6 Power Series 幂级数

本页译自 OpenStax《Calculus Volume 2》第 6 章 Power Series（幂级数）：6.1–6.4 四节（幂级数与其函数／幂级数的性质／泰勒与麦克劳林级数／泰勒级数的应用）+ Chapter Review / Key Terms / Key Equations / Key Concepts / Review Exercises 全译，段段对照。公式经本地 MathJax 渲染，自定义宏已注入。

6.1 Power Series and Functions 6.1 幂级数与其函数

• 6.1.1 Identify a power series and provide examples of them.
• 6.1.2 Determine the radius of convergence and interval of convergence of a power series.
• 6.1.3 Use a power series to represent a function.

• 6.1.1 识别幂级数并举例说明。
• 6.1.2 确定幂级数的收敛半径与收敛区间。
• 6.1.3 用幂级数表示一个函数。

A power series is a type of series with terms involving a variable. More specifically, if the variable is *x*, then all the terms of the series involve powers of *x*. As a result, a power series can be thought of as an infinite polynomial. Power series are used to represent common functions and also to define new functions. In this section we define power series and show how to determine when a power series converges and when it diverges. We also show how to represent certain functions using power series.

Form of a Power Series 幂级数的形式

A series of the form

$${\sum\limits_{n = 0}^{\infty}{c_{n}x^{n}}} = c_{0} + c_{1}x + c_{2}x^{2} + \text{⋯},$$

where *x* is a variable and the coefficients *c_n* are constants, is known as a power series. The series

$$1 + x + x^{2} + \text{⋯} = {\sum\limits_{n = 0}^{\infty}x^{n}}$$

is an example of a power series. Since this series is a geometric series with ratio $r = x,$ we know that it converges if $|x| < 1$ and diverges if $|x| \geq 1.$

A series of the form

$${\sum\limits_{n = 0}^{\infty}{c_{n}x^{n}}} = c_{0} + c_{1}x + c_{2}x^{2} + \text{⋯}$$ (6.1)

is a power series centered at $x = 0.$ A series of the form

$${\sum\limits_{n = 0}^{\infty}{c_{n}\left( {x - a} \right)^{n}}} = c_{0} + c_{1}\left( {x - a} \right) + c_{2}\left( {x - a} \right)^{2} + \text{⋯}$$ (6.2)

is a power series centered at $x = a.$

To make this definition precise, we stipulate that $x^{0} = 1$ and $\left( {x - a} \right)^{0} = 1$ even when $x = 0$ and $x = a,$ respectively.

The series

$${\sum\limits_{n = 0}^{\infty}\frac{x^{n}}{n\text{!}}} = 1 + x + \frac{x^{2}}{2\text{!}} + \frac{x^{3}}{3\text{!}} + \text{⋯}$$

and

$${\sum\limits_{n = 0}^{\infty}{n\text{!}x^{n}}} = 1 + x + 2\text{!}x^{2} + 3\text{!}x^{3} + \text{⋯}$$

are both power series centered at $x = 0.$ The series

$${\sum\limits_{n = 0}^{\infty}\frac{\left( {x - 2} \right)^{n}}{\left( {n + 1} \right)3^{n}}} = 1 + \frac{x - 2}{2 \cdot 3} + \frac{\left( {x - 2} \right)^{2}}{3 \cdot 3^{2}} + \frac{\left( {x - 2} \right)^{3}}{4 \cdot 3^{3}} + \text{⋯}$$

is a power series centered at $x = 2.$

Convergence of a Power Series 幂级数的收敛性

Since the terms in a power series involve a variable *x*, the series may converge for certain values of *x* and diverge for other values of *x*. For a power series centered at $x = a,$ the value of the series at $x = a$ is given by $c_{0}.$ Therefore, a power series always converges at its center. Some power series converge only at that value of *x*. Most power series, however, converge for more than one value of *x*. In that case, the power series either converges for all real numbers *x* or converges for all *x* in a finite interval. For example, the geometric series $\sum\limits_{n = 0}^{\infty}x^{n}$ converges for all *x* in the interval $\left( {-1,1} \right),$ but diverges for all *x* outside that interval. We now summarize these three possibilities for a general power series.

Convergence of a Power Series 幂级数的收敛性

Consider the power series ${\sum\limits_{n = 0}^{\infty}{c_{n}\left( {x - a} \right)^{n}}}.$ The series satisfies exactly one of the following properties:

1. The series converges at $x = a$ and diverges for all $x \neq a.$
2. The series converges for all real numbers *x*.
3. There exists a real number $R > 0$ such that the series converges if $\left| {x - a} \right| < R$ and diverges if $\left| {x - a} \right| > R.$ At the values *x* where $\left| {x - a} \right| = R,$ the series may converge or diverge.

1. 级数在 $x = a$ 处收敛，而对所有 $x \neq a$ 发散。
2. 级数对所有实数 *x* 收敛。
3. 存在实数 $R > 0$，使得当 $\left| {x - a} \right| < R$ 时级数收敛，当 $\left| {x - a} \right| > R$ 时级数发散。在 $\left| {x - a} \right| = R$ 的 *x* 值处，级数可能收敛也可能发散。

Proof 证明

Suppose that the power series is centered at $a = 0.$ (For a series centered at a value of *a* other than zero, the result follows by letting $y = x - a$ and considering the series ${\sum\limits_{n = 1}^{\infty}{c_{n}y^{n}}}.)$ We must first prove the following fact:

If there exists a real number $d \neq 0$ such that $\sum\limits_{n = 0}^{\infty}{c_{n}d^{n}}$ converges, then the series $\sum\limits_{n = 0}^{\infty}{c_{n}x^{n}}$ converges absolutely for all *x* such that $|x| < |d|.$

Since $\sum\limits_{n = 0}^{\infty}{c_{n}d^{n}}$ converges, the *n*th term $c_{n}d^{n}\rightarrow 0$ as $n\rightarrow\infty.$ Therefore, there exists an integer *N* such that $\left| {c_{n}d^{n}} \right| \leq 1$ for all $n \geq N.$ Writing

$$\left| {c_{n}x^{n}} \right| = \left| {c_{n}d^{n}} \right|\left| \frac{x}{d} \right|^{n},$$

we conclude that, for all $n \geq N,$

$$\left| {c_{n}x^{n}} \right| \leq \left| \frac{x}{d} \right|^{n}.$$

The series

$$\sum\limits_{n = N}^{\infty}\left| \frac{x}{d} \right|^{n}$$

is a geometric series that converges if $\left| \frac{x}{d} \right| < 1.$ Therefore, by the comparison test, we conclude that $\sum\limits_{n = N}^{\infty}{c_{n}x^{n}}$ also converges for $|x| < |d|.$ Since we can add a finite number of terms to a convergent series, we conclude that $\sum\limits_{n = 0}^{\infty}{c_{n}x^{n}}$ converges for $|x| < |d|.$

With this result, we can now prove the theorem. Consider the series

$$\sum\limits_{n = 0}^{\infty}{a_{n}x^{n}}$$

and let *S* be the set of real numbers for which the series converges. Suppose that the set $S = \left\{ 0 \right\}.$ Then the series falls under case i. Suppose that the set *S* is the set of all real numbers. Then the series falls under case ii. Suppose that $S \neq \left\{ 0 \right\}$ and *S* is not the set of real numbers. Then there exists a real number $x* \neq 0$ such that the series does not converge. Thus, the series cannot converge for any *x* such that $|x| > \left| {x*} \right|.$ Therefore, the set *S* must be a bounded set, which means that it must have a smallest upper bound. (This fact follows from the Least Upper Bound Property for the real numbers, which is beyond the scope of this text and is covered in real analysis courses.) Call that smallest upper bound *R*. Since $S \neq \left\{ 0 \right\},$ the number $R > 0.$ Therefore, the series converges for all *x* such that $|x| < R,$ and the series falls into case iii.

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If a series $\sum\limits_{n = 0}^{\infty}{c_{n}\left( {x - a} \right)^{n}}$ falls into case iii. of Convergence of a Power Series, then the series converges for all *x* such that $\left| {x - a} \right| < R$ for some $R > 0,$ and diverges for all *x* such that $\left| x - a \middle| > R. \right.$ The series may converge or diverge at the values *x* where $\left| {x - a} \right| = R.$ The set of values *x* for which the series $\sum\limits_{n = 0}^{\infty}{c_{n}\left( {x - a} \right)^{n}}$ converges is known as the interval of convergence. Since the series diverges for all values *x* where $\left| {x - a} \right| > R,$ the length of the interval is 2*R*, and therefore, the radius of the interval is *R*. The value *R* is called the radius of convergence. For example, since the series $\sum\limits_{n = 0}^{\infty}x^{n}$ converges for all values *x* in the interval $\left( {-1,1} \right)$ and diverges for all values *x* such that $|x| \geq 1,$ the interval of convergence of this series is $\left( {-1,1} \right).$ Since the length of the interval is 2, the radius of convergence is 1.

Consider the power series ${\sum\limits_{n = 0}^{\infty}{c_{n}\left( {x - a} \right)^{n}}}.$ The set of real numbers *x* where the series converges is the interval of convergence. If there exists a real number $R > 0$ such that the series converges for $\left| {x - a} \right| < R$ and diverges for $\left| {x - a} \right| > R,$ then *R* is the radius of convergence. If the series converges only at $x = a,$ we say the radius of convergence is $R = 0.$ If the series converges for all real numbers *x*, we say the radius of convergence is $R = \infty$ (Figure 6.2).

To determine the interval of convergence for a power series, we typically apply the ratio test. In Example 6.1, we show the three different possibilities illustrated in Figure 6.2.

Finding the Interval and Radius of Convergence 求收敛区间与收敛半径

For each of the following series, find the interval and radius of convergence.

1. $\sum\limits_{n = 0}^{\infty}\frac{x^{n}}{n\text{!}}$

2. $\sum\limits_{n = 0}^{\infty}{n\text{!}x^{n}}$

3. $\sum\limits_{n = 0}^{\infty}\frac{\left( {x - 2} \right)^{n}}{\left( {n + 1} \right)3^{n}}$

Solution 解答

1. To check for convergence, apply the ratio test. We have

$$\begin{array}{cl} \rho & {= \underset{n\rightarrow\infty}{\text{lim}}\left| \frac{\frac{x^{n + 1}}{\left( {n + 1} \right)\text{!}}}{\frac{x^{n}}{n\text{!}}} \right|} \\ & {= \underset{n\rightarrow\infty}{\text{lim}}\left| {\frac{x^{n + 1}}{\left( {n + 1} \right)\text{!}} \cdot \frac{n\text{!}}{x^{n}}} \right|} \\ & {= \underset{n\rightarrow\infty}{\text{lim}}\left| {\frac{x^{n + 1}}{\left( {n + 1} \right) \cdot n\text{!}} \cdot \frac{n\text{!}}{x^{n}}} \right|} \\ & {= \underset{n\rightarrow\infty}{\text{lim}}\left| \frac{x}{n + 1} \right|} \\ & {= |x|\underset{n\rightarrow\infty}{\text{lim}}\frac{1}{n + 1}} \\ & {= 0 < 1} \end{array}$$

for all values of *x*. Therefore, the series converges for all real numbers *x*. The interval of convergence is $\left( {\text{−}\infty,\infty} \right)$ and the radius of convergence is $R = \infty.$

2. Apply the ratio test. For $x \neq 0,$ we see that

$$\begin{matrix} \rho & {= \underset{n\rightarrow\infty}{\text{lim}}\left| \frac{\left( {n + 1} \right)\text{!}x^{n + 1}}{n\text{!}x^{n}} \right|} \\ & {= \underset{n\rightarrow\infty}{\text{lim}}\left| {\left( {n + 1} \right)x} \right|} \\ & {= |x|\underset{n\rightarrow\infty}{\text{lim}}\left( {n + 1} \right)} \\ & {= \infty.} \end{matrix}$$

Therefore, the series diverges for all $x \neq 0.$ Since the series is centered at $x = 0,$ it must converge there, so the series converges only for $x = 0.$ The interval of convergence is the single value $x = 0$ and the radius of convergence is $R = 0.$

3. In order to apply the ratio test, consider

$$\begin{array}{cl} \rho & {= \underset{n\rightarrow\infty}{\text{lim}}\left| \frac{\frac{\left( {x - 2} \right)^{n + 1}}{\left( {n + 2} \right)3^{n + 1}}}{\frac{\left( {x - 2} \right)^{n}}{\left( {n + 1} \right)3^{n}}} \right|} \\ & {= \underset{n\rightarrow\infty}{\text{lim}}\left| {\frac{\left( {x - 2} \right)^{n + 1}}{\left( {n + 2} \right)3^{n + 1}} \cdot \frac{\left( {n + 1} \right)3^{n}}{\left( {x - 2} \right)^{n}}} \right|} \\ & {= \underset{n\rightarrow\infty}{\text{lim}}\left| \frac{\left( {x - 2} \right)\left( {n + 1} \right)}{3\left( {n + 2} \right)} \right|} \\ & {= \frac{|x - 2|}{3}.} \end{array}$$

The ratio $\rho < 1$ if $\left| {x - 2} \right| < 3.$ Since $\left| {x - 2} \right| < 3$ implies that $-3 < x - 2 < 3,$ the series converges absolutely if $-1 < x < 5.$ The ratio $\rho > 1$ if $\left| {x - 2} \right| > 3.$ Therefore, the series diverges if $x < -1$ or $x > 5.$ The ratio test is inconclusive if $\rho = 1.$ The ratio $\rho = 1$ if and only if $x = -1$ or $x = 5.$ We need to test these values of *x* separately. For $x = -1,$ the series is given by

$${\sum\limits_{n = 0}^{\infty}\frac{(-1)^{n}}{n + 1}} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \text{⋯}.$$

Since this is the alternating harmonic series, it converges. Thus, the series converges at $x = -1.$ For $x = 5,$ the series is given by

$${\sum\limits_{n = 0}^{\infty}\frac{1}{n + 1}} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \text{⋯}.$$

This is the harmonic series, which is divergent. Therefore, the power series diverges at $x = 5.$ We conclude that the interval of convergence is $\left\lbrack {-1,5} \right)$ and the radius of convergence is $R = 3.$

Find the interval and radius of convergence for the series ${\sum\limits_{n = 1}^{\infty}\frac{x^{n}}{\sqrt{n}}}.$

Representing Functions as Power Series 用幂级数表示函数

Being able to represent a function by an “infinite polynomial” is a powerful tool. Polynomial functions are the easiest functions to analyze, since they only involve the basic arithmetic operations of addition, subtraction, multiplication, and division. If we can represent a complicated function by an infinite polynomial, we can use the polynomial representation to differentiate or integrate it. In addition, we can use a truncated version of the polynomial expression to approximate values of the function. So, the question is, when can we represent a function by a power series?

Consider again the geometric series

$$1 + x + x^{2} + x^{3} + \text{⋯} = {\sum\limits_{n = 0}^{\infty}x^{n}}.$$ (6.3)

Recall that the geometric series

$$a + ar + ar^{2} + ar^{3} + \text{⋯}$$

converges if and only if $|r| < 1.$ In that case, it converges to $\frac{a}{1 - r}.$ Therefore, if $|x| < 1,$ the series in Example 6.3 converges to $\frac{1}{1 - x}$ and we write

$$1 + x + x^{2} + x^{3} + \text{⋯} = \frac{1}{1 - x}\ \text{for}\ |x| < 1.$$

As a result, we are able to represent the function $f(x) = \frac{1}{1 - x}$ by the power series

$$1 + x + x^{2} + x^{3} + \text{⋯}\ \text{when}\ |x| < 1.$$

We now show graphically how this series provides a representation for the function $f(x) = \frac{1}{1 - x}$ by comparing the graph of *f* with the graphs of several of the partial sums of this infinite series.

Graphing a Function and Partial Sums of its Power Series 绘制函数及其幂级数部分和的图像

Sketch a graph of $f(x) = \frac{1}{1 - x}$ and the graphs of the corresponding partial sums $S_{N}(x) = {\sum\limits_{n = 0}^{N}x^{n}}$ for $N = 2,4,6$ on the interval $\left( {-1,1} \right).$ Comment on the approximation $S_{N}$ as *N* increases.

Solution

From the graph in Figure 6.3 you see that as *N* increases, $S_{N}$ becomes a better approximation for $f(x) = \frac{1}{1 - x}$ for *x* in the interval $\left( {-1,1} \right).$

Sketch a graph of $f(x) = \frac{1}{1 - x^{2}}$ and the corresponding partial sums $S_{N}(x) = {\sum\limits_{n = 0}^{N}x^{2n}}$ for $N = 2,4,6$ on the interval $\left( {-1,1} \right).$

Next we consider functions involving an expression similar to the sum of a geometric series and show how to represent these functions using power series.

Representing a Function with a Power Series 用幂级数表示函数

Use a power series to represent each of the following functions $f.$ Find the interval of convergence.

1. $f(x) = \frac{1}{1 + x^{3}}$

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

Solution

1. You should recognize this function *f* as the sum of a geometric series, because

$$\frac{1}{1 + x^{3}} = \frac{1}{1 - \left( {\text{−}x^{3}} \right)}.$$

Using the fact that, for $|r| < 1,\frac{a}{1 - r}$ is the sum of the geometric series

$${\sum\limits_{n = 0}^{\infty}a}r^{n} = a + ar + ar^{2} + \text{⋯},$$

we see that, for $\left| {\text{−}x^{3}} \right| < 1,$

$$\begin{array}{cl} \frac{1}{1 + x^{3}} & {= \frac{1}{1 - \left( {\text{−}x^{3}} \right)}} \\ & {= {\sum\limits_{n = 0}^{\infty}\left( {\text{−}x^{3}} \right)^{n}}} \\ & {= 1 - x^{3} + x^{6} - x^{9} + \text{⋯}.} \end{array}$$

Since this series converges if and only if $\left| {\text{−}x^{3}} \right| < 1,$ the interval of convergence is $\left( {-1,1} \right),$ and we have

$$\frac{1}{1 + x^{3}} = 1 - x^{3} + x^{6} - x^{9} + \text{⋯}\ \text{for}\ |x| < 1.$$

2. This function is not in the exact form of a sum of a geometric series. However, with a little algebraic manipulation, we can relate *f* to a geometric series. By factoring 4 out of the two terms in the denominator, we obtain

$$\begin{array}{cl} \frac{x^{2}}{4 - x^{2}} & {= \frac{x^{2}}{4\left( {1 - \frac{x^{2}}{4}} \right)}} \\ & {= \frac{x^{2}}{4\left( {1 - \left( \frac{x}{2} \right)^{2}} \right)}.} \end{array}$$

Therefore, we have

$$\begin{array}{cl} \frac{x^{2}}{4 - x^{2}} & {= \frac{x^{2}}{4\left( {1 - \left( \frac{x}{2} \right)^{2}} \right)}} \\ & {= \frac{\frac{x^{2}}{4}}{1 - \left( \frac{x}{2} \right)^{2}}} \\ & {= {\sum\limits_{n = 0}^{\infty}{\frac{x^{2}}{4}\left( \frac{x}{2} \right)^{2n}}}.} \end{array}$$

The series converges as long as $\left| \left( \frac{x}{2} \right)^{2} \right| < 1$ (note that when $\left| \left( \frac{x}{2} \right)^{2} \right| = 1$ the series does not converge). Solving this inequality, we conclude that the interval of convergence is $\left( {-2,2} \right)$ and

$$\begin{array}{cl} \frac{x^{2}}{4 - x^{2}} & {= {\sum\limits_{n = 0}^{\infty}\frac{x^{2n + 2}}{4^{n + 1}}}} \\ & {= \frac{x^{2}}{4} + \frac{x^{4}}{4^{2}} + \frac{x^{6}}{4^{3}} + \text{⋯}} \end{array}$$

for $|x| < 2.$

Represent the function $f(x) = \frac{x^{3}}{2 - x}$ using a power series and find the interval of convergence.

In the remaining sections of this chapter, we will show ways of deriving power series representations for many other functions, and how we can make use of these representations to evaluate, differentiate, and integrate various functions.

Section 6.1 Exercises 6.1 节习题

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

1.

If $\sum\limits_{n = 1}^{\infty}{a_{n}x^{n}}$ converges, then $a_{n}x^{n}\rightarrow 0$ as $n\rightarrow\infty.$

2\.

$\sum\limits_{n = 1}^{\infty}{a_{n}x^{n}}$ converges at $x = 0$ for any real numbers $a_{n}.$

3.

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

4\.

If $\sum\limits_{n = 1}^{\infty}{a_{n}x^{n}}$ has radius of convergence $R > 0$ and if $\left| b_{n} \right| \leq \left| a_{n} \right|$ for all *n*, then the radius of convergence of $\sum\limits_{n = 1}^{\infty}{b_{n}x^{n}}$ is greater than or equal to *R*.

5.

Suppose that $\sum\limits_{n = 0}^{\infty}{a_{n}\left( {x - 3} \right)^{n}}$ converges at $x = 6.$ At which of the following points might the series diverge? Use the fact that if $\sum{a_{n}\left( {x - c} \right)^{n}}$ converges at *x*, then it converges at any point closer to *c* than *x*.

1. $x = 1$

2. $x = 2$

3. $x = 3$

4. $x = 0$

5. $x = 5.99$

6. $x = 0.000001$

6\.

Suppose that $\sum\limits_{n = 0}^{\infty}{a_{n}\left( {x + 1} \right)^{n}}$ converges at $x = -2.$ At which of the following points must the series also converge? Use the fact that if $\sum{a_{n}\left( {x - c} \right)^{n}}$ converges at *x*, then it converges at any point closer to *c* than *x*.

1. $x = 2$

2. $x = -1$

3. $x = -3$

4. $x = 0$

5. $x = 0.99$

6. $x = 0.000001$

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

7.

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

8\.

$\sum\limits_{n = 0}^{\infty}\frac{a_{n}x^{n}}{2^{n}}$

9.

$\sum\limits_{n = 0}^{\infty}\frac{a_{n}\pi^{n}x^{n}}{e^{n}}$

10\.

$\sum\limits_{n = 0}^{\infty}\frac{a_{n}(-1)^{n}x^{n}}{10^{n}}$

11.

$\sum\limits_{n = 0}^{\infty}{a_{n}(-1)^{n}x^{2n}}$

12\.

$\sum\limits_{n = 0}^{\infty}{a_{n}(-4)^{n}x^{2n}}$

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

13.

$\sum\limits_{n = 1}^{\infty}\frac{\left( {2x} \right)^{n}}{n}$

14\.

$\sum\limits_{n = 1}^{\infty}{(-1)^{n}\frac{x^{n}}{\sqrt{n}}}$

15.

$\sum\limits_{n = 1}^{\infty}\frac{nx^{n}}{2^{n}}$

16\.

$\sum\limits_{n = 1}^{\infty}\frac{nx^{n}}{e^{n}}$

17.

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

18\.

$\sum\limits_{k = 1}^{\infty}\frac{k^{e}x^{k}}{e^{k}}$

19.

$\sum\limits_{k = 1}^{\infty}\frac{\pi^{k}x^{k}}{k^{\pi}}$

20\.

$\sum\limits_{n = 1}^{\infty}\frac{x^{n}}{n\text{!}}$

21.

$\sum\limits_{n = 1}^{\infty}\frac{10^{n}x^{n}}{n\text{!}}$

22\.

${\sum\limits_{n = 1}^{\infty}(-1)^{n}}\frac{x^{n}}{\text{ln}\mspace{2mu}\left( {2n} \right)}$

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

23.

$\sum\limits_{k = 1}^{\infty}\frac{\left( {k\text{!}} \right)^{2}x^{k}}{\left( {2k} \right)\text{!}}$

24\.

$\sum\limits_{n = 1}^{\infty}\frac{\left( {2n} \right)\text{!}x^{n}}{n^{2n}}$

25.

${\sum\limits_{k = 1}^{\infty}\frac{k\text{!}}{1 \cdot 3 \cdot 5\text{⋯}\left( {2k - 1} \right)}}x^{k}$

26\.

$\sum\limits_{k = 1}^{\infty}{\frac{2 \cdot 4 \cdot 6\text{⋯}2k}{\left( {2k} \right)\text{!}}x^{k}}$

27.

$\sum\limits_{n = 1}^{\infty}\frac{x^{n}}{\begin{pmatrix} {2n} \\ n \end{pmatrix}}$ where $\begin{pmatrix} n \\ k \end{pmatrix} = \frac{n\text{!}}{k\text{!}\left( {n - k} \right)\text{!}}$

28\.

$\sum\limits_{n = 1}^{\infty}{\text{sin}^{2}nx^{n}}$

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

29.

$\sum\limits_{n = 1}^{\infty}{\frac{\left( {n\text{!}} \right)^{3}}{\left( {3n} \right)\text{!}}x^{n}}$

30\.

$\sum\limits_{n = 1}^{\infty}{\frac{2^{3n}\left( {n\text{!}} \right)^{3}}{\left( {3n} \right)\text{!}}x^{n}}$

31.

$\sum\limits_{n = 1}^{\infty}{\frac{n\text{!}}{n^{n}}x^{n}}$

32\.

$\sum\limits_{n = 1}^{\infty}{\frac{\left( {2n} \right)\text{!}}{n^{2n}}x^{n}}$

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

33.

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

34\.

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

35.

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

36\.

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

37.

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

38\.

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

39.

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

40\.

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

41.

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

42\.

$f(x) = \frac{x^{2}}{5 - 4x + x^{2}};a = 2$

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

43.

