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Mathematics LibreTexts

10.2E: Exercises for Section 10.2

  • Gilbert Strang & Edwin “Jed” Herman
  • OpenStax

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1) If f(x)=n=0xnn! and g(x)=n=0(1)nxnn!, find the power series of 12(f(x)+g(x)) and of 12(f(x)g(x)).

Answer
12(f(x)+g(x))=n=0x2n(2n)! and 12(f(x)g(x))=n=0x2n+1(2n+1)!.

2) If C(x)=n=0x2n(2n)! and S(x)=n=0x2n+1(2n+1)!, find the power series of C(x)+S(x) and of C(x)S(x).

In exercises 3 - 6, use partial fractions to find the power series of each function.

3) 4(x3)(x+1)

Answer
4(x3)(x+1)=1x31x+1=13(1x3)11(x)=13n=0(x3)nn=0(1)nxn=n=0((1)n+113n+1)xn

4) 3(x+2)(x1)

5) 5(x2+4)(x21)

Answer
5(x2+4)(x21)=1x211411+(x2)2=n=0x2n14n=0(1)n(x2)n=n=0((1)+(1)n+112n+2)x2n

6) 30(x2+1)(x29)

In exercises 7 - 10, express each series as a rational function.

7) n=11xn

Answer
1xn=01xn=1x111x=1x1

8) n=11x2n

9) n=11(x3)2n1

Answer
1x3111(x3)2=x3(x3)21

10) n=1(1(x3)2n11(x2)2n1)

Exercises 11 - 16 explore applications of annuities.

11) Calculate the present values P of an annuity in which $10,000 is to be paid out annually for a period of 20 years, assuming interest rates of r=0.03,r=0.05, and r=0.07.

Answer
P=P1++P20 where Pk=10,0001(1+r)k. Then P=10,00020k=11(1+r)k=10,0001(1+r)20r. When r=0.03,P10,000×14.8775=148,775. When r=0.05,P10,000×12.4622=124,622. When r=0.07,P105,940.

12) Calculate the present values P of annuities in which $9,000 is to be paid out annually perpetually, assuming interest rates of r=0.03,r=0.05 and r=0.07.

13) Calculate the annual payouts C to be given for 20 years on annuities having present value $100,000 assuming respective interest rates of r=0.03,r=0.05, and r=0.07.

Answer
In general, P=C(1(1+r)N)r for N years of payouts, or C=Pr1(1+r)N. For N=20 and P=100,000, one has C=6721.57 when r=0.03;C=8024.26 when r=0.05; and C9439.29 when r=0.07.

14) Calculate the annual payouts C to be given perpetually on annuities having present value $100,000 assuming respective interest rates of r=0.03,r=0.05, and r=0.07.

15) Suppose that an annuity has a present value P=1 million dollars. What interest rate r would allow for perpetual annual payouts of $50,000?

Answer
In general, P=Cr. Thus, r=CP=5×104106=0.05.

16) Suppose that an annuity has a present value P=10 million dollars. What interest rate r would allow for perpetual annual payouts of $100,000?

In exercises 17 - 20, express the sum of each power series in terms of geometric series, and then express the sum as a rational function.

17) x+x2x3+x4+x5x6+ (Hint: Group powers x3k,x3k1, and x3k2.)

Answer
(x+x2x3)(1+x3+x6+)=x+x2x31x3

18) x+x2x3x4+x5+x6x7x8+ (Hint: Group powers x4k,x4k1, etc.)

19) xx2x3+x4x5x6+x7 (Hint: Group powers x3k,x3k1, and x3k2.)

Answer
(xx2x3)(1+x3+x6+)=xx2x31x3

20) x2+x24x38+x416+x532x664+ (Hint: Group powers (x2)3k,(x2)3k1, and (x2)3k2.)

In exercises 21 - 24, find the power series of f(x)g(x) given f and g as defined.

21) f(x)=2n=0xn,g(x)=n=0nxn

Answer
an=2,bn=n so cn=nk=0bkank=2nk=0k=(n)(n+1) and f(x)g(x)=n=1n(n+1)xn

22) f(x)=n=1xn,g(x)=n=11nxn. Express the coefficients of f(x)g(x) in terms of Hn=nk=11k.

