Content check — Calculus Vol. 2, Chapter 6: Power Series

Every problem in the chapter beside the book’s answer, a chapter at a time as you scroll.

Chapter 6: Power Series

161 problems · 12 figures

Checkpoint 6.1 §6.1 source ↗

Find the interval and radius of convergence for the series

\sum_{n = 1}^{\infty} \frac{x^{n}}{\sqrt{n}} .

Answer

The interval of convergence is

[−1, 1) .

The radius of convergence is

R = 1 .

Checkpoint 6.2 §6.1 source ↗

Sketch a graph of

f (x) = \frac{1}{1 - x^{2}}

and the corresponding partial sums

S_{N} (x) =

\sum_{n = 0}^{N} x^{2 n}

for

N = 2, 4, 6

on the interval

(−1, 1) .

Answer

This figure is the graph of y = 1/(1-x^2), which is a curve concave up, symmetrical about the y axis. Also on this graph are three partial sums of the function, S sub 6, S sub 4, and S sub 2. These curves, in order, gradually become flatter.

Checkpoint 6.3 §6.1 source ↗

Represent the function

f (x) = \frac{x^{3}}{2 - x}

using a power series and find the interval of convergence.

Answer

\sum_{n = 0}^{\infty} \frac{x^{n + 3}}{2^{n + 1}}

with interval of convergence

(−2, 2)

6-1 §6.1 source ↗

In the following exercises, state whether each statement is true, or give an example to show that it is false.

If $\sum_{n = 1}^{\infty} a_{n} x^{n}$ converges, then $a_{n} x^{n} \to 0$ as $n \to \infty .$

Answer

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

6-3 §6.1 source ↗

In the following exercises, state whether each statement is true, or give an example to show that it is false.

Given any sequence $a_{n},$ there is always some $R > 0,$ possibly very small, such that $\sum_{n = 1}^{\infty} a_{n} x^{n}$ converges on $(− R, R) .$

Answer

False. It would imply that

a_{n} x^{n} \to 0

for

| x | < R .

a_{n} = n^{n},

then

a_{n} x^{n} = {(n x)}^{n}

does not tend to zero for any

x \neq 0 .

6-5 §6.1 source ↗

In the following exercises, state whether each statement is true, or give an example to show that it is false.

Suppose that $\sum_{n = 0}^{\infty} a_{n} {(x - 3)}^{n}$ converges at $x = 6 .$ At which of the following points might the series diverge? Use the fact that if $\sum a_{n} {(x - c)}^{n}$ 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;

x = 2;

x = 3;

x = 0;

x = 5.99;

and f.

x = 0.000001 .

6-7 §6.1 source ↗

In the following exercises, suppose that $| \frac{a_{n + 1}}{a_{n}} | \to 1$ as $n \to \infty .$ Find the radius of convergence for each series.

$\sum_{n = 0}^{\infty} a_{n} 2^{n} x^{n}$

Answer

| \frac{a_{n + 1} 2^{n + 1} x^{n + 1}}{a_{n} 2^{n} x^{n}} | =

2 | x | | \frac{a_{n + 1}}{a_{n}} | \to

2 | x |

R = \frac{1}{2}

6-9 §6.1 source ↗

In the following exercises, suppose that $| \frac{a_{n + 1}}{a_{n}} | \to 1$ as $n \to \infty .$ Find the radius of convergence for each series.

$\sum_{n = 0}^{\infty} \frac{a_{n} π^{n} x^{n}}{e^{n}}$

Answer

| \frac{a_{n + 1} {(\frac{π}{e})}^{n + 1} x^{n + 1}}{a_{n} {(\frac{π}{e})}^{n} x^{n}} | =

\frac{π | x |}{e} | \frac{a_{n + 1}}{a_{n}} | \to

\frac{π | x |}{e}

R = \frac{e}{π}

6-11 §6.1 source ↗

In the following exercises, suppose that $| \frac{a_{n + 1}}{a_{n}} | \to 1$ as $n \to \infty .$ Find the radius of convergence for each series.

$\sum_{n = 0}^{\infty} a_{n} {(−1)}^{n} x^{2 n}$

Answer

| \frac{a_{n + 1} {(−1)}^{n + 1} x^{2 n + 2}}{a_{n} {(−1)}^{n} x^{2 n}} | =

| x^{2} | | \frac{a_{n + 1}}{a_{n}} | \to

| x^{2} |

R = 1

6-13 §6.1 source ↗

In the following exercises, find the radius of convergence R and interval of convergence for $\sum a_{n} x^{n}$ with the given coefficients $a_{n} .$

$\sum_{n = 1}^{\infty} \frac{{(2 x)}^{n}}{n}$

Answer

a_{n} = \frac{2^{n}}{n}

\frac{a_{n + 1} x}{a_{n}} \to 2 x .

R = \frac{1}{2} .

When

x = \frac{1}{2}

the series is harmonic and diverges. When

x = - \frac{1}{2}

the series is alternating harmonic and converges. The interval of convergence is

I = [- \frac{1}{2}, \frac{1}{2}) .

6-15 §6.1 source ↗

In the following exercises, find the radius of convergence R and interval of convergence for $\sum a_{n} x^{n}$ with the given coefficients $a_{n} .$

$\sum_{n = 1}^{\infty} \frac{n x^{n}}{2^{n}}$

Answer

a_{n} = \frac{n}{2^{n}}

\frac{a_{n + 1} x}{a_{n}} \to \frac{x}{2}

R = 2 .

When

x = ± 2

the series diverges by the divergence test. The interval of convergence is

I = (−2, 2) .

6-17 §6.1 source ↗

In the following exercises, find the radius of convergence R and interval of convergence for $\sum a_{n} x^{n}$ with the given coefficients $a_{n} .$

$\sum_{n = 1}^{\infty} \frac{n^{2} x^{n}}{2^{n}}$

Answer

a_{n} = \frac{n^{2}}{2^{n}}

R = 2 .

When

x = ± 2

the series diverges by the divergence test. The interval of convergence is

I = (−2, 2) .

6-19 §6.1 source ↗

In the following exercises, find the radius of convergence R and interval of convergence for $\sum a_{n} x^{n}$ with the given coefficients $a_{n} .$

$\sum_{k = 1}^{\infty} \frac{π^{k} x^{k}}{k^{π}}$

Answer

a_{k} = \frac{π^{k}}{k^{π}}

R = \frac{1}{π} .

When

x = ± \frac{1}{π}

the series is an absolutely convergent p-series. The interval of convergence is

I = [- \frac{1}{π}, \frac{1}{π}] .

6-21 §6.1 source ↗

In the following exercises, find the radius of convergence R and interval of convergence for $\sum a_{n} x^{n}$ with the given coefficients $a_{n} .$

$\sum_{n = 1}^{\infty} \frac{10^{n} x^{n}}{n!}$

Answer

a_{n} =

\frac{10^{n}}{n!},

\frac{a_{n + 1} x}{a_{n}} =

\frac{10 x}{n + 1} \to

0 <

1

so the series converges for all x by the ratio test and

I = (− \infty, \infty) .

6-23 §6.1 source ↗

In the following exercises, find the radius of convergence of each series.

$\sum_{k = 1}^{\infty} \frac{{(k!)}^{2} x^{k}}{(2 k)!}$

Answer

a_{k} =

\frac{{(k!)}^{2}}{(2 k)!}

\frac{a_{k + 1}}{a_{k}} =

\frac{{(k + 1)}^{2}}{(2 k + 2) (2 k + 1)} \to

\frac{1}{4}

R = 4

6-25 §6.1 source ↗

In the following exercises, find the radius of convergence of each series.

$\sum_{k = 1}^{\infty} \frac{k!}{1 \cdot 3 \cdot 5 ⋯ (2 k - 1)} x^{k}$

Answer

a_{k} =

\frac{k!}{1 \cdot 3 \cdot 5 ⋯ (2 k - 1)}

\frac{a_{k + 1}}{a_{k}} =

\frac{k + 1}{2 k + 1} \to

\frac{1}{2}

R = 2

6-27 §6.1 source ↗

In the following exercises, find the radius of convergence of each series.

$\sum_{n = 1}^{\infty} \frac{x^{n}}{(\begin{matrix} 2 n \\ n \end{matrix})}$ where $(\begin{matrix} n \\ k \end{matrix}) = \frac{n!}{k! (n - k)!}$

Answer

a_{n} = \frac{1}{(\begin{matrix} 2 n \\ n \end{matrix})}

\frac{a_{n + 1}}{a_{n}} =

\frac{{((n + 1)!)}^{2}}{(2 n + 2)!} \frac{2 n!}{{(n!)}^{2}} =

\frac{{(n + 1)}^{2}}{(2 n + 2) (2 n + 1)} \to

\frac{1}{4}

R = 4

6-29 §6.1 source ↗

In the following exercises, use the ratio test to determine the radius of convergence of each series.

$\sum_{n = 1}^{\infty} \frac{{(n!)}^{3}}{(3 n)!} x^{n}$

Answer

\frac{a_{n + 1}}{a_{n}} =

\frac{{(n + 1)}^{3}}{(3 n + 3) (3 n + 2) (3 n + 1)} \to

\frac{1}{27}

R = 27

6-31 §6.1 source ↗

In the following exercises, use the ratio test to determine the radius of convergence of each series.

$\sum_{n = 1}^{\infty} \frac{n!}{n^{n}} x^{n}$

Answer

a_{n} = \frac{n!}{n^{n}}

\frac{a_{n + 1}}{a_{n}} =

\frac{(n + 1)!}{n!} \frac{n^{n}}{{(n + 1)}^{n + 1}} =

{(\frac{n}{n + 1})}^{n} \to

\frac{1}{e}

R = e

6-33 §6.1 source ↗

In the following exercises, given that $\frac{1}{1 - x} =$ $\sum_{n = 0}^{\infty} x^{n}$ with convergence in $(−1, 1),$ find the power series for each function with the given center a, and identify its interval of convergence.