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

44\.

$\sum\limits_{n = 1}^{\infty}\frac{x^{n}}{n^{n}}$

45.

$\sum\limits_{k = 1}^{\infty}{\left( \frac{k - 1}{2k + 3} \right)^{k}x^{k}}$

46\.

$\sum\limits_{k = 1}^{\infty}{\left( \frac{2k^{2} - 1}{k^{2} + 3} \right)^{k}x^{k}}$

47.

${\sum\limits_{n = 1}^{\infty}a_{n}} = \left( {n^{1\text{/}n} - 1} \right)^{n}x^{n}$

48\.

Suppose that $p(x) = {\sum\limits_{n = 0}^{\infty}{a_{n}x^{n}}}$ such that $a_{n} = 0$ if *n* is even. Explain why $p(x) = - p\left( {\text{−}x} \right).$

49.

Suppose that $p(x) = {\sum\limits_{n = 0}^{\infty}{a_{n}x^{n}}}$ such that $a_{n} = 0$ if *n* is odd. Explain why $p(x) = p\left( {\text{−}x} \right).$

50\.

Suppose that $p(x) = {\sum\limits_{n = 0}^{\infty}{a_{n}x^{n}}}$ converges on $\left( {-1,1} \right\rbrack.$ Find the interval of convergence of $p\left( {Ax} \right).$

51.

Suppose that $p(x) = {\sum\limits_{n = 0}^{\infty}{a_{n}x^{n}}}$ converges on $\left( {-1,1} \right\rbrack.$ Find the interval of convergence of $p\left( {2x - 1} \right).$

In the following exercises, suppose that $p(x) = {\sum\limits_{n = 0}^{\infty}{a_{n}x^{n}}}$ satisfies $\underset{n\rightarrow\infty}{\text{lim}}\frac{a_{n + 1}}{a_{n}} = 1$ where $a_{n} \geq 0$ for each *n*. State whether each series converges on the full interval $\left( {-1,1} \right),$ or if there is not enough information to draw a conclusion. Use the comparison test when appropriate.

52\.

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

53.

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

54\.

${\sum\limits_{n = 0}^{\infty}{a_{2n}x^{n}}}\ \left( {Hint\text{:}\ x = \text{±}\sqrt{x^{2}}} \right)$

55.

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

56\.

Suppose that $p(x)$ is a polynomial of degree *N*. Find the radius and interval of convergence of ${\sum\limits_{n = 1}^{\infty}{p(n)x^{n}}}.$

57.

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

58\.

\[T\] Plot the graphs of $\text{−}\text{ln}\mspace{2mu}\left( {1 - x} \right)$ and of the partial sums $S_{N} = {\sum\limits_{n = 1}^{N}\frac{x^{n}}{n}}$ for $n = 10,50,100$ on the interval $\left\lbrack {-0.99,0.99} \right\rbrack.$ Comment on the behavior of the sums near $x = -1$ and near $x = 1$ as *N* increases.

59.

\[T\] Plot the graphs of the partial sums $S_{n} = {\sum\limits_{n = 1}^{N}\frac{x^{n}}{n^{2}}}$ for $n = 10,50,100$ on the interval $\left\lbrack {-0.99,0.99} \right\rbrack.$ Comment on the behavior of the sums near $x = -1$ and near $x = 1$ as *N* increases.

60\.

\[T\] Plot the graphs of the partial sums $S_{N} = {\sum\limits_{n = 1}^{N}{\text{sin}\mspace{2mu} nx^{n}}}$ for $n = 10,50,100$ on the interval $\left\lbrack {-0.99,0.99} \right\rbrack.$ Comment on the behavior of the sums near $x = -1$ and near $x = 1$ as *N* increases.

61.

\[T\] Plot the graphs of the partial sums $S_{N} = {\sum\limits_{n = 0}^{N}{(-1)^{n}\frac{x^{2n + 1}}{\left( {2n + 1} \right)\text{!}}}}$ for $n = 3,5,10$ on the interval $\left\lbrack {-2\pi,2\pi} \right\rbrack.$ Comment on how these plots approximate $\text{sin}\mspace{2mu} x$ as *N* increases.

62\.

\[T\] Plot the graphs of the partial sums $S_{N} = {\sum\limits_{n = 0}^{N}{(-1)^{n}\frac{x^{2n}}{\left( {2n} \right)\text{!}}}}$ for $n = 3,5,10$ on the interval $\left\lbrack {-2\pi,2\pi} \right\rbrack.$ Comment on how these plots approximate $\text{cos}\mspace{2mu} x$ as *N* increases.

6.2 Properties of Power Series 6.2 幂级数的性质

• 6.2.1 Combine power series by addition or subtraction.
• 6.2.2 Create a new power series by multiplication by a power of the variable or a constant, or by substitution.
• 6.2.3 Multiply two power series together.
• 6.2.4 Differentiate and integrate power series term-by-term.

• 6.2.1 通过加法或减法组合幂级数。
• 6.2.2 通过对变量的幂或常数相乘，或通过代换，构造新的幂级数。
• 6.2.3 将两个幂级数相乘。
• 6.2.4 对幂级数逐项求导与逐项积分。

In the preceding section on power series and functions we showed how to represent certain functions using power series. In this section we discuss how power series can be combined, differentiated, or integrated to create new power series. This capability is particularly useful for a couple of reasons. First, it allows us to find power series representations for certain elementary functions, by writing those functions in terms of functions with known power series. For example, given the power series representation for $f(x) = \frac{1}{1 - x},$ we can find a power series representation for $f^{\prime}(x) = \frac{1}{\left( {1 - x} \right)^{2}}.$ Second, being able to create power series allows us to define new functions that cannot be written in terms of elementary functions. This capability is particularly useful for solving differential equations for which there is no solution in terms of elementary functions.

Combining Power Series 幂级数的组合

If we have two power series with the same interval of convergence, we can add or subtract the two series to create a new power series, also with the same interval of convergence. Similarly, we can multiply a power series by a power of *x* or evaluate a power series at $x^{m}$ for a positive integer *m* to create a new power series. Being able to do this allows us to find power series representations for certain functions by using power series representations of other functions. For example, since we know the power series representation for $f(x) = \frac{1}{1 - x},$ we can find power series representations for related functions, such as

$$y = \frac{3x}{1 - x^{2}}\ \text{and}\ y = \frac{1}{\left( {x - 1} \right)\left( {x - 3} \right)}.$$

In Combining Power Series we state results regarding addition or subtraction of power series, composition of a power series, and multiplication of a power series by a power of the variable. For simplicity, we state the theorem for power series centered at $x = 0.$ Similar results hold for power series centered at $x = a.$

Combining Power Series 幂级数的组合

Suppose that the two power series $\sum\limits_{n = 0}^{\infty}{c_{n}x^{n}}$ and $\sum\limits_{n = 0}^{\infty}{d_{n}x^{n}}$ converge to the functions *f* and *g*, respectively, on a common interval *I*.

1. The power series $\sum\limits_{n = 0}^{\infty}\left( {c_{n}x^{n} \pm d_{n}x^{n}} \right)$ converges to $f \pm g$ on *I*.

2. For any integer $m \geq 0$ and any real number *b*, the power series $\sum\limits_{n = 0}^{\infty}{bx^{m}c_{n}x^{n}}$ converges to $bx^{m}f(x)$ on *I*.

3. For any integer $m \geq 0$ and any real number *b*, the series $\sum\limits_{n = 0}^{\infty}{c_{n}\left( {bx^{m}} \right)^{n}}$ converges to $f\left( {bx^{m}} \right)$ for all *x* such that $bx^{m}$ is in *I*.

Proof 证明

We prove i. in the case of the series ${\sum\limits_{n = 0}^{\infty}\left( {c_{n}x^{n} + d_{n}x^{n}} \right)}.$ Suppose that $\sum\limits_{n = 0}^{\infty}{c_{n}x^{n}}$ and $\sum\limits_{n = 0}^{\infty}{d_{n}x^{n}}$ converge to the functions *f* and *g*, respectively, on the interval *I*. Let *x* be a point in *I* and let $S_{N}(x)$ and $T_{N}(x)$ denote the *N*th partial sums of the series $\sum\limits_{n = 0}^{\infty}{c_{n}x^{n}}$ and ${\sum\limits_{n = 0}^{\infty}{d_{n}x^{n}}},$ respectively. Then the sequence $\left\{ {S_{N}(x)} \right\}$ converges to $f(x)$ and the sequence $\left\{ {T_{N}(x)} \right\}$ converges to $g(x).$ Furthermore, the *N*th partial sum of $\sum\limits_{n = 0}^{\infty}\left( {c_{n}x^{n} + d_{n}x^{n}} \right)$ is

$$\begin{array}{cl} {\sum\limits_{n = 0}^{N}\left( {c_{n}x^{n} + d_{n}x^{n}} \right)} & {= {\sum\limits_{n = 0}^{N}{c_{n}x^{n}}} + {\sum\limits_{n = 0}^{N}{d_{n}x^{n}}}} \\ & {= S_{N}(x) + T_{N}(x).} \end{array}$$

Because

$$\begin{array}{cl} {\underset{N\rightarrow\infty}{\text{lim}}\left( {S_{N}(x) + T_{N}(x)} \right)} & {= \underset{N\rightarrow\infty}{\text{lim}}S_{N}(x) + \underset{N\rightarrow\infty}{\text{lim}}T_{N}(x)} \\ & {= f(x) + g(x),} \end{array}$$

we conclude that the series $\sum\limits_{n = 0}^{\infty}\left( {c_{n}x^{n} + d_{n}x^{n}} \right)$ converges to $f(x) + g(x)~\text{on}~I.$

□

We examine products of power series in a later theorem. First, we show several applications of Combining Power Series and how to find the interval of convergence of a power series given the interval of convergence of a related power series.

Combining Power Series 幂级数的组合

Suppose that $\sum\limits_{n = 0}^{\infty}{a_{n}x^{n}}$ is a power series whose interval of convergence is $\left( {-1,1} \right),$ and suppose that $\sum\limits_{n = 0}^{\infty}{b_{n}x^{n}}$ is a power series whose interval of convergence is $\left( {-2,2} \right).$

1. Find the interval of convergence of the series ${\sum\limits_{n = 0}^{\infty}\left( {a_{n}x^{n} + b_{n}x^{n}} \right)}.$

2. Find the interval of convergence of the series ${\sum\limits_{n = 0}^{\infty}{a_{n}3^{n}x^{n}}}.$

Solution 解答

1. Since the interval $\left( {-1,1} \right)$ is a common interval of convergence of the series $\sum\limits_{n = 0}^{\infty}{a_{n}x^{n}}$ and ${\sum\limits_{n = 0}^{\infty}{b_{n}x^{n}}},$ the interval of convergence of the series $\sum\limits_{n = 0}^{\infty}\left( {a_{n}x^{n} + b_{n}x^{n}} \right)$ is $\left( {-1,1} \right).$

2. Since $\sum\limits_{n = 0}^{\infty}{a_{n}x^{n}}$ is a power series centered at zero with radius of convergence 1, it converges for all *x* in the interval $\left( {-1,1} \right).$ By Combining Power Series, the series

$${\sum\limits_{n = 0}^{\infty}{a_{n}3^{n}x^{n}}} = {\sum\limits_{n = 0}^{\infty}{a_{n}\left( {3x} \right)^{n}}}$$

converges if 3*x* is in the interval $\left( {-1,1} \right).$ Therefore, the series converges for all *x* in the interval $\left( {- \frac{1}{3},\frac{1}{3}} \right).$

Suppose that $\sum\limits_{n = 0}^{\infty}{a_{n}x^{n}}$ has an interval of convergence of $\left( {-1,1} \right).$ Find the interval of convergence of ${\sum\limits_{n = 0}^{\infty}{a_{n}\left( \frac{x}{2} \right)^{n}}}.$

In the next example, we show how to use Combining Power Series and the power series for a function *f* to construct power series for functions related to *f*. Specifically, we consider functions related to the function $f(x) = \frac{1}{1 - x}$ and we use the fact that

$$\frac{1}{1 - x} = {\sum\limits_{n = 0}^{\infty}x^{n}} = 1 + x + x^{2} + x^{3} + \text{⋯}$$

for $|x| < 1.$

Constructing Power Series from Known Power Series 由已知幂级数构造幂级数

Use the power series representation for $f(x) = \frac{1}{1 - x}$ combined with Combining Power Series to construct a power series for each of the following functions. Find the interval of convergence of the power series.

1. $f(x) = \frac{3x}{1 + x^{2}}$

2. $f(x) = \frac{1}{\left( {x - 1} \right)\left( {x - 3} \right)}$

Solution 解答

1. First write $f(x)$ as

$$f(x) = 3x\left( \frac{1}{1 - \left( {\text{−}x^{2}} \right)} \right).$$

Using the power series representation for $f(x) = \frac{1}{1 - x}$ and parts ii. and iii. of Combining Power Series, we find that a power series representation for *f* is given by

$${\sum\limits_{n = 0}^{\infty}{3x\left( {\text{−}x^{2}} \right)^{n}}} = {\sum\limits_{n = 0}^{\infty}{3(-1)^{n}x^{2n + 1}}}.$$

Since the interval of convergence of the series for $\frac{1}{1 - x}$ is $\left( {-1,1} \right),$ the interval of convergence for this new series is the set of real numbers *x* such that $\left| x^{2} \right| < 1.$ Therefore, the interval of convergence is $\left( {-1,1} \right).$

2. To find the power series representation, use partial fractions to write $f(x) = \frac{1}{\left( {x - 1} \right)\left( {x - 3} \right)}$ as the sum of two fractions. We have

$$\begin{array}{cl} \frac{1}{\left( {x - 1} \right)\left( {x - 3} \right)} & {= \frac{\text{−}{1\text{/}2}}{x - 1} + \frac{1\text{/}2}{x - 3}} \\ & {= \frac{1\text{/}2}{1 - x} - \frac{1\text{/}2}{3 - x}} \\ & {= \frac{1\text{/}2}{1 - x} - \frac{1\text{/}6}{1 - \frac{x}{3}}.} \end{array}$$

First, using part ii. of Combining Power Series, we obtain

$$\frac{1\text{/}2}{1 - x} = {\sum\limits_{n = 0}^{\infty}{\frac{1}{2}x^{n}}}\ \text{for}\ |x| < 1.$$

Then, using parts ii. and iii. of Combining Power Series, we have

$$\frac{1\text{/}6}{1 - {x\text{/}3}} = {\sum\limits_{n = 0}^{\infty}{\frac{1}{6}\left( \frac{x}{3} \right)^{n}}}\ \text{for}\ |x| < 3.$$

Since we are combining these two power series, the interval of convergence of the difference must be the smaller of these two intervals. Using this fact and part i. of Combining Power Series, we have

$$\frac{1}{\left( {x - 1} \right)\left( {x - 3} \right)} = {\sum\limits_{n = 0}^{\infty}{\left( {\frac{1}{2} - \frac{1}{6 \cdot 3^{n}}} \right)x^{n}}}$$

where the interval of convergence is $\left( {-1,1} \right).$

Use the series for $f(x) = \frac{1}{1 - x}$ on $|x| < 1$ to construct a series for $\frac{1}{\left( {1 - x} \right)\left( {x - 2} \right)}.$ Determine the interval of convergence.

In Example 6.5, we showed how to find power series for certain functions. In Example 6.6 we show how to do the opposite: given a power series, determine which function it represents.

Finding the Function Represented by a Given Power Series 求给定幂级数所表示的函数

Consider the power series ${\sum\limits_{n = 0}^{\infty}{2^{n}x^{n}}}.$ Find the function *f* represented by this series. Determine the interval of convergence of the series.

Solution 解答

Writing the given series as

$${\sum\limits_{n = 0}^{\infty}{2^{n}x^{n}}} = {\sum\limits_{n = 0}^{\infty}\left( {2x} \right)^{n}},$$

we can recognize this series as the power series for

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

Since this is a geometric series, the series converges if and only if $\left| {2x} \right| < 1.$ Therefore, the interval of convergence is $\left( {- \frac{1}{2},\frac{1}{2}} \right).$

Find the function represented by the power series ${\sum\limits_{n = 0}^{\infty}{\frac{1}{3^{n}}x^{n}}}.$ Determine its interval of convergence.

Recall the questions posed in the chapter opener about which is the better way of receiving payouts from lottery winnings. We now revisit those questions and show how to use series to compare values of payments over time with a lump sum payment today. We will compute how much future payments are worth in terms of today's dollars, assuming we have the ability to invest winnings and earn interest. The value of future payments in terms of today's dollars is known as the *present value* of those payments.

Chapter Opener: Present Value of Future Winnings 章首问题：未来奖金的现值

Suppose you win the lottery and are given the following three options: (1) Receive 20 million dollars today; (2) receive 1.5 million dollars per year over the next 20 years; or (3) receive 1 million dollars per year indefinitely (being passed on to your heirs). Which is the best deal, assuming that the annual interest rate is 5%? We answer this by working through the following sequence of questions.

1. How much is the 1.5 million dollars received annually over the course of 20 years worth in terms of today's dollars, assuming an annual interest rate of 5%?

2. Use the answer to part a. to find a general formula for the present value of payments of *C* dollars received each year over the next *n* years, assuming an average annual interest rate *r*.

3. Find a formula for the present value if annual payments of *C* dollars continue indefinitely, assuming an average annual interest rate *r*.

4. Use the answer to part c. to determine the present value of 1 million dollars paid annually indefinitely.

5. Use your answers to parts a. and d. to determine which of the three options is best.

Solution 解答

1. Consider the payment of 1.5 million dollars made at the end of the first year. If you were able to receive that payment today instead of one year from now, you could invest that money and earn 5% interest. Therefore, the present value of that money *P*₁ satisfies $P_{1}\left( {1 + 0.05} \right) = 1.5\ \text{million dollars}.$ We conclude that

$$P_{1} = \frac{1.5}{1.05} = \text{\$}1.429\ \text{million dollars}\text{.}$$

Similarly, consider the payment of 1.5 million dollars made at the end of the second year. If you were able to receive that payment today, you could invest that money for two years, earning 5% interest, compounded annually. Therefore, the present value of that money *P*₂ satisfies $P_{2}\left( {1 + 0.05} \right)^{2} = 1.5\ \text{million dollars}.$ We conclude that

$$P_{2} = \frac{1.5}{(1.05)^{2}} = \text{\$}1.361\ \text{million dollars}\text{.}$$

The value of the future payments today is the sum of the present values $P_{1},P_{2},\text{…},P_{20}$ of each of those annual payments. The present value *P_k* satisfies

$$P_{k} = \frac{1.5}{(1.05)^{k}}.$$

Therefore,

$$\begin{array}{cl} P & {= \frac{1.5}{1.05} + \frac{1.5}{(1.05)^{2}} + \text{⋯} + \frac{1.5}{(1.05)^{20}}} \\ & {= \text{\$}18.693\ \text{million dollars}\text{.}} \end{array}$$

2. Using the result from part a. we see that the present value *P* of *C* dollars paid annually over the course of *n* years, assuming an annual interest rate *r*, is given by

$$P = \frac{C}{1 + r} + \frac{C}{\left( {1 + r} \right)^{2}} + \text{⋯} + \frac{C}{\left( {1 + r} \right)^{n}}\ \text{dollars}\text{.}$$

3. Using the result from part b. we see that the present value of an annuity that continues indefinitely is given by the infinite series

$$P = {\sum\limits_{n = 0}^{\infty}\frac{C}{\left( {1 + r} \right)^{n + 1}}}.$$

We can view the present value as a power series in *r*, which converges as long as $\left| \frac{1}{1 + r} \right| < 1.$ Since $r > 0,$ this series converges. Rewriting the series as

$$P = \frac{C}{\left( {1 + r} \right)}{\sum\limits_{n = 0}^{\infty}\left( \frac{1}{1 + r} \right)^{n}},$$

we recognize this series as the power series for

$$f(r) = \frac{1}{1 - \left( \frac{1}{1 + r} \right)} = \frac{1}{\left( \frac{r}{1 + r} \right)} = \frac{1 + r}{r}.$$

We conclude that the present value of this annuity is

$$P = \frac{C}{1 + r} \cdot \frac{1 + r}{r} = \frac{C}{r}.$$

4. From the result to part c. we conclude that the present value *P* of $C = 1\ \text{million dollars}$ paid out every year indefinitely, assuming an annual interest rate $r = 0.05,$ is given by

$$P = \frac{1}{0.05} = 20\ \text{million dollars}\text{.}$$

5. From part a. we see that receiving \$1.5 million dollars over the course of 20 years is worth \$18.693 million dollars in today's dollars. From part d. we see that receiving \$1 million dollars per year indefinitely is worth \$20 million dollars in today's dollars. Therefore, either receiving a lump-sum payment of \$20 million dollars today or receiving \$1 million dollars indefinitely have the same present value.

Multiplication of Power Series 幂级数的乘法

We can also create new power series by multiplying power series. Being able to multiply two power series provides another way of finding power series representations for functions.

The way we multiply them is similar to how we multiply polynomials. For example, suppose we want to multiply

$${\sum\limits_{n = 0}^{\infty}{c_{n}x^{n}}} = c_{0} + c_{1}x + c_{2}x^{2} + \text{⋯}$$

and

$${\sum\limits_{n = 0}^{\infty}{d_{n}x^{n}}} = d_{0} + d_{1}x + d_{2}x^{2} + \text{⋯}.$$

It appears that the product should satisfy

$$\begin{array}{cl} {\left( {\sum\limits_{n = 0}^{\infty}{c_{n}x^{n}}} \right)\left( {\sum\limits_{n = -0}^{\infty}{d_{n}x^{n}}} \right)} & {= \left( {c_{0} + c_{1}x + c_{2}x^{2} + \text{⋯}} \right) \cdot \left( {d_{0} + d_{1}x + d_{2}x^{2} + \text{⋯}} \right)} \\ & {= c_{0}d_{0} + \left( {c_{1}d_{0} + c_{0}d_{1}} \right)\mspace{2mu} x + \left( {c_{2}d_{0} + c_{1}d_{1} + c_{0}d_{2}} \right)\mspace{2mu} x^{2} + \text{⋯}.} \end{array}$$

In Multiplying Power Series, we state the main result regarding multiplying power series, showing that if $\sum\limits_{n = 0}^{\infty}{c_{n}x^{n}}$ and $\sum\limits_{n = 0}^{\infty}{d_{n}x^{n}}$ converge on a common interval *I*, then we can multiply the series in this way, and the resulting series also converges on the interval *I*.

Multiplying Power Series 幂级数的乘法

Suppose that the power series $\sum\limits_{n = 0}^{\infty}{c_{n}x^{n}}$ and $\sum\limits_{n = 0}^{\infty}{d_{n}x^{n}}$ converge to *f* and *g*, respectively, on a common interval *I*. Let

$$\begin{array}{cl} e_{n} & {= c_{0}d_{n} + c_{1}d_{n - 1} + c_{2}d_{n - 2} + \text{⋯} + c_{n - 1}d_{1} + c_{n}d_{0}} \\ & {= {\sum\limits_{k = 0}^{n}{c_{k}d_{n - k}}}.} \end{array}$$

Then

$$\left( {\sum\limits_{n = 0}^{\infty}{c_{n}x^{n}}} \right)\left( {\sum\limits_{n = 0}^{\infty}{d_{n}x^{n}}} \right) = {\sum\limits_{n = 0}^{\infty}{e_{n}x^{n}}}$$

and

$${\sum\limits_{n = 0}^{\infty}{e_{n}x^{n}}}\ \text{converges to}\ f(x) \cdot g(x)\ \text{on}\ I.$$

The series ${\sum\limits_{n = 0}^{\infty}e_{n}}x^{n}$ is known as the Cauchy product of the series $\sum\limits_{n = 0}^{\infty}{c_{n}x^{n}}$ and ${\sum\limits_{n = 0}^{\infty}{d_{n}x^{n}}}.$

We omit the proof of this theorem, as it is beyond the level of this text and is typically covered in a more advanced course. We now provide an example of this theorem by finding the power series representation for

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

using the power series representations for

$$y = \frac{1}{1 - x}\ \text{and}\ y = \frac{1}{1 - x^{2}}.$$

Multiplying Power Series 幂级数的乘法

Multiply the power series representation

$$\begin{array}{cl} \frac{1}{1 - x} & {= {\sum\limits_{n = 0}^{\infty}x^{n}}} \\ & {= 1 + x + x^{2} + x^{3} + \text{⋯}} \end{array}$$

for $\left| x \middle| < 1 \right.$ with the power series representation

$$\begin{array}{cl} \frac{1}{1 - x^{2}} & {= {\sum\limits_{n = 0}^{\infty}\left( x^{2} \right)^{n}}} \\ & {= 1 + x^{2} + x^{4} + x^{6} + \text{⋯}} \end{array}$$

for $|x| < 1$ to construct a power series for $f(x) = \frac{1}{\left( {1 - x} \right)\left( {1 - x^{2}} \right)}$ on the interval $\left( {-1,1} \right).$

Solution 解答

We need to multiply

$$\left( {1 + x + x^{2} + x^{3} + \text{⋯}} \right)\left( {1 + x^{2} + x^{4} + x^{6} + \text{⋯}} \right).$$

Writing out the first several terms, we see that the product is given by

$$\begin{array}{l} {\left( {1 + x^{2} + x^{4} + x^{6} + \text{⋯}} \right) + \left( {x + x^{3} + x^{5} + x^{7} + \text{⋯}} \right) + \left( {x^{2} + x^{4} + x^{6} + x^{8} + \text{⋯}} \right) + \left( {x^{3} + x^{5} + x^{7} + x^{9} + \text{⋯}} \right)} \\ {= 1 + x + \left( {1 + 1} \right)x^{2} + \left( {1 + 1} \right)x^{3} + \left( {1 + 1 + 1} \right)x^{4} + \left( {1 + 1 + 1} \right)x^{5} + \text{⋯}} \\ {= 1 + x + 2x^{2} + 2x^{3} + 3x^{4} + 3x^{5} + \text{⋯}.} \end{array}$$

Since the series for $y = \frac{1}{1 - x}$ and $y = \frac{1}{1 - x^{2}}$ both converge on the interval $\left( {-1,1} \right),$ the series for the product also converges on the interval $\left( {-1,1} \right).$

Multiply the series $\frac{1}{1 - x} = {\sum\limits_{n = 0}^{\infty}x^{n}}$ by itself to construct a series for $\frac{1}{\left( {1 - x} \right)\left( {1 - x} \right)}.$

Differentiating and Integrating Power Series 幂级数的求导与积分

Consider a power series ${\sum\limits_{n = 0}^{\infty}{c_{n}x^{n}}} = c_{0} + c_{1}x + c_{2}x^{2} + \text{⋯}$ that converges on some interval *I*, and let $f$ be the function defined by this series. Here we address two questions about $f.$

• Is $f$ differentiable, and if so, how do we determine the derivative $f^{\prime}?$
• How do we evaluate the indefinite integral ${\int{f(x)\mspace{2mu} dx}}?$

• $f$ 是否可导？若可导，我们如何求其导数 $f^{\prime}?$
• 我们如何求不定积分 ${\int{f(x)\mspace{2mu} dx}}?$

We know that, for a polynomial with a finite number of terms, we can evaluate the derivative by differentiating each term separately. Similarly, we can evaluate the indefinite integral by integrating each term separately. Here we show that we can do the same thing for convergent power series. That is, if

$$f(x) = {\sum\limits_{n = 0}^{\infty}{c_{n}x^{n}}} = c_{0} + c_{1}x + c_{2}x^{2} + \text{⋯}$$

converges on some interval *I*, then

$$f^{\prime}(x) = c_{1} + 2c_{2}x + 3c_{3}x^{2} + \text{⋯}$$

and

$${\int{f(x)\mspace{2mu} dx}} = C + c_{0}x + c_{1}\frac{x^{2}}{2} + c_{2}\frac{x^{3}}{3} + \text{⋯}$$

converges on *I*. As noted below, behavior at the endpoints of the interval must be investigated individually.

