Bismillah hir Rehman nir Raheem Assalat o Wasalam o Allika Ya RasoolALLAH
Calculus (11
th
Ed. Text Book)
Thomas, Weir, Hass, Giordano
Ch Ch
Published By: Muhammad Hassan Riaz Yousufi
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
11
INFINITE SEQUENCES AND SERIES
You
Chapter
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OVERVIEW While everyone knows how to add together two numbers, or even several, how to add together infinitely many numbers is not so clear. In this chapter we study such questions, the subject of the theory of infinite series. Infinite series sometimes have a finite sum, as in 1 1 1 1 + + + + Á = 1. 2 4 8 16
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This sum is represented geometrically by the areas of the repeatedly halved unit square shown here. The areas of the small rectangles add together to give the area of the unit square, which they fill. Adding together more and more terms gets us closer and closer to the total.
1/8 1/16
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1/2
1/4
Mu
ham
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Other infinite series do not have a finite sum, as with 1 + 2 + 3 + 4 + 5 + Á.
The sum of the first few terms gets larger and larger as we add more and more terms. Taking enough terms makes these sums larger than any prechosen constant. With some infinite series, such as the harmonic series 1 +
1 1 1 1 1 + + + + + Á 5 2 3 4 6
it is not obvious whether a finite sum exists. It is unclear whether adding more and more terms gets us closer to some sum, or gives sums that grow without bound. As we develop the theory of infinite sequences and series, an important application gives a method of representing a differentiable function ƒ(x) as an infinite sum of powers of x. With this method we can extend our knowledge of how to evaluate, differentiate, and integrate polynomials to a class of functions much more general than polynomials. We also investigate a method of representing a function as an infinite sum of sine and cosine functions. This method will yield a powerful tool to study functions.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.1
i
Sequences A sequence is a list of numbers
HISTORICAL ESSAY
a1, a2 , a3 , Á , an , Á
Sequences and Series
747
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11.1
Sequences
You
in a given order. Each of a1, a2, a3 and so on represents a number. These are the terms of the sequence. For example the sequence 2, 4, 6, 8, 10, 12, Á , 2n, Á
has first term a1 = 2, second term a2 = 4 and nth term an = 2n. The integer n is called the index of an , and indicates where an occurs in the list. We can think of the sequence
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a1, a2 , a3 , Á , an , Á
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as a function that sends 1 to a1 , 2 to a2 , 3 to a3 , and in general sends the positive integer n to the nth term an . This leads to the formal definition of a sequence.
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DEFINITION Infinite Sequence An infinite sequence of numbers is a function whose domain is the set of positive integers.
The function associated to the sequence 2, 4, 6, 8, 10, 12, Á , 2n, Á
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sends 1 to a1 = 2, 2 to a2 = 4, and so on. The general behavior of this sequence is described by the formula an = 2n.
12, 14, 16, 18, 20, 22 Á
is described by the formula an = 10 + 2n. It can also be described by the simpler formula bn = 2n, where the index n starts at 6 and increases. To allow such simpler formulas, we let the first index of the sequence be any integer. In the sequence above, 5an6 starts with a1 while 5bn6 starts with b6 . Order is important. The sequence 1, 2, 3, 4 Á is not the same as the sequence 2, 1, 3, 4 Á . Sequences can be described by writing rules that specify their terms, such as
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We can equally well make the domain the integers larger than a given number n0 , and we allow sequences of this type also. The sequence
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an = 2n, 1 bn = s 1dn + 1 n , cn =
n  1 n ,
dn = s 1dn + 1
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 11: Infinite Sequences and Series
5an6 =
E 21, 22, 23, Á , 2n, Á F
i
or by listing terms,
1 1 1 1 5bn6 = e 1,  , ,  , Á , s 1dn + 1 n , Á f 2 3 4
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n  1 1 2 3 4 5cn6 = e 0, , , , , Á , n , Á f 2 3 4 5
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748
5dn6 = 51, 1, 1, 1, 1, 1, Á , s 1dn + 1, Á 6. We also sometimes write 5an6 =
E 2n F n = 1. . q
nR
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Figure 11.1 shows two ways to represent sequences graphically. The first marks the first few points from a1, a2 , a3 , Á , an , Á on the real axis. The second method shows the graph of the function defining the sequence. The function is defined only on integer inputs, and the graph consists of some points in the xyplane, located at s1, a1 d, s2, a2 d, Á , sn, an d, Á . an
a2 a3 a4 a5
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a1 1
0
2
an 兹n
a3 a 2
a1
0
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1
Mu
ham
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a2 a4
a5 a3
1 an n
Diverges
3 2 1
0 an
Converges to 0
1 0
1
2
an a1
0
3
4
5
n
Converges to 0
1
1 an (1) n1 1n
n
1 2 3 4 5
0
n
FIGURE 11.1 Sequences can be represented as points on the real line or as points in the plane where the horizontal axis n is the index number of the term and the vertical axis an is its value.
Convergence and Divergence Sometimes the numbers in a sequence approach a single value as the index n increases. This happens in the sequence 1 1 1 1 e 1, , , , Á , n , Á f 2 3 4 whose terms approach 0 as n gets large, and in the sequence
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1 1 2 3 4 e 0, , , , , Á , 1  n , Á f 2 3 4 5
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.1
Sequences
749
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E 21, 22, 23, Á , 2n, Á F
i
whose terms approach 1. On the other hand, sequences like
have terms that get larger than any number as n increases, and sequences like 51, 1, 1, 1, 1, 1, Á , s 1dn + 1, Á 6
0
a2 a3
aN
a1
L L
an
an
bounce back and forth between 1 and 1, never converging to a single value. The following definition captures the meaning of having a sequence converge to a limiting value. It says that if we go far enough out in the sequence, by taking the index n to be larger then some value N, the difference between an and the limit of the sequence becomes less than any preselected number P 7 0.
You
L
N
n
n
n 7 N
FIGURE 11.2 an : L if y = L is a horizontal asymptote of the sequence of points 5sn, an d6 . In this figure, all the an’s after aN lie within P of L.
Nicole Oresme (ca. 1320–1382)
ƒ an  L ƒ 6 P.
If no such number L exists, we say that 5an6 diverges. If 5an6 converges to L, we write limn: q an = L, or simply an : L, and call L the limit of the sequence (Figure 11.2).
The definition is very similar to the definition of the limit of a function ƒ(x) as x tends to q (limx: q ƒsxd in Section 2.4). We will exploit this connection to calculate limits of sequences.
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HISTORICAL BIOGRAPHY
Q
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1 2 3
DEFINITIONS Converges, Diverges, Limit The sequence 5an6 converges to the number L if to every positive number P there corresponds an integer N such that for all n,
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L (N, aN)
0
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L (n, an)
L
EXAMPLE 1
Applying the Definition
Show that
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1 (a) lim n = 0 n: q
(b) lim k = k n: q
sany constant kd
Mu
ham
Solution
(a) Let P 7 0 be given. We must show that there exists an integer N such that for all n, n 7 N
Q
1 ` n  0 ` 6 P.
This implication will hold if s1>nd 6 P or n 7 1>P. If N is any integer greater than 1>P, the implication will hold for all n 7 N. This proves that limn: q s1>nd = 0. (b) Let P 7 0 be given. We must show that there exists an integer N such that for all n, n 7 N
Q
ƒ k  k ƒ 6 P.
Since k  k = 0, we can use any positive integer for N and the implication will hold. This proves that limn: q k = k for any constant k.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 11: Infinite Sequences and Series
EXAMPLE 2
A Divergent Sequence
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Show that the sequence 51, 1, 1, 1, 1, 1, Á , s 1dn + 1, Á 6 diverges.
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Suppose the sequence converges to some number L. By choosing P = 1>2 in the definition of the limit, all terms an of the sequence with index n larger than some N must lie within P = 1>2 of L. Since the number 1 appears repeatedly as every other term of the sequence, we must have that the number 1 lies within the distance P = 1>2 of L. It follows that ƒ L  1 ƒ 6 1>2, or equivalently, 1>2 6 L 6 3>2. Likewise, the number 1 appears repeatedly in the sequence with arbitrarily high index. So we must also have that ƒ L  s 1d ƒ 6 1>2, or equivalently, 3>2 6 L 6 1>2. But the number L cannot lie in both of the intervals (1> 2, 3> 2) and s 3>2, 1>2d because they have no overlap. Therefore, no such limit L exists and so the sequence diverges. Note that the same argument works for any positive number P smaller than 1, not just 1> 2.
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Solution
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The sequence {1n} also diverges, but for a different reason. As n increases, its terms become larger than any fixed number. We describe the behavior of this sequence by writing lim 2n = q . n: q
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In writing infinity as the limit of a sequence, we are not saying that the differences between the terms an and q become small as n increases. Nor are we asserting that there is some number infinity that the sequence approaches. We are merely using a notation that captures the idea that an eventually gets and stays larger than any fixed number as n gets large.
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DEFINITION Diverges to Infinity The sequence 5an6 diverges to infinity if for every number M there is an integer N such that for all n larger than N, an 7 M. If this condition holds we write lim an = q
n: q
or
an : q .
Mu
ham
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Similarly if for every number m there is an integer N such that for all n 7 N we have an 6 m, then we say 5an6 diverges to negative infinity and write lim an =  q
n: q
or
an :  q .
A sequence may diverge without diverging to infinity or negative infinity. We saw this in Example 2, and the sequences 51, 2, 3, 4, 5, 6, 7, 8, Á 6 and 51, 0, 2, 0, 3, 0, Á 6 are also examples of such divergence.
Calculating Limits of Sequences If we always had to use the formal definition of the limit of a sequence, calculating with P’s and N’s, then computing limits of sequences would be a formidable task. Fortunately we can derive a few basic examples, and then use these to quickly analyze the limits of many more sequences. We will need to understand how to combine and compare sequences. Since sequences are functions with domain restricted to the positive integers, it is not too surprising that the theorems on limits of functions given in Chapter 2 have versions for sequences.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.1
Sequences
751
5.
Quotient Rule:
limn: q san + bn d = A + B limn: q san  bn d = A  B limn: q san # bn d = A # B limn: q sk # bn d = k # B sAny number kd an A limn: q = if B Z 0 B bn
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Sum Rule: Difference Rule: Product Rule: Constant Multiple Rule:
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1. 2. 3. 4.
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THEOREM 1 Let 5an6 and 5bn6 be sequences of real numbers and let A and B be real numbers. The following rules hold if limn: q an = A and limn: q bn = B.
The proof is similar to that of Theorem 1 of Section 2.2, and is omitted.
EXAMPLE 3
Applying Theorem 1
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By combining Theorem 1 with the limits of Example 1, we have: 1 1 (a) lim a n b = 1 # lim n = 1 # 0 = 0 n: q n: q n: q
(c)
n  1 1 1 lim a1  n b = lim 1  lim n = 1  0 = 1 n b = n: q n: q n: q
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(b) lim a
Constant Multiple Rule and Example 1a
lim
n: q
5 1 = 5 # lim n n: q n2
#
1 lim n = 5 # 0 # 0 = 0
n: q
s4>n 6 d  7 0  7 4  7n 6 = lim = = 7. 1 + 0 n: q n 6 + 3 n: q 1 + s3>n 6 d
Product Rule
Sum and Quotient Rules
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(d) lim
Difference Rule and Example 1a
Mu
ham
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Be cautious in applying Theorem 1. It does not say, for example, that each of the sequences 5an6 and 5bn6 have limits if their sum 5an + bn6 has a limit. For instance, 5an6 = 51, 2, 3, Á 6 and 5bn6 = 51, 2, 3, Á 6 both diverge, but their sum 5an + bn6 = 50, 0, 0, Á 6 clearly converges to 0. One consequence of Theorem 1 is that every nonzero multiple of a divergent sequence 5an6 diverges. For suppose, to the contrary, that 5can6 converges for some number c Z 0. Then, by taking k = 1>c in the Constant Multiple Rule in Theorem 1, we see that the sequence 1 e c # can f = 5an6
converges. Thus, 5can6 cannot converge unless 5an6 also converges. If 5an6 does not converge, then 5can6 does not converge. The next theorem is the sequence version of the Sandwich Theorem in Section 2.2. You are asked to prove the theorem in Exercise 95.
THEOREM 2 The Sandwich Theorem for Sequences Let 5an6, 5bn6, and 5cn6 be sequences of real numbers. If an … bn … cn holds for all n beyond some index N, and if limn: q an = limn: q cn = L, then limn: q bn = L also.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 752
Chapter 11: Infinite Sequences and Series
EXAMPLE 4
Applying the Sandwich Theorem
Since 1>n : 0, we know that cos n n :0
1 :0 2n 1 (c) s 1dn n : 0
cos n 1 1  n … n … n;
because
0 …
because
1 1 1  n … s 1dn n … n .
1 1 … n; 2n
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(b)
because
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(a)
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An immediate consequence of Theorem 2 is that, if ƒ bn ƒ … cn and cn : 0, then bn : 0 because cn … bn … cn . We use this fact in the next example.
The application of Theorems 1 and 2 is broadened by a theorem stating that applying a continuous function to a convergent sequence produces a convergent sequence. We state the theorem without proof (Exercise 96). y
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y 2x (1, 2)
THEOREM 3 The Continuous Function Theorem for Sequences Let 5an6 be a sequence of real numbers. If an : L and if ƒ is a function that is continuous at L and defined at all an , then ƒsan d : ƒsLd.
1 , 21/2 2 1 , 21/3 3
1
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2
EXAMPLE 5
Applying Theorem 3
Show that 2sn + 1d>n : 1.
We know that sn + 1d>n : 1. Taking ƒsxd = 1x and L = 1 in Theorem 3 gives 1sn + 1d>n : 11 = 1.
0
1 3
1 2
1
x
EXAMPLE 6
The Sequence 521>n6
The sequence 51>n6 converges to 0. By taking an = 1>n, ƒsxd = 2x , and L = 0 in Theorem 3, we see that 21>n = ƒs1>nd : ƒsLd = 20 = 1. The sequence 521>n6 converges to 1 (Figure 11.3).
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FIGURE 11.3 As n : q , 1>n : 0 and 21>n : 20 (Example 6).
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Solution
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Using l’Hôpital’s Rule The next theorem enables us to use l’Hôpital’s Rule to find the limits of some sequences. It formalizes the connection between limn: q an and limx: q ƒsxd.
THEOREM 4 Suppose that ƒ(x) is a function defined for all x Ú n0 and that 5an6 is a sequence of real numbers such that an = ƒsnd for n Ú n0 . Then lim ƒsxd = L
x: q
Q
lim an = L.
n: q
Proof Suppose that limx: q ƒsxd = L. Then for each positive number P there is a number M such that for all x, x 7 M
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Q
ƒ ƒsxd  L ƒ 6 P.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.1
Sequences
753
EXAMPLE 7
an = ƒsnd
and
ƒ an  L ƒ = ƒ ƒsnd  L ƒ 6 P.
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Q
n 7 N
i
Let N be an integer greater than M and greater than or equal to n0 . Then
Applying L’Hôpital’s Rule
lim
n: q
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Show that ln n n = 0.
The function sln xd>x is defined for all x Ú 1 and agrees with the given sequence at positive integers. Therefore, by Theorem 5, limn: q sln nd>n will equal limx: q sln xd>x if the latter exists. A single application of l’Hôpital’s Rule shows that
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Solution
1>x 0 ln x = lim = = 0. x 1 x: q x: q 1
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lim
We conclude that limn: q sln nd>n = 0.
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When we use l’Hôpital’s Rule to find the limit of a sequence, we often treat n as a continuous real variable and differentiate directly with respect to n. This saves us from having to rewrite the formula for an as we did in Example 7.
EXAMPLE 8
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Find
Applying L’Hôpital’s Rule
By l’Hôpital’s Rule (differentiating with respect to n),
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Solution
ham
EXAMPLE 9
Mu
2n . 5n n: q lim
2n 2n # ln 2 = lim 5 n: q 5n n: q q = . lim
Applying L’Hôpital’s Rule to Determine Convergence
Does the sequence whose nth term is an = a
n + 1 b n  1
n
converge? If so, find limn: q an . q The limit leads to the indeterminate form 1 . We can apply l’Hôpital’s Rule if # we first change the form to q 0 by taking the natural logarithm of an :
Solution
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ln an = ln a
n + 1 b n  1
= n ln a
n
n + 1 b. n  1
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 754
Chapter 11: Infinite Sequences and Series
= lim
ln a
n: q
= lim
n + 1 b n  1
n + 1 b n  1 1>n
2>sn 2  1d
1>n 2 2n 2 = 2. = lim 2 n: q n  1 n: q
q #0
0 0
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n: q
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lim ln an = lim n ln a
n: q
i
Then,
l’Hôpital’s Rule
The sequence 5an6 converges to e 2 .
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Since ln an : 2 and ƒsxd = e x is continuous, Theorem 4 tells us that an = e ln an : e 2 .
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Commonly Occurring Limits
The next theorem gives some limits that arise frequently.
1.
3. 4.
n: q
ln n n = 0 n
lim 2n = 1
n: q
lim x 1>n = 1
5.
ham
e n (rounded)
n!
Mu
n 1 5 10 20
3 148 22,026 4.9 * 10 8
1 120 3,628,800 2.4 * 10 18
6.
sx 7 0d
n: q
lim x n = 0
s ƒ x ƒ 6 1d
n: q
n
x lim a1 + n b = e x n: q
ma
Factorial Notation The notation n! (“n factorial”) means the product 1 # 2 # 3 Á n of the integers from 1 to n. Notice that sn + 1d! = sn + 1d # n! . Thus, 4! = 1 # 2 # 3 # 4 = 24 and 5! = 1 # 2 # 3 # 4 # 5 = 5 # 4! = 120 . We define 0! to be 1. Factorials grow even faster than exponentials, as the table suggests.
lim
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2.
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THEOREM 5 The following six sequences converge to the limits listed below:
xn = 0 n: q n! lim
sany xd
sany xd
In Formulas (3) through (6), x remains fixed as n : q .
Proof The first limit was computed in Example 7. The next two can be proved by taking logarithms and applying Theorem 4 (Exercises 93 and 94). The remaining proofs are given in Appendix 3.
EXAMPLE 10 (a)
Applying Theorem 5
ln sn 2 d 2 ln n = n :2#0 = 0 n n
(b) 2n 2 = n 2>n = sn 1/n d2 : s1d2 = 1 n
(c) 23n = 3 sn d : 1 # 1 = 1 1>n
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1/n
Formula 1 Formula 2 Formula 3 with x = 3 and Formula 2
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
n
1 (d) a b : 0 2 n
100 n :0 n!
Formula 5 with x = 2
Formula 6 with x = 100
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(f)
n
n  2 2 2 n b = a1 + n b : e
1 2
755
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(e) a
Formula 4 with x = 
Sequences
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11.1
Recursive Definitions
So far, we have calculated each an directly from the value of n. But sequences are often defined recursively by giving
EXAMPLE 11
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The value(s) of the initial term or terms, and A rule, called a recursion formula, for calculating any later term from terms that precede it.
Sequences Constructed Recursively
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1. 2.
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(a) The statements a1 = 1 and an = an  1 + 1 define the sequence 1, 2, 3, Á , n, Á of positive integers. With a1 = 1, we have a2 = a1 + 1 = 2, a3 = a2 + 1 = 3, and so on. (b) The statements a1 = 1 and an = n # an  1 define the sequence 1, 2, 6, 24, Á , n!, Á of factorials. With a1 = 1, we have a2 = 2 # a1 = 2, a3 = 3 # a2 = 6, a4 = 4 # a3 = 24, and so on. (c) The statements a1 = 1, a2 = 1, and an + 1 = an + an  1 define the sequence 1, 1, 2, 3, 5, Á of Fibonacci numbers. With a1 = 1 and a2 = 1, we have a3 = 1 + 1 = 2, a4 = 2 + 1 = 3, a5 = 3 + 2 = 5, and so on. (d) As we can see by applying Newton’s method, the statements x0 = 1 and xn + 1 = xn  [ssin xn  x n2 d>scos xn  2xn d] define a sequence that converges to a solution of the equation sin x  x 2 = 0.
Bounded Nondecreasing Sequences
Mu
ham
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The terms of a general sequence can bounce around, sometimes getting larger, sometimes smaller. An important special kind of sequence is one for which each term is at least as large as its predecessor.
DEFINITION Nondecreasing Sequence A sequence 5an6 with the property that an … an + 1 for all n is called a nondecreasing sequence.
EXAMPLE 12
Nondecreasing Sequences
(a) The sequence 1, 2, 3, Á , n, Á of natural numbers n 1 2 3 ,Á (b) The sequence , , , Á , 2 3 4 n + 1 (c) The constant sequence 536
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 756
Chapter 11: Infinite Sequences and Series
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There are two kinds of nondecreasing sequences—those whose terms increase beyond any finite bound and those whose terms do not.
EXAMPLE 13
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DEFINITIONS Bounded, Upper Bound, Least Upper Bound A sequence 5an6 is bounded from above if there exists a number M such that an … M for all n. The number M is an upper bound for 5an6. If M is an upper bound for 5an6 but no number less than M is an upper bound for 5an6, then M is the least upper bound for 5an6.
Applying the Definition for Boundedness
(a) The sequence 1, 2, 3, Á , n, Á has no upper bound.
n 1 2 3 (b) The sequence , , , Á , , Á is bounded above by M = 1. 2 3 4 n + 1
nR
y
No number less than 1 is an upper bound for the sequence, so 1 is the least upper bound (Exercise 113).
yL
L
(8, a 8 )
(5, a 5) (1, a 1)
0
1
2
3
4
5
6
7
8
x
(a) an … L for all values of n and (b) given any P 7 0, there exists at least one integer N for which aN 7 L  P. The fact that 5an6 is nondecreasing tells us further that
Mu
ham
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FIGURE 11.4 If the terms of a nondecreasing sequence have an upper bound M, they have a limit L … M .
A nondecreasing sequence that is bounded from above always has a least upper bound. This is the completeness property of the real numbers, discussed in Appendix 4. We will prove that if L is the least upper bound then the sequence converges to L. Suppose we plot the points s1, a1 d, s2, a2 d, Á , sn, an d, Á in the xyplane. If M is an upper bound of the sequence, all these points will lie on or below the line y = M (Figure 11.4). The line y = L is the lowest such line. None of the points sn, an d lies above y = L, but some do lie above any lower line y = L  P, if P is a positive number. The sequence converges to L because
ass a
M
dH
yM
an Ú aN 7 L  P
for all n Ú N.
Thus, all the numbers an beyond the Nth number lie within P of L. This is precisely the condition for L to be the limit of the sequence {an}. The facts for nondecreasing sequences are summarized in the following theorem. A similar result holds for nonincreasing sequences (Exercise 107).
THEOREM 6 The Nondecreasing Sequence Theorem A nondecreasing sequence of real numbers converges if and only if it is bounded from above. If a nondecreasing sequence converges, it converges to its least upper bound.
Theorem 6 implies that a nondecreasing sequence converges when it is bounded from above. It diverges to infinity if it is not bounded from above.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.1 Sequences
Finding Terms of a Sequence
23. an = 2 + s0.1dn
1  n n2 s 1dn + 1 3. an = 2n  1
2. an =
an + 1 = an + s1>2n d an + 1 = an>sn + 1d
an + 1 = s 1dn + 1an>2
an + 1 = nan>sn + 1d
11. a1 = a2 = 1, 12. a1 = 2,
29. an =
n 2  2n + 1 n  1
an + 2 = an + 1 + an
a2 = 1,
an + 2 = an + 1>an
Finding a Sequence’s Formula
15. The sequence 1, 4, 9, 16, 25, Á
1’s with alternating signs
Squares of the positive integers; with alternating signs
Reciprocals of squares of the positive integers, with alternating signs
ma
1 1 1 1 16. The sequence 1,  , ,  , , Á 4 9 16 25
1’s with alternating signs
Squares of the positive integers diminished by 1
18. The sequence 3, 2, 1, 0, 1, Á
Integers beginning with 3
ham
17. The sequence 0, 3, 8, 15, 24, Á
19. The sequence 1, 5, 9, 13, 17, Á
Every other odd positive integer Every other even positive integer
21. The sequence 1, 0, 1, 0, 1, Á
Alternating 1’s and 0’s
Mu
20. The sequence 2, 6, 10, 14, 18, Á
22. The sequence 0, 1, 1, 2, 2, 3, 3, 4, Á
Each positive integer repeated
Finding Limits
Which of the sequences 5an6 in Exercises 23–84 converge, and which diverge? Find the limit of each convergent sequence.
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28. an =
n + 3 n 2 + 5n + 6
30. an =
1  n3 70  4n 2
1 32. an = s 1dn a1  n b
33. an = a
34. an = a2 
1 1 b a3 + n b 2n 2 n
n + 1 1 b a1  n b 2n
35. an =
s 1dn + 1 2n  1
1 36. an = a b 2
37. an =
2n An + 1
38. an =
39. an = sin a
dH
14. The sequence 1, 1, 1, 1, 1, Á
2n + 1
1  32n
31. an = 1 + s 1dn
In Exercises 13–22, find a formula for the nth term of the sequence. 13. The sequence 1, 1, 1, 1, 1, Á
n + s 1dn n
iaz
2n  1 6. an = 2n
26. an =
nR
10. a1 = 2,
1  5n 4 n 4 + 8n 3
ass a
9. a1 = 2,
27. an =
1 n!
Each of Exercises 7–12 gives the first term or two of a sequence along with a recursion formula for the remaining terms. Write out the first ten terms of the sequence. 8. a1 = 1,
1  2n 1 + 2n
4. an = 2 + s 1dn
2n 5. an = n + 1 2
7. a1 = 1,
25. an =
24. an =
You
Each of Exercises 1–6 gives a formula for the nth term an of a sequence 5an6 . Find the values of a1, a2 , a3 , and a4 . 1. an =
suf
i
EXERCISES 11.1
757
p 1 + nb 2
1 s0.9dn
40. an = np cos snpd
41. an =
sin n n
42. an =
sin2 n 2n
43. an =
n 2n
44. an =
3n n3
46. an =
ln n ln 2n
45. a n =
ln sn + 1d 2n
47. an = 81>n
48. an = s0.03d1>n
7 49. an = a1 + n b n
51. an = 210n 3 53. an = a n b 55. an =
n! nn
n
n
52. an = 2n 2 54. an = sn + 4d1>sn + 4d 56. an = ln n  ln sn + 1d
57. an = 24nn 59. an =
1 50. an = a1  n b
1>n
ln n n 1>n n
n
n
58. an = 232n + 1
(Hint: Compare with 1> n.)
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 758
Chapter 11: Infinite Sequences and Series
1 64. an = ln a1 + n b n b 66. an = a n + 1
n
65. an = a
n
xn b 67. an = a 2n + 1 n
68. an = a1 
1 b n2 s10>11dn 70. an = s9/10dn + s11/12dn 72. an = sinh sln nd
i
,
x 7 0
Do the sequences converge? If so, to what value? In each case, begin by identifying the function ƒ that generates the sequence.
3n # 6n 2n # n!
a. x0 = 1,
71. an = tanh n n2 1 sin n 2n  1
1 1 77. an = a b + 3 22n sln nd200 79. an = n
tan1 n
2
sln nd5
81. an = n  2n  n 2
2n 1
2n 2  1  2n 2 + n L1
n
1 x dx
84. an =
Theory and Examples
1 p dx, L1 x
p 7 1
dH
n
ƒsxn d . ƒ¿sxn d
xn + 1 = xn 
x n2  2 xn 1 = + xn 2xn 2 tan xn  1
b. x0 = 1,
xn + 1 = xn 
c. x0 = 1,
xn + 1 = xn  1
sec2 xn
88. a. Suppose that ƒ(x) is differentiable for all x in [0, 1] and that ƒs0d = 0 . Define the sequence 5an6 by the rule an = nƒs1>nd . Show that lim n: q an = ƒ¿s0d . Use the result in part (a) to find the limits of the following sequences 5an6.
1 b. an = n tan1 n
c. an = nse 1>n  1d
ass a
2n
1 83. an = n
xn + 1 = xn 
1>n
75. an = tan1 n
78. an = 2n + n
82. an =
87. Newton’s method The following sequences come from the recursion formula for Newton’s method,
n
n
1 n
80. an =
69. an =
73. an =
1 74. an = n a1  cos n b 76. an =
3n + 1 b 3n  1
suf
n! 2n # 3n
You
62. an =
b. The fractions rn = xn>yn approach a limit as n increases. What is that limit? (Hint: Use part (a) to show that r n2  2 = ;s1>yn d2 and that yn is not less than n.)
n! 10 6n 1>sln nd 1 63. an = a n b 61. an =
iaz
s 4dn n!
nR
60. an =
85. The first term of a sequence is x1 = 1 . Each succeeding term is the sum of all those that come before it:
2 d. an = n ln a1 + n b
89. Pythagorean triples A triple of positive integers a, b, and c is called a Pythagorean triple if a 2 + b 2 = c 2 . Let a be an odd positive integer and let b = j
a2 k 2
and
c = l
a2 m 2
be, respectively, the integer floor and ceiling for a 2>2 .
ma
xn + 1 = x1 + x2 + Á + xn .
Write out enough early terms of the sequence to deduce a general formula for xn that holds for n Ú 2 . 86. A sequence of rational numbers is described as follows:
ham
a a + 2b 1 3 7 17 , , , ,Á, , ,Á. 1 2 5 12 b a + b
Here the numerators form one sequence, the denominators form a second sequence, and their ratios form a third sequence. Let xn and yn be, respectively, the numerator and the denominator of the nth fraction rn = xn>yn .
Mu
a. Verify that x 12  2y 12 = 1, x 22  2y 22 = +1 and, more generally, that if a 2  2b 2 = 1 or +1 , then sa + 2bd2  2sa + bd2 = +1
or
a 2 2
a 2 2
a
a. Show that a 2 + b 2 = c 2 . (Hint: Let a = 2n + 1 and express b and c in terms of n.)
1 ,
respectively.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.1 Sequences
106. The first term of a sequence is x1 = cos s1d . The next terms are x2 = x1 or cos (2), whichever is larger; and x3 = x2 or cos (3), whichever is larger (farther to the right). In general,
a2 m 2
90. The nth root of n! a. Show that limn: q s2npd1>s2nd = 1 and hence, using Stirling’s approximation (Chapter 8, Additional Exercise 50a), that n n 2n! L e T
for large values of n .
b. Test the approximation in part (a) for n = 40, 50, 60, Á , as far as your calculator will allow.
91. a. Assuming that limn: q s1>n c d = 0 if c is any positive constant, show that ln n = 0 nc
b. Prove that limn: q s1>n c d = 0 if c is any positive constant. (Hint: If P = 0.001 and c = 0.04 , how large should N be to ensure that ƒ 1>n c  0 ƒ 6 P if n 7 N ?)
dH
92. The zipper theorem Prove the “zipper theorem” for sequences: If 5an6 and 5bn6 both converge to L, then the sequence a1, b1, a2 , b2 , Á , an , bn , Á n
93. Prove that limn: q 2n = 1 .
ma
94. Prove that limn: q x 1>n = 1, sx 7 0d . 95. Prove Theorem 2.
96. Prove Theorem 3.
In Exercises 97–100, determine if the sequence is nondecreasing and if it is bounded from above.
99. an =
3n + 1 n + 1
s2n + 3d! sn + 1d! 2 1 100. an = 2  n  n 2
ham
97. an =
2n3n n!
98. an =
Which of the sequences in Exercises 101–106 converge, and which diverge? Give reasons for your answers.
Mu
1 101. an = 1  n 103. an =
2n  1 2n
108. an =
n + 1 n
109. an =
1 + 22n
2n 4n + 1 + 3n 111. an = 4n
1  4n 110. an = 2n 112. a1 = 1,
if c is any positive constant.
converges to L.
(Continuation of Exercise 107.) Using the conclusion of Exercise 107, determine which of the sequences in Exercises 108–112 converge and which diverge.
an + 1 = 2an  3
113. The sequence 5n>sn + 1d6 has a least upper bound of 1 Show that if M is a number less than 1, then the terms of 5n>sn + 1d6 eventually exceed M. That is, if M 6 1 there is an integer N such that n>sn + 1d 7 M whenever n 7 N . Since n>sn + 1d 6 1 for every n, this proves that 1 is a least upper bound for 5n>sn + 1d6 .
ass a
lim
n: q
107. Nonincreasing sequences A sequence of numbers 5an6 in which an Ú an + 1 for every n is called a nonincreasing sequence. A sequence 5an6 is bounded from below if there is a number M with M … an for every n. Such a number M is called a lower bound for the sequence. Deduce from Theorem 6 that a nonincreasing sequence that is bounded from below converges and that a nonincreasing sequence that is not bounded from below diverges.
You
l
xn + 1 = max 5xn , cos sn + 1d6 .
.
iaz
a: q
a2 k 2
nR
lim
j
suf
i
b. By direct calculation, or by appealing to the figure here, find
759
1 102. an = n  n 104. an =
2n  1 3n
n + 1 105. an = ss 1d + 1d a n b n
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114. Uniqueness of least upper bounds Show that if M1 and M2 are least upper bounds for the sequence 5an6 , then M1 = M2 . That is, a sequence cannot have two different least upper bounds. 115. Is it true that a sequence 5an6 of positive numbers must converge if it is bounded from above? Give reasons for your answer.
116. Prove that if 5an6 is a convergent sequence, then to every positive number P there corresponds an integer N such that for all m and n, m 7 N
and
n 7 N
Q
ƒ am  an ƒ 6 P .
117. Uniqueness of limits Prove that limits of sequences are unique. That is, show that if L1 and L2 are numbers such that an : L1 and an : L2 , then L1 = L2 . 118. Limits and subsequences If the terms of one sequence appear in another sequence in their given order, we call the first sequence a subsequence of the second. Prove that if two subsequences of a sequence 5an6 have different limits L1 Z L2 , then 5an6 diverges. 119. For a sequence 5an6 the terms of even index are denoted by a2k and the terms of odd index by a2k + 1 . Prove that if a2k : L and a2k + 1 : L , then an : L . 120. Prove that a sequence 5an6 converges to 0 if and only if the sequence of absolute values 5ƒ an ƒ6 converges to 0.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 760
Calculator Explorations of Limits
i
b. Graph ƒsxd = 7.25s0.94dx and use Trace to find where the graph crosses the line y = 3.5 .
122. ƒ 2n  1 ƒ 6 10 3 124. 2n>n! 6 10 7
125. Sequences generated by Newton’s method Newton’s method, applied to a differentiable function ƒ(x), begins with a starting value x0 and constructs from it a sequence of numbers 5xn6 that under favorable circumstances converges to a zero of ƒ. The recursion formula for the sequence is xn + 1 = xn 
ƒsxn d . ƒ¿sxn d
a. Show that the recursion formula for ƒsxd = x 2  a, a 7 0 , can be written as xn + 1 = sxn + a>xn d>2 . b. Starting with x0 = 1 and a = 3 , calculate successive terms of the sequence until the display begins to repeat. What number is being approximated? Explain.
COMPUTER EXPLORATIONS
Use a CAS to perform the following steps for the sequences in Exercises 129–140. a. Calculate and then plot the first 25 terms of the sequence. Does the sequence appear to be bounded from above or below? Does it appear to converge or diverge? If it does converge, what is the limit L? b. If the sequence converges, find an integer N such that ƒ an  L ƒ … 0.01 for n Ú N . How far in the sequence do you have to get for the terms to lie within 0.0001 of L?
dH
127. A recursive definition of P/ 2 If you start with x1 = 1 and define the subsequent terms of 5xn6 by the rule xn = xn  1 + cos xn  1 , you generate a sequence that converges rapidly to p>2 . a. Try it. b. Use the accompanying figure to explain why the convergence is so rapid. y
cos xn 1
ma
1
xn 1
xn 1
x
1
ham
0
Mu
128. According to a frontpage article in the December 15, 1992, issue of the Wall Street Journal, Ford Motor Company used about 7 14 hours of labor to produce stampings for the average vehicle, down from an estimated 15 hours in 1980. The Japanese needed only about 3 12 hours. Ford’s improvement since 1980 represents an average decrease of 6% per year. If that rate continues, then n years from 1992 Ford will use about n
Sn = 7.25s0.94d
hours of labor to produce stampings for the average vehicle. Assuming that the Japanese continue to spend 3 12 hours per vehicle,
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0.5 130. an = a1 + n b
n
129. an = 2n
an + 1 = an +
132. a1 = 1,
an + 1 = an + s 2dn 1 134. an = n sin n
133. an = sin n 135. an =
sin n n
ln n 136. an = n
137. an = s0.9999dn 139. an =
8n n!
n
1 5n
131. a1 = 1,
ass a
126. (Continuation of Exercise 125.) Repeat part (b) of Exercise 125 with a = 2 in place of a = 3 .
You
n
121. ƒ 20.5  1 ƒ 6 10 3 123. s0.9dn 6 10 3
iaz
n
T
suf
In Exercises 121–124, experiment with a calculator to find a value of N that will make the inequality hold for all n 7 N . Assuming that the inequality is the one from the formal definition of the limit of a sequence, what sequence is being considered in each case and what is its limit?
how many more years will it take Ford to catch up? Find out two ways: a. Find the first term of the sequence 5Sn6 that is less than or equal to 3.5.
nR
T
Chapter 11: Infinite Sequences and Series
138. an = 1234561>n 140. an =
n 41 19n
141. Compound interest, deposits, and withdrawals If you invest an amount of money A0 at a fixed annual interest rate r compounded m times per year, and if the constant amount b is added to the account at the end of each compounding period (or taken from the account if b 6 0), then the amount you have after n + 1 compounding periods is r An + 1 = a1 + m bAn + b .
(1)
a. If A0 = 1000, r = 0.02015, m = 12 , and b = 50 , calculate and plot the first 100 points sn, An d . How much money is in your account at the end of 5 years? Does 5An6 converge? Is 5An6 bounded?
b. Repeat part (a) with A0 = 5000, r = 0.0589, m = 12 , and b = 50 . c. If you invest 5000 dollars in a certificate of deposit (CD) that pays 4.5% annually, compounded quarterly, and you make no further investments in the CD, approximately how many years will it take before you have 20,000 dollars? What if the CD earns 6.25%?
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.1 Sequences
142. Logistic difference equation The recursive relation an + 1 = rans1  an d is called the logistic difference equation, and when the initial value a0 is given the equation defines the logistic sequence 5an6 . Throughout this exercise we choose a0 in the interval 0 6 a0 6 1 , say a0 = 0.3 .
i
e. The situation gets even more interesting. There is actually an increasing sequence of bifurcation values 3 6 3.45 6 3.54 6 Á 6 cn 6 cn + 1 Á such that for cn 6 r 6 cn + 1 the logistic sequence 5an6 eventually oscillates steadily among 2n values, called an attracting 2ncycle. Moreover, the bifurcation sequence 5cn6 is bounded above by 3.57 (so it converges). If you choose a value of r 6 3.57 you will observe a 2ncycle of some sort. Choose r = 3.5695 and plot 300 points. f. Let us see what happens when r 7 3.57 . Choose r = 3.65 and calculate and plot the first 300 terms of 5an6 . Observe how the terms wander around in an unpredictable, chaotic fashion. You cannot predict the value of an + 1 from previous values of the sequence.
ass a
a. Choose r = 3>4 . Calculate and plot the points sn, an d for the first 100 terms in the sequence. Does it appear to converge? What do you guess is the limit? Does the limit seem to depend on your choice of a0 ?
suf
For the values of the constants A0 , r, m, and b given in part (a), validate this assertion by comparing the values of the first 50 terms of both sequences. Then show by direct substitution that the terms in Equation (2) satisfy the recursion formula in Equation (1).
You
(2)
iaz
k
mb mb r Ak = a1 + m b aA0 + r b  r .
d. Next explore the behavior for r values near the endpoints of each of the intervals 3.45 6 r 6 3.54 and 3.54 6 r 6 3.55 . Plot the first 200 terms of the sequences. Describe in your own words the behavior observed in your plots for each interval. Among how many values does the sequence appear to oscillate for each interval? The values r = 3.45 and r = 3.54 (rounded to two decimal places) are also called bifurcation values because the behavior of the sequence changes as r crosses over those values.
nR
d. It can be shown that for any k Ú 0 , the sequence defined recursively by Equation (1) satisfies the relation
b. Choose several values of r in the interval 1 6 r 6 3 and repeat the procedures in part (a). Be sure to choose some points near the endpoints of the interval. Describe the behavior of the sequences you observe in your plots.
g. For r = 3.65 choose two starting values of a0 that are close together, say, a0 = 0.3 and a0 = 0.301 . Calculate and plot the first 300 values of the sequences determined by each starting value. Compare the behaviors observed in your plots. How far out do you go before the corresponding terms of your two sequences appear to depart from each other? Repeat the exploration for r = 3.75 . Can you see how the plots look different depending on your choice of a0 ? We say that the logistic sequence is sensitive to the initial condition a0 .
Mu
ham
ma
dH
c. Now examine the behavior of the sequence for values of r near the endpoints of the interval 3 6 r 6 3.45 . The transition value r = 3 is called a bifurcation value and the new behavior of the sequence in the interval is called an attracting 2cycle. Explain why this reasonably describes the behavior.
