Calculus single and multivariable 5th edition hughes hallett solutions manual

Calculus Single And Multivariable 5th Edition Hughes-Hallett Solutions Manual Visit to download the full and correct content document: https://testbankdeal.com/dow nload/calculus-single-and-multivariable-5th-edition-hughes-hallett-solutions-manual/

10.1 SOLUTIONS

CHAPTER TEN Solutions for Section 10.1 Exercises 1. Let f (x) =

1 = (1 − x)−1 . Then f (0) = 1.

1−x f 0 (x) = 1!(1 − x)−2 00

f (x) = 2!(1 − x)

−3

f 0 (0) = 1!, f 00 (0) = 2!,

f 000 (x) = 3!(1 − x)−4 f 000 (0) = 3!,

f (4) (x) = 4!(1 − x)−5 f (4) (0) = 4!,

f (5) (x) = 5!(1 − x)−6 f (5) (0) = 5!,

f (6) (x) = 6!(1 − x)−7 f (6) (0) = 6!,

f (7) (x) = 7!(1 − x)−8 f (7) (0) = 7!.

P3 (x) = 1 + x + x2 + x3 , P5 (x) = 1 + x + x2 + x3 + x4 + x5 , P7 (x) = 1 + x + x2 + x3 + x4 + x5 + x6 + x7 .

2. Let

1 = (1 + x)−1 . Then f (0) = 1.

1+x f 0 (x) = −1!(1 + x)−2 00

f (x) = 2!(1 + x)

−3

f 0 (0) = −1,

f 00 (0) = 2!,

f 000 (x) = −3!(1 + x)−4 f 000 (0) = −3!,

f (4) (x) = 4!(1 + x)−5 f

(5)

(x) = −5!(1 + x)

−6

(x) = −7!(1 + x)

−8

f (6) (x) = 6!(1 + x)−7 f

(7)

f (8) (x) = 8!(1 + x)−9

f (4) (0) = 4!,

f (5) (0) = −5!, f (6) (0) = 6!,

f (7) (0) = −7!, f (8) (0) = 8!.

P4 (x) = 1 − x + x2 − x3 + x4 ,

P6 (x) = 1 − x + x2 − x3 + x4 − x5 + x6 ,

P8 (x) = 1 − x + x2 − x3 + x4 − x5 + x6 − x7 + x8 .

3. Let f (x) =

√ 1 + x = (1 + x)1/2 . Then f (0) = 1, and f 0 (x) = 12 (1 + x)−1/2

f 0 (0) = 12 ,

f 00 (x) = − 14 (1 + x)−3/2

f 00 (0) = − 41 ,

000

f (x) = 83 (1 + x)−5/2 (1 + x)−7/2 f (4) (x) = − 15 16

f 000 (0) = 38 ,

f (4) (0) = − 15 .

16

783

784

Chapter Ten /SOLUTIONS

Thus, 1 1 x − x2 , 2 8 1 1 2 1 3 P3 (x) = 1 + x − x + x , 2 8 16 1 1 1 3 5 4 P4 (x) = 1 + x − x2 + x − x .

2 8 16 128 P2 (x) = 1 +

√ 4. Let f (x) = 3 1 − x = (1 − x)1/3 . Then f (0) = 1, and f 0 (x) = − 13 (1 − x)−2/3 00

f (x) = − 322 (1 − x)−5/3 f 000 (x) = − 3103 (1 − x)−8/3 f (4) (x) = − 3804 (1 − x)−11/3

f 0 (0) = − 13 ,

f 00 (0) = − 322 ,

f 000 (0) = − 3103 ,

f (4) (0) = − 3804 .

Then, 1 1 2 2 1 1 x− x = 1 − x − x2 , 3 2! 32 3 9

1 10 1 5 3 1 3 P3 (x) = P2 (x) − x , x = 1 − x − x2 − 3! 33 3 9 81 1 80 4 1 1 5 3 10 4 P4 (x) = P3 (x) − x = 1 − x − x2 − x − x .

4! 34 3 9 81 243 P2 (x) = 1 −

5. Let f (x) = cos x. Then f (0) = cos(0) = 1, and f 0 (x) = − sin x

f 00 (x) = − cos x 000

f 0 (0) = 0, f 00 (0) = −1,

f (x) = sin x

f 000 (0) = 0,

f (4) (x) = cos x

f (4) (0) = 1,

f

(5)

(x) = − sin x f (5) (0) = 0,

f (6) (x) = − cos x f (6) (0) = −1.

Thus, x2 , 2!

2 x4 x + , P4 (x) = 1 − 2!

4! x2 x4 x6 P6 (x) = 1 − + − .

2!

4!

6!

P2 (x) = 1 −

6. Let f (x) = ln(1 + x). Then f (0) = ln 1 = 0, and f 0 (x) = (1 + x)−1

f 0 (0) = 1,

f 00 (x) = (−1)(1 + x)−2

f 00 (0) = −1,

000

f (x) = 2(1 + x)

−3

f 000 (0) = 2,

f (4) (x) = −3!(1 + x)−4

f (4) (0) = −3!,

f (6) (x) = −5!(1 + x)−6

f (6) (0) = −5!,

f (8) (x) = −7!(1 + x)−8

f (8) (0) = −7!,

f

f

f

(5)

(7)

(9)

(x) = 4!(1 + x)

(x) = 6!(1 + x)

(x) = 8!(1 + x)

−5

−7

−9

f (5) (0) = 4!,

f (7) (0) = 6!,

f (9) (0) = 8!.

10.1 SOLUTIONS

So, x2 x3 x4 x5 + − + , 2 3 4 5 2 3 4 5 x x x x6 x7 x P7 (x) = x − + − + − + , 2 3 4 5 6 7 2 3 4 5 6 7 x x x x x x x8 x9 P9 (x) = x − + − + − + − + .

2 3 4 5 6 7 8 9 P5 (x) = x −

7. Let f (x) = arctan x. Then f (0) = arctan 0 = 0, and f 0 (x) = 1/(1 + x2 ) = (1 + x2 )−1

f 0 (0) = 1,

00

f 00 (0) = 0,

2 −2

f (x) = (−1)(1 + x )

2x

f 000 (x) = 2!(1 + x2 )−3 22 x2 + (−1)(1 + x2 )−2 2 f

(4)

2 −4 3 3

(x) = −3!(1 + x )

2 −3 3

2 x + 2!(1 + x )

2 x

2 −3 2

+ 2!(1 + x )

f 000 (0) = −2, f (4) (0) = 0.

2 x

Therefore,

1 3 x .

3

P3 (x) = P4 (x) = x − 8. Let f (x) = tan x. So f (0) = tan 0 = 0, and f 0 (x) = 1/ cos2 x

f 0 (0) = 1,

00

f 00 (0) = 0,

f (x) = 2 sin x/ cos3 x 000

2

2

4

f (x) = (2/ cos x) + (6 sin x/ cos x) f

(4)

f 000 (0) = 2,

(x) = (16 sin x/ cos3 x) + (24 sin3 x/ cos5 x) f (4) (0) = 0.

Thus, P3 (x) = P4 (x) = x +

x3 .

3

1 9. Let f (x) = √ = (1 + x)−1/2 . Then f (0) = 1.

1+x f 0 (x) = − 12 (1 + x)−3/2

f 0 (0) = − 12 ,

f 00 (x) = 232 (1 + x)−5/2

f 00 (0) = 232 ,

000

f 000 (0) = − 3·5 , 23

f (x) = − 3·5 (1 + x)−7/2 23 (1 + x)−9/2 f (4) (x) = 3·5·7 24

f (4) (0) = 3·5·7 24

Then, 1 3 2 1 3 1 x+ x = 1 − x + x2 , 2 2! 22 2 8 1 3 5 3 1 3·5 3 x = 1 − x + x2 − x , P3 (x) = P2 (x) − 3! 23 2 8 16 1 3·5·7 4 1 3 5 3 35 4 P4 (x) = P3 (x) + x = 1 − x + x2 − x + x .

4! 24 2 8 16 128

P2 (x) = 1 −

10. Let f (x) = (1 + x)p . (a) Suppose that p = 0. Then f (x) = 1 and f (k) (x) = 0 for any k ≥ 1. Thus P2 (x) = P3 (x) = P4 (x) = 1. (b) If p = 1 then f (x) = 1 + x, so f (0) = 1, f 0 (x) = 1, f (k) (x) = 0 Thus P2 (x) = P3 (x) = P4 (x) = 1 + x.

k ≥ 2.

785

786

Chapter Ten /SOLUTIONS

(c) In general: f (x) = (1 + x)p , f 0 (x) = p(1 + x)p−1 , f 00 (x) = p(p − 1)(1 + x)p−2 ,

f 000 (x) = p(p − 1)(p − 2)(1 + x)p−3 ,

f (4) (x) = p(p − 1)(p − 2)(p − 3)(1 + x)p−4 . f (0) = 1, f 0 (0) = p, f 00 (0) = p(p − 1),

f 000 (0) = p(p − 1)(p − 2),

f (4) (0) = p(p − 1)(p − 2)(p − 3).

p(p − 1) 2 x , 2 p(p − 1) 2 p(p − 1)(p − 2) 3 x + x , P3 (x) = 1 + px + 2 6 p(p − 1) 2 p(p − 1)(p − 2) 3 P4 (x) = 1 + px + x + x 2 6 p(p − 1)(p − 2)(p − 3) 4 + x .

24

P2 (x) = 1 + px +

√ 11. Let f (x) = 1 − x = (1 − x)1/2 . Then f 0 (x) = − 12 (1 − x)−1/2 , f 00 (x) = − 14 (1 − x)−3/2 , f 000 (x) = − 38 (1 − x)−5/2 .

So f (0) = 1, f 0 (0) = − 12 , f 00 (0) = − 41 , f 000 (0) = − 83 , and 1 1 2 3 1 3 1 x− x − x 2 4 2!

8 3!

2 3 x x x = 1− − − .

2 8 16

P3 (x) = 1 −

12. Let f (x) = ex . Since f (k) (x) = ex = f (x) for all k ≥ 1, the Taylor polynomial of degree 4 for f (x) = e x about x = 1 is e1 e1 e1 (x − 1)2 + (x − 1)3 + (x − 1)4 2!

3!

4! h i 1 1 1 2 3 = e 1 + (x − 1) + (x − 1) + (x − 1) + (x − 1)4 .

2 6 24

P4 (x) = e1 + e1 (x − 1) +

√ 1 + x = (1 + x)1/2 .

1 3 1 0 Then f (x) = (1 + x)−1/2 , f 00 (x) = − (1 + x)−3/2 , and f 000 (x) = (1 + x)−5/2 . The Taylor polynomial of degree 2 4 8 three about x = 1 is thus

13. Let f (x) =

1

P3 (x) = (1 + 1)1/2 +

− (1 + 1) 1 (1 + 1)−1/2 (x − 1) + 4 2 2!

(1 + 1)−5/2 (x − 1)3 3!

√ (x − 1)2 (x − 1)3 x−1 = 2 1+ − + .

4 32 128 3

+8

−3/2

(x − 1)2

10.1 SOLUTIONS

787

14. Let f (x) = cos x. f ( π2 ) = 0. f 0 (x) = − sin x

f 0 ( π2 ) = −1,

00

f 00 ( π2 ) = 0,

f (x) = − cos x

f 000 (x) = sin x

f 000 ( π2 ) = 1,

(4)

f (4) ( π2 ) = 0.

f

(x) = cos x

So,

1 π x− 3!

2

π 1 π 3 = − x− . + x− 2 3!

2

P4 (x) = 0 − x −

π 2

+0+

3

15. Let f (x) = sin x.

Then f 0 (x) = cos x, f 00 (x) = − sin x, and f 000 (x) = − cos x, so the Taylor polynomial for sin x of degree three about x = −π/4 is

π π π + cos − x+ 4 4 4

π

− sin − π4 π 2 − cos − 4 π 3 + x+ x+ + 2!

4 3!

4 √

2

1 2 π 1 π π 3 . − = −1 + x + + x+ x+ 2 4 2 4 6 4

P3 (x) = sin −

16. Let f (x) = ln(x2 ). Then ln(12 ) = ln 1 = 0.

Then f 0 (x) = 2x−1 , f 00 (x) = −2x−2 , f 000 (x) = 4x−3 , and f (4) (x) = −12x−4 .

The Taylor polynomial of degree 4 for f (x) = ln(x2 ) about x = 1 is −2 · 1−2 4 · 1−3 −12 · 1−4 (x − 1)2 + (x − 1)3 + (x − 1)4 2!

3!

4!

12 4 (x − 1)4 = 0 + 2(x − 1) − (x − 1)2 + (x − 1)3 − 6 24 2 1 = 2(x − 1) − (x − 1)2 + (x − 1)3 − (x − 1)4 .

3 2

P4 (x) = ln(12 ) + 2 · 1−1 (x − 1) +

Problems 17. Using the fact that

f 00 (0) 2 f 000 (0) 3 x + x 2!