$f (x) = \frac{1}{x}; a = 1$ (Hint: $\frac{1}{x} =$ $\frac{1}{1 - (1 - x)})$

Answer

f (x) =

\sum_{n = 0}^{\infty} {(1 - x)}^{n}

I = (0, 2)

6-35 §6.1 source ↗

$f (x) =$ $\frac{x}{1 - x^{2}};$ $a =$ $0$

Answer

\sum_{n = 0}^{\infty} x^{2 n + 1}

I = (−1, 1)

6-37 §6.1 source ↗

$f (x) =$ $\frac{x^{2}}{1 + x^{2}};$ $a =$ $0$

Answer

\sum_{n = 0}^{\infty} {(−1)}^{n} x^{2 n + 2}

I = (−1, 1)

6-39 §6.1 source ↗

$f (x) =$ $\frac{1}{1 - 2 x};$ $a =$ $0 .$

Answer

\sum_{n = 0}^{\infty} 2^{n} x^{n}

(- \frac{1}{2}, \frac{1}{2})

6-41 §6.1 source ↗

$f (x) =$ $\frac{x^{2}}{1 - 4 x^{2}};$ $a =$ $0$

Answer

\sum_{n = 0}^{\infty} 4^{n} x^{2 n + 2}

(- \frac{1}{2}, \frac{1}{2})

6-43 §6.1 source ↗

Use the next exercise to find the radius of convergence of the given series in the subsequent exercises.

Explain why, if ${| a_{n} |}^{1 / n} \to$ $r >$ $0,$ then ${| a_{n} x^{n} |}^{1 / n} \to$ $| x | r <$ $1$ whenever $| x | < \frac{1}{r}$ and, therefore, the radius of convergence of $\sum_{n = 1}^{\infty} a_{n} x^{n}$ is $R = \frac{1}{r} .$

Answer

{| a_{n} x^{n} |}^{1 / n} =

{| a_{n} |}^{1 / n} | x | \to

| x | r

n \to \infty

and

| x | r < 1

when

| x | < \frac{1}{r} .

Therefore,

\sum_{n = 1}^{\infty} a_{n} x^{n}

converges when

| x | < \frac{1}{r}

by the nth root test.

6-45 §6.1 source ↗

Use the next exercise to find the radius of convergence of the given series in the subsequent exercises.

$\sum_{k = 1}^{\infty} {(\frac{k - 1}{2 k + 3})}^{k} x^{k}$

Answer

a_{k} =

{(\frac{k - 1}{2 k + 3})}^{k}

{(a_{k})}^{1 / k} \to

\frac{1}{2} <

1

R = 2

6-47 §6.1 source ↗

Use the next exercise to find the radius of convergence of the given series in the subsequent exercises.

$\sum_{n = 1}^{\infty} a_{n} =$ ${(n^{1 / n} - 1)}^{n} x^{n}$

Answer

a_{n} =

{(n^{1 / n} - 1)}^{n}

{(a_{n})}^{1 / n} \to 0

R = \infty

6-49 §6.1 source ↗

Use the next exercise to find the radius of convergence of the given series in the subsequent exercises.

Suppose that $p (x) =$ $\sum_{n = 0}^{\infty} a_{n} x^{n}$ such that $a_{n} = 0$ if n is odd. Explain why $p (x) = p (− x) .$

Answer

We can rewrite

p (x) =

\sum_{n = 0}^{\infty} a_{2 n + 1} x^{2 n + 1}

and

p (x) = p (− x)

since only even powers of

x

remain,

p (x)

is an even function, for which, by definition

p (x) = p (–x)

6-51 §6.1 source ↗

Use the next exercise to find the radius of convergence of the given series in the subsequent exercises.

Suppose that $p (x) =$ $\sum_{n = 0}^{\infty} a_{n} x^{n}$ converges on $(−1, 1] .$ Find the interval of convergence of $p (2 x - 1) .$

Answer

x \in [0, 1],

then

y =

2 x -

1 \in [−1, 1]

p (2 x - 1) =

p (y) =

\sum_{n = 0}^{\infty} a_{n} y^{n}

converges.

6-53 §6.1 source ↗

In the following exercises, suppose that $p (x) =$ $\sum_{n = 0}^{\infty} a_{n} x^{n}$ satisfies $lim_{n \to \infty} \frac{a_{n + 1}}{a_{n}} =$ $1$ where $a_{n} \geq 0$ 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.

$\sum_{n = 0}^{\infty} a_{2 n} x^{2 n}$

Answer

Converges on

(−1, 1)

by the ratio test

6-55 §6.1 source ↗

$\sum_{n = 0}^{\infty} a_{n^{2}} x^{n^{2}}$ (Hint: Let $b_{k} = a_{k}$ if $k = n^{2}$ for some n, otherwise $b_{k} = 0 .)$

Answer

Consider the series

\sum b_{k} x^{k}

where

b_{k} = a_{k}

k = n^{2}

and

b_{k} = 0

otherwise. Then

b_{k} \leq a_{k}

and so the series converges on

(−1, 1)

by the comparison test.

6-57 §6.1 source ↗

[T] Plot the graphs of $\frac{1}{1 - x}$ and of the partial sums $S_{N} = \sum_{n = 0}^{N} x^{n}$ for $n = 10, 20, 30$ on the interval $[−0.99, 0.99] .$ Comment on the approximation of $\frac{1}{1 - x}$ by $S_{N}$ near $x = −1$ and near $x = 1$ as N increases.

Answer

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

The approximation is more accurate near $x = −1 .$ The partial sums follow $\frac{1}{1 - x}$ more closely as N increases but are never accurate near $x = 1$ since the series diverges there.

6-59 §6.1 source ↗

[T] Plot the graphs of the partial sums $S_{n} =$ $\sum_{n = 1}^{N} \frac{x^{n}}{n^{2}}$ 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

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

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

6-61 §6.1 source ↗

[T] Plot the graphs of the partial sums $S_{N} =$ $\sum_{n = 0}^{N} {(−1)}^{n} \frac{x^{2 n + 1}}{(2 n + 1)!}$ for $n = 3, 5, 10$ on the interval $[−2 π, 2 π] .$ Comment on how these plots approximate $sin x$ as N increases.

Answer

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.

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

Checkpoint 6.4 §6.2 source ↗

Suppose that

\sum_{n = 0}^{\infty} a_{n} x^{n}

has an interval of convergence of

(−1, 1) .

Find the interval of convergence of

\sum_{n = 0}^{\infty} a_{n} {(\frac{x}{2})}^{n} .

Answer

Interval of convergence is

(−2, 2) .

Checkpoint 6.5 §6.2 source ↗

Use the series for

f (x) = \frac{1}{1 - x}

| x | < 1

to construct a series for

\frac{1}{(1 - x) (x - 2)} .

Determine the interval of convergence.

Answer

\sum_{n = 0}^{\infty} (−1 + \frac{1}{2^{n + 1}}) x^{n} .

The interval of convergence is

(−1, 1) .

Checkpoint 6.6 §6.2 source ↗

Find the function represented by the power series

\sum_{n = 0}^{\infty} \frac{1}{3^{n}} x^{n} .

Determine its interval of convergence.

Answer

f (x) = \frac{3}{3 - x} .

The interval of convergence is

(−3, 3) .

Checkpoint 6.7 §6.2 source ↗

Multiply the series

\frac{1}{1 - x} =

\sum_{n = 0}^{\infty} x^{n}

by itself to construct a series for

\frac{1}{(1 - x) (1 - x)} .

Answer

1 +

2 x +

3 x^{2} +

4 x^{3} +

⋯

Checkpoint 6.8 §6.2 source ↗

Differentiate the series

\frac{1}{{(1 - x)}^{2}} =

\sum_{n = 0}^{\infty} (n + 1) x^{n}

term-by-term to find a power series representation for

\frac{2}{{(1 - x)}^{3}}

on the interval

(−1, 1) .

Answer

\sum_{n = 0}^{\infty} (n + 2) (n + 1) x^{n}

Checkpoint 6.9 §6.2 source ↗

Integrate the power series

ln (1 + x) =

\sum_{n = 1}^{\infty} {(−1)}^{n + 1} \frac{x^{n}}{n}

term-by-term to evaluate

\int ln (1 + x) d x .

Answer

\sum_{n = 2}^{\infty} \frac{{(−1)}^{n} x^{n}}{n (n - 1)}

6-63 §6.2 source ↗

f (x) =

\sum_{n = 0}^{\infty} \frac{x^{n}}{n!}

and

g (x) =

\sum_{n = 0}^{\infty} {(−1)}^{n} \frac{x^{n}}{n!},

find the power series of

\frac{1}{2} (f (x) + g (x))

and of

\frac{1}{2} (f (x) - g (x)) .

Answer

\frac{1}{2} (f (x) + g (x)) =

\sum_{n = 0}^{\infty} \frac{x^{2 n}}{(2 n)!}

and

\frac{1}{2} (f (x) - g (x)) =

\sum_{n = 0}^{\infty} \frac{x^{2 n + 1}}{(2 n + 1)!} .

6-65 §6.2 source ↗

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

$\frac{4}{(x - 3) (x + 1)}$

Answer

\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}^{\infty} {(\frac{x}{3})}^{n} -

\sum_{n = 0}^{\infty} {(−1)}^{n} x^{n} =

\sum_{n = 0}^{\infty} ({(−1)}^{n + 1} - \frac{1}{3^{n + 1}}) x^{n}

6-67 §6.2 source ↗

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

$\frac{5}{(x^{2} + 4) (x^{2} - 1)}$

Answer

\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}^{\infty} x^{2 n} -

\frac{1}{4} \sum_{n = 0}^{\infty} {(−1)}^{n} {(\frac{x}{2})}^{2 n} =

\sum_{n = 0}^{\infty} (- 1 + (−1) \cdot^{n + 1} \frac{1}{2^{n + 2}}) x^{2 n}

6-69 §6.2 source ↗

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

$\sum_{n = 1}^{\infty} \frac{1}{x^{n}}$

Answer

\frac{1}{x} \sum_{n = 0}^{\infty} \frac{1}{x^{n}} =

\frac{1}{x} \frac{1}{1 - \frac{1}{x}} =

\frac{1}{x - 1}

6-71 §6.2 source ↗

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

$\sum_{n = 1}^{\infty} \frac{1}{{(x - 3)}^{2 n - 1}}$

Answer

\frac{1}{x - 3} \frac{1}{1 - \frac{1}{{(x - 3)}^{2}}} =

\frac{x - 3}{{(x - 3)}^{2} - 1}

6-73 §6.2 source ↗

The following exercises explore applications of annuities.

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 \frac{1}{{(1 + r)}^{k}} .

Then

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 \approx

10,000 \times 14.8775 =

148,775 .

