10.2E: Exercises for Section 10.2

1) If $\displaystyle f(x)=\sum_{n=0}^∞\frac{x^n}{n!}$ and $\displaystyle g(x)=\sum_{n=0}^∞(−1)^n\frac{x^n}{n!}$, find the power series of $\frac{1}{2}\big(f(x)+g(x)\big)$ and of $\frac{1}{2}\big(f(x)−g(x)\big)$.

Answer

$\displaystyle \frac{1}{2}\big(f(x)+g(x)\big)=\sum_{n=0}^∞\frac{x^{2n}}{(2n)!}$ and $\displaystyle \frac{1}{2}\big(f(x)−g(x)\big)=\sum_{n=0}^∞\frac{x^{2n+1}}{(2n+1)!}$.

2) If $\displaystyle C(x)=\sum_{n=0}^∞\frac{x^{2n}}{(2n)!}$ and $\displaystyle S(x)=\sum_{n=0}^∞\frac{x^{2n+1}}{(2n+1)!}$, 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) $\dfrac{4}{(x−3)(x+1)}$

Answer

$\displaystyle \frac{4}{(x−3)(x+1)}=\frac{1}{x−3}−\frac{1}{x+1}=−\frac{1}{3(1−\frac{x}{3})}−\frac{1}{1−(−x)}=−\frac{1}{3}\sum_{n=0}^∞\left(\frac{x}{3}\right)^n−\sum_{n=0}^∞(−1)^nx^n=\sum_{n=0}^∞\left((−1)^{n+1}−\frac{1}{3^{n+1}}\right)x^n$

4) $\dfrac{3}{(x+2)(x−1)}$

5) $\dfrac{5}{(x^2+4)(x^2−1)}$

Answer

$\displaystyle \frac{5}{(x^2+4)(x^2−1)}=\frac{1}{x^2−1}−\frac{1}{4}\frac{1}{1+\left(\frac{x}{2}\right)^2}=−\sum_{n=0}^∞x^{2n}−\frac{1}{4}\sum_{n=0}^∞(−1)^n\left(\frac{x}{2}\right)^n=\sum_{n=0}^∞\left((−1)+(−1)^{n+1}\frac{1}{2^{n+2}}\right)x^{2n}$

6) $\dfrac{30}{(x^2+1)(x^2−9)}$

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

7) $\displaystyle \sum_{n=1}^∞\frac{1}{x^n}$

Answer

$\displaystyle \frac{1}{x}\sum_{n=0}^∞\frac{1}{x^n}=\frac{1}{x}\cdot \frac{1}{1−\frac{1}{x}}=\frac{1}{x−1}$

8) $\displaystyle \sum_{n=1}^∞\frac{1}{x^{2n}}$

9) $\displaystyle \sum_{n=1}^∞\frac{1}{(x−3)^{2n−1}}$

Answer

$\displaystyle \frac{1}{x−3}\cdot \frac{1}{1−\frac{1}{(x−3)^2}}=\frac{x−3}{(x−3)^2−1}$

10) $\displaystyle \sum_{n=1}^∞\left(\frac{1}{(x−3)^{2n−1}}−\frac{1}{(x−2)^{2n−1}}\right)$

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

$P=P_1+⋯+P_{20}$ where $P_k=10,000\dfrac{1}{(1+r)^k}$. Then $\displaystyle P=10,000\sum_{k=1}^{20}\frac{1}{(1+r)^k}=10,000\frac{1−(1+r)^{−20}}{r}$. When $r=0.03, \,P≈10,000×14.8775=148,775.$ When $r=0.05, \,P≈10,000×12.4622=124,622.$ When $r=0.07, \, P≈105,940$.

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, $P=\dfrac{C(1−(1+r)^{−N})}{r}$ for $N$ years of payouts, or $C=\dfrac{Pr}{1−(1+r)^{−N}}$. For $N=20$ and $P=100,000$, one has $C=6721.57$ when $r=0.03; \, C=8024.26$ when $r=0.05$; and $C≈9439.29$ 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, $P=\dfrac{C}{r}.$ Thus, $r=\dfrac{C}{P}=5×\frac{10^4}{10^6}=0.05.$

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) $x+x^2−x^3+x^4+x^5−x^6+⋯$ (Hint: Group powers $x^{3k}, \, x^{3k−1},$ and $x^{3k−2}$.)