3! and identifying coefficients with those given for P3 (x), we obtain the following: f (x) ≈ P3 (x) = f (0) + f 0 (0)x +

(a) f (0) = constant term which equals 2, so f (0) = 2. (b) f 0 (0) = coefficient of x which equals −1, so f 0 (0) = −1. f 00 (0) 2 so f 00 (0) = −2/3. (c) 2! = coefficient of x which equals −1/3, 000

(d) f 3!(0) = coefficient of x3 which equals 2,

so f 000 (0) = 12.

18. f (x) = 4x2 − 7x + 2

f 0 (x) = 8x − 7

f 00 (x) = 8

f (0) = 2 f 0 (0) = −7

f 00 (0) = 8,

so P2 (x) = 2 + (−7)x + 82 x2 = 4x2 − 7x + 2. We notice that f (x) = P2 (x) in this case.

788

Chapter Ten /SOLUTIONS

19. f 0 (x) = 3x2 + 14x − 5, f 00 (x) = 6x + 14, f 000 (x) = 6. Thus, about a = 0, 14 2 6 −5 x+ x + x3 1!

2!

3! = 1 − 5x + 7x2 + x3

P3 (x) = 1 +

= f (x).

20. (a) We’ll make the following conjecture: “If f (x) is a polynomial of degree n, i.e.

f (x) = a0 + a1 x + a2 x2 + · · · + an−1 xn−1 + an xn , then Pn (x), the nth degree Taylor polynomial for f (x) about x = 0, is f (x) itself.” (b) All we need to do is to calculate Pn (x), the nth degree Taylor polynomial for f about x = 0 and see if it is the same as f (x). f (0) = a0 ; f 0 (0) = (a1 + 2a2 x + · · · + nan xn−1 ) x=0 = a1 ;

f 00 (0) = (2a2 + 3 · 2a3 x + · · · + n(n − 1)an xn−2 ) x=0 = 2!a2 .

If we continue doing this, we’ll see in general f (k) (0) = k!ak ,

k = 1, 2, 3, · · · , n.

Therefore, f 00 (0) 2 f (n) (0) n f 0 (0) x+ x + ··· + x 1!

2! n!

2 n = a0 + a1 x + a2 x + · · · + a n x

Pn (x) = f (0) +

= f (x).

21. Since P2 (x) is the second degree Taylor polynomial for f (x) about x = 0, P 2 (0) = f (0), which says a = f (0). Since d P2 (x) = f 0 (0), dx x=0 b = f 0 (0); and since d2 = f 00 (0), P2 (x) dx2 x=0 2c = f 00 (0). In other words, a is the y-intercept of f (x), b is the slope of the tangent line to f (x) at x = 0 and c tells us the concavity of f (x) near x = 0. So c < 0 since f is concave down; b > 0 since f is increasing; a > 0 since f (0) > 0.

22. As we can see from Problem 21, a is the y-intercept of f (x), b is the slope of the tangent line to f (x) at x = 0 and c tells us the concavity of f (x) near x = 0.

So a > 0, b < 0 and c < 0.

23. As we can see from Problem 21, a is the y-intercept of f (x), b is the slope of the tangent line to f (x) at x = 0 and c tells us the concavity of f (x) near x = 0.

So a < 0, b > 0 and c > 0.

24. As we can see from Problem 21, a is the y-intercept of f (x), b is the slope of the tangent line to f (x) at x = 0 and c tells us the concavity of f (x) near x = 0.

So a < 0, b < 0 and c > 0.

25. x3

x − 3! sin x x2 = lim = lim 1 − x→0 x→0 x→0 x x 3! lim

= 1.

10.1 SOLUTIONS

789

26. x2

x4

1 − (1 − 2! + 4! ) 1 − cos x = lim = lim x→0 x→0 x→0 x2 x2 lim

27. For f (h) = eh , P4 (h) = 1 + h + (a)

x2 1 − 2 4!

=

1 .

2

h2 h3 h4 + + . So 2 3!

4!

P4 (h) − 1 − h eh − 1 − h = lim h→0 h→0 h2 h2 lim

3 4 h2 + h3! + h4!

2 = lim h→0 h2

= lim

h→0

=

1 .

2

1 h h2 + + 2 3!

4!

(b) 2

2

eh − 1 − h − h2 P4 (h) − 1 − h − h2 = lim 3 h→0 h→0 h h3 lim

4 h3 + h4!

3! h→0 h3

= lim

= lim

h→0

1 1 = . = 3!

6

1 h + 3!

4!

Using Taylor polynomials of higher degree would not have changed the results since the terms with higher powers of h all go to zero as h → 0.

28. (a) We use the Taylor polynomial of degree two for f and h about x = 2.

f 00 (2) 3 (x − 2)2 = (x − 2)2 2!

2 h00 (2) 7 2 0 (x − 2) = (x − 2)2 h(x) ≈ h(2) + h (2)(x − 2) + 2!

2

f (x) ≈ f (2) + f 0 (2)(x − 2) +

Thus, using the fact that near x = 2 we can approximate a function by Taylor polynomials 3 (x − 2)2 f (x) 3 = lim 27 = . x→2 (x − 2)2 x→2 h(x) 7 2

lim

(b) We use the Taylor polynomial of degree two for f and g about x = 2. f 00 (2) 3 (x − 2)2 = (x − 2)2 2!

2 g 00 (2) 5 g(x) ≈ g(2) + g 0 (2)(x − 2) + (x − 2)2 = 22(x − 2) + (x − 2)2 .

2!

2

f (x) ≈ f (2) + f 0 (2)(x − 2) +

Thus, f (x) = lim x→2 g(x) x→2 lim

3 (x − 2)2 2

22(x − 2) + 5(x − 2)2

= lim

x→2

3 (x − 2) 2

22 + 5(x − 2)

=

0 = 0.

22

29. (a) Since the coefficient of the x-term of each f is 1, we know f 10 (0) = f20 (0) = f30 (0) = 1. Thus, each of the f s slopes upward near 0, and are in the second figure.

The coefficient of the x-term in g1 and in g2 is 1, so g10 (0) = g20 (0) = 1. For g3 however, g30 (0) = −1. Thus, g1 and g2 slope up near 0, but g3 slopes down. The gs are in the first figure. (b) Since g1 (0) = g2 (0) = g3 (0) = 1, the point A is (0, 1).

Since f1 (0) = f2 (0) = f3 (0) = 2, the point B is (0, 2).

790

Chapter Ten /SOLUTIONS

(c) Since g3 slopes down, g3 is I. Since the coefficient of x2 for g1 is 2, we know g100 (0) =2 2!

so

g100 (0) = 4.

By similar reasoning g200 (0) = 2. Since g1 and g2 are concave up, and g1 has a larger second derivative, g1 is III and g2 is II. Calculating the second derivatives of the f s from the coefficients x2 , we find f100 (0) = 4

f200 (0) = −2

f300 (0) = 2.

Thus, f1 and f3 are concave up, with f1 having the larger second derivative, so f1 is III and f3 is II. Then f2 is concave down and is I. 30. Let f (x) be a function that has derivatives up to order n at x = a. Let Pn (x) = C0 + C1 (x − a) + · · · + Cn (x − a)n be the polynomial of degree n that approximates f (x) about x = a. We require that P n (x) and all of its first n derivatives agree with those of the function f (x) at x = a, i.e., we want f (a) = Pn (a), f 0 (a) = Pn0 (a), f 00 (a) = Pn00 (a), .. . f (n) (a) = Pn(n) (a).

When we substitute x = a in Pn (x), all the terms except the first drop out, so f (a) = C0 .

Now differentiate Pn (x): Pn0 (x) = C1 + 2C2 (x − a) + 3C3 (x − a)2 + · · · + nCn (x − a)n−1 .

Substitute x = a again, which yields

f 0 (a) = Pn0 (a) = C1 .

Differentiate Pn0 (x): Pn00 (x) = 2C2 + 3 · 2C3 (x − a) + · · · + n(n − 1)Cn (x − a)n−2

and substitute x = a again:

f 00 (a) = Pn00 (a) = 2C2 .

Differentiating and substituting again gives f 000 (a) = Pn000 (a) = 3 · 2C3 .

Similarly,

f (k) (a) = Pn (k) (a) = k!Ck .

000

00

So, C0 = f (a), C1 = f 0 (a), C2 = f 2!(a) , C3 = f 3!(a) , and so on.

If we adopt the convention that f (0) (a) = f (a) and 0! = 1, then Ck =

f (k) (a) , k = 0, 1, 2, · · · , n. k!

Therefore, f (x) ≈ Pn (x) = C0 + C1 (x − a) + C2 (x − a)2 · · · + Cn (x − a)n = f (a) + f 0 (a)(x − a) +

f 00 (a) f (n) (a) (x − a)2 + · · · + (x − a)n .

2! n!

10.1 SOLUTIONS

791

2

31. (a) f (x) = ex .

2 2 2 f 0 (x) = 2xex , f 00 (x) = 2(1 + 2x2 )ex , f 000 (x) = 4(3x + 2x3 )ex , 2 2 f (4) (x) = 4(3 + 6x2 )ex + 4(3x + 2x3 )2xex .

The Taylor polynomial about x = 0 is 0 2 0 12 4 x + x2 + x3 + x 1!

2!

3!

4!

1 = 1 + x 2 + x4 .

2

P4 (x) = 1 +

(b) f (x) = ex . The Taylor polynomial of degree 2 is Q2 (x) = 1 +

x x2 1 + = 1 + x + x2 .

1!

2!

2

If we substitute x2 for x in the Taylor polynomial for ex of degree 2, we will get P4 (x), the Taylor polynomial for 2 ex of degree 4: 1 2 2 (x ) 2 1 = 1 + x 2 + x4 2 = P4 (x).

Q2 (x2 ) = 1 + x2 +

(c) Let Q10 (x) = 1 +

x x2 x10 + + ··· + be the Taylor polynomial of degree 10 for ex about x = 0. Then 1!

2!

10!

P20 (x) = Q10 (x2 ) (x2 )2 (x2 )10 x2 + + ··· + 1!

2!

10! x2 x4 x20 = 1+ + + ··· + .

1!

2!

10! = 1+

5

x + · · · + x5! . Then (d) Let ex ≈ Q5 (x) = 1 + 1!

e−2x ≈ Q5 (−2x)

(−2x)2 (−2x)3 (−2x)4 (−2x)5 −2x + + + + 1!

2!

3!

4!

5!

4 3 2 4 4 5 2 = 1 − 2x + 2x − x + x − x .

3 3 15

= 1+

t3

32. (a)

t − 3! sin t t2 ≈ =1− t t 6

Z 1 0

t3

(b)

t2 1− 6

dt = t −

t3 = 0.94444 · · · 18 0

t4 t2 + 1− 6 120

dt = t −

t3 t5 = 0.94611 · · · + 18 600 0

sin t dt ≈ t

Z 1 0

1

t5

t − 3! + 5! t2 t4 sin t ≈ =1− + t t 6 120

Z 1 0

sin t dt ≈ t

Z 1 0

1

33. (a) The equation sin x = 0.2 has one solution near x = 0 and infinitely many others, one near each multiple of π. See x3 Figure 10.1. The equation x − = 0.2 has three solutions, one near x = 0 and two others. See Figure 10.2.

3!

792

Chapter Ten /SOLUTIONS y

y

y = 0.2 x

y = 0.2 x

3

Figure 10.2: Graph of y = x − x3! and y = 0.2

Figure 10.1: Graph of y = sin x and y = 0.2

(b) Near x = 0, the cubic Taylor polynomial x − x3 /3! ≈ sin x. Thus, the solutions to the two equations near x = 0 are approximately equal. The other solutions are not close. The reason is that x − x 3 /3! only approximates sin x near x = 0 but not further away. See Figure 10.3. y

sin x y = 0.2

x

x − x3 /3!

Figure 10.3 34. Changing sin θ into θ makes sense if the two values are almost equal. If we measure θ in radians, this is true for values of θ close to zero. (Recall the first degree Taylor polynomial: sin θ ≈ θ.) In other words, the switch is justified when the pendulum does not swing very far from the vertical.

35. (a) The graphs of y = cos x and y = 1 − 0.1x cross at x = 0 and for another x-value just to the right of x = 0. (There are other crossings much further to the right.) (b) Since x2 cos x ≈ 1 − 2 the equation becomes 1−

x2 = 1 − 0.1x 2 2 x = 0.1x 2 x = 0, 0.2.

The solution x = 0 is an exact solution to the original equation; x = 0.2 is an approximate solution to the original equation.

Solutions for Section 10.2 Exercises 1. Differentiating (1 + x)3/2 : f (x) = (1 + x)3/2 0

f (x) = (3/2)(1 + x) 00

f (0) = 1, f 0 (0) = 23 ,

1/2

f (x) = (1/2)(3/2)(1 + x)

−1/2

= (3/4)(1 + x)

−1/2

f 00 (0) = 43 ,

f 000 (x) = (−1/2)(3/4)(1 + x)−3/2 = (−3/8)(1 + x)−3/2 f 000 (0) = − 38 .

10.2 SOLUTIONS

793

(3/4)x2 (−3/8)x3 3 ·x+ + + ··· 2 2!

3!

