Chaotic Dynamics Fractals, Tilings, and Substitutions (Complete Instructor's Resources with Solution

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SUPPLEMENTARY PROBLEMS We split the problems into two sections, Section A and Section B. Section A problems are fairly standard, whilst Section B problems are a little more demanding.

As usual, more difficult problems are denoted with an asterisk. In Section A, the first number denotes the chapter of the text the problem refers to. (So Problem 2.3 refers to a problem supplementary to Chapter 2). Double starred questions are problems that the author has not yet fully considered, and may be difficult.

Section A 1.1 You pay into an annuity a sum of $P dollars. This annuity pays you $α per month. The interest is r% (calculated as simple interest on the remaining balance at the end of each month). If A(n) is the amount remaining at the end of the nth month, A(0) = P , write down A(n + 1) in terms of A(n), and deduce a closed formula for A(n).

If P = $100, 000, α = $500 and the interest rate is 4% per month, find out how long the annuity will last.

1.2 Let g(x) = x2 + 1/4. Use induction to show that: (a) g n (x) → 1/2 as n → ∞, if −1/2 < x < 1/2, (b) g n (x) → ∞ as n → ∞, if |x| > 1/2.

Deduce that x = 1/2 is a fixed point of g that is stable on the left and unstable on the right.

1.3 If p(x) is a polynomial of degree n, what is the degree of (i) p2 (x), (ii) pk (x)?

2.1 Let f : [0, 1] → [0, 1] be continuous on [0, 1] and differentiable on (0, 1) (a C 1 function), with |f ′ (x)| < 1 for all x ∈ (0, 1). (a) Prove that f (x) has a unique fixed point p in [0, 1] (See 1.4 # 2). (b) Prove that f (x) cannot have a point of period 2 in [0, 1] (Hint: Use the Mean Value Theorem). (c) Prove that f n (x) → p as n → ∞, for all x ∈ (0, 1).

2.2 Suppose that f is an odd function, f (α) ̸= 0 and Nf (α) = −α. (a) Show that f n (x) − x and Nf (x) are odd functions.

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(b) Show that {−α, α} is a 2-cycle for Nf . Use this to find the 2-cycles for Nf when (i) f (x) = x3 − 4x, (ii) f (x) = sin(x). What happens with f (x) = x3 + 2x. (c) Show that if {α, β} is a 2-cycle for the odd function f , then so is {−α, −β}.

Give an example of an odd function f with a 2-cycle not of the form {−α, α} (Hint: Consider f (x) = x3 − λx for some λ > 2).

2.3 (a) Let f (x) be a polynomial that is also an odd function. Note that g(x) = ([f (x)]2 − x2 )/x is a polynomial. Show that g(x) divides f 2 (x) − x. (Hint: Note that some 2-cycles arise from solving f (x) = −x). (b) Check the above for (i) f (x) = x3 − x, (ii) f (x) = x3 − λx for 0 ≤ λ < 3. Find all the two cycles for f (x) = x3 − 3x/2, and deduce that there are some that don’t arise from solving f (x) = −x.

2.4 Show that the fixed points of T (the standard tent map) are in the first and sixth positions of the fixed points of T 3 (we are using the ordering on the line real line).

2.5 (a) Show that if a function f : R → R has a 2-cycle {α, β}, then the line y = −x+a intersects the graph in the two points (α, β) and (β, α), for some a ∈ R. Conversely, show that if the line y = −x + a intersects the graph of f in two points equidistant from the line y = x, then f has a 2-cycle. Illustrate the situation graphically. √ { 2 x; 1√≤ x ≤ 10 (b) Graph the function f : [1, 10] → [1, 10] defined by f (x) = , x2 /10; 10 < x ≤ 10 and use (a) to show that f has a 2-cycle. Find the 2-cycle. (c)∗∗ Does the function f from (b) have any other { periodic orbits? What about the 2x; 1 ≤ x ≤ 5 function g : [1, 10] → [1, 10] defined by g(x) = . Does g have x/5; 5 < x ≤ 10 any periodic points? Is g transitive. { 2.6 fa : [0, 1] → [0, 1] is defined by fa (x) =

ax2 ; 0 ≤ x ≤ 1/2 . a(1 − x)2 ; 1/2 < x ≤ 1

(a) Find the fixed points for each a ∈ [0, 4] and determine their stability. (b) For the fixed point x = 0, find the immediate basin of attraction. Deduce that fa is not chaotic for a ∈ [0, 4]. (c)∗∗ What is the basin of attraction of x = 0. Investigate the dynamics of fa as a increases (use Mathematica).

3.1 Let f : R → R have a period-16 point. What other periodic points must f have?

What if instead f has a period-18 point?

3.2 Show that if f : R → R is a continuous function for which there exists y ∈ R with f 2 (y) < y < f (y), then f has a fixed point. Give an example to show that f need not have a 2-cycle (See Exercises 3.2 # 12, where the requirement that f : [a, b] → [a, b] should be included).

4.1 (a) Let f : X → X be a continuous function on a metric space X. f is a periodic function if there exists p ∈ Z+ satisfying f p (x) = x for all x ∈ X (the minimum such p is the period of f ). Show that such a function is a homeomorphism and give two examples of (non-trivial) periodic functions of period p > 1. (b) Let f : [0, 1] → [0, 1] be a continuous periodic function. What can you say about p?

5.1 Write down the definition of sensitive dependence on initial conditions as a quantified statement, and use it to write down its negation.

6.1 Show that if f : X → X is a continuous function on a metric space X for which f 2 is chaotic, then f is chaotic. Is the converse true?

7.1 f : R → R is defined by f (x) = x2 − 2x + 2. Show that f is linearly conjugate to g : R → R, g(x) = x2 . {

−2x + 1; 0 ≤ x ≤ 1/2 . Show 2x − 1; 1/2 < x ≤ 1 that f is conjugate to the tent map by constructing a conjugacy.

7.2 A map f : [0, 1] → [0, 1] is defined by f (x) =

8.1 If x = c is an attracting fixed point of a polynomial p(x) of degree n > 1, we know that the immediate basin of attraction is an open interval. Why must the interval be bounded?

8.2 (a) Let p(x) = (x−a)(x−b)(x−c) be a polynomial of degree 3 having three distinct real roots. Find the Schwarzian derivative and show directly that it is negative.

(b) If p is a polynomial of degree n > 1 having all roots real and distinct, show that p has a negative Schwarzian derivative.

8.3 Let p(x) be a polynomial of degree n > 1. (a) If p′ has n − 1 real roots (some of which may be repeated), show that p has a negative Schwarzian derivative. (b) If p(x) has n real roots (not necessarily distinct), show that p has a negative Schwarzian derivative. (c) Give an example of a degree 3 polynomial not having negative Schwarzian derivative everywhere.

8.4 Let p(x) = x3 + ax2 + bx + c, a ̸= 0, where p′ (x) > 0 for all x ∈ R. Show that the Schwarzian derivative Sp(x) is positive on an interval. { 9.1 fa : [0, 1] → [0, 1], a > 1, is defined by fa (x) =

ax; 0 ≤ x ≤ 1/a . −a(x − 1)/(a − 1); 1/a < x ≤ 1

fa is a skewed tent map. Sketch the graph of fa . (a) Show that fa is conjugate to the tent map. (Hint: Use the methods of Section 9.1). (b) Show that if h is the conjugacy, then h(0) = 0, and that h cannot be linear near 0. (Hint: Consider h−1 ◦ f ◦ h near 0).

9.2 The polynomial pk (µ) = µk−1 −µk−2 −µk−3 +µk−4 −µk−5 +·√· ·+µ−1, k√≥ 5 (odd), arises in the solution to Exercise 9.2 # 3(iii). Show that pk ( 2) = 1 − 2 < 0 for k = 5, 7, 9 . . .. Deduce that T√2 (Tent map) cannot have points having odd periods (see also Section 2.8).

10.1 (a) Give a description of points in the Sierpinski Carpet in an analogous way to the ternary representation of points in the Cantor Set (as a first step in the construction, take the interval [0, 1] × [0, 1]). Deduce that the Sierpinski Carpet is invariant under the Iterated Function System defined by the maps

( fi

x y

)

( =

x/3 + ai y/3 + bi

) ,

(

) ( ) ( ) ( ) ( ) ai 0 0 0 1/3 i = 1, . . . , 8, where are the points , , , , b 0 1/3 2/3 0 ( ) ( ) ( i ) ( ) 1/3 2/3 2/3 2/3 , , and .

2/3 0 1/3 2/3 (b) Give a description of Menger sponge in terms of ternary expansions.

10.2 (a) Show that the Sierpinski triangle has topological dimension zero. (b) Show that the Sierpinski triangle is invariant under the Iterated Function System given in Example 10.5.9.

10.3 For each n ≥ 1, set An = {(k/n, ℓ/n) : 0 ≤ k ≤ n, 0 ≤ ℓ ≤ n}, a discrete subset of B = [0, 1] × [0, 1]. If D denotes the Hausdorff metric on the compact subsets of R2 , show that limn→∞ An = B with respect to the Hausdorff metric.

10.4 Let F1 , F2 : R → R be an iterated function system given by F1 (x) = x/4 and F2 (x) = x/2 + 1/2. Describe the attractor and find its fractal dimension.

14.1 f : C → C is defined by f (z) = z̄. Show that f is not differentiable at any point in C. 14.2 A function f : C → C is defined by f (z) = z 8 . Find the fixed points of f .

Use your calculations to find the real linear and quadratic factors of the polynomial p(z) = z 7 − 1.

14.3. Prove that if fc (z) = z 2 + c has an attracting periodic point, then c ∈ M, the Mandelbrot set. (Hint: Use any appropriate results from Section 14.5).

14.4. Show that p(z) = az + b, a ̸= 1 on C, is conjugate to q(z) = az.

14.5. Let fc (z) = z 2 + c. Find the values of c so that z = i is a period-2 point. Find the fixed points in each case and determine their stability. Does c ∈ M? (Hint: It is useful to use Mathematica to find the square roots of complex numbers).

z−i gives a conjugacy between the Newton z+i map Nf1 , where f1 (z) = z 2 + 1, and the function f0 (z) = z 2 . Deduce the Julia set of Nf1 , and that it is chaotic on its Julia set. (Hint: Find H −1 (z) and set z = eiθ ).

14.6 Show that the function H(z) =

p2 (z) − z = p(z) + z? If so, find all p(z) − z such polynomials and their period-2 points. What does this say about the periodic points of the polynomial?

14.7 Is it possible for a polynomial p(z) to satisfy

(b) What about

p2 (z) − z = p(z) − z? p(z) − z

Section B 1. Consider the family of functions fλ : R → R, fλ (x) = x3 − λx, for a parameter λ ∈ R. (a) Find all the fixed points and determine their nature and where they are created as λ varies. (b) Find where a two cycle is created (you may use a computer algebra facility), and give the graph of where this happens. Determine the stability of the hyperbolic 2-cycles. (c) Use a computer algebra facility to find an approximate value of λ where the 3-cycle is created. Give the graph of this situation.

2. Let f (x) = ax3 + bx + c where a and b satisfy a/b > 0. Denote by Nf the corresponding Newton function. (a) Show that Nf has a unique fixed point. (b) Show that Nf cannot have any period-2 points. (c) Why does it follow that Nf has no points of period n, n > 2?

3. This question is related to Exercise 4.2 # 8, parts (a), (b) and (c). (a) Show that the function f (x) = −1/(x + 1) has the property that f 3 (x) = x for all x ̸= −1, x ̸= 0.

(b) Let f : R → R be a function defined on a set I, with f 3 (x) = x for all x ∈ I? Set g(x) = f 2 (x). Show that g 3 (x) = x for all x ∈ I. Deduce a different function from the one in (a) with this property. (c) In general, show that such a function cannot have a 2-cycle. (d) Deduce that a function f : R → R with the property f 3 (x) = x cannot be continuous (i.e., there must be some points of discontinuity for f , where we may not have f 3 (x) = x). (e) Show that f is one-to-one. Need f be onto? Show that f is strictly decreasing and investigate any symmetry. What if f : [−∞, ∞] → [−∞, ∞] - consider the example in (a) above? (f) If f ′ (x) exists for all x ∈ I, show that the 3-cycles are non-hyperbolic, (assume that f (x) is not the function y = x) . ax + b satisfies f 3 (x) = x, show that if f (x) is not the cx + d function y = x, and a ̸= d then (g) Suppose that f (x) =

a2 + bc + ad + d2 = 0.

(i) Show that if ad − bc > 0, then such a function cannot have any fixed points. (ii) Deduce two other functions with the property that f 3 (x) = x wherever defined, one having fixed points and one with no fixed points.

4.∗∗ (check). If f : D → D has the property |f ′ (z)| < 1 for all z ∈ D \ S1 , show that |f (z) − f (w)| < |z − w| for all z, w ∈ D. Does it follow that f is a contraction mapping (consider some examples)? Show that f has a unique fixed point α. Why can we deduce that Bf (α) = D? (D = {z ∈ C : |z| ≤ 1}). az + b 5. There is a correspondence between functions of the form f (z) = and cz + d [ ] a b matrices , when a, b, c, d ∈ R, ad − bc ̸= 0. To be more precise, show that c d there is a group homomorphism Φ : GL2 (R) → F , where GL2 (R) is the multiplicative group of 2 × 2 matrices having real entries and non-zero determinant, and F is az + b , the group (under composition) of linear fractional transformations f (z) = cz + d ( ) a b (ad − bc ̸= 0), where Φ is defined by Φ = f (z). Is the map Φ one-to-one or c d onto?

6. (a) Show that fc : R → R, fc (x) = x2 + c has a 3-cycle for c ≤ −7/4. (b) Deduce that Lµ : R → R, Lµ (x) = µx(1 − x) has a 3-cycle for µ ≤ 1 −

√ 8.

7.∗ Use the following steps to show that the Koch curve is continuous, i.e., there is a continuous function f : [0, 1] → R2 , whose range is the top part of the Koch curve (see Section 10.1): (a) Denote by fn : [0, 1] → R2 the piecewise defined function which is linear on each of the 4n segments defined at the nth stage of the construction of the Koch curve.

Show that |fn (t) − fn+1 (t)| ≤ 3−n for all t ∈ [0, 1]. (b) Deduce that (fn (t)) is a Cauchy sequence for each t ∈ [0, 1], so that f (t) = limn→∞ fn (t) exists for all t ∈ [0, 1]. (c) Show that[ convergence of the sequence of functions (fn ) to f is uniform on [0, 1] ] (i.e., limn→∞ sup0≤t≤1 |fn (t) − f (t)| = 0). (d) Now complete the proof using the fact that the uniform limit of a sequence of continuous functions on an interval [a, b] is a continuous function. That f is not differentiable anywhere can be deduced from the fact that the length of the graph of f is infinite on any arbitrarily small interval.

SUPPLEMENTARY PROBLEM ANSWERS Section A 1.1 You pay into an annuity a sum of $P dollars. This annuity pays you $α per month. The interest is r% (calculated as simple interest on the remaining balance at the end of each month). If A(n) is the amount remaining at the end of the nth month, A(0) = P , write down A(n + 1) in terms of A(n), and deduce a closed formula for A(n).

If P = $100, 000, α = $500 and the interest rate is 4% per month, find out how long the annuity will last.

Answer. This gives rise to an affine difference equation: A(n + 1) = A(n) + A(n)r/1200 − α = (1 + r/1200)A(n) − α,

A(0) = P.

Set a = 1 + r/1200, b = −α and x0 = P , to give −α −α A(n) = (P + )(1 + r/1200)n + 1 + r/1200 − 1 1 − 1 − r/1200 = (P − 1200α/r)(1 + r/1200)n + 1200α/r.

Now set P = $ 100,000, α = $ 500 and r = 4, so that we find when A(n) = 0 as this is when all the money is paid off (after n months), i.e., A(n) = (100000 − 1200 × 500/4)(1 + 4/1200)n + 1200 × 500/4 = 0.

Solving gives n = months, so our income ceases after years and months.

1.2 Let g(x) = x2 + 1/4. Use induction to show that: (a) g n (x) → 1/2 as n → ∞, if −1/2 < x < 1/2, (b) g n (x) → ∞ as n → ∞, if |x| > 1/2.

Deduce that x = 1/2 is a fixed point of g that is stable on the left and unstable on the right.

Answer. (a) Let −1/2 < x < 1/2, then g(x) = x2 + 1/4 < 1/4 + 1/4 = 1/2, so 0 ≤ g(x) < 1/2. Suppose that 0 ≤ g n (x) < 1/2, then g n+1 (x) = [g n (x)]2 + 1/4 < 1/4 + 1/4 = 1/2, so 0 ≤ g n+1 (x) < 1/2. It follows by induction that this is true for all n.

We can check that g(x) > x for −1/2 < x < 1/2, so suppose that g n (x) > g n−1 (x), then g n+1 (x) = [g n (x)]2 + 1/2 > [g n−1 (x)]2 + 1/4 = g n (x), so this is true for all n. Thus for each x ∈ (−1/2, 1/2), (g n (x)) is an increasing sequence bounded above, so is convergent to the only fixed point, x = 1/2. (b) In a similar way, if x > 1/2, then we show that (g n (x)) is an increasing sequence which must converge to infinity since there are no fixed points of g in (1/2, ∞).

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1.3 If p(x) is a polynomial of degree n, what is the degree of (i) p2 (x), (ii) pk (x)?

Answer. The degree of p2 (x) is n2 , and the degree of pk (x) is nk .

2.1 Let f : [0, 1] → [0, 1] be continuous on [0, 1] and differentiable on (0, 1) (a C 1 function), with |f ′ (x)| < 1 for all x ∈ (0, 1). (a) Prove that f (x) has a unique fixed point p in [0, 1] (See 1.4 # 2). (b) Prove that f (x) cannot have a point of period 2 in [0, 1] (Hint: Use the Mean Value Theorem). (c) Prove that f n (x) → p as n → ∞, for all x ∈ (0, 1).

Answer. (a) Suppose that there are two distinct fixed points p and q in [0, 1], say f (p) − f (q) = p < q. By the Mean Value Theorem there exists a ∈ (p, q) with f ′ (a) = p−q p−q = 1, a contradiction. p−q (b) Suppose there is a 2-cycle {a, b}, so f (a) = b, f (b) = a with a < b. By the Mean f (a) − f (b) b−a Value Theorem there exists α ∈ (a, b) with f ′ (α) = = = −1, which a−b a−b again contradicts |f ′ (α)| < 1. (c) We show that limn→∞ f n (x) = p for all x ∈ (0, 1). The basin of attraction of p is open in [0, 1]. Suppose that it is an interval of the form (a, b), where 0 < a < b < 1.

Then we have seen that this implies that either a or b is a fixed point, or {a, b} is a 2-cycle. Clearly none of these can happen, so we must have a = 0 and b = 1. Thus the basin of attraction must contain (0, 1). (It must be equal to [0, 1] for otherwise these would also be fixed etc.).

2.2 Suppose that f is an odd function, f (α) ̸= 0 and Nf (α) = −α where α ̸= 0. (a) Show that f n (x) − x and Nf (x) are odd functions. (b) Show that {−α, α} is a 2-cycle for Nf . Use this to find the 2-cycles for Nf when (i) f (x) = x3 − 4x, (ii) f (x) = sin(x). What happens with f (x) = x3 + 2x. (c) Show that if {α, β} is a 2-cycle for the odd function f , then so is {−α, −β}.

Give an example of an odd function f with a 2-cycle not of the form {−α, α} (Hint: Consider fλ (x) = x3 − λx for some λ > 2).

Answer. (a) f (−x) = −f (x), so suppose that f n−1 (−x) = −f n−1 (x), then f n (−x) = f (f n−1 (−x)) = f (−f n−1 (x)) = −f n (x), so by induction, f n (x) is odd for all n ∈ Z+ .

It follows that f n (x) − x is an odd function. Now f ′ (x) is an even function, so f (x) Nf (x) = x − ′ is an odd function. f (x) (b) Nf2 (α) = Nf (−α) = −Nf (α) = α, so {−α, α} is a 2-cycle. (i) If f (x) = x3 − 4x, x3 − 4x 2x3 then Nf (x) = x − 2 = (fixed points are x = 0, ±2). If we solve 3x − 4 √ 3x2 − 4 √ √ Nf (x) = −x, we get x = ±2/ 5, so {−2/ 5, 2/ 5} is a 2-cycle. (ii) If f (x) = sin(x), Nf (x) = x − tan(x), so if Nf (x) = −x, tan(x) = 2x. This has many solutions of the form {−α, α}.

3 For f (x) = x3 +2x, Nf (x) = 3x2x2 +2 , and Nf (x) = −x has no non-trivial solutions, so there is no 2-cycle of this form. (c) f (α) = β and f (β) = α, so that f (−α) = −f (α) = −β and f (−β) = −f (β) = −α, so that {−α, −β} is also a 2-cycle. √ √ If fλ (x) = x3 − λx, setting fλ (x) = −x gives the 2-cycle {− λ − 1, λ − 1}, (λ > 1), which is of the above form. To find a 2-cycle not of this form, we solve fλ2 (x) = x, or (x3 − λx)3 − λx − x = 0, and use the fact that x(x2 − λ − 1)(x2 − λ + 1) is a factor. We see that f 2 (x) − x =√x(x2 − λ − 1)(x2 − λ + 1)(x4 − λx2 + 1). Solving x4 − λx2 + 1 = 0 gives x2 = (λ ± λ2 − 4)/2, so for λ > 2 we have two additional period-2 points, for example when λ = 3 we get two 2-cycles {−1.61803, 0.618034} and {−0.618034, 1.61803}. which are not of the form {−α, α}, 2.3 (a) Let f (x) be a polynomial that is also an odd function. Note that g(x) = ([f (x)]2 − x2 )/x is a polynomial. Show that g(x) divides f 2 (x) − x. (Hint: Note that some 2-cycles arise from solving f (x) = −x). (b) Check the above for (i) f (x) = x3 − x, (ii) f (x) = x3 − λx for 0 ≤ λ < 3. Find all the two cycles for f (x) = x3 − 3x/2, and deduce that there are some that don’t arise from solving f (x) = −x.

Answer. (a) g(x) = ([f (x)]2 − x2 )/x = (f (x) + x)(f (x) − x)/x and since f (x) is odd, x = 0 is a fixed point of f (x), so (f (x) − x)/x is a polynomial. Now both f (x) + x and f (x) − x are odd with x being odd, so g(x) is odd.

If (x − α) is a factor of g(x), then either (i) α = 0, then since f (x) is odd, x is a factor of f 2 (x) − x, or (ii) If α ̸= 0 and f (α) = α, then f ( α) = α, so x − α is a factor of f 2 (x) − x, or (iii) If α ̸= 0 and f (α) = −α, then f 2 (α) = α, so x − α is a factor of f 2 (x) − x. The result follows. (b) Set f (x) = x3 − x. Then g(x) = ([f (x)]2 − x2 )/x = x3 (x2 − 1), f 2 (x) − x = (x3 − x)3 − x3 and (f 2 (x) − x)/g(x) = x4 − x2 + 1.

Now write f (x) = x3 − λx, g(x) = ([f (x)]2 − x2 )/x = x(x4 − 2λx2 + λ2 − 1) = −x(x2 − λ + 1)(x2 + λ + 1), f 2 (x) − x = (x3 − λx)3 − λ(x3 − λx) − x and (f 2 (x) − x)/g(x) = x4 − λx2 + 1. We can now find the 2-cycles when λ = 3/2.

2.4 Show that the fixed points of T (the standard tent map) are in the first and sixth positions of the fixed points of T 3 (we are using the ordering on the real line).

Answer This actually follows from considerations of the truncated tent map (see the answer to Exercise 3.1 # 14, and Exercise 12.3 # 11).

2.5 Show that if a function f : R → R has a 2-cycle {α, β}, then the line y = −x + a intersects the graph in the two points (α, β) and (β, α), for some a ∈ R. Conversely, show that if the line y = −x + a intersects the graph of f in two points equidistant from the line y = x, then f has a 2-cycle. √ { 2 x; 1√≤ x ≤ 10 , (b) Graph the function f : [1, 10] → [1, 10] defined by f (x) = x2 /10; 10 < x ≤ 10 and use (a) to show that f has a 2-cycle. Find the 2-cycle. (c)∗∗ Does the function f from (b) have any other { periodic orbits? What about the 2x; 1 ≤ x ≤ 5 function g : [1, 10] → [1, 10] defined by g(x) = . Does g have x/5; 5 < x ≤ 10 any periodic points? Is g transitive.

Answer. (a) We have f (α) = β and f (β) = α, where α ̸= β. Then the line joining (α, β) to (β, α) has equation y = −x + α + β, intersecting the line y = x at (α + β)/2, (α + β)/2) (the mid-point). On the other hand, if the line y = −x + a intersects the graph of f in two points equidistant from the line y = x, the symmetry tells us that the points are of the form (α, β) and (β, α), so must give rise to a 2-cycle. (b) Take (α, β) on the graph of y = x2 /10 and (β, α) on the graph of y = x2 , then β = α2 /10 and α = β 2 , which gives β = β 4 /10, so β = 101/3 , so the 2-cycle is {101/3 , 102/3 } . (c) f has other periodic cycles, for example {101/7 , 102/7 , 104/7 } is a 3-cycle. Clearly g has no fixed points or period-2 points. In fact g cannot have any periodic points since g n (x) = 2k x/5m for some k, m (where at least of k or m is greater than one), and this cannot be equal to x. { 2.6 fa : [0, 1] → [0, 1] is defined by fa (x) =

ax2 ; 0 ≤ x ≤ 1/2 .

2 a(1 − x) ; 1/2 < x ≤ 1

(a) Find the fixed points for each a ∈ [0, 4] and determine their stability. (b) For the fixed point x = 0, find the immediate basin of attraction. Deduce that fa is not chaotic for a ∈ [0, 4]. (c)∗∗ What is the basin of attraction of x = 0. Investigate the dynamics of fa as a increases (use Mathematica).

Answer. (a) Solving fa (x) = x gives ax2 = x, 0 ≤ x ≤ 1/2 or a(1√− x)2 = x, 1/2 < x ≤ 1. This gives x = 0, x = 1/a for 2 < a ≤ 4, and x = (2a + 1 − 4a + 1)/2a (since x ≥ 1/2 we need a ≥ 2 for these fixed point to exist: if a < 2, 1/a > 1/2 etc.).

Now clearly x = 0 is attracting and fa′ (x) = 2ax for 0 < x <√1/2, so fa′ (1/a) = 2 gives 1/a as repelling (2 < a ≤ 4). Similarly, for x = (2a + 1 − 4a + 1)/2a, we see it is repelling. (b) If 0 < a < 2, and 0 ≤ x ≤ 1/2, then fa (x) = ax2 = ax · x < x, so fan (x) is a decreasing sequence, converging to 0, and therefore [0, 1/2] is contained in the basin of attraction of x = 0. Similarly the basin of attraction contains [1/2, 1], so it is equal to [0, 1]. For 2 ≤ a ≤ 4, the fixed point 1/a will not be in the basin of attraction of x = 0 and we see that the immediate basin of attraction is [0, 1/a). It follows that fa cannot be chaotic, for example, it is not transitive. (c) For 0 < a < 4 this question seems to be complicated, the basin of attraction of 0 is possibly equal to a Cantor set, depending on the value of a. Use Mathematica to investigate this further.

3.1 Let f : R → R have a period-16 point. What other periodic points must f have?

What if instead f has a period-18 point?

Answer. 16 = 24 , so f must also have points of period 23 = 8, 22 = 4, 21 = 2 and a fixed point.

18 = 2 × 9, so f must also have points of period 2m, for all m ∈ Z+ \ {3, 5, 7}, together with a fixed point.

3.2 Show that if f : R → R is a continuous function for which there exists y ∈ R with f 2 (y) < y < f (y), then f has a fixed point. Give an example to show that f need not have a 2-cycle (See Exercises 3.2 # 12, where the requirement that f : [a, b] → [a, b] should be included).

Answer. Set g(x) = f (x)−x, then g(y) = f (y)−y > 0, and g(f (y)) = f 2 (y)−f (y) < 0, so by the IVT there exists c ∈ (y, f (y)) with g(c) = 0, or f (c) = c.

Set f (x) = −2x+2, and y = 0, then x = 2/3 is a fixed point and f (y) = f (0) = 2, f (y) = f (2) = −2, so that f 2 (y) < y < f (y). f is a continuous function and f 2 (x) = 4x − 2, whose only fixed point is x = 2/3, so there are no period-2 points.

2

4.1 (a) Let f : X → X be a continuous function on a metric space X. f is a periodic function if there exists p ∈ Z+ satisfying f p (x) = x for all x ∈ X (the minimum such p is the period of f ). Show that such a function is a homeomorphism and give two examples of (non-trivial) periodic functions of period p > 1.

(b) Let f : [0, 1] → [0, 1] be a continuous periodic function. What can you say about p?

Answer. (a) f : X → X is continuous, f (f p−1 (x)) = x, so f is onto and if f (x) = f (y) then f p (x) = f p (y), or x = y, so f is one-to-one. Clearly f is a homeomorphism.

Define f [0, 1] → [0, 1], by f (x) = −x, then f is periodic of period 2. Define 1 g : S → S1 , by g(z) = az, where a = e2πi/p , p > 1, then g is periodic of period p. (b) f : [0, 1] → [0, 1] is continuous and periodic, so it is a homeomorphism. Thus it is either increasing or decreasing. If f is increasing it cannot have any period-2 points, so we must have f (x) = x for all x ∈ [0, 1]. If it is decreasing, we must have f 2 (x) = x for all x ∈ [0, 1]. We conclude that p = 1 or p = 2. An example of the second case is f (x) = 1 − x. In fact any map having its graph symmetric with respect to y = x has this property, e.g., f (x) = (1 − 1/a)x + 1 for 0 ≤ x ≤ a, f (x) = a(x − 1)/(a − 1) for a < x ≤ 1, where 0 < a < 1.

5.1 Write down the definition of sensitive dependence on initial conditions as a quantified statement, and use it to write down its negation.

Answer. Let f : X → X be a function defined on the metric space (X, d). Then f has sensitive dependence on initial conditions if: ∃ δ > 0, ∀ x ∈ X, ∀ open U ⊂ X (x ∈ U ), ∃ y ∈ U, ∃ n ∈ Z+ , d(f n (x), f n (y)) > δ.

Then the negation is: ∀ δ > 0, ∃ x ∈ X, ∃ open U ⊂ X (x ∈ U ), ∀ y ∈ U, ∀ n ∈ Z+ , d(f n (x), f n (y)) ≤ δ.

6.1 Show that if f : X → X is a continuous function on a metric space X for which f 2 is chaotic, then f is chaotic. Is the converse true?

Answer. Suppose that f 2 is chaotic, then the periodic points of f 2 are dense in X, so the periodic points of f are dense in X. f 2 has a transitive point, say x0 , so clearly x0 is a transitive point for f . The continuity on a metric space now implies that f is chaotic.

On the other hand 7.1 f : R → R is defined by f (x) = x2 − 2x + 2. Show that f is linearly conjugate to g : R → R, g(x) = x2 .

Answer. Set h(x) = x − 1, then we can check that h ◦ f (x) = g ◦ h(x) for all x ∈ R.

{

−2x + 1; 0 ≤ x ≤ 1/2 . Show 2x − 1; 1/2 < x ≤ 1 that f is conjugate to the tent map by constructing a conjugacy.

7.2 A map f : [0, 1] → [0, 1] is defined by f (x) =

Answer. Set k(x) = −x + 1, where k : [0, 1) → [0, 1) (we choose this since it sends fixed points of T to fixed points of f ). Then we can check that k ◦ T (x) = f ◦ k(x) for all x ∈ [0, 1). k is clearly a homeomorphism.

8.1 If x = c is an attracting fixed point of a polynomial p(x) of degree n > 1, we know that the immediate basin of attraction is an open interval. Why must the interval be bounded?

Answer. We are assuming c ̸= ±∞. However, we have seen that ±∞ are attracting “fixed points” for a polynomial, so their basins of attraction include intervals of the form (−∞, a) and (b, ∞). It follows that the basins of attraction of any other attracting fixed point must be bounded.

8.2 (a) Let p(x) = (x−a)(x−b)(x−c) be a polynomial of degree 3 having three distinct real roots. Find the Schwarzian derivative and show directly that it is negative. (b) If p is a polynomial of degree n > 1 having all roots real and distinct, show that p has a negative Schwarzian derivative.

Answer. (a) The Schwarzian derivative is 6(a2 + b2 + c2 + b(c − 4x) + a(b + c − 4x) − 4cx + 6x2 ) , (b(c − 2x) + a(b + c − 2x) + x(−2c + 3x))2 where we have a quadratic in the numerator, whose discriminant is Sp(x) = −

16(a + b + c)2 − 24(a2 + b2 + c2 + ab + bc + ca) = 4[(a − b)2 + (b − c)2 + (c − a)2 ] ≥ 0, so that Sp(x) is always negative. (b) If p(x) is a polynomial of degree n > 1 having all of its roots real and distinct, then p(x) has n − 1 critical points which are real and distinct (use Rolle’s Theorem).

It follows from Lemma 8.1.1 that the Schwarzian derivative of p is negative.

8.3 Let p(x) be a polynomial of degree n > 1. (a) If p′ has n − 1 real roots (some of which may be repeated), show that p has a negative Schwarzian derivative. (b) If p(x) has n real roots (not necessarily distinct), show that p has a negative Schwarzian derivative. (c) Give an example of a degree 3 polynomial not having negative Schwarzian derivative everywhere.

Answer. (a) Generalize the proof of Lemma 8.1.1 in the obvious way. Similarly for (b) (as in 8.2(b)).

6 − 36x2 (c) Set p(x) = x3 + x, then we can check that Sp(x) = , which is positive (3x2 + 1)2 √ √ on the interval (−1/ 6, 1/ 6).

8.4 Let p(x) = x3 + ax2 + bx + c, a ̸= 0, where p′ (x) > 0 for all x ∈ R. Show that the Schwarzian derivative Sp(x) is positive on an interval. −6(6x2 + 4ax + a2 − b) .

(3x2 + 2ax + b)2 Looking at the discriminant of the numerator, it is positive when a2 < 3b, giving an interval on which Sp(x) is positive. Suppose p′ (x) = 3x2 + 2ax + b > 0 for all x ∈ R, then its discriminant satisfies: 4a2 − 12b < 0, or a2 < 3b. We conclude that if p′ (x) < 0, then Sp(x) is positive on an interval.

Answer. The Schwarzian derivative of p(x) is Sp(x) =

{ 9.1 fa : [0, 1] → [0, 1], a > 1, is defined by fa (x) =

ax; 0 ≤ x ≤ 1/a . −a(x − 1)/(a − 1); 1/a < x ≤ 1

fa is skewed tent map. Sketch the graph of fa . (a) Show that fa is conjugate to the tent map. (Hint: Use the methods of Section 9.1). (b) Show that if h is the conjugacy, then h(0) = 0, and that h cannot be linear near 0. (Hint: Consider h−1 ◦ f ◦ h near 0).

Answer. (a) Using the method of 9.1, start by selecting x0 ∈ (0, a) arbitrarily, and form a fundamental domain 9.2 The polynomial pk (µ) = µk−1 − µk−2 − µk−3 + µk−4 − µk−5 + ·√· · + µ − 1,√k ≥ 5 (odd), arises in the solution to Exercise 9.2 # 3(iii). Show that pk ( 2) = 1 − 2 < 0 for k = 5, 7, 9 . . .. Deduce that T√2 (tent map) cannot have points having odd periods (see also Section 2.8).

Answer. We can check that

pk (µ) = µk−1 − µk−2 − µk−3 + µk−4 − µk−5 + · · · + µ − 1 = 0 = µk−1 − µk−2 −

1 + µk−2 , 1+µ

√ √ so pk ( 2) = 1 − 2 < 0, for all k = 5, 7, 9 . . .. It follows that T√2 cannot have points of odd period greater than one (see the answers to 9.2 # 3(iii)).

10.1 (a) Give a description of points in the Sierpinski Carpet in an analogous way to the ternary representation of points in the Cantor Set (as a first step in the construction, take the interval [0, 1] × [0, 1]). Deduce that the Sierpinski Carpet is invariant under the Iterated Function System defined by the maps ( fi

x y

)

( =

x/3 + ai y/3 + bi

) ,

(

) ( ) ( ) ( ) ( ) ai 0 0 0 1/3 i = 1, . . . , 8, where are the points , , , , b 0 1/3 2/3 0 ( ) ( ) ( i ) ( ) 1/3 2/3 2/3 2/3 , , and .

2/3 0 1/3 2/3 (b) Give a description of Menger sponge in terms of ternary expansions.

Answer. (a) Define A0 = [0, 1] × [0, 1] as a first step in the construction of the Sierpinski Carpet. Now we remove open middle third squares to obtain A1 , A2 , . . ., and set S = ∩∞ n=0 An , the Sierpinski Carpet. In a similar way to the Cantor set, we see that (x, y) ∈ S if and only if both x and y have ternary expansions x = ·x1 x2 x3 . . ., y = ·y1 y2 y3 . . . where (xi , yi ) ̸= (1, 1) for every i. This is because the points removed at the nth stage of the construction are of the form (x, y), where the ith digit in the expansions of x and y are both 1. We can now check that S is invariant under the iterated function system. (b) The points (x, y, z) ∈ M , the Menger sponge, if and only if (x, y, z) = (·x1 x2 x3 . . . , ·y1 y2 y3 . . . , ·z1 z2 z3 . . .) (ternary expansions), such that (xi , yi ), (xi , zi ), (yi , zi ) ̸= (1, 1) for every i.

10.2 (a) Show that the Sierpinski triangle has topological dimension one. (b) Show that the Sierpinski triangle is invariant under the Iterated Function System given in Section 10.5.9.

Answer. (a) Given a point in the Sierpinski triangle, we can surround that point by a circle that intersects exactly three points of the triangle (vertices of a sub-triangle).

From the self similarity, we can find arbitrary small circles with this property. (b) It is clear that the set S = {(x, y) : both x and y have a ternary expansion in terms of 0’s and 2’s alone} is invariant under the iterated function system. UNFINISHED

10.3 For each n ≥ 1, set An = {(k/n, ℓ/n) : 0 ≤ k ≤ n, 0 ≤ ℓ ≤ n}, a discrete subset of B = [0, 1] × [0, 1]. If D denotes the Hausdorff metric on the compact subsets of R2 , show that limn→∞ An = B with respect to the Hausdorff metric.

Answer.

10.4 Let F1 , F2 : R → R be an iterated function system given by F1 (x) = x/4 and F2 (x) = x/2 + 1/2. Describe the attractor and find its fractal dimension.

14.1 f : C → C is defined by f (z) = z̄. Show that f is not differentiable at any point in C. z̄ − ā Answer. If we look at lim as z approaches a from different directions, we get z→a z − a different answers, so the limit does not exist.

14.2 A function f : C → C is defined by f (z) = z 8 . Find the fixed points of f .

Use your calculations to find the real linear and quadratic factors of the polynomial p(z) = z 7 − 1.

Answer. f (z) = z 8 , so solving z 8 = z gives z = 0, or z 7 = 1. Set z = eiθ , then e7iθ = e2kπi , for k = 0, . . . , 6 so θ = 2kπ/7, so z = 0 and z = e2kπi/7 , k = 0, 1, . . . 6 are the fixed points of f .

We deduce that z 7 − 1 = (z − 1)(z − e2πi/7 )(z − e−2πi/7 )(z − e4πi/7 )(z − e−4πi/7 )(z − e6πi/7 )(z − e6πi/7 ), or z 7 − 1 = (z − 1)(z 2 − 2z cos(2π/7) + 1)(z 2 − 2z cos(4π/7) + 1)(z 2 − 2z cos(6π/7) + 1)

14.3. Prove that if fc (z) = z 2 + c has an attracting periodic point, then c ∈ M, the Mandelbrot set. (Hint: Use any appropriate results from Section 14.5).

Answer. If z0 is an attracting fixed point of fc , then it must attract a critical point.

The only critical point of fc is z = 0. It follows that z = 0 has a bounded orbit under fc , or c ∈ M, the Mandelbrot set.

14.4. Show that p(z) = az + b, a ̸= 1 on C, is conjugate to q(z) = az.

Answers. Set h(z) = z − b/(1 − a), then we can check that h ◦ p(z) = q ◦ h(z) for all z.

14.5. Let fc (z) = z 2 + c. Find the values of c so that z = i is a period-2 point. Find the fixed points in each case and determine their stability. Does c ∈ M? (Hint: It is useful to use Mathematica to find the square roots of complex numbers). √ Answer. Solving fc (z) = z gives z 2 − z + c = 0, or z = (1 ± 1 − 4c)/2 as the 2 fixed points. Now fc2 (z) = (z + c)2 + c, so setting √ z = i gives (−1 + c)2 + c = i, √ c2 − c + 1 − i = 0, so c = (1 ± 1 − 4(1 − i))/2 = (1 ± 4i − 3)/2 = (1 ± (1 + 2i))/2 = 1 + i, or −i. We can check that f−i (i) = −1 − i and f−i (−1 − i) = i, and similarly f1+i (i) = −1 + 1 + i = i, so z = i is actually a fixed point of f1+i . We can now check√that the 2-cycle is repelling. Setting c = −i now gives the fixed points as (1 ± 1 + 4i)/2.

3 We can check that 0 is eventually periodic for f−i (f−i (0) = i), so i ∈ M. z−i gives a conjugacy between the Newton z+i 2 map Nf1 , where f1 (z) = z + 1, and the function f0 (z) = z 2 . Deduce the Julia set of Nf1 , and that it is chaotic on its Julia set. (Hint: Find H −1 (z) and set z = eiθ ).

14.6 Show that the function H(z) =

z2 − 1 , then Answer. Set Nf1 (z) = N1 (z) = 2z ( ) ( )2 ( 2 ) ( )2 z−i z −1 z−i z−i f0 ◦ H(z) = f0 = , H ◦ N1 (z) = H = , z+i z+i 2z z+i giving a conjugacy. sin(θ) −iz − i , and setting z = eiθ gives H −1 (eiθ ) = . It Now H −1 (z) = z−1 cos(θ) − 1 follows that the Julia set of N1 is the image of S1 under H −1 , which we can now see is the real axis. p2 (z) − z 14.7 Is it possible for a polynomial p(z) to satisfy = p(z) + z? If so, find all p(z) − z such polynomials and their period-2 points. What does this say about the periodic points of the polynomial? (b) What about

p2 (z) − z = p(z) − z? p(z) − z

p2 (z) − z = p(z) + z, where p has degree n. Then p2 (z) − z = p(z) − z [p(z)]2 − z 2 , where the right hand side had degree 2n, whilst the left hand side has degree n2 , so we must have n = 2.

Set p(z) = az 2 +bz+c, then we can see that a = 1 and z 2 (1+b)+z(b2 −1)c+bc = 0, 2 so b = −1 and c is arbitrary. Thus √ p(z) = z − z +√c. To find the fixed points we see 2 that z − 2z + c = 0, z = (2 ± 4 − 4c)/2 √ = 1 ± 1 − c. The period-2 points must 2 satisfy p(z) + z = 0, so z + c = 0 or z = ± −c.

Answer. Suppose that

(b) In this case p2 (z)−z = (p(z)−z)2 . Again p(z) has to be a quadratic. On equating coefficients, we see there is no solution.

Section B 1. Consider the family of functions fλ : R → R, fλ (x) = x3 − λx, for a parameter λ ∈ R. (a) Find all the fixed points and determine their nature and where they are created as λ varies. (b) Find where a two cycle is created (you may use a computer algebra facility), and give the graph of where this happens. Determine the stability of the hyperbolic 2-cycles. (c) Use a computer algebra facility to find an approximate value of λ where the 3-cycle is created. Give the graph of this situation.

Answer. Set fλ (x) = x3 − λx, λ ∈ R.

√ (a) fλ (x) = x, when x3 − λx = x, so x = 0 or x2 = 1 + λ, x = ± 1 + λ. This is created, giving 2 new fixed points, when λ > −1.

Now fλ′ (x) = 3x2 − λ, so fλ′ (0) = −λ, so x = 0 is stable when |λ| < 1. When λ = −1, f ′ (0) = 1, fλ′′ (x) = 6x, fλ′′ (0) = 0, fλ′′′ (0) = 6 > 0, so x = 0 is unstable.

When λ = 1, fλ′ (0) = −1, and Sf (0) = −f ′′′ (0) − 3/2[f ′′ (0)]2 = 6 − 0 < 0, so x = 0 is stable. For all other √ values of λ, x = 0 is unstable.

Furthermore, f (± 1 + λ) = 3(1 + λ) − 1 = 3λ + 2. However, |3λ + 2| < 1 if and only if −2 < λ < −1, and this cannot happen since λ > −1 always for these fixed points to exist. Thus these fixed points are always repelling.

2 (b) Using Mathematica, we can see that fλ2 (x) − x = (x3 − λx − x)(x − λ +√1)(x4 − √ λx2 + 1. Ignoring the fixed points we see that there is a 2-cycle {− λ − 1, λ − 1}, √ √ created when λ > 1, and another 2-cycle given by x = ± (λ + λ2 − 4)/2, created when λ > 2. We have a hyperbolic 2-cycle {−1, 1} created when λ = 2, which is semi-stable.

2. Let f (x) = ax3 + bx + c where a and b satisfy a/b > 0. Denote by Nf the corresponding Newton function. (a) Show that Nf has a unique fixed point. (b) Show that Nf cannot have any period-2 points. (c) Why does it follow that Nf has no points of period n, n > 2? f (x) = ax3 + bx + c,

where a/b > 0, so a and b are both greater than zero, or are both less than zero.

Answer. (a) Thus f ′ (x) = 3ax2 + b is either strictly positive (f is increasing) or strictly negative (f is decreasing), so in either case there is exactly one point where f (x) = 0 (f is a cubic). It follows that the Newton function must have exactly one fixed point. ax3 + bx + c 2ax3 − c = . Suppose that {x0 , x1 } is a 2-cycle 3ax2 + b 3ax2 + b for Nf (so x0 ̸= x1 ), then (b) Now Nf (x) = x −

2ax30 − c 2ax31 − c = x , and = x0 , 1 3ax20 + b 3ax21 + b so that 2ax30 − c = 3ax20 x1 + bx1 , and 2ax31 − c = 3ax21 x0 + bx0 .

Subtracting gives 2a(x30 − x31 ) = 3ax20 x1 − 3ax21 x0 + b(x1 − x0 ), 2a(x20 + x0 x1 + x21 ) = 3ax0 x1 − b, 2x20 − x0 x1 + 2x21 = −b/a, 2x20 + 4x0 x1 + 2x21 = −b/a + 5x0 x1 , 2(x0 + x1 )2 = −b/a + 5x0 x1 ≥ 0, so x0 x1 ≥ b/5a.

In addition 2x20 −4x0 x1 +2x21 = −b/a−3x0 x1 , so 2(x0 −x1 )2 = −b/a−3x0 x1 > 0, x0 x1 < −b/3a.

It follows that b/5a ≤ x0 x1 < −b/3a, and this is impossible for a/b > 0. (c) Since the denominator of Nf is never zero, Nf is a continuous function, having no period-2 points. It follows from Sharkovsky’s Theorem that Nf cannot have points of any period n > 1.

3. This question is related to Exercise 4.2 # 8, parts (a), (b) and (c). (a) Show that the function f (x) = −1/(x + 1) has the property that f 3 (x) = x for all x ̸= −1, x ̸= 0. (b) Let f : R → R be a function defined on a set I, with f 3 (x) = x for all x ∈ I? Set g(x) = f 2 (x). Show that g 3 (x) = x for all x ∈ I. Deduce a different function from the one in (a) with this property. (c) In general, show that such a function cannot have a 2-cycle. (d) Deduce that a function f : R → R with the property f 3 (x) = x cannot be continuous (i.e., there must be some points of discontinuity for f , where we may not have f 3 (x) = x).

(e) Show that f is one-to-one. Need f be onto? Show that f is strictly decreasing and investigate any symmetry. What if f : [−∞, ∞] → [−∞, ∞] - consider the example in (a) above? (f) If f ′ (x) exists for all x ∈ I, show that the 3-cycles are non-hyperbolic, (assume that f (x) is not the function y = x) . ax + b satisfies f 3 (x) = x, show that if f (x) is not the cx + d function y = x, and a ̸= d then

(g) Suppose that f (x) =

a2 + bc + ad + d2 = 0.

(i) Show that if ad − bc > 0, then such a function cannot have any fixed points. (ii) Deduce two other functions with the property that f 3 (x) = x wherever defined, one having fixed points and one with no fixed points.

Answer. (a) If f (x) = −1/(x+1), then f 2 (x) = −1/(−1/(x+1)+1) = −(x+1)/x = −1 − 1/x, and so f 3 (x) = −1 + (x + 1) = x for all x ̸= 0, −1. (b) If f 3 (x) = x for all x ∈ I, then if g(x) = f 2 (x), g 3 (x) = f 6 (x) = f 3 (x) = x. Thus we see that the function g(x) = −1 − 1/x also has the property that g 3 (x) = x for all x ̸= 0. − 1. (c) and (d) If f (x) = x for all x ∈ R, then clearly f 3 (x) = x everywhere, a continuous function. Suppose that f : R → R with f 3 (x) = x for all x, where f is not the identity function. There exists points c with f (c) ̸= c. Suppose there is a 2-cycle {c0 , c1 } with c0 ̸= c1 , then f 3 (c0 ) = c1 , leading to a contradiction. Since there are points of period-3, the continuity of f and Sharkovsky’s Theorem imply that f has a 2-cycle, giving a contradiction. It follows that f cannot be continuous. We see from the example in (a), that f need not have a fixed point. (e) If f (x) = f (y), then f 3 (x) = f 3 (y), so x = y and f is one-to-one. If z is in the range of f , then f (f 2 (z)) = z, so if f 3 (x) = x for all x ∈ R, then by setting x = f 2 (z) we see that f would be onto. However there is a difficulty with this argument because of the lack of continuity of f , such a function cannot exist (besides y = x). If we allow f to be a function f : [−∞, ∞] → [−∞, ∞], we can make this argument work (for example, in (a) we can think of {0, −1, ∞} as a 3-cycle). (f) On the other hand, suppose that c is a fixed point for f . Set g(x) = f 3 (x), then g ′ (x) = f ′ (f 2 (x))f ′ (f (x))f ′ (x) = 1, so g ′ (c) = [f ′ (c)]3 = 1 and f ′ (c) = 1, so c is non-hyperbolic of the appropriate type. Clearly any 3-cycle is hyperbolic. (a2 + bc + ad + d2 )(b − x(−a + d + cx)) (g) We have f (x) − x = 2 = 0. This is a cx + d2 (d + cx) + bc(2d + cx) + ac(b + dx) only possible if b(a2 + bc + ad + d2 ) = 0, c(a2 + bc + ad + d2 ) = 0 and (a − d)(a2 + 3

bc + ad + d2 ) = 0, so if a ̸= d, then (a2 + bc + ad + d2 ) = 0 (the condition a ̸= d rules out the function y = x. x+3 We can check that the function f (x) = has f 3 (x) = x for all x in the −x + 1 domain. f has no fixed points (here a = 1, b = 3, c = −1 and d = 1). ax + b = x gives x = (a − d ± cx + d √ (a − d)2 + 4bc)/2c, so in order for fixed points, we need (a−d)2 +4bc ≥ 0. However, Now suppose that ad − bc > 0. Solving

0 = a2 + bc + ad + d2 = (a − d)2 + 3ad + bc = (a − d)2 + 4bc + 3(ad − bc), so (a − d)2 + 4bc = −3(ad − bc). It follows that if ad − bc > 0, there can be no fixed points.

4 (check). If f : D → D has the property |f ′ (z)| < 1 for all z ∈ D \ S1 , show that |f (z)−f (w)| < |z −w| for all z, w ∈ D. Does it follow that f is a contraction mapping (consider some examples)? Show that f has a unique fixed point α. Why can we deduce that Bf (α) = D? (D = {z ∈ C : |z| ≤ 1}).

Answer. az + b 5. There is a correspondence between functions of the form f (z) = and cz + d [ ] a b matrices , when a, b, c, d ∈ R, ad − bc ̸= 0. To be more precise, show that c d there is a group homomorphism Φ : GL2 (R) → F , where GL2 (R) is the multiplicative group of 2 × 2 matrices having real entries and non-zero determinant, and F is az + b the group (under composition) of linear fractional transformations f (z) = , cz + d ( ) a b (ad − bc ̸= 0), where Φ is defined by Φ = f (z). Is the map Φ one-to-one or c d onto? ( ) a b Answer. The map Φ is defined by Φ = f (z) - clearly an onto function. c d a2 z + b 2 a1 z + b 1 , and f2 (z) = , then Set f1 (z) = c1 z + d 1 c2 z + d2 [( ) ( )] [( )] a1 b 1 a2 b 2 a1 a2 + b 1 c 2 a1 b 2 + b 1 d 2 Φ · =Φ c1 d1 c2 d 2 c1 a2 + c2 d1 b2 c1 + d1 d2

=

(a1 a2 + b1 c2 )z + a1 b2 + d2 b1 = f1 (f2 (z)), (a2 c1 + c2 d1 )z + c1 b2 + d1 d2

so the homomorphism condition is satisfied. Φ is not one-to-one, for example ( Φ

1 0 1 1

)

( =Φ

2 0 2 2

) .

6. (a) Show that fc : R → R, fc (x) = x2 + c has a 3-cycle for c ≤ −7/4. (b) Deduce that Lµ : R → R, Lµ (x) = µx(1 − x) has a 3-cycle for µ ≤ 1 −

√ 8.

2µ − µ2 (see Section 4 7.3). √ Setting c = −7/4 and solving√gives µ2 − 2µ − 7 = 0, or µ = 1 ± 8, etc. This also show that Lµ , when µ =√1 − 8 is conjugate to this fc , so has a 3-cycle. There will be a 3-cycle for µ ≤ 1 − 8 (using the conjugacy).

Answer. We have seen that fc and Lµ are conjugate when c =

7.∗ Use the following steps to show that the Koch curve is continuous, i.e., there is a continuous function f : [0, 1] → R2 , whose range is the top part of the Koch curve (see Section 10.1): (a) Denote by fn : [0, 1] → R2 the piecewise defined function which is linear on each of the 4n segments defined at the nth stage of the construction of the Koch curve.

Show that |fn (t) − fn+1 (t)| ≤ 3−n for all t ∈ [0, 1]. (b) Deduce that (fn (t)) is a Cauchy sequence for each t ∈ [0, 1], so that f (t) = limn→∞ fn (t) exists for all t ∈ [0, 1]. (c) Show that[ convergence of the sequence of functions (fn ) to f is uniform on [0, 1] ] (i.e., limn→∞ sup0≤t≤1 |fn (t) − f (t)| = 0). (d) Now complete the proof using the fact that the uniform limit of a sequence of continuous functions is a continuous function. That f is not differentiable anywhere can be deduced from the fact that the length of the graph of f is infinite on any arbitrarily small interval.

Answer. (a) fn and fn+1 agree on all points in the nth stage of the construction that do not change when we construct the (n + 1)th stage, so f and fn+1 only differ on the newly constructed intervals. The largest distance between points on these new intervals is 3−n . (b) Use the triangle inequality in the usual way to see that (fn (t)) is a Cauchy sequence for each t, so converges to some f (t). (c) Now we can see that |f (t) − fn (t)| ≤ 2 · 3−n for each t ∈ [0, 1], so we have the uniform convergence of fn to f .

(d) Since each fn is continuous, it follows that f is continuous.

ANSWERS TO CHAOTIC DYNAMICS TEXT: PART 2, CHAPTERS 11-14 Geoffrey R. Goodson Section 11.4 1. If r ∈ (0, 1) has a finite binary representation, then we can write r = ·a1 a2 . . . an , for some ai ∈ {0, 1}. Then r = a1 /2 + a2 /22 + · · · + an /2n = (2n−1 a1 + 2n−2 a2 + · · · + an )/2n = k/2n , for some k, n ∈ Z+ .

Conversely, if r ∈ (0, 1) where r = k/2n , k, n ∈ Z+ , then we can write the binary expansion of k as k = an + 2an−1 + · · · + 2n−2 a2 + 2n−1 an , where ai ∈ {0, 1}, so that r = (an + 2an−1 + · · · + 2n−2 a2 + 2n−1 a1 )/2n = a1 /2 + a2 /22 + · · · + an /2n = ·a1 a2 . . . an , and the result follows.

2. We have 5×63/(212 −1) = 5×32 ×7/(5×32 ×7×13) = 1/13. Our results tell us that we have a repeating binary representation having period 12. Also, ·000100111011 = (1/24 +1/27 +1/28 +1/29 +1/211 +1/212 )(1/(1−1/212 )) = 32 ×5×7/(212 −1) = 1/13.

3. Suppose that r ∈ (0, 1) has an infinite repeating binary expansion, say r = ·a1 a2 . . . an a1 a2 . . ., then r = a1 /2 + a2 /22 + · · · + an /2n + a1 /2n+1 + · · · = (2n−1 a1 + 2n−2 a2 +· · ·+an )(1/2n +1/22n +· · · ) = (2n−1 a1 +2n−2 a2 +· · ·+an )(1/2n /(1−1/2n )) = (2n−1 a1 + 2n−2 a2 + · · · + an ))/(2n − 1) = t/q, where t, q ∈ Z+ and q is odd.