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761
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
i f u s u o Y iaz 11.2 Infinite Series
11.2
Infinite Series
R n ssa
761
An infinite series is the sum of an infinite sequence of numbers
a H d a m m a h u M
a1 + a2 + a3 + Á + an + Á
The goal of this section is to understand the meaning of such an infinite sum and to develop methods to calculate it. Since there are infinitely many terms to add in an infinite series, we cannot just keep adding to see what comes out. Instead we look at what we get by summing the first n terms of the sequence and stopping. The sum of the first n terms
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sn = a1 + a2 + a3 + Á + an
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 762
Chapter 11: Infinite Sequences and Series
1 1 1 1 + + + + Á 2 4 8 16
You
1 +
suf
i
is an ordinary finite sum and can be calculated by normal addition. It is called the nth partial sum. As n gets larger, we expect the partial sums to get closer and closer to a limiting value in the same sense that the terms of a sequence approach a limit, as discussed in Section 11.1. For example, to assign meaning to an expression like
iaz
We add the terms one at a time from the beginning and look for a pattern in how these partial sums grow.
Suggestive expression for partial sum
s1 = 1
First:
nR
Partial sum
2  1 1 2 2 1 2 4 o
1 s2 = 1 + 2 1 1 s3 = 1 + + 2 4 o 1 1 1 sn = 1 + + + Á + n  1 2 4 2
Second:
ass a
Third: o
dH
nth:
2 
1 2n  1
Value 1 3 2 7 4 o n 2  1 2n  1
Indeed there is a pattern. The partial sums form a sequence whose nth term is sn = 2 
1 . 2n  1
ma
This sequence of partial sums converges to 2 because limn: q s1>2n d = 0. We say 1 1 1 + + Á + n  1 + Á is 2.” 2 4 2
Is the sum of any finite number of terms in this series equal to 2? No. Can we actually add an infinite number of terms one by one? No. But we can still define their sum by defining it to be the limit of the sequence of partial sums as n : q , in this case 2 (Figure 11.5). Our knowledge of sequences and limits enables us to break away from the confines of finite sums.
1
1/4
0
1
1/2
Mu
ham
“the sum of the infinite series 1 +
1/8
2
FIGURE 11.5 As the lengths 1, 12 , 14 , 18 , Á are added one by one, the sum approaches 2.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.2
763
i
DEFINITIONS Infinite Series, nth Term, Partial Sum, Converges, Sum Given a sequence of numbers 5an6, an expression of the form a1 + a2 + a3 + Á + an + Á
suf
HISTORICAL BIOGRAPHY
Infinite Series
Blaise Pascal (1623–1662)
s1 = a1 s2 = a1 + a2 o
You
is an infinite series. The number an is the nth term of the series. The sequence 5sn6 defined by
n
iaz
sn = a1 + a2 + Á + an = a ak k=1
o
nR
is the sequence of partial sums of the series, the number sn being the nth partial sum. If the sequence of partial sums converges to a limit L, we say that the series converges and that its sum is L. In this case, we also write q
a1 + a2 + Á + an + Á = a an = L. n=1
ass a
If the sequence of partial sums of the series does not converge, we say that the series diverges.
dH
When we begin to study a given series a1 + a2 + Á + an + Á , we might not know whether it converges or diverges. In either case, it is convenient to use sigma notation to write the series as q
a an ,
n=1
A useful shorthand when summation from 1 to q is understood
q
a ak ,
or
k=1
a an
ma
Geometric Series
Mu
ham
Geometric series are series of the form q
a + ar + ar 2 + Á + ar n  1 + Á = a ar n  1 n=1
in which a and r are fixed real numbers and a Z 0. The series can also be written as q g n = 0 ar n . The ratio r can be positive, as in 1 +
1 1 1 + + Á + a b 2 4 2
n1
+ Á,
or negative, as in 1 
1 1 1 +  Á + a b 3 9 3
n1
+ Á.
If r = 1, the nth partial sum of the geometric series is sn = a + as1d + as1d2 + Á + as1dn  1 = na,
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 764
Chapter 11: Infinite Sequences and Series
as1  r n d , 1  r
sn =
sr Z 1d.
Multiply sn by r. Subtract rsn from sn . Most of the terms on the right cancel. Factor.
You
sn = a + ar + ar 2 + Á + ar n  1 rsn = ar + ar 2 + Á + ar n  1 + ar n sn  rsn = a  ar n sns1  rd = as1  r n d
suf
i
and the series diverges because limn: q sn = ; q , depending on the sign of a. If r = 1, the series diverges because the nth partial sums alternate between a and 0. If ƒ r ƒ Z 1, we can determine the convergence or divergence of the series in the following way:
We can solve for sn if r Z 1 .
iaz
If ƒ r ƒ 6 1, then r n : 0 as n : q (as in Section 11.1) and sn : a>s1  rd. If ƒ r ƒ 7 1, then ƒ r n ƒ : q and the series diverges.
nR
If ƒ r ƒ 6 1, the geometric series a + ar + ar 2 + Á + ar n  1 + Á converges to a>s1  rd: q
a n1 , = a ar 1  r n=1
ƒ r ƒ 6 1.
ass a
If ƒ r ƒ Ú 1, the series diverges.
dH
We have determined when a geometric series converges or diverges, and to what value. Often we can determine that a series converges without knowing the value to which it converges, as we will see in the next several sections. The formula a>s1  rd for the sum of a geometric series applies only when the summation index begins with n = 1 in the exq q pression g n = 1 ar n  1 (or with the index n = 0 if we write the series as g n = 0 ar n).
EXAMPLE 1
Index Starts with n = 1
ma
The geometric series with a = 1>9 and r = 1>3 is
Mu
ham
EXAMPLE 2
1 1 1 1 1 + + + Á = a a b 9 27 81 n=1 9 3 q
n1
=
1>9 1 = . 6 1  s1>3d
Index Starts with n = 0
The series
s 1dn5 5 5 5 = 5  + + Á a 4n 4 16 64 n=0 q
is a geometric series with a = 5 and r = 1>4. It converges to 5 a = = 4. 1  r 1 + s1>4d
EXAMPLE 3
A Bouncing Ball
You drop a ball from a meters above a flat surface. Each time the ball hits the surface after falling a distance h, it rebounds a distance rh, where r is positive but less than 1. Find the total distance the ball travels up and down (Figure 11.6).
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.2 Solution
765
The total distance is
i
a
Infinite Series
This sum is 2ar>s1  rd. ar
s = 6 ar 3
1 + s2>3d 5>3 = 6a b = 30 m. 1>3 1  s2>3d
Repeating Decimals
iaz
EXAMPLE 4
You
If a = 6 m and r = 2>3, for instance, the distance is ar 2
suf
2ar 1 + r s = a + 2ar + 2ar 2 + 2ar 3 + Á = a + = a . 1 r 1  r (''''''')'''''''*
Express the repeating decimal 5.232323 Á as the ratio of two integers. (a)
23 23 23 + + + Á 100 s100d2 s100d3
nR
Solution
5.232323 Á = 5 +
ass a
= 5 +
2
23 1 1 + a b + Áb a1 + 100 100 100
a = 1, r = 1>100
('''''''')''''''''* 1>s1  0.01d
= 5 +
23 518 23 1 b = 5 + = a 100 0.99 99 99
dH
Unfortunately, formulas like the one for the sum of a convergent geometric series are rare and we usually have to settle for an estimate of a series’ sum (more about this later). The next example, however, is another case in which we can find the sum exactly.
EXAMPLE 5
A Nongeometric but Telescoping Series q
(b)
ma
1 Find the sum of the series a . n = 1 nsn + 1d
Mu
ham
FIGURE 11.6 (a) Example 3 shows how to use a geometric series to calculate the total vertical distance traveled by a bouncing ball if the height of each rebound is reduced by the factor r. (b) A stroboscopic photo of a bouncing ball.
We look for a pattern in the sequence of partial sums that might lead to a formula for sk . The key observation is the partial fraction decomposition
Solution
1 1 1 = n , n + 1 nsn + 1d
so k
k
1 1 1 a nsn + 1d = a a n  n + 1 b n=1 n=1 and sk = a
1 1 1 1 1 1 1 1  b + a  b + a  b + Á + a b. 1 2 2 3 3 4 k k + 1
Removing parentheses and canceling adjacent terms of opposite sign collapses the sum to sk = 1 
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1 . k + 1
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 766
Chapter 11: Infinite Sequences and Series
i
We now see that sk : 1 as k : q . The series converges, and its sum is 1: q
suf
1 a nsn + 1d = 1. n=1
You
Divergent Series
One reason that a series may fail to converge is that its terms don’t become small.
EXAMPLE 6
Partial Sums Outgrow Any Number
(a) The series q
iaz
2 Á + n2 + Á an = 1 + 4 + 9 +
n=1
q
a
3 n + 1 2 4 Á + n + 1 + Á n = 1 + 2 + 3 + n
ass a
n=1
nR
diverges because the partial sums grow beyond every number L. After n = 1, the partial sum sn = 1 + 4 + 9 + Á + n 2 is greater than n 2 . (b) The series
diverges because the partial sums eventually outgrow every preassigned number. Each term is greater than 1, so the sum of n terms is greater than n.
The nthTerm Test for Divergence
dH
Observe that limn: q an must equal zero if the series g n = 1 an converges. To see why, let S represent the series’ sum and sn = a1 + a2 + Á + an the nth partial sum. When n is large, both sn and sn  1 are close to S, so their difference, an , is close to zero. More formally, an = sn  sn  1
q
:
S  S = 0.
Difference Rule for sequences
ma
This establishes the following theorem.
Mu
ham
Caution q Theorem 7 does not say that g n = 1 an converges if an : 0 . It is possible for a series to diverge when an : 0 .
THEOREM 7 q
If a an converges, then an : 0. n=1
Theorem 7 leads to a test for detecting the kind of divergence that occurred in Example 6.
The nthTerm Test for Divergence q
lim an fails to exist or is different from zero. a an diverges if n: q
n=1
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.2
Applying the nthTerm Test
q
n=1 q
n + 1 n + 1 n diverges because n : 1
You
(b) a
suf
(a) a n 2 diverges because n 2 : q n=1 q
767
i
EXAMPLE 7
Infinite Series
(c) a s 1dn + 1 diverges because limn: q s 1dn + 1 does not exist n=1 q
n n 1 =  Z 0. (d) a diverges because limn: q 2n + 5 2 n = 1 2n + 5
an : 0 but the Series Diverges
iaz
EXAMPLE 8 The series
1 1 1 1 1 1 1 1 1 + + + + + + Á + n + n + Á + n + Á 2 2 4 4 4 4 2 2 2
nR
1 +
(')'* 2 terms
('''')''''* 4 terms
(''''')'''''* 2n terms
ass a
diverges because the terms are grouped into clusters that add to 1, so the partial sums increase without bound. However, the terms of the series form a sequence that converges to 0. Example 1 of Section 11.3 shows that the harmonic series also behaves in this manner.
Combining Series
dH
Whenever we have two convergent series, we can add them term by term, subtract them term by term, or multiply them by constants to make new convergent series.
ma
THEOREM 8 If gan = A and gbn = B are convergent series, then 1.
Mu
ham
2. 3.
Sum Rule: Difference Rule: Constant Multiple Rule:
gsan + bn d = gan + gbn = A + B gsan  bn d = gan  gbn = A  B sAny number kd. gkan = kgan = kA
Proof The three rules for series follow from the analogous rules for sequences in Theorem 1, Section 11.1. To prove the Sum Rule for series, let An = a1 + a2 + Á + an ,
Bn = b1 + b2 + Á + bn .
Then the partial sums of gsan + bn d are
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sn = sa1 + b1 d + sa2 + b2 d + Á + san + bn d = sa1 + Á + an d + sb1 + Á + bn d = An + Bn .
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 768
Chapter 11: Infinite Sequences and Series
suf
i
Since An : A and Bn : B, we have sn : A + B by the Sum Rule for sequences. The proof of the Difference Rule is similar. To prove the Constant Multiple Rule for series, observe that the partial sums of gkan form the sequence sn = ka1 + ka2 + Á + kan = ksa1 + a2 + Á + an d = kAn ,
1. 2.
You
which converges to kA by the Constant Multiple Rule for sequences. As corollaries of Theorem 8, we have
Every nonzero constant multiple of a divergent series diverges. If gan converges and gbn diverges, then gsan + bn d and gsan  bn d both diverge.
We omit the proofs.
iaz
CAUTION Remember that gsan + bn d can converge when gan and gbn both diverge. For example, gan = 1 + 1 + 1 + Á and gbn = s 1d + s 1d + s 1d + Á diverge, whereas gsan + bn d = 0 + 0 + 0 + Á converges to 0. EXAMPLE 9
nR
Find the sums of the following series.
3n  1  1 1 1 = a a n1  n1 b n1 6 6 n=1 n=1 2 q
q
(a) a
q
q
ass a
1 1 = a n1  a n1 n=1 2 n=1 6 1 1 = 1  s1>2d 1  s1>6d
dH
= 2 =
6 5
q
4 1 a n = 4 a 2n n=0 2 n=0
ma ham Mu
Geometric series with a = 1 and r = 1>2, 1>6
4 5
q
(b)
Difference Rule
= 4a
1 b 1  s1>2d
Constant Multiple Rule
Geometric series with a = 1, r = 1>2
= 8
Adding or Deleting Terms We can add a finite number of terms to a series or delete a finite number of terms without altering the series’ convergence or divergence, although in the case of convergence this will q q usually change the sum. If g n = 1 an converges, then g n = k an converges for any k 7 1 and q
q
Á + ak  1 + a an = a1 + a2 + a an .
n=1
Conversely, if
q g n = k an
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n=k
converges for any k 7 1, then q
q g n = 1 an
converges. Thus,
q
1 1 1 1 1 a 5n = 5 + 25 + 125 + a 5n n=4
n=1
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.2 Infinite Series
769
1 1 1 1 1 a 5n = a a 5n b  5  25  125 . n=4 n=1 q
suf
q
Richard Dedekind (1831–1916)
You
Reindexing HISTORICAL BIOGRAPHY
i
and
As long as we preserve the order of its terms, we can reindex any series without altering its convergence. To raise the starting value of the index h units, replace the n in the formula for an by n  h: q
q
n=1
Á. a an  h = a1 + a2 + a3 +
n=1+h
iaz
a an =
To lower the starting value of the index h units, replace the n in the formula for an by n + h: q
q
Á. a an + h = a1 + a2 + a3 +
nR
a an =
n=1
n=1h
ass a
It works like a horizontal shift. We saw this in starting a geometric series with the index n = 0 instead of the index n = 1, but we can use any other starting index value as well. We usually give preference to indexings that lead to simple expressions.
EXAMPLE 10
Reindexing a Geometric Series
We can write the geometric series q
dH
1 1 1 Á a 2n  1 = 1 + 2 + 4 + n=1
as
q
1 a n, n=0 2
q
1
a n  5, n=5 2
q
or even
1 a 2n + 4 .
n = 4
Mu
ham
ma
The partial sums remain the same no matter what indexing we choose.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
i f u s u o
11.2 Infinite Series
EXERCISES 11.2 Finding nth Partial Sums
1. 2 +
H d a
2 2 2 2 + + + Á + n1 + Á 3 9 27 3
9 9 9 9 Á + + Á + + 2. n + 2 3 100 100 100 100
m m a
1 1 1 1 +  + Á + s 1dn  1 n  1 + Á 2 4 8 2 4. 1  2 + 4  8 + Á + s 1dn  1 2n  1 + Á 3. 1 
h u M
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Y z a i R
5.
1 1 1 1 + # + # + Á + + Á 2#3 3 4 4 5 sn + 1dsn + 2d
6.
5 5 5 5 + # + # + Á + + Á 1#2 2 3 3 4 nsn + 1d
n a s as
In Exercises 1–6, find a formula for the nth partial sum of each series and use it to find the series’ sum if the series converges.
769
Series with Geometric Terms In Exercises 7–14, write out the first few terms of each series to show how the series starts. Then find the sum of the series. s 1dn 4n n=0 q
7. a
q
1 8. a n n=2 4
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 770
Chapter 11: Infinite Sequences and Series q
5 1 12. a a n  n b 3 n=0 2
41. a s 1dnx n
n
s 1d 1 b 13. a a n + 5n n=0 2
q n=0 q
43. a 3 a
2 14. a a n b 5 n=0 q
n+1
n=0
Telescoping Series Use partial fractions to find the sum of each series in Exercises 15–22. q
q
6 16. a n = 1 s2n  1ds2n + 1d
4 15. a n = 1 s4n  3ds4n + 1d
q
x  1 b 2
suf
q
i
Then express the inequality ƒ r ƒ 6 1 in terms of x and find the values of x for which the inequality holds and the series converges.
n=0 q
5 1 11. a a n + n b 3 n=0 2 q
5 10. a s 1dn n 4
42. a s 1dnx 2n n=0 q
n s 1dn 1 a b 2 3 + sin x n=0
n
44. a
You
q
7 9. a n n=1 4
In Exercises 45–50, find the values of x for which the given geometric series converges. Also, find the sum of the series (as a function of x) for those values of x. q
q
45. a 2nx n
46. a s 1dnx 2n
n=0 q
n=0 q
n
2n + 1 18. a 2 2 n = 1 n sn + 1d
47. a s 1dnsx + 1dn
1 48. a a b sx  3dn 2 n=0
19. a a
1 1 20. a a 1>n  1>sn + 1d b 2 n=1 2
49. a sinn x
50. a sln xdn
q
1
n=1
2n

1 2n + 1
b
q
n=0 q
q
22. a stan1 snd  tan1 sn + 1dd
51. 0.23 = 0.23 23 23 Á
n=1
52. 0.234 = 0.234 234 234 Á
ass a
Convergence or Divergence
53. 0.7 = 0.7777 Á
Which series in Exercises 23–40 converge, and which diverge? Give reasons for your answers. If a series converges, find its sum. 22
q
25. a s 1d
n+1
n=1 q
n
3 2n
26. a s 1d
n+1
n=1 q n=0 q
q
57. 1.24123 = 1.24 123 123 123 Á
n
cos np 5n
ma
n=0 q
1 32. a n , n=0 x
q
ƒxƒ 7 1
1 34. a a1  n b q
n
ham
2n  1 3n n=0
nn 36. a n = 1 n!
n b 37. a ln a n + 1 n=1
n b 38. a ln a 2n + 1 n=1
n
Mu
q
Theory and Examples 59. The series in Exercise 5 can also be written as q
1 a sn + 1dsn + 2d n=1
n=0
q
1 a sn + 3dsn + 4d . n = 1
and
Write it as a sum beginning with (a) n = 2 , (b) n = 0 , (c) n = 5 . 60. The series in Exercise 6 can also be written as
n=1 q
n! 35. a n n = 0 1000
e 39. a a p b
58. 3.142857 = 3.142857 142857 Á
n=1 q
2 31. a n n = 1 10
where d is a digit
55. 0.06 = 0.06666 Á
1 30. a ln n
29. a e 2n
q
54. 0.d = 0.dddd Á ,
56. 1.414 = 1.414 414 414 Á
q
28. a
n=0
q
q
n=0
27. a cos np
33. a
24. a A 22 B n
dH
n=0
b
n=0
Express each of the numbers in Exercises 51–58 as the ratio of two integers.
q
1
n=0
nR
q
q
q
Repeating Decimals
1 1 b 21. a a ln sn + 2d ln sn + 1d n=1
23. a a
iaz
40n 17. a 2 2 n = 1 s2n  1d s2n + 1d
q
q
5 a n = 1 nsn + 1d
q
and
5 . a n = 0 sn + 1dsn + 2d
q
Write it as a sum beginning with (a) n = 1 , (b) n = 3 , (c) n = 20 .
q
61. Make up an infinite series of nonzero terms whose sum is
e np 40. a ne n=0 p
Geometric Series
In each of the geometric series in Exercises 41–44, write out the first few terms of the series to find a and r, and find the sum of the series.
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a. 1
b. 3
c. 0.
62. (Continuation of Exercise 61.) Can you make an infinite series of nonzero terms that converges to any number you want? Explain.
63. Show by example that gsan>bn d may diverge even though gan and gbn converge and no bn equals 0.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 771
11.2 Infinite Series
1/8
65. Show by example that gsan>bn d may converge to something other than A> B even when A = gan, B = gbn Z 0 , and no bn equals 0.
67. What happens if you add a finite number of terms to a divergent series or delete a finite number of terms from a divergent series? Give reasons for your answer.
a. a = 2
that converges to the number
1/2
77. Helga von Koch’s snowflake curve Helga von Koch’s snowflake is a curve of infinite length that encloses a region of finite area. To see why this is so, suppose the curve is generated by starting with an equilateral triangle whose sides have length 1.
iaz
69. Make up a geometric series gar 5 if
n1
You
1/4
66. If gan converges and an 7 0 for all n, can anything be said about gs1>an d ? Give reasons for your answer.
68. If gan converges and gbn diverges, can anything be said about their termbyterm sum gsan + bn d ? Give reasons for your answer.
suf
i
64. Find convergent geometric series A = gan and B = gbn that illustrate the fact that gan bn may converge without being equal to AB.
b. a = 13>2 .
a. Find the length Ln of the nth curve Cn and show that limn: q Ln = q .
1 + e b + e 2b + e 3b + Á = 9 .
nR
70. Find the value of b for which
b. Find the area An of the region enclosed by Cn and calculate limn: q An .
71. For what values of r does the infinite series 1 + 2r + r 2 + 2r 3 + r 4 + 2r 5 + r 6 + Á
ass a
converge? Find the sum of the series when it converges.
72. Show that the error sL  sn d obtained by replacing a convergent geometric series with one of its partial sums sn is ar n>s1  rd .
Curve 1 Curve 2
dH
73. A ball is dropped from a height of 4 m. Each time it strikes the pavement after falling from a height of h meters it rebounds to a height of 0.75h meters. Find the total distance the ball travels up and down.
74. (Continuation of Exercise 73.) Find the total number of seconds the ball in Exercise 73 is traveling. (Hint: The formula s = 4.9t 2 gives t = 2s>4.9 .)
Mu
ham
ma
75. The accompanying figure shows the first five of a sequence of squares. The outermost square has an area of 4 m2 . Each of the other squares is obtained by joining the midpoints of the sides of the squares before it. Find the sum of the areas of all the squares.
76. The accompanying figure shows the first three rows and part of the fourth row of a sequence of rows of semicircles. There are 2n semicircles in the nth row, each of radius 1>2n . Find the sum of the areas of all the semicircles.
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Curve 3
Curve 4
78. The accompanying figure provides an informal proof that q g n = 1 s1>n 2 d is less than 2. Explain what is going on. (Source: “Convergence with Pictures” by P. J. Rippon, American Mathematical Monthly, Vol. 93, No. 6, 1986, pp. 476–478.)
1 72
1 32 1
1 1 22
1
1 62 1 52
…
1 42
1 2
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1 4
…
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 11: Infinite Sequences and Series
The Integral Test
i
11.3
suf
772
Given a series gan , we have two questions: Does the series converge? If it converges, what is its sum?
You
1. 2.
nR
iaz
Much of the rest of this chapter is devoted to the first question, and in this section we answer that q question by making a connection to the convergence of the improper integral 11 ƒsxd dx. However, as a practical matter the second question is also important, and we will return to it later. In this section and the next two, we study series that do not have negative terms. The reason for this restriction is that the partial sums of these series form nondecreasing sequences, and nondecreasing sequences that are bounded from above always converge (Theorem 6, Section 11.1). To show that a series of nonnegative terms converges, we need only show that its partial sums are bounded from above. It may at first seem to be a drawback that this approach establishes the fact of convergence without producing the sum of the series in question. Surely it would be better to compute sums of series directly from formulas for their partial sums. But in most cases such formulas are not available, and in their absence we have to turn instead to the twostep procedure of first establishing convergence and then approximating the sum.
ass a
Nondecreasing Partial Sums
Suppose that g n = 1 an is an infinite series with an Ú 0 for all n. Then each partial sum is greater than or equal to its predecessor because sn + 1 = sn + an : s1 … s2 … s3 … Á … sn … sn + 1 … Á . q
dH
Since the partial sums form a nondecreasing sequence, the Nondecreasing Sequence Theorem (Theorem 6, Section 11.1) tells us that the series will converge if and only if the partial sums are bounded from above.
ma
Corollary of Theorem 6 q A series g n = 1 an of nonnegative terms converges if and only if its partial sums are bounded from above.
EXAMPLE 1
ham
HISTORICAL BIOGRAPHY
The Harmonic Series
The series
Mu
Nicole Oresme (1320–1382)
q
1 1 1 Á + 1 + Á a n = 1 + 2 + 3 + n n=1 is called the harmonic series. The harmonic series is divergent, but this doesn’t follow from the nthTerm Test. The nth term 1> n does go to zero, but the series still diverges. The reason it diverges is because there is no upper bound for its partial sums. To see why, group the terms of the series in the following way: 1 +
1 1 1 1 1 1 1 1 1 1 + a + b + a + + + b + a + + Á + b + Á. 5 7 2 3 4 6 8 9 10 16 ('')''* 7
2 4
To Read it Online & Download:
=
1 2
(''''')'''''* 7
4 8
=
1 2
('''''')''''''* 7
8 16
=
1 2
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.3
The Integral Test
773
You
suf
i
The sum of the first two terms is 1.5. The sum of the next two terms is 1>3 + 1>4, which is greater than 1>4 + 1>4 = 1>2. The sum of the next four terms is 1>5 + 1>6 + 1>7 + 1>8, which is greater than 1>8 + 1>8 + 1>8 + 1>8 = 1>2. The sum of the next eight terms is 1>9 + 1>10 + 1>11 + 1>12 + 1>13 + 1>14 + 1>15 + 1>16, which is greater than 8>16 = 1>2. The sum of the next 16 terms is greater than 16>32 = 1>2, and so on. In general, the sum of 2n terms ending with 1>2n + 1 is greater than 2n>2n + 1 = 1>2. The sequence of partial sums is not bounded from above: If n = 2k , the partial sum sn is greater than k >2. The harmonic series diverges.
The Integral Test
EXAMPLE 2
y
iaz
We introduce the Integral Test with a series that is related to the harmonic series, but whose nth term is 1>n 2 instead of 1> n. Does the following series converge? q
nR
1 1 1 1 Á + 1 + Á a n 2 = 1 + 4 + 9 + 16 + n2
(1, f(1))
n=1
Graph of f(x) 12 x (2, f(2)) 1 32 (3, f(3))
1 42
1 22 0
q 2 11 s1>x d
1
2
3
4 …
1 n2
(n, f(n))
n1 n…
x
ma ham
Mu
Caution The series and integral need not have the same value in the convergent case. As we q noted in Example 2, g n = 1s1>n 2 d = q p2>6 while 11 s1>x 2 d dx = 1 .
1 1 1 1 + 2 + 2 + Á + 2 12 2 3 n = ƒs1d + ƒs2d + ƒs3d + Á + ƒsnd
sn =
dH
FIGURE 11.7 The sum of the areas of the rectangles under the graph of f (x) = 1>x 2 is less than the area under the graph (Example 2).
q
ass a
1 12
We determine the convergence of g n = 1s1>n 2 d by comparing it with dx. To carry out the comparison, we think of the terms of the series as values of the function ƒsxd = 1>x 2 and interpret these values as the areas of rectangles under the curve y = 1>x 2 . As Figure 11.7 shows, Solution
n
6 ƒs1d +
1 dx 2 x L1 q
1 dx x2 6 1 + 1 = 2. 6 1 +
L1
As in Section 8.8, Example 3, q 2 11 s1>x d dx = 1 .
Thus the partial sums of g n = 11>n 2 are bounded from above (by 2) and the series converges. The sum of the series is known to be p2>6 L 1.64493. (See Exercise 16 in Section 11.11.) q
THEOREM 9 The Integral Test Let 5an6 be a sequence of positive terms. Suppose that an = ƒsnd, where ƒ is a continuous, positive, decreasing function of x for all x Ú N (N a positive inteq q ger). Then the series g n = N an and the integral 1N ƒsxd dx both converge or both diverge. Proof We establish the test for the case N = 1. The proof for general N is similar. We start with the assumption that ƒ is a decreasing function with ƒsnd = an for every n. This leads us to observe that the rectangles in Figure 11.8a, which have areas
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 774
Chapter 11: Infinite Sequences and Series
a1, a2 , Á , an , collectively enclose more area than that under the curve y = ƒsxd from x = 1 to x = n + 1. That is,
suf
i
y
n+1
L1
y f (x) a2 an
0
1
n
3
2
n1
x
In Figure 11.8b the rectangles have been faced to the left instead of to the right. If we momentarily disregard the first rectangle, of area a1 , we see that n
(a)
a2 + a3 + Á + an …
y
L1
If we include a1 , we have
2
an n1 n
3
x
Combining these results gives n+1
(b)
L1
n
ƒsxd dx. L1 These inequalities hold for each n, and continue to hold as n : q . q If 11 ƒsxd dx is finite, the righthand inequality shows that gan is finite. If q 11 ƒsxd dx is infinite, the lefthand inequality shows that gan is infinite. Hence the series and the integral are both finite or both infinite. L1
FIGURE 11.8 Subject to the conditions of q the Integral Test, the series g n = 1 an and q the integral 11 ƒsxd dx both converge or both diverge.
ƒsxd dx.
iaz
a3
ƒsxd dx … a1 + a2 + Á + an … a1 +
nR
1
a1 + a2 + Á + an … a1 +
ass a
0
a2
ƒsxd dx.
n
y f (x)
a1
You
a1
ƒsxd dx … a1 + a2 + Á + an .
EXAMPLE 3
The pSeries
Show that the pseries
q
dH
1 1 1 1 Á + 1p + Á a n p = 1p + 2p + 3p + n n=1
( p a real constant) converges if p 7 1, and diverges if p … 1. If p 7 1, then ƒsxd = 1>x p is a positive decreasing function of x. Since
Mu
ham
ma
Solution
q
L1
b
x p + 1 1 x p dx = lim c d p dx = x b: q p + 1 1 L1 q
=
1 1 lim a  1b 1  p b: q b p  1
=
1 1 s0  1d = , 1  p p  1
b p  1 : q as b : q because p  1 7 0.
the series converges by the Integral Test. We emphasize that the sum of the pseries is not 1>s p  1d. The series converges, but we don’t know the value it converges to. If p 6 1, then 1  p 7 0 and q
L1
1 1 dx = lim sb 1  p  1d = q . 1  p b: q xp
The series diverges by the Integral Test.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.3 The Integral Test
775
1 1 1 + + Á + n + Á. 2 3
suf
1 +
i
If p = 1, we have the (divergent) harmonic series
We have convergence for p 7 1 but divergence for every other value of p.
EXAMPLE 4
iaz
You
The pseries with p = 1 is the harmonic series (Example 1). The pSeries Test shows that the harmonic series is just barely divergent; if we increase p to 1.000000001, for instance, the series converges! The slowness with which the partial sums of the harmonic series approaches infinity is impressive. For instance, it takes about 178,482,301 terms of the harmonic series to move the partial sums beyond 20. It would take your calculator several weeks to compute a sum with this many terms. (See also Exercise 33b.)
A Convergent Series
The series
q
nR
1 a n2 + 1
n=1
converges by the Integral Test. The function ƒsxd = 1>sx 2 + 1d is positive, continuous, and decreasing for x Ú 1, and b 1 dx = lim C arctan x D 1 q b: x + 1 = lim [arctan b  arctan 1]
ass a
q
b: q
=
p p p = . 2 4 4
dH
L1
2
Again we emphasize that p>4 is not the sum of the series. The series converges, but we do not know the value of its sum.
Mu
ham
ma
Convergence of the series in Example 4 can also be verified by comparison with the series g1>n 2 . Comparison tests are studied in the next section.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
i f su
u o zY
11.3 The Integral Test
EXERCISES 11.3 Determining Convergence or Divergence
a s as
Which of the series in Exercises 1–30 converge, and which diverge? Give reasons for your answers. (When you check an answer, remember that there may be more than one way to determine the series’ convergence or divergence.) q
1 1. a n n = 1 10 q
q
2. a e n n=1 q
m m a h u M
5 4. a n=1 n + 1 q
H ad
q
1 7. a  n 8 n=1
5. a
3
2n 8 8. a n n=1 q n=1
n 3. a n=1 n + 1 q
6. a
a i R n
q
2
n2n ln n 9. a n n=1 q
10. a
2n 2 13. a n=0 n + 1 n=2 q
q
5n 12. a n n=1 4 + 3
q
2n 15. a n + 1 n=1
2n 11. a n n=1 3
ln n
1 14. a n = 1 2n  1
q
q
n 2n 1 17. a 18. a a1 + n b n = 1 2n A 2n + 1 B n = 2 ln n n=1 q q 1 1 19. a 20. a n n n = 1 sln 2d n = 1 sln 3d q q s1>nd 1 21. a 22. a 2 n = 3 sln nd2ln2 n  1 n = 1 ns1 + ln nd q
16. a
1
q
q
n=2
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 776
Chapter 11: Infinite Sequences and Series q
e 25. a 2n n=1 1 + e q
2 26. a n n=1 1 + e
i
a. Show that
n=1 q
n
q
L2
dx xsln xd p
s p a positive constantd
q
1
8 tan n 27. a 2 n=1 1 + n
n 28. a 2 n=1 n + 1
q
converges if and only if p 7 1 .
b. What implications does the fact in part (a) have for the convergence of the series
q
29. a sech n
30. a sech2 n
n=1
n=1
You
n=1 q
39. Logarithmic pseries
1 24. a n tan n
suf
q
1 23. a n sin n
q
1 p ? a n = 2 nsln nd
Theory and Examples 2a 1 b 32. a a n + 1 n=3 n  1
q
q
33. a. Draw illustrations like those in Figures 11.7 and 11.8 to show that the partial sums of the harmonic series satisfy the inequalities L1
1 b. a 1.01 n = 2 nsln nd
q
1 1 Á + 1 x dx … 1 + 2 + n n
… 1 +
q
1 a. a n = 2 nsln nd
L1
1 x dx = 1 + ln n.
dH
T b. There is absolutely no empirical evidence for the divergence of the harmonic series even though we know it diverges. The partial sums just grow too slowly. To see what we mean, suppose you had started with s1 = 1 the day the universe was formed, 13 billion years ago, and added a new term every second. About how large would the partial sum sn be today, assuming a 365day year?
34. Are there any values of x for which g n = 1s1>snxdd converges? Give reasons for your answer. q
q g n = 1 an
q
1 1 c. a d. a 3 n ln sn d nsln nd3 n=2 n=2 41. Euler’s constant Graphs like those in Figure 11.8 suggest that as n increases there is little change in the difference between the sum
ass a
n+1
ln sn + 1d =
q
iaz
a 1 b 31. a a n + 4 n=1 n + 2
Give reasons for your answer.
40. (Continuation of Exercise 39.) Use the result in Exercise 39 to determine which of the following series converge and which diverge. Support your answer in each case.
nR
For what values of a, if any, do the series in Exercises 31 and 32 converge?
1 1 + Á + n 2
1 +
and the integral n
ln n =
L1
1 x dx .
To explore this idea, carry out the following steps. a. By taking ƒsxd = 1>x in the proof of Theorem 9, show that
or
36. (Continuation of Exercise 35.) Is there a “largest” convergent series of positive numbers? Explain.
0 6 ln sn + 1d  ln n … 1 +
ham
ma
35. Is it true that if is a divergent series of positive numbers q then there is also a divergent series g n = 1 bn of positive numbers with bn 6 an for every n? Is there a “smallest” divergent series of positive numbers? Give reasons for your answers.
37. The Cauchy condensation test The Cauchy condensation test says: Let 5an6 be a nonincreasing sequence (an Ú an + 1 for all n) of positive terms that converges to 0. Then gan converges if and only if g2na2n converges. For example, gs1>nd diverges because g2n # s1>2n d = g1 diverges. Show why the test works. 38. Use the Cauchy condensation test from Exercise 37 to show that q
Mu
1 a. a diverges; n = 2 n ln n q
1 b. a p converges if p 7 1 and diverges if p … 1 . n=1 n
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ln sn + 1d … 1 +
1 1 + Á + n … 1 + ln n 2
1 1 + Á + n  ln n … 1 . 2
Thus, the sequence an = 1 +
1 1 + Á + n  ln n 2
is bounded from below and from above. b. Show that 1 6 n + 1 Ln
n+1
1 x dx = ln sn + 1d  ln n ,
and use this result to show that the sequence 5an6 in part (a) is decreasing.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.3 The Integral Test
42. Use the integral test to show that
1 1 1 + + Á + n  ln n : g . 2
q
ae
You
converges.
Mu
ham
ma
dH
ass a
nR
iaz
The number g , whose value is 0.5772 Á , is called Euler’s constant. In contrast to other special numbers like p and e, no other
n2
n=0
i
expression with a simple law of formulation has ever been found for g .
suf
Since a decreasing sequence that is bounded from below converges (Exercise 107 in Section 11.1), the numbers an defined in part (a) converge:
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
Comparison Tests
usu
11.4
777
fi
11.4 Comparison Tests
Yo
We have seen how to determine the convergence of geometric series, pseries, and a few others. We can test the convergence of many more series by comparing their terms to those of a series whose convergence is known.
iaz
THEOREM 10 The Comparison Test Let gan be a series with no negative terms.
(a) gan converges if there is a convergent series gcn with an … cn for all n 7 N, for some integer N.
nR
(b) gan diverges if there is a divergent series of nonnegative terms gdn with an Ú dn for all n 7 N, for some integer N.
Ha ssa
Proof In Part (a), the partial sums of gan are bounded above by HISTORICAL BIOGRAPHY
q
M = a1 + a2 + Á + aN +
Albert of Saxony (ca. 1316–1390)
a cn .
n=N+1
ad
They therefore form a nondecreasing sequence with a limit L … M. In Part (b), the partial sums of gan are not bounded from above. If they were, the partial sums for gdn would be bounded by q
M * = d1 + d2 + Á + dN +
a an
n=N+1
Mu
ha
mm
and gdn would have to converge instead of diverge.
EXAMPLE 1
Applying the Comparison Test
(a) The series q
5 a 5n  1 n=1 diverges because its nth term 5 = 5n  1
1 7 n 1 n 5 is greater than the nth term of the divergent harmonic series.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 778
Chapter 11: Infinite Sequences and Series
i
(b) The series 1 1 1 1 Á a n! = 1 + 1! + 2! + 3! + n=0
suf
q
You
converges because its terms are all positive and less than or equal to the corresponding terms of q
1 1 1 1 + a n = 1 + 1 + + 2 + Á. 2 2 n=0 2 The geometric series on the left converges and we have q
iaz
1 1 1 + a n = 1 + = 3. 1  s1>2d n=0 2
The fact that 3 is an upper bound for the partial sums of g n = 0 s1>n!d does not mean that the series converges to 3. As we will see in Section 11.9, the series converges to e. (c) The series
nR
2 1 1 1 1 1 + + 1 + + + + Á + + Á n 7 3 2 + 21 4 + 22 8 + 23 2 + 2n
ass a
5 +
q
converges. To see this, we ignore the first three terms and compare the remaining terms q with those of the convergent geometric series g n = 0 s1>2n d. The term 1>s2n + 2nd of the truncated sequence is less than the corresponding term 1>2n of the geometric series. We see that term by term we have the comparison, 1
dH 1 +
2 + 21
+
1
4 + 22
+
1
1 1 1 + Á … 1 + + + + Á 2 4 8 8 + 23
So the truncated series and the original series converge by an application of the Comparison Test.
ma
The Limit Comparison Test
Mu
ham
We now introduce a comparison test that is particularly useful for series in which an is a rational function of n.
THEOREM 11 Limit Comparison Test Suppose that an 7 0 and bn 7 0 for all n Ú N (N an integer). an = c 7 0, then gan and gbn both converge or both diverge. n: q bn
1.
If lim
2.
If lim
an = 0 and gbn converges, then gan converges. bn
3.
If lim
an = q and gbn diverges, then gan diverges. bn
n: q
n: q
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.4 Comparison Tests
779
n 7 N Q `
suf
i
Proof We will prove Part 1. Parts 2 and 3 are left as Exercises 37(a) and (b). Since c>2 7 0, there exists an integer N such that for all n Limit definition with P = c>2, L = c , and an replaced by an>bn
an c  c` 6 . 2 bn

You
Thus, for n 7 N, an c c 6  c 6 , 2 2 bn
iaz
an 3c c 6 6 , 2 2 bn
3c c a bbn 6 an 6 a bbn . 2 2
Using the Limit Comparison Test
ass a
EXAMPLE 2
nR
If gbn converges, then gs3c>2dbn converges and gan converges by the Direct Comparison Test. If gbn diverges, then gsc>2dbn diverges and gan diverges by the Direct Comparison Test.
