
# 1.2E: Exercises

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###### Exercise $$\PageIndex{1}$$

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 $$\displaystyle \frac{1}{2}(f(x)+g(x))$$ and of $$\displaystyle \frac{1}{2}(f(x)−g(x))$$.

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

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 $$\displaystyle C(x)+S(x)$$ and of $$\displaystyle C(x)−S(x)$$.

###### Exercise $$\PageIndex{2}$$

In the following exercises, use partial fractions to find the power series of each function.

1. $$\displaystyle \frac{4}{(x−3)(x+1)}$$

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

2. $$\displaystyle \frac{3}{(x+2)(x−1)}$$

3. $$\displaystyle \frac{5}{(x^2+4)(x^2−1)}$$

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

4. $$\displaystyle \frac{30}{(x^2+1)(x^2−9)}$$

###### Exercise $$\PageIndex{3}$$

In the following exercises, express each series as a rational function.

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

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

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

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

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

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

###### Exercise $$\PageIndex{4}$$

The following exercises explore applications of annuities.

1. 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 $$\displaystyle r=0.03,r=0.05$$, and $$\displaystyle r=0.07$$. Answer $$\displaystyle P=P_1+⋯+P_{20}$$ where $$\displaystyle P_k=10,000\frac{1}{(1+r)^k}$$. Then $$\displaystyle P=10,000\sum_{k=1}^{20}\frac{1}{(1+r)^k}=10,000\frac{1−(1+r)^{−20}}{r}$$. When $$\displaystyle r=0.03,P≈10,000×14.8775=148,775.$$ When $$\displaystyle r=0.05,P≈10,000×12.4622=124,622.$$ When $$\displaystyle r=0.07,P≈105,940$$. 2. Calculate the present values P of annuities in which$9,000 is to be paid out annually perpetually, assuming interest rates of $$\displaystyle r=0.03,r=0.05$$ and $$\displaystyle r=0.07$$.

3. Calculate the annual payouts C to be given for 20 years on annuities having present value $100,000 assuming respective interest rates of $$\displaystyle r=0.03,r=0.05,$$ and $$\displaystyle r=0.07.$$ Answer In general, $$\displaystyle P=\frac{C(1−(1+r)^{−N})}{r}$$ for N years of payouts, or $$\displaystyle C=\frac{Pr}{1−(1+r)^{−N}}$$. For $$\displaystyle N=20$$ and $$\displaystyle P=100,000$$, one has $$\displaystyle C=6721.57$$ when $$\displaystyle r=0.03;C=8024.26$$ when $$\displaystyle r=0.05$$; and $$\displaystyle C≈9439.29$$ when $$\displaystyle r=0.07$$. 4. Calculate the annual payouts C to be given perpetually on annuities having present value$100,000 assuming respective interest rates of $$\displaystyle r=0.03,r=0.05,$$ and $$\displaystyle r=0.07$$.

5. Suppose that an annuity has a present value $$\displaystyle P=1$$ million dollars. What interest rate r would allow for perpetual annual payouts of $50,000? Answer In general, $$\displaystyle P=\frac{C}{r}.$$ Thus, $$\displaystyle r=\frac{C}{P}=5×\frac{10^4}{10^6}=0.05.$$ 6. Suppose that an annuity has a present value $$\displaystyle P=10$$ million dollars. What interest rate r would allow for perpetual annual payouts of$100,000?

###### Exercise $$\PageIndex{5}$$

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

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

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

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

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

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

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

###### Exercise $$\PageIndex{6}$$

In the following exercises, find the power series of $$\displaystyle f(x)g(x)$$ given f and g as defined.

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

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

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

3. $$\displaystyle f(x)=g(x)=\sum_{n=1}^∞(\frac{x}{2})^n$$

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

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

###### Exercise $$\PageIndex{7}$$

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

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

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

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

In the following exercises, integrate the given series expansion of $$\displaystyle f$$ term-by-term from zero to x to obtain the corresponding series expansion for the indefinite integral of $$\displaystyle f$$.

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

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

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

###### Exercise $$\PageIndex{8}$$

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

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

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

2. Evaluate $$\displaystyle \sum_{n=1}^∞\frac{n}{3^n}$$ as $$\displaystyle f′(\frac{1}{3})$$ where $$\displaystyle f(x)=\sum_{n=0}^∞x6n$$.

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

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

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

###### Exercise $$\PageIndex{9}$$

In the following exercises, 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.