Evaluating the derivative and indefinite integral in this way is called term-by-term differentiation of a power series and term-by-term integration of a power series, respectively. The ability to differentiate and integrate power series term-by-term also allows us to use known power series representations to find power series representations for other functions. For example, given the power series for $f(x) = \frac{1}{1 - x},$ we can differentiate term-by-term to find the power series for $f^{\prime}(x) = \frac{1}{\left( {1 - x} \right)^{2}}.$ Similarly, using the power series for $g(x) = \frac{1}{1 + x},$ we can integrate term-by-term to find the power series for $G(x) = \text{ln}\mspace{2mu}\left( {1 + x} \right),$ an antiderivative of *g*. We show how to do this in Example 6.9 and Example 6.10. First, we state Term-by-Term Differentiation and Integration for Power Series, which provides the main result regarding differentiation and integration of power series.

Term-by-Term Differentiation and Integration for Power Series 幂级数的逐项求导与逐项积分

Suppose that the power series $\sum\limits_{n = 0}^{\infty}{c_{n}\left( {x - a} \right)^{n}}$ converges on the interval $\left( {a - R,a + R} \right)$ for some $R > 0.$ Let *f* be the function defined by the series

$$\begin{array}{cl} {f(x)} & {= {\sum\limits_{n = 0}^{\infty}{c_{n}\left( {x - a} \right)^{n}}}} \\ & {= c_{0} + c_{1}\left( {x - a} \right) + c_{2}\left( {x - a} \right)^{2} + c_{3}\left( {x - a} \right)^{3} + \text{⋯}} \end{array}$$

for $\left| {x - a} \right| < R.$ Then *f* is differentiable on the interval $\left( {a - R,a + R} \right)$ and we can find $f^{\prime}$ by differentiating the series term-by-term:

$$\begin{array}{cl} {f^{\prime}(x)} & {= {\sum\limits_{n = 1}^{\infty}{nc_{n}\left( {x - a} \right)^{n - 1}}}} \\ & {= c_{1} + 2c_{2}\left( {x - a} \right) + 3c_{3}\left( {x - a} \right)^{2} + \text{⋯}} \end{array}$$

for $\left| {x - a} \right| < R.$ Also, to find ${\int{f(x)\mspace{2mu} dx}},$ we can integrate the series term-by-term. The resulting series converges on $\left( {a - R,a + R} \right),$ and we have

$$\begin{array}{cl} {\int{f(x)\mspace{2mu} dx}} & {= C + {\sum\limits_{n = 0}^{\infty}{c_{n}\frac{\left( {x - a} \right)^{n + 1}}{n + 1}}}} \\ & {= C + c_{0}\left( {x - a} \right) + c_{1}\frac{\left( {x - a} \right)^{2}}{2} + c_{2}\frac{\left( {x - a} \right)^{3}}{3} + \text{⋯}} \end{array}$$

for $\left| {x - a} \right| < R.$

The proof of this result is beyond the scope of the text and is omitted. Note that although Term-by-Term Differentiation and Integration for Power Series guarantees the same radius of convergence when a power series is differentiated or integrated term-by-term, it says nothing about what happens at the endpoints. It is possible that the differentiated and integrated power series have different behavior at the endpoints than does the original series. We see this behavior in the next examples.

Differentiating Power Series 幂级数的求导

1. Use the power series representation

$$\begin{array}{cl} {f(x)} & {= \frac{1}{1 - x}} \\ & {= {\sum\limits_{n = 0}^{\infty}x^{n}}} \\ & {= 1 + x + x^{2} + x^{3} + \text{⋯}} \end{array}$$

for $|x| < 1$ to find a power series representation for

$$g(x) = \frac{1}{\left( {1 - x} \right)^{2}}$$

on the interval $\left( {-1,1} \right).$ Determine whether the resulting series converges at the endpoints.

2. Use the result of part a. to evaluate the sum of the series ${\sum\limits_{n = 0}^{\infty}\frac{n + 1}{4^{n}}}.$

Solution 解答

1. Since $g(x) = \frac{1}{\left( {1 - x} \right)^{2}}$ is the derivative of $f(x) = \frac{1}{1 - x},$ we can find a power series representation for *g* by differentiating the power series for *f* term-by-term. The result is

$$\begin{array}{cl} {g(x)} & {= \frac{1}{\left( {1 - x} \right)^{2}}} \\ & {= \frac{d}{dx}\left( \frac{1}{1 - x} \right)} \\ & {= {\sum\limits_{n = 0}^{\infty}{\frac{d}{dx}\left( x^{n} \right)}}} \\ & {= \frac{d}{dx}\left( {1 + x + x^{2} + x^{3} + \text{⋯}} \right)} \\ & {= 0 + 1 + 2x + 3x^{2} + 4x^{3} + \text{⋯}} \\ & {= {\sum\limits_{n = 0}^{\infty}{\left( {n + 1} \right)x^{n}}}} \end{array}$$

for $|x| < 1.$ Term-by-Term Differentiation and Integration for Power Series does not guarantee anything about the behavior of this series at the endpoints. Testing the endpoints by using the divergence test, we find that the series diverges at both endpoints $x = \text{±}1.$ Note that this is the same result found in Example 6.8.

2. From part a. we know that

$${\sum\limits_{n = 0}^{\infty}{\left( {n + 1} \right)x^{n}}} = \frac{1}{\left( {1 - x} \right)^{2}}.$$

Therefore,

$$\begin{array}{cl} {\sum\limits_{n = 0}^{\infty}\frac{n + 1}{4^{n}}} & {= {\sum\limits_{n = 0}^{\infty}{\left( {n + 1} \right)\left( \frac{1}{4} \right)^{n}}}} \\ & {= \frac{1}{\left( {1 - \frac{1}{4}} \right)^{2}}} \\ & {= \frac{1}{\left( \frac{3}{4} \right)^{2}}} \\ & {= \frac{16}{9}.} \end{array}$$

Differentiate the series $\frac{1}{\left( {1 - x} \right)^{2}} = {\sum\limits_{n = 0}^{\infty}{\left( {n + 1} \right)x^{n}}}$ term-by-term to find a power series representation for $\frac{2}{\left( {1 - x} \right)^{3}}$ on the interval $\left( {-1,1} \right).$

Integrating Power Series 幂级数的积分

For each of the following functions *f*, find a power series representation for *f* by integrating the power series for $f^{\prime}$ and find its interval of convergence.

1. $f(x) = \text{ln}\mspace{2mu}\left( {1 + x} \right)$

2. $f(x) = \text{tan}^{-1}x$

Solution 解答

1. For $f(x) = \text{ln}\mspace{2mu}\left( {1 + x} \right),$ the derivative is $f^{\prime}(x) = \frac{1}{1 + x}.$ We know that

$$\begin{array}{cl} \frac{1}{1 + x} & {= \frac{1}{1 - \left( {\text{−}x} \right)}} \\ & {= {\sum\limits_{n = 0}^{\infty}\left( {\text{−}x} \right)^{n}}} \\ & {= 1 - x + x^{2} - x^{3} + \text{⋯}} \end{array}$$

for $|x| < 1.$ To find a power series for $f(x) = \text{ln}\mspace{2mu}\left( {1 + x} \right),$ we integrate the series term-by-term.

$$\begin{array}{cl} {\int{f^{\prime}(x)\mspace{2mu} dx}} & {= {\int{\left( {1 - x + x^{2} - x^{3} + \text{⋯}} \right)\mspace{2mu} dx}}} \\ & {= C + x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + \text{⋯}} \end{array}$$

Since $f(x) = \text{ln}\mspace{2mu}\left( {1 + x} \right)$ is an antiderivative of $\frac{1}{1 + x},$ it remains to solve for the constant *C*. Since $\text{ln}\mspace{2mu}\left( {1 + 0} \right) = 0,$ we have $C = 0.$ Therefore, a power series representation for $f(x) = \text{ln}\mspace{2mu}\left( {1 + x} \right)$ is

$$\begin{array}{cl} {\text{ln}\mspace{2mu}\left( {1 + x} \right)} & {= x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \frac{x^{4}}{4} + \text{⋯}} \\ & {= {\sum\limits_{n = 1}^{\infty}{(-1)^{n + 1}\frac{x^{n}}{n}}}} \end{array}$$

for $|x| < 1.$ Term-by-Term Differentiation and Integration for Power Series does not guarantee anything about the behavior of this power series at the endpoints. However, checking the endpoints, we find that at $x = 1$ the series is the alternating harmonic series, which converges. Also, at $x = -1,$ the series is the harmonic series, which diverges. It is important to note that, even though this series converges at $x = 1,$ Term-by-Term Differentiation and Integration for Power Series does not guarantee that the series actually converges to $\text{ln}\mspace{2mu}(2).$ In fact, the series does converge to $\text{ln}\mspace{2mu}(2),$ but showing this fact requires more advanced techniques. (Abel’s theorem, covered in more advanced texts, deals with this more technical point.) The interval of convergence is $\left( {-1,1} \right\rbrack.$

2. The derivative of $f(x) = \text{tan}^{-1}x$ is $f^{\prime}(x) = \frac{1}{1 + x^{2}}.$ We know that

$$\begin{array}{cl} \frac{1}{1 + x^{2}} & {= \frac{1}{1 - \left( {\text{−}x^{2}} \right)}} \\ & {= {\sum\limits_{n = 0}^{\infty}\left( {\text{−}x^{2}} \right)^{n}}} \\ & {= 1 - x^{2} + x^{4} - x^{6} + \text{⋯}} \end{array}$$

for $|x| < 1.$ To find a power series for $f(x) = \text{tan}^{-1}x,$ we integrate this series term-by-term.

$$\begin{array}{cl} {\int{f^{\prime}(x)\mspace{2mu} dx}} & {= {\int{\left( {1 - x^{2} + x^{4} - x^{6} + \text{⋯}} \right)\mspace{2mu} dx}}} \\ & {= C + x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \frac{x^{7}}{7} + \text{⋯}} \end{array}$$

Since $\text{tan}^{-1}(0) = 0,$ we have $C = 0.$ Therefore, a power series representation for $f(x) = \text{tan}^{-1}x$ is

$$\begin{array}{cl} {\text{tan}^{-1}x} & {= x - \frac{x^{3}}{3} + \frac{x^{5}}{5} - \frac{x^{7}}{7} + \text{⋯}} \\ & {= {\sum\limits_{n = 0}^{\infty}{(-1)^{n}\frac{x^{2n + 1}}{2n + 1}}}} \end{array}$$

for $|x| < 1.$ Again, Term-by-Term Differentiation and Integration for Power Series does not guarantee anything about the convergence of this series at the endpoints. However, checking the endpoints and using the alternating series test, we find that the series converges at $x = 1$ and $x = -1.$ As discussed in part a., using Abel’s theorem, it can be shown that the series actually converges to $\text{tan}^{-1}(1)$ and $\text{tan}^{-1}(-1)$ at $x = 1$ and $x = -1,$ respectively. Thus, the interval of convergence is $\left\lbrack {-1,1} \right\rbrack.$

Integrate the power series $\text{ln}\mspace{2mu}\left( {1 + x} \right) = {\sum\limits_{n = 1}^{\infty}{(-1)^{n + 1}\frac{x^{n}}{n}}}$ term-by-term to evaluate ${\int{\text{ln}\mspace{2mu}\left( {1 + x} \right)\mspace{2mu} dx}}.$

Up to this point, we have shown several techniques for finding power series representations for functions. However, how do we know that these power series are unique? That is, given a function *f* and a power series for *f* at *a*, is it possible that there is a different power series for *f* at *a* that we could have found if we had used a different technique? The answer to this question is no. This fact should not seem surprising if we think of power series as polynomials with an infinite number of terms. Intuitively, if

$$c_{0} + c_{1}x + c_{2}x^{2} + \text{⋯} = d_{0} + d_{1}x + d_{2}x^{2} + \text{⋯}$$

for all values *x* in some open interval *I* about zero, then the coefficients *c_n* should equal *d_n* for $n \geq 0.$ We now state this result formally in Uniqueness of Power Series.

Uniqueness of Power Series 幂级数的唯一性

Let $\sum\limits_{n = 0}^{\infty}{c_{n}\left( {x - a} \right)^{n}}$ and $\sum\limits_{n = 0}^{\infty}{d_{n}\left( {x - a} \right)^{n}}$ be two convergent power series such that

$${\sum\limits_{n = 0}^{\infty}{c_{n}\left( {x - a} \right)^{n}}} = {\sum\limits_{n = 0}^{\infty}{d_{n}\left( {x - a} \right)^{n}}}$$

for all *x* in an open interval containing *a*. Then $c_{n} = d_{n}$ for all $n \geq 0.$

Proof 证明

Let

$$\begin{array}{cl} {f(x)} & {= c_{0} + c_{1}\left( {x - a} \right) + c_{2}\left( {x - a} \right)^{2} + c_{3}\left( {x - a} \right)^{3} + \text{⋯}} \\ & {= d_{0} + d_{1}\left( {x - a} \right) + d_{2}\left( {x - a} \right)^{2} + d_{3}\left( {x - a} \right)^{3} + \text{⋯}.} \end{array}$$

Then $f(a) = c_{0} = d_{0}.$ By Term-by-Term Differentiation and Integration for Power Series, we can differentiate both series term-by-term. Therefore,

$$\begin{array}{cl} {f^{\prime}(x)} & {= c_{1} + 2c_{2}\left( {x - a} \right) + 3c_{3}\left( {x - a} \right)^{2} + \text{⋯}} \\ & {= d_{1} + 2d_{2}\left( {x - a} \right) + 3d_{3}\left( {x - a} \right)^{2} + \text{⋯},} \end{array}$$

and thus, $f^{\prime}(a) = c_{1} = d_{1}.$ Similarly,

$$\begin{array}{cl} {f^{''}(x)} & {= 2c_{2} + 3 \cdot 2c_{3}\left( {x - a} \right) + \text{⋯}} \\ & {= 2d_{2} + 3 \cdot 2d_{3}\left( {x - a} \right) + \text{⋯}} \end{array}$$

implies that $f^{''}(a) = 2c_{2} = 2d_{2},$ and therefore, $c_{2} = d_{2}.$ More generally, for any integer $n \geq 0,f^{(n)}(a) = n\text{!}c_{n} = n\text{!}d_{n},$ and consequently, $c_{n} = d_{n}$ for all $n \geq 0.$

□

In this section we have shown how to find power series representations for certain functions using various algebraic operations, differentiation, or integration. At this point, however, we are still limited as to the functions for which we can find power series representations. Next, we show how to find power series representations for many more functions by introducing Taylor series.

Section 6.2 Exercises 6.2 节习题

63.

If $f(x) = {\sum\limits_{n = 0}^{\infty}\frac{x^{n}}{n\text{!}}}$ and $g(x) = {\sum\limits_{n = 0}^{\infty}{(-1)^{n}\frac{x^{n}}{n\text{!}}}},$ find the power series of $\frac{1}{2}\left( {f(x) + g(x)} \right)$ and of $\frac{1}{2}\left( {f(x) - g(x)} \right).$

64\.

If $C(x) = {\sum\limits_{n = 0}^{\infty}\frac{x^{2n}}{\left( {2n} \right)\text{!}}}$ and $S(x) = {\sum\limits_{n = 0}^{\infty}\frac{x^{2n + 1}}{\left( {2n + 1} \right)\text{!}}},$ find the power series of $C(x) + S(x)$ and of $C(x) - S(x).$

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

65.

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

66\.

$\frac{3}{\left( {x + 2} \right)\left( {x - 1} \right)}$

67.

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

68\.

$\frac{30}{\left( {x^{2} + 1} \right)\left( {x^{2} - 9} \right)}$

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

69.

$\sum\limits_{n = 1}^{\infty}\frac{1}{x^{n}}$

70\.

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

71.

$\sum\limits_{n = 1}^{\infty}\frac{1}{\left( {x - 3} \right)^{2n - 1}}$

72\.

$\sum\limits_{n = 1}^{\infty}\left( {\frac{1}{\left( {x - 3} \right)^{2n - 1}} - \frac{1}{\left( {x - 2} \right)^{2n - 1}}} \right)$

The following exercises explore applications of annuities.

73.

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 $r = 0.07.$

74\.

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 $r = 0.07.$

75.

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 $r = 0.07.$

76\.

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 $r = 0.07.$

77.

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

78\.

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

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.

79.

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

80\.

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

81.

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

82\.

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

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

83.

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

84\.

$f(x) = {\sum\limits_{n = 1}^{\infty}x^{n}},g(x) = {\sum\limits_{n = 1}^{\infty}{\frac{1}{n}x^{n}}}.$ Express the coefficients of $f(x)g(x)$ in terms of $H_{n} = {\sum\limits_{k = 1}^{n}\frac{1}{k}}.$

85.

$f(x) = g(x) = {\sum\limits_{n = 1}^{\infty}\left( \frac{x}{2} \right)^{n}}$

86\.

$f(x) = g(x) = {\sum\limits_{n = 1}^{\infty}{nx^{n}}}$

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*.

87.

$f(x) = \frac{1}{1 + x} = {\sum\limits_{n = 0}^{\infty}{(-1)^{n}x^{n}}}$

88\.

$f(x) = \frac{1}{1 - x^{2}} = {\sum\limits_{n = 0}^{\infty}x^{2n}}$

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 $f.$

89.

$f(x) = \frac{2x}{\left( {1 + x^{2}} \right)^{2}} = {\sum\limits_{n = 1}^{\infty}{(-1)^{n}\left( {2n} \right)x^{2n - 1}}}$

90\.

$f(x) = \frac{2x}{1 + x^{2}} = 2{\sum\limits_{n = 0}^{\infty}{(-1)^{n}x^{2n + 1}}}$

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

91.

Evaluate $\sum\limits_{n = 1}^{\infty}\frac{n}{2^{n}}$ as $f^{\prime}\left( \frac{1}{2} \right)$ where $f(x) = {\sum\limits_{n = 0}^{\infty}x^{n}}.$

92\.

Evaluate $\sum\limits_{n = 1}^{\infty}\frac{n}{3^{n}}$ as $f^{\prime}\left( \frac{1}{3} \right)$ where $f(x) = {\sum\limits_{n = 0}^{\infty}x^{n}}.$

93.

Evaluate $\sum\limits_{n = 2}^{\infty}\frac{n\left( {n - 1} \right)}{2^{n}}$ as $f^{''}\left( \frac{1}{2} \right)$ where $f(x) = {\sum\limits_{n = 0}^{\infty}x^{n}}.$

94\.

Evaluate $\sum\limits_{n = 0}^{\infty}\frac{(-1)^{n}}{2n + 1}$ as $\int_{0}^{1}{f(t)\mspace{2mu} dt}$ where $f(x) = {\sum\limits_{n = 0}^{\infty}{(-1)^{n}x^{2n}}} = \frac{1}{1 + x^{2}}.$

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

95.

$f(x) = \text{ln}\mspace{2mu} x$ centered at $x = 1$ (*Hint:* $x = 1 - \left( {1 - x} \right))$

96\.

$\text{ln}\mspace{2mu}\left( {1 - x} \right)$ at $x = 0$

97.

$\text{ln}\mspace{2mu}\left( {1 - x^{2}} \right)$ at $x = 0$

98\.

$f(x) = \frac{2x}{\left( {1 - x^{2}} \right)^{2}}$ at $x = 0$

99.

$f(x) = \text{tan}^{-1}\left( x^{2} \right)$ at $x = 0$

100\.

$f(x) = \text{ln}\mspace{2mu}\left( {1 + x^{2}} \right)$ at $x = 0$

101.

$f(x) = {\int_{0}^{x}{\text{ln}\mspace{2mu} tdt}}$ where $\text{ln}\mspace{2mu}(x) = {\sum\limits_{n = 1}^{\infty}{(-1)^{n - 1}\frac{\left( {x - 1} \right)^{n}}{n}}}$

102\.

\[T\] Evaluate the power series expansion $\text{ln}\mspace{2mu}\left( {1 + x} \right) = {\sum\limits_{n = 1}^{\infty}{(-1)^{n - 1}\frac{x^{n}}{n}}}$ at $x = 1$ to show that $\text{ln}\mspace{2mu}(2)$ 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 $\text{ln}\mspace{2mu}(2)$ accurate to within 0.001, and find such an approximation.

103.

\[T\] Subtract the infinite series of $\text{ln}\mspace{2mu}\left( {1 - x} \right)$ from $\text{ln}\mspace{2mu}\left( {1 + x} \right)$ to get a power series for $\text{ln}\mspace{2mu}\left( \frac{1 + x}{1 - x} \right).$ Evaluate at $x = \frac{1}{3}.$ What is the smallest *N* such that the *N*th partial sum of this series approximates $\text{ln}\mspace{2mu}(2)$ with an error less than 0.001?

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.

104\.

$\sum\limits_{k = 0}^{\infty}\left( {x^{k} - x^{2k + 1}} \right)$

105.

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

106\.

$\sum\limits_{k = 1}^{\infty}\left( {1 + x^{2}} \right)^{\text{−}k}$ using $y = \frac{1}{1 + x^{2}}$

107.

$\sum\limits_{k = 1}^{\infty}2^{\text{−}kx}$ using $y = 2^{\text{−}x}$

108\.

Show that, up to powers *x*³ and *y*³, $E(x) = {\sum\limits_{n = 0}^{\infty}\frac{x^{n}}{n\text{!}}}$ satisfies $E\left( {x + y} \right) = E(x)E(y).$

109.

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

110\.

Show that if $f(x) = {\sum\limits_{n = 0}^{\infty}{a_{n}x^{n}}}$ is a sum of even powers, that is, $a_{n} = 0$ if *n* is odd, then $F = {\int_{0}^{x}{f(t)\mspace{2mu} dt}}$ is a sum of odd powers, while if *f* is a sum of odd powers, then *F* is a sum of even powers.

111.

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

112\.

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

113.

\[T\] Given the power series expansion $\text{ln}\mspace{2mu}\left( {1 + x} \right) = {\sum\limits_{n = 1}^{\infty}{(-1)^{n - 1}\frac{x^{n}}{n}}},$ determine how many terms *N* of the sum evaluated at $x = -1\text{/}2$ are needed to approximate $\text{ln}\mspace{2mu}(2)$ accurate to within 1/1000. Evaluate the corresponding partial sum ${\sum\limits_{n = 1}^{N}{(-1)^{n - 1}\frac{x^{n}}{n}}}.$

114\.

\[T\] Given the power series expansion $\text{tan}^{-1}(x) = {\sum\limits_{k = 0}^{\infty}{(-1)^{k}\frac{x^{2k + 1}}{2k + 1}}},$ use the alternating series test to determine how many terms *N* of the sum evaluated at $x = 1$ are needed to approximate $\text{tan}^{-1}(1) = \frac{\pi}{4}$ accurate to within 1/1000. Evaluate the corresponding partial sum ${\sum\limits_{k = 0}^{N}{(-1)^{k}\frac{x^{2k + 1}}{2k + 1}}}.$

115.

\[T\] Recall that $\text{tan}^{-1}\left( \frac{1}{\sqrt{3}} \right) = \frac{\pi}{6}.$ Assuming an exact value of $\left( \frac{1}{\sqrt{3}} \right),$ estimate $\frac{\pi}{6}$ by evaluating partial sums $S_{N}\left( \frac{1}{\sqrt{3}} \right)$ of the power series expansion $\text{tan}^{-1}(x) = {\sum\limits_{k = 0}^{\infty}{(-1)^{k}\frac{x^{2k + 1}}{2k + 1}}}$ 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?