2 3 3x x 3x + − + ··· = 1+ 2 8 16

f (x) = (1 + x)3/2 = 1 +

√ 2. Differentiating 4 x + 1: √ f (x) = 4 x + 1 = (x + 1)1/4 0

f (x) = (1/4)(x + 1)

f (0) = 1,

−3/4

00

f (x) = (−3/4)(1/4)(x + 1)

f 0 (0) = 41 , −7/4

= (−3/16)(x + 1)

−7/4

3 f 00 (0) = − 16 ,

21 f 000 (x) = (−7/4)(−3/16)(x + 1)−11/4 = (21/64)(x + 1)−11/4 f 000 (0) = 64 .

√ (−3/16)x2 (21/64)x3 1 f (x) = 4 x + 1 = 1 + · x + + + ··· 4 2!

3!

2 3 x 3x 7x = 1+ − + − ···.

4 32 128 3. Differentiating sin(−x): f (x) = sin(−x)

f (0) = 0,

f 0 (x) = cos(−x)(−1) = − cos(−x)

f 0 (0) = −1,

f 00 (x) = −(− sin(−x))(−1) = − sin(−x)

f 00 (0) = 0,

000

f (x) = − cos(−x)(−1) = cos(−x)

f (4) (x) = − sin(−x)(−1) = sin(−x)

f

(5)

f 000 (0) = 1 f (4) (0) = 0, f (5) (0) = −1,

(x) = cos(−x)(−1) = − cos(−x)

f (6) (x) = −(− sin(−x))(−1) = − sin(−x) f (6) (0) = 0,

f (7) (x) = − cos(−x)(−1) = cos(−x)

f (7) (0) = 1.

1x3 0x4 −1x5 0x6 1x7 0x2 + + + + + + ··· 2!

3!

4!

5!

6!

7! x3 x5 x7 = −x + − + + ···.

3!

5!

7!

f (x) = sin(−x) = 0 − 1 · x +

Notice that the series for sin(−x) is obtained from the series for sin x by changing the signs. This is expected since sin(−x) = − sin x.

4. Differentiating ln(1 − x)

f (x) = ln(1 − x)

f (0) = 0,

1 (−1) = −(1 − x)−1 f (x) = 1−x 00 −2

f 0 (0) = −1,

0

f (x) = −(−(1 − x)

)(−1) = −(1 − x)

(4)

−4

−2

f 000 (x) = −2(−(1 − x)−3 )(−1) = −2(1 − x)−3

f

(x) = −3(−2(1 − x)

)(−1) = −6(1 − x)

f 00 (0) = −1,

−4

f 000 (0) = −2

f (4) (0) = −6.

(−1)x2 (−2)x3 (−6)x4 + + + ··· 2!

3!

4! x2 x3 x4 = −x − − − + ···.

2 3 4

f (x) = ln(1 − x) = 0 − 1 · x +

5.

1 = (1 − x)−1 f (x) = 1−x 0

f (x) = −(1 − x) 00

−2

f (x) = −2(1 − x)

(−1) = (1 − x)

−3

f (0) = 1, −2

(−1) = 2(1 − x)

−3

f 0 (0) = 1, f 00 (0) = 2,

f 000 (x) = −6(1 − x)−4 (−1) = 6(1 − x)−4 f 000 (0) = 6.

794

Chapter Ten /SOLUTIONS

f (x) =

1 2x2 6x3 = 1+1·x+ + + ··· 1−x 2!

3!

2 3 = 1 + x + x + x + ···

6.

1

1 = (1 + x)− 2 f (x) = √1+x 3 f (x) = − 21 (1 + x)− 2 5 f 00 (x) = 43 (1 + x)− 2 7 f 000 (x) = − 15 (1 + x)− 2 8

f (0) = 1 f 0 (0) = − 12

0

f 00 (0) = 34

f 000 (0) = − 15 8

( 3 )x2 (− 15 )x3 1 1 8 f (x) = √ x+ 4 = 1+ − + + ··· 2 2!

3!

1+x x 3x2 5x3 = 1− + − + ··· 2 8 16 7. f (x) =

√ 1 1 + x = (1 + x) 2

1 f (x) = 21 (1 + x)− 2 3 f 00 (x) = − 14 (1 + x)− 2 5 f 000 (x) = 83 (1 + x)− 2

0

f (x) =

8.

f (0) = 1 f 0 (0) = 21 f 00 (0) = − 14

f 000 (0) = 83

√ (− 14 )x2 ( 3 )x3 1 + 8 + ··· 1+x = 1+ x+ 2 2!

3! x2 x3 x = 1+ − + + ··· 2 8 16 √ 1 f (y) = 3 1 − y = (1 − y) 3

f (0) = 1

2 2 f (y) = 31 (1 − y)− 3 (−1) = − 13 (1 − y)− 3 5 5 f 00 (y) = 92 (1 − y)− 3 (−1) = − 29 (1 − y)− 3 8 8 10 f 000 (y) = 27 (1 − y)− 3 (−1) = − 10 (1 − y)− 3 27

0

f 0 (0) = − 13

f 00 (0) = 29

f 000 (0) = − 10 27

(− 29 )y 2 )y 3 (− 10 1 27 y+ + + ··· 3 2!

3! y2 5y 3 y − − ··· = 1− − 3 9 81

p

f (y) = 3 1 − y = 1 + −

9. f (x) = sin x 0

f (x) = cos x f 00 (x) = − sin x

f 000 (x) = − cos x

f ( π4 ) =

√

2 , √2 0 π f ( 4 ) = 22 , √ f 00 ( π4 ) = − 22 , √ f 000 ( π4 ) = − 22 .

√ √ √ (x − π )2 √ (x − π )3 2 2 2 2 π 4 4 + x− − − − ··· 2 2 4 2 2!

2 3! √ √ √ 2 √ 3 2 2 π 2 π 2 π + − − ··· = x− − x− x− 2 2 4 4 4 12 4

sin x =

10.2 SOLUTIONS

10. f (θ) = cos θ 0

f (θ) = − sin θ

f 00 (θ) = − cos θ

f 000 (θ) = sin θ

11. Differentiating gives

√

f ( π4 ) =

2 , 2√ 0 π f ( 4 ) = − 22 , √ f 00 ( π4 ) = − 22 , √ f 000 ( π4 ) = 22 .

√ √ √ (θ − π )2 √ (θ − π )3 π 2 2 2 2 4 4 θ− − cos θ = − + − ··· 2 2 4 2 2!

2 3! √ √ √ 2 √ 3 π π π 2 2 2 2 = − θ− − θ− θ− + − ··· 2 2 4 4 4 12 4 f (t) = cos t f 0 (t) = − sin t

f 00 (t) = − cos t

f 000 (t) = sin t

√

3 , 2 0 π 1 f ( 6 ) = −2, √ f 00 ( π6 ) = − 23 , f 000 ( π6 ) = 12 .

f ( π6 ) =

√

√ (t − π )2 π 3 π 1 1 (t − 6 ) 3 3 6 t− − − + − ··· 2 2 6 2 2!

2 3! √ √

3 π 3 π 2 π 3 1 1 t− − t− t− = − + − ··· 2 2 6 4 6 12 6

cos t =

12.

f (θ) = sin θ f 0 (θ) = cos θ 00

f (θ) = − sin θ

f 000 (θ) = − cos θ

√

f (− π4 ) = − 22 ,

f 0 (− π4 ) =

√

2 , √2 π f (− 4 ) = 22 , √ f 000 (− π4 ) = − 22 .

00

√ √ (θ + π )2 √ (θ + π )3 2 2 π 2 2 4 4 + θ+ + − + ··· 2 2 4 2 2!

2 3! √ √ √ 2 √ 3 2 2 2 2 π π π + − + ···. =− θ+ + θ+ θ+ 2 2 4 4 4 12 4

sin θ = −

√

13. f (x) = tan x

f ( π4 ) = 1,

f 0 (x) = cos12 x

f 0 ( π4 ) = 2,

sin x) sin x f (x) = −2(− = 2cos 3x cos3 x x(− sin x) f 000 (x) = −6 sincos + cos22 x 4x 00

f 00 ( π4 ) = 4, f 000 ( π4 ) = 16.

3

x − π4 (x − π4 )2 π +4 + 16 + ··· 4 2!

3!

2

3 π π π 8 = 1+2 x− +2 x− x− + + ··· 4 4 3 4

tan x = 1 + 2 x −

14. f (x) = x1

f (1) = 1

0

f 0 (1) = −1

f (x) = − x12 f 00 (x) = x23 f 000 (x) = − x64

f 00 (1) = 2

f 000 (1) = −6

795

796

Chapter Ten /SOLUTIONS

2(x − 1)2 6(x − 1)3 1 = 1 − (x − 1) + − + ··· x 2!

3! = 1 − (x − 1) + (x − 1)2 − (x − 1)3 + · · · .

15. Again using the derivatives found in Problem 14, we have f (2) =

1 , 2

1 f 0 (2) = − , 4

f 00 (2) =

1 , 4

3 f 000 (2) = − .

8

(x − 2)2 3(x − 2)3 1 1 x−2 = − + − + ··· x 2 4 4 · 2!

8 · 3! (x − 2) (x − 2)2 (x − 2)3 1 = − + − + ··· 2 4 8 16 16. Using the derivatives from Problem 14, we have f (−1) = −1,

f 0 (−1) = −1,

f 00 (−1) = −2,

f 000 (−1) = −6.

Hence, 2(x + 1)2 6(x + 1)3 1 = −1 − (x + 1) − − − ··· x 2!

3! = −1 − (x + 1) − (x + 1)2 − (x + 1)3 − · · · 17. The general term can be written as xn for n ≥ 0.

18. The general term can be written as (−1)n xn for n ≥ 0.

19. The general term can be written as −xn /n for n ≥ 1.

20. The general term can be written as (−1)n−1 xn /n for n ≥ 1.

21. The general term can be written as (−1)k x2k+1 /(2k + 1)! for k ≥ 0.

22. The general term can be written as (−1)k x2k+1 /(2k + 1) for k ≥ 0.

23. The general term can be written as x2k /k! for k ≥ 0.

24. The general term can be written as (−1)k x4k+2 /(2k)! for k ≥ 0.

Problems 25. By looking at Figure 10.4 we can that the Taylor polynomials are reasonable approximations for the function f (x) = √1 between x = −1 and x = 1. Thus a good guess is that the interval of convergence is −1 < x < 1.

1+x 4 3

P10 (x)

P4 (x)

2

P2 (x)

1

1 f (x) = √1+x

x −1

1

2

Figure 10.4

3

10.2 SOLUTIONS

797

26. By √ looking at Figure 10.5, we see that the Taylor polynomials are reasonable approximations for the function f (x) = 1 + x between x = −1 and x = 1. Thus a good guess is that the interval of convergence is −1 < x < 1. f (x) =

2

√

x+1 P2 (x)

1 P10 (x)

P4 (x) x

−1

1

2

3

Figure 10.5 27. The graph suggests that the Taylor polynomials converge to f (x) = Since

1 on the interval −1 < x < 1. See Figure 10.6.

1−x

1 = 1 + x + x 2 + x3 · · · .

1−x

the ratio test gives

|an+1 | |xn+1 | = lim = |x|. n→∞ |an | n→∞ |xn | Thus, the series converges if |x| < 1|; that is, −1 < x < 1. lim

1 f (x) = 1−x

-

P7 (x)

−1

x

1

-

P5 (x) P3 (x)-

Figure 10.6 28. The Taylor series for ln(1 − x) is ln(1 − x) = −x −

so

x3 xn x2 − − ··· − − ···, 2 3 n

1/(n + 1) |an+1 | n = |x| lim = |x| lim = |x|. n→∞ n→∞ n + 1 |an | 1/n Thus the series converges for |x| < 1, and the radius of convergence is 1. Note: This series can be obtained from the series for ln(1 + x) by replacing x by −x and has the same radius of convergence as the series for ln(1 + x). lim

n→∞

29. (a) We have shown that the series is

1 + px +

p(p − 1) 2 p(p − 1)(p − 2) 3 x + x + ··· 2!

3!

so the general term is (b) We use the ratio test lim

n→∞

p(p − 1) . . . (p − (n − 1)) n x . n!

|an+1 | p(p − 1) . . . (p − (n − 1))(p − n) · n! p−n . = |x| lim = |x| lim n→∞ n→∞ n + 1 |an | (n + 1)!p(p − 1) . . . (p − (n − 1))

Since p is fixed, we have

lim

n→∞

p−n = 1, n+1

so

R = 1.

798

Chapter Ten /SOLUTIONS

30. The second coefficient of the Taylor expansion is g 00 (0) = 1, 2

so

g 00 (0) = 2.

g 000 (0) =0 3!

so

g 000 (0) = 0.

g (10) (0) 1 = 10!

5!

so

g (10) (0) =

Similarly, the third coefficient is Finally, the tenth coefficient is

10! .

5!

31. Let Cn be the coefficient of the nth term in the series. Note that 0 = C1 =

d 2 x2 , (x e ) dx x=0

and since 1 = C6 = 2 we have

2 d6 (x2 ex ) dx6

x=0

6!