Conversely suppose r ∈ (0, 1) where r = t/q, t, q ∈ Z+ , and q is odd. We must show that r has a repeating binary expansion. It suffices to prove this for r = 1/q, where q is odd. Suppose that ai ∈ {0, 1} with ) 1 a1 a2 an an+1 a1 a2 an 1 ( an+1 an+2 = + 2 + · · · + n + n+1 + · · · = + 2 +···+ n + n + 2 + ··· , q 2 2 2 2 2 2 2 2 2 2 where n is the first time an+1 = a1 Then

(a ) an+2 n+1 2n − q(2n−1 a1 + · · · + an ) = q + 2 + ··· .

2 2 (a ) an+2 n+1 + 2 + · · · is an integer, or It follows that q 2 2 an+1 an+2 m + 2 + ··· = , 2 2 q 1

2

for some m ∈ Z+ . Since an+1 = a1 , we must have m = 1, and the result follows.

4. Set f (x) = ax2 +bx+c and g(x) = x2 −α, α ̸= 0. Then Nf (x) = x−

ax2 + bx + c = 2ax + b

ax2 − c x2 − α x2 + α x α , and Ng (x) = x − = = + . Set h(x) = 2ax + b, then 2ax + b 2x 2x 2 2x ( 2 ) ( 2 ) ax − c ax − c 2a2 x2 − 2ac + 2abx + b2 h ◦ Nf (x) = h = 2a +b= , 2ax + b 2ax + b 2ax + b and Ng ◦ h(x) = Ng (2ax + b) = =

2ax + b α + 2 2(2ax + b)

(2ax + b)2 + α 2a2 x2 + 2abx + b2 − 2ac = , 2(2ax + b) 2ax + b

and these are equal.

Now set f (x) = (x+1)2 and g(x) = x2 . In this case α = 0, and b2 −4ac = 4−4 = 0, so by the above, Nf and Ng are conjugate. However, f and g are not conjugate since g has two fixed points, but f has no fixed points. x3 + 2x2 + x/5 + 2/5 2x3 + 2x2 − 2/5 = , 3x2 + 4x + 1/5 3x2 + 4x + 1/5 2/1000 + 2/100 − 2/5 2 + 20 − 400 378 and so Nf (1/10) = = =− = −3/5. Also, 3/100 + 4/10 + 1/5 30 + 400 + 200 630 −54 + 90 − 50 14 1 −54/125 + 18/25 − 2/5 = = = , so we have a Nf (−3/5) = 27/25 − 12/5 + 1/5 135 − 300 + 25 140 10 2-cycle {1/10, −3/5}.

5. If f (x) = (x+2)(x2 +1/5), then Nf (x) = x−

x2 − 1 x 1 = + . Note that x = ±1 are 2x 2 2x x0 1 (super-attracting) fixed points for Ng . Let x0 ∈ (1, ∞), then 1 < Ng (x0 ) = + < 2 2x0 n x0 . Continuing this argument, we see that the sequence (Ng (x0 )) is decreasing and bounded below by 1, so must converge to the fixed point x = 1. We use a similar argument when 0 < x0 < 1 to obtain an increasing sequence bounded above. We deduce in this way that the basin of attraction of 1 is (0, ∞). Similarly, the basin of attraction of x = −1 is (−∞, 0).

6. Let g(x) = x2 − 1, then Ng (x) = x −

3

√ x α 7. Let g(x) = x2 − α, where α > 0, then Ng (x) = + has fixed points ± α. We √ 2 2x √ follow the steps of Exercise # 6, letting x0 ∈ ( α, ∞), then as before α < Ng (x0 ) < √ √ x0 , and continue as previously (using x20 − 2x0 α + α = (x0 − α)2 ≥ 0).

8. Suppose that f and g are linearly conjugate via h(x) = ax + b and Nf and Ng are also linearly conjugate via h. Say h(g(x)) = f (h(x)), then h′ (g(x))g ′ (x) = f ′ (h(x))h′ (x), or ag ′ (x) = f ′ (ax + b) · a, and g ′ (x) = f ′ (ax + b). Now f (h(x)) h(g(x)) ag(x) + b Nf (h(x)) = h(x) − ′ = h(x) − ′ = ax + b − . f (h(x)) g (x) g ′ (x) Also ( ) ( ) g(x) g(x) ag(x) h ◦ Ng (x) = h x − ′ =a x− ′ + b = ax + b − ′ . g (x) g (x) g (x) It follows that

ag(x) + b ag(x) = ′ , or b = 0. ′ g (x) g (x)

4

Section 12.3 1. f : [a, b] → [a, b] is a continuous function having no 2-cycles. Let c ∈ [a, b] with f (c) > c (which we are assuming to exist), then we cannot have f 2 (c) < c < f (c) as this implies the existence of a 2-cycle. It follows that f 2 (c) > c.

2. f : [0, 1] → [0, 1] is a continuous function which is also onto, so has at least one fixed point. Since f is onto, there exists c0 , c1 ∈ [0, 1] with f (c0 ) = 0 and f (c1 ) = 1.

If g(x) = f (x) − x, then g(c0 ) = f (c0 ) − c0 = −c0 ≤ 0, g(c1 ) = f (c1 ) − c1 = 1 − c1 ≥ 0, so there is a fixed point p for f lying between c0 and c1 .

Case 1. Suppose that 0 ≤ c0 < c1 ≤ 1. Set g(x) = f (x) − x, then g(0) = f (0) − 0 = f (0) ≥ 0, and g(1) = f (1) − 1 ≤ 0. It follows that there exists q1 ∈ [0, c0 ) with f (q1 ) = q1 , p ∈ (c0 , c1 ) with f (p) = p and q2 ∈ [c1 , 1) with f (q2 ) = q2 , so the result follows in this case.

Case 2. Suppose that 0 ≤ c1 < c0 ≤ 1 and set h(x) = f 2 (x) − x, then as before, we have h(0) = f 2 (0) − 0 = f 2 (0) ≥ 0 and h(1) = f 2 (1) − 1 ≤ 0 so we now consider: Case 2a. Suppose that 0 ≤ c1 < c0 ≤ 1 and f (1) < c1 , then as above there is a fixed point p ∈ (c1 , c0 ). Also, h(0) ≥ 0 and h(c1 ) = f 2 (c1 ) − c1 = f (1) − c1 < 0, so there exists q ∈ [0, c1 ) with h(q) = 0, or f 2 (q) = q and our result follows Case 2b. Suppose that 0 ≤ c1 < c0 ≤ 1 and f (1) > c1 . In this case we can find α ∈ [c0 , 1], with f (α) = c1 (using IVT, since c1 < c0 < 1). We have h(α) = f 2 (α) − α = f (c1 ) − α = 1 − α > 0. Since h(1) ≤ 0, there exists q ∈ (α, 1] with f 2 (q) = q, and the result follows.

The function f : [0, 1] → [0, 1], f (x) = x2 is continuous and onto with two fixed points, but no period-2 points.

3. f : [a, b] → [a, b] is a continuous function having no points of period two with a unique fixed point x = c. By Coppel’s Theorem, limn→∞ f n (x) exists for all x ∈ [a, b] and is therefore a fixed point for each such x. It follows that limn→∞ f n (x) = c for all x ∈ [a, b].

5

4. f m (x0 ) = x0 . Suppose that f k (x0 ) = x0 , where k is the least such positive integer.

By the Division Algorithm, we may write m = ka + r where a and r are integers with 0 ≤ r < k. Then f m (x0 ) = f ka+r (x0 ) = f r (f ka (x0 )) = f r (x0 ), so that f r (x0 ) = x0 , which is a contradiction, so we must have r = 0.

5. f has a 3-cycle: {a, b, c} with f (a) = b, f (b) = c, f (c) = a. Set g(x) = f 2 (x) − x, then g(a) = f 2 (a) − a = c − a > 0, and g(b) = f 2 (b) − b = a − b < 0. It follows that there exists x ∈ (a, b) with g(x) = 0, or f 2 (x) = x. However, it may be that f (x) = x. Set p = max{x ∈ (a, b) : f 2 (x) = x}.

Then (p, c) = (p, f (b)) ⊆ f (p, b), so there exists q ∈ (p, b) with f (q) = b. Thus g(b) = f 2 (b) − b = a − c < 0, g(q) = f 2 (q) − q = f (b) − q = c − q > 0.

It follows that there exists r ∈ (q, b) with f 2 (r) = r, contradicting p being the maximum such, since r > p. Therefore we must have f 2 (p) = p and f (p) ̸= p.

6. We give an example of a continuous function f : [a, b] → [a, b] for which there exists c ∈ [a, b] with c < f 2 (c) < f (c), but no period-2 point.

Set f (x) = 3x(1 − x), x ∈ [0, 1], a map having two fixed points but no period-2 points. Set c = ·6, the f (c) = 3 × 6/10 × 4/10 = ·72, f (·72) = 3 × 72/100 × 28/100 = ·6408, and ·6 < ·6408 < ·72, or c < f 2 (c) < f (c).

7(a). Suppose that f 2 has a period-m point, then f 2m (x0 ) = x0 , for some x0 . We use kn 2m Lemma 12.2.3(c), setting n = 2 and k = m, then f has a point of period = , s s kn where s|n (s divides n). It follows that s = 1, or s = 2 and (s, k) = 1, so = m, or s 2m. (b) Suppose that f has period-m points, where m is odd. Set n = 2k in Lemma 12.2.3(b), k ≥ 1, then x0 is a period point of f n with period m/d where d = (m, 2k ) = k 1, so f 2 has period-m points.

6

(c) Suppose that f has a period-(2k · m) point with m odd, k ≥ 1. Set n = 2k 2k m in Lemma 12.2.3(b), then f n has period points where d = (2k m, 2k ) = 2k , so d 2k m 2k m 2k = k = m, so f has points of period m. d 2 8.(a) f is a C 1 -function, with fixed point p for which f ′ (p) < −1. We wish to show that f has a 2-cycle. Now f (p) = p. From the continuity of f ′ (x) there exists δ > 0 for which f ′ (x) < −1, for x ∈ (p − δ, p + δ). Let c ∈ (p − δ, p), then from the Mean Value Theorem, there exist α ∈ (c, p) with f ′ (α) =

f (p) − f (c) p − f (c) < −1, or < −1. p−c p−c

It follows that f (c) > p, so f (c) > c. In a similar manner, there exists β ∈ (c, f (c)) with f ′ (β) =

f 2 (c) − f (c) < −1, f 2 (c) − f (c) < −f (c) + c, f (c) − c

or f 2 (c) < c.

It follows that f 2 (c) < c < f (c), and an application of Lemma 12.1.3 gives the result. (b) Let Lµ (x) = µx(1−x) for µ > 3. x = 1−1/µ is a fixed point, and L′µ (x) = µ−2µx, so that L′µ (1 − 1/µ) = µ − 2µ(1 − 1/µ) = 2 − µ < −1. The result follows from (a). (c) Let fλ (x) = x3 − λx. We have a fixed point at x = 0. fλ′ (x) = 3x2 − λ, so fλ′ (0) = −λ < −1, for λ > 0, so a 2-cycle is created. Also, we have f2′ (2) = 3x2 − 2x = −1 when

9. (a) Eλ (x) = λex , then E−e (−1) = −ee−1 = −1, so x = −1 is a fixed point. Also, ′ (−1) = −ee−1 = −1, so from the above, we have a period doubling Eλ′ (x) = λex , E−e bifurcation. (b) If fµ (x) = µxe−x , we set µ = e2 and p = 2, so that fe2 (x) = e2 xe−x , and fe2 (2) = e2 2e−2 = 2, so x = 2 is a fixed point. Now fµ′ (x) = µ(e−x − xe−x ), so fµ′ (2) = µ(e−2 − 2e−2 ) and fe′2 (2) = −e2 e−2 = −1.

7

10. (a) We give an alternate proof to Lemmas 12.1.3 and 12.1.4 for a continuous function f on a compact interval I = [α, β]: We assume there exists a, b ∈ [α, β] with f (b) < a < b ≤ f (a). Set g(x) = f (x)−x, then g(a) = f (a) − a > 0 and g(b) = f (b) − b < 0, so f has a fixed point z ∈ [a, b].

Now we have [z, f (a)] ⊆ f ([a, z]), so there exists v ∈ [a, z] with f (v) = b > z. Set u = α if f (x) > z for all x ∈ [α, v], otherwise set u = max{x ∈ [α, v] : f (x) = z} (so if u < x < v, we have f (x) > z). If h(x) = f 2 (x) − x, then h(u) = f 2 (u) − u = f (z) − u = z − u > 0, and h(v) = f 2 (v) − v = f (b) − v < 0, so there exists y ∈ [u, v] with f 2 (y) = y. Since f (x) > z for all x ∈ [u, v], we must have f (y) ̸= y, so y is a period-2 point. (b) Suppose there exists c ∈ I and n ≥ 2 with f n (c) < c < f (c).

Set A = {f k (c) : 0 ≤ k ≤ n − 1}, a = max{x ∈ A : c ≤ x and f (x) > x}, and choose b ∈ {x ∈ A : a < x ≤ f (a)} with f (b) < a. We first show that b exists and hence that f (b) < a < b ≤ f (a). This will then imply the existence of a fixed point and a period-2 point (from (a) above). Now a ̸= f n−1 (c) since in this case f (a) = f n (c) < c ≤ a. The set {x ∈ A : a < x ≤ f (a)} is non-empty since firstly X = {x ∈ A : a < x} ̸= ∅ (since a < f (a), so f (a) ∈ X), and d = min{x ∈ A : a < x} has the property: either d = f (a) or d < f (a).

We conclude that there exists b ∈ {x ∈ A : a < x ≤ f (a)} with f (b) < a, for if not, for all x ∈ A with a < x ≤ f (a), we have f (x) > a. This is impossible since d cannot have this property as f (d) < d, so f (d) ≤ a.

Now if f (d) = a, then cannot have f (a) = d, so must have f (a) = e > d, f (e) < e, so either f (e) < a or f (e) = d, but this gives a 3-cycle, completing the proof.

11. We are looking at the truncated tent map Th (x) = min{h, T (x)}, where T is the usual tent map.

Claim 1. If O = {x1 , x2 , . . . xm }, an m-cycle for T and hO = xm , we set a = min{hO }.

We show that a is the largest member of Q, a unique m-cycle for the the truncated tent map Ta , say Q = {z1 , z2 , . . . zm }, where a = zm .

The minimum is taken over a finite set, so a exists and a = zm for some m-cycle Q = {z1 , z2 , . . . zm } (in increasing order). Since each member of Q is less than or equal to a = zm , the orbit of z1 under Ta is the same as the orbit of z1 under T . This is not true for any other m-cycle of T since there are members having value greater

8

than a. It follows that Ta has exactly one m-cycle and zm = a (any other m-cycle with this property would coincide with Q).

Claim 2. We must have T (a) = T (zm ) = z1 because T (a) = 2(1 − a), and any zk < 2(1 − a) must be the image of some zk > zm , giving a contradiction, or else it is the image of some zp < zk and continuing this argument leads to a contradiction.

Claim 3. It now directly follows that every member of Q lies in the interval [T (a), a].

We claim that every periodic orbit of Ta enters [T (a), a] and is therefore contained in [T (a), a]. If we have another periodic orbit, say {a1 , a2 , . . . , ak } (written in ascending order) must have ak < a, for otherwise ak would not be in the range of Ta , contradicting ak being on a k-cycle. As before, T (ak ) = a1 > T (a), so {a1 , a2 , . . . , ak } ⊂ [T (a), a].

Claim 4. Suppose k ◃ m and Ta has a k-cycle R = {a1 , a2 , . . . , ak } (recall that 3 ◃ m for all m), then R is contained in [T (a), a] = [T (zm ), zm ]. By Sharkovsky’s Theorem, Ta has an m-cycle contained in [T (ak ), ak ], which therefore must be different to Q (defined above), a contradiction since the m-cycle is unique, so R cannot exist. This proves the first part of the converse of Sharkovsky’s Theorem.

Claim 5. Choose a = a2n , so that Ta2n has a unique 2n -cycle, but no k-cycle with k ◃ 2n . Set a = limn→∞ a2n . Clearly (a2n ) is an increasing sequence bounded above, so this limit exists. Since a > a2n for all n, we see that Ta has 2n -cycles for all n ≥ 0.

To see that Ta has no m-cycle (m ̸= 2k for any k), with m ◃ 2n for all n, suppose to the contrary that such an m exists with m-cycle {z1 , z2 , . . . , zm } (in increasing order). Then [T (zm ), zm ] ⊂ [T (a), a] (using the fact that the range of Ta contains the m-cycle), and in particular zm < a (if zm = a, using Sharkovsky’s Theorem, we could choose a different cycle with this property). It follows that there exists n ∈ Z+ with zm < a2n < a. This implies that Ta2n has a unique 2n -cycle, no 2k -cycles for k > n and an m-cycle, contradicting Sharkovsky’s Theorem. It follows that no such m-cycle exists.

9

Section 13.1 1. We determine the dynamics of the dynamical systems whose matrices are: [ ] 2 0 (i) . Since both eigenvalues are real and greater than one, x = 0 is a repelling 0 3 fixed point. [ ] 1 0 (ii) . Both eigenvalues have absolute value equal to one, so we have the 0 −1 non-hyperbolic case. We see that in this case the matrix acts as a reflection in the x-axis. [ ] .5 0 (iii) . Both eigenvalues are real, lying between zero and one, so the fixed 0 .3 point x = 0 is an attractor. ] [ 1/2 0 . The fixed point x = 0 is a saddle, with some trajectories close to (iv) 0 2 zero moving away from zero, and some moving toward zero.

2. We diagonalize each of the following matrices A, and find An , and we deduce the dynamics linear transformation. [ of the induced ] 1 1/2 (i) . The characteristic equation is |A − λI| = λ2 − 9λ/4 − 7/8 = 3/4 5/4 [ ] [ ] 1/2 0 1 2 (λ − 1/2)(λ − 7/4), so Λ = , and we can check that P = , and 0 7/4 −1 3 [ ] [ ][ ][ ] 1/5 2/5 1 2 1/2n 0 1/5 2/5 −1 n n −1 P = . Then A = P Λ P = . −1/5 3/5 −1 3 0 (7/4)n −1/5 3/5 We see that we have a saddle at x = 0. ] [ −5/3 3 . |A − λI| = λ2 − 5λ/3 + 4/9 = (λ − 1/3)(λ − 4/3), so Λ = (ii) −2 10/3 [ ] [ ] [ ] 1/3 0 3 1 1 −1 −1 , and P = , P = . Then An = P Λn P −1 = 0 4/3 2 1 −2 3 ] [ ][ ][ 3 1 1/3n 0 1 −1 . Again we have a saddle at x = 0.

2 1 0 (4/3)n −2 3

10

[

] [ ] −5 −7 2 0 2 (iii) . |A−λI| = λ −λ/2−3 = (λ−2)(λ+3/2), so Λ = , 7/2 11/2 0 −3/2 [ ] [ ][ n ][ ] −2 1 −2 1 2 0 −1 1 and P = . An = P Λn P −1 = . n 1 −1 1 −1 0 (−3/2) 1 −2 In this case, x = 0 is a repelling fixed point. [

] a b 3. We write each matrix in the form QΛQ−1 (where Λ = , for some −b a a, b ∈ R), and hence deduce the dynamics of the corresponding linear map. [ ] −7 15 (i) A = , so |A−λI| = (−7−λ)(11−λ)+90 = λ2 −4λ+13, so solving for −6 11 the eigenvalues gives λ = 2±3i, and [ we set a] = 2 and b = 3. If α is an eigenvector, [ ]we 2 3 3 −1 can take α = (3 − i, 2)t then Λ = , and Q = [Re(α), Im(α)] = . −3 2 2 0 [ ] 1/6 2/3 (ii) B = . |B − λI| = (1/6 − λ)(5/6 − λ) + 2/9 = λ2 − λ + 13/36, −1/3 5/6 [ ] 1/2 1/3 giving eigenvalues 1/2 ± i/3, and eigenvector α = (1 + i, 1), so Λ = , −1/3 1/2 [ ] 1 1 and Q = [Re(α), Im(α)] = .

1 0 [

a b −b a

]

, for a, b ∈ C. |Ia,b − λI| = ] [ a + ib 0 2 2 , (a − λ) + b = 0 when a − λ = ±ib, so λ = a ± ib. Set Λ = 0 a − ib t t then [the corresponding ] [eigenvectors can ] be chosen to be (i, 1) and (−i, 1) , so i −i −i/2 1/2 P = , P −1 = with Ia,b = P ΛP −1 .

1 1 i/2 −1/2 [ ][ ] [ ] a b c d ac − bd ad + bc (b) Ia,b Ic,d = = = Ic,d Ia,b . −b a −d c −ad − bc ac − bd [ ] i 1 2 2 (c) det(Ia,b ) = a + b ̸= 0 if a ̸= 0 or b ̸= 0. If A = , then det(A) = −1 i i2 + 12 = 0.

4. (a) We diagonalize the matrix Ia,b =

11

5. (a) Let v1 ̸= 0 be an eigenvector of A corresponding to the eigenvalue λ. Choose v2 ̸= 0 with (A − λI)v2 = v1 and set P = [v1 v2 ]. Then AP = A [v1 v2 ] = [Av1 Av2 ], and ] [ ] [ [ ][ ] λ 1 p11 p12 λ 1 λp11 p11 + λp12 P = = , 0 λ p21 p22 0 λ λp21 p21 + λp22 so AP = P Λ. P is invertible since v1 and v2 are linearly dependent. [ n ] λ nλn−1 n (b) We can now check that A = P P −1 .

0 λn (c) If F (x) = Ax, we determine the long term behavior of F . If |λ| > 1 then we have a repelling fixed point, and if |λ| < 1 it is attracting. When |λ| = 1, we have the non-hyperbolic case. ] 5 −4 6. A = , then |A − λI| = (5 − λ)(−7 − λ) + 36 = λ2 + 2λ + 1 = (λ + 1)2 , so 9 −7 λ = −1 is an eigenvalue of algebraic multiplicity two, geometric multiplicity one. We find v[1 = (2 ]3)t , and solve (A + I)v2 = (2 3)t to find v2 . We can choose v2 = (1 1), so 2 1 P = , and we can check that AP = P Λ. Since the eigenvalue has absolute 3 1 value one, we have the non-hyperbolic case. [

2 7. (a) We assume that [ A ∈]M2 (R) has characteristic equation λ −τ λ+∆ = 0. On the a b other hand, if A = , then |A−λI| = (a−λ)(d−λ)−bc = λ2 −λ(a+d)+ad−bc, c d so on equating coefficients, we see that τ = a + d = trace of A, and ∆ = ad − bc = determinant of A.

(b) Suppose that ∆ < 1, ∆ > −τ −1 and ∆ > τ −1, then λ1 λ2 < 1 (note that although the matrix is real, the eigenvalues need not be real, however, both τ = λ1 + λ2 and ∆ = λ2 λ2 are real).

Case 1. Suppose that the eigenvalues are not purely real, then they are given by √ √ τ ± τ 2 − 4∆ τ ± i 4∆ − τ 2 = λ= 2 2 since τ 2 − 4∆ < 0. Hence |λ|2 = (τ 2 + 4∆ − τ 2 )/4 = ∆ < 1, and the result follows.

Case 2. Suppose the eigenvalues are real, then we obtain −1 < λ1 λ2 < 1, and

− λ1 λ2 − 1 < λ1 + λ2 < λ1 λ2 + 1.

12

We can write λ1 λ2 = ∆ for ∆ ∈ (−1, 1), then λ2 = ∆/λ1 so that −∆ − 1 < λ1 + ∆/λ1 < ∆ + 1, λ1 − ∆ − 1 + ∆/λ1 < 0, and λ1 + ∆ + 1 + ∆/λ1 > 0, giving λ21 − λ1 (∆ + 1) + ∆ λ21 + λ1 (∆ + 1) + ∆ < 0, and > 0, λ1 ∆1 (λ1 + 1)(λ1 + ∆) (λ1 − 1)(λ1 − ∆) < 0 and > 0. If λ1 > 0 this gives ∆ < λ1 < 1 λ1 λ1 and −∆ < λ1 and we conclude that |λ1 | < 1. If λ1 < 0, then λ1 < ∆ or λ1 > 1, and −1 < λ1 < −∆, and again we get |λ1 | < 1. Similarly for λ2 . or

Conversely, if |λi | < 1 for i = 1, 2, then |∆| = |λ1 λ2 | < 1, so −1 < ∆ < 1. In the case that λ1 = α + iβ, λ2 = α − iβ are not purely real, we must show that ∆ > τ − 1, or λ1 λ2 > λ1 + λ2 − 1, or α2 + β 2 > 2α − 1. If we can show that 2αβ > 2α − 1 then we are done. This can be done, for example using Mathematica (subject to α2 + β 2 < 1).

For the real case, reverse the above steps for the real case, again splitting into different cases (using −1 < λi < 1), for example assume λ1 > 0 and we may assume one of λ > ∆ etc. (c) If ∆ = τ − 1, then our equation becomes λ2 − τ λ + τ − 1 = (λ − 1)(λ − τ + 1) = 0, when λ = 1, or λ = τ −1. If ∆ = −τ −1, then λ2 −τ λ−τ −1 = (λ+1)(λ−τ −1) = 0, when λ = 1, or λ = τ + 1. (d) We have seen in (b) that if 4∆ > τ 2 , the eigenvalues are non-real (NOTE: There is an error in the statement of this question.) If ∆ = 1, then λ2 = λ¯1 and each eigenvalue has absolute value equal to one.

8. (a) F : R2 → R2 , is a linear map, so F (0) = 0. We first show that if F is continuous at zero, then it is continuous everywhere: Suppose that given z ∈ R2 and ϵ > 0 there exists δ > 0 such that if d(z, 0) < δ, then d(F (z), 0) < ϵ.

Fix x ∈ R2 and choose y ∈ R2 with d(x, y) < δ, or d(x − y, 0) < δ (since we are using the standard distance metric on R2 ). It follows that d(F (x − y), 0) < ϵ, or d(F (x), F (y)) < ϵ (since F is linear), and hence F is continuous on all of R2 .

13

We can represent F as multiplication by a matrix A (which we can assume is non-zero): ( )[ ]( ) ( ) x1 a b x1 ax1 + bx2 F = . x2 c d x2 cx1 + dx2 It is now most convenient to use the Cauchy-Schwartz inequality in the form (ax1 + bx2 )2 ≤ (x21 + x22 )(a2 + b2 ), (which can be verified directly). Then (( ) ( ))2 ax1 + bx2 0 d , = (ax1 +bx2 )2 +(cx1 +dx2 )2 ≤ (x21 +x22 )(a2 +b2 +c2 +d2 ), cx1 + dx2 0 so given ϵ > 0, we can take δ = ϵ/(a2 + b2 + c2 + d2 )1/2 to show continuity at zero. (b) F has eigenvalues λ1 , λ2 with |λ1 | < 1 and |λ2 |[ < 1. If ]these eigenvalues are λ1 0 distinct we can diagonalize A = P ΛP −1 where Λ = It is then clear that 0 λ2 ( ) ( ) x1 0 , as n → ∞. If the matrix is not diagonalizable, we can use the → Fn x2 0 Jordan form in a similar way to see that this result holds. (c) H(x) = P · x where P ∈ Mn (Rn ). If P is invertible, H −1 (x) = P −1 (x), so both H and H −1 are continuous. Since H −1 exists, H is one-to-one and onto, so it is a homeomorphism.

14

Section 13.2 1. We find the fixed points and the stability of the maps F : R2 → R2 : ( ) ( 2 ) x x − y2 (i) F = . We set x2 − y 2 = x and 2xy = y and solve simultaney 2xy ously. We see(that)the [fixed points]are (0, 0) and ( (1, )0). [ ] x 2x −2y 0 0 0 Now JF = , and JF = , and we have stay 2y 2x 0 0 0 ( ) [ ] 1 2 0 2−λ 0 bility at (0, 0). Also JF = , where = (2 − λ)2 = 0, 0 0 2 0 2−λ when λ = 2, giving an unstable fixed point. ( ) ( 2 ) x x − y2 + x − 1 (ii) F = . Solving x2 − y 2 + x − 1 = x and 2xy + y = y y 2xy + y ( ) [ ] ( ) x 2x + 1 −2y 1 gives fixed points (±1, 0). Now JF = , and JF = y 2y 2x + 1 0 [ ] ( ) [ ] 3 0 −1 −1 0 , so we have instability in this case. JF = , so λ = −1 0 3 0 0 −1 is a repeated eigenvalue, giving the non-hyperbolic case..

2. We find a and b for which the map F (has )a fixed ( point, and we) determine y x . We set = the stability when a = 6 and b = 1: F ay − bx − y 3 y ) ( ) ( x y = , so that x = y and then −by + ay − y 3 = y. Thus 3 y ay − bx − y y = 0, or y 2 = a − b − 1 and in particular, we require a − b ≥ 1. Thus the fixed points √ √ are (0, 0)t and (± a − b − 1, ± a − b ]− 1)t . ] [ ( ) [ ( ) 0 1 x 0 1 0 t , so = Now JF = , so at (0, 0) , JF 0 −b a −b a − 3y 2 y when = 6 [and b =] 1, the fixed points are (0, 0)t , (2, 2)t and (−2, −2)t , and ( a) √ 0 0 1 JF = . This gives rise to eigenvalues λ = 3 ± 2 2, so we have 0 −1 6 instability. ( ) [ ] √ 2 0 1 t At (2, 2) , JF = , with eigenvalues λ = −3 ± 2 2, and simi2 −1 −6 larly at (−2, −2), again giving instability.

15

(

)

) 1 − ax2 + y 3. (a) The Henon map is defined by H = , where we are bx assuming a ̸= 0 and 0 < b < 1. ( ) ( ) x1 x2 Now H is one-to-one since clearly H = H implies x1 = x2 and y y ( 1 ) ( )2 x c y1 = y2 . If (c, d)t ∈ R2 , then solving H = gives x = d/b and y = y d ( ) ( ) x y/b 2 2 −1 c−1+ad /b , so H is onto. The inverse is given by H = . y x − 1 + ay 2 /b2 x y

(

(

) 2 1 − ax + y (b) Set b = 1 and K(x, y)t = (−y, −x)t , then K ◦ H =K = bx ( ) ( ) ( ) ( ) −bx x −y −x/b and H −1 ◦K = H −1 = , and 2 −1 + ax − y y −x −y − 1 + ax2 /b2 these are equal when b = 1. x y

)

(

] ) [ −2ax 1 x . Here we need = 4. (a) The Jacobian of the Henon map is JH b 0 y to use the fact that the area of the unit square is altered by a factor of |det(A)|, for a matrix A. Now |det(JH(x, y)y )| = b, so JH acts as a contraction of the area, preserving the area when b = 1. (

(b) We can check that H = H1 ◦ H2 ◦ H3 , where Hi are the given maps, and H1 is clearly a reflection, H2 is a contraction (since 0 < b < 1) and H3 is area preserving (as in (a) with b = 1.

5. (a) We show that the fixed point√(x− , y − ) is unstable (assuming (1 − b)2 + 4a > 0, b − 1 − (1 − b)2 + 4a a ̸= 0), where x = x− = . The eigenvalues of the Jacobian 2a √ satisfy λ2 + 2axλ − b = or λ± = −ax ± a2 x2 + b. We will show that λ+ = −ax + √ a2 x2 + b > 1. √ Now 2ax = b−1− (1 − b)2 + 4a < b−1 < 0, so 2ax+1 < b and a2 x2 +2ax+1 < a2 x2 + b, or ax + 1 ≤ |ax + 1| < It follows that 1 < −ax +

√

√

a2 x2 + b.

a2 x2 + b = λ+ , so we have instability.

16

(b) We show that the fixed point in (a) is a saddle. Now we have (1 −√ b)2 + 4a > 0, so from (a) |λ+ | > 1, so it suffices to show that |λ− | < 1. Note that −ax− (ax)2 + b < 0 since b > 0, so λ− < 0. Also from above, ax < 0, so √ √ √ λ− = −ax − (ax)2 + b ≥ −ax − |ax| − b = − b > −1, √ √ √ (using 0 < b < 1 and s + t ≤ s + t for s, t ≥ 0). It follows that −1 < λ− < 0, so we have a saddle. (Note. We have not used the fact that a < 3(1 − b)2 /4). (

) ( ) ( ) ( ) x 1 − ax2 + y x 1 − a(1 − ax2 + y)2 + bx 2 6. We have H = , so H = , y bx y b(1 − ax2 + y) so we set y = (abx2 − b)/(b − 1) and substitute into 1 − a(1 − ax2 + y)2 + bx = x.

After simplifying (use a CAS) we get (−a + 2a2 x2 − a3 x4 + (b − 1)2 (1 + (b − 1)x))/(b − 1)2 = 0, √ giving the required√ equation. This has solutions x = (1 − b ± 4a − 3(1 − b)2 )/(2a) and x = (−1 + b ± (1 − b)2 + 4a)/(2a). Disregarding the fixed point, we see that a period-2 point will appear when a > 3(1 − b)2 /4.

17

Section 13.4 1. T (x) = ax (mod 1) on [0, 1) and T ′ (eix ) = eiax on S1 are conjugate via ϕ : [0, 1) → S1 , ϕ(x) = e2πix , (where we are identifying 0 and 1, as mentioned in Section 13.3). In this case, we have seen that ϕ is a homeomorphism. Now ϕ ◦ T (x) = ϕ(ax) = e2πiax , and T ′ ◦ ϕ(x) = T ′ (e2πix ) = e2πiax , so these maps are conjugate. (

) ax1 + bx2 2. We consider when T : [0, 1) → [0, 1) defined by T = cx1 + dx2 ) ( x1 (mod 1) is well defined. Suppose first that a, b, c, d ∈ Z, clearly T ∈ [0, 1)2 . x2 Suppose that we are given two points in [0, 1)2 with (x1 , x2 )t = (y1 , y2 )t , then x1 = y1 , x2 = y2 (mod 1), so since a, b are integers, ax1 +bx2 = ay1 +by2 (mod 1), and similarly cx1 + dx2 = cy1 + dy2 (mod 1), so T is well defined. A similar argument gives the converse.

2

2

x1 x2

)

(

(

) ( ) x1 ax1 + bx2 3. As above, T : [0, 1) → [0, 1) is defined by T = (mod 1), x2 cx1 + dx2 ( ix ) ( i(ax +bx ) ) e 1 e 1 2 ′ 1 1 1 1 ′ and also T : S × S → S × S is defined by T = . Define ix2 e ei(cx1 +dx2 ) Φ : [0, 1)2 → S1 × S1 , by Φ(x, y)t = (e2πix , e2πiy )t . We can check (as in Exercise 1) that Φ is a homeomorphism, and Φ ◦ T = T ′ ◦ Φ, so T and T ′ are conjugate.

2

2

(

) ( ) x1 ax1 + bx2 4. We show that T : [0, 1) → [0, 1) defined by T = (mod x2 cx1 + dx2 1), [ is onto ] when det(A) ̸= 0, and one-to-one if and only if det(A) = ±1 (where A = a b ). We are assuming that a, b, c, d ∈ Z, so T is well defined. Solving A(x, y)t = c d [ ] z b (z, w)t we get x = det(A1 )/ det(A) and y = det(A2 )/ det(A) where A1 = w d [ ] a y and A2 = (Kramer’s Rule), having a solution for every (z, w)t ∈ R2 if and c w only if det(A) ̸= 0, so will certainly have a solution for (z, w)t ∈ [0, 1)2 .

2

2

18

Now suppose that ad−bc = ±1, then if T (x1 , x2 )t = T (y1 , y2 )t , we have ax1 +bx2 = ay1 + by2 (mod 1) and cx1 + dx2 = cy1 + dy2 (mod 1). Multiplying the first equation by c and the second by a and subtracting (where we may assume a, c ̸= 0) gives (x2 − y2 )(ad − bc) = 0 (mod 1), so since ad − bc = ±1, x2 = y2 (mod 1) and similarly x1 = y1 (mod 1), so T is one-to-one. The converse can be obtained by reversing the argument. (

) ( ) x 2x 5. T : T → T is the endomorphism defined by T = (mod 1). y 2y We( show that T)is chaotic on T2 . )Write x = ·x1 x2 x3 . . . and y = ·y1 y2 y3 . . ., then ( ·x1 x2 x3 . . . ·x2 x3 x4 . . .

T = , so for example (·1010 . . . , ·0101)t = (2/3, 1/3)t ·y1 y2 y3 . . . ·y2 y3 y4 . . . is a period-2 point (as is (1/3, 2/3), and we see that there are four distinct period-2 points. Similarly, (·100100100 . . . , ·0100100100 . . .)t = (4/7, 2/7)t is a period-3 point, and in a similar way to the one-dimensional case, we see that the period points are dense in T2 . The fact that T has sensitive dependence on initial conditions is similar to the one-dimensional case (if (x1 , x2 )t ̸= (y1 , y2 ), the components will differ in some term of their binary expansions, and then they will move widely apart under iteration

by T ).

To show that T is transitive, we construct a point having a dense orbit. This can be done by modifying the construction for the one-dimensional case. We define a point (x0 , y0 )t ∈ T2 2

2

(

x0 y0

)

( =

·00 11 00 00 00 00 01 . . . ·01 01 00 01 10 11 00 . . .

) ,

where we list all possible “pairs” of 1-blocks, all possible “pairs” of 2-blocks, etc. An argument similar to the one-dimensional case shows that his is a transitive point. { 6. B : [0, 1] → [0, 1] is defined by B(x, y) = 2

2

(2x, y/2) ; 0 ≤ x < 1/2 .

(2x − 1, y/2 + 1/2) ; 1/2 ≤ x ≤ 1

(a) The Baker’s transformation acts on the unit square by mapping the left hand rectangle: [0, 1/2) × [0, 1], onto the bottom half of the unit square: [0, 1] × [0, 1/2).

In a similar way, [1/2, 1] × [0, 1] gets mapped onto [0, 1] × [1/2, 1]. This resembles kneading dough (giving rise to its name).

19

(b) Let x, y ∈ [0, 1] and set x = ·x1 x2 x3 . . ., y = ·y1 y2 y3 . . ., then B(x, y) = B(·x1 x2 x3 . . . , ·y1 y2 y3 . . .) = (·x2 x3 x4 . . . , ·x1 y1 y2 . . .).

It is clear that B is invertible, with inverse B −1 (·x1 x2 x3 . . . , ·y1 y2 y3 . . .) = (·y1 x1 x2 x3 x4 . . . , ·y2 y3 y4 . . .).

The fixed points are (0, 0) and (1, 1). Now B 2 (= ·x1 x2 x3 . . . , ·y1 y2 y3 . . .) = (·x3 x4 . . . , ·x2 x1 y1 y2 . . .), so for period-2 points we require x1 = x3 , x2 = x4 , x3 = x5 , . . ., and y1 = x2 , y2 = x1 , y3 = y1 , . . ., so for example (·01010 . . . , 10101 . . .) = (2/3, 1/3) is a period-2 point, with {(2/3, 1/3), (1/3, 2/3)} being the 3-cycle. (c) Now B(x, y) = (x, y) when (x, y) = (0, 0) or (1, 1). B 2 (x, y) = (x, y) when x1 = x3 , x2 = x4 , y1 = x2 y1 = x2 , y1 = y3 etc, so (x, y) = (·010101 . . . , ·101010 . . .) = (1/3, 2/3) etc. (d) We must show that the periodic points of B are dense in [0, 1] × [0, 1] and B is transitive. The easiest way to do this is to define a map σ : Σ → Σ, where Σ = {(. . . x−2 x−1 x¯1 x2 x3 . . .) : xi ∈ {0, 1}}, is the set of all doubly infinite sequences of zeros and ones, where the “bar” indicates the zeroth coordinate. Then σ(. . . x−2 x−1 x¯1 x2 x3 . . .) = (. . . x−2 x−1 x1 x¯2 x3 x4 . . .).

If we define a map ϕ : [0, 1]2 → Σ by ϕ(·x1 x2 x3 . . . , ·y1 y2 y3 . . .) = (. . . y2 y1 x¯1 x2 x3 . . .) then we can check that ϕ is one-to-one and onto (except for those points that do not have a unique binary representation), with ϕ ◦ B = σ ◦ ϕ. (For example, (. . . 1001̄001001 . . .) is period-3 for σ, so that (·100100100 . . . , ·00100100 . . .) is period3 for B). Thus to find the period points of B, we find those of σ, and to find a point of transitivity for B we find one for σ. In the same way as for the one sided shift, we see that the periodic points are dense in Σ (with the usual metric on Σ), and we can transfer these points to [0, 1]2 via ϕ to see that the corresponding points are dense in [0, 1]2 .

To show transitivity, take a one sided sequence ω0′ consisting of all possible blocks (as in the one-sided case). Set ω0 = ω˜0′ ω0′ ∈ Σ (where ω˜0′ is ω0′ , but written in reverse). By iterating ω0 sufficiently, we can approximate any sequence ω = (. . . y−N y−N +1 . . . y−1 x1 x2 x3 . . . yN . . .) using σ N (ω0 ). We can then transfer this point to [0, 1]2 , to give the required result, where we approximate the point (x, y) =

20

(·x1 x2 . . . xN . . . , ·y1 y2 y3 . . . yN . . .) by B N (·ω0′ , ·ω0′ ). Since B is a continuous map on a metric space which has dense periodic points and is transitive, it is chaotic.

7. T : T2 → T2 is an endomorphism of the torus induced by a matrix A with det(A) ̸= 0 and has no eigenvalues that are roots of unity. We show that T is transitive if and only if A is a matrix having no eigenvalues that are roots of unity.

Suppose that T is transitive, then it is onto and det(A) ̸= 0. If the eigenvalues λ1 and λ2 , are roots of unity, then |λ1 | = λ2 | = 1 and det(A) = λ1 λ2 = 1 and λ2 = λ¯1 , det(A) = 1. Now we can follow the steps of the solution to Exercise 13.5 # 4, to see that T cannot be transitive.

For the converse we assume that T is transitive and look at different cases.

Case 1. If det(A) = ±1, where neither eigenvalue has absolute value equal to one, then we have a hyperbolic toral automorphism, so we can use the result of Theorem 13.5.4 to conclude that T is transitive.

Case 2. det(A) = ±1, where both eigenvalues have absolute value equal to one, then we see that this case cannot happen (also from Exercise 13.5 # 4).

Case 3. det(A) ̸= ±1 and det(A) ̸= 0. In this case the eigenvalues cannot have absolute value one, so cannot be roots of unity: The eigenvalues are either both real or both have an imaginary part, occurring in conjugate pairs. If they are not real, say λ1 and λ̄1 , with |λ1 | = 1, then also |λ̄1 | = 1, so det(A) = ±1, which is a contradiction.

It remains only to show that T is transitive in this case. In the real case, the proof of Theorem 13.5.4 shows that the eigenvalues are irrational. Notice that ad − bc is an integer, so we must have |λ1 λ2 | = | det(A)| ≥ 2, so at least one of the eigenvalues has absolute value greater than 1, say |λ1 | > 1. Then there exists an eigenvector x0 ∈ T2 with An (x0 ) = λn1 x0 → ∞ as n → ∞. It now suffices to show that {T n (x0 ) : n ∈ Z+ } is dense in T2 - we omit the proof as it is beyond the scope of this text.

21

Section 13.5 1. We ask which of the following matrices induce hyperbolic toral automorphisms?

All of these matrices except (iv) have integer entries, so (iv) does not give a hyperbolic toral automorphism. We check to see if det(A) = ±1 and that neither eigenvalue is of absolute value equal to one. [ ] 3 2 (i) A = . det(A) = 9 − 8 = 1, and det(A − λI) = λ2 − 6λ + 1 = 0 when 4 3 √ λ = 3 ± 2 2, so it gives a hyperbolic toral automorphism [ ] 1 −2 (ii) A = . det(A) = −3 + 6 ̸= 1, so it does not give a hyperbolic toral 3 −3 automorphism ] [ 5 7 (iii) A = . det(A) = 15 − 14 = 1 and det(A − λI) = λ2 − 8λ + 1 = 0 when 2 3 √ λ = 4 ± 15, so it is a hyperbolic toral automorphism [ ] 2 1/2 (iv) A = . Since we do not have integer entries it does not give a 2 1 hyperbolic toral automorphism ] [ 3 1 . det(A) = −6 + 7 = −1 and det(A − λI) = λ2 − λ + 1 = 0 (v) A = −7 −2 √ when λ = (1 ± i 3)/2. These eigenvalues have absolute value equal to one, so it does not give a hyperbolic toral automorphism. [

] 2 1 2. (a) We have seen in Example 13.5.2 (i) that A = define a hyperbolic 1 1 toral automorphism. The only fixed point is (0, 0)2 . We have seen that the period points are of the form (p1 /q1 , p2 /q2 )t , and we can check that {(1/5, 2/5)t , (4/5, 3/5)t } and {(2/5, 4/5)t , (3/5, 1/5)t } are the 2-cycles, and the 3-cycles are {(1/2, 1/2)t , (1/2, 0)t , (0, 1/2)t }, {(1/4, 1/4)t , (3/4, 1/2)t , (0, 1/4)t }, {(3/4, 0)t , (1/2, 3/4)t , (3/4, 1/4)t }, {(3/4, 3/4)t , (1/4, 1/2)t , (0, 3/4)t }. [

] F2n F2n−1 (b) We can use induction to show that A = , where Fn is the nth F2n−1 F2n−2 term of the Fibonacci sequence (F0 = 1, F1 = 1). n

22

√ (c) The eigenvalues of A are (3 ± 5)/2 and we can check [ that the eigenvectors ] can √ √ 2 √2 be taken to be (2, 5 − 1)t and (2, − 5 − 1)t . Set P = √ , then 5−1 − 5−1 √ [ ] [(3 + 5)/2]n 0 √ An = P Λn P −1 , where Λn = . By calculating P −1 , 0 [(3 − 5)/2]n it is now easy to equate the entries in the (1, 2) position to deduce ( ( √ )n √ )n 1 3+ 5 3− 5 1 −√ .

F2n−1 = √ 2 2 5 5

2 2 3. Let [ T : T ] → T be a hyperbolic toral automorphism induced by the matrix a−λ b a b a b = (a − λ)(d − = ad − bc = ±1, and A= . Then c d−λ c d c d λ) − bc = λ2 − λ(a + d) + ad − bc, so the eigenvalues are √ √ a + d ± (a + d)2 − 4(ad − bc) a + d ± (a − d)2 + 4bc λ= = .

2 2 √ a + d ± i −(a − d)2 − 4bc 2 Suppose that (a − d) + 4bc < 0, then λ = , so 2

|λ|2 =

(a + d)2 + (−(a − d)2 − 4bc) = ad − bc = ±1, 4

contradicting T being a hyperbolic toral automorphism.

That the eigenspaces are lines with irrational slopes now follows from Theorem 13.5.4 (ii). [

] a b 4. We have T : T → T a torus automorphism with matrix A = . It follows c d that a, b, c, d ∈ Z and that T is one-to-one and onto, so detA = ±1. Thus if T fails to be a hyperbolic toral automorphism, then the eigenvalues of A must be of absolute value ±1.

2

2

(a) Suppose the eigenvalues are λ1 = 1 and λ2 = −1, then the characteristic equation is det(A − λI) = λ2 − (a + d)λ + ad − bc, and it is also equal to (λ − λ1 )(λ − λ2 ) = λ2 − (λ1 + λ2 )λ + λ1 λ2 = λ2 − 1, so that a + d = 0 and detA = −1.

23

[

] a b (b) d = −a (from (a)), so A = . We deduce that T is not chaotic as c −a [ ] 1 0 −1 follows: the matrix A can be diagonalized a A = P ΛP where Λ = , so 0 −1 that A2 = I, the identity matrix. It follows that T cannot be chaotic. (c) We are supposing that λ1 and λ2 are not purely real and have absolute value equal to one with det(A) = λ1 λ2 = 1. Then λ1 and λ2 are roots of the equation λ2 − (a + d)λ + ad − bc having real (integer) coefficients. It follows that λ2 = λ¯1 .

Set λ1 = α − iβ, then also √ √ λ1 = (a + d ± (a + d)2 − 4(ad − bc))/2 = (a + d ± (a + d)2 − 4)/2.

Since this is non-real, we must have (a + d)2 − 4 < 0, or −2 < a + d < 2. Since a and d are integers, we must have either: Case (i) a + d = 0, so α = (a + d)/2 = 0 and λ1 = iβ. It follows that β = ±1 and λ1 = ±i. Case (ii) a + d = 1, then √ √ √ λ1 = (1 ± 1 − 4)/2 = (1 ± 3)/2, or Case (iii) a + d = −1 and λ1 = (−1 ± 3)/2. [ ] √ ω 0 (d) In a similar way to (b), with Λ = , where ω = (−1 + 3)/2, ω 3 = 1, we 0 ω̄ [ ] −ω 0 3 see that A = I, and similarly in the other case where Λ = .

0 −ω̄

24

Section 14.2 1. To prove the triangle inequality: |z + w|2 = (z + w)(z̄ + w̄) = z z̄ + z w̄ + z̄w + ww̄ = |z|2 + 2Re(z w̄) + |w|2 ≤ |z|2 + 2|z w̄| + |w|2 = |z|2 + 2|z||w| + |w|2 = (|z| + |w|)2 , and since everything is positive, |z + w| ≤ |z| + |w|.

Replacing z by z − w gives |z| ≤ |z − w| + |w|, or |z| − |w| ≤ |z − w|, and similarly |w| − |z| ≤ |z − w|, and the result follows.

2. We have fc : C → C, fc (z) = z 2 + c. Recall that z0 is an attracting fixed point if |f ′ (z0 )| < 1. If |f ′ (z0 )| = 1, it is neutral and if |f ′ (z0 )| > 1 it is repelling. (a) Suppose that z0 and z1 are fixed points of fc , then they are solutions of z 2 −z +c = √ 0, so zi = (1 ± 1 − 4c)/2, i = 0, 1. Adding the two solutions gives z0 + z1 = 1. Now f ′ (z) = 2z, so f ′ (z0 ) + f ′ (z1 ) = 2z0 + 2z1 = 2. Furthermore |f ′ (z0 )| = 2|z0 |, and |f ′ (z1 )| = 2|z1 |, so that if |f ′ (z0 )| = 2|z0 | ≤ 1, then |z0 | ≤ 1/2, so 1 = |z0 + z1 | ≤ |z0 | + |z1 | ≤ 1/2 + |z1 |, |z1 | ≥ 1/2, |f ′ (z1 )| ≥ 1, so z1 won’t be attracting.

If c = 1/4, then z0 = z1 = 1/2 and |f (z0 )| = 1, so is not attracting. (b) z0 is super-attracting if f ′ (z0 ) = 0, so z0 = 0. Since fc (z0 ) = z0 , we must have c = 0. (c) If z0 is on a super-attracting 2-cycle of fc , then (fc2 )′ (z0 ) = 0, so fc′ (z0 )fc′ (fc (z0 )) = 0, so we may assume z0 = 0. Then fc2 (0) = 0, so fc (c) = 0 and c2 + c = 0, c = −1. (d) If z0 is a period-n point which is super-attracting, then (fcn )′ (z0 ) = 0. This implies one of the members of the n-cycle is 0, so we may assume z0 = 0. Conversely, if z0 = 0 is on an n-cycle, since f ′ (z0 ) = 0, we have (f n )′ (z0 ) = 0, so the cycle is super-attracting. (e) In a similar manner to (d), z0 = 0 must be on the 3-cycle. Then fc (0) = c, fc2 (c) = c2 + c, so the 3-cycle is {0, c, c2 + c}, and since f 3 (0) = 0, we must have (c2 + c)2 + c = 0, or c3 + 2c2 + c + 1 = 0 (disregarding c = 0). (f) If fc (z) = z 2 + c with c = −1.75488 . . ., then we have fc (0) = c, fc2 (0) = z1 (say), f 3 (0) = z2 and f 4 (0) = c, i.e., z = c is an eventually periodic point with {c, z1 , z2 } being a 3-cycle which is not super-attracting.

25

z n − an f (z) − f (a) = lim z→a z→a z − a z−a n−1 n−2 (z − a)(z + z a + · · · + zan−2 + an−1 ) = lim z→a z−a n−1 n−2 = lim (z + z a + · · · + zan−2 + an−1 ) = nan−1 ,

3. If f (z) = z n , then f ′ (a) = lim

z→a

and the result follows.

4. f : C → C is continuous and O(z0 ) = {z0 , f (z0 ), f 2 (z0 ), . . .} has a unique limit point α. It follows that there is a sequence of positive integers (nk ) with f nk (z0 ) → α as k → ∞. From the continuity, f (f nk )(z0 ) → f (α), as k → ∞, or f nk +1 (z0 ) → f (α), so that f (α) is also a limit point of the orbit. The uniqueness then implies that f (α) = α.

If there are p limit points, say A = {α1 , α2 , . . . , αp }, then arguing as above, there is a sequence (nk ) with f nk (z0 ) → α1 as k → ∞, so that f nk +ℓ (z0 ) → f ℓ (α1 ) as k → ∞, and each f ℓ (α) is also a limit point. Since the set A has only p members, α1 is either periodic with some period m ≤ p, or is eventually periodic. Similarly for α2 , α3 , . . . , αp .

5. We first show that if limz→a f (z) = L, then limz→a |f (z)| = |L|. Let ϵ > 0, then there exists δ > 0 such that if 0 < |z − a| < δ, then |f (z) − L| < ϵ. But in this case ||f (z)| − |L|| ≤ |f (z) − L| < ϵ, so the result follows.

Now suppose that |L| > 0, then there exists δ > 0 such that if 0 < |z − a| < δ, then ||f (z)| − |L|| < |L|/2, or |L|/2 < |f (z)| < 3|L|/2, from which the result follows.

26

Section 14.3 1. p(z) is a polynomial having degree at least 2. We show that p(∞) = ∞. Suppose that p(z) = an z n +an−1 z n−1 +· · ·+a1 z +a0 , where a0 , a1 , . . . , an are complex constants and an ̸= 0. Clearly p(z)/z n → an as z → ∞, i.e., given ϵ > 0 there exists M ∈ R+ such that if |z| > M , then |p(z)/z n − an | < ϵ, so |an | − ϵ < |p(z)/z n | < |an | + ϵ (we are assuming an ̸= 0, and can choose ϵ < |an |). Thus |p(z)| > (|an | − ϵ)|z n | → ∞ as z → ∞, so we have p(∞) = ∞.

Let h(z) = 1/z, then set q(z) = h−1 ◦ p ◦ h(z) = h−1 ◦ p(1/z) = h−1 (an /z n + an−1 /z n−1 + · · · + a1 /z + a0 ) = z n /(an + an−1 z + · · · + a1 z n−1 + a0 z n ).

Then q(0) = 0, and q ′ (z) =

(a0 z n + a1 z n−1 + · · · + an )nz n−1 − z n (na0 z n−1 + · · · + an−1 ) , so q ′ (0) = 0. (a0 z n + a1 z n−1 + · · · + an )2

Thus p′ (∞) = 0 and p has a super-attracting fixed point at ∞.

If p(z) = az + b where a ̸= 0, then we still have p(∞) = ∞ and q(z) = z/(a + bz), so q ′ (z) = a/(a + bz)2 and q ′ (0) = 1/a. Thus if |a| > 1, ∞ is attracting for p, and if |a| < 1 it is repelling. This fits in with what we know about the real case.

2. p(z) = z 2 − z. We first find the fixed points: Solving p(z) = z gives z 2 − 2z = 0, so z = 0 or z = 2. If p2 (z) − z = 0, then (z 2 − z)2 − z 2 + z = z or z 4 − 2z 3 = 0, giving no additional solutions, so there are no period-2 points.’

3. As before, p(z) is a polynomial having degree at least 2, and we may write p(z) = an z n + an−1 z n−1 + · · · + a1 z + a0 , where a0 , a1 , . . . , an are complex constants. an z n + an−1 z n−1 + · · · + a1 z + a0 p(z) =z− Now Np (z) = z − ′ p (z) nan z n−1 + (n − 1)an−1 z n−2 + · · · + a1 =

(n − 1)an z n + (n − 2)an−1 z n−1 + · · · + a2 z 2 − a0 , nan z n−1 + · · · + 2a2 z + a1

so clearly Np (∞) = ∞.

27

( −1

−1

Thus g(z) = h ◦Np ◦h(z) = h

(n − 1)an + (n − 2)an−1 z + · · · + a2 z n−2 − a0 z n nan z + · · · + 2a2 z n−1 + a1 z n

)

nan z + · · · + 2a2 z n−1 + a1 z n . Then (n − 1)an + (n − 2)an−1 z + · · · + a2 z n−2 − a0 z n g ′ (z) =

((n − 1)an + · · · − a0 z n )(nan + · · · ) − (nan z + · · · + a1 z n )((n − 2)an−1 + · · · ) , ((n − 1)an + (n − 2)an−1 z + · · · + a2 z n−2 − a0 z n )2

so Np′ (∞) = g ′ (0) =

n(n − 1)a2n n = > 1, so that ∞ is a repelling fixed point.