6.3 Taylor and Maclaurin Series 6.3 泰勒与麦克劳林级数

• 6.3.1 Describe the procedure for finding a Taylor polynomial of a given order for a function.
• 6.3.2 Explain the meaning and significance of Taylor’s theorem with remainder.
• 6.3.3 Estimate the remainder for a Taylor series approximation of a given function.

• 6.3.1 描述求函数在给定阶数下泰勒多项式的步骤。
• 6.3.2 解释带余项的泰勒定理的含义与意义。
• 6.3.3 估计给定函数泰勒级数逼近的余项。

In the previous two sections we discussed how to find power series representations for certain types of functions––specifically, functions related to geometric series. Here we discuss power series representations for other types of functions. In particular, we address the following questions: Which functions can be represented by power series and how do we find such representations? If we can find a power series representation for a particular function $f$ and the series converges on some interval, how do we prove that the series actually converges to $f?$

Overview of Taylor/Maclaurin Series 泰勒/麦克劳林级数概述

Consider a function $f$ that has a power series representation at $x = a.$ Then the series has the form

$${\sum\limits_{n = 0}^{\infty}{c_{n}\left( {x - a} \right)^{n}}} = c_{0} + c_{1}(x - a) + c_{2}{(x - a)}^{2} + \text{⋯}.$$ (6.4)

What should the coefficients be? For now, we ignore issues of convergence, but instead focus on what the series should be, if one exists. We return to discuss convergence later in this section. If the series Equation 6.4 is a representation for $f$ at $x = a,$ we certainly want the series to equal $f(a)$ at $x = a.$ Evaluating the series at $x = a,$ we see that

$$\begin{array}{cl} {\sum\limits_{n = 0}^{\infty}{c_{n}\left( {x - a} \right)^{n}}} & {= c_{0} + c_{1}\left( {a - a} \right) + c_{2}\left( {a - a} \right)^{2} + \text{⋯}} \\ & {= c_{0}.} \end{array}$$

Thus, the series equals $f(a)$ if the coefficient $c_{0} = f(a).$ In addition, we would like the first derivative of the power series to equal $f^{\prime}(a)$ at $x = a.$ Differentiating Equation 6.4 term-by-term, we see that

$$\frac{d}{dx}\left( {\sum\limits_{n = 0}^{\infty}{c_{n}\left( {x - a} \right)^{n}}} \right) = c_{1} + 2c_{2}\left( {x - a} \right) + 3c_{3}\left( {x - a} \right)^{2} + \text{⋯}.$$

Therefore, at $x = a,$ the derivative is

$$\begin{array}{cl} & \\ & \\ {\frac{d}{dx}\left( {\sum\limits_{n = 0}^{\infty}{c_{n}\left( {x - a} \right)^{n}}} \right)} & {= c_{1} + 2c_{2}\left( {a - a} \right) + 3c_{3}\left( {a - a} \right)^{2} + \text{⋯}} \\ & {= c_{1}.} \end{array}$$

Therefore, the derivative of the series equals $f^{\prime}(a)$ if the coefficient $c_{1} = f^{\prime}(a).$ Continuing in this way, we look for coefficients *c_n* such that all the derivatives of the power series Equation 6.4 will agree with all the corresponding derivatives of $f$ at $x = a.$ The second and third derivatives of Equation 6.4 are given by

$$\frac{d^{2}}{dx^{2}}\left( {\sum\limits_{n = 0}^{\infty}{c_{n}\left( {x - a} \right)^{n}}} \right) = 2c_{2} + 3 \cdot 2c_{3}\left( {x - a} \right) + 4 \cdot 3c_{4}\left( {x - a} \right)^{2} + \text{⋯}$$

and

$$\frac{d^{3}}{dx^{3}}\left( {\sum\limits_{n = 0}^{\infty}{c_{n}\left( {x - a} \right)^{n}}} \right) = 3 \cdot 2c_{3} + 4 \cdot 3 \cdot 2c_{4}\left( {x - a} \right) + 5 \cdot 4 \cdot 3c_{5}\left( {x - a} \right)^{2} + \text{⋯}.$$

Therefore, at $x = a,$ the second and third derivatives

$$\begin{array}{cl} {\frac{d^{2}}{dx^{2}}\left( {\sum\limits_{n = 0}^{\infty}{c_{n}\left( {x - a} \right)^{n}}} \right)} & {= 2c_{2} + 3 \cdot 2c_{3}\left( {a - a} \right) + 4 \cdot 3c_{4}\left( {a - a} \right)^{2} + \text{⋯}} \\ & {= 2c_{2}} \end{array}$$

and

$$\begin{array}{cl} {\frac{d^{3}}{dx^{3}}\left( {\sum\limits_{n = 0}^{\infty}{c_{n}\left( {x - a} \right)^{n}}} \right)} & {= 3 \cdot 2c_{3} + 4 \cdot 3 \cdot 2c_{4}\left( {a - a} \right) + 5 \cdot 4 \cdot 3c_{5}\left( {a - a} \right)^{2} + \text{⋯}} \\ & {= 3 \cdot 2c_{3}} \end{array}$$

equal $f^{''}(a)$ and $f\text{'''}(a),$ respectively, if $c_{2} = \frac{f^{''}(a)}{2}$ and $c_{3} = \frac{f\text{'''}(a)}{3 \cdot 2}.$ More generally, we see that if $f$ has a power series representation at $x = a,$ then the coefficients should be given by $c_{n} = \frac{f^{(n)}(a)}{n\text{!}}.$ That is, the series should be

$${\sum\limits_{n = 0}^{\infty}\frac{f^{(n)}(a)}{n\text{!}}}\left( {x - a} \right)^{n} = f(a) + f^{\prime}(a)\left( {x - a} \right) + \frac{f^{''}(a)}{2\text{!}}\left( {x - a} \right)^{2} + \frac{f\text{'''}(a)}{3\text{!}}\left( {x - a} \right)^{3} + \text{⋯}.$$

This power series for $f$ is known as the Taylor series for $f$ at $a.$ If $a = 0,$ then this series is known as the Maclaurin series for $f.$

If $f$ has derivatives of all orders at $x = a,$ then the Taylor series for the function $f$ at $a$ is

$${\sum\limits_{n = 0}^{\infty}\frac{f^{(n)}(a)}{n\text{!}}}\left( {x - a} \right)^{n} = f(a) + f^{\prime}(a)\left( {x - a} \right) + \frac{f^{''}(a)}{2\text{!}}\left( {x - a} \right)^{2} + \text{⋯} + \frac{f^{(n)}(a)}{n\text{!}}\left( {x - a} \right)^{n} + \text{⋯}.$$ (6.5)

The Taylor series for $f$ at 0 is known as the Maclaurin series for $f.$

Later in this section, we will show examples of finding Taylor series and discuss conditions under which the Taylor series for a function will converge to that function. Here, we state an important result. Recall from Uniqueness of Power Series that power series representations are unique. Therefore, if a function $f$ has a power series at $a,$ then it must be the Taylor series for $f$ at $a.$

Uniqueness of Taylor Series 泰勒级数的唯一性

If a function $f$ has a power series at *a* that converges to $f$ on some open interval containing *a*, then that power series is the Taylor series for $f$ at *a*.

The proof follows directly from Uniqueness of Power Series.

To determine if a Taylor series converges, we need to look at its sequence of partial sums. These partial sums are finite polynomials, known as Taylor polynomials.

Visit the MacTutor History of Mathematics archive to read brief biographies of Brook Taylor and Colin Maclaurin and how they developed the concepts named after them.

Taylor Polynomials 泰勒多项式

The *n*th partial sum of the Taylor series for a function $f$ at $a$ is known as the *n*th Taylor polynomial. For example, the 0th, 1st, 2nd, and 3rd partial sums of the Taylor series are given by

$$\begin{array}{l} {p_{0}(x) = f(a),} \\ {p_{1}(x) = f(a) + f^{\prime}(a)\left( {x - a} \right),} \\ {p_{2}(x) = f(a) + f^{\prime}(a)\left( {x - a} \right) + \frac{f^{''}(a)}{2\text{!}}\left( {x - a} \right)^{2},} \\ {p_{3}(x) = f(a) + f^{\prime}(a)\left( {x - a} \right) + \frac{f^{''}(a)}{2\text{!}}\left( {x - a} \right)^{2} + \frac{f\text{'''}(a)}{3\text{!}}\left( {x - a} \right)^{3},} \end{array}$$

respectively. These partial sums are known as the 0th, 1st, 2nd, and 3rd Taylor polynomials of $f$ at $a,$ respectively. If $a = 0,$ then these polynomials are known as Maclaurin polynomials for $f.$ We now provide a formal definition of Taylor and Maclaurin polynomials for a function $f.$

If $f$ has *n* derivatives at $x = a,$ then the *n*th Taylor polynomial for $f$ at $a$ is

$$p_{n}(x) = f(a) + f^{\prime}(a)\left( {x - a} \right) + \frac{f^{''}(a)}{2\text{!}}\left( {x - a} \right)^{2} + \frac{f\text{'''}(a)}{3\text{!}}\left( {x - a} \right)^{3} + \text{⋯} + \frac{f^{(n)}(a)}{n\text{!}}\left( {x - a} \right)^{n}.$$

The *n*th Taylor polynomial for $f$ at 0 is known as the *n*th Maclaurin polynomial for $f.$

We now show how to use this definition to find several Taylor polynomials for $f(x) = \text{ln}\mspace{2mu} x$ at $x = 1.$

Finding Taylor Polynomials 求泰勒多项式

Find the Taylor polynomials $p_{0},p_{1},p_{2}$ and $p_{3}$ for $f(x) = \text{ln}\mspace{2mu} x$ at $x = 1.$ Use a graphing utility to compare the graph of $f$ with the graphs of $p_{0},p_{1},p_{2}$ and $p_{3}.$

Solution 解答

To find these Taylor polynomials, we need to evaluate $f$ and its first three derivatives at $x = 1.$

$$\begin{array}{rllccrll} {f(x)} & = & {\text{ln}\mspace{2mu} x} & & & {f(1)} & = & 0 \\ {f^{\prime}(x)} & = & \frac{1}{x} & & & {f^{\prime}(1)} & = & 1 \\ {f^{''}(x)} & = & {- \frac{1}{x^{2}}} & & & {f^{''}(1)} & = & -1 \\ {f\text{'''}(x)} & = & \frac{2}{x^{3}} & & & {f\text{'''}(1)} & = & 2 \end{array}$$

Therefore,

$$\begin{array}{cll} {p_{0}(x)} & = & {f(1) = 0,} \\ {p_{1}(x)} & = & {f(1) + f^{\prime}(1)\left( {x - 1} \right) = x - 1,} \\ {p_{2}(x)} & = & {f(1) + f^{\prime}(1)\left( {x - 1} \right) + \frac{f^{''}(1)}{2}\left( {x - 1} \right)^{2} = \left( {x - 1} \right) - \frac{1}{2}\left( {x - 1} \right)^{2},} \\ {p_{3}(x)} & = & {f(1) + f^{\prime}(1)\left( {x - 1} \right) + \frac{f^{''}(1)}{2}\left( {x - 1} \right)^{2} + \frac{f\text{'''}(1)}{3\text{!}}\left( {x - 1} \right)^{3}} \\ & = & {\left( {x - 1} \right) - \frac{1}{2}\left( {x - 1} \right)^{2} + \frac{1}{3}\left( {x - 1} \right)^{3}.} \end{array}$$

The graphs of $y = f(x)$ and the first three Taylor polynomials are shown in Figure 6.5.

Find the Taylor polynomials $p_{0},p_{1},p_{2}$ and $p_{3}$ for $f(x) = \frac{1}{x^{2}}$ at $x = 1.$

We now show how to find Maclaurin polynomials for *e^x*, $\text{sin}\mspace{2mu} x,$ and $\text{cos}\mspace{2mu} x.$ As stated above, Maclaurin polynomials are Taylor polynomials centered at zero.

Finding Maclaurin Polynomials 求麦克劳林多项式

For each of the following functions, find formulas for the Maclaurin polynomials $p_{0},p_{1},p_{2}$ and $p_{3}.$ Find a formula for the *n*th Maclaurin polynomial and write it using sigma notation. Use a graphing utilty to compare the graphs of $p_{0},p_{1},p_{2}$ and $p_{3}$ with $f.$

1. $f(x) = e^{x}$

2. $f(x) = \text{sin}\mspace{2mu} x$

3. $f(x) = \text{cos}\mspace{2mu} x$

Solution 解答

1. Since $f(x) = e^{x},$ we know that $f(x) = f^{\prime}(x) = f^{''}(x) = \text{⋯} = f^{(n)}(x) = e^{x}$ for all positive integers *n*. Therefore,

$$f(0) = f^{\prime}(0) = f^{''}(0) = \text{⋯} = f^{(n)}(0) = 1$$

for all positive integers *n*. Therefore, we have

$$\begin{array}{cll} {p_{0}(x)} & = & {f(0) = 1,} \\ {p_{1}(x)} & = & {f(0) + f^{\prime}(0)x = 1 + x,} \\ {p_{2}(x)} & = & {f(0) + f^{\prime}(0)x + \frac{f^{''}(0)}{2\text{!}}x^{2} = 1 + x + \frac{1}{2}x^{2},} \\ {p_{3}(x)} & = & {f(0) + f^{\prime}(0)x + \frac{f^{''}(0)}{2}x^{2} + \frac{f\text{'''}(0)}{3\text{!}}x^{3}} \\ & = & {1 + x + \frac{1}{2}x^{2} + \frac{1}{3\text{!}}x^{3},} \\ {p_{n}(x)} & = & {f(0) + f^{\prime}(0)x + \frac{f^{''}(0)}{2}x^{2} + \frac{f\text{'''}(0)}{3\text{!}}x^{3} + \text{⋯} + \frac{f^{(n)}(0)}{n\text{!}}x^{n}} \\ & = & {1 + x + \frac{x^{2}}{2\text{!}} + \frac{x^{3}}{3\text{!}} + \text{⋯} + \frac{x^{n}}{n\text{!}}} \\ & = & {{\sum\limits_{k = 0}^{n}\frac{x^{k}}{k\text{!}}}.} \end{array}$$

The function and the first three Maclaurin polynomials are shown in Figure 6.6.

2. For $f(x) = \text{sin}\mspace{2mu} x,$ the values of the function and its first four derivatives at $x = 0$ are given as follows:

$$\begin{array}{rllccrll} {f(x)} & = & {\text{sin}\mspace{2mu} x} & & & {f(0)} & = & 0 \\ {f^{\prime}(x)} & = & {\text{cos}\mspace{2mu} x} & & & {f^{\prime}(0)} & = & 1 \\ {f^{''}(x)} & = & {\text{−}\text{sin}\mspace{2mu} x} & & & {f^{''}(0)} & = & 0 \\ {f\text{'''}(x)} & = & {\text{−}\text{cos}\mspace{2mu} x} & & & {f\text{'''}(0)} & = & -1 \\ {f^{(4)}(x)} & = & {\text{sin}\mspace{2mu} x} & & & {f^{(4)}(0)} & = & 0. \end{array}$$

Since the fourth derivative is $\text{sin}\mspace{2mu} x,$ the pattern repeats. That is, $f^{({2m})}(0) = 0$ and $f^{({2m + 1})}(0) = (-1)^{m}$ for $m \geq 0.$ Thus, we have

$$\begin{array}{l} {p_{0}(x) = 0,} \\ {p_{1}(x) = 0 + x = x,} \\ {p_{2}(x) = 0 + x + 0 = x,} \\ {p_{3}(x) = 0 + x + 0 - \frac{1}{3\text{!}}x^{3} = x - \frac{x^{3}}{3\text{!}},} \\ {p_{4}(x) = 0 + x + 0 - \frac{1}{3\text{!}}x^{3} + 0 = x - \frac{x^{3}}{3\text{!}},} \\ {p_{5}(x) = 0 + x + 0 - \frac{1}{3\text{!}}x^{3} + 0 + \frac{1}{5\text{!}}x^{5} = x - \frac{x^{3}}{3\text{!}} + \frac{x^{5}}{5\text{!}},} \end{array}$$

and for $m \geq 0,$

$$\begin{array}{cl} {p_{2m + 1}(x)} & {= p_{2m + 2}(x)} \\ & {= x - \frac{x^{3}}{3\text{!}} + \frac{x^{5}}{5\text{!}} - \text{⋯} + (-1)^{m}\frac{x^{2m + 1}}{\left( {2m + 1} \right)\text{!}}} \\ & {= {\sum\limits_{k = 0}^{m}{(-1)^{k}\frac{x^{2k + 1}}{\left( {2k + 1} \right)\text{!}}}}.} \end{array}$$

Graphs of the function and its Maclaurin polynomials are shown in Figure 6.7.

3. For $f(x) = \text{cos}\mspace{2mu} x,$ the values of the function and its first four derivatives at $x = 0$ are given as follows:

$$\begin{array}{rllccrll} {f(x)} & = & {\text{cos}\mspace{2mu} x} & & & {f(0)} & = & 1 \\ {f^{\prime}(x)} & = & {\text{−}\text{sin}\mspace{2mu} x} & & & {f^{\prime}(0)} & = & 0 \\ {f^{''}(x)} & = & {\text{−}\text{cos}\mspace{2mu} x} & & & {f^{''}(0)} & = & -1 \\ {f\text{'''}(x)} & = & {\text{sin}\mspace{2mu} x} & & & {f\text{'''}(0)} & = & 0 \\ {f^{(4)}(x)} & = & {\text{cos}\mspace{2mu} x} & & & {f^{(4)}(0)} & = & 1. \end{array}$$

Since the fourth derivative is $\text{cos}\mspace{2mu} x,$ the pattern repeats. In other words, $f^{({2m})}(0) = (-1)^{m}$ and $f^{({2m + 1})} = 0$ for $m \geq 0.$ Therefore,

$$\begin{array}{l} {p_{0}(x) = 1,} \\ {p_{1}(x) = 1 + 0 = 1,} \\ {p_{2}(x) = 1 + 0 - \frac{1}{2\text{!}}x^{2} = 1, } \\ {p_{3}(x) = 1 + 0 - \frac{1}{2\text{!}}x^{2} + 0 = 1 - \frac{x^{2}}{2\text{!}},} \\ {p_{4}(x) = 1 + 0 - \frac{1}{2\text{!}}x^{2} + 0 + \frac{1}{4\text{!}}x^{4} = 1 - \frac{x^{2}}{2\text{!}} + \frac{x^{4}}{4\text{!}},} \\ {p_{5}(x) = 1 + 0 - \frac{1}{2\text{!}}x^{2} + 0 + \frac{1}{4\text{!}}x^{4} + 0 = 1 - \frac{x^{2}}{2\text{!}} + \frac{x^{4}}{4\text{!}},} \end{array}$$

and for $n \geq 0,$

$$\begin{array}{cl} {p_{2m}(x)} & {= p_{2m + 1}(x)} \\ & {= 1 - \frac{x^{2}}{2\text{!}} + \frac{x^{4}}{4\text{!}} - \text{⋯} + (-1)^{m}\frac{x^{2m}}{\left( {2m} \right)\text{!}}} \\ & {= {\sum\limits_{k = 0}^{m}(-1)^{k}}\frac{x^{2k}}{\left( {2k} \right)\text{!}}.} \end{array}$$

Graphs of the function and the Maclaurin polynomials appear in Figure 6.8.

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 *n*th Maclaurin polynomial. Write your answer using sigma notation.

Taylor's Theorem with Remainder 带余项的泰勒定理

Recall that the *n*th Taylor polynomial for a function $f$ at *a* is the *n*th partial sum of the Taylor series for $f$ at *a*. Therefore, to determine if the Taylor series converges, we need to determine whether the sequence of Taylor polynomials $\left\{ p_{n} \right\}$ converges. However, not only do we want to know if the sequence of Taylor polynomials converges, we want to know if it converges to $f.$ To answer this question, we define the remainder $R_{n}(x)$ as

$$R_{n}(x) = f(x) - p_{n}(x).$$

$$R_{n}(x) = f(x) - p_{n}(x).$$

For the sequence of Taylor polynomials to converge to $f,$ we need the remainder *R_n* to converge to zero. To determine if *R_n* converges to zero, we introduce Taylor's theorem with remainder. Not only is this theorem useful in proving that a Taylor series converges to its related function, but it will also allow us to quantify how well the *n*th Taylor polynomial approximates the function.

Here we look for a bound on $\left| R_{n} \right|.$ Consider the simplest case: $n = 0.$ Let *p*₀ be the 0th Taylor polynomial at *a* for a function $f.$ The remainder *R*₀ satisfies

$$\begin{array}{cl} {R_{0}(x)} & {= f(x) - p_{0}(x)} \\ & {= f(x) - f(a).} \end{array}$$

$$\begin{array}{cl} {R_{0}(x)} & {= f(x) - p_{0}(x)} \\ & {= f(x) - f(a).} \end{array}$$

If $f$ is differentiable on an interval *I* containing *a* and *x*, then by the Mean Value Theorem there exists a real number *c* between *a* and *x* such that $f(x) - f(a) = f^{\prime}(c)\left( {x - a} \right).$ Therefore,

$$R_{0}(x) = f^{\prime}(c)\left( {x - a} \right).$$

$$R_{0}(x) = f^{\prime}(c)\left( {x - a} \right).$$

Using the Mean Value Theorem in a similar argument, we can show that if $f$ is *n* times differentiable on an interval *I* containing *a* and *x*, then the *n*th remainder *R_n* satisfies

$$R_{n}(x) = \frac{f^{({n + 1})}(c)}{\left( {n + 1} \right)\text{!}}\left( {x - a} \right)^{n + 1}$$

$$R_{n}(x) = \frac{f^{({n + 1})}(c)}{\left( {n + 1} \right)\text{!}}\left( {x - a} \right)^{n + 1}$$

for some real number *c* between *a* and *x*. It is important to note that the value *c* in the numerator above is not the center *a*, but rather an unknown value *c* between *a* and *x*. This formula allows us to get a bound on the remainder *R_n*. If we happen to know that $\left| {f^{({n + 1})}(x)} \right|$ is bounded by some real number *M* on this interval *I*, then

$$\left| {R_{n}(x)} \right| \leq \frac{M}{\left( {n + 1} \right)\text{!}}\left| {x - a} \right|^{n + 1}$$

$$\left| {R_{n}(x)} \right| \leq \frac{M}{\left( {n + 1} \right)\text{!}}\left| {x - a} \right|^{n + 1}$$

for all *x* in the interval *I*.

We now state Taylor's theorem, which provides the formal relationship between a function $f$ and its *n*th degree Taylor polynomial $p_{n}(x).$ This theorem allows us to bound the error when using a Taylor polynomial to approximate a function value, and will be important in proving that a Taylor series for $f$ converges to $f.$

Taylor's Theorem with Remainder 带余项的泰勒定理

Let $f$ be a function that can be differentiated $n + 1$ times on an interval *I* containing the real number *a*. Let *p_n* be the *n*th Taylor polynomial of $f$ at *a* and let

$$R_{n}(x) = f(x) - p_{n}(x)$$

$$R_{n}(x) = f(x) - p_{n}(x)$$

be the *n*th remainder. Then for each *x* in the interval *I*, there exists a real number *c* between *a* and *x* such that

$$R_{n}(x) = \frac{f^{({n + 1})}(c)}{\left( {n + 1} \right)\text{!}}\left( {x - a} \right)^{n + 1}.$$

$$R_{n}(x) = \frac{f^{({n + 1})}(c)}{\left( {n + 1} \right)\text{!}}\left( {x - a} \right)^{n + 1}.$$

If there exists a real number *M* such that $\left| {f^{({n + 1})}(x)} \right| \leq M$ for all $x \in I,$ then

$$\left| {R_{n}(x)} \right| \leq \frac{M}{\left( {n + 1} \right)\text{!}}\left| {x - a} \right|^{n + 1}$$

$$\left| {R_{n}(x)} \right| \leq \frac{M}{\left( {n + 1} \right)\text{!}}\left| {x - a} \right|^{n + 1}$$

for all *x* in *I*.