Answer

$(x+x^2−x^3)(1+x^3+x^6+⋯)=\dfrac{x+x^2−x^3}{1−x^3}$

18) $x+x^2−x^3−x^4+x^5+x^6−x^7−x^8+⋯$ (Hint: Group powers $x^{4k}, \, x^{4k−1},$ etc.)

19) $x−x^2−x^3+x^4−x^5−x^6+x^7−⋯$ (Hint: Group powers $x^{3k}, \, x^{3k−1}$, and $x^{3k−2}$.)

Answer

$(x−x^2−x^3)(1+x^3+x^6+⋯)=\dfrac{x−x^2−x^3}{1−x^3}$

20) $\displaystyle \frac{x}{2}+\frac{x^2}{4}−\frac{x^3}{8}+\frac{x^4}{16}+\frac{x^5}{32}−\frac{x^6}{64}+⋯$ (Hint: Group powers $\left(\dfrac{x}{2}\right)^{3k}, \, \left(\dfrac{x}{2}\right)^{3k−1},$ and $\left(\dfrac{x}{2}\right)^{3k−2}$.)

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

21) $\displaystyle f(x)=2\sum_{n=0}^∞x^n,g(x)=\sum_{n=0}^∞nx^n$

Answer

$a_n=2, \, b_n=n$ so $\displaystyle c_n=\sum_{k=0}^nb_ka_{n−k}=2\sum_{k=0}^nk=(n)(n+1)$ and $\displaystyle f(x)g(x)=\sum_{n=1}^∞n(n+1)x^n$

22) $\displaystyle f(x)=\sum_{n=1}^∞x^n,\; g(x)=\sum_{n=1}^∞\frac{1}{n}x^n$. Express the coefficients of $f(x)g(x)$ in terms of $\displaystyle H_n=\sum_{k=1}^n\frac{1}{k}$.

23) $\displaystyle f(x)=g(x)=\sum_{n=1}^∞\left(\frac{x}{2}\right)^n$

Answer

$a_n=b_n=2^{−n}$ so $\displaystyle c_n=\sum_{k=1}^nb_ka_{n−k}=2^{−n}\sum_{k=1}^n1=\frac{n}{2^n}$ and $\displaystyle f(x)g(x)=\sum_{n=1}^∞n\left(\frac{x}{2}\right)^n$

24) $\displaystyle f(x)=g(x)=\sum_{n=1}^∞nx^n$

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) $\displaystyle f(x)=\frac{1}{1+x}=\sum_{n=0}^∞(−1)^nx^n$

Answer

The derivative of $f$ is $\displaystyle −\frac{1}{(1+x)^2}=−\sum_{n=0}^∞(−1)^n(n+1)x^n$.

26) $\displaystyle f(x)=\frac{1}{1−x^2}=\sum_{n=0}^∞x^{2n}$

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) $\displaystyle f(x)=\frac{2x}{(1+x^2)^2}=\sum_{n=1}^∞(−1)^n(2n)x^{2n−1}$

Answer

The indefinite integral of $f$ is $\displaystyle \frac{1}{1+x^2}=\sum_{n=0}^∞(−1)^nx^{2n}$.

28) $\displaystyle f(x)=\frac{2x}{1+x^2}=2\sum_{n=0}^∞(−1)^nx^{2n+1}$

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

29) Evaluate $\displaystyle \sum_{n=1}^∞\frac{n}{2^n}$ as $f′\left(\frac{1}{2}\right)$ where $\displaystyle f(x)=\sum_{n=0}^∞x^n$.

Answer

$\displaystyle f(x)=\sum_{n=0}^∞x^n=\frac{1}{1−x}; \; f′(\frac{1}{2})=\sum_{n=1}^∞\frac{n}{2^{n−1}}=\frac{d}{dx}(1−x)^{−1}\Big|_{x=1/2}=\frac{1}{(1−x)^2}\Big|_{x=1/2}=4$ so $\displaystyle \sum_{n=1}^∞\frac{n}{2^n}=2.$

30) Evaluate $\displaystyle \sum_{n=1}^∞\frac{n}{3^n}$ as $f′\left(\frac{1}{3}\right)$ where $\displaystyle f(x)=\sum_{n=0}^∞x^n$.