2 2 (n − 1) an n−1

If p(z) = az + b, then Np (z) = z −

b az + b = − , so ∞ is not a fixed point of Np . a a

az + b = z, where (a − d)2 + 4bc = 0, then az + b = cz + d √ a − d ± (a − d)2 + 4bc 2 2 cz + dz, cz + z(d − a) − b = 0, so z = . It follows that if 2c (a − d)2 + 4bc = 0, then f (z) has a unique fixed point.

In Section 14.3.3 we mentioned [ that ]there is a correspondence between the maps a b f (z) and matrices of the form . The eigenvalues of this matrix are λ = c d √ a + d ± (a − d)2 + 4bc , which coincide when (a − d)2 + 4bc = 0. In this case the 2 matrix has a single repeated eigenvalue, say λ0 . The matrix is not diagonalizable since the eigenspace is one-dimensional. [However,] it follows from Section 13.1 that λ0 1 the matrix is similar to one of the form (the Jordan form). This implies 0 λ0 λ0 z + 1 that f conjugate to a transformation of the form g(z) = , which is of the form λ0 g(z) = z + α.

4. Suppose that f (z) =

5. (a) As in the last example, if f (z) has two distinct eigenvalues, [ ]say λ1 and λ2 , then λ1 0 the matrix is diagonalizable (similar to one of the form ), so f is conjugate 0 λ2 λ1 to a map of the form g(z) = z = αz. λ2 (b) To obtain a closed form for f n , we should find the eigenvalues λ1 and λ2 explicitly, and find the corresponding eigenvectors. Writing the eigenvectors as the columns of a

=

28

matrix P , we then have A = P ΛP −1 (where Λ has the eigenvalues down the diagonal).

Then An = P Λn P −1 , from which we can read off f n . z−1 . If z belongs to the imaginary axis, then we can write z = iα z+1 for some α ∈ R. Then

6. Set T (z) =

T (iα) =

(iα − 1)(−iα + 1) α2 − 1 + 2iα iα − 1 = = . iα + 1 (iα + 1)(−iα + 1) α2 + 1

α2 − 1 + 2iα (α2 − 1)2 + 4α2 (α2 + 1)2 2 If w = , then |w| = = 2 = 1, so w ∈ S1 , and 2 2 2 2 α +1 (α + 1) (α + 1) by varying α we get every member of S1 .

7. A straight line from the North Pole N (0, 0, 1) to a point P on the complex plane, intersects the Riemann sphere S at S(z). If the coordinates of P are (x0 , y0 , 0) (z = x0 + iy0 ), then the coordinates of S(z) are of the form w = (cx0 , cy0 , cz0 ) for some c ∈ R to be determined. The sphere has equation x2 + y 2 + (z − 1/2)2 = 1/4, so that c2 x20 + c2 y02 + (cz0 − 1/2)2 = 1/4, or c2 (x20 + y02 ) = z0 (1 − z0 ). √ √ x20 + y02 c2 x20 + c2 y02 = Also, using similar triangles, , so c = 1 − z0 . Combining 1 1 − z0 1 these two equations gives c = 2 , and x0 + y02 + 1 ) ( y0 x20 + y02 x0 , , . w= x20 + y02 + 1 x20 + y02 + 1 x20 + y02 + 1

8. We have a metric ρ(w1 , w2 ) on the Riemann sphere S, giving the shortest distance between the points w1 and w2 on the sphere. A distance d is defined on the complex plane by d(z1 , z2 ) = ρ(S(z1 ), S(z2 )). We show that d is a metric on the extended b Clearly d(z1 , z2 ) ≥ 0 always, and if z1 = z2 , then S(z1 ) = S(z2 ), complex plane C. so ρ(S(z1 ), S(z2 )) = 0, or d(z1 , z2 ) = 0. Also if d(z1 , z2 ) = 0, then z1 = z2 (where we b → S is one-to-one and onto). Therefore we only have to are using the fact that S : C show that the triangle inequality holds. Here we use the fact that ρ is a metric on S: d(z1 , z3 ) = ρ(S(z1 ), S(z3 )) ≤ ρ(S(z1 ), S(z2 )) + ρ(S(z2 ), S(z3 )) = d(z1 , z2 ) + d(z2 , z3 ).

29

b : d(z, 0) > r} (it is assumed that ∞ ∈ Dr ). Let Let r > 0 and set Dr = {z ∈ C z0 ∈ Dr with z0 ̸= ∞, then S(z0 ) ∈ S and S(z0 ) ̸= N the North Pole. It is clear we can find an open disc centered on S(z0 ) on S, contained in {w ∈ S : ρ(w, 0) > r} and that this projects to an open set of C contained in Dr . If z0 = ∞, then similar considerations may be used. b →: C, b defined by h(z) = 1/z, h(0) = ∞ and h(∞) = 0. h is 9. We have h : C clearly one-to-one and onto. We show that h is continuous, so is a homeomorphism b and S are defined, the (since h−1 = h). Clearly from the way the metrics on C b → S is a homeomorphism. We sketch a solution: Suppose that projection map S : C b then z0 = x0 + iy0 and z = x + iy are close in C, ) ( y0 x20 + y02 x0 , , , S(z0 ) = x20 + y02 + 1 x20 + y02 + 1 x20 + y02 + 1 and similarly for S(z), and these are close. We can check that ) ( ( ) ( ) −y0 1 x0 1 x0 −y0 1 , , = S = , , , z0 x20 + y02 + 1 x20 + y02 + 1 x20 + y02 + 1 |z0 |2 + 1 |z0 |2 + 1 |z0 |2 + 1 ( ) 1 and this will be close to S . z

30

Section 14.4 1. fc (z) = z 2 + c, so f−6 (2) = 4 − 6 = −2 and f−6 (−2) = 4 − 6 = −2. Since the orbit of z = 2 is bounded, 2 ∈ K−6 . Clearly also −2 ∈ K−6 .

If we find a fixed point of f−1+3i , we solve z 2 − 1 + 3i = z, or z 2 − z − 1 + 3i = 0. √ This gives z = (1 ± 5 − 12i)/2 = (1 ± (3 − 2i))/2 = 2 − i or −1 + i, so these both lie in K−1+3i .

2. fi (z) = z 2 + i. Now fi (0) = i, fi (i) = −1 + i, fi (−1 + i) = 1 − 2i − 1 + i = −i, fi (−i) = −1 + i, so that {−1 + i, −i} is a 2-cycle and z = 0 is eventually periodic.

First note that the fixed points satisfy the equation z 2 − z + i = 0. Solving fi2 (z) = z, we obtain z 4 + 2iz 2 − z − 1 + i = 0. The polynomial z 2 − z + i must be a factor, leaving only two roots to find. These must be z = −i and z = −1 + i since {−i, −1 + i} is a 2-cycle (we can check that z = −i is a solution of the equation fi2 (z) − z = 0).

3. We have seen that z = −i is a period 2-point for fi . To check its stability, we calculate |fi′ (−i)fi′ (−1 + i)| = |(−2i)(−2 + 2i)| > 1, so z = −i is repelling, and hence belongs to the Julia set J(fi ).

4. We have seen that if z0 is a period-n point, then (f n )′ (z0 ) = f ′ (z0 )f ′ (f (z0 ))f ′ (f 2 (z0 )) . . . f ′ (f n−1 (z0 )) = f ′ (z0 )f ′ (z1 ) . . . f ′ (zn−1 ), where {z0 , z1 , . . . , zn−1 } is the n-cycle. It follows that the same value is obtained no matter at which period n-point we start with as it will just permute the product.

Suppose that |(f n )′ (z0 )| = λ, then as above, |(f nk )′ (z0 )| = |f ′ (z0 )f ′ (f (z0 )) . . . f ′ (f n−1 (z0 )) . . . f ′ (z0 )f ′ (f (z0 )) . . . f ′ (f n−1 (z0 ))|, (nk-times), so |(f nk )′ (z0 )| = λk .

5. We are considering f0 : S1 → S1 , f0 (z) = z 2 , f−2 : [−2, 2] → [−2, 2], f−2 (z) = z 2 −2, and h : S1 → [−2, 2], h(z) = z+1/z. Now h◦f0 (z) = h(z 2 ) = z 2 +1/z 2 and f−2 ◦h(z) = f−2 (z + 1/z) = (z + 1/z)2 − 2 = z 2 + 1/z 2 , so these are equal. We need to show that h is well defined and onto. Let z = a + ib ∈ S1 (a and b real), then −1 ≤ a ≤ 1

31

h(z) = 2a, so −2 ≤ h(z) ≤ 2, so h is well defined. Now let w ∈ [−2, 2], then solving √ √ h(z) = w gives z 2 − zw + 1 = 0, or z = (w ± w2 − 4)/2 = w/2 ± i 4 − w2 /2, where |z|2 = w2 /4 + (4 − w2 )/4 = 1. In other words, for w ∈ [−2, 2] we get two solutions, both in S1 (one solution when w = ±2), so h is onto, but not one-to-one. I.e., h is a factor map but not a conjugation.

6. Suppose that fc has a bounded attracting periodic orbit, say z0 . Then the basin of attraction Bfc (z0 ) of z0 is an open set in K(fc ). J(fc ) is the closure of the repelling periodic points, so Bfc (z0 ) cannot contain any such points, i.e., K(fc ) ̸= J(fc ).

Consider the map f1/4 (z) = z 2 + 1/4, having a unique neutral fixed point z = 1/2 ′ (so |f1/4 (1/2)| = 1). We claim that 0 ∈ K(f1/4 ) \ J(f1/4 ). This follows from the fact n that limn→∞ f1/4 (0) = 1/2, which was proved for the real function f1/4 in Chapter 2. This shows that 0 ∈ K(f1/4 ), so it suffices to show that there is an open ball surrounding z = 0, say Bδ (0) whose points are contained in K(f1//4 ). In fact we can check that it suffices to take δ = 1/2.

Furthermore, it follows that f1/4 has no points having a bounded attracting periodic orbit, since any such orbit must attract a critical point (see Theorem 15.5.4).

Now z = 0 is the only critical point, and 0 is attracted to 1/2. (In the next section we will see that c = 1/4 lies on the boundary of the Mandelbrot set. In fact if fc has an attracting periodic orbit, then c lies in the interior of the Mandelbrot set).

7. We have a map f : X → X and C ⊂ X. (a) Suppose that both f (C) ⊆ C and f −1 (C) ⊆ C. To show f −1 (C) = C, it suffices to show that C ⊆ f −1 (C). Let x ∈ C, then f (x) ∈ f (C), so f (x) ∈ C and x ∈ f −1 (C) and the result follows. (b) Suppose that f is onto with f −1 (C) = C, then we show that f (C) = C. Let x ∈ C, then x ∈ f −1 (C), so f (x) ∈ C. It follows that f (C) ⊆ C. If f (C) ̸= C, there exists y ∈ C with f (x) ̸= y for all x ∈ C. Since f is onto, there exists z ∈ X with f (z) = y ∈ C. Thus z ∈ f −1 (C) = C, a contradiction.

8. (a) To show that p(z) = z 2 − z is conjugate to q(z) = z 2 − 3/4, we can use Example 7.3.2 with a = 1, b = −1, c = 0, r = 1, s = 0 and t = −3/4, and check that

32

c = (b2 − s2 + 2s − 2b + 4rt)/4a, or 0 = (1 − 0 + 0 + 2 + 4(−3/4))/2, which is indeed that case. The conjugacy is given by h(z) = z − 1/2. (b) p(z) = z 2 + c. Set q(z) = p(z) − z, then q(p(z)) = p2 (z) − p(z) and q(p(z)) + q(z) = p2 (z) − p(z) + p(z) − z = p2 (z) − z.

Suppose p(α) = α, then q(α) = 0 and p2 (α) − α = q(p(α)) + q(α) = q(α) + 0 = 0, so any factor of p(z) − z is also a factor of p2 (z) − z, so p(z) − z divides p2 (z) − z. p2 (z) − z is a quadratic and p(z) − z p2 (z) − z we may assume that p(z) is of the form p(z) = z 2 + c. Suppose we have = p(z) − z p(z)−z, or (z 2 +c)2 −z +c = (z 2 −z +c)2 , then we see this is not possible since it gives a cubic term on the right, but no cubic term on the left. Another possibility is that p2 (z) − z are repeated, and coincide with one of the roots of p(z) − z. In the roots of p(z) − z √ p2 (z) − z this case we have = z 2 +z +c+1, so that z = (−1± 1 − 4 − 4c)/2, so if the p(z) − z p2 (z) − z roots are repeated, c = −3/4. We can check that = z 2 +z+1/4 = (z+1/2)2 , p(z) − z so has a single root: z = −1/2, and p(z) − z = z 2 − z − 3/4 = (z − 3/2)(z + 1/2), so the fixed points are z = 3/2 and z = −1/2. No other situation can arise.

Suppose that p(z) has no points of period-2. Then

(c) We show that p(z) = Az 3 + Bz 2 + Cz + D (A ̸= 0), is linearly conjugate to a polynomial of the form q(z) = z 3 + az + b. This can be checked if we set √ B B2 , b=D A+ √ a=C− 3A 3 A and h(z) = αz + β where α =

√

) ( 2B 2 , 1−C + 9A

B A and β = √ , then h ◦ p(z) = q ◦ h(z).

3 A

9. Let f (z) = az + b, then f (z) = z when z = b/(1 − a), provided a ̸= 1 (if a = 1 and b = 0, every point is a fixed point, but none is repelling). Now f ′ (z) = a, so the fixed point is repelling only when |a| > 1. Since there is now a single fixed point, J(f ) = {a}. In the other cases, J(f ) = ∅.

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10. f and g are conjugate polynomials via a conjugation h. If A is the set of repelling periodic points of f , then h(A) is the set of repelling periodic points of g. Now J(f ) = A, and J(g) = h(A), and since h is homeomorphism, J(g) = h(A) = h(J(f )), which completes the proof.

11. Let p(z) be a polynomial of degree d > 1, and k ∈ Z+ . We have seen that K(p) = p−1 (A) ∩ p−2 (A) ∩ . . . ∩ p−n (A) ∩ . . . , so K(pk ) = p−k (A) ∩ p−2k (A) ∩ . . . ∩ p−nk (A) ∩ . . . , where the set A is defined in 14.4.11. These are nested sequences of sets and clearly K(p) ⊆ K(pk ). Because of the nested property we see that the reverse is also true, so K(pk ) = K(p). Since these sets must have the same boundary, we have J(pk ) = J(p).

12. fc (z) = z 2 + c, c > 2 real and A = B rc (0) = B c (0), the closure of the open ball of radius c centered on 0.

Now f −1 (A) = {z ∈ C : fc (z) ∈ A}, and its boundary is the set {z : |z 2 + c| = c}.

Consider (z 2 + c)(z̄ 2 + c̄) = c2 , or |z 4 | + z 2 c̄ + z̄ 2 c = 0.

Set z = reiθ , then r4 + 2ℜ(r2 e2iθ c) = 0, so r2 = −2c cos(2θ).

13. Fc (z) = z d + c, where d ≥ 2 is an integer. Then the basin of attraction of infinity is the set Bc (∞) = {z ∈ C : F n (z) → ∞, as n → ∞}.

We show that Bc (∞) is an open set. Let z0 ∈ Bc (∞), then we can find an integer k > 0 with |Fck (z0 )| > max{|c|, 2}+1.

Since Fck is a continuous function, there exists δ > 0 such that |Fck (z)−Fck (z0 )| < 1 whenever |z − z0 | < δ. Now let z ∈ Bδ (z0 ), then we obtain from above and using the triangle inequality that |Fck (z)| > |Fck (z0 )| − 1 > max{|c|, 2}, so by the escape criterion, Fcn (Fck (z)) → ∞ as n → ∞. I.e., z ∈ Bc (∞). It follows that Bc (∞) is an open set.

34 2 2 14. f−1 (z) = z 2 − 1 and f−1 (z) = (z 2 − 1)2 − 1 = z 4 − 2z 2 . Then f−1 (0) = 0, 2 2 f−1 (−1) = −1 and f−1 (∞) = ∞. Also f−1 (0) = −1, so we have a 2-cycle which is clearly attracting, and we can check that ∞ is an attracting fixed point. It follows that each of these points is centered on disjoint regions making up different components of the Fatou set, so the Fatou set has a least three components. It follows that the Fatou set has infinitely many components.

15. We have seen that for a polynomial p(z), there are always points z ∈ C with pn (z) → ∞, so Bp (∞) is non-empty (in fact we have seen that ∞ is a super-attracting fixed point for p - Exercises 14.3 # 1). Bp (∞) is an invariant set, for if z ∈ Bp (∞), then pn (z) → ∞, so pn (p(z)) → ∞ or p(z) ∈ Bp (∞). (Here we are using the metric d arising from the metric ρ on the Riemann sphere - see Exercises 14.3 # 8). To show that Bp (∞) is an open set we can use the method of the proof of Theorem 4.3.6 on b d) and the fact that p is continuous on (C, b d). the metric space (C, Let U be an open set in C (where of course we assume U ̸= C and U ̸= ∅). Then there is a straight line L in C for which U ∩ L ̸= ∅. Consequently we may assume without loss of generality that U ⊂ R, U ̸= R. If the set U is bounded above or below, then it is clear that sup(U ) (respectively inf(U )) belongs to the boundary of U . If U is unbounded, there exist p ∈ R \ U . Then if p does not belong to the boundary of U (for example suppose that U = R \ C, where C is the Cantor set), we can set α = sup{x ∈ U : x < p}, then α is a boundary point of U .

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Section 14.5 1. (a) Suppose |c| > 2, then fc (0) = c has absolute value greater than 2, and greater than or equal to |c|, so that fcn (fc )(0) → ∞ as n → ∞. This proves (a). It follows that c ∈ / M, so (b) follows. (b) If |c| > 2, then from (a) the orbit of zero under fc is unbounded. It follows that M ⊆ {z ∈ C : |z| ≤ 2}. (c) Fc (z) = z 3 + c. We follow the proof of Proposition 14.4. (Note: The text has a √ typo, saying Proposition 14.5.5). Suppose |z| > max{|c|, 2}, then |Fc (z)| = |z 3 + c| ≥ |z 3 | − |c| = |z|3 − |c| > |z|3 − |z| = |z|(|z|2 − 1).

Since |z|2 > 2, |z|2 − 1 > 2 − 1 = 1. Set |z|2 − 1 = α > 1, then |Fc (z)| > α|z|, now continue in the same way as in the proof of Proposition 14.4.4 to see that |Fcn (z)| is an increasing sequence unbounded above.

2. fc (z) = z 2 + c, so fc (0) = c, fc2 (0) = c2 + c and fc3 (0) = (c2 + c)2 + c. Suppose that fc3 (c2 +c) = c2 +c, then (((c2 +c)2 +c)2 +c)2 +c = c2 +c, so that (((c2 +c)2 +c)2 +c)2 −c2 = 0. This can be seen to factor as (use a CAS) c3 (c + 2)(c3 + 2c2 + c + 1)2 (c6 + 2c5 + 2c4 + 2c3 + c2 + 1) = 0. c = −2 is the smallest real value in the Mandelbrot set. Solutions to c3 + 2c2 + c + 1 = 0 make c so that fc has a super-attracting orbit of period-3.

The last factor has a root: c = −1 · 239255 − i0 · 4126021816 . . ., and its Julia set is a “dendrite”.

All these c-values lie in the Mandelbrot set because they lie on the orbit of 0. For example, f−2 (0) = −2 → 2 → 2 · · · , which is bounded.

Note that if fc [fc2 (0)] = fc2 (0), then fc (c2 + c) = c2 + c, and simplifying, this gives c(c + 2) = 0. These values of c make c2 + c a fixed point of fc .

3. If M is the Mandelbrot set, we must show that if c ∈ M, then c̄ ∈ M. In this case, the orbit O(0) = {fcn (0) : n ∈ Z+ }, is bounded, i.e., the set {0, c, c2 + c, (c2 + c)2 + c, . . .}, is bounded. Taking the conjugate gives the orbit {fc̄n (0) : n ∈ Z+ } as being bounded, and this shows that c̄ ∈ M.

36

4. Let c ∈ M lie on the boundary of the primary bulb of the Mandelbrot set.

From the proof of Proposition 14.5.6, these c lie in the boundary of the set {c ∈ √ √ C : |1 − 1 − 4c|/2 < 1/2}, or in other words satisfy |1 − 1 − 4c| = 1. Suppose that c satisfies this equation, then the fixed points of fc are given by fc (z) = z, or √ z 2 − z + c = 0, so the roots are zi = (1 ± 1 − 4c)/2, i = 1, 2, and one of these (say √ z1 ), satisfies |fc′ (z1 )| = |2z1 | = |1 − 1 − 4c| = 1.

37

Section 14.6 1. We show that if f (z) = az 2 + bz + c and q(z) = z 2 − α where α = b2 − 4ac, then the corresponding Newton functions Nf and Nq are conjugate. Set h(z) = 2az + b, then we can check that h ◦ Nf (x) = Nq ◦ h(x), for all x in the domain of Nf .

2. p(z) is a polynomial of degree d > 1 with Newton function Np (z) = z − p(z)/p′ (z). (a) We have seen previously that Np′ (z) = p(z)p′′ (z)/(p′ (z))2 . (b) We have p(z) = z m q(z) where q(0) ̸= 0 (and we are assuming m > 0 is an integer).

Then Np (z) = z − z m q(z)/(mz m−1 q(z) + z m q ′ (z)) = z − zq(z)/(mq(z) + zq ′ (z)), so Np (0) = 0. From (a) we have Np′ (z) =

z m q(z)(m(m − 1)z m−2 q(z) + mz m−1 q ′ (z) + mz m−1 q ′ (z) + z m q ′′ (z)) , (mz m−1 q(z) + z m q ′ (z))2

and this gives Np′ (0) =

m−1 , m

so z = 0 is super-attracting only when m = 1. (c) This problem was solved in Exercises 14.3 # 3. (d) We can now check that if p(z) = z 3 − z 2 = z 2 (z − 1) then m = 2 and Np (z) = 2z 2 − z 6z 2 − 8z + 2 , Np (0) = 0. Also Np′ (z) = , so Np′ (0) = 1/2.

3z − 2 (3z − 2)2 3z − 2z 2 (2 − z 2 )(3 − 4z) − (3z − 2z 2 )(−2z) ′ Also, g(z) = h◦Np ◦h(z) = , so g (z) = , 2 − z2 (2 − z 2 )2 so g ′ (0) = Np′ (∞) = 3/2. z 2 − α2 z 2 + α2 z−α 3. (a) We have f0 (z) = z 2 and Nα = z − = and T (z) = .

2z ( 2z z+α ( )2 ) z−α z 2 + α2 z 2 − 2αz + α2 Then f0 (T (z)) = , and T (Nα (z)) = T = 2 , and z+α 2z z + 2αz + α2 the result follows.

38

(b) If α = i, then since the Julia set of f0 (z) = z 2 is the unit circle S1 , the Julia iz + i set of Nα will be T −1 (S1 ), where T −1 (z) = . Also, notice that T −1 (0) = i, −z + 1 T −1 (1) = ∞ and T −1 (∞) = −i, giving the fixed points of Ni . z = ±i are attracting sin(θ) and ∞ is repelling. We can check that T −1 (eiθ ) = , so belongs to the real 1 − cos(θ) axis, and that every member of R can be written in this form, so the Julia set of Nα is R. The interior of the circle maps to the upper half plane, so this is the basin of attraction of z = i, and the exterior maps to the lower half plane, giving the basin of attraction on z = −i.

Similar analysis for the situation where α = 1 shows that the imaginary axis will now be the Julia set.

4. g(z) = (z − α)m h(z) where m > 1 is an integer and h(α) ̸= 0. (a) Set f (z) = (g(z))1/m and Nm (z) = Nf (z) = z − z−

f (z) m(g(z))1/m = z − = f ′ (z) (g(z))(1−m)/m g ′ (z)

mg(z) . g ′ (z)

(b) Now Nm (z) = z −

m(z − α)m h(z) , so Nm (α) = α, m(z − α)m−1 h(z) + (z − α)m h′ (z)

so z = α is a fixed point for Nm . We can check that ′ Nm (z) = 1 − m +

mg(z)g ′′ (z) (g ′ (z))2

(

m(m − 1)h(z) + 2m(z − α)h′ (z) + (z − α)2 h′′ (z) = 1 − m + m(z − α) h(z) (z − α)2m−2 [mh(z) + (z − α)h′ (z)]2 ′ It follows that Nm (α) = 0, so z = α is super-attracting for Nm .

2m−2

) .

(c) g(z) = (z − a)2 (z − b), so we find N2 and from (a) we have N2 (z) = z −

2(z − a)2 (z − b) z 2 + az − 2ab . = 2(z − a)(z − b) + (z − a)2 3z − a − 2b

We now we use a calculation to show that p = h ◦ N2 ◦ h−1 , where p(z) = z 2 − 3/4 and 3z + a − 4b h(z) = , to give the conjugacy. We can check using Mathematica that 2(z − a) p ◦ h = h ◦ N2 to give our result (see Mathematica Labs).

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5. p(z) = z 3 − 1 and Np (z) = (2z 3 + 1)/3z 2 . (a) It is straightforward to show by direct calculation that BNp (ω) = ωBNp (1), where ω ̸= 1 is a cube root of unity. Similarly for BNp (ω 2 ) = ω 2 BNp (1). (b) If Npn (z) → 1 as n → ∞, then clearly Npn (z̄) → 1 as n → ∞, so we have symmetry with respect to the real axis. (c) For x ∈ R+ , Np (x) = (2x3 + 1)/3x2 . In this case we can check that Npn (x) → 1 as n → ∞, so R+ ⊂ BNp (1) (use the fact that Np (x) < x for x > 1, so that Npn (x) is a decreasing sequence which must converge to the fixed point 1, and similarly Np (x) > x for 0 < x < 1).

6. Let p(z) be a polynomial having degree d > 1 with Newton function Np (z) = z − p(z)/p′ (z). (z − α)p′ (z) − p(z) p(z) − α = . Hence g(z) = [Np (z) − α]p′ (z) is p′ (z) p′ (z) a polynomial with g(α) = 0, so (z − α) is a factor. Also, g ′ (z) = (z − α)p′′ (z) + p′ (z) − p′ (z), so g ′ (α) = 0. It follows that (z − α)2 is a factor of g(z), i.e., g(z) = (z − α)2 q(z) for some polynomial q(z).

Thus (Np (z) − α)p′ (z) = (z − α)2 q(z), so that

(a) Np (z) − α = z −

q(z) Np (z) − α = (z − α)2 ′ . p (z)

(b) Now g ′′ (z) = (z − α)p′′′ (z) + p′′ (z), so if p′′ (α) = 0, then g ′′ (α) = 0, and the result follows as before. p(z) z 2 (2z − a − 1) = p′ (z) a − 2az + z(−2 + 3z) 2 2z(z − 1)(a + a − 4az + z(−1 + 3z)) Np′ (z) = = 0 when z = 0, z = 1, z = a and (a − 2az + z(−2 + 3z))2 z = (a + 1)/3. (c) pα (z) = z(z − 1)(z − α). Np (z) = z −

(z − a)2 (2z − 1) , so that when a = 1/2, (z − 1/2)3 is a ((2 − 3z)z + a(−1 + 2z)) factor of Np (z) − α.

(d) Np (z) − α = −

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7. Suppose that p(z) = r(z − a)(z − b)(z − b) where a, b and c are distinct, so p is a polynomial having three distinct roots. We also have pα (z) = z(z − 1)(z − α).

To show that there exists α such that Np and Npα = Nα are conjugate, it suffices to show there exists h : C → C, h(x) = Az + B, with pα (h(z)) = p(z) for some α a−z a−c (see Proposition 11.3.1). We can check that if we set h(z) = and α = a−b a−b then h(a) = 0, h(b) = 1 and h(c) = α then Np and Nα are conjugate. (h is mapping fixed points of Np (zeros of p) to fixed points of Nα (zeros of pα ), so must be a conjugacy). We can check that p(z) = pα (Az + B). Note: To find h we can use a standard cross ratio method to find a linear map which sends the zeros of one cubic to those of another - consider the ratios (z, a, b, c) = (w, 0, 1, α), by which we mean (z − a)(b − c) (w − 0)(1 − α) = . Solving for w gives (z − c)(b − a) (w − α)(1 − 0) (a − z)(b − c)α h(z) = w = . z(−a + c + aα − bα) + ab − ac − αac + αbc a−z a−c . This gives h(z) = , To simplify this we set −a + c + aα − bα = 0, or α = a−b a−b having the required properties.

The problem can be completed by using a result of Lucas (1874) that says the critical points of the polynomial p(z) are contained in the convex hull of the roots of p(z) (see P. Blanchard [17], where more detail is given).

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Section 14.7 1. Let z, w ∈ C. (i) Set z = x + iy and w = u + iv where x, y, u, v are real, then ez+w = ex+u+i(y+v) = ex+u (cos(y + v) + i sin(y + v)) = ex eu (cos(y) cos(v) − sin(y) sin(v) + i sin(y) cos(v) + i cos(y) sin(v)) = ex (cos(y) + i sin(y))eu (cos(v) + i sin(v)). (ii) (ez )n = enx (cos(y) + i sin(y))n = enx (cos(ny) + i sin(ny)) = enz . (iii) e2kπi = cos(2kπ) + i sin(2kπi) = 1 + i × 0 = 1.

2. To see that (ez )w ̸= ezw in general, set z = iπ/2 and w = i, then ez = i, ezw = e−π/2 and (ez )w = ii = ei log(i) = ei(iπ/2+2kπi) = e−π/2−2kπ , for k ∈ Z.

3. (i) Using the same notation as in Exercise 1, sin(z + w) = (ei(z+w) − e−i(z+w) )/2i = (eiz eiw − e−iz e−iw )/2i.

On the other hand sin(z) cos(w) + cos(z) sin(w) = (eiz − e−iz )(eiw + e−iw )/4i + (eiz + e−iz )(eiw − e−iw )/4i = (2eiz eiw − 2e−iz e−iw )/4i and these are equal, so the result follows.

(ii) A similar calculation shows that

cos(z + w) = cos(z) cos(w) − sin(z) sin(w). (iii) tan(z + w) =

sin(z) cos(w) + cos(z) sin(w) tan(z) + tan(w) sin(z + w) = = . cos(z + w) cos(z) cos(w) − sin(z) sin(w) 1 − tan(z) tan(w)

(iv) sin(z + 2π) = (eiz+2πi − e−iz−2πi )/2i = (eiz − e−iz )/2i = sin(z) and similarly for cos(z + 2π). Now sin(z + π) = (eiz+πi − e−iz−πi )/2i = − sin(z) and similarly cos(z + π) = − cos(z), so tan(z + π) = tan(z).

4. Consider f : S → C \ {0}, where S = {z ∈ C : z = x + iy, x, y ∈ R, −π < y ≤ π} and f (z) = ez . We show that f is one-to-one and onto.

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Suppose that ez1 = ez2 where z1 = x1 + iy1 and z2 = x2 + iy2 , then equating real and imaginary parts, ex1 cos(y1 ) = ex2 cos(y2 ), and ex1 sin(y1 ) = ex2 sin(y2 ).

Squaring and adding gives e2x1 = e2x2 , so x1 = x2 . Then we see that cos(y1 ) = cos(y2 ) and sin(y1 ) = sin(y2 ), together with the restriction on y gives y1 = y2 , so z1 = z2 , and f is one-to-one.

Given w ∈ C \ {0}, say w = u + iv, we find z ∈ S with ez = w, or ex cos(y) = u, ex sin(y) = v. Clearly we can find x with e2x = u2 + v 2 (since w ̸= 0). Also we can find −π < y ≤ π with tan(y) = v/u, so f is onto.

5. Now f : T → C \ {±1}, where T = {z ∈ C : z = x + iy, x, y ∈ R, −π/2 < y < π/2} and f (z) = tan(z). From what was said in 14.7.9 and in a similar manner to the previous solution, we see that f is one-to-one and onto.

6. We show that tan(2 arctan(tan z)) = tan(2z) for z ∈ C. We know that arctan(tan z) = z for all z ∈ T = {z ∈ C : −π/2 < ℜ(z) < π/2}. If ℜ(z) does not lie between −π/2 and π/2, there exists n ∈ Z with −π/2 < ℜ(z +nπ) < π/2, so that tan(z) = tan(z + nπ), and then arctan(tan(z)) = arctan(tan(z + nπ)) = nπ + z, so tan(2(nπ + z)) = tan(2z) and the result follows.

7. To show f (z) = ez is continuous at z0 ∈ C, let ϵ > 0. We must find δ > 0 such that if z ∈ C with |z − z0 | < δ, then |ez − ez0 | < ϵ.

From the definition |ez − ez0 |2 = |ex (cos(y) + i sin(y)) − ex0 (cos(y0 ) + i sin(y0 ))|2 = |(ex cos(y)−ex0 cos(y0 ))+i(ex0 sin(y) −ex0 sin(y0 )|2 = (ex −ex0 )2 +2ex+x0 (1−cos(y−y0 )).

Now we can use the continuity of the real functions ex and cos(y), to finish the proof.

8. f (z) = sin(z). Set z = ix where x ∈ R, then f (ix) = (e−x − ex )/2i, so if g(x) = |f (ix)|, then for x > 0 we have g(x) = (ex − e−x )/2 = sinh(x) > x. This is a

43

function whose iterates increase, having no fixed points, so we must have g n (x) → ∞ as n → ∞. It follows that |f n (z)| → ∞, for z purely imaginary, as n → ∞. x 2 − a2 . We have 9. f (x) = x2 + a2 , and so the Newton function is Nf (x) = 2x { 2x + 1; −1 ≤ x < 0 C(x) = and h : (−1, 1) → R, h(x) = a tan(πx/2). We then 2x − 1; 0 ≤ x < 1 have tan2 (πx/2) − 1 Nf ◦ h(x) = Nf (a tan(πx/2)) = a = −a cot(πx).

2 tan(πx/2 Also h ◦ C(x) = a tan(π(2x ± 1)/2) = a tan(πx ± π/2) = −a cot(πx), so that Nf ◦ h = h ◦ C. We now check that C ◦ k(x) = k ◦ D, so that C and D are conjugate. We have k : [−1, { 1) → [0, 1), k(x) = (x + 1)/2, clearly a linear x + 1; −1 ≤ x < 0 homeomorphism with k(C(x)) = = D(k(x)), and conjugacy x: 0≤x<1 follows. It now follows that Nf and D are conjugate.

10. We know that D : [0, 1) → [0, 1), D(x) = 2x (mod 1) is chaotic. By the previous exercise D is conjugate to Nf : R → R, Nf (x) = x/2 − a2 /(2x). It follows that Nf is chaotic on R. 11. If g = f /f ′ , then

( )( ) [f ′ (z)]2 f (z)f ′ (z) f (z) Ng (z) = z−g(z)/g (z) = z− ′ = z− . f (z) f ′ (z)f ′ (z) − f (z)f ′′ (z) [f ′ (z)]2 − f (z)f ′′ (z) Differentiating gives [ ′ ] (f (z))4 − [f (z)f ′′ (z)]2 − [f ′ (z)f ′′ (z) − f (z)f ′′′ (z)]f (z)f ′ (z) ′ Ng (z) = 1 − . [(f ′ (z))2 − f (z)f ′′ (z)]2 ′

f (z)[(f ′ (z))2 f ′′ (z) − 2f (z)(f ′′ (z))2 + f (z)f ′ (z)f ′′′ (z)] [[f ′ (z)]2 − f (z)f ′′ (z)]2 Substituting z = α clearly gives Ng′ (α) = 0. f ′′ (α) We use Mathematica to show that Ng′′ (α) = − ′ . (Note: There is an error f (α) in the question, where it is not correct that Ng′′ (α) = 0). This can be done by setting =−

P[z_]:= z-(f[z]/f’[z])/D[(f[z]/f’[z]),z],

44

and then finding P’[z], P’’[z].

To substitute P (α) = 0, use the Mathematica substitution command: P’’[z]/.{P[a]->0}.

We are using P in place of N since Mathematica reserves N for a different command. Now we can find Ng′′′ (α) by finding P ′′′ [z] and then using the substitution.

P’’’[z]/.{P[a]->0}.

This gives Ng′′′ (α) = −4

(

[ ]2 ) f ′′′ (α) 3 f ′′ (α) − , which is not quite the same as f ′ (α) 4 f ′ (α)

Sf (α)!

Note If we set g = f /(f ′ )1/2 , then we can check that Ng′ (α) = Ng′′ (α) = 0 and that Ng′′′ (α) = −Sf (α).

ANSWERS TO CHAOTIC DYNAMICS TEXT: PART 1, CHAPTERS 1-14 Please contact me at ggoodson@towson.edu, to ask questions or let me know about mistakes/typos. Any comments, suggestions for improvement or new problems are most welcome.

Many thanks, Geoff Goodson.

Section 1.1 1. Lµ (x) = µx(1 − x), so that L2µ (x) = µ(µx(1 − x))(1 − µ(1 − x)) = −µ3 x4 + +2µ3 x3 + µ2 (µ − 1)x2 + µ2 x, and we can check (use a computer algebra system) that L3µ (x) = µ3 (1 − x)x(1 − µ(1 − x)x)(1 − µ2 (1 − x)x(1 − µ(1 − x)x)) = −µ7 x8 + 4µ7 x7 − 6µ7 x6 − 2µ6 x6 + 4µ7 x5 + 6µ6 x5 − µ7 x4 −6µ6 x4 − µ5 x4 − µ4 x4 + 2µ6 x3 + 2µ5 x3 + 2µ4 x3 − µ5 x2 − µ4 x2 − µ3 x2 + µ3 x.

2. (i) x0 = 2 and xn+1 = xn /3 + 2. We apply Example 1.1.3 with a = 1/3, b = 2 2 2 and x0 = 2, to give the closed form solution xn = (2 + )(1/3)n + = 1/3 − 1 1 − 1/3 −1/3n + 3. (ii) In this case xn+1 = −3xn + 4, so that a = −3, b = 4, where x0 = −1, so that 4 4 xn = (−1 + )(−3)n + = (−2)(−3)n + 1. −3 − 1 1+3 3. The logistic difference equation is xn+1 = µxn (1 − xn ). (i) Set µ = 2 and xn = (1 − yn )/2 then the logistic equation becomes 2(1 − yn ) (1 − yn ) (1 − yn+1 ) = (1 − ).

2 2 2 n Simplifying gives yn+1 = yn2 , so that yn = y02 . Now (1 − y0 )/2 = x0 , so y0 = 1 − 2x0 n 1 − (1 − 2x0 )2 and xn = is the closed form solution.

2 1

2

(ii) Set µ = 4, and xn = sin2 (θn ) and the logistic equation becomes sin2 (θn+1 ) = 4 sin2 (θn )(1 − sin2 (θn )). Simplifying gives sin2 (θn+1 ) = 4 sin2 (θn ) cos2 (θn ) = (2 sin(θn ) cos(θn ))2 = (sin(2θn ))2 .

We conclude that θn+1 = 2θn , a linear difference equation that is easy to solve. We see that θn = 2n θ0 , so substituting back (using x0 = sin2 (θ0 )), gives √ xn = sin2 (2n arcsin( x0 )).

4. This gives rise to an affine difference equation:

A(n + 1) = A(n) + A(n)r/1200 − M = (1 + r/1200)A(n) − M,

A(0) = P.

Set a = 1 + r/1200, b = −M and x0 = P , to give A(n) = (P +

−M −M )(1 + r/1200)n + 1 + r/1200 − 1 1 − 1 − r/1200

= (P − 1200M/r)(1 + r/1200)n +

1200M . r

Now set P = $ 100,000, M = $ 1,000 and r = 4, so that we find when A(n) = 0 as this is when the loan will be paid off (after n months), i.e., A(n) = (100000 − 1200 × 1000/4)(1 + 4/1200)n + 1200 × 1000/4 = 0.

Solving gives n = 121.842 months, so we can pay off after 10 years and 2 months.

5. Set A = A(0), then A(1) = A+A.r1 /100−A.r2 /100 = A−A.r/100 = A(1−r/100), where r = r2 − r1 , and generally A(n) = A(1 − r/100)n , a linear dynamical system.

When r1 = 3, r2 = 3.65, r = 0.65%, so that A(10) = 500, 000(1 − .65/100)10 =, A(20) = 500, 000(1 − .65/100)20 =, A(30) = 500, 000(1 − .65/100)30 = 411, 155.54.

3

{

µx; 0 ≤ x < 1/2 . If µ ∈ (0, 2], and x ∈ [0, 1/2), then 0 ≤ µx < µ(1 − x); 1/2 ≤ x ≤ 1 1. Also if x ∈ [1/2, 1], then 1 − x ∈ [0, 1/2], so µ(1 − x) ∈ [0, 1]. Therefore we have Tµ (x) ∈ [0, 1] for all x ∈ [0, 1], so Tµ is a dynamical system on [0, 1].

6. Tµ (x) =

7. f (x) = x2 + bx + c where f : [0, 1] → [0, 1], so f (0) = c and f (1) = 1 + b + c, so we require 0 ≤ c ≤ 1 and 0 ≤ 1 + b + c ≤ 1. The maximum for f will occur at one of these points. The minimum for f will occur where f ′ (x) = 0. Now f ′ (x) = 2x + b = 0 when x = −b/2. f (−b/2) = c − b2 /4, so in addition we require 0 ≤ c − b2 /4 ≤ 1.

Using this information, we can plot a region in the b, c-plane for which points (b, c) gives ([0, 1], f ) as a dynamical system.

8. (a) f (x) = x3 − 3x, I = [−1, 1]. f (1) = −2, f (−1) = −1 + 3 = 2, so this is not a dynamical system on [−1, 1] f ′ (x) = 3x2 − 3 = 0 when x = ±1, and f (1), f (−1) lie within [−2, 2] (using the fact the the maxima/minima of such a function occur either at the end points or where the derivative is zero), we see that f is a dynamical system on I = [−2, 2]. (b) f (x) = 2x3 − 6x, (i) I = [−1, 1]. In this case f (−1) = −2 + 6 = 4, so it is not a dynamical system √ on [−1, √ 1]. (ii) I = [− 7/2, 7/2] Although the end points are fixed, ±1 lie within the interval, with f (±1) = ∓4, so f does not define a dynamical system. (

) √ 1−x 9. Is fµ (x) = µx , (0 < µ < (5 5 + 11)/2), a dynamical system on [0, 1]?

1+x ( ) 1 − x − x2 ′ Now fµ (0) = 0 and fµ (1) = 0. Also fµ (x) = 2µx = 0 when (1 + x)2 √ x = ( 5 − 1)/2 (ignoring the negative root). Now 2

√ fµ (( 5 − 1)/2) = µ

) ( √ )2 ( ) (√ √ 5 5 − 11 5−1 1 − ( 5 − 1)/2 √ =µ , 2 2 1 + ( 5 − 1)/2

√ 5 5 + 11 . so we require 0 < µ ≤ 2

4 n

10. Find f n (x) for (i) f (x) = x2 . Since f 2 (x) = x4 , f 3 (x) = x8 , we have f n (x) = x2 . (ii) f (x) = |x + 1|, f 2 (x) = ||x + 1| + 1| = |x + 1| + 1, and generally f n (x) = |x + 1| + n − 1, n ≥ 1.  22 x; 0 ≤ x < 1/4   {  2 2x; 0 ≤ x < 1/2 2 x − 3/4; 1/4 ≤ x < 1/2 (ii) f (x) = . We see that f 2 (x) = , 2 2x − 1; 1/2 ≤ x < 1 2 x − 1/2; 1/2 ≤ x < 3/4    22 x − 1; 3/4 ≤ x < 1 and we continue in this way to find f n (x).

11. f (x) =

f

n+1

2 . The statement is true for n = 1. Suppose it holds for n, then x+1

( n ) 2 (x + 2) + (−1)n (2x − 2) 2n+1 (x + 2) + (−1)n (2x − 2) (x) = f = , 2n (x + 2) − (−1)n (x − 1) 2n+1 (x + 1) − (−1)n+1 (x − 1)

so by induction, it is true for all n ≥ 1.

12. (a) The first 10 terms are 0, 0, 1, 1, 2, 4, 7, 13, 24, 44, . . .. [ ][ ] 1 1 Fn+1 (b) F · vn = = vn+1 .

1 0 Fn   1 1 1 (c) If T =  1 1 0 , then wn+1 = T · wn .

1 0 0 13. We have Lµ (x) = µx(1 − x), and we wish to find a closed form solution to the difference equation xn+1 = Lµ (xn ), when µ = −2. Set xn = (1 − f (θn ))/2, where f is to be chosen later. Since xn+1 = −2xn (1 − xn ), we have: 1 − f (θn+1 ) = −2 2

(

1 − f (θn )) 2

)(

(1 − f (θn )) 1− 2

and simplifying gives f (θn+1 ) = −f (θn )2 + 2.

) = (f (θn ) − 1)(f (θn ) + 1)/2,

5

√ ) π − 3θ Set f (θ) = 2 cos , and substitute into the last equation to give 3 ( ) ( ) √ √ π − 3θn+1 π − 3θ n 2 cos = −4 cos2 +2 3 3 ( ) ) ( √ √ π − 3θn 2π − 2 3θn 2 = 2(1 − 2 cos ) = −2 cos , 3 3 (

so that

) ( ) ( ) √ √ √ 2π − 2 3θn π + 2 3θn 3θn+1 = − cos = cos . cos 3 3 3 √ √ Hence (π − 3θn+1 )/3 = (π + 2 3θn )/3, or θn+1 = −2θn , so that in general θn = (−2)n θ0 . It follows that [ ( )] √ 1 − f (θn ) 1 − f ((−2)n θ0 ) 1 π − 3(−2)n θ0 xn = = = 1 − 2 cos , 2 2 2 3 (

π−

where f (θ0 ) = 1 − 2x0 , θ0 = f −1 (1 − 2x0 ).

6

Section 1.2 1. f (x) = x2 + bx + c where f : R → R has a fixed point when f (x) = x, or x2 + bx + c = x, x2 + x(b − 1) + c = 0. Solving this quadratic gives x=

1−b±

√

(b − 1)2 − 4c .

2

Consequently the requirement for f to have a fixed point is (b − 1)2 ≥ 4c. In the case that fc (x) = x2 + c, b = 0, so we require 1 ≥ 4c.

2. Let f (x) = ax3 + bx2 + cx + d where a ̸= 0 and g(x) = f (x) − x. If a > 0, then limx→∞ g(x) = ∞ and limx→−∞ g(x) = −∞, so there are both positive and negative values of g(x). The Intermediate Value Theorem then implies that g has at least one zero, which must be a fixed point of f . Similarly if a < 0.

3. If f (x) = 1 − |x|, then f (1/2) = 1/2 and f (n/2) = n/2 for n ∈ Z odd. We see that f 2 (x) = |x| if |x| ≤ 1 and f 2 (x) = 2 − |x| if |x| > 1, so that f 2 (x) = x for 0 ≤ x ≤ 1.

The points −1 ≤ x < 0 are eventually periodic.

4 f n (c) = c where c is unique. Suppose f (c) = c1 ̸= c, then f n (c1 ) = f n+1 (c) = f (c) = c1 , contradicting the uniqueness of c, so we must have f (c) = c.

5. Use a computer algebra system.

6. Consider Lµ : [0, 1] → [0, 1], Lµ (x) = µx(1 − x). (a) If Lµ (x) = 0, then clearly x = 0 or x = 1.√If Lµ (x) = 1, then µx(1 − x) = 1, so µx2 − µx + 1 = 0. Solving gives x = (µ ± µ2 − 4µ)/2µ, which is not real for 0 < µ < 4, since in this case µ(µ − 4) < 0. (b) If Lµ (x) = 1 − 1/µ, then µx2 − µx + 1 − 1/µ = 0, so that x = (µ ±

√ √ µ2 − 4µ(1 − 1/µ)/2µ = (µ ± µ2 − 4µ − 4)/2µ = (µ ± (µ − 2))/2µ.

7

This simplifies to 1/µ or 1 − 1/µ for √1 < µ ≤ 2. If we now solve Lµ (x) = 1/µ we get √ 2 x = (µ ± µ − 4µ1/µ)/2µ = (µ ± µ2 − 4)/2µ, which has no solution for this range of values of µ. However, if µ > 2, then there are additional solutions, thus giving the answer to (c). (d) Now set µ = 5/2, then solving 5/2x2 − 5/2x = x gives x = 0 or x = 3/5. Now solve L5/2 (x) = 3/5: 25x2 − 25x + 6 = 0 to give x = 3/5 or x = 2/5. We can continue in this way indefinitely. √ 7. (a) f (x) = (1 + x)−1 . Solving f (x) = x gives x2 + x − 1 = 0, or x = (−1 ± 5)/2 as the fixed points. Solving f 2 (x) = x also gives x2 + x − 1 = 0, so there are no additional solutions, so no period-2 points. (b) Clearly x1 = f (x0 ), and ( ) Fn−1 x0 + Fn Fn x0 + Fn+1 1 f (xn ) = f = , = Fn−1 x0 +Fn Fn + Fn+1 Fn+1 x0 + Fn+2 1 + F x +F n 0

n+1

so the result follows by induction.

8. f : S → S where S = {a1 , a2 , . . . , ap }, a finite set. Consider the set {a1 , f (a1 ), f 2 (a1 ), . . .}, a finite set, so it must eventually start to repeat itself, so the points of S must either be periodic or eventually periodic.

9. The function f (x) = x + 1/2, 0 ≤ x < 1/2 and f (x) = x − 1/2, 1/2 ≤ x ≤ 1, is one-to-one and onto (but not continuous on [0, 1]), and has no fixed point.

8

Section 1.4 6 −1 = x gives 6−x = x2 , or x2 +x−6 = 0, (x+3)(x−2) = 0, so the x 6 fixed points are x = 2 and x = −3. Now f ′ (x) = − 2 , so that f ′ (2) = −6/4 = −3/2 x and x = 2 is repelling. f ′ (−3) = −6/9, and x = −3 is attracting.

1. Solving f (x) =

2. f : R → R is differentiable everywhere, and f ′ (x) ̸= 1 for all x. Suppose that there are two distinct fixed points c1 and c2 , say c1 < c2 . By the Mean Value Theorem f (c1 ) − f (c2 ) c1 − c2 there exists a ∈ (c1 , c2 ) with f ′ (a) = = = 1, a contradiction. c1 − c2 c1 − c2 3. Use a computer algebra system.

4. (a) We have Sµ (x) = µ sin(x). When µ = π/2, π/2 sin(π/2) = π/2 · 1 = π/2, ′ ′ so that c = π/2 is a fixed point. Now Sπ/2 (x) = π/2 cos(x), and so Sπ/2 (π/2) = π/2 cos(π/2) = 0, so the fixed point is super-attracting. (b) Cµ (x) = µ cos(x). Here we see that if µ = −π, then C−π (x) = −π cos(x) and ′ C−π (π) = −π cos(π) = −π(−1) = π, so that π is a fixed point. Also C−π (x) = ′ π sin(x), so that C−π (π) = 0, and x = π is a super-attracting fixed point.

1 has no points of period n > 1 for if f n (x) = x, then from Exercises 1+x 1.1 # 11, we have 5. f (x) =

2n+1 (x + 2) + (−1)n (2x − 2) = x, 2n+1 (x + 1) − (−1)n+1 (x − 1) The only solutions this has are the fixed points of f .

6. f (x) = x + 1/x > x for x > 0. In a similar way, f 2 (x) > f (x) and the sequence (f n (x)) is increasing. If it is bounded above, it must converge, to c (say). Then c must be a fixed point, contradicting the fact that f clearly has no fixed points.

9

x2 + 1 x 1 = − . Then 2x 2 2x 1 x 1 1 x 1 1 Nf2 (x) = ( − ) − x = − − 1 2 2 2x 4 4x x − x1 2( 2 − 2x )

7. f (x) = x2 + 1 and so Nf (x) = x −

=

x 1 x x4 − 6x2 + 1 − − 2 = .

4 4x x − 1 4x(x2 − 1)

−6x +1 = x, so that x4 − 6x2 + 1 = 4x4 − 4x2 , or Solving Nf2 (x) = x, we set x4x(x 2 −1) √ 3x4 + 2x2 − 1 = 0, (3x2 − 1)(x2 + 1) = 0, so the only solutions are x = ±1/ 3, giving two period-2 points.

4

2

f (x) f ′ (x) = c − lim = c − 0 = c (using L’Hospital’s rule). x→c f ′ (x) x→c f ′′ (x)

8.(a) lim Nf (x) = c − lim x→c

f (x)f ′′ (x) ′ , which is of the form 0/0, so we can apply (b) We have lim Nf (x) = lim x→c x→c [f ′ (x)]2 L’Hospital’s rule to give 1 f (x)f ′′′ (x) f ′ (x)f ′′ (x) + f (x)f ′′′ (x) ′ = + lim ′ Nf (c) = lim x→c 2f ′ (x)f ′′ (x) 2 x→c 2f (x)f ′′ (x) =

f (x) f ′′′ (x) 1 1 1 1 + lim ′ lim ′′ = +0= .

2 2 x→ c f (x) x→c f (x) 2 2

(Where we use the fact that limx→ c ff′(x) = 0, f ′′′ (x) is continuous at x = c and (x) f ′′ (c) ̸= 0). (c) f (x) = x3 − x2 , f (0) = 0 and Nf (x) = x − Nf′ (x) =

6x2 − 8x + 2 1 → as x → 0.

2 (3x − 2) 2

x3 − x2 x2 − x 2x2 − x = x − = , so 3x2 − 2x 3x − 2 3x − 2

9. We show that if Nf′ has a removable discontinuity at x = c, then it can be removed (k − 1) by setting Nf′ (c) = . k We are assuming that f is a polynomial of the form f (x) = (x − c)k h(x), where h(c) ̸= 0 and k > 1. We know that Nf′ (x) =

f (x)f ′′ (x) [f ′ (x)]2

10

=

(x − c)k h(x)[k(k − 1)(x − c)k−2 h(x) + 2k(x − c)k−1 h′ (x) + (x − c)k h′′ (c)] [k(x − c)k−1 h(x) + (x − c)k h′ (x)]2 =

as x → c.

(x − c)2k−2 h(x)[k(k − 1)h(x) + 2k(x − c)h′ (x) + (x − c)2 h′′ (x)] (x − c)2k−2 [kh(x) + (x − c)h′ (x)]2 k(k − 1) k−1 h(c)[k(k − 1)h(c)] = = → 2 2 2 k (h(c)) k k

10. We apply Newton’s method to the function g(x) =

f (x) . The Newton function f ′ (x)

for g is

f (x)f ′ (x) , [f ′ (x)]2 − f (x)f ′′ (x) so that f (x) = 0 if and only if Ng (x) = x. Differentiating gives [ ′ ] (f (x))4 − [f (x)f ′′ (x)]2 − [f ′ (x)f ′′ (x) − f (x)f ′′′ (x)]f (x)f ′ (x) ′ Ng (x) = 1 − , [(f ′ (x))2 − f (x)f ′′ (x)]2 Ng (x) = x −

and substituting x = c gives Ng′ (c) = 0, so that c is a super-attracting fixed point for Ng .  0 ≤ x < 1/4  4x; 2 − 4x; 1/4 ≤ x < 1/2 . Clearly x = 0 is a repelling fixed point. If 11. f (x) =  0; 1/2 ≤ x < 1 x ∈ [1/2, 1], then f (x) = 0, so f n (x) → 0 as n → ∞. Now x = 2/5 is a fixed point of f , so cannot be attracted to 0, i.e., x = 0 is not globally attracting. It can be seen that if 1/8 < x < 1/4, then f (x) > 1/2 and then f 2 (x) = 0. Similarly if 1/2 < x < 3/8, f (x) > 1/2 etc. If we set x = ·a1 a2 a3 . . ., ai  ∈ {0, 1, 2, 3}, the 4-nary expansion  ·a2 a3 a4 . . . ; a1 = 0 ·a′ a′ a′ . . . ; a2 = 1 , of x ∈ [0, 1], then we can check that f (x) =  2 3 4 0; a2 = 2, or a2 = 3 where a′i = 3 if ai = 0, a′i = 2 if ai = 2 and a′i = 0 if ai = 3. It is then seen that ·12 12 12 12 . . . = 2/5 is the fixed point. Points that are not in the set {x ∈ [0, 1] : f n (x) → 0} have a 4-nary expansion of the form: ·000 . . . 00 1 2 000 . . . 00 1 2 000 . . . i.e., starting with n zeros for some n ≥ 0, with the possibility of n = 0, followed by a single 1, followed by a single 2, then some m ≥ 0, 0’s, i.e., the expansion only involves 0’s, followed by 12, followed by 0’s, followed by 12 and continues in this way. In fact ·012 012 012 . . . = 2/21 is a period-3 point.

