Working with Taylor Series

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

\int e^{− x^{2}} d x,

The Binomial Series

Our first goal in this section is to determine the Maclaurin series for the function

f (x) = {(1 + x)}^{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) = {(1 + x)}^{r}

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,

cannot be written as a finite polynomial. However, we can find a power series for

f .

is a nonnegative integer, then [link] for the coefficients agrees with [link] for the coefficients, and the formula for the binomial series agrees with [link] for the finite binomial expansion. More generally, to denote the binomial coefficients for any real number

r,

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

we conclude that the interval of convergence for the binomial series is

(−1, 1) .

the series diverges at both endpoints. The binomial series does converge to

{(1 + x)}^{r}

We can use this definition to find the binomial series for

f (x) = \sqrt{1 + x}

Common Functions Expressed as Taylor Series

At this point, we have derived Maclaurin series for exponential, trigonometric, and logarithmic functions, as well as functions of the form

f (x) = {(1 + x)}^{r} .

In [link], we summarize the results of these series. We remark that the convergence of the Maclaurin series for

f (x) = ln (1 + x)

relies on a more advanced theorem than we present here. (Refer to Abel’s theorem for a discussion of this more technical point.)

Maclaurin Series for Common Functions
Function	Maclaurin Series	Interval of Convergence
$f (x) = \frac{1}{1 - x}$	$\sum_{n = 0}^{\infty} x^{n}$	$−1 < x < 1$
$f (x) = e^{x}$	$\sum_{n = 0}^{\infty} \frac{x^{n}}{n!}$	$− \infty < x < \infty$
$f (x) = sin x$	$\sum_{n = 0}^{\infty} {(−1)}^{n} \frac{x^{2 n + 1}}{(2 n + 1)!}$	$− \infty < x < \infty$
$f (x) = cos x$	$\sum_{n = 0}^{\infty} {(−1)}^{n} \frac{x^{2 n}}{(2 n)!}$	$− \infty < x < \infty$
$f (x) = ln (1 + x)$	$\sum_{n = 1}^{\infty} {(−1)}^{n + 1} \frac{x^{n}}{n}$	$−1 < x \leq 1$
$f (x) = {tan}^{−1} x$	$\sum_{n = 0}^{\infty} {(−1)}^{n} \frac{x^{2 n + 1}}{2 n + 1}$	$−1 < x \leq 1$
$f (x) = {(1 + x)}^{r}$	$\sum_{n = 0}^{\infty} (\begin{matrix} r \\ n \end{matrix}) x^{n}$	$−1 < x < 1$

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 [link], to create Maclaurin series for other functions.

We also showed previously in this chapter how power series can be differentiated term by term to create a new power series. In [link], we differentiate the binomial series for

\sqrt{1 + x}

directly from the definition, but differentiating the binomial series for

\sqrt{1 + x}

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

Recall that this is a first-order separable equation and its solution is

y = C e^{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_{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^{'} = y

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

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.

Evaluating Nonelementary Integrals

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

\int e^{− x^{2}} d x .

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} - 3 x} + e^{x^{3}} - sin (5 x + 4)

is an elementary function, although not a particularly simple-looking function. Any integral of the form

\int f (x) d x

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

\int e^{− x^{2}} d x .

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

μ

Another application in which a nonelementary integral arises involves the period of a pendulum. The integral is

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.

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.

Key Concepts

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

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

{(1 + x^{2})}^{−1 / 3}

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

{(1 - x)}^{1.01}

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

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

In the following exercises, use the substitution ${(b + x)}^{r} = {(b + a)}^{r} {(1 + \frac{x - a}{b + a})}^{r}$

in the binomial expansion to find the Taylor series of each function with the given center.

\sqrt{x + 2}

at $a = 0$

\sqrt{x^{2} + 2}

at $a = 0$

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

\sqrt{x + 2}

at $a = 1$

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

at $a = 1$

(Hint: $2 x - x^{2} = 1 - {(x - 1)}^{2})$

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

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

{(x - 8)}^{1 / 3}

at $a = 9$

\sqrt{x}

at $a = 4$

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

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

x^{1 / 3}

at $a = 27$

\sqrt{x}

at $x = 9$

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

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

[T] ${(15)}^{1 / 4}$

using ${(16 - x)}^{1 / 4}$

[T] ${(1001)}^{1 / 3}$

using ${(1000 + x)}^{1 / 3}$

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

Using, for example, a fourth-degree estimate at $x = 1$

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

whereas ${(1001)}^{1 / 3} = 10.00332222839093 ....$

Two terms would suffice for three-digit accuracy.

