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## Applications and Examples Using term-by-term differentiation and integration,
\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 $\lvert x\rvert<1$. -----------------------------------------------------------------------------
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 $C=0$.
-----------------------------------------------------------------------------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}
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)&=&\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}$ $$\tan^{-1}(x)= \sum_{n=0}^\infty(-1)^n\frac{x^{2n+1}}{2n+1}=x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots$$ ----------------------------------------------------------------------------- |