When

r =

0.05,

P \approx

10,000 \times 12.4622 =

124,622 .

When

r =

0.07,

P \approx

105,940 .

6-75 §6.2 source ↗

The following exercises explore applications of annuities.

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 =

\frac{C (1 - {(1 + r)}^{− N})}{r}

for N years of payouts, or

C =

\frac{P r}{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 \approx 9439.29

when

r = 0.07 .

6-77 §6.2 source ↗

The following exercises explore applications of annuities.

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 = \frac{C}{r} .

Thus,

r =

\frac{C}{P} =

5 \times \frac{10^{4}}{10^{6}} =

0.05 .

6-79 §6.2 source ↗

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.

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

Answer

(x + x^{2} - x^{3}) (1 + x^{3} + x^{6} + ⋯) =

\frac{x + x^{2} - x^{3}}{1 - x^{3}}

6-81 §6.2 source ↗

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.

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

Answer

(x - x^{2} - x^{3}) (1 + x^{3} + x^{6} + ⋯) =

\frac{x - x^{2} - x^{3}}{1 - x^{3}}

6-83 §6.2 source ↗

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

$f (x) =$ $2 \sum_{n = 0}^{\infty} x^{n},$ $g (x) =$ $\sum_{n = 0}^{\infty} n x^{n}$

Answer

a_{n} = 2, b_{n} = n

c_{n} =

\sum_{k = 0}^{n} b_{k} a_{n - k} =

2 \sum_{k = 0}^{n} k =

(n) (n + 1)

and

f (x) g (x) =

\sum_{n = 1}^{\infty} n (n + 1) x^{n}

6-85 §6.2 source ↗

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

$f (x) =$ $g (x) =$ $\sum_{n = 1}^{\infty} {(\frac{x}{2})}^{n}$

Answer

a_{n} = b_{n} = 2^{− n}

c_{n} =

\sum_{k = 1}^{n - 1} b_{k} a_{n - k} =

2^{− n} \sum_{k = 1}^{n - 1} 1 =

\frac{n - 1}{2^{n}}

and

f (x) g (x) =

\sum_{n = 2}^{\infty} (n - 1) {(\frac{x}{2})}^{n}

6-87 §6.2 source ↗

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.

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

Answer

The derivative of

f

-

\frac{1}{{(1 + x)}^{2}} =

− \sum_{n = 0}^{\infty} {(−1)}^{n} (n + 1) x^{n} .

6-89 §6.2 source ↗

In the following exercises, 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 $f .$

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

Answer

The indefinite integral of

f

- \frac{1}{1 + x^{2}} =

\sum_{n = 0}^{\infty} {(−1)}^{n} x^{2 n} .

6-91 §6.2 source ↗

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

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

Answer

f (x) =

\sum_{n = 0}^{\infty} x^{n} =

\frac{1}{1 - x};

f^{'} (\frac{1}{2}) =

\sum_{n = 1}^{\infty} \frac{n}{2^{n - 1}} =

{\frac{d}{d x} {(1 - x)}^{−1} |}_{x = 1 / 2} =

{\frac{1}{{(1 - x)}^{2}} |}_{x = 1 / 2} =

4

\sum_{n = 1}^{\infty} \frac{n}{2^{n}} = 2 .

6-93 §6.2 source ↗

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

Evaluate $\sum_{n = 2}^{\infty} \frac{n (n - 1)}{2^{n}}$ as $f^{″} (\frac{1}{2})$ where $f (x) =$ $\sum_{n = 0}^{\infty} x^{n} .$

Answer

f (x) =

\sum_{n = 0}^{\infty} x^{n} =

\frac{1}{1 - x};

f^{″} (\frac{1}{2}) =

\sum_{n = 2}^{\infty} \frac{n (n - 1)}{2^{n - 2}} =

{\frac{d^{2}}{d x^{2}} {(1 - x)}^{−1} |}_{x = 1 / 2} =

{\frac{2}{{(1 - x)}^{3}} |}_{x = 1 / 2} =

16

\sum_{n = 2}^{\infty} \frac{n (n - 1)}{2^{n}} =

4 .

6-95 §6.2 source ↗

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

$f (x) = ln x$ centered at $x = 1$ (Hint: $x = 1 - (1 - x))$

Answer

\int \sum {(1 - x)}^{n} d x =

\int \sum {(−1)}^{n} {(x - 1)}^{n} d x =

\sum \frac{{(−1)}^{n} {(x - 1)}^{n + 1}}{n + 1}

6-97 §6.2 source ↗

$ln (1 - x^{2})$ at $x = 0$

Answer

− \int_{t = 0}^{x^{2}} \frac{1}{1 - t} d t =

− \sum_{n = 0}^{\infty} \int_{0}^{x^{2}} t^{n} d x -

\sum_{n = 0}^{\infty} \frac{x^{2 (n + 1)}}{n + 1} =

− \sum_{n = 1}^{\infty} \frac{x^{2 n}}{n}

6-99 §6.2 source ↗

$f (x) =$ ${tan}^{−1} (x^{2})$ at $x = 0$

Answer

\int_{0}^{x^{2}} \frac{d t}{1 + t^{2}} =

\sum_{n = 0}^{\infty} {(−1)}^{n} \int_{0}^{x^{2}} t^{2 n} d t =

{\sum_{n = 0}^{\infty} {(−1)}^{n} \frac{t^{2 n + 1}}{2 n + 1} |}_{t = 0}^{x^{2}} =

\sum_{n = 0}^{\infty} {(−1)}^{n} \frac{x^{4 n + 2}}{2 n + 1}

6-101 §6.2 source ↗

$f (x) =$ $\int_{0}^{x} ln t d t$ where $ln (x) =$ $\sum_{n = 1}^{\infty} {(−1)}^{n - 1} \frac{{(x - 1)}^{n}}{n}$

Answer

Term-by-term integration gives

\int_{0}^{x} ln t d t =

\sum_{n = 1}^{\infty} {(−1)}^{n - 1} \frac{{(x - 1)}^{n + 1}}{n (n + 1)} =

\sum_{n = 1}^{\infty} {(−1)}^{n - 1} (\frac{1}{n} - \frac{1}{n + 1}) {(x - 1)}^{n + 1} =

(x - 1) ln x +

\sum_{n = 2}^{\infty} {(−1)}^{n} \frac{{(x - 1)}^{n}}{n} =

x ln x -

x .

6-103 §6.2 source ↗

[T] Subtract the infinite series of $ln (1 - x)$ from $ln (1 + x)$ to get a power series for $ln (\frac{1 + x}{1 - x}) .$ Evaluate at $x = \frac{1}{3} .$ 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 (1 - x) =

− \sum_{n = 1}^{\infty} \frac{x^{n}}{n}

ln (1 + x) =

\sum_{n = 1}^{\infty} {(−1)}^{n - 1} \frac{x^{n}}{n} .

Thus,

ln (\frac{1 + x}{1 - x}) =

\sum_{n = 1}^{\infty} (1 + {(−1)}^{n - 1}) \frac{x^{n}}{n} =

2 \sum_{n = 1}^{\infty} \frac{x^{2 n - 1}}{2 n - 1} .

When

x = \frac{1}{3}

we obtain

ln (2) =

2 \sum_{n = 1}^{\infty} \frac{1}{3^{2 n - 1} (2 n - 1)} .

We have

2 \sum_{n = 1}^{3} \frac{1}{3^{2 n - 1} (2 n - 1)} =

0.69300 …,

while

2 \sum_{n = 1}^{4} \frac{1}{3^{2 n - 1} (2 n - 1)} =

0.69313 …

and

ln (2) =

0.69314 …;

therefore,

N = 4 .

6-105 §6.2 source ↗

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.

$\sum_{k = 1}^{\infty} \frac{x^{3 k}}{6 k}$

Answer

\sum_{k = 1}^{\infty} \frac{x^{k}}{k} =

− ln (1 - x)

\sum_{k = 1}^{\infty} \frac{x^{3 k}}{6 k} =

-

\frac{1}{6} ln (1 - x^{3}) .

The radius of convergence is equal to 1 by the ratio test.

6-107 §6.2 source ↗

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.

$\sum_{k = 1}^{\infty} 2^{− k x}$ using $y = 2^{− x}$

Answer

y = 2^{− x},

then

\sum_{k = 1}^{\infty} y^{k} =

\frac{y}{1 - y} =

\frac{2^{− x}}{1 - 2^{− x}} =

\frac{1}{2^{x} - 1} .

a_{k} = 2^{− k x},

then

\frac{a_{k + 1}}{a_{k}} =

2^{− x} <

1

when

x > 0 .

So the series converges for all

x > 0 .

6-109 §6.2 source ↗

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.

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

Answer

Answers will vary.

6-111 §6.2 source ↗

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.

[T] Suppose that the coefficients a_n of the series $\sum_{n = 0}^{\infty} a_{n} x^{n}$ are defined by the recurrence relation $a_{n} =$ $\frac{a_{n - 1}}{n} +$ $\frac{a_{n - 2}}{n (n - 1)} .$ For $a_{0} = 0$ and $a_{1} = 1,$ compute and plot the sums $S_{N} =$ $\sum_{n = 0}^{N} a_{n} x^{n}$ for $N = 2, 3, 4, 5$ on $[−1, 1] .$

Answer

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.

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

6-113 §6.2 source ↗

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.

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

Answer

When

x =

-

\frac{1}{2},

− ln (2) =

ln (\frac{1}{2}) =

− \sum_{n = 1}^{\infty} \frac{1}{n 2^{n}} .

Since

\sum_{n = 11}^{\infty} \frac{1}{n 2^{n}} <

\sum_{n = 11}^{\infty} \frac{1}{2^{n}} =

\frac{1}{2^{10}},

one has

\sum_{n = 1}^{10} \frac{1}{n 2^{n}} =

0.69306 …

whereas

ln (2) =

0.69314 …;

therefore,

N = 10 .

6-115 §6.2 source ↗

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.

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

Answer

6 S_{N} (\frac{1}{\sqrt{3}}) =

2 \sqrt{3} \sum_{n = 0}^{N} {(−1)}^{n} \frac{1}{3^{n} (2 n + 1)} .

