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Find the Maclaurin series of each of the following functions by using one of the series listed in [link] .
Find the Maclaurin series for $\text{sin}\left({x}^{2}\right).$
$\sum _{n=0}^{\infty}\frac{{\left(\mathrm{-1}\right)}^{n}{x}^{4n+2}}{\left(2n+1\right)\text{!}}$
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.
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
so the binomial series for $\frac{1}{\sqrt{1+x}}$ is given by
Find the binomial series for $f\left(x\right)=\frac{1}{{\left(1+x\right)}^{3\text{/}2}}$
$\sum _{n=1}^{\infty}\frac{{\left(\mathrm{-1}\right)}^{n}}{n\text{!}}\phantom{\rule{0.2em}{0ex}}\frac{1\xb73\xb75\text{\cdots}\left(2n-1\right)}{{2}^{n}}{x}^{n}$
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.
Consider the differential equation
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 $y={\displaystyle \sum _{n=0}^{\infty}{c}_{n}{x}^{n}}$ 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.
Use power series to solve the initial-value problem
Suppose that there exists a power series solution
Differentiating this series term by term, we obtain
If y satisfies the differential equation, then
Using [link] on the uniqueness of power series representations, we know that these series can only be equal if their coefficients are equal. Therefore,
Using the initial condition $y\left(0\right)=3$ combined with the power series representation
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
Therefore,
You might recognize
as the Taylor series for ${e}^{x}.$ Therefore, the solution is $y=3{e}^{x}.$
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