.
Math 263C, WINTER 2007, Solution of HW 11.10
Attn. If f (x) has a power series representation at a, this means, if
f(x) =
1
X
n=0
cn (x ¡ a)n ; jx ¡ aj < R
then its coe±cients are given by
cn =
f (n) (a)
:
n!
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9. Find a power series representation for the function
f (x) = sinhx:
Then ¯nd the radius of convergence.
ANSWER. Write sinhx =
1
P
cn xn . Now we ¯nd the coe±cients cn (here
n=0
a = 0). f (n) (x) = coshx if n is odd, and f (n) (x) = sinhx if n is even. Hence
1
.
f (n) (0) = 1 if n is odd, and f (n) (0) = 0 if n is even. Thus c2k = 0; c2k+1 = (2k+1)!
Substitute this for cn in the formula of sinhx we have:
1
X
x2n+1
:
sinhx =
(2n + 1)!
n=0
Now we use the Ratio Test to ¯nd the convergence radius R. Set an =
x2n+1
(2n+1)!
2n+3
x
1
then lim j aan+1
j = lim j (2n+3)!
: (2n+1)!
lim jx2 j (2n+2)(2n+3)
= x2 :0 = 0 < 1.
x2n+1 j = n!1
n
n!1
n!1
Hence R = 1:
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15. This problem was presented in class.
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23. Obtain a power series for cos¼x, and give the radius of convergence.
Typeset by AMS-TEX
1
2
ANSWER. By formula (16) in page 766, cosx =
cos¼x =
1
P
n=0
1
P
n=0
2n
(¡1)n (¼x)
(2n)! :
2n
x
. It follows
(¡1)n (2n)!
Similar to # 9 above we ¯nd R = 1.
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25. Obtain a power series for xtan¡1 x, and give the radius of convergence.
ANSWER. By a formula on page 767, we have tan¡1 x =
1
P
n=0
1; I = [¡1; 1]: Hence
¡1
xtan
1
X
2n+1
(¡1)n x2n+1 ; R =
1
X
x2n+1
x2n+2
x=x
=
:
(¡1)
(¡1)n
2n
+
1
2n
+
1
n=0
n=0
n
2n+4
2n+1
To ¯nd R and I we use the limit ratio test. lim j aan+1
j = lim j x2n+3 : x2n+1 j =
n
2n+1
n!1 2n+3
jx3 j lim
n!1
n!1
= jx3 j < 1 ) jxj < 1 ) ¡1 < x < 1: Now we check the endpoints
x = ¡1; x = 1, and see (prove it!) that the series of the function xtan¡1 x is convergent
at x = 1 and x = ¡1. Hence R = 1; I = [¡1; 1]:
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27. Use the series of ex , then multiply the series with x2 we will get a power series
for f(x) = x2 ex with R = 1; I = (¡1; 1).
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29. Obtain a power series for sin2 x, and give the radius of convergence.
SOLUTION. Use the formula sin2 x = 12 (1 ¡ cos2x) we have:
1
1
1
2n
2n
X
X
1
1
1 X
(2x)2n
n (2x)
n (2x)
sin x = [1 ¡
(¡1)
] = [1 ¡ 1 ¡
(¡1)
]= [
(¡1)n+1
]
2
(2n)!
2
(2n)!
2
(2n)!
n=0
n=1
n=1
2
=
1
X
(¡1)n+1
n=1
22n¡1 x2n
:
(2n)!
Use the limit ratio test to show that R = 1 and so I = (¡1; 1). (You should do
this for more practice!)
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31. By a formula on page 767, sinx =
1
P
n=0
2n+1
x
(¡1)n (2n+1)!
: Hence we have:
3
1
1
2n+1
X
sinx
1 X
x2n
n x
= :
=
:
(¡1)
(¡1)n
x
x n=0
(2n + 1)! n=0
(2n + 1)!
This series includes the case x = 0, in this case the series is equal to 1, as given in the
problem 39. Using the limit ratio test we see R = 1.
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R
39. Using in¯nite series to evaluate the in¯nite integral xcos(x3 )dx.
ANSWER. By the formula (16) on page 766 we have:
1
X
1
1
3 2n
X
X
x2n
x6n+1
3
n (x )
3
cosx =
(¡1)
) cos(x ) =
(¡1)
) xcos(x ) =
(¡1)n
(2n)!
(2n)!
(2n)!
n=0
n=0
n=0
)
Z
n
Z X
1
1
6n+1
X
x6n+2
nx
n
xcos(x )dx = [
(¡1)
]dx = C +
(¡1)
:
(2n)!
(6n + 2)(2n)!
n=0
n=0
3
Where C is a constant. Use the limit ratio test we will ¯nd R = 1: (Check it!!)