,

6! d 6 2 x2 = (x e ) = 360. dx6 2 x=0

32. (a) From the coefficients of the (x − 1) terms of the f s, we see that f10 (1) = 1,

f20 (1) = −1

f30 (1) = −2.

f200 (1) = 1, 2!

f300 (1) = 1, 2!

From the (x − 1)2 terms of the f s, we see that f100 (1) = −1, 2!

so f100 (1) = −2, f200 (1) = 2, f300 (1) = 2.

Thus, f1 slopes up at x = 1 and f2 and f3 slope down; f3 slopes down more steeply than f2 . This means that the f s are in the first figure, since graphs II and III in the second figure have the same negative slope at point B. By a similar argument, we find g10 (4) = −1,

g20 (4) = −1,

g30 (4) = 1, and g100 (4) = −2,

g200 (4) = 2,

g300 (4) = 2.

Thus, two of the gs slope down, one of which is concave up and one is concave down; the third g slopes up and is concave up. This confirms that the gs are in the second figure. (b) Since f1 (1) = f2 (1) = f3 (1) = 3, the point A is (1, 3).

Since g1 (4) = g2 (4) = g3 (4) = 5, the point B is (4, 5). (c) In the first figure, graph I is f1 since it slopes up. Graph II is f2 since it slopes down, but less steeply than graph III, which is f3 .

In the second figure, graph I is g3 , since it slopes up. Graph II is g2 since it slopes down and is concave up.

Graph III is g1 since it slopes down and is concave down.

33. We define eiθ to be

(iθ)3 (iθ)4 (iθ)5 (iθ)6 (iθ)2 + + + + + ··· 2!

3!

4!

5!

6!

Suppose we consider the expression cos θ + i sin θ, with cos θ and sin θ replaced by their Taylor series: eiθ = 1 + iθ +

cos θ + i sin θ =

θ4 θ6 θ2 1− + − + ··· 2!

4!

6!

θ3 θ5 +i θ− + − ··· 3!

5!

Reordering terms, we have cos θ + i sin θ = 1 + iθ −

θ2 iθ3 θ4 iθ5 θ6 − + + − − ··· 2!

3!

4!

5!

6!

10.3 SOLUTIONS

799

Using the fact that i2 = −1, i3 = −i, i4 = 1, i5 = i, · · ·, we can rewrite the series as (iθ)2 (iθ)3 (iθ)4 (iθ)5 (iθ)6 + + + + + ··· 2!

3!

4!

5!

6!

cos θ + i sin θ = 1 + iθ +

Amazingly enough, this series is the Taylor series for ex with iθ substituted for x. Therefore, we have shown that cos θ + i sin θ = eiθ .

34. This is the series for ex with x replaced by 2, so the series converges to e2 .

35. This is the series for sin x with x replaced by 1, so the series converges to sin 1.

36. This is the series for 1/(1 − x) with x replaced by 1/4, so the series converges to 1/(1 − (1/4)) = 4/3.

37. This is the series for cos x with x replaced by 10, so the series converges to cos 10.

38. This is the series for ln(1 + x) with x replaced by 1/2, so the series converges to ln(3/2).

39. The Taylor series for f (x) = 1/(1 + x) is 1 = 1 − x + x 2 − x3 + · · · .

1+x Substituting x = 0.1 gives 1 − 0.1 + (0.1)2 − (0.1)3 + · · · =

1 1 = .

1 + 0.1 1.1

Alternatively, this is a geometric series with a = 1, x = −0.1.

40. This is the series for ex with x = 3 substituted. Thus 1+3+

9 27 81 32 33 34 + + + ··· = 1 + 3 + + + + · · · = e3 .

2!

3!

4!

2!

3!

4!

41. This is the series for cos x with x = 1 substituted. Thus 1−

1 1 1 + − + · · · = cos 1.

2!

4!

6!

42. This is the series for ex with −0.1 substituted for x, so 1 − 0.1 + 43. Since 1 + x + x2 + x3 + · · · = 44. Since x −

0.001 0.01 − + · · · = e−0.1 .

2!

3!

1 1 1 4 , a geometric series, we solve = 5 giving = 1 − x, so x = .

1−x 1−x 5 5

1 2 1 3 x + x + · · · = ln(1 + x), we solve ln(1 + x) = 0.2, giving 1 + x = e0.2 , so x = e0.2 − 1.

2 3

Solutions for Section 10.3 Exercises 2

3

1. Substitute y = −x into ey = 1 + y + y2! + y3! + · · ·. We get (−x)2 (−x)3 + + ··· 2!

3! x2 x3 = 1−x+ − + ···.

2!

3!

e−x = 1 + (−x) +

800

Chapter Ten /SOLUTIONS

2. We’ll use

p

1 −1 y 2 1 y+ 2 2 2 2!

3 1 −1 −3 y + + ··· 2 2 2 3!

2 3 y y y = 1+ − + − ···.

2 8 16

1

1 + y = (1 + y) 2 = 1 +

Substitute y = −2x. √ (−2x) (−2x)2 (−2x)3 1 − 2x = 1 + − + − ··· 2 8 16 x3 x2 − − ··· = 1−x− 2 2 3. Substitute x = θ 2 into series for cos x: (θ2 )4 (θ2 )6 (θ2 )2 + − + ··· 2!

4!

6! θ8 θ12 θ4 + − + ···. = 1− 2!

4!

6!

cos (θ 2 ) = 1 −

2

3

4

4. Substituting x = −2y into ln(1 + x) = x − x2 + x3 − x4 + · · · gives (−2y)3 (−2y)4 (−2y)2 + − + ··· 2 3 4 8 = −2y − 2y 2 − y 3 − 4y 4 − · · · .

3

ln(1 − 2y) = (−2y) −

d (arcsin x) = √ 1 5. Since dx

1−x2

5 6 x + · · ·, integrating gives = 1 + 12 x2 + 83 x4 + 16

arcsin x = c + x +

1 3 3 5 5 7 x + x + x + ···.

6 40 112

Since arcsin 0 = 0, c = 0.

6. We substitute 3t into the series for sin x and multiply by t. Since sin x = x −

x5 x7 x3 + − + ···, 3!

5!

7!

substituting 3t gives (3t)3 (3t)5 (3t)7 + − + ··· 3!

5!

7! −9 3 81 5 −243 7 = 3t + t + t + t + ···, 2 40 560

sin(3t) = (3t) −

so t sin(3t) = 3t2 −

9 4 81 6 243 8 t + t − t + ···.

2 40 560

1

5 3 1 x + · · · gives 7. Substituting x = −z 2 into √1+x = (1 + x)− 2 = 1 − 21 x + 38 x2 − 16

3(−z 2 )2 5(−z 2 )3 (−z 2 ) 1 √ + − + ··· = 1− 2 2 8 16 1−z 1 3 5 6 = 1 + z2 + z4 + z + ···.

2 8 16

10.3 SOLUTIONS

8. z −z 2 =z 2 = ze z e

(−z 2 )3 (−z 2 )2 + + ··· 1 + (−z ) + 2!

3!

2

z5 z7 − + ··· 2!

3!

= z − z3 + 9. φ3 cos(φ2 ) = φ3

1−

= φ3 −

(φ2 )4 (φ2 )6 (φ2 )2 + − + ··· 2!

4!

6!

φ7 φ11 φ15 + − + ··· 2!

4!

6!

801

5

3

10. Substituting the series for sin θ = θ − θ3! + θ5! − · · · into

p

gives

1+y =1+

√ 1 1 + sin θ = 1 + 2 1 + 16

1 1 1 3 y − y2 + y − ··· 2 8 16

θ3 θ5 + − ··· 3!

5!

θ−

θ3 θ5 θ− + − ··· 3!

5!

3

−

(1 + t) sin t =

Multiplying and collecting terms yields

p

1+

θ−

+ ···

t t2 t3 − + − ··· 2 8 16

θ3 θ5 + − ··· 3!

5!

t−

t3 t5 + − ··· 3!

5!

et cos t =

1+t+

t2 t3 t4 + + + ··· 2!

3!

4!

Multiplying out and collecting terms gives t

e cos t = 1 + t + = 1+t−

2

t t2 − 2!

2!

+

3

t4 t3 − + ···.

3 6

1−

t2 t4 t6 + − + ··· 2!

4!

6!

t3 t3 t4 t4 t2 (1 + t) sin t = t + − + + − 2 3!

8 16 12 7 3 1 4 1 t − t + ···. = t + t2 − 2 24 48

12.

t t3 − 3!

2!

2

− ···

θ2 θ θ 1 θ− + − 2 8 16 2 · 3!

1 1 1 3 = 1 + θ − θ2 − θ + ··· 2 8 48

p

3

3

= 1+

11.

1 8

+

4

+ ···

t t4 t4 + − 4!

4!

(2!)2

+ ···

13. Multiplying out gives (1 + x)3 = 1 + 3x + 3x2 + x3 . Since this polynomial equals the original function for all x, it must be the Taylor series. The general term is 0 · xn for n ≥ 4.

802

Chapter Ten /SOLUTIONS

14. Substituting t2 into the series for sin x gives (t2 )5 (−1)k (t2 )2k+1 (t2 )3 + + ··· + + ··· 3!

5!

(2k + 1)!

sin(t2 ) = t2 −

(−1)k t4k+2 t6 t10 + + ··· + + ··· 3!

5!

(2k + 1)!

= t2 − Therefore 2

3

t sin(t ) − t =

(−1)k t4k+3 t7 t11 t − + + ··· + + ··· 3!

5!

(2k + 1)!

=−

3

(−1)k t4k+3 t11 t7 + + ··· + + ··· 3!

5!

(2k + 1)!

− t3

for k ≥ 1.

15. Using the Binomial theorem: 1 √ 1−x

= (1 − x)−1/2

= 1+ −

n

1

(−1/2)(−3/2) · · · (− 2 − n + 1)(−x) (−1/2)(−3/2)(−x)2 1 (−x) + + ··· + + · · · for n ≥ 1.

2 2! n!

Substituting y 2 for x:

p

1 1 − y2

= (1 − y 2 )−1/2 1

(1/2)(3/2) · · · ( 2 + n − 1)y 1 2 3 4 y + y + ··· + 2 8 n!

= 1+

16.

1 1 1 x = = 1+ 2+x 2(1 + x2 ) 2 2

1 2

=

1−

x + 2

2 x 2

−

−1

3 x 2

+ ···

2n

+ · · · for n ≥ 1.

17. Using the binomial expansion for (1 + x)1/2 with x = h/T :

1/2 1/2 √

√ h T h 1/2 = T 1+ = T 1+ T +h = T + h T T T

(1/2)(−1/2) 2 (1/2)(−1/2)(−3/2) 3 √ h h h = T 1 + (1/2) + + ··· T 2!

T 3!

T =

√ T

1+

1 2

h T

−

1 8

2 h T

+

1 16

3

18. Using the binomial expansion for (1 + x)−1 with x = −r/a: 1 1 = a−r a−a 1 = a = =

1 a 1 a

r a

=

1+

r a

r a

··· .

−1

1 1 r = 1+ − a a a 1 − ar

r 1 + (−1) − a 1− −

+

r a

2 r a

(−1)(−2) r + − 2! a

+ −

2

+

− −

3 r a

r a

h T

3

2

+ ···

+ ··· .

(−1)(−2)(−3) r − + 3! a

3

+ ···

10.3 SOLUTIONS

803

19. Using the binomial expansion for (1 + x)−2 with x = r/a: 1 1 1 1 = 2 = 2 = 2 r r (a + r)2 a a+a a a 1+ a =

1 a2

=

1 a2

1 + (−2) 1−2

r a

(−2)(−3) 2!

r a

+

+3

2 r a

−4

2 r a

3 r a

−2 r a

1+ +

(−2)(−3)(−4) 3!

3 r a

+ ···

+ ··· .

20.

1/3 √

1/3 √ t 1/3 t t 3 3 P +t = P +P = P 1+ = P 1+ P P P

(1/3)(−2/3) 2 (1/3)(−2/3)(−5/3) 3 √ t t t 3 = P 1 + (1/3) + + ··· P 2!

P 3!

P =

√ 3 P

1+

1 3

t P

−

1 9

2 t P

+

5 81

3

··· .

− 21

t P

21. a a √ = = 2 1 a2 + x 2 a(1 + xa2 ) 2

= 1+ −

1 2

1 1 + − 3!

2 = 1−

1 2

x2 1 1 + − a2 2!

2

2 x a

x2 1+ 2 a

3 − 2

+

3 8

5 − 2

4 x a

−

3 2

2 2

2 3 x a2

−

5 16

x a2

+ ···

6 x a

+ ···

Problems 22. (a) Since the Taylor series for ex and e−x are given by x2 x3 x4 + + + ··· 2!

3!

4! x2 x3 x4 e−x = 1 − x + − + + ···, 2!

3!

4! ex = 1 + x +

we have

x3 x4 x4 x2 +0 +2 + · · · = 2 + x2 + + ···.

2!

3!

4!