One has

π -

6 S_{4} (\frac{1}{\sqrt{3}}) =

0.00101 …

and

π -

6 S_{5} (\frac{1}{\sqrt{3}}) =

0.00028 …

N = 5

is the smallest partial sum with accuracy to within 0.001. Also,

π -

6 S_{7} (\frac{1}{\sqrt{3}}) =

0.00002 …

while

π -

6 S_{8} (\frac{1}{\sqrt{3}}) =

−0.000007 …

N = 8

is the smallest N to give accuracy to within 0.00001.

Checkpoint 6.10 §6.3 source ↗

Find the Taylor polynomials

p_{0}, p_{1}, p_{2}

and

p_{3}

for

f (x) = \frac{1}{x^{2}}

x = 1 .

Answer

p_{0} (x) =

1;

p_{1} (x) =

1 -

2 (x - 1);

p_{2} (x) =

1 -

2 (x - 1) +

3 {(x - 1)}^{2};

p_{3} (x) =

1 -

2 (x - 1) +

3 {(x - 1)}^{2} -

4 {(x - 1)}^{3}

Checkpoint 6.11 §6.3 source ↗

Find formulas for the Maclaurin polynomials

p_{0}, p_{1}, p_{2}

and

p_{3}

for

f (x) = \frac{1}{1 + x} .

Find a formula for the nth Maclaurin polynomial. Write your answer using sigma notation.

Answer

p_{0} (x) =

1;

p_{1} (x) =

1 -

x;

p_{2} (x) =

1 -

x +

x^{2};

p_{3} (x) =

1 -

x +

x^{2} -

x^{3};

p_{n} (x) =

1 -

x +

x^{2} -

x^{3} +

⋯ +

{(−1)}^{n} x^{n} =

\sum_{k = 0}^{n} {(−1)}^{k} x^{k}

Checkpoint 6.12 §6.3 source ↗

Find the first and second Taylor polynomials for

f (x) = \sqrt{x}

x = 4 .

Use these polynomials to estimate

\sqrt{6} .

Use Taylor’s theorem to bound the error.

Answer

$p_{1} (x) =$ $2 +$ $\frac{1}{4} (x - 4);$ $p_{2} (x) =$ $2 +$ $\frac{1}{4} (x - 4) -$ $\frac{1}{64} {(x - 4)}^{2};$ $p_{1} (6) =$ $2.5;$ $p_{2} (6) =$ $2.4375;$

$| R_{1} (6) | \leq 0.0625;$ $| R_{2} (6) | \leq 0.015625$

Checkpoint 6.13 §6.3 source ↗

Use the fourth Maclaurin polynomial for

cos x

to approximate

cos (\frac{π}{12}) .

Answer

0.96593

Checkpoint 6.14 §6.3 source ↗

Find the Taylor series for

f (x) = \frac{1}{2 x}

x = 2

and determine its interval of convergence.

Answer

\sum_{n = 0}^{\infty} \frac{{2 - x}^{n}}{2^{n + 1}} .

The interval of convergence is

(0, 4) .

Checkpoint 6.15 §6.3 source ↗

Find the Maclaurin series for

f (x) = cos x .

Use the ratio test to show that the interval of convergence is

(− \infty, \infty) .

Show that the Maclaurin series converges to

cos x

for all real numbers x.

Answer

$\sum_{n = 0}^{\infty} \frac{{(−1)}^{n} x^{2 n}}{(2 n)!}$

By the ratio test, the interval of convergence is $(− \infty, \infty) .$ Since $| R_{n} (x) | \leq \frac{{| x |}^{n + 1}}{(n + 1)!},$ the series converges to $cos x$ for all real x.

6-117 §6.3 source ↗

In the following exercises, find the Taylor polynomials of degree two approximating the given function centered at the given point.

$f (x) = 1 + x + x^{2}$ at $a = −1$

Answer

f (−1) =

1;

f^{'} (−1) =

−1;

f^{″} (−1) =

2;

f (x) =

1 -

(x + 1) +

{(x + 1)}^{2}

6-119 §6.3 source ↗

In the following exercises, find the Taylor polynomials of degree two approximating the given function centered at the given point.

$f (x) =$ $sin (2 x)$ at $a = \frac{π}{2}$

Answer

f^{'} (x) =

2 cos (2 x);

f^{″} (x) =

−4 sin (2 x);

p_{2} (x) =

−2 (x - \frac{π}{2})

6-121 §6.3 source ↗

In the following exercises, find the Taylor polynomials of degree two approximating the given function centered at the given point.

$f (x) = ln x$ at $a = 1$

Answer

f^{'} (x) =

\frac{1}{x};

f^{″} (x) =

-

\frac{1}{x^{2}};

p_{2} (x) =

0 +

(x - 1) -

\frac{1}{2} {(x - 1)}^{2}

6-123 §6.3 source ↗

In the following exercises, find the Taylor polynomials of degree two approximating the given function centered at the given point.

$f (x) = e^{x}$ at $a = 1$

Answer

p_{2} (x) =

e +

e (x - 1) +

\frac{e}{2} {(x - 1)}^{2}

6-125 §6.3 source ↗

In the following exercises, verify that the given choice of n in the remainder estimate $| R_{n} | \leq \frac{M}{(n + 1)!} {(x - a)}^{n + 1},$ where M is the maximum value of $| f^{(n + 1)} (z) |$ on the interval between a and the indicated point, yields $| R_{n} | \leq \frac{1}{1000} .$ Find the value of the Taylor polynomial p_n of $f$ at the indicated point. If $R n$ is not less than $1 / 1000$ , determine what it is.

[T] ${(28)}^{1 / 3};$ $a =$ $27,$ $n =$ $1$

Answer

\frac{d^{2}}{d x^{2}} x^{1 / 3} =

-

\frac{2}{9 x^{5 / 3}} \geq −0.00092 …

when

x \geq 28

so the remainder estimate applies to the linear approximation

x^{1 / 3} \approx

p_{1} (27) =

3 +

\frac{x - 27}{27},

which gives

{(28)}^{1 / 3} \approx

3 +

\frac{1}{27} =

3. \bar{037},

while

{(28)}^{1 / 3} \approx

3.03658 .

6-127 §6.3 source ↗

[T] e²; $a = 0, n = 9$

Answer

Using the estimate

\frac{2^{10}}{10!} <

0.000283

we can use the Taylor expansion of order 9 to estimate e^x at

x = 2 .

e^{2} \approx

p_{9} (2) =

1 +

2 +

\frac{2^{2}}{2} +

\frac{2^{3}}{6} +

⋯ +

\frac{2^{9}}{9!} =

7.3887 …

whereas

e^{2} \approx 7.3891 .

6-129 §6.3 source ↗

[T] $ln (2);$ $a =$ $1,$ $n =$ $1000$

Answer

Since

\frac{d^{n}}{d x^{n}} (ln x) =

{(−1)}^{n - 1} \frac{(n - 1)!}{x^{n}},

R_{1000} \approx

\frac{1}{1001} .

One has

p_{1000} (1) =

\sum_{n = 1}^{1000} \frac{{(−1)}^{n - 1}}{n} \approx

0.6936

whereas

ln (2) \approx

0.6931 ⋯ .

6-131 §6.3 source ↗

Integrate the approximation $e^{x} \approx$ $1 +$ $x +$ $\frac{x^{2}}{2} +$ $⋯ +$ $\frac{x^{6}}{720}$ evaluated at −x² to approximate $\int_{0}^{1} e^{− x^{2}} d x .$

Answer

$\int_{0}^{1} (1 - x^{2} + \frac{x^{4}}{2} - \frac{x^{6}}{6} + \frac{x^{8}}{24} - \frac{x^{10}}{120} + \frac{x^{12}}{720}) d x$

$=$ $1 -$ $\frac{1^{3}}{3} +$ $\frac{1^{5}}{10} -$ $\frac{1^{7}}{42} +$ $\frac{1^{9}}{9 \cdot 24} -$ $\frac{1^{11}}{120 \cdot 11} +$ $\frac{1^{13}}{720 \cdot 13} \approx$ $0.74683$ whereas $\int_{0}^{1} e^{− x^{2}} d x \approx$ $0.74682 .$

6-133 §6.3 source ↗

In the following exercises, find the smallest value of n such that the remainder estimate $| R_{n} | \leq \frac{M}{(n + 1)!} {(x - a)}^{n + 1},$ where M is the maximum value of $| f^{(n + 1)} (z) |$ on the interval between a and the indicated point, yields $| R_{n} | \leq \frac{1}{1000}$ on the indicated interval.

$f (x) = cos x$ on $[- \frac{π}{2}, \frac{π}{2}],$ $a =$ $0$

Answer

Since

f^{(n + 1)} (z)

sin z

cos z,

we have

M = 1 .

Since

| x - 0 | \leq \frac{π}{2},

we seek the smallest n such that

\frac{π^{n + 1}}{2^{n + 1} (n + 1)!} \leq 0.001 .

The smallest such value is

n = 7 .

The remainder estimate is

R_{7} \leq 0.00092 .

6-135 §6.3 source ↗

$f (x) = e^{− x}$ on $[−3, 3], a = 0$

Answer

Since

f^{(n + 1)} (z) =

± e^{− z}

one has

M = e^{3} .

Since

| x - 0 | \leq 3,

one seeks the smallest n such that

\frac{3^{n + 1} e^{3}}{(n + 1)!} \leq 0.001 .

The smallest such value is

n = 14 .

The remainder estimate is

R_{14} \leq 0.000220 .

6-137 §6.3 source ↗

In the following exercises, the maximum of the right-hand side of the remainder estimate $| R_{1} | \leq \frac{max | f^{″} (z) |}{2} R^{2}$ on $[a - R, a + R]$ occurs at a or $a \pm R .$ Estimate the maximum value of R such that $\frac{max | f^{″} (z) |}{2} R^{2} \leq 0.1$ on $[a - R, a + R]$ by plotting this maximum as a function of R.

[T] $sin x$ approximated by x, $a = 0$

Answer

This graph has a horizontal line at y=0.2. It also has a curve starting at the origin and concave up. The curve and the line intersect at the ordered pair (0.5966, 0.2).

Since $sin x$ is increasing for small x and since $si n^{″} x =$ $− sin x,$ the estimate applies whenever $R^{2} sin (R) \leq 0.2,$ which applies up to $R = 0.596 .$

6-139 §6.3 source ↗

[T] $cos x$ approximated by $1, a = 0$

Answer

Since the second derivative of $cos x$ is $− cos x$ and since $cos x$ is decreasing away from $x = 0,$ the estimate applies when $R^{2} cos R \leq 0.2$ or $R \leq 0.447 .$

6-141 §6.3 source ↗

In the following exercises, find the Taylor series of the given function centered at the indicated point.