11

We see that f has a 2-cycle {2/17, 8/17} and two 3-cycles: {2/21, 8/21, 10/21} and {2/65, 8/65, 32/65}, so the set {x ∈ [0, 1] : f n (x) → 0} excludes all such points and is actually quite complicated. It is instructive to graph f (x) and look at the iterates using a computer algebra system.

12. Definition 1.4.1(i) of a stable fixed point can be stated as saying c is a stable fixed point of f : I → I if: ∀ ϵ > 0, ∃ δ > 0, ∀ x ∈ I, (|x − c| < δ, −→ ∀ n ∈ Z+ , |f n (x) − c| < ϵ).

The negation is then: ∃ ϵ > 0, ∀ δ > 0, ∃ x ∈ I, (|x − c| < δ) ∧ (∃ n ∈ Z+ , |f n (x) − c| ≥ ϵ). f (x) > f ′ (x) x since f (x)/f ′ (x) < 0 for all x. Inductively, we see that Nfn (x) > Nfn−1 (x) > Nfn−2 (x) > · · · > x. It follows that we cannot have Nfn (x) = x for any x ∈ R. 13. Let f : R → (0, ∞) with f ′ (x) < 0 for all x ∈ R. Then Nf (x) = x −

12

Section 1.5 x3 x 3x2 1 + = x, gives x = 0 or x = ±1. Now f ′ (x) = + , 2 2 2 2 ′ so |f (0)| = 1/2 giving stability, whilst |f (±1)| = 2 > 1, so these are unstable.

1. (i) Solving f (x) = x, or

1 (ii) If f (x) = arctan(x) = x, then x = 0 is the only fixed point. f ′ (x) = , 1 + x2 −2x , so f ′′ (0) = 0. We f ′ (0) = 1 giving a non-hyperbolic fixed point. f ′′ (x) = (1 + x2 )2 can check that f ′′′ (0) < 0, and by Theorem 1.5.3 (iii) we have a stable fixed point. (iii) If f (x) = x3 + x2 + x = x, then x = 0 and x = −1 are the fixed points. f ′ (x) = 3x2 + 2x + 1, so f ′ (0) = 1 gives x = 0 a non-hyperbolic fixed point. f ′′ (x) = 6x + 2, so f ′′ (0) ̸= 0 and by Theorem 1.5.3 (i) x = 0 is semi-stable. Since f ′ (−1) = 3−2+1 = 2, x = −1 is unstable. (iv) If f (x) = x3 − x2 + x = x, then x = 0 and x = 1 are the fixed points. f ′ (x) = 3x2 −2x+1, so f ′ (0) = 1 giving x = 0 a non-hyperbolic fixed point. f ′′ (x) = 6x−2, so f ′′ (0) = −2 and by Theorem 1.5.3 (i) x = 0 is semi-stable. Since f ′ (1) = 3−2+1 = 2, x = 1 is unstable. { 3x/4; x ≤ 1/2 (v) If f (x) = , then x = 0 is the only fixed point. Since 3(1 − x)/4; x > 1/2 f ′ (0) = 3/4, x = 0 is stable.

1±

√

1 − 4c .

2 We have seen that this has two solution for c < 1/4, which are both hyperbolic fixed points. Set 2. (a) fc (x) = x + c. Solving x + c = x, or x − x + c = 0 gives x = 2

2

x0 =

1+

2

√

√ 1 − 1 − 4c 1 − 4c , and x1 = .

2 2 (

′

′

′

1+

√

1 − 4c 2

)

Now f (x) = 2x and so for the stability of x0 |f (x0 )| = |f |= √ |1 + 1 − 4c| < 1. However, this can never so )x0 is unstable. √ ( happen √ 1 − 1 − 4c For the stability of x1 |f ′ (x1 )| = |f ′ | = |1 − 1 − 4c| < 1. This 2 √ is equivalent to −1 < 1 − 1 − 4c < 1, or c > −3/4. We conclude that x1 is a stable fixed point for −3/4 < c < 1/4.

13 ′ (b) When c = 1/4 we have a hyperbolic fixed point at c = 1/2. f1/4 (1/2) = 1 and ′′ f1/2 (1/4) = 2, so we have semi-stability.

3. (a) f (x) = −2x3 + 2x2 + x = x gives x = 0 and x = 1. f ′ (x) = −6x2 + 4x + 1, so f ′ (0) = 1 and f ′ (1) = −1, so we have two non-hyperbolic fixed points. f ′′ (x) = −12x + 4, so f ′′ (0) = 4, giving semi-stability. Also, f ′′′ (x) = −12, so Sf (1) = 12 − 3/2[−8]2 < 0, so x = 1 is stable. (b) f (x) = ax3 + bx2 + cx + d where f (0) = 0 and f (1) = 1. It follows that d = 0 and a + b + c = 1. There are four cases to consider.

Case 1. f ′ (0) = 1 and f ′ (1) = 1. Then since f ′ (x) = 3ax2 + 2bx + c, c = 1 and 3a + 2b + 1 = 1 or 3a + 2b = 0, and this, together with a + b = 0 show that a = b = 0, so only a trivial solution in this case.

Case 2. f ′ (0) = 1 and f ′ (1) = −1. Then c = 1 as before, but 3a + 2b + 1 = −1, so we solve a + b = 0 and 3a + 2b = −2, to give a = −2, b = 2, c = 1 and d = 0: f (x) = −2x3 + 2x2 + x, which is just (a) above.

Case 3. f ′ (0) = −1 and f ′ (1) = 1. Then c = −1 and 3a + 2b − 1 = 1, so we solve a + b = 2 and 3a + 2b = 2 to give a = −2, b = 4 to give f (x) = −2x3 + 4x2 − x.

Case 4. f ′ (0) = −1 and f ′ (1) = −1, then c = −1 and 3a + 2b − 1 = −1, so we solve a + b = 2 and 3a + 2b = −0, to give a = −4 and b = 6, so f (x) = −4x3 + 6x2 − x. (c) We have already seen in (a) the stability of the fixed points of f (x) = −2x3 + 2x2 + x.

For f (x) = −2x3 + 4x2 − x, we have f ′ (x) = −6x2 + 8x − 1, f ′ (0) = −1, f ′ (1) = 1, f ′′ (x) = −12x + 8, f ′′ (0) = 8, f ′′ (1) = −4, so x = 1 is semi-stable. f ′′′ (x) = −12 and Sf (0) = 12 − 3/2[8]2 < 0, so we have stability.

Now consider f (x) = −4x3 + 6x2 − x, f ′ (x) = −18x2 + 16x − 1, so we have f ′ (0) = −1, f ′′ (x) = −36x + 16, so f ′′ (0) = 16, f ′′′ (0) = −36. It follows that Sf (0) = 36 − 3/2[16]2 < 0, so x = 0 is stable.

4. (a) f (x) = x + µ cos(x). Solving f (x) = x gives µ cos(x) = 0, which has infinitely many solutions. These are x = (2n + 1)π/2, for n ∈ Z.

14

Now f ′ (x) = 1 − µ sin(x), so ′

{

f ((2n + 1)π/2) = 1 − µ sin((2n + 1)π/2) = 1 − µ(−1) = n

1 + µ; n odd , 1 − µ; n even

so that for n odd and µ > 0, xn = (2n + 1)π/2 is unstable. For n even xn is stable when |1 − µ| < 1, or 0 < µ < 2 (also answering (b)). (c) When µ = 1, f ′ (x) = −1 in the latter case, so we look at the Schwarzian derivative. f ′′ (x) = −µ cos(x), so f ′′ ((2n + 1)π/2) = 0, f ′′′ (x) = µ sin(x) and f ′′′ ((2n + 1)π/2) = µ = 2. so we have Sf ((2n + 1)π/2) = −2 < 0 (all for n even), so we have stability.

5. f (x) = x2 + ax + b, so that p2 + ap + b = p. (i) f ′ (x) = 2x + a, so 2p + a = 1, p = (1−a)/2. It follows that (1−a)2 /4+a(1−a)/2+b = (1−a)/2, or a2 −2a+1 = 4b, (a − 1)2 = 4b. (ii) If f ′ (p) = 1, then we have f ′′ (p) = 2 > 0, so we have stability on the left. (iii) If f ′ (p) = −1, then since f ′′ (x) = 2 > 0 and f ′′′ (x) = 0, we have Sf (p) < 0, so p is asymptotically stable.

6. If f (x) = ex , then f ′ (x) = f ′′ (x) = f ′′′ (x) = ex , so that Sf (x) = f ′′′ (x)/f ′ (x) − 3/2 [f ′′ (x)/f ′ (x)]2 = 1 − 3/2 = −1/2, so is always negative.

If f (x) = sin(x), then f ′ (x) = cos(x), f ′′ (x) = − sin(x) and f ′′′ (x) = − cos(x).

Hence ( )2 − cos(x) 3 − sin(x) − = −1 − 3 tan2 (x)/2 < 0.

Sf (x) = cos(x) 2 cos(x)

[ ]2 ax + b f ′′′ (x) 3 f ′′ (x) 7. Let f (x) = = 0 wherever , then we show that Sf (x) = ′ − cx + d f (x) 2 f ′ (x) it is defined. We can check that f ′ (x) = so that

ad − bc −2c(ad − bc) 6c2 (ad − bc) ′′ ′′′ , f (x) = , and f (x) = , (cx + d)2 (cx + d)3 (cx + d)4

[ ]2 6c2 (ad − bc) (cx + d)2 3 −2c(ad − bc)(cx + d)2 Sf (x) = − (cx + d)4 (ad − bc) 2 (cx + d)3 (ad − bc)

15

[ ] 6c2 3 4c2 = − = 0. (cx + d)2 2 (cx + d)2

{

x sin(1/x); x ̸= 0 . (a) f (x) = x, when x = 0, or sin(1/x) = 1. The 0; x=0 2 latter equation gives x = , n ∈ Z. Now f ′ (x) = sin(1/x) + x cos(1/x) · (4n + 1)π 1 (−1/x2 ) = sin(1/x) − cos(1/x), so that f ′ (2/(4n + 1)π) = sin((4n + 1)π/2) − 0 = 1, x so the fixed point is non-hyperbolic.

8. f (x) =

(b) x = 0 is a fixed point, but fixed points of the form 2/(4n + 1)π become arbitrarily close to zero, for large n, so x = 0 is not an isolated fixed point.

9. f (x) is a polynomial with fixed point f (c) = c. (i) If f ′ (c) = 1 and g(x) = f (x) − x, then g(c) = 0 and g ′ (x) = f ′ (x) − 1, so g ′ (c) = 0.

It follows that (x − c)2 is a factor of g(x). (ii) If |f ′ (c)| = 1 and h(x) = f 2 (x) − x, then h(c) = 0 as in (i) and h′ (x) = f ′ (f (x)) · f ′ (x)−1, so h′ (c) = f ′ (f (c))f ′ (c)−1 = f ′ (c)f ′ (c)−1 = 0 when f ′ (c) = ±1. Therefore (x − c)2 is a factor of h(x). (iii) If f ′ (c) = −1 and h(x) = f 2 (x) − x, then (ii) shows that (x − c)2 is a factor of h(x). It therefore suffices to show that h′′ (c) = 0. Now h′′ (x) = f ′′ (f (x)) · [f ′ (x)]2 + f ′ (f (x)) · f ′′ (x), so h′′ (c) = f ′′ (c)[f ′ (c)]2 + f ′ (c)f ′′ (c) = f ′′ (c) − f ′′ (c) = 0, and the result follows. (iv) L3 (x) = 3x(1 − x), where we have L3 (2/3) = 2/3 and L′3 (2/3) = −1. Then L23 (x) − x = 3(3x(1 − x)(1 − 3x(1 − x)) = −x(3x − 2)3 , so (3x − 2)3 is a factor of L23 (x) − x as was to be verified. (v) f (x) = −2x3 + 2x2 + x, has f (0) = 0, f ′ (0) = 1, f (1) = 1 and f ′ (1) = −1. Now f 2 (x) − x = 2(−2x3 + 2x2 + x) + 2 − 2x3 + 2x2 + x) + (−2x3 + 2x2 + x). We can check that f 2 (x) − x = 4x2 (x − 1)3 (4x4 − 6x2 − 4x − 1), so x2 and (x − 1)3 are factors of f 2 (x) − x as was to be verified.

16

10. Again f (x) is a polynomial with fixed point f (c) = c. (i) Suppose that (x − c)2 is a factor of h(x) = f 2 (x) − x, then h(c) = 0 and h′ (c) = 0.

But h′ (x) = f ′ (f (x))f ′ (x) − 1, so h′ (c) = f ′ (f (c))f ′ (c) − 1 = [f ′ (c)]2 − 1 = 0 that f ′ (c) = ±1 and the fixed point is non-hyperbolic. (ii) If (x − c)3 is not a factor of h(x), then h′′ (c) ̸= 0. Since h′′ (x) = f ′′ (f (x))[f ′ (x)]2 + f ′ (f (x))f ′′ (x), h′′ (c) = f ′′ (c) − [f ′ (c)]2 + f ′ (c)f ′′ (c) ̸= 0, and since [f ′ (c)]2 = 1, we must have have f ′ (c) = 1. (iii) If (x − c)3 is a factor of h(x), and f ′′ (c) ̸= 0, then h′′ (c) = 0 implies f ′ (c) = −1.

11. (a) If g(x) = f (x) − x = cos(x) − x, then g(0) = 1 and g(π/2) = −π/2, so by the Intermediate Value Theorem, g(c) = 0 for some c ∈ (0, π/2), or f (c) = c. (b) If f (x) = cos(x) and x ∈ R, consider f n (x), n ∈ Z+ . We may assume x ∈ [−1, 1], since after the first iteration, this is the case. In particular, we must have c < 1.

Suppose that x ̸= c with x ∈ (0, 1], then by the Mean Value Theorem, there exists a f (x) − f (c) between x and c satisfying f ′ (a) = , so that |f (x) − c| = | sin(a)||x − c|. x−c We can assume that 0 < sin(a) < 1 for 0 < a ≤ 1, so we can find λ, 0 < λ < 1 with |f (x)−c| < λ|x−c|. We now repeat this argument to show that |f n (x)−c| < λn |x−c|, so that f n (x) → c as n → ∞. A similar argument may be used for x ∈ [−1, 0). (c) f (x) can have eventual fixed points. For example, f (−c) = cos(−c) = cos(c) = c, so that −c is eventually fixed. Similarly, f (c + 2nπ) = c, for n ∈ Z. (d) f (x) can have no points p with f 2 (p) = p other than c, since there existence would contradict limn→∞ f n (p) = c.

12. (a) f : R → R has the property that there are two fixed points x0 < x1 with f ′ (x0 ) < 1 and f ′ (x1 ) < 1. We show that there is another fixed point for f lying between x0 and x1 .

Since f ′ (x) is continuous, there exists δ > 0 with f ′ (x) < 1 for all x ∈ (x0 − δ, x0 + δ). Set g(x) = f (x) − x, then g ′ (x) = f ′ (x) − 1 < 0 for x ∈ (x0 − δ, x0 + δ), so g(x) is strictly decreasing on this interval.

17

Hence there exists α ∈ (x0 , x0 + δ) with g(α) < g(x0 ) = 0. Similarly there exists δ , β ∈ (x1 − δ ′ , x1 ) with g(β) > 0. It follows by the IVT that there exists c ∈ (α, β) with g(c) = 0. Now x0 < α < c < β < x1 with f (c) = c. ′

(b) If c is unique, then we show f ′ (c) ≥ 1. Suppose instead that f ′ (c) < 1, then if g(x) = f (x) − x, g ′ (x) = f ′ (x) − 1, and g ′ (c) = f ′ (c) − 1 < 0, so g is decreasing on an interval surrounding c.

However, if c is the unique fixed point, from (a) we must have f (x) < x for x < c and f (x) > x for x > c (if not, then f will have other fixed points). This contradicts g being decreasing near c. (c) Set f (x) = −4x3 + 6x2 − x, then f (0) = 0, f (1) = 1, f ′ (x) = −12x2 + 12x − 1, so f ′ (0) = −1, f ′ (1) = −1. Solving f (x) = x gives 2x(x − 1)(2x − 1) = 0, so we have a third fixed point between x = 0 and x = 1, at x = 1/2.

18

Section 2.2 1. Find the basin of attraction of the fixed points of (i) f (x) = x3 . The fixed points are x = 0, x = ±1. Only x = 0 is attracting, and clearly Bf (0) = (−1, 1). (ii) f (x) = x3 − 3x. Solving f (x) = x gives x3 − 4x = 0, x = 0 or x = ±2. f ′ (x) = 3x2 − 3, f ′ (0) = −3, f ′ (±2) = 9, so no point is attracting. (iii) f (x) = x/2 + 1/2x. Solving f (x) = x gives x = ±1. f ′ (x) = 1/2 − 1/2x2 and f ′ (±1) = 0. We claim that Bf (1) = (0, ∞) and Bf (−1) = (−∞, 0). In fact if x > 1, x/2 + 1/2x < x/2 + x/2 = x, or f (x) < x. Inductively, we see that (f n (x)) is a decreasing sequence bounded below by 1, so must converge to the only fixed point: 1. Similarly if 0 < x < 1, so Bf (1) = (0, ∞). The case of the fixed point x = −1 is similar.

2. Lµ (x) = µx(1 − x) has fixed point x0 = 1 − 1/µ. Now L′µ (x) = µ − 2µx, so L′µ (1 − 1/µ) = 2 − µ = 0, when µ = 2, x0 = 1/2.

3. If Lµ (a) = a and Lµ (b) = a, then L2µ (a) = a, so that a and b satisfy the equation L2µ (x) − x = x[µ2 x2 − µ(µ + 1)x + µ + 1](µx − µ + 1) = 0. We can disregard the linear factor (when a ̸= B), since that gives rise to the fixed point.

4. Lµ (x) = µx(1 − x) and L′µ (x) = µ − 2µx = 0 when x = 1/2. Now (L2µ )′ (x) = L′µ (Lµ (x))L′µ (x) = 0 when L′µ (x) = 0, or when L′µ (Lµ (x)) = 0. The first equation gives x = 1/2, and the second gives µ − 2µ(µx(1 −√x)) = 0, or 2µx2 − 2µx + 1 = 0. √ Solving gives x = (2µ ± 4µ2 − 8µ)/2µ = 1/2 ± µ2 − 2µ/2µ. There are no new solution for µ ≤ 2, but for µ > 2, we get two new critical points.

Consider the map L2µ restricted to the interval [1/µ, 1 − 1/µ]. The maximum and minimum values occur either at an end point or where (L2µ )′ (x) = 0. Both end points √ are mapped to 1−1/µ by L2µ . We can check that 1/µ > 1/2− µ2 − 2µ/2µ, and that √ 1 − 1/µ < 1/2 + µ2 − 2µ/2µ, so the only critical point in the interval is x = 1/2.

L2µ (1/2) = Lµ (µ/4) = µ · µ/4(1 − µ/4) = µ2 /4 − µ3 /16 ≥ 1/2 (check), so L2µ maps [1/µ, 1 − 1/µ] into the interval [1/2, 1 − 1/µ].

19

5. f : I → I is a continuous function on an interval I. J ⊂ I is the immediate basin of attraction of some attracting fixed point c ∈ J. Suppose that J = (a, b), a < b and f (J) = (α, β), α < β, then J ∩ f (J) ̸= ∅ since c belongs to both sets. Suppose α < a and let z ∈ (α, a), then z lies in the basin of attraction of f . Thus the set (α, b) is an interval contained in the basin of attraction of c, contradicting the fact that J is the largest such interval. It follows that a ≤ α and similarly β ≤ b, so f (J) ⊆ J. 6. Lµ (x) = µx(1−x), where 1 < µ ≤ 2, so that 1−1/µ ≤ 1/2, and Lµ (1/µ) = 1−1/µ.

Suppose that 0 < x < 1 − 1/µ, then Lµ (x) = µx(1 − x) > x, and Lµ (x) < 1 − 1/µ (check). We see that in general x < Lµ (x) < L2µ (x) < · · · < Lnµ (x) < 1 − 1/µ, so we have an increasing sequence which is bounded above by 1 − 1/µ, so must converge to the fixed point 1 − 1/µ.

Now suppose that 1 − 1/µ < x < 1/2, then 1 − 1/µ < Lµ (x) = µx(1 − x) < x, and we now obtain a decreasing sequence (Lnµ (x)) which must converge to the same fixed point, 1 − 1/µ.

If 1/2 < x ≤ 1/µ, then 1 − 1/µ < Lµ (x) < 1/2, so we now apply the second case to obtain a decreasing sequence that converges to 1 − 1/µ.

If 1/µ < x < 1, 0 < lµ (x) < 1 − 1/µ, and we are in the first case. In this way, we see that the basin of attraction of 1 − 1/µ is (0, 1).

7. f : R → R with two fixed points α and β, R = Bf (α) ∪ {p} ∪ Bf (β) = (−∞, p) ∪ {p} ∪ (p, ∞). We must have f (p) = p, since it cannot lie in either basin of attraction.

In other words, p is a repelling fixed point.

8. If 0 < µ < 1, then we know that c = 0 is an attracting fixed point, and there are no other fixed points. Suppose that the basin of attraction is an ”open” set of the form [0, a), a < 1. Given any sequence (xn ) ⊂ [0, a), with limn→∞ xn = a, we must have limn→∞ f (xn ) = f (a). It follows that f (a) = a since f (a) must be a limit point of [0, a), which does not lie in a. This contradicts the fact that x = 0 is the only fixed point. It follows that a = 1 and since f (1) = 0, the basin of attraction is [0, 1].

20

9. f : [0, 1] → [0, 1] continuous with f (0) = 0 = f (1), f (x) > 0 for x ∈ (0, 1). We know that x = 0 is a repelling fixed point and there is a unique attracting fixed point c ∈ (0, 1), and also no period-2 points. Suppose that the basin of attraction of c is an interval of the form (a, b), 0 < a < b < 1. The points a and b are either fixed, a 2-cycle or one is fixed and the other is eventually fixed (see the argument in 2.2.3).

None of these can happen (by hypothesis), so we must have a = 0 and b = 1 with basin of attraction for c equal to (0, 1).

10. Lµ (x) = µx(1 − x), where 2 < µ ≤ 3, so that if the fixed point is p = 1 − 1/µ, then 1/µ < 1/2 < p, and as before Lµ (1/µ) = p. Again we split the analysis into a number of different cases to show that the basin of attraction of p is (0, 1). Set q = 1/µ and note that 0 < q < 1/2 < p < 1, Lµ (q) = p = Lµ (p) and the maximum value of Lµ (x) is µ/4, when x = 1/2.

Let x ∈ (0, q), then Lµ (x) > x. If (Lnµ (x)) were an increasing sequence contained in (0, q), then it would converge to a fixed point, and since this cannot happen, there exists m > 0 with Lm µ (x) > q Now suppose that x ∈ (q, p], then p ≤ Lµ (x) ≤ µ/4. Now using the fact that q < Lµ (µ/4) and that Lµ is decreasing on [1/2, 1], it follows that if p < x ≤ µ/4, then q < Lµ (µ/4) ≤ Lµ (x) < p. Furthermore, since p < µ/4, and Lµ is decreasing on [p, 1] we obtain 0 < Lµ (x) < p, for x ∈ (µ/4, 1).

It follows that if x ∈ (0, 1), then x has an iterate in the interval (q, p). In fact we see that if x ∈ (q, p), then so is the sequence (L2n µ (x)) and if x ∈ (p, µ/4), then so is 2n the sequence (Lµ (x)). We showed earlier that L2µ (x) − x = 0 only has two solutions (i.e., L2µ has only two fixed points, namely 0 and p for 2 < µ ≤ 3). We can complete the proof by showing that if x ∈ (q, p) then the sequence (L2n µ (x)) converges to p.

2 Using our remarks above, Lµ (x) − x has no roots in the interval (q, p), so does not change sign on that interval. To find the sign, we calculate L2µ (1/2) − 1/2, and we can check that this is always positive on (q, p). It follows that the sequence (L2n µ (x)) is increasing in (q, p) so converges to a fixed point, which can only be p. It clearly follows that Lnµ (x) → p as n → ∞, and this must hold for all x ∈ (0, 1). This completes the proof.

21

Section 2.3 π 1. (i) f (x) = cos(x), f (0) = π/2 and f (π/2) = 0, so {0, π/2} is a 2-cycle. Now 2 π ′ f (x) = − sin(x), so |f ′ (0) · f ′ (π/2)| = 0, and the 2-cycle is super-attracting.

2 1 3 (ii) If g(x) = − x3 − x2 + 1, then g(0) = 1, g(1) = −1 and g(−1) = 0, so {0, 1, −1} 2 2 3 ′ is a 3-cycle for g. g (x) = − x2 − 3x, so g ′ (0) = 0 and this is enough to tell us that 2 the 3-cycle is super-attracting.

2. If fc (x) = x2 + c = x, then fc (x) − x = x2 − x + c = 0, and this will be a factor of fc2 (x) − x. Now fc2 (x) − x = (x2 + c)2 + c − x = x4 + 2cx2 + c2 − x + c = (x2 − x + c)(x2 + x + c + 1). √ −1 ± 1 − 4(c + 1) = Ignoring the fixed points, we solve x2 +x+c+1 = 0 giving x = 2 √ −1 ± −3 − 4c , so there is a 2-cycle when c < −3/4.

2 ′ Now fc (x) = 2x, so denoting the 2-cycle by {x1 , x2 }, for asymptotic stability we require |fc′ (x1 ) · fc′ (x2 )| < 1, i.e., √ √ |(−1 − −3 − 4c)(−1 + −3 − 4c)| < 1, or − 1 < 1 − (−4c − 3) < 1.

Thus the requirement for the 2-cycle to be attracting is −5/4 < c < −3/4.

3. (a) If f (x) = ax + b, a ̸= 0. Suppose that {x0 , x1 } is a 2-cycle, then ax0 + b = x1 , and ax1 + b = x0 , subtracting gives a(x0 − x1 ) = x1 − x0 , so as a ̸= 0 and x0 ̸= x1 we must have a = −1, b = x0 + x1 .

Set f (x) = −x + b, then f 2 (x) = −(−x + b) + b = x, so every x gives rise to a 2-cycle, except x = b/2, which is a fixed point, and this is the only way 2-cycles can arise. (b) If f (x) = ax2 − x + c, a, c > 0, then again suppose that {x0 , x1 } is a 2-cycle, then ax20 − x0 + c = x1 , and ax21 − x1 + c = x0 ,

22

Subtracting gives a(x20 −x21 )−(x0 −x1 ) = x1 −x0 , so a(x0 +x1 )−1 = −1, a(x0 +x1 ) = 0.

If a ̸= 0then x0 = −x1 and ax20 − x0 + c = −x0 so ax20 + c = 0. This is impossible for a, c > 0, so there are no period-2 points. b b (c) f (x) = a − , a ̸= 0, b ̸= 0. Solving a − = x gives x2 − ax + b = 0, so x x √ a ± a2 − 4b x= , and there is a fixed point when a2 ≥ 4b.

2 b If f 2 (x) = x, then a − = x, giving a(a − b/x) − b = x(a − b/x), or a − b/x x2 − ax + b = 0. This has no new solutions besides the fixed points, so there are no period-2 points (d) f (x) =

ax + b 2 , a + bc ̸= 0. Solving f 2 (x) = x, we have cx − a ax+b a( cx−a )+b ax+b c( cx−a )−a

=

a2 x + bcx a(ax + b) + b(cx + a) = = x, c(ax + b) − a(cx − a) bc + a2

provided a2 + bc ̸= 0. Consequently, every point in the domain of f is of period two, √ except the fixed points, which we can check are x = (a ± a2 + bc)/c.

4. (a) f : R → R is continuous with a two-cycle {x1 , x2 } say. Suppose x1 < x2 and set g(x) = f (x) − x, also a continuous function. Then g(x1 ) = f (x1 ) − x1 = x2 − x1 > 0, and g(x2 ) = f (x2 ) − x2 = x1 − x2 < 0, so by the Intermediate Value Theorem, there exists c ∈ (x1 , x2 ) where g(c) = 0, or f (c) = c. (b) As in (a), we have a < b < c, a 3-cycle. If g(x) = f (x) − x, g(b) = f (b) − b = c − b > 0, and g(c) = f (c) − c = a − c < 0, so there exists x1 ∈ (b, c) with g(x1 ) = 0, or f (x1 ) = x1 .

Now set h(x) = f 2 (x) − x, then h(a) = f 2 (a) − a = c − a > 0, and h(b) = f 2 (b) − b = a − b < 0, so there exists x0 ∈ (a, b) with h(x0 ) = 0, or f 2 (x0 ) = 0. Note that at this stage we cannot say that x0 is a period-2 point, but later we shall see that a period-2 point has to exist.

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5. {0, 1} is a non-hyperbolic 2-cycle for f (x) = ax2 + bx + c, so that f (0) = 1 and f (1) = 0. These give c = 1 and a + b + 1 = 0, a = −b − 1. f ′ (x) = 2ax + b.

Case 1. f ′ (0)f ′ (1) = 1 implies b(2a + b) = 1, b(−2b − 2 + b) = 1, b2 + 2b + 1 = 0, so b = −1 and a = 0 (this simply gives f (x) = −x + 1 for which every point is on a 2-cycle).

Case 2. f ′ (0)f ′ (1) = −1, so b(2a + b) = −1, b(−2b − 2 + b) = −1, b2 + 2b − 1 = 0, so √ √ √ b = (−2 ± 8)/2 = −1 ± 2, a = ∓ 2. We deduce √ √ √ √ (i) f (x) = − 2x2 + ( 2 − 1)x + 1, or (ii) f (x) = 2x2 − ( 2 + 1)x + 1.

In sub-case (i) Let g(x) = f 2 (x). We can use Mathematica to show that g ′′ (0) = √ √ √ √ −4(−3 + 2), g ′′ (1) = −4(3 + 2), g ′′′ (0) = 24(−1 + 2), g ′′′ (1) = 24(−1 − 2), √ √ Sg(0) = 120(−2 + 2) < 0, so we have stability, and Sg(1) = −120(2 + 2) < 0, which is also stable.

Sub-case (ii) is done similarly.

6. (a) f (x) = x3 − 6x2 + 7x + 2, so that the Newton function is Nf (x) = x − x3 − 6x2 + 7x + 2 2x3 − 6x2 − 2 2−6−2 = , so that N (1) = = 3, Nf (3) = f 3x2 − 12x + 7 3x2 − 12x + 7 3 − 12 + 7 f (x)f ′′ (x) 54 − 54 − 2 = 1. Now f ′ (x) = 3x2 −12x+7, f ′′ (x) = 6x−12, and Nf′ (x) = , 27 − 36 + 7 [f ′ (x)]2 so 4 × (−6) (−4) × 6 f (1)f ′′ (1) f (3)f ′′ (3) = × = 36 > 1, (Nf2 )′ (1) = Nf′ (1)Nf′ (3) = [f ′ (1)]2 [f ′ (3)]2 (−2)2 (−2)2 so the 2-cycle is unstable.

7. Let f (x) = ax3 +bx+1, a ̸= 0, then f (0) = 1, and f (1) = a+b+1 = 0, so a+b = −1.

If the 2-cycle {0, 1} is non-hyperbolic, then |f ′ (0)f ′ (1)| = 1. Now f ′ (x) = 3ax2 + b, so f ′ (0) = b and f ′ (1) = 3a + b, so b(3a + b) = ±1, so b(3(−1 − b) + b) = ±1, or √ 2b2 + 3b ± 1 = 0, b = (−3 ± 9 ± 8)/4. It follows that we have four cases (i) b = −1, a = −b − 1 = 0, so this case does not give rise to a cubic (we have the linear map f (x) = −x + 1 studied earlier). x3 x (ii) b = −1/2, a = −1/2, so f (x) = − − + 1.

2 2( ( √ √ ) √ ) √ −1 + 17 −1 + 17 −3 − 17 −3 − 17 ,a= , f (x) = x3 + x + 1. (iii) b = 4 4 4 4

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( ( √ √ ) √ ) √ −3 + 17 −1 − 17 −1 − 17 −3 + 17 (iv) b = ,a= , f (x) = x3 + x + 1.

4 4 4 4 In case (ii), using f ′ (x) = −3x2 /2−1/2, we see that f ′ (0)f ′ (1) = (−1/2)(−2) = 1, so it is non-hyperbolic, so we attempt to apply Theorem 1.5.3. Using Mathematica, if we set g(x) = f 2 (x), then g ′ (0) = 1 and g ′′ (0) = −3/4 < 0, so by Theorem 1.5.3, x = 0 is a semi-stable fixed point for g, with stability on the left. It follows that the 2-cycle {0, 1} of f is semi-stable, with stability on the left.

In case (iii), again set g(x) = f 2 (x), then since g ′ (0) = −1 we will need Theorem 1.5.7, using the Schwarzian derivative. In this case we can check that Sg(0) = √ −9 (1191 + 241 17) < 0, so x = 0 is asymptotically stable for g and {0, 1} is therefore 64 stable for f .

In case (iv), with g(x) = f 2 (x), g ′ (0) = −1 so apply Theorem 1.5.7 again. In √ 9 (−1191 + 241 17) < 0 so again x = 0 is this case we can check that Sg(0) = 64 asymptotically stable for g and {0, 1} is stable for f .

8. (a) Cµ (x) = µ cos(x). Set µ = λ, then Cλ (0) = λ, and in order for this to give rise to a 3-cycle, we require Cλ (λ) = π/2, since we then have Cλ (π/2) = λ cos(π/2) = 0.

In other words, if λ satisfies λ cos(λ) = π/2, then {0.λ, π/2} will be a 3-cycle for Cλ .

Since Cλ′ (0) = −λ sin(0) = 0, the 3-cycle will be super-attracting. Note that we can check that the equation λ cos(λ) = π/2 has a solution. (b) Sµ (x) = µ sin(x). Since Sµ′ (x) = µ cos(x) = 0 when x = π/2, we create a 2-cycle {π/2, λ}, so we require Sλ (π/2) = λ sin(π/2) = λ, and Sλ (λ) = λ sin(λ) = π/2.

Again we can check that there is a λ satisfying this equation, so {π/2, λ} is a superattracting 2-cycle. We find a super-attracting 3-cycle {π/2, λ, µ} in a similar way. ′ (c) Sπ/2 (π/2) = π/2 sin(π/2) = π/2, and Sπ/2 (π/2) = π/2 cos(π/2) = 0, so π/2 is a super attracting fixed point of Sπ/2 . We know that the immediate basin of attraction is an open interval contained in the basin of attraction. We need to show that it is distinct from the basin of attraction. Note that Sπ/2 (0) = 0, and Sπ/2 (π) = 0 and these are not in the immediate basin of attraction of π/2, so BSπ/2 (π/2) ⊂ (0, π). On the other hand, Sπ/2 (5π/2) = π/2, so 5π/2 belongs to the basin of attraction of π/2, so the immediate basin of attraction is not equal to the basin of attraction.

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If x ∈ (0, π/2), then Sπ/2 (x) > x (show that sin(x)/x > 2/π for x > 0), we see n that (Sπ/2 (x)) is an increasing sequence bounded above by π/2, so must converge to the fixed point π/2. Also if x ∈ (π/2, π), then Sπ/2 (x) is mapped to (0, π/2). It follows that the immediate basin of attraction is (0, π). (d) Use a computer algebra system.

9. Let Fµ be a family of maps. It is often the case in our examples, that if x = c gives rise to a super attracting fixed point, then c occurs in other super attracting cycles that arise, for example with the logistic map Lµ (x) = µx(1 − x), x = 1/2 is a fixed point when µ = 1/2, and since L′µ (x) = µ − 2µx, L′2 (1/2) = 0, and also x = 1/2 √ is part of a super-attracting 2-cycle when µ = 1 + 5 and super-attracting 3-cycle when µ = 3 · 83187 . . . (see Section 2.6.3). Similar phenomena happens with various trigonometric functions.

In these examples, when solving (Fµ )′ (c) = 0, or Fµ′ (c0 )Fµ′ (c1 ) . . . Fµ′ (cn−1 ) = 0, there is a solution, say c0 , independent of µ. In general this phenomena won’t arise if the critical point is dependent on µ.

10. (a) We have f (x) = xep−x , for x > 0 and p > 0. If f (x) = x, then x = 0 or ep−x = 1, so x = 0 and x = p are the fixed points. f ′ (x) = ep−x + xep−x (−1) = ep−x (1 − x), so f ′ (0) = ep and f ′ (p) = 1 − p. This tells us that x = 0 is unstable, and and x = p is unstable for p > 2. (b) Note that (a) tells us that we have fixed points at x = 0 and x = p with 0 < p.

These are fixed points for f 2 (x), and we have (f 2 )′ (0) = e2p > 1 and (f 2 )′ (p) = (1 − p)2 > 1 when p > 2. It follows that Exercise 1.5 # 12 tells us that f 2 (x) has a fixed point between 0 and p, which must be a period-2 point for f , since f has no other fixed points.

11. {x0 , x1 , . . . , xn−1 } is an n-cycle for f , then we have seen that (f n )′ (x0 ) = f ′ (x0 )f ′ (x1 ) · · · f ′ (xn−1 ), and since the terms in the product may be permuted in any way, we see that this is equal to (f n )′ (xi ), for every 0 ≤ i ≤ n − 1.

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12. n is the smallest positive integer with f n (x0 ) = x0 (the period of x0 ). We have f m (x0 ) = x0 . By the Division Algorithm, we can write m = qn + r, for some r, q ∈ N, with 0 ≤ r < n. Then f m (x0 ) = f qn+r (x0 ) = f r (f qn (x0 )) = f r (x0 ). If r > 0, we have a contradiction, since the period of f is n > r, so we must have r = 0, so m = nq.

13. f p (x) = x and f q (x) = x and n is the highest common factor of p and q. This means that (i) n|p and m|q, and (ii) if also n|p and n|q, then n|m. Suppose that x has period m, then f m (x) = x, so by the previous exercise m|p and m|q. It follows that n|m, or m = kn for some k ≥ 1, so f m (x) = x.

14. We have f : R → R an odd function: f (−x) = −f (x) for all x ∈ R. (a) Now in this case, f (0) = f (−0) = −f (0), so f (0) = 0 and 0 is a fixed point of f .

If the graph of y = f (x) intersects the graph y = −x at x = c, then f (c) = −c, so f 2 (c) = f (−c) = −f (c) = c, so x = c is a period-2 point when c ̸= 0. (b) f (x) = x3 − 3x/2 is an odd function. Solving f (x) = −x gives x3 − 3x/2 = −x, √ √ or x3 − x/2 = 0, x(x2 − 1/2) = 0. x = 0 is a fixed point, and {−1/ 2, 1/ 2} gives a 2-cycle. Note: There may be other 2-cycles that don’t arise directly in this way, but this is not the case for f (x) = x3 − 3x/2. (c) f : R → R an even function: f (−x) = f (x) for all x ∈ R. This time, if the graph of y = f (x) intersects the graph y = −x at x = c, then f (c) = −c, so f 2 (c) = f (−c) = f (c) = −c, so −c is a fixed point of f , and c is therefore an eventual fixed point (when c ̸= 0).

If f (x) = cos(x), then f is an even function, so solving f (x) = −x gives cos(x) = −x. We have seen that cos(x) = x when x = ·739085 . . ., so from symmetry we have c = − · 739085 . . . is an eventual fixed point. {

x sin(1/x); x ̸= 0 We saw in Exercises 1.5 # 8, f (x) = x, when 0; x=0 2 x = 0, x = , n ∈ Z. and the non-zero fixed points are non-hyperbolic.

(4n + 1)π 2 Since f (x) is an even function, there are eventual fixed points at x = − , (4n + 1)π n ∈ Z. 15.(a) f (x) =

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(b) We have seen that the non-zero fixed points are non-hyperbolic, and can be seen to have one-sided stability. Note that 0 < f (x) ≤ x for all x > 0 and that f maps points to points having lesser or equal absolute value. It follows that fixed points and period-2 points can only happen when | sin(1/x)| = 1. If sin(1/x) = 1, the result is a fixed point. Suppose that sin(1/x) = −1, for example when x = −2/(3π), we have f (−2/(3π)) = −2/(3π) sin(−3π/2) = 2/(3π) sin(3π/2) = −2/(3π), an eventual fixed point (as in 14(c)). In this way we see that f has no period two points. { x cos(1/x); x ̸= 0 (c) If g(x) = , then g is an odd function. Solving g(x) = x gives 0; x=0 1 cos(1/x) = 1 for x ̸= 0, so that 1/x = 2kπ, k = ±1, ±2, . . ., or x = . Together 2kπ with x = 0, these are the fixed points of g.

Now the intersection of the graph of g with the line y = −x gives rise to period-2 points. We solve g(x) = −x, or cos(1/x) = −1, so 1/x = (2k + 1)π, k ∈ Z, or 1 , so for example we see that {1/π, −1/π} is a 2-cycle. x= (2k + 1)π 16. We have f : R → R a continuously differentiable function, with exactly two distinct fixed points x0 and x1 , which satisfy |f ′ (x0 )| > 1, |f ′ (x1 )| > 1. It follows from Exercise 1.5 # 12, that either f ′ (x0 ) < −1 and f ′ (x1 ) > 1 or vice-versa, for otherwise f would have another fixed point between x0 and x1 . Assume the former holds and set g(x) = f 2 (x), then g has fixed points at x0 and x1 , and g ′ (x) = f ′ (f (x))f ′ (x), g ′ (x0 ) = f ′ (x0 )f ′ (x0 ) = [f (x0 )]2 > 1, and similarly g ′ (x1 ) > 1. Again using Exercise 1.5 # 12, it follows that g has a fixed point between x0 and x1 . This fixed point of f 2 cannot be a fixed point of f , so must be a period-2 point of f .

17. Suppose that f (x) = ax2 + bx + c (a ̸= 0), has a 2-cycle {x0 , x1 } (so x0 ̸= x1 ), which is hyperbolic of the type where f ′ (x0 )f ′ (x1 ) = 1. We obtain a contradiction to show that this cannot happen.

We have ax20 +bx0 +c = x1 and ax21 +bx1 +c = x0 . Subtracting these two equations gives a(x20 − x21 ) + b(x0 − x1 ) = x1 − x0 , or a(x0 + x1 ) + b = −1, b = −1 − a(x0 + x1 ).

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Now f ′ (x0 )f ′ (x1 ) = (2ax0 + b)(2ax1 + b) = 1, so substituting for b gives (2ax0 − ax0 − ax1 − 1)(2ax1 − ax0 − ax1 − 1) = 1 This reduces to (ax0 − ax1 − 1)(ax1 − ax0 − 1) = 1, or

− [(ax0 − ax1 ) − 1][(ax0 − ax1 ) + 1] = 1,

so −(ax0 − ax1 )2 + 1 = 1, (ax0 − ax1 )2 = 0, so if a ̸= 0, we have x0 = x1 , contradicting the fact that we have a 2-cycle.

18. f (x) is a polynomial with g(x) = f 2 (x) − x having a repeated root at x0 (where f (x0 ) = x1 ̸= x0 ). We must show that {x0 , x1 } is a non-hyperbolic 2-cycle for f of the type where f ′ (x0 )f ′ (x1 ) = 1.

We have g(x) = f 2 (x) − x = (x − x0 )2 h(x) for some polynomial h. Clearly f 2 (x0 ) = x0 , so if x1 = f (x0 ), {x0 , x1 } is a 2-cycle.

Now g ′ (x) = f ′ (f (x))f ′ (x)−1, g ′ (x0 ) = f ′ (f (x0 ))f ′ (x0 )−1 = f ′ (x1 )f ′ (x0 )−1 = 0, so the 2-cycle is non-hyperbolic of the required type.

Conversely, suppose that f (x) is a polynomial having a 2-cycle {x0 , x1 } and g(x) = f (x) − x, then g(x0 ) = f 2 (x0 ) − x0 = 0, so (x − x0 ) is a factor of g(x). g ′ (x) = f ′ (f (x))f ′ (x) − 1, so g ′ (x0 ) = f ′ (x1 )f ′ (x0 ) − 1 = 0, and it follows that (x − x0 )2 is also a factor of g(x).

2

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Section 2.4 √ 1. (i) fc (x) = x2 + c. Solving fc (x) = x gives x2 − x + c = 0, so x = (1 ± 1 − 4c)/2. √ Now f ′ (x) = 2x, so for the fixed point to be super-attracting, we require 1± 1 − 4c = 0, which give c = 0 and x = 0.

Now fc2 (x) = (x2 + c)2 + c = x4 + 2cx2 + c2 + c and solving fc2 (x) − x = x4 + 2cx2 − x + c2 + c = (x2 − x + c)(x2 + x + c + 1) = 0 √ gives x = (−1 ± −4c − 3)/2. f ′ (x) = 2x, so for the 2-cycle to be super-attracting, √ √ √ we require f ′ ((−1 − −4c − 3)/2)f ′ ((−1 + −4c − 3)/2) = (−1 − −4c − 3))(−1 + √ √ −4c − 3) = 0 or 4 + 4c = 0, c = −1. This gives x = (−1 ± 4 − 3)/2 = 0 or −1.

We can now check that {0, −1} is a super-attracting 2-cycle for f−1 (x) = x2 − 1. (ii) If we have a super-attracting 3-cycle, say {x1 , x2 , x3 }, we require f ′ (x1 )f ′ (x2 )f ′ (x3 ) = 0, so must have f ′ (x1 ) = 2x1 = 0 (say), so x1 = 0. Now fc (0) = c and f 2 (0) = c2 + c, so that 3-cycle is {0, c, c2 + c}. Since we must have x = 0 satisfying the equation f 3 (x) = x we have f 3 (x) − x = ((x2 + c)2 + c)2 + c − x = 0 is satisfied by x = 0, so (c2 + c)2 + c = 0, or c3 + 2c2 + c + 1 = 0.

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Section 2.6 1. L4 (x) = 4x(1 − x), so that L4 (sin2 (π/5)) = 4 sin2 (π/5)(1 − sin2 (π/5)) = 4 sin2 (π/5) cos2 (π/5) = (2 sin(π/5) cos(π/5))2 = sin2 (2π/5), and similarly L4 (sin2 (2π/5)) = sin2 (4π/5) = sin2 (π/5), so we have a 2-cycle.

In a similar way we check that {sin2 (π/7), sin2 (2π/7), sin2 (3π/7)} and {sin2 (π/9), sin2 (2π/9), sin2 (4π/9)} are 3-cycles for L4 .

2. The 4-cycles of L4 are found by using the formula in Remark 2.6.6: Per4 (L4 ) = {sin2 (kπ/15) : 0 ≤ k < 8} ∪ {sin2 (kπ/17) : 0 < k ≤ 8}.

This set will include the fixed points and period-2 points, which are {0, sin2 (π/3)} and {sin2 (π/5), sin2 (2π/5)} respectively. This leaves {sin2 (π/15), sin2 (2π/15), sin2 (4π/15), sin2 (7π/15)}, {sin2 (3π/17), sin2 (6π/17), sin2 (5π/17), sin2 (7π/17)} {sin2 (π/17), sin2 (2π/17), sin2 (4π/17), sin2 (8π/17)} as 4-cycles. The 5-cycles are found in a similar way.

3, 4, 5, 6 and 7 can be solved using a computer algebra system. (x − x2 ) (1 − 2x − x2 ) ′ 8. (a) We have gµ : [0, 1] → [0, 1], gµ (x) = µ , so that gµ (x) = µ = 1+x (1 + x)2 √ 0, when x = 2 − 1 > 0.

The maximum/minimum occur where gµ′ (x) = 0 or at the end points. Now √ √ √ gµ (0) = 0 = gµ (1) and gµ ( 2 − 1) = µ( 2 − 1)2 = µ(3 − 2 2), giving the maximum value for gµ . √ (b) It follows that gµ is a dynamical system for 0 ≤ µ(3 − 2 2) ≤ 1, so we require √ 0 ≤ µ ≤ 3 + 2 2.

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(c) Solving gµ (x) = x gives x = 0 or µ − µx = 1 + x, or x = (µ − 1)/(µ + 1) and these are the fixed points of gµ . (d) From (a) gµ′ (0) = µ, so we have stability for 0 ≤ µ < 1. ( ) ( ) µ−1 µ−1 2 1 − 2 − ( ) µ − 1 µ2 − 2µ − 1 µ+1 µ+1 gµ′ =µ = − . µ+1 2µ [1 + µ−1 ]2 µ+1 √ 2 Thus we require | µ −2µ−1 | < 1, and this gives 1 < µ < 2 + 5 (can check that 2µ √ √ 2 + 5 < 3 + 2 2).

9. (a) We wish to show that the logistic map L4 (x) = 4x(1 − x) cannot have a super-attracting cycle. For a contradiction, suppose that {c0 , c1 , . . . , cn } is a super attracting n-cycle for L4 , then (Ln4 )′ (c0 ) = L′4 (c0 )L′4 (c1 ) . . . L′4 (cn−1 ) = 0, so that (4 − 8c0 )(4 − 8c1 ) · · · (4 − 8cn−1 ) = 0.

It follows that ci = 1/2 for some 0 ≤ i < n and so 1/2 lies on an n-cycle. But L4 (1/2) = 4(1/2)(1 − 1/2) = 1, and L4 (1) = 0, does not give an n-cycle. (b) We find a point x0 ∈ (0, 1), which is not a periodic point for L4 .

Recall that if xn+1 = 4xn (1 − xn ) is the corresponding difference equation, we can give a closed form solution by setting xn = sin2 (θn ), so that xn = sin2 (2n θ0 ), where x0 = sin2 (θ0 ). Consequently, if we can choose θ0 so that all the terms of the the sequence (sin2 (2n θ0 )) are all distinct, then this will give rise to an orbit of x0 for L4 which is infinite, so is non-periodic and not eventually periodic. It suffices to take θ0 √ √ so that it is not a rational multiple of π, e.g. θ0 = 2π, so that x0 = sin2 ( 2π).

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Section 2.8 1. If µ > 1, then the derivative of Tµ has the property that |(Tµ )′ (x)| > 1 for every x ̸= 1/2. It clearly follows that every periodic point will be unstable.

2. We may assume that µ > 1. If c = 1/2 has period-3 under Tµ , then since Tµ (1/2) = µ/2 > 1/2, so Tµ (µ/2) = µ(1 − µ/2) and we must have Tµ (µ(1 − µ/2)) = 1/2. Now µ(1 − µ/2) < 1/2 as this is equivalent to (µ − 1)2 > 0. The requirement for µ is therefore µ2 (1 − µ/2) = 1/2, or µ3 − 2µ2 + 1 = (µ − 1)(µ2 − µ − 1) = 0, so we require √ µ = (1 + 5)/2.

3. We wish to show that for µ ≥ (1 + {

√

5)/2 we have a 3-cycle for Tµ :

µ µ2 µ3 , , }.

1 + µ3 1 + µ3 1 + µ3

1 µ3 > . In addition, 1 + µ3 2 √ µ2 1 2 ≤ if and only if (µ − 1)(µ − µ +1) > 0.

This will hold when µ > (1+ 5)/2.

1 + µ3 2 It follows that ) ) ) ( ) ( ( ( µ3 µ µ µ2 µ3 3 2 , Tµ = Tµ = Tµ =µ 1− = 3 3 3 3 1+µ 1+µ 1+µ 1+µ 1 + µ3

We may assume that µ > 1. Notice that in this case

so we have a 3-cycle.

4. We now show that for µ3 − µ2 − µ − 1 > 0 we have a 4-cycle for Tµ , namely: { } µ µ2 µ3 µ + µ2 + µ3 , , , .

1 + µ + µ2 + µ3 1 + µ + µ2 + µ3 1 + µ + µ2 + µ3 1 + µ + µ2 + µ3 2

µ 3 2 Now 1+µ+µ 2 +µ3 < 1/2 is equivalent to f (µ) = µ − µ + µ + 1 > 0. We can check that f ′ (µ) > 0 always, so f is increasing with f (1) = 2, so our inequality must hold. µ2 3 2 The requirement 1+µ+µ 2 +µ3 < 1/2 is equivalent to µ − µ − µ − 1 > 0, which we are 2

3

µ+µ +µ assuming holds. Finally, we can check that 1+µ+µ 2 +µ3 > 1/2 is true for µ > 1.

Then we simply check that ( ) ( ) ( ) µ µ2 µ3 4 3 2 Tµ = Tµ = Tµ 1 + µ + µ2 + µ3 1 + µ + µ2 µ3 1 + µ + µ2 + µ3

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( = Tµ µ(1 −

( ) ) µ3 µ + µ2 + µ3 µ ) = Tµ = .

2 3 2 3 1+µ+µ +µ 1+µ+µ +µ 1 + µ + µ2 + µ3

5. Lµ (x) = µx(1 − x) = µx − µx2 , so L′µ (x) = µx − 2µx and L′µ (1/2) = 0. It follows that if c0 = 1/2 is a period-n point for f , with corresponding n-cycle {c0 , c1 , . . . , cn−1 }, then (Lnµ )′ (1/2) = L′µ (1/2)L′µ (c1 ) . . . L′µ (cn−1 ) = 0, so the n-cycle is super-attracting.

6. Use a computer algebra system.

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Section 3.2 1. f (x) = −1 − 1/x = −(x + 1)/x, then f 2 (x) = −1 + x/(x + 1) = −1/(x + 1) and f 3 (x) = −1/(1 − 1 − 1/x) = x, and we see that every point is of period-3. There are no fixed points since f (x) = x gives x2 + x + 1 = 0, having no real solution. Note that f is not continuous, so we conclude that continuity is essential in Sharkovsky’s Theorem

2. In the Sharkovsky ordering of the numbers 25 to 35: 25 > 27 > 29 > 31 > 33 > 35 > 2 × 13 > 2 × 15 > 2 × 17 > 22 × 7 > 26 .

3. If m > n in the Sharkovsky ordering, suppose that m = 2k p and n = 2l q, p and q odd. From Sharkovsky’s Theorem we have two cases. Either k < l, so k + 1 < l + 1 and 2m > 2n or k = l and p < q, so again 2m > 2n. Conversely, if 2m > 2n, then we argue in a similar way that m > n.

4. Use two copies of a the interval containing a and b, so that f ([a, b]) ⊆ [a, b], so there exists c ∈ [a, b] with f (c) = c. c is a fixed point of f .

5. If f (x) = (x − 1/x)/2 = x, then x2 = −1, which has ( no real solutions, so there ) 1 x 1 1 2 are no fixed points. On the other hand, if f (x) = − − = 2 2 2x x/2 − 1/(2x) √ x4 − 6x2 + 1 (x2 +1)(x2 −3) 2 .

Solving f (x) = x gives = 0, so x = ±1/ 3, giving a 2 4x(x −1) 4x(x2 − 1) 2-cycle.

If we plot f 3 (x) using Mathematica, we see that there are 3-cycles. This does not contradict Sharkovsky’s Theorem since f is not a continuous function.

6. f : [1, 7] → [1, 7] is defined so that f (1) = 4, f (2) = 7, f (3) = 6, f (4) = 5, f (5) = 3, f (6) = 2, f (7) = 1, and the corresponding points are joined so that f is piecewise linear and continuous.

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We see that {1, 4, 5, 3, 6, 2, 7} is a 7-cycle. Now we can check that f 5 [1, 2] = [2, 7], f 5 [5, 6] = [1, 5], f 5 [6, 7] = [1, 6], so that no period-5 points lie in these intervals.

Now f 5 [4, 5] = [1, 7] ⊃ [4, 5], so that f 5 has a fixed point in [4, 5], say α. It follows that f (α) ∈ [3, 5], so if f (α) ∈ [3, 4], then f 2 (α) ∈ [5, 6] f 3 (α) ∈ [2, 3], f 4 (α) ∈ [6, 7], α = f 5 (α) ∈ [1, 2], which is impossible, so we must have f (α) ∈ [4, 5], so f 2 (α) ∈ [3, 5] etc. Continuing in this way we see that

f 2 (α) ∈ [4, 5], f 3 (α) ∈ [4, 5], f 4 (α) ∈ [4, 5],

so the orbit of α is contained in the interval [4, 5].

On [4, 5], we can check that f (x) = −2x + 13, so f (x) = x when x = 13/3, the fixed point of f . We can also check that f 5 (x) = −32x + 143 on [4, 5], so that f 5 (x) = x when x = 143/33 = 13/3, so there are no additional fixed points for f 5 (x).

7. Suppose that F : R → R is a continuous increasing function. If {a1 , a2 } is a 2-cycle, with a1 < a2 , then f (a1 ) < f (a2 ), so a1 < a2 = f (a1 ) < f (a2 ) = a1 , a contradiction.

Since f cannot have any 2-cycles, it cannot have any m-cycles, for m > 2.

8. Follow the steps in the proof of the Li-Yorke Theorem: There is an interval B1 ⊂ I1 with f (B1 ) = I1 . There is an interval B2 ⊂ I2 with f (B2 ) = B1 and there is an interval B3 ⊂ I2 with f (B3 ) = B1 . It follows that f 3 (B3 ) = f 2 (B2 ) = f (B1 ) = I2 ⊃ B3 , so f 3 has a fixed point in B3 , which cannot be a fixed point of f , so is a period-3 point of f.