In the following exercises, use the binomial approximation $\sqrt{1 - x} \approx 1 - \frac{x}{2} - \frac{x^{2}}{8} - \frac{x^{3}}{16} - \frac{5 x^{4}}{128} - \frac{7 x^{5}}{256}$

for $\| x \| < 1$

to approximate each number. Compare this value to the value given by a scientific calculator.

[T] $\frac{1}{\sqrt{2}}$

using $x = \frac{1}{2}$

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

[T] $\sqrt{5} = 5 \times \frac{1}{\sqrt{5}}$

using $x = \frac{4}{5}$

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

The approximation is $2.3152;$

the CAS value is $2.23 … .$

[T] $\sqrt{3} = \frac{3}{\sqrt{3}}$

using $x = \frac{2}{3}$

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

[T] $\sqrt{6}$

using $x = \frac{5}{6}$

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

The approximation is $2.583 …;$

the CAS value is $2.449 … .$

Integrate the binomial approximation of $\sqrt{1 - x}$

to find an approximation of $\int_{0}^{x} \sqrt{1 - t} d t .$

[T] Recall that the graph of $\sqrt{1 - x^{2}}$

is an upper semicircle of radius $1 .$

Integrate the binomial approximation of $\sqrt{1 - x^{2}}$

up to order $8$

from $x = −1$

to $x = 1$

to estimate $\frac{π}{2} .$

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

Thus

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

whereas $\frac{π}{2} = 1.570 ...$

In the following exercises, use the expansion ${(1 + x)}^{1 / 3} = 1 + \frac{1}{3} x - \frac{1}{9} x^{2} + \frac{5}{81} x^{3} - \frac{10}{243} x^{4} + ⋯$

to write the first five terms (not necessarily a quartic polynomial) of each expression.

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

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

{(1 + x)}^{4 / 3} = (1 + x) (1 + \frac{1}{3} x - \frac{1}{9} x^{2} + \frac{5}{81} x^{3} - \frac{10}{243} x^{4} + ⋯) = 1 + \frac{4 x}{3} + \frac{2 x^{2}}{9} - \frac{4 x^{3}}{81} + \frac{5 x^{4}}{243} + ⋯

{(3 + 2 x)}^{1 / 3}; a = −1

{(x^{2} + 6 x + 10)}^{1 / 3}; a = −3

{(1 + {(x + 3)}^{2})}^{1 / 3} = 1 + \frac{1}{3} {(x + 3)}^{2} - \frac{1}{9} {(x + 3)}^{4} + \frac{5}{81} {(x + 3)}^{6} - \frac{10}{243} {(x + 3)}^{8} + ⋯

Use ${(1 + x)}^{1 / 3} = 1 + \frac{1}{3} x - \frac{1}{9} x^{2} + \frac{5}{81} x^{3} - \frac{10}{243} x^{4} + ⋯$

with $x = 1$

to approximate $2^{1 / 3} .$

Use the approximation ${(1 - x)}^{2 / 3} = 1 - \frac{2 x}{3} - \frac{x^{2}}{9} - \frac{4 x^{3}}{81} - \frac{7 x^{4}}{243} - \frac{14 x^{5}}{729} + ⋯$

for $\| x \| < 1$

to approximate $2^{1 / 3} = {2.2}^{−2 / 3} .$

Twice the approximation is $1.260 …$

whereas $2^{1 / 3} = 1.2599. ...$

Find the $25 th$

derivative of $f (x) = {(1 + x^{2})}^{13}$

at $x = 0 .$

Find the $99$

th derivative of $f (x) = {(1 + x^{4})}^{25} .$

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

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

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

f (x) = 2^{x}

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

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

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

For $x > 0, sin (\sqrt{x}) = \sum_{n = 0}^{\infty} {(−1)}^{n} \frac{x^{(2 n + 1) / 2}}{\sqrt{x} (2 n + 1)!} = \sum_{n = 0}^{\infty} {(−1)}^{n} \frac{x^{n}}{(2 n + 1)!} .$

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

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

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

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

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

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

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

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

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

by integrating the Maclaurin series of $f$

term by term. If $f$

is not strictly defined at zero, you may substitute the value of the Maclaurin series at zero.

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

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

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

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

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

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

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

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

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

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

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

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

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

f (x) = sin (x + \frac{π}{4}) = sin x cos (\frac{π}{4}) + cos x sin (\frac{π}{4})

f (x) = tan x

x + \frac{x^{3}}{3} + \frac{2 x^{5}}{15} + ⋯

f (x) = ln (cos x)

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

1 + x - \frac{x^{3}}{3} - \frac{x^{4}}{6} + ⋯

f (x) = e^{sin x}

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

1 + x^{2} + \frac{2 x^{4}}{3} + \frac{17 x^{6}}{45} + ⋯

f (x) = tanh x

f (x) = \frac{tan \sqrt{x}}{\sqrt{x}}

(see expansion for $tan x)$

Using the expansion for $tan x$

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

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

ln (1 + x)

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

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

so $R = 1$

by the ratio test.