12 x −x (b) For x near 0, we can approximate e +e by its second degree Taylor polynomial, P2 (x), whose graph is a parabola: ex + e−x = 2 + 0x + 2

ex + e−x ≈ P2 (x) = 2 + x2 .

23. (a) Since the Taylor series for ex and e−x are given by x2 x3 x4 x5 + + + + ··· 2!

3!

4!

5! x2 x3 x4 x5 e−x = 1 − x + − + − + ···, 2!

3!

4!

5! ex = 1 + x +

we have ex − e−x = 0 + 2x + 0

x2 x3 x4 x5 x3 x5 +2 +0 + 2 · · · = 2x + + + ···.

2!

3!

4!

5!

3 60

804

Chapter Ten /SOLUTIONS

(b) For x near 0, we can approximate ex − e−x by its third degree Taylor polynomial, P3 (x): ex − e−x ≈ P3 (x) = 2x +

x3 .

3

The function P3 (x) is a cubic polynomial whose graph is symmetric about the origin.

24. Notice that

P

pxp−1 , is the derivative, term-by-term, of a geometric series: ∞ X p=1

pxp−1 = 1 · x0 + 2 · x1 + 3 · x2 + · · · =

d (x + x2 + x3 + · · ·). {z } dx | Geometric series

For |x| < 1, the sum of the geometric series with first term x and common ratio x is x + x2 + x3 + · · · = Differentiating gives

∞ X

pxp−1 =

p=1

d dx

x 1−x

=

25. From the series for ln(1 + y), we get

ln(1 + y) = y −

y6 y8 y4 + − + ··· 2 3 4

sin y = y −

y5 y7 y3 + − + ··· 3!

5!

7!

sin y 2 = y 2 −

y 10 y 14 y6 + − + ··· 3!

5!

7!

cos y = 1 −

y2 y4 y6 + − + ··· 2!

4!

6!

So The Taylor series for cos y is

So

1(1 − x) − x(−1) 1 = .

(1 − x)2 (1 − x)2

y2 y3 y4 + − + ···, 2 3 4

ln(1 + y 2 ) = y 2 − The Taylor series for sin y is

x .

1−x

y2 y4 y6 − + + ··· 2!

4!

6!

Near y = 0, we can drop terms beyond the fourth degree in each expression: 1 − cos y =

ln(1 + y 2 ) ≈ y 2 −

y4 2

sin y 2 ≈ y 2 y2 y4 1 − cos y ≈ − .

2!

4!

(Note: These functions are all even, so what holds for negative y will hold for positive y.)

Clearly 1 − cos y is smallest, because the y 2 term has a factor of 12 . Thus, for small y, y2 y4 y4 − < y2 − < y2 2!

4!

2 so

1 − cos y < ln(1 + y 2 ) < sin(y 2 ).

10.3 SOLUTIONS

26. The Taylor series about 0 for y =

1 is 1 − x2

805

y = 1 + x 2 + x4 + x6 + · · · .

The series for y = (1 + x)1/4 is, using the binomial expansion, y =1+ The series for y =

q

1+

1 3 x2 + − 2!

4 4

·

−

1 3 1 x+ − 4 4 4

x2 1 1 − + 8 2 2

−

7 4

x3 + ···.

3!

x x = (1 + )1/2 is, again using the binomial expansion, 2 2 y =1+

1 x 1 1 − · + 2 2 2 2

1 Similarly for y = √ = (1 − x)−(1/2) , 1−x

y =1+ −

1 1 (−x) + − 2 2

3 2

Near 0, let’s truncate these series after their x2 terms:

·

x2 1 + − 2!

2

−

−

3 2

3 2

·

x3 + ···.

48

−

5 2

·

−x3 + ···.

3!

1 ≈ 1 + x2 , 1 − x2 3 2 1 x , (1 + x)1/4 ≈ 1 + x − 4 32 q 1 1 2 x x , 1+ ≈ 1+ x− 2 4 32 1 3 1 √ ≈ 1 + x + x2 .

2 8 1−x

II.

1 Thus 1−x 2 looks like a parabola opening upward near the origin, with y-axis as the axis of symmetry, so (a) = I. 1 Now √1−x has the largest positive slope ( 21 ), and is concave up (because the coefficient of x2 is positive). So (d) = 1

The last two both have positive slope ( 41 ) and are concave down. Since (1 + x) 4 has the smallest second derivative (i.e., the most negative coefficient of x2 ), (b) = IV and therefore (c) = III. 27. Since it does not depend on n, we can factor out e−k , giving ∞ X k n−1 n=1

(n − 1)!

e−k = e−k

∞ X k n−1 n=1

= e−k

|

(n − 1)!

k k2 k3 1 + + + + ··· 0!

1!

2!

3!

{z

This is the series for ek

= e−k · ek

}

= 1.

28. (a) The Taylor approximation to f (x) = cosh x about x = 0 is of the form cosh x ≈ cosh(0) + f 0 (0)x +

f 00 (0)x2 f (n) (0)xn + ... + .

2! n!

We have the following results: f (x) = cosh x so f (0) = 1, f 0 (x) = sinh x f 00 (x) =

so f 0 (0) = 0,

d (sinh x) = cosh x so f 00 (0) = 1, dx f 000 (x) = sinh x so f 000 (0) = 0.

806

Chapter Ten /SOLUTIONS

The derivatives continue to alternate between cosh x and sinh x, so their values at 0 continue to alternate between 0 and 1. Therefore x2 x3 x4 cosh x ≈ 1 + 0 · x + 1 · +0· +1· + ···, 2!

3!

4! so the degree 8 Taylor approximation is given by cosh x ≈ 1 +

x4 x6 x8 x2 + + + .

2!

4!

6!

8!

(b) We use the polynomial obtained from part (a) to estimate cosh 1, cosh 1 ≈ 1 +

1 1 1 1 + + + = 1.543080357.

2!

4!

6!

8!

Compared to the actual value of cosh 1 = 1.543080635 . . ., the error is less than 10 −6 . d (c) Since (cosh x) = sinh x, we have dx sinh x ≈

d dx

1+

x2 x4 x6 x8 + + + 2!

4!

6!

8!

2x 4x3 6x5 8x7 + + + 2!

4!

6!

8! x3 x5 x7 = x+ + + .

3!

5!

7!

=

29. Since ex =

∞ X xn n=0

n!

and sinh 2x = (e2x − e−2x )/2, the Taylor expansion for sinh 2x is ∞ X (2x)n

1 sinh 2x = 2

n=0

n!

−

∞ X (−2x)n n=0

n!

!

1 = 2 =

∞ X n=0

(2x)n (1 − (−1) ) n! n

∞ X (2x)2m+1

m=0

(2m + 1)!

!

.

Since cosh 2x = (e2x + e−2x )/2, we have 1 cosh 2x = 2

∞ X (2x)n n=0

n!

+

∞ X (−2x)n n=0

n!

!

1 = 2 =

∞ X n=0

(2x)n (1 + (−1) ) n!

∞ X (2x)2m

m=0

(2m)!

n

!

.

30. (a) The Taylor series for 1/(1 − x) = 1 + x + x2 + x3 + . . ., so 1 1 = = 1 + (0.02) + (0.02)2 + (0.02)3 + . . .

0.98 1 − 0.02 = 1.020408 . . . (b) Since d/dx(1/(1 − x)) = (1/(1 − x))2 , the Taylor series for 1/(1 − x)2 is d (1 + x + x2 + x3 + . . .) = 1 + 2x + 3x2 + 4x3 + · · · dx Thus 1 1 = = 1 + 2(0.01) + 3(0.0001) + 4(0.000001) + · · · (0.99)2 (1 − 0.01)2 = 1.0203040506 . . .

10.3 SOLUTIONS

f (x) = (1 + ax)(1 + bx)−1 = (1 + ax) 1 − bx + (bx)2 − (bx)3 + · · ·

31. (a)

= 1 + (a − b)x + (b2 − ab)x2 + · · ·

807

2

(b) ex = 1 + x + x2 + · · · Equating coefficients: a − b = 1, 1 b − ab = .

2 2

Solving gives a = 12 , b = − 12 .

32. (a) If φ = 0,

left side = b(1 + 1 + 1) = 3b ≈ 0 so the equation is almost satisfied and there could be a solution near φ = 0. (b) We have φ5 φ3 + − ··· 3!

5! φ2 φ4 cos φ = 1 − + − ··· 2!

4! sin φ = φ −

So cos2 φ =

1−

φ2 φ4 + − ··· 2!

4!

Neglecting terms of order φ2 and higher, we get

1−

φ2 φ4 + − ··· .

2!

4!

sin φ ≈ φ

cos φ ≈ 1

cos2 φ ≈ 1.

So φ + b(1 + 1 + 1) ≈ 0, whence φ ≈ −3b. mM 33. (a) µ = . m+M (b)

mM = m.

M

If M >> m, then the denominator m + M ≈ M , so µ ≈

M µ=m m+M

=m

1 M M M m + M M

=m

m + M

2

3

m We can use the binomial expansion since M < 1.

µ=m 1−

m M

−

m M

1 m 1+ M

+ ···

1 1 m ≈ 1836 ≈ 0.000545.

M, then M 1836 So a first order approximation to µ would give µ = m(1 − 0.000545). The percentage difference from µ = m is −0.0545%.

(c) If m ≈

34. (a) Solving for ω0 gives

1 2 =0 ω0 C 1 ω0 L − =0 ω0 C

ω0 L −

1 ω0 C 1 ω02 = LC

ω0 L =

ω0 =

r

1 .

LC

(Note: We discarded the negative root, because we need positive frequencies.)

808

Chapter Ten /SOLUTIONS

(b) Set ω = ω0 + ∆ω and get

1 ωL − ωC

2

1 (ω0 + ∆ω)L − (ω0 + ∆ω)C

=

We need to find a Taylor expansion for the term

.

1 . Using the binomial expansion we have C(ω0 + ∆ω)

∆ω −1 1 1 1+ = C(ω0 + ∆ω) Cω0 ω0

∆ω 1 1− = + ··· Cω0 ω0

So our expression becomes:

2

ωL −

1 ωC

2

= ω0 L + L∆ω − =

r

r

1 Cω0

1−

1 1 · L + L∆ω − LC C

L + L∆ω − = C = 2L∆ω − · · · .

So

r

r

∆ω + ··· ω0

LC ∆ω LC + · 1 C 1

L + L∆ω − · · · C

1 2 ≈ 4L2 (∆ω)2 . ωC Notice that if ∆ω = 0 in the above expression, we get 0, which is what we expected, since in this case ω = ω 0 . ωL −

35. (a) Factoring the expression for t1 − t2 , we get 2l2 2l1 2l1 2l2 + − p − p c(1 − v 2 /c2 ) c(1 − v 2 /c2 ) c 1 − v 2 /c2 c 1 − v 2 /c2

∆t = t1 − t2 =

2(l1 + l2 ) 2(l1 + l2 ) − p c(1 − v 2 /c2 ) c 1 − v 2 /c2

=

2(l1 + l2 ) = c

1 1 −p 1 − v 2 /c2 1 − v 2 /c2

!

.

Expanding the two terms within the parentheses in terms of v 2 /c2 gives

v2 1− 2 c

−1

(−1)(−2) v2 = 1+ 2 + c 2! = 1+

v2 1− 2 c

−1/2

−1 2

2!

−3 2

2

(−1)(−2)(−3) + 3!

−v 2 c2

2

−v 2 c2

v2 v4 v6 + 4 + 6 + ··· 2 c c c

1 v2 + = 1+ 2 c2 = 1+

3 v4 5 v6 1 v2 + + + ··· 2 c2 8 c4 16 c6

+

−1 2

−3 2

3!

−v 2 c2

−5 2

3

+ ···

−v 2 c2

3

+ ···

Thus, we have ∆t =

2(l1 + l2 ) c

=

2(l1 + l2 ) c

∆t ≈

(l1 + l2 ) c

1+

v2 v4 v6 1 v2 3 v4 5 v6 + 4 + 6 + ··· − 1 − − − − ··· 2 2 4 c c c 2c 8c 16 c6

1 v2 5 v4 11 v 6 + + + ··· 2 4 2c 8c 16 c6

2

v 5 v4 + 2 c 4 c4

.

10.3 SOLUTIONS

809

(b) For small v. we can neglect all but the first nonzero term, so ∆t ≈

(l1 + l2 ) 2 (l1 + l2 ) v 2 · 2 = v . c c c3

Thus, ∆t is proportional to v 2 with constant of proportionality (l1 + l2 )/c3 .

2

36. (a) Since the expression under the square root sign, 1 − vc2 must be positive in order to give a real value of m, we have v2 >0 c2 2 v <1 c2 2 v < c2 ,

1−

so

− c < v < c.

In other words, the object can never travel faster that the speed of light. (b)

m

m0

(c) Notice that m = m0

1−

v2 c2

binomial expansion to get:

−1/2

v

c

−c 2

. If we substitute u = − vc2 , we get m = m0 (1 + u)−1/2 and we can use the

m = m0 = m0

1−

(−1/2)(−3/2) 2 1 u+ u + ··· 2 2!

1+

1 v2 3 v4 + + ··· .