1. $$\displaystyle f(x)=lnx$$ centered at $$\displaystyle x=1$$ (Hint: $$\displaystyle x=1−(1−x)$$)

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

2. $$\displaystyle ln(1−x)$$ at $$\displaystyle x=0$$

3. $$\displaystyle ln(1−x^2)$$ at $$\displaystyle x=0$$

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

4. $$\displaystyle f(x)=\frac{2x}{(1−x^2)^2}$$ at $$\displaystyle x=0$$

5. $$\displaystyle f(x)=tan^{−1}(x^2)$$ at $$\displaystyle x=0$$

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

6. $$\displaystyle f(x)=ln(1+x^2)$$ at $$\displaystyle x=0$$

7. $$\displaystyle f(x)=∫^x_0lntdt$$ where $$\displaystyle ln(x)=\sum_{n=1}^∞(−1)^{n−1}\frac{(x−1)^n}{n}$$

Term-by-term integration gives $$\displaystyle ∫^x_0lntdt=\sum_{n=1}^∞(−1)^{n−1}\frac{(x−1)^{n+1}}{n(n+1)}=\sum_{n=1}^∞(−1)^{n−1}(\frac{1}{n}−\frac{1}{n+1})(x−1)^{n+1}=(x−1)lnx+\sum_{n=2}^∞(−1)^n\frac{(x−1)^n}{n}=xlnx−x.$$

###### Exercise $$\PageIndex{10}$$

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

1. $$\displaystyle \sum_{k=0}^∞(x^k−x^{2k+1})$$

2. $$\displaystyle \sum_{k=1}^∞\frac{x^{3k}}{6k}$$

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

3. $$\displaystyle \sum_{k=1}^∞(1+x^2)^{−k}$$ using $$\displaystyle y=\frac{1}{1+x^2}$$

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

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

###### Exercise $$\PageIndex{11}$$

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

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

3. Show that if $$\displaystyle f(x)=\sum_{n=0}^∞a_nx^n$$ is a sum of even powers, that is, $$\displaystyle a_n=0$$ if $$\displaystyle n$$ 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.

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

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

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

6. 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 $$\displaystyle x=−1/2$$ are needed to approximate $$\displaystyle ln(2)$$ accurate to within 1/1000. Evaluate the corresponding partial sum $$\displaystyle \sum_{n=1}^N(−1)^{n−1}\frac{x^n}{n}$$.

When $$\displaystyle x=−\frac{1}{2},−ln(2)=ln(\frac{1}{2})=−\sum_{n=1}^∞\frac{1}{n2^n}$$. 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 $$\displaystyle ln(2)=0.69314…;$$ therefore, $$\displaystyle N=10.$$
7. 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 $$\displaystyle x=1$$ are needed to approximate $$\displaystyle tan^{−1}(1)=\frac{π}{4}$$ accurate to within 1/1000. Evaluate the corresponding partial sum $$\displaystyle \sum_{k=0}^N(−1)^k\frac{x^{2k+1}}{2k+1}$$.
8. Recall that $$\displaystyle tan^{−1}(\frac{1}{\sqrt{3}})=\frac{π}{6}.$$ Assuming an exact value of $$\displaystyle \frac{1}{\sqrt{3}})$$, estimate $$\displaystyle \frac{π}{6}$$ by evaluating partial sums $$\displaystyle S_N(\frac{1}{\sqrt{3}})$$ of the power series expansion $$\displaystyle tan^{−1}(x)=\sum_{k=0}^∞(−1)^k\frac{x^{2k+1}}{2k+1}$$ at $$\displaystyle x=\frac{1}{\sqrt{3}}$$. What is the smallest number $$\displaystyle N$$ such that $$\displaystyle 6S_N(\frac{1}{\sqrt{3}})$$ approximates $$\displaystyle π$$ accurately to within 0.001? How many terms are needed for accuracy to within 0.00001?
$$\displaystyle 6S_N(\frac{1}{\sqrt{3}})=2\sqrt{3}\sum_{n=0}^N(−1)^n\frac{1}{3^n(2n+1).}$$ One has $$\displaystyle π−6S_4(\frac{1}{\sqrt{3}})=0.00101…$$ and $$\displaystyle π−6S_5(\frac{1}{\sqrt{3}})=0.00028…$$ so $$\displaystyle N=5$$ is the smallest partial sum with accuracy to within 0.001. Also, $$\displaystyle π−6S_7(\frac{1}{\sqrt{3}})=0.00002…$$ while $$\displaystyle π−6S_8(\frac{1}{\sqrt{3}})=−0.000007…$$ so $$\displaystyle N=8$$ is the smallest N to give accuracy to within 0.00001.