23) f(x)=g(x)=n=1(x2)n

Answer
an=bn=2n so cn=nk=1bkank=2nnk=11=n2n and f(x)g(x)=n=1n(x2)n

24) f(x)=g(x)=n=1nxn

In exercises 25 - 26, differentiate the given series expansion of f term-by-term to obtain the corresponding series expansion for the derivative of f.

25) f(x)=11+x=n=0(1)nxn

Answer
The derivative of f is 1(1+x)2=n=0(1)n(n+1)xn.

26) f(x)=11x2=n=0x2n

In exercises 27 - 28, integrate the given series expansion of f term-by-term from zero to x to obtain the corresponding series expansion for the indefinite integral of f.

27) f(x)=2x(1+x2)2=n=1(1)n(2n)x2n1

Answer
The indefinite integral of f is 11+x2=n=0(1)nx2n.

28) f(x)=2x1+x2=2n=0(1)nx2n+1

In exercises 29 - 32, evaluate each infinite series by identifying it as the value of a derivative or integral of geometric series.

29) Evaluate n=1n2n as f(12) where f(x)=n=0xn.

Answer
f(x)=n=0xn=11x;f(12)=n=1n2n1=ddx(1x)1|x=1/2=1(1x)2|x=1/2=4 so n=1n2n=2.

30) Evaluate n=1n3n as f(13) where f(x)=n=0xn.

31) Evaluate n=2n(n1)2n as f(12) where f(x)=n=0xn.

Answer
f(x)=n=0xn=11x;f(12)=n=2n(n1)2n2=d2dx2(1x)1|x=1/2=2(1x)3|x=1/2=16 so n=2n(n1)2n=4.

32) Evaluate n=0(1)nn+1 as 10f(t)dt where f(x)=n=0(1)nx2n=11+x2.

In exercises 33 - 39, given that 11x=n=0xn, use term-by-term differentiation or integration to find power series for each function centered at the given point.

33) f(x)=lnx centered at x=1 (Hint: x=1(1x))

Answer
(1x)ndx=(1)n(x1)ndx=(1)n(x1)n+1n+1

34) ln(1x) at x=0

35) ln(1x2) at x=0

Answer
x2t=011tdt=n=0x20tndxn=0x2(n+1)n+1=n=1x2nn

36) f(x)=2x(1x2)2 at x=0

37) f(x)=tan1(x2) at x=0

Answer
x20dt1+t2=n=0(1)nx20t2ndt=n=0(1)nt2n+12n+1x2t=0=n=0(1)nx4n+22n+1

38) f(x)=ln(1+x2) at x=0

39) f(x)=x0lntdt where ln(x)=n=1(1)n1(x1)nn

Answer
Term-by-term integration gives x0lntdt=n=1(1)n1(x1)n+1n(n+1)=n=1(1)n1(1n1n+1)(x1)n+1=(x1)lnx+n=2(1)n(x1)nn=xlnxx.

40) [T] Evaluate the power series expansion ln(1+x)=n=1(1)n1xnn at x=1 to show that ln(2) is the sum of the alternating harmonic series. Use the alternating series test to determine how many terms of the sum are needed to estimate ln(2) accurate to within 0.001, and find such an approximation.

41) [T] Subtract the infinite series of ln(1x) from ln(1+x) to get a power series for ln(1+x1x). Evaluate at x=13. What is the smallest N such that the Nth partial sum of this series approximates ln(2) with an error less than 0.001?

Answer
We have ln(1x)=n=1xnn so ln(1+x)=n=1(1)n1xnn. Thus, ln(1+x1x)=n=1(1+(1)n1)xnn=2n=1x2n12n1. When x=13 we obtain ln(2)=2n=1132n1(2n1). We have 23n=1132n1(2n1)=0.69300, while 24n=1132n1(2n1)=0.69313 and ln(2)=0.69314; therefore, N=4.

In exercises 42 - 45, using a substitution if indicated, express each series in terms of elementary functions and find the radius of convergence of the sum.