Which of the following series converge, and which diverge? q
(a)
q
5 7 9 3 2n + 1 2n + 1 + + + + Á = a = a 2 2 4 9 16 25 sn + 1d n + 2n + 1 n=1 n=1 q
1 1 1 1 1 + + + + Á = a n 7 1 3 15 2  1 n=1
(c)
1 + 3 ln 3 1 + n ln n 1 + 4 ln 4 1 + 2 ln 2 + + + Á = a 2 9 14 21 n=2 n + 5
dH (b)
ma
q
Solution
Mu
ham
(a) Let an = s2n + 1d>sn 2 + 2n + 1d. For large n, we expect an to behave like 2n>n 2 = 2>n since the leading terms dominate for large n, so we let bn = 1>n. Since q
q
n=1
n=1
1 a bn = a n diverges
and an 2n 2 + n = lim 2 = 2, n: q bn n: q n + 2n + 1 lim
gan diverges by Part 1 of the Limit Comparison Test. We could just as well have taken bn = 2>n, but 1> n is simpler.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 780
Chapter 11: Infinite Sequences and Series
q
q
1 a bn = a 2n converges n=1 n=1
You
and
suf
i
(b) Let an = 1>s2n  1d. For large n, we expect an to behave like 1>2n , so we let bn = 1>2n . Since
an 2n = lim n n: q bn n: q 2  1 lim
= lim
n: q
1 1  s1>2n d
iaz
= 1,
nR
gan converges by Part 1 of the Limit Comparison Test. (c) Let an = s1 + n ln nd>sn 2 + 5d. For large n, we expect an to behave like sn ln nd>n 2 = sln nd>n, which is greater than 1> n for n Ú 3, so we take bn = 1>n. Since q
q
n=2
n=2
and
ass a
1 a bn = a n diverges
an n + n 2 ln n = lim n: q bn n: q n2 + 5 = q, lim
dH
gan diverges by Part 3 of the Limit Comparison Test. q
EXAMPLE 3
ln n converge? 3>2 n=1 n
Does a
Because ln n grows more slowly than n c for any positive constant c (Section 11.1, Exercise 91), we would expect to have
Mu
ham
ma
Solution
ln n n 1>4 1 6 3>2 = 5>4 3>2 n n n
for n sufficiently large. Indeed, taking an = sln nd>n 3>2 and bn = 1>n 5>4 , we have lim
n: q
an ln n = lim 1>4 bn n: q n 1>n = lim n: q s1>4dn 3>4
l’Hôpital’s Rule
4 = 0. n 1>4 Since gbn = gs1>n 5>4 d (a pseries with p 7 1) converges, gan converges by Part 2 of the Limit Comparison Test.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.4 Comparison Tests
Determining Convergence or Divergence
n
q
8. a
n=1 q
q
1 10. a 2 n = 2 sln nd q
13. a
1
2n ln n 1 16. a 2 n = 1 s1 + ln nd n=2 q
q
n=1 q
14. a
n 2n 2  1 n + 2n 22. a 2 n n=1 n 2 n=2 q
6. a
n + 1
40. Prove that if gan is a convergent series of nonnegative terms, then ga n2 converges.
n 2 2n 1 9. a n = 3 ln sln nd n=1 q
1 2n + 2 sln nd2 3
n sln nd2
q
12. a
sln nd3 n
n=1 q
n 3>2 ln sn + 1d 17. a n + 1 n=2 n=1 q
2n 20. a 2 n=1 n + 1
3
1 15. a n = 1 1 + ln n q
1 18. a 2 n = 1 s1 + ln nd q
1  n 21. a n n = 1 n2
q
1 23. a n  1 + 1 n=1 3
q
sin n 2n
n=1 q
3
q
1
19. a
11. a
3. a
q
3n  1 + 1 3n n=1
24. a
q
1 25. a sin n
1 26. a tan n
10n + 1 27. a nsn + 1dsn + 2d n=1
5n 3  3n 28. a 2 n sn  2dsn 2 + 5d n=3
29. a
n=1 q
32. a
n=1 q
q
1
tan n n 1.1 tanh n n2
30. a
n=1 q
33. a
n=1
1
sec n n 1.3
q
31. a
n=1 q
coth n n2 n
2n 34. a 2 n=1 n
1 n
n 2n
ma
q
n=1 q
dH
n=1 q
1 35. a 1 + 2 + 3 + Á + n n=1
You
n + 2n 2n 5. a n = 1 3n  1 n=1 q
COMPUTER EXPLORATION
41. It is not yet known whether the series
iaz
n b 7. a a n = 1 3n + 1 q
3
2. a
3 2 2n + 2 n 1 + cos n 4. a n2 n=1
n=1 q
39. Suppose that an 7 0 and bn 7 0 for n Ú N (N an integer). If lim n: q san>bn d = q and gan converges, can anything be said about g bn ? Give reasons for your answer.
q
1 a 3 2 n = 1 n sin n
converges or diverges. Use a CAS to explore the behavior of the series by performing the following steps.
nR
1
1. a
2
q
a. Define the sequence of partial sums k
1 sk = a 3 2 . n = 1 n sin n
ass a
q
38. If g n = 1 an is a convergent series of nonnegative numbers, can q anything be said about g n = 1san>nd ? Explain. q
Which of the series in Exercises 1–36 converge, and which diverge? Give reasons for your answers. q
suf
i
EXERCISES 11.4
781
What happens when you try to find the limit of sk as k : q ? Does your CAS find a closed form answer for this limit?
b. Plot the first 100 points sk, sk d for the sequence of partial sums. Do they appear to converge? What would you estimate the limit to be? c. Next plot the first 200 points sk, sk d . Discuss the behavior in your own words.
d. Plot the first 400 points sk, sk d . What happens when k = 355 ? Calculate the number 355> 113. Explain from your calculation what happened at k = 355 . For what values of k would you guess this behavior might occur again?
q
1 36. a 2 2 1 + 2 + 3 + Á + n2 n=1
Theory and Examples
You will find an interesting discussion of this series in Chapter 72 of Mazes for the Mind by Clifford A. Pickover, St. Martin’s Press, Inc., New York, 1992.
Mu
ham
37. Prove (a) Part 2 and (b) Part 3 of the Limit Comparison Test.
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i f u s u o Y
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
11.5 The Ratio and Root Tests
11.5
The Ratio and Root Tests
z a i R n a
s s a dH
781
The Ratio Test measures the rate of growth (or decline) of a series by examining the ratio an + 1>an . For a geometric series gar n , this rate is a constant ssar n + 1 d>sar n d = rd, and the series converges if and only if its ratio is less than 1 in absolute value. The Ratio Test is a powerful rule extending that result. We prove it on the next page using the Comparison Test.
Mu
a m m a h
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
THEOREM 12 The Ratio Test Let gan be a series with positive terms and suppose that an + 1 = r. n: q an Then
iaz
(a) the series converges if r 6 1, (b) the series diverges if r 7 1 or r is infinite, (c) the test is inconclusive if r = 1.
You
lim
i
Chapter 11: Infinite Sequences and Series
suf
782
Proof
nR
(a) R<1. Let r be a number between r and 1. Then the number P = r  r is positive. Since an + 1 an : r,
aN + 1 6 raN , aN + 2 6 raN + 1 6 r 2aN , aN + 3 6 raN + 2 6 r 3aN , o aN + m 6 raN + m  1 6 r maN .
dH
That is,
ass a
an + 1>an must lie within P of r when n is large enough, say for all n Ú N. In particular an + 1 when n Ú N. an 6 r + P = r,
Mu
ham
ma
These inequalities show that the terms of our series, after the Nth term, approach zero more rapidly than the terms in a geometric series with ratio r 6 1. More precisely, consider the series gcn , where cn = an for n = 1, 2, Á , N and cN + 1 = raN , cN + 2 = r 2aN , Á , cN + m = r maN , Á . Now an … cn for all n, and q
Á + aN  1 + aN + raN + r 2aN + Á a cn = a1 + a2 +
n=1
= a1 + a2 + Á + aN  1 + aN s1 + r + r 2 + Á d. The geometric series 1 + r + r 2 + Á converges because ƒ r ƒ 6 1, so gcn converges. Since an … cn , gan also converges. (b) 1<R ◊ ˆ . From some index M on, an + 1 an 7 1
and
aM 6 aM + 1 6 aM + 2 6 Á .
The terms of the series do not approach zero as n becomes infinite, and the series diverges by the nthTerm Test.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.5
783
The Ratio and Root Tests
1 a n
q
and
n=1
1 a n2 n=1
suf
q
i
(c) R = 1. The two series
show that some other test for convergence must be used when r = 1. q
1>sn + 1d an + 1 n : 1. = an = n + 1 1>n
You
1 For a n : n=1
2 1>sn + 1d2 an + 1 n = = a b : 12 = 1. an n + 1 1>n 2
q
1 For a 2 : n=1 n
iaz
In both cases, r = 1, yet the first series diverges, whereas the second converges. The Ratio Test is often effective when the terms of a series contain factorials of expressions involving n or expressions raised to a power involving n.
Applying the Ratio Test
nR
EXAMPLE 1
Investigate the convergence of the following series. q
2n + 5 3n n=0
q s2nd! (b) a n = 1 n!n!
ass a
(a) a
q
4nn!n! (c) a n = 1 s2nd!
Solution
(a) For the series g n = 0 s2n + 5d>3n , q
s2n + 1 + 5d>3n + 1 an + 1 1 an = s2n + 5d>3n = 3
#
2n + 1 + 5 1 = 3 2n + 5
1#2 # a 2 + 5 ## 2n n b : 1 + 5 2
3 1
=
2 . 3
dH
The series converges because r = 2>3 is less than 1. This does not mean that 2> 3 is the sum of the series. In fact, n
2n + 5 5 5 21 2 1 a 3n = a a 3 b + a 3n = 1  s2>3d + 1  s1>3d = 2 . n=0 n=0 n=0
ma
q
q
s2n + 2d! s2nd! and , then an + 1 = n!n! sn + 1d!sn + 1d! n!n!s2n + 2ds2n + 1ds2nd! an + 1 an = sn + 1d!sn + 1d!s2nd! =
s2n + 2ds2n + 1d 4n + 2 : 4. = n + 1 sn + 1dsn + 1d
The series diverges because r = 4 is greater than 1. (c) If an = 4nn!n!>s2nd!, then
Mu
ham
(b) If an =
q
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4n + 1sn + 1d!sn + 1d! # s2nd! an + 1 an = s2n + 2ds2n + 1ds2nd! 4nn!n! =
2sn + 1d 4sn + 1dsn + 1d : 1. = 2n + 1 s2n + 2ds2n + 1d
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 784
Chapter 11: Infinite Sequences and Series
suf
i
Because the limit is r = 1, we cannot decide from the Ratio Test whether the series converges. When we notice that an + 1>an = s2n + 2d>s2n + 1d, we conclude that an + 1 is always greater than an because s2n + 2d>s2n + 1d is always greater than 1. Therefore, all terms are greater than or equal to a1 = 2, and the nth term does not approach zero as n : q . The series diverges.
You
The Root Test
The convergence tests we have so far for gan work best when the formula for an is relatively simple. But consider the following.
Solution
Let an = e
n>2n, 1>2n,
n odd n even.
Does gan converge?
iaz
EXAMPLE 2
We write out several terms of the series: q
=
nR
3 5 7 1 1 1 1 Á a an = 21 + 22 + 23 + 24 + 25 + 26 + 27 + n=1 3 5 7 1 1 1 1 + + + + + + + Á. 2 4 8 16 32 64 128
ass a
Clearly, this is not a geometric series. The nth term approaches zero as n : q , so we do not know if the series diverges. The Integral Test does not look promising. The Ratio Test produces 1 , 2n
n odd
n + 1 , 2
n even.
dH
an + 1 an = d
As n : q , the ratio is alternately small and large and has no limit. A test that will answer the question (the series converges) is the Root Test.
ma
THEOREM 13 The Root Test Let gan be a series with an Ú 0 for n Ú N, and suppose that n
lim 2an = r.
n: q
Mu
ham
Then
(a) the series converges if r 6 1, (b) the series diverges if r 7 1 or r is infinite, (c) the test is inconclusive if r = 1.
Proof n
n
(a) R<1. Choose an P 7 0 so small that r + P 6 1. Since 2an : r, the terms 2an eventually get closer than P to r. In other words, there exists an index M Ú N such that
To Read it Online & Download:
n
2an 6 r + P
when n Ú M.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.5
The Ratio and Root Tests
785
an 6 sr + Pdn
suf
for n Ú M.
i
Then it is also true that Now, g n = M sr + Pdn , a geometric series with ratio sr + Pd 6 1, converges. By q comparison, g n = M an converges, from which it follows that q
q
q
You
Á + aM  1 + a an = a1 + a an
n=1
n=M
iaz
converges. n (b) 1<R ◊ ˆ . For all indices beyond some integer M, we have 2an 7 1, so that an 7 1 for n 7 M. The terms of the series do not converge to zero. The series diverges by the nthTerm Test. (c) R = 1. The series g n = 1 s1>nd and g n = 1 s1>n 2 d show that the test is not conclusive when r = 1. The first series diverges and the second converges, but in both cases n 2an : 1.
EXAMPLE 3
q
nR
q
Applying the Root Test
Which of the following series converges, and which diverges? q
q
n2 (a) a n n=1 2
1 b (c) a a n=1 1 + n q
ass a
2n (b) a 2 n=1 n
Solution
n
2 2n 2 n2 n n (a) a n converges because n = n n = B2 n=1 2 22
dH
q
q
n 2n n 2 (b) a 2 diverges because = A n2 n=1 n
n
n nB2 A2
2
:
1 6 1. 2
2 2 : 7 1. n 2 1 2 n A B n
n
1 n 1 1 b converges because a b = : 0 6 1. (c) a a 1 + n B 1 + n n=1 1 + n
ma
q
Mu
ham
EXAMPLE 2 Let an = e Solution
n>2n, 1>2n,
Revisited n odd n even.
Does gan converge?
We apply the Root Test, finding that n 2n>2, n 2an = e 1>2,
n odd n even.
Therefore, n
2n 1 n … 2an … . 2 2 n
n
Since 2n : 1 (Section 11.1, Theorem 5), we have limn: q 2an = 1>2 by the Sandwich Theorem. The limit is less than 1, so the series converges by the Root Test.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 786
Chapter 11: Infinite Sequences and Series
Determining Convergence or Divergence
n
q
n 1. a n n=1 2
2. a n 2e n n=1 q
q
3. a n!e n
n! 4. a n n = 1 10
n 10 5. a n n = 1 10
6. a a
2 + s 1dn 7. a 1.25n n=1
8. a
n=1 q
q
n=1 q
n
1 b 10. a a1 3n n=1 q
n=1 q
n
q
q
q
ln n 15. a n
16. a
n=1 q
sn + 1dsn + 2d n! n=1
n=1 q
n ln n 2n
18. a e nsn 3 d
sn + 3d! 19. a n n = 1 3!n!3
20. a
dH
17. a
n=1 q
n
n2nsn + 1d! 3nn! n=1
q
q
q
n! 21. a n = 1 s2n + 1d!
n! 22. a n n=1 n
q
q
q
2
ham
q
an + 1 =
28. a1 = 1,
an + 1 =
1 , 3
an + 1 =
30. a1 = 3,
an + 1 =
31. a1 = 2,
an + 1 =
q
40. a
sn!dn 2
n sn d nn 42. a n 2 n = 1 s2 d n=1 q
2sn d 1 # 3 # Á # s2n  1d 43. a 4n2nn! n=1 q 1 # 3 # Á # s2n  1d 44. a # # Á # s2nd]s3n + 1d n = 1 [2 4 n=1 q
Theory and Examples 45. Neither the Ratio nor the Root Test helps with pseries. Try them on q
1 a np
n=1
q
3n 26. a 3 n n=1 n 2
Which of the series g n = 1 an defined by the formulas in Exercises 27–38 converge, and which diverge? Give reasons for your answers.
and show that both tests fail to provide information about convergence. 46. Show that neither the Ratio Test nor the Root Test provides information about the convergence of
1 + sin n an n
1 + tan1 n an n 3n  1 a 2n + 5 n n a n + 1 n 2 n an
Mu
29. a1 =
nn
24. a
n! ln n 25. a n = 1 nsn + 2d!
27. a1 = 2,
41. a
n sn>2d n = 2 sln nd
ma
n 23. a n n = 2 sln nd
q
ass a
1 1 14. a a n  2 b n n=1
s3nd! n!sn + 1d!sn + 2d!
q sn!dn 39. a n 2 n = 1 sn d
sln ndn nn n=1
q
1 , an + 1 2 1 n 35. a1 = , an + 1 = 2an 3 1 36. a1 = , an + 1 = san dn + 1 2 2nn!n! 37. an = s2nd! 34. a1 =
Which of the series in Exercises 39–44 converge, and which diverge? Give reasons for your answers.
12. a
1 1 13. a a n  2 b n n=1
an + 1
38. an =
q
ln n 11. a 3 n=1 n
33. a1 = 1,
nR
q
n
s 2dn 3n n=1
q
3 9. a a1  n b
n  2 n b
an + 1 =
You
22
2n a 2 n 1 + ln n = an n n + ln n = a n + 10 n
32. a1 = 5,
iaz
Which of the series in Exercises 1–26 converge, and which diverge? Give reasons for your answers. (When checking your answers, remember there may be more than one way to determine a series’ convergence or divergence.) q
suf
i
EXERCISES 11.5
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q
1 a sln nd p
s p constantd .
n=2
n>2n, if n is a prime number 1>2n, otherwise. Does gan converge? Give reasons for your answer.
47. Let an = e
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
Alternating Series, Absolute and Conditional Convergence
787
suf
11.6
Alternating Series, Absolute and Conditional Convergence
i
11.6
A series in which the terms are alternately positive and negative is an alternating series. Here are three examples: s 1dn + 1 1 1 1 1 + Á +  +  Á + n 5 2 3 4
2 + 1 
You
1 
s 1dn4 1 1 1 +  + Á + + Á 2 4 8 2n
1  2 + 3  4 + 5  6 + Á + s 1dn + 1n + Á
(1) (2) (3)
THEOREM 14 The series
nR
iaz
Series (1), called the alternating harmonic series, converges, as we will see in a moment. Series (2) a geometric series with ratio r = 1>2, converges to 2>[1 + s1>2d] = 4>3. Series (3) diverges because the nth term does not approach zero. We prove the convergence of the alternating harmonic series by applying the Alternating Series Test.
The Alternating Series Test (Leibniz’s Theorem)
ass a
q
a s 1d
n+1
un = u1  u2 + u3  u4 + Á
n=1
converges if all three of the following conditions are satisfied: The un’s are all positive. un Ú un + 1 for all n Ú N, for some integer N. un : 0.
dH
1. 2. 3.
Mu
ham
ma
Proof If n is an even integer, say n = 2m, then the sum of the first n terms is s2m = su1  u2 d + su3  u4 d + Á + su2m  1  u2m d = u1  su2  u3 d  su4  u5 d  Á  su2m  2  u2m  1 d  u2m .
The first equality shows that s2m is the sum of m nonnegative terms, since each term in parentheses is positive or zero. Hence s2m + 2 Ú s2m , and the sequence 5s2m6 is nondecreasing. The second equality shows that s2m … u1 . Since 5s2m6 is nondecreasing and bounded from above, it has a limit, say lim s2m = L.
m: q
(4)
If n is an odd integer, say n = 2m + 1, then the sum of the first n terms is s2m + 1 = s2m + u2m + 1 . Since un : 0, lim u2m + 1 = 0
m: q
and, as m : q , s2m + 1 = s2m + u2m + 1 : L + 0 = L.
(5)
Combining the results of Equations (4) and (5) gives lim sn = L (Section 11.1, Exern: q cise 119).
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 11: Infinite Sequences and Series
The alternating harmonic series
i
EXAMPLE 1
q
a s 1d
n+1
n=1
1 1 1 1 Á n = 1  2 + 3  4 +
suf
788
satisfies the three requirements of Theorem 14 with N = 1; it therefore converges.
You
u4
s2
s4
L
s3
s1
x
FIGURE 11.9 The partial sums of an alternating series that satisfies the hypotheses of Theorem 14 for N = 1 straddle the limit from the beginning.
iaz
u2 u3
0
A graphical interpretation of the partial sums (Figure 11.9) shows how an alternating series converges to its limit L when the three conditions of Theorem 14 are satisfied with N = 1. (Exercise 63 asks you to picture the case N 7 1.) Starting from the origin of the xaxis, we lay off the positive distance s1 = u1 . To find the point corresponding to s2 = u1  u2 , we back up a distance equal to u2 . Since u2 … u1 , we do not back up any farther than the origin. We continue in this seesaw fashion, backing up or going forward as the signs in the series demand. But for n Ú N, each forward or backward step is shorter than (or at most the same size as) the preceding step, because un + 1 … un . And since the nth term approaches zero as n increases, the size of step we take forward or backward gets smaller and smaller. We oscillate across the limit L, and the amplitude of oscillation approaches zero. The limit L lies between any two successive sums sn and sn + 1 and hence differs from sn by an amount less than un + 1 . Because
nR
u1
ass a
for n Ú N, ƒ L  sn ƒ 6 un + 1 we can make useful estimates of the sums of convergent alternating series.
dH
THEOREM 15 The Alternating Series Estimation Theorem q If the alternating series g n = 1 s 1dn + 1un satisfies the three conditions of Theorem 14, then for n Ú N, sn = u1  u2 + Á + s 1dn + 1un
ma
approximates the sum L of the series with an error whose absolute value is less than un + 1 , the numerical value of the first unused term. Furthermore, the remainder, L  sn , has the same sign as the first unused term.
Mu
EXAMPLE 2
We try Theorem 15 on a series whose sum we know:
q
1 1 1 1 1 1 1 n 1 a s 1d 2n = 1  2 + 4  8 + 16  32 + 64  128 n=0

ham
We leave the verification of the sign of the remainder for Exercise 53.
+
1  Á. 256
The theorem says that if we truncate the series after the eighth term, we throw away a total that is positive and less than 1> 256. The sum of the first eight terms is 0.6640625. The sum of the series is 1 2 1 = . = 3 3>2 1  s 1>2d The difference, s2>3d  0.6640625 = 0.0026041666 Á , is positive and less than s1>256d = 0.00390625.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.6
Alternating Series, Absolute and Conditional Convergence
789
suf
i
Absolute and Conditional Convergence
You
DEFINITION Absolutely Convergent A series gan converges absolutely (is absolutely convergent) if the corresponding series of absolute values, g ƒ an ƒ , converges.
The geometric series
1 1 1 +  + Á 2 4 8
iaz
1 
converges absolutely because the corresponding series of absolute values 1 1 1 + + + Á 2 4 8
nR
1 +
ass a
converges. The alternating harmonic series does not converge absolutely. The corresponding series of absolute values is the (divergent) harmonic series.
DEFINITION Conditionally Convergent A series that converges but does not converge absolutely converges conditionally.
ma
dH
The alternating harmonic series converges conditionally. Absolute convergence is important for two reasons. First, we have good tests for convergence of series of positive terms. Second, if a series converges absolutely, then it converges. That is the thrust of the next theorem.
THEOREM 16
The Absolute Convergence Test
q
q
Mu
ham
If a ƒ an ƒ converges, then a an converges. n=1
n=1
Proof For each n,  ƒ an ƒ … an … ƒ an ƒ , q gn=1
so
0 … an + ƒ an ƒ … 2 ƒ an ƒ .
q gn=1
If 2 ƒ an ƒ converges and, by the Direct Comparison Test, ƒ an ƒ converges, then q the nonnegative series g n = 1 san + ƒ an ƒ d converges. The equality an = san + ƒ an ƒ d  ƒ an ƒ q now lets us express g n = 1 an as the difference of two convergent series: q
q
q
q
a an = a san + ƒ an ƒ  ƒ an ƒ d = a san + ƒ an ƒ d  a ƒ an ƒ .
Therefore,
n=1
n=1
q g n = 1 an
converges.
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n=1
n=1
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 790
Chapter 11: Infinite Sequences and Series
EXAMPLE 3
Applying the Absolute Convergence Test
q
suf
i
CAUTION We can rephrase Theorem 16 to say that every absolutely convergent series converges. However, the converse statement is false: Many convergent series do not converge absolutely (such as the alternating harmonic series in Example 1).
You
1 1 1 1 (a) For a s 1dn + 1 2 = 1  + + Á , the corresponding series of absolute 4 9 16 n n=1 values is the convergent series q
1 1 1 1 Á. a n 2 = 1 + 4 + 9 + 16 +
n=1
iaz
The original series converges because it converges absolutely. q
nR
sin n sin 1 sin 2 sin 3 (b) For a 2 = + + + Á , the corresponding series of absolute 1 4 9 n=1 n values is q ƒ sin 1 ƒ ƒ sin 2 ƒ sin n + + Á, a ` n2 ` = 1 4 n=1 which converges by comparison with g n = 1 s1>n 2 d because ƒ sin n ƒ … 1 for every n. The original series converges absolutely; therefore it converges.
ass a
q
EXAMPLE 4
Alternating pSeries
If p is a positive constant, the sequence 51>n p6 is a decreasing sequence with limit zero. Therefore the alternating pseries s 1dn  1 1 1 1 = 1  p + p  p + Á, a np 2 3 4 n=1
dH
q
p 7 0
converges. If p 7 1, the series converges absolutely. If 0 6 p … 1, the series converges conditionally.
ma
Conditional convergence: Absolute convergence:
1 1 1 + + Á 22 23 24 1 1 1 1  3>2 + 3>2  3>2 + Á 2 3 4 1 
Mu
ham
Rearranging Series
THEOREM 17
The Rearrangement Theorem for Absolutely Convergent Series q If g n = 1 an converges absolutely, and b1, b2 , Á , bn , Á is any arrangement of the sequence 5an6, then gbn converges absolutely and q
q
a bn = a an .
n=1
n=1
(For an outline of the proof, see Exercise 60.)
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.6
791
Applying the Rearrangement Theorem
i
EXAMPLE 5
Alternating Series, Absolute and Conditional Convergence
1 
suf
As we saw in Example 3, the series
1 1 1 1 + + Á + s 1dn  1 2 + Á 4 9 16 n
1 1 1 1 1 1 1 1 1 + + + + Á. 4 16 9 25 49 36 64 100 144
iaz
1 
You
converges absolutely. A possible rearrangement of the terms of the series might start with a positive term, then two negative terms, then three positive terms, then four negative terms, and so on: After k terms of one sign, take k + 1 terms of the opposite sign. The first ten terms of such a series look like this:
(See Exercise 61.)
nR
The Rearrangement Theorem says that both series converge to the same value. In this example, if we had the second series to begin with, we would probably be glad to exchange it for the first, if we knew that we could. We can do even better: The sum of either series is also equal to q q 1 1 . a a 2 2 n = 1 s2n  1d n = 1 s2nd
ass a
If we rearrange infinitely many terms of a conditionally convergent series, we can get results that are far different from the sum of the original series. Here is an example.
EXAMPLE 6
Rearranging the Alternating Harmonic Series
The alternating harmonic series
dH
1 1 1 1 1 1 1 1 1 1 1  +  +  +  + +  Á 5 7 1 2 3 4 6 8 9 10 11
can be rearranged to diverge or to reach any preassigned sum. (a) Rearranging g n = 1 s 1dn + 1>n to diverge. The series of terms g[1>s2n  1d] diverges to + q and the series of terms gs 1>2nd diverges to  q . No matter how far out in the sequence of oddnumbered terms we begin, we can always add enough positive terms to get an arbitrarily large sum. Similarly, with the negative terms, no matter how far out we start, we can add enough consecutive evennumbered terms to get a negative sum of arbitrarily large absolute value. If we wished to do so, we could start adding oddnumbered terms until we had a sum greater than +3, say, and then follow that with enough consecutive negative terms to make the new total less than 4. We could then add enough positive terms to make the total greater than +5 and follow with consecutive unused negative terms to make a new total less than 6, and so on. In this way, we could make the swings arbitrarily large in either direction. q (b) Rearranging g n = 1 s 1dn + 1>n to converge to 1. Another possibility is to focus on a particular limit. Suppose we try to get sums that converge to 1. We start with the first term, 1> 1, and then subtract 1> 2. Next we add 1> 3 and 1> 5, which brings the total back to 1 or above. Then we add consecutive negative terms until the total is less than 1. We continue in this manner: When the sum is less than 1, add positive terms until the total is 1 or more; then subtract (add negative) terms until the total is again less than 1. This process can be continued indefinitely. Because both the oddnumbered
Mu
ham
ma
q
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 792
Chapter 11: Infinite Sequences and Series
suf
i
terms and the evennumbered terms of the original series approach zero as n : q , the amount by which our partial sums exceed 1 or fall below it approaches zero. So the new series converges to 1. The rearranged series starts like this:
You
1 1 1 1 1 1 1 1 1 1 1 1 1 1  + +  + +  + +  + + 5 7 1 2 3 4 9 6 11 13 8 15 17 10 1 1 1 1 1 1 1 1 + + + + + + Á 19 21 12 23 25 14 27 16
5.
Mu
ham
ma
dH
6.
nR
2. 3. 4.
The nthTerm Test: Unless an : 0, the series diverges. Geometric series: gar n converges if ƒ r ƒ 6 1; otherwise it diverges. pseries: g1>n p converges if p 7 1; otherwise it diverges. Series with nonnegative terms: Try the Integral Test, Ratio Test, or Root Test. Try comparing to a known series with the Comparison Test. Series with some negative terms: Does g ƒ an ƒ converge? If yes, so does gan , since absolute convergence implies convergence. Alternating series: gan converges if the series satisfies the conditions of the Alternating Series Test.
ass a
1.
iaz
The kind of behavior illustrated by the series in Example 6 is typical of what can happen with any conditionally convergent series. Therefore we must always add the terms of a conditionally convergent series in the order given. We have now developed several tests for convergence and divergence of series. In summary:
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 792
i f su
Chapter 11: Infinite Sequences and Series
EXERCISES 11.6 Determining Convergence or Divergence
q
Which of the alternating series in Exercises 1â€“10 converge, and which diverge? Give reasons for your answers. q
q
1 1. a s 1dn + 1 2 n n=1 q
3. a s 1d
n+1
n=1 q
2. a s 1dn + 1
n a b 10
n
n=1 q
4. a s 1d
n+1
n=1 q
n
3>2
10 n n 10
H d
ln n 6. a s 1dn + 1 n
1 5. a s 1dn + 1 ln n n=2 q
n=1 q
a m m a h
ln n 7. a s 1dn + 1 ln n 2 n=2 q
2n + 1 9. a s 1dn + 1 n + 1 n=1
1 8. a s 1dn ln a1 + n b n=1 q
10. a s 1dn + 1 n=1
32n + 1 2n + 1
Absolute Convergence
u M
Which of the series in Exercises 11â€“44 converge absolutely, which converge, and which diverge? Give reasons for your answers.
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q
11. a s 1dn + 1s0.1dn n=1 q
13. a s 1dn
n a s s a
1
u o Y z a i R
n=1 q
1
2n
15. a s 1d
n+1
n=1 q
n n3 + 1
1 17. a s 1dn n + 3 n=1 q
3 + n 19. a s 1dn + 1 5 + n n=1 q
21. a s 1dn + 1 n=1 q
1 + n n2
12. a s 1dn + 1 n=1 q
14. a
s0.1dn n
s 1dn
1 + 2n n! 16. a s 1dn + 1 n 2 n=1 n=1 q
q
sin n 18. a s 1dn 2 n n=1 q
1 20. a s 1dn ln sn 3 d n=2 q
s 2dn + 1 22. a n n=1 n + 5
23. a s 1dnn 2s2>3dn
24. a s 1dn + 1 A 210 B
tan1 n 25. a s 1d 2 n + 1 n=1
1 26. a s 1dn + 1 n ln n
q
n=1 q
n
n
n=1 q n=2
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4100 AWL/Thomas_ch11p746847 8/25/04 2:41 PM Page 793
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.6 Alternating Series, Absolute and Conditional Convergence
s 100dn n! n=1 q
q
29. a
30. a s 5dn ln n b 32. a s 1dn a ln n 2 n=2
s 1d 31. a 2 n = 1 n + 2n + 1 33. a
to estimate L. Compute
q
cos np
cos np 34. a n
n 2n s 1dnsn + 1dn 35. a s2ndn n=1 n=1 q
n=1 q
n+1
s20 +
2
s 1d sn!d s2nd! n=1
36. a
sn!d2 3n 38. a s 1dn s2n + 1d! n=1
s2nd! 37. a s 1dn n 2 n!n n=1 q
q
39. a s 1dn A 2n + 1  2n B 40. a s 1dn A 2n 2 + n  n B q
q
n=1 q
n=1
n=1 q
42. a
n=1 q
s 1dn
q
43. a s 1dn sech n
2n + 2n + 1
as an approximation to the sum of the alternating harmonic series. The exact sum is ln 2 = 0.6931. Á 53. The sign of the remainder of an alternating series that satisfies the conditions of Theorem 14 Prove the assertion in Theorem 15 that whenever an alternating series satisfying the conditions of Theorem 14 is approximated with one of its partial sums, then the remainder (sum of the unused terms) has the same sign as the first unused term. (Hint: Group the remainder’s terms in consecutive pairs.) 54. Show that the sum of the first 2n terms of the series
n=1
44. a s 1dn csch n
1 
ass a
n=1
Error Estimation
q
1 45. a s 1dn + 1 n
dH
It can be shown that the sum is ln 2.
n=1 q
1 46. a s 1dn + 1 n 10 n=1 q
s0.01dn 47. a s 1dn + 1 n
ma
q
1 = a s 1dnt n, 48. 1 + t n=0
0 6 t 6 1
ham
T Approximate the sums in Exercises 49 and 50 with an error of magnitude less than 5 * 10 6 . n=0 q
1 50. a s 1dn n! n=0
1 1 1 1 1 + # + # + # + # + Á. 1#2 2 3 3 4 4 5 5 6
Do these series converge? What is the sum of the first 2n + 1 terms of the first series? If the series converge, what is their sum?
55. Show that if g n = 1 an diverges, then g n = 1 ƒ an ƒ diverges. q
56. Show that if
q gn=1
As you will see in Section 11.7, the sum is ln (1.01).
n=1
q
1 1 1 1 1 1 1 1 1 +  +  +  +  + Á 5 5 2 2 3 3 4 4 6
is the same as the sum of the first n terms of the series
In Exercises 45–48, estimate the magnitude of the error involved in using the sum of the first four terms to approximate the sum of the entire series.
1 49. a s 1dn s2nd!
1 21
#
nR
41. a s 1dn A 2n + 1n  2n B
1 2
iaz
q
sn + sn + 1 1 = sn + s 1dn + 2an + 1 2 2
n
You
n=1 q
n1
q
T 52. The limit L of an alternating series that satisfies the conditions of Theorem 14 lies between the values of any two consecutive partial sums. This suggests using the average
i
q
ln n 28. a s 1dn n  ln n n=1
suf
q
n 27. a s 1dn n + 1 n=1
793
As you will see in Section 11.9, the sum is cos 1, the cosine of 1 radian.
As you will see in Section 11.9, the sum is e1 .
Mu
Theory and Examples 51. a. The series
1 1 1 1 1 1 1 1  +  +  + Á + n  n + Á 3 2 9 4 27 8 3 2
does not meet one of the conditions of Theorem 14. Which one?
b. Find the sum of the series in part (a).
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q
an converges absolutely, then
` a an ` … a ƒ an ƒ . q
q
n=1
n=1
57. Show that if g n = 1 an and g n = 1 bn both converge absolutely, then so does q
q
q
q
a. a san + bn d
b. a san  bn d
n=1 q
c. a kan
n=1
(k any number)
n=1
58. Show by example that g n = 1 an bn may diverge even if g n = 1 an q and g n = 1 bn both converge. q
q
T 59. In Example 6, suppose the goal is to arrange the terms to get a new series that converges to 1>2 . Start the new arrangement with the first negative term, which is 1>2 . Whenever you have a sum that is less than or equal to 1>2, start introducing positive terms, taken in order, until the new total is greater than 1>2 . Then add negative terms until the total is less than or equal to 1>2 again. Continue this process until your partial sums have
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 11: Infinite Sequences and Series In other words, if a series converges absolutely, its positive terms form a convergent series, and so do its negative terms. Furthermore,
i
been above the target at least three times and finish at or below it. If sn is the sum of the first n terms of your new series, plot the points sn, sn d to illustrate how the sums are behaving. 60. Outline of the proof of the Rearrangement Theorem (Theorem 17) a. Let P be a positive real number, let L = g n = 1 an , and let k sk = g n = 1 an . Show that for some index N1 and for some index N2 Ú N1 ,
q n=1
1 2 1 2 1 2 1 2 1 2  +  +  +  +  + Á. 5 7 5 3 2 3 4 9 11 6
… a ƒ ak ƒ + ƒ sN2  L ƒ 6 P. k = N1
b. The argument in part (a) shows that if g n = 1 an converges q
absolutely then g n = 1 bn converges and g n = 1 bn = g n = 1 an . Now show that because converges to g n = 1 ƒ an ƒ .
ƒ an ƒ converges,
q
61. Unzipping absolutely convergent series ƒ an ƒ converges and
bn = e then
q gn=1
an , 0,
if an Ú 0 if an 6 0,
q gn=1
ƒ bn ƒ
dH
a. Show that if
q gn=1
q
bn converges.
Collect terms with the same denominator, as the arrows indicate, to arrive at
ass a
q
q gn=1
by 2 to get
2S = 2  1 +
q
q
iaz
k=1
` a bk  L ` … ` a bk  sN2 ` + ƒ sN2  L ƒ
1 1 1 1 1 +  +  + 5 2 3 4 6 1 1 1 1 1 1  + + + Á 7 8 9 10 11 12
nR
k=1
You
P ƒ sN2  L ƒ 6 2 .
Since all the terms a1, a2 , Á , aN 2 appear somewhere in the sequence 5bn6 , there is an index N3 Ú N2 such that if n n Ú N3 , then A g k = 1 bk B  sN2 is at most a sum of terms am with m Ú N1 . Therefore, if n Ú N3 , n
n=1
because bn = san + ƒ an ƒ d>2 and cn = san  ƒ an ƒ d>2 . 62. What is wrong here?:
S = 1 
n
n=1
Multiply both sides of the alternating harmonic series
q
and
q
a an = a bn + a cn
q
P a ƒ an ƒ 6 2 n = N1
q
suf
794
2S = 1 
1 1 1 1 1 +  +  + Á. 5 2 3 4 6
The series on the righthand side of this equation is the series we started with. Therefore, 2S = S , and dividing by S gives 2 = 1 . (Source: “Riemann’s Rearrangement Theorem” by Stewart Galanor, Mathematics Teacher, Vol. 80, No. 8, 1987, pp. 675–681.)
63. Draw a figure similar to Figure 11.9 to illustrate the convergence of the series in Theorem 14 when N 7 1 .
b. Use the results in part (a) to show likewise that if g n = 1 ƒ an ƒ converges and
ma
q
cn = e
0, an ,
if an Ú 0 if an 6 0,
then g n = 1 cn converges.
Mu
ham
q
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i f u s u o Y
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 794
Chapter 11: Infinite Sequences and Series
11.7
z a i R n a
Power Series
s s a dH
Now that we can test infinite series for convergence we can study the infinite polynomials mentioned at the beginning of this chapter. We call these polynomials power series because they are defined as infinite series of powers of some variable, in our case x. Like polynomials, power series can be added, subtracted, multiplied, differentiated, and integrated to give new power series.
Mu
a m m a h
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.7
795
Power Series
i
Power Series and Convergence
suf
We begin with the formal definition.
You
DEFINITIONS Power Series, Center, Coefficients A power series about x 0 is a series of the form q
n 2 Á + cn x n + Á . a cn x = c0 + c1 x + c2 x +
n=0
(1)
A power series about x a is a series of the form q
iaz
n 2 Á + cnsx  adn + Á a cnsx  ad = c0 + c1sx  ad + c2sx  ad +
n=0
(2)
nR
in which the center a and the coefficients c0, c1, c2, Á , cn, Á are constants.
Equation (1) is the special case obtained by taking a = 0 in Equation (2).
EXAMPLE 1
A Geometric Series
ass a
Taking all the coefficients to be 1 in Equation (1) gives the geometric power series q
n 2 Á + xn + Á . ax = 1 + x + x +
n=0
dH
This is the geometric series with first term 1 and ratio x. It converges to 1>s1  xd for ƒ x ƒ 6 1. We express this fact by writing 1 = 1 + x + x 2 + Á + x n + Á, 1  x
1 6 x 6 1.
(3)
Mu
ham
ma
Up to now, we have used Equation (3) as a formula for the sum of the series on the right. We now change the focus: We think of the partial sums of the series on the right as polynomials Pnsxd that approximate the function on the left. For values of x near zero, we need take only a few terms of the series to get a good approximation. As we move toward x = 1, or 1, we must take more terms. Figure 11.10 shows the graphs of ƒsxd = 1>s1  xd, and the approximating polynomials yn = Pnsxd for n = 0, 1, 2, and 8. The function ƒsxd = 1>s1  xd is not continuous on intervals containing x = 1, where it has a vertical asymptote. The approximations do not apply when x Ú 1.
EXAMPLE 2
A Geometric Series
The power series n
1 
1 1 1 sx  2d + sx  2d2 + Á + a b sx  2dn + Á 2 4 2
(4)
matches Equation (2) with a = 2, c0 = 1, c1 = 1>2, c2 = 1>4, Á , cn = s 1>2dn . This x  2 is a geometric series with first term 1 and ratio r = . The series converges for 2
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 796
Chapter 11: Infinite Sequences and Series
1 1x
i
y
suf
y 9 8 7
y8 1 x x 2 x 3 x 4 x 5 x 6 x7 x 8
You
5 4
y2 1 x x2
3
y1 1 x
2 1
y0 1 x
iaz
–1
0
1
`
nR
FIGURE 11.10 The graphs of ƒsxd = 1>s1  xd and four of its polynomial approximations (Example 1).
x  2 ` 6 1 or 0 6 x 6 4. The sum is 2
y
so
y0 1
0
1
Mu
ham
FIGURE 11.11 The graphs of ƒsxd = 2>x and its first three polynomial approximations (Example 2).
0 6 x 6 4.
Series (4) generates useful polynomial approximations of ƒsxd = 2>x for values of x near 2: P0sxd = 1
ma
1
2 y2 3 3x x 4 2 (2, 1) y 2x y1 2 x 2 x 3 2
n sx  2d sx  2d2 1 2 +  Á + a b sx  2dn + Á, x = 1 2 4 2
dH
2
1 2 = x, x  2 1 + 2
ass a
1 = 1  r
P1sxd = 1 
x 1 sx  2d = 2 2 2
P2sxd = 1 
3x x2 1 1 + , sx  2d + sx  2d2 = 3 2 4 2 4
and so on (Figure 11.11).