{tan}^{−1} x

ln (1 + x^{2})

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

so $R = 1$

by the ratio test.

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

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

Add series of $e^{x}$

and $e^{− x}$

term by term. Odd terms cancel and $cosh x = \sum_{n = 0}^{\infty} \frac{x^{2 n}}{(2 n)!} .$

Differentiate term by term the Maclaurin series of $sinh x$

and compare the result with the Maclaurin series of $cosh x .$

[T] Let $S_{n} (x) = \sum_{k = 0}^{n} {(−1)}^{k} \frac{x^{2 k + 1}}{(2 k + 1)!}$

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

denote the respective Maclaurin polynomials of degree $2 n + 1$

of $sin x$

and degree $2 n$

of $cos x .$

Plot the errors $\frac{S_{n} (x)}{C_{n} (x)} - tan x$

for $n = 1, .., 5$

and compare them to $x + \frac{x^{3}}{3} + \frac{2 x^{5}}{15} + \frac{17 x^{7}}{315} - tan x$

on $(- \frac{π}{4}, \frac{π}{4}) .$

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

The ratio $\frac{S_{n} (x)}{C_{n} (x)}$

approximates $tan x$

better than does $p_{7} (x) = x + \frac{x^{3}}{3} + \frac{2 x^{5}}{15} + \frac{17 x^{7}}{315}$

for $N \geq 3 .$

The dashed curves are $\frac{S_{n}}{C_{n}} - tan$

for $n = 1, 2 .$

The dotted curve corresponds to $n = 3,$

and the dash-dotted curve corresponds to $n = 4 .$

The solid curve is $p_{7} - tan x .$

Use the identity $2 sin x cos x = sin (2 x)$

to find the power series expansion of ${sin}^{2} x$

at $x = 0 .$

(Hint: Integrate the Maclaurin series of $sin (2 x)$

term by term.)

If $y = \sum_{n = 0}^{\infty} a_{n} x^{n},$

find the power series expansions of $x y^{'}$

and $x^{2} y ″ .$

By the term-by-term differentiation theorem, $y^{'} = \sum_{n = 1}^{\infty} n a_{n} x^{n - 1}$

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

whereas $y^{'} = \sum_{n = 2}^{\infty} n (n - 1) a_{n} x^{n - 2}$

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

[T] Suppose that $y = \sum_{k = 0}^{\infty} a_{k} x^{k}$

satisfies $y^{'} = −2 x y$

and $y (0) = 0 .$

Show that $a_{2 k + 1} = 0$

for all $k$

and that $a_{2 k + 2} = \frac{− a_{2 k}}{k + 1} .$

Plot the partial sum $S_{20}$

of $y$

on the interval $[−4, 4] .$

[T] Suppose that a set of standardized test scores is normally distributed with mean $μ = 100$

and standard deviation $σ = 10 .$

Set up an integral that represents the probability that a test score will be between $90$

and $110$

and use the integral of the degree $10$

Maclaurin polynomial of $\frac{1}{\sqrt{2 π}} e^{− x^{2} / 2}$

to estimate this probability.

The probability is $p = \frac{1}{\sqrt{2 π}} \int_{(a - μ) / σ}^{(b - μ) / σ} e^{− x^{2} / 2} d x$

where $a = 90$

and $b = 100,$

that is, $p = \frac{1}{\sqrt{2 π}} \int_{−1}^{1} e^{− x^{2} / 2} d x = \frac{1}{\sqrt{2 π}} \int_{−1}^{1} \sum_{n = 0}^{5} {(−1)}^{n} \frac{x^{2 n}}{2^{n} n!} d x = \frac{2}{\sqrt{2 π}} \sum_{n = 0}^{5} {(−1)}^{n} \frac{1}{(2 n + 1) 2^{n} n!} \approx 0.6827 .$

[T] Suppose that a set of standardized test scores is normally distributed with mean $μ = 100$

and standard deviation $σ = 10 .$

Set up an integral that represents the probability that a test score will be between $70$

and $130$

and use the integral of the degree $50$

Maclaurin polynomial of $\frac{1}{\sqrt{2 π}} e^{− x^{2} / 2}$

to estimate this probability.