42) k=0(xkx2k+1)

43) k=1x3k6k

Answer
k=1xkk=ln(1x) so k=1x3k6k=16ln(1x3). The radius of convergence is equal to 1 by the ratio test.

44) k=1(1+x2)k using y=11+x2

45) k=12kx using y=2x

Answer
If y=2x, then k=1yk=y1y=2x12x=12x1. If ak=2kx, then ak+1ak=2x<1 when x>0. So the series converges for all x>0.

46) Show that, up to powers x3 and y3, E(x)=n=0xnn! satisfies E(x+y)=E(x)E(y).

47) Differentiate the series E(x)=n=0xnn! term-by-term to show that E(x) is equal to its derivative.

Answer
Answers will vary.

48) Show that if f(x)=n=0anxn is a sum of even powers, that is, an=0 if n is odd, then F=x0f(t)dt is a sum of odd powers, while if I is a sum of odd powers, then F is a sum of even powers.

49) [T] Suppose that the coefficients an of the series n=0anxn are defined by the recurrence relation an=an1n+an2n(n1). For a0=0 and a1=1, compute and plot the sums SN=Nn=0anxn for N=2,3,4,5 on [1,1].

Answer

The solid curve is S5. The dashed curve is S2, dotted is S3, and dash-dotted is S4

This is a graph of three curves. They are all increasing and become very close as the curves approach x = 0. Then they separate as x moves away from 0.

50) [T] Suppose that the coefficients an of the series n=0anxn are defined by the recurrence relation an=an1nan2n(n1). For a0=1 and a1=0, compute and plot the sums SN=Nn=0anxn for N=2,3,4,5 on [1,1].

51) [T] Given the power series expansion ln(1+x)=n=1(1)n1xnn, determine how many terms N of the sum evaluated at x=1/2 are needed to approximate ln(2) accurate to within 1/1000. Evaluate the corresponding partial sum Nn=1(1)n1xnn.

Answer
When x=12,ln(2)=ln(12)=n=11n2n. Since n=111n2n<n=1112n=1210, one has 10n=11n2n=0.69306 whereas ln(2)=0.69314; therefore, N=10.

52) [T] Given the power series expansion tan1(x)=k=0(1)kx2k+12k+1, use the alternating series test to determine how many terms N of the sum evaluated at x=1 are needed to approximate \tan^{−1}(1)=\frac{π}{4} accurate to within 1/1000. Evaluate the corresponding partial sum \displaystyle \sum_{k=0}^N(−1)^k\frac{x^{2k+1}}{2k+1}.

53) [T] Recall that \tan^{−1}\left(\frac{1}{\sqrt{3}}\right)=\frac{π}{6}. Assuming an exact value of \frac{1}{\sqrt{3}}), estimate \frac{π}{6} by evaluating partial sums S_N\left(\frac{1}{\sqrt{3}}\right) of the power series expansion \displaystyle \tan^{−1}(x)=\sum_{k=0}^∞(−1)^k\frac{x^{2k+1}}{2k+1} at x=\frac{1}{\sqrt{3}}. What is the smallest number N such that 6S_N\left(\frac{1}{\sqrt{3}}\right) approximates π accurately to within 0.001? How many terms are needed for accuracy to within 0.00001?

Answer
\displaystyle 6S_N\left(\frac{1}{\sqrt{3}}\right)=2\sqrt{3}\sum_{n=0}^N(−1)^n\frac{1}{3^n(2n+1).} One has π−6S_4\left(\frac{1}{\sqrt{3}}\right)=0.00101… and π−6S_5\left(\frac{1}{\sqrt{3}}\right)=0.00028… so N=5 is the smallest partial sum with accuracy to within 0.001. Also, π−6S_7\left(\frac{1}{\sqrt{3}}\right)=0.00002… while π−6S_8\left(\frac{1}{\sqrt{3}}\right)=−0.000007… so N=8 is the smallest N to give accuracy to within 0.00001.

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 10.2E: Exercises for Section 10.2 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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10.2: Properties of Power Series
10.3: Taylor and Maclaurin Series