9. f : R → R has a 5-cycle {a1 , a2 , a3 , a4 , a5 }. (a) Showing that f has a 7-cycle is a special case of the general proof of Sharkovsky’s Theorem (that if f has a period-m point for m odd, then f has a period m + 2 point - see 12.2.5 and 12.2.6. This is a non-trivial fact). We present this special case as follows: Set P = {a1 , a2 , a3 , a4 , a5 }, the 5-cycle for f , where a1 < a2 < a3 < a4 < a5 . Then we show there exists v: f (v) ∈ P with f 2 (v) < v < f (v).

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Write A = {x ∈ P : f (x) > x}, a = max(A) and b = min{x ∈ P : a < x}.

Neither f (b) > a or f (a) < b can happen, so we must have f (b) ≤ a < b ≤ f (a).

Now choose v ∈ [a, b] so that f (v) = b. This can be done since [a, b] ⊂ f [a, b]. Since b is a period-5 point b ̸= v. (In fact v may not belong to P, but it is possible that v = a). Then we have shown there exists v ∈ [a, b) with a1 ≤ f 2 (v) < v < f (v) = b ∈ P. If g(x) = f (x) − x, then g(v) = f (v) − v = b − v > 0, and g(b) = f (b) − b = f 2 (v) − f (v) < 0. Thus, g has a zero z in [v, b], which must be a fixed point of f .

In a similar way, since f 2 (a1 ) > a1 and f 2 (v) < v, there are points x ∈ [a1 , v] with f 2 (x) = x. It follows that y = max{x : a1 ≤ x ≤ v, f 2 (x) = x}, exists.

Note that f (x) > z, x ∈ [y, v], for if there is an x with f (x) < z, then [z, f (v)] ⊆ [f (x), f (v)] ⊆ f [x, v] ⊆ f [y, v], and then we can find p ∈ [y, v] with f (p) = z. If h(x) = f 2 (x) − x, then h(p) = f 2 (p) − p = z − p > 0 (f (z) = z), and h(v) < 0, consequently there must exist w ∈ (y, v) with f 2 (w) = w. This contradicts the maximality of y.

Therefore, f (x) > z > x on [y, v] (and also f 2 (x) < x on (y, v], so y must be a period-2 point).

We have shown that f (x) > z > x > f 2 (x), for x ∈ (y, v].

Since f 7 (y) = f (y) > y and f 7 (v) = f 2 (v) = f (b) < v, the point q = min{x : y ≤ x ≤ v, f 7 (x) = x} exists.

Now q cannot be a fixed point of f because f has no fixed points in (y, v). It follows that q has period 7, and the result follows. (b) Use four copies of the real line and note that f ([a4 , a5 ]) ⊇ [a1 , a5 ] and f ([a3 , a4 ]) ⊃ [a4 , a5 ].

Find B1 ⊂ [a4 , a5 ] with f (B1 ) = [a3 , a4 ] (look at the second and third copies).

Find B2 ⊂ [a4 , a5 ] with f (B2 ) = B1 (think of B2 as being in the first copy). Set

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B = B2 , f 3 (B) = f 3 (B2 ) = f (B2 ) ⊃ f ([a3 , a4 ]) ⊃ [a4 , a5 ] ⊃ B. Thus there is a fixed point in B for f 3 , which cannot be a fixed point of f , so must be a period-3 point of f.

10. f : R → R has a 4-cycle {a, b, c, d} with a < b < c < d. The possibilities are f (a) = a, f (b) = c, f (c) = d, f (d) = a, with its mirror image f (a) = d, f (d) = c, f (c) = b, f (b) = a. each of these gives rise to a 3-cycle (using Proposition 3.1.6).

Other possibilities are f (a) = c, f (c) = d, f (d) = b, f (b) = a and its mirror image f (a) = b, f (b) = d, f (d) = c, f (c) = a.

The final possibility is f (a) = d, f (d) = b, f (b) = c, f (c) = a and its mirror image f (a) = c, f (c) = b, f (b) = d, f (d) = a. We can list these as follows, where they come in pairs as the cycle and its mirror image. (i) and (ii) give rise to a 3-cycle. (i) a → b → c → d → a (ii) a → d → c → b → a (iii) a → c → d → b → a (iv) a → b → d → c → a (v) a → d → b → c → a (vi) a → c → b → d → a.

To see that a map satisfying (v) need not have a 3-cycle, consider the map f : [1, 4] → [1, 4] with f (1) = 4, f (2) = 3, f (3) = 1 and f (4) = 2, and extended piecewise linearly. f has a 4-cycle {1, 4, 2, 3} and can be seen to have a single fixed point. We can check that f has no period-3 points (use a computer algebra system for example).

Doing the same type of construction with (iii), we get an function having a 3-cycle.

11. f : [a, b] → [a, b] is continuous with limn→∞ f n (x) existing for each x ∈ [a, b]. Suppose that f has a point of period n > 1, then from Sharkovsky’s Theorem, f has a 2cycle, say {x0 , x1 } with x0 ̸= x1 . The sequence (f n (x0 )) is just {x0 , x1 , x0 , x1 , x0 , , . . .}, oscillating between x0 and x1 , so that limn→∞ f n (x0 ) does not exist, contradicting our hypothesis. It follows that there can be no periodic points of period n > 1.

12. Note: We need to assume that f : [a, b] → [a, b], a continuous map. (a) There exists c ∈ [a, b] with a ≤ f 2 (c) < c < f (c) ≤ b, and we show that f has a period-2 point. Set g(x) = f 2 (x) − x, then g(a) = f 2 (a) − a ≥ 0 and

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g(c) = f 2 (c) − c < 0, so by IVT there exists p ∈ [a, c] with f 2 (p) = p. We may assume that p = max{x ∈ [a, c] : f 2 (x) = x}. If f (p) ̸= p, we are done, so suppose that f (p) = p. Then (p, f (c)) ⊂ f (p, c), so c ∈ f (p, c). Hence there exists q ∈ (p, c) with f (q) = c.

Then since g(c) = f 2 (c) − c < 0 and g(q) = f 2 (q) − q = f (c) − q > f (c) − c > 0, we deduce that f 2 (r) = r for some r ∈ [q, c]. This is a contradiction, since it gives a point r ∈ [a, c] with f 2 (r) = r and r > p. Consequently, p must be a period-2 point. (b) We now suppose there exists c ∈ [a, b] with f 3 (c) < c < f (c), and again show that f has a period-2 point. There are 4 different possibilities that arise: (i) a ≤ f 2 (c) < f 3 (c) < c < f (c) ≤ b (ii) a ≤ f 3 (c) < f 2 (c) < c < f (c) ≤ b . If cases (i), (ii) or (iii) hold, then it follows (iii) a ≤ f 3 (c) < c < f (c) < f 2 (c) ≤ b (iv) a ≤ f 3 (c) < c < f 2 (c) < f (c) ≤ b from (a) that there is a period- point (for (iii), set d = f (c), then f 2 (d) < d < f (d)).

For case (iv), set g(x) = f 3 (x) − x, then g(a) = f 3 (a) − a ≥ 0 and g(c) = f 3 (c) − c < 0, so by IVT there exists p ∈ [a, c] with f 3 (p) = p. We may assume that p = max{x ∈ [a, c] : f 3 (x) = x}.

Hence there exists q ∈ (p, c) with f (q) = c. Then since g(c) = f 3 (c) − c < 0 and g(q) = f 3 (q) − q = f 2 (c) − q > 0 (using (iv)), we deduce that f 3 (r) = r for some r ∈ [q, f (c)]. This is a contradiction, since it gives a point r ∈ [a, f (c)] with f 3 (r) = r and r > p. Consequently, p must be a period-3 point. It now follows by the Li-Yorke Theorem that f has period-2 points.

13. Find a period-2 point for the map of Example 3.2.3. Note that f has a unique fixed point in the interval [3, 4].

Now f [1, 2] = [3, 5], and f [3, 5] = [1, 4], so f 2 [1, 2] = [1, 4] ⊇ [1, 2], so there exists x0 ∈ [1, 2] with f 2 (x0 ) = x0 . x0 must be a period-2 point since there is no fixed point in [1, 2].

On [1, 2], f (x) = 2x + 1. Since f [1, 2] = [3, 5] and f [3, 4] = [2, 4] we must have f (x0 ) ∈ [4, 5], and the equation of f on [4, 5] is f (x) = −x + 6. Composing these gives f 2 on [1, 2], i. e., f 2 (x) = −(2x + 1) + 6 = −2x + 5. Solving f 2 (x) = x gives x = 5/3, and we can check that this is the period-2 point, the 2-cycle being {5/3, 13/3}.

14. Th : [0, 1] → [0, 1] is the truncated tent map: Th (x) = min{h, T (x)}, where T is the tent map.

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(a) Now T2/3 (2/3) = 2/3, so T2/3 has fixed points at 0 and 2/3. Suppose that {a, b} is a 2-cycle for T2/3 , then both a < 2/3 and b < 2/3. We may assume that a < 1/2 and b > 1/2, for otherwise we cannot have a 2-cycle.

If 0 < a ≤ 1/3, then T2/3 (a) = b = 2a ≤ 2/3, so T2/3 (b) = 2(1 − 2a) = 2 − 4a = a, which is impossible, as this gives a = 2/5, b = 4/5 > 2/3. If 1/3 < a ≤ 1/2, then T2/3 (a) = 2/3 and we do not have a 2-cycle. (b) T4/5 (4/5) = 2(1−4/5) = 2/5, and T4/5 (2/5) = 2(1−2/5) = 4/5, so that {2/5, 4/5} is a 2-cycle for T4/5 . Suppose we have a 4-cycle, say {a, b, c, d}. These numbers are either less than 2/5, or greater than 3/5, for otherwise they get mapped to 4/5 and we do not have a 4-cycle. On the other hand at least one of these must be less than 1/2 and one must be greater than 1/2, for otherwise we don’t have a 4-cycle. It follows that one of these numbers (say a < 2/5), gets mapped to b > 3/5. Now T4/5 (a) = 2a > 3/5, so a > 3/10. Also T4/5 (2a) = 2(1 − 2a) = 2 − 4a > 2 − 8/5 = 2/5, which leads to a contradiction, unless 2 − 4a > 3/5, or 4a < 7/5, a < 7/20.

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Section 4.2 1. Let Bϵ (a) ⊆ X be an open ball. We show that it is an open set. Let x ∈ Bϵ (a) and set δ = min{d(a, x), ϵ − d(a, x)}, then Bδ (x) ⊆ Bϵ (a). Since every x ∈ Bϵ (a) can be surrounded by an open ball centered on x, entirely contained in Bϵ (a), it is an open set.

If A = {a1 , a2 , . . . , an } is a finite set in X, then we show X \ A is open. Let x ∈ X \ A, and set ϵ = min{d(x, ai ), 1 ≤ i ≤ n}, then Bϵ (x) ⊆ (X \ A), so X \ A is open and A is closed.

2. Let xn ∈ B ϵ (a) be a sequence contained in the closed ball, with limn→∞ xn = x, then it suffices to show that x ∈ B ϵ (a).

Now d(xn , a) ≤ ϵ for n ≥ 1. Let δ > 0, then there exists N ∈ Z+ such that for all n ≥ N , d(x, xn ) < δ. It follows from the triangle inequality that d(x, a) ≤ d(x, xn ) + d(xn , a) < δ + ϵ.

Since this is true for all δ > 0, we must have d(x, a) ≤ ϵ, so x ∈ B ϵ (a).

Now set A = {0, 1} with the discrete metric. If B1 (0) = {x ∈ A : d(0, x) < 1}, then B 1 (0) = B1 (0) = {0}, whereas the closed ball is {x ∈ A : d(0, x) ≤ 1} = {0, 1}.

3. (i) d1 satisfies 1, 2 and 3 of definition 4.1.1. We show that the triangle inequality does not hold. Set x = 4, y = 3 and z = 2, then d1 (x, z) = (4 − 2)2 = 4, d1 (x, y) = (4 − 3)2 = 1 and d1 (y, z) = (3 − 2)2 = 1, so d1 (x, z) > d1 (x, y) + d1 (y, z). (ii) d2 (x, y) ̸= d2 (y, x) so (3) of 1.4.1 does not hold. (iii) For example, d3 (2, −2) = 0, but 2 ̸= −2, so it is not a metric. (iv) d4 (1, 1) = 1 ̸= 0, so it is not a metric. (v) We show that d5 is a metric. First note that (1), (2), and (3) of the definition hold, so we only have to show that the triangle inequality holds. Consider the function f (x) = x/(1+x). f ′ (x) > 0, so this function is increasing. Now |x−z| ≤ |x−y|+|y−z|, so that |x − y| + |y − z| |x − y| |y − z| |x − z| ≤ = + 1 + |x − z| 1 + |x − y| + |y − z| 1 + |x − y| + |y − z| 1 + |x − y| + |y − z|

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|x − y| |y − z| + , 1 + |x − y| 1 + |y − z| and this is the triangle inequality. ≤

4. x = (x1 , x2 ), y = (y1 , y2 ) in R2 . (i) d′ (x, y) = max{|x1 − y1 |, |x2 − y2 |}, so clearly (1), (2) and (3) of 1.4.1 hold. To show that (4) holds, note that |x1 − z1 | ≤ |x1 − y1 | + |y1 − z1 | ≤ max{|x1 − y1 |, |x2 − y2 |} + max{|y1 − z1 |, |y2 − z2 |} and similarly |x2 − z2 | ≤ |x2 − y2 | + |y2 − z2 | ≤ max{|x1 − y1 |, |x2 − y2 |} + max{|y1 − z1 |, |y2 − z2 |}, and the result follows. (ii) d′′ (x, y) = |x1 − y1 | + |x2 − y2 |, so also (1), (2) and (3) of 1.4.1 are easily seen to hold. Clearly |x1 − z1 | ≤ |x1 − y1 | + |y1 − z1 |, and |x2 − z2 | ≤ |x2 − y2 | + |y2 − z2 |, so adding gives the triangle inequality in this case.

Consider Bϵ (x) = {y ∈ R2 : d′ (x, y) < ϵ} = {(y1 , y2 ) : max{|x1 − y1 |, |x2 − y2 |} < ϵ} = {(y1 , y2 ) : |x1 − y1 | < ϵ, |x2 − y2 | < ϵ}, an open square centered centered on (x1 , y1 ).

For Bϵ (x) = {y ∈ R2 : d′′ (x, y) < ϵ} = {(y1 , y2 ) : |x1 − y1 | + |x2 − y2 | < ϵ}, also a square centered on (x1 , y1 ), but the sides of the square parallel to y = ±x.

5. Let A = ∩ni=1 Ai , be the finite union of open sets. Let a ∈ A, then a ∈ Ai for all i = 1, 2, . . . , n. Since each Ai is open, for each 1 ≤ i ≤ n there is an open ball Bϵi (a) ⊆ Ai .

Set ϵ = min{ϵi : 1 ≤ i ≤ n}, Then Bϵ (a) ⊆ Bϵi (a) ⊆ Ai for i = 1, 2, . . . , n, so Bϵ (a) ⊆ ∩ni=1 Ai = A, and so A is open.

6. Set ω1 = (s1 , s2 , . . .), ω2 = (t1 , t2 , . . .) and ω3 = (u1 , u2 , . . .), where si , ti , ui ∈ {0, 1}. ∞ ∑ |sk − tk | If d(ω1 , ω2 ) = , then we show that d defines a metric on the sequence 2k k=1

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space of 0’s and 1’s. Clearly always d(ω1 , ω2 ) ≥ 0, d(ω1 , ω2 ) = d(ω2 , ω1 ), and if d(ω1 , ω2 ) = 0, then sk = tk for all k, so ω1 = ω2 . The triangle inequality follows from |sk − uk | ≤ |sl − tk | + |tk − uk |. Divide by 2k and summing gives the required result.

1 1 3 + + 0 + ··· = .

0 2 2 2 1 1 1 (ii) ω1 = (0, 1, 0, 1, . . .) and ω2 = (1, 0, 1, 0, . . .), so d(ω1 , ω2 ) = 0 + + 2 + · · · = 2.

2 2 2 (i) ω1 = (0, 1, 1, . . .) and ω2 = (1, 0, 1, 1, . . .), so d(ω1 , ω2 ) =

7. f : I → I is a continuous function on an interval I. (a) If f 2 (x) = x for all x in a dense set, then f 2 (x) = x for all x ∈ I (suppose that (xn ) is a sequence in the dense set which converges to x ∈ R, then f 2 (xn ) = xn for all n, and use the continuity of f ). (b) If f (x) = f (y), then f 2 (x) = f 2 (y), so x = y and f is one-to-one. If z ∈ I, then f 2 (z) = z, so if we set x = f (z), f (x) = z, so f is onto. It follows that f −1 exists, and since f 2 (x) = x for all x ∈ I we must have f −1 (x) = f (x) (this tells us that the graph of f is symmetric with respect to the line y = x).

Suppose there exists a ∈ I with f (a) ̸= a, say f (a) > a, then if g(x) = f (x) − x, g(a) = f (a) − a > 0 and g(f (a)) = f 2 (a) − f (a) = a − f (a) < 0. It follows by the Intermediate Value Theorem that there exists c ∈ I with g(c) = 0, or f (c) = c.

Similarly for the case where f (a) < a. Obviously if there is no such a, then every point is a fixed point. (c) Suppose that f is increasing on I with f ̸= idI , the identity function on I. Then there exists α ∈ I with f (α) ̸= α. Suppose that β > α, with β = f (α) then the point (α, β) lies on the graph of f , and so does the point (β, α) (since the graph is symmetric about the line y = x). Thus β = f (α) < f (β) = α, a contradiction.

In a similar way, if β < α, we deduce α = f (β) < f (α) = β and again we have a contradiction. We deduce that f must be decreasing on I. (d) If f ′ (x) exists everywhere and g(x) = f 2 (x) = x, then g ′ (x) = f ′ (f (x))f ′ (x) = 1 for all x ∈ I. If f (c0 ) = c1 and f (c1 ) = c0 , we have g ′ (c0 ) = f ′ (c1 )f ′ (c0 ) = 1, so any 2-cycle is non-hyperbolic. If f (c) = c, then f ′ (f (c))f ′ (c) = 1, so f ′ (c) = ±1. If f is decreasing, we must have f ′ (c) = −1.

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(d) Define f : (0, ∞) → (0, ∞) by f (x) = 1/x, then f is continuous with f 2 (x) = x for all x ∈ (0, ∞). We have seen that the fixed point x = 1 is hyperbolic of the required type and similarly for all the 2-cycles.

8. (a) If f (x) = −1/(x + 1), then f 2 (x) = −1/(−1/(x + 1) + 1) = −(x + 1)/x = −1 − 1/x, and so f 3 (x) = −1 + (x + 1) = x for all x ̸= 0, −1. (b) If f 3 (x) = x for all x ∈ I, then if g(x) = f 2 (x), g 3 (x) = f 6 (x) = f 3 (x) = x. Thus we see that the function g(x) = −1 − 1/x also has the property that g 3 (x) = x for all x ̸= 0. − 1. (c) If f (x) = x for all x ∈ R, then clearly f 3 (x) = x everywhere, a continuous function. Suppose that f : R → R with f 3 (x) = x for all x, where f is not the identity function. There exists points c with f (c) ̸= c. Suppose there is a 2-cycle {c0 , c1 } with c0 ̸= c1 , then f 3 (c0 ) = c1 , leading to a contradiction. Since there are points of period-3, the continuity of f and Sharkovsky’s Theorem imply that f has a 2-cycle, giving a contradiction. It follows that f cannot be continuous. We see from the example in (a), that f need not have a fixed point.

On the other hand, suppose that c is a fixed point for f . Set g(x) = f 3 (x), then g ′ (x) = f ′ (f 2 (x))f ′ (f (x))f ′ (x) = 1, so g ′ (c) = [f ′ (c)]3 = 1 and f ′ (c) = 1, so c is non-hyperbolic of the appropriate type. Clearly any 3-cycle is hyperbolic.

9. Pern (f ) = {x ∈ R : f n (x) = x} where f is continuous. Let (xk ) be a sequence in Pern (f ) with limk→∞ xk = x. Then f n (xk ) = xk for all n and k, so x = limk→∞ xk = limk→∞ f n (xk ) = f n (x), from the continuity of f n , so the set is closed.

10. If f : (0, ∞) → (0, ∞) is defined by f (x) = 1/x, then the period-2 points are the set (0, 1) ∪ (1, ∞), x = 1 being a fixed point. This is not a closed set.

11. C[a, b] is the set of continuous functions f : [a, b] → R. (i) We show that d1 (f, g) = max |f (x)−g(x)| defines a metric on C[a, b]. If d1 (f, g) = x∈[a,b]

0, then maxx∈[a,b] |f (x) − g(x)| = 0, so f (x) = g(x) for all x ∈ [a, b]. Clearly d1 (f, g) ≥ 0 always, d1 (f, f ) = 0 and d1 (f, g) = d1 (g, f ). The triangle inequality follows from

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the inequalities max |f (x) − h(x)| ≤ max (|f (x) − g(x)| + |g(x) − h(x)|)

x∈[a,b]

x∈[a,b]

≤ max |f (x) − g(x)| + max |g(x) − h(x)|. x∈[a,b]

x∈[a,b]

∫ b |f (x) − g(x)| dx. As before, d2 (f, g) ≥ 0, d2 (f, f ) = 0 and d2 (f, g) =

(ii) d2 (f, g) = a

d2 (g, f ). If d2 (f, g) = 0, since |f − g| is a continuous function, if it were greater than zero at x = c, then it must be greater than zero at some interval (c − δ, c + δ), from which it is easy to see that the integral would be non-zero. For the triangle inequality, use |f (x) − h(x)| ≤ |f (x) − g(x)| + |g(x) − h(x)|, and then integrate.

12. We show that the periodic points of the tent map T : [0, 1] → [0, 1] are dense in [0, 1].

Notice that T (0) = 0, T (1/2) = 1, and T (1) = 0 and similarly T 2 (0) = 0, T 2 (1/4) = 1, T 2 (1/2) = 0, T 2 (3/4) = 1 and T 2 (1) = 0. We continue this inductively to show that T n (0) = 0, T n (1/2n ) = 1, T n (2/2n ) = 0, . . . , T n ((2k − 1)/2n ) = 1, T n (2k/2n ) = 0, . . . , T n (1) = 0, It follows that T n ([(2k − 1)/2n , 2k/2n ]) = [0, 1] ⊃ [(2k − 1)/2n , 2k/2n ], for 1 ≤ k ≤ 2n−1 , so there is a fixed point of T n in the interval [(2k −1)/2n , 2k/2n ]. Since these intervals can be made arbitrarily small, each containing a periodic point, the periodic points are dense in [0, 1].

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Section 4.3 x .

1 + |x| (i) f is a well defined function (−1 − |x| < x < 1 + |x| for all x ∈ R). (ii) f is one-to-one since we can check that f ′ (x) = 1/(1 + x)2 > 0 if x > 0, and f ′ (x) = 1/(1 − x)2 > 0 if x < 0. (iii) f is onto, for if z ∈ [0, 1), f (z/(1 − z)) = z, and if z ∈ (−1, 0), then f (z/(1 + z)) = z. { x/(1 − x) 0 ≤ x < 1 In this way we see that the inverse map is defined by f (x) = , x/(1 + x) −1 < x < 0 clearly a continuous function. Thus f is a homeomorphism.

1. (a) f : R → (−1, 1) is defined by f (x) =

(b) We extend f to be a function f : [−∞, ∞] → [−∞, ∞] by setting f (−∞) = −1 and f (∞) = 1. Set d(x, y) = |f (x)−f (y)| for x, y ∈ [−∞, ∞], We show that d defines a metric on [−∞, ∞].

It is clear that properties (1), (2), (3) and (4) hold for x, y ∈ R, so consider d(∞, ∞) = |f (∞) − f (∞)| = 0, d(∞, −∞) = |f (∞) − f (−∞)| = 2, and we can check that all the other required properties hold.

2. f : X → X satisfies d(f (x), f (y)) ≤ αd(x, y) for all x, y ∈ X, a metric space (α ∈ (0, ∞)). Let ϵ > 0 and set δ = ϵ/α, then if d(x, y) < δ, then d(f (x), f (y)) ≤ αd(x, y) < αδ = αϵ/α = ϵ, so f is continuous (actually we have shown uniform continuity, which is slightly stronger that continuity).

3. Continuity is similar to Exercise 2. Suppose that f (x) = f (y), then 0 = d(f (x), f (y)) = d(x, y), so x = y, showing that f is one-to-one. The isometries of R preserve distance, so that |f (x) − f (y)| = |x − y|. Thus |f (x) − f (0)| = |x|, and if we set f (0) = b, then we deduce f (x) = ±x + b. A continuity argument can be used to show that f cannot oscillate between ±x, so f (x) = x + b, or f (x) = −x + b for some b ∈ R. 4. f : [a, b] → [a, b] is a homeomorphism, so it is continuous, one-to-one and onto.

The results of Section 4.3 show that f is either increasing or decreasing.

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Case 1. f is increasing. If f (a) > a, let α ∈ (a, f (a)), then there is no x ∈ [a, b] with f (x) = α, since if x > a, f (x) > f (a) > α, contradicting f being onto. So we must have f (a) = a. Similarly f (b) = b.

Case 2. f is decreasing. If α < β in [a, b], then f (α) > f (β) and similarly f 2 (α) < f 2 (β). In other words, f 2 is increasing (and of course is also a homeomorphism), so f 2 (a) = a and f 2 (b) = b. Suppose that f (a) = r ̸= b. Let α ∈ (r, b), then if x > a, f (x) < f (a) < α, so we again conclude that f is not onto, so we must have f (a) = b, giving a 2-cycle.

5. f : X → X is continuous on (X, d). (a) Let a ∈ X and (xn ) be a sequence in X with limn→∞ xn = a. Let ϵ > 0, then there exists δ > 0 such that if x ∈ X, then d(x, a) < δ implies d(f (x), f (a)) < ϵ. We can find N ∈ N such that if n > N , then d(xn , a) < δ. Hence if n > N , d(f (xn ), f (a)) < ϵ, so that limn→∞ f (xn ) = f (a). (b) x0 ∈ X with limn→∞ f n (x0 ) = p, then from (a) lim f n+1 (x0 ) = lim f (f n (x0 )) = f (p).

n→∞

n→∞

But the former limit is equal to p since (f n+1 (x0 )) is the same sequence as (f n (x0 )), but with a different starting point. It follows that f (p) = p. (c) If O(x0 ) = {x0 , f (x0 ), f 2 (x0 ), . . .} has exactly one limit point, say p, then it is easy to see that there is a sequence (f nk (x0 )) with d(f nk (x0 ), p) < 1/k for k = 1, 2, . . ., and then limk→∞ f nk (x0 ) = p. The continuity of f implies that limk→∞ f nk +1 (x0 ) = f (p).

This says that f (p) is also a limit point of the orbit, so the uniqueness implies that f (p) = p.

6. (a) f : X → Y is a function and U ⊂ X, V ⊂ Y . Suppose that f (U ) ⊂ V and let x ∈ U , then f (x) ∈ V , so by definition of f −1 (V ), x ∈ f −1 (V ), so U ⊂ f −1 (V ).

Now suppose U ⊂ f −1 (V ) and y ∈ f (U ), then there exists x ∈ U with y = f (x).

Thus x ∈ f −1 (V ), so y = f (x) ∈ V . (b) Let x ∈ f −1 (Y \ V ), (where we may assume this set is non-empty). Then f (x) ∈ Y \ V , so f (x) ∈ / V , x ∈ X, but x ∈ / f −1 (V ), so that x ∈ X \ f −1 (V ). The other direction is similar.

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(c) Suppose that f is continuous and C is a closed subset of Y . The Y \ C is open in Y , so f −1 (Y \ C) = X \ f −1 (C) is open in X. It follows that f −1 (C) is closed in X. Let U be open in Y , then Y \ U is closed, so if f −1 (C) is closed in X for all closed sets C in Y , then f −1 (Y \ U ) = X \ f −1 (U ) is closed in X. It follows that f −1 (U ) is open in X, so f is continuous.

7. (a) f : R → R is a homeomorphism with f 2 (x) = x for all x. It follows that f −1 = f , so if (α, β) lies on the graph of f , so does (β, α), and the graph is symmetric about the line y = x. ax + b , where ad − bc ̸= 0, a well defined and continuous function. Let cx + d ax + b ay + b = , and (ad − bc)(x − y) = 0, so x, y ̸= −d/c, then if f (x) = f (y), then cx + d cy + d ( ) dz − b x = y and f is one-to-one. If z ∈ R \ {a/c}, then we can check that f = z, a − cz dx − b so f is onto and the inverse is f −1 (x) = , which is also continuous, so f is a a − cx homeomorphism. (b) f (x) =

(c) If f (x) = x, then

ax + b = x, so cx2 + x(d − a) − b = 0 and cx + d √ a − d ± (a − d)2 + 4bc . x= 2c

It follows that there will be a single fixed point when (a − d)2 + 4bc = 0 and two fixed points when (a − d)2 + 4bc > 0. ax + b a(ax + b)/(cx − a) + b x(a2 + bc) , then f 2 (x) = = = x, so f is cx − a c(ax + b)/(cx − a) − a a2 + bc an involution. ax + b (d) If f (x) = , we see that if the point (x, y) lies on the graph, then so does cx + a a(−y)+b ax+b the point (−x, −y) since y = cx+a if and only if −x = c(−y)+a . As an example, set f (x) = −1/x, where a = 0, b = −1, c = 1. (d) f (x) =

8. We have f : R → R continuous with a fixed point f (c) = c. The basin of attraction of c is the interval (a, b).

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Let (xn ) be a sequence in (a, b) with limn→∞ xn = a. Since f is continuous, we must have limn→∞ f (xn ) = f (a). But the points f (xn ) must lie in (a, b) as it is an invariant set, so f (a) is a limit point of (a, b). f (a) cannot lie in (a, b) as this would imply that a lies in the basin of attraction of c. It follows that either f (a) = a or f (a) = b. In a similar way we have either f (b) = b or f (b) = a. Putting the different possibilities together gives us that one of the cases (a), (b) or (c) must hold.

9. f : X → X is continuous with at least 2 attracting fixed points. X is a metric space that is not the union of two open sets. Set A = ∪α∈Λ Bf (α), the union of the basins of attractions of all attracting fixed points. Let C = X \ A, a closed set since each basin of attraction is open. It is an invariant set, for if x ∈ C and f (x) ∈ / C, then f (x) ∈ Bf (α) for some α ∈ Λ. It follows that α ∈ Bf (α), giving a contradiction.

We are assuming that some Bf (α) is not both open and closed (if it were, then X would not be connected). Let x ∈ Bf (α) \ Bf (α), then x ∈ / Bf (β), for any other β ∈ Λ, for if it were, since Bf (β) is open, there would be a ball surrounding x entirely contained in Bf (β), contradicting x ∈ Bf (α). It follows that C is non-empty.

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Section 4.4 1. We have seen that f : R → R, f (x) = x3 is a homeomorphism. It is not a diffeomorphism since f ′ (0) = 0 (the inverse map is not differentiable at x = 0). (i) For a diffeomorphism f : R → R with exactly one fixed point, we could take f (x) = 2x for example. (ii) Set f (x) = x3 − x2 + x, then clearly f is onto. Now f ′ (x) = 3x2 − 2x + 1, and this is never zero since the discriminant is b2 − 4ac = 4 − 12 = −8 < 0, so f ′ (x) > 0 always. It follows that f is increasing so is one-to-one and is clearly a diffeomorphism. f (x) = x when x2 (x − 1) = 0, so x = 0 or x = 1. (iii) If f (x) = x3 + x/2, then f ′ (x) = 3x2 + 1/2 > 0 always, so f is a diffeomorphism. √ If f (x) = x, then x3 = x/2, x = 0 or x = ±1/ 2, so f has exactly three fixed points.

2. We have f : R → R. For what values of λ is f (a) a homeomorphism, (d) a diffeomorphism. (i) f (x) = λx + 3 will be a diffeomorphism if λ ̸= 0, since f ′ (x) = λ and f will either be strictly increasing or strictly decreasing. (ii) f (x) = λx2 . f is never one-to-one, so cannot be a homeomorphism. (iii) f (x) = x3 + λx. f ′ (x) = 3x2 + λ > 0 if λ > 0, so that f is increasing and f is one to one. Being a cubic, f is onto and continuous, so is a homeomorphism and since f ′ (x) > 0 everywhere, f is a diffeomorphism. If λ = 0, we have f (x) = x3 , which we have seen is a homeomorphism, but not a diffeomorphism. If λ < 0, f is not one-to-one, so is not a homeomorphism. (iv) f (x) = x + λ sin(x), f ′ (x) = 1 + λ cos(x) > 0 always, when |λ| < 1, and f will be a diffeomorphism. This is also the case when λ = ±1. If |λ| > 1, then f will not be one-to-one. (v) f (x) = λx + sin(x), f ′ (x) = λ + cos(x) > 0 if λ > 1. f is clearly onto in this case, so is a diffeomorphism. If λ = 1, there are points where f ′ (x) = 0, so f will not be a diffeomorphism, but is still a homeomorphism. If λ < 1, f will not be one-to-one.

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(vi) f (x) = λx + arctan(x), f ′ (x) = λ + 1/(1 + x2 ) > 0 when λ > 0, and we have a diffeomorphism. If λ = 0, f will not be onto (we have a diffeomorphism onto (π/2, π/2)). If λ ≤ −1, then f ′ (x) ≤ −1 + 1/(1 + x2 ) ≤ 0, so f is decreasing and is therefore one-to-one. f is not a diffeomorphism for −1 < λ < 0.

3. Use a computer algebra system.

4. (a) Let fn (x) = 1 + x + x2 /2! + x3 /3! + · · · + x2n+1 /(2n + 1)!, for n = 0, 1, 2, . . .., ′ then f0 (x) = 1 + x and f0′ (x) = 1 > 0. Suppose that fn−1 (x) > 0 for all x ∈ R, then ′ (x) + x2n /(2n)! > 0. fn′ (x) = 1 + x + x2 /2! + x3 /3! + · · · + x2n /(2n)! = fn−1

It follows from the Principle of Mathematical Induction that fn (x) > 0, for all x ∈ R, and all n ≥ 0. In particular, fn (x) is a one-to-one function whose derivative is never zero. Since fn is an odd degree polynomial, limx→∞ fn (x) = ∞ and limx→−∞ fn (x) = −∞, so the range of fn is all of R, and fn is an onto function. It now follows that fn : R → R is an order preserving diffeomorphism. Consequently, fn cannot have points of period greater than one. Note that for x > 0, fn (x) > x, so fn (x) cannot have any fixed points in (0, ∞).

Set hn (x) = fn (x) − x, then as above limx→∞ hn (x) = ∞ and limx→−∞ hn (x) = −∞, so hn (cn ) = 0, or fn (cn ) = cn for some cn ∈ R. Now hn (x) = 1 + x2 /2! + x3 /3! + · · · + x2n+1 /(2n + 1)! and hn (0) = 1, h′n (x) = x + x2 /2! + · · · + x2n /(2n)!, h′′n (x) = 1 + x + x2 /2! + · · · + x2n−1 /(2n − 1)!) = fn−1 (x), where fn−1 is strictly increasing with a a single zero, say qn−1 < 0 (since each fn is positive for x > 0). Thus hn (x) is concave down for x < qn−1 and concave up for x > qn−1 with a local minimum at x = 0, hn (0) = 1. It follows that the point cn where hn (cn ) = 0 is unique, and in particular, the fixed point of fn is unique (if hn had another zero, the concavity would have to change at some other point). (b) If gn (x) = 1 + x + x2 /2! + x3 /3! + · · · + x2n /(2n)!, then the above argument shows that gn (x) > 0 for all x ∈ R. In addition, gn′′ (x) > 0 from the same argument, so gn is concave up with a single critical point.

Now gn (x) > x for x > 0, so gn (x) cannot have any fixed points in (0, ∞).

Furthermore, since gn (x) > 0 for all x, there cannot be fixed points in (−∞, 0], so gn (x) does not have any fixed points. Since gn (x) is continuous, Sharkovsky’s Theorem implies that gn (x) has no periodic points.

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5. Let f : [a, b] → [a, b] be a continuous increasing function. Here we need to assume that f (x) ̸= Id[a,b] , the identity on the set [a, b]. We have seen that such a function cannot have points of period n > 1, so assume that the fixed points are dense, i.e., f (x) = x for all x ∈ [a, b]. A continuity argument then shows that f (x) = Id[a,b] .

6. To prove the Intermediate Value Theorem for f : [a, b] → R, a continuous function with f (a) < 0 and f (b) > 0, we proceed as follows: Set a1 = (a + b)/2 and b1 = b, if f ((a + b)/2) < 0, and set a = a1 , b1 = (a + b)/2 if f ((a + b)/2) > 0 (we may assume thatf ((a + b)/2) ̸= 0, for otherwise we are done).

Now set a2 = (a1 + b1 )/2 and b2 = b1 if f ((a1 + b1 )/2) < 0, otherwise set a2 = a1 , b2 = (a1 + b1 )/2 if f ((a1 + b1 )/2) > 0 (again we may assume that f ((a1 + b1 )/2) ̸= 0).

Continue in this way inductively, so having defined an and bn , we set an+1 = (an +bn )/2 and bn+1 = bn if f ((an + bn )/2) < 0, otherwise set an+1 = an , bn+1 = (an + bn )/2 if f ((an + bn )/2) > 0. We obtain sequences (an ) and (bn ) for which a1 ≤ a2 ≤ a3 · · · ≤ an ≤ an+1 ≤ · · · ≤ bn+1 ≤ bn ≤ · · · ≤ b3 ≤ b2 ≤ b1 .

Thus (an ) is an increasing sequence bounded above, and (bn ) is a decreasing sequence bounded below, so both limits exists, and α = limn→∞ an ≤ limn→∞ bn = β. Now we always have f (an ) < 0 and f (bn ) > 0 for all n, (unless the algorithm terminated at a zero). Continuity of f then implies that f (α) ≤ 0 and f (β) ≥ 0.

We claim that α = β, and this completes the proof since then f (α) = f (β) = 0.

Suppose not, then β − α = ϵ > 0, so we can find n so that an is within ϵ/2 of α and bn within ϵ/2 of β, for which (an + bn )/2 lies almost midway between α and β, contradicting their construction.

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Section 5.2 1. f : N2 → N is defined by f (m, n) = 2m 3n . f is well defined, and if f (m1 , n1 ) = f (m2 , n2 ), then 2m1 3n1 = 2m2 3n2 , so 2m1 −m2 = 3n2 −n1 , and this implies m1 = m2 and n1 = n2 , so f is one-to-one. If A = range of f , then f : N2 → A is one-to-one and onto a subset of N. It is easy to see that A is a countable set, so N2 is a countable set.

If we define g : N2 → Q+ by f (m, n) = m/n, we see that g is one-to-one and onto, so Q+ is a countable set.

2. (a) Suppose that B ⊆ A, where A is a set of measure zero. Since we can cover A by a collection of open intervals whose total length can be made arbitrarily small, and B ⊆ A, so we can use the same collection to cover B. It follows that B is also a set of measure zero. (b) The same proof that the rationals are of measure zero shows that any countable set has measure. (c) Let An be countable sets, for n = 1, 2, . . ., and set A = ∪∞ n=1 . We can enumerate each An as An = {an1 , an2 , an3 , . . .}, and then we can either use a proof similar to that which showed the rationals are countable, or we can define a map f : N2 → A by setting f (n, m) = anm (where we can assume that all the members of A are distinct).

Then f is one-to-one and onto, so the result follows from Exercise 1.

3. Set Σ = {(a1 , a2 , a3 , . . .) : ai ∈ {0.1}}. Just as we proved that the reals are not countable, we again assume that we have a countable set, so we can list the elements as a sequence, say (ωm ), where ωm = (am1 , am2 , am3 , . . . , amn , . . .). Now define ω ∈ Σ by ω = (a′11 , a′22 , . . . , a′n,n , . . . , ) where a′ii = 0 if aii = 1, and a′ii = 1, if aii = 0. Then ω differs from ωn in the nth position, so does not appear in the sequence (ωm ), so does not belong to Σ, which is a contradiction, so Σ is not countable.

4. (a) [0, ∞) is not compact since for example the sets An = (−1, n) are open and ∪∞ n=1 An = [−1, ∞), so the sets An form an open cover of [0, ∞), but clearly no finite sub-collection of these sets can contain [0, ∞).

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If B = {1, 1/2, 1/3, . . . , 1/n, . . .}, set An = (0, 1/n), then B ⊂ ∪∞ n=1 An , but no finite subset of the An ’s has this property. (b) We have seen that the interval [a, b], a < b is compact. If K is a closed and bounded subset of R, then K ⊂ [a, b] for some a < b. Any open cover of K can be extended in an obvious way to [a, b], and since [a, b] is compact, there is a finite sub-cover of [a, b], which is a finite sub-cover of K.

5. We have seen that an interval cannot have measure zero, and a subset of a set of measure zero has measure zero. The result follows.

6. Two sets A and B have the same cardinality if there is a one-to-one onto map f : A → B. (a) Let f : A → P(A) be a function, where P(A) is the power set of A. We show that f cannot be onto. Set B = {x ∈ A : x ∈ / f (x)}. B is a subset of A, so suppose that there exists x ∈ A with f (x) = B. There are two possibilities, either (i) x ∈ B, or (ii) x∈ / B. If (i) holds then x ∈ / f (x) = B, a contradiction, If (ii) holds, x ∈ f (x) = B, another contradiction. We conclude that there is no x ∈ A with f (x) = B, so f is not onto. (b) It follows that N and P(N) have different cardinalities. In particular, since P(N) is an infinite set that cannot be put into a one-to-one correspondence with N, it is not a countable set. In a similar way the power set of P(N) has different cardinality to both those of N and P(N). This process can be continued indefinitely to give infinitely many sets having different cardinalities. (c) To show that R and R2 have the same cardinality, we first show that (0, 1) × (0, 1) and (0, 1) have the same cardinality.

Define F : (0, 1) × (0, 1) → (0, 1) by F (·x1 x2 x3 . . . , ·y1 y2 y3 . . .) = ·x1 y1 x2 y2 x2 y2 x3 y3 . . . (where we have used the unique infinite binary representations of x = ·x1 x2 x3 . . . and y = ·y1 y2 y3 . . . in (0, 1). F is clearly well defined and onto. Suppose that F (x, y) = F (a, b), then ·x1 y1 x2 y2 x2 y2 x3 y3 . . . = ·a1 b1 a2 b2 a2 b2 a3 b3 . . ., and from the uniqueness of representation x = a, y = b, so F is one-to-one.

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7. Let f : [a, b] → R be a continuous function. We show that f is bounded. Given c ∈ [a, b], and ϵ = 1 > 0, there exists δc > 0 such that if x ∈ [a, b] and |x − c| < δc , then |f (x) − f (c)| < 1, or f (c) − 1 < f (x) < f (c) + 1 (here we are using the obvious interpretation at the end points). Set Ic = (c − δc , c + δc ), then the set {Ic : c ∈ [a, b]} is an open cover of [a, b]. By the Heine-Borel Theorem, there is a finite sub-cover, say {Ic1 , Ic2 , . . . , Icn }. Set M = max{f (ci ) + 1 : 1 ≤ i ≤ n} and m = min{f (ci ) − 1 : 1 ≤ i ≤ n}, then it is clear that m < f (x) < M for all x ∈ [a, b], so f is bounded on [a, b].

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Section 5.3 1. (a) Find the ternary expansion of x = 1/2, 1/4, 2/9 ∈ [0, 1].

Since 1/2 lies exactly in the center of [0, 1], every term in the ternary expansion will be a 1, i.e., 1/2 = ·11111 . . . = 1/3 + 1/32 = 1/33 + · · · = (1/3)/(1 − 1/3) = 1/2.

It does not belong to the Cantor set since it does not have a ternary expansion using only 0’s or 2’s (clearly it does not have a terminating ternary expansion).

Consider ·020202 . . . = 2/32 + 2/34 = 2/36 + · · · = (2/32 )/(1 − 1/32 ) = 1/4 ∈ C, since it has a ternary expansion in terms of 0’s and 2’s.

2/9 has ternary expansion ·02000 . . ., so belongs to C. (b) C ′ = {x ∈ [0, 1] : x does not have a ternary expansion involving 1}. Then 2/9 ∈ / C ′ since 2/9 = ·001111 . . .

2. Let C(α) = ∩∞ n=1 Sn (α), where each Sn (α) is a closed set constructed in the following way. Let α ∈ (0, 1) and an = 2n−1 α/3n , and S0 (α) = [0, 1]. S1 (α) is constructed by removing an open interval of length a1 = α/3 from the center of S0 (α).

S1 (α) is the union of two disjoint closed intervals. From each of these intervals we remove an open interval of length a2 /2 to obtain a set S2 (α) which is the union of 22 = 4 disjoint closed intervals of equal length. We continue the construction in this way. Now |S0 (α)| = 1, |S1 (α)| = 1 − α/3, |S2 (α)| = 1 − α/3 − 2α/32 , and generally |Sn (α)| = 1 − α/3 − 2α/32 − · · · − 2n−1 α/3n . In total, we remove a set of length α/3/(1 − 2/3) = α < 1 from [0, 1]. Clearly C(α) is a closed non-empty set which is not of measure zero.

We claim that C(α) has empty interior (contains no open intervals). Suppose that U ⊂ C(α) is an open interval. Each of the intervals removed is centered on a dyadic rational (a number of the form k/2n ), and all such dyadic rationals have intervals surrounding them removed. It is not difficult to see that any open interval contains dyadic rationals, so part of U must be removed, contradicting U ⊂ C(α).

3. We have x ∈ [0, 1] with ternary expansion x = ·x1 x2 x3 . . . (xi ∈ {0, 1, 2}, i = 1, 2, 3, . . .). Set N = ∞ if xn ̸= 1 for all n ∈ Z+ , otherwise set N = min{n ∈ Z+ : xn = 1}. Let yn = xn /2 for n < N and yN = 1.

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∑ n (a) We must show that if x has more than one ternary expansion, then N n=1 yn /2 is independent of which ternary expansion is used.

The numbers x ∈ [0, 1] which have a non-unique ternary expansion are the triadic rationals - numbers of the form x = k/3n , where 0 < k < 3n−1 , where we may assume that k = a1 3n−1 + a2 3n−2 + · · · + an−1 3 + an , ai ∈ {0, 1, 2} and a1 ̸= 0 and an ̸= 0.

Then we can write x = ·a1 a2 . . . an−1 an 0 0 0 0 . . . = ·a1 a2 . . . an−1 bn 2 2 2 . . ., where bn = an − 1.

Now if ai = 1, but aj ̸= 1 for 1 ≤ j < i, then i = N , and clearly κ will take the same value with either representation. If n = N , an = 1 and bn = 0, so again κ will take the same value (use ·0 0 . . . 0 2 2 2 . . . = ·0 0 . . . 1 0 0 0 . . .. These are the only ∑ n possibilities that we need to consider. This shows that the map κ(x) = N n=1 yn /2 is well defined. (b) We now show that the function κ : [0, 1] → [0, 1] defined by κ(x) =

N ∑ yn

is 2n continuous, onto, and increasing. First note that it is well defined from (a). Suppose that x, z ∈ [0, 1] with x < z, then there exists xi and zi in their ternary expansion with xi < zi (use the infinite ternary expansion and we may assume this is the first position that xj ̸= zj ), if N < i, then κ(x) = κ(z). If xi = 0 and zi =1, then again we can check that κ(x) ≤ κ(z). Similarly if xi = 0 and zi = 2, or if xi = 1 and zi = 2, so that κ is increasing (but takes constant values on certain intervals, as we see in (c)).

To see that κ is continuous, let ϵ > 0 and x ∈ (0, 1). Choose M ∈ Z+ so large that 1/2M < ϵ, and set δ = 1/3M . If y ∈ [0, 1] with |x − y| < δ, with x and y lying in the same triadic interval (k/3M , (k + 1)/3M ), then xi = yi , i = 1, 2, . . . , M (ternary expansion), then it follows that |κ(x) − κ(y)| = 0 if N ≤ M , whereas if ∞ ∑ |xn − yn | 1 M > N , we have |κ(x) − κ(y)| ≤ ≤ M < ϵ. On the other hand, if n+1 2 2 n=M +1 n=1

x ∈ ((k − 1)/3M , k/3M ) and y ∈ (k/3M , (k + 1)/3M ), different triadic intervals, then we can check that κ(x) and κ(y) are still close and continuity follows (from the way κ is defined). Continuity at the end points is dealt with similarly.

Note that κ(1) = κ(·2 2 2 2 . . .) = 2/22 + 2/23 + 2/24 + · · · = 1, and κ(0) = 0, so since κ is continuous and increasing with κ(1) = 1, κ(0) = 0, κ must be onto. (c) We show that κ is constant on each interval contained in the complement of the Cantor set (for example, κ(x) = 1/2 for all x ∈ (1/3, 2/3)), and κ′ (x) = 0 everywhere except on a set of measure zero. If 1/3 < x < 2/3, then x = ·1a2 a3 . . ., where some ai ̸= 0, 2, so that N = 1 and κ(x) = y/2 = 1/2 since y = 1. If 1/9 < x < 2/9, then

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x = ·0 1 a3 a4 . . ., where ai ̸= 0, 2, so N = 2 and then κ(x) = y1 /2+y2 /4 = x1 /4+1/4 = 0/4 + 1/4 = 1/4. On the other hand, if 7/9 < x < 8/9, then x = ·2 1 a3 a4 . . ., so N = 2 and κ(x) = 2/4 + 1/4 = 3/4. Continuing in this way, we see that κ is constant on each interval contained in the complement of the Cantor set. If Sn is the set obtained at the nth stage of the construction of the Cantor set, then κ(x) is constant on the intervals making up An = [0, 1] \ Sn . It follows that κ′ (x) = 0 for all x ∈ An .

Therefore the set on which κ(x) is not zero is contained in Sn for all n. Since the sets Sn have lengths that can be made arbitrarily small, this is a set of measure zero.

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Section 6.5 1. T = T2 is the standard tent map. {We have seen that if x ∈ [0, 1] has binary ·a2 a3 . . . : a1 = 0 , where bi = 0 if ai = 1 expansion x = ·a1 a2 a3 . . ., then T (x) = ·b2 b3 . . . ; a1 = 1 and bi = 1 if ai = 0.

We give a proof by induction. The result we wish to{ prove is true for n = 1. ·ak ak+1 . . . : ak−1 = 0 Suppose that it is true for n = k − 1, then T k−1 (x) = . ·bk bk+1 . . . ; ak−1 = 1 Thus there are 4 cases to consider: Case 1: ak−1 = 0, ak = 0, then T k (x) = ·ak+1 ak+2 . . ..

Case 2: ak−1 = 0, ak = 1, then T k (x) = ·bk+1 bk+2 . . ..

Case 3: ak−1 = 1, ak = 0, then bk = 1, so T k (x) = ·ak+1 ak+2 . . ..

Case 4: ak−1 = 1, ak = 1, then{bk = 0, so T k (x) = ·bk+1 bk+2 . . .. ·ak+1 ak+2 . . . : ak = 0 In all cases we have T k (x) = . The result now follows ·bk+1 bk+2 . . . ; ak = 1 by induction.

We can check that x = ·01110 01110 01110 . . . gives a period-5 point, giving the 5cycle {14/31, 28/31, 6/31, 12/31, 24/31}, and the other 5-cycles can be found similarly.

2(a). We define a map f : Σ → C where Σ = {(a1 , a2 , . . .), ai = 0 or 1} and C is the Cantor set, by f (a1 , a2 , . . .) = .b1 b2 . . . , where bi = 0 if ai = 0 and bi = 2, if ai = 1.

Clearly f is well defined, (unlike the case of f : [0, 1] → C, there is no ambiguity with sequence such as (1, 0, 0, 0, . . .) and (0, 1, 1, 1, . . .) as these represent quite different members of Σ). We start by showing that f is continuous. Let ϵ > 0. We will show that if ω1 = (s1 , s2 , . . .) and ω2 = (t1 , t2 , . . .), there exists δ > 0 such that if d(ω1 , ω2 ) < δ, then |f (ω1 ) − f (ω2 )| < ϵ.

Choose n so large that 1/3n < ϵ, and set δ = 1/2n . Then if d(ω1 , ω2 ) < δ, s1 = t1 , s2 = ∑∞ k t2 , . . . sn = tn , so that |f (ω1 ) − f (ω2 )| = |.00 . . . 0cn+1 cn+2 . . . | ≤ k=n+1 2/3 = n 1/3 < ϵ.

It is clear that f is onto since every possible ternary expansion involving 0’s and 2’s arises. f is one-to-one, since we are using the unique ternary expansion of y ∈ C involving only 0’s and 2’s in its expansion. The proof that the inverse is continuous is similar to the above proof.

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(b) We define f : [0, 1] → C in the natural way, by setting f (·a1 a2 a3 . . .) = ·b1 b2 b3 . . ., where bi = 0 when ai = 0 and bi = 2 when ai = 1, mapping binary expansion to ternary expansion, (in the non-unique case, the binary expansion that gives a member of C may be used). f is not continuous, for example, points such as x1 = ·100000001000 . . . and x2 = ·011111111000 . . . are close together (near 1/2), but their images: f (x1 ) = ·.2000000002000 . . . (near 2/3), and f (x2 ) = ·022222222000 . . . (near 1/3), are far apart. It can be shown that [0, 1] and the Cantor set have different “topological properties”, so cannot be homeomorphic.

3. We have a transitive map f : I → I, where I is an interval. U and V are nonempty open sets in I. Let x0 ∈ I be a transitive point for f , i.e., O(x0 ) = I. Then there exists r, s ∈ Z+ (say r > s, r = s + m), with y = f r (x0 ) ∈ U , f s (x0 ) ∈ V , so that y = f r (x0 ) = f s+m (x0 ) = f m (f s (x0 )) ∈ f m (V ).

In other words, y ∈ U ∩ f m (V ), so U ∩ f m (V ) ̸= ∅.

4. F : [0, 1) → [0, 1) is the tripling map, F (x) = 3x (mod 1). Let x ∈ [0, 1) have ternary expansion x = ·a1 a2 a3 . . . = a1 /3 + a2 /32 + a3 /33 · · · , ai ∈ {0, 1, 2}, then we easily see that F (x) = .a2 a3 a4 . . ., so F acts as a shift on the ternary expansion (care needs to be taken at certain numbers having a non-unique ternary expansion such as x = 1/3 and x = 2/3).

We can now check that x0 = 0 1 2 0 0 0 1 0 2 1 0 1 1 1 2 2 0 2 1 2 2 0 0 0 0 0 1 . . . , where we have listed all possible one blocks (using 0, 1 and 2), all possible 2-blocks, all possible 3-blocks, and we continue this indefinitely. Again we check that x0 is a transitive point for F , using the same argument as for the doubling map. If x = ·a1 a2 . . . an an+1 . . ., set xn = ·a1 a2 . . . an a1 a2 . . . an a1 . . ., a point of period n which we can use to approximate x arbitrarily closely.

5. D : [0, 1) → [0, 1) is the doubling map D(x) = 2x (mod 1). It is straightforward to check that Dn (x) = 2n (x − k/2n ) for k/2n ≤ x < (k + 1)/2n , 0 ≤ k ≤ 2n − 1, so that Dn ([k/2n , (k + 1)/2n )) = [0, 1). It follows that Dn has a unique fixed point in

60

[k/2n , (k + 1)/2n ) and that Pern (D) = 2n − 1 (where we exclude x = 1 since it is not in the domain).

6. fλ : [0, 1] → [0, 1] is defined by fλ (x) = 1 − |2x − 1|λ for 0 < λ ≤ 4. (a) If λ = 1, we have f1 (x) = 1 − |2x − 1| = T2 , the tent map. f2 (x) = 1 − (2x − 1)2 = 1 − 4x2 + 4x − 1 = 4x(1 − x) = L4 (x). (b) Suppose that λ < 1/2, then we show that x = 0 is an attracting fixed point, for then it follows that fλ cannot be chaotic. If 0 ≤ x < 1/2, fλ (x) = 1 − (1 − 2x)λ , and then fλ′ (x) = λ(1 − 2x)λ−1 (−2), |fλ′ (0)| = | − 2λ| = 2λ < 1/2, so x = 0 is attracting. (c) Use a computer algebra system.

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Section 6.6 {

3x; x < 1/2 , restricted to the 3(1 − x); x ≥ 1/2 { ·a2 a3 a4 . . . ; ; 0 ≤ x ≤ 1/3 , Cantor set. We can check that (see Section 5.4) T3 (x) = ·b2 b3 b4 . . . ; 2/3 ≤ x ≤ 1 where bi = 0 if ai = 2, bi = 1 if ai = 1 and bi = 2 if ai = 0. Therefore, if we restrict to the Cantor set (ai = 0 or 2 only), then we have the required formula for n = 1.

Induction now proceeds in the same way as in Exercise 6.5 # 1(a).