2 2c 8 c4

(d) We would expect this series to converge only for values of the original function that exist, namely when |v| < c.

37. (a) To find when V takes on its minimum values, set dV = 0. So dr −V0

d dr

2

r0 r

6

−

r0 r

12

=0

−V0 −12r06 r−7 + 12r012 r−13 = 0

12r06 r−7 = 12r012 r−13 r06 = r6

6

12

r = r0 .

r0 12r06 V0 1− , we see that V 0 (r) > 0 for r > r0 and V 0 (r) < 0 for r < r0 . Thus, r7 r V = −V0 (2(1)6 − (1)12 ) = −V0 is a minimum.

(Note: We discard the negative root −r0 since the distance r must be positive.)

Rewriting V 0 (r) as

(b)

V (r) = −V0

2

r0 r

6

−

r0 r

V 0 (r) = −V0 (−12r06 r−7 + 12r012 r−13 )

V

00

(r) = −V0 (84r06 r−8 − 156r012 r−14 )

V (r0 ) = −V0

V 0 (r0 ) = 0

V 00 (r0 ) = 72V0 r0−2

810

Chapter Ten /SOLUTIONS

The Taylor series is thus: V (r) = −V0 + 72V0 r0−2 · (r − r0 )2 · (c) The difference between V and its minimum value −V0 is V − (−V0 ) = 36V0

1 + ··· 2

(r − r0 )2 + ··· r02

which is approximately proportional to (r − r0 )2 since terms containing higher powers of (r − r0 ) have relatively small values for r near r0 . (d) From part (a) we know that dV /dr = 0 when r = r0 , hence F = 0 when r = r0 . Since, if we discard powers of (r − r0 ) higher than the second,

(r − r0 )2 V (r) ≈ −V0 1 − 36 r02 giving

r − r0 r − r0 dV ≈ 72 · (−V0 ) = −72V0 . dr r02 ro2 So F is approximately proportional to (r − r0 ).

F =−

38. Solving for P gives

P = Expanding P in terms of 1/V gives

n2 a nRT − 2.

V − nb V

nb −1 n2 a nRT 1− − 2 V V V

n2 a nb nRT + ··· − 2 1+ P = V V V 2 2 nRT n bRT n a P = + − 2 V V2 V

2 1 1 2 P = nRT + n (bRT − a) + ···.

V V

P =

Solutions for Section 10.4 Exercises 1. The error bound in approximating e01 using the Taylor polynomial of degree 3 for f (x) = ex about x = 0 is: |E3 | = |f (0.1) − P3 (0.1)| ≤

M (0.1)4 M · |0.1 − 0|4 = , 4!

24

where |f (4) (x)| ≤ M for 0 ≤ x ≤ 0.1. Now, f (4) (x) = ex . Since ex is increasing for all x, we see that |f (4) (x)| is maximized for x between 0 and 0.1 when x = 0.1. Thus, |f (4) | ≤ e0.1 , so The Taylor polynomial of degree 3 is

|E3 | ≤

e0.1 · (0.1)4 = 0.00000460.

24

P3 (x) = 1 + x + The approximation is P3 (0.1), so the actual error is

1 2 1 x + x3 .

2!

3!

E3 = e0.1 − P3 (0.1) = 1.10517092 − 1.10516667 = 0.00000425, which is slightly less than the bound.

10.4 SOLUTIONS

811

2. The error bound in approximating sin(0.2) using the Taylor polynomial of degree 3 for f (x) = sin x about x = 0 is: |E3 | = |f (0.2) − P3 (0.2)| ≤

M · |0.2 − 0|4 M (0.2)4 = , 4!

24

where |f (4) (x)| ≤ M for 0 ≤ x ≤ 0.2. Now, f (4) (x) = sin x. By looking at the graph of sin x, we see that |f (4) (x)| is maximized for x between 0 and 0.2 when x = 0.2. Thus, |f (4) | ≤ sin(0.2), so The Taylor polynomial of degree 3 is

|E3 | ≤

sin(0.2) · (0.2)4 = 0.0000132.

24 P3 (x) = x −

1 3 x .

3!

The approximation is P3 (0.1), so the actual error is

E3 = sin(0.2) − P3 (0.2) = 0.19866933 − 0.19866667 = 0.00000266, which is much less than the bound.

3. The error bound in approximating cos(−0.3) using the Taylor polynomial of degree 3 for f (x) = cos x about x = 0 is: |E3 | = |f (−0.3) − P3 (−0.3)| ≤

M · | − 0.3 − 0|4 M (−0.3)4 = , 4!

24

where |f (4) (x)| ≤ M for 0 ≥ x ≥ −0.3. Now, f (4) (x) = cos x, which has its largest value of 1 at x = 0. Thus, |f (4) | ≤ cos 0 = 1, so The Taylor polynomial of degree 3 is

|E3 | ≤

1 · (−0.3)4 = 0.000338.

24

P3 (x) = 1 −

1 2 x .

2!

The approximation is P3 (−0.3), so the actual error is

E3 = cos(−0.3) − P (−0.3) = 0.955336 − 0.955000 = 0.000336, which is slightly less than the bound. √ √ 0.9 using the Taylor polynomial of degree three for f (x) = 1 + x about x = 0 is:

4. The error bound in approximating

|E3 | = |f (−0.1) − P3 (−0.1)| ≤

M · (−0.1)4 M · | − 0.1 − 0|4 = , 4!

24

where |f (4) | ≤ M for 0 ≥ x ≥ −0.1. Since f (4) (x) = −

15 (1 + x)−7/2 , 16

we see that if x is between 0 and −0.1, the maximum is at x = −0.1. Thus |f (4) x)| ≤ (15/16)(1 − 0.1)−7/2 , so |E3 | ≤

(−0.1)4 15 (1 − 0.1)−7/2 · = 0.00000565.

16 24

The Taylor polynomial of degree 3 is 1 1 1 x2 1 1 3 x3 x + (− ) + (− )(− ) 2 2 2 2!

2 2 2 3!

1 2 1 3 1 x . = 1+ x− x + 2 8 16

P3 (x) = 1 +

The approximation is P3 (−0.1), so the actual error is √ E3 = 0.9 − P3 (−0.1) = 0.94868330 − 0.94868750 = −0.00000420, which is slightly less, in absolute value, than the bound.

812

Chapter Ten /SOLUTIONS

5. The error bound in approximating ln(1.5) using the Taylor polynomial of degree 3 for f (x) = ln(1 + x) about x = 0 is: |E4 | = |f (0.5) − P3 (0.5)| ≤

M (0.5)4 M · |0.5 − 0|4 = , 4!

24

where |f (4) (x)| ≤ M for 0 ≤ x ≤ 0.5. Since f (4) (x) =

−3!

(1 + x)4

and the denominator attains its minimum when x = 0, we have |f (4) (x)| ≤ 3!, so |E4 | ≤

3! (0.5)4 = 0.0156.

24

The Taylor polynomial of degree 3 is P3 (x) = 0 + x + (−1) = x−

x3 x2 + (−1)(−2) 2!

3!

1 2 1 3 x + x .

2 3

The approximation is P3 (0.5), so the actual error is E3 = ln(1.5) − P3 (0.5) = 0.4055 − 0.4167 = −0.0112 which is slightly less, in absolute value, than the bound. √ 6. The error bound in approximating 1/ 3 using the Taylor polynomial of degree three for f (x) = (1 + x)−1/2 about x = 0 is: M · |2 − 0|4 M · 24 |E3 | = |f (2) − P3 (2)| ≤ = , 4!

24 where |f (4) | ≤ M for 0 ≤ x ≤ 2. Since 105 f (4) (x) = (1 + x)−9/2 , 16 we see that if x is between 0 and 2, then |f (4) x)| ≤ 105 . Thus, 16 |E3 | ≤

105 105 24 · = = 4.375.

16 24 24

This is not a very helpful bound on the error, but that is to be expected as the Taylor series does not converge at x = 2.

(At x = 2, we are outside the interval of convergence.)

The Taylor polynomial of degree 3 is 1 1 3 x2 1 3 5 x3 P3 (x) = 1 + ( )x + (− )(− ) + (− )(− )(− ) 2 2 2 2!

2 2 2 3!

1 3 5 3 = 1 − x + x2 − x .

2 8 16 The approximation is P3 (2), so the actual error is 1 E3 = √ − P3 (2) = 0.577 − (−1.0) = 1.577, 3 which is much less than the bound, but still very large.

7. The error bound in approximating tan(1) using the Taylor polynomial of degree three for f (x) = tan x about x = 0 is: |E3 | = |f (1) − P3 (x)| ≤

M · |1 − 0|4 M = 4!

24

where |f (4) (x)| ≤ M for 0 ≤ x ≤ 1. Now, f (4) (x) =

16 sin x 24 sin3 x + . cos3 x cos5 x

10.4 SOLUTIONS

813

From a graph of f (4) (x), we see that f (4) (x) is increasing for x between 0 and 1. Thus, |f (4) (x)| ≤ |f (4) (1)| = 396, so

396 = 16.5.

24 This is not a very helpful error bound! The reason the error bound is so huge is that x = 1 is getting near the vertical asymptote of the tangent graph, and the fourth derivative is enormous there.

Observing that any term in f 00 (x) or f 000 (x) involving tan x is zero at x = 0, we calculate the Taylor polynomial of degree 3: |E3 | ≤

P3 (x) = 0 + x + (0) = x+

x3 x2 + (2) 2!

3!

1 3 x .

3

The approximation is P3 (1), so the actual error is E3 = tan 1 − P3 (1) = 1.557 − 1.333 = 0.224, which is much less than the bound.

8. The error bound in approximating 0.51/3 using the Taylor polynomial of degree 3 for f (x) = (1 − x)1/3 about x = 0 is: |E3 | = |f (0.5) − P3 (0.5)| ≤

M · |0.5 − 0|4 M (0.5)4 = , 4!

24

where |f (4) (x)| ≤ M for 0 ≤ x ≤ 0.5. Now, f (4) (x) = −

80 (1 − x)−11/3 .

81

By looking at the graph of (1 − x)−11/3 , we see that |f (4) (x)| is maximized for x between 0 and 0.5 when x = 0.5.

Thus,

80 11/3 80 1 −11/3 = ·2 , |f (4) | ≤ 81 2 81 so 80 · 211/3 · (0.5)4 |E3 | ≤ = 0.033.

81 · 24 The Taylor polynomial of degree 3 is 1 −2 (−x)2 1 −2 −5 (−x)3 1 (−x) + + 3 3 3 2!

3 3 3 3!

5 3 1 2 1 x . = 1− x− x − 3 9 81

P3 (x) = 1 +

The approximation is P3 (0.5), so the actual error is E3 = (0.5)1/3 − P3 (0.5) = 0.79370 − 0.79784 = −0.00414, which is much less than the bound.

Problems √ √ 9. Let f (x) = 1 + x. We use a Taylor polynomial with x = 1 to approximate 2. The error bound for the Taylor √ approximation of degree three for f (x) = 2 about x = 0 is: |E3 | = |f (1) − P3 (1)| ≤ where |f (4) (x)| ≤ M for 0 ≤ x ≤ 1.

Now, f (4) (x) = −

M · |1 − 0|4 M = , 4!

24

−15 15 (1 + x)−7/2 = .

16 16(1 + x)7/2

814

Chapter Ten /SOLUTIONS

Since 1 ≤ (1 + x)7/2 for x between 0 and 1, we see that 15

|f (4) (x)| = for x between 0 and 1. Thus,

16(1 + x)

≤

7 2

15 16

15 < 0.039 16 · 24

|E3 | ≤

10. (a) θ is the first degree approximation of f (θ) = sin θ; it is also the second degree approximation, since the next term in the Taylor expansion is 0.

P1 (θ) = θ is an overestimate for 0 < θ ≤ 1, and is an underestimate for −1 ≤ θ < 0. (This can be seen easily from a graph.) (b) Using the second degree error bound, if |f (3) (θ)| ≤ M for −1 ≤ θ ≤ 1, then |E2 | ≤

M · |θ|3 M ≤ .

3!

6

For what value of M is |f (3) (θ)| ≤ M for −1 ≤ θ ≤ 1? Well, |f (3) (θ)| = | − cos θ| ≤ 1. So |E2 | ≤ 16 = 0.17.

θ3 is the third degree Taylor approximation of f (θ) = sin θ; it is also the fourth degree approximation, since the 11. (a) θ − 3! next term in the Taylor expansion is 0.

P3 (θ) is an underestimate for 0 < θ ≤ 1, and is an overestimate for −1 ≤ θ < 0. (This can be checked with a calculator.) (b) Using the fourth degree error bound, if |f (5) (θ)| ≤ M for −1 ≤ θ ≤ 1, then |E4 | ≤

M · |θ|5 M ≤ .

5!

120

For what value of M is |f (5) (θ)| ≤ M for −1 ≤ θ ≤ 1? Since f (5) (θ) = cos θ and | cos θ| ≤ 1, we have |E4 | ≤

1 ≤ 0.0084.