31) Evaluate $\displaystyle \sum_{n=2}^∞\frac{n(n−1)}{2^n}$ as $f''\left(\frac{1}{2}\right)$ where $\displaystyle f(x)=\sum_{n=0}^∞x^n$.

Answer

$\displaystyle f(x)=\sum_{n=0}^∞x^n=\frac{1}{1−x}; \; f''\left(\frac{1}{2}\right)=\sum_{n=2}^∞\frac{n(n−1)}{2^{n−2}}=\frac{d^2}{dx^2}(1−x)^{−1}\Big|_{x=1/2}=\frac{2}{(1−x)^3}\Big|_{x=1/2}=16$ so $\displaystyle \sum_{n=2}^∞n\frac{(n−1)}{2^n}=4.$

32) Evaluate $\displaystyle \sum_{n=0}^∞\frac{(−1)^n}{n+1}$ as $\displaystyle ∫^1_0f(t) \, dt$ where $\displaystyle f(x)=\sum_{n=0}^∞(−1)^nx^{2n}=\frac{1}{1+x^2}$.

In exercises 33 - 39, given that $\displaystyle \frac{1}{1−x}=\sum_{n=0}^∞x^n$, use term-by-term differentiation or integration to find power series for each function centered at the given point.

33) $f(x)=\ln x$ centered at $x=1$ (Hint: $x=1−(1−x)$)

Answer

$\displaystyle ∫\sum(1−x)^n\,dx=∫\sum(−1)^n(x−1)^n\,dx=\sum \frac{(−1)^n(x−1)^{n+1}}{n+1}$

34) $\ln(1−x)$ at $x=0$

35) $\ln(1−x^2)$ at $x=0$

Answer

$\displaystyle −∫^{x^2}_{t=0}\frac{1}{1−t}dt=−\sum_{n=0}^∞∫^{x^2}_0t^ndx−\sum_{n=0}^∞\frac{x^{2(n+1)}}{n+1}=−\sum_{n=1}^∞\frac{x^{2n}}{n}$

36) $f(x)=\dfrac{2x}{(1−x^2)^2}$ at $x=0$

37) $f(x)=\tan^{−1}(x^2)$ at $x=0$

Answer

$\displaystyle ∫^{x^2}_0\frac{dt}{1+t^2}=\sum_{n=0}^∞(−1)^n∫^{x^2}_0t^{2n}dt=\sum_{n=0}^∞(−1)^n\frac{t^{2n+1}}{2n+1}∣^{x^2}_{t=0}=\sum_{n=0}^∞(−1)^n\frac{x^{4n+2}}{2n+1}$

38) $f(x)=\ln(1+x^2)$ at $x=0$

39) $\displaystyle f(x)=∫^x_0\ln t\,dt$ where $\displaystyle \ln(x)=\sum_{n=1}^∞(−1)^{n−1}\frac{(x−1)^n}{n}$

Answer

Term-by-term integration gives $\displaystyle ∫^x_0\ln t\,dt=\sum_{n=1}^∞(−1)^{n−1}\frac{(x−1)^{n+1}}{n(n+1)}=\sum_{n=1}^∞(−1)^{n−1}\left(\frac{1}{n}−\frac{1}{n+1}\right)(x−1)^{n+1}=(x−1)\ln x+\sum_{n=2}^∞(−1)^n\frac{(x−1)^n}{n}=x\ln x−x.$

40) [T] Evaluate the power series expansion $\displaystyle \ln(1+x)=\sum_{n=1}^∞(−1)^{n−1}\frac{x^n}{n}$ 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 $\ln(1−x)$ from $\ln(1+x)$ to get a power series for $\ln\left(\dfrac{1+x}{1−x}\right)$. Evaluate at $x=\frac{1}{3}$. What is the smallest $N$ such that the $N^{\text{th}}$ partial sum of this series approximates $\ln(2)$ with an error less than $0.001$?

