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Mathematics LibreTexts

10.1E: Exercises for Section 10.1

  • Gilbert Strang & Edwin “Jed” Herman
  • OpenStax

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In exercises 1 - 4, state whether each statement is true, or give an example to show that it is false.

1) If n=1anxn converges, then anxn0 as n.

Answer
True. If a series converges then its terms tend to zero.

2) n=1anxn converges at x=0 for any real numbers an.

3) Given any sequence an, there is always some R>0, possibly very small, such that n=1anxn converges on (R,R).

Answer
False. It would imply that anxn0 for |x|<R. If an=nn, then anxn=(nx)n does not tend to zero for any x0.

4) If n=1anxn has radius of convergence R>0 and if |bn||an| for all n, then the radius of convergence of n=1bnxn is greater than or equal to R.

5) Suppose that n=0an(x3)n converges at x=6. At which of the following points must the series also converge? Use the fact that if an(xc)n converges at x, then it converges at any point closer to c than x.

a. x=1

b. x=2

c. x=3

d. x=0

e. x=5.99

f. x=0.000001

Answer
It must converge on (0,6] and hence at: a. x=1; b. x=2; c. x=3; d. x=0; e. x=5.99; and f. x=0.000001.

6) Suppose that n=0an(x+1)n converges at x=2. At which of the following points must the series also converge? Use the fact that if an(xc)n converges at x, then it converges at any point closer to c than x.

a. x=2

b. x=1

c. x=3

d. x=0

e. x=0.99

f. x=0.000001

In the following exercises, suppose that |an+1an|1 as n. Find the radius of convergence for each series.

7) n=0an2nxn

Answer
|an+12n+1xn+1an2nxn|=2|x||an+1an|2|x| so R=12

8) n=0anxn2n

9) n=0anπnxnen

Answer
|an+1(πe)n+1xn+1an(πe)nxn|=π|x|e|an+1an|π|x|e so R=eπ

10) n=0an(1)nxn10n

11) n=0an(1)nx2n

Answer
|an+1(1)n+1x2n+2an(1)nx2n|=|x2||an+1an||x2| so R=1

12) n=0an(4)nx2n

In exercises 13 - 22, find the radius of convergence R and interval of convergence for anxn with the given coefficients an.

13) n=1(2x)nn

Answer
an=2nn so an+1xan2x. so R=12. When x=12 the series is harmonic and diverges. When x=12 the series is alternating harmonic and converges. The interval of convergence is I=[12,12).

14) n=1(1)nxnn

15) n=1nxn2n

Answer
an=n2n so an+1xanx2 so R=2. When x=±2 the series diverges by the divergence test. The interval of convergence is I=(2,2).

16) n=1nxnen

17) n=1n2xn2n

Answer
an=n22n so R=2. When x=±2 the series diverges by the divergence test. The interval of convergence is I=(2,2).

18) k=1kexkek

19) k=1πkxkkπ

Answer
ak=πkkπ so R=1π. When x=±1π the series is an absolutely convergent p-series. The interval of convergence is I=[1π,1π].

20) n=1xnn!

21) n=110nxnn!

Answer
an=10nn!,an+1xan=10xn+10<1 so the series converges for all x by the ratio test and I=(,).

22) n=1(1)nxnln(2n)

In exercises 23 - 28, find the radius of convergence of each series.

23) k=1(k!)2xk(2k)!

Answer
ak=(k!)2(2k)! so ak+1ak=(k+1)2(2k+2)(2k+1)14 so R=4

24) n=1(2n)!xnn2n

25) k=1k!135(2k1)xk

Answer
ak=k!135(2k1) so ak+1ak=k+12k+112 so R=2

26) k=12462k(2k)!xk

27) n=1xn(2nn) where (nk)=n!k!(nk)!

Answer
an=1(2nn) so an+1an=((n+1)!)2(2n+2)!2n!(n!)2=(n+1)2(2n+2)(2n+1)14 so R=4

28) n=1sin2nxn

In exercises 29 - 32, use the ratio test to determine the radius of convergence of each series.

29) n=1(n!)3(3n)!xn

Answer
an+1an=(n+1)3(3n+3)(3n+2)(3n+1)127 so R=27

30) n=123n(n!)3(3n)!xn

31) n=1n!nnxn

Answer
an=n!nn so an+1an=(n+1)!n!nn(n+1)n+1=(nn+1)n1e so R=e

32) n=1(2n)!n2nxn

In the following exercises, given that 11x=n=0xn with convergence in (1,1), find the power series for each function with the given center a, and identify its interval of convergence.

33) f(x)=1x;a=1 (Hint: 1x=11(1x))

Answer
f(x)=n=0(1x)n on I=(0,2)

34) f(x)=11x2;a=0

35) f(x)=x1x2;a=0

Answer
n=0x2n+1 on I=(1,1)

36) f(x)=11+x2;a=0

37) f(x)=x21+x2;a=0

Answer
n=0(1)nx2n+2 on I=(1,1)

38) f(x)=12x;a=1

39) f(x)=112x;a=0.