$1 + x + x^{2} + x^{3}$ at $a = −1$

Answer

{(x + 1)}^{3} -

2 {(x + 1)}^{2} +

2 (x + 1)

6-143 §6.3 source ↗

In the following exercises, find the Taylor series of the given function centered at the indicated point.

$cos x$ at $a = 2 π$

Answer

Values of derivatives are the same as for

x = 0

cos x =

{\sum_{n = 0}^{\infty} (−1)}^{n} \frac{{(x - 2 π)}^{2 n}}{(2 n)!}

6-145 §6.3 source ↗

In the following exercises, find the Taylor series of the given function centered at the indicated point.

$cos x$ at $x = \frac{π}{2}$

Answer

cos (\frac{π}{2}) =

0,

− sin (\frac{π}{2}) =

−1

cos x =

\sum_{n = 0}^{\infty} {(−1)}^{n + 1} \frac{{(x - \frac{π}{2})}^{2 n + 1}}{(2 n + 1)!},

which is also

− cos (x - \frac{π}{2}) .

6-147 §6.3 source ↗

In the following exercises, find the Taylor series of the given function centered at the indicated point.

$e^{x}$ at $a = 1$

Answer

The derivatives are

f^{(n)} (1) = e

e^{x} =

e \sum_{n = 0}^{\infty} \frac{{(x - 1)}^{n}}{n!} .

6-149 §6.3 source ↗

In the following exercises, find the Taylor series of the given function centered at the indicated point.

$\frac{1}{{(x - 1)}^{3}}$ at $a = 0$

Answer

\frac{1}{{(x - 1)}^{3}} =

− (\frac{1}{2}) \frac{d^{2}}{d x^{2}} \frac{1}{1 - x} =

− \sum_{n = 0}^{\infty} (\frac{(n + 2) (n + 1) x^{n}}{2})

6-151 §6.3 source ↗

In the following exercises, compute the Taylor series of each function around $x = 1 .$

$f (x) = 2 - x$

Answer

2 - x = 1 - (x - 1)

6-153 §6.3 source ↗

In the following exercises, compute the Taylor series of each function around $x = 1 .$

$f (x) = {(x - 2)}^{2}$

Answer

{((x - 1) - 1)}^{2} =

{(x - 1)}^{2} -

2 (x - 1) +

1

6-155 §6.3 source ↗

In the following exercises, compute the Taylor series of each function around $x = 1 .$

$f (x) = \frac{1}{x}$

Answer

\frac{1}{1 - (1 - x)} =

\sum_{n = 0}^{\infty} {(−1)}^{n} {(x - 1)}^{n}

6-157 §6.3 source ↗

In the following exercises, compute the Taylor series of each function around $x = 1 .$

$f (x) =$ $\frac{x}{4 x - 2 x^{2} - 1}$

Answer

x \sum_{n = 0}^{\infty} 2^{n} {(1 - x)}^{2 n} = \sum_{n = 0}^{\infty} 2^{n} {(x - 1)}^{2 n + 1} + \sum_{n = 0}^{\infty} 2^{n} {(x - 1)}^{2 n}

6-159 §6.3 source ↗

In the following exercises, compute the Taylor series of each function around $x = 1 .$

$f (x) = e^{2 x}$

Answer

e^{2 x} =

e^{2 (x - 1) + 2} =

e^{2} \sum_{n = 0}^{\infty} \frac{2^{n} {(x - 1)}^{n}}{n!}

6-161 §6.3 source ↗

[T] In the following exercises, identify the value of x such that the given series $\sum_{n = 0}^{\infty} a_{n}$ is the value of the Maclaurin series of $f (x)$ at $x .$ Approximate the value of $f (x)$ using $S_{10} =$ $\sum_{n = 0}^{10} a_{n} .$

$\sum_{n = 0}^{\infty} \frac{2^{n}}{n!}$

Answer

x =

e^{2};

S_{10} =

\frac{34,913}{4725} \approx

7.3889947

6-163 §6.3 source ↗

$\sum_{n = 0}^{\infty} \frac{{(−1)}^{n} {(2 π)}^{2 n + 1}}{(2 n + 1)!}$

Answer

sin (2 π) =

0;

S_{10} =

8.27 \times 10^{−5}

6-165 §6.3 source ↗

The following exercises make use of the functions $S_{5} (x) =$ $x -$ $\frac{x^{3}}{6} +$ $\frac{x^{5}}{120}$ and $C_{4} (x) =$ $1 -$ $\frac{x^{2}}{2} +$ $\frac{x^{4}}{24}$ on $[− π, π] .$

[T] Plot ${cos}^{2} x -$ ${(C_{4} (x))}^{2}$ on $[− π, π] .$ Compare the maximum difference with the square of the Taylor remainder estimate for $cos x .$

Answer

This graph has a concave up curve that is symmetrical about the y axis. The lowest point of the graph is the origin with the rest of the curve above the x-axis.

The difference is small on the interior of the interval but approaches $1$ near the endpoints. The remainder estimate is $| R_{4} | =$ $\frac{π^{5}}{120} \approx$ $2.552 .$

6-167 §6.3 source ↗

The following exercises make use of the functions $S_{5} (x) =$ $x -$ $\frac{x^{3}}{6} +$ $\frac{x^{5}}{120}$ and $C_{4} (x) =$ $1 -$ $\frac{x^{2}}{2} +$ $\frac{x^{4}}{24}$ on $[− π, π] .$

[T] Compare $\frac{S_{5} (x)}{C_{4} (x)}$ on $[−1, 1]$ to $tan x .$ Compare this with the Taylor remainder estimate for the approximation of $tan x$ by $x +$ $\frac{x^{3}}{3} +$ $\frac{2 x^{5}}{15} .$

Answer

This graph has two curves. The solid curve is very flat and close to the x-axis. It passes through the origin. The second curve, a broken line, is concave down and symmetrical about the y-axis. It is very close to the x-axis between -3 and 3.

The difference is on the order of $10^{−4}$ on $[−1, 1]$ while the Taylor approximation error is around $0.1$ near $\pm 1 .$ The top curve is a plot of ${tan}^{2} x -$ ${(\frac{S_{5} (x)}{C_{4} (x)})}^{2}$ and the lower dashed plot shows $t^{2} - {(\frac{S_{5}}{C_{4}})}^{2} .$

6-169 §6.3 source ↗

The following exercises make use of the functions $S_{5} (x) =$ $x -$ $\frac{x^{3}}{6} +$ $\frac{x^{5}}{120}$ and $C_{4} (x) =$ $1 -$ $\frac{x^{2}}{2} +$ $\frac{x^{4}}{24}$ on $[− π, π] .$

(Taylor approximations and root finding.) Recall that Newton’s method $x_{n + 1} =$ $x_{n} -$ $\frac{f (x_{n})}{f' (x_{n})}$ approximates solutions of $f (x) = 0$ near the input $x_{0} .$

a. If $f$ and $g$ are inverse functions, explain why a solution of $g (x) = a$ is the value $f (a) of f .$

b. Let $p_{N} (x)$ be the $N th$ degree Maclaurin polynomial of $e^{x} .$ Use Newton’s method to approximate solutions of $p_{N} (x) - 2 = 0$ for $N = 4, 5, 6 .$

c. Explain why the approximate roots of $p_{N} (x) - 2 = 0$ are approximate values of $ln (2) .$

Answer

a. Answers will vary. b. The following are the

x_{n}

values after

10

iterations of Newton’s method to approximation a root of

p_{N} (x) - 2 = 0 :

for

N =

4,

x =

0.6939...;

for

N =

5,

x =

0.6932...;

for

N =

6,

x =

0.69315...;

.

(Note:

ln (2) =

0.69314 ...)

c. Answers will vary.

6-171 §6.3 source ↗

In the following exercises, use the fact that if $q (x) =$ $\sum_{n = 1}^{\infty} a_{n} {(x - c)}^{n}$ converges in an interval containing $c,$ then $lim_{x \to c} q (x) =$ $a_{0}^{}$ to evaluate each limit using Taylor series.

$lim_{x \to 0} \frac{ln (1 - x^{2})}{x^{2}}$

Answer

\frac{ln (1 - x^{2})}{x^{2}} \to

− 1

6-173 §6.3 source ↗

$lim_{x \to 0^{+}} \frac{cos (\sqrt{x}) - 1}{2 x}$

Answer

\frac{cos (\sqrt{x}) - 1}{2 x} \approx

\frac{(1 - \frac{x}{2} + \frac{x^{2}}{4!} - ⋯) - 1}{2 x} \to

-

\frac{1}{4}

Checkpoint 6.16 §6.4 source ↗

Find the binomial series for

f (x) =

\frac{1}{{(1 + x)}^{2}} .

Answer

\sum_{n = 0}^{\infty} {(−1)}^{n} (n + 1) x^{n}

Checkpoint 6.17 §6.4 source ↗

Find the Maclaurin series for

sin (x^{2}) .

Answer

\sum_{n = 0}^{\infty} \frac{{(−1)}^{n} x^{4 n + 2}}{(2 n + 1)!}

Checkpoint 6.18 §6.4 source ↗

Find the binomial series for

f (x) =

\frac{1}{{(1 + x)}^{3 / 2}}

Answer

\sum_{n = 1}^{\infty} \frac{{(−1)}^{n}}{n!} \frac{1 \cdot 3 \cdot 5 ⋯ (2 n - 1)}{2^{n}} x^{n}

Checkpoint 6.19 §6.4 source ↗

Use power series to solve

y^{'} =

2 y,

y (0) =

5 .

Answer

y = 5 e^{2 x}

Checkpoint 6.20 §6.4 source ↗

Use power series to solve

y^{″} + x^{2} y = 0

with the initial condition

y (0) = a

and

y^{'} (0) = b .

Answer

y =

a (1 - \frac{x^{4}}{3 \cdot 4} + \frac{x^{8}}{3 \cdot 4 \cdot 7 \cdot 8} - ⋯) +

b (x - \frac{x^{5}}{4 \cdot 5} + \frac{x^{9}}{4 \cdot 5 \cdot 8 \cdot 9} - ⋯)

Checkpoint 6.21 §6.4 source ↗

Express

\int cos \sqrt{x} d x

as an infinite series. Evaluate

\int_{0}^{1} cos \sqrt{x} d x

to within an error of

0.01 .