120

12. (a) The Taylor polynomial of degree 0 about t = 0 for f (t) = e t is simply P0 (x) = 1. Since et ≥ 1 on [0, 0.5], the approximation is an underestimate. (b) Using the zero degree error bound, if |f 0 (t)| ≤ M for 0 ≤ t ≤ 0.5, then |E0 | ≤ M · |t| ≤ M (0.5).

Since |f 0 (t)| = |et | = et is increasing on [0, 0.5], |f 0 (t)| ≤ e0.5 <

√

4 = 2.

Therefore |E0 | ≤ (2)(0.5) = 1.

th

(Note: By looking at a graph of f (t) and its 0 degree approximation, it is easy to see that the greatest error occurs when t = 0.5, and the error is e0.5 − 1 ≈ 0.65 < 1. So our error bound works.)

13. (a) The second-degree Taylor polynomial for f (t) = et is P2 (t) = 1 + t + t2 /2. Since the full expansion of et = 1 + t + t2 /2 + t3 /6 + t4 /24 + · · · is clearly larger than P2 (t) for t > 0, P2 (t) is an underestimate on [0, 0.5]. (b) Using the second-degree error bound, if |f (3) (t)| ≤ M for 0 ≤ t ≤ 0.5, then |E2 | ≤

M (0.5)3 M · |t|3 ≤ .

3!

6

Since |f (3) (t)| = et , and et is increasing on [0, 0.5], f (3) (t) ≤ e0.5 < So |E2 | ≤

√

4 = 2.

(2)(0.5)3 < 0.047.

6

10.4 SOLUTIONS

14. (a)

815

(i) The vertical distance between the graph of y = cos x and y = P 10 (x) at x = 6 is no more than 4, so |Error in P10 (6)| ≤ 4.

Since at x = 6 the cos x and P20 (x) graphs are indistinguishable in this figure, the error must be less than the smallest division we can see, which is about 0.2 so, |Error in P20 (6)| ≤ 0.2. (ii) The maximum error occurs at the ends of the interval, that is, at x = −9, x = 9. At x = 9, the graphs of y = cos x and y = P20 (x) are no more than 1 apart, so Maximum error in P20 (x) ≤ 1. for −9 ≤ x ≤ 9

(b) We are looking for the largest x-interval on which the graphs of y = cos x and y = P 10 (x) are indistinguishable.

This is hard to estimate accurately from the figure, though −4 ≤ x ≤ 4 certainly satisfies this condition.

15. The maximum possible error for the nth degree Taylor polynomial about x = 0 approximating cos x is |En | ≤ M ·|x−0|n+1 , where | cos(n+1) x| ≤ M for 0 ≤ x ≤ 1. Now the derivatives of cos x are simply cos x, sin x, − cos x, and (n+1)! n+1

1 − sin x. The largest magnitude these ever take is 1, so | cos(n+1) (x)| ≤ 1, and thus |En | ≤ |x| ≤ (n+1)! . The same (n+1)! argument works for sin x.

1 16. By the results of Problem 15, if we approximate cos 1 using the n th degree polynomial, the error is at most (n+1)! .

For the answer to be correct to four decimal places, the error must be less than 0.00005. Thus, the first n such that 1 1 1 < 0.00005 will work. In particular, when n = 7, 8! = 40370 < 0.00005, so the 7th degree Taylor polynomial (n+1)!

1 will give the desired result. For six decimal places, we need (n+1)! < 0.0000005. Since n = 9 works, the 9th degree Taylor polynomial is sufficient.

17. (a) Table 10.1 E1 = sin x − x

Table 10.2 E1 = sin x − x

x

sin x

E

x

sin x

−0.5

−0.4794

0.0206

0

0

−0.3

−0.2955 −0.0998

−0.4 −0.2 −0.1

−0.3894

0.0106

0.1

0.0998

0.0045

0.2

0.1987

−0.1987

0.0013

0.3

0.2955

0.0002

0.4

0.3894

0.5

0.4794

(b) See answer to part (a) above. (c)

E 0 −0.0002 −0.0013 −0.0045 −0.0106 −0.0206

0.03 E1 x −0.5

0.5 −0.03

The fact that the graph of E1 lies between the horizontal lines at ±0.03 shows that |E1 | < 0.03 for −0.5 ≤ x ≤ 0.5.

18. (a) See Figure 10.7. The graph of E1 looks like a parabola. Since the graph of E1 is sandwiched between the graph of y = x2 and the x axis, we have |E1 | ≤ x2 for |x| ≤ 0.1.

816

Chapter Ten /SOLUTIONS y 0.01

y y = x2

y = x3

E1 −0.1

E2 x 0.1

−0.1

x 0.1 −0.01

−0.001

Figure 10.7

Figure 10.8

(b) See Figure 10.8. The graph of E2 looks like a cubic, sandwiched between the graph of y = x3 and the x axis, so |E2 | ≤ x3

for |x| ≤ 0.1.

(c) Using the Taylor expansion

x3 x2 + + ··· 2!

3!

ex = 1 + x + we see that E1 = ex − (1 + x) =

x3 x4 x2 + + + ···.

2!

3!

4!

Thus for small x, the x2 /2! term dominates, so E1 ≈ and so E1 is approximately a quadratic.

Similarly E2 = ex − (1 + x +

x2 , 2!

x2 x3 x4 )= + + ···.

2 3!

4!

Thus for small x, the x3 /3! term dominates, so E2 ≈

x3 3!

and so E2 is approximately a cubic.

19.

0.01 −0.1

0.1

x

E0 −0.01

The graph of E0 looks like a parabola, and the graph shows |E0 | < 0.01

for |x| ≤ 0.1.

(In fact |E0 | < 0.005 on this interval.) Since x2 x4 x6 + − + ···, 2!

4!

6! x2 x4 x6 E0 = cos x − 1 = − + − + ···.

2!

4!

6!

cos x = 1 −

So, for small x, E0 ≈ − and therefore the graph of E0 is parabolic.

x2 , 2

10.4 SOLUTIONS

817

20. Since f (x) = ex , the (n + 1)st derivative f (n+1) (x) is also ex , no matter what n is. Now fix a number x and let M = ex , then |f (n+1) (t)| ≤ et ≤ ex on the interval 0 ≤ t ≤ x. (This works for x ≥ 0; if x < 0 then we can take M = 1.) The important observation is that for any x the same number M bounds all the higher derivatives f (n+1) (x).

By the error bound formula, we now have |En (x)| = |ex − Pn (x)| ≤

M |x|n+1 (n + 1)!

for every n.

To show that the errors go to zero, we must show that for a fixed x and a fixed number M , M |x|n+1 → 0 (n + 1)!

as n → ∞.

1 |x|n+1 → 0 (n + 1)!

as n → ∞.

Since M is fixed, we need only show that

21.

This was shown in the text on page 528. Therefore, the Taylor series 1 + x + x 2 /2! + · · · does converge to ex . x5 x3 + − ··· 3!

5!

Write the error in approximating sin x by the Taylor polynomial of degree n = 2k + 1 as E n so that sin x = x −

sin x = x −

x3 x5 x2k+1 + − · · · (−1)k + En .

3!

5!

(2k + 1)!

(Notice that (−1)k = 1 if k is even and (−1)k = −1 if k is odd.) We want to show that if x is fixed, En → 0 as k → ∞.

Since f (x) = sin x, all the derivatives of f (x) are ± sin x or ± cos x, so we have for all n and all x |f (n+1) (x)| ≤ 1.

Using the bound on the error given in the text on page 528, we see that |En | ≤

1 |x|2k+2 .

(2k + 2)!

By the argument in the text on page 528, we know that for all x, |x|2k+2 |x|n+1 = →0 (2k + 2)! (n + 1)!

as n = 2k + 1 → ∞.

Thus the Taylor series for sin x does converge to sin x for every x.

22. (a) Since 4 arctan 1 = π, we approximate π by approximating 4 arctan x by Taylor polynomials with x = 1. Let f (x) = 4 arctan x. We find the Taylor polynomial of f about x = 0. f (0) = 0 4 so 1 + x2 8x f 00 (x) = − (1 + x2 )2 f 0 (x) =

f 000 (x) = −

f 0 (0) = 4 so

f 00 (0) = 0

8 32x2 + 2 2 (1 + x ) (1 + x2 )3

so

f 000 (0) = −8.

Thus, the third degree Taylor polynomial for f is F3 (x) =

4x 8 4 − x3 = 4x − x3 .

1!

3!

3

In particular,

8 4 = ≈ 2.67.

3 3 Note: If you already have the Taylor series for 1/(1 + x2 ), the Taylor polynomial for arctan x can also be found by integration.

F3 (1) = 4 −

818

Chapter Ten /SOLUTIONS

(b) We now approximate π by looking at g(x) = 2 arcsin x about x = 0 and substituting x = 1. g(0) = 0 g 0 (x) = √ g 00 (x) = g 000 (x) =

2 1 − x2 2x

so

g 0 (0) = 2

so

g 00 (0) = 0

3

(1 − x2 ) 2 2

+

3 2

(1 − x2 )

6x2 5

(1 − x2 ) 2

g 000 (0) = 2.

so

Thus, the third degree Taylor polynomial for g is G3 (x) =

2x 2x3 1 + = 2x + x3 .

1!

3!

3

In particular,

7 G3 (1) = ≈ 2.33.

3 √ Note: If you already have the Taylor series for 1/ 1 − x2 , the Taylor polynomial for arcsin x can also be found by integration. (c) To estimate the maximum possible error, |E3 |, in the approximation using the arctangent, we need a bound on the fourth derivative of f (x) = arctan x on 0 ≤ x ≤ 1. Since f (4) (x) = −

192x3 96x + , (1 + x2 )4 (1 + x2 )3

now use a graphing calculator to see that the maximum value of |f (4) (x)|, on 0 ≤ x ≤ 1 is about 18.6. Thus, |E3 | ≤

18.6 ≈ 0.78.

4!

(Notice that π ≈ 3.14 is within 0.78 of 2.67.) (d) To estimate the maximum possible error, |En |, in an approximation using the arcsine, we need a bound on the derivatives of g(x) = arcsin x on 0 ≤ x ≤ 1. The derivatives of arcsin x contain terms of the form (1 − x 2 )−a , for some positive a. As x gets close to 1, the value of (1 − x2 )−a approaches ∞. Thus, we cannot get a bound on the derivatives of arcsin x, so the error formula does not give us a bound on |E n |.

Solutions for Section 10.5 Exercises 1. No, a Fourier series has terms of the form cos nx, not cosn x.

2. Not a Fourier series because terms are not of the form sin nx.

3. Yes. Terms are of the form sin nx and cos nx.

4. Yes. This is a Fourier series where the cos nx terms all have coefficients of zero.

5. a0 =

1 a1 = π

1 2π

Z 0

1 f (x) cos x dx = π −π

Z 0

1 2π

Z π

f (x) dx =

−π

Z π

=

−π

"

−π

−1 dx +

Z π 0

− cos x dx + 0

1 dx = 0

Z π

cos x dx

0

# π

1 − sin x + sin x π −π 0

= 0.

10.5 SOLUTIONS

Similarly, a2 and a3 are both 0.

Rπ (In fact, notice f (x) cos nx is an odd function, so −π f (x) cos nx = 0.)

1 b1 = π

Z π

1 f (x) sin x dx = π −π =

1 b2 = π

Z π

1 b3 = π

Z π

− sin x dx + 0

sin x dx

0

π

Z 0

−π

− sin 2x dx + 0

Z π

=

Z 0

sin 2x dx

0

−π

− sin 3x dx + 0

Z π

4 π

π

1 1 1 cos 2x + (− cos 2x) π 2 2 −π 0

1 f (x) sin 3x dx = π −π =

−π

Z π

1 cos x + (− cos x) π −π 0

1 f (x) sin 2x dx = π −π =

Z 0

π

1 1 1 + (− cos 3x) cos 3x π 3 3 −π 0

= 0.

sin 3x dx

0

819

=

4 .

3π

4 sin 3x.

Thus, F1 (x) = F2 (x) = π4 sin x and F3 (x) = π4 sin x + 3π

1

1

−π π

−π

x

−1

−1

F1 (x) = F2 (x) = π4 sin x

6. First, a0 =

1 2π

Z π

f (x) dx =

−π

1 2π

4 F3 (x) = π4 sin x + 3π sin 3x

Z 0

−π

−x dx +

To find the ai ’s, we use the integral table. For n ≥ 1, an =

1 π

Z π

x

π

f (x) cos(nx) dx =

−π

1 π

1 = π

Z 0

−π

Z π

x dx =

0

0

π

1 x2 x2 − + 2π 2 −π 2 0

−x cos(nx) dx +

Z π

x cos(nx) dx

0

=

x 1 − sin(nx) − 2 cos(nx) n n

+

x 1 sin(nx) + 2 cos(nx) n n

−

1 1 1 1 + 2 cos(−nπ) + 2 cos(nπ) − 2 n2 n n n

=

1 π

=

2 (cos nπ − 1) πn2

π .

2

0

−π

π 0

4 . To find the bi ’s, note that f (x) is even, so for n ≥ 1, f (x) sin(nx) is odd.

