Calculus II Quiz 10: Convergence of Series and Power Series Representation of arctan x, Exercises of Calculus

The solutions to quiz 10 for math 106a,b - calculus ii, winter 2008. It includes the calculation of the interval of convergence for a given series and the derivation of the power series representation of arctan x. The document also shows how to use the power series representation to evaluate the limit of arctan x as x approaches 0.

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MATH 106A,B - CALCULUS II WINTER 2008
QUIZ 10
NAME:
Show ALL your work CAREFULLY.
(a) Find the interval of convergence of the following infinite series [Do not forget to check the endpoints!].
โˆž
X
k=1
(4x)k
k2
Consider
lim
kโ†’โˆž
|4x|k+1
(k+ 1)2ยทk2
|4x|k= lim
kโ†’โˆž
|4x|๎˜’k
k+ 1๎˜“2
=|4x|lim
kโ†’โˆž ๎˜’k
k+ 1๎˜“2
=|4x|.
By the Ratio Test, the series converges if |4x|<1, i.e., โˆ’1<4x < 1or โˆ’1
4<x< 1
4. When
x=1
4, the series becomes P1
k2which is known to converge. When x=โˆ’1
4, the series becomes
P(โˆ’1)k
k2which converges since the associated series of absolute values P1
k2converges. Hence,
the interval of convergence is [โˆ’1
4,1
4].
(b) Recall that 1
1โˆ’t= 1 + t+t2+t3+t4+... for |t|<1.
Use this to obtain a power series representation of arctanx. [What is the derivative of arctanx?]
By letting t=โˆ’x2, we have
1
1 + x2= 1 โˆ’x2+x4โˆ’x6+... for |x|<1.
It follows that
arctan x=Z1
1 + x2dx =xโˆ’x3
3+x5
5โˆ’x7
7+...
=X
n=1
(โˆ’1)n+1x2nโˆ’1
2nโˆ’1for |x|<1.
(c) Use (b) to evaluate limxโ†’0arctanx
x.
Using the power series representation of arctan xfrom (b), we have
lim
xโ†’0
arctan x
x= lim
xโ†’0
1
x๎˜’xโˆ’x3
3+x5
5โˆ’x7
7+...๎˜“
= lim
xโ†’0๎˜’1โˆ’x2
3+x4
5โˆ’x6
7+...๎˜“= 1.
Date: April 4, 2008.
1

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MATH 106A,B - CALCULUS II WINTER 2008

QUIZ 10

NAME:

Show ALL your work CAREFULLY.

(a) Find the interval of convergence of the following infinite series [Do not forget to check the endpoints!]. โˆ‘^ โˆž k=

(4x)k k^2

Consider

k^ limโ†’โˆž

| 4 x|k+ (k + 1)^2

k^2 | 4 x|k^ = (^) klimโ†’โˆž | 4 x|

k k + 1

= | 4 x| (^) klimโ†’โˆž

k k + 1

= | 4 x|.

By the Ratio Test, the series converges if | 4 x| < 1 , i.e., โˆ’ 1 < 4 x < 1 or โˆ’ 14 < x < 14. When x = 14 , the series becomes

k^2 which is known to converge. When^ x^ =^ โˆ’^ 1 โˆ‘ 4 , the series becomes (โˆ’1)k k^2 which converges since the associated series of absolute values^

k^2 converges. Hence, the interval of convergence is [โˆ’ 14 , 14 ].

(b) Recall that 1 1 โˆ’ t = 1 + t + t^2 + t^3 + t^4 + ... for |t| < 1.

Use this to obtain a power series representation of arctan x. [What is the derivative of arctan x?]

By letting t = โˆ’x^2 , we have 1 1 + x^2 = 1^ โˆ’^ x

(^2) + x (^4) โˆ’ x (^6) + ... for |x| < 1.

It follows that

arctan x =

1 + x^2 dx = x โˆ’ x^3 3

x^5 5

x^7 7

n=

(โˆ’1)n+1x^2 nโˆ’^1 2 n โˆ’ 1 for^ |x|^ <^1.

(c) Use (b) to evaluate limxโ†’ 0 arctanx x.

Using the power series representation of arctan x from (b), we have

x^ limโ†’ 0

arctan x x = lim xโ†’ 0

x

x โˆ’ x^3 3

x^5 5

x^7 7

= lim xโ†’ 0

1 โˆ’ x

2 3

  • x

4 5 โˆ’ x

6 7

Date: April 4, 2008. 1