[T] Suppose that $\sum_{n = 0}^{\infty} a_{n} x^{n}$

converges to a function $f (x)$

such that $f (0) = 1, f^{'} (0) = 0,$

and $f ″ (x) = − f (x) .$

Find a formula for $a_{n}$

and plot the partial sum $S_{N}$

for $N = 20$

on $[−5, 5] .$

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

As in the previous problem one obtains $a_{n} = 0$

if $n$

is odd and $a_{n} = − (n + 2) (n + 1) a_{n + 2}$

if $n$

is even, so $a_{0} = 1$

leads to $a_{2 n} = \frac{{(−1)}^{n}}{(2 n)!} .$

[T] Suppose that $\sum_{n = 0}^{\infty} a_{n} x^{n}$

converges to a function $f (x)$

such that $f (0) = 0, f^{'} (0) = 1,$

and $f ″ (x) = − f (x) .$

Find a formula for $a_{n}$

and plot the partial sum $S_{N}$

for $N = 10$

on $[−5, 5] .$

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

converges to a function $y$

such that $y ″ - y^{'} + y = 0$

where $y (0) = 1$

and $y' (0) = 0 .$

Find a formula that relates $a_{n + 2}, a_{n + 1},$

and $a_{n}$

and compute $a_{0}, ..., a_{5} .$

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

and $y^{'} = \sum_{n = 0}^{\infty} (n + 1) a_{n + 1} x^{n}$

so $y ″ - y^{'} + y = 0$

implies that $(n + 2) (n + 1) a_{n + 2} - (n + 1) a_{n + 1} + a_{n} = 0$

or $a_{n} = \frac{a_{n - 1}}{n} - \frac{a_{n - 2}}{n (n - 1)}$

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

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

so $a_{2} = \frac{1}{2}, a_{3} = \frac{1}{6}, a_{4} = 0,$

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

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

converges to a function $y$

such that $y ″ - y^{'} + y = 0$

where $y (0) = 0$

and $y^{'} (0) = 1 .$

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 $\int_{a}^{b} f (t) d t$

by that of a Taylor approximation $\int_{a}^{b} P_{n} (t) d t$

is at most $\int_{a}^{b} R_{n} (t) d t .$

In the following exercises, the Taylor remainder estimate $R_{n} \leq \frac{M}{(n + 1)!} {\| x - a \|}^{n + 1}$

guarantees that the integral of the Taylor polynomial of the given order approximates the integral of $f$

with an error less than $\frac{1}{10} .$

Evaluate the integral of the appropriate Taylor polynomial and verify that it approximates the CAS value with an error less than $\frac{1}{100} .$
Compare the accuracy of the polynomial integral estimate with the remainder estimate.

[T] $\int_{0}^{π} \frac{sin t}{t} d t; P_{s} = 1 - \frac{x^{2}}{3!} + \frac{x^{4}}{5!} - \frac{x^{6}}{7!} + \frac{x^{8}}{9!}$

(You may assume that the absolute value of the ninth derivative of $\frac{sin t}{t}$

is bounded by $0.1 .)$

a. (Proof) b. We have $R_{s} \leq \frac{0.1}{(9)!} π^{9} \approx 0.0082 < 0.01 .$

We have $\int_{0}^{π} (1 - \frac{x^{2}}{3!} + \frac{x^{4}}{5!} - \frac{x^{6}}{7!} + \frac{x^{8}}{9!}) d x = π - \frac{π^{3}}{3 \cdot 3!} + \frac{π^{5}}{5 \cdot 5!} - \frac{π^{7}}{7 \cdot 7!} + \frac{π^{9}}{9 \cdot 9!} = 1.852...,$

whereas $\int_{0}^{π} \frac{sin t}{t} d t = 1.85194...,$

so the actual error is approximately $0.00006 .$

[T] $\int_{0}^{2} e^{− x^{2}} d x; p_{11} = 1 - x^{2} + \frac{x^{4}}{2} - \frac{x^{6}}{3!} + ⋯ - \frac{x^{22}}{11!}$

(You may assume that the absolute value of the $23 rd$

derivative of $e^{− x^{2}}$

is less than $2 \times 10^{14} .)$

The following exercises deal with Fresnel integrals.