Thus, Z a1 = − π4 , a2 = 0, and a3 = − 9π π

Thus,

−π

4 cos 3x. f (x) sin(nx) = 0, so all the bi ’s are 0. F1 = F2 = π2 − π4 cos x, F3 = π2 − π4 cos x − 9π

820

Chapter Ten /SOLUTIONS π

π

−π

π

−π

x

π

x 4 cos 3x F3 (x) = π2 − π4 cos x − 9π

F1 (x) = F2 (x) = π2 − π4 cos x

7. The energy of the function f (x) is E= =

1 π

Z π

1 π

(f (x))2 dx =

−π

Z π

3

π

x2 dx =

−π

1 3 x 3π −π

2π 2 1 3 (π − (−π 3 )) = = π 2 = 6.57974.

3π 3π 3

4 . Therefore the energy in the From Problem 6, we know all the bi ’s are 0 and a0 = π2 , a1 = − π4 , a2 = 0 , a3 = − 9π constant term and first three harmonics is

A20 + A21 + A22 + A23 = 2a20 + a21 + a22 + a23 =2 which means that they contain 8. First, we find a0 .

2 π 4

+

6.57596 = 0.99942 ≈ 99.942% of the total energy.

6.57974 a0 =

1 2π

Z π

x2 dx =

−π

1 2π

To find an , n ≥ 1, we use the integral table (III-15 and III-16).

1 an = π

16 16 +0+ = 6.57596 π2 81π 2

Z π

3 π x 3

=

−π

π2 3

1 x2 2x 2 x cos nx dx = sin(nx) + 2 cos(nx) − 3 sin(nx) π n n n −π 2

2π 1 2π cos(nπ) + 2 cos(−nπ) = π n2 n 4 = 2 cos(nπ) n

π

−π

Again, cos(nπ) = (−1)n for all integers n, so an = (−1)n n42 . Note that bn =

1 π

Z π

x2 sin nx dx.

−π

x2 is an even function, and sin nx is odd, so x2 sin nx is odd. Thus We deduce that the nth Fourier polynomial for f (where n ≥ 1) is

i=1

In particular, we have the graphs in Figure 10.9.

−π

x2 sin nx dx = 0, and bn = 0 for all n.

4 π2 X (−1)i 2 cos(ix). + 3 i n

Fn (x) =

Rπ

10.5 SOLUTIONS

i

π2

F3 (x) f (x) = x2 F2 (x)

π

π

F1 (x)

x

Figure 10.9 9.

1 a0 = 2π

Z π

1 h(x) dx = 2π −π

Z π

x dx =

0

π 4

As in Problem 10, we use the integral table (III-15 and III-16) to find formulas for a n and bn . an =

1 π

Z π

1 π

h(x) cos(nx) dx =

−π

Z π

x cos nx dx =

0

1 π

1 = π =

1 n2 π

1 x sin(nx) + 2 cos(nx) n n 1 1 cos(nπ) − 2 n2 n

π 0

cos(nπ) − 1 .

Note that since cos(nπ) = (−1)n , an = 0 if n is even and an = − n22π if n is odd.

Z π

1 π

bn =

−π

1 π

=

h(x) cos(nx) dx =

1 = π =−

−

π − cos(nπ) n

1 cos(nπ) n

π X + 4 n

i=1

This can also be written as

where

2

x sin x dx

π 0

cos(iπ) − 1

· cos(ix) +

(−1)i+1 sin(ix) . i

[ n2 ] n π X (−1)i+1 sin(ix) X −2 Hn (x) = + + cos((2i − 1)x) 4 i (2i − 1)2 π i=1

n

1 i2 π

0

if n ≥ 1

We have that the nth Fourier polynomial for h (for n ≥ 1) is Hn (x) =

Z π

x 1 cos(nx) + 2 sin(nx) n n

1 (−1)n+1 n

=

1 π

i=1

denotes the biggest integer smaller than or equal to n2 . In particular, we have the graphs in Figure 10.10.

821

822

Chapter Ten /SOLUTIONS

h(x)

π

H3 (x)

H2 (x) H1 (x)

x

π

−π

Figure 10.10 10. To find the nth Fourier polynomial, we must come up with a general formula for an and bn . First, we find a0 .

1 a0 = 2π

Z π

Z π

π

Now we use the integral table (III-15 and III-16) to find an and bn for n ≥ 1.

1 an = π

Z π

1 x cos nx dx = π −π 1 π

=

1 x sin(nx) + 2 cos(nx) n n

bn =

1 π

x sin nx dx =

−π

1 π

1 = π

π

−π

1 1 cos(nπ) − 2 cos(−nπ) n2 n

(Note that since x cos nx is odd, we could have deduced that

Z π

1 x2 =0 x dx = 2π 2 −π −π

1 g(x) dx = 2π −π

Rπ

−

−π

=0

x cos nx = 0.)

1 x cos(nx) + 2 sin(nx) n n

π

π π − cos(nπ) − cos(−nπ) n n

2 cos(nπ) n Notice that cos(nπ) = (−1)n for all integers n, so bn = (−1)n+1 ( n2 ).

Thus the nth Fourier polynomial for g is

−π

=−

Gn (x) =

n X

(−1)i+1

i=1

In particular, we have the graphs in Figure 10.11.

2 sin(ix). i

g(x)

π

−π π

−π

Figure 10.11

G3 (x) G2 (x) G1 (x) x

10.5 SOLUTIONS

Problems 11. (a)

(i) The graph of y = sin x + 13 sin 3x is in Figure 10.12. y 1

y = sin x + 31 sin 3x

−π

−2π

2π

π

x

−1

Figure 10.12 (ii) The graph of y = sin x + 31 sin 3x + 51 sin 5x is in Figure 10.12. y

y = sin x + 31 sin 3x + 51 sin 5x

1 −π

−2π

2π

π

x

−1

Figure 10.13 (b) Following the pattern, we add the term 71 sin 7x to get Figure 10.14. y

y = sin x + 13 sin 3x + 51 sin 5x + 71 sin 7x

1

−2π

−π

2π

π

x

−1

Figure 10.14 (c) The equation is

f (x) =

 .. ..   . .      1 − 2π ≤ x < −π    −1

−π ≤x<0

.

π ≤ x < 2π .. .

 1      −1     ..

0≤x<π

The square wave function is not continuous at x = 0, ± π, ± 2π, . . .. See Figure 10.15. y 1

−3π −2π

π

−π −1

Figure 10.15

x 2π

3π

823

824

Chapter Ten /SOLUTIONS

12. (a) The graph of g(x) is 1

−π

g(x)

−π/2

x

π

π/2

First find the Fourier coefficients: a0 is the average value of g on [−π, π] so from the graph, it is clear that a0 =

1 1 (π × 1) = , 2π 2

or analytically, 1 a0 = 2π

Z π

1 g(x) dx = 2π −π

1 1 = (π) = , 2π 2

1 ak = π

Z π

Z π/2

cos kx dx =

=

2 sin

1 g(x) sin kx dx = π −π

1 =− kπ So,

kπ kπ − sin − sin 2 2

Z π

π/2

1 1 = x 2π −π/2 2π

1 dx =

−π/2

1 g(x) cos kx dx = π −π

1 = kπ

1 bk = π

Z π/2

kπ kπ cos − cos − 2 2

1 kπ

−π/2

kπ 2

1 sin kx kπ −π/2

,

π/2

sin kx dx = −

.=−

1 (0) = 0 kπ

1 cos kx kπ −π/2

π 2 1 2 sin = , π 2 π

1 2π a2 = 2 sin = 0, 2π 2

1 3π 2 a3 = 2 sin =− , 3π 2 3π a1 =

which gives

F3 (x) =

1 2 2 + cos x − cos 3x.

2 π 3π

1

−π

π/2

π π − − 2 2

π/2

−π/2

Z π/2

F3 (x) g(x)

π/2

π

x

(b) There are cosines instead of sines (but the energy spectrum remains the same).

10.5 SOLUTIONS

825

13. We have f (x) = x, 0 ≤ x < 1. Let t = 2πx − π. Notice that as x varies from 0 to 1, t varies from −π to π. Thus if we rewrite the function in terms of t, we can find the Fourier series in terms of t in the usual way. To do this, let g(t) = f (x) = x = t+π on −π ≤ t < π. We now find the fourth degree Fourier polynomial for g.

2π ao =

1 2π

Z π

g(t) dt =

−π

1 2π

Z π

t+π 1 dt = 2π (2π)2 −π

Notice, a0 is the average value of both f and g. For n ≥ 1, 1 π

an =

Z π

t+π 1 cos(nt)dt = 2 2π 2π −π

1 2π 2

=

= 0.

1 π

bn =

h

Z π

2

t+π 1 sin(nt) dt = 2 2π 2π −π

h

Z π

=

−π

1 2

i π

−π

(t sin(nt) + π sin(nt)) dt

−π

t 1 π 1 − cos(nt) + 2 sin(nt) − cos(nt) 2π 2 n n n

=

π

(t cos(nt) + π cos(nt))dt

−π

1 π t sin(nt) + 2 cos(nt) + sin(nt) n n n

Z π

t + πt 2

i π

−π

1 4π 2 2 = (− cos(πn)) = − cos(πn) = (−1)n+1 .

2π 2 n πn πn We get the integrals for an and bn using the integral table (formulas III-15 and III-16).

Thus, the Fourier polynomial of degree 4 for g is: G4 (t) =

1 2 1 2 1 + sin t − sin 2t + sin 3t − sin 4t.

2 π π 3π 2π

Now, since g(t) = f (x), the Fourier polynomial of degree 4 for f can be found by replacing t in terms of x again. Thus, F4 (x) =

2 1 2 1 1 + sin(2πx − π) − sin(4πx − 2π) + sin(6πx − 3π) − sin(8πx − 4π).

2 π π 3π 2π

Now, using the fact that sin(x − π) = − sin x and sin(x − 2π) = sin x, etc., we have: F4 (x) =

1 2 1 2 1 − sin(2πx) − sin(4πx) − sin(6πx) − sin(8πx).

2 π π 3π 2π

f (x) = x

1

F4 (x) x 1

14. Since the period is 2, we make the substitution t = πx − π. Thus, x = t+π . We find the Fourier coefficients. Notice π that all of the integrals are the same as in Problem 13 except for an extra factor of 2. Thus, a 0 = 1, an = 0, and 4 (−1)n+1 , so: bn = πn 4 2 4 1 G4 (t) = 1 + sin t − sin 2t + sin 3t − sin 4t. π π 3π π Again, we substitute back in to get a Fourier polynomial in terms of x: 2 4 sin(πx − π) − sin(2πx − 2π) π π 1 4 sin(3πx − 3π) − sin(4πx − 4π) + 3π π 4 2 4 1 = 1 − sin(πx) − sin(2πx) − sin(3πx) − sin(4πx). π π 3π π

F4 (x) = 1 +

826

Chapter Ten /SOLUTIONS 3 2

f (x) = x

1

F4 (x) x 1

2

−1

Notice in this case, the terms in our series are sin(nπx), not sin(2πnx), as in Problem 13. In general, the terms will be sin(n 2π x), where b is the period. b 15. The signal received on earth is in the form of a periodic function h(t), which can be expanded in a Fourier series h(t) = a0 + a1 cos t + a2 cos 2t + a3 cos 3t + · · · + b1 sin t + b2 sin 2t + b3 sin 3t + · · ·

If the periodic noise consists of only the second and higher harmonics of the Fourier series, then the original signal contributed the fundamental harmonic plus the constant term, i.e., + a1 cos t + b1 sin t =

a

0 |{z}

{z

|

constant term

}

A cos }t | {z

.

original signal

fundamental harmonic

In order to find A, we need to find a0 , a1 , and b1 . Looking at the graph of h(t), we see 1 (Area above the x-axis – Area below the x-axis) 2π h

i π π π π π 1 80 − 50 + 30 + 30 + 50 = 2π 2 4 4 4 4 h i π π 1 1 · 0, = 0 = 80 − 80 = 2π 2 2 2π

a0 = average value of h(t) =

a1 =

1 π

1 = π

Z π

Z −3π/4 −π

+

=

h(t) cos t dt

−π

"

Z π/4

−50 cos t dt +

80 cos t dt +

−π/4

Z −π/2

1 −50 sin t π −π

Z π/2

− 30 sin t

π/4

+80 sin t −π/4

−30 cos t dt + −π/4

π/4

π

− 50 sin t

√

√

3π/4

π/2

#

−30 cos t dt

0 cos t dt +

Z π

3π/4

−50 cos t dt

√ √ 2 2 2 2 1 − 0 − 30 − − (−1) + 80 − − −50 − π 2 2 2 2

√ √ 2 2 −30 1 − − 50 0 − 2 2 √ √ √ √ √ √ 1 = [25 2 + 15 2 − 30 + 40 2 + 40 2 − 30 + 15 2 + 25 2] π √ 1 = [160 2 − 60] = 52.93, π =

Z 3π/4

−π/2

π/2

− 30 sin t

−π/2

−3π/4

π/4

−3π/4

0 cos t dt +

Z −π/4