# 6.4 Working with taylor series  (Page 3/11)

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## Deriving maclaurin series from known series

Find the Maclaurin series of each of the following functions by using one of the series listed in [link] .

1. $f\left(x\right)=\text{cos}\sqrt{x}$
2. $f\left(x\right)=\text{sinh}\phantom{\rule{0.1em}{0ex}}x$
1. Using the Maclaurin series for $\text{cos}\phantom{\rule{0.1em}{0ex}}x$ we find that the Maclaurin series for $\text{cos}\sqrt{x}$ is given by
$\begin{array}{cc}\hfill \sum _{n=0}^{\infty }\frac{{\left(-1\right)}^{n}{\left(\sqrt{x}\right)}^{2n}}{\left(2n\right)\text{!}}& =\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^{n}{x}^{n}}{\left(2n\right)\text{!}}\hfill \\ & =1-\frac{x}{2\text{!}}+\frac{{x}^{2}}{4\text{!}}-\frac{{x}^{3}}{6\text{!}}+\frac{{x}^{4}}{8\text{!}}-\text{⋯}.\hfill \end{array}$

This series converges to $\text{cos}\sqrt{x}$ for all $x$ in the domain of $\text{cos}\sqrt{x};$ that is, for all $x\ge 0.$
2. To find the Maclaurin series for $\text{sinh}\phantom{\rule{0.1em}{0ex}}x,$ we use the fact that
$\text{sinh}\phantom{\rule{0.1em}{0ex}}x=\frac{{e}^{x}-{e}^{\text{−}x}}{2}.$

Using the Maclaurin series for ${e}^{x},$ we see that the $n\text{th}$ term in the Maclaurin series for $\text{sinh}\phantom{\rule{0.1em}{0ex}}x$ 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 $\frac{2{x}^{n}}{n\text{!}}.$ Therefore, the Maclaurin series for $\text{sinh}\phantom{\rule{0.1em}{0ex}}x$ has only odd-order terms and is given by
$\sum _{n=0}^{\infty }\frac{{x}^{2n+1}}{\left(2n+1\right)\text{!}}=x+\frac{{x}^{3}}{3\text{!}}+\frac{{x}^{5}}{5\text{!}}+\text{⋯}.$

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

$\sum _{n=0}^{\infty }\frac{{\left(-1\right)}^{n}{x}^{4n+2}}{\left(2n+1\right)\text{!}}$

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 $\sqrt{1+x}$ 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

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

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

$\begin{array}{cc}\hfill \frac{1}{\sqrt{1+x}}& =2\frac{d}{dx}\sqrt{1+x}\hfill \\ & =1+\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n}}{n\text{!}}\phantom{\rule{0.2em}{0ex}}\frac{1·3·5\text{⋯}\left(2n-1\right)}{{2}^{n}}{x}^{n}.\hfill \end{array}$

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

$\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n}}{n\text{!}}\phantom{\rule{0.2em}{0ex}}\frac{1·3·5\text{⋯}\left(2n-1\right)}{{2}^{n}}{x}^{n}$

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

Consider the differential equation

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

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 ${y}^{\prime }=y$ to illustrate the technique.

## Power series solution of a differential equation

Use power series to solve the initial-value problem

${y}^{\prime }=y,\phantom{\rule{0.5em}{0ex}}y\left(0\right)=3.$

Suppose that there exists a power series solution

$y\left(x\right)=\sum _{n=0}^{\infty }{c}_{n}{x}^{n}={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+{c}_{3}{x}^{3}+{c}_{4}{x}^{4}+\text{⋯}.$

Differentiating this series term by term, we obtain

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

If y satisfies the differential equation, then

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

Using [link] on the uniqueness of power series representations, we know that these series can only be equal if their coefficients are equal. Therefore,

$\begin{array}{c}{c}_{0}={c}_{1},\hfill \\ {c}_{1}=2{c}_{2},\hfill \\ {c}_{2}=3{c}_{3},\hfill \\ {c}_{3}=4{c}_{4},\hfill \\ \hfill \text{⋮}.\hfill \end{array}$

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

$y\left(x\right)={c}_{0}+{c}_{1}x+{c}_{2}{x}^{2}+{c}_{3}{x}^{3}+\text{⋯},$

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

$\begin{array}{}\\ \\ {c}_{1}={c}_{0}=3=\frac{3}{1\text{!}},\hfill \\ {c}_{2}=\frac{{c}_{1}}{2}=\frac{3}{2}=\frac{3}{2\text{!}},\hfill \\ {c}_{3}=\frac{{c}_{2}}{3}=\frac{3}{3·2}=\frac{3}{3\text{!}},\hfill \\ {c}_{4}=\frac{{c}_{3}}{4}=\frac{3}{4·3·2}=\frac{3}{4\text{!}}.\hfill \end{array}$

Therefore,

$\begin{array}{cc}\hfill y& =3\left[1+\frac{1}{1\text{!}}x+\frac{1}{2\text{!}}{x}^{2}+\frac{1}{3\text{!}}{x}^{3}\frac{1}{4\text{!}}{x}^{4}+\text{⋯}\right]\hfill \\ & =3\sum _{n=0}^{\infty }\frac{{x}^{n}}{n\text{!}}.\hfill \end{array}$

You might recognize

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

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

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