Answer

C +

\sum_{n = 1}^{\infty} {(−1)}^{n + 1} \frac{x^{n}}{n (2 n - 2)!}

The definite integral is approximately

0.514

to within an error of

0.01 .

Checkpoint 6.22 §6.4 source ↗

Use the first five terms of the Maclaurin series for

e^{− x^{2} / 2}

to estimate the probability that a randomly selected test score is between

100

and

150 .

Use the alternating series test to determine the accuracy of this estimate.

Answer

The estimate is approximately

0.3414 .

This estimate is accurate to within

0.0000094 .

6-175 §6.4 source ↗

In the following exercises, use appropriate substitutions to write down the Maclaurin series for the given binomial.

${(1 + x^{2})}^{−1 / 3}$

Answer

{(1 + x^{2})}^{−1 / 3} = \sum_{n = 0}^{\infty} (\begin{matrix} - \frac{1}{3} \\ n \end{matrix}) x^{2 n}

6-177 §6.4 source ↗

In the following exercises, use appropriate substitutions to write down the Maclaurin series for the given binomial.

${(1 - 2 x)}^{2 / 3}$

Answer

{(1 - 2 x)}^{2 / 3} = \sum_{n = 0}^{\infty} {(−1)}^{n} 2^{n} (\begin{matrix} \frac{2}{3} \\ n \end{matrix}) x^{n}

6-179 §6.4 source ↗

In the following exercises, use the substitution ${(b + x)}^{r} =$ ${(b + a)}^{r} {(1 + \frac{x - a}{b + a})}^{r}$ in the binomial expansion to find the Taylor series of each function with the given center.

$\sqrt{x^{2} + 2}$ at $a = 0$

Answer

\sqrt{2 + x^{2}} = \sum_{n = 0}^{\infty} 2^{(1 / 2) - n} (\begin{matrix} \frac{1}{2} \\ n \end{matrix}) x^{2 n}; (| x^{2} | < 2)

6-181 §6.4 source ↗

$\sqrt{2 x - x^{2}}$ at $a = 1$ (Hint: $2 x -$ $x^{2} =$ $1 -$ ${(x - 1)}^{2})$

Answer

\sqrt{2 x - x^{2}} =

\sqrt{1 - {(x - 1)}^{2}}

\sqrt{2 x - x^{2}} = \sum_{n = 0}^{\infty} {(−1)}^{n} (\begin{matrix} \frac{1}{2} \\ n \end{matrix}) {(x - 1)}^{2 n}

6-183 §6.4 source ↗

$\sqrt{x}$ at $a = 4$

Answer

\sqrt{x} = 2 \sqrt{1 + \frac{x - 4}{4}}

\sqrt{x} = \sum_{n = 0}^{\infty} 2^{1 - 2 n} (\begin{matrix} \frac{1}{2} \\ n \end{matrix}) {(x - 4)}^{n}

6-185 §6.4 source ↗

$\sqrt{x}$ at $α = 9$

Answer

\sqrt{x} =

\sum_{n = 0}^{\infty} 3^{1 - 2 n} (\frac{1}{2} n) {(x - 9)}^{n}

6-187 §6.4 source ↗

In the following exercises, use the binomial theorem to estimate each number, computing enough terms to obtain an estimate accurate to an error of at most $1 / 1000 .$

[T] ${(1001)}^{1 / 3}$ using ${(1000 + x)}^{1 / 3}$

Answer

10 {(1 + \frac{x}{1000})}^{1 / 3} = \sum_{n = 0}^{\infty} 10^{1 - 3 n} (\begin{matrix} \frac{1}{3} \\ n \end{matrix}) x^{n} .

Using, for example, a fourth-degree estimate at

x = 1

gives

\begin{matrix} {(1001)}^{1 / 3} & \approx 10 (1 + (\begin{matrix} \frac{1}{3} \\ 1 \end{matrix}) 10^{−3} + (\begin{matrix} \frac{1}{3} \\ 2 \end{matrix}) 10^{−6} + (\begin{matrix} \frac{1}{3} \\ 3 \end{matrix}) 10^{−9} + (\begin{matrix} \frac{1}{3} \\ 4 \end{matrix}) 10^{−12}) \\ = 10 (1 + \frac{1}{{3.10}^{3}} - \frac{1}{{9.10}^{6}} + \frac{5}{{81.10}^{9}} - \frac{10}{{243.10}^{12}}) = 10.00333222... \end{matrix}

whereas

{(1001)}^{1 / 3} =

10.00332222839093 ....

Two terms would suffice for three-digit accuracy.

6-189 §6.4 source ↗

In the following exercises, use the binomial approximation $\sqrt{1 - x} \approx$ $1 -$ $\frac{x}{2} -$ $\frac{x^{2}}{8} -$ $\frac{x^{3}}{16} -$ $\frac{5 x^{4}}{128} -$ $\frac{7 x^{5}}{256}$ for $| x | < 1$ to approximate each number. Compare this value to the value given by a scientific calculator.

[T] $\sqrt{5} = 5 \times \frac{1}{\sqrt{5}}$ using $x = \frac{4}{5}$ in ${(1 - x)}^{1 / 2}$

Answer

The approximation is

2.3152;

the CAS value is

2.23 … .

6-191 §6.4 source ↗

[T] $\sqrt{6}$ using $x = \frac{5}{6}$ in ${(1 - x)}^{1 / 2}$

Answer

The approximation is

2.583 …;

the CAS value is

2.449 … .

6-193 §6.4 source ↗

[T] Recall that the graph of $\sqrt{1 - x^{2}}$ is an upper semicircle of radius $1 .$ Integrate the binomial approximation of $\sqrt{1 - x^{2}}$ up to order $8$ from $x = −1$ to $x = 1$ to estimate $\frac{π}{2} .$

Answer

$\sqrt{1 - x^{2}} =$ $1 -$ $\frac{x^{2}}{2} -$ $\frac{x^{4}}{8} -$ $\frac{x^{6}}{16} -$ $\frac{5 x^{8}}{128} +$ $⋯ .$ Thus

$\int_{−1}^{1} \sqrt{1 - x^{2}} d x =$ $x -$ $\frac{x^{3}}{6} -$ $\frac{x^{5}}{40} -$ $\frac{x^{7}}{7 \cdot 16} -$ $\frac{5 x^{9}}{9 \cdot 128} +$ $⋯ |_{−1}^{1} \approx$ $2 -$ $\frac{1}{3} -$ $\frac{1}{20} -$ $\frac{1}{56} -$ $\frac{10}{9 \cdot 128} +$ $error =$ $1.590 ...$ whereas $\frac{π}{2} = 1.570 ...$

6-195 §6.4 source ↗

In the following exercises, use the expansion ${(1 + x)}^{1 / 3} =$ $1 +$ $\frac{1}{3} x -$ $\frac{1}{9} x^{2} +$ $\frac{5}{81} x^{3} -$ $\frac{10}{243} x^{4} +$ $⋯$ to write the first five terms (not necessarily a quartic polynomial) of each expression.

${(1 + 4 x)}^{4 / 3};$ $a =$ $0$

Answer

\begin{array}{l} {(1 + 4 x)}^{4 / 3} = (1 + 4 x) {(1 + 4 x)}^{1 / 3} \\ = (1 + 4 x) (1 + \frac{4 x}{3} - \frac{16 x^{3}}{9} + \frac{320 x^{3}}{81} - \frac{2560 x^{4}}{243}) \\ = 1 + \frac{16}{3} x + \frac{32}{9} x^{2} - \frac{256}{81} x^{3} + \frac{1280}{243} x^{4} - \frac{10240}{243} x^{5} \end{array}

6-197 §6.4 source ↗

${(x^{2} + 6 x + 10)}^{1 / 3};$ $a =$ $−3$

Answer

{(1 + {(x + 3)}^{2})}^{1 / 3} =

1 +

\frac{1}{3} {(x + 3)}^{2} -

\frac{1}{9} {(x + 3)}^{4} +

\frac{5}{81} {(x + 3)}^{6} -

\frac{10}{243} {(x + 3)}^{8} +

⋯

6-199 §6.4 source ↗

Use the approximation ${(1 - x)}^{2 / 3} =$ $1 -$ $\frac{2 x}{3} -$ $\frac{x^{2}}{9} -$ $\frac{4 x^{3}}{81} -$ $\frac{7 x^{4}}{243} -$ $\frac{14 x^{5}}{729} +$ $⋯$ for $| x | < 1$ to approximate $2^{1 / 3} =$ ${2.2}^{−2 / 3} .$

Answer

Twice the approximation is

1.260 …

whereas

2^{1 / 3} =

1.2599. ...

6-201 §6.4 source ↗

Find the $99$ th derivative at $x = 0$ of $f (x) =$ ${(1 + x^{4})}^{25} .$

Answer

f^{(99)} (0) = 0

6-203 §6.4 source ↗

In the following exercises, find the Maclaurin series of each function.

$f (x) = 2^{x}$

Answer

\sum_{n = 0}^{\infty} \frac{{(ln (2) x)}^{n}}{n!}

6-205 §6.4 source ↗

In the following exercises, find the Maclaurin series of each function.

$f (x) =$ $\frac{sin (\sqrt{x})}{\sqrt{x}},$ $(x > 0),$

Answer

For

x >

0,

sin (\sqrt{x}) =

\sum_{n = 0}^{\infty} {(−1)}^{n} \frac{x^{(2 n + 1) / 2}}{\sqrt{x} (2 n + 1)!} =

\sum_{n = 0}^{\infty} {(−1)}^{n} \frac{x^{n}}{(2 n + 1)!} .

6-207 §6.4 source ↗

In the following exercises, find the Maclaurin series of each function.

$f (x) = e^{x^{3}}$

Answer

e^{x^{3}} =

\sum_{n = 0}^{\infty} \frac{x^{3 n}}{n!}

6-209 §6.4 source ↗

In the following exercises, find the Maclaurin series of each function.