Proof 证明

Fix a point $x \in I$ and introduce the function *g* such that

$$g(t) = f(x) - f(t) - f^{\prime}(t)\left( {x - t} \right) - \frac{f^{''}(t)}{2\text{!}}\left( {x - t} \right)^{2} - \text{⋯} - \frac{f^{(n)}(t)}{n\text{!}}\left( {x - t} \right)^{n} - R_{n}(x)\frac{\left( {x - t} \right)^{n + 1}}{\left( {x - a} \right)^{n + 1}}.$$

$$g(t) = f(x) - f(t) - f^{\prime}(t)\left( {x - t} \right) - \frac{f^{''}(t)}{2\text{!}}\left( {x - t} \right)^{2} - \text{⋯} - \frac{f^{(n)}(t)}{n\text{!}}\left( {x - t} \right)^{n} - R_{n}(x)\frac{\left( {x - t} \right)^{n + 1}}{\left( {x - a} \right)^{n + 1}}.$$

We claim that *g* satisfies the criteria of Rolle's theorem. Since *g* is a polynomial function (in *t*), it is a differentiable function. Also, *g* is zero at $t = a$ and $t = x$ because

$$\begin{array}{cll} {g(a)} & = & {f(x) - f(a) - f^{\prime}(a)\left( {x - a} \right) - \frac{f^{''}(a)}{2\text{!}}\left( {x - a} \right)^{2} + \text{⋯} + \frac{f^{(n)}(a)}{n\text{!}}\left( {x - a} \right)^{n} - R_{n}(x)} \\ & = & {f(x) - p_{n}(x) - R_{n}(x)} \\ & = & {0,} \\ {g(x)} & = & {f(x) - f(x) - 0 - \text{⋯} - 0} \\ & = & 0. \end{array}$$

$$\begin{array}{cll} {g(a)} & = & {f(x) - f(a) - f^{\prime}(a)\left( {x - a} \right) - \frac{f^{''}(a)}{2\text{!}}\left( {x - a} \right)^{2} + \text{⋯} + \frac{f^{(n)}(a)}{n\text{!}}\left( {x - a} \right)^{n} - R_{n}(x)} \\ & = & {f(x) - p_{n}(x) - R_{n}(x)} \\ & = & {0,} \\ {g(x)} & = & {f(x) - f(x) - 0 - \text{⋯} - 0} \\ & = & 0. \end{array}$$

Therefore, *g* satisfies Rolle's theorem, and consequently, there exists *c* between *a* and *x* such that $g^{\prime}(c) = 0.$ We now calculate $g^{\prime}.$ Using the product rule, we note that

$$\frac{d}{dt}\left\lbrack {\frac{f^{(n)}(t)}{n\text{!}}\left( {x - t} \right)^{n}} \right\rbrack = \frac{\text{−}f^{(n)}(t)}{\left( {n - 1} \right)\text{!}}\left( {x - t} \right)^{n - 1} + \frac{f^{({n + 1})}(t)}{n\text{!}}\left( {x - t} \right)^{n}.$$

$$\frac{d}{dt}\left\lbrack {\frac{f^{(n)}(t)}{n\text{!}}\left( {x - t} \right)^{n}} \right\rbrack = \frac{\text{−}f^{(n)}(t)}{\left( {n - 1} \right)\text{!}}\left( {x - t} \right)^{n - 1} + \frac{f^{({n + 1})}(t)}{n\text{!}}\left( {x - t} \right)^{n}.$$

Consequently,

$$\begin{array}{cl} {g^{\prime}(t)} & {= \text{−}f^{\prime}(t) + \left\lbrack {f^{\prime}(t) - f^{''}(t)\left( {x - t} \right)} \right\rbrack + \left\lbrack {f^{''}(t)\left( {x - t} \right) - \frac{f\text{'''}(t)}{2\text{!}}\left( {x - t} \right)^{2}} \right\rbrack + \text{⋯}} \\ & {\mspace{25mu} + \left\lbrack {\frac{f^{(n)}(t)}{\left( {n - 1} \right)\text{!}}\left( {x - t} \right)^{n - 1} - \frac{f^{({n + 1})}(t)}{n\text{!}}\left( {x - t} \right)^{n}} \right\rbrack + \left( {n + 1} \right)R_{n}(x)\frac{\left( {x - t} \right)^{n}}{\left( {x - a} \right)^{n + 1}}.} \end{array}$$

$$\begin{array}{cl} {g^{\prime}(t)} & {= \text{−}f^{\prime}(t) + \left\lbrack {f^{\prime}(t) - f^{''}(t)\left( {x - t} \right)} \right\rbrack + \left\lbrack {f^{''}(t)\left( {x - t} \right) - \frac{f\text{'''}(t)}{2\text{!}}\left( {x - t} \right)^{2}} \right\rbrack + \text{⋯}} \\ & {\mspace{25mu} + \left\lbrack {\frac{f^{(n)}(t)}{\left( {n - 1} \right)\text{!}}\left( {x - t} \right)^{n - 1} - \frac{f^{({n + 1})}(t)}{n\text{!}}\left( {x - t} \right)^{n}} \right\rbrack + \left( {n + 1} \right)R_{n}(x)\frac{\left( {x - t} \right)^{n}}{\left( {x - a} \right)^{n + 1}}.} \end{array}$$

Notice that there is a telescoping effect. Therefore,

$$g^{\prime}(t) = - \frac{f^{({n + 1})}(t)}{n\text{!}}\left( {x - t} \right)^{n} + \left( {n + 1} \right)R_{n}(x)\frac{\left( {x - t} \right)^{n}}{\left( {x - a} \right)^{n + 1}}.$$

$$g^{\prime}(t) = - \frac{f^{({n + 1})}(t)}{n\text{!}}\left( {x - t} \right)^{n} + \left( {n + 1} \right)R_{n}(x)\frac{\left( {x - t} \right)^{n}}{\left( {x - a} \right)^{n + 1}}.$$

By Rolle's theorem, we conclude that there exists a number *c* between *a* and *x* such that $g^{\prime}(c) = 0.$ Since

$$g^{\prime}(c) = - \frac{f^{({n + 1})}(c)}{n\text{!}}\left( {x - c} \right)^{n} + \left( {n + 1} \right)R_{n}(x)\frac{\left( {x - c} \right)^{n}}{\left( {x - a} \right)^{n + 1}}$$

$$g^{\prime}(c) = - \frac{f^{({n + 1})}(c)}{n\text{!}}\left( {x - c} \right)^{n} + \left( {n + 1} \right)R_{n}(x)\frac{\left( {x - c} \right)^{n}}{\left( {x - a} \right)^{n + 1}}$$

we conclude that

$$- \frac{f^{({n + 1})}(c)}{n\text{!}}\left( {x - c} \right)^{n} + \left( {n + 1} \right)R_{n}(x)\frac{\left( {x - c} \right)^{n}}{\left( {x - a} \right)^{n + 1}} = 0.$$

$$- \frac{f^{({n + 1})}(c)}{n\text{!}}\left( {x - c} \right)^{n} + \left( {n + 1} \right)R_{n}(x)\frac{\left( {x - c} \right)^{n}}{\left( {x - a} \right)^{n + 1}} = 0.$$

Adding the first term on the left-hand side to both sides of the equation and dividing both sides of the equation by $\frac{(n + 1)(x - c)^{n}}{(x - a)^{n + 1}},$ we conclude that

$$R_{n}(x) = \frac{f^{({n + 1})}(c)}{\left( {n + 1} \right)\text{!}}\left( {x - a} \right)^{n + 1}$$

$$R_{n}(x) = \frac{f^{({n + 1})}(c)}{\left( {n + 1} \right)\text{!}}\left( {x - a} \right)^{n + 1}$$

as desired. From this fact, it follows that if there exists *M* such that $\left| {f^{({n + 1})}(x)} \right| \leq M$ for all *x* in *I*, then

$$\left| {R_{n}(x)} \right| \leq \frac{M}{\left( {n + 1} \right)\text{!}}\left| {x - a} \right|^{n + 1}.$$

$$\left| {R_{n}(x)} \right| \leq \frac{M}{\left( {n + 1} \right)\text{!}}\left| {x - a} \right|^{n + 1}.$$

□

Not only does Taylor's theorem allow us to prove that a Taylor series converges to a function, but it also allows us to estimate the accuracy of Taylor polynomials in approximating function values. We begin by looking at linear and quadratic approximations of $f(x) = \sqrt[3]{x}$ at $x = 8$ and determine how accurate these approximations are at estimating $\sqrt[3]{11}.$

Using Linear and Quadratic Approximations to Estimate Function Values 用线性与二次逼近估计函数值

Consider the function $f(x) = \sqrt[3]{x}.$

1. Find the first and second Taylor polynomials for $f$ at $x = 8.$ Use a graphing utility to compare these polynomials with $f$ near $x = 8.$

2. Use these two polynomials to estimate $\sqrt[3]{11}.$

3. Use Taylor's theorem to bound the error.

Solution 解答

1. For $f(x) = \sqrt[3]{x},$ the values of the function and its first two derivatives at $x = 8$ are as follows:

$$\begin{array}{rllccrll} {f(x)} & = & \sqrt[3]{x} & & & {f(8)} & = & 2 \\ {f^{\prime}(x)} & = & \frac{1}{3x^{2\text{/}3}} & & & {f^{\prime}(8)} & = & \frac{1}{12} \\ {f^{''}(x)} & = & \frac{-2}{9x^{5\text{/}3}} & & & {f^{''}(8)} & = & {- \frac{1}{144}.} \end{array}$$

$$\begin{array}{rllccrll} {f(x)} & = & \sqrt[3]{x} & & & {f(8)} & = & 2 \\ {f^{\prime}(x)} & = & \frac{1}{3x^{2\text{/}3}} & & & {f^{\prime}(8)} & = & \frac{1}{12} \\ {f^{''}(x)} & = & \frac{-2}{9x^{5\text{/}3}} & & & {f^{''}(8)} & = & {- \frac{1}{144}.} \end{array}$$

Thus, the first and second Taylor polynomials at $x = 8$ are given by

$$\begin{array}{cll} {p_{1}(x)} & = & {f(8) + f^{\prime}(8)\left( {x - 8} \right)} \\ & = & {2 + \frac{1}{12}\left( {x - 8} \right)} \\ {p_{2}(x)} & = & {f(8) + f^{\prime}(8)\left( {x - 8} \right) + \frac{f^{''}(8)}{2\text{!}}\left( {x - 8} \right)^{2}} \\ & = & {2 + \frac{1}{12}\left( {x - 8} \right) - \frac{1}{288}\left( {x - 8} \right)^{2}.} \end{array}$$

$$\begin{array}{cll} {p_{1}(x)} & = & {f(8) + f^{\prime}(8)\left( {x - 8} \right)} \\ & = & {2 + \frac{1}{12}\left( {x - 8} \right)} \\ {p_{2}(x)} & = & {f(8) + f^{\prime}(8)\left( {x - 8} \right) + \frac{f^{''}(8)}{2\text{!}}\left( {x - 8} \right)^{2}} \\ & = & {2 + \frac{1}{12}\left( {x - 8} \right) - \frac{1}{288}\left( {x - 8} \right)^{2}.} \end{array}$$

The function and the Taylor polynomials are shown in Figure 6.9.

2. Using the first Taylor polynomial at $x = 8,$ we can estimate

$$\sqrt[3]{11} \approx p_{1}(11) = 2 + \frac{1}{12}\left( {11 - 8} \right) = 2.25.$$

$$\sqrt[3]{11} \approx p_{1}(11) = 2 + \frac{1}{12}\left( {11 - 8} \right) = 2.25.$$

Using the second Taylor polynomial at $x = 8,$ we obtain

$$\sqrt[3]{11} \approx p_{2}(11) = 2 + \frac{1}{12}\left( {11 - 8} \right) - \frac{1}{288}\left( {11 - 8} \right)^{2} = 2.21875.$$

$$\sqrt[3]{11} \approx p_{2}(11) = 2 + \frac{1}{12}\left( {11 - 8} \right) - \frac{1}{288}\left( {11 - 8} \right)^{2} = 2.21875.$$

3. By Taylor's Theorem with Remainder, there exists a *c* in the interval $\left( {8,11} \right)$ such that the remainder when approximating $\sqrt[3]{11}$ by the first Taylor polynomial satisfies

$$R_{1}(11) = \frac{f^{''}(c)}{2\text{!}}\left( {11 - 8} \right)^{2}.$$

$$R_{1}(11) = \frac{f^{''}(c)}{2\text{!}}\left( {11 - 8} \right)^{2}.$$

We do not know the exact value of *c*, so we find an upper bound on $R_{1}(11)$ by determining the maximum value of $f^{''}$ on the interval $\left( {8,11} \right).$ Since $f^{''}(x) = - \frac{2}{9x^{5\text{/}3}},$ the largest value for $\left| {f^{''}(x)} \right|$ on that interval occurs at $x = 8.$ Using the fact that $f^{''}(8) = - \frac{1}{144},$ we obtain

$$\left| {R_{1}(11)} \right| \leq \frac{1}{144 \cdot 2\text{!}}\left( {11 - 8} \right)^{2} = 0.03125.$$

$$\left| {R_{1}(11)} \right| \leq \frac{1}{144 \cdot 2\text{!}}\left( {11 - 8} \right)^{2} = 0.03125.$$

Similarly, to estimate $R_{2}(11),$ we use the fact that

$$R_{2}(11) = \frac{f\text{'''}(c)}{3\text{!}}\left( {11 - 8} \right)^{3}.$$

$$R_{2}(11) = \frac{f\text{'''}(c)}{3\text{!}}\left( {11 - 8} \right)^{3}.$$

Since $f\text{'''}(x) = \frac{10}{27x^{8\text{/}3}},$ the maximum value of $f\text{'''}$ on the interval $\left( {8,11} \right)$ is $f\text{'''}(8) \approx 0.0014468.$ Therefore, we have

$$\left| {R_{2}(11)} \right| \leq \frac{0.0011468}{3\text{!}}\left( {11 - 8} \right)^{3} \approx 0.0065104.$$

$$\left| {R_{2}(11)} \right| \leq \frac{0.0011468}{3\text{!}}\left( {11 - 8} \right)^{3} \approx 0.0065104.$$

Find the first and second Taylor polynomials for $f(x) = \sqrt{x}$ at $x = 4.$ Use these polynomials to estimate $\sqrt{6}.$ Use Taylor's theorem to bound the error.

Approximating sin *x* Using Maclaurin Polynomials 用麦克劳林多项式近似 sin *x*

From Example 6.12b., the Maclaurin polynomials for $\text{sin}\mspace{2mu} x$ are given by

$$\begin{array}{cl} {p_{2m + 1}(x)} & {= p_{2m + 2}(x)} \\ & {= x - \frac{x^{3}}{3\text{!}} + \frac{x^{5}}{5\text{!}} - \frac{x^{7}}{7\text{!}} + \text{⋯} + (-1)^{m}\frac{x^{2m + 1}}{\left( {2m + 1} \right)\text{!}}} \end{array}$$

$$\begin{array}{cl} {p_{2m + 1}(x)} & {= p_{2m + 2}(x)} \\ & {= x - \frac{x^{3}}{3\text{!}} + \frac{x^{5}}{5\text{!}} - \frac{x^{7}}{7\text{!}} + \text{⋯} + (-1)^{m}\frac{x^{2m + 1}}{\left( {2m + 1} \right)\text{!}}} \end{array}$$

for $m = 0,1,2,\text{…}.$

1. Use the fifth Maclaurin polynomial for $\text{sin}\mspace{2mu} x$ to approximate $\text{sin}\left( \frac{\pi}{18} \right)$ and bound the error.

2. For what values of *x* does the fifth Maclaurin polynomial approximate $\text{sin}\mspace{2mu} x$ to within 0.0001?

Solution 解答

1. The fifth Maclaurin polynomial is

$$p_{5}(x) = x - \frac{x^{3}}{3\text{!}} + \frac{x^{5}}{5\text{!}}.$$

$$p_{5}(x) = x - \frac{x^{3}}{3\text{!}} + \frac{x^{5}}{5\text{!}}.$$

Using this polynomial, we can estimate as follows:

$$\begin{array}{cl} {\text{sin}\left( \frac{\pi}{18} \right)} & {\approx p_{5}\left( \frac{\pi}{18} \right)} \\ & {= \frac{\pi}{18} - \frac{1}{3\text{!}}\left( \frac{\pi}{18} \right)^{3} + \frac{1}{5\text{!}}\left( \frac{\pi}{18} \right)^{5}} \\ & {\approx 0.173648.} \end{array}$$

$$\begin{array}{cl} {\text{sin}\left( \frac{\pi}{18} \right)} & {\approx p_{5}\left( \frac{\pi}{18} \right)} \\ & {= \frac{\pi}{18} - \frac{1}{3\text{!}}\left( \frac{\pi}{18} \right)^{3} + \frac{1}{5\text{!}}\left( \frac{\pi}{18} \right)^{5}} \\ & {\approx 0.173648.} \end{array}$$

To estimate the error, use the fact that the sixth Maclaurin polynomial is $p_{6}(x) = p_{5}(x)$ and calculate a bound on $R_{6}\left( \frac{\pi}{18} \right).$ By Uniqueness of Taylor Series, the remainder is

$$R_{6}\left( \frac{\pi}{18} \right) = \frac{f^{(7)}(c)}{7\text{!}}\left( \frac{\pi}{18} \right)^{7}$$

$$R_{6}\left( \frac{\pi}{18} \right) = \frac{f^{(7)}(c)}{7\text{!}}\left( \frac{\pi}{18} \right)^{7}$$

for some *c* between 0 and $\frac{\pi}{18}.$ Using the fact that $\left| {f^{(7)}(x)} \right| \leq 1$ for all *x*, we find that the magnitude of the error is at most

$$\frac{1}{7\text{!}} \cdot \left( \frac{\pi}{18} \right)^{7} \leq 9.8\ \times \ 10^{-10}.$$

$$\frac{1}{7\text{!}} \cdot \left( \frac{\pi}{18} \right)^{7} \leq 9.8\ \times \ 10^{-10}.$$

2. We need to find the values of *x* such that

$$\frac{1}{7\text{!}}|x|^{7} \leq 0.0001.$$

$$\frac{1}{7\text{!}}|x|^{7} \leq 0.0001.$$

Solving this inequality for *x*, we have that the fifth Maclaurin polynomial gives an estimate to within 0.0001 as long as $|x| < 0.907.$

Use the fourth Maclaurin polynomial for $\text{cos}\mspace{2mu} x$ to approximate $\text{cos}\left( \frac{\pi}{12} \right).$

Now that we are able to bound the remainder $R_{n}(x),$ we can use this bound to prove that a Taylor series for $f$ at *a* converges to $f.$

Representing Functions with Taylor and Maclaurin Series 用泰勒级数与麦克劳林级数表示函数

We now discuss issues of convergence for Taylor series. We begin by showing how to find a Taylor series for a function, and how to find its interval of convergence.

Finding a Taylor Series 求泰勒级数

Find the Taylor series for $f(x) = \frac{1}{x}$ at $x = 1.$ Determine the interval of convergence.

Solution 解答

For $f(x) = \frac{1}{x},$ the values of the function and its first four derivatives at $x = 1$ are

$$\begin{array}{rllccrll} {f(x)} & = & \frac{1}{x} & & & {f(1)} & = & 1 \\ {f^{\prime}(x)} & = & {- \frac{1}{x^{2}}} & & & {f^{\prime}(1)} & = & -1 \\ {f^{''}(x)} & = & \frac{2}{x^{3}} & & & {f^{''}(1)} & = & {2\text{!}} \\ {f\text{'''}(x)} & = & {- \frac{3 \cdot 2}{x^{4}}} & & & {f\text{'''}(1)} & = & {-3\text{!}} \\ {f^{(4)}(x)} & = & \frac{4 \cdot 3 \cdot 2}{x^{5}} & & & {f^{(4)}(1)} & = & {4\text{!.}} \end{array}$$

That is, we have $f^{(n)}(1) = (-1)^{n}n\text{!}$ for all $n \geq 0.$ Therefore, the Taylor series for $f$ at $x = 1$ is given by

$${\sum\limits_{n = 0}^{\infty}{\frac{f^{(n)}(1)}{n\text{!}}\left( {x - 1} \right)^{n}}} = {\sum\limits_{n = 0}^{\infty}{(-1)^{n}\left( {x - 1} \right)^{n}}}.$$

To find the interval of convergence, we use the ratio test. We find that

$$\frac{\left| a_{n + 1} \right|}{\left| a_{n} \right|} = \frac{\left| {(-1)^{n + 1}\left( {x - 1} \right)^{n + 1}} \right|}{\left| {(-1)^{n}\left( {x - 1} \right)^{n}} \right|} = \left| {x - 1} \right|.$$

Thus, the series converges if $\left| {x - 1} \right| < 1.$ That is, the series converges for $0 < x < 2.$ Next, we need to check the endpoints. At $x = 2,$ we see that

$${\sum\limits_{n = 0}^{\infty}{(-1)^{n}\left( {2 - 1} \right)^{n}}} = {\sum\limits_{n = 0}^{\infty}(-1)^{n}}$$

diverges by the divergence test. Similarly, at $x = 0,$

$${\sum\limits_{n = 0}^{\infty}{(-1)^{n}\left( {0 - 1} \right)^{n}}} = {\sum\limits_{n = 0}^{\infty}(-1)^{2n}} = {\sum\limits_{n = 0}^{\infty}1}$$

diverges. Therefore, the interval of convergence is $\left( {0,2} \right).$

Find the Taylor series for $f(x) = \frac{1}{2x}$ at $x = 2$ and determine its interval of convergence.

We know that the Taylor series found in this example converges on the interval $\left( {0,2} \right),$ but how do we know it actually converges to $f?$ We consider this question in more generality in a moment, but for this example, we can answer this question by writing

$$f(x) = \frac{1}{x} = \frac{1}{1 - \left( {1 - x} \right)}.$$

That is, $f$ can be represented by the geometric series ${\sum\limits_{n = 0}^{\infty}\left( {1 - x} \right)^{n}}.$ Since this is a geometric series, it converges to $\frac{1}{x}$ as long as $\left| {1 - x} \right| < 1.$ Therefore, the Taylor series found in Example 6.15 does converge to $f(x) = \frac{1}{x}$ on $\left( {0,2} \right).$

We now consider the more general question: if a Taylor series for a function $f$ converges on some interval, how can we determine if it actually converges to $f?$ To answer this question, recall that a series converges to a particular value if and only if its sequence of partial sums converges to that value. Given a Taylor series for $f$ at *a*, the *n*th partial sum is given by the *n*th Taylor polynomial *p_n*. Therefore, to determine if the Taylor series converges to $f,$ we need to determine whether

$$\underset{n\rightarrow\infty}{\text{lim}}p_{n}(x) = f(x).$$

Since the remainder $R_{n}(x) = f(x) - p_{n}(x),$ the Taylor series converges to $f$ if and only if

$$\underset{n\rightarrow\infty}{\text{lim}}R_{n}(x) = 0.$$

We now state this theorem formally.

Convergence of Taylor Series 泰勒级数的收敛性

Suppose that $f$ has derivatives of all orders on an interval *I* containing *a*. Then the Taylor series

$$\sum\limits_{n = 0}^{\infty}{\frac{f^{(n)}(a)}{n\text{!}}\left( {x - a} \right)^{n}}$$

converges to $f(x)$ for all *x* in *I* if and only if

$$\underset{n\rightarrow\infty}{\text{lim}}R_{n}(x) = 0$$

for all *x* in *I*.

With this theorem, we can prove that a Taylor series for $f$ at *a* converges to $f$ if we can prove that the remainder $R_{n}(x)\rightarrow 0.$ To prove that $R_{n}(x)\rightarrow 0,$ we typically use the bound

$$\left| {R_{n}(x)} \right| \leq \frac{M}{\left( {n + 1} \right)\text{!}}\left| {x - a} \right|^{n + 1}$$

from Taylor’s theorem with remainder.

In the next example, we find the Maclaurin series for *e^x* and $\text{sin}\mspace{2mu} x$ and show that these series converge to the corresponding functions for all real numbers by proving that the remainders $R_{n}(x)\rightarrow 0$ for all real numbers *x*.

Finding Maclaurin Series 求麦克劳林级数

For each of the following functions, find the Maclaurin series and its interval of convergence. Use Taylor’s Theorem with Remainder to prove that the Maclaurin series for $f$ converges to $f$ on that interval.