EXAMPLE 3
Testing for Convergence Using the Ratio Test
For what values of x do the following power series converge? q
xn x2 x3 +  Á (a) a s 1dn  1 n = x 2 3 n=1 q
x3 x5 x 2n  1 = x +  Á (b) a s 1dn  1 5 2n 1 3 n=1 q
xn x2 x3 = 1 + x + + + Á (c) a 2! 3! n = 0 n! q
(d) a n!x n = 1 + x + 2!x 2 + 3!x 3 + Á n=0
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.7
Power Series
797
Apply the Ratio Test to the series g ƒ un ƒ , where un is the nth term of the series
Solution
suf
i
in question. un + 1 n (a) ` un ` = x : x. n + 1ƒ ƒ ƒ ƒ
–1
0
x
1
iaz
un + 1 2n  1 2 (b) ` un ` = x : x2. 2n + 1
You
The series converges absolutely for ƒ x ƒ 6 1. It diverges if ƒ x ƒ 7 1 because the nth term does not converge to zero. At x = 1, we get the alternating harmonic series 1  1>2 + 1>3  1>4 + Á , which converges. At x = 1 we get 1  1>2 1>3  1>4  Á , the negative of the harmonic series; it diverges. Series (a) converges for 1 6 x … 1 and diverges elsewhere.
ass a
nR
The series converges absolutely for x 2 6 1. It diverges for x 2 7 1 because the nth term does not converge to zero. At x = 1 the series becomes 1  1>3 + 1>5  1>7 + Á , which converges by the Alternating Series Theorem. It also converges at x = 1 because it is again an alternating series that satisfies the conditions for convergence. The value at x = 1 is the negative of the value at x = 1. Series (b) converges for 1 … x … 1 and diverges elsewhere. –1
0
1
x
un + 1 ƒxƒ x n + 1 # n! (c) ` un ` = ` : 0 for every x. n` = n + 1 x sn + 1d!
dH
The series converges absolutely for all x. 0
x
sn + 1d!x n + 1 un + 1 (d) ` un ` = ` ` = sn + 1d ƒ x ƒ : q unless x = 0. n!x n
Mu
ham
ma
The series diverges for all values of x except x = 0. x
0
Example 3 illustrates how we usually test a power series for convergence, and the possible results.
THEOREM 18
The Convergence Theorem for Power Series q
If the power series a an x n = a0 + a1 x + a2 x 2 + Á converges for n=0
x = c Z 0, then it converges absolutely for all x with ƒ x ƒ 6 ƒ c ƒ . If the series diverges for x = d, then it diverges for all x with ƒ x ƒ 7 ƒ d ƒ .
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 798
Chapter 11: Infinite Sequences and Series
Proof Suppose the series g n = 0 an c n converges. Then limn: q an c n = 0. Hence, there is an integer N such that ƒ an c n ƒ 6 1 for all n Ú N. That is, 1 ƒ c ƒn
for n Ú N.
Now take any x such that ƒ x ƒ 6 ƒ c ƒ and consider
(5)
You
ƒ an ƒ 6
suf
i
q
ƒ a0 ƒ + ƒ a1 x ƒ + Á + ƒ aN  1x N  1 ƒ + ƒ aN x N ƒ + ƒ aN + 1 x N + 1 ƒ + Á .
There are only a finite number of terms prior to ƒ aN x N ƒ , and their sum is finite. Starting with ƒ aN x N ƒ and beyond, the terms are less than N
x + `c`
N+1
x + `c`
N+2
+ Á
iaz
x `c`
(6)
ass a
nR
because of Inequality (5). But Series (6) is a geometric series with ratio r = ƒ x>c ƒ , which is less than 1, since ƒ x ƒ 6 ƒ c ƒ . Hence Series (6) converges, so the original series converges absolutely. This proves the first half of the theorem. The second half of the theorem follows from the first. If the series diverges at x = d and converges at a value x0 with ƒ x0 ƒ 7 ƒ d ƒ , we may take c = x0 in the first half of the theorem and conclude that the series converges absolutely at d. But the series cannot converge absolutely and diverge at one and the same time. Hence, if it diverges at d, it diverges for all x with ƒ x ƒ 7 ƒ d ƒ .
dH
To simplify the notation, Theorem 18 deals with the convergence of series of the form gan x n . For series of the form gansx  adn we can replace x  a by x¿ and apply the results to the series gansx¿dn .
The Radius of Convergence of a Power Series
Mu
ham
ma
The theorem we have just proved and the examples we have studied lead to the conclusion that a power series gcnsx  adn behaves in one of three possible ways. It might converge only at x = a, or converge everywhere, or converge on some interval of radius R centered at x = a. We prove this as a Corollary to Theorem 18.
COROLLARY TO THEOREM 18 The convergence of the series gcnsx  adn is described by one of the following three possibilities: 1.
2. 3.
There is a positive number R such that the series diverges for x with ƒ x  a ƒ 7 R but converges absolutely for x with ƒ x  a ƒ 6 R. The series may or may not converge at either of the endpoints x = a  R and x = a + R. The series converges absolutely for every x sR = q d. The series converges at x = a and diverges elsewhere sR = 0d.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.7
Power Series
799
nR
iaz
You
suf
i
Proof We assume first that a = 0, so that the power series is centered at 0. If the series converges everywhere we are in Case 2. If it converges only at x = 0 we are in Case 3. Otherwise there is a nonzero number d such that gcn d n diverges. The set S of values of x for which the series gcn x n converges is nonempty because it contains 0 and a positive number p as well. By Theorem 18, the series diverges for all x with ƒ x ƒ 7 ƒ d ƒ , so ƒ x ƒ … ƒ d ƒ for all x H S, and S is a bounded set. By the Completeness Property of the real numbers (see Appendix 4) a nonempty, bounded set has a least upper bound R. (The least upper bound is the smallest number with the property that the elements x H S satisfy x … R.) If ƒ x ƒ 7 R Ú p, then x x S so the series gcn x n diverges. If ƒ x ƒ 6 R, then ƒ x ƒ is not an upper bound for S (because it’s smaller than the least upper bound) so there is a number b H S such that b 7 ƒ x ƒ . Since b H S, the series gcn b n converges and therefore the series gcn ƒ x ƒ n converges by Theorem 18. This proves the Corollary for power series centered at a = 0. For a power series centered at a Z 0, we set x¿ = sx  ad and repeat the argument with x¿ . Since x¿ = 0 when x = a, a radius R interval of convergence for gcnsx¿dn centered at x¿ = 0 is the same as a radius R interval of convergence for gcnsx  adn centered at x = a. This establishes the Corollary for the general case.
dH
ass a
R is called the radius of convergence of the power series and the interval of radius R centered at x = a is called the interval of convergence. The interval of convergence may be open, closed, or halfopen, depending on the particular series. At points x with ƒ x  a ƒ 6 R, the series converges absolutely. If the series converges for all values of x, we say its radius of convergence is infinite. If it converges only at x = a, we say its radius of convergence is zero.
How to Test a Power Series for Convergence
ma
1.
Mu
ham
2.
3.
Use the Ratio Test (or nthRoot Test) to find the interval where the series converges absolutely. Ordinarily, this is an open interval ƒx  aƒ 6 R
or
a  R 6 x 6 a + R.
If the interval of absolute convergence is finite, test for convergence or divergence at each endpoint, as in Examples 3a and b. Use a Comparison Test, the Integral Test, or the Alternating Series Test. If the interval of absolute convergence is a  R 6 x 6 a + R, the series diverges for ƒ x  a ƒ 7 R (it does not even converge conditionally), because the nth term does not approach zero for those values of x.
TermbyTerm Differentiation A theorem from advanced calculus says that a power series can be differentiated term by term at each interior point of its interval of convergence.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 800
Chapter 11: Infinite Sequences and Series
suf
i
THEOREM 19 The TermbyTerm Differentiation Theorem If gcnsx  adn converges for a  R 6 x 6 a + R for some R 7 0, it defines a function ƒ: q
ƒsxd = a cnsx  adn,
a  R 6 x 6 a + R.
You
n=0
Such a function ƒ has derivatives of all orders inside the interval of convergence. We can obtain the derivatives by differentiating the original series term by term: q
ƒ¿sxd = a ncnsx  adn  1 n=1 q n=2
iaz
ƒ–sxd = a nsn  1dcnsx  adn  2 ,
EXAMPLE 4
nR
and so on. Each of these derived series converges at every interior point of the interval of convergence of the original series.
Applying TermbyTerm Differentiation
Find series for ƒ¿sxd and ƒ–sxd if
1 = 1 + x + x2 + x3 + x4 + Á + xn + Á 1  x
ass a ƒsxd =
q
= a x n,
1 6 x 6 1
n=0
dH
Solution
ƒ¿sxd =
1 = 1 + 2x + 3x 2 + 4x 3 + Á + nx n  1 + Á s1  xd2 q
= a nx n  1,
1 6 x 6 1
ma
n=1
Mu
ham
ƒ–sxd =
2 = 2 + 6x + 12x 2 + Á + nsn  1dx n  2 + Á s1  xd3 q
= a nsn  1dx n  2,
1 6 x 6 1
n=2
CAUTION
Termbyterm differentiation might not work for other kinds of series. For example, the trigonometric series q sin sn!xd a n2 n=1 converges for all x. But if we differentiate term by term we get the series q n!cos sn!xd , a n2 n=1 which diverges for all x. This is not a power series, since it is not a sum of positive integer powers of x.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.7
Power Series
801
i
TermbyTerm Integration
THEOREM 20 Suppose that
You
suf
Another advanced calculus theorem states that a power series can be integrated term by term throughout its interval of convergence.
The TermbyTerm Integration Theorem q
ƒsxd = a cnsx  adn n=0
iaz
converges for a  R 6 x 6 a + R sR 7 0d. Then q
a cn
n =0
(x  a) n+1 n + 1
nR
converges for a  R 6 x 6 a + R and q
L
ƒsxd dx = a cn n=0
sx  adn + 1 + C n + 1
ass a
for a  R 6 x 6 a + R.
EXAMPLE 5
A Series for tan1 x, 1 … x … 1
dH
Identify the function
x5 x3 +  Á, 5 3
1 … x … 1.
We differentiate the original series term by term and get
ma
Solution
ƒsxd = x 
ƒ¿sxd = 1  x 2 + x 4  x 6 + Á,
1 6 x 6 1.
Mu
ham
This is a geometric series with first term 1 and ratio x 2 , so ƒ¿sxd =
1 1 = . 1  s x 2 d 1 + x2
We can now integrate ƒ¿sxd = 1>s1 + x 2 d to get L
ƒ¿sxd dx =
dx = tan1 x + C. 2 L1 + x
The series for ƒ(x) is zero when x = 0, so C = 0. Hence ƒsxd = x 
x3 x5 x7 + + Á = tan1 x, 5 7 3
1 6 x 6 1.
In Section 11.10, we will see that the series also converges to tan1 x at x = ;1.
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(7)
4100 AWL/Thomas_ch11p746847 8/25/04 2:41 PM Page 802
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 802
Chapter 11: Infinite Sequences and Series
EXAMPLE 6
suf
i
Notice that the original series in Example 5 converges at both endpoints of the original interval of convergence, but Theorem 20 can guarantee the convergence of the differentiated series only inside the interval.
A Series for ln s1 + xd, 1 6 x … 1
You
The series
1 = 1  t + t2  t3 + Á 1 + t converges on the open interval 1 6 t 6 1. Therefore,
x
x
t2 t3 t4 1 dt = t + + Ád 2 3 4 0 L0 1 + t x2 x3 x4 + + Á, 2 3 4
Theorem 20
1 6 x 6 1.
nR
= x 
iaz
ln s1 + xd =
ass a
It can also be shown that the series converges at x = 1 to the number ln 2, but that was not guaranteed by the theorem.
USING TECHNOLOGY
Study of Series
ma
dH
Series are in many ways analogous to integrals. Just as the number of functions with explicit antiderivatives in terms of elementary functions is small compared to the number of integrable functions, the number of power series in x that agree with explicit elementary functions on xintervals is small compared to the number of power series that converge on some xinterval. Graphing utilities can aid in the study of such series in much the same way that numerical integration aids in the study of definite integrals. The ability to study power series at particular values of x is built into most Computer Algebra Systems. If a series converges rapidly enough, CAS exploration might give us an idea of the q sum. For instance, in calculating the early partial sums of the series g k = 1 [1>s2k  1 d] (Section 11.4, Example 2b), Maple returns Sn = 1.6066 95152 for 31 … n … 200. This suggests that the sum of the series is 1.6066 95152 to 10 digits. Indeed,
Mu
ham
q
q
q
1 1 1 1 = a k1 6 a k  1 = 199 6 1.25 * 10 60 . a k 2 s2  s1>2k  1 dd k = 201 2  1 k = 201 2 k = 201 2
The remainder after 200 terms is negligible. However, CAS and calculator exploration cannot do much for us if the series converges or diverges very slowly, and indeed can be downright misleading. For example, q try calculating the partial sums of the series g k = 1 [1>s10 10kd]. The terms are tiny in comparison to the numbers we normally work with and the partial sums, even for hundreds of terms, are miniscule. We might well be fooled into thinking that the series conq verges. In fact, it diverges, as we can see by writing it as s1>10 10 dg k = 1 s1>kd, a constant times the harmonic series. We will know better how to interpret numerical results after studying error estimates in Section 11.9.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.7
Power Series
803
i
Multiplication of Power Series
suf
Another theorem from advanced calculus states that absolutely converging power series can be multiplied the way we multiply polynomials. We omit the proof.
You
THEOREM 21 The Series Multiplication Theorem for Power Series q q If Asxd = g n = 0 an x n and Bsxd = g n = 0 bn x n converge absolutely for ƒ x ƒ 6 R, and n
cn = a0 bn + a1 bn  1 + a2 bn  2 + Á + an  1b1 + an b0 = a ak bn  k , k=0
converges absolutely to A(x)B(x) for ƒ x ƒ 6 R:
iaz
then
q g n = 0 cn x n
a a an x n b # a a bn x n b = a cn x n . q
q
n=0
n=0
nR
n=0
q
EXAMPLE 7
Multiply the geometric series
q
for ƒ x ƒ 6 1,
ass a
n 2 Á + xn + Á = 1 , ax = 1 + x + x + 1  x n=0
by itself to get a power series for 1>s1  xd2 , for ƒ x ƒ 6 1. Solution
Let
q
dH
Asxd = a an x n = 1 + x + x 2 + Á + x n + Á = 1>s1  xd n=0 q
Bsxd = a bn x n = 1 + x + x 2 + Á + x n + Á = 1>s1  xd n=0
Mu
ham
ma
and
Á + ak bn  k + Á + an b0 cn = a(''''''''''')''''''''''''* 0 bn + a1 bn  1 + n + 1 terms
= ('''')''''* 1 + 1 + Á + 1 = n + 1. n + 1 ones
Then, by the Series Multiplication Theorem, q
q
Asxd # Bsxd = a cn x n = a sn + 1dx n n=0
n=0
= 1 + 2x + 3x 2 + 4x 3 + Á + sn + 1dx n + Á is the series for 1>s1  xd2 . The series all converge absolutely for ƒ x ƒ 6 1. Notice that Example 4 gives the same answer because
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d 1 1 a b = . dx 1  x s1  xd2
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 804
Chapter 11: Infinite Sequences and Series
In Exercises 1–32, (a) find the series’ radius and interval of convergence. For what values of x does the series converge (b) absolutely, (c) conditionally?
q
4. a
sx  2dn 5. a 10 n n=0
6. a s2xdn n=0 q
q
nx n 7. a n=0 n + 2
8. a
n=1 q
xn
10. a
n 2n 3n s 1dnx n 11. a n! n=0 n=1
q
q
16. a
2n 2 + 3 q nsx + 3dn 17. a 5n n=0 n=0
q
19. a
n=0 q
n=0
s 1dnx n
q
q
n
n=1 q
1 21. a a1 + n b x n
22. a sln ndx n
23. a n nx n
24. a n!sx  4dn
n=1 q
n=1
n=0
s 1dn + 1sx + 2dn n2n n=1 q
Mu 29. a
n=1 q
31. a
n=1
s4x  5d2n + 1 n 3>2
n
sx + pd 2n
s3x + 1dn + 1 30. a 2n + 2 n=1 q
q
32. a
n=0
n
x3 x5 x7 x9 x 11 + + + Á 3! 5! 7! 9! 11!
converges to sin x for all x. a. Find the first six terms of a series for cos x. For what values of x should the series converge? b. By replacing x by 2x in the series for sin x, find a series that converges to sin 2x for all x. c. Using the result in part (a) and series multiplication, calculate the first six terms of a series for 2 sin x cos x. Compare your answer with the answer in part (b). 42. The series ex = 1 + x +
x2 x3 x4 x5 + + + + Á 2! 3! 4! 5!
converges to e x for all x.
Get the information you need about a 1>(n ln n) from Section 11.3, Exercise 38.
n
x 28. a n = 2 n ln n q
n=0
Get the information you need about 2 a 1>(n(ln n) ) from Section 11.3, Exercise 39.
xn 27. a 2 n = 2 nsln nd q
q
26. a s 2dnsn + 1dsx  1dn
ham
25. a
q
ma
n=1 q
n
40. If you integrate the series in Exercise 39 term by term, what new series do you get? For what values of x does the new series converge, and what is another name for its sum?
sin x = x 
20. a 2ns2x + 5dn n
n=0
x2  1 b 2
41. The series
2n 2 + 3
nx n 18. a n 2 n = 0 4 sn + 1d
2nx n 3n
38. a a
converge? What is its sum? What series do you get if you differentiate the given series term by term? For what values of x does the new series converge? What is its sum?
14. a
xn
n=0 q
1 1 1 sx  3d + sx  3d2 + Á + a b sx  3dn + Á 2 4 2
ass a
q
2n n n
q
x 2n + 1 n = 0 n!
15. a
sx  1dn
s2x + 3d2n + 1 n! n=0
q
n
39. For what values of x does the series 1 
3x 12. a n = 0 n!
13. a
x + 1 b 3 2
Theory and Examples
s 1dnsx + 2dn n
n=1
q
n=0
dH
q
9. a
n=1 q
q
s3x  2d n
3. a s 1dns4x + 1dn n=0 q
37. a a
n
q
36. a sln xdn
nR
n=0 q
sx + 1d2n 9n n=0 q
34. a
n 2x 35. a a  1b 2 n=0
q
2. a sx + 5dn
n=0 q
sx  1d2n 4n n=0 q
33. a
iaz
q
1. a x n
In Exercises 33–38, find the series’ interval of convergence and, within this interval, the sum of the series as a function of x.
You
Intervals of Convergence
suf
i
EXERCISES 11.7
A x  22 B
2n + 1
2n
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a. Find a series for sd>dxde x . Do you get the series for e x ? Explain your answer. b. Find a series for 1 e x dx . Do you get the series for e x ? Explain your answer. c. Replace x by x in the series for e x to find a series that converges to e x for all x. Then multiply the series for e x and e x to find the first six terms of a series for e x # e x .
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.7 Power Series
17x 7 62x 9 x3 2x 5 + + + + Á 3 15 315 2835
45. Uniqueness of convergent power series
i
tan x = x +
c. Check your result in part (b) by multiplying the series for sec x by the series given for tan x in Exercise 43.
suf
43. The series
a. Show that if two power series g n = 0 an x n and g n = 0 bn x n are convergent and equal for all values of x in an open interval s c, cd , then an = bn for every n. (Hint: Let q q ƒsxd = g n = 0 an x n = g n = 0 bn x n . Differentiate term by term to show that an and bn both equal f snds0d>sn!d .)
a. Find the first five terms of the series for ln ƒ sec x ƒ . For what values of x should the series converge? b. Find the first five terms of the series for sec2 x . For what values of x should this series converge?
q
You
q
converges to tan x for p>2 6 x 6 p>2 .
805
b. Show that if g n = 0 an x n = 0 for all x in an open interval s c, cd , then an = 0 for every n. q
converges to sec x for p>2 6 x 6 p>2 .
46. The sum of the series g n = 0 sn2>2n d To find the sum of this series, express 1>s1  xd as a geometric series, differentiate both sides of the resulting equation with respect to x, multiply both sides of the result by x, differentiate again, multiply by x again, and set x equal to 1> 2. What do you get? (Source: David E. Dobbs’ letter to the editor, Illinois Mathematics Teacher, Vol. 33, Issue 4, 1982, p. 27.)
a. Find the first five terms of a power series for the function ln ƒ sec x + tan x ƒ . For what values of x should the series converge?
47. Convergence at endpoints Show by examples that the convergence of a power series at an endpoint of its interval of convergence may be either conditional or absolute.
sec x = 1 +
61 6 277 8 Á 5 4 x2 x + x + x + + 2 24 720 8064
48. Make up a power series whose interval of convergence is a. s 3, 3d
b. s 2, 0d
c. (1, 5).
Mu
ham
ma
dH
ass a
b. Find the first four terms of a series for sec x tan x. For what values of x should the series converge?
iaz
44. The series
q
nR
c. Check your result in part (b) by squaring the series given for sec x in Exercise 44.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
i f su
11.8 Taylor and Maclaurin Series
11.8
Taylor and Maclaurin Series
805
u o Y
This section shows how functions that are infinitely differentiable generate power series called Taylor series. In many cases, these series can provide useful polynomial approximations of the generating functions.
Series Representations
n a ss
z a i R
We know from Theorem 19 that within its interval of convergence the sum of a power series is a continuous function with derivatives of all orders. But what about the other way around? If a function ƒ(x) has derivatives of all orders on an interval I, can it be expressed as a power series on I? And if it can, what will its coefficients be? We can answer the last question readily if we assume that ƒ(x) is the sum of a power series
a uh
M
m m
ad
a H
q
ƒsxd = a ansx  adn n=0
= a0 + a1sx  ad + a2sx  ad2 + Á + ansx  adn + Á
with a positive radius of convergence. By repeated termbyterm differentiation within the interval of convergence I we obtain ƒ¿sxd = a1 + 2a2sx  ad + 3a3sx  ad2 + Á + nansx  adn  1 + Á ƒ–sxd = 1 # 2a2 + 2 # 3a3sx  ad + 3 # 4a4sx  ad2 + Á ƒ‡sxd = 1 # 2 # 3a3 + 2 # 3 # 4a4sx  ad + 3 # 4 # 5a5sx  ad2 + Á ,
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Chapter 11: Infinite Sequences and Series
i
with the nth derivative, for all n, being
suf
f sndsxd = n!an + a sum of terms with sx  ad as a factor. Since these equations all hold at x = a, we have
and, in general, ƒsndsad = n!an .
You
ƒ¿sad = a1, # ƒ–sad = 1 2a2 , ƒ‡sad = 1 # 2 # 3a3 ,
These formulas reveal a pattern in the coefficients of any power series g n = 0 ansx  adn that converges to the values of ƒ on I (“represents ƒ on I”). If there is such a series (still an open question), then there is only one such series and its nth coefficient is
iaz
q
ƒsndsad . n!
nR
an =
If ƒ has a series representation, then the series must be
ass a
ƒsxd = ƒsad + ƒ¿sadsx  ad + + Á +
ƒ–sad sx  ad2 2!
ƒsndsad sx  adn + Á . n!
(1)
dH
But if we start with an arbitrary function ƒ that is infinitely differentiable on an interval I centered at x = a and use it to generate the series in Equation (1), will the series then converge to ƒ(x) at each x in the interior of I? The answer is maybe—for some functions it will but for other functions it will not, as we will see.
Taylor and Maclaurin Series
HISTORICAL BIOGRAPHIES Brook Taylor (1685–1731)
ma
DEFINITIONS Taylor Series, Maclaurin Series Let ƒ be a function with derivatives of all orders throughout some interval containing a as an interior point. Then the Taylor series generated by ƒ at x = a is
Mu
ham
Colin Maclaurin (1698–1746)
q
a
k=0
ƒ skdsad ƒ–sad sx  adk = ƒsad + ƒ¿sadsx  ad + sx  ad2 2! k! + Á +
ƒ sndsad sx  adn + Á . n!
The Maclaurin series generated by ƒ is ƒ–s0d 2 ƒsnds0d n ƒ skds0d k Á + x = ƒs0d + ƒ¿s0dx + x + x + Á, a k! 2! n! k=0 q
the Taylor series generated by ƒ at x = 0.
The Maclaurin series generated by ƒ is often just called the Taylor series of ƒ.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.8
807
Finding a Taylor Series
i
EXAMPLE 1
Taylor and Maclaurin Series
We need to find ƒs2d, ƒ¿s2d, ƒ–s2d, Á . Taking derivatives we get ƒsxd = x 1,
ƒs2d = 21 =
ƒ¿sxd = x 2,
ƒ¿s2d = 
1 , 22
ƒ–s2d 1 = 23 = 3 , 2! 2
iaz
ƒ–sxd = 2!x 3,
1 , 2
You
Solution
suf
Find the Taylor series generated by ƒsxd = 1>x at a = 2. Where, if anywhere, does the series converge to 1> x?
nR
ƒ‡sxd = 3!x 4,
ƒ‡s2d 1 =  4, 3! 2
o
ass a
ƒ sndsxd = s 1dnn!x sn + 1d,
o ƒ snds2d s 1dn = n+1 . n! 2
The Taylor series is
dH
ƒs2d + ƒ¿s2dsx  2d +
=
ƒ–s2d ƒsnds2d sx  2d2 + Á + sx  2dn + Á 2! n!
n sx  2d2 sx  2d 1 Á + s 1dn sx  2d + Á . + 2 22 23 2n + 1
This is a geometric series with first term 1> 2 and ratio r = sx  2d>2. It converges absolutely for ƒ x  2 ƒ 6 2 and its sum is
ma
1>2 1 1 = = x. 1 + sx  2d>2 2 + sx  2d
Mu
ham
In this example the Taylor series generated by ƒsxd = 1>x at a = 2 converges to 1> x for ƒ x  2 ƒ 6 2 or 0 6 x 6 4.
Taylor Polynomials The linearization of a differentiable function ƒ at a point a is the polynomial of degree one given by P1sxd = ƒsad + ƒ¿sadsx  ad. In Section 3.8 we used this linearization to approximate ƒ(x) at values of x near a. If ƒ has derivatives of higher order at a, then it has higherorder polynomial approximations as well, one for each available derivative. These polynomials are called the Taylor polynomials of ƒ.
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Chapter 11: Infinite Sequences and Series
+
ƒ–sad sx  ad2 + Á 2!
You
Pnsxd = ƒsad + ƒ¿sadsx  ad +
suf
i
DEFINITION Taylor Polynomial of Order n Let ƒ be a function with derivatives of order k for k = 1, 2, Á , N in some interval containing a as an interior point. Then for any integer n from 0 through N, the Taylor polynomial of order n generated by ƒ at x = a is the polynomial
ƒskdsad ƒsndsad sx  adk + Á + sx  adn . n! k!
EXAMPLE 2 y
nR
iaz
We speak of a Taylor polynomial of order n rather than degree n because ƒsndsad may be zero. The first two Taylor polynomials of ƒsxd = cos x at x = 0, for example, are P0sxd = 1 and P1sxd = 1. The firstorder Taylor polynomial has degree zero, not one. Just as the linearization of ƒ at x = a provides the best linear approximation of ƒ in the neighborhood of a, the higherorder Taylor polynomials provide the best polynomial approximations of their respective degrees. (See Exercise 32.)
Finding Taylor Polynomials for e x
3.0
y e x y P3(x)
Solution
we have
ƒ¿sxd = e x,
ƒs0d = e 0 = 1,
dH
2.0
Since
ƒsxd = e x,
y P2(x)
2.5
ass a
Find the Taylor series and the Taylor polynomials generated by ƒsxd = e x at x = 0.
y P1(x)
ƒ¿s0d = 1,
ƒs0d + ƒ¿s0dx +
ma
1.0
0
0.5
1.0
FIGURE 11.12 The graph of ƒsxd = e x and its Taylor polynomials P1sxd = 1 + x P2sxd = 1 + x + sx 2>2!d P3sxd = 1 + x + sx 2>2!d + sx 3>3!d. Notice the very close agreement near the center x = 0 (Example 2).
Mu
Á,
ƒsnds0d = 1,
Á,
.Á
ƒsnds0d n ƒ–s0d 2 x + Á + x + Á 2! n! = 1 + x +
x2 xn + Á + + Á 2 n!
q
xk = a . k = 0 k!
x
ham
–0.5
ƒsndsxd = e x,
The Taylor series generated by ƒ at x = 0 is
1.5
0.5
Á,
This is also the Maclaurin series for e x . In Section 11.9 we will see that the series converges to e x at every x. The Taylor polynomial of order n at x = 0 is Pnsxd = 1 + x +
x2 xn + Á + . 2 n!
See Figure 11.12.
EXAMPLE 3
Finding Taylor Polynomials for cos x
Find the Taylor series and Taylor polynomials generated by ƒsxd = cos x at x = 0.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.8
809
ƒsxd =
i
The cosine and its derivatives are sin x,
ƒ¿sxd =
cos x,
suf
Solution
Taylor and Maclaurin Serie
s3d
ƒ sxd =
ƒ–sxd = cos x, o ƒs2ndsxd = s 1dn cos x,
sin x,
You
o ƒs2n + 1dsxd = s 1dn + 1 sin x.
At x = 0, the cosines are 1 and the sines are 0, so ƒs2nds0d = s 1dn,
ƒs2n + 1ds0d = 0.
The Taylor series generated by ƒ at 0 is
ƒ–s0d 2 ƒ‡s0d 3 ƒsnds0d n x + x + Á + x + Á 2! 3! n!
= 1 + 0#x q
x2 x4 x 2n + 0 # x3 + + Á + s 1dn + Á 2! 4! s2nd!
s 1dkx 2k . s2kd!
nR
= a
iaz
ƒs0d + ƒ¿s0dx +
k=0
ass a
This is also the Maclaurin series for cos x. In Section 11.9, we will see that the series converges to cos x at every x. Because ƒs2n + 1ds0d = 0, the Taylor polynomials of orders 2n and 2n + 1 are identical: P2nsxd = P2n + 1sxd = 1 
x2 x4 x 2n +  Á + s 1dn . 2! 4! s2nd!
dH
Figure 11.13 shows how well these polynomials approximate ƒsxd = cos x near x = 0. Only the righthand portions of the graphs are given because the graphs are symmetric about the yaxis.
y
Mu
ham
ma
2
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1
P4
P0
P8
P12
P16 y cos x
0
1
2
3
4
5
6
7
8
9
x
–1 P2
P6
P10
P14
P18
–2
FIGURE 11.13 The polynomials n
P2nsxd = a
k=0
s 1dkx 2k s2kd!
converge to cos x as n : q . We can deduce the behavior of cos x arbitrarily far away solely from knowing the values of the cosine and its derivatives at x = 0 (Example 3).
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Chapter 11: Infinite Sequences and Series
A Function ƒ Whose Taylor Series Converges at Every x but Converges to ƒ(x) Only at x = 0
suf
i
EXAMPLE 4
It can be shown (though not easily) that 0, 2 e 1>x ,
x = 0 x Z 0
You
ƒsxd = e
(Figure 11.14) has derivatives of all orders at x = 0 and that ƒsnds0d = 0 for all n. This means that the Taylor series generated by ƒ at x = 0 is ƒs0d + ƒ¿s0dx +
ƒ–s0d 2 ƒsnds0d n x + Á + x + Á 2! n!
iaz
= 0 + 0 # x + 0 # x2 + Á + 0 # xn + Á = 0 + 0 + Á + 0 + Á.
The series converges for every x (its sum is 0) but converges to ƒ(x) only at x = 0. 0 , x0 y –1/x2 e , x 0
nR
y
ass a
1
–4
–3
–2
–1
0
1
2
3
4
x
dH
FIGURE 11.14 The graph of the continuous extension of 2 y = e 1>x is so flat at the origin that all of its derivatives there are zero (Example 4).
Two questions still remain. 1.
ma
2.
For what values of x can we normally expect a Taylor series to converge to its generating function? How accurately do a function’s Taylor polynomials approximate the function on a given interval?
Mu
ham
The answers are provided by a theorem of Taylor in the next section.
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Chapter 11: Infinite Sequences and Series
EXERCISES 11.8
R n ssa
Finding Taylor Polynomials
a H d
In Exercises 1–8, find the Taylor polynomials of orders 0, 1, 2, and 3 generated by ƒ at a. 1. ƒsxd = ln x,
a = 1
2. ƒsxd = ln s1 + xd,
a = 0
3. ƒsxd = 1>x,
a = 2
4. ƒsxd = 1>sx + 2d,
a = 0
a m m a h u M
5. ƒsxd = sin x,
a = p>4
6. ƒsxd = cos x,
7. ƒsxd = 2x,
a = 4
8. ƒsxd = 2x + 4,
a = p>4
a = 0
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i f u s u o Y iaz
Finding Taylor Series at x = 0 (Maclaurin Series)
Find the Maclaurin series for the functions in Exercises 9–20. 9. e x
11.
1 1 + x
13. sin 3x
10. e x>2 1 1  x x 14. sin 2 12.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.8 Taylor and Maclaurin Series
e + e 2
18. sinh x =
19. x 4  2x 3  5x + 4
e  e 2
x
20. sx + 1d2
Finding Taylor Series
32. Of all polynomials of degree ◊ n, the Taylor polynomial of order n gives the best approximation Suppose that ƒ(x) is differentiable on an interval centered at x = a and that gsxd = b0 + b1sx  ad + Á + bnsx  adn is a polynomial of degree n with constant coefficients b0, Á , bn . Let Esxd = ƒsxd  gsxd . Show that if we impose on g the conditions
i
17. cosh x =
x
x
2
23. ƒsxd = x + x + 1,
x:a
a = 1
then
a = 2
24. ƒsxd = 3x 5  x 4 + 2x 3 + x 2  2,
gsxd = ƒsad + ƒ¿sadsx  ad +
a = 1
a = 1
26. ƒsxd = x>s1  xd, 27. ƒsxd = e x,
a = 2
28. ƒsxd = 2x,
a = 1
a = 0
ƒ–sad sx  ad2 + Á 2! +
ƒ sndsad sx  adn . n!
Thus, the Taylor polynomial Pnsxd is the only polynomial of degree less than or equal to n whose error is both zero at x = a and negligible when compared with sx  adn .
Theory and Examples 29. Use the Taylor series generated by e x at x = a to show that sx  ad2 e = e c1 + sx  ad + + Á d. 2! a
Quadratic Approximations
The Taylor polynomial of order 2 generated by a twicedifferentiable function ƒ(x) at x = a is called the quadratic approximation of ƒ at x = a . In Exercises 33–38, find the (a) linearization (Taylor polynomial of order 1) and (b) quadratic approximation of ƒ at x = 0 .
ass a
x
The error is negligible when compared to sx  adn .
iaz
25. ƒsxd = 1>x 2,
Esxd = 0, sx  adn
nR
4
b. lim
a = 2
22. ƒsxd = 2x 3 + x 2 + 3x  8,
The approximation error is zero at x = a .
a. Esad = 0
In Exercises 21–28, find the Taylor series generated by ƒ at x = a . 21. ƒsxd = x 3  2x + 4,
suf
16. 5 cos px x
You
15. 7 cos s xd
811
30. (Continuation of Exercise 29.) Find the Taylor series generated by e x at x = 1 . Compare your answer with the formula in Exercise 29.
34. ƒsxd = e sin x
33. ƒsxd = ln scos xd 35. ƒsxd = 1> 21  x 37. ƒsxd = sin x
2
36. ƒsxd = cosh x 38. ƒsxd = tan x
Mu
ham
ma
dH
31. Let ƒ(x) have derivatives through order n at x = a . Show that the Taylor polynomial of order n and its first n derivatives have the same values that ƒ and its first n derivatives have at x = a .
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
i f u s u o
11.9 Convergence of Taylor Series; Error Estimates
Y z a i R
Convergence of Taylor Series; Error Estimates
11.9
n a s as
811
This section addresses the two questions left unanswered by Section 11.8: 1. 2.
H d a m
m a h
Mu
When does a Taylor series converge to its generating function? How accurately do a function’s Taylor polynomials approximate the function on a given interval?
Taylor’s Theorem We answer these questions with the following theorem.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 812
Chapter 11: Infinite Sequences and Series
ƒsndsad ƒsn + 1dscd sb  adn + sb  adn + 1 . n! sn + 1d!
iaz
+
ƒ–sad sb  ad2 + Á 2!
You
ƒsbd = ƒsad + ƒ¿sadsb  ad +
suf
i
THEOREM 22 Taylor’s Theorem If ƒ and its first n derivatives ƒ¿, ƒ–, Á , ƒsnd are continuous on the closed interval between a and b, and ƒsnd is differentiable on the open interval between a and b, then there exists a number c between a and b such that
nR
Taylor’s Theorem is a generalization of the Mean Value Theorem (Exercise 39). There is a proof of Taylor’s Theorem at the end of this section. When we apply Taylor’s Theorem, we usually want to hold a fixed and treat b as an independent variable. Taylor’s formula is easier to use in circumstances like these if we change b to x. Here is a version of the theorem with this change.
ass a
Taylor’s Formula If ƒ has derivatives of all orders in an open interval I containing a, then for each positive integer n and for each x in I, ƒsxd = ƒsad + ƒ¿sadsx  ad +
dH
+
ƒ–sad sx  ad2 + Á 2!
ƒsndsad sx  adn + Rnsxd, n!
(1)
where
ma
Rnsxd =
f sn + 1dscd sx  adn + 1 sn + 1d!
for some c between a and x.
(2)
Mu
ham
When we state Taylor’s theorem this way, it says that for each x H I, ƒsxd = Pnsxd + Rnsxd.
The function Rnsxd is determined by the value of the sn + 1dst derivative ƒsn + 1d at a point c that depends on both a and x, and which lies somewhere between them. For any value of n we want, the equation gives both a polynomial approximation of ƒ of that order and a formula for the error involved in using that approximation over the interval I. Equation (1) is called Taylor’s formula. The function Rnsxd is called the remainder of order n or the error term for the approximation of ƒ by Pnsxd over I. If Rnsxd : 0 as n : q for all x H I, we say that the Taylor series generated by ƒ at x = a converges to ƒ on I, and we write q
ƒsxd = a
k=0
ƒskdsad sx  adk . k!
Often we can estimate Rn without knowing the value of c, as the following example illustrates.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.9
813
The Taylor Series for e x Revisited
i
EXAMPLE 1
Convergence of Taylor Series; Error Estimates
suf
Show that the Taylor series generated by ƒsxd = e x at x = 0 converges to ƒ(x) for every real value of x. The function has derivatives of all orders throughout the interval I = s  q , q d. Equations (1) and (2) with ƒsxd = e x and a = 0 give ex = 1 + x +
You
Solution
xn x2 + Á + + Rnsxd 2! n!
and ec xn+1 sn + 1d!
for some c between 0 and x.
iaz
Rnsxd =
Polynomial from Section 11.8, Example 2
nR
Since e x is an increasing function of x, e c lies between e 0 = 1 and e x . When x is negative, so is c, and e c 6 1. When x is zero, e x = 1 and Rnsxd = 0. When x is positive, so is c, and e c 6 e x . Thus, ƒ Rnsxd ƒ …
ass a
and
ƒ x ƒn+1 sn + 1d!
ƒ Rnsxd ƒ 6 e x
xn+1 sn + 1d!
when x … 0,
when x 7 0.
Finally, because
xn+1 = 0 n: q sn + 1d!
dH
lim
for every x,
Section 11.1
lim Rnsxd = 0, and the series converges to e x for every x. Thus,
n: q
q
ma
x2 xk xk ex = a = 1 + x + + Á. + Á + 2! k! k = 0 k!
(3)
Estimating the Remainder
Mu
ham
It is often possible to estimate Rnsxd as we did in Example 1. This method of estimation is so convenient that we state it as a theorem for future reference.
THEOREM 23 The Remainder Estimation Theorem If there is a positive constant M such that ƒ ƒsn + 1dstd ƒ … M for all t between x and a, inclusive, then the remainder term Rnsxd in Taylor’s Theorem satisfies the inequality ƒ Rnsxd ƒ … M
ƒ x  a ƒ n+1 . sn + 1d!
If this condition holds for every n and the other conditions of Taylor’s Theorem are satisfied by ƒ, then the series converges to ƒ(x).
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 814
Chapter 11: Infinite Sequences and Series
The Taylor Series for sin x at x = 0
You
EXAMPLE 2
suf
i
We are now ready to look at some examples of how the Remainder Estimation Theorem and Taylor’s Theorem can be used together to settle questions of convergence. As you will see, they can also be used to determine the accuracy with which a function is approximated by one of its Taylor polynomials.
Show that the Taylor series for sin x at x = 0 converges for all x. The function and its derivatives are ƒsxd =
sin x,
ƒ¿sxd =
cos x,
ƒ–sxd =
 sin x,
ƒ‡sxd =
 cos x,
o
iaz
Solution
o
ƒ(2k + 1)sxd = s 1dk cos x,
nR
ƒ(2k)sxd = s 1dk sin x, so
f s2kds0d = 0
and
f s2k + 1ds0d = s 1dk .
ass a
The series has only oddpowered terms and, for n = 2k + 1, Taylor’s Theorem gives sin x = x 
s 1dkx 2k + 1 x3 x5 +  Á + + R2k + 1sxd. 3! 5! s2k + 1d!
dH
All the derivatives of sin x have absolute values less than or equal to 1, so we can apply the Remainder Estimation Theorem with M = 1 to obtain ƒ R2k + 1sxd ƒ … 1 #
ƒ x ƒ 2k + 2 . s2k + 2d!
Mu
ham
ma
Since s ƒ x ƒ 2k + 2>s2k + 2d!d : 0 as k : q , whatever the value of x, R2k + 1sxd : 0, and the Maclaurin series for sin x converges to sin x for every x. Thus,
EXAMPLE 3
q s 1dkx 2k + 1 x3 x5 x7 sin x = a = x + + Á. 3! 5! 7! k = 0 s2k + 1d!