The Fresnel integrals are defined by $C (x) = \int_{0}^{x} cos (t^{2}) d t$

and $S (x) = \int_{0}^{x} sin (t^{2}) d t .$

Compute the power series of $C (x)$

and $S (x)$

and plot the sums $C_{N} (x)$

and $S_{N} (x)$

of the first $N = 50$

nonzero terms on $[0, 2 π] .$

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

Since $cos (t^{2}) = \sum_{n = 0}^{\infty} {(−1)}^{n} \frac{t^{4 n}}{(2 n)!}$

and $sin (t^{2}) = \sum_{n = 0}^{\infty} {(−1)}^{n} \frac{t^{4 n + 2}}{(2 n + 1)!},$

one has $S (x) = \sum_{n = 0}^{\infty} {(−1)}^{n} \frac{x^{4 n + 3}}{(4 n + 3) (2 n + 1)!}$

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

The sums of the first $50$

nonzero terms are plotted below with $C_{50} (x)$

the solid curve and $S_{50} (x)$

the dashed curve.

[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 $(C (t), S (t)) .$

Plot the curve $(C_{50}, S_{50})$

for $0 \leq t \leq 2 π,$

the coordinates of which were computed in the previous exercise.

Estimate $\int_{0}^{1 / 4} \sqrt{x - x^{2}} d x$

by approximating $\sqrt{1 - x}$

using the binomial approximation $1 - \frac{x}{2} - \frac{x^{2}}{8} - \frac{x^{3}}{16} - \frac{5 x^{4}}{2128} - \frac{7 x^{5}}{256} .$

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

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

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

[T] Use Newton’s approximation of the binomial $\sqrt{1 - x^{2}}$

to approximate $π$

as follows. The circle centered at $(\frac{1}{2}, 0)$

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 $(\frac{1}{4}, \frac{\sqrt{3}}{4})$

corresponds to $\frac{1}{6}$

of the circle and has area $\frac{π}{24} .$

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 $y = \sqrt{x} \sqrt{1 - x} = \sqrt{x} \times (binomial expansion of \sqrt{1 - x})$

and integrate term by term. Use this approach with the binomial approximation from the previous exercise to estimate $π .$

Use the approximation $T \approx 2 π \sqrt{\frac{L}{g}} (1 + \frac{k^{2}}{4})$

to approximate the period of a pendulum having length $10$

meters and maximum angle $θ_{max} = \frac{π}{6}$

where $k = sin (\frac{θ_{max}}{2}) .$

Compare this with the small angle estimate $T \approx 2 π \sqrt{\frac{L}{g}} .$

T \approx 2 π \sqrt{\frac{10}{9.8}} (1 + \frac{{sin}^{2} (θ / 12)}{4}) \approx 6.453

seconds. The small angle estimate is $T \approx 2 π \sqrt{\frac{10}{9.8} \approx 6.347} .$

The relative error is around $2$

percent.

Suppose that a pendulum is to have a period of $2$

seconds and a maximum angle of $θ_{max} = \frac{π}{6} .$

Use $T \approx 2 π \sqrt{\frac{L}{g}} (1 + \frac{k^{2}}{4})$

to approximate the desired length of the pendulum. What length is predicted by the small angle estimate $T \approx 2 π \sqrt{\frac{L}{g}} ?$

Evaluate $\int_{0}^{π / 2} {sin}^{4} θ d θ$

in the approximation $T = 4 \sqrt{\frac{L}{g}} \int_{0}^{π / 2} (1 + \frac{1}{2} k^{2} {sin}^{2} θ + \frac{3}{8} k^{4} {sin}^{4} θ + ⋯) d θ$

to obtain an improved estimate for $T .$

\int_{0}^{π / 2} {sin}^{4} θ d θ = \frac{3 π}{16} .

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

[T] An equivalent formula for the period of a pendulum with amplitude $θ_{max}$

is $T (θ_{max}) = 2 \sqrt{2} \sqrt{\frac{L}{g}} \int_{0}^{θ_{max}} \frac{d θ}{\sqrt{cos θ} - cos (θ_{max})}$

where $L$

is the pendulum length and $g$

is the gravitational acceleration constant. When $θ_{max} = \frac{π}{3}$

we get $\frac{1}{\sqrt{cos t - 1 / 2}} \approx \sqrt{2} (1 + \frac{t^{2}}{2} + \frac{t^{4}}{3} + \frac{181 t^{6}}{720}) .$

Integrate this approximation to estimate $T (\frac{π}{3})$

in terms of $L$

and $g .$

Assuming $g = 9.806$

meters per second squared, find an approximate length $L$

such that $T (\frac{π}{3}) = 2$

seconds.

Chapter Review Exercises

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

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

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.

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

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?

In the following exercises, find the Maclaurin series for

F (x) = \int_{0}^{x} f (t) d t

Working with Taylor Series

The Binomial Series

Common Functions Expressed as Taylor Series

Solving Differential Equations with Power Series

Evaluating Nonelementary Integrals

Key Concepts

Chapter Review Exercises

Glossary