1. (a) T3 : C → C is the tent map T3 (x) =

We can now check that {12/13, 3/13, 9/13} (using 12/13 = ·220 220 220 . . .) is a 3cycle, and the other 3-cycles are found in the same way. x = ·02220 02220 02220 . . . = 111/364, gives a period-5 point. (b) The denseness of the periodic points can be shown in a similar way to that for the standard tent map. (c) Again we can construct an explicit transitive points for T3 , using all possible 1blocks, 2-blocks etc. It is convenient to place an additional zero between each block to make sure it appears in the desired form, when iterated. (d) The continuity of T3 restricted to C simply follows from the hint, since T3 is continuous on R. Since T3 is a continuous map on a metric space having a dense set of periodic points and is transitive, we can use Theorem 6.6.1 to deduce that it has sensitive dependence on initial conditions, so is chaotic. { 2. f : [0, 1] → [0, 1] is defined by f (x) =

x + 1/2; 0 ≤ x < 1/2 .

2(1 − x); 1/2 ≤ x ≤ 1

(a) If x = ·a1 a2 a3 . . ., where a1 = 0, then f (x) = x + 1/2 = 1/2 + a2 /22 + a3 /23 + · · · = ·1a2 a3 . . .. If a1 = 1, we have seen that f (x) = ·b2 b3 . . ., where bi = 1 if ai = 0 and bi = 0 if ai = 1. (b) f (0) = 1/2, f (1/2) = 1 and f (1) = 0, so {0, 1/2, 1} is a 3-cycle. If x = ·101010 . . . = 2/3, then x is a fixed point since f (2/3) = 2(1 − 2/3) = 2/3. If x = ·010101 . . . = 1/3, then f (x) = ·1101010 . . ., and f 2 (x) = ·010101 . . . giving the 2-cycle {1/3, 5/6} for f . (c) By calculating f 4 (·a1 a2 a3 . . .) for different cases and equating to ·a1 a2 . . ., we see that x = ·10001 1 10001 1 10001 . . . = 35/63 is a period-4 point for f . This gives

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the 4-cycle as {35/63, 56/63, 14/63, 91/126}. We can check that points of the form x = ·0a1 a2 . . . a2k 1 a′1 a′2 . . . a′2k are of period 2 + 2k exactly, (where aj = 0, j odd, and aj = 1, j even), and points of the form x = ·00a1 a2 . . . a2k , where aj = 1 if j is odd, aj = 0, if j is even, are of period 3 + 2k.

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Section 7.1 1. (i) f : X → X is conjugate to itself via the identity map 1X (x) = x, for all x ∈ X. (ii) If f is conjugate to g via h, then h ◦ f = g ◦ h, so h−1 ◦ g = f ◦ h−1 , so g is conjugate to f via h−1 . (iii) f is conjugate to g via h and g is conjugate to k via q, then we have h ◦ f = g ◦ h and q ◦ g = k ◦ q. It follows that (q ◦ h) ◦ f = q ◦ (h ◦ f ) = q ◦ (g ◦ h) = (q ◦ g) ◦ h = (k ◦ q) ◦ h = k ◦ (q ◦ h), where q ◦ h is also a homeomorphism. It follows that f is conjugate to k via q ◦ h. It now follows that conjugacy is an equivalence relation. { √ x; x≥0 √ 2. fa : R → R is defined by fa (x) = ax. Set h(x) = . Then h is − −x; x < 0 clearly a homeomorphism. { { √ √ f1/2 ( √ x) x/2; x≥0 √ f1/2 ◦ h(x) = = . f1/2 (− −x) − −x/2; x < 0 { √ x/4; x≥0 √ , h ◦ f1/4 = h(x/4) = − −x/4; x < 0 and these are equal, so f1/2 and f1/4 are conjugate. (b) Suppose that fa , fb : [0, ∞) → [0, ∞), 0 < a, b < 1, and h(x) = xp , p > 0. Then fa ◦ h(x) = fa (xp ) = axp , and h ◦ fb (x) = h(bx) = bp xp , so we require a = bp , or ln(a) = p ln(b), p = ln(a)/ ln(b) > 0. With this value of p, h will be a homeomorphism, giving a conjugacy between fa and fb . A similar argument works when a > 1, b > 1.

If a > 1 and 0 < b < 1, then p < 0, for example, suppose a = 1/2, b = 2, p = −1, h(x) = 1/x. In this case x = 0 is an attracting fixed point for f1/2 but not for f2 , so they are not conjugate (f1/2 ◦ h(x) = 1/2x = h ◦ f2 (x), and we can consider them conjugate if we allow x = ∞ as an attracting fixed point).

3. We have f : X → X, g : Y → Y conjugate via a homeomorphism h : X → Y , with h ◦ f = g ◦ h. Suppose that g is one-to-one and f (x) = f (y). Then h(f (x)) = h(f (y))

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and so g(h(x)) = g(h(y)), h(x) = h(y) and so x = y. Similarly if f is one-to-one, then g is one-to-one.

Suppose that g is onto, and let z ∈ X. Then there exists w ∈ Y with g(w) = h(z), as g is onto. Since h is onto, there exists x ∈ X with h(x) = w, so that g(h(x)) = g(w) = h(z), and then h(f (x)) = h(z), so f (x) = z since h is one-to-one. This shows that f is onto.

4. This time we are implicitly assuming that f : I → I, g : J → J are conjugate via a homeomorphism h : I → J, with h ◦ f = g ◦ h, where I and J are sub-intervals of R. Consequently, h is either increasing or decreasing.

Case 1. Suppose that h is increasing, and that x0 is a local maximum for f . Then there exists ϵ > 0 for which f is increasing on (x0 −ϵ, x0 ] and decreasing on [x0 , x0 +ϵ).

We show that g is increasing on h(x0 −ϵ, x0 ] = (α0 , y0 ], and decreasing on h[x0 , x0 + ϵ) = [y0 , β0 ).

Let z, w ∈ (α0 , y0 ], z < w, then there exists x1 , x2 ∈ [x0 , x0 + ϵ) with h(x1 ) = z, h(x2 ) = w, with x1 < x2 , since h is increasing. Hence f (x1 ) < f (x2 ), so h(f (x1 )) < h(f (x2 )), or g(h(x1 )) < g(h(x2 )) and g(z) < g(w), so g is increasing on (α0 , y0 ].

Similarly, g is decreasing on [y0 , β0 ), so g has a local maximum at y0 , where y0 = h(x0 ).

The other cases are similar.

5. We have f : [0, 1] → [0, 1], g : [0, 1] → [0, 1] conjugate via a homeomorphism h : [0, 1] → [0, 1], with h ◦ f = g ◦ h, where f (0) = f (1) = 0 g(0) = g(1) = 0. We show that h is increasing.

We know that h is either increasing or decreasing. Suppose that h is decreasing, then we must have h(0) = 1 and h(1) = 0. Since h(f (0)) = g(h(0) this gives h(0) = g(1), or 1 = 0, which is impossible, so h must be increasing.

6. (a) f : X → X is a homeomorphism conjugate to its inverse f −1 via a homeomorphism g, so f ◦ g = g ◦ f −1 . Suppose that g is an involution, then g −1 is an involution and f = (g ◦ f −1 ) ◦ g −1 = h ◦ k where k = g −1 is an involution, and h = g ◦ f −1 so

h2 = (g ◦ f −1 ) ◦ (g ◦ f −1 ) = (f ◦ g) ◦ (g ◦ f −1 ) = f ◦ g 2 ◦ f −1 = f ◦ f −1 = identity.

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Conversely suppose that f = h ◦ k, where both h and k are involutions. Then f ◦ k = h ◦ k 2 = h, and k ◦ f −1 = k ◦ k −1 ◦ h−1 = k ◦ k ◦ h = k 2 ◦ h = h, so f ◦ k = k ◦ f −1 . Note that since k is an involution, it is one-to one and onto. (b) f (x) = 1/(3 − x). Set h(x) = 1/x and k(x) = 3 − x. h and k are both involutions.

Also, h ◦ k(x) = h(3 − x) = 1/(3 − x) = f (x), so f is the product of two involutions, so is conjugate to f −1 .

7. Here the question should read: Tn : [−1, 1] → [−1, 1], Tn (x) = cos(n arccos(x)), the nth Chebyshev polynomial, and Λn : [0, 1] → [0, 1] is a piecewise linear continuous function defined by joining (0, 0), (1/n, 1), (2/n, 0), (3/n, 1) . . . with straight lines. We must show that Tn and Λn are conjugate via h : [0, 1] → [−1, 1], h(x) = cos(πx). We can check that { 2k ≤ x ≤ 2k n( n − x); 2k−1 n n , Λ(x) = 2k 2k n(x − n ); n ≤ x ≤ 2k+1 n for k = 0, 1, . . . , n/2, n even, k = 0, 1, . . . (n + 1)/2, n odd.

Then Tn ◦ h(x) = Tn (cos(πx)) = cos(n arccos(cos(πx))) = cos(nπx), and h ◦ 2k )) = cos(π(nx − 2kπ) = cos(πnx), for 2k ≤ x ≤ 2k+1 , and Λn (x) = h(n(x − n n n 2k h ◦ Λn (x) = h(n( − x)) = cos(π(2kπ − nx) = cos(πnx), for 2k−1 ≤ x ≤ 2k , so the n n n result follows.

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Section 7.2 1. (a) h : S1 → [0, 1] defined by h(eix ) = sin2 (x) is well defined, for if eix = eiy , then x = y + 2kπ for some k ∈ Z, but then sin2 (y + 2kπ) = sin2 (y), so sin2 (y) = sin2 (x). h is onto since sine takes every value in [−1, 1] for 0 ≤ x < 2π (but clearly h is not one-to-one). The continuity of h is clear for z ∈ S1 , z ̸= 1. The continuity when z = 1 follows from the fact that if x is close to zero, eix is close to 1, and sin2 (x) is close to 0 (e.g., if ϵ > 0 is small, then sin2 (ϵ) and sin2 (2π − ϵ) are close in value). (b) h : Σ → S1 , h(a1 , a2 , a3 , . . .) = e2π(·a1 a2 a3 ...) . We have seen that h is onto, but not one-to-one. For continuity, let ϵ > 0, and choose n so large that 1/2n < ϵ. Set δ = 1/2n , then if d(ω1 , ω2 ) < δ, the first n coordinates of ωi are equal. The metric d′ on S1 gives the distance around the arc of the circle, which is just the difference in the arguments, i.e., d′ (h(ω1 ), h(ω2 )) = | · a1 a2 a3 . . . an an+1 . . . − ·a1 a2 . . . an a′n+1 . . . | = k+1 = 1/2n < ϵ, so h is continuous. | · 000 · · · 00 a′n+1 · · · | < Σ∞ k=n+1 1/2

2. We show that f is a factor of D as it is not true that D is a factor of f . Here D : [0, 1) → [0, 1) is the doubling map D(x) = 2x (mod 1), and f : S1 → S1 is the angle doubling map f (z) = z 2 . We show that there exists a continuous onto map h : [0, 1) → S1 such that h ◦ D = f ◦ h.

Define h : [0, 1) → S1 by h(x) = e2πix , for 0 ≤ x < 1. h is continuous, one-toone and onto. (Note: The inverse of h is not continuous at z = 1, since h−1 (1) = h−1 (e2π0 ) = 0, and if ϵ > 0 is small, e2π(1−ϵ) is close to 1, but h−1 (e2π(1−ϵ) ) = 1 − ϵ, which is not close to 0, so h is not a homeomorphism). h ◦ D(x) = h(2x) = e4πix , and f ◦ h(x) = f (e2πix ) = e2πi2x = e4πix , so f is factor of D. 3. (a) If g : S1 → S1 , g(z) = z 3 , then g(eiθ ) = e3iθ , so that g triples the angle θ. (b) If g n (z) = z for some n ∈ Z+ , then z 3 = z or z 3 −1 = 1. Write z = eiθ , then we want to find the (3n − 1)th roots of unity. Then: n

n

e(3 −1)iθ = e2kπi , for some k ∈ Z+ , n

giving the 3n − 1 distinct roots: zk = e2kπi/(3 −1) , k = 0, 1, 2, . . . , 3n − 2, showing that n

Pern (g) = {e2kπi/(3 −1) : 0 ≤ k < 3n − 1}, n ∈ Z+ . n

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These points are equally spaced around the circle, a distance 2π/(3n − 1) apart, which can be made arbitrarily small by taking n large enough. It follows that the periodic points of g are dense in S1 . (c) Again we show instead, that g is a factor of F . We need to show that there is a continuous onto map h : [0, 1) → S1 with h ◦ F = g ◦ h. Now set h(x) = e2πix , for 0 ≤ x < 1, then h is continuous, onto and one-to-one, then as for h in Exercise 2, we obtain the result (again, h is not a homeomorphism since its inverse in not continuous at z = 1 as mentioned in Exercise 2).

4. We wish to show that T3 : C → C, the member of the tent family with µ = 3, but restricted to the Cantor set, is conjugate to the shift map σ : Σ → Σ. Define h : C → Σ by { 0; if T3n (x) ∈ [0, 1/3] h(x) = (a1 , a2 , a3 , . . .), where an = .

1; if T3n (x) ∈ [2/3, 1] h is one-to-one, onto and continuous, with a continuous inverse. Now σ ◦ h(·x1 x2 x3 . . .) = σ(a1 , a2 , a3 , . . .) = (a2 , a3 , . . .), (as above). Also, h ◦ T3 (·x1 x2 x3 . . .) = (z1 , z2 , z3 , . . .) (say), where { { 0; if T3n+1 (x) ∈ [0, 1/3] 0; if T3n (T3 (x)) ∈ [0, 1/3] = zn = , 1; if T3n (T3 (x)) ∈ [2/3, 1] 1; if T3n+1 (x) ∈ [2/3, 1] and this is just the definition of an+1 , so σ ◦ h(·x1 x2 x3 . . .) = h ◦ T3 (·x1 x2 x3 . . .), and the result follows.

5. In order for the shift map σ : Σ → Σ and the doubling map D : [0, 1) → [0, 1) to be conjugate, we need the metric spaces Σ and [0, 1) to be homeomorphic, and this is not the case.

6. U : [−1, 1] → [−1, 1] is defined by U (x) = 1−2x2 . T : [0, 1] → [0, 1] is the tent map and h : [0, 1] → [−1, 1] is defined by h(x) = − cos(πx). h is continuous, one-to-one and onto with continuous inverse, so h is a homeomorphism.

2 2 { Now U ◦ h(x) = U (− cos(πx)) { = 12 − 2 cos (πx) = sin (2πx) and h ◦ T (x) = h(2x); 0 ≤ x < 1/2 sin (2x); 0 ≤ x < 1/2 = = sin2 (2x). The h(2(1 − x)); 1/2 ≤ x ≤ 1 sin2 (2(1 − x)); 1/2 ≤ x ≤ 1 conjugacy now follows.

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7. We have h◦f (x) = g◦h(x), where h is a diffeomorphism. Then differentiating gives h′ (f (x)) · f ′ (x) = g ′ (h(x)) · h′ (x), so if f (p) = p then h′ (p)f ′ (p) = g ′ (h(p))h′ (p), then since h′ (p) ̸= 0 (we have a diffeomorphism), then f ′ (p) = g ′ (h(p)), so the derivative at the fixed point is preserved.

For periodic points, if f n (p) = p, in the same way we get (f n )′ (p) = (g n )′ (h(p)).

3 8. G : [−1, 1]  → [−1, 1] is defined by G(x) = 4x −3x, and F : [0, 1] → [0, 1] is defined 0 ≤ x < 1/3  3x; −3x + 2; 1/3 ≤ x < 2/3 . If h(x) = cos(πx), then h : [0, 1] → [−1, 1]

by F (x) =  3x − 2; 2/3 ≤ x ≤ 1 is one-to-one, onto and continuous, with continuous inverse, so is a homeomorphism.

3  G ◦ h(x) = G(cos(πx)) = 4 cos (πx) − 3 cos(πx) = cos(3πx), and h ◦ F (x) = 0 ≤ x < 1/3  cos(3πx); cos(−3πx + 2π); 1/3 ≤ x < 2/3 = cos(3πx), so F and G are conjugate.  cos(3πx − 2π); 2/3 ≤ x ≤ 1

9. If instead of Y = h(X) in the hypothesis of Theorem 7.2.4 we have Y = h(X), then we get h(X) = h(Per(f )) ⊆ h(Per(f )) ⊆ Per(g) ⊆ Y, and taking the closure gives the denseness of the periodic points. Similarly for the transitivity.

10. We show that stable fixed points are preserved by conjugacy. Let f be a function defined on a metric space (X, d) with fixed point p. Recall that p is a stable fixed point for the dynamical system (X, f ), if given ϵ > 0, there exists δ > 0 such that if x ∈ Bδ (p), then f n (x) ∈ Bϵ (p) for n = 1, 2, . . .. Note that f is necessarily continuous at x = p.

Suppose that h◦f = g◦h, where (Y, g) is another dynamical system and h : X → Y is a homeomorphism. Since f (p) = p, then g(h(p)) = h(p). Let ϵ > 0, then we show that there exists δ > 0 such that if y ∈ Bδ (h(p)), then g n (y) ∈ Bϵ (h(p)) for n = 1, 2, . . . .

Since h is continuous, h−1 (Bϵ (h(p))) is open in X, so there exists ϵ′ > 0 with Bϵ′ (p) ⊆ h−1 (Bϵ (h(p))). From the stability of p, there exists δ ′ > 0 such that x ∈ Bδ′ (p) implies f n (x) ∈ Bϵ′ (p) for n = 1, 2, . . ..

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Thus if h(x) ∈ h(Bδ′ (p)), then h ◦ f n (x) ∈ h(Bϵ′ (p)), n = 1, 2, . . . , so g n ◦ h(x) ∈ Bϵ (h(p)), n = 1, 2, . . . . Hence if y ∈ h(Bδ′ (p)) then g n (y) ∈ Bϵ (h(p)), n = 1, 2, . . ..

Since h−1 is continuous, there exists δ > 0 with Bδ (h(p)) ⊆ h(Bδ′ (p)), so y ∈ Bδ (h(p)) implies g n (y) ∈ Bϵ (h(p)), n = 1, 2, . . ., and the fixed point h(p) is stable for g.

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Section 7.3 1. (X, f ) and Y, g) are conjugate dynamical systems with f ◦ h = g ◦ h, and A ⊂ X is invariant under f . We need to show that h(A) is invariant under g. Let y ∈ h(A), then y = h(x) for some x ∈ A, so f (x) ∈ A and h(f (x)) ∈ h(A), or g(h(x)) ∈ h(A), i.e., g(y) ∈ h(A), so the invariance follows.

2. fc (x) = x2 + c, with c = (2µ − µ2 )/4. fc′ (x) = 2x = 0 when x = 0. fc′′ (x) = 2 so we have a minimum at x = 0 with fc (0) = c. In addition, fc (µ/2) = µ2 /4 + c = fc (−µ/2), so we simply have to check that −µ/2 ≤ c < µ2 /4 + c ≤ µ/2. This is just −µ/2 < c = (2µ − µ2 )/4 < µ2 /4 + c = (2µ − µ2 )/4 < µ/2. The first inequality is equivalent to µ(µ − 4) ≤ 0, which certainly holds for 0 < µ ≤ 4, and the other inequalities are straightforward.

3. fc (x) = x2 + c and Lµ (x) = µx(1 − x) are considered as dynamical systems on R. (a) Suppose c > 1/4, then fc has no fixed points, but Lµ always has the fixed point x = 0, so they cannot be conjugate for any µ.

1 2µ − µ2 = , so (µ − 1)2 = 0, and µ = 1. It follows that f1/4 (b) If c = 1/4, then 4 4 and L1 are linearly conjugate. √ √ 2µ − µ2 2 ± 4 − 16c 2 (c) If c = , then µ − 2µ + 4c = 0, so that µ = = 1 ± 1 − 4c.

4 2 √ This is defined for c < 1/4, giving two values of µ, say µ1 = 1 + 1 − 4c and √ µ2 = 1 − 1 − 4c with fc conjugate to both Lµ1 and Lµ2 (so for example, when c = −2, µ1 = 4 and µ2 = −2, so that L4 and L−2 are conjugate). Note that here we are talking about conjugacy as functions from R to R. √ √ (d) If c = 2/9, then µ1 = 1 + 1 − 8/9 = 4/3, and µ2 = 1 − 1 − 8/9 = 2/3, so that L2/3 and L4/3 are conjugate. This does not contradict what we know about the logistic maps as we are looking at these maps as functions defined on the whole real line. The fixed points of Lµ occur at x = 0 and x = 1 − 1/µ. L4/3 has fixed points at x = 0 and x = 1/4, and L2/3 has fixed points at x = 0 and x = −1/2, i.e., both maps have two fixed points.

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(e) If c < 0, µ1 > 0, but µ2 < 0, so there is a unique µ > 0 corresponding to c < 0.

(Note: there is an error in the statement of this question, replace c > 0 by c < 0).

4. (a) fa (x) = ax and fb (x) = bx for x ∈ R. Suppose that h(x) = cx + d, c ̸= 0, and h ◦ fa (x) = fb ◦ h(x), then h(ax) = fb (cx + d), or acx + d = bcx + bd for all x. It follows that a = b, so fa and fb can only be linearly conjugate if they are the same map. (b) Differentiating h ◦ fa (x) = fb ◦ h(x), or rather h(ax) = b · h(x) gives a · h′ (ax) = b · h′ (x), so if a ̸= b, h′ (0) = 0, so h is not a diffeomorphism. (c) Suppose that 0 < a, b < 1 and h : [0, 1] → [0, 1] is a conjugacy between fa and fb , so that h ◦ fa (x) = fb ◦ h(x). Then h(ax) = b · h(x), so h(0) = b · h(0) and h(0) = 0.

Since h is a homeomorphism of [0, 1], we must also have h(1) = 1.

We therefore have h(a) = b · h(1) = b, h(a2 ) = b · h(a) = b2 , and inductively, h(an ) = bn for n ∈ Z+ 5. Set p(x) = ax2 +bx+d, and fc′ (x) = x2 +c′ , then we apply Example 7.3.2 with a = b2 − 0 + 0 − 2b + 4c′ a, b = b, c = d, r = 1, s = 0 and t = c′ . Then we require d = , or 4a 2 2 ′ 4ad − b + 2b b − 2b + 4c . In other words, we choose c′ = , to give the required d= 4a 4 conjugacy. The uniqueness of c′ follows from the proof of 7.3.2.

6. We have L4 : [0, 1] → [0, 1], L4 (x) = 4x(1 − x) and F : [−1, 1] → [−1, 1], F (x) = 2x2 − 1. To show they are conjugate we use Example 7.3.2. Set a = −4, b = 4, c = 0 and r = 2, s = 0 and t = −1. We need to check that c = (b2 − s2 + 2s − 2b + 4rt)/4a, or that 0 = (16 − 0 + 0 − 8 − 8)/(−16), and this is indeed the case. In this situation, the conjugating map is h(x) = −2x + 1, and we see that h(0) = 1, and h(1) = −1, so h is a homeomorphism that maps [0, 1] onto [−1, 1].

7. (a) We consider the logistic maps Lµ : [0, 1 → [0, 1], Lµ (x) = µx(1 − x) and the 1−µ 1 , 2−µ ] → [ 1−µ , 1 ], Fµ (x) = (2 − µ)x(1 − x). We claim logistic type map Fµ : [ 2−µ 2−µ 2−µ 1−µ µ 1 these are conjugate for µ ̸= 2, via the map h : [0, 1] → [ 2−µ , 2−µ ], h(x) = 2−µ x + 1−µ .

2−µ

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We check that h ◦ Lµ (x) = Fµ ◦ h(x) for x ∈ [0, 1]. In fact h(Lµ (x)) =

µ 2 x − µ 2 x2 − µ + 1 = Fµ (h(x)), 2−µ

1 as is readily checked. Note that h(0) = 1−µ , and h(1) = 2−µ , and h is a linear 2−µ homeomorphism between the two intervals.

(b) Set µ = 4, then (a) says that L4 : [0, 1] → [0, 1], L4 (x) = 4x(1 − x) is conjugate to F4 : [−1/2, 3/2] → [−1/2, 3/2], F4 (x) = −2x(1 − x) = L−2 . (c) Set µ = 4/3, then L4/3 is conjugate to F4/3 = 2/3x(1 − x) = L2/3 as a map on [−1/2, 3/2]. (d) We have seen that the closed form solution to the difference equation xn+1 = √ 4xn (1 − xn ) is xn = sin2 (2n arcsin( x0 )) (from Exercises 1.1 # 3(ii)). We use the conjugacy between L4 and L−2 to obtain the closed form solution to xn+1 = −2xn (1− xn ).

Set xn = −2yn + 3/2 and substitute into xn+1 = −2xn (1 − xn ) to give −2yn+1 + 3/2 = −2(−2yn + 3/2)(1 + 2yn − 3/2) = (4yn − 3)(2yn − 1/2), √ to give yn+1 = 4yn (1 − yn ). It follows that yn = sin2 (2n arcsin( y0 )) , but since √ yn = −xn /2 + 3/4, this√ gives −xn /2 + 3/4 = sin2 (2n arcsin( −x0 /2 + 3/4)), so that xn = −2 sin2 (2n arcsin( x0 /2 + 3/4)) + 3/2.

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Section 8.2 1. If F (x) = x4 − 2x2 − 3, then F ′ (x) = 4x3 − 4x = 4x(x2 − 1) = 4x(x − 1)(x + 1).

Since all the roots of the derivative of F are real and distinct, it follows from Lemma 8.1.1 that the Schwarzian derivative of F is negative.

2. F (x) =

f ′′ (x) f ′ (x)f ′′′ (x) − [f ′′ (x)]2 ′ , and , so F (x) = f ′ (x) [f ′ (x)]2 [ ′′ ]2 [ ]2 1 f ′′′ (x) f (x) 1 f ′′ (x) ′ 2 F (x) − [F (x)] = ′ − − = Sf (x) 2 f (x) f ′ (x) 2 f ′ (x)

3. If c is an attracting fixed point of a C 1 function f , then we cannot have |f ”′ (c)| > 1, as this implies c is repelling, so we must have |f ”′ (c)| ≤ 1.

4. Let f (x) = x − x5 , then solving f (x) = x gives x = 0 as the unique fixed point. f is an odd function, so we can find period-2 points by solving f (x) = −x. This gives x5 − 2x = 0, so x = 0, or x = ±21/4 . Now f ′ (x) = 1 − 5x4 and f ′ (x) = 0, when x = 1/51/4 . For the stability, note that f ′ (0) = 1, so x = 0 is a non-hyperbolic fixed point. f ′′ (x) = 20x3 and f ′′ (0) = 0, f ′′′ (0) = 60x2 , so f ′′′ (0) = 0, so Theorem 1.5.3 is inconclusive. However, f (x) = x − x5 < x for x > 0 and f (x) > x for x < 0, so our usual arguments show that x = 0 is an attracting fixed point. Suppose that basin of attraction is an interval (a, b) with 0 ∈ (a, b). As usual, a and b are either fixed, eventually fixed, or give a 2-cycle. This is only possible if a = −21/4 and b = 21/4 .

The resulting interval contains the critical point x = 1/51/4 .

5. (a) Suppose that p(x) is a polynomial of degree n. The equation p(x) = x, or p(x) − x = 0 can have at most n distinct solutions. We are assuming there are exactly n distinct solutions, say {x1 , x2 , . . . , xn }, all fixed points of p. First note that if x = c is an attracting fixed point of a polynomial, then the immediate basin of attraction is a bounded open interval. This is because there are always points that are attracted to ∞ (we can think of ∞ as an attracting fixed point: p(∞) = ∞, and for x large (positive or negative), pn (x) → ∞ as n → ∞). If all x1 , x2 , . . . xn ,

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are attracting fixed points, then each of the immediate basins of attraction (say the open intervals {I1 , I2 , . . . , In }), contains a critical point ci , i = 1, 2, . . . , n. Now p′ (x) is a polynomial of degree n − 1, so can have at most n − 1 critical points, giving a contradiction. Thus there can be at most n − 1 attracting fixed points. (b) We know that any quadratic f (x) = ax2 + bx + c (a ̸= 0), has negative Schwarzian derivative: f ′ (x) = 2ax + b, f ′′ (x) = 2a and f ′′′ (x) = 0. It follows that Sf (x) = −3[2a/(2ax + b)]2 /2 < 0. √ If Lµ (x) = µx(1 − x) where µ > 2 + 5, then L′µ (x) = µ − 2µx.

Now any periodic point c must satisfy 0 ≤ c ≤ 1 (since we are looking at the equation Lnµ (x) = x for 0 ≤ x ≤ 1). If c has the property that |L′µ (c)| > 1, then it will be a repelling periodic point. Now L′µ (x) achieves its minimum value when √ µ2 − 4µ)/2µ, i = 1, or Lµ (x) = 1. This is when µx2 − µx + 1 = 0, or x = (µ ± i √ √ i = 2. Then L′µ (xi ) = µ − 2µ(µ ± µ2 − 4µ)/2µ = ∓ µ2 − 4µ This will be repelling √ when µ2 − 4µ > 1, µ2 − µ − 1 > 0, or µ > (1 + 5)/2.

6. (a) p(x) = x3 + ax2 + bx + c, p′ (x) = 3x2 + 2ax + b, p′′ (x) = 6x + 2a and p′′′ (x) = 6, so that [ ]2 6 3 6x + 2a Sp(x) = 2 − .

3x + 2ax + b 2 3x2 + 2ax + b This simplifies to give Sp(x) =

−6[6x2 + 4ax + a2 − b] , (3x2 + 2ax + b)2

so we require the quadratic in the numerator to have no roots. This happens when a2 − 3b > 0. (b) If a2 − 3b < 0, then the above quadratic has two solutions, say x1 < x2 . The quadratic is negative for x1 < x < x2 , so the Schwarzian derivative will be positive for these values of x. Conversely, if Sp(x) > 0 on an interval, we must have a2 − 3b < 0. (c) We show that there is an interval on which the Schwarzian derivative is positive, if and only if p′ (x) > 0 for all x (this is slightly different to what is claimed in the problem). If p(x) = x3 + ax2 + bx + c, then p′ (x) = 3x2 + 2ax + b > 0 everywhere, if and only if 4a2 −12b < 0 (the discriminant of the quadratic), if and only if a2 −3b < 0.

Consequently from (b), p′ (x) > 0 if and only if there is an interval on which p(x) has a positive Schwarzian derivative. (NOTE: For example, p(x) = x3 has Schwarzian

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derivative which is not negative on any interval, and is strictly increasing on R with p′ (x) ≥ 0 for all x. The requirement that p′ (x) > 0 is essential for the conclusion). (d) If p(x) = x3 − x2 + x/2, then a2 − 3b = 1 − 3/2 = −1/2 < 0, so there is an interval on which Sp(x) is positive. This interval is given by x1 and x2 above, where √ √ xi = (−4a ± 2 6b − 2a2 )/12 = (4 ± 2 3 − 2)/12 = 1/2, or 1/6, so the interval on which Sp(x) > 0 is (1/6, 1/2). p(0) = 0 and p′ (x) = 3x2 − 2x + 1/2, so p′ (0) = 1/2 and x = 0 is an attracting √ fixed point. Solving p′ (x) = 0 gives x = (2 ± 4 − 6)/6, so there are no real solutions, so p(x) has no critical points.

7. f (x) = ax2 + bx + c (a ̸= 0), f has negative Schwarzian derivative, and has exactly one critical point, so f can have at most one attracting cycle.

8. G(x) = λ arctan x, G′ (x) = λ/(1 + x2 ), and this is never zero, so G has no critical points.

G′′ (x) = −2xλ/(1 + x2 )2 and G′′′ (x) = (6λx2 − 2λ)/(1 + x2 )3 , so that [ ] [ ]2 6λx2 − 2λ (1 + x2 ) 3 −2xλ (1 + x2 ) −2 SG(x) = − . = 2 3 2 2 (1 + x ) λ 2 (1 + x ) λ (1 + x2 )2 (a) G(0) = 0, so x = 0 is a fixed point. G′ (0) = λ, and it is stable for |λ| < 1. Since there are no critical points, the basin of attraction must be unbounded, and from the symmetry, it must be equal to (−∞, ∞). (b) If λ > 1, x = 0 is no longer attracting. If g(x) = x − λ arctan(x), then g(x) > 0 for x large, and g(x) < 0 for x small (e.g. set x = 1/2), so g has a zero x1 > 0, and similarly a zero x2 < 0. x1 and x2 are fixed points of G. Since G′ (x) = λ, these fixed points are attracting. Again the basins of attraction must be unbounded intervals, which we can argue must be (0, ∞) and (−∞, 0) (cannot be of the form (a, ∞) as then a would have to be another fixed point). (c) Suppose λ < −1. We use the fact that G(x) is an odd function and a solution to G(x) = −x gives a period-2 point. If g(x) = x + λ arctan x, then g(x) > 0 for x large.

Also g ′ (x) > 0 for λ ≥ −1. For λ < −1, g has two critical points, g(0) = 0 and we see that g(x1 ) = 0 for some x1 > 0, and g(−x1 ) = 0. Thus {−x1 , x1 } is a 2-cycle for

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G. Again we argue that the basin of attraction of x1 is unbounded, excluding 0, so must be (−∞, 0) ∪ (0, ∞). (d) We have used Theorem 8.1.4 to show that the basins of attraction must be unbounded.

9.(a) f (x) = 2x3 + bx2 + cx + d. x = 0 is a period-2 point, f 2 (1) = f (1), and both x = 0 and x = 1 are critical points of f .

Now f ′ (x) = 6x2 + 2bx + c, so f ′ (0) = c = 0 and f ′ (1) = 6 + 2b = 0, so b = −3, and f (x) = 2x3 − 3x2 + d. f (0) = d and since f (d) = 0 we have 2d3 − 3d2 + d = 0. We cannot have d = 0, so 2d2 − 3d + 1 = 0, or (2d − 1)(d − 1) = 0. If d = 1, then f (1) = 0 and f 2 (1) = f (0) = 1, so f (x) = 2x3 − 3x2 + 1 does not satisfy all of the conditions. If d = 1/2, then f (1) = 2 − 3 + 1/2 = −1/2, and f (−1/2) = −2/8 − 3/4 + 1/2 = −1/2, so d = 1/2, and f (x) = 2x3 − 3x2 + 1/2. (b) Since the critical points of f are real and distinct, f has negative Schwarzian derivative. (c) Theorem 8.1.4 tells us that f can have at most two attracting cycles. [ ]2 f ′′′ (x) 3 f ′′ (x) − 10. Sf (x) = ′ = 0, so that 2(f ′ (x))2 f ′′′ (x) = 3f ′ (x)(f ′′ (x))2 . Note f (x) 2 f ′ (x) that

[ ] f ′′ (x)[2f ′ (x)f ′′′ (x) − 3(f ′′ (x))2 ] d (f ′′ (x))2 = = 0, dx (f ′ (x))3 [f ′ (x)]4

(y ′ )2 = c′ , a constant, or y ′ = cy 3/2 , a y3 4 separable differential equation. Solving gives −2/y 1/2 = cx + d, or f ′ (x) = , (cx + d)2 −1 + e, which is of the required form. so f (x) = cx + d from above. Set y = f ′ (x), then this says that

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11. f is a C 4 function with Newton function Nf (x) = x − f (x)f ′′ (x) . Thus [f ′ (x)]2 Nf′′ (x) =

f (x) and Nf′ (x) = f ′ (x)

[f ′ (x)]2 (f ′ (x)f ′′ (x) + f (x)f ′′′ (x)) − f (x)f ′′ (x)2f ′ (x)f ′′ (x) [f ′ (x)]4

[f ′ (x)]2 f ′′ (x) + f (x)f ′ (x)f ′′′ (x) − 2f (x)[f (x)]2 , [f ′ (x)]2 and (noting that f (c) = 0), { [ Nf′′′ (x) = [f ′ (x)]3 2f ′ (x)[f ′′ (x)]2 + [f ′ (x)]2 f ′′′ (x) + f (x)[f ′ (x)f ′′′ (x)]′ + f ′ (x)[f ′ (x)f ′′′ (x)] −2f (x)([f ′′ (x)]2 )′ − 2f ′ (x)[f ′′ (x)]2 ] − 3[f ′ (x)]2 f ′′′ (x) [[f ′ (x)]2 f ′′ (x) + other terms]} /[f ′ (x)]6 =

N ′′′ f (c) =

[f ′ (c)]4 [2[f ′′ (c)]2 + f ′ (c)f ′′′ (c) + f ′ (c)f ′′′ (c) − 2[f ′′ (c)]2 ] − 3[f ′ (c)]4 [f ′′ (c)]2 [f ′ (c)]6 [ ′′ ]2 f ′′ (c) f (c) 2f ′ (c)f ′′′ (c) − 3[f ′′ (c)]2 =2 ′ −3 ′ = = 2Sf (c). ′ 2 [f (c)] f (c) f (c)

12. Suppose p(x) = an xn + an−1 xn−1 + · · · + a4 x4 + a3 x3 + a1 x + a0 , where a3 /a1 > 0.

Then [ ]2 [ ]2 p′′′ (0) 3 p′′ (0) 6a3 0 6a3 Sp(0) = ′ − = − = > 0. ′ p (0) 2 p (0) a1 a1 a1 Since p is a polynomial, all its derivatives are continuous, so Sp(x) is also continuous where it is defined. p′ (x) ̸= 0 near x = 0, so Sp(x) is continuous near x = 0, so as Sp(0) > 0 we must have Sp(x) > 0 in an interval containing 0.

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Section 9.1 1. (a) 0 < a, b < 1, fa (x) = ax, fb (x) = bx, x ∈ [0, 1]. Then fa (x) < x and fb (x) < x for all x ∈ (0, 1). Set x0 = 1 and take I = (fa (x0 ), x0 ] = (a, 1] as a fundamental domain for fa , and set J = (b, 1]. Define h : I → J by setting h(a) = b, h(1) = 1 and linearly on I so that h is an increasing bijection. n ∪∞ n=0 fa (I) = (0, 1] is a disjoint union, giving a partition of (0, 1]. Set h(0) = 0.

Now if y ∈ f n (I), then fa−n (y) ∈ I, so we set h(y) = fbn (h(fa−n (y))), and we can show in the usual way that h is a homeomorphism on [0, 1] with h ◦ fa = fb ◦ h. (b) Proceed exactly as in part (a), but replace fb by g, and write J = (g(1), 1] 2. Let 0 < λ, µ < 1. h : [0, 1] → [0, 1] is an orientation preserving homeomorphism with h ◦ Lµ (x) = Lλ ◦ h(x) for all x ∈ [0, 1].

Now Lλ (1−h(x)) = λ(1−h(x))(h(x))) = Lλ (h(x)) = h(Lµ (x)) = h(Lµ (1−x)) = Lλ ◦h(1−x).

It follows that 1 − h(x) = h(1 − x) or 1 − h(x) = 1 − h(1 − x). The latter cannot happen as h is one-to-one, so h(x) + h(1 − x) = 1, 2h(1/2) = 1, h(1/2) = 1/2.

Also h(Lµ (1/2)) = Lλ (h(1/2)), so h(µ/4) = λ/4.

3. f, g : [0, 1] → [0, 1] are homeomorphisms. (a) If f is order preserving and g is order reversing, then we have seen that f has no points of period two, but g does have a 2-cycle, so they cannot be conjugate. (b) Suppose that f (x) > x and g(x) < x for all x ∈ (0, 1). In this case f and g are increasing on [0, 1] with the same fixed points: f (0) = 0 = g(0) and f (1) = 1 = g(1).

Now x = 0 is an attracting fixed point for g and repelling for f , whilst x = 1 is attracting for f and repelling for g. Thus any homeomorphism with h ◦ f = g ◦ h must have h(0) = 1 and h(1) = 0 (mapping attracting fixed point to attracting fixed point, etc.). Fix x0 ∈ (0, 1) and write J = (f (x0 ), x0 ], a fundamental domain for f .

Continuing in this way, we can show that f and g are conjugate.

4. We show that Sλ is conjugate to Sµ , on [0, π] when λ, µ ∈ (0, 1) and Sλ (x) = λ sin(x). We follow the steps in Example 9.1.2 # 2. x = 0 is an attracting fixed

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point. Consider the maps restricted to the interval [0, π/2], where they are both increasing. Set I = (Sµ (π/2), π/2] = (µ, π/2] as a fundamental domain for Sµ , and similarly J = (λ, π/2] as a fundamental domain for Sλ . Define h : (µ, π/2] → (λ, π/2] so that h(µ) = λ and h(π/2) = π/2, and linearly elsewhere, so it is a homeomorphism. n It can be seen that (0, π/2] = ∪∞ n=0 Sµ (I) disjointly, using the fact that x = 0 is an attracting fixed point. We can extend the definition of h to (0, π/2] by setting h(x) = Sλm ◦h◦Sµ−n (x), x ∈ (0, π/2]. Set h(0) = 0, then h is increasing and continuous on [0, π/2]. Define h on (π/2, π] by setting h(x) = 1 − h(x) for x ∈ [0, π/2]. We can check that h has the required properties.

5. Ra : S1 → S1 is defined by Ra (z) = az, a rotation of the circle. (a) Ra and Rb can be conjugate, for example if b = ā = 1/a, then if h(z) = z̄ we have h(Ra (z)) = h(az) = āz̄, and Rā (h(z)) = Rā (z̄) = āz̄. This shows that Ra is conjugate to its inverse Rā . Other methods (see Chapter 17) can be used to show that generally, they cannot be conjugate). (b) Tα : [0, 1) → [0, 1) is defined by Tα (x) = x + α (mod 1), and Ra (z) = az, where a = e2πiα . Then if ϕ : [0, 1) → S1 , ϕ(x) = e2πix , ϕ ◦ Tα (x) = e2πi(x+α) = e2πiα e2πix = Ra (e2πix ) = Ra ◦ϕ(x). This shows that Ra is a factor of Tα , but they are not conjugate as ϕ is not a homeomorphism.

6. Tα : [0, 1] → [0, 1] is defined by Tα (x) = x + α (mod 1). The inverse is the map Tα−1 = 1 − α (mod 1). Set h(x) = −x (mod 1), for some β ∈ [0, 1). Then h is a homeomorphism of [0, 1] and h(Tα (x) = −x − α (mod 1), and Tα−1 (h(x)) = −x − α (mod 1), so that Tα and Tα−1 are conjugate.

Tα and Tα2 cannot be conjugate when α is irrational. This follows from results in Chapter 17, concerning the eigenvalues of rotations.

7. The proof that for 0 < µ ≤ 2 the map Lµ is conjugate to L2µ is very similar to the proof of Lµ being conjugate to Lλ for 1 < µ < λ < 2 (see Examples 9.1.2 # 3).

Suppose instead that µ > 2, then Lµ has a single critical point at x = 1/2.

However, L2µ (x) = µ(µx(1−x))(1−µ(1−x)) = −µ3 x4 ++2µ3 x3 +µ2 (µ−1)x2 +µ2 x, and √ µ ± −2µ + µ2 2 ′ 2 (Lµ ) (x) = µ (−1 + 2x)(1 + 2µ(−1 + x)x) = 0, when x = 1/2, x = , 2µ

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so that L2µ has three critical points for µ > 2. It follows that Lµ and L2µ cannot be conjugate.

For the tent family Tµ we see in a similar way that Tµ is conjugate to Tµ2 for 0 < µ ≤ 1, but for similar reasons to the logistic maps, they cannot be conjugate for µ > 1.

8. (i) It is straightforward to check that L4 ◦k −1 = k −1 ◦T2 , where k −1 (x) = sin2 (πx/2), as we have seen previously. (ii) Suppose that h ◦ L4 (x) = T2 ◦ h(x) for all x ∈ [0, 1], where h : [0, 1] → [0, 1] is a homeomorphism. We know that either h(0) = 0 or h(0) = 1. Now h(0) = T2 (h(0)), so if h(0) = 1, 1 = h(0) = T2 (1) = 0, a contradiction, so h(0) = 0, i.e., h is an order preserving homeomorphism of [0, 1]. (iii) We have seen that h maps critical points to critical points, and since it is order preserving, it must map maxima to maxima etc. (iv) Now {1/2n , 2/2n , . . . , (2n − 1)/2n } are the critical points of T2n , so suppose the critical points of Ln4 are {a1 , a2 , . . . , a2n −1 }, appearing in that order. For the conjugation h between L4 and T2 we have Ln4 ◦ h−1 = h−1 ◦ T2n , and T2n (k/2n ) = 1, so Ln4 ◦ h−1 (k/2n ) = h−1 ◦ T2n (k/2n ). This implies Ln4 ◦ h−1 (k/2n ) = h−1 (1) = 1. The fact that h−1 is order preserving implies that h−1 (k/2n ) = ak , for k = 1, 2, . . . , 2n − 1.

The same must be true for any homeomorphism, so h−1 and k −1 must be equal on the critical points. (v) The critical points of T2 occur at {k/2n : k = 1, 2, . . . , 2n − 1, n ∈ Z+ }, and these are dense in [0, 1]. It follows that the critical points of L4 are dense in [0, 1].

Since h(x) = k(x) at all local maxima and minima and h(x) and k(x) are continuous functions that agree on dense sets, then they are equal for all x ∈ [0, 1]. In other words the conjugacy between L4 and T2 is unique. √ 2 (vi) We have shown that the only conjugacy is k(x) = arcsin( x). This function π 1 ′ is not differentiable at x = 0 or at x = 1: k (x) = √ . π x − x2

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9. We know that conjugacies map critical points to critical points. The critical points of T2n occur at k/2n , k = 1, 2, . . . , 2n − 1. Since k ◦ L4 = T2 ◦ k, or L4 ◦ k −1 = k −1 ◦ T2 , where k −1 (x) = sin2 (πx/2), the critical points occur at sin2 (πk/2n+1 ).

10. Suppose that ϕ : [0, 1] → [0, 1] is a homeomorphism with L4 ◦ ϕ(x) = ϕ ◦ L4 (x) √ 2 for all x ∈ [0, 1]. We have shown that if k(x) = arcsin( x), then k ◦ L4 = T2 ◦ k, π and that k is the unique map with this property. Then L4 ◦ k −1 = k −1 ◦ T2 , and so ϕ ◦ L4 ◦ k −1 = ϕ ◦ k −1 ◦ T2 , and L4 ◦ ϕ ◦ k −1 = ϕ ◦ k −1 ◦ T2 . The uniqueness implies that ϕ ◦ k −1 = k −1 , so ϕ is the identity map.

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Section 9.2 ( ) ( ) µ3 µ3 µ4 1 µ4 1 1. If µ > 1 and < , then Tµ = > , so that Tµ = 1 + µ4 2 1 + µ4 1 + µ4 ) 2 1 + µ4 ) ( ( µ µ3 1 µ µ2 µ3 1 µ4 µ 1− = < < , so T = < < , µ 4 4 4 4 4 4 1+µ ) 1+µ 1+µ ( 2 1+µ 1+µ 1+µ 2 ) ( 3 3 µ3 µ µ µ2 = = and Tµ . I.e., Tµ4 and 1 + µ4 1 + µ4 1 + µ4 1 + µ4 { } µ3 µ4 µ µ2 , , , 1 + µ4 1 + µ4 1 + µ4 1 + µ4 is a 4-cycle. This happens when 1 + µ4 > 2µ3 , or µ4 − 2µ3 + 1 > 0, i.e., (µ − 1)(µ3 − µ2 − µ − 1) > 0. We have shown that Tµ has 2n cycles if µ > 1. These must be distinct from the above cycle, so there is no contradiction. ) ( µ5 1 µ4 µ4 = > , (when < (b) For a 5-cycles, we need µ > 1 and Tµ 1 + µ5 1 + µ5 2 1 + µ5 1 ). For this we require µ4 − µ3 + µ2 − µ + 1 > 0.

2 ( n−1 ) µ 1 µn−1 ≤ , then T (c) Generalizing the above argument, suppose that = µ 1 + µn 2 1 + µn 1 µn > , so that 1 + µn 2 ( ) ( ) µn−1 1 µn µn µ Tµ < < , = µ 1 − = n n n n 1+µ 1+µ 1+µ 1+µ 2 ( ) µ µk µn−1 1 µ2 and Tµ .

Continuing in this way (since ≤ < = 1 + µn 1 + µn 1 + µn 1 + µn 2 n−1 n−1 µ µ 1 for k = 1, 2, . . . n − 1), we eventually get back to , so that if ≤ , or n n 1+µ 1+µ 2 equivalently µn−1 − µn−2 − µn−3 − · · · − 1 ≥ 0, we have an n-cycle.

2. Let µ > 1. We show that

µ − µ2 + µ4 is a period-4 point for Tµ . First note that 1 + µ4

µ − µ2 + µ4 1 > as this is equivalent to µ4 − 2µ2 + 2µ − 1 > 0, or (µ − 1)(µ3 + µ2 − 4 1+µ 2 µ + 1) > 0. For µ > 1, it is clear that this holds. Thus ( ) µ − µ2 + µ4 µ − µ2 + µ3 1 Tµ = < , 4 4 1+µ 1+µ 2

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since (µ − 1)2 (µ2 + 1) > 0. Thus ( ) µ − µ2 + µ3 µ2 − µ3 + µ4 1 Tµ = > , 4 4 1+µ 1+µ 2 as this is equivalent to (µ − 1)(µ3 − µ2 + µ + 1) > 0. Hence ( 2 ) 1 µ − µ3 + µ4 µ − µ3 + µ4 > , Tµ = 4 4 1+µ 1+µ 2 since this is equivalent to µ4 − 2µ3 + 2µ − 1 > 0, or (µ − 1)3 (µ + 1) > 0. Consequently ) ( µ − µ2 + µ4 µ − µ3 + µ4 = , Tµ 1 + µ4 1 + µ4 so we have a 4-cycle. √ 3. Consider the tent map T√2 : [0, 1] → [0, 1], T√2 (x) = 2x, 0 ≤ x ≤ 1/2, T√2 (x) = √ 2(1 − x), 1/2 ≤ x ≤ 1. √ √ √ √ √ (i) T√4 2 (1/2) = T√3 2 ( 2/2) = T√2 2 ( 2(1 − 2/2)) = T√2 2 ( 2 − 1) = T√2 (2 − 2) = √ √ √ √ 2(1 − 2 + 2) = 2 − 2, where 2 − 2 is a fixed point of T√2 . (ii) We first show that there is a sub-interval of [0, 1] on which T√2 is chaotic. √ √ Note that 2− 2 and 2−1 are a fixed point and an eventual fixed point (they are √ √ √ √ fixed points of T√2 2 ). Consider T√2 : [ 2−1, 1/ 2] → [ 2−1, 1/ 2]. We can see that √ √ √ T√2 is a dynamical system on this sub-interval since T√2 ( 2 − 1) = 2 − 2 < 1/ 2, √ √ √ √ T√2 (1/ 2) = 2 − 1 < 1/ 2 and T√2 (1/2) = 1/ 2 < 1 (see Exercise # 5 below with √ µ = 2). We claim that T√2 is chaotic on this interval. To prove this, let U and V be any open intervals in the sub-interval. If we can show that there exists n ∈ Z+ with V ∩ T√n2 (U ) ̸= ∅, then it follows that T√2 is transitive, and hence is chaotic (using the fact that a continuous transitive map on a compact interval is chaotic). Using the ideas of Exercise # 6, we can find p, q ∈ Z+ with 1/2 ∈ T p (V ) ∩ T q (U ), from which the result follows.

We know that T2 : [0, 1] → [0, 1] is chaotic and has points of all possible periods. √ √ √ √ Corollary 9.2.3 tells us that T√2 2 : [ 2 − 1, 2 − 2] → [ 2 − 1, 2 − 2] is conjugate √ √ to T2 on [0, 1] (so T√2 2 is chaotic on [ 2 − 1, 2 − 2]). Let n ∈ Z+ , then there exists x0 ∈ [0, 1] with T2n (x0 ) = x0 , a point of period n. It follows that there exists √ √ y0 ∈ [ 2 − 1, 2 − 2] with T√2n2 (y0 ) = y0 (the image of x0 under the conjugation map).

It follows that y0 is a point of period n for T√2 2 and hence of period n or 2n for T√2 .

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We can rule out period n when n is odd (see (iii) below), so that T√2 has points of all even periods. (iii) We know that T√2 does not have a point of period 3, and we show that it has no periodic points of odd period greater than 1. We use a result from Section 2.8, which says that the smallest µ for which Tµ has a point of period k (k > 3 odd), is the largest root of the equation: pk (µ) = µk−1 − µk−2 − µk−3 + µk−4 − µk−5 + · · · + µ − 1 = 0 = µk−1 − µk−2 −

1 + µk−2 .

1+µ

√ √ √ When we substitute µ = 2 into this equation, we get pk ( 2) = 1 − 2 < 0, for all √ k = 5, 7, 9 . . . i.e., it is negative, showing that 2 is smaller than this value of µ for every k ≥ 5. It follows that T√2 cannot have points of odd period greater than one.

4. To show that Tµ2 ([1/(1 + µ), µ/(1 + µ)]) ⊆ [1/(1 + µ), µ/(1 + µ)], for 1 < µ ≤ √ 2, we use that fact that the extrema of Tµ2 occur either at the end points of the interval, or at critical points. The critical points of Tµ2 occur at x = 1/2, x = 1/2µ and x = 1 − 1/(2µ). However, 1/2µ and 1 − 1/2µ do not lie in the interval under consideration. Now Tµ2 (1/(1 + µ)) = Tµ (µ/(1 + µ)) = µ(1 − µ/(1 + µ)) = 1/(1 + µ), a fixed point of Tµ2 for µ > 1, and similarly for µ/(1 + µ). Also, Tµ2 (1/2)) = Tµ (µ/2) = µ(1 − µ/2) = µ − µ2 /2 ∈ [1/(1 + µ), µ/(1 + µ)] (use the fact that (µ − 1)(µ2 − 2) < 0 √ for 1 < µ ≤ 2). The result now follows.

5. Let 1 < µ < 2. We show that Tµ : [µ − µ2 /2, µ/2] → [µ − µ2 /2, µ/2] defines a dynamical system. Note that µ−µ2 /2 < 1/2 < µ/2. Now Tµ (µ−µ2 /2) = µ2 −µ3 /2 ∈ [µ − µ2 /2, µ/2], and Tµ (µ/2) = µ(1 − µ/2) = µ2 − µ2 /2. Also Tµ (1/2) = µ/2. Since the maxima and minima occur either at an end point or at a critical point, it follows that Tµ is a well defined dynamical system.

Now µ−µ2 < 1/(1+µ) < µ/(1+µ) (use µ3 −µ+1 > 0 for the first inequality, which clearly holds for µ > 1). Also µ/2 > µ/(1+µ) as is easily checked, so the result follows.

Note: The problem should read: Show that [1/(1 + µ), µ/(1 + µ)] ⊂ [µ − µ2 /2, µ/2]. √ 6. Tµ : [0, 1] → [0, 1] is the tent map with µ > 2. Let U ⊂ [0, 1] be an open interval and suppose that 1/2 ∈ / U , then Tµ (U ) is an open interval whose length is increased by a factor of µ, so that |Tµ (U )| > µ|U |. Continuing this argument, as long as 1/2

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does not belong to Tµn (U ), we have |Tµn (U )| > µ|Tµn−1 (U )| > µn |U |, so in particular |Tµm+2 (U )| > µm+2 |U |. The latter term will eventually exceed 1. As this is impossible, it follows that 1/2 ∈ Tµm+2 (U ) for some m ≥ 1.

Now if µ > 1/2, then Tµ (1/2) = µ/2 > 1/2, and Tµ2 (1/2) = µ(1−µ/2) < 1/2, since (µ − 1)2 > 0. We therefore have [Tµ2 (1/2), Tµ (1/2)] = [µ(1 − µ/2), µ/2]. Now T n (U ) is always an interval, and we may suppose that 1/2 ∈ T m (U ) and 1/2 ∈ T m+1 (U ) (for example, if this is not the case, we get a sequence of adjacent iterations, only one of which contains 1/2, again contradicting the fact that [0, 1] is a bounded set). Then Tµ2 (1/2) ∈ T m+2 (U ) and Tµ (1/2) ∈ T m+2 (U ). In particular, µ/2 ∈ T m+2 (U ) so the maximum of Tµ appears in T m+2 (U ). The Intermediate Value Theorem now implies that [Tµ2 (1/2), Tµ (1/2)] ⊂ Tµm+2 (U ), and the result follows.

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Section 10.3 1. Let the length of the curve at the n stage of the construction be Ln , with L0 = 3 as we started with an equilateral triangle of side length 1. Each side of the triangle is increased in length by a factor of 4/3 (we remove one line of length 1/3 and add two such lines), so L1 = 3 · 4/3 = 4 This procedure is continued at each stage of the construction, so that each line appearing in the triangle is increased in length by a factor of 4/3, so that Ln = 4Ln−1 /3. It follows that Ln = 4n /3n−1 , so that Ln → ∞ as n → ∞. We conclude that the Koch Snow Flake is a curve of infinite length.