(4)
The Taylor Series for cos x at x = 0 Revisited
Show that the Taylor series for cos x at x = 0 converges to cos x for every value of x. We add the remainder term to the Taylor polynomial for cos x (Section 11.8, Example 3) to obtain Taylor’s formula for cos x with n = 2k:
Solution
cos x = 1 
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x2 x4 x 2k +  Á + s 1dk + R2ksxd. 2! 4! s2kd!
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.9
Convergence of Taylor Series; Error Estimates
815
ƒ R2ksxd ƒ … 1 #
ƒ x ƒ 2k + 1 . s2k + 1d!
suf
i
Because the derivatives of the cosine have absolute value less than or equal to 1, the Remainder Estimation Theorem with M = 1 gives
q
cos x = a
k=0
s 1dkx 2k x2 x4 x6 = 1 + + Á. 2! 4! 6! s2kd!
(5)
Finding a Taylor Series by Substitution
iaz
EXAMPLE 4
You
For every value of x, R2k : 0 as k : q . Therefore, the series converges to cos x for every value of x. Thus,
Find the Taylor series for cos 2x at x = 0.
q
cos 2x = a
k=0
nR
Solution We can find the Taylor series for cos 2x by substituting 2x for x in the Taylor series for cos x:
s2xd4 s2xd6 s 1dks2xd2k s2xd2 + + Á = 1 2! 4! 6! s2kd! 22x 2 24x 4 26x 6 + + Á 2! 4! 6!
ass a
= 1 
Equation (5) with 2x for x
q
22kx 2k = a s 1dk . s2kd! k=0
dH
Equation (5) holds for  q 6 x 6 q , implying that it holds for  q 6 2x 6 q , so the newly created series converges for all x. Exercise 45 explains why the series is in fact the Taylor series for cos 2x.
EXAMPLE 5
Finding a Taylor Series by Multiplication
ma
Find the Taylor series for x sin x at x = 0. We can find the Taylor series for x sin x by multiplying the Taylor series for sin x (Equation 4) by x:
Mu
ham
Solution
x sin x = x ax = x2 
x3 x5 x7 + + Áb 3! 5! 7!
x6 x8 x4 + + Á. 3! 5! 7!
The new series converges for all x because the series for sin x converges for all x. Exercise 45 explains why the series is the Taylor series for x sin x.
Truncation Error The Taylor series for e x at x = 0 converges to e x for all x. But we still need to decide how many terms to use to approximate e x to a given degree of accuracy. We get this information from the Remainder Estimation Theorem.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 816
Chapter 11: Infinite Sequences and Series
Calculate e with an error of less than 10 6 .
We can use the result of Example 1 with x = 1 to write e = 1 + 1 +
1 1 + Á + + Rns1d, 2! n!
You
Solution
suf
i
EXAMPLE 6
with Rns1d = e c
1 sn + 1d!
for some c between 0 and 1.
iaz
For the purposes of this example, we assume that we know that e 6 3. Hence, we are certain that
nR
3 1 6 Rns1d 6 sn + 1d! sn + 1d!
because 1 6 e c 6 3 for 0 6 c 6 1. By experiment we find that 1>9! 7 10 6 , while 3>10! 6 10 6 . Thus we should take sn + 1d to be at least 10, or n to be at least 9. With an error of less than 10 6 , 1 1 1 + + Á + L 2.718282. 2 3! 9!
ass a
e = 1 + 1 +
For what values of x can we replace sin x by x  sx 3>3!d with an error of magnitude no greater than 3 * 10 4 ?
EXAMPLE 7
Here we can take advantage of the fact that the Taylor series for sin x is an alternating series for every nonzero value of x. According to the Alternating Series Estimation Theorem (Section 11.6), the error in truncating sin x = x 
Mu
ham
ma
after sx 3>3!d is no greater than
`
x3 x5 +  Á 3! 5! 
dH
Solution
ƒ x ƒ5 x5 ` = . 120 5!
Therefore the error will be less than or equal to 3 * 10 4 if ƒ x ƒ5 6 3 * 10 4 120
or
5 ƒ x ƒ 6 2360 * 10 4 L 0.514.
Rounded down, to be safe
The Alternating Series Estimation Theorem tells us something that the Remainder Estimation Theorem does not: namely, that the estimate x  sx 3>3!d for sin x is an underestimate when x is positive because then x 5>120 is positive. Figure 11.15 shows the graph of sin x, along with the graphs of a number of its approximating Taylor polynomials. The graph of P3sxd = x  sx 3>3!d is almost indistinguishable from the sine curve when 1 … x … 1.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.9
Convergence of Taylor Series; Error Estimates
817
2 P1
P5
P9
P13
P17
5
6
7
y sin x
1
1
2
3
4
–1 P3
P7
8
9
x
You
0
suf
i
y
P11
–2
P19
iaz
FIGURE 11.15 The polynomials
P15
n s 1dkx 2k + 1 P2n + 1sxd = a k = 0 s2k + 1d!
nR
converge to sin x as n : q . Notice how closely P3sxd approximates the sine curve for x 6 1 (Example 7).
ass a
You might wonder how the estimate given by the Remainder Estimation Theorem compares with the one just obtained from the Alternating Series Estimation Theorem. If we write sin x = x 
x3 + R3 , 3!
then the Remainder Estimation Theorem gives ƒ x ƒ4 ƒ x ƒ4 ƒ R3 ƒ … 1 # 4! = 24 ,
dH
which is not as good. But if we recognize that x  sx 3>3!d = 0 + x + 0x 2 sx 3/3!d + 0x 4 is the Taylor polynomial of order 4 as well as of order 3, then
ma
sin x = x 
x3 + 0 + R4 , 3!
Mu
ham
and the Remainder Estimation Theorem with M = 1 gives ƒ x ƒ5 ƒ x ƒ5 ƒ R4 ƒ … 1 # 5! = 120 .
This is what we had from the Alternating Series Estimation Theorem.
Combining Taylor Series On the intersection of their intervals of convergence, Taylor series can be added, subtracted, and multiplied by constants, and the results are once again Taylor series. The Taylor series for ƒsxd + gsxd is the sum of the Taylor series for ƒ(x) and g(x) because the nth derivative of f + g is f snd + g snd , and so on. Thus we obtain the Taylor series for s1 + cos 2xd>2 by adding 1 to the Taylor series for cos 2x and dividing the combined results by 2, and the Taylor series for sin x + cos x is the termbyterm sum of the Taylor series for sin x and cos x.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 818
Chapter 11: Infinite Sequences and Series
i
Euler’s Identity
i 3 = i 2i = i,
i 4 = i 2i 2 = 1,
and so on, to simplify the result, we obtain e iu = 1 +
i 5 = i 4i = i,
You
i 2 = 1,
suf
As you may recall, a complex number is a number of the form a + bi, where a and b are real numbers and i = 21. If we substitute x = iu (u real) in the Taylor series for e x and use the relations
i 2u2 i 3u3 i 4u4 i 5u5 i 6u6 iu + + + + + + Á 1! 2! 3! 4! 5! 6!
u4 u6 u3 u5 u2 + + Á b + i au +  Á b = cos u + i sin u. 2! 4! 6! 3! 5!
iaz
= a1 
DEFINITION
nR
This does not prove that e iu = cos u + i sin u because we have not yet defined what it means to raise e to an imaginary power. Rather, it says how to define e iu to be consistent with other things we know.
For any real number u, e iu = cos u + i sin u.
ass a
(6)
Equation (6), called Euler’s identity, enables us to define e a + bi to be e a # e bi for any complex number a + bi. One consequence of the identity is the equation
dH
e ip = 1.
When written in the form e ip + 1 = 0, this equation combines five of the most important constants in mathematics.
A Proof of Taylor’s Theorem
ma
We prove Taylor’s theorem assuming a 6 b. The proof for a 7 b is nearly the same. The Taylor polynomial
Mu
ham
Pnsxd = ƒsad + ƒ¿sadsx  ad +
ƒ–sad f sndsad sx  ad2 + Á + sx  adn 2! n!
and its first n derivatives match the function ƒ and its first n derivatives at x = a. We do not disturb that matching if we add another term of the form Ksx  adn + 1 , where K is any constant, because such a term and its first n derivatives are all equal to zero at x = a. The new function fnsxd = Pnsxd + Ksx  adn + 1 and its first n derivatives still agree with ƒ and its first n derivatives at x = a. We now choose the particular value of K that makes the curve y = fnsxd agree with the original curve y = ƒsxd at x = b. In symbols, ƒsbd = Pnsbd + Ksb  adn + 1,
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or
K =
ƒsbd  Pnsbd sb  adn + 1
.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.9 Convergence of Taylor Series; Error Estimates
819
i
With K defined by Equation (7), the function
suf
Fsxd = ƒsxd  fnsxd
F¿sc1 d = 0
You
measures the difference between the original function ƒ and the approximating function fn for each x in [a, b]. We now use Rolle’s Theorem (Section 4.2). First, because Fsad = Fsbd = 0 and both F and F¿ are continuous on [a, b], we know that for some c1 in sa, bd.
Next, because F¿sad = F¿sc1 d = 0 and both F¿ and F– are continuous on [a, c1], we know that for some c2 in sa, c1 d.
iaz
F–sc2 d = 0
Rolle’s Theorem, applied successively to F–, F‡, Á , F sn  1d implies the existence of in sa, c2 d
such that F‡sc3 d = 0,
c4
in sa, c3 d
such that F s4dsc4 d = 0,
nR
c3
o
in sa, cn  1 d
cn
such that F sndscn d = 0.
ass a
Finally, because F snd is continuous on [a, cn] and differentiable on sa, cn d, and F sndsad = F sndscn d = 0, Rolle’s Theorem implies that there is a number cn + 1 in sa, cn d such that F sn + 1dscn + 1 d = 0. n+1
If we differentiate Fsxd = ƒsxd  Pnsxd  Ksx  ad
(8)
a total of n + 1 times, we get
dH
F sn + 1dsxd = ƒsn + 1dsxd  0  sn + 1d!K.
(9)
Equations (8) and (9) together give K =
ƒsn + 1dscd sn + 1d!
for some number c = cn + 1 in sa, bd.
ma
Equations (7) and (10) give ƒsbd = Pnsbd +
ƒsn + 1dscd sb  adn + 1 . sn + 1d!
Mu
ham
This concludes the proof.
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i f u s u o
11.9 Convergence of Taylor Series; Error Estimates
Y z a i R n a s as
EXERCISES 11.9 Taylor Series by Substitution
More Taylor Series
H d a
Use substitution (as in Example 4) to find the Taylor series at x = 0 of the functions in Exercises 1–6. 1. e
m m ha
5x
px 4. sin a b 2
2. e
x>2
5. cos 2x + 1
Mu
819
3. 5 sin s xd
6. cos A x 3>2> 22 B
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Find Taylor series at x = 0 for the functions in Exercises 7–18. 7. xe x
10. sin x  x +
8. x 2 sin x
x3 3!
11. x cos px
9.
x2  1 + cos x 2
12. x 2 cos sx 2 d
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 820
Chapter 11: Infinite Sequences and Series
1 s1  xd2
18.
16. x ln s1 + 2xd 2 s1  xd3
i
x2 1  2x
Finding and Identifying Maclaurin Series
Error Estimates 19. For approximately what values of x can you replace sin x by x  sx 3>6d with an error of magnitude no greater than 5 * 10 4 ? Give reasons for your answer. 20. If cos x is replaced by 1  sx 2>2d and ƒ x ƒ 6 0.5 , what estimate can be made of the error? Does 1  sx 2>2d tend to be too large, or too small? Give reasons for your answer.
Recall that the Maclaurin series is just another name for the Taylor series at x = 0 . Each of the series in Exercises 31–34 is the value of the Maclaurin series of a function ƒ(x) at some point. What function and what point? What is the sum of the series? 31. s0.1d 32. 1 
21. How close is the approximation sin x = x when ƒ x ƒ 6 10 3 ? For which of these values of x is x 6 sin x ?
33.
22. The estimate 21 + x = 1 + sx>2d is used when x is small. Estimate the error when ƒ x ƒ 6 0.01 .
34. p 
s 1dkspd2k p4 p2 + 4  Á + 2k + Á 4 # 2! 4 # 4! 4 # s2k!d 2
s 1dkp2k + 1 p5 p p3 + 5  Á + 2k + 1 + Á  3 3 3 #5 3 #3 3 s2k + 1d pk p2 p3 +  Á + s 1dk  1 + Á 2 3 k
35. Multiply the Maclaurin series for e x and sin x together to find the first five nonzero terms of the Maclaurin series for e x sin x . 36. Multiply the Maclaurin series for e x and cos x together to find the first five nonzero terms of the Maclaurin series for e x cos x .
ass a
24. (Continuation of Exercise 23.) When x 6 0 , the series for e x is an alternating series. Use the Alternating Series Estimation Theorem to estimate the error that results from replacing e x by 1 + x + sx 2>2d when 0.1 6 x 6 0 . Compare your estimate with the one you obtained in Exercise 23.
s0.1d3 s0.1d5 s 1dks0.1d2k + 1 +  Á + + Á 3! 5! s2k + 1d!
nR
23. The approximation e x = 1 + x + sx 2>2d is used when x is small. Use the Remainder Estimation Theorem to estimate the error when ƒ x ƒ 6 0.1 .
You
17.
15.
iaz
14. sin2 x
T b. Graph ƒsxd = s1  cos xd>x 2 together with y = s1>2d  sx 2>24d and y = 1>2 for 9 … x … 9 . Comment on the relationships among the graphs.
suf
13. cos2 x (Hint: cos2 x = s1 + cos 2xd>2 .)
dH
25. Estimate the error in the approximation sinh x = x + sx 3>3!d when ƒ x ƒ 6 0.5 . (Hint: Use R4 , not R3 .) 26. When 0 … h … 0.01 , show that e h may be replaced by 1 + h with an error of magnitude no greater than 0.6% of h. Use e 0.01 = 1.01 . 27. For what positive values of x can you replace ln s1 + xd by x with an error of magnitude no greater than 1% of the value of x?
ma
28. You plan to estimate p>4 by evaluating the Maclaurin series for tan1 x at x = 1 . Use the Alternating Series Estimation Theorem to determine how many terms of the series you would have to add to be sure the estimate is good to two decimal places.
ham
29. a. Use the Taylor series for sin x and the Alternating Series Estimation Theorem to show that 1 
sin x x2 6 x 6 1, 6
x Z 0.
Mu
T b. Graph ƒsxd = ssin xd>x together with the functions y = 1  sx 2>6d and y = 1 for 5 … x … 5 . Comment on the relationships among the graphs. 30. a. Use the Taylor series for cos x and the Alternating Series Estimation Theorem to show that 1  cos x x2 1 1 6 , 6 2 24 2 x2
x Z 0.
(This is the inequality in Section 2.2, Exercise 52.)
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37. Use the identity sin2 x = s1  cos 2xd>2 to obtain the Maclaurin series for sin2 x . Then differentiate this series to obtain the Maclaurin series for 2 sin x cos x. Check that this is the series for sin 2x.
38. (Continuation of Exercise 37.) Use the identity cos2 x = cos 2x + sin2 x to obtain a power series for cos2 x .
Theory and Examples 39. Taylor’s Theorem and the Mean Value Theorem Explain how the Mean Value Theorem (Section 4.2, Theorem 4) is a special case of Taylor’s Theorem. 40. Linearizations at inflection points Show that if the graph of a twicedifferentiable function ƒ(x) has an inflection point at x = a , then the linearization of ƒ at x = a is also the quadratic approximation of ƒ at x = a . This explains why tangent lines fit so well at inflection points. 41. The (second) second derivative test ƒsxd = ƒsad + ƒ¿sadsx  ad +
Use the equation ƒ–sc2 d sx  ad2 2
to establish the following test. Let ƒ have continuous first and second derivatives and suppose that ƒ¿sad = 0 . Then a. ƒ has a local maximum at a if ƒ– … 0 throughout an interval whose interior contains a; b. ƒ has a local minimum at a if ƒ– Ú 0 throughout an interval whose interior contains a.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.9 Convergence of Taylor Series; Error Estimates
i
T 48. a. Graph the curves y = s1>3d  sx 2 d>5 and y = sx  tan1 xd>x 3 together with the line y = 1>3 . b. Use a Taylor series to explain what you see. What is
43. a. Use Taylor’s formula with n = 2 to find the quadratic approximation of ƒsxd = s1 + xdk at x = 0 (k a constant).
a. Let P be an approximation of p accurate to n decimals. Show that P + sin P gives an approximation correct to 3n decimals. (Hint: Let P = p + x .)
Euler’s Identity
49. Use Equation (6) to write the following powers of e in the form a + bi .
cos u =
and x4 x5 x e = x + x + + + Á, 2! 3! 2
3
dH
2 x
e iu + e iu 2
obtained by multiplying Taylor series by powers of x, as well as series obtained by integration and differentiation of convergent power series, are themselves the Taylor series generated by the functions they represent.
ham
ma
46. Taylor series for even functions and odd functions (Continuaq tion of Section 11.7, Exercise 45.) Suppose that ƒsxd = g n = 0 an x n converges for all x in an open interval s c, cd . Show that a. If ƒ is even, then a1 = a3 = a5 = Á = 0 , i.e., the Taylor series for ƒ at x = 0 contains only even powers of x. b. If ƒ is odd, then a0 = a2 = a4 = Á = 0 , i.e., the Taylor series for ƒ at x = 0 contains only odd powers of x. 47. Taylor polynomials of periodic functions a. Show that every continuous periodic function ƒsxd,  q 6 x 6 q , is bounded in magnitude by showing that there exists a positive constant M such that ƒ ƒsxd ƒ … M for all x.
Mu
b. Show that the graph of every Taylor polynomial of positive degree generated by ƒsxd = cos x must eventually move away from the graph of cos x as ƒ x ƒ increases. You can see this in Figure 11.13. The Taylor polynomials of sin x behave in a similar way (Figure 11.15).
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and
sin u =
e iu  e iu . 2i
nR
51. Establish the equations in Exercise 50 by combining the formal Taylor series for e iu and e iu . 52. Show that
a. cosh iu = cos u ,
b. sinh iu = i sin u .
53. By multiplying the Taylor series for e x and sin x, find the terms through x 5 of the Taylor series for e x sin x . This series is the imaginary part of the series for
ass a
x4 x6 x8 + + Á 3! 5! 7!
c. e ip>2
50. Use Equation (6) to show that
q
x sin x = x 2 
b. e ip>4
a. e ip
T b. Try it with a calculator. 45. The Taylor series generated by ƒsxd = g n = 0 an x n is q q g n = 0 an x n A function defined by a power series g n = 0 an x n with a radius of convergence c 7 0 has a Taylor series that converges to the function at every point of s c, cd . Show this by q showing that the Taylor series generated by ƒsxd = g n = 0 an x n is q n the series g n = 0 an x itself. An immediate consequence of this is that series like
You
44. Improving approximations to P
x  tan1 x ? x:0 x3 lim
iaz
b. If k = 3 , for approximately what values of x in the interval [0, 1] will the error in the quadratic approximation be less than 1> 100?
suf
42. A cubic approximation Use Taylor’s formula with a = 0 and n = 3 to find the standard cubic approximation of ƒsxd = 1>s1  xd at x = 0 . Give an upper bound for the magnitude of the error in the approximation when ƒ x ƒ … 0.1 .
821
e x # e ix = e s1 + idx .
Use this fact to check your answer. For what values of x should the series for e x sin x converge?
54. When a and b are real, we define e sa + ibdx with the equation e sa + ibdx = e ax # e ibx = e axscos bx + i sin bxd .
Differentiate the righthand side of this equation to show that d sa + ibdx e = sa + ibde sa + ibdx . dx Thus the familiar rule sd>dxde kx = ke kx holds for k complex as well as real. 55. Use the definition of e iu to show that for any real numbers u, u1 , and u2 , a. e iu1e iu2 = e isu1 + u2d,
b. e iu = 1>e iu .
56. Two complex numbers a + ib and c + id are equal if and only if a = c and b = d . Use this fact to evaluate L
e ax cos bx dx
and
L
e ax sin bx dx
from L
e sa + ibdx dx =
a  ib sa + ibdx e + C, a2 + b2
where C = C1 + iC2 is a complex constant of integration.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 11: Infinite Sequences and Series
a. For what values of x can the function be replaced by each approximation with an error less than 10 2 ? b. What is the maximum error we could expect if we replace the function by each approximation over the specified interval? Using a CAS, perform the following steps to aid in answering questions (a) and (b) for the functions and intervals in Exercises 57–62. Step 1: Plot the function over the specified interval. Step 2: Find the Taylor polynomials P1sxd, P2sxd, and P3sxd at x = 0.
i
Step 5: Compare your estimated error with the actual error Ensxd = ƒ ƒsxd  Pnsxd ƒ by plotting Ensxd over the specified interval. This will help answer question (b).
Step 6: Graph the function and its three Taylor approximations together. Discuss the graphs in relation to the information discovered in Steps 4 and 5. 57. ƒsxd =
1 21 + x
,
3 ƒxƒ … 4
58. ƒsxd = s1 + xd3>2,

1 … x … 2 2
x , ƒxƒ … 2 x2 + 1 60. ƒsxd = scos xdssin 2xd, ƒ x ƒ … 2 61. ƒsxd = e x cos 2x, ƒ x ƒ … 1 59. ƒsxd =
62. ƒsxd = e x>3 sin 2x,
ƒxƒ … 2
Mu
ham
ma
dH
ass a
Step 3: Calculate the sn + 1dst derivative ƒsn + 1dscd associated with the remainder term for each Taylor polynomial. Plot the derivative as a function of c over the specified interval and estimate its maximum absolute value, M.
suf
Taylor’s formula with n = 1 and a = 0 gives the linearization of a function at x = 0 . With n = 2 and n = 3 we obtain the standard quadratic and cubic approximations. In these exercises we explore the errors associated with these approximations. We seek answers to two questions:
Step 4: Calculate the remainder Rnsxd for each polynomial. Using the estimate M from Step 3 in place of ƒsn + 1dscd, plot Rnsxd over the specified interval. Then estimate the values of x that answer question (a).
You
Linear, Quadratic, and Cubic Approximations
iaz
COMPUTER EXPLORATIONS
nR
822
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 822
i f su
Chapter 11: Infinite Sequences and Series
u o zY
Applications of Power Series
11.10
a i R
This section introduces the binomial series for estimating powers and roots and shows how series are sometimes used to approximate the solution of an initial value problem, to evaluate nonelementary integrals, and to evaluate limits that lead to indeterminate forms. We provide a selfcontained derivation of the Taylor series for tan1 x and conclude with a reference table of frequently used series.
n a s s a
The Binomial Series for Powers and Roots
H d a m am
The Taylor series generated by ƒsxd = s1 + xdm , when m is constant, is 1 + mx +
M
uh
msm  1dsm  2d 3 Á msm  1d 2 x + x + 2! 3! +
msm  1dsm  2d Á sm  k + 1d k Á x + . k!
(1)
This series, called the binomial series, converges absolutely for ƒ x ƒ 6 1. To derive the
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.10
Applications of Power Series
823
ƒsxd = s1 + xdm ƒ¿sxd = ms1 + xdm  1 ƒ–sxd = msm  1ds1 + xdm  2
You
ƒ‡sxd = msm  1dsm  2ds1 + xdm  3
suf
i
series, we first list the function and its derivatives:
o skd
ƒ sxd = msm  1dsm  2d Á sm  k + 1ds1 + xdm  k .
nR
iaz
We then evaluate these at x = 0 and substitute into the Taylor series formula to obtain Series (1). If m is an integer greater than or equal to zero, the series stops after sm + 1d terms because the coefficients from k = m + 1 on are zero. If m is not a positive integer or zero, the series is infinite and converges for ƒ x ƒ 6 1. To see why, let uk be the term involving x k . Then apply the Ratio Test for absolute convergence to see that uk + 1 m  k ` uk ` = ` x` :ƒxƒ k + 1
as k : q .
ass a
Our derivation of the binomial series shows only that it is generated by s1 + xdm and converges for ƒ x ƒ 6 1. The derivation does not show that the series converges to s1 + xdm . It does, but we omit the proof.
dH
The Binomial Series For 1 6 x 6 1,
q m s1 + xdm = 1 + a a b x k , k k=1
ma
where we define m a b = m, 1
msm  1d m a b = , 2! 2
Mu
ham
and msm  1dsm  2d Á sm  k + 1d m a b = k! k
EXAMPLE 1
for k Ú 3.
Using the Binomial Series
If m = 1, a
1 b = 1, 1
a
1s 2d 1 b = = 1, 2! 2
and a
1s 2ds 3d Á s 1  k + 1d 1 k! b = = s 1dk a b = s 1dk . k! k! k
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Chapter 11: Infinite Sequences and Series
suf
i
With these coefficient values and with x replaced by x, the binomial series formula gives the familiar geometric series q
s1 + xd1 = 1 + a s 1dkx k = 1  x + x 2  x 3 + Á + s 1dkx k + Á . k=1
Using the Binomial Series
You
EXAMPLE 2
We know from Section 3.8, Example 1, that 21 + x L 1 + sx>2d for ƒ x ƒ small. With m = 1>2, the binomial series gives quadratic and higherorder approximations as well, along with error estimates that come from the Alternating Series Estimation Theorem:
iaz
s1 + xd1>2
3 1 1 1 1 a b a b a b a b a b 2 2 2 2 2 x = 1 + + x2 + x3 2 2! 3!
= 1 +
nR
3 5 1 1 a b a b a b a b 2 2 2 2 + x4 + Á 4!
5x 4 x x2 x3 + + Á. 2 8 16 128
ass a
Substitution for x gives still other approximations. For example, 21  x 2 L 1 
1 1 1 1  x L 1 2x 8x 2
for ƒ x 2 ƒ small
1 for ` x ` small, that is, ƒ x ƒ large.
dH
A
x2 x4 2 8
Power Series Solutions of Differential Equations and Initial Value Problems
Mu
ham
ma
When we cannot find a relatively simple expression for the solution of an initial value problem or differential equation, we try to get information about the solution in other ways. One way is to try to find a power series representation for the solution. If we can do so, we immediately have a source of polynomial approximations of the solution, which may be all that we really need. The first example (Example 3) deals with a firstorder linear differential equation that could be solved with the methods of Section 9.2. The example shows how, not knowing this, we can solve the equation with power series. The second example (Example 4) deals with an equation that cannot be solved analytically by previous methods.
EXAMPLE 3
Series Solution of an Initial Value Problem
Solve the initial value problem y¿  y = x, Solution
ys0d = 1.
We assume that there is a solution of the form y = a0 + a1 x + a2 x 2 + Á + an  1x n  1 + an x n + Á .
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(2)
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.10
Applications of Power Series
825
(3)
suf
y¿ = a1 + 2a2 x + 3a3 x 2 + Á + nan x n  1 + Á
i
Our goal is to find values for the coefficients ak that make the series and its first derivative
satisfy the given differential equation and initial condition. The series y¿  y is the difference of the series in Equations (2) and (3):
You
y¿  y = sa1  a0 d + s2a2  a1 dx + s3a3  a2 dx 2 + Á + snan  an  1 dx n  1 + Á .
(4)
If y is to satisfy the equation y¿  y = x, the series in Equation (4) must equal x. Since power series representations are unique (Exercise 45 in Section 11.7), the coefficients in Equation (4) must satisfy the equations Constant terms
3a3  a2 = 0
Coefficients of x2
nR
iaz
a1  a0 = 0 2a2  a1 = 1
nan  an  1
Coefficients of x
o
o = 0
Coefficients of xn  1
o
o
ass a
We can also see from Equation (2) that y = a0 when x = 0, so that a0 = 1 (this being the initial condition). Putting it all together, we have a0 = 1,
a3 =
a1 = a0 = 1,
1 + a1 1 + 1 2 = = , 2 2 2 an  1 2 an = n = , Á n! a2 =
a2 2 2 = # = ,Á, 3 3 2 3!
Mu
ham
ma
dH
Substituting these coefficient values into the equation for y (Equation (2)) gives y = 1 + x + 2#
x2 x3 xn + 2# + Á + 2# + Á 2! 3! n!
x2 x3 xn = 1 + x + 2 a + + Á + + Áb 2! 3! n! ('''''')''''''* the Taylor series for ex  1  x
= 1 + x + 2se x  1  xd = 2e x  1  x.
The solution of the initial value problem is y = 2e x  1  x. As a check, we see that ys0d = 2e 0  1  0 = 2  1 = 1 and y¿  y = s2e x  1d  s2e x  1  xd = x.
EXAMPLE 4
Solving a Differential Equation
Find a power series solution for
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y– + x 2y = 0.
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(5)
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 11: Infinite Sequences and Series
We assume that there is a solution of the form y = a0 + a1 x + a2 x 2 + Á + an x n + Á ,
i
Solution
(6)
suf
826
and find what the coefficients ak have to be to make the series and its second derivative y– = 2a2 + 3 # 2a3 x + Á + nsn  1dan x n  2 + Á
(7)
You
satisfy Equation (5). The series for x 2y is x 2 times the righthand side of Equation (6): x 2y = a0 x 2 + a1 x 3 + a2 x 4 + Á + an x n + 2 + Á .
(8)
The series for y– + x 2y is the sum of the series in Equations (7) and (8):
y– + x 2y = 2a2 + 6a3 x + s12a4 + a0 dx 2 + s20a5 + a1 dx 3
iaz
+ Á + snsn  1dan + an  4 dx n  2 + Á .
(9)
2a2 = 0,
nR
Notice that the coefficient of x n  2 in Equation (8) is an  4 . If y and its second derivative y– are to satisfy Equation (5), the coefficients of the individual powers of x on the righthand side of Equation (9) must all be zero: 6a3 = 0,
and for all n Ú 4,
12a4 + a0 = 0,
20a5 + a1 = 0,
ass a
nsn  1dan + an  4 = 0.
(10)
(11)
We can see from Equation (6) that
a0 = ys0d,
a1 = y¿s0d.
dH
In other words, the first two coefficients of the series are the values of y and y¿ at x = 0. Equations in (10) and the recursion formula in Equation (11) enable us to evaluate all the other coefficients in terms of a0 and a1 . The first two of Equations (10) give a2 = 0,
a3 = 0.
ma
Equation (11) shows that if an  4 = 0, then an = 0; so we conclude that a6 = 0,
a7 = 0,
a10 = 0,
a11 = 0,
Mu
ham
and whenever n = 4k + 2 or 4k + 3, an is zero. For the other coefficients we have an  4 an = nsn  1d so that a0 a0 a4 , a8 = # = # # # 4#3 8 7 3 4 7 8 a8 a0 = = # # # # # 11 # 12 3 4 7 8 11 12
a4 = a12 and
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a5 = a13
a5 a1 a1 , a9 = # = # # # # 5 4 9 8 4 5 8 9 a9 a1 = = # # # # # . 12 # 13 4 5 8 9 12 13
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.10
Applications of Power Series
827
x4 x8 x 12 + + Áb 3#4 3#4#7#8 3 # 4 # 7 # 8 # 11 # 12
+ a1 ax 
x5 x9 x 13 + # # #  # # # # # + Áb . # 4 5 4 5 8 9 4 5 8 9 12 13
You
y = a0 a1 
suf
i
The answer is best expressed as the sum of two separate series—one multiplied by a0 , the other by a1 :
Both series converge absolutely for all x, as is readily seen by the Ratio Test.
Evaluating Nonelementary Integrals
iaz
Taylor series can be used to express nonelementary integrals in terms of series. Integrals like 1 sin x 2 dx arise in the study of the diffraction of light. Express 1 sin x 2 dx as a power series.
EXAMPLE 5
From the series for sin x we obtain
nR
Solution
sin x 2 = x 2 
x3 x 11 x 15 x 10 x7  # + +  Á. # # 3 7 3! 11 5! 15 7! 19 # 9!
ass a
Therefore,
x6 x 10 x 14 x 18 + +  Á. 3! 5! 7! 9!
sin x 2 dx = C +
L
EXAMPLE 6
dx with an error of less than 0.001.
dH
Estimate
Estimating a Definite Integral
1 2 10 sin x
From the indefinite integral in Example 5,
Solution
1
L0
sin x 2 dx =
1 1 1 1 1  # + +  Á. 3 7 3! 11 # 5! 15 # 7! 19 # 9!
Mu
ham
ma
The series alternates, and we find by experiment that 1 L 0.00076 11 # 5!
is the first term to be numerically less than 0.001. The sum of the preceding two terms gives 1
L0
sin x 2 dx L
1 1 L 0.310. 3 42
With two more terms we could estimate 1
L0
sin x 2 dx L 0.310268
with an error of less than 10 6 . With only one term beyond that we have 1
L0
sin x 2 dx L
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 828
Chapter 11: Infinite Sequences and Series
suf
i
with an error of about 1.08 * 10 9 . To guarantee this accuracy with the error formula for the Trapezoidal Rule would require using about 8000 subintervals.
Arctangents
You
In Section 11.7, Example 5, we found a series for tan1 x by differentiating to get d 1 tan1 x = = 1  x2 + x4  x6 + Á dx 1 + x2 and integrating to get
x7 x3 x5 + Á. + 5 7 3
iaz
tan1 x = x 
However, we did not prove the termbyterm integration theorem on which this conclusion depended. We now derive the series again by integrating both sides of the finite formula
nR
n + 1 2n + 2 1 2 4 6 Á + s 1dnt 2n + s 1d t = 1 t + t t + , 1 + t2 1 + t2
(12)
ass a
in which the last term comes from adding the remaining terms as a geometric series with first term a = s 1dn + 1t 2n + 2 and ratio r = t 2 . Integrating both sides of Equation (12) from t = 0 to t = x gives tan1 x = x 
where
x5 x3 x7 x 2n + 1 + + Rnsxd, + Á + s 1dn 5 7 3 2n + 1
s 1dn + 1t 2n + 2 dt. 1 + t2 L0 x
dH
Rnsxd =
The denominator of the integrand is greater than or equal to 1; hence ƒxƒ
ƒ Rnsxd ƒ …
L0
t 2n + 2 dt =
ƒ x ƒ 2n + 3 . 2n + 3
Mu
ham
ma
If ƒ x ƒ … 1, the right side of this inequality approaches zero as n : q . Therefore limn: q Rnsxd = 0 if ƒ x ƒ … 1 and s 1dnx 2n + 1 , 2n + 1 n=0 q
tan1 x = a tan
1
ƒ x ƒ … 1.
x5 x7 x3 + + Á, x = x 5 7 3
(13)
ƒxƒ … 1
We take this route instead of finding the Taylor series directly because the formulas for the higherorder derivatives of tan1 x are unmanageable. When we put x = 1 in Equation (13), we get Leibniz’s formula: s 1dn p 1 1 1 1 = 1  +  +  Á + + Á. 5 7 4 3 9 2n + 1 Because this series converges very slowly, it is not used in approximating p to many decimal places. The series for tan1 x converges most rapidly when x is near zero. For that reason, people who use the series for tan1 x to compute p use various trigonometric identities.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.10
Applications of Power Series
829
1 b = tan1 , 3
suf
1 2
a = tan1
and
You
then tan sa + bd =
i
For example, if
1 tan a + tan b 2 + = 1  tan a tan b 1 
and
1 3 1 6
= 1 = tan
p 4
iaz
p 1 1 = a + b = tan1 + tan1 . 4 2 3
nR
Now Equation (13) may be used with x = 1>2 to evaluate tan1 (1>2) and with x = 1>3 to give tan1 (1>3). The sum of these results, multiplied by 4, gives p.
Evaluating Indeterminate Forms
ass a
We can sometimes evaluate indeterminate forms by expressing the functions involved as Taylor series.
EXAMPLE 7
Limits Using Power Series
dH
Evaluate
lim
x:1
ln x . x  1
We represent ln x as a Taylor series in powers of x  1. This can be accomplished by calculating the Taylor series generated by ln x at x = 1 directly or by replacing x by x  1 in the series for ln (1 + x) in Section 11.7, Example 6. Either way, we obtain
Mu
ham
ma
Solution
ln x = sx  1d 
1 sx  1d2 + Ă , 2
from which we find that ln x 1 = lim a1  sx  1d + Ă b = 1. x 1 2 x:1 x:1 lim
EXAMPLE 8
Limits Using Power Series
Evaluate
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lim
x:0
sin x  tan x . x3
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 11: Infinite Sequences and Series
The Taylor series for sin x and tan x, to terms in x 5 , are sin x = x 
x3 x5 +  Á, 3! 5!
tan x = x +
x3 2x 5 + + Á. 3 15
sin x  tan x = 
You
Hence,
i
Solution
suf
830
x3 x5 x2 1  Á = x 3 a  Áb 2 8 2 8
and
sin x  tan x x2 1 = lim a Áb 2 8 x:0 x:0 x3
iaz
lim
nR
1 =  . 2
If we apply series to calculate limx:0 ss1>sin xd  s1/xdd, we not only find the limit successfully but also discover an approximation formula for csc x.
Find lim a x:0
1 1  xb. sin x
dH
Solution
Approximation Formula for csc x
ass a
EXAMPLE 9
Mu
ham
ma
x  sin x 1 1  x = = sin x x sin x x3 a
x  ax x # ax 
x3 x5 +  Áb 3! 5! x3 x5 +  Áb 3! 5!
x2 1 + Áb 3! 5!
x2 1 + Á 3! 5! = = x . x2 x2 2 Á Á 1 + + b x a1 3! 3!
Therefore, x2 1 + Á 3! 5! 1 1 lim a  x b = lim § x ¥ = 0. x:0 sin x x:0 x2 1 + Á 3! From the quotient on the right, we can see that if ƒ x ƒ is small, then x 1 1 1  x L x# = 6 3! sin x
To Read it Online & Download:
or
x 1 csc x L x + . 6
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.10 Applications of Power Series
suf
i
TABLE 11.1 Frequently used Taylor series q
1 = 1 + x + x 2 + Á + x n + Á = a x n, 1  x n=0
ƒxƒ 6 1
q
1 = 1  x + x 2  Á + s xdn + Á = a s 1dnx n, 1 + x n=0
cos x = 1 
q s 1dnx 2n + 1 x3 x5 x 2n + 1 + Á = a , +  Á + s 1dn 3! 5! s2n + 1d! n = 0 s2n + 1d! q s 1dnx 2n x2 x4 x 2n + Á = a , +  Á + s 1dn 2! 4! s2nd! s2nd! n=0
ƒxƒ 6 q
ƒxƒ 6 q
q s 1dn  1x n x2 x3 xn +  Á + s 1dn  1 n + Á = a , n 2 3 n=1
nR
ln s1 + xd = x 
1 6 x … 1
x 2n + 1 1 + x x3 x5 x 2n + 1 = 2 tanh1 x = 2 ax + + + Á + + Áb = 2 a , 5 1  x 3 2n + 1 n = 0 2n + 1
tan1 x = x 
q s 1dnx 2n + 1 x3 x5 x 2n + 1 +  Á + s 1dn + Á = a , 5 3 2n + 1 2n + 1 n=0
Binomial Series s1 + xdm = 1 + mx +
q
ƒxƒ 6 1
ƒxƒ … 1
ass a
ln
ƒxƒ 6 q
iaz
sin x = x 
x2 xn xn + Á + + Á = a , 2! n! n = 0 n!
You
ƒxƒ 6 1
q
ex = 1 + x +
831
msm  1dsm  2dx 3 msm  1dsm  2d Á sm  k + 1dx k msm  1dx 2 + + Á + + Á 2! 3! k!
where m a b = m, 1
ƒ x ƒ 6 1,
dH
q m = 1 + a a bx k, k k=1
msm  1d m , a b = 2! 2
msm  1d Á sm  k + 1d m a b = k! k
for k Ú 3.
Mu
ham
ma
m Note: To write the binomial series compactly, it is customary to define a b to be 1 and to take x 0 = 1 (even in the usually 0 m q excluded case where x = 0), yielding s1 + xdm = g k = 0 a bx k . If m is a positive integer, the series terminates at x m and the k result converges for all x.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
Y z a i R n a s as
i f u s u o
11.10 Applications of Power Series
EXERCISES 11.10 Binomial Series
H d ma
Find the first four terms of the binomial series for the functions in Exercises 1–10. 1. s1 + xd1>2
M
3. s1  xd1>2
x 5. a1 + b 2
x 6. a1  b 2
m a uh
1>2
4. s1  2xd
2. s1 + xd1>3
2
2
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7. s1 + x 3 d1>2 1 9. a1 + x b
1>2
8. s1 + x 2 d1>3
2 10. a1  x b
1>3
Find the binomial series for the functions in Exercises 11–14. 11. s1 + xd4
12. s1 + x 2 d3
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831
4100 AWL/Thomas_ch11p746847 8/25/04 2:41 PM Page 832
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 832
Chapter 11: Infinite Sequences and Series x
4
44. Fsxd =
2
t 2e t dt,
L0
[0, 1]
i
x b 2
x
45. Fsxd =
Find series solutions for the initial value problems in Exercises 15–32. 16. y¿  2y = 0,
17. y¿  y = 1,
y s0d = 0
18. y¿ + y = 1,
19. y¿  y = x,
y s0d = 0
20. y¿ + y = 2x,
23. s1  xdy¿  y = 0,
y s0d = 1
22. y¿  x y = 0,
y s0d = 1
y s0d = 2
24. s1 + x 2 dy¿ + 2xy = 0,
49. lim
27. y– + y = x,
y¿s0d = 1 and y s0d = 2
28. y–  y = x,
y¿s0d = 2 and y s0d = 1
51. lim
y¿s0d = b and y s0d = a
31. y– + x 2y = x,
y¿s0d = b and y s0d = a
32. y–  2y¿ + y = 0,
0.2
0.2
L0 0.1
35.