Answer

We have $\displaystyle \ln(1−x)=−\sum_{n=1}^∞\frac{x^n}{n}$ so $\displaystyle \ln(1+x)=\sum_{n=1}^∞(−1)^{n−1}\frac{x^n}{n}$. Thus, $\displaystyle \ln\left(\frac{1+x}{1−x}\right)=\sum_{n=1}^∞\big(1+(−1)^{n−1}\big)\frac{x^n}{n}=2\sum_{n=1}^∞\frac{x^{2n−1}}{2n−1}$. When $x=\frac{1}{3}$ we obtain $\displaystyle \ln(2)=2\sum_{n=1}^∞\frac{1}{3^{2n−1}(2n−1)}$. We have $\displaystyle 2\sum_{n=1}^3\frac{1}{3^{2n−1}(2n−1)}=0.69300…$, while $\displaystyle 2\sum_{n=1}^4\frac{1}{3^{2n−1}(2n−1)}=0.69313…$ 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) $\displaystyle \sum_{k=0}^∞(x^k−x^{2k+1})$

43) $\displaystyle \sum_{k=1}^∞\frac{x^{3k}}{6k}$

Answer

$\displaystyle \sum_{k=1}^∞\frac{x^k}{k}=−\ln(1−x)$ so $\displaystyle \sum_{k=1}^∞\frac{x^{3k}}{6k}=−\frac{1}{6}\ln(1−x^3)$. The radius of convergence is equal to $1$ by the ratio test.

44) $\displaystyle \sum_{k=1}^∞(1+x^2)^{−k}$ using $y=\dfrac{1}{1+x^2}$

45) $\displaystyle \sum_{k=1}^∞2^{−kx}$ using $y=2^{−x}$

Answer

If $y=2^{−x}$, then $\displaystyle \sum_{k=1}^∞y^k=\frac{y}{1−y}=\frac{2^{−x}}{1−2^{−x}}=\frac{1}{2^x−1}$. If $a_k=2^{−kx}$, then $\dfrac{a_{k+1}}{a_k}=2^{−x}<1$ when $x>0$. So the series converges for all $x>0$.

46) Show that, up to powers $x^3$ and $y^3$, $\displaystyle E(x)=\sum_{n=0}^∞\frac{x^n}{n!}$ satisfies $E(x+y)=E(x)E(y)$.

47) Differentiate the series $\displaystyle E(x)=\sum_{n=0}^∞\frac{x^n}{n!}$ term-by-term to show that $E(x)$ is equal to its derivative.

Answer

Answers will vary.

48) Show that if $\displaystyle f(x)=\sum_{n=0}^∞a_nx^n$ is a sum of even powers, that is, $a_n=0$ if $n$ is odd, then $\displaystyle F=∫^x_0f(t)\, dt$ 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 $\displaystyle \sum_{n=0}^∞a_nx^n$ are defined by the recurrence relation $a_n=\dfrac{a_{n−1}}{n}+\dfrac{a_{n−2}}{n(n−1)}$. For $a_0=0$ and $a_1=1$, compute and plot the sums $\displaystyle S_N=\sum_{n=0}^Na_nx^n$ for $N=2,3,4,5$ on $[−1,1].$

Answer

The solid curve is $S_5$. The dashed curve is $S_2$, dotted is $S_3$, and dash-dotted is $S_4$

50) [T] Suppose that the coefficients an of the series $\displaystyle \sum_{n=0}^∞a_nx^n$ are defined by the recurrence relation $a_n=\dfrac{a_{n−1}}{\sqrt{n}}−\dfrac{a_{n−2}}{\sqrt{n(n−1)}}$. For $a_0=1$ and $a_1=0$, compute and plot the sums $\displaystyle S_N=\sum_{n=0}^Na_nx^n$ for $N=2,3,4,5$ on $[−1,1]$.

51) [T] Given the power series expansion $\displaystyle \ln(1+x)=\sum_{n=1}^∞(−1)^{n−1}\frac{x^n}{n}$, 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 $\displaystyle \sum_{n=1}^N(−1)^{n−1}\frac{x^n}{n}$.

Answer

When $\displaystyle x=−\frac{1}{2}, \;−\ln(2)=\ln\left(\frac{1}{2}\right)=−\sum_{n=1}^∞\frac{1}{n2^n}$. Since $\displaystyle \sum^∞_{n=11}\frac{1}{n2^n}<\sum_{n=11}^∞\frac{1}{2^n}=\frac{1}{2^{10}},$ one has $\displaystyle \sum_{n=1}^{10}\frac{1}{n2^n}=0.69306…$ whereas $\ln(2)=0.69314…;$ therefore, $N=10.$

52) [T] Given the power series expansion $\displaystyle \tan^{−1}(x)=\sum_{k=0}^∞(−1)^k\frac{x^{2k+1}}{2k+1}$, 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.