To find the area enclosed by the curve, we are starting with an equilateral triangle having side length 1. Denote by An the area at the nth stage of the construction. Then √ √ A1 = 1/2×base×height = 1/2×1× 3/2 = 3/4. We add three equilateral triangles √ of side length 1/3, whose total area is therefore 3(1/2×1/3×1/3× 3/2). We now add two triangle on the face of each new triangle, and two more triangles adjacent to each new triangle, so the total number of new triangles has increased by a factor of 4. At the same time the length of the sides of the new triangle has decreased by a factor of 3, and the area therefore decreases by a factor of 32 . This procedure continues to obtain a √ √ geometric series having ratio r = 4/32 . In other words A2 = 3/4+3(1/3·1/3· 3/4).

We now add 4 × 3 = 12 equilateral triangles of side length 1/32 with total area √ √ √ √ 4 · 3(1/32 · 1/32 · 3/4), so A3 = 3/4 + 3(1/3 · 1/3 · 3/4) + 4 · 3(1/32 · 1/32 · 3/4).

Summing this infinite geometric series gives √ √ √ √ √ 3 3(1/3 · 1/3 · 3/4) 3 3(1/3 · 1/3 · 3/4) 2 3 + = + = .

4 1 − 4/32 4 5/9 5

2. Taking intervals (boxes) of length (size) 1/4, the first step can be covered by using 2 intervals, so N1/4 (K) = 2. At the second stage of the construction we have 4 intervals, each of length 1/42 , so we need 22 intervals to cover the set K, i.e., N1/42 (K) = 22 .

Continuing in this way, at the kth stage of the construction we have 2k intervals, each of length 1/4k , so that 1 log 2k log 2 = . = k k→∞ log 4 log 4 2

D = lim

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3. (a) If A = Q ∩ [0, 1], set δ = 1/n, n = 1, 2, 3, . . .. If δ = 1, then Nδ (A) = 1. If δ = 1/2, then Nδ (A) = 2, and if δ = 1/3, then Nδ (A) = 3. Continuing in this way, if log n δ = 1/n, then Nδ (A) = n. Thus dim(A) = limn→∞ log = 1. n (b) For A = [0, 1] × [0, 1], again set δ = 1/n, n = 1, 2, 3, . . .. If δ = 1, then Nδ (A) = 1.

If δ = 1/2, then Nδ (A) = 4, and if δ = 1/3, then Nδ (A) = 32 , and continuing, if n2 δ = 1/n, then Nδ (A) = n2 . Thus dim(A) = limn→∞ log = 2. log n

4. In the construction of the Sierpinski Carpet S, we start with the unit square [0, 1] × [0, 1], partitioned into 9 equal sub-squares, and remove the open middle third square: (1/3, 2/3) × (1/3, 2/3). We continue in this way removing open middle third squares from the remaining eight squares. (a) Let An be the area left at the nth stage of the construction, so A1 = 1, A2 = 1 − 1/32 , A3 = 1 − 1/32 − 8/34 , A4 = 1 − 1/32 − 8/34 − 82 /36 , A4 = 1 − 1/32 − 8/34 − 82 /36 − 83 /38 , so ultimately, the area left is A = 1 − 1/32 − 8/34 − 82 /36 − 83 /38 − · · · = 1 −

1/32 = 0, 1 − 8/32

so we have removed 1 square unit. (b) If δ = 1/3, we need Nδ (S) = 8 boxes to cover S. When δ = 1/32 , we need 82 boxes. log 8 log 8n n n = , is Generally when δ = 1/3 , we need 8 boxes. Thus dim(S) = lim n→∞ log 3n log 3 the box counting dimension of S. (c) Points of the form (k/3n , l/3m ) remain in S (we can think of S as the intersection of the sets S1 = [0, 1] × [0, 1], S2 = [0, 1] × [0, 1] \ (1/3, 2/3) × (1/3, 2/3), S3 etc., and such points belong to each Sn ). In fact S is a type of 2-dimensional Cantor set, and consists of points whose coordinates have a ternary expansion consisting only of 0’s and 2’s. However, it is not hard to see that every open ball contains points having ternary expansions which cannot be written only in terms of 0’s and 2’s so cannot be contained in the Sierpinski Carpet. Therefore S has no interior.

5. Start with the unit cube and subdivide it into 27 equal cubes having side length 1/3. Remove such sub-cubes from each face and also remove the central sub-cube.

We have removed 7 cubes having a total volume of 7/33 , so the volume remaining is 1 − 7/33 .

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From the remaining 20 cubes, repeat this process, removing 7 cubes having side length 1/32 and volume 7/36 from each of the 20 sub-cubes, so the volume remaining is 1 − 7/33 − 7 · 20/(32 )3 . We continue in this way so that at the next stage we have 1 − 7/33 − 7 · 20/(33 )2 − 7 · 202 /(33 )3 , and continuing indefinitely gives a volume of

1−

∞ ∑ n=1

7·

7/33 20n−1 = 1 − =0 (33 )n 1 − 20/33

To find the box counting dimension of M : We can cover M with 20 cubes of side length 1/3.

We can cover M with 202 cubes of side length 1/32 , We can cover M with 20k cubes of side length 1/3k , so

log 20n log 20 = . n n→∞ − log 1/3 log 3

D = dim(M ) = lim

6. Set A = {0, 1, 1/2, 1/3, 1/4, . . .}. If δ = 1, then Nδ (A) = 1. If δ = 1/2, then Nδ (A) = 2, and if δ = 1/3, then Nδ (A) = 3. If we continue in this way, with δ = 1/n, and Nδ (A) = n, we will be using unneeded boxes. Rather set δ = 1/n(n − 1), n = 2, 3, . . .. Then we cover the points of A in the interval [0, 1/n] with n − 1 boxes.

There are n − 1 remaining points: {1/(n − 1), 1/(n − 2), . . . 1/(n − (n − 1)) = 1} that are uncovered, and we cover these with n − 1 boxes. Thus δ = 1/n(n − 1) and log(2n − 2) = 1/2. (This suggests that the Nδ (A) = 2n − 2. Thus dim(A) = lim n→∞ log(n(n − 1)) definition of box counting dimension does not work well for all types of sets.)

7. This figure is a self-similar fractal. We see that five copies one third the size of the log c log 5 original appear. i.e., c = 5 and a = 3, so d = = . log a log 3

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

 1 1 1 1 1 1 1 1 1  1 0 1 1 0 1 1 0 1     1 1 1 1 1 1 1 1 1      1 1 1  1 1 1 0 0 0 1 1 1    8. We have A1 =  1 0 1  , so that A2 = A1 ⊗A1 =  1 0 1 0 0 0 1 0 1  ,   1 1 1  1 1 1 0 0 0 1 1 1   1 1 1 1 1 1 1 1 1     1 0 1 1 0 1 1 0 1  1 1 1 1 1 1 1 1 1 and A3 = A1 ⊗A1 ⊗A1 is constructed similarly. It is clear that coloring the unit square according to the matrix An will give rise to the n approximation to the Sierpinski carpet.   1 0 1 0 0 0 1 0 1  1 1 0 0 0 0 1 1 0     0 1 1 0 0 0 0 1 1      1 0 1  1 0 1 1 0 1 0 0 0    (a) G1 =  1 1 0  , and G2 = G1 ⊗ G1 =  1 1 0 1 1 0 0 0 0  . To   0 1 1  0 1 1 0 1 1 0 0 0   0 0 0 1 0 1 1 0 1     0 0 0 1 1 0 1 1 0  0 0 0 0 1 1 0 1 1 find the box counting dimension of the resulting fractal, we need 6 boxes of side length 1/3 to cover the fractal, 62 boxes of side length 1/32 , 6n boxes of side length 1/3n , so the dimension is D = lim n → ∞ log(6n )/ log(3n ) = log(6)/ log(3). ( ) (b) To generate the Cantor set C, use the matrix S = 1 0 1 , so that S ⊗ S = ( ) 1 0 1 0 0 0 1 0 1 . For the Sierpinski  triangle, for a possible approxima0 0 0 1 0 0 0  0 0 1 1 1 0 0   tion could be given by:   0 1 0 0 0 1 0 .

1 1 1 0 1 1 1

9. Suppose that (rn ) is a decreasing sequence with rn > 0 for all n, limn→∞ rn = 0, log rn+1 log N (rn ) log N (r) lim = 1 and lim = α, then we wish to show lim+ = α. n→∞ log rn n→∞ − log rn r→0 − log r

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If rn+1 < r ≤ rn , then log N (rn ) ≤ log N (r) ≤ log N (rn+1 ), and − log rn ≤ − log r < − log rn+1 . It follows that log N (r) log N (rn+1 ) log N (rn ) ≤ ≤ , − log rn+1 − log r − log rn where it can be seen that the first and last term go to α as n → ∞. log N (rn ) Now let ϵ > 0, and choose M large enough that for all n > M , −α < − log rn+1 log N (rn+1 ) ϵ, and similarly we get − α < ϵ. − log rn Let δ = rM and 0 < r < δ, then there exists n with rn+1 < r ≤ rn < δ, so the log N (r) − α < ϵ for 0 < r < δ, and the above inequality holds. It follows that − log r result follows.

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Section 10.4 1. We show that the following sequences are Cauchy in R: (i) xn = (n + 1)/n. Let ϵ > 0 and choose N ∈ Z+ with N > 1/ϵ. Suppose that m > n > N , then |xm − xn | = |1/n − 1/m| = (m − n)/(mn) < 1/n < 1/N < ϵ, so the sequence is Cauchy. (ii) xn = 1 + 1/2 + 1/4 + · · · + 1/2n . Let ϵ > 0 and choose N so large that 2N > 1/ϵ (N > ln(1/ϵ)/ ln(2)).

Suppose that m > n > N , then |xm − xn | = |1/2n+1 + 1/2n+2 + · · · + 1/2m | < 1/2n < 1/2N < ϵ, and the result follows. ∑ (iii) xn = nk=1 (−1)k /k. Suppose that m > n, then ( ) ( ) 1 1 (−1)m (−1)n 1 1 |xm − xn | = −1 + − + · · · + − −1 + − + · · · + 2 3 m 2 3 n (−1)n+1 (−1)m 1 1 (−1)n+m+1 1 + ··· + = − + − ... + . n+1 m n+1 n+2 n+3 m We split this into two cases. Case 1: m + n even, then this becomes =

1 1 1 1 1 − + − ··· + − n+1 n+2 n+3 m−1 m ( ) ( ) ( ) 1 1 1 1 1 1 = − + − + ··· + − , n+1 n+2 n+3 n+4 m−1 m (since each term in parentheses is positive). It follows that ( ) ( ) 1 1 1 1 1 1 1 |xm − xn | = − − − < . − ··· − − n+1 n+2 n+3 m m+1 m+1 n+1 A similar inequality is obtained when m + n is odd. It now follows that (xn ) is a Cauchy sequence.

2. (a) The real sequence (xn ) satisfies |xn+1 − xn | ≤ k|xn − xn−1 | for n = 2, 3, . . ., where 0 < k < 1.

Now |xn+1 − xn | ≤ k|xn − xn−1 | ≤ k 2 |xn−1 − xn−2 | ≤ · · · ≤ k n−1 |x2 − x1 |. It follows that if m > n > 1, then |xm − xn | ≤ |xm − xm−1 | + |xm−1 − xm−2 | + · · · + |xn+1 − xn | ≤ k m−2 |x2 − x1 | + kn − 1 |x2 −x1 |. Since this can be made arbitrarily k m−3 |x2 −x1 |+· · ·+k n−1 |x2 −x1 | ≤ k−1 small, the sequence is Cauchy.

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(b) If xn = an for 0 < a < 1, then |xn+1 − xn | = |an+1 − an | = a|an − an−1 | = |a|xn − xn−1 |, so from (a) the sequence is Cauchy. ( )2 ( )2 xn 1 xn 1 xn 1 2 (c) x1 = 1 and xn+1 = + . Now (xn+1 ) −2 = + −2 = − ≥ 2 xn 2 xn 2 xn 0. Also |xn+1 − xn | =

=

xn − xn−1 xn−1 − xn |xn − xn−1 | 2 + = 1− 2 xn xn−1 2 xn xn−1 |xn − xn−1 | 2 |xn − xn−1 | (1 − )≤ , 2 xn xn−1 2

since xn xn−1 − 2 ≥ 0 and xn xn−1 > 0. The result now follows from (a).

3. xn = (−1)n . Set ϵ = 1, and fix N ∈ Z+ . Set m = 2N and n = 2N + 1, then |xm − xn | = |(−1)2N − (−1)2N +1 | = |1 + 1| = 2 > ϵ, so the sequence is not Cauchy.

Section 10.5 1. We show that the Cantor Set is invariant under the map F , given by f1 (x) = x/3 and f2 (x) = x/3 + 2/3, so that C = F (C) = f1 (C) ∪ f2 (C).

Let y ∈ f1 (C) ∪ f2 (C), then suppose first that y ∈ f1 (C), then y = f (x) for some x ∈ C, so that x = ·x1 x2 x3 . . . = x1 /3 + x2 /32 + x3 /33 + · · · , has a ternary expansion using only 0’s and 2’s. Then f1 (x) = x1 /32 + x2 /33 + x3 /34 + · · · = ·0x1 x2 x3 . . ., so again f1 (x) ∈ C. Similarly if y ∈ f2 (C), we again obtain y ∈ C. This shows that f1 (C) ∪ f2 (C) ⊆ C. Now let x ∈ C. In the first instance, suppose that x ∈ C ∩ [0, 1/3], then x = ·0x1 x2 x3 . . ., for some xi ∈ {0, 2}. Set z = ·x1 x2 x3 . . . ∈ C, then f1 (z) = ·0x1 x2 x3 . . . = x, so x ∈ f1 (C). Similarly for the case where x ∈ C ∩ [2/3, 1]. This shows that f1 (C) ∪ f2 (C) ⊇ C, so we must have f1 (C) ∪ f2 (C) = C.

2. To find the box counting dimension of the Sierpinski Triangle K, we use triangles as our “boxes”. When δ = 1/2, Nδ (K) = 3. When δ = 1/22 , Nδ (K) = 32 , and in log 3 log(3n ) = . general, when δ = 2n , then Nδ (K) = 3n , so that dim(K) = lim n n→∞ log(2 ) log 2

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3. We have f : I → I a function defined on an non-empty closed interval I, satisfying |f (x) − f (y)| < |x − y| for all x ̸= y. (a) Suppose that f has two fixed points α ̸= β, then |α − β| = |f (α) − f (β)| < |α − β|, which is a contradiction. (b) Suppose that I = [a, b], and that inf{|f (x) − x| : x ∈ [a, b]} = α > 0. We know that a continuous function (such as |f (x) − x|) attains its minimum, so there exists x0 ∈ [a, b] with |f (x0 ) − x0 | = α. Since the minimum is attained at x0 , at f (x0 ) we have |f 2 (x0 ) − f (x0 )| ≥ α. But by hypothesis |f 2 (x0 ) − f (x0 )| < |f (x0 ) − x0 | = α, which is a contradiction, so we must have α = 0 and f (x0 ) = x0 . (c) Let g(x) = x + 1/x, then |g(x) − g(y)| = |x + 1/x − y − 1/y| = |x − y||1 − 1/(xy)| < |x − y| since xy > 1 for x ̸= y. (d) Let h(x) = [x + sin(x)]/2, then |h(x) − h(y)| = |[x + sin(x)]/2 − [y + sin(y)]/2| = |(x − y)/2 + (sin(x) − sin(y))/2| = |(x − y)/2 + cos(x + y)/2 sin(x − y)/2| < |(x − y)/2 + (x − y)/2| = |x − y|, when x ̸= y. Clearly x = 0 is a fixed point.

4. f (x) = x2 − a, and so Nf (x) = x −

x2 − a x a = + , so that 2x 2 2x

x a y a |x − y| a + − − = 1− .

2 2x 2 2y 2 xy a a Note that 1 − < 1 since a/(xy) < 1, and also 1 − > −2, since a/(xy) < 3 (the xy xy maximum of a/(xy) occurs when a = 3, x = y = 1), so we get |Nf (x) − Nf (y)| < a 1 . α|x − y|, where α = 1 − 2 xy |Nf (x) − Nf (y)| =

5. We have f : D → D, where D = {x ∈ Rn : d(x, 0) ≤ 1}, where d(f (x), f (y)) ≤ d(x, y) for all x, y ∈ D. 1 (k − 1) Set gk (x) = (1 − )f (x) = f (x), for some k ∈ Z+ then k k (k − 1) k−1 d(f (x), f (y)) ≤ d(x, y). k k It follows that gk is a contraction mapping having a fixed point xk , k = 1, 2, . . .. The sequence (xk ) has a convergent sub-sequence, say (xkn ), with limn→∞ xkn = x0 . An d(gk (x), gk (y)) =

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approximation argument can now be used to show that x0 is a fixed point of f : d(f (x0 ), x0 ) ≤ d(f (x0 ), f (xkn )) + d(f (xkn ), gkn (xkn )) + d(gkn (xkn ), x0 ).

The first term can be made small using the continuity of f , the second term can be made small using the definition of gk , and the third term is just d(xnk , x0 ), which can be made small using the definition of x0 .

6. (a) If f k is a contraction for some k > 1, then f k has a fixed point say x = p. p is unique, so it cannot be a period-k point of f (otherwise f (p) would also be a fixed point of f distinct from p). Thus x = p must be the (unique) fixed point of f . Now for each x ∈ X we have f kn (x) = (f k )n (x) → p as n → ∞. Thus f ((f k )n )(x) → f (p), or f kn+1 (x) → f (p) = p, and similarly f kn+j (x) → f j (p) = p for all x ∈ X, j = 0, 1, . . . , k − 1, as n → ∞. It follows that f n (x) → p as n → ∞ for all x ∈ X. { 0; 0 ≤ x ≤ 1/2 (b) If f : [0, 1] → [0, 1] is defined by f (x) = , then f is not con1/2; 1/2 < x ≤ 1 tinuous, but f 2 (x) = 0 for all x ∈ [0, 1], so is continuous and is clearly a contraction. (c) Let f : R → R, f (x) = e−x , then |f (x) − f (y)| = |e−x − e−y |. Now suppose that x < y < 0, then by the MVT there exists c ∈ (x, y) with f ′ (c) = (f (x)−f (y))/(x−y), so that |f (x) − f (y)| = e−c |x − y| > |x − y|, since e−c > 1. Now if x < y, then by the MVT there exists c ∈ (x, y) with f ′ (f (c))f ′ (c) = (f 2 (x) − f 2 (y))/(x − y), so −c −c that |f 2 (x) − f 2 (y)| = e−c e−e |x − y| = e−(c+e ) |x − y|. We can now check that −c e−(c+e ) < 1/e, and the result follows.

7. We wish to find the distance between the sets A = {0, 1/n, 2/n, . . . , (n − 1)/n, 1} and the interval B = [0, 1]. Recall that Uδ (B) = {x ∈ R : |x − y| ≤ δ, for some y ∈ [0, 1]} = [−δ, 1 + δ]. Also, if δ ≥ 1/(2n), then Uδ (A) = {x ∈ R : |x − y| ≤ δ, for some y ∈ A} = [−δ, 1 + δ], and if δ < 1/(2n), then Uδ (A) will be the union of n closed intervals which cannot contain [0, 1].

It follows from the definition of the Hausdorff metric, D(A, B) = inf{δ > 0 : A ⊆ [−δ, 1 + δ] and [0, 1] ⊆ Uδ (A)},

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that D(A, B) = 1/(2n), so the distance between a finite set and an infinite set can be made arbitrarily small by taking n large enough.

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Section 11.4 1. If r ∈ (0, 1) has a finite binary representation, then we can write r = ·a1 a2 . . . an , for some ai ∈ {0, 1}. Then r = a1 /2 + a2 /22 + · · · + an /2n = (2n−1 a1 + 2n−2 a2 + · · · + an )/2n = k/2n , for some k, n ∈ Z+ .

Conversely, if r ∈ (0, 1) where r = k/2n , k, n ∈ Z+ , then we can write the binary expansion of k as k = an + 2an−1 + · · · + 2n−2 a2 + 2n−1 an , where ai ∈ {0, 1}, so that r = (an + 2an−1 + · · · + 2n−2 a2 + 2n−1 a1 )/2n = a1 /2 + a2 /22 + · · · + an /2n = ·a1 a2 . . . an , and the result follows.

2. We have 5×63/(212 −1) = 5×32 ×7/(5×32 ×7×13) = 1/13. Our results tell us that we have a repeating binary representation having period 12. Also, ·000100111011 = (1/24 +1/27 +1/28 +1/29 +1/211 +1/212 )(1/(1−1/212 )) = 32 ×5×7/(212 −1) = 1/13.

3. Suppose that r ∈ (0, 1) has an infinite repeating binary expansion, say r = ·a1 a2 . . . an a1 a2 . . ., then r = a1 /2 + a2 /22 + · · · + an /2n + a1 /2n+1 + · · · = (2n−1 a1 + 2n−2 a2 +· · ·+an )(1/2n +1/22n +· · · ) = (2n−1 a1 +2n−2 a2 +· · ·+an )(1/2n /(1−1/2n )) = (2n−1 a1 + 2n−2 a2 + · · · + an ))/(2n − 1) = t/q, where t, q ∈ Z+ and q is odd.

Conversely suppose r ∈ (0, 1) where r = t/q, t, q ∈ Z+ , and q is odd. We must show that r has a repeating binary expansion. It suffices to prove this for r = 1/q, where q is odd. Suppose that ai ∈ {0, 1} with ) 1 a1 a2 an an+1 a1 a2 an 1 ( an+1 an+2 = + 2 + · · · + n + n+1 + · · · = + 2 +···+ n + n + 2 + ··· , q 2 2 2 2 2 2 2 2 2 2 where n is the first time an+1 = a1 Then 2n − q(2n−1 a1 + · · · + an ) = q It follows that q

(a

n+1

2

(a

n+1

2

+

) an+2 + · · · .

22

) an+2 + 2 + · · · is an integer, or 2 an+1 an+2 m + 2 + ··· = , 2 2 q

for some m ∈ Z+ . Since an+1 = a1 , we must have m = 1, and the result follows.

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4. Set f (x) = ax2 +bx+c and g(x) = x2 −α, α ̸= 0. Then Nf (x) = x−

ax2 + bx + c = 2ax + b

ax2 − c x2 − α x2 + α x α , and Ng (x) = x − = = + . Set h(x) = 2ax + b, then 2ax + b 2x 2x 2 2x ( 2 ) ( 2 ) ax − c ax − c 2a2 x2 − 2ac + 2abx + b2 h ◦ Nf (x) = h = 2a +b= , 2ax + b 2ax + b 2ax + b and Ng ◦ h(x) = Ng (2ax + b) =

=

2ax + b α + 2 2(2ax + b)

(2ax + b)2 + α 2a2 x2 + 2abx + b2 − 2ac = , 2(2ax + b) 2ax + b

and these are equal.

Now set f (x) = (x+1)2 and g(x) = x2 . In this case α = 0, and b2 −4ac = 4−4 = 0, so by the above, Nf and Ng are conjugate. However, f and g are not conjugate since g has two fixed points, but f has no fixed points.

2x3 + 2x2 − 2/5 x3 + 2x2 + x/5 + 2/5 = , 3x2 + 4x + 1/5 3x2 + 4x + 1/5 2/1000 + 2/100 − 2/5 2 + 20 − 400 378 and so Nf (1/10) = = =− = −3/5. Also, 3/100 + 4/10 + 1/5 30 + 400 + 200 630 −54 + 90 − 50 14 1 −54/125 + 18/25 − 2/5 = = = , so we have a Nf (−3/5) = 27/25 − 12/5 + 1/5 135 − 300 + 25 140 10 2-cycle {1/10, −3/5}.

5. If f (x) = (x+2)(x2 +1/5), then Nf (x) = x−

x2 − 1 x 1 = + . Note that x = ±1 are 2x 2 2x x0 1 (super-attracting) fixed points for Ng . Let x0 ∈ (1, ∞), then 1 < Ng (x0 ) = + < 2 2x0 n x0 . Continuing this argument, we see that the sequence (Ng (x0 )) is decreasing and bounded below by 1, so must converge to the fixed point x = 1. We use a similar argument when 0 < x0 < 1 to obtain an increasing sequence bounded above. We deduce in this way that the basin of attraction of 1 is (0, ∞). Similarly, the basin of attraction of x = −1 is (−∞, 0).

6. Let g(x) = x2 − 1, then Ng (x) = x −

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√ x α 7. Let g(x) = x2 − α, where α > 0, then Ng (x) = + has fixed points ± α. We √ 2 2x √ follow the steps of Exercise # 6, letting x0 ∈ ( α, ∞), then as before α < Ng (x0 ) < √ √ x0 , and continue as previously (using x20 − 2x0 α + α = (x0 − α)2 ≥ 0).

8. Suppose that f and g are linearly conjugate via h(x) = ax + b and Nf and Ng are also linearly conjugate via h. Say h(g(x)) = f (h(x)), then h′ (g(x))g ′ (x) = f ′ (h(x))h′ (x), or ag ′ (x) = f ′ (ax + b) · a, and g ′ (x) = f ′ (ax + b). Now f (h(x)) h(g(x)) ag(x) + b Nf (h(x)) = h(x) − ′ = h(x) − ′ = ax + b − . f (h(x)) g (x) g ′ (x) Also ( ) ( ) g(x) g(x) ag(x) h ◦ Ng (x) = h x − ′ =a x− ′ + b = ax + b − ′ . g (x) g (x) g (x) It follows that

ag(x) + b ag(x) = ′ , or b = 0. ′ g (x) g (x)

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Section 12.3 1. f : [a, b] → [a, b] is a continuous function having no 2-cycles. Let c ∈ [a, b] with f (c) > c (which we are assuming to exist), then we cannot have f 2 (c) < c < f (c) as this implies the existence of a 2-cycle. It follows that f 2 (c) > c.

2. f : [0, 1] → [0, 1] is a continuous function which is also onto, so has at least one fixed point. Since f is onto, there exists c0 , c1 ∈ [0, 1] with f (c0 ) = 0 and f (c1 ) = 1.

If g(x) = f (x) − x, then g(c0 ) = f (c0 ) − c0 = −c0 ≤ 0, g(c1 ) = f (c1 ) − c1 = 1 − c1 ≥ 0, so there is a fixed point p for f lying between c0 and c1 .

Case 1. Suppose that 0 ≤ c0 < c1 ≤ 1. Set g(x) = f (x) − x, then g(0) = f (0) − 0 = f (0) ≥ 0, and g(1) = f (1) − 1 ≤ 0. It follows that there exists q1 ∈ [0, c0 ) with f (q1 ) = q1 , p ∈ (c0 , c1 ) with f (p) = p and q2 ∈ [c1 , 1) with f (q2 ) = q2 , so the result follows in this case.

Case 2. Suppose that 0 ≤ c1 < c0 ≤ 1 and set h(x) = f 2 (x) − x, then as before, we have h(0) = f 2 (0) − 0 = f 2 (0) ≥ 0 and h(1) = f 2 (1) − 1 ≤ 0 so we now consider: Case 2a. Suppose that 0 ≤ c1 < c0 ≤ 1 and f (1) < c1 , then as above there is a fixed point p ∈ (c1 , c0 ). Also, h(0) ≥ 0 and h(c1 ) = f 2 (c1 ) − c1 = f (1) − c1 < 0, so there exists q ∈ [0, c1 ) with h(q) = 0, or f 2 (q) = q and our result follows Case 2b. Suppose that 0 ≤ c1 < c0 ≤ 1 and f (1) > c1 . In this case we can find α ∈ [c0 , 1], with f (α) = c1 (using IVT, since c1 < c0 < 1). We have h(α) = f 2 (α) − α = f (c1 ) − α = 1 − α > 0. Since h(1) ≤ 0, there exists q ∈ (α, 1] with f 2 (q) = q, and the result follows.

The function f : [0, 1] → [0, 1], f (x) = x2 is continuous and onto with two fixed points, but no period-2 points.

3. f : [a, b] → [a, b] is a continuous function having no points of period two with a unique fixed point x = c. By Coppel’s Theorem, limn→∞ f n (x) exists for all x ∈ [a, b] and is therefore a fixed point for each such x. It follows that limn→∞ f n (x) = c for all x ∈ [a, b].

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4. f m (x0 ) = x0 . Suppose that f k (x0 ) = x0 , where k is the least such positive integer.

By the Division Algorithm, we may write m = ka + r where a and r are integers with 0 ≤ r < k. Then f m (x0 ) = f ka+r (x0 ) = f r (f ka (x0 )) = f r (x0 ), so that f r (x0 ) = x0 , which is a contradiction, so we must have r = 0.

5. f has a 3-cycle: {a, b, c} with f (a) = b, f (b) = c, f (c) = a. Set g(x) = f 2 (x) − x, then g(a) = f 2 (a) − a = c − a > 0, and g(b) = f 2 (b) − b = a − b < 0. It follows that there exists x ∈ (a, b) with g(x) = 0, or f 2 (x) = x. However, it may be that f (x) = x. Set p = max{x ∈ (a, b) : f 2 (x) = x}.

Then (p, c) = (p, f (b)) ⊆ f (p, b), so there exists q ∈ (p, b) with f (q) = b. Thus g(b) = f 2 (b) − b = a − c < 0, g(q) = f 2 (q) − q = f (b) − q = c − q > 0.

It follows that there exists r ∈ (q, b) with f 2 (r) = r, contradicting p being the maximum such, since r > p. Therefore we must have f 2 (p) = p and f (p) ̸= p.

6. We give an example of a continuous function f : [a, b] → [a, b] for which there exists c ∈ [a, b] with c < f 2 (c) < f (c), but no period-2 point.

Set f (x) = 3x(1 − x), x ∈ [0, 1], a map having two fixed points but no period-2 points. Set c = ·6, the f (c) = 3 × 6/10 × 4/10 = ·72, f (·72) = 3 × 72/100 × 28/100 = ·6408, and ·6 < ·6408 < ·72, or c < f 2 (c) < f (c).

7(a). Suppose that f 2 has a period-m point, then f 2m (x0 ) = x0 , for some x0 . We use kn 2m Lemma 12.2.3(c), setting n = 2 and k = m, then f has a point of period = , s s kn where s|n (s divides n). It follows that s = 1, or s = 2 and (s, k) = 1, so = m, or s 2m. (b) Suppose that f has period-m points, where m is odd. Set n = 2k in Lemma 12.2.3(b), k ≥ 1, then x0 is a period point of f n with period m/d where d = (m, 2k ) = k 1, so f 2 has period-m points.

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(c) Suppose that f has a period-(2k · m) point with m odd, k ≥ 1. Set n = 2k 2k m in Lemma 12.2.3(b), then f n has period points where d = (2k m, 2k ) = 2k , so d 2k m 2k m 2k = k = m, so f has points of period m. d 2 8.(a) f is a C 1 -function, with fixed point p for which f ′ (p) < −1. We wish to show that f has a 2-cycle. Now f (p) = p. From the continuity of f ′ (x) there exists δ > 0 for which f ′ (x) < −1, for x ∈ (p − δ, p + δ). Let c ∈ (p − δ, p), then from the Mean Value Theorem, there exist α ∈ (c, p) with f ′ (α) =

f (p) − f (c) p − f (c) < −1, or < −1. p−c p−c

It follows that f (c) > p, so f (c) > c. In a similar manner, there exists β ∈ (c, f (c)) with f ′ (β) =

f 2 (c) − f (c) < −1, f 2 (c) − f (c) < −f (c) + c, f (c) − c

or f 2 (c) < c.

It follows that f 2 (c) < c < f (c), and an application of Lemma 12.1.3 gives the result. (b) Let Lµ (x) = µx(1−x) for µ > 3. x = 1−1/µ is a fixed point, and L′µ (x) = µ−2µx, so that L′µ (1 − 1/µ) = µ − 2µ(1 − 1/µ) = 2 − µ < −1. The result follows from (a). (c) Let fλ (x) = x3 − λx. We have a fixed point at x = 0. fλ′ (x) = 3x2 − λ, so fλ′ (0) = −λ < −1, for λ > 0, so a 2-cycle is created. Also, we have f2′ (2) = 3x2 − 2x = −1 when

9. (a) Eλ (x) = λex , then E−e (−1) = −ee−1 = −1, so x = −1 is a fixed point. Also, ′ (−1) = −ee−1 = −1, so from the above, we have a period doubling Eλ′ (x) = λex , E−e bifurcation. (b) If fµ (x) = µxe−x , we set µ = e2 and p = 2, so that fe2 (x) = e2 xe−x , and fe2 (2) = e2 2e−2 = 2, so x = 2 is a fixed point. Now fµ′ (x) = µ(e−x − xe−x ), so fµ′ (2) = µ(e−2 − 2e−2 ) and fe′2 (2) = −e2 e−2 = −1.

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10. (a) We give an alternate proof to Lemmas 12.1.3 and 12.1.4 for a continuous function f on a compact interval I = [α, β]: We assume there exists a, b ∈ [α, β] with f (b) < a < b ≤ f (a). Set g(x) = f (x)−x, then g(a) = f (a) − a > 0 and g(b) = f (b) − b < 0, so f has a fixed point z ∈ [a, b].

Now we have [z, f (a)] ⊆ f ([a, z]), so there exists v ∈ [a, z] with f (v) = b > z. Set u = α if f (x) > z for all x ∈ [α, v], otherwise set u = max{x ∈ [α, v] : f (x) = z} (so if u < x < v, we have f (x) > z). If h(x) = f 2 (x) − x, then h(u) = f 2 (u) − u = f (z) − u = z − u > 0, and h(v) = f 2 (v) − v = f (b) − v < 0, so there exists y ∈ [u, v] with f 2 (y) = y. Since f (x) > z for all x ∈ [u, v], we must have f (y) ̸= y, so y is a period-2 point. (b) Suppose there exists c ∈ I and n ≥ 2 with f n (c) < c < f (c).

Set A = {f k (c) : 0 ≤ k ≤ n − 1}, a = max{x ∈ A : c ≤ x and f (x) > x}, and choose b ∈ {x ∈ A : a < x ≤ f (a)} with f (b) < a. We first show that b exists and hence that f (b) < a < b ≤ f (a). This will then imply the existence of a fixed point and a period-2 point (from (a) above). Now a ̸= f n−1 (c) since in this case f (a) = f n (c) < c ≤ a. The set {x ∈ A : a < x ≤ f (a)} is non-empty since firstly X = {x ∈ A : a < x} ̸= ∅ (since a < f (a), so f (a) ∈ X), and d = min{x ∈ A : a < x} has the property: either d = f (a) or d < f (a).

We conclude that there exists b ∈ {x ∈ A : a < x ≤ f (a)} with f (b) < a, for if not, for all x ∈ A with a < x ≤ f (a), we have f (x) > a. This is impossible since d cannot have this property as f (d) < d, so f (d) ≤ a.

Now if f (d) = a, then cannot have f (a) = d, so must have f (a) = e > d, f (e) < e, so either f (e) < a or f (e) = d, but this gives a 3-cycle, completing the proof.

11. We are looking at the truncated tent map Th (x) = min{h, T (x)}, where T is the usual tent map.

Claim 1. If O = {x1 , x2 , . . . xm }, an m-cycle for T and hO = xm , we set a = min{hO }.

We show that a is the largest member of Q, a unique m-cycle for the the truncated tent map Ta , say Q = {z1 , z2 , . . . zm }, where a = zm .

The minimum is taken over a finite set, so a exists and a = zm for some m-cycle Q = {z1 , z2 , . . . zm } (in increasing order). Since each member of Q is less than or equal to a = zm , the orbit of z1 under Ta is the same as the orbit of z1 under T . This is not true for any other m-cycle of T since there are members having value greater

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than a. It follows that Ta has exactly one m-cycle and zm = a (any other m-cycle with this property would coincide with Q).

Claim 2. We must have T (a) = T (zm ) = z1 because T (a) = 2(1 − a), and any zk < 2(1 − a) must be the image of some zk > zm , giving a contradiction, or else it is the image of some zp < zk and continuing this argument leads to a contradiction.

Claim 3. It now directly follows that every member of Q lies in the interval [T (a), a].

We claim that every periodic orbit of Ta enters [T (a), a] and is therefore contained in [T (a), a]. If we have another periodic orbit, say {a1 , a2 , . . . , ak } (written in ascending order) must have ak < a, for otherwise ak would not be in the range of Ta , contradicting ak being on a k-cycle. As before, T (ak ) = a1 > T (a), so {a1 , a2 , . . . , ak } ⊂ [T (a), a].

Claim 4. Suppose k ◃ m and Ta has a k-cycle R = {a1 , a2 , . . . , ak } (recall that 3 ◃ m for all m), then R is contained in [T (a), a] = [T (zm ), zm ]. By Sharkovsky’s Theorem, Ta has an m-cycle contained in [T (ak ), ak ], which therefore must be different to Q (defined above), a contradiction since the m-cycle is unique, so R cannot exist. This proves the first part of the converse of Sharkovsky’s Theorem.

Claim 5. Choose a = a2n , so that Ta2n has a unique 2n -cycle, but no k-cycle with k ◃ 2n . Set a = limn→∞ a2n . Clearly (a2n ) is an increasing sequence bounded above, so this limit exists. Since a > a2n for all n, we see that Ta has 2n -cycles for all n ≥ 0.

To see that Ta has no m-cycle (m ̸= 2k for any k), with m ◃ 2n for all n, suppose to the contrary that such an m exists with m-cycle {z1 , z2 , . . . , zm } (in increasing order). Then [T (zm ), zm ] ⊂ [T (a), a] (using the fact that the range of Ta contains the m-cycle), and in particular zm < a (if zm = a, using Sharkovsky’s Theorem, we could choose a different cycle with this property). It follows that there exists n ∈ Z+ with zm < a2n < a. This implies that Ta2n has a unique 2n -cycle, no 2k -cycles for k > n and an m-cycle, contradicting Sharkovsky’s Theorem. It follows that no such m-cycle exists.

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Section 13.1 1. We determine the dynamics of the dynamical systems whose matrices are: [ ] 2 0 (i) . Since both eigenvalues are real and greater than one, x = 0 is a repelling 0 3 fixed point. [ ] 1 0 (ii) . Both eigenvalues have absolute value equal to one, so we have the 0 −1 non-hyperbolic case. We see that in this case the matrix acts as a reflection in the x-axis. [ ] .5 0 (iii) . Both eigenvalues are real, lying between zero and one, so the fixed 0 .3 point x = 0 is an attractor. ] [ 1/2 0 . The fixed point x = 0 is a saddle, with some trajectories close to (iv) 0 2 zero moving away from zero, and some moving toward zero.

2. We diagonalize each of the following matrices A, and find An , and we deduce the dynamics linear transformation. [ of the induced ] 1 1/2 (i) . The characteristic equation is |A − λI| = λ2 − 9λ/4 − 7/8 = 3/4 5/4 [ ] [ ] 1/2 0 1 2 (λ − 1/2)(λ − 7/4), so Λ = , and we can check that P = , and 0 7/4 −1 3 [ ] [ ][ ][ ] 1/5 2/5 1 2 1/2n 0 1/5 2/5 −1 n n −1 P = . Then A = P Λ P = . −1/5 3/5 −1 3 0 (7/4)n −1/5 3/5 We see that we have a saddle at x = 0. ] [ −5/3 3 . |A − λI| = λ2 − 5λ/3 + 4/9 = (λ − 1/3)(λ − 4/3), so Λ = (ii) −2 10/3 [ ] [ ] [ ] 1/3 0 3 1 1 −1 −1 , and P = , P = . Then An = P Λn P −1 = 0 4/3 2 1 −2 3 ] [ ][ ][ 3 1 1/3n 0 1 −1 . Again we have a saddle at x = 0.

2 1 0 (4/3)n −2 3

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[

] [ ] −5 −7 2 0 2 (iii) . |A−λI| = λ −λ/2−3 = (λ−2)(λ+3/2), so Λ = , 7/2 11/2 0 −3/2 [ ] [ ][ n ][ ] −2 1 −2 1 2 0 −1 1 and P = . An = P Λn P −1 = . n 1 −1 1 −1 0 (−3/2) 1 −2 In this case, x = 0 is a repelling fixed point. [

] a b 3. We write each matrix in the form QΛQ−1 (where Λ = , for some −b a a, b ∈ R), and hence deduce the dynamics of the corresponding linear map. [ ] −7 15 (i) A = , so |A−λI| = (−7−λ)(11−λ)+90 = λ2 −4λ+13, so solving for −6 11 the eigenvalues gives λ = 2±3i, and [ we set a] = 2 and b = 3. If α is an eigenvector, [ ]we 2 3 3 −1 can take α = (3 − i, 2)t then Λ = , and Q = [Re(α), Im(α)] = . −3 2 2 0 [ ] 1/6 2/3 (ii) B = . |B − λI| = (1/6 − λ)(5/6 − λ) + 2/9 = λ2 − λ + 13/36, −1/3 5/6 [ ] 1/2 1/3 giving eigenvalues 1/2 ± i/3, and eigenvector α = (1 + i, 1), so Λ = , −1/3 1/2 [ ] 1 1 and Q = [Re(α), Im(α)] = .

1 0 [

a b −b a

]

, for a, b ∈ C. |Ia,b − λI| = ] [ a + ib 0 2 2 , (a − λ) + b = 0 when a − λ = ±ib, so λ = a ± ib. Set Λ = 0 a − ib t t then [the corresponding ] [eigenvectors can ] be chosen to be (i, 1) and (−i, 1) , so i −i −i/2 1/2 P = , P −1 = with Ia,b = P ΛP −1 .

1 1 i/2 −1/2 [ ][ ] [ ] a b c d ac − bd ad + bc (b) Ia,b Ic,d = = = Ic,d Ia,b . −b a −d c −ad − bc ac − bd [ ] i 1 2 2 (c) det(Ia,b ) = a + b ̸= 0 if a ̸= 0 or b ̸= 0. If A = , then det(A) = −1 i i2 + 12 = 0.

4. (a) We diagonalize the matrix Ia,b =

106

5. (a) Let v1 ̸= 0 be an eigenvector of A corresponding to the eigenvalue λ. Choose v2 ̸= 0 with (A − λI)v2 = v1 and set P = [v1 v2 ]. Then AP = A [v1 v2 ] = [Av1 Av2 ], and ] [ ] [ [ ][ ] λ 1 p11 p12 λ 1 λp11 p11 + λp12 P = = , 0 λ p21 p22 0 λ λp21 p21 + λp22 so AP = P Λ. P is invertible since v1 and v2 are linearly dependent. [ n ] λ nλn−1 n (b) We can now check that A = P P −1 .

0 λn (c) If F (x) = Ax, we determine the long term behavior of F . If |λ| > 1 then we have a repelling fixed point, and if |λ| < 1 it is attracting. When |λ| = 1, we have the non-hyperbolic case. ] 5 −4 6. A = , then |A − λI| = (5 − λ)(−7 − λ) + 36 = λ2 + 2λ + 1 = (λ + 1)2 , so 9 −7 λ = −1 is an eigenvalue of algebraic multiplicity two, geometric multiplicity one. We find v[1 = (2 ]3)t , and solve (A + I)v2 = (2 3)t to find v2 . We can choose v2 = (1 1), so 2 1 P = , and we can check that AP = P Λ. Since the eigenvalue has absolute 3 1 value one, we have the non-hyperbolic case. [

2 7. (a) We assume that [ A ∈]M2 (R) has characteristic equation λ −τ λ+∆ = 0. On the a b other hand, if A = , then |A−λI| = (a−λ)(d−λ)−bc = λ2 −λ(a+d)+ad−bc, c d so on equating coefficients, we see that τ = a + d = trace of A, and ∆ = ad − bc = determinant of A.

(b) Suppose that ∆ < 1, ∆ > −τ −1 and ∆ > τ −1, then λ1 λ2 < 1 (note that although the matrix is real, the eigenvalues need not be real, however, both τ = λ1 + λ2 and ∆ = λ2 λ2 are real).

Case 1. Suppose that the eigenvalues are not purely real, then they are given by √ √ τ ± τ 2 − 4∆ τ ± i 4∆ − τ 2 = λ= 2 2 since τ 2 − 4∆ < 0. Hence |λ|2 = (τ 2 + 4∆ − τ 2 )/4 = ∆ < 1, and the result follows.

Case 2. Suppose the eigenvalues are real, then we obtain −1 < λ1 λ2 < 1, and

− λ1 λ2 − 1 < λ1 + λ2 < λ1 λ2 + 1.

107

We can write λ1 λ2 = ∆ for ∆ ∈ (−1, 1), then λ2 = ∆/λ1 so that −∆ − 1 < λ1 + ∆/λ1 < ∆ + 1, λ1 − ∆ − 1 + ∆/λ1 < 0, and λ1 + ∆ + 1 + ∆/λ1 > 0, giving λ21 − λ1 (∆ + 1) + ∆ λ21 + λ1 (∆ + 1) + ∆ < 0, and > 0, λ1 ∆1 (λ1 + 1)(λ1 + ∆) (λ1 − 1)(λ1 − ∆) < 0 and > 0. If λ1 > 0 this gives ∆ < λ1 < 1 λ1 λ1 and −∆ < λ1 and we conclude that |λ1 | < 1. If λ1 < 0, then λ1 < ∆ or λ1 > 1, and −1 < λ1 < −∆, and again we get |λ1 | < 1. Similarly for λ2 . or

Conversely, if |λi | < 1 for i = 1, 2, then |∆| = |λ1 λ2 | < 1, so −1 < ∆ < 1. In the case that λ1 = α + iβ, λ2 = α − iβ are not purely real, we must show that ∆ > τ − 1, or λ1 λ2 > λ1 + λ2 − 1, or α2 + β 2 > 2α − 1. If we can show that 2αβ > 2α − 1 then we are done. This can be done, for example using Mathematica (subject to α2 + β 2 < 1).

For the real case, reverse the above steps for the real case, again splitting into different cases (using −1 < λi < 1), for example assume λ1 > 0 and we may assume one of λ > ∆ etc. (c) If ∆ = τ − 1, then our equation becomes λ2 − τ λ + τ − 1 = (λ − 1)(λ − τ + 1) = 0, when λ = 1, or λ = τ −1. If ∆ = −τ −1, then λ2 −τ λ−τ −1 = (λ+1)(λ−τ −1) = 0, when λ = 1, or λ = τ + 1. (d) We have seen in (b) that if 4∆ > τ 2 , the eigenvalues are non-real (NOTE: There is an error in the statement of this question.) If ∆ = 1, then λ2 = λ¯1 and each eigenvalue has absolute value equal to one.

8. (a) F : R2 → R2 , is a linear map, so F (0) = 0. We first show that if F is continuous at zero, then it is continuous everywhere: Suppose that given z ∈ R2 and ϵ > 0 there exists δ > 0 such that if d(z, 0) < δ, then d(F (z), 0) < ϵ.

Fix x ∈ R2 and choose y ∈ R2 with d(x, y) < δ, or d(x − y, 0) < δ (since we are using the standard distance metric on R2 ). It follows that d(F (x − y), 0) < ϵ, or d(F (x), F (y)) < ϵ (since F is linear), and hence F is continuous on all of R2 .

108

We can represent F as multiplication by a matrix A (which we can assume is non-zero): ( )[ ]( ) ( ) x1 a b x1 ax1 + bx2 F = . x2 c d x2 cx1 + dx2 It is now most convenient to use the Cauchy-Schwartz inequality in the form (ax1 + bx2 )2 ≤ (x21 + x22 )(a2 + b2 ), (which can be verified directly). Then (( ) ( ))2 ax1 + bx2 0 d , = (ax1 +bx2 )2 +(cx1 +dx2 )2 ≤ (x21 +x22 )(a2 +b2 +c2 +d2 ), cx1 + dx2 0 so given ϵ > 0, we can take δ = ϵ/(a2 + b2 + c2 + d2 )1/2 to show continuity at zero. (b) F has eigenvalues λ1 , λ2 with |λ1 | < 1 and |λ2 |[ < 1. If ]these eigenvalues are λ1 0 distinct we can diagonalize A = P ΛP −1 where Λ = It is then clear that 0 λ2 ( ) ( ) x1 0 , as n → ∞. If the matrix is not diagonalizable, we can use the → Fn x2 0 Jordan form in a similar way to see that this result holds. (c) H(x) = P · x where P ∈ Mn (Rn ). If P is invertible, H −1 (x) = P −1 (x), so both H and H −1 are continuous. Since H −1 exists, H is one-to-one and onto, so it is a homeomorphism.

109

Section 13.2 1. We find the fixed points and the stability of the maps F : R2 → R2 : ( ) ( 2 ) x x − y2 (i) F = . We set x2 − y 2 = x and 2xy = y and solve simultaney 2xy ously. We see(that)the [fixed points]are (0, 0) and ( (1, )0). [ ] x 2x −2y 0 0 0 Now JF = , and JF = , and we have stay 2y 2x 0 0 0 ( ) [ ] 1 2 0 2−λ 0 bility at (0, 0). Also JF = , where = (2 − λ)2 = 0, 0 0 2 0 2−λ when λ = 2, giving an unstable fixed point. ( ) ( 2 ) x x − y2 + x − 1 (ii) F = . Solving x2 − y 2 + x − 1 = x and 2xy + y = y y 2xy + y ( ) [ ] ( ) x 2x + 1 −2y 1 gives fixed points (±1, 0). Now JF = , and JF = y 2y 2x + 1 0 [ ] ( ) [ ] 3 0 −1 −1 0 , so we have instability in this case. JF = , so λ = −1 0 3 0 0 −1 is a repeated eigenvalue, giving the non-hyperbolic case..

2. We find a and b for which the map F (has )a fixed ( point, and we) determine y x . We set = the stability when a = 6 and b = 1: F ay − bx − y 3 y ) ( ) ( x y = , so that x = y and then −by + ay − y 3 = y. Thus 3 y ay − bx − y y = 0, or y 2 = a − b − 1 and in particular, we require a − b ≥ 1. Thus the fixed points √ √ are (0, 0)t and (± a − b − 1, ± a − b ]− 1)t . ] [ ( ) [ ( ) 0 1 x 0 1 0 t , so = Now JF = , so at (0, 0) , JF 0 −b a −b a − 3y 2 y when = 6 [and b =] 1, the fixed points are (0, 0)t , (2, 2)t and (−2, −2)t , and ( a) √ 0 0 1 JF = . This gives rise to eigenvalues λ = 3 ± 2 2, so we have 0 −1 6 instability. ( ) [ ] √ 2 0 1 t At (2, 2) , JF = , with eigenvalues λ = −3 ± 2 2, and simi2 −1 −6 larly at (−2, −2), again giving instability.

110

(

)

) 1 − ax2 + y 3. (a) The Henon map is defined by H = , where we are bx assuming a ̸= 0 and 0 < b < 1. ( ) ( ) x1 x2 Now H is one-to-one since clearly H = H implies x1 = x2 and y y ( 1 ) ( )2 x c y1 = y2 . If (c, d)t ∈ R2 , then solving H = gives x = d/b and y = y d ( ) ( ) x y/b 2 2 −1 c−1+ad /b , so H is onto. The inverse is given by H = . y x − 1 + ay 2 /b2 x y

(

(

) 2 1 − ax + y (b) Set b = 1 and K(x, y)t = (−y, −x)t , then K ◦ H =K = bx ( ) ( ) ( ) ( ) −bx x −y −x/b and H −1 ◦K = H −1 = , and 2 −1 + ax − y y −x −y − 1 + ax2 /b2 these are equal when b = 1. x y

)

(

] ) [ −2ax 1 x . Here we need = 4. (a) The Jacobian of the Henon map is JH b 0 y to use the fact that the area of the unit square is altered by a factor of |det(A)|, for a matrix A. Now |det(JH(x, y)y )| = b, so JH acts as a contraction of the area, preserving the area when b = 1. (

(b) We can check that H = H1 ◦ H2 ◦ H3 , where Hi are the given maps, and H1 is clearly a reflection, H2 is a contraction (since 0 < b < 1) and H3 is area preserving (as in (a) with b = 1.

5. (a) We show that the fixed point√(x− , y − ) is unstable (assuming (1 − b)2 + 4a > 0, b − 1 − (1 − b)2 + 4a a ̸= 0), where x = x− = . The eigenvalues of the Jacobian 2a √ satisfy λ2 + 2axλ − b = or λ± = −ax ± a2 x2 + b. We will show that λ+ = −ax + √ a2 x2 + b > 1. √ Now 2ax = b−1− (1 − b)2 + 4a < b−1 < 0, so 2ax+1 < b and a2 x2 +2ax+1 < a2 x2 + b, or ax + 1 ≤ |ax + 1| < It follows that 1 < −ax +

√

√

a2 x2 + b.

a2 x2 + b = λ+ , so we have instability.

111

(b) We show that the fixed point in (a) is a saddle. Now we have (1 −√ b)2 + 4a > 0, so from (a) |λ+ | > 1, so it suffices to show that |λ− | < 1. Note that −ax− (ax)2 + b < 0 since b > 0, so λ− < 0. Also from above, ax < 0, so √ √ √ λ− = −ax − (ax)2 + b ≥ −ax − |ax| − b = − b > −1, √ √ √ (using 0 < b < 1 and s + t ≤ s + t for s, t ≥ 0). It follows that −1 < λ− < 0, so we have a saddle. (Note. We have not used the fact that a < 3(1 − b)2 /4). (

) ( ) ( ) ( ) x 1 − ax2 + y x 1 − a(1 − ax2 + y)2 + bx 2 6. We have H = , so H = , y bx y b(1 − ax2 + y) so we set y = (abx2 − b)/(b − 1) and substitute into 1 − a(1 − ax2 + y)2 + bx = x.

After simplifying (use a CAS) we get (−a + 2a2 x2 − a3 x4 + (b − 1)2 (1 + (b − 1)x))/(b − 1)2 = 0, √ giving the required√ equation. This has solutions x = (1 − b ± 4a − 3(1 − b)2 )/(2a) and x = (−1 + b ± (1 − b)2 + 4a)/(2a). Disregarding the fixed point, we see that a period-2 point will appear when a > 3(1 − b)2 /4.

112

Section 13.4 1. T (x) = ax (mod 1) on [0, 1) and T ′ (eix ) = eiax on S1 are conjugate via ϕ : [0, 1) → S1 , ϕ(x) = e2πix , (where we are identifying 0 and 1, as mentioned in Section 13.3). In this case, we have seen that ϕ is a homeomorphism. Now ϕ ◦ T (x) = ϕ(ax) = e2πiax , and T ′ ◦ ϕ(x) = T ′ (e2πix ) = e2πiax , so these maps are conjugate. (

) ax1 + bx2 2. We consider when T : [0, 1) → [0, 1) defined by T = cx1 + dx2 ) ( x1 (mod 1) is well defined. Suppose first that a, b, c, d ∈ Z, clearly T ∈ [0, 1)2 . x2 Suppose that we are given two points in [0, 1)2 with (x1 , x2 )t = (y1 , y2 )t , then x1 = y1 , x2 = y2 (mod 1), so since a, b are integers, ax1 +bx2 = ay1 +by2 (mod 1), and similarly cx1 + dx2 = cy1 + dy2 (mod 1), so T is well defined. A similar argument gives the converse.

2

2

x1 x2

)

(

(

) ( ) x1 ax1 + bx2 3. As above, T : [0, 1) → [0, 1) is defined by T = (mod 1), x2 cx1 + dx2 ( ix ) ( i(ax +bx ) ) e 1 e 1 2 ′ 1 1 1 1 ′ and also T : S × S → S × S is defined by T = . Define ix2 e ei(cx1 +dx2 ) Φ : [0, 1)2 → S1 × S1 , by Φ(x, y)t = (e2πix , e2πiy )t . We can check (as in Exercise 1) that Φ is a homeomorphism, and Φ ◦ T = T ′ ◦ Φ, so T and T ′ are conjugate.

2

2

(

) ( ) x1 ax1 + bx2 4. We show that T : [0, 1) → [0, 1) defined by T = (mod x2 cx1 + dx2 1), [ is onto ] when det(A) ̸= 0, and one-to-one if and only if det(A) = ±1 (where A = a b ). We are assuming that a, b, c, d ∈ Z, so T is well defined. Solving A(x, y)t = c d [ ] z b (z, w)t we get x = det(A1 )/ det(A) and y = det(A2 )/ det(A) where A1 = w d [ ] a y and A2 = (Kramer’s Rule), having a solution for every (z, w)t ∈ R2 if and c w only if det(A) ̸= 0, so will certainly have a solution for (z, w)t ∈ [0, 1)2 .

2

2

113

Now suppose that ad−bc = ±1, then if T (x1 , x2 )t = T (y1 , y2 )t , we have ax1 +bx2 = ay1 + by2 (mod 1) and cx1 + dx2 = cy1 + dy2 (mod 1). Multiplying the first equation by c and the second by a and subtracting (where we may assume a, c ̸= 0) gives (x2 − y2 )(ad − bc) = 0 (mod 1), so since ad − bc = ±1, x2 = y2 (mod 1) and similarly x1 = y1 (mod 1), so T is one-to-one. The converse can be obtained by reversing the argument. (

) ( ) x 2x 5. T : T → T is the endomorphism defined by T = (mod 1). y 2y We( show that T)is chaotic on T2 . )Write x = ·x1 x2 x3 . . . and y = ·y1 y2 y3 . . ., then ( ·x1 x2 x3 . . . ·x2 x3 x4 . . .