1. *e^x*

2. $\text{sin}\mspace{2mu} x$

Solution 解答

1. Using the *n*th Maclaurin polynomial for *e^x* found in Example 6.12a., we find that the Maclaurin series for *e^x* is given by

$${\sum\limits_{n = 0}^{\infty}\frac{x^{n}}{n\text{!}}}.$$

To determine the interval of convergence, we use the ratio test. Since

$$\frac{\left| a_{n + 1} \right|}{\left| a_{n} \right|} = \frac{|x|^{n + 1}}{\left( {n + 1} \right)\text{!}} \cdot \frac{n\text{!}}{|x|^{n}} = \frac{|x|}{n + 1},$$

we have

$$\underset{n\rightarrow\infty}{\text{lim}}\frac{\left| a_{n + 1} \right|}{\left| a_{n} \right|} = \underset{n\rightarrow\infty}{\text{lim}}\frac{|x|}{n + 1} = 0$$

for all *x*. Therefore, the series converges absolutely for all *x*, and thus, the interval of convergence is $\left( {\text{−}\infty,\infty} \right).$ To show that the series converges to *e^x* for all *x*, we use the fact that $f^{(n)}(x) = e^{x}$ for all $n \geq 0$ and *e^x* is an increasing function on $\left( {\text{−}\infty,\infty} \right).$ Therefore, for any real number *b*, the maximum value of *e^x* for all $|x| \leq b$ is *e^b*. Thus,

$$\left| {R_{n}(x)} \right| \leq \frac{e^{b}}{\left( {n + 1} \right)\text{!}}|x|^{n + 1}.$$

Since we just showed that

$$\sum\limits_{n = 0}^{\infty}\frac{|x|^{n}}{n\text{!}}$$

converges for all *x*, by the divergence test, we know that

$$\underset{n\rightarrow\infty}{\text{lim}}\frac{|x|^{n + 1}}{\left( {n + 1} \right)\text{!}} = 0$$

for any real number *x*. By combining this fact with the squeeze theorem, the result is $\underset{n\rightarrow\infty}{\text{lim}}R_{n}(x) = 0.$

2. Using the *n*th Maclaurin polynomial for $\text{sin}\mspace{2mu} x$ found in Example 6.12b., we find that the Maclaurin series for $\text{sin}\mspace{2mu} x$ is given by

$${\sum\limits_{n = 0}^{\infty}{(-1)^{n}\frac{x^{2n + 1}}{\left( {2n + 1} \right)\text{!}}}}.$$

In order to apply the ratio test, consider

$$\frac{\left| a_{n + 1} \right|}{\left| a_{n} \right|} = \frac{|x|^{2n + 3}}{\left( {2n + 3} \right)\text{!}} \cdot \frac{\left( {2n + 1} \right)\text{!}}{|x|^{2n + 1}} = \frac{|x|^{2}}{\left( {2n + 3} \right)\left( {2n + 2} \right)}.$$

Since

$$\underset{n\rightarrow\infty}{\text{lim}}\frac{|x|^{2}}{\left( {2n + 3} \right)\left( {2n + 2} \right)} = 0$$

for all *x*, we obtain the interval of convergence as $\left( {\text{−}\infty,\infty} \right).$ To show that the Maclaurin series converges to $\text{sin}\mspace{2mu} x,$ look at $R_{n}(x).$ For each *x* there exists a real number *c* between 0 and *x* such that

$$R_{n}(x) = \frac{f^{({n + 1})}(c)}{\left( {n + 1} \right)\text{!}}x^{n + 1}.$$

Since $\left| {f^{({n + 1})}(c)} \right| \leq 1$ for all integers *n* and all real numbers *c*, we have

$$\left| {R_{n}(x)} \right| \leq \frac{|x|^{n + 1}}{\left( {n + 1} \right)\text{!}}$$

for all real numbers *x*. Using the same idea as in part a., the result is $\underset{n\rightarrow\infty}{\text{lim}}R_{n}(x) = 0$ for all *x*, and therefore, the Maclaurin series for $\text{sin}\mspace{2mu} x$ converges to $\text{sin}\mspace{2mu} x$ for all real *x*.

Find the Maclaurin series for $f(x) = \text{cos}\mspace{2mu} x.$ Use the ratio test to show that the interval of convergence is $\left( {\text{−}\infty,\infty} \right).$ Show that the Maclaurin series converges to $\text{cos}\mspace{2mu} x$ for all real numbers *x*.

Proving that *e* is Irrational 证明 *e* 是无理数

In this project, we use the Maclaurin polynomials for *e^x* to prove that *e* is irrational. The proof relies on supposing that *e* is rational and arriving at a contradiction. Therefore, in the following steps, we suppose $e = {r\text{/}s}$ for some integers *r* and *s* where $s \neq 0.$

1. Write the Maclaurin polynomials $p_{0}(x),p_{1}(x),p_{2}(x),p_{3}(x),p_{4}(x)$ for *e^x*. Evaluate $p_{0}(1),p_{1}(1),p_{2}(1),p_{3}(1),p_{4}(1)$ to estimate *e*.

2. Let $R_{n}(x)$ denote the remainder when using $p_{n}(x)$ to estimate *e^x*. Therefore, $R_{n}(x) = e^{x} - p_{n}(x),$ and $R_{n}(1) = e - p_{n}(1).$ Assuming that $e = \frac{r}{s}$ for integers *r* and *s*, evaluate $R_{0}(1),R_{1}(1),R_{2}(1),R_{3}(1),R_{4}(1).$

3. Using the results from part 2, show that for each remainder $R_{0}(1),R_{1}(1),R_{2}(1),R_{3}(1),R_{4}(1),$ we can find an integer *k* such that $kR_{n}(1)$ is an integer for $n = 0,1,2,3,4.$

4. Write down the formula for the *n*th Maclaurin polynomial $p_{n}(x)$ for *e^x* and the corresponding remainder $R_{n}(x).$ Show that $sn\text{!}R_{n}(1)$ is an integer.

5. Use Taylor’s theorem to write down an explicit formula for $R_{n}(1).$ Conclude that $R_{n}(1) \neq 0,$ and therefore, $sn\text{!}R_{n}(1) \neq 0.$

6. Use Taylor’s theorem to find an estimate on $R_{n}(1).$ Use this estimate combined with the result from part 5 to show that $\left| {sn\text{!}R_{n}(1)} \right| < \frac{se}{n + 1}.$ Conclude that if *n* is large enough, then $\left| {sn\text{!}R_{n}(1)} \right| < 1.$ Therefore, $sn\text{!}R_{n}(1)$ is an integer with magnitude less than 1. Thus, $sn\text{!}R_{n}(1) = 0.$ But from part 5, we know that $sn\text{!}R_{n}(1) \neq 0.$ We have arrived at a contradiction, and consequently, the original supposition that *e* is rational must be false.

Section 6.3 Exercises 6.3 节习题

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

116\.

$f(x) = 1 + x + x^{2}$ at $a = 1$

117.

$f(x) = 1 + x + x^{2}$ at $a = -1$

118\.

$f(x) = \text{cos}\left( {2x} \right)$ at $a = \pi$

119.

$f(x) = \text{sin}\left( {2x} \right)$ at $a = \frac{\pi}{2}$

120\.

$f(x) = \sqrt{x}$ at $a = 4$

121.

$f(x) = \text{ln}\mspace{2mu} x$ at $a = 1$

122\.

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

123.

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

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

124\.

\[T\] $\sqrt{10};a = 9,n = 3$

125.

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

126\.

\[T\] $\text{sin}(6);a = 2\pi,n = 5$

127.

\[T\] *e*²; $a = 0,n = 9$

128\.

\[T\] $\text{cos}\left( \frac{\pi}{5} \right);a = 0,n = 4$

129.

\[T\] $\text{ln}\mspace{2mu}(2);a = 1,n = 1000$

130\.

Integrate the approximation $\text{sin}\mspace{2mu} t \approx t - \frac{t^{3}}{6} + \frac{t^{5}}{120} - \frac{t^{7}}{5040}$ evaluated at *πt* to approximate ${\int_{0}^{1}{\frac{\text{sin}\mspace{2mu}\pi t}{\pi t}dt}}.$

131.

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

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

132\.

$f(x) = \text{sin}\mspace{2mu} x$ on $\left\lbrack {\text{−}\pi,\pi} \right\rbrack,a = 0$

133.

$f(x) = \text{cos}\mspace{2mu} x$ on $\left\lbrack {- \frac{\pi}{2},\frac{\pi}{2}} \right\rbrack,a = 0$

134\.

$f(x) = e^{-2x}$ on $\left\lbrack {-1,1} \right\rbrack,a = 0$

135.

$f(x) = e^{\text{−}x}$ on $\left\lbrack {-3,3} \right\rbrack,a = 0$

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

136\.

\[T\] *e^x* approximated by $1 + x,a = 0$

137.

\[T\] $\text{sin}\mspace{2mu} x$ approximated by *x*, $a = 0$

138\.

\[T\] $\text{ln}\mspace{2mu} x$ approximated by $x - 1,a = 1$

139.

\[T\] $\text{cos}\mspace{2mu} x$ approximated by $1,a = 0$

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

140\.

$x^{4}$ at $a = -1$

141.

$1 + x + x^{2} + x^{3}$ at $a = -1$

142\.

$\text{sin}\mspace{2mu} x$ at $a = \pi$

143.

$\text{cos}\mspace{2mu} x$ at $a = 2\pi$

144\.

$\text{sin}\mspace{2mu} x$ at $x = \frac{\pi}{2}$

145.

$\text{cos}\mspace{2mu} x$ at $x = \frac{\pi}{2}$

146\.

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

147.

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

148\.

$\frac{1}{\left( {x - 1} \right)^{2}}$ at $a = 0$ (*Hint:* Differentiate $\frac{1}{1 - x}.)$

149.

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

150\.

$F(x) = {\int_{0}^{x}{\text{cos}\left( \sqrt{t} \right)\mspace{2mu} dt}};f(t) = {\sum\limits_{n = 0}^{\infty}{(-1)^{n}\frac{t^{n}}{\left( {2n} \right)\text{!}}}}$ at $a = 0$ (*Note*: $f$ is the Taylor series of $\text{cos}\left( \sqrt{t} \right).)$

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

151.

$f(x) = 2 - x$

152\.

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

153.

$f(x) = \left( {x - 2} \right)^{2}$

154\.

$f(x) = \text{ln}\mspace{2mu} x$

155.

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

156\.

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

157.

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

158\.

$f(x) = e^{\text{−}x}$

159.

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

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

160\.

$\sum\limits_{n = 0}^{\infty}\frac{1}{n\text{!}}$

161.

$\sum\limits_{n = 0}^{\infty}\frac{2^{n}}{n\text{!}}$

162\.

$\sum\limits_{n = 0}^{\infty}\frac{(-1)^{n}\left( {2\pi} \right)^{2n}}{\left( {2n} \right)\text{!}}$

163.

$\sum\limits_{n = 0}^{\infty}\frac{(-1)^{n}\left( {2\pi} \right)^{2n + 1}}{\left( {2n + 1} \right)\text{!}}$

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 $\lbrack\text{−}\pi,\pi\rbrack.$

164\.

\[T\] Plot $\text{sin}^{2}x - \left( {S_{5}(x)} \right)^{2}$ on $\lbrack\text{−}\pi,\pi\rbrack.$ Compare the maximum difference with the square of the Taylor remainder estimate for $\text{sin}\mspace{2mu} x.$

165.

\[T\] Plot $\text{cos}^{2}x - \left( {C_{4}(x)} \right)^{2}$ on $\lbrack\text{−}\pi,\pi\rbrack.$ Compare the maximum difference with the square of the Taylor remainder estimate for $\text{cos}\mspace{2mu} x.$

166\.

\[T\] Plot $\left| {2S_{5}(x)C_{4}(x) - \text{sin}\left( {2x} \right)} \right|$ on $\lbrack\text{−}\pi,\pi\rbrack.$

167.

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

168\.

\[T\] Plot $e^{x} - e_{4}(x)$ where $e_{4}(x) = 1 + x + \frac{x^{2}}{2} + \frac{x^{3}}{6} + \frac{x^{4}}{24}$ on $\lbrack 0,2\rbrack.$ Compare the maximum error with the Taylor remainder estimate.

169.

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

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

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

3. Explain why the approximate roots of $p_{N}(x) - 2 = 0$ are approximate values of $\text{ln}\mspace{2mu}(2).$

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

170\.

$\underset{x\rightarrow 0}{\text{lim}}\frac{\text{cos}\mspace{2mu} x - 1}{x^{2}}$

171.

$\underset{x\rightarrow 0}{\text{lim}}\frac{\text{ln}\mspace{2mu}\left( {1 - x^{2}} \right)}{x^{2}}$

172\.

$\underset{x\rightarrow 0}{\text{lim}}\frac{e^{x^{2}} - x^{2} - 1}{x^{4}}$

173.

$\underset{x\rightarrow 0^{+}}{\text{lim}}\frac{\text{cos}\left( \sqrt{x} \right) - 1}{2x}$

6.4 Working with Taylor Series 6.4 泰勒级数的应用

• 6.4.1 Write the terms of the binomial series.
• 6.4.2 Recognize the Taylor series expansions of common functions.
• 6.4.3 Recognize and apply techniques to find the Taylor series for a function.
• 6.4.4 Use Taylor series to solve differential equations.
• 6.4.5 Use Taylor series to evaluate nonelementary integrals.

• 6.4.1 写出二项级数的各项。
• 6.4.2 识别常用函数的泰勒级数展开式。
• 6.4.3 识别并运用相关技巧求函数的泰勒级数。
• 6.4.4 用泰勒级数求解微分方程。
• 6.4.5 用泰勒级数计算非初等积分。

In the preceding section, we defined Taylor series and showed how to find the Taylor series for several common functions by explicitly calculating the coefficients of the Taylor polynomials. In this section we show how to use those Taylor series to derive Taylor series for other functions. We then present two common applications of power series. First, we show how power series can be used to solve differential equations. Second, we show how power series can be used to evaluate integrals when the antiderivative of the integrand cannot be expressed in terms of elementary functions. In one example, we consider $\left. \int e^{\text{−}x^{2}} \right.dx,$ an integral that arises frequently in probability theory.

The Binomial Series 二项级数

Our first goal in this section is to determine the Maclaurin series for the function $f(x) = \left( {1 + x} \right)^{r}$ for all real numbers $r.$ The Maclaurin series for this function is known as the binomial series. We begin by considering the simplest case: $r$ is a nonnegative integer. We recall that, for $r = 0,1,2,3,4,f(x) = \left( {1 + x} \right)^{r}$ can be written as

$$\begin{array}{l} \\ {f(x) = \left( {1 + x} \right)^{0} = 1,} \\ {f(x) = \left( {1 + x} \right)^{1} = 1 + x,} \\ {f(x) = \left( {1 + x} \right)^{2} = 1 + 2x + x^{2},} \\ {f(x) = \left( {1 + x} \right)^{3} = 1 + 3x + 3x^{2} + x^{3},} \\ {f(x) = \left( {1 + x} \right)^{4} = 1 + 4x + 6x^{2} + 4x^{3} + x^{4}.} \end{array}$$

The expressions on the right-hand side are known as binomial expansions and the coefficients are known as binomial coefficients. More generally, for any nonnegative integer $r,$ the binomial coefficient of $x^{n}$ in the binomial expansion of $\left( {1 + x} \right)^{r}$ is given by

$$\left( \begin{array}{l} r \\ n \end{array} \right) = \frac{r\text{!}}{n\text{!}\left( {r - n} \right)\text{!}}$$ (6.6)

and

$$\begin{array}{cl} {f(x)} & {= \left( {1 + x} \right)^{r}} \\ & {= \left( \begin{array}{l} r \\ 0 \end{array} \right)1 + \left( \begin{array}{l} r \\ 1 \end{array} \right)x + \left( \begin{array}{l} r \\ 2 \end{array} \right)x^{2} + \left( \begin{array}{l} r \\ 3 \end{array} \right)x^{3} + \text{⋯} + \left( \begin{array}{l} r \\ {r - 1} \end{array} \right)x^{r - 1} + \left( \begin{array}{l} r \\ r \end{array} \right)x^{r}} \\ & {= {\sum\limits_{n = 0}^{r}{\left( \begin{array}{l} r \\ n \end{array} \right)x^{n}}}.} \end{array}$$ (6.7)

For example, using this formula for $r = 5,$ we see that

$$\begin{array}{cl} {f(x)} & {= \left( {1 + x} \right)^{5}} \\ & {= \left( \begin{array}{l} 5 \\ 0 \end{array} \right)1 + \left( \begin{array}{l} 5 \\ 1 \end{array} \right)x + \left( \begin{array}{l} 5 \\ 2 \end{array} \right)x^{2} + \left( \begin{array}{l} 5 \\ 3 \end{array} \right)x^{3} + \left( \begin{array}{l} 5 \\ 4 \end{array} \right)x^{4} + \left( \begin{array}{l} 5 \\ 5 \end{array} \right)x^{5}} \\ & {= \frac{5\text{!}}{0\text{!}5\text{!}}1 + \frac{5\text{!}}{1\text{!}4\text{!}}x + \frac{5\text{!}}{2\text{!}3\text{!}}x^{2} + \frac{5\text{!}}{3\text{!}2\text{!}}x^{3} + \frac{5\text{!}}{4\text{!}1\text{!}}x^{4} + \frac{5\text{!}}{5\text{!}0\text{!}}x^{5}} \\ & {= 1 + 5x + 10x^{2} + 10x^{3} + 5x^{4} + x^{5}.} \end{array}$$

We now consider the case when the exponent $r$ is any real number, not necessarily a nonnegative integer. If $r$ is not a nonnegative integer, then $f(x) = \left( {1 + x} \right)^{r}$ cannot be written as a finite polynomial. However, we can find a power series for $f.$ Specifically, we look for the Maclaurin series for $f.$ To do this, we find the derivatives of $f$ and evaluate them at $x = 0.$

$$\begin{array}{rllccrll} {f(x)} & = & \left( {1 + x} \right)^{r} & & & {f(0)} & = & 1 \\ {f^{\prime}(x)} & = & {r\left( {1 + x} \right)^{r - 1}} & & & {f\prime(0)} & = & r \\ {f^{''}(x)} & = & {r\left( {r - 1} \right)\left( {1 + x} \right)^{r - 2}} & & & {f^{''}(0)} & = & {r\left( {r - 1} \right)} \\ {f\text{'''}(x)} & = & {r\left( {r - 1} \right)\left( {r - 2} \right)\left( {1 + x} \right)^{r - 3}} & & & {f\text{'''}(0)} & = & {r\left( {r - 1} \right)\left( {r - 2} \right)} \\ {f^{(n)}(x)} & = & {r\left( {r - 1} \right)\left( {r - 2} \right)\text{⋯}\left( {r - n + 1} \right)\left( {1 + x} \right)^{r - n}} & & & {f^{(n)}(0)} & = & {r\left( {r - 1} \right)\left( {r - 2} \right)\text{⋯}\left( {r - n + 1} \right)} \end{array}$$

We conclude that the coefficients in the binomial series are given by

$$\frac{f^{(n)}(0)}{n\text{!}} = \frac{r\left( {r - 1} \right)\left( {r - 2} \right)\text{⋯}\left( {r - n + 1} \right)}{n\text{!}}.$$ (6.8)

We note that if $r$ is a nonnegative integer, then the $\left( {r + 1} \right)\text{st}$ derivative $f^{({r + 1})}$ is the zero function, and the series terminates. In addition, if $r$ is a nonnegative integer, then Equation 6.8 for the coefficients agrees with Equation 6.6 for the coefficients, and the formula for the binomial series agrees with Equation 6.7 for the finite binomial expansion. More generally, to denote the binomial coefficients for any real number $r,$ we define

$$\left( \begin{array}{l} r \\ n \end{array} \right) = \frac{r\left( {r - 1} \right)\left( {r - 2} \right)\text{⋯}\left( {r - n + 1} \right)}{n\text{!}}.$$

With this notation, we can write the binomial series for $\left( {1 + x} \right)^{r}$ as

$$\sum\limits_{n = 0}^{\infty}{\left( \begin{array}{l} r \\ n \end{array} \right)x^{n} = 1 + rx + \frac{r\left( {r - 1} \right)}{2\text{!}}x^{2} + \text{⋯} + \frac{r\left( {r - 1} \right)\text{⋯}\left( {r - n + 1} \right)}{n\text{!}}x^{n} + \text{⋯}.}$$ (6.9)

We now need to determine the interval of convergence for the binomial series Equation 6.9. We apply the ratio test. Consequently, we consider

$$\begin{array}{cl} \frac{\left| a_{n + 1} \right|}{\left| a_{n} \right|} & {= \frac{\left| {r\left( {r - 1} \right)\left( {r - 2} \right)\text{⋯}\left( {r - n} \right)} \right||x|^{n + 1}}{\left( {n + 1} \right)\text{!}} \cdot \frac{n\text{!}}{\left| {r\left( {r - 1} \right)\left( {r - 2} \right)\text{⋯}\left( {r - n + 1} \right)} \right||x|^{n}}} \\ & {= \frac{\left| {r - n} \right||x|}{\left| {n + 1} \right|}.} \end{array}$$

Since

$$\underset{n\rightarrow\infty}{\text{lim}}\frac{\left| a_{n + 1} \right|}{\left| a_{n} \right|} = |x| < 1$$

if and only if $|x| < 1,$ we conclude that the interval of convergence for the binomial series is $\left( {-1,1} \right).$ The behavior at the endpoints depends on $r.$ It can be shown that for $r \geq 0$ the series converges at both endpoints; for $-1 < r < 0,$ the series converges at $x = 1$ and diverges at $x = -1;$ and for $r < -1,$ the series diverges at both endpoints. The binomial series does converge to $\left( {1 + x} \right)^{r}$ in $\left( {-1,1} \right)$ for all real numbers $r,$ but proving this fact by showing that the remainder $R_{n}(x)\rightarrow 0$ is difficult.

For any real number $r,$ the Maclaurin series for $f(x) = \left( {1 + x} \right)^{r}$ is the binomial series. It converges to $f$ for $|x| < 1,$ and we write

$$\begin{array}{cl} \left( {1 + x} \right)^{r} & {= {\sum\limits_{n = 0}^{\infty}{\left( \begin{array}{l} r \\ n \end{array} \right)x^{n}}}} \\ & {= 1 + rx + \frac{r\left( {r - 1} \right)}{2\text{!}}x^{2} + \text{⋯} + \frac{r\left( {r - 1} \right)\text{⋯}\left( {r - n + 1} \right)}{n\text{!}}x^{n} + \text{⋯}} \end{array}$$

for $|x| < 1.$

We can use this definition to find the binomial series for $f(x) = \sqrt{1 + x}$ and use the series to approximate $\sqrt{1.5}.$

Finding Binomial Series 求二项级数

1. Find the binomial series for $f(x) = \sqrt{1 + x}.$

2. Use the third-order Maclaurin polynomial $p_{3}(x)$ to estimate $\sqrt{1.5}.$ Use Taylor’s theorem to bound the error. Use a graphing utility to compare the graphs of $f$ and $p_{3}.$

Solution 解答

1. Here $r = \frac{1}{2}.$ Using the definition for the binomial series, we obtain

$$\begin{array}{cl} \sqrt{1 + x} & {= 1 + \frac{1}{2}x + \frac{\left( {1\text{/}2} \right)\left( {\text{−}{1\text{/}2}} \right)}{2\text{!}}x^{2} + \frac{\left( {1\text{/}2} \right)\left( {\text{−}{1\text{/}2}} \right)\left( {\text{−}{3\text{/}2}} \right)}{3\text{!}}x^{3} + \text{⋯}} \\ & {= 1 + \frac{1}{2}x - \frac{1}{2\text{!}}\ \frac{1}{2^{2}}x^{2} + \frac{1}{3\text{!}}\ \frac{1 \cdot 3}{2^{3}}x^{3} - \text{⋯} + \frac{(-1)^{n + 1}}{n\text{!}}\ \frac{1 \cdot 3 \cdot 5\text{⋯}\left( {2n - 3} \right)}{2^{n}}x^{n} + \text{⋯}} \\ & {= 1 + {\sum\limits_{n = 1}^{\infty}{\frac{(-1)^{n + 1}}{n\text{!}}\ \frac{1 \cdot 3 \cdot 5\text{⋯}\left( {2n - 3} \right)}{2^{n}}x^{n}.}}} \end{array}$$

2. From the result in part a. the third-order Maclaurin polynomial is

$$p_{3}(x) = 1 + \frac{1}{2}x - \frac{1}{8}x^{2} + \frac{1}{16}x^{3}.$$

Therefore,

$$\begin{array}{cl} \sqrt{1.5} & {= \sqrt{1 + 0.5}} \\ & {\approx 1 + \frac{1}{2}(0.5) - \frac{1}{8}(0.5)^{2} + \frac{1}{16}(0.5)^{3}} \\ & {\approx 1.2266.} \end{array}$$

From Taylor’s theorem, the error satisfies

$$R_{3}(0.5) = \frac{f^{(4)}(c)}{4\text{!}}(0.5)^{4}$$

for some $c$ between $0$ and $0.5.$ Since $f^{(4)}(x) = - \frac{15}{2^{4}\left( {1 + x} \right)^{7\text{/}2}},$ and the maximum value of $\left| {f^{(4)}(x)} \right|$ on the interval $\left( {0,0.5} \right)$ occurs at $x = 0,$ we have

$$\left| {R_{3}(0.5)} \right| \leq \frac{15}{4\text{!}2^{4}}(0.5)^{4} \approx 0.00244.$$

The function and the Maclaurin polynomial $p_{3}$ are graphed in Figure 6.10.

Find the binomial series for $f(x) = \frac{1}{\left( {1 + x} \right)^{2}}.$

Common Functions Expressed as Taylor Series 用泰勒级数表示常用函数

At this point, we have derived Maclaurin series for exponential, trigonometric, and logarithmic functions, as well as functions of the form $f(x) = \left( {1 + x} \right)^{r}.$ In Table 6.1, we summarize the results of these series. We remark that the convergence of the Maclaurin series for $f(x) = \text{ln}\mspace{2mu}\left( {1 + x} \right)$ at the endpoint $x = 1$ and the Maclaurin series for $f(x) = \text{tan}^{-1}x$ at the endpoints $x = 1$ and $x = -1$ relies on a more advanced theorem than we present here. (Refer to Abel’s theorem for a discussion of this more technical point.)

Table 6.1 Maclaurin Series for Common Functions

Earlier in the chapter, we showed how you could combine power series to create new power series. Here we use these properties, combined with the Maclaurin series in Table 6.1, to create Maclaurin series for other functions.

Deriving Maclaurin Series from Known Series 由已知级数推导麦克劳林级数

Find the Maclaurin series of each of the following functions by using one of the series listed in Table 6.1.