$f (x) = {sin}^{2} x$ using the identity ${sin}^{2} x =$ $\frac{1}{2} -$ $\frac{1}{2} cos (2 x)$

Answer

{sin}^{2} x =

− \sum_{k = 1}^{\infty} \frac{{(−1)}^{k} 2^{2 k - 1} x^{2 k}}{(2 k)!}

6-211 §6.4 source ↗

In the following exercises, find the Maclaurin series of $F (x) =$ $\int_{0}^{x} f (t) d t$ by integrating the Maclaurin series of $f$ term by term. If $f$ is not strictly defined at zero, you may substitute the value of the Maclaurin series at zero.

$F (x) =$ ${tan}^{−1} x;$ $f (t) =$ $\frac{1}{1 + t^{2}} =$ $\sum_{n = 0}^{\infty} {(−1)}^{n} t^{2 n}$

Answer

{tan}^{−1} x =

\sum_{k = 0}^{\infty} \frac{{(−1)}^{k} x^{2 k + 1}}{2 k + 1}

6-213 §6.4 source ↗

$F (x) = {sin}^{−1} x; f (t) = \frac{1}{\sqrt{1 - t^{2}}} = \sum_{k = 0}^{\infty} (\begin{matrix} \frac{1}{2} \\ k \end{matrix}) \frac{t^{2 k}}{k!}$

Answer

{sin}^{−1} x = \sum_{n = 0}^{\infty} (\begin{matrix} \frac{1}{2} \\ n \end{matrix}) \frac{x^{2 n + 1}}{(2 n + 1) n!}

6-215 §6.4 source ↗

$F (x) =$ $\int_{0}^{x} cos (\sqrt{t}) d t;$ $f (t) =$ $\sum_{n = 0}^{\infty} {(−1)}^{n} \frac{x^{n}}{(2 n)!}$

Answer

F (x) =

\sum_{n = 0}^{\infty} {(−1)}^{n} \frac{x^{n + 1}}{(n + 1) (2 n)!}

6-217 §6.4 source ↗

$F (x) =$ $\int_{0}^{x} \frac{ln (1 + t)}{t} d t;$ $f (t) =$ $\sum_{n = 0}^{\infty} {(−1)}^{n} \frac{t^{n}}{n + 1}$

Answer

F (x) =

\sum_{n = 1}^{\infty} {(−1)}^{n + 1} \frac{x^{n}}{n^{2}}

6-219 §6.4 source ↗

In the following exercises, compute at least the first three nonzero terms (not necessarily a quadratic polynomial) of the Maclaurin series of $f .$

$f (x) = tan x$

Answer

x +

\frac{x^{3}}{3} +

\frac{2 x^{5}}{15} +

⋯

6-221 §6.4 source ↗

In the following exercises, compute at least the first three nonzero terms (not necessarily a quadratic polynomial) of the Maclaurin series of $f .$

$f (x) = e^{x} cos x$

Answer

1 +

x -

\frac{x^{3}}{3} -

\frac{x^{4}}{6} +

⋯

6-223 §6.4 source ↗

In the following exercises, compute at least the first three nonzero terms (not necessarily a quadratic polynomial) of the Maclaurin series of $f .$

$f (x) = {sec}^{2} x$

Answer

1 +

x^{2} +

\frac{2 x^{4}}{3} +

\frac{17 x^{6}}{45} +

⋯

6-225 §6.4 source ↗

In the following exercises, compute at least the first three nonzero terms (not necessarily a quadratic polynomial) of the Maclaurin series of $f .$

$f (x) = \frac{tan \sqrt{x}}{\sqrt{x}}$ (see expansion for $tan x)$

Answer

Using the expansion for

tan x

gives

1 + \frac{x}{3} + \frac{2 x^{2}}{15} .

6-227 §6.4 source ↗

In the following exercises, find the radius of convergence of the Maclaurin series of each function.

$\frac{1}{1 + x^{2}}$

Answer

\frac{1}{1 + x^{2}} =

\sum_{n = 0}^{\infty} {(−1)}^{n} x^{2 n}

R = 1

by the ratio test.

6-229 §6.4 source ↗

In the following exercises, find the radius of convergence of the Maclaurin series of each function.

$ln (1 + x^{2})$

Answer

ln (1 + x^{2}) =

\sum_{n = 1}^{\infty} \frac{{(−1)}^{n - 1}}{n} x^{2 n}

R = 1

by the ratio test.

6-231 §6.4 source ↗

In the following exercises, find the radius of convergence of the Maclaurin series of each function.

Find the Maclaurin series of $cosh x =$ $\frac{e^{x} + e^{− x}}{2} .$

Answer

Add series of

e^{x}

and

e^{− x}

term by term. Odd terms cancel and

cosh x =

\sum_{n = 0}^{\infty} \frac{x^{2 n}}{(2 n)!} .

6-233 §6.4 source ↗

In the following exercises, find the radius of convergence of the Maclaurin series of each function.

[T] Let $S_{n} (x) =$ $\sum_{k = 0}^{n} {(−1)}^{k} \frac{x^{2 k + 1}}{(2 k + 1)!}$ and $C_{n} (x) =$ $\sum_{n = 0}^{n} {(−1)}^{k} \frac{x^{2 k}}{(2 k)!}$ denote the respective Maclaurin polynomials of degree $2 n + 1$ of $sin x$ and degree $2 n$ of $cos x .$ Plot the errors $\frac{S_{n} (x)}{C_{n} (x)} -$ $tan x$ for $n = 1, .., 5$ and compare them to $x +$ $\frac{x^{3}}{3} +$ $\frac{2 x^{5}}{15} +$ $\frac{17 x^{7}}{315} -$ $tan x$ on $(- \frac{π}{4}, \frac{π}{4}) .$

Answer

This graph has two curves. The first one is a decreasing function passing through the origin. The second is a broken line which is an increasing function passing through the origin. The two curves are very close around the origin.

The ratio $\frac{S_{n} (x)}{C_{n} (x)}$ approximates $tan x$ better than does $p_{7} (x) =$ $x +$ $\frac{x^{3}}{3} +$ $\frac{2 x^{5}}{15} +$ $\frac{17 x^{7}}{315}$ for $N \geq 3 .$ The dashed curves are $\frac{S_{n}}{C_{n}} - tan$ for $n = 1, 2 .$ The dotted curve corresponds to $n = 3,$ and the dash-dotted curve corresponds to $n = 4 .$ The solid curve is $p_{7} - tan x .$

6-235 §6.4 source ↗

In the following exercises, find the radius of convergence of the Maclaurin series of each function.

If $y =$ $\sum_{n = 0}^{\infty} a_{n} x^{n},$ find the power series expansions of $x y^{'}$ and $x^{2} y^{″} .$

Answer

By the term-by-term differentiation theorem,

y^{'} =

\sum_{n = 1}^{\infty} n a_{n} x^{n - 1}

y^{'} =

\sum_{n = 1}^{\infty} n a_{n} x^{n - 1} x y^{'} =

\sum_{n = 1}^{\infty} n a_{n} x^{n},

whereas

y^{'} =

\sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2}

x y^{″} =

\sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n} .

6-237 §6.4 source ↗

In the following exercises, find the radius of convergence of the Maclaurin series of each function.

[T] Suppose that a set of standardized test scores is normally distributed with mean $μ = 100$ and standard deviation $σ = 10 .$ Set up an integral that represents the probability that a test score will be between $90$ and $110$ and use the integral of the degree $10$ Maclaurin polynomial of $\frac{1}{\sqrt{2 π}} e^{− x^{2} / 2}$ to estimate this probability.

Answer

The probability is

p =

\frac{1}{\sqrt{2 π}} \int_{(a - μ) / σ}^{(b - μ) / σ} e^{− x^{2} / 2} d x

where

a = 90

and

b = 100,

that is,

p =

\frac{1}{\sqrt{2 π}} \int_{−1}^{1} e^{− x^{2} / 2} d x =

\frac{1}{\sqrt{2 π}} \int_{−1}^{1} \sum_{n = 0}^{5} {(−1)}^{n} \frac{x^{2 n}}{2^{n} n!} d x =

\frac{2}{\sqrt{2 π}} \sum_{n = 0}^{5} {(−1)}^{n} \frac{1}{(2 n + 1) 2^{n} n!} \approx

0.6827 .

6-239 §6.4 source ↗

In the following exercises, find the radius of convergence of the Maclaurin series of each function.

[T] Suppose that $\sum_{n = 0}^{\infty} a_{n} x^{n}$ converges to a function $f (x)$ such that $f (0) =$ $1,$ $f^{'} (0) =$ $0,$ and $f^{″} (x) =$ $− f (x) .$ Find a formula for $a_{n}$ and plot the partial sum $S_{N}$ for $N = 20$ on $[−5, 5] .$

Answer

This graph is a wave curve symmetrical about the origin. It has a peak at y = 1 above the origin. It has lowest points at -3 and 3.

As in the previous problem one obtains $a_{n} = 0$ if $n$ is odd and $a_{n} =$ $− (n + 2) (n + 1) a_{n + 2}$ if $n$ is even, so $a_{0} = 1$ leads to $a_{2 n} =$ $\frac{{(−1)}^{n}}{(2 n)!} .$

6-241 §6.4 source ↗

In the following exercises, find the radius of convergence of the Maclaurin series of each function.

Suppose that $\sum_{n = 0}^{\infty} a_{n} x^{n}$ converges to a function $y$ such that $y^{″} - y^{'} + y = 0$ where $y (0) = 1$ and $y' (0) = 0 .$ Find a formula that relates $a_{n + 2}, a_{n + 1},$ and $a_{n}$ and compute $a_{0}, ..., a_{5} .$

Answer

y^{″} =

\sum_{n = 0}^{\infty} (n + 2) (n + 1) a_{n + 2} x^{n}

and

y^{'} =

\sum_{n = 0}^{\infty} (n + 1) a_{n + 1} x^{n}

y^{″} - y^{'} + y = 0

implies that

(n + 2) (n + 1) a_{n + 2} -

(n + 1) a_{n + 1} +

a_{n} =

0

a_{n} =

\frac{a_{n - 1}}{n} -

\frac{a_{n - 2}}{n (n - 1)}

for all

n \cdot y (0) = a_{0} = 1

and

y^{'} (0) = a_{1} = 0,

a_{2} =

\frac{1}{2},

a_{3} =

\frac{1}{6},

a_{4} =

0,

and

a_{5} = - \frac{1}{120} .