L0
34.
L0
4
dx
36.
3 2 1 + x 2 dx
L0
T Use series to approximate the values of the integrals in Exercises 37–40 with an error of magnitude less than 10 8 . L0
0.1
sin x x dx
38.
0.1
39.
L0
ma
37.
40.
2
e x dx
L0
1
21 + x 4 dx
L0
1  cos x dx x2
ham
41. Estimate the error if cos t 2 is approximated by 1 1
integral 10 cos t 2 dt .
42. Estimate the error if cos 2t is approximated by 1 
t4 t8 + in the 2 4! t t2 t3 + 2 4! 6!
Mu
In Exercises 43–46, find a polynomial that will approximate F(x) throughout the given interval with an error of magnitude less than 10 3 . x
L0
u5
tan1 y  sin y
y:0
y 3 cos y
54. lim sx + 1d sin x: q
1 x + 1
x2  4 x:2 ln sx  1d
56. lim
Theory and Examples 57. Replace x by x in the Taylor series for ln s1 + xd to obtain a series for ln s1  xd . Then subtract this from the Taylor series for ln s1 + xd to show that for ƒ x ƒ 6 1 , ln
1 + x x3 x5 = 2 ax + + + Á b. 5 1  x 3
58. How many terms of the Taylor series for ln s1 + xd should you add to be sure of calculating ln (1.1) with an error of magnitude less than 10 8 ? Give reasons for your answer. 59. According to the Alternating Series Estimation Theorem, how many terms of the Taylor series for tan1 1 would you have to add to be sure of finding p>4 with an error of magnitude less than 10 3 ? Give reasons for your answer. 60. Show that the Taylor series for ƒsxd = tan1 x diverges for ƒ x ƒ 7 1. T 61. Estimating Pi About how many terms of the Taylor series for tan1 x would you have to use to evaluate each term on the righthand side of the equation
1
in the integral 10 cos 2t dt .
43. Fsxd =
sin u  u + su3>6d
0.25
1 21 + x
0.1
y3
ln s1 + x 2 d x:0 1  cos x
e x  1 dx x
50. lim 52. lim
55. lim
dH
sin x 2 dx
y  tan1 y
2
T In Exercises 33–36, use series to estimate the integrals’ values with an error of magnitude less than 10 3 . (The answer section gives the integrals’ values rounded to five decimal places.)
e x  e x x x:0 u :0
x: q
Approximations and Nonelementary Integrals
33.
t
4
53. lim x 2se 1>x  1d
y¿s0d = 1 and y s0d = 0
(b) [0, 1]
48. lim
nR
30. y–  x y = 0,
2
1  cos t  st 2>2d
y:0
y¿s2d = 2 and y s2d = 0
2
x
ass a
29. y–  y = x,
e x  s1 + xd
t: 0
y¿s0d = 0 and y s0d = 1
ln s1 + td dt, (a) [0, 0.5] t
Use series to evaluate the limits in Exercises 47–56. x:0
y s0d = 3
26. y– + y = 0,
L0
(b) [0, 1]
Indeterminate Forms 47. lim
y¿s0d = 1 and y s0d = 0
25. y–  y = 0,
x
46. Fsxd =
y s0d = 2
2
y s0d = 1
21. y¿  xy = 0,
y s0d = 1
iaz
y s0d = 1
15. y¿ + y = 0,
tan1 t dt, (a) [0, 0.5]
L0
You
Initial Value Problems
suf
14. a1 
13. s1  2xd3
sin t 2 dt,
[0, 1]
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p = 48 tan1
1 1 1 + 32 tan1  20 tan1 57 18 239
with an error of magnitude less than 10 6 ? In contrast, the conq vergence of g n = 1s1>n 2 d to p2>6 is so slow that even 50 terms will not yield twoplace accuracy. 62. Integrate the first three nonzero terms of the Taylor series for tan t from 0 to x to obtain the first three nonzero terms of the Taylor series for ln sec x.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.10 Applications of Power Series 63. a. Use the binomial series and the fact that
suf
i
69. Series for sin1 x Integrate the binomial series for s1  x 2 d1>2 to show that for ƒ x ƒ 6 1 ,
d sin1 x = s1  x 2 d1>2 dx
1 # 3 # 5 # Á # s2n  1d x 2n + 1 . 2 # 4 # 6 # Á # s2nd 2n + 1 n=1 q
sin1 x = x + a
to generate the first four nonzero terms of the Taylor series for sin1 x . What is the radius of convergence?
70. Series for tan1 x for ƒ x ƒ 7 1 Derive the series
b. Series for cos x Use your result in part (a) to find the first five nonzero terms of the Taylor series for cos1 x . 64. a. Series for sinh x Taylor series for
Find the first four nonzero terms of the
You
1
1
p 1 1 1  x + + Á, x 7 1 2 5x 5 3x 3 p 1 1 1 + Á, x 6 1 , tan1 x =   x + 2 5x 5 3x 3 tan1 x =
x
dt . L0 21 + t 2
T b. Use the first three terms of the series in part (a) to estimate sinh1 0.25 . Give an upper bound for the magnitude of the estimation error. 65. Obtain the Taylor series for 1>s1 + xd2 from the series for 1>s1 + xd . 2
66. Use the Taylor series for 1>s1  x d to obtain a series for 2x>s1  x 2 d2 .
2#4#4#6#6#8# Á p = # # # # # # Á . 4 3 3 5 5 7 7 Find p to two decimal places with this formula.
1 , 9
#
1 1 1 1 1 = 2  4 + 6  8 + Á 1 + s1>t 2 d t t t t
in the first case from x to q and in the second case from  q to x.
71. The value of g n = 1 tan1s2>n2 d q
a. Use the formula for the tangent of the difference of two angles to show that tan stan1 sn + 1d  tan1 sn  1dd =
2 n2
b. Show that N
p 1 2 1 1 a tan n 2 = tan sN + 1d + tan N  4 . n=1
c. Find the value of g n = 1 tan1 q
2 . n2
1 . 13
Mu
ham
ma
1 , 5
dH
T 68. Construct a table of natural logarithms ln n for n = 1, 2, 3, Á , 10 by using the formula in Exercise 57, but taking advantage of the relationships ln 4 = 2 ln 2, ln 6 = ln 2 + ln 3, ln 8 = 3 ln 2, ln 9 = 2 ln 3 , and ln 10 = ln 2 + ln 5 to reduce the job to the calculation of relatively few logarithms by series. Start by using the following values for x in Exercise 57: 1 , 3
1 1 = 2 1 + t2 t
ass a
T 67. Estimating Pi The English mathematician Wallis discovered the formula
iaz
by integrating the series
nR
sinh1 x =
833
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
i f u s u o
11.11 Fourier Series
11.11
Fourier Series
HISTORICAL BIOGRAPHY JeanBaptiste Joseph Fourier
We have seen how Taylor series can be used to approximate a function ƒ by polynomials. The Taylor polynomials give a close fit to ƒ near a particular point x = a, but the error in the approximation can be large at points that are far away. There is another method that often gives good approximations on wide intervals, and often works with discontinuous functions for which Taylor polynomials fail. Introduced by Joseph Fourier, this method approximates functions with sums of sine and cosine functions. It is well suited for analyzing periodic functions, such as radio signals and alternating currents, for solving heat transfer problems, and for many other problems in science and engineering.
d a mm
(1766–1830)
a h Mu
n a s s a H
Y z a i R
833
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 834
Chapter 11: Infinite Sequences and Series
suf
i
Suppose we wish to approximate a function ƒ on the interval [0, 2p] by a sum of sine and cosine functions, ƒnsxd = a0 + sa1 cos x + b1 sin xd + sa2 cos 2x + b2 sin 2xd + Á + san cos nx + bn sin nxd
You
or, in sigma notation, n
ƒnsxd = a0 + a sak cos kx + bk sin kxd. k=1
(1)
2. 3.
ƒnsxd and ƒ(x) give the same value when integrated from 0 to 2p. ƒnsxd cos kx and ƒ(x) cos kx give the same value when integrated from 0 to 2p sk = 1, Á , nd. ƒnsxd sin kx and ƒ(x) sin kx give the same value when integrated from 0 to 2p sk = 1, Á , nd.
nR
1.
iaz
We would like to choose values for the constants a0 , a1, a2 , Á an and b1, b2 , Á , bn that make ƒnsxd a “best possible” approximation to ƒ(x). The notion of “best possible” is defined as follows:
Altogether we impose 2n + 1 conditions on ƒn : 2p
ƒsxd dx,
L0
ass a
L0
2p
ƒnsxd dx =
2p
L0
2p
ƒnsxd cos kx dx =
L0
2p
k = 1, Á , n,
ƒsxd sin kx dx,
k = 1, Á , n.
2p
ƒnsxd sin kx dx =
dH
L0
ƒsxd cos kx dx,
L0
It is possible to choose a0 , a1, a2 , Á an and b1, b2 , Á , bn so that all these conditions are satisfied, by proceeding as follows. Integrating both sides of Equation (1) from 0 to 2p gives 2p
ma ham Mu
ƒnsxd dx = 2pa0
L0
since the integral over [0, 2p] of cos kx equals zero when k Ú 1, as does the integral of sin kx. Only the constant term a 0 contributes to the integral of ƒn over [0, 2p]. A similar calculation applies with each of the other terms. If we multiply both sides of Equation (1) by cos x and integrate from 0 to 2p then we obtain 2p
ƒnsxd cos x dx = pa1 .
L0 This follows from the fact that
2p
L0
cos px cos px dx = p
and 2p
L0
2p
cos px cos qx dx =
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L0
2p
cos px sin mx dx =
L0
sin px sin qx dx = 0
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.11
Fourier Series
835
2p
ƒnsxd sin x dx = pb1 .
L0
You
Proceeding in a similar fashion with
suf
i
whenever p, q and m are integers and p is not equal to q (Exercises 9–13). If we multiply Equation (1) by sin x and integrate from 0 to 2p we obtain
cos 2x, sin 2x, Á , cos nx, sin nx
we obtain only one nonzero term each time, the term with a sinesquared or cosinesquared term. To summarize,
L0 2p
ƒnsxd dx = 2pa0
ƒnsxd cos kx dx = pak,
k = 1, Á , n
nR
L0
iaz
2p
2p
ƒnsxd sin kx dx = pbk,
ass a
L0
k = 1, Á , n
We chose ƒn so that the integrals on the left remain the same when ƒn is replaced by ƒ, so we can use these equations to find a0 , a1, a2 , Á an and b1, b2 , Á , bn from ƒ: 2p
Mu
ham
ma
dH
a0 =
1 ak = p
1 ƒsxd dx 2p L0
(2)
2p
L0
1 bk = p
ƒsxd cos kx dx,
k = 1, Á , n
(3)
k = 1, Á , n
(4)
2p
L0
ƒsxd sin kx dx,
The only condition needed to find these coefficients is that the integrals above must exist. If we let n : q and use these rules to get the coefficients of an infinite series, then the resulting sum is called the Fourier series for ƒ(x), q
a0 + a sak cos kx + bk sin kxd.
(5)
k=1
EXAMPLE 1
Finding a Fourier Series Expansion
Fourier series can be used to represent some functions that cannot be represented by Taylor series; for example, the step function ƒ shown in Figure 11.16a.
To Read it Online & Download:
ƒsxd = e
1, 2,
if 0 … x … p if p 6 x … 2p.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 836
Chapter 11: Infinite Sequences and Series
y 2
1
1
0
x
2
–2
–
(a)
You
2
suf
i
y
0
2
3
4
x
(b)
FIGURE 11.16 (a) The step function 1, 2,
0 … x … p p 6 x … 2p
iaz
ƒsxd = e
nR
(b) The graph of the Fourier series for ƒ is periodic and has the value 3>2 at each point of discontinuity (Example 1).
The coefficients of the Fourier series of ƒ are computed using Equations (2), (3), and (4). 2p
1 ƒsxd dx 2p L0
ass a
a0 =
p
=
dH
1 ak = p
2p
ma
2p
p
cos kx dx +
L0
sin kx 1 = p ac d k
1 bk = p
ham Mu
ƒsxd cos kx dx
L0
1 = p a
p 0
+ c
Lp
2 cos kx dxb 2p
2 sin kx d b = 0, k p
k Ú 1
2p
L0
1 = p a
ƒsxd sin kx dx 2p
p
L0
sin kx dx +
cos kx 1 = p a cd k =
2p
3 1 a 1 dx + 2 dxb = 2p L0 2 Lp
p 0
Lp
+ c
2 sin kx dxb 2p
2 cos kx d b k p
s 1dk  1 cos kp  1 = . kp kp
So
To Read it Online & Download:
a0 =
3 , 2
a1 = a2 = Á = 0,
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.11
Fourier Series
837
b2 = 0,
b3 = 
2 , 3p
b4 = 0,
b5 = 
2 , 5p
b6 = 0, Á
suf
2 b1 =  p ,
i
and
The Fourier series is
You
3 sin 3x sin 5x 2 + Áb .  p asin x + + 5 2 3
nR
iaz
Notice that at x = p, where the function ƒ(x) jumps from 1 to 2, all the sine terms vanish, leaving 3> 2 as the value of the series. This is not the value of ƒ at p, since ƒspd = 1. The Fourier series also sums to 3> 2 at x = 0 and x = 2p. In fact, all terms in the Fourier series are periodic, of period 2p, and the value of the series at x + 2p is the same as its value at x. The series we obtained represents the periodic function graphed in Figure 11.16b, with domain the entire real line and a pattern that repeats over every interval of width 2p. The function jumps discontinuously at x = np, n = 0, ;1, ;2, Á and at these points has value 3> 2, the average value of the onesided limits from each side. The convergence of the Fourier series of ƒ is indicated in Figure 11.17.
y
ass a
y
f
2 1.5
2
1.5
f1
1
dH
(a)
x
(b)
0
ma
2
1.5
f3
ham
2
1.5
FIGURE 11.17
x
(c)
0
2
x
y
f
f
2 1.5
f9
1
0
f5
1
2
y
f
2
1
0
Mu
f
y
f15
1
(d)
2
x
0
(e)
2
The Fourier approximation functions ƒ1, ƒ3, ƒ5, ƒ9 , and ƒ15 of the function ƒsxd = e
To Read it Online & Download:
x
1, 2,
0 … x … p in Example 1. p 6 x … 2p
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 838
Chapter 11: Infinite Sequences and Series
i
Convergence of Fourier Series
You
suf
Taylor series are computed from the value of a function and its derivatives at a single point x = a, and cannot reflect the behavior of a discontinuous function such as ƒ in Example 1 past a discontinuity. The reason that a Fourier series can be used to represent such functions is that the Fourier series of a function depends on the existence of certain integrals, whereas the Taylor series depends on derivatives of a function near a single point. A function can be fairly “rough,” even discontinuous, and still be integrable. The coefficients used to construct Fourier series are precisely those one should choose to minimize the integral of the square of the error in approximating ƒ by ƒn . That is, 2p
L0
[ƒsxd  ƒnsxd]2 dx
nR
iaz
is minimized by choosing a0 , a1, a2 , Á an and b1, b2 , Á , bn as we did. While Taylor series are useful to approximate a function and its derivatives near a point, Fourier series minimize an error which is distributed over an interval. We state without proof a result concerning the convergence of Fourier series. A function is piecewise continuous over an interval I if it has finitely many discontinuities on the interval, and at these discontinuities onesided limits exist from each side. (See Chapter 5, Additional Exercises 11–18.)
dH
ass a
THEOREM 24 Let ƒ(x) be a function such that ƒ and ƒ¿ are piecewise continuous on the interval [0, 2p]. Then ƒ is equal to its Fourier series at all points where ƒ is continuous. At a point c where ƒ has a discontinuity, the Fourier series converges to ƒsc + d + ƒsc  d 2
Mu
ham
ma
where ƒsc + d and ƒsc  d are the right and lefthand limits of ƒ at c.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 838
Chapter 11: Infinite Sequences and Series
u o zY
EXERCISES 11.11 Finding Fourier Series In Exercises 1–8, find the Fourier series associated with the given functions. Sketch each function. 1. ƒsxd = 1
0 … x … 2p .
1, 2. ƒsxd = e 1,
0 … x … p p 6 x … 2p
3. ƒsxd = e
x, x  2p,
4. ƒsxd = e
x 2, 0,
5. ƒsxd = e x
d a m
0 … x … p p 6 x … 2p
m a h
0 … x … p p 6 x … 2p
u M
0 … x … 2p .
a i R an
6. ƒsxd = e
s s Ha
To Read it Online & Download:
e x, 0,
7. ƒsxd = e
cos x, 0,
8. ƒsxd = e
2, x,
i f su
0 … x … p p 6 x … 2p
0 … x … p p 6 x … 2p
0 … x … p p 6 x … 2p
Theory and Examples Establish the results in Exercises 9–13, where p and q are positive integers. 2p
9.
L0
cos px dx = 0 for all p .
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 11.11 Fourier Series 2p
2p
11.
cos px cos qx dx = e
0, p,
if p Z q . if p = q
L0 sHint: cos A cos B = s1>2d[cossA + Bd + cossA  Bd].d 2p
12.
sin px sin qx dx = e
0, p,
if p Z q . if p = q
L0 sHint: sin A sin B = s1>2d[cos sA  Bd  cos sA + Bd].d 2p
sin px cos qx dx = 0 for all p and q . L0 sHint: sin A cos B = s1>2d[sin sA + Bd + sin sA  Bd].d
a. Use Theorem 24 to verify that the Fourier series for ƒsxd in Exercise 3 converges to ƒsxd for 0 6 x 6 2p .
b. Although ƒ¿sxd = 1 , show that the series obtained by termbyterm differentiation of the Fourier series in part (a) diverges. 16. Use Theorem 24 to find the Value of the Fourier series determined q p2 1 in Exercise 4 and show that = a 2. 6 n=1 n
Mu
ham
ma
dH
ass a
nR
iaz
13.
15. Termbyterm differentiation
suf
L0
14. Fourier series of sums of functions If ƒ and g both satisfy the conditions of Theorem 24, is the Fourier series of ƒ + g on [0, 2p] the sum of the Fourier series of ƒ and the Fourier series of g? Give reasons for your answer.
i
sin px dx = 0 for all p .
You
10.
839
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
Chapter 11
Chapter 11
i f u s u o
Questions to Guide Your Review
Questions to Guide Your Review
Y z
839
1. What is an infinite sequence? What does it mean for such a sequence to converge? To diverge? Give examples.
15. When do pseries converge? Diverge? How do you know? Give examples of convergent and divergent pseries.
2. What is a nondecreasing sequence? Under what circumstances does such a sequence have a limit? Give examples.
16. What are the Direct Comparison Test and the Limit Comparison Test? What is the reasoning behind these tests? Give examples of their use.
3. What theorems are available for calculating limits of sequences? Give examples.
iR a
4. What theorem sometimes enables us to use l’Hôpital’s Rule to calculate the limit of a sequence? Give an example.
17. What are the Ratio and Root Tests? Do they always give you the information you need to determine convergence or divergence? Give examples.
5. What six sequence limits are likely to arise when you work with sequences and series?
18. What is an alternating series? What theorem is available for determining the convergence of such a series?
6. What is an infinite series? What does it mean for such a series to converge? To diverge? Give examples.
19. How can you estimate the error involved in approximating the sum of an alternating series with one of the series’ partial sums? What is the reasoning behind the estimate?
n a sa s H d
7. What is a geometric series? When does such a series converge? Diverge? When it does converge, what is its sum? Give examples.
8. Besides geometric series, what other convergent and divergent series do you know? 9. What is the nthTerm Test for Divergence? What is the idea behind the test?
a m m
10. What can be said about termbyterm sums and differences of convergent series? About constant multiples of convergent and divergent series?
20. What is absolute convergence? Conditional convergence? How are the two related? 21. What do you know about rearranging the terms of an absolutely convergent series? Of a conditionally convergent series? Give examples. 22. What is a power series? How do you test a power series for convergence? What are the possible outcomes? 23. What are the basic facts about a. termbyterm differentiation of power series?
11. What happens if you add a finite number of terms to a convergent series? A divergent series? What happens if you delete a finite number of terms from a convergent series? A divergent series?
b. termbyterm integration of power series?
12. How do you reindex a series? Why might you want to do this?
Give examples.
a h u M
c. multiplication of power series?
13. Under what circumstances will an infinite series of nonnegative terms converge? Diverge? Why study series of nonnegative terms?
24. What is the Taylor series generated by a function ƒ(x) at a point x = a ? What information do you need about ƒ to construct the series? Give an example.
14. What is the Integral Test? What is the reasoning behind it? Give an example of its use.
25. What is a Maclaurin series?
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 840
Chapter 11: Infinite Sequences and Series 31. How can you sometimes use power series to estimate the values of nonelementary definite integrals?
27. What are Taylor polynomials? Of what use are they?
32. What are the Taylor series for 1>s1  xd, 1>s1 + xd, e x, sin x, cos x, ln s1 + xd, ln [s1 + xd>s1  xd] , and tan1 x ? How do you estimate the errors involved in replacing these series with their partial sums?
suf
28. What is Taylor’s formula? What does it say about the errors involved in using Taylor polynomials to approximate functions? In particular, what does Taylor’s formula say about the error in a linearization? A quadratic approximation?
i
26. Does a Taylor series always converge to its generating function? Explain.
30. How can you sometimes use power series to solve initial value problems?
34. State the theorem on convergence of the Fourier series for ƒ(x) when ƒ and ƒ¿ are piecewise continuous on [0, 2p] .
Mu
ham
ma
dH
ass a
nR
iaz
You
29. What is the binomial series? On what interval does it converge? How is it used?
33. What is a Fourier series? How do you calculate the Fourier coefficients a0 , a1, a2 , Á and b1, b2 , Á for a function ƒ(x) defined on the interval [0, 2p] ?
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 840
i f su
Chapter 11: Infinite Sequences and Series
Chapter 11
Practice Exercises
u o Y z
Convergent or Divergent Sequences
Convergent or Divergent Series
Which of the sequences whose nth terms appear in Exercises 1–18 converge, and which diverge? Find the limit of each convergent sequence.
Which of the series in Exercises 25–40 converge absolutely, which converge conditionally, and which diverge? Give reasons for your answers.
1. an = 1 +
s 1dn n
1  2n 3. an = 2n np 5. an = sin 2
2. an =
7. an = 9. an =
ln sn 2 d n
11. an = a 13. an =
n  5 n b
H d
1>n
a m
s 4dn n!
Find the sums of the series in Exercises 19–24. q
h u
1 19. a n = 3 s2n  3ds2n  1d q
9 21. a n = 1 s3n  1ds3n + 2d q
23. a e n=0
n
M
q
n2n 2 + 1
2 20. a n = 2 nsn + 1d
8 22. a n = 3 s4n  3ds4n + 1d 3 24. a s 1d n 4 n=1 n
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1 30. a 2 n = 2 n sln nd
q
ln n 32. a n = 3 ln sln nd q s 1dn 3n 2 34. a 3 n=1 n + 1
s 3dn n! n=1
2n 3n 38. a n n=1 n
q
37. a q
39. a
n=1
1 2nsn + 1dsn + 2d
q
q
40. a
n=2
1 n2n 2  1
Power Series In Exercises 41–50, (a) find the series’ radius and interval of convergence. Then identify the values of x for which the series converges (b) absolutely and (c) conditionally. sx + 4dn n3n n=1 q
41. a 43. a
s 1dn  1s3x  1dn
n=1 q
q
2n
q
q s 1dnsn 2 + 1d 36. a 2 n = 1 2n + n  1
q
q
n=1
n
s 1dn
n + 1 n! n=1
q
35. a
m a
Convergent Series
sa
27. a
s 1d 29. a n = 1 ln sn + 1d q
n sa
n=1
n
16. an = 22n + 1 18. an =
n=1
s 1dn
q
33. a
ln s2n 3 + 1d n
n
sn + 1d! n!
2n
q
ln s2n + 1d n
iR a
q
5 26. a n
q
ln n 31. a 3 n=1 n
3 14. an = a n b
n n 3 An
n=1
6. an = sin np
1 12. an = a1 + n b
15. an = ns21>n  1d 17. an =
2n
q
1
1 28. a 3 n = 1 2n
10. an = n
q
25. a
4. an = 1 + s0.9dn
8. an =
n + ln n n
1  s 1dn
xn 45. a n n n=1
n2
q sx  1d2n  2 42. a n = 1 s2n  1d!
sn + 1ds2x + 1dn s2n + 1d2n n=0 q
44. a q
46. a
n=1
xn 2n
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 11 s 1dnsx  1d2n + 1 2n + 1 n=0
Nonelementary Integrals
q
i
48. a
Use series to approximate the values of the integrals in Exercises 77–80 with an error of magnitude less than 10 8 . (The answer section gives the integrals’ values rounded to 10 decimal places.)
q
49. a scsch ndx
50. a scoth ndx
n
n=1
n
n=1
1>2
Maclaurin Series
77.
Each of the series in Exercises 51–56 is the value of the Taylor series at x = 0 of a function ƒ(x) at a particular point. What function and what point? What is the sum of the series?
79.
L0
tan1 x x dx
In Exercises 81–86:
8 4 2 2 +  Á + s 1dn  1 n + Á 3 18 81 n3
81. lim
54. 1 
p2n p2 p4 + Á +  Á + s 1dn 2n # # 9 2! 81 4! 3 s2nd!
83. lim a t: 0
n
sln 2d sln 2d + Á + + Á 2! n!
1

9 23
1
+
45 23
+ s 1dn  1
85. lim
z:0
1
+ Á
Find Taylor series at x = 0 for the functions in Exercises 57–64. 1 58. 1 + x3
59. sin px
60. sin
61. cos sx 5>2 d
62. cos 25x
63. e spx>2d
64. e x
2x 3
ma
at
x = 2
67. ƒsxd = 1>sx + 1d
at
x = 3
68. ƒsxd = 1>x
at
x = a 7 0
Mu
Use power series to solve the initial value problems in Exercises 69–76. 69. y¿ + y = 0,
ys0d = 1
70. y¿  y = 0,
ys0d = 3
71. y¿ + 2y = 0,
ys0d = 3
72. y¿ + y = 1,
ys0d = 0
ys0d = 1 74. y¿ + y = x,
ys0d = 0
75. y¿  y = x,
ys0d = 1
ssin hd>h  cos h h2
h:0
2
86. lim
y:0
y cos y  cosh y
76. y¿  y = x,
lim a
x:0
sin 3x r + 2 + sb = 0 . x3 x
89. a. Show that the series 1 1 a asin 2n  sin 2n + 1 b q
converges.
Initial Value Problems
73. y¿  y = 3x,
1  cos2 z ln s1  zd + sin z
84. lim
n=1
ham
66. ƒsxd = 1>s1  xd
1 1  2b 2  2 cos t t
Theory and Examples
In Exercises 65–68, find the first four nonzero terms of the Taylor series generated by ƒ at x = a . x = 1
e u  e u  2u u  sin u u:0
T b. Compare the accuracies of the approximations sin x L x and sin x L 6x>s6 + x 2 d by comparing the graphs of ƒsxd = sin x  x and gsxd = sin x  s6x>s6 + x 2 dd . Describe what you find.
2
at
dx
88. a. Show that the approximation csc x L 1>x + x>6 in Section 11.10, Example 9, leads to the approximation sin x L 6x>s6 + x 2 d .
dH
1 57. 1  2x
65. ƒsxd = 23 + x 2
2x
87. Use a series representation of sin 3x to find values of r and s for which
 Á
s2n  1d A 23 B 2n  1
Taylor Series
L0
tan1 x
82. lim
nR
1 23
7 sin x x:0 e 2x  1
ass a
56.
1>64
80.
a. Use power series to evaluate the limit.
p3 p2n + 1 p5 +  Á + s 1dn + Á 3! 5! s2n + 1d!
55. 1 + ln 2 +
x sin sx 3 d dx
T b. Then use a grapher to support your calculation.
53. p 
2
L0
Indeterminate Forms
1 1 1 +  Á + s 1dn n + Á 4 16 4 n
52.
1>2
78.
iaz
51. 1 
1
3
e x dx
L0
suf
q
841
You
sn + 1dx 2n  1 3n n=0 q
47. a
Practice Exercises
ys0d = 2
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T b. Estimate the magnitude of the error involved in using the sum of the sines through n = 20 to approximate the sum of the series. Is the approximation too large, or too small? Give reasons for your answer. q 1 1 90. a. Show that the series a atan  tan b converges. 2n 2n + 1 n=1 T b. Estimate the magnitude of the error in using the sum of the tangents through tan s1>41d to approximate the sum of the series. Is the approximation too large, or too small? Give reasons for your answer.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 11: Infinite Sequences and Series
91. Find the radius of convergence of the series
Show that the series
2 # 5 # 8 # Á # s3n  1d n x . 2 # 4 # 6 # Á # s2nd n=1 q
a3 a1 a2 + + + Á 1 2 3
92. Find the radius of convergence of the series
diverges. q
95. a. Find the interval of convergence of the series 1 3 1 6 Á x + x + 6 180 1 # 4 # 7 # Á # s3n  2d 3n + x + Á. s3nd!
y = 1 +
and find the values of the constants a and b.
96. a. Find the Maclaurin series for the function x 2>s1 + xd . b. Does the series converge at x = 1 ? Explain.
97. If g n = 1 an and g n = 1 bn are convergent series of nonnegative q numbers, can anything be said about g n = 1 an bn ? Give reasons for your answer.
dH
q
98. If g n = 1 an and g n = 1 bn are divergent series of nonnegative q numbers, can anything be said about g n = 1 an bn ? Give reasons for your answer.
ma
99. Prove that the sequence 5xn6 and the series g k= 1 sxk + 1  xk d both converge or both diverge. q
100. Prove that g n = 1 san>s1 + an dd converges if an 7 0 for all n and q g n = 1 an converges.
ham
q
1
2 x 10 x e dx. There are several ways to do this. 1 a. Use the Trapezoidal Rule with n = 2 to estimate 10 x 2e x dx. b. Write out the first three nonzero terms of the Taylor series at x = 0 for x 2e x to obtain the fourth Taylor polynomial P(x) 1 for x 2e x . Use 10 Psxd dx to obtain another estimate for
nR
2 x 10 x e dx. c. The second derivative of ƒsxd = x 2e x is positive for all x 7 0 . Explain why this enables you to conclude that the Trapezoidal Rule estimate obtained in part (a) is too large. (Hint: What does the second derivative tell you about the graph of a function? How does this relate to the trapezoidal approximation of the area under this graph?)
ass a
= x ay + b
q
diverges.
104. Suppose you wish to obtain a quick estimate for the value of
1
b. Show that the function defined by the series satisfies a differential equation of the form
q
1 1 + a n = 2 n ln n
iaz
93. Find a closedform formula for the nth partial sum of the series q g n = 2 ln s1  s1>n 2 dd and use it to determine the convergence or divergence of the series. q 94. Evaluate g k= 2 s1>sk 2  1dd by finding the limits as n : q of the series’ nth partial sum.
q
You
103. Use the result in Exercise 102 to show that
3 # 5 # 7 # Á # s2n + 1d n a 4 # 9 # 14 # Á # s5n  1d sx  1d . n=1 q
dx 2
suf
a
d 2y
i
842
1
e. Use integration by parts to evaluate 10 x 2e x dx.
Fourier Series Find the Fourier series for the functions in Exercises 105–108. Sketch each function. 105. ƒsxd = e
0, 1,
0 … x … p p 6 x … 2p
106. ƒsxd = e
x, 1,
0 … x … p p 6 x … 2p
107. ƒsxd = e
p  x, x  2p,
108. ƒsxd = ƒ sin x ƒ ,
0 … x … p p 6 x … 2p
0 … x … 2p
Mu
101. (Continuation of Section 4.7, Exercise 27.) If you did Exercise 27 in Section 4.7, you saw that in practice Newton’s method stopped too far from the root of ƒsxd = sx  1d40 to give a useful estimate of its value, x = 1 . Prove that nevertheless, for any starting value x0 Z 1 , the sequence x0 , x1, x2 , Á , xn , Á of approximations generated by Newton’s method really does converge to 1.
d. All the derivatives of ƒsxd = x 2e x are positive for x 7 0 . Explain why this enables you to conclude that all Maclaurin polynomial approximations to ƒ(x) for x in [0, 1] will be too small. (Hint: ƒsxd = Pnsxd + Rnsxd .)
102. Suppose that a1, a2 , a3 , Á , an are positive numbers satisfying the following conditions: i. a1 Ú a2 Ú a3 Ú Á ; ii. the series a2 + a4 + a8 + a16 + Á diverges.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 11
Chapter 11
q stan1 nd2 2. a 2 n=1 n + 1 q logn sn!d 4. a n3
1 n + s1>2d n = 1 s3n  2d q
3. a s 1dn tanh n
+ 17. Evaluate
6. a1 = a2 = 7,
an + 1 =
n an sn  1dsn + 1d
7. a1 = a2 = 1,
an + 1 =
1 1 + an
ass a
an = n>3n if n is even
a. Use the figure to show that g n = 1 An 6 s1>2dsƒs1d  ƒs2dd .
dH
ƒsn + 1dscd ƒsndsad sx  adn + sx  adn + 1 n! sn + 1d!
x = 0.4
x = 69
12. ln x
near near
14. tan1 x
n
lim c a ƒskd n: q k=1
n
L1
ƒsxd dx d .
If ƒsxd = 1>x , the limit in part (c) is Euler’s constant (Section 11.3, Exercise 41). (Source: “Convergence with Pictures” by P. J. Rippon, American Mathematical Monthly, Vol. 93, No. 6, 1986, pp. 476–478.)
x = 6.3 x = 1.3
near
x = 2
Mu
near
10. sin x
n
c. Then show the existence of
ma
ham near
x = 1
n
1 ƒsxd dx d . lim c a ƒskd  sƒs1d + ƒsndd 2 n: q k=1 L1
expresses the value of ƒ at x in terms of the values of ƒ and its derivatives at x = a . In numerical computations, we therefore need a to be a point where we know the values of ƒ and its derivatives. We also need a to be close enough to the values of ƒ we are interested in to make sx  adn + 1 so small we can neglect the remainder. In Exercises 9–14, what Taylor series would you choose to represent the function near the given value of x? (There may be more than one good answer.) Write out the first four nonzero terms of the series you choose.
13. cos x
q
b. Then show the existence of
ƒ–sad ƒsxd = ƒsad + ƒ¿sadsx  ad + sx  ad2 + Á 2!
11. e
nx n a sn + 1ds2x + 1dn n=1
19. Generalizing Euler’s constant The accompanying figure shows the graph of a positive twicedifferentiable decreasing function ƒ whose second derivative is positive on s0, q d . For each n, the number An is the area of the lunar region between the curve and the line segment joining the points (n, ƒ(n)) and sn + 1, ƒsn + 1dd .
if n Ú 2
Taylor’s formula
near
1 dx . 1 + x2
converges absolutely.
if n Ú 2
Choosing Centers for Taylor Series
x
Ln
q
(Hint: Write out several terms, see which factors cancel, and then generalize.)
9. cos x
n=0
18. Find all values of x for which
nsn + 1d an sn + 2dsn + 3d
if n is odd,
a
iaz
an + 1 =
n+1
q
Which of the series defined by the formulas in Exercises 5–8 converge, and which diverge? Give reasons for your answers.
+
7 3 + + Á. 10 8 10 9
n=2
q g n = 1 an
8. an = 1>3n
7 3 7 3 2 2 2 + + + + + + 10 10 5 10 7 10 2 10 3 10 4 10 6
nR
n=1
1 +
You
Which of the series defined by the formulas in Exercises 1–4 converge, and which diverge? Give reasons for your answers. q
suf
16. Find the sum of the infinite series
q g n = 1 an
1. a
843
i
Additional and Advanced Exercises
Convergence or Divergence
5. a1 = 1,
Additional and Advanced Exercises
Theory and Examples 15. Let a and b be constants with 0 6 a 6 b . Does the sequence 5sa n + b n d1>n6 converge? If it does converge, what is the limit?
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 844
Chapter 11: Infinite Sequences and Series
y
suf
i
23. Find a value for the constant b that will make the radius of convergence of the power series q
b nx n a ln n n=2 A1
equal to 5.
A2 A3
You
…
24. How do you know that the functions sin x, ln x, and e x are not polynomials? Give reasons for your answer. 25. Find the value of a for which the limit
f(1)
A2
lim
sin saxd  sin x  x x3
x:0
A3
2
1
3
26. Find values of a and b for which
… 4
iaz
f(4) 0
is finite and evaluate the limit.
y f(x)
f(3)
x
5
lim
2b
ass a
2b
2b
2b
2b
2b • • •
ƒsnd un C un + 1 = 1 + n + n 2 ,
a constant ,
ham
cos sa>nd n lim a1 b , n n: q
appear to depend on the value of a? If so, how? b. Does the value of lim a1 
n: q
cos sa>nd n b , bn
a and b constant, b Z 0 ,
(1)
where ƒ ƒsnd ƒ 6 K for n Ú N , then g n = 1 un converges if C 7 1 and diverges if C … 1 . Show that the results of Raabe’s test agree with what you q q know about the series g n = 1 s1>n 2 d and g n = 1 s1>nd . q
28. (Continuation of Exercise 27.) Suppose that the terms of g n = 1 un are defined recursively by the formulas q
u1 = 1,
2b
ma
2b
21. a. Does the value of
un + 1 =
s2n  1d2 un . s2nds2n + 1d
Apply Raabe’s test to determine whether the series converges. 29. If g n = 1 an converges, and if an Z 1 and an 7 0 for all n, q
a. Show that g n = 1 a n2 converges. q
b. Does
an >s1  an d converge? Explain.
q gn=1
30. (Continuation of Exercise 29.) If g n = 1 an converges, and if q 1 7 an 7 0 for all n, show that g n = 1 ln s1  an d converges. q
(Hint: First show that ƒ ln s1  an d ƒ … an>s1  an d .) 31. Nicole Oresme’s Theorem Prove Nicole Oresme’s Theorem that
appear to depend on the value of b? If so, how?
Mu
T
dH
c. Is every point on the original triangle removed? Explain why or why not.
= 1 .
27. Raabe’s (or Gauss’s) test The following test, which we state without proof, is an extension of the Ratio Test. q Raabe’s test: If g n = 1 un is a series of positive constants and there exist constants C, K, and N such that
b. Find the sum of this infinite series and hence find the total area removed from the original triangle.
2b
2x 2
x:0
20. This exercise refers to the “right side up” equilateral triangle with sides of length 2b in the accompanying figure. “Upside down” equilateral triangles are removed from the original triangle as the sequence of pictures suggests. The sum of the areas removed from the original triangle forms an infinite series. a. Find this infinite series.
cos saxd  b
nR
f(2)
c. Use calculus to confirm your findings in parts (a) and (b).
22. Show that if
q g n = 1 an
converges, then
a a q
n=1
1 + sin san d n b 2
1 +
1 2
#
2 +
1 4
#
3 + Á +
n + Á = 4. 2n  1
(Hint: Differentiate both sides of the equation 1>s1  xd = q 1 + g n = 1 x n .)
converges.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 11 32. a. Show that
i
q
nsn + 1d 2x 2 = n x sx  1d3 n=1
suf
for ƒ x ƒ 7 1 by differentiating the identity x2 1  x
c. As an engineer applying statistical control to an industrial operation, you inspect items taken at random from the assembly line. You classify each sampled item as either “good” or “bad.” If the probability of an item’s being good is p and of an item’s being bad is q = 1  p , the probability that the first bad item found is the nth one inspected is p n  1q . The average number inspected up to and including the first q bad item found is g n = 1 np n  1q . Evaluate this sum, assuming 0 6 p 6 1.
You
n=1
twice, multiplying the result by x, and then replacing x by 1> x.
b. Use part (a) to find the real solution greater than 1 of the equation q
nsn + 1d . xn n=1
x = a
33. A fast estimate of P /2 As you saw if you did Exercise 127 in Section 11.1, the sequence generated by starting with x0 = 1 and applying the recursion formula xn + 1 = xn + cos xn converges rapidly to p>2 . To explain the speed of the convergence, let Pn = sp>2d  xn . (See the accompanying figure.) Then p  xn  cos xn 2
= Pn  cos a
a. pk = 2k
p  Pn b 2
c. pk =
= Pn  sin Pn
Use this equality to show that
1 sP d3 . 6 n
0 6 Pn + 1 6
where C0 = the change in concentration achievable by a single dose (mg> mL), k = the elimination constant sh1 d , and t0 = time between doses (h). See the accompanying figure.
⑀n
C
xn
ham
1
cos xn
xn
0
1
x
q gn=1
Mu
34. If an is a convergent series of positive numbers, can anyq thing be said about the convergence of g n = 1 ln s1 + an d ? Give reasons for your answer. 35. Quality control
a. Differentiate the series 1 = 1 + x + x2 + Á + xn + Á 1  x to obtain a series for 1>s1  xd2 .