1. $f(x) = \text{cos}\sqrt{x}$

2. $f(x) = \text{sinh}\mspace{2mu} x$

Solution 解答

1. Using the Maclaurin series for $\text{cos}\mspace{2mu} x$ we find that the Maclaurin series for $\text{cos}\sqrt{x}$ is given by

$$\begin{array}{cl} {\sum\limits_{n = 0}^{\infty}\frac{(-1)^{n}\left( \sqrt{x} \right)^{2n}}{\left( {2n} \right)\text{!}}} & {= {\sum\limits_{n = 0}^{\infty}\frac{(-1)^{n}x^{n}}{\left( {2n} \right)\text{!}}}} \\ & {= 1 - \frac{x}{2\text{!}} + \frac{x^{2}}{4\text{!}} - \frac{x^{3}}{6\text{!}} + \frac{x^{4}}{8\text{!}} - \text{⋯}.} \end{array}$$

This series converges to $\text{cos}\sqrt{x}$ for all $x$ in the domain of $\text{cos}\sqrt{x};$ that is, for all $x \geq 0.$

2. To find the Maclaurin series for $\text{sinh}\mspace{2mu} x,$ we use the fact that

$$\text{sinh}\mspace{2mu} x = \frac{e^{x} - e^{\text{−}x}}{2}.$$

Using the Maclaurin series for $e^{x},$ we see that the $n\text{th}$ term in the Maclaurin series for $\text{sinh}\mspace{2mu} x$ is given by

$$\frac{x^{n}}{n\text{!}} - \frac{\left( {\text{−}x} \right)^{n}}{n\text{!}}.$$

For $n$ even, this term is zero. For $n$ odd, this term is $\frac{2x^{n}}{n\text{!}}.$ Therefore, the Maclaurin series for $\text{sinh}\mspace{2mu} x$ has only odd-order terms and is given by

$${\sum\limits_{n = 0}^{\infty}\frac{x^{2n + 1}}{\left( {2n + 1} \right)\text{!}}} = x + \frac{x^{3}}{3\text{!}} + \frac{x^{5}}{5\text{!}} + \text{⋯}.$$

Find the Maclaurin series for $\text{sin}\left( x^{2} \right).$

We also showed previously in this chapter how power series can be differentiated term by term to create a new power series. In Example 6.19, we differentiate the binomial series for $\sqrt{1 + x}$ term by term to find the binomial series for $\frac{1}{\sqrt{1 + x}}.$ Note that we could construct the binomial series for $\frac{1}{\sqrt{1 + x}}$ directly from the definition, but differentiating the binomial series for $\sqrt{1 + x}$ is an easier calculation.

Differentiating a Series to Find a New Series 对级数求导以得到新级数

Use the binomial series for $\sqrt{1 + x}$ to find the binomial series for $\frac{1}{\sqrt{1 + x}}.$

Solution 解答

The two functions are related by

$$\frac{d}{dx}\sqrt{1 + x} = \frac{1}{2\sqrt{1 + x}},$$

so the binomial series for $\frac{1}{\sqrt{1 + x}}$ is given by

$$\begin{array}{cl} \frac{1}{\sqrt{1 + x}} & {= 2\frac{d}{dx}\sqrt{1 + x}} \\ & {= 1 + {\sum\limits_{n = 1}^{\infty}{\frac{(-1)^{n}}{n\text{!}}\ \frac{1 \cdot 3 \cdot 5\text{⋯}\left( {2n - 1} \right)}{2^{n}}x^{n}}}.} \end{array}$$

Find the binomial series for $f(x) = \frac{1}{\left( {1 + x} \right)^{3\text{/}2}}$

In this example, we differentiated a known Taylor series to construct a Taylor series for another function. The ability to differentiate power series term by term makes them a powerful tool for solving differential equations. We now show how this is accomplished.

Solving Differential Equations with Power Series 用幂级数求解微分方程

Consider the differential equation

$$y^{\prime}(x) = y.$$

Recall that this is a first-order separable equation and its solution is $y = Ce^{x}.$ This equation is easily solved using techniques discussed earlier in the text. For most differential equations, however, we do not yet have analytical tools to solve them. Power series are an extremely useful tool for solving many types of differential equations. In this technique, we look for a solution of the form $y = {\sum\limits_{n = 0}^{\infty}{c_{n}x^{n}}}$ and determine what the coefficients would need to be. In the next example, we consider an initial-value problem involving $y^{\prime} = y$ to illustrate the technique.

Power Series Solution of a Differential Equation 微分方程的幂级数解

Use power series to solve the initial-value problem

$$y^{\prime} = y,\mspace{9mu} y(0) = 3.$$

Solution 解答

Suppose that there exists a power series solution

$$y(x) = {\sum\limits_{n = 0}^{\infty}{c_{n}x^{n}}} = c_{0} + c_{1}x + c_{2}x^{2} + c_{3}x^{3} + c_{4}x^{4} + \text{⋯}.$$

Differentiating this series term by term, we obtain

$$y^{\prime} = c_{1} + 2c_{2}x + 3c_{3}x^{2} + 4c_{4}x^{3} + \text{⋯}.$$

If *y* satisfies the differential equation, then

$$c_{0} + c_{1}x + c_{2}x^{2} + c_{3}x^{3} + \text{⋯} = c_{1} + 2c_{2}x + 3c_{3}x^{2} + 4c_{3}x^{3} + \text{⋯}.$$

Using Uniqueness of Power Series on the uniqueness of power series representations, we know that these series can only be equal if their coefficients are equal. Therefore,

$$\begin{matrix} {c_{0} = c_{1},} \\ {c_{1} = 2c_{2},} \\ {c_{2} = 3c_{3},} \\ {c_{3} = 4c_{4},} \\ {\text{⋮}.} \end{matrix}$$

Using the initial condition $y(0) = 3$ combined with the power series representation

$$y(x) = c_{0} + c_{1}x + c_{2}x^{2} + c_{3}x^{3} + \text{⋯},$$

we find that $c_{0} = 3.$ We are now ready to solve for the rest of the coefficients. Using the fact that $c_{0} = 3,$ we have

$$\begin{array}{l} \\ \\ {c_{1} = c_{0} = 3 = \frac{3}{1\text{!}},} \\ {c_{2} = \frac{c_{1}}{2} = \frac{3}{2} = \frac{3}{2\text{!}},} \\ {c_{3} = \frac{c_{2}}{3} = \frac{3}{3 \cdot 2} = \frac{3}{3\text{!}},} \\ {c_{4} = \frac{c_{3}}{4} = \frac{3}{4 \cdot 3 \cdot 2} = \frac{3}{4\text{!}}.} \end{array}$$

Therefore,

$$\begin{array}{cl} y & {= 3\left\lbrack {1 + \frac{1}{1\text{!}}x + \frac{1}{2\text{!}}x^{2} + \frac{1}{3\text{!}}x^{3} + \frac{1}{4\text{!}}x^{4} + \text{⋯}} \right\rbrack} \\ & {= 3{\sum\limits_{n = 0}^{\infty}\frac{x^{n}}{n\text{!}}}.} \end{array}$$

You might recognize

$$\sum\limits_{n = 0}^{\infty}\frac{x^{n}}{n\text{!}}$$

as the Taylor series for $e^{x}.$ Therefore, the solution is $y = 3e^{x}.$

Use power series to solve $y^{\prime} = 2y,\mspace{9mu} y(0) = 5.$

We now consider an example involving a differential equation that we cannot solve using previously discussed methods. This differential equation

$$y^{''} - xy = 0$$

is known as Airy’s equation. It has many applications in mathematical physics, such as modeling the diffraction of light. Here we show how to solve it using power series.

Power Series Solution of Airy’s Equation Airy 方程的幂级数解

Use power series to solve

$$y^{''} - xy = 0$$

with the initial conditions $y(0) = a$ and $y\prime(0) = b.$

Solution 解答

We look for a solution of the form

$$y = {\sum\limits_{n = 0}^{\infty}{c_{n}x^{n} = c_{0} + c_{1}x + c_{2}x^{2} + c_{3}x^{3} + c_{4}x^{4} + \text{⋯}.}}$$

Differentiating this function term by term, we obtain

$$\begin{array}{rll} y^{\prime} & = & {c_{1} + 2c_{2}x + 3c_{3}x^{2} + 4c_{4}x^{3} + \text{⋯},} \\ y^{''} & = & {2 \cdot 1c_{2} + 3 \cdot 2c_{3}x + 4 \cdot 3c_{4}x^{2} + \text{⋯}.} \end{array}$$

If *y* satisfies the equation $y^{''} = xy,$ then

$$2 \cdot 1c_{2} + 3 \cdot 2c_{3}x + 4 \cdot 3c_{4}x^{2} + \text{⋯} = x\left( {c_{0} + c_{1}x + c_{2}x^{2} + c_{3}x^{3} + \text{⋯}} \right).$$

Using Uniqueness of Power Series on the uniqueness of power series representations, we know that coefficients of the same degree must be equal. Therefore,

$$\begin{matrix} {2 \cdot 1c_{2} = 0,} \\ {3 \cdot 2c_{3} = c_{0},} \\ {4 \cdot 3c_{4} = c_{1},} \\ {5 \cdot 4c_{5} = c_{2},} \\ {\text{⋮}.} \end{matrix}$$

More generally, for $n \geq 3,$ we have $n \cdot \left( {n - 1} \right)c_{n} = c_{n - 3}.$ In fact, all coefficients can be written in terms of $c_{0}$ and $c_{1}.$ To see this, first note that $c_{2} = 0.$ Then

$$\begin{array}{l} \\ \\ {c_{3} = \frac{c_{0}}{3 \cdot 2},} \\ {c_{4} = \frac{c_{1}}{4 \cdot 3}.} \end{array}$$

For $c_{5},c_{6},c_{7},$ we see that

$$\begin{array}{l} \\ \\ {c_{5} = \frac{c_{2}}{5 \cdot 4} = 0,} \\ {c_{6} = \frac{c_{3}}{6 \cdot 5} = \frac{c_{0}}{6 \cdot 5 \cdot 3 \cdot 2},} \\ {c_{7} = \frac{c_{4}}{7 \cdot 6} = \frac{c_{1}}{7 \cdot 6 \cdot 4 \cdot 3}.} \end{array}$$

Therefore, the series solution of the differential equation is given by

$$y = c_{0} + c_{1}x + 0 \cdot x^{2} + \frac{c_{0}}{3 \cdot 2}x^{3} + \frac{c_{1}}{4 \cdot 3}x^{4} + 0 \cdot x^{5} + \frac{c_{0}}{6 \cdot 5 \cdot 3 \cdot 2}x^{6} + \frac{c_{1}}{7 \cdot 6 \cdot 4 \cdot 3}x^{7} + \text{⋯}.$$

The initial condition $y(0) = a$ implies $c_{0} = a.$ Differentiating this series term by term and using the fact that $y^{\prime}(0) = b,$ we conclude that $c_{1} = b.$ Therefore, the solution of this initial-value problem is

$$y = a\left( {1 + \frac{x^{3}}{3 \cdot 2} + \frac{x^{6}}{6 \cdot 5 \cdot 3 \cdot 2} + \text{⋯}} \right) + b\left( {x + \frac{x^{4}}{4 \cdot 3} + \frac{x^{7}}{7 \cdot 6 \cdot 4 \cdot 3} + \text{⋯}} \right).$$

Use power series to solve $y^{''} + x^{2}y = 0$ with the initial condition $y(0) = a$ and $y^{\prime}(0) = b.$

Evaluating Nonelementary Integrals 非初等积分的求值

Solving differential equations is one common application of power series. We now turn to a second application. We show how power series can be used to evaluate integrals involving functions whose antiderivatives cannot be expressed using elementary functions.

One integral that arises often in applications in probability theory is $\left. \int{e^{\text{−}x^{2}}dx} \right..$ Unfortunately, the antiderivative of the integrand $e^{\text{−}x^{2}}$ is not an elementary function. By elementary function, we mean a function that can be written using a finite number of algebraic combinations or compositions of exponential, logarithmic, trigonometric, or power functions. We remark that the term “elementary function” is not synonymous with noncomplicated function. For example, the function $f(x) = \sqrt{x^{2} - 3x} + e^{x^{3}} - \text{sin}\left( {5x + 4} \right)$ is an elementary function, although not a particularly simple-looking function. Any integral of the form $\left. \int{f(x)\mspace{2mu} dx} \right.$ where the antiderivative of $f$ cannot be written as an elementary function is considered a nonelementary integral.

Nonelementary integrals cannot be evaluated using the basic integration techniques discussed earlier. One way to evaluate such integrals is by expressing the integrand as a power series and integrating term by term. We demonstrate this technique by considering $\left. \int{e^{\text{−}x^{2}}dx} \right..$

Using Taylor Series to Evaluate a Definite Integral 用泰勒级数求定积分

1. Express $\left. \int{e^{\text{−}x^{2}}dx} \right.$ as an infinite series.

2. Evaluate $\int_{0}^{1}{e^{\text{−}x^{2}}dx}$ to within an error of $0.01.$

Solution 解答

1. The Maclaurin series for $e^{\text{−}x^{2}}$ is given by

$$\begin{array}{cl} e^{\text{−}x^{2}} & {= {\sum\limits_{n = 0}^{\infty}\frac{\left( {\text{−}x^{2}} \right)^{n}}{n\text{!}}}} \\ & {= 1 - x^{2} + \frac{x^{4}}{2\text{!}} - \frac{x^{6}}{3\text{!}} + \text{⋯} + (-1)^{n}\frac{x^{2n}}{n\text{!}} + \text{⋯}} \\ & {= {\sum\limits_{n = 0}^{\infty}{(-1)^{n}\frac{x^{2n}}{n\text{!}}.}}} \end{array}$$

Therefore,

$$\begin{array}{cl} \left. \int{e^{\text{−}x^{2}}dx} \right. & {= \left. \int{\left( {1 - x^{2} + \frac{x^{4}}{2\text{!}} - \frac{x^{6}}{3\text{!}} + \text{⋯} + (-1)^{n}\frac{x^{2n}}{n\text{!}} + \text{⋯}} \right)\mspace{2mu} dx} \right.} \\ & {= C + x - \frac{x^{3}}{3} + \frac{x^{5}}{5 \cdot 2\text{!}} - \frac{x^{7}}{7 \cdot 3\text{!}} + \text{⋯} + (-1)^{n}\frac{x^{2n + 1}}{\left( {2n + 1} \right)n\text{!}} + \text{⋯}.} \end{array}$$

2. Using the result from part a. we have

$${\int_{0}^{1}{e^{\text{−}x^{2}}dx}} = 1 - \frac{1}{3} + \frac{1}{10} - \frac{1}{42} + \frac{1}{216} - \text{⋯}.$$

The sum of the first four terms is approximately $0.74.$ By the alternating series test, this estimate is accurate to within an error of less than $\frac{1}{216} \approx 0.0046296 < 0.01.$

Express $\left. \int{\text{cos}\sqrt{x}dx} \right.$ as an infinite series. Evaluate $\int_{0}^{1}{\text{cos}\sqrt{x}dx}$ to within an error of $0.01.$

As mentioned above, the integral $\left. \int{e^{\text{−}x^{2}}dx} \right.$ arises often in probability theory. Specifically, it is used when studying data sets that are normally distributed, meaning the data values lie under a bell-shaped curve. For example, if a set of data values is normally distributed with mean $\mu$ and standard deviation $\sigma,$ then the probability that a randomly chosen value lies between $x = a$ and $x = b$ is given by

$$\frac{1}{\sigma\sqrt{2\pi}}{\int_{a}^{b}{e^{\text{−}{({x - \mu})}^{2}\text{/}{({2\sigma^{2}})}}dx}}.$$ (6.10)

(See Figure 6.11.)

To simplify this integral, we typically let $z = \frac{x - \mu}{\sigma}.$ This quantity $z$ is known as the $z$ score of a data value. With this simplification, integral Equation 6.10 becomes

$$\frac{1}{\sqrt{2\pi}}{\int_{{({a - \mu})}\text{/}\sigma}^{{({b - \mu})}\text{/}\sigma}e^{\text{−}z^{2}\text{/}2}}dz.$$ (6.11)

In Example 6.23, we show how we can use this integral in calculating probabilities.

Using Maclaurin Series to Approximate a Probability 用麦克劳林级数近似概率

Suppose a set of standardized test scores are normally distributed with mean $\mu = 100$ and standard deviation $\sigma = 50.$ Use Equation 6.11 and the first six terms in the Maclaurin series for $e^{\text{−}x^{2}\text{/}2}$ to approximate the probability that a randomly selected test score is between $x = 100$ and $x = 200.$ Use the alternating series test to determine how accurate your approximation is.

Solution 解答

Since $\mu = 100,\sigma = 50,$ and we are trying to determine the area under the curve from $a = 100$ to $b = 200,$ integral Equation 6.11 becomes

$$\frac{1}{\sqrt{2\pi}}{\int_{0}^{2}e^{\text{−}z^{2}\text{/}2}}dz.$$

The Maclaurin series for $e^{\text{−}x^{2}\text{/}2}$ is given by

$$\begin{array}{cl} e^{\text{−}x^{2}\text{/}2} & {= {\sum\limits_{n = 0}^{\infty}\frac{\left( {- \frac{x^{2}}{2}} \right)^{n}}{n\text{!}}}} \\ & {= 1 - \frac{x^{2}}{2^{1} \cdot 1\text{!}} + \frac{x^{4}}{2^{2} \cdot 2\text{!}} - \frac{x^{6}}{2^{3} \cdot 3\text{!}} + \text{⋯} + (-1)^{n}\frac{x^{2n}}{2^{n} \cdot n\text{!}} + \text{⋯}} \\ & {= {\sum\limits_{n = 0}^{\infty}{(-1)^{n}\frac{x^{2}{}^{n}}{2^{n} \cdot n\text{!}}}}.} \end{array}$$

Therefore,

$$\begin{array}{cll} {\frac{1}{\sqrt{2\pi}}\left. \int{e^{\text{−}z^{2}\text{/}2}dz} \right.} & = & {\frac{1}{\sqrt{2\pi}}\left. \int{\left( {1 - \frac{z^{2}}{2^{1} \cdot 1\text{!}} + \frac{z^{4}}{2^{2} \cdot 2\text{!}} - \frac{z^{6}}{2^{3} \cdot 3\text{!}} + \text{⋯} + (-1)^{n}\frac{z^{2n}}{2^{n} \cdot n\text{!}} + \text{⋯}} \right)\mspace{2mu} dz} \right.} \\ & = & {\frac{1}{\sqrt{2\pi}}\left( {C + z - \frac{z^{3}}{3 \cdot 2^{1} \cdot 1\text{!}} + \frac{z^{5}}{5 \cdot 2^{2} \cdot 2\text{!}} - \frac{z^{7}}{7 \cdot 2^{3} \cdot 3\text{!}} + \text{⋯} + (-1)^{n}\frac{z^{2n + 1}}{\left( {2n + 1} \right)2^{n} \cdot n\text{!}} + \text{⋯}} \right)} \\ {\frac{1}{\sqrt{2\pi}}{\int_{0}^{2}{e^{\text{−}z^{2}\text{/}2}dz}}} & = & {\frac{1}{\sqrt{2\pi}}\left( {2 - \frac{8}{6} + \frac{32}{40} - \frac{128}{336} + \frac{512}{3456} - \frac{2^{11}}{11 \cdot 2^{5} \cdot 5\text{!}} + \text{⋯}} \right).} \end{array}$$

Using the first five terms, we estimate that the probability is approximately $0.4922.$ By the alternating series test, we see that this estimate is accurate to within

$$\frac{1}{\sqrt{2\pi}}\ \frac{2^{13}}{13 \cdot 2^{6} \cdot 6\text{!}} \approx 0.00546.$$

Analysis 分析

If you are familiar with probability theory, you may know that the probability that a data value is within two standard deviations of the mean is approximately $95\text{\%}.$ Here we calculated the probability that a data value is between the mean and two standard deviations above the mean, so the estimate should be around $47.5\text{\%}.$ The estimate, combined with the bound on the accuracy, falls within this range.

Use the first five terms of the Maclaurin series for $e^{\text{−}x^{2}\text{/}2}$ 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.

Another application in which a nonelementary integral arises involves the period of a pendulum. The integral is

$${\int_{0}^{\pi\text{/}2}\frac{d\theta}{\sqrt{1 - k^{2}\text{sin}^{2}\theta}}}.$$

An integral of this form is known as an elliptic integral of the first kind. Elliptic integrals originally arose when trying to calculate the arc length of an ellipse. We now show how to use power series to approximate this integral.

Period of a Pendulum 单摆的周期

The period of a pendulum is the time it takes for a pendulum to make one complete back-and-forth swing. For a pendulum with length $L$ that makes a maximum angle $\theta_{\text{max}}$ with the vertical, its period $T$ is given by

$$T = 4\sqrt{\frac{L}{g}}{\int_{0}^{\pi\text{/}2}\frac{d\theta}{\sqrt{1 - k^{2}\text{sin}^{2}\theta}}}$$

where $g$ is the acceleration due to gravity and $k = \text{sin}\left( \frac{\theta_{\text{max}}}{2} \right)$ (see Figure 6.12). (We note that this formula for the period arises from a non-linearized model of a pendulum. In some cases, for simplification, a linearized model is used and $\text{sin}\mspace{2mu}\theta$ is approximated by $\theta.)$ Use the binomial series

$$\frac{1}{\sqrt{1 + x}} = 1 + {\sum\limits_{n = 1}^{\infty}\frac{(-1)^{n}}{n\text{!}}}\ \frac{1 \cdot 3 \cdot 5\text{⋯}\left( {2n - 1} \right)}{2^{n}}x^{n}$$

to estimate the period of this pendulum. Specifically, approximate the period of the pendulum if

1. you use only the first term in the binomial series, and

2. you use the first two terms in the binomial series.

Solution 解答

We use the binomial series, replacing $x$ with $\text{−}k^{2}\text{sin}^{2}\theta.$ Then we can write the period as

$$T = 4\sqrt{\frac{L}{g}}{\int_{0}^{\pi\text{/}2}{\left( {1 + \frac{1}{2}k^{2}\text{sin}^{2}\theta + \frac{1 \cdot 3}{2\text{!}2^{2}}k^{4}\text{sin}^{4}\theta + \text{⋯}} \right)\mspace{2mu} d\theta}}.$$

1. Using just the first term in the integrand, the first-order estimate is

$$T \approx 4\sqrt{\frac{L}{g}}{\int_{0}^{\pi\text{/}2}{d\theta = 2\pi\sqrt{\frac{L}{g}}.}}$$

If $\theta_{\text{max}}$ is small, then $k = \text{sin}\left( \frac{\theta_{\text{max}}}{2} \right)$ is small. We claim that when $k$ is small, this is a good estimate. To justify this claim, consider

$${\int_{0}^{\pi\text{/}2}{\left( {1 + \frac{1}{2}k^{2}\text{sin}^{2}\theta + \frac{1 \cdot 3}{2\text{!}2^{2}}k^{4}\text{sin}^{4}\theta + \text{⋯}} \right)\mspace{2mu} d\theta}}.$$

Since $\left| {\text{sin}\mspace{2mu} x} \right| \leq 1,$ this integral is bounded by

$${\int_{0}^{\pi\text{/}2}{\left( {\frac{1}{2}k^{2} + \frac{1.3}{2\text{!}2^{2}}k^{4} + \text{⋯}} \right)\mspace{2mu} d\theta}} < \frac{\pi}{2}\left( {\frac{1}{2}k^{2} + \frac{1 \cdot 3}{2\text{!}2^{2}}k^{4} + \text{⋯}} \right).$$

Furthermore, it can be shown that each coefficient on the right-hand side is less than $1$ and, therefore, that this expression is bounded by

$$\frac{\pi k^{2}}{2}\left( {1 + k^{2} + k^{4} + \text{⋯}} \right) = \frac{\pi k^{2}}{2} \cdot \frac{1}{1 - k^{2}},$$

which is small for $k$ small.

2. For larger values of $\theta_{\text{max}},$ we can approximate $T$ by using more terms in the integrand. By using the first two terms in the integral, we arrive at the estimate

$$\begin{array}{cl} T & {\approx 4\sqrt{\frac{L}{g}}{\int_{0}^{\pi\text{/}2}{\left( {1 + \frac{1}{2}k^{2}\text{sin}^{2}\theta} \right)d}}\theta} \\ & {= 2\pi\sqrt{\frac{L}{g}}\left( {1 + \frac{k^{2}}{4}} \right).} \end{array}$$

The applications of Taylor series in this section are intended to highlight their importance. In general, Taylor series are useful because they allow us to represent known functions using polynomials, thus providing us a tool for approximating function values and estimating complicated integrals. In addition, they allow us to define new functions as power series, thus providing us with a powerful tool for solving differential equations.

Section 6.4 Exercises 6.4 节习题

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

174\.

$\left( {1 - x} \right)^{1\text{/}3}$

175.

$\left( {1 + x^{2}} \right)^{-1\text{/}3}$

176\.

$\left( {1 - x} \right)^{1.01}$

177.

$\left( {1 - 2x} \right)^{2\text{/}3}$

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

178\.

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

179.

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

180\.

$\sqrt{x + 2}$ at $a = 1$

181.

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

182\.

$\left( {x - 8} \right)^{1\text{/}3}$ at $a = 9$

183.

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

184\.

$x^{1\text{/}3}$ at $a = 27$

185.