6-243 §6.4 source ↗

The error in approximating the integral $\int_{a}^{b} f (t) d t$ by that of a Taylor approximation $\int_{a}^{b} P_{n} (t) d t$ is at most $\int_{a}^{b} R_{n} (t) d t .$ In the following exercises, the Taylor remainder estimate $R_{n} \leq \frac{M}{(n + 1)!} {| x - a |}^{n + 1}$ guarantees that the integral of the Taylor polynomial of the given order approximates the integral of $f$ with an error less than $\frac{1}{10} .$

[T] $\int_{0}^{π} \frac{sin t}{t} d t;$ $P_{s} =$ $1 -$ $\frac{x^{2}}{3!} +$ $\frac{x^{4}}{5!} -$ $\frac{x^{6}}{7!} +$ $\frac{x^{8}}{9!}$ (You may assume that the absolute value of the ninth derivative of $\frac{sin t}{t}$ is bounded by $0.1 .)$

Answer

a. (Proof) b. We have

R_{s} \leq \frac{0.1}{(9)!} π^{9} \approx

0.0082 <

0.01 .

We have

\int_{0}^{π} (1 - \frac{x^{2}}{3!} + \frac{x^{4}}{5!} - \frac{x^{6}}{7!} + \frac{x^{8}}{9!}) d x =

π -

\frac{π^{3}}{3 \cdot 3!} +

\frac{π^{5}}{5 \cdot 5!} -

\frac{π^{7}}{7 \cdot 7!} +

\frac{π^{9}}{9 \cdot 9!} =

1.852...,

whereas

\int_{0}^{π} \frac{sin t}{t} d t =

1.85194...,

so the actual error is approximately

0.00006 .

6-245 §6.4 source ↗

The following exercises deal with Fresnel integrals.

The Fresnel integrals are defined by $C (x) =$ $\int_{0}^{x} cos (t^{2}) d t$ and $S (x) =$ $\int_{0}^{x} sin (t^{2}) d t .$ Compute the power series of $C (x)$ and $S (x)$ and plot the sums $C_{N} (x)$ and $S_{N} (x)$ of the first $N = 50$ nonzero terms on $[0, 2 π] .$

Answer

This graph has two curves. The first one is a solid curve labeled Csub50(x). It begins at the origin and is a wave that gradually decreases in amplitude. The highest it reaches is y = 1. The second curve is labeled Ssub50(x). It is a wave that gradually decreases in amplitude. The highest it reaches is 0.9. It is very close to the pattern of the first curve with a slight shift to the right.

Since $cos (t^{2}) =$ $\sum_{n = 0}^{\infty} {(−1)}^{n} \frac{t^{4 n}}{(2 n)!}$ and $sin (t^{2}) =$ $\sum_{n = 0}^{\infty} {(−1)}^{n} \frac{t^{4 n + 2}}{(2 n + 1)!},$ one has $S (x) =$ $\sum_{n = 0}^{\infty} {(−1)}^{n} \frac{x^{4 n + 3}}{(4 n + 3) (2 n + 1)!}$ and $C (x) =$ $\sum_{n = 0}^{\infty} {(−1)}^{n} \frac{x^{4 n + 1}}{(4 n + 1) (2 n)!} .$ The sums of the first $50$ nonzero terms are plotted below with $C_{50} (x)$ the solid curve and $S_{50} (x)$ the dashed curve.

6-247 §6.4 source ↗

The following exercises deal with Fresnel integrals.

Estimate $\int_{0}^{1 / 4} \sqrt{x - x^{2}} d x$ by approximating $\sqrt{1 - x}$ using the binomial approximation $1 -$ $\frac{x}{2} -$ $\frac{x^{2}}{8} -$ $\frac{x^{3}}{16} -$ $\frac{5 x^{4}}{2128} -$ $\frac{7 x^{5}}{256} .$

Answer

$\int_{0}^{1 / 4} \sqrt{x} (1 - \frac{x}{2} - \frac{x^{2}}{8} - \frac{x^{3}}{16} - \frac{5 x^{4}}{128} - \frac{7 x^{5}}{256}) d x$

$=$ $\frac{2}{3} 2^{−3} -$ $\frac{1}{2} \frac{2}{5} 2^{−5} -$ $\frac{1}{8} \frac{2}{7} 2^{−7} -$ $\frac{1}{16} \frac{2}{9} 2^{−9} -$ $\frac{5}{128} \frac{2}{11} 2^{−11} -$ $\frac{7}{256} \frac{2}{13} 2^{−13} =$ $0.0767732 ...$

whereas $\int_{0}^{1 / 4} \sqrt{x - x^{2}} d x =$ $0.076773 .$

6-249 §6.4 source ↗

The following exercises deal with Fresnel integrals.

Use the approximation $T \approx$ $2 π \sqrt{\frac{L}{g}} (1 + \frac{k^{2}}{4})$ to approximate the period of a pendulum having length $10$ meters and maximum angle $θ_{max} = \frac{π}{6}$ where $k =$ $sin (\frac{θ_{max}}{2}) .$ Compare this with the small angle estimate $T \approx 2 π \sqrt{\frac{L}{g}} .$

Answer

T \approx

2 π \sqrt{\frac{10}{9.8}} (1 + \frac{{sin}^{2} (θ / 12)}{4}) \approx

6.453

seconds. The small angle estimate is

T \approx

2 π \sqrt{\frac{10}{9.8} \approx 6.347} .

The relative error is around

2

percent.

6-251 §6.4 source ↗

The following exercises deal with Fresnel integrals.

Evaluate $\int_{0}^{π / 2} {sin}^{4} θ d θ$ in the approximation $T =$ $4 \sqrt{\frac{L}{g}} \int_{0}^{π / 2} (1 + \frac{1}{2} k^{2} {sin}^{2} θ + \frac{3}{8} k^{4} {sin}^{4} θ + ⋯) d θ$ to obtain an improved estimate for $T .$

Answer

\int_{0}^{π / 2} {sin}^{4} θ d θ =

\frac{3 π}{16} .

Hence

T \approx

2 π \sqrt{\frac{L}{g}} (1 + \frac{k^{2}}{4} + \frac{9}{256} k^{4}) .

6-253 §6.R source ↗

True or False? In the following exercises, justify your answer with a proof or a counterexample.

If the radius of convergence for a power series $\sum_{n = 0}^{\infty} a_{n} x^{n}$ is $5,$ then the radius of convergence for the series $\sum_{n = 1}^{\infty} n a_{n} x^{n - 1}$ is also $5 .$

Answer

True

6-255 §6.R source ↗

True or False? In the following exercises, justify your answer with a proof or a counterexample.

For small values of $x, sin x \approx x .$

Answer

True

6-257 §6.R source ↗

In the following exercises, find the radius of convergence and the interval of convergence for the given series.

$\sum_{n = 0}^{\infty} n^{2} {(x - 1)}^{n}$

Answer

ROC:

1;

IOC:

(0, 2)

6-259 §6.R source ↗

In the following exercises, find the radius of convergence and the interval of convergence for the given series.

$\sum_{n = 0}^{\infty} \frac{3 n x^{n}}{12^{n}}$

Answer

ROC:

12;

IOC:

(−12, 12)

6-261 §6.R source ↗

In the following exercises, find the power series representation for the given function. Determine the radius of convergence and the interval of convergence for that series.

$f (x) = \frac{x^{2}}{x + 3}$

Answer

\sum_{n = 0}^{\infty} \frac{{(−1)}^{n}}{3^{n + 1}} x^{n + 2};

ROC:

3;

IOC:

(−3, 3)

6-263 §6.R source ↗

In the following exercises, find the power series for the given function using term-by-term differentiation or integration.

$f (x) =$ ${tan}^{−1} (2 x)$

Answer

integration:

\sum_{n = 0}^{\infty} \frac{{(−1)}^{n}}{2 n + 1} {(2 x)}^{2 n + 1}

6-265 §6.R source ↗

In the following exercises, evaluate the Taylor series expansion of degree four for the given function at the specified point. What is the error in the approximation?

$f (x) =$ $x^{3} -$ $2 x^{2} +$ $4,$ $a =$ $−3$

Answer

p_{4} (x) =

{(x + 3)}^{3} -

11 {(x + 3)}^{2} +

39 (x + 3) -

41;

exact

6-267 §6.R source ↗

In the following exercises, find the Maclaurin series for the given function.

$f (x) =$ $cos (3 x)$

Answer

\sum_{n = 0}^{\infty} \frac{{(−1)}^{n} {(3 x)}^{2 n}}{2 n!}

6-269 §6.R source ↗

In the following exercises, find the Taylor series at the given value.

$f (x) =$ $sin x,$ $a =$ $\frac{π}{2}$

Answer

\sum_{n = 0}^{\infty} \frac{{(−1)}^{n}}{(2 n)!} {(x - \frac{π}{2})}^{2 n}

6-271 §6.R source ↗

In the following exercises, find the Maclaurin series for the given function.

$f (x) = e^{− x^{2}} - 1$

Answer

\sum_{n = 1}^{\infty} \frac{{(−1)}^{n}}{n!} x^{2 n}

6-273 §6.R source ↗

In the following exercises, find the Maclaurin series for $F (x) =$ $\int_{0}^{x} f (t) d t$ by integrating the Maclaurin series of $f (x)$ term by term.

$f (x) = \frac{sin x}{x}$

Answer

F (x) =

\sum_{n = 0}^{\infty} \frac{{(−1)}^{n}}{(2 n + 1) (2 n + 1)!} x^{2 n + 1}

6-275 §6.R source ↗

In the following exercises, find the Maclaurin series for $F (x) =$ $\int_{0}^{x} f (t) d t$ by integrating the Maclaurin series of $f (x)$ term by term.

Use power series to prove Euler’s formula: $e^{i x} =$ $cos x +$ $i sin x$

Answer

Answers may vary.

6-277 §6.R source ↗

The following exercises consider problems of annuity payments.

A lottery winner has an annuity that has a present value of $$ 10$ million. What interest rate would they need to live on perpetual annual payments of $$ 250,000 ?$

Answer

2.5 %

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Problems and answers are adapted from Calculus Volume 2 by Gilbert Strang and Edwin “Jed” Herman, OpenStax, and are used under the CC BY-NC-SA 4.0 licence. Access for free at https://openstax.org/books/calculus-volume-2/pages/1-introduction. Illustrations: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. This adaptation is released under the same licence, is non-commercial, and is neither affiliated with nor endorsed by OpenStax or Rice University.