Applications and Examples

Using term-by-term differentiation and integration, we can compute the power series of more functions, as in the following examples.

Example 1:  Find a series representation for $\displaystyle \frac{1}{(1-x)^2}$. 

Solution 1:  Notice that we cannot just do algebra to $\displaystyle \frac{1}{1-x}$ to solve this.  But, given that  $\displaystyle \frac{1}{1-x}=\sum_{n=0}^\infty x^n $ when $\lvert x\rvert<1$, and given that our function is the derivative of $\displaystyle \frac{1}{1-x}$ (check this!  Why isn't there a negative?),

$\begin{eqnarray}
\frac{1}{(1-x)^2}&=&\frac{d}{dx} \left(\frac{1}{1-x}\right)\\
&=&\frac{d}{dx} \left(\sum_{n=0}^\infty x^n \right)\\
&=&\sum_{n=1}^\infty n\, x^{n-1}\\
&=&1+2x+3x^2+4x^3+\cdots,
\end{eqnarray}$

and we have our series representation when $\lvert x\rvert<1$.
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Example 2:  Find a series representation for $\ln(1+x)$. 

Solution 2:  To do this, we must find a series that we know, for which $\ln(1+x)$ is the derivative or the antiderivative.  At this point, we don't know that many series; really all we know is the standard geometric series and variants of it.  (Any ideas?)  After some thought, we realize $\displaystyle \frac{d}{dx}\ln (1+x)=\frac{1}{1+x}$ and $\displaystyle\frac{1}{1+x}=\sum_{n=0}^\infty(-1)^nx^n$ (from our previous work).  So, when $\lvert x\rvert<1$,
$\begin{eqnarray}
\ln(1+x)&=&\int\frac{1}{1+x}\,dx\\
&=&\int\left(\sum_{n=0}^\infty(-1)^nx^n\right)\,dx\\
&=&\left(\sum_{n=0}^\infty(-1)^n\frac{x^{n+1}}{n+1}\right)+C\\
&=&\sum_{n=0}^\infty(-1)^n\frac{x^{n+1}}{n+1}\\
&=&x - \frac{x^2}{2} + \frac{x^3}{3} -\frac{x^4}{4}+ \cdots.
\end{eqnarray}$
To see why $C=0$, plug $x=0$ into $\ln(1+x)$.  Since $\ln(1+0)=0$, our series is 0 when $x=0$, so $C=0$.
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Example 3:  Use the fact that $\displaystyle\frac{d}{dx} \tan^{-1}(x)=\frac{1}{1+x^2}$ to find a series representation for $\tan^{-1}(x)$. 
DO this before reading further.






Solution 3:  We know how to compute $\displaystyle\frac{1}{1+x^2}=\sum_{n=0}^\infty(-x^2)^n=\sum_{n=0}^\infty(-1)^nx^{2n}$.  So, when $\lvert x\rvert<1$,

$\begin{eqnarray}
\tan^{-1}(x)&=&\int\frac{1}{1+x^2}\,dx\\
&=&\int\left(\sum_{n=0}^\infty(-1)^nx^{2n}\right)\,dx\\
&=&\int \left(1-x^2+x^4-x^6+x^8-x^{10}+\cdots\right)\,dx\\
&=&C+x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots\\
&=&\left(\sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1}\right)+C\\
\end{eqnarray}$
To solve for $C$, plug $x=0$ into $\tan^{-1}(x)$.  We get $\tan^{-1}(0)=0$, so our series at $x=0$ must be $0$, and hence $C=0$.  we have
$$\tan^{-1}(x)= \sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1}=x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots$$
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Most calculator and computer approximations are done via series, so if we can find a series to represent a hard-to-compute function, we are happy, since series are easy to compute (especially for a computer) to any degree of accuracy you wish.  The video will go through some of these examples, and will demonstrate why this is so important.