$\sqrt{x}$ at $\alpha = 9$

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 $1\text{/}1000.$

186\.

\[T\] $(15)^{1\text{/}4}$ using $\left( {16 - x} \right)^{1\text{/}4}$

187.

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

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

188\.

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

189.

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

190\.

\[T\] $\sqrt{3} = \frac{3}{\sqrt{3}}$ using $x = \frac{2}{3}$ in $\left( {1 - x} \right)^{1\text{/}2}$

191.

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

192\.

Integrate the binomial approximation of $\sqrt{1 - x}$ to find an approximation of ${\int_{0}^{x}\sqrt{1 - t}}dt.$

193.

\[T\] 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 $\frac{\pi}{2}.$

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

194\.

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

195.

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

196\.

$\left( {3 + 2x} \right)^{1\text{/}3};a = -1$

197.

$\left( {x^{2} + 6x + 10} \right)^{1\text{/}3};a = -3$

198\.

Use $\left( {1 + x} \right)^{1\text{/}3} = 1 + \frac{1}{3}x - \frac{1}{9}x^{2} + \frac{5}{81}x^{3} - \frac{10}{243}x^{4} + \text{⋯}$ with $x = 1$ to approximate $2^{1\text{/}3}.$

199.

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

200\.

Find the $25\text{th}$ derivative of $f(x) = \left( {1 + x^{2}} \right)^{13}$ at $x = 0.$

201.

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

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

202\.

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

203.

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

204\.

$f(x) = \frac{\text{sin}\mspace{2mu} x}{x}$

205.

$f(x) = \frac{\text{sin}\left( \sqrt{x} \right)}{\sqrt{x}},\mspace{9mu}\left( {x > 0} \right),$

206\.

$f(x) = \text{sin}\left( x^{2} \right)$

207.

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

208\.

$f(x) = \text{cos}^{2}x$ using the identity $\text{cos}^{2}x = \frac{1}{2} + \frac{1}{2}\mspace{2mu}\text{cos}\mspace{2mu}\left( {2x} \right)$

209.

$f(x) = \text{sin}^{2}x$ using the identity $\text{sin}^{2}x = \frac{1}{2} - \frac{1}{2}\mspace{2mu}\text{cos}\mspace{2mu}\left( {2x} \right)$

In the following exercises, find the Maclaurin series of $F(x) = {\int_{0}^{x}{f(t)\mspace{2mu} dt}}$ 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.

210\.

$F(x) = {\int_{0}^{x}e^{\text{−}t^{2}}}dt;f(t) = e^{\text{−}t^{2}} = {\sum\limits_{n = 0}^{\infty}(-1)^{n}}\frac{t^{2n}}{n\text{!}}$

211.

$F(x) = \text{tan}^{-1}x;\mspace{9mu} f(t) = \frac{1}{1 + t^{2}} = {\sum\limits_{n = 0}^{\infty}{(-1)^{n}t^{2n}}}$

212\.

$F(x) = \text{tanh}^{-1}x;\mspace{9mu} f(t) = \frac{1}{1 - t^{2}} = {\sum\limits_{n = 0}^{\infty}t^{2n}}$

213.

$F(x) = \text{sin}^{-1}x;\mspace{9mu} f(t) = \frac{1}{\sqrt{1 - t^{2}}} = {\sum\limits_{k = 0}^{\infty}{\left( \begin{array}{l}\frac{1}{2} \\k\end{array} \right)\frac{t^{2k}}{k\text{!}}}}$

214\.

$F(x) = {\int_{0}^{x}{\frac{\text{sin}\mspace{2mu} t}{t}dt}};\mspace{9mu} f(t) = \frac{\text{sin}\mspace{2mu} t}{t} = {\sum\limits_{n = 0}^{\infty}{(-1)^{n}\frac{t^{2n}}{\left( {2n + 1} \right)\text{!}}}}$

215.

$F(x) = {\int_{0}^{x}{\text{cos}\left( \sqrt{t} \right)\mspace{2mu} dt}};\mspace{9mu} f(t) = {\sum\limits_{n = 0}^{\infty}{(-1)^{n}\frac{x^{n}}{\left( {2n} \right)\text{!}}}}$

216\.

$F(x) = {\int_{0}^{x}{\frac{1 - \text{cos}\mspace{2mu} t}{t^{2}}dt}};\mspace{9mu} f(t) = \frac{1 - \text{cos}\mspace{2mu} t}{t^{2}} = {\sum\limits_{n = 0}^{\infty}{(-1)^{n}\frac{t^{2n}}{\left( {2n + 2} \right)\text{!}}}}$

217.

$F(x) = {\int_{0}^{x}{\frac{\text{ln}\mspace{2mu}\left( {1 + t} \right)}{t}dt}};\mspace{9mu} f(t) = {\sum\limits_{n = 0}^{\infty}{(-1)^{n}\frac{t^{n}}{n + 1}}}$

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

218\.

$f(x) = \text{sin}\left( {x + \frac{\pi}{4}} \right) = \text{sin}\mspace{2mu} x\mspace{2mu}\text{cos}\left( \frac{\pi}{4} \right) + \text{cos}\mspace{2mu} x\mspace{2mu}\text{sin}\left( \frac{\pi}{4} \right)$

219.

$f(x) = \text{tan}\mspace{2mu} x$

220\.

$f(x) = \text{ln}\mspace{2mu}\left( {\text{cos}\mspace{2mu} x} \right)$

221.

$f(x) = e^{x}\text{cos}\mspace{2mu} x$

222\.

$f(x) = e^{\text{sin}\mspace{2mu} x}$

223.

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

224\.

$f(x) = \text{tanh}\mspace{2mu} x$

225.

$f(x) = \frac{\text{tan}\sqrt{x}}{\sqrt{x}}$ (see expansion for $\text{tan}\mspace{2mu} x)$

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

226\.

$\text{ln}\mspace{2mu}\left( {1 + x} \right)$

227.

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

228\.

$\text{tan}^{-1}x$

229.

$\text{ln}\mspace{2mu}\left( {1 + x^{2}} \right)$

230\.

Find the Maclaurin series of $\text{sinh}\mspace{2mu} x = \frac{e^{x} - e^{\text{−}x}}{2}.$

231.

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

232\.

Differentiate term by term the Maclaurin series of $\text{sinh}\mspace{2mu} x$ and compare the result with the Maclaurin series of $\text{cosh}\mspace{2mu} x.$

233.

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

234\.

Use the identity $2\mspace{2mu}\text{sin}\mspace{2mu} x\mspace{2mu}\text{cos}\mspace{2mu} x = \text{sin}\left( {2x} \right)$ to find the power series expansion of $\text{sin}^{2}x$ at $x = 0.$ (*Hint:* Integrate the Maclaurin series of $\text{sin}\left( {2x} \right)$ term by term.)

235.

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

236\.

\[T\] Suppose that $y = {\sum\limits_{k = 0}^{\infty}{a_{k}x^{k}}}$ satisfies $y^{\prime} = -2xy$ and $y(0) = 0.$ Show that $a_{2k + 1} = 0$ for all $k$ and that $a_{2k + 2} = \frac{\text{−}a_{2k}}{k + 1}.$ Plot the partial sum $S_{20}$ of $y$ on the interval $\lbrack-4,4\rbrack.$

237.

\[T\] Suppose that a set of standardized test scores is normally distributed with mean $\mu = 100$ and standard deviation $\sigma = 10.$ 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 $\frac{1}{\sqrt{2\pi}}e^{\text{−}x^{2}\text{/}2}$ to estimate this probability.

238\.

\[T\] Suppose that a set of standardized test scores is normally distributed with mean $\mu = 100$ and standard deviation $\sigma = 10.$ Set up an integral that represents the probability that a test score will be between $70$ and $130$ and use the integral of the degree $50$ Maclaurin polynomial of $\frac{1}{\sqrt{2\pi}}e^{\text{−}x^{2}\text{/}2}$ to estimate this probability.

239.

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

240\.

\[T\] Suppose that $\sum\limits_{n = 0}^{\infty}{a_{n}x^{n}}$ converges to a function $f(x)$ such that $f(0) = 0,\mspace{9mu} f^{\prime}(0) = 1,$ and $f^{''}(x) = \text{−}f(x).$ Find a formula for $a_{n}$ and plot the partial sum $S_{N}$ for $N = 10$ on $\lbrack-5,5\rbrack.$

241.

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

242\.

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

The error in approximating the integral $\int_{a}^{b}{f(t)\mspace{2mu} dt}$ by that of a Taylor approximation $\int_{a}^{b}{P_{n}(t)\mspace{2mu} dt}$ is at most ${\int_{a}^{b}{R_{n}(t)\mspace{2mu} dt}}.$ In the following exercises, the Taylor remainder estimate $R_{n} \leq \frac{M}{\left( {n + 1} \right)\text{!}}\left| {x - a} \right|^{n + 1}$ guarantees that the integral of the Taylor polynomial of the given order approximates the integral of $f$ with an error less than $\frac{1}{10}.$

1. Evaluate the integral of the appropriate Taylor polynomial and verify that it approximates the CAS value with an error less than $\frac{1}{100}.$

2. Compare the accuracy of the polynomial integral estimate with the remainder estimate.

243.

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

244\.

\[T\] ${\int_{0}^{2}e^{\text{−}x^{2}}}dx;p_{11} = 1 - x^{2} + \frac{x^{4}}{2} - \frac{x^{6}}{3\text{!}} + \text{⋯} - \frac{x^{22}}{11\text{!}}$ (You may assume that the absolute value of the $23\text{rd}$ derivative of $e^{\text{−}x^{2}}$ is less than $2\ \times \ 10^{14}.)$

The following exercises deal with Fresnel integrals.

245.

The Fresnel integrals are defined by $C(x) = {\int_{0}^{x}{\text{cos}\left( t^{2} \right)\mspace{2mu} dt}}$ and $S(x) = {\int_{0}^{x}{\text{sin}\left( t^{2} \right)\mspace{2mu} dt}}.$ 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 $\lbrack 0,2\pi\rbrack.$

246\.

\[T\] The Fresnel integrals are used in design applications for roadways and railways and other applications because of the curvature properties of the curve with coordinates $\left( {C(t),S(t)} \right).$ Plot the curve $\left( {C_{50},S_{50}} \right)$ for $0 \leq t \leq 2\pi,$ the coordinates of which were computed in the previous exercise.

247.

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

248\.

\[T\] Use Newton's approximation of the binomial $\sqrt{1 - x^{2}}$ to approximate $\pi$ as follows. The circle centered at $\left( {\frac{1}{2},0} \right)$ with radius $\frac{1}{2}$ has upper semicircle $y = \sqrt{x}\sqrt{1 - x}.$ The sector of this circle bounded by the $x$-axis between $x = 0$ and $x = \frac{1}{2}$ and by the line joining $\left( {\frac{1}{4},\frac{\sqrt{3}}{4}} \right)$ corresponds to $\frac{1}{6}$ of the circle and has area $\frac{\pi}{24}.$ This sector is the union of a right triangle with height $\frac{\sqrt{3}}{4}$ and base $\frac{1}{4}$ and the region below the graph between $x = 0$ and $x = \frac{1}{4}.$ To find the area of this region you can write $y = \sqrt{x}\sqrt{1 - x} = \sqrt{x}\ \times \ \left( {\text{binomial expansion of}\sqrt{1 - x}} \right)$ and integrate term by term. Use this approach with the binomial approximation from the previous exercise to estimate $\pi.$

249.

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

250\.

Suppose that a pendulum is to have a period of $2$ seconds and a maximum angle of $\theta_{\text{max}} = \frac{\pi}{6}.$ Use $T \approx 2\pi\sqrt{\frac{L}{g}}\left( {1 + \frac{k^{2}}{4}} \right)$ to approximate the desired length of the pendulum. What length is predicted by the small angle estimate $T \approx 2\pi\sqrt{\frac{L}{g}}?$

251.

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

252\.

\[T\] An equivalent formula for the period of a pendulum with amplitude $\theta_{\text{max}}$ is $T\left( \theta_{\text{max}} \right) = 2\sqrt{2}\sqrt{\frac{L}{g}}{\int_{0}^{\theta_{\text{max}}}\frac{d\theta}{\sqrt{\text{cos}\mspace{2mu}\theta} - \text{cos}\left( \theta_{\text{max}} \right)}}$ where $L$ is the pendulum length and $g$ is the gravitational acceleration constant. When $\theta_{\text{max}} = \frac{\pi}{3}$ we get $\frac{1}{\sqrt{\text{cos}\mspace{2mu} t - 1\text{/}2}} \approx \sqrt{2}\left( {1 + \frac{t^{2}}{2} + \frac{t^{4}}{3} + \frac{181t^{6}}{720}} \right).$ Integrate this approximation to estimate $T\left( \frac{\pi}{3} \right)$ in terms of $L$ and $g.$ Assuming $g = 9.806$ meters per second squared, find an approximate length $L$ such that $T\left( \frac{\pi}{3} \right) = 2$ seconds.

Key Terms 关键术语

binomial series

the Maclaurin series for $f(x) = \left( {1 + x} \right)^{r};$ it is given by

$\left( {1 + x} \right)^{r} = {\sum\limits_{n = 0}^{\infty}{\left( \begin{array}{l}

r \\

n

\end{array} \right)x^{n}}} = 1 + rx + \frac{r\left( {r - 1} \right)}{2\text{!}}x^{2} + \text{⋯} + \frac{r\left( {r - 1} \right)\text{⋯}\left( {r - n + 1} \right)}{n\text{!}}x^{n} + \text{⋯}$ for $|x| < 1$

interval of convergence

the set of real numbers *x* for which a power series converges

Maclaurin polynomial

a Taylor polynomial centered at 0; the *n*th Taylor polynomial for $f$ at 0 is the *n*th Maclaurin polynomial for $f$

Maclaurin series

a Taylor series for a function $f$ at $x = 0$ is known as a Maclaurin series for $f$

nonelementary integral

an integral for which the antiderivative of the integrand cannot be expressed as an elementary function

power series

a series of the form $\sum\limits_{n = 0}^{\infty}{c_{n}x^{n}}$ is a power series centered at $x = 0\text{;}$ a series of the form $\sum\limits_{n = 0}^{\infty}{c_{n}\left( {x - a} \right)^{n}}$ is a power series centered at $x = a$

radius of convergence

if there exists a real number $R > 0$ such that a power series centered at $x = a$ converges for $\left| {x - a} \right| < R$ and diverges for $\left| {x - a} \right| > R,$ then *R* is the radius of convergence; if the power series only converges at $x = a,$ the radius of convergence is $R = 0\text{;}$ if the power series converges for all real numbers *x*, the radius of convergence is $R = \infty$

Taylor polynomials

the *n*th Taylor polynomial for $f$ at $x = a$ is $p_{n}(x) = f(a) + f^{\prime}(a)\left( {x - a} \right) + \frac{f^{''}(a)}{2\text{!}}\left( {x - a} \right)^{2} + \text{⋯} + \frac{f^{(n)}(a)}{n\text{!}}\left( {x - a} \right)^{n}$

Taylor series

a power series at *a* that converges to a function $f$ on some open interval containing *a*

Taylor’s theorem with remainder

for a function $f$ and the *n*th Taylor polynomial for $f$ at $x = a,$ the remainder $R_{n}(x) = f(x) - p_{n}(x)$ satisfies $R_{n}(x) = \frac{f^{({n + 1})}(c)}{\left( {n + 1} \right)\text{!}}\left( {x - a} \right)^{n + 1}$

for some *c* between *x* and *a*; if there exists an interval *I* containing *a* and a real number *M* such that $\left| {f^{({n + 1})}(x)} \right| \leq M$ for all *x* in *I*, then $\left| {R_{n}(x)} \right| \leq \frac{M}{\left( {n + 1} \right)\text{!}}\left| {x - a} \right|^{n + 1}$

term-by-term differentiation of a power series

a technique for evaluating the derivative of a power series $\sum\limits_{n = 0}^{\infty}{c_{n}\left( {x - a} \right)^{n}}$ by evaluating the derivative of each term separately to create the new power series $\sum\limits_{n = 1}^{\infty}{nc_{n}\left( {x - a} \right)^{n - 1}}$

term-by-term integration of a power series

a technique for integrating a power series $\sum\limits_{n = 0}^{\infty}{c_{n}\left( {x - a} \right)^{n}}$ by integrating each term separately to create the new power series $C + {\sum\limits_{n = 0}^{\infty}{c_{n}\frac{\left( {x - a} \right)^{n + 1}}{n + 1}}}$

Key Equations 关键方程

Key Concepts 核心概念

6.1 Power Series and Functions 6.1 幂级数与其函数

• For a power series centered at $x = a,$ one of the following three properties hold:

• 对于以 $x = a$ 为中心的幂级数，下列三种性质之一必成立：

1. The power series converges only at $x = a.$ In this case, we say that the radius of convergence is $R = 0.$

2. The power series converges for all real numbers *x*. In this case, we say that the radius of convergence is $R = \infty.$

3. There is a real number *R* such that the series converges for $\left| {x - a} \right| < R$ and diverges for $\left| {x - a} \right| > R.$ In this case, the radius of convergence is *R*.

• If a power series converges on a finite interval, the series may or may not converge at the endpoints.
• The ratio test may often be used to determine the radius of convergence.
• The geometric series ${\sum\limits_{n = 0}^{\infty}x^{n}} = \frac{1}{1 - x}$ for $|x| < 1$ allows us to represent certain functions using geometric series.

• 若一个幂级数在一个有限区间上收敛，则该级数在端点处可能收敛，也可能发散。
• 比值判别法常可用来确定收敛半径。
• 对于 $|x| < 1$，几何级数 ${\sum\limits_{n = 0}^{\infty}x^{n}} = \frac{1}{1 - x}$ 使我们能够利用几何幂级数表示某些函数。

6.2 Properties of Power Series 6.2 幂级数的性质

• Given two power series $\sum\limits_{n = 0}^{\infty}{c_{n}x^{n}}$ and $\sum\limits_{n = 0}^{\infty}{d_{n}x^{n}}$ that converge to functions *f* and *g* on a common interval *I*, the sum and difference of the two series converge to $f \pm g,$ respectively, on *I*. In addition, for any real number *b* and integer $m \geq 0,$ the series $\sum\limits_{n = 0}^{\infty}{bx^{m}c_{n}x^{n}}$ converges to $bx^{m}f(x)$ and the series $\sum\limits_{n = 0}^{\infty}{c_{n}\left( {bx^{m}} \right)^{n}}$ converges to $f\left( {bx^{m}} \right)$ whenever *bx^m* is in the interval *I*.
• Given two power series that converge on an interval $\left( {\text{−}R,R} \right),$ the Cauchy product of the two power series converges on the interval $\left( {\text{−}R,R} \right).$
• Given a power series that converges to a function *f* on an interval $\left( {\text{−}R,R} \right),$ the series can be differentiated term-by-term and the resulting series converges to $f^{\prime}$ on $\left( {\text{−}R,R} \right).$ The series can also be integrated term-by-term and the resulting series converges to $\int{f(x)\mspace{2mu} dx}$ on $\left( {\text{−}R,R} \right).$

• 给定两个幂级数 $\sum\limits_{n = 0}^{\infty}{c_{n}x^{n}}$ 与 $\sum\limits_{n = 0}^{\infty}{d_{n}x^{n}}$，它们在公共区间 *I* 上分别收敛到函数 *f* 与 *g*，则这两个级数的和与差在 *I* 上分别收敛到 $f \pm g$。此外，对任意实数 *b* 与整数 $m \geq 0,$ 级数 $\sum\limits_{n = 0}^{\infty}{bx^{m}c_{n}x^{n}}$ 收敛到 $bx^{m}f(x)$，且级数 $\sum\limits_{n = 0}^{\infty}{c_{n}\left( {bx^{m}} \right)^{n}}$ 收敛到 $f\left( {bx^{m}} \right)$，只要 *bx^m* 落在区间 *I* 内。
• 给定两个在区间 $\left( {\text{−}R,R} \right)$ 上收敛的幂级数，则这两个幂级数的柯西乘积在同一区间 $\left( {\text{−}R,R} \right)$ 上收敛。
• 给定在区间 $\left( {\text{−}R,R} \right)$ 上收敛到函数 *f* 的幂级数，该级数可逐项求导，所得级数在 $\left( {\text{−}R,R} \right)$ 上收敛到 $f^{\prime}$。该级数也可逐项积分，所得级数在 $\left( {\text{−}R,R} \right)$ 上收敛到 $\int{f(x)\mspace{2mu} dx}$。

6.3 Taylor and Maclaurin Series 6.3 泰勒级数与麦克劳林级数

• Taylor polynomials are used to approximate functions near a value $x = a.$ Maclaurin polynomials are Taylor polynomials at $x = 0.$
• The *n*th degree Taylor polynomials for a function $f$ are the partial sums of the Taylor series for $f.$
• If a function $f$ has a power series representation at $x = a,$ then it is given by its Taylor series at $x = a.$
• A Taylor series for $f$ converges to $f$ if and only if $\underset{n\rightarrow\infty}{\text{lim}}R_{n}(x) = 0$ where $R_{n}(x) = f(x) - p_{n}(x).$
• The Taylor series for *e^x*, $\text{sin}\mspace{2mu} x,$ and $\text{cos}\mspace{2mu} x$ converge to the respective functions for all real *x*.

• 泰勒多项式用于在 $x = a$ 附近逼近函数。麦克劳林多项式是取 $x = 0$ 处的泰勒多项式。
• 函数 $f$ 的 *n* 次泰勒多项式是 $f$ 的泰勒级数的部分和。
• 若函数 $f$ 在 $x = a$ 处有幂级数表示，则该表示即由其在 $x = a$ 处的泰勒级数给出。
• $f$ 的泰勒级数收敛到 $f$ 当且仅当 $\underset{n\rightarrow\infty}{\text{lim}}R_{n}(x) = 0$，其中 $R_{n}(x) = f(x) - p_{n}(x)$。
• *e^x*、$\text{sin}\mspace{2mu} x$ 与 $\text{cos}\mspace{2mu} x$ 的泰勒级数对所有实数 *x* 都收敛到各自的函数。

6.4 Working with Taylor Series 6.4 泰勒级数的应用

• The binomial series is the Maclaurin series for $f(x) = \left( {1 + x} \right)^{r}.$ It converges for $|x| < 1.$
• Taylor series for functions can often be derived by algebraic operations with a known Taylor series or by differentiating or integrating a known Taylor series.
• Power series can be used to solve differential equations.
• Taylor series can be used to help approximate integrals that cannot be evaluated by other means.

• 二项级数是函数 $f(x) = \left( {1 + x} \right)^{r}$ 的麦克劳林级数。它在 $|x| < 1$ 时收敛。
• 函数的泰勒级数常可通过对已知泰勒级数进行代数运算，或对已知泰勒级数求导、积分而得到。
• 幂级数可用于求解微分方程。
• 泰勒级数可用于帮助近似计算那些无法用其他方法求值的积分。

Review Exercises 复习题

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

253.

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

254\.

Power series can be used to show that the derivative of $e^{x}\ \text{is}\ e^{x}.$ (*Hint:* Recall that $e^{x} = {\sum\limits_{n = 0}^{\infty}\frac{1}{n\text{!}}}x^{n}.)

255.

For small values of $x,{\text{sin}\mspace{2mu} x \approx x.}$

256\.

The radius of convergence for the Maclaurin series of $f(x) = 3^{x}$ is $3.$

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

257.

${\sum\limits_{n = 0}^{\infty}n^{2}}\left( {x - 1} \right)^{n}$

258\.

$\sum\limits_{n = 0}^{\infty}\frac{x^{n}}{n^{n}}$

259.

$\sum\limits_{n = 0}^{\infty}\frac{3nx^{n}}{12^{n}}$

260\.

${\sum\limits_{n = 0}^{\infty}\frac{2^{n}}{e^{n}}}\left( {x - e} \right)^{n}$

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.

261.

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

262\.

$f(x) = \frac{8x + 2}{2x^{2} - 3x + 1}$

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

263.

$f(x) = \text{tan}^{-1}\left( {2x} \right)$

264\.

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

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?

265.

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

266\.

$f(x) = e^{1\text{/}{({4x})}},a = 4$

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

267.

$f(x) = \text{cos}\left( {3x} \right)$

268\.

$f(x) = \text{ln}\mspace{2mu}\left( {x + 1} \right)$

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

269.

$f(x) = \text{sin}\mspace{2mu} x,a = \frac{\pi}{2}$

270\.

$f(x) = \frac{3}{x},a = 1$

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

271.

$f(x) = e^{\text{−}x^{2}} - 1$

272\.

$f(x) = \text{cos}\mspace{2mu} x - x\mspace{2mu}\text{sin}\mspace{2mu} x$

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

273.

$f(x) = \frac{\text{sin}\mspace{2mu} x}{x}$

274\.

$f(x) = 1 - e^{x}$

275.

Use power series to prove Euler’s formula: $e^{ix} = \text{cos}\mspace{2mu} x + i\mspace{2mu}\text{sin}\mspace{2mu} x$

The following exercises consider problems of annuity payments.

276\.

For annuities with a present value of $\text{\$}1$ million, calculate the annual payouts given over $25$ years assuming interest rates of $1\text{\%},5\text{\%},\ \text{and}\ 10\text{\%}.$

277.

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

278\.

Calculate the necessary present value of an annuity in order to support annual payouts of $\text{\$}15,000$ given over $25$ years assuming interest rates of $1\text{\%},5\text{\%},\ \text{and}\ 10\text{\%}.$