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1 1 1 = k k + 1 ksk + 1d
5k  1 6k
Rn = C0 e kt0 + C0 e 2k t0 + Á + C0 e nk t0 ,
ma
y
b. pk =
T 37. Safe and effective dosage The concentration in the blood resulting from a single dose of a drug normally decreases with time as the drug is eliminated from the body. Doses may therefore need to be repeated periodically to keep the concentration from dropping below some particular level. One model for the effect of repeated doses gives the residual concentration just before the sn + 1dst dose as
dH
1 1 A P B 3  5! A Pn B 5 + Á . 3! n
=
36. Expected value Suppose that a random variable X may assume the values 1, 2, 3, Á , with probabilities p1, p2, p3, Á , where pk is the probability that X equals k sk = 1, 2, 3, Á d . Suppose also q that pk Ú 0 and that g k = 1 pk = 1 . The expected value of X, deq noted by E(X), is the number g k = 1 k pk , provided the series conq verges. In each of the following cases, show that g k = 1 pk = 1 and find E(X) if it exists. (Hint: See Exercise 35.)
ass a
Pn + 1 =
iaz
=
Concentration (mg/mL)
n+1
nR
q
845
b. In one throw of two dice, the probability of getting a roll of 7 is p = 1>6 . If you throw the dice repeatedly, the probability that a 7 will appear for the first time at the nth throw is q n  1p , where q = 1  p = 5>6 . The expected number of q throws until a 7 first appears is g n = 1 nq n  1p . Find the sum of this series.
a
ax
Additional and Advanced Exercises
C1 C0 C0 e– k t 0
Cn1
C2 Rn C0
0
R2 R1 C0 e– k t 0
R3 t
t0 Time (h)
a. Write Rn in closed form as a single fraction, and find R = limn: q Rn .
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 11: Infinite Sequences and Series is said to converge if the series q
a ln s1 + an d ,
c. If k = 0.01 h1 and t0 = 10 h , find the smallest n such that Rn 7 s1>2dR .
38. Time between drug doses (Continuation of Exercise 37.) If a drug is known to be ineffective below a concentration CL and harmful above some higher concentration CH , one needs to find values of C0 and t0 that will produce a concentration that is safe (not above CH ) but effective (not below CL). See the accompanying figure. We therefore want to find values for C0 and t0 for which
n=1
obtained by taking the natural logarithm of the product, converges. Prove that the product converges if an 7 1 for every n q and if g n = 1 ƒ an ƒ converges. (Hint: Show that
You
(Source: Prescribing Safe and Effective Dosage, B. Horelick and S. Koont, COMAP, Inc., Lexington, MA.)
i
b. Calculate R1 and R10 for C0 = 1 mg>mL, k = 0.1 h1 , and t0 = 10 h . How good an estimate of R is R10 ?
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846
ƒ ln s1 + an d ƒ …
ƒ an ƒ … 2 ƒ an ƒ 1  ƒ an ƒ
when ƒ an ƒ 6 1>2 .) 40. If p is a constant, show that the series q
and
1 1 + a # p # n = 3 n ln n [ln sln nd]
iaz
R = CL
C0 + R = CH .
a. converges if p 7 1 , b. diverges if p … 1 . In general, if ƒ1sxd = x, ƒn + 1sxd = ln sƒnsxdd , and n takes on the values 1, 2, 3, Á , we find that ƒ2sxd = ln x, ƒ3sxd = ln sln xd , and so on. If ƒnsad 7 1 , then
Highest safe level
CH
nR
Concentration in blood
C
C0
q
CL
Lowest effective level
La
0
t
Time
ass a
t0
1 CH ln . k CL
dH
Thus C0 = CH  CL . When these values are substituted in the equation for R obtained in part (a) of Exercise 37, the resulting equation simplifies to t0 =
dx ƒ1sxdƒ2sxd Á ƒnsxdsƒn + 1sxddp
ma
To reach an effective level rapidly, one might administer a “loading” dose that would produce a concentration of CH mg>mL . This could be followed every t0 hours by a dose that raises the concentration by C0 = CH  CL mg>mL . a. Verify the preceding equation for t0 .
ham
b. If k = 0.05 h1 and the highest safe concentration is e times the lowest effective concentration, find the length of time between doses that will assure safe and effective concentrations. c. Given CH = 2 mg>mL, CL = 0.5 mg/mL , and k = 0.02 h1 , determine a scheme for administering the drug.
Mu
d. Suppose that k = 0.2 h1 and that the smallest effective concentration is 0.03 mg> mL. A single dose that produces a concentration of 0.1 mg> mL is administered. About how long will the drug remain effective?
39. An infinite product The infinite product q
Á q s1 + an d = s1 + a1 ds1 + a2 ds1 + a3 d
converges if p 7 1 and diverges if p … 1 .
41. a. Prove the following theorem: If 5cn6 is a sequence of numbers n such that every sum tn = g k = 1 ck is bounded, then the series q q g n = 1 cn >n converges and is equal to g n = 1 tn>(n(n + 1)) . Outline of proof: Replace c1 by t1 and cn by tn  tn  1 2n + 1 for n Ú 2. If s2n + 1 = g k = 1 ck > k, show that s2n + 1 = t1 a1 
1 1 1 b + t2 a  b 2 2 3
t2n + 1 1 1 b + + Á + t2n a 2n 2n + 1 2n + 1 2n t2n + 1 tk + . = a 2n + 1 k = 1 k sk + 1d
Because ƒ tk ƒ 6 M for some constant M, the series q
tk a k sk + 1d
k=1
converges absolutely and s2n + 1 has a limit as n : q . 2n Finally, if s2n = g k = 1 ck >k, then s2n + 1  s2n = c2n + 1>s2n + 1d approaches zero as n : q because ƒ c2n + 1 ƒ = ƒ t2n + 1  t2n ƒ 6 2M. Hence the sequence of partial sums of the series gck >k converges and the limit is q g k = 1 tk >sk sk + 1dd.
n=1
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 11 Additional and Advanced Exercises b. Show how the foregoing theorem applies to the alternating harmonic series
L0
xn+2 . n + 2
suf
x
ƒ Rn + 1 ƒ …
t n + 1 dt =
a Hint: As t varies from 0 to x,
1 1 1 1 1 1 1   + +   + Á. 5 7 2 3 4 6
1 + t Ú 1 and
converges. (After the first term, the signs are two negative, two positive, two negative, two positive, and so on in that pattern.)
`
to ln s1 + xd for
t n + 1>s1 + td … t n + 1 ,
You
c. Show that the series
and
x
ƒstd dt ` …
L0
x
L0
ƒ ƒstd ƒ dt. b
iaz
d. If 1 6 x 6 0 , show that
a. Show by long division or otherwise that
ƒ Rn + 1 ƒ
s 1dn + 1t n + 1 1 = 1  t + t 2  t 3 + Á + s 1dnt n + . 1 + t 1 + t b. By integrating the equation of part (a) with respect to t from 0 to x, show that
+ s 1dn
xn+1 + Rn + 1 n + 1
where x
`
n+1
ƒtƒ tn+1 .b ` … 1 + t 1  ƒxƒ
e. Use the foregoing results to prove that the series x 
s 1dnx n + 1 x2 x3 x4 + + Á + + Á 2 3 4 n + 1
converges to ln s1 + xd for 1 6 x … 1 .
Mu
ham
ma
dH
tn+1 dt . Rn + 1 = s 1dn + 1 L0 1 + t
n+2 ƒxƒ tn+1 dt ` = . sn + 2ds1  ƒ x ƒ d L0 1  ƒ x ƒ x
a Hint: If x 6 t … 0, then ƒ 1 + t ƒ Ú 1  ƒ x ƒ and
ass a
x2 x3 x4 ln s1 + xd = x + + Á 2 3 4
… `
nR
42. The convergence of 1 6 x … 1
i
c. If x Ú 0 , show that
1 1 1 1 1 1  +  +  + Á. 5 2 3 4 6
q g n = 1 [s 1dn  1x n]>n
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847
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 11
i
Technology Application Projects
Mathematica / Maple Module Bouncing Ball The model predicts the height of a bouncing ball, and the time until it stops bouncing.
You
Mathematica / Maple Module
Mu
ham
ma
dH
ass a
nR
iaz
Taylor Polynomial Approximations of a Function A graphical animation shows the convergence of the Taylor polynomials to functions having derivatives of all orders over an interval in their domains.
To Read it Online & Download:
847
suf
Chapter 11
Technology Application Projects
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suf
i
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
12
VECTORS AND THE GEOMETRY OF SPACE
You
Chapter
ass a
nR
iaz
OVERVIEW To apply calculus in many realworld situations and in higher mathematics, we need a mathematical description of threedimensional space. In this chapter we introduce threedimensional coordinate systems and vectors. Building on what we already know about coordinates in the xyplane, we establish coordinates in space by adding a third axis that measures distance above and below the xyplane. Vectors are used to study the analytic geometry of space, where they give simple ways to describe lines, planes, surfaces, and curves in space. We use these geometric ideas in the rest of the book to study motion in space and the calculus of functions of several variables, with their many important applications in science, engineering, economics, and higher mathematics.
12.1
ThreeDimensional Coordinate Systems
dH ma
z
z = constant
ham
(0, 0, z)
(0, y, z)
(x, 0, z)
0
P(x, y, z)
(0, y, 0)
y
Mu
(x, 0, 0) x
x = constant
To locate a point in space, we use three mutually perpendicular coordinate axes, arranged as in Figure 12.1. The axes shown there make a righthanded coordinate frame. When you hold your right hand so that the fingers curl from the positive xaxis toward the positive yaxis, your thumb points along the positive zaxis. So when you look down on the xyplane from the positive direction of the zaxis, positive angles in the plane are measured counterclockwise from the positive xaxis and around the positive zaxis. (In a lefthanded coordinate frame, the zaxis would point downward in Figure 12.1 and angles in the plane would be positive when measured clockwise from the positive xaxis. This is not the convention we have used for measuring angles in the xyplane. Righthanded and lefthanded coordinate frames are not equivalent.) The Cartesian coordinates (x, y, z) of a point P in space are the numbers at which the planes through P perpendicular to the axes cut the axes. Cartesian coordinates for space are also called rectangular coordinates because the axes that define them meet at right angles. Points on the xaxis have y and zcoordinates equal to zero. That is, they have coordinates of the form (x, 0, 0). Similarly, points on the yaxis have coordinates of the form (0, y, 0), and points on the zaxis have coordinates of the form (0, 0, z). The planes determined by the coordinates axes are the xyplane, whose standard equation is z = 0; the yzplane, whose standard equation is x = 0; and the xzplane, whose standard equation is y = 0. They meet at the origin (0, 0, 0) (Figure 12.2). The origin is also identified by simply 0 or sometimes the letter O. The three coordinate planes x = 0, y = 0, and z = 0 divide space into eight cells called octants. The octant in which the point coordinates are all positive is called the first octant; there is no conventional numbering for the other seven octants.
y = constant
(x, y, 0)
FIGURE 12.1 The Cartesian coordinate system is righthanded.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.1
849
ThreeDimensional Coordinate Systems
You
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i
The points in a plane perpendicular to the xaxis all have the same xcoordinate, this being the number at which that plane cuts the xaxis. The y and zcoordinates can be any numbers. Similarly, the points in a plane perpendicular to the yaxis have a common ycoordinate and the points in a plane perpendicular to the zaxis have a common zcoordinate. To write equations for these planes, we name the common coordinate’s value. The plane x = 2 is the plane perpendicular to the xaxis at x = 2. The plane y = 3 is the plane perpendicular to the yaxis at y = 3. The plane z = 5 is the plane perpendicular to the zaxis at z = 5. Figure 12.3 shows the planes x = 2, y = 3, and z = 5, together with their intersection point (2, 3, 5). z
z
(0, 0, 5)
yzplane: x 0 x yplane: z 0
Line y 3, z 5 Plane z 5 Line x 2, z 5
Plane x 2
nR
Origin
(2, 3, 5)
iaz
xzplane: y 0
Plane y 3 0
(0, 3, 0)
(2, 0, 0)
(0, 0, 0)
ass a
x
y
FIGURE 12.2 The planes x = 0, y = 0 , and z = 0 divide space into eight octants.
y x Line x 2, y 3
FIGURE 12.3 The planes x = 2, y = 3 , and z = 5 determine three lines through the point (2, 3, 5).
ma
dH
The planes x = 2 and y = 3 in Figure 12.3 intersect in a line parallel to the zaxis. This line is described by the pair of equations x = 2, y = 3. A point (x, y, z) lies on the line if and only if x = 2 and y = 3. Similarly, the line of intersection of the planes y = 3 and z = 5 is described by the equation pair y = 3, z = 5. This line runs parallel to the xaxis. The line of intersection of the planes x = 2 and z = 5, parallel to the yaxis, is described by the equation pair x = 2, z = 5. In the following examples, we match coordinate equations and inequalities with the sets of points they define in space.
Mu
ham
EXAMPLE 1
Interpreting Equations and Inequalities Geometrically
(a) z Ú 0 (b) x = 3 (c) z = 0, x … 0, y Ú 0 (d) x Ú 0, y Ú 0, z Ú 0 (e) 1 … y … 1 (f) y = 2, z = 2
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The halfspace consisting of the points on and above the xyplane. The plane perpendicular to the xaxis at x = 3. This plane lies parallel to the yzplane and 3 units behind it. The second quadrant of the xyplane. The first octant. The slab between the planes y = 1 and y = 1 (planes included). The line in which the planes y = 2 and z = 2 intersect. Alternatively, the line through the point s0, 2, 2d parallel to the xaxis.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 850
Chapter 12: Vectors and the Geometry of Space
EXAMPLE 2
The circle x 2 y 2 4, z 3
Graphing Equations
i
z
(0, 2, 3)
x2 + y2 = 4
and
suf
What points P(x, y, z) satisfy the equations
(2, 0, 3)
z = 3?
The points lie in the horizontal plane z = 3 and, in this plane, make up the circle x 2 + y 2 = 4. We call this set of points “the circle x 2 + y 2 = 4 in the plane z = 3” or, more simply, “the circle x 2 + y 2 = 4, z = 3 ” (Figure 12.4).
The plane z3
Distance and Spheres in Space
(0, 2, 0) (2, 0, 0)
You
Solution
y
The formula for the distance between two points in the xyplane extends to points in space.
x
FIGURE 12.4 The circle x 2 + y 2 = 4 in the plane z = 3 (Example 2).
iaz
x 2 y 2 4, z 0
The Distance Between P1sx1 , y1 , z1 d and P2sx2 , y2 , z2 d is
nR
ƒ P1 P2 ƒ = 2sx2  x1 d2 + s y2  y1 d2 + sz2  z1 d2
ass a
Proof We construct a rectangular box with faces parallel to the coordinate planes and the points P1 and P2 at opposite corners of the box (Figure 12.5). If Asx2 , y1 , z1 d and Bsx2 , y2 , z1 d are the vertices of the box indicated in the figure, then the three box edges P1 A, AB, and BP2 have lengths z P1(x1, y1, z 1)
P2(x 2, y 2, z 2 )
ƒ P1 A ƒ = ƒ x2  x1 ƒ ,
ƒ AB ƒ = ƒ y2  y1 ƒ ,
ƒ BP2 ƒ = ƒ z2  z1 ƒ .
Because triangles P1 BP2 and P1 AB are both rightangled, two applications of the Pythagorean theorem give
0 A(x2, y1, z 1)
B(x 2, y 2, z 1) y
Mu
ham
FIGURE 12.5 We find the distance between P1 and P2 by applying the Pythagorean theorem to the right triangles P1 AB and P1 BP2 .
and
ƒ P1 B ƒ 2 = ƒ P1 A ƒ 2 + ƒ AB ƒ 2
(see Figure 12.5). So
ƒ P1 P2 ƒ 2 = ƒ P1 B ƒ 2 + ƒ BP2 ƒ 2
ma
x
dH
ƒ P1 P2 ƒ 2 = ƒ P1 B ƒ 2 + ƒ BP2 ƒ 2
= ƒ P1 A ƒ 2 + ƒ AB ƒ 2 + ƒ BP2 ƒ 2
Substitute 2 2 2 ƒ P1 B ƒ = ƒ P1 A ƒ + ƒ AB ƒ .
= ƒ x2  x1 ƒ 2 + ƒ y2  y1 ƒ 2 + ƒ z2  z1 ƒ 2 = sx2  x1 d2 + sy2  y1 d2 + sz2  z1 d2
Therefore ƒ P1 P2 ƒ = 2sx2  x1 d2 + sy2  y1 d2 + sz2  z1 d2
EXAMPLE 3
Finding the Distance Between Two Points
The distance between P1s2, 1, 5d and P2s 2, 3, 0d is
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ƒ P1 P2 ƒ = 2s 2  2d2 + s3  1d2 + s0  5d2 = 216 + 4 + 25 = 245 L 6.708.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.1 z
851
i
We can use the distance formula to write equations for spheres in space (Figure 12.6). A point P(x, y, z) lies on the sphere of radius a centered at P0sx0 , y0 , z0 d precisely when ƒ P0 P ƒ = a or
P(x, y, z)
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P0(x 0 , y0 , z 0 )
ThreeDimensional Coordinate Systems
sx  x0 d2 + sy  y0 d2 + sz  z0 d2 = a 2 .
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a
The Standard Equation for the Sphere of Radius a and Center sx0 , y0 , z0 d
0
sx  x0 d2 + sy  y0 d2 + sz  z0 d2 = a 2
y x
EXAMPLE 4
Finding the Center and Radius of a Sphere
iaz
FIGURE 12.6 The standard equation of the sphere of radius a centered at the point sx0 , y0 , z0 d is
Find the center and radius of the sphere
x 2 + y 2 + z 2 + 3x  4z + 1 = 0.
nR
sx  x0 d2 + s y  y0 d2 + sz  z0 d2 = a 2 .
ass a
Solution We find the center and radius of a sphere the way we find the center and radius of a circle: Complete the squares on the x, y, and zterms as necessary and write each quadratic as a squared linear expression. Then, from the equation in standard form, read off the center and radius. For the sphere here, we have
x 2 + y 2 + z 2 + 3x  4z + 1 = 0 sx 2 + 3xd + y 2 + sz 2  4zd = 1
2
2
2
dH
3 3 4 4 ax 2 + 3x + a b b + y 2 + az 2  4z + a b b = 1 + a b + a b 2 2 2 2 ax +
2
2
3 9 21 b + y 2 + sz  2d2 = 1 + + 4 = . 2 4 4
ma
From this standard form, we read that x0 = 3>2, y0 = 0, z0 = 2, and a = 221>2. The center is s 3>2, 0, 2d. The radius is 221>2.
Mu
ham
EXAMPLE 5
Interpreting Equations and Inequalities
(a) x 2 + y 2 + z 2 6 4 (b) x 2 + y 2 + z 2 … 4
(c) x 2 + y 2 + z 2 7 4 (d) x 2 + y 2 + z 2 = 4, z … 0
The interior of the sphere x 2 + y 2 + z 2 = 4. The solid ball bounded by the sphere x 2 + y 2 + z 2 = 4. Alternatively, the sphere x 2 + y 2 + z 2 = 4 together with its interior. The exterior of the sphere x 2 + y 2 + z 2 = 4. The lower hemisphere cut from the sphere x 2 + y 2 + z 2 = 4 by the xyplane (the plane z = 0).
Just as polar coordinates give another way to locate points in the xyplane (Section 10.5), alternative coordinate systems, different from the Cartesian coordinate system developed here, exist for threedimensional space. We examine two of these coordinate systems in Section 15.6.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 852
Chapter 12: Vectors and the Geometry of Space
Sets, Equations, and Inequalities
25. The line through the point s1, 3, 1d parallel to the
2
2
2
z = 0
4. x = 1,
5. x + y = 4, 7. x + z = 4,
y = 0
2
2
z = 2
2
2
x = 0
6. x + y = 4,
z = 0
8. y + z = 1,
y = 0
9. x 2 + y 2 + z 2 = 1,
z = 0
x = 0
10. x 2 + y 2 + z 2 = 25,
12. x 2 + s y  1d2 + z 2 = 4,
29. The slab bounded by the planes z = 0 and z = 1 (planes included)
y = 0
z = 0
b. x Ú 0,
b. 0 … x … 1,
14. a. 0 … x … 1 c. 0 … x … 1,
0 … y … 1,
15. a. x 2 + y 2 + z 2 … 1 z = 0
2
2
no restriction on z
2
2
b. x + y + z … 1, 18. a. x = y,
2
2
b. x + y … 1,
z = 3
z Ú 0
dH
c. x + y … 1,
17. a. x 2 + y 2 + z 2 = 1,
0 … y … 1
b. x 2 + y 2 + z 2 7 1
2
16. a. x + y … 1,
z = 0
0 … z … 1
2
2
y … 0,
z Ú 0
z = 0
b. x = y,
30. The solid cube in the first octant bounded by the coordinate planes and the planes x = 2, y = 2 , and z = 2 31. The halfspace consisting of the points on and below the xyplane 32. The upper hemisphere of the sphere of radius 1 centered at the origin 33. The (a) interior and (b) exterior of the sphere of radius 1 centered at the point (1, 1, 1)
ass a
y Ú 0,
28. The set of points in space that lie 2 units from the point (0, 0, 1) and, at the same time, 2 units from the point s0, 0, 1d Write inequalities to describe the sets in Exercises 29–34.
z = 0
In Exercises 13–18, describe the sets of points in space whose coordinates satisfy the given inequalities or combinations of equations and inequalities. 13. a. x Ú 0,
27. The circle in which the plane through the point (1, 1, 3) perpendicular to the zaxis meets the sphere of radius 5 centered at the origin
y = 4
11. x 2 + y 2 + sz + 3d2 = 25,
c. zaxis
iaz
2
2. x = 1,
b. yaxis
nR
3. y = 0,
y = 3
a. xaxis
26. The set of points in space equidistant from the origin and the point (0, 2, 0)
You
In Exercises 1–12, give a geometric description of the set of points in space whose coordinates satisfy the given pairs of equations. 1. x = 2,
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EXERCISES 12.1
no restriction on z
34. The closed region bounded by the spheres of radius 1 and radius 2 centered at the origin. (Closed means the spheres are to be included. Had we wanted the spheres left out, we would have asked for the open region bounded by the spheres. This is analogous to the way we use closed and open to describe intervals: closed means endpoints included, open means endpoints left out. Closed sets include boundaries; open sets leave them out.)
Distance
19. The plane perpendicular to the
35. P1s1, 1, 1d,
P2s3, 3, 0d
36. P1s 1, 1, 5d,
P2s2, 5, 0d
37. P1s1, 4, 5d,
P2s4, 2, 7d
38. P1s3, 4, 5d,
P2s2, 3, 4d
39. P1s0, 0, 0d,
P2s2, 2, 2d
40. P1s5, 3, 2d,
P2s0, 0, 0d
a. xaxis at (3, 0, 0) c. zaxis at s0, 0, 2d
ma
In Exercises 19–28, describe the given set with a single equation or with a pair of equations. b. yaxis at s0, 1, 0d
20. The plane through the point s3, 1, 2d perpendicular to the b. yaxis
c. zaxis
ham
a. xaxis
21. The plane through the point s3, 1, 1d parallel to the a. xyplane
b. yzplane
c. xzplane
22. The circle of radius 2 centered at (0, 0, 0) and lying in the a. xyplane
b. yzplane
c. xzplane
Mu
23. The circle of radius 2 centered at (0, 2, 0) and lying in the a. xyplane
b. yzplane
c. plane y = 2
24. The circle of radius 1 centered at s 3, 4, 1d and lying in a plane parallel to the a. xyplane
b. yzplane
In Exercises 35–40, find the distance between points P1 and P2.
c. xzplane
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Spheres Find the centers and radii of the spheres in Exercises 41–44. 41. sx + 2d2 + y 2 + sz  2d2 = 8 42. ax +
2
2
2
1 1 1 21 b + ay + b + az + b = 2 2 2 4
43. A x  22 B 2 + A y  22 B 2 + A z + 22 B 2 = 2 44. x 2 + ay +
2
2
29 1 1 b + az  b = 3 3 9
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.1 ThreeDimensional Coordinate Systems
Radius
45. (1, 2, 3)
214
46. s0, 1, 5d
2
47. s 2, 0, 0d
23
48. s0, 7, 0d
7
Theory and Examples
2
i
53. Find a formula for the distance from the point P(x, y, z) to the a. xaxis
b. yaxis
c. zaxis
54. Find a formula for the distance from the point P(x, y, z) to the a. xyplane
Find the centers and radii of the spheres in Exercises 49â€“52. 2
52. 3x 2 + 3y 2 + 3z 2 + 2y  2z = 9
You
Center
51. 2x 2 + 2y 2 + 2z 2 + x + y + z = 9
suf
Find equations for the spheres whose centers and radii are given in Exercises 45â€“48.
853
2
49. x + y + z + 4x  4z = 0
b. yzplane
c. xzplane
55. Find the perimeter of the triangle with vertices As 1, 2, 1d, Bs1, 1, 3d , and C(3, 4, 5). 56. Show that the point P(3, 1, 2) is equidistant from the points As2, 1, 3d and B(4, 3, 1).
Mu
ham
ma
dH
ass a
nR
iaz
50. x 2 + y 2 + z 2  6y + 8z = 0
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
i f u s u o 12.2 Vectors
12.2
Vectors
853
Some of the things we measure are determined simply by their magnitudes. To record mass, length, or time, for example, we need only write down a number and name an appropriate unit of measure. We need more information to describe a force, displacement, or velocity. To describe a force, we need to record the direction in which it acts as well as how large it is. To describe a body’s displacement, we have to say in what direction it moved as well as how far. To describe a body’s velocity, we have to know where the body is headed as well as how fast it is going.
Y z
Component Form Terminal point B
AB
Initial point A
FIGURE 12.7 1 AB .
The directed line segment
A quantity such as force, displacement, or velocity is called a vector and is represented by a directed line segment (Figure 12.7). The arrow points in the direction of the action and its length gives the magnitude of the action in terms of a suitably chosen unit. For example, a force vector points in the direction in which the force acts; its length is a measure of the force’s strength; a velocity vector points in the direction of motion and its length is the speed of the moving object. Figure 12.8 displays the velocity vector v at a specific location for a particle moving along a path in the plane or in space. (This application of vectors is studied in Chapter 13.)
a m m
a h u M
n a sa s H d
iR a
y
z v
v
0
x
0
(a) two dimensions
x
y
(b) three dimensions
FIGURE 12.8 The velocity vector of a particle moving along a path (a) in the plane (b) in space. The arrowhead on the path indicates the direction of motion of the particle.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 854
Chapter 12: Vectors and the Geometry of Space y B
i
DEFINITIONS Vector, Initial and Terminal Point, Length 1 A vector in the plane is a directed line segment. The directed line segment AB 1 has initial point A and terminal point B; its length is denoted by ƒ AB ƒ . Two vectors are equal if they have the same length and direction.
D C P
0
Q(x 2 , y 2 , z 2 )
Position vector of PQ v 具v1, v 2, v3 典 v2
x
(v1, v 2 , v3)
v3 v1
y
ass a
z
nR
FIGURE 12.9 The four arrows in the plane (directed line segments) shown here have the same length and direction. They therefore represent the same vector, and we 1 1 1 1 write AB = CD = OP = EF .
The arrows we use when we draw vectors are understood to represent the same vector if they have the same length, are parallel, and point in the same direction (Figure 12.9) regardless of the initial point. In textbooks, vectors are usually written in lowercase, boldface letters, for example u, v, and w. Sometimes we use uppercase boldface letters, such as F, to denote a force vector. u, In handwritten form, it is customary to draw small arrows above the letters, for example s s. y, w s s , and F We need a way to represent vectors algebraically so that we can be more precise about the direction of a vector. 1 1 Let v = PQ . There is one directed line segment equal to PQ whose initial point is the origin (Figure 12.10). It is the representative of v in standard position and is the vector we normally use to represent v. We can specify v by writing the coordinates of its terminal point sv1, v2 , v3 d when v is in standard position. If v is a vector in the plane its terminal point sv1, v2 d has two coordinates.
iaz
E
P(x 1, y1, z 1)
You
x F
DEFINITION Component Form If v is a twodimensional vector in the plane equal to the vector with initial point at the origin and terminal point sv1, v2 d, then the component form of v is
dH
O
suf
A
v = 8v1, v29.
If v is a threedimensional vector equal to the vector with initial point at the origin and terminal point sv1, v2, v3 d, then the component form of v is v = 8v1, v2 , v39.
Mu
ham
ma
1 FIGURE 12.10 A vector PQ in standard position has its initial point at the origin. 1 The directed line segments PQ and v are parallel and have the same length.
So a twodimensional vector is an ordered pair v = 8v1, v29 of real numbers, and a threedimensional vector is an ordered triple v = 8v1, v2 , v39 of real numbers. The numbers v1, v2 , and v3 are called the components of v. 1 Observe that if v = 8v1, v2 , v39 is represented by the directed line segment PQ , where the initial point is Psx1, y1, z1 d and the terminal point is Qsx2 , y2 , z2 d, then x1 + v1 = x2 , y1 + v2 = y2 , and z1 + v3 = z2 (see Figure 12.10). Thus, v1 = x2  x1, v2 = y2  y1 , and 1 v3 = z2  z1 are the components of PQ . In summary, given the points Psx1, y1, z1 d and Qsx2 , y2 , z2 d, the standard position 1 vector v = 8v1, v2 , v39 equal to PQ is v = 8x2  x1, y2  y1, z2 , z19.
If v is twodimensional with Psx1, y1 d and Qsx2 , y2 d as points in the plane, then v = 8x2  x1, y2  y19. There is no third component for planar vectors. With this understanding, we will develop the algebra of threedimensional vectors and simply drop the third component when the vector is twodimensional (a planar vector).
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.2
Vectors
855
You
suf
i
Two vectors are equal if and only if their standard position vectors are identical. Thus 8u1, u2 , u39 and 8v1, v2 , v39 are equal if and only if u1 = v1, u2 = v2 , and u3 = v3 . 1 The magnitude or length of the vector PQ is the length of any of its equivalent directed line segment representations. In particular, if v = 8x2  x1, y2  y1, z2  z19 is the 1 standard position vector for PQ , then the distance formula gives the magnitude or length of v, denoted by the symbol ƒ v ƒ or ƒ ƒ v ƒ ƒ .
1 The magnitude or length of the vector v = PQ is the nonnegative number
(See Figure 12.10.)
iaz
ƒ v ƒ = 2v12 + v22 + v32 = 2sx2  x1 d2 + s y2  y1 d2 + sz2  z1 d2
EXAMPLE 1
nR
The only vector with length 0 is the zero vector 0 = 80, 09 or 0 = 80, 0, 09. This vector is also the only vector with no specific direction.
Component Form and Length of a Vector
ass a
Find the (a) component form and (b) length of the vector with initial point Ps 3, 4, 1d and terminal point Qs 5, 2, 2d. Solution
1 (a) The standard position vector v representing PQ has components
dH
v1 = x2  x1 = 5  s 3d = 2,
v2 = y2  y1 = 2  4 = 2,
and
v3 = z2  z1 = 2  1 = 1.
ma
1 The component form of PQ is v = 82, 2, 19.
1 (b) The length or magnitude of v = PQ is
ham
y
F = 具a, b典
Mu
45°
x
FIGURE 12.11 The force pulling the cart forward is represented by the vector F of magnitude 20 (pounds) making an angle of 45° with the horizontal ground (positive xaxis) (Example 2).
ƒ v ƒ = 2s 2d2 + s 2d2 + s1d2 = 29 = 3.
EXAMPLE 2
Force Moving a Cart
A small cart is being pulled along a smooth horizontal floor with a 20lb force F making a 45° angle to the floor (Figure 12.11). What is the effective force moving the cart forward? Solution
The effective force is the horizontal component of F = 8a, b9, given by 22 a = ƒ F ƒ cos 45° = s20d a b L 14.14 lb. 2
Notice that F is a twodimensional vector.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 856
Chapter 12: Vectors and the Geometry of Space
i
Vector Algebra Operations
You
suf
Two principal operations involving vectors are vector addition and scalar multiplication. A scalar is simply a real number, and is called such when we want to draw attention to its differences from vectors. Scalars can be positive, negative, or zero.
DEFINITIONS Vector Addition and Multiplication of a Vector by a Scalar Let u = 8u1, u2 , u39 and v = 8v1, v2 , v39 be vectors with k a scalar.
iaz
Addition: u + v = 8u1 + v1, u2 + v2 , u3 + v39 Scalar multiplication: ku = 8ku1, ku2 , ku39
y
ass a
nR
We add vectors by adding the corresponding components of the vectors. We multiply a vector by a scalar by multiplying each component by the scalar. The definitions apply to planar vectors except there are only two components, 8u1, u29 and 8v1, v29. The definition of vector addition is illustrated geometrically for planar vectors in Figure 12.12a, where the initial point of one vector is placed at the terminal point of the other. Another interpretation is shown in Figure 12.12b (called the parallelogram law of addition), where the sum, called the resultant vector, is the diagonal of the parallelogram. In physics, forces add vectorially as do velocities, accelerations, and so on. So the force acting on a particle subject to electric and gravitational forces is obtained by adding the two force vectors. y
dH
具u1 v1, u 2 v 2 典
u+v
v2
v
u
u+v v
v1
u
ma
u2
Mu
1.5u
ham
0
u
2u
(a)
x
0 (b)
FIGURE 12.12 (a) Geometric interpretation of the vector sum. (b) The parallelogram law of vector addition.
Figure 12.13 displays a geometric interpretation of the product ku of the scalar k and vector u. If k 7 0, then ku has the same direction as u; if k 6 0, then the direction of ku is opposite to that of u. Comparing the lengths of u and ku, we see that 2 2 2 ƒ ku ƒ = 2sku1 d2 + sku2 d2 + sku3 d2 = 2k 2su 1 + u 2 + u 3 d
= 2k 2 2u 12 + u 22 + u 32 = ƒ k ƒ ƒ u ƒ .
–2u
FIGURE 12.13 Scalar multiples of u.
x
u1
The length of ku is the absolute value of the scalar k times the length of u. The vector s 1du = u has the same length as u but points in the opposite direction.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.2
Vectors
857
If u = 8u1, u2 , u39 and v = 8v1, v2 , v39, then
suf
u  v = u + s vd.
v
i
By the difference u  v of two vectors, we mean
uv
(a)
You
u  v = 8u1  v1, u2  v2, u3  v39.
u
Note that su  vd + v = u, so adding the vector su  vd to v gives u (Figure 12.14a). Figure 12.14b shows the difference u  v as the sum u + s vd.
EXAMPLE 3
Performing Operations on Vectors
Let u = 81, 3, 19 and v = 84, 7, 09. Find u
(a) 2u + 3v
–v u (–v)
(b) u  v
Solution
(c) `
1 u` . 2
iaz
v
(b) u  v = 81, 3, 19  84, 7, 09 = 81  4, 3  7, 1  09 = 85, 4, 19 (c)
`
2
2
2
3 1 1 3 1 1 1 1 u ` = ` h , , i ` = a b + a b + a b = 211 . 2 2 2 2 2 2 2 2 C
ass a
FIGURE 12.14 (a) The vector u  v, when added to v, gives u. (b) u  v = u + s vd .
nR
(a) 2u + 3v = 281, 3, 19 + 384, 7, 09 = 82, 6, 29 + 812, 21, 09 = 810, 27, 29
(b)
dH
Vector operations have many of the properties of ordinary arithmetic. These properties are readily verified using the definitions of vector addition and multiplication by a scalar.
Properties of Vector Operations Let u, v, w be vectors and a, b be scalars.
ma
1.
Mu
ham
3. 5. 7. 9.
u + v = v + u u + 0 = u 0u = 0 asbud = sabdu sa + bdu = au + bu
2. su + vd + w = u + sv + wd 4. u + s ud = 0 6. 1u = u 8. asu + vd = au + av
An important application of vectors occurs in navigation.
EXAMPLE 4
Finding Ground Speed and Direction
A Boeing® 767® airplane, flying due east at 500 mph in still air, encounters a 70mph tailwind blowing in the direction 60° north of east. The airplane holds its compass heading due east but, because of the wind, acquires a new ground speed and direction. What are they?
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 858
Chapter 12: Vectors and the Geometry of Space
If u = the velocity of the airplane alone and v = the velocity of the tailwind, then ƒ u ƒ = 500 and ƒ v ƒ = 70 (Figure 12.15). The velocity of the airplane with respect to the ground is given by the magnitude and direction of the resultant vector u + v. If we let the positive xaxis represent east and the positive yaxis represent north, then the component forms of u and v are
Solution
500
u = 8500, 09
E
u
v = 870 cos 60°, 70 sin 60°9 = 835, 35239.
and
Therefore,
NOT TO SCALE
u + v = 8535, 35239
FIGURE 12.15 Vectors representing the velocities of the airplane u and tailwind v in Example 4.
You
uv
ƒ u + v ƒ = 25352 + s3513d2 L 538.4 and
3523 L 6.5°. 535
Figure 12.15
nR
u = tan1
iaz
v 30˚ 70
suf
i
N
The new ground speed of the airplane is about 538.4 mph, and its new direction is about 6.5° north of east.
Unit Vectors
ass a
A vector v of length 1 is called a unit vector. The standard unit vectors are i = 81, 0, 09,
j = 80, 1, 09,
and
k = 80, 0, 19.
Any vector v = 8v1, v2 , v39 can be written as a linear combination of the standard unit vectors as follows:
z
= v181, 0, 09 + v280, 1, 09 + v380, 0, 19
OP2 x 2 i y2 j z 2 k
k
We call the scalar (or number) v1 the icomponent of the vector v, v2 the jcomponent, and v3 the kcomponent. In component form, the vector from P1sx1, y1, z1 d to P2sx2 , y2 , z2 d is 1 P1 P2 = sx2  x1 di + s y2  y1 dj + sz2  z1 dk
P1P2
j
ham
i
y
x
= v1 i + v2 j + v3 k.
ma
P2(x2, y 2 , z 2 )
O
dH
v = 8v1, v2 , v39 = 8v1, 0, 09 + 80, v2 , 09 + 80, 0, v39
P1(x 1, y 1 , z 1 )
(Figure 12.16). Whenever v Z 0, its length ƒ v ƒ is not zero and
`
OP1 x 1i y 1 j z 1k
Mu
FIGURE 12.16 The vector from P1 to P2 1 is P1 P2 = sx2  x1 di + s y2  y1 dj + sz2  z1 dk.
1 1 v` = ƒ v ƒ = 1. v v ƒ ƒ ƒ ƒ
That is, v> ƒ v ƒ is a unit vector in the direction of v, called the direction of the nonzero vector v.
EXAMPLE 5
Finding a Vector’s Direction
Find a unit vector u in the direction of the vector from P1s1, 0, 1d to P2s3, 2, 0d.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.2
859
1 We divide P1 P2 by its length: 1 P1 P2 = s3  1di + s2  0dj + s0  1dk = 2i + 2j  k
suf
i
Solution
Vectors
EXAMPLE 6
You
1 ƒ P1 P2 ƒ = 2s2d2 + s2d2 + s 1d2 = 24 + 4 + 1 = 29 = 3 1 2i + 2j  k P1 P2 2 2 1 = i + j  k. u = 1 = 3 3 3 3 ƒ P1 P2 ƒ 1 The unit vector u is the direction of P1 P2 .
Expressing Velocity as Speed Times Direction
Solution
Speed is the magnitude (length) of v:
ƒ v ƒ = 2s3d2 + s 4d2 = 29 + 16 = 5.
nR
HISTORICAL BIOGRAPHY
iaz
If v = 3i  4j is a velocity vector, express v as a product of its speed times a unit vector in the direction of motion.
The unit vector v> ƒ v ƒ has the same direction as v:
Hermann Grassmann (1809–1877)
3i  4j v 3 4 = = i  j. 5 5 5 ƒvƒ
ass a
So
3 4 v = 3i  4j = 5 a i  jb . 5 5 (')'* Length (speed)
Direction of motion
dH
In summary, we can express any nonzero vector v in terms of its two important features, v . length and direction, by writing v = ƒ v ƒ ƒvƒ
Mu
ham
ma
If v Z 0, then v 1. is a unit vector in the direction of v; ƒvƒ v 2. the equation v = ƒ v ƒ expresses v in terms of its length and direction. ƒvƒ
EXAMPLE 7
A Force Vector
A force of 6 newtons is applied in the direction of the vector v = 2i + 2j  k. Express the force F as a product of its magnitude and direction. v Solution The force vector has magnitude 6 and direction , so ƒvƒ 2i + 2j  k 2i + 2j  k v F = 6 = 6 = 6 2 2 2 3 ƒvƒ 22 + 2 + s 1d
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2 2 1 = 6 a i + j  kb . 3 3 3
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 860
Chapter 12: Vectors and the Geometry of Space
i
Midpoint of a Line Segment
suf
Vectors are often useful in geometry. For example, the coordinates of the midpoint of a line segment are found by averaging.
P1(x 1, y1, z 1)
x1 + x2 y1 + y2 z1 + z2 , , b. 2 2 2
iaz
a
You
The midpoint M of the line segment joining points P1sx1, y1, z1 d and P2sx2 , y2 , z2 d is the point
To see why, observe (Figure 12.17) that
1 1 1 1 1 1 1 1 OM = OP1 + sP1 P2 d = OP1 + sOP2  OP1 d 2 2
x x 2 y1 y 2 z1 z 2 , , M 1 2 2 2
ass a
P2(x 2, y 2 , z 2 )
EXAMPLE 8
O
Finding Midpoints
The midpoint of the segment joining P1s3, 2, 0d and P2s7, 4, 4d is a
3 + 7 2 + 4 0 + 4 , , b = s5, 1, 2d. 2 2 2
Mu
ham
ma
dH
FIGURE 12.17 The coordinates of the midpoint are the averages of the coordinates of P1 and P2 .
nR
1 1 1 sOP1 + OP2 d 2 y1 + y2 x1 + x2 z1 + z2 = i + j + k. 2 2 2 =
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 860
i f su
Chapter 12: Vectors and the Geometry of Space
u o zY
EXERCISES 12.2 Vectors in the Plane
In Exercises 1–8, let u = 83, 29 and v = 82, 59 . Find the (a) component form and (b) magnitude (length) of the vector. 1. 3u
2. 2v
3. u + v
4. u  v
5. 2u  3v
6. 2u + 5v
7.
