Representing Functions as Power Series

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Representing Functions as Power Series

Functions as Power Series
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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.