T = , so for example (·1010 . . . , ·0101)t = (2/3, 1/3)t ·y1 y2 y3 . . . ·y2 y3 y4 . . . is a period-2 point (as is (1/3, 2/3), and we see that there are four distinct period-2 points. Similarly, (·100100100 . . . , ·0100100100 . . .)t = (4/7, 2/7)t is a period-3 point, and in a similar way to the one-dimensional case, we see that the period points are dense in T2 . The fact that T has sensitive dependence on initial conditions is similar to the one-dimensional case (if (x1 , x2 )t ̸= (y1 , y2 ), the components will differ in some term of their binary expansions, and then they will move widely apart under iteration

by T ).

To show that T is transitive, we construct a point having a dense orbit. This can be done by modifying the construction for the one-dimensional case. We define a point (x0 , y0 )t ∈ T2 2

2

(

x0 y0

)

( =

·00 11 00 00 00 00 01 . . . ·01 01 00 01 10 11 00 . . .

) ,

where we list all possible “pairs” of 1-blocks, all possible “pairs” of 2-blocks, etc. An argument similar to the one-dimensional case shows that his is a transitive point. { 6. B : [0, 1] → [0, 1] is defined by B(x, y) = 2

2

(2x, y/2) ; 0 ≤ x < 1/2 .

(2x − 1, y/2 + 1/2) ; 1/2 ≤ x ≤ 1

(a) The Baker’s transformation acts on the unit square by mapping the left hand rectangle: [0, 1/2) × [0, 1], onto the bottom half of the unit square: [0, 1] × [0, 1/2).

In a similar way, [1/2, 1] × [0, 1] gets mapped onto [0, 1] × [1/2, 1]. This resembles kneading dough (giving rise to its name).

114

(b) Let x, y ∈ [0, 1] and set x = ·x1 x2 x3 . . ., y = ·y1 y2 y3 . . ., then B(x, y) = B(·x1 x2 x3 . . . , ·y1 y2 y3 . . .) = (·x2 x3 x4 . . . , ·x1 y1 y2 . . .).

It is clear that B is invertible, with inverse B −1 (·x1 x2 x3 . . . , ·y1 y2 y3 . . .) = (·y1 x1 x2 x3 x4 . . . , ·y2 y3 y4 . . .).

The fixed points are (0, 0) and (1, 1). Now B 2 (= ·x1 x2 x3 . . . , ·y1 y2 y3 . . .) = (·x3 x4 . . . , ·x2 x1 y1 y2 . . .), so for period-2 points we require x1 = x3 , x2 = x4 , x3 = x5 , . . ., and y1 = x2 , y2 = x1 , y3 = y1 , . . ., so for example (·01010 . . . , 10101 . . .) = (2/3, 1/3) is a period-2 point, with {(2/3, 1/3), (1/3, 2/3)} being the 3-cycle. (c) Now B(x, y) = (x, y) when (x, y) = (0, 0) or (1, 1). B 2 (x, y) = (x, y) when x1 = x3 , x2 = x4 , y1 = x2 y1 = x2 , y1 = y3 etc, so (x, y) = (·010101 . . . , ·101010 . . .) = (1/3, 2/3) etc. (d) We must show that the periodic points of B are dense in [0, 1] × [0, 1] and B is transitive. The easiest way to do this is to define a map σ : Σ → Σ, where Σ = {(. . . x−2 x−1 x¯1 x2 x3 . . .) : xi ∈ {0, 1}}, is the set of all doubly infinite sequences of zeros and ones, where the “bar” indicates the zeroth coordinate. Then σ(. . . x−2 x−1 x¯1 x2 x3 . . .) = (. . . x−2 x−1 x1 x¯2 x3 x4 . . .).

If we define a map ϕ : [0, 1]2 → Σ by ϕ(·x1 x2 x3 . . . , ·y1 y2 y3 . . .) = (. . . y2 y1 x¯1 x2 x3 . . .) then we can check that ϕ is one-to-one and onto (except for those points that do not have a unique binary representation), with ϕ ◦ B = σ ◦ ϕ. (For example, (. . . 1001̄001001 . . .) is period-3 for σ, so that (·100100100 . . . , ·00100100 . . .) is period3 for B). Thus to find the period points of B, we find those of σ, and to find a point of transitivity for B we find one for σ. In the same way as for the one sided shift, we see that the periodic points are dense in Σ (with the usual metric on Σ), and we can transfer these points to [0, 1]2 via ϕ to see that the corresponding points are dense in [0, 1]2 .

To show transitivity, take a one sided sequence ω0′ consisting of all possible blocks (as in the one-sided case). Set ω0 = ω˜0′ ω0′ ∈ Σ (where ω˜0′ is ω0′ , but written in reverse). By iterating ω0 sufficiently, we can approximate any sequence ω = (. . . y−N y−N +1 . . . y−1 x1 x2 x3 . . . yN . . .) using σ N (ω0 ). We can then transfer this point to [0, 1]2 , to give the required result, where we approximate the point (x, y) =

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(·x1 x2 . . . xN . . . , ·y1 y2 y3 . . . yN . . .) by B N (·ω0′ , ·ω0′ ). Since B is a continuous map on a metric space which has dense periodic points and is transitive, it is chaotic.

7. T : T2 → T2 is an endomorphism of the torus induced by a matrix A with det(A) ̸= 0 and has no eigenvalues that are roots of unity. We show that T is transitive if and only if A is a matrix having no eigenvalues that are roots of unity.

Suppose that T is transitive, then it is onto and det(A) ̸= 0. If the eigenvalues λ1 and λ2 , are roots of unity, then |λ1 | = λ2 | = 1 and det(A) = λ1 λ2 = 1 and λ2 = λ¯1 , det(A) = 1. Now we can follow the steps of the solution to Exercise 13.5 # 4, to see that T cannot be transitive.

For the converse we assume that T is transitive and look at different cases.

Case 1. If det(A) = ±1, where neither eigenvalue has absolute value equal to one, then we have a hyperbolic toral automorphism, so we can use the result of Theorem 13.5.4 to conclude that T is transitive.

Case 2. det(A) = ±1, where both eigenvalues have absolute value equal to one, then we see that this case cannot happen (also from Exercise 13.5 # 4).

Case 3. det(A) ̸= ±1 and det(A) ̸= 0. In this case the eigenvalues cannot have absolute value one, so cannot be roots of unity: The eigenvalues are either both real or both have an imaginary part, occurring in conjugate pairs. If they are not real, say λ1 and λ̄1 , with |λ1 | = 1, then also |λ̄1 | = 1, so det(A) = ±1, which is a contradiction.

It remains only to show that T is transitive in this case. In the real case, the proof of Theorem 13.5.4 shows that the eigenvalues are irrational. Notice that ad − bc is an integer, so we must have |λ1 λ2 | = | det(A)| ≥ 2, so at least one of the eigenvalues has absolute value greater than 1, say |λ1 | > 1. Then there exists an eigenvector x0 ∈ T2 with An (x0 ) = λn1 x0 → ∞ as n → ∞. It now suffices to show that {T n (x0 ) : n ∈ Z+ } is dense in T2 - we omit the proof as it is beyond the scope of this text.

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Section 13.5 1. We ask which of the following matrices induce hyperbolic toral automorphisms?

All of these matrices except (iv) have integer entries, so (iv) does not give a hyperbolic toral automorphism. We check to see if det(A) = ±1 and that neither eigenvalue is of absolute value equal to one. [ ] 3 2 (i) A = . det(A) = 9 − 8 = 1, and det(A − λI) = λ2 − 6λ + 1 = 0 when 4 3 √ λ = 3 ± 2 2, so it gives a hyperbolic toral automorphism [ ] 1 −2 (ii) A = . det(A) = −3 + 6 ̸= 1, so it does not give a hyperbolic toral 3 −3 automorphism ] [ 5 7 (iii) A = . det(A) = 15 − 14 = 1 and det(A − λI) = λ2 − 8λ + 1 = 0 when 2 3 √ λ = 4 ± 15, so it is a hyperbolic toral automorphism [ ] 2 1/2 (iv) A = . Since we do not have integer entries it does not give a 2 1 hyperbolic toral automorphism ] [ 3 1 . det(A) = −6 + 7 = −1 and det(A − λI) = λ2 − λ + 1 = 0 (v) A = −7 −2 √ when λ = (1 ± i 3)/2. These eigenvalues have absolute value equal to one, so it does not give a hyperbolic toral automorphism. [

] 2 1 2. (a) We have seen in Example 13.5.2 (i) that A = define a hyperbolic 1 1 toral automorphism. The only fixed point is (0, 0)2 . We have seen that the period points are of the form (p1 /q1 , p2 /q2 )t , and we can check that {(1/5, 2/5)t , (4/5, 3/5)t } and {(2/5, 4/5)t , (3/5, 1/5)t } are the 2-cycles, and the 3-cycles are {(1/2, 1/2)t , (1/2, 0)t , (0, 1/2)t }, {(1/4, 1/4)t , (3/4, 1/2)t , (0, 1/4)t }, {(3/4, 0)t , (1/2, 3/4)t , (3/4, 1/4)t }, {(3/4, 3/4)t , (1/4, 1/2)t , (0, 3/4)t }. [

] F2n F2n−1 (b) We can use induction to show that A = , where Fn is the nth F2n−1 F2n−2 term of the Fibonacci sequence (F0 = 1, F1 = 1). n

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√ (c) The eigenvalues of A are (3 ± 5)/2 and we can check [ that the eigenvectors ] can √ √ 2 √2 be taken to be (2, 5 − 1)t and (2, − 5 − 1)t . Set P = √ , then 5−1 − 5−1 √ [ ] [(3 + 5)/2]n 0 √ An = P Λn P −1 , where Λn = . By calculating P −1 , 0 [(3 − 5)/2]n it is now easy to equate the entries in the (1, 2) position to deduce ( ( √ )n √ )n 1 3+ 5 3− 5 1 −√ .

F2n−1 = √ 2 2 5 5

2 2 3. Let [ T : T ] → T be a hyperbolic toral automorphism induced by the matrix a−λ b a b a b = (a − λ)(d − = ad − bc = ±1, and A= . Then c d−λ c d c d λ) − bc = λ2 − λ(a + d) + ad − bc, so the eigenvalues are √ √ a + d ± (a + d)2 − 4(ad − bc) a + d ± (a − d)2 + 4bc λ= = .

2 2 √ a + d ± i −(a − d)2 − 4bc 2 Suppose that (a − d) + 4bc < 0, then λ = , so 2

|λ|2 =

(a + d)2 + (−(a − d)2 − 4bc) = ad − bc = ±1, 4

contradicting T being a hyperbolic toral automorphism.

That the eigenspaces are lines with irrational slopes now follows from Theorem 13.5.4 (ii). [

] a b 4. We have T : T → T a torus automorphism with matrix A = . It follows c d that a, b, c, d ∈ Z and that T is one-to-one and onto, so detA = ±1. Thus if T fails to be a hyperbolic toral automorphism, then the eigenvalues of A must be of absolute value ±1.

2

2

(a) Suppose the eigenvalues are λ1 = 1 and λ2 = −1, then the characteristic equation is det(A − λI) = λ2 − (a + d)λ + ad − bc, and it is also equal to (λ − λ1 )(λ − λ2 ) = λ2 − (λ1 + λ2 )λ + λ1 λ2 = λ2 − 1, so that a + d = 0 and detA = −1.

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[

] a b (b) d = −a (from (a)), so A = . We deduce that T is not chaotic as c −a [ ] 1 0 −1 follows: the matrix A can be diagonalized a A = P ΛP where Λ = , so 0 −1 that A2 = I, the identity matrix. It follows that T cannot be chaotic. (c) We are supposing that λ1 and λ2 are not purely real and have absolute value equal to one with det(A) = λ1 λ2 = 1. Then λ1 and λ2 are roots of the equation λ2 − (a + d)λ + ad − bc having real (integer) coefficients. It follows that λ2 = λ¯1 .

Set λ1 = α − iβ, then also √ √ λ1 = (a + d ± (a + d)2 − 4(ad − bc))/2 = (a + d ± (a + d)2 − 4)/2.

Since this is non-real, we must have (a + d)2 − 4 < 0, or −2 < a + d < 2. Since a and d are integers, we must have either: Case (i) a + d = 0, so α = (a + d)/2 = 0 and λ1 = iβ. It follows that β = ±1 and λ1 = ±i. Case (ii) a + d = 1, then √ √ √ λ1 = (1 ± 1 − 4)/2 = (1 ± 3)/2, or Case (iii) a + d = −1 and λ1 = (−1 ± 3)/2. [ ] √ ω 0 (d) In a similar way to (b), with Λ = , where ω = (−1 + 3)/2, ω 3 = 1, we 0 ω̄ [ ] −ω 0 3 see that A = I, and similarly in the other case where Λ = .

0 −ω̄

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Section 14.2 1. To prove the triangle inequality: |z + w|2 = (z + w)(z̄ + w̄) = z z̄ + z w̄ + z̄w + ww̄ = |z|2 + 2Re(z w̄) + |w|2 ≤ |z|2 + 2|z w̄| + |w|2 = |z|2 + 2|z||w| + |w|2 = (|z| + |w|)2 , and since everything is positive, |z + w| ≤ |z| + |w|.

Replacing z by z − w gives |z| ≤ |z − w| + |w|, or |z| − |w| ≤ |z − w|, and similarly |w| − |z| ≤ |z − w|, and the result follows.

2. We have fc : C → C, fc (z) = z 2 + c. Recall that z0 is an attracting fixed point if |f ′ (z0 )| < 1. If |f ′ (z0 )| = 1, it is neutral and if |f ′ (z0 )| > 1 it is repelling. (a) Suppose that z0 and z1 are fixed points of fc , then they are solutions of z 2 −z +c = √ 0, so zi = (1 ± 1 − 4c)/2, i = 0, 1. Adding the two solutions gives z0 + z1 = 1. Now f ′ (z) = 2z, so f ′ (z0 ) + f ′ (z1 ) = 2z0 + 2z1 = 2. Furthermore |f ′ (z0 )| = 2|z0 |, and |f ′ (z1 )| = 2|z1 |, so that if |f ′ (z0 )| = 2|z0 | ≤ 1, then |z0 | ≤ 1/2, so 1 = |z0 + z1 | ≤ |z0 | + |z1 | ≤ 1/2 + |z1 |, |z1 | ≥ 1/2, |f ′ (z1 )| ≥ 1, so z1 won’t be attracting.

If c = 1/4, then z0 = z1 = 1/2 and |f (z0 )| = 1, so is not attracting. (b) z0 is super-attracting if f ′ (z0 ) = 0, so z0 = 0. Since fc (z0 ) = z0 , we must have c = 0. (c) If z0 is on a super-attracting 2-cycle of fc , then (fc2 )′ (z0 ) = 0, so fc′ (z0 )fc′ (fc (z0 )) = 0, so we may assume z0 = 0. Then fc2 (0) = 0, so fc (c) = 0 and c2 + c = 0, c = −1. (d) If z0 is a period-n point which is super-attracting, then (fcn )′ (z0 ) = 0. This implies one of the members of the n-cycle is 0, so we may assume z0 = 0. Conversely, if z0 = 0 is on an n-cycle, since f ′ (z0 ) = 0, we have (f n )′ (z0 ) = 0, so the cycle is super-attracting. (e) In a similar manner to (d), z0 = 0 must be on the 3-cycle. Then fc (0) = c, fc2 (c) = c2 + c, so the 3-cycle is {0, c, c2 + c}, and since f 3 (0) = 0, we must have (c2 + c)2 + c = 0, or c3 + 2c2 + c + 1 = 0 (disregarding c = 0). (f) If fc (z) = z 2 + c with c = −1.75488 . . ., then we have fc (0) = c, fc2 (0) = z1 (say), f 3 (0) = z2 and f 4 (0) = c, i.e., z = c is an eventually periodic point with {c, z1 , z2 } being a 3-cycle which is not super-attracting.

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z n − an f (z) − f (a) = lim z→a z→a z − a z−a n−1 n−2 (z − a)(z + z a + · · · + zan−2 + an−1 ) = lim z→a z−a n−1 n−2 = lim (z + z a + · · · + zan−2 + an−1 ) = nan−1 ,

3. If f (z) = z n , then f ′ (a) = lim

z→a

and the result follows.

4. f : C → C is continuous and O(z0 ) = {z0 , f (z0 ), f 2 (z0 ), . . .} has a unique limit point α. It follows that there is a sequence of positive integers (nk ) with f nk (z0 ) → α as k → ∞. From the continuity, f (f nk )(z0 ) → f (α), as k → ∞, or f nk +1 (z0 ) → f (α), so that f (α) is also a limit point of the orbit. The uniqueness then implies that f (α) = α.

If there are p limit points, say A = {α1 , α2 , . . . , αp }, then arguing as above, there is a sequence (nk ) with f nk (z0 ) → α1 as k → ∞, so that f nk +ℓ (z0 ) → f ℓ (α1 ) as k → ∞, and each f ℓ (α) is also a limit point. Since the set A has only p members, α1 is either periodic with some period m ≤ p, or is eventually periodic. Similarly for α2 , α3 , . . . , αp .

5. We first show that if limz→a f (z) = L, then limz→a |f (z)| = |L|. Let ϵ > 0, then there exists δ > 0 such that if 0 < |z − a| < δ, then |f (z) − L| < ϵ. But in this case ||f (z)| − |L|| ≤ |f (z) − L| < ϵ, so the result follows.

Now suppose that |L| > 0, then there exists δ > 0 such that if 0 < |z − a| < δ, then ||f (z)| − |L|| < |L|/2, or |L|/2 < |f (z)| < 3|L|/2, from which the result follows.

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Section 14.3 1. p(z) is a polynomial having degree at least 2. We show that p(∞) = ∞. Suppose that p(z) = an z n +an−1 z n−1 +· · ·+a1 z +a0 , where a0 , a1 , . . . , an are complex constants and an ̸= 0. Clearly p(z)/z n → an as z → ∞, i.e., given ϵ > 0 there exists M ∈ R+ such that if |z| > M , then |p(z)/z n − an | < ϵ, so |an | − ϵ < |p(z)/z n | < |an | + ϵ (we are assuming an ̸= 0, and can choose ϵ < |an |). Thus |p(z)| > (|an | − ϵ)|z n | → ∞ as z → ∞, so we have p(∞) = ∞.

Let h(z) = 1/z, then set q(z) = h−1 ◦ p ◦ h(z) = h−1 ◦ p(1/z) = h−1 (an /z n + an−1 /z n−1 + · · · + a1 /z + a0 ) = z n /(an + an−1 z + · · · + a1 z n−1 + a0 z n ).

Then q(0) = 0, and q ′ (z) =

(a0 z n + a1 z n−1 + · · · + an )nz n−1 − z n (na0 z n−1 + · · · + an−1 ) , so q ′ (0) = 0. (a0 z n + a1 z n−1 + · · · + an )2

Thus p′ (∞) = 0 and p has a super-attracting fixed point at ∞.

If p(z) = az + b where a ̸= 0, then we still have p(∞) = ∞ and q(z) = z/(a + bz), so q ′ (z) = a/(a + bz)2 and q ′ (0) = 1/a. Thus if |a| > 1, ∞ is attracting for p, and if |a| < 1 it is repelling. This fits in with what we know about the real case.

2. p(z) = z 2 − z. We first find the fixed points: Solving p(z) = z gives z 2 − 2z = 0, so z = 0 or z = 2. If p2 (z) − z = 0, then (z 2 − z)2 − z 2 + z = z or z 4 − 2z 3 = 0, giving no additional solutions, so there are no period-2 points.’

3. As before, p(z) is a polynomial having degree at least 2, and we may write p(z) = an z n + an−1 z n−1 + · · · + a1 z + a0 , where a0 , a1 , . . . , an are complex constants. an z n + an−1 z n−1 + · · · + a1 z + a0 p(z) =z− Now Np (z) = z − ′ p (z) nan z n−1 + (n − 1)an−1 z n−2 + · · · + a1 =

(n − 1)an z n + (n − 2)an−1 z n−1 + · · · + a2 z 2 − a0 , nan z n−1 + · · · + 2a2 z + a1

so clearly Np (∞) = ∞.

122

( −1

−1

Thus g(z) = h ◦Np ◦h(z) = h

(n − 1)an + (n − 2)an−1 z + · · · + a2 z n−2 − a0 z n nan z + · · · + 2a2 z n−1 + a1 z n

)

nan z + · · · + 2a2 z n−1 + a1 z n . Then (n − 1)an + (n − 2)an−1 z + · · · + a2 z n−2 − a0 z n g ′ (z) =

((n − 1)an + · · · − a0 z n )(nan + · · · ) − (nan z + · · · + a1 z n )((n − 2)an−1 + · · · ) , ((n − 1)an + (n − 2)an−1 z + · · · + a2 z n−2 − a0 z n )2

so Np′ (∞) = g ′ (0) =

n(n − 1)a2n n = > 1, so that ∞ is a repelling fixed point.

2 2 (n − 1) an n−1

If p(z) = az + b, then Np (z) = z −

b az + b = − , so ∞ is not a fixed point of Np . a a

az + b = z, where (a − d)2 + 4bc = 0, then az + b = cz + d √ a − d ± (a − d)2 + 4bc 2 2 cz + dz, cz + z(d − a) − b = 0, so z = . It follows that if 2c (a − d)2 + 4bc = 0, then f (z) has a unique fixed point.

In Section 14.3.3 we mentioned [ that ]there is a correspondence between the maps a b f (z) and matrices of the form . The eigenvalues of this matrix are λ = c d √ a + d ± (a − d)2 + 4bc , which coincide when (a − d)2 + 4bc = 0. In this case the 2 matrix has a single repeated eigenvalue, say λ0 . The matrix is not diagonalizable since the eigenspace is one-dimensional. [However,] it follows from Section 13.1 that λ0 1 the matrix is similar to one of the form (the Jordan form). This implies 0 λ0 λ0 z + 1 that f conjugate to a transformation of the form g(z) = , which is of the form λ0 g(z) = z + α.

4. Suppose that f (z) =

5. (a) As in the last example, if f (z) has two distinct eigenvalues, [ ]say λ1 and λ2 , then λ1 0 the matrix is diagonalizable (similar to one of the form ), so f is conjugate 0 λ2 λ1 to a map of the form g(z) = z = αz. λ2 (b) To obtain a closed form for f n , we should find the eigenvalues λ1 and λ2 explicitly, and find the corresponding eigenvectors. Writing the eigenvectors as the columns of a

=

123

matrix P , we then have A = P ΛP −1 (where Λ has the eigenvalues down the diagonal).

Then An = P Λn P −1 , from which we can read off f n . z−1 . If z belongs to the imaginary axis, then we can write z = iα z+1 for some α ∈ R. Then

6. Set T (z) =

T (iα) =

(iα − 1)(−iα + 1) α2 − 1 + 2iα iα − 1 = = . iα + 1 (iα + 1)(−iα + 1) α2 + 1

α2 − 1 + 2iα (α2 − 1)2 + 4α2 (α2 + 1)2 2 If w = , then |w| = = 2 = 1, so w ∈ S1 , and 2 2 2 2 α +1 (α + 1) (α + 1) by varying α we get every member of S1 .

7. A straight line from the North Pole N (0, 0, 1) to a point P on the complex plane, intersects the Riemann sphere S at S(z). If the coordinates of P are (x0 , y0 , 0) (z = x0 + iy0 ), then the coordinates of S(z) are of the form w = (cx0 , cy0 , cz0 ) for some c ∈ R to be determined. The sphere has equation x2 + y 2 + (z − 1/2)2 = 1/4, so that c2 x20 + c2 y02 + (cz0 − 1/2)2 = 1/4, or c2 (x20 + y02 ) = z0 (1 − z0 ). √ √ x20 + y02 c2 x20 + c2 y02 = Also, using similar triangles, , so c = 1 − z0 . Combining 1 1 − z0 1 these two equations gives c = 2 , and x0 + y02 + 1 ) ( y0 x20 + y02 x0 , , . w= x20 + y02 + 1 x20 + y02 + 1 x20 + y02 + 1

8. We have a metric ρ(w1 , w2 ) on the Riemann sphere S, giving the shortest distance between the points w1 and w2 on the sphere. A distance d is defined on the complex plane by d(z1 , z2 ) = ρ(S(z1 ), S(z2 )). We show that d is a metric on the extended b Clearly d(z1 , z2 ) ≥ 0 always, and if z1 = z2 , then S(z1 ) = S(z2 ), complex plane C. so ρ(S(z1 ), S(z2 )) = 0, or d(z1 , z2 ) = 0. Also if d(z1 , z2 ) = 0, then z1 = z2 (where we b → S is one-to-one and onto). Therefore we only have to are using the fact that S : C show that the triangle inequality holds. Here we use the fact that ρ is a metric on S: d(z1 , z3 ) = ρ(S(z1 ), S(z3 )) ≤ ρ(S(z1 ), S(z2 )) + ρ(S(z2 ), S(z3 )) = d(z1 , z2 ) + d(z2 , z3 ).

124

b : d(z, 0) > r} (it is assumed that ∞ ∈ Dr ). Let Let r > 0 and set Dr = {z ∈ C z0 ∈ Dr with z0 ̸= ∞, then S(z0 ) ∈ S and S(z0 ) ̸= N the North Pole. It is clear we can find an open disc centered on S(z0 ) on S, contained in {w ∈ S : ρ(w, 0) > r} and that this projects to an open set of C contained in Dr . If z0 = ∞, then similar considerations may be used. b →: C, b defined by h(z) = 1/z, h(0) = ∞ and h(∞) = 0. h is 9. We have h : C clearly one-to-one and onto. We show that h is continuous, so is a homeomorphism b and S are defined, the (since h−1 = h). Clearly from the way the metrics on C b → S is a homeomorphism. We sketch a solution: Suppose that projection map S : C b then z0 = x0 + iy0 and z = x + iy are close in C, ) ( y0 x20 + y02 x0 , , , S(z0 ) = x20 + y02 + 1 x20 + y02 + 1 x20 + y02 + 1 and similarly for S(z), and these are close. We can check that ) ( ( ) ( ) −y0 1 x0 1 x0 −y0 1 , , = S = , , , z0 x20 + y02 + 1 x20 + y02 + 1 x20 + y02 + 1 |z0 |2 + 1 |z0 |2 + 1 |z0 |2 + 1 ( ) 1 and this will be close to S . z

125

Section 14.4 1. fc (z) = z 2 + c, so f−6 (2) = 4 − 6 = −2 and f−6 (−2) = 4 − 6 = −2. Since the orbit of z = 2 is bounded, 2 ∈ K−6 . Clearly also −2 ∈ K−6 .

If we find a fixed point of f−1+3i , we solve z 2 − 1 + 3i = z, or z 2 − z − 1 + 3i = 0. √ This gives z = (1 ± 5 − 12i)/2 = (1 ± (3 − 2i))/2 = 2 − i or −1 + i, so these both lie in K−1+3i .

2. fi (z) = z 2 + i. Now fi (0) = i, fi (i) = −1 + i, fi (−1 + i) = 1 − 2i − 1 + i = −i, fi (−i) = −1 + i, so that {−1 + i, −i} is a 2-cycle and z = 0 is eventually periodic.

First note that the fixed points satisfy the equation z 2 − z + i = 0. Solving fi2 (z) = z, we obtain z 4 + 2iz 2 − z − 1 + i = 0. The polynomial z 2 − z + i must be a factor, leaving only two roots to find. These must be z = −i and z = −1 + i since {−i, −1 + i} is a 2-cycle (we can check that z = −i is a solution of the equation fi2 (z) − z = 0).

3. We have seen that z = −i is a period 2-point for fi . To check its stability, we calculate |fi′ (−i)fi′ (−1 + i)| = |(−2i)(−2 + 2i)| > 1, so z = −i is repelling, and hence belongs to the Julia set J(fi ).

4. We have seen that if z0 is a period-n point, then (f n )′ (z0 ) = f ′ (z0 )f ′ (f (z0 ))f ′ (f 2 (z0 )) . . . f ′ (f n−1 (z0 )) = f ′ (z0 )f ′ (z1 ) . . . f ′ (zn−1 ), where {z0 , z1 , . . . , zn−1 } is the n-cycle. It follows that the same value is obtained no matter at which period n-point we start with as it will just permute the product.

Suppose that |(f n )′ (z0 )| = λ, then as above, |(f nk )′ (z0 )| = |f ′ (z0 )f ′ (f (z0 )) . . . f ′ (f n−1 (z0 )) . . . f ′ (z0 )f ′ (f (z0 )) . . . f ′ (f n−1 (z0 ))|, (nk-times), so |(f nk )′ (z0 )| = λk .

5. We are considering f0 : S1 → S1 , f0 (z) = z 2 , f−2 : [−2, 2] → [−2, 2], f−2 (z) = z 2 −2, and h : S1 → [−2, 2], h(z) = z+1/z. Now h◦f0 (z) = h(z 2 ) = z 2 +1/z 2 and f−2 ◦h(z) = f−2 (z + 1/z) = (z + 1/z)2 − 2 = z 2 + 1/z 2 , so these are equal. We need to show that h is well defined and onto. Let z = a + ib ∈ S1 (a and b real), then −1 ≤ a ≤ 1

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h(z) = 2a, so −2 ≤ h(z) ≤ 2, so h is well defined. Now let w ∈ [−2, 2], then solving √ √ h(z) = w gives z 2 − zw + 1 = 0, or z = (w ± w2 − 4)/2 = w/2 ± i 4 − w2 /2, where |z|2 = w2 /4 + (4 − w2 )/4 = 1. In other words, for w ∈ [−2, 2] we get two solutions, both in S1 (one solution when w = ±2), so h is onto, but not one-to-one. I.e., h is a factor map but not a conjugation.

6. Suppose that fc has a bounded attracting periodic orbit, say z0 . Then the basin of attraction Bfc (z0 ) of z0 is an open set in K(fc ). J(fc ) is the closure of the repelling periodic points, so Bfc (z0 ) cannot contain any such points, i.e., K(fc ) ̸= J(fc ).

Consider the map f1/4 (z) = z 2 + 1/4, having a unique neutral fixed point z = 1/2 ′ (so |f1/4 (1/2)| = 1). We claim that 0 ∈ K(f1/4 ) \ J(f1/4 ). This follows from the fact n that limn→∞ f1/4 (0) = 1/2, which was proved for the real function f1/4 in Chapter 2. This shows that 0 ∈ K(f1/4 ), so it suffices to show that there is an open ball surrounding z = 0, say Bδ (0) whose points are contained in K(f1//4 ). In fact we can check that it suffices to take δ = 1/2.

Furthermore, it follows that f1/4 has no points having a bounded attracting periodic orbit, since any such orbit must attract a critical point (see Theorem 15.5.4).

Now z = 0 is the only critical point, and 0 is attracted to 1/2. (In the next section we will see that c = 1/4 lies on the boundary of the Mandelbrot set. In fact if fc has an attracting periodic orbit, then c lies in the interior of the Mandelbrot set).

7. We have a map f : X → X and C ⊂ X. (a) Suppose that both f (C) ⊆ C and f −1 (C) ⊆ C. To show f −1 (C) = C, it suffices to show that C ⊆ f −1 (C). Let x ∈ C, then f (x) ∈ f (C), so f (x) ∈ C and x ∈ f −1 (C) and the result follows. (b) Suppose that f is onto with f −1 (C) = C, then we show that f (C) = C. Let x ∈ C, then x ∈ f −1 (C), so f (x) ∈ C. It follows that f (C) ⊆ C. If f (C) ̸= C, there exists y ∈ C with f (x) ̸= y for all x ∈ C. Since f is onto, there exists z ∈ X with f (z) = y ∈ C. Thus z ∈ f −1 (C) = C, a contradiction.

8. (a) To show that p(z) = z 2 − z is conjugate to q(z) = z 2 − 3/4, we can use Example 7.3.2 with a = 1, b = −1, c = 0, r = 1, s = 0 and t = −3/4, and check that

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c = (b2 − s2 + 2s − 2b + 4rt)/4a, or 0 = (1 − 0 + 0 + 2 + 4(−3/4))/2, which is indeed that case. The conjugacy is given by h(z) = z − 1/2. (b) p(z) = z 2 + c. Set q(z) = p(z) − z, then q(p(z)) = p2 (z) − p(z) and q(p(z)) + q(z) = p2 (z) − p(z) + p(z) − z = p2 (z) − z.

Suppose p(α) = α, then q(α) = 0 and p2 (α) − α = q(p(α)) + q(α) = q(α) + 0 = 0, so any factor of p(z) − z is also a factor of p2 (z) − z, so p(z) − z divides p2 (z) − z. p2 (z) − z is a quadratic and p(z) − z p2 (z) − z we may assume that p(z) is of the form p(z) = z 2 + c. Suppose we have = p(z) − z p(z)−z, or (z 2 +c)2 −z +c = (z 2 −z +c)2 , then we see this is not possible since it gives a cubic term on the right, but no cubic term on the left. Another possibility is that p2 (z) − z are repeated, and coincide with one of the roots of p(z) − z. In the roots of p(z) − z √ p2 (z) − z this case we have = z 2 +z +c+1, so that z = (−1± 1 − 4 − 4c)/2, so if the p(z) − z p2 (z) − z roots are repeated, c = −3/4. We can check that = z 2 +z+1/4 = (z+1/2)2 , p(z) − z so has a single root: z = −1/2, and p(z) − z = z 2 − z − 3/4 = (z − 3/2)(z + 1/2), so the fixed points are z = 3/2 and z = −1/2. No other situation can arise.

Suppose that p(z) has no points of period-2. Then

(c) We show that p(z) = Az 3 + Bz 2 + Cz + D (A ̸= 0), is linearly conjugate to a polynomial of the form q(z) = z 3 + az + b. This can be checked if we set √ B B2 , b=D A+ √ a=C− 3A 3 A and h(z) = αz + β where α =

√

) ( 2B 2 , 1−C + 9A

B A and β = √ , then h ◦ p(z) = q ◦ h(z).

3 A

9. Let f (z) = az + b, then f (z) = z when z = b/(1 − a), provided a ̸= 1 (if a = 1 and b = 0, every point is a fixed point, but none is repelling). Now f ′ (z) = a, so the fixed point is repelling only when |a| > 1. Since there is now a single fixed point, J(f ) = {a}. In the other cases, J(f ) = ∅.

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10. f and g are conjugate polynomials via a conjugation h. If A is the set of repelling periodic points of f , then h(A) is the set of repelling periodic points of g. Now J(f ) = A, and J(g) = h(A), and since h is homeomorphism, J(g) = h(A) = h(J(f )), which completes the proof.

11. Let p(z) be a polynomial of degree d > 1, and k ∈ Z+ . We have seen that K(p) = p−1 (A) ∩ p−2 (A) ∩ . . . ∩ p−n (A) ∩ . . . , so K(pk ) = p−k (A) ∩ p−2k (A) ∩ . . . ∩ p−nk (A) ∩ . . . , where the set A is defined in 14.4.11. These are nested sequences of sets and clearly K(p) ⊆ K(pk ). Because of the nested property we see that the reverse is also true, so K(pk ) = K(p). Since these sets must have the same boundary, we have J(pk ) = J(p).

12. fc (z) = z 2 + c, c > 2 real and A = B rc (0) = B c (0), the closure of the open ball of radius c centered on 0.

Now f −1 (A) = {z ∈ C : fc (z) ∈ A}, and its boundary is the set {z : |z 2 + c| = c}.

Consider (z 2 + c)(z̄ 2 + c̄) = c2 , or |z 4 | + z 2 c̄ + z̄ 2 c = 0.

Set z = reiθ , then r4 + 2ℜ(r2 e2iθ c) = 0, so r2 = −2c cos(2θ).

13. Fc (z) = z d + c, where d ≥ 2 is an integer. Then the basin of attraction of infinity is the set Bc (∞) = {z ∈ C : F n (z) → ∞, as n → ∞}.

We show that Bc (∞) is an open set. Let z0 ∈ Bc (∞), then we can find an integer k > 0 with |Fck (z0 )| > max{|c|, 2}+1.

Since Fck is a continuous function, there exists δ > 0 such that |Fck (z)−Fck (z0 )| < 1 whenever |z − z0 | < δ. Now let z ∈ Bδ (z0 ), then we obtain from above and using the triangle inequality that |Fck (z)| > |Fck (z0 )| − 1 > max{|c|, 2}, so by the escape criterion, Fcn (Fck (z)) → ∞ as n → ∞. I.e., z ∈ Bc (∞). It follows that Bc (∞) is an open set.

129 2 2 14. f−1 (z) = z 2 − 1 and f−1 (z) = (z 2 − 1)2 − 1 = z 4 − 2z 2 . Then f−1 (0) = 0, 2 2 f−1 (−1) = −1 and f−1 (∞) = ∞. Also f−1 (0) = −1, so we have a 2-cycle which is clearly attracting, and we can check that ∞ is an attracting fixed point. It follows that each of these points is centered on disjoint regions making up different components of the Fatou set, so the Fatou set has a least three components. It follows that the Fatou set has infinitely many components.

15. We have seen that for a polynomial p(z), there are always points z ∈ C with pn (z) → ∞, so Bp (∞) is non-empty (in fact we have seen that ∞ is a super-attracting fixed point for p - Exercises 14.3 # 1). Bp (∞) is an invariant set, for if z ∈ Bp (∞), then pn (z) → ∞, so pn (p(z)) → ∞ or p(z) ∈ Bp (∞). (Here we are using the metric d arising from the metric ρ on the Riemann sphere - see Exercises 14.3 # 8). To show that Bp (∞) is an open set we can use the method of the proof of Theorem 4.3.6 on b d) and the fact that p is continuous on (C, b d). the metric space (C, Let U be an open set in C (where of course we assume U ̸= C and U ̸= ∅). Then there is a straight line L in C for which U ∩ L ̸= ∅. Consequently we may assume without loss of generality that U ⊂ R, U ̸= R. If the set U is bounded above or below, then it is clear that sup(U ) (respectively inf(U )) belongs to the boundary of U . If U is unbounded, there exist p ∈ R \ U . Then if p does not belong to the boundary of U (for example suppose that U = R \ C, where C is the Cantor set), we can set α = sup{x ∈ U : x < p}, then α is a boundary point of U .

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Section 14.5 1. (a) Suppose |c| > 2, then fc (0) = c has absolute value greater than 2, and greater than or equal to |c|, so that fcn (fc )(0) → ∞ as n → ∞. This proves (a). It follows that c ∈ / M, so (b) follows. (b) If |c| > 2, then from (a) the orbit of zero under fc is unbounded. It follows that M ⊆ {z ∈ C : |z| ≤ 2}. (c) Fc (z) = z 3 + c. We follow the proof of Proposition 14.4. (Note: The text has a √ typo, saying Proposition 14.5.5). Suppose |z| > max{|c|, 2}, then |Fc (z)| = |z 3 + c| ≥ |z 3 | − |c| = |z|3 − |c| > |z|3 − |z| = |z|(|z|2 − 1).

Since |z|2 > 2, |z|2 − 1 > 2 − 1 = 1. Set |z|2 − 1 = α > 1, then |Fc (z)| > α|z|, now continue in the same way as in the proof of Proposition 14.4.4 to see that |Fcn (z)| is an increasing sequence unbounded above.

2. fc (z) = z 2 + c, so fc (0) = c, fc2 (0) = c2 + c and fc3 (0) = (c2 + c)2 + c. Suppose that fc3 (c2 +c) = c2 +c, then (((c2 +c)2 +c)2 +c)2 +c = c2 +c, so that (((c2 +c)2 +c)2 +c)2 −c2 = 0. This can be seen to factor as (use a CAS) c3 (c + 2)(c3 + 2c2 + c + 1)2 (c6 + 2c5 + 2c4 + 2c3 + c2 + 1) = 0. c = −2 is the smallest real value in the Mandelbrot set. Solutions to c3 + 2c2 + c + 1 = 0 make c so that fc has a super-attracting orbit of period-3.

The last factor has a root: c = −1 · 239255 − i0 · 4126021816 . . ., and its Julia set is a “dendrite”.

All these c-values lie in the Mandelbrot set because they lie on the orbit of 0. For example, f−2 (0) = −2 → 2 → 2 · · · , which is bounded.

Note that if fc [fc2 (0)] = fc2 (0), then fc (c2 + c) = c2 + c, and simplifying, this gives c(c + 2) = 0. These values of c make c2 + c a fixed point of fc .

3. If M is the Mandelbrot set, we must show that if c ∈ M, then c̄ ∈ M. In this case, the orbit O(0) = {fcn (0) : n ∈ Z+ }, is bounded, i.e., the set {0, c, c2 + c, (c2 + c)2 + c, . . .}, is bounded. Taking the conjugate gives the orbit {fc̄n (0) : n ∈ Z+ } as being bounded, and this shows that c̄ ∈ M.

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4. Let c ∈ M lie on the boundary of the primary bulb of the Mandelbrot set.

From the proof of Proposition 14.5.6, these c lie in the boundary of the set {c ∈ √ √ C : |1 − 1 − 4c|/2 < 1/2}, or in other words satisfy |1 − 1 − 4c| = 1. Suppose that c satisfies this equation, then the fixed points of fc are given by fc (z) = z, or √ z 2 − z + c = 0, so the roots are zi = (1 ± 1 − 4c)/2, i = 1, 2, and one of these (say √ z1 ), satisfies |fc′ (z1 )| = |2z1 | = |1 − 1 − 4c| = 1.

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Section 14.6 1. We show that if f (z) = az 2 + bz + c and q(z) = z 2 − α where α = b2 − 4ac, then the corresponding Newton functions Nf and Nq are conjugate. Set h(z) = 2az + b, then we can check that h ◦ Nf (x) = Nq ◦ h(x), for all x in the domain of Nf .

2. p(z) is a polynomial of degree d > 1 with Newton function Np (z) = z − p(z)/p′ (z). (a) We have seen previously that Np′ (z) = p(z)p′′ (z)/(p′ (z))2 . (b) We have p(z) = z m q(z) where q(0) ̸= 0 (and we are assuming m > 0 is an integer).

Then Np (z) = z − z m q(z)/(mz m−1 q(z) + z m q ′ (z)) = z − zq(z)/(mq(z) + zq ′ (z)), so Np (0) = 0. From (a) we have Np′ (z) =

z m q(z)(m(m − 1)z m−2 q(z) + mz m−1 q ′ (z) + mz m−1 q ′ (z) + z m q ′′ (z)) , (mz m−1 q(z) + z m q ′ (z))2

and this gives Np′ (0) =

m−1 , m

so z = 0 is super-attracting only when m = 1. (c) This problem was solved in Exercises 14.3 # 3. (d) We can now check that if p(z) = z 3 − z 2 = z 2 (z − 1) then m = 2 and Np (z) = 2z 2 − z 6z 2 − 8z + 2 , Np (0) = 0. Also Np′ (z) = , so Np′ (0) = 1/2.

3z − 2 (3z − 2)2 3z − 2z 2 (2 − z 2 )(3 − 4z) − (3z − 2z 2 )(−2z) ′ Also, g(z) = h◦Np ◦h(z) = , so g (z) = , 2 − z2 (2 − z 2 )2 so g ′ (0) = Np′ (∞) = 3/2. z 2 − α2 z 2 + α2 z−α 3. (a) We have f0 (z) = z 2 and Nα = z − = and T (z) = .

2z ( 2z z+α ( )2 ) z−α z 2 + α2 z 2 − 2αz + α2 Then f0 (T (z)) = , and T (Nα (z)) = T = 2 , and z+α 2z z + 2αz + α2 the result follows.

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(b) If α = i, then since the Julia set of f0 (z) = z 2 is the unit circle S1 , the Julia iz + i set of Nα will be T −1 (S1 ), where T −1 (z) = . Also, notice that T −1 (0) = i, −z + 1 T −1 (1) = ∞ and T −1 (∞) = −i, giving the fixed points of Ni . z = ±i are attracting sin(θ) and ∞ is repelling. We can check that T −1 (eiθ ) = , so belongs to the real 1 − cos(θ) axis, and that every member of R can be written in this form, so the Julia set of Nα is R. The interior of the circle maps to the upper half plane, so this is the basin of attraction of z = i, and the exterior maps to the lower half plane, giving the basin of attraction on z = −i.

Similar analysis for the situation where α = 1 shows that the imaginary axis will now be the Julia set.

4. g(z) = (z − α)m h(z) where m > 1 is an integer and h(α) ̸= 0. (a) Set f (z) = (g(z))1/m and Nm (z) = Nf (z) = z − z−

f (z) m(g(z))1/m = z − = f ′ (z) (g(z))(1−m)/m g ′ (z)

mg(z) . g ′ (z)

(b) Now Nm (z) = z −

m(z − α)m h(z) , so Nm (α) = α, m(z − α)m−1 h(z) + (z − α)m h′ (z)

so z = α is a fixed point for Nm . We can check that ′ Nm (z) = 1 − m +

mg(z)g ′′ (z) (g ′ (z))2

(

m(m − 1)h(z) + 2m(z − α)h′ (z) + (z − α)2 h′′ (z) = 1 − m + m(z − α) h(z) (z − α)2m−2 [mh(z) + (z − α)h′ (z)]2 ′ It follows that Nm (α) = 0, so z = α is super-attracting for Nm .

2m−2

) .

(c) g(z) = (z − a)2 (z − b), so we find N2 and from (a) we have N2 (z) = z −

2(z − a)2 (z − b) z 2 + az − 2ab . = 2(z − a)(z − b) + (z − a)2 3z − a − 2b

We now we use a calculation to show that p = h ◦ N2 ◦ h−1 , where p(z) = z 2 − 3/4 and 3z + a − 4b h(z) = , to give the conjugacy. We can check using Mathematica that 2(z − a) p ◦ h = h ◦ N2 to give our result (see Mathematica Labs).

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5. p(z) = z 3 − 1 and Np (z) = (2z 3 + 1)/3z 2 . (a) It is straightforward to show by direct calculation that BNp (ω) = ωBNp (1), where ω ̸= 1 is a cube root of unity. Similarly for BNp (ω 2 ) = ω 2 BNp (1). (b) If Npn (z) → 1 as n → ∞, then clearly Npn (z̄) → 1 as n → ∞, so we have symmetry with respect to the real axis. (c) For x ∈ R+ , Np (x) = (2x3 + 1)/3x2 . In this case we can check that Npn (x) → 1 as n → ∞, so R+ ⊂ BNp (1) (use the fact that Np (x) < x for x > 1, so that Npn (x) is a decreasing sequence which must converge to the fixed point 1, and similarly Np (x) > x for 0 < x < 1).

6. Let p(z) be a polynomial having degree d > 1 with Newton function Np (z) = z − p(z)/p′ (z). (z − α)p′ (z) − p(z) p(z) − α = . Hence g(z) = [Np (z) − α]p′ (z) is p′ (z) p′ (z) a polynomial with g(α) = 0, so (z − α) is a factor. Also, g ′ (z) = (z − α)p′′ (z) + p′ (z) − p′ (z), so g ′ (α) = 0. It follows that (z − α)2 is a factor of g(z), i.e., g(z) = (z − α)2 q(z) for some polynomial q(z).

Thus (Np (z) − α)p′ (z) = (z − α)2 q(z), so that (a) Np (z) − α = z −

q(z) Np (z) − α = (z − α)2 ′ . p (z)

(b) Now g ′′ (z) = (z − α)p′′′ (z) + p′′ (z), so if p′′ (α) = 0, then g ′′ (α) = 0, and the result follows as before. p(z) z 2 (2z − a − 1) = p′ (z) a − 2az + z(−2 + 3z) 2 2z(z − 1)(a + a − 4az + z(−1 + 3z)) Np′ (z) = = 0 when z = 0, z = 1, z = a and (a − 2az + z(−2 + 3z))2 z = (a + 1)/3. (c) pα (z) = z(z − 1)(z − α). Np (z) = z −

(z − a)2 (2z − 1) , so that when a = 1/2, (z − 1/2)3 is a ((2 − 3z)z + a(−1 + 2z)) factor of Np (z) − α.

(d) Np (z) − α = −

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7. Suppose that p(z) = r(z − a)(z − b)(z − b) where a, b and c are distinct, so p is a polynomial having three distinct roots. We also have pα (z) = z(z − 1)(z − α).

To show that there exists α such that Np and Npα = Nα are conjugate, it suffices to show there exists h : C → C, h(x) = Az + B, with pα (h(z)) = p(z) for some α a−z a−c (see Proposition 11.3.1). We can check that if we set h(z) = and α = a−b a−b then h(a) = 0, h(b) = 1 and h(c) = α then Np and Nα are conjugate. (h is mapping fixed points of Np (zeros of p) to fixed points of Nα (zeros of pα ), so must be a conjugacy). We can check that p(z) = pα (Az + B). Note: To find h we can use a standard cross ratio method to find a linear map which sends the zeros of one cubic to those of another - consider the ratios (z, a, b, c) = (w, 0, 1, α), by which we mean (z − a)(b − c) (w − 0)(1 − α) = . Solving for w gives (z − c)(b − a) (w − α)(1 − 0) (a − z)(b − c)α h(z) = w = . z(−a + c + aα − bα) + ab − ac − αac + αbc a−z a−c . This gives h(z) = , To simplify this we set −a + c + aα − bα = 0, or α = a−b a−b having the required properties.

The problem can be completed by using a result of Lucas (1874) that says the critical points of the polynomial p(z) are contained in the convex hull of the roots of p(z) (see P. Blanchard [17], where more detail is given).

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Section 14.7 1. Let z, w ∈ C. (i) Set z = x + iy and w = u + iv where x, y, u, v are real, then ez+w = ex+u+i(y+v) = ex+u (cos(y + v) + i sin(y + v)) = ex eu (cos(y) cos(v) − sin(y) sin(v) + i sin(y) cos(v) + i cos(y) sin(v)) = ex (cos(y) + i sin(y))eu (cos(v) + i sin(v)). (ii) (ez )n = enx (cos(y) + i sin(y))n = enx (cos(ny) + i sin(ny)) = enz . (iii) e2kπi = cos(2kπ) + i sin(2kπi) = 1 + i × 0 = 1.

2. To see that (ez )w ̸= ezw in general, set z = iπ/2 and w = i, then ez = i, ezw = e−π/2 and (ez )w = ii = ei log(i) = ei(iπ/2+2kπi) = e−π/2−2kπ , for k ∈ Z.

3. (i) Using the same notation as in Exercise 1, sin(z + w) = (ei(z+w) − e−i(z+w) )/2i = (eiz eiw − e−iz e−iw )/2i.

On the other hand sin(z) cos(w) + cos(z) sin(w) = (eiz − e−iz )(eiw + e−iw )/4i + (eiz + e−iz )(eiw − e−iw )/4i = (2eiz eiw − 2e−iz e−iw )/4i and these are equal, so the result follows.

(ii) A similar calculation shows that

cos(z + w) = cos(z) cos(w) − sin(z) sin(w). (iii) tan(z + w) =

sin(z) cos(w) + cos(z) sin(w) tan(z) + tan(w) sin(z + w) = = . cos(z + w) cos(z) cos(w) − sin(z) sin(w) 1 − tan(z) tan(w)

(iv) sin(z + 2π) = (eiz+2πi − e−iz−2πi )/2i = (eiz − e−iz )/2i = sin(z) and similarly for cos(z + 2π). Now sin(z + π) = (eiz+πi − e−iz−πi )/2i = − sin(z) and similarly cos(z + π) = − cos(z), so tan(z + π) = tan(z).

4. Consider f : S → C \ {0}, where S = {z ∈ C : z = x + iy, x, y ∈ R, −π < y ≤ π} and f (z) = ez . We show that f is one-to-one and onto.

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Suppose that ez1 = ez2 where z1 = x1 + iy1 and z2 = x2 + iy2 , then equating real and imaginary parts, ex1 cos(y1 ) = ex2 cos(y2 ), and ex1 sin(y1 ) = ex2 sin(y2 ).

Squaring and adding gives e2x1 = e2x2 , so x1 = x2 . Then we see that cos(y1 ) = cos(y2 ) and sin(y1 ) = sin(y2 ), together with the restriction on y gives y1 = y2 , so z1 = z2 , and f is one-to-one.

Given w ∈ C \ {0}, say w = u + iv, we find z ∈ S with ez = w, or ex cos(y) = u, ex sin(y) = v. Clearly we can find x with e2x = u2 + v 2 (since w ̸= 0). Also we can find −π < y ≤ π with tan(y) = v/u, so f is onto.

5. Now f : T → C \ {±1}, where T = {z ∈ C : z = x + iy, x, y ∈ R, −π/2 < y < π/2} and f (z) = tan(z). From what was said in 14.7.9 and in a similar manner to the previous solution, we see that f is one-to-one and onto.

6. We show that tan(2 arctan(tan z)) = tan(2z) for z ∈ C. We know that arctan(tan z) = z for all z ∈ T = {z ∈ C : −π/2 < ℜ(z) < π/2}. If ℜ(z) does not lie between −π/2 and π/2, there exists n ∈ Z with −π/2 < ℜ(z +nπ) < π/2, so that tan(z) = tan(z + nπ), and then arctan(tan(z)) = arctan(tan(z + nπ)) = nπ + z, so tan(2(nπ + z)) = tan(2z) and the result follows.

7. To show f (z) = ez is continuous at z0 ∈ C, let ϵ > 0. We must find δ > 0 such that if z ∈ C with |z − z0 | < δ, then |ez − ez0 | < ϵ.

From the definition |ez − ez0 |2 = |ex (cos(y) + i sin(y)) − ex0 (cos(y0 ) + i sin(y0 ))|2 = |(ex cos(y)−ex0 cos(y0 ))+i(ex0 sin(y) −ex0 sin(y0 )|2 = (ex −ex0 )2 +2ex+x0 (1−cos(y−y0 )).

Now we can use the continuity of the real functions ex and cos(y), to finish the proof.

8. f (z) = sin(z). Set z = ix where x ∈ R, then f (ix) = (e−x − ex )/2i, so if g(x) = |f (ix)|, then for x > 0 we have g(x) = (ex − e−x )/2 = sinh(x) > x. This is a

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function whose iterates increase, having no fixed points, so we must have g n (x) → ∞ as n → ∞. It follows that |f n (z)| → ∞, for z purely imaginary, as n → ∞. x 2 − a2 . We have 9. f (x) = x2 + a2 , and so the Newton function is Nf (x) = 2x { 2x + 1; −1 ≤ x < 0 C(x) = and h : (−1, 1) → R, h(x) = a tan(πx/2). We then 2x − 1; 0 ≤ x < 1 have tan2 (πx/2) − 1 Nf ◦ h(x) = Nf (a tan(πx/2)) = a = −a cot(πx).

2 tan(πx/2 Also h ◦ C(x) = a tan(π(2x ± 1)/2) = a tan(πx ± π/2) = −a cot(πx), so that Nf ◦ h = h ◦ C. We now check that C ◦ k(x) = k ◦ D, so that C and D are conjugate. We have k : [−1, { 1) → [0, 1), k(x) = (x + 1)/2, clearly a linear x + 1; −1 ≤ x < 0 homeomorphism with k(C(x)) = = D(k(x)), and conjugacy x: 0≤x<1 follows. It now follows that Nf and D are conjugate.

10. We know that D : [0, 1) → [0, 1), D(x) = 2x (mod 1) is chaotic. By the previous exercise D is conjugate to Nf : R → R, Nf (x) = x/2 − a2 /(2x). It follows that Nf is chaotic on R. 11. If g = f /f ′ , then

( )( ) [f ′ (z)]2 f (z)f ′ (z) f (z) Ng (z) = z−g(z)/g (z) = z− ′ = z− . f (z) f ′ (z)f ′ (z) − f (z)f ′′ (z) [f ′ (z)]2 − f (z)f ′′ (z) Differentiating gives [ ′ ] (f (z))4 − [f (z)f ′′ (z)]2 − [f ′ (z)f ′′ (z) − f (z)f ′′′ (z)]f (z)f ′ (z) ′ Ng (z) = 1 − . [(f ′ (z))2 − f (z)f ′′ (z)]2 ′

f (z)[(f ′ (z))2 f ′′ (z) − 2f (z)(f ′′ (z))2 + f (z)f ′ (z)f ′′′ (z)] [[f ′ (z)]2 − f (z)f ′′ (z)]2 Substituting z = α clearly gives Ng′ (α) = 0. f ′′ (α) We use Mathematica to show that Ng′′ (α) = − ′ . (Note: There is an error f (α) in the question, where it is not correct that Ng′′ (α) = 0). This can be done by setting =−

P[z_]:= z-(f[z]/f’[z])/D[(f[z]/f’[z]),z],

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and then finding P’[z], P’’[z].

To substitute P (α) = 0, use the Mathematica substitution command: P’’[z]/.{P[a]->0}.

We are using P in place of N since Mathematica reserves N for a different command. Now we can find Ng′′′ (α) by finding P ′′′ [z] and then using the substitution.

P’’’[z]/.{P[a]->0}.

This gives Ng′′′ (α) = −4

(

[ ]2 ) f ′′′ (α) 3 f ′′ (α) − , which is not quite the same as f ′ (α) 4 f ′ (α)

Sf (α)!

Note If we set g = f /(f ′ )1/2 , then we can check that Ng′ (α) = Ng′′ (α) = 0 and that Ng′′′ (α) = −Sf (α).