8. 
5 12 u + v 13 13
In Exercises 9–16, find the component form of the vector. 1 9. The vector PQ , where P = s1, 3d and Q = s2, 1d 1 10. The vector OP where O is the origin and P is the midpoint of segment RS, where R = s2, 1d and S = s 4, 3d
h u M
11. The vector from the point A = s2, 3d to the origin 1 1 12. The sum of AB and CD , where A = s1, 1d, B = s2, 0d, C = s 1, 3d , and D = s 2, 2d
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14. The unit vector that makes an angle u = 3p>4 with the positive xaxis
n a s s a
H d a m am
3 4 u + v 5 5
a i R
13. The unit vector that makes an angle u = 2p>3 with the positive xaxis
15. The unit vector obtained by rotating the vector 80, 19 120° counterclockwise about the origin 16. The unit vector obtained by rotating the vector 81, 09 135° counterclockwise about the origin
Vectors in Space In Exercises 17–22, express each vector in the form v = v1 i + v2 j + v3 k . 1 17. P1 P2 if P1 is the point s5, 7, 1d and P2 is the point s2, 9, 2d 1 18. P1 P2 if P1 is the point (1, 2, 0) and P2 is the point s 3, 0, 5d 1 19. AB if A is the point s 7, 8, 1d and B is the point s 10, 8, 1d 1 20. AB if A is the point (1, 0, 3) and B is the point s 1, 4, 5d
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.2 Vectors 21. 5u  v if u = 81, 1, 19 and v = 82, 0, 39
861
Length
Geometry and Calculation
Direction
a. 7
In Exercises 23 and 24, copy vectors u, v, and w head to tail as needed to sketch the indicated vector.
j 3 4  i  k b. 22 5 5 13 3 4 12 c. i j k 12 13 13 13 1 1 1 d. a 7 0 i + j k 22 23 26 33. Find a vector of magnitude 7 in the direction of v = 12i  5k .
You
23.
suf
22. 2u + 3v if u = 81, 0, 29 and v = 81, 1, 19
i
32. Find the vectors whose lengths and directions are given. Try to do the calculations without writing.
b. u + v + w
c. u  v
d. u  w
34. Find a vector of magnitude 3 in the direction opposite to the direction of v = s1>2di  s1>2dj  s1>2dk .
iaz
a. u + v
Vectors Determined by Points; Midpoints
24.
In Exercises 35–38, find
1
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a. the direction of P1 P2 and
b. the midpoint of line segment P1 P2 . P2s2, 5, 0d
35. P1s 1, 1, 5d 36. P1s1, 4, 5d
P2s4, 2, 7d
b. u  v + w
c. 2u  v
d. u + v + w
Length and Direction
dH
a. u  v
ass a
P2s2, 3, 4d 37. P1s3, 4, 5d P2s2, 2, 2d 38. P1s0, 0, 0d 1 39. If AB = i + 4j  2k and B is the point (5, 1, 3), find A. 1 40. If AB = 7i + 3j + 8k and A is the point s 2, 3, 6d , find B.
In Exercises 25–30, express each vector as a product of its length and direction. 26. 9i  2j + 6k 3 4 27. 5k 28. i + k 5 5 j 1 i 1 1 k i j k + + 29. 30. 26 26 26 23 23 23 31. Find the vectors whose lengths and directions are given. Try to do the calculations without writing.
ham
ma
25. 2i + j  2k
Length
Direction
a. 2
i
k 3 j + 5 6 i 7
Mu
b. 23 1 c. 2 d. 7
4 k 5 3 2 j + k 7 7
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Theory and Applications 41. Linear combination Let u = 2i + j, v = i + j , and w = i  j . Find scalars a and b such that u = av + bw .
42. Linear combination Let u = i  2j , v = 2i + 3j , and w = i + j . Write u = u1 + u2 , where u1 is parallel to v and u2 is parallel to w. (See Exercise 41.) 43. Force vector You are pulling on a suitcase with a force F (pictured here) whose magnitude is ƒ F ƒ = 10 lb . Find the i and jcomponents of F. F 30°
44. Force vector A kite string exerts a 12lb pull s ƒ F ƒ = 12d on a kite and makes a 45° angle with the horizontal. Find the horizontal and vertical components of F.
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4100 AWL/Thomas_ch12p848905 8/25/04 2:43 PM Page 862
Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 12: Vectors and the Geometry of Space c. Find the coordinates of the point in which the medians of ¢ABC intersect. According to Exercise 29, Section 6.4, this point is the plate’s center of mass.
i
45° F
z
You
C(1, 1, 3)
45. Velocity An airplane is flying in the direction 25° west of north at 800 km> h. Find the component form of the velocity of the airplane, assuming that the positive xaxis represents due east and the positive yaxis represents due north.
a. At what point is the tree located? b. At what point is the telephone pole?
y
B(1, 3, 0)
M
iaz
x
A(4, 2, 0)
50. Find the vector from the origin to the point of intersection of the medians of the triangle whose vertices are As1, 1, 2d,
Bs2, 1, 3d,
and
Cs 1, 2, 1d .
51. Let ABCD be a general, not necessarily planar, quadrilateral in space. Show that the two segments joining the midpoints of opposite sides of ABCD bisect each other. (Hint: Show that the segments have the same midpoint.)
ass a
47. Location A bird flies from its nest 5 km in the direction 60° north of east, where it stops to rest on a tree. It then flies 10 km in the direction due southeast and lands atop a telephone pole. Place an xycoordinate system so that the origin is the bird’s nest, the xaxis points east, and the yaxis points north.
c.m.
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46. Velocity An airplane is flying in the direction 10° east of south at 600 km> h. Find the component form of the velocity of the airplane, assuming that the positive xaxis represents due east and the positive yaxis represents due north.
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862
48. Use similar triangles to find the coordinates of the point Q that divides the segment from P1sx1, y1, z1 d to P2sx2 , y2 , z2 d into two lengths whose ratio is p>q = r .
dH
49. Medians of a triangle Suppose that A, B, and C are the corner points of the thin triangular plate of constant density shown here. a. Find the vector from C to the midpoint M of side AB.
53. Suppose that A, B, and C are vertices of a triangle and that a, b, and c are, respectively, the midpoints of the opposite sides. Show 1 1 1 that Aa + Bb + Cc = 0 .
54. Unit vectors in the plane Show that a unit vector in the plane can be expressed as u = scos udi + ssin udj , obtained by rotating i through an angle u in the counterclockwise direction. Explain why this form gives every unit vector in the plane.
Mu
ham
ma
b. Find the vector from C to the point that lies twothirds of the way from C to M on the median CM.
52. Vectors are drawn from the center of a regular nsided polygon in the plane to the vertices of the polygon. Show that the sum of the vectors is zero. (Hint: What happens to the sum if you rotate the polygon about its center?)
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 862
12.3
The Dot Product
F
v
m a h u M
FIGURE 12.18 The magnitude of the force F in the direction of vector v is the length ƒ F ƒ cos u of the projection of F onto v.
a i R an
If a force F is applied to a particle moving along a path, we often need to know the magnitude of the force in the direction of motion. If v is parallel to the tangent line to the path at the point where F is applied, then we want the magnitude of F in the direction of v. Figure 12.18 shows that the scalar quantity we seek is the length ƒ F ƒ cos u, where u is the angle between the two vectors F and v. In this section, we show how to calculate easily the angle between two vectors directly from their components. A key part of the calculation is an expression called the dot product. Dot products are also called inner or scalar products because the product results in a scalar, not a vector. After investigating the dot product, we apply it to finding the projection of one vector onto another (as displayed in Figure 12.18) and to finding the work done by a constant force acting through a displacement.
d a m
Length F cos
i f su
u o zY
Chapter 12: Vectors and the Geometry of Space
s s Ha
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.3
863
The Dot Product
i
Angle Between Vectors
v
u
You
suf
When two nonzero vectors u and v are placed so their initial points coincide, they form an angle u of measure 0 … u … p (Figure 12.19). If the vectors do not lie along the same line, the angle u is measured in the plane containing both of them. If they do lie along the same line, the angle between them is 0 if they point in the same direction, and p if they point in opposite directions. The angle u is the angle between u and v. Theorem 1 gives a formula to determine this angle.
FIGURE 12.19 The angle between u and v.
iaz
THEOREM 1 Angle Between Two Vectors The angle u between two nonzero vectors u = 8u1, u2 , u39 8v1, v2 , v39 is given by
v =
u1 v1 + u2 v2 + u3 v3 b. ƒuƒ ƒvƒ
nR
u = cos1 a
and
ass a
Before proving Theorem 1 (which is a consequence of the law of cosines), let’s focus attention on the expression u1 v1 + u2 v2 + u3 v3 in the calculation for u.
dH
DEFINITION Dot Product The dot product u # v s“u dot v”d of vectors u = 8u1, u2 , u39 and v = 8v1, v2 , v39 is
ma
EXAMPLE 1
u # v = u1 v1 + u2 v2 + u3 v3 .
Finding Dot Products
ham
(a) 81, 2, 19 # 86, 2, 39 = s1ds 6d + s 2ds2d + s 1ds 3d = 6  4 + 3 = 7
w
Mu
u
v
1 1 (b) a i + 3j + kb # s4i  j + 2kd = a bs4d + s3ds 1d + s1ds2d = 1 2 2 The dot product of a pair of twodimensional vectors is defined in a similar fashion: 8u1, u29 # 8v1, v29 = u1 v1 + u2 v2 . Proof of Theorem 1 Applying the law of cosines (Equation (6), Section 1.6) to the triangle in Figure 12.20, we find that
FIGURE 12.20 The parallelogram law of addition of vectors gives w = u  v.
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ƒ w ƒ 2 = ƒ u ƒ 2 + ƒ v ƒ 2  2 ƒ u ƒ ƒ v ƒ cos u 2 ƒ u ƒ ƒ v ƒ cos u = ƒ u ƒ 2 + ƒ v ƒ 2  ƒ w ƒ 2 .
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Law of cosines
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 864
Chapter 12: Vectors and the Geometry of Space
ƒvƒ2 =
A 2v12 + v22 + v32 B 2 = v12 + v22 + v32
A 2su1  v1 d2 + su2  v2 d2 + su3  v3 d2 B 2
You
ƒwƒ2 =
A 2u 12 + u 22 + u 32 B 2 = u 12 + u 22 + u 32
suf
ƒuƒ2 =
i
Because w = u  v, the component form of w is 8u1  v1, u2  v2 , u3  v39. So
= su1  v1 d2 + su2  v2 d2 + su3  v3 d2
= u 12  2u1v1 + v12 + u 22  2u2v2 + v22 + u 32  2u3v3 + v32 and
iaz
ƒ u ƒ 2 + ƒ v ƒ 2  ƒ w ƒ 2 = 2su1 v1 + u2v2 + u3 v3). Therefore,
2 ƒ u ƒ ƒ v ƒ cos u = ƒ u ƒ 2 + ƒ v ƒ 2  ƒ w ƒ 2 = 2su1 v1 + u2 v2 + u3 v3 d
nR
ƒ u ƒ ƒ v ƒ cos u = u1 v1 + u2 v2 + u3 v3 cos u =
ass a
So
u1 v1 + u2 v2 + u3 v3 ƒuƒ ƒvƒ
u = cos1 a
u1 v1 + u2 v2 + u3 v3 b ƒuƒ ƒvƒ
With the notation of the dot product, the angle between two vectors u and v can be written as
dH
u = cos1 a
EXAMPLE 2
u#v b. ƒuƒ ƒvƒ
Finding the Angle Between Two Vectors in Space
ma
Find the angle between u = i  2j  2k and v = 6i + 3j + 2k.
Mu
ham
Solution
We use the formula above: u # v = s1ds6d + s 2ds3d + s 2ds2d = 6  6  4 = 4 ƒ u ƒ = 2s1d2 + s 2d2 + s 2d2 = 29 = 3 ƒ v ƒ = 2s6d2 + s3d2 + s2d2 = 249 = 7 u = cos1 a = cos1 a
u#v b ƒuƒ ƒvƒ 4 b L 1.76 radians. s3ds7d
The angle formula applies to twodimensional vectors as well.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.3 y
EXAMPLE 3
865
Finding an Angle of a Triangle
i
B(3, 5)
The Dot Product
C(5, 2)
A
x
1
FIGURE 12.21 Example 3.
1 1 The angle u is the angle between the vectors CA and CB . The component forms of these two vectors are 1 1 CA = 85, 29 and CB = 82, 39.
Solution
1
The triangle in
You
suf
Find the angle u in the triangle ABC determined by the vertices A = s0, 0d, B = s3, 5d, and C = s5, 2d (Figure 12.21).
First we calculate the dot product and magnitudes of these two vectors. 1 1 CA # CB = s 5ds 2d + s 2ds3d = 4
iaz
1 ƒ CA ƒ = 2s 5d2 + s 2d2 = 229 1 ƒ CB ƒ = 2s 2d2 + s3d2 = 213 Then applying the angle formula, we have
nR
1 1 CA # CB u = cos1 £ 1 1 ≥ ƒ CA ƒ ƒ CB ƒ
ass a
= cos1 £
4
A 229 B A 213 B
L 78.1°
or
≥
1.36 radians.
dH
Perpendicular (Orthogonal) Vectors
ma
Two nonzero vectors u and v are perpendicular or orthogonal if the angle between them is p>2. For such vectors, we have u # v = 0 because cos sp>2d = 0. The converse is also true. If u and v are nonzero vectors with u # v = ƒ u ƒ ƒ v ƒ cos u = 0, then cos u = 0 and u = cos1 0 = p>2.
Mu
ham
DEFINITION Orthogonal Vectors Vectors u and v are orthogonal (or perpendicular) if and only if u # v = 0.
EXAMPLE 4
Applying the Definition of Orthogonality
(a) u = 83, 29 and v = 84, 69 are orthogonal because u # v = s3ds4d + s 2ds6d = 0. (b) u = 3i  2j + k and v = 2j + 4k are orthogonal because u # v = s3ds0d + s 2ds2d + s1ds4d = 0. (c) 0 is orthogonal to every vector u since
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0 # u = 80, 0, 09 # 8u1, u2, u39 = s0dsu1 d + s0dsu2 d + s0dsu3 d = 0.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 866
Chapter 12: Vectors and the Geometry of Space
i
Dot Product Properties and Vector Projections
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The dot product obeys many of the laws that hold for ordinary products of real numbers (scalars).
HISTORICAL BIOGRAPHY
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You
Properties of the Dot Product If u, v, and w are any vectors and c is a scalar, then 1. u # v = v # u 2. scud # v = u # scvd = csu # vd 3. u # sv + wd = u # v + u # w 4. u # u = ƒ u ƒ 2 5. 0 # u = 0.
Proofs of Properties 1 and 3 The properties are easy to prove using the definition. For instance, here are the proofs of Properties 1 and 3.
Carl Friedrich Gauss (1777–1855) Q
u # v = u1 v1 + u2 v2 + u3 v3 = v1 u1 + v2 u2 + v3 u3 = v # u u # sv + wd = 8u1, u2 , u39 # 8v1 + w1, v2 + w2 , v3 + w39
ass a
1. 3.
= u1sv1 + w1 d + u2sv2 + w2 d + u3sv3 + w3 d
u
= u1 v1 + u1 w1 + u2 v2 + u2 w2 + u3 v3 + u3 w3
v P
R
S
= su1 v1 + u2 v2 + u3 v3 d + su1 w1 + u2 w2 + u3 w3 d
u v P
S
We now return to the problem of projecting one vector onto another, posed in the 1 1 opening to this section. The vector projection of u = PQ onto a nonzero vector v = PS 1 (Figure 12.22) is the vector PR determined by dropping a perpendicular from Q to the line PS. The notation for this vector is
ma
R
dH
= u#v + u#w
Q
ham
FIGURE 12.22 The vector projection of u onto v.
Force u
projv u
s“the vector projection of u onto v”d.
If u represents a force, then projv u represents the effective force in the direction of v (Figure 12.23). If the angle u between u and v is acute, projv u has length ƒ u ƒ cos u and direction v> ƒ v ƒ (Figure 12.24). If u is obtuse, cos u 6 0 and projv u has length  ƒ u ƒ cos u and direction v> ƒ v ƒ . In both cases, projv u = s ƒ u ƒ cos ud
Mu
v
FIGURE 12.23 If we pull on the box with force u, the effective force moving the box forward in the direction v is the projection of u onto v.
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= a
u#v v b ƒvƒ ƒvƒ
= a
u#v bv. ƒvƒ2
v ƒvƒ ƒ u ƒ cos u =
ƒ u ƒ ƒ v ƒ cos u u#v = ƒvƒ ƒvƒ
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH The Dot Product
867
u
u
proj v u
proj v u
v
v
You
suf
i
12.3
Length u cos (a)
Length –u cos
(b)
iaz
FIGURE 12.24 The length of projv u is (a) ƒ u ƒ cos u if cos u Ú 0 and (b)  ƒ u ƒ cos u if cos u 6 0 .
nR
The number ƒ u ƒ cos u is called the scalar component of u in the direction of v. To summarize,
Vector projection of u onto v:
projv u = a
u#v bv ƒvƒ2
(1)
ass a
Scalar component of u in the direction of v: ƒ u ƒ cos u =
v u#v = u# v v ƒ ƒ ƒ ƒ
(2)
dH
Note that both the vector projection of u onto v and the scalar component of u onto v depend only on the direction of the vector v and not its length (because we dot u with v> ƒ v ƒ , which is the direction of v).
ma
EXAMPLE 5
Finding the Vector Projection
Find the vector projection of u = 6i + 3j + 2k onto v = i  2j  2k and the scalar component of u in the direction of v.
Mu
ham
Solution
We find projv u from Equation (1): u#v 6  6  4 projv u = v # v v = si  2j  2kd 1 + 4 + 4 = 
8 8 4 4 si  2j  2kd =  i + j + k. 9 9 9 9
We find the scalar component of u in the direction of v from Equation (2): ƒ u ƒ cos u = u #
v 1 2 2 = s6i + 3j + 2kd # a i  j  kb 3 3 3 ƒvƒ
= 2  2 
4 4 =  . 3 3
Equations (1) and (2) also apply to twodimensional vectors.
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Chapter 12: Vectors and the Geometry of Space
Finding Vector Projections and Scalar Components
i
EXAMPLE 6
projv F = ¢ =
F#v ≤v ƒvƒ2
You
The vector projection is
5  6 1 si  3jd = si  3jd 1 + 9 10
= 
3 1 i + j. 10 10
iaz
Solution
suf
Find the vector projection of a force F = 5i + 2j onto v = i  3j and the scalar component of F in the direction of v.
The scalar component of F in the direction of v is
Work
D
Q
F cos
FIGURE 12.25 The work done by a constant force F during a displacement D is s ƒ F ƒ cos ud ƒ D ƒ .
ass a
P
In Chapter 6, we calculated the work done by a constant force of magnitude F in moving an object through a distance d as W = Fd. That formula holds only if the force is directed 1 along the line of motion. If a force F moving an object through a displacement D = PQ has some other direction, the work is performed by the component of F in the direction of D. If u is the angle between F and D (Figure 12.25), then scalar component of F Work = ain the direction of D bslength of Dd
dH
F
5  6 F#v 1 = = . ƒvƒ 21 + 9 210
nR
ƒ F ƒ cos u =
= s ƒ F ƒ cos ud ƒ D ƒ = F # D.
Mu
ham
ma
DEFINITION Work by Constant Force 1 The work done by a constant force F acting through a displacement D = PQ is W = F # D = ƒ F ƒ ƒ D ƒ cos u,
where u is the angle between F and D.
EXAMPLE 7
Applying the Definition of Work
If ƒ F ƒ = 40 N (newtons), ƒ D ƒ = 3 m, and u = 60°, the work done by F in acting from P to Q is
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Work = = = =
ƒ F ƒ ƒ D ƒ cos u s40ds3d cos 60° s120ds1>2d 60 J s joulesd.
Definition Given values
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.3
The Dot Product
869
suf
i
We encounter more challenging work problems in Chapter 16 when we learn to find the work done by a variable force along a path in space.
Writing a Vector as a Sum of Orthogonal Vectors
u = u1 i + u2 j
u = u1 i + su2 j + u3 kd
u proj v u
v proj v u
FIGURE 12.26 Writing u as the sum of vectors parallel and orthogonal to v.
nR
iaz
(since i # j = i # k = j # k = 0). Sometimes, however, it is more informative to express u as a different sum. In mechanics, for instance, we often need to write a vector u as a sum of a vector parallel to a given vector v and a vector orthogonal to v. As an example, in studying the motion of a particle moving along a path in the plane (or space), it is desirable to know the components of the acceleration vector in the direction of the tangent to the path (at a point) and of the normal to the path. (These tangential and normal components of acceleration are investigated in Section 13.4.) The acceleration vector can then be expressed as the sum of its (vector) tangential and normal components (which reflect important geometric properties about the nature of the path itself, such as curvature). Velocity and acceleration vectors are studied in the next chapter. Generally, for vectors u and v, it is easy to see from Figure 12.26 that the vector
ass a
u
or
You
We know one way to write a vector u = 8u1, u29 or u = 8u1, u2, u39 as a sum of two orthogonal vectors:
u  projv u
is orthogonal to the projection vector projv u (which has the same direction as v). The following calculation verifies this observation:
ma
dH
su  projv ud # projv u = ¢ u  ¢ = ¢ =
u#v u#v v # ≤v 2≤ ≤ ¢ ƒvƒ ƒvƒ2 2
u#v u#v su # vd  ¢ ≤ sv # vd 2≤ ƒvƒ ƒvƒ2
su # vd2 su # vd2 ƒvƒ2 ƒvƒ2
Equation (1)
Dot product properties 2 and 3 v # v = ƒ v ƒ 2 cancels
= 0.
u = projv u + su  projv ud
expresses u as a sum of orthogonal vectors.
How to Write u as a Vector Parallel to v Plus a Vector Orthogonal to v
Mu
ham
So the equation
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u = projv u + su  projv ud u#v u#v v + ¢u  ¢ ≤v≤ 2≤ ƒvƒ ƒvƒ2 (')'* ('')''*
= ¢
Parallel to v
Orthogonal to v
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 870
Chapter 12: Vectors and the Geometry of Space
Force on a Spacecraft
i
EXAMPLE 8
suf
A force F = 2i + j  3k is applied to a spacecraft with velocity vector v = 3i  j. Express F as a sum of a vector parallel to v and a vector orthogonal to v.
F = projv F + sF  projv Fd F#v F#v = v # v v + aF  v # v vb
=
6  1 6  1 bv + aF  a bvb 9 + 1 9 + 1
iaz
= a
You
Solution
5 5 s3i  jd + a2i + j  3k s3i  jdb 10 10
nR
3 3 1 1 = a i  jb + a i + j  3kb . 2 2 2 2
ass a
The force s3>2di  s1/2dj is the effective force parallel to the velocity v. The force s1>2di + s3/2dj  3k is orthogonal to v. To check that this vector is orthogonal to v, we find the dot product:
Mu
ham
ma
dH
3 3 3 1 a i + j  3kb # s3i  jd =  = 0. 2 2 2 2
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 870
i f su
Chapter 12: Vectors and the Geometry of Space
EXERCISES 12.3 Dot Product and Projections
T
In Exercises 1–8, find a. ƒuƒ b. the cosine of the angle between v and u
u = 2i + 4j  25k
2. v = s3>5di + s4>5dk,
u = 5i + 12j
3. v = 10i + 11j  2k,
u = 3j + 4k
4. v = 2i + 10j  11k,
u = 2i + 2j + k
5. v = 5j  3k,
d a m m a h
u = 2i + 217j
8. v = h
1
i,
u M 1
,
a H
u = 22i + 23j + 2k
7. v = 5i + j,
22 23
u = h
1
22
,
1
23
v = i + 2j  k
10. u = 2i  2j + k,
u = i + j + k
6. v = i + j,
n a ss
9. u = 2i + j,
c. the scalar component of u in the direction of v 1. v = 2i  4j + 25k,
Angles Between Vectors
Find the angles between the vectors in Exercises 9–12 to the nearest hundredth of a radian.
v # u, ƒ v ƒ ,
d. the vector projv u .
u o Y z a i R
i
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11. u = 23i  7j,
v = 3i + 4k
v = 23i + j  2k
12. u = i + 22j  22k,
v = i + j + k
13. Triangle Find the measures of the angles of the triangle whose vertices are A = s 1, 0d, B = s2, 1d , and C = s1, 2d . 14. Rectangle Find the measures of the angles between the diagonals of the rectangle whose vertices are A = s1, 0d, B = s0, 3d, C = s3, 4d , and D = s4, 1d .
15. Direction angles and direction cosines The direction angles a, b , and g of a vector v = ai + bj + ck are defined as follows: a is the angle between v and the positive xaxis s0 … a … pd b is the angle between v and the positive yaxis s0 … b … pd g is the angle between v and the positive zaxis s0 … g … pd .
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.3 The Dot Product z
871
v2
␥
v
v1 v 2
You
0
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two vectors to be orthogonal to their difference? Give reasons for your answer.

␣
v1
y x
–v 2
v1 v 2
b cos b = , ƒvƒ
c cos g = , ƒvƒ
and cos2 a + cos2 b + cos2 g = 1 . These cosines are called the direction cosines of v. b. Unit vectors are built from direction cosines Show that if v = ai + bj + ck is a unit vector, then a, b, and c are the direction cosines of v.
East
rth
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No
Decomposing Vectors
In Exercises 17–19, write u as the sum of a vector parallel to v and a vector orthogonal to v. v = i + j
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17. u = 3j + 4k, 18. u = j + k,
v = i + j
19. u = 8i + 4j  12k,
A
v
–u
O
B
u
23. Diagonals of a rhombus Show that the diagonals of a rhombus (parallelogram with sides of equal length) are perpendicular. 24. Perpendicular diagonals Show that squares are the only rectangles with perpendicular diagonals.
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C
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16. Water main construction A water main is to be constructed with a 20% grade in the north direction and a 10% grade in the east direction. Determine the angle u required in the water main for the turn from north to east.
22. Orthogonality on a circle Suppose that AB is the diameter of a circle with center O and that C is a point on one of the two arcs 1 1 joining A and B. Show that CA and CB are orthogonal.
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a cos a = , ƒvƒ
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a. Show that
v = i + 2j  k
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20. Sum of vectors u = i + sj + kd is already the sum of a vector parallel to i and a vector orthogonal to i. If you use v = i , in the decomposition u = projv u + su  projv ud , do you get projv u = i and su  projv ud = j + k ? Try it and find out.
25. When parallelograms are rectangles Prove that a parallelogram is a rectangle if and only if its diagonals are equal in length. (This fact is often exploited by carpenters.) 26. Diagonal of parallelogram Show that the indicated diagonal of the parallelogram determined by vectors u and v bisects the angle between u and v if ƒ u ƒ = ƒ v ƒ . K L
27. Projectile motion A gun with muzzle velocity of 1200 ft> sec is fired at an angle of 8° above the horizontal. Find the horizontal and vertical components of the velocity. 28. Inclined plane Suppose that a box is being towed up an inclined plane as shown in the figure. Find the force w needed to make the component of the force parallel to the inclined plane equal to 2.5 lb.
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Geometry and Examples 21. Sums and differences In the accompanying figure, it looks as if v1 + v2 and v1  v2 are orthogonal. Is this mere coincidence, or are there circumstances under which we may expect the sum of
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH Chapter 12: Vectors and the Geometry of Space
Theory and Examples
Work u#v =
i
872
43. Work along a line Find the work done by a force F = 5i (magnitude 5 N) in moving an object along the line from the origin to the point (1, 1) (distance in meters).
b. Under what circumstances, if any, does ƒ u # v ƒ equal ƒ u ƒ ƒ v ƒ ? Give reasons for your answer.
44. Locomotive The union Pacific’s Big Boy locomotive could pull 6000ton trains with a tractive effort (pull) of 602,148 N (135,375 lb). At this level of effort, about how much work did Big Boy do on the (approximately straight) 605km journey from San Francisco to Los Angeles?
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30. Copy the axes and vector shown here. Then shade in the points (x, y) for which sxi + yjd # v … 0 . Justify your answer.
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29. a. CauchySchwartz inequality Use the fact that ƒ u ƒ ƒ v ƒ cos u to show that the inequality ƒ u # v ƒ … ƒ u ƒ ƒ v ƒ holds for any vectors u and v.
45. Inclined plane How much work does it take to slide a crate 20 m along a loading dock by pulling on it with a 200 N force at an angle of 30° from the horizontal?
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0
31. Orthogonal unit vectors If u1 and u2 are orthogonal unit vectors and v = au1 + bu2 , find v # u1 .
Equations for Lines in the Plane
60° 1000 lb magnitude force
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32. Cancellation in dot products In realnumber multiplication, if uv1 = uv2 and u Z 0 , we can cancel the u and conclude that v1 = v2 . Does the same rule hold for the dot product: If u # v1 = u # v2 and u Z 0 , can you conclude that v1 = v2 ? Give reasons for your answer.
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46. Sailboat The wind passing over a boat’s sail exerted a 1000lb magnitude force F as shown here. How much work did the wind perform in moving the boat forward 1 mi? Answer in footpounds.
x
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33. Line perpendicular to a vector Show that the vector v = ai + bj is perpendicular to the line ax + by = c by establishing that the slope of v is the negative reciprocal of the slope of the given line.
Angles Between Lines in the Plane The acute angle between intersecting lines that do not cross at right angles is the same as the angle determined by vectors normal to the lines or by the vectors parallel to the lines. n1
n2
L2 L2
34. Line parallel to a vector Show that the vector v = ai + bj is parallel to the line bx  ay = c by establishing that the slope of the line segment representing v is the same as the slope of the given line.
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F
v2
v1 L1
L1
Use this fact and the results of Exercise 33 or 34 to find the acute angles between the lines in Exercises 47–52.
35. Ps2, 1d,
47. 3x + y = 5,
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In Exercises 35–38, use the result of Exercise 33 to find an equation for the line through P perpendicular to v. Then sketch the line. Include v in your sketch as a vector starting at the origin. v = i + 2j
36. Ps 1, 2d,
v = 2i  j
37. Ps 2, 7d, 38. Ps11, 10d,
v = 2i + j
Mu 41. Ps1, 2d,
v = i  j
v = i  2j
40. Ps0, 2d, 42. Ps1, 3d,
y =  23x + 2
49. 23x  y = 2,
v = 2i  3j
50. x + 23y = 1,
In Exercises 39–42, use the result of Exercise 34 to find an equation for the line through P parallel to v. Then sketch the line. Include v in your sketch as a vector starting at the origin. 39. Ps 2, 1d,
2x  y = 4
48. y = 23x  1,
v = 2i + 3j v = 3i  2j
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51. 3x  4y = 3, 52. 12x + 5y = 1,
x  23y = 1
A 1  23 B x + A 1 + 23 B y = 8
x  y = 7
2x  2y = 3
Angles Between Differentiable Curves The angles between two differentiable curves at a point of intersection are the angles between the curves’ tangent lines at these points. Find
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.3 The Dot Product
y = x2
stwo points of intersectiond
x = y2 2
56. y = x ,
stwo points of intersectiond
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55. y = x 3,
y = 1x
stwo points of intersectiond
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53. y = s3>2d  x 2,
54. x = s3>4d  y 2, x = y 2  s3>4d stwo points of intersectiond
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the angles between the curves in Exercises 53â€“56. Note that if v = ai + bj is a vector in the plane, then the vector has slope b> a provided a Z 0 .
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873
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH
12.4
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12.4 The Cross Product
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The Cross Product
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In studying lines in the plane, when we needed to describe how a line was tilting, we used the notions of slope and angle of inclination. In space, we want a way to describe how a plane is tilting. We accomplish this by multiplying two vectors in the plane together to get a third vector perpendicular to the plane. The direction of this third vector tells us the “inclination” of the plane. The product we use to multiply the vectors together is the vector or cross product, the second of the two vector multiplication methods we study in calculus. Cross products are widely used to describe the effects of forces in studies of electricity, magnetism, fluid flows, and orbital mechanics. This section presents the mathematical properties that account for the use of cross products in these fields.
The Cross Product of Two Vectors in Space uv
an
We start with two nonzero vectors u and v in space. If u and v are not parallel, they determine a plane. We select a unit vector n perpendicular to the plane by the righthand rule. This means that we choose n to be the unit (normal) vector that points the way your right thumb points when your fingers curl through the angle u from u to v (Figure 12.27). Then the cross product u * v (“u cross v”) is the vector defined as follows.
v
n
ass
u
DEFINITION
ad H
u * v = s ƒ u ƒ ƒ v ƒ sin ud n
The construction of
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FIGURE 12.27 u * v.
Cross Product
Unlike the dot product, the cross product is a vector. For this reason it’s also called the vector product of u and v, and applies only to vectors in space. The vector u * v is orthogonal to both u and v because it is a scalar multiple of n. Since the sines of 0 and p are both zero, it makes sense to define the cross product of two parallel nonzero vectors to be 0. If one or both of u and v are zero, we also define u * v to be zero. This way, the cross product of two vectors u and v is zero if and only if u and v are parallel or one or both of them are zero.
Parallel Vectors Nonzero vectors u and v are parallel if and only if u * v = 0 .
The cross product obeys the following laws.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 874
Chapter 12: Vectors and the Geometry of Space
You
u
vu
To visualize Property 4, for example, notice that when the fingers of a right hand curl through the angle u from v to u, the thumb points the opposite way and the unit vector we choose in forming v * u is the negative of the one we choose in forming u * v (Figure 12.28). Property 1 can be verified by applying the definition of cross product to both sides of the equation and comparing the results. Property 2 is proved in Appendix 6. Property 3 follows by multiplying both sides of the equation in Property 2 by 1 and reversing the order of the products using Property 4. Property 5 is a definition. As a rule, cross product multiplication is not associative so su * vd * w does not generally equal u * sv * wd. (See Additional Exercise 15.) When we apply the definition to calculate the pairwise cross products of i, j, and k, we find (Figure 12.29)
k i j –( j i)
–i –j
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FIGURE 12.28 The construction of v * u.
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Properties of the Cross Product If u, v, and w are any vectors and r, s are scalars, then 1. srud * ss vd = srsdsu * vd 2. u * sv + wd = u * v + u * w 3. sv + wd * u = v * u + w * u 4. v * u = su * vd 5. 0 * u = 0
v
j k i –(i k)
–k
i
k * i = si * kd = j
Diagram for recalling these products
and
i * i = j * j = k * k = 0.
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FIGURE 12.29 The pairwise cross products of i, j, and k.
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j * k = sk * jd = i
x i j k –(k j)
j
i * j = sj * id = k
y
i
k
ƒ u * v ƒ Is the Area of a Parallelogram
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Because n is a unit vector, the magnitude of u * v is
Area base ⋅ height u ⋅ vsin u × v
v
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h v sin
u
FIGURE 12.30 The parallelogram determined by u and v.
ƒ u * v ƒ = ƒ u ƒ ƒ v ƒ ƒ sin u ƒ ƒ n ƒ = ƒ u ƒ ƒ v ƒ sin u.
This is the area of the parallelogram determined by u and v (Figure 12.30), ƒ u ƒ being the base of the parallelogram and ƒ v ƒ ƒ sin u ƒ the height.
Determinant Formula for u * v Our next objective is to calculate u * v from the components of u and v relative to a Cartesian coordinate system.
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.4
u * v = su1 i + u2 j + u3 kd * sv1 i + v2 j + v3 kd = u1 v1 i * i + u1 v2 i * j + u1 v3 i * k
 a2 `
a3 b2 b3 3 = a1 ` c2 c3 b1 c1
+ u2 v1 j * i + u2 v2 j * j + u2 v3 j * k
+ u3 v1 k * i + u3 v2 k * j + u3 v3 k * k
= su2 v3  u3 v2 di  su1 v3  u3 v1 dj + su1 v2  u2 v1 dk.
b3 ` c3
b3 b1 ` + a3 ` c3 c1
The terms in the last line are the same as the terms in the expansion of the symbolic determinant
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a2 b2 c2
c1
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1 ` = s2ds3d  s1ds 4d 3 = 6 + 4 = 10
b2 ` c2
i 3 u1
EXAMPLE
v1
3 1 3
k u3 3 . v3
We therefore have the following rule. 1 ` 3
ass a
1 1 1 1 3 = s 5d ` ` 3 1 1 2 1 2  s3d ` ` + s1d ` 4 1 4 = 5s1  3d  3s2 + 4d + 1s6 + 4d = 10  18 + 10 = 2
5 3 2 4
j u2 v2
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a1 3 b1
i
Then the distributive laws and the rules for multiplying i, j, and k tell us that
b ` = ad  bc d
EXAMPLE 2 4
v = v1 i + v2 j + v3 k.
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u = u1 i + u2 j + u3 k,
2 * 2 and 3 * 3 determinants are evaluated as follows:
`
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Suppose that
Determinants
a ` c
The Cross Product
Calculating Cross Products Using Determinants If u = u1 i + u2 j + u3 k and v = v1 i + v2 j + v3 k, then i u * v = 3 u1 v1
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(For more information, see the Web site at www.awbc.com/thomas.)
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EXAMPLE 1
j u2 v2
k u3 3 . v3
Calculating Cross Products with Determinants
Find u * v and v * u if u = 2i + j + k and v = 4i + 3j + k.
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Solution
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R(–1, 1, 2)
i 3 u * v = 2 4
0
Mu
P(1, –1, 0)
x
j 1 3
k 1 13 = p 3 1
1 2 pi  p 1 4
1 2 pj + p 1 4
1 pk 3
= 2i  6j + 10k y
Q(2, 1, –1)
FIGURE 12.31 The area of triangle PQR 1 1 is half of ƒ PQ * PR ƒ (Example 2).
v * u = su * vd = 2i + 6j  10k
EXAMPLE 2
Finding Vectors Perpendicular to a Plane
Find a vector perpendicular to the plane of Ps1, 1, 0d, Qs2, 1, 1d, and Rs 1, 1, 2d (Figure 12.31).
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 876
Chapter 12: Vectors and the Geometry of Space
1 1 The vector PQ * PR is perpendicular to the plane because it is perpendicular to both vectors. In terms of components, 1 PQ = s2  1di + s1 + 1dj + s 1  0dk = i + 2j  k 1 PR = s 1  1di + s1 + 1dj + s2  0dk = 2i + 2j + 2k j 2 2
k 2 1 3 = ` 2 2
1 1 `i  ` 2 2
1 1 `j + ` 2 2
2 `k 2
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= 6i + 6k.
EXAMPLE 3
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i 1 1 3 PQ * PR = 1 2
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Solution
Finding the Area of a Triangle
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Find the area of the triangle with vertices Ps1, 1, 0d, Qs2, 1, 1d, and Rs 1, 1, 2d (Figure 12.31). The area of the parallelogram determined by P, Q, and R is 1 1 Values from Example 2. ƒ PQ * PR ƒ = ƒ 6i + 6k ƒ
Solution
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= 2s6d2 + s6d2 = 22 # 36 = 622 .
The triangle’s area is half of this, or 322 .
EXAMPLE 4
Finding a Unit Normal to a Plane
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Find a unit vector perpendicular to the plane of Ps1, 1, 0d, Qs2, 1, 1d, and Rs 1, 1, 2d.
1 1 Since PQ * PR is perpendicular to the plane, its direction n is a unit vector perpendicular to the plane. Taking values from Examples 2 and 3, we have 1 1 PQ * PR 6i + 6k 1 1 n = 1 = i + k. 1 = 6 22 22 22 ƒ PQ * PR ƒ
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Solution
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Torque
For ease in calculating the cross product using determinants, we usually write vectors in the form v = v1 i + v2 j + v3 k rather than as ordered triples v = 8v1, v2 , v39.
r
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Component of F perpendicular to r. Its length is F sin .
F
FIGURE 12.32 The torque vector describes the tendency of the force F to drive the bolt forward.
Torque When we turn a bolt by applying a force F to a wrench (Figure 12.32), the torque we produce acts along the axis of the bolt to drive the bolt forward. The magnitude of the torque depends on how far out on the wrench the force is applied and on how much of the force is perpendicular to the wrench at the point of application. The number we use to measure the torque’s magnitude is the product of the length of the lever arm r and the scalar component of F perpendicular to r. In the notation of Figure 12.32,
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Magnitude of torque vector = ƒ r ƒ ƒ F ƒ sin u,
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Bismillah hir Rehman nir Raheem****Asalat o Wasalam o alika Ya RasoolALLAH 12.4
The Cross Product
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or ƒ r * F ƒ . If we let n be a unit vector along the axis of the bolt in the direction of the torque, then a complete description of the torque vector is r * F, or Torque vector = s ƒ r ƒ ƒ F ƒ sin ud n.
EXAMPLE 5
70° F
20 lb magnitude force
FIGURE 12.33 The magnitude of the torque exerted by F at P is about 56.4 ftlb (Example 5).
Finding the Magnitude of a Torque
The magnitude of the torque generated by force F at the pivot point P in Figure 12.33 is 1 1 ƒ PQ * F ƒ = ƒ PQ ƒ ƒ F ƒ sin 70° L s3ds20ds0.94d L 56.4 ftlb.
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3 ft bar P
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Recall that we defined u * v to be 0 when u and v are parallel. This is consistent with the torque interpretation as well. If the force F in Figure 12.32 is parallel to the wrench, meaning that we are trying to turn the bolt by pushing or pulling along the line of the wrench’s handle, the torque produced is zero.
Triple Scalar or Box Product
The product su * vd # w is called the triple scalar product of u, v, and w (in that order). As you can see from the formula
ass a
ƒ su * vd # w ƒ = ƒ u * v ƒ ƒ w ƒ ƒ cos u ƒ ,
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the absolute value