Answer
n=02nxn on (12,12)

40) f(x)=114x2;a=0

41) f(x)=x214x2;a=0

Answer
n=04nx2n+2 on (12,12)

42) f(x)=x254x+x2;a=2

Use the result of exercise 43 to find the radius of convergence of the given series in the subsequent exercises (44 - 47).

43) Explain why, if |an|1/nr>0, then |anxn|1/n|x|r<1 whenever |x|<1r and, therefore, the radius of convergence of n=1anxn is R=1r.

Answer
|anxn|1/n=|an|1/n|x||x|r as n and |x|r<1 when |x|<1r. Therefore, n=1anxn converges when |x|<1r by the nth root test.

44) n=1xnnn

45) k=1(k12k+3)kxk

Answer
ak=(k12k+3)k so (ak)1/k12<1 so R=2

46) k=1(2k21k2+3)kxk

47) n=1an=(n1/n1)nxn

Answer
an=(n1/n1)n so (an)1/n0 so R=

48) Suppose that p(x)=n=0anxn such that an=0 if n is even. Explain why p(x)=p(x).

49) Suppose that p(x)=n=0anxn such that an=0 if n is odd. Explain why p(x)=p(x).

Answer
We can rewrite p(x)=n=0a2n+1x2n+1 and p(x)=p(x) since x2n+1=(x)2n+1.

50) Suppose that p(x)=n=0anxn converges on (1,1]. Find the interval of convergence of p(Ax).

51) Suppose that p(x)=n=0anxn converges on (1,1]. Find the interval of convergence of p(2x1).

Answer
If x[0,1], then y=2x1[1,1] so p(2x1)=p(y)=n=0anyn converges.

In the following exercises, suppose that p(x)=n=0anxn satisfies limnan+1an=1 where an0 for each n. State whether each series converges on the full interval (1,1), or if there is not enough information to draw a conclusion. Use the comparison test when appropriate.

52) n=0anx2n

53) n=0a2nx2n

Answer
Converges on (1,1) by the ratio test

54) n=0a2nxn (Hint:x=±x2)

55) n=0an2xn2 (Hint: Let bk=ak if k=n2 for some n, otherwise bk=0.)

Answer
Consider the series bkxk where bk=ak if k=n2 and bk=0 otherwise. Then bkak and so the series converges on (1,1) by the comparison test.

56) Suppose that p(x) is a polynomial of degree N. Find the radius and interval of convergence of n=1p(n)xn.

57) [T] Plot the graphs of 11x and of the partial sums SN=Nn=0xn for n=10,20,30 on the interval [0.99,0.99]. Comment on the approximation of 11x by SN near x=1 and near x=1 as N increases.

Answer

The approximation is more accurate near x=1. The partial sums follow 11x more closely as N increases but are never accurate near x=1 since the series diverges there.

This figure is the graph of y = 1/(1-x), which is an increasing curve with vertical asymptote at 1.

58) [T] Plot the graphs of ln(1x) and of the partial sums SN=Nn=1xnn for n=10,50,100 on the interval [0.99,0.99]. Comment on the behavior of the sums near x=1 and near x=1 as N increases.

59) [T] Plot the graphs of the partial sums Sn=Nn=1xnn2 for n=10,50,100 on the interval [0.99,0.99]. Comment on the behavior of the sums near x=1 and near x=1 as N increases.

Answer

The approximation appears to stabilize quickly near both x=±1.

This figure is the graph of y = -ln(1-x) which is an increasing curve passing through the origin.

60) [T] Plot the graphs of the partial sums SN=Nn=1(sinn)xn for n=10,50,100 on the interval [0.99,0.99]. Comment on the behavior of the sums near x=1 and near x=1 as N increases.

61) [T] Plot the graphs of the partial sums SN=Nn=0(1)nx2n+1(2n+1)! for n=3,5,10 on the interval [2π,2π]. Comment on how these plots approximate sinx as N increases.

Answer

The polynomial curves have roots close to those of sinx up to their degree and then the polynomials diverge from sinx.

This figure is the graph of the partial sums of (-1)^n times x^(2n+1) divided by (2n+1)! For n=3,5,10. The curves approximate the sine curve close to the origin and then separate as the curves move away from the origin.

62) [T] Plot the graphs of the partial sums SN=Nn=0(1)nx2n(2n)! for n=3,5,10 on the interval [2π,2π]. Comment on how these plots approximate cosx as N increases.

Contributors

Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.


This page titled 10.1E: Exercises for Section 10.1 is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Gilbert Strang & Edwin “Jed” Herman (OpenStax) via source content that was edited to the style and standards of the LibreTexts platform.

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