Represent each of the following functions by a power series centered at $0$:

Solution We use the function $f(x) = \dfrac{1}{1-x}$, $-1< x < 1$, represented by the geometric series $\sum\limits_{k\,=\,0}^{\infty }x^{k}$.

(a) In the geometric series for $f(x) =\dfrac{1}{1-x} ,$ we replace $x$ by $2x^{2}$. This series converges if $\vert 2x^{2}\vert < 1$, or equivalently if $-\dfrac{\sqrt{2}}{2} < x < \dfrac{\sqrt{2}}{2}$. Then on the open interval $\left( -\dfrac{\sqrt{2}}{2},\,\dfrac{\sqrt{2}}{2}\right)$, the function $h\,(x) =\dfrac{1}{ 1-2x^{2}}$ is represented by the power series $$\begin{eqnarray*} h(x) &=& \dfrac{1}{1-2x^{2}}=1+(2x^{2}) + (2x^{2}) ^{2}+ (2x^{2})^{3}+ \cdots \\ &=& 1+2x^{2}+4x^{4}+8x^{6}+\cdots+2^{n}x^{2n} +\cdots=\sum\limits_{k\,=\,0}^{ \infty }(2x^{2}) ^{k}= \sum\limits_{k\,=\,0}^{\infty }2^{k}x^{2k} \end{eqnarray*}$$

(b) We begin by writing $$g(x) =\dfrac{1}{3+x}=\dfrac{1}{3}\left(\dfrac{1}{1+\dfrac{x}{3}} \right) =\dfrac{1}{3}\left[ \dfrac{1}{1-\left( -\dfrac{x}{3}\right) }\right]$$

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Now in the geometric series for $f(x) =\dfrac{1}{1-x}$, replace $x$ by $-\dfrac{x}{3}$. This series converges if $\left\vert -\dfrac{x}{3} \right\vert < 1$, or equivalently if $-3< x < 3$. Then in the open interval $(-3,3)$, $g(x) =\dfrac{1}{3+x}$ is represented by the power series $$\begin{eqnarray*} g\,(x) &=& \dfrac{1}{3+x}=\dfrac{1}{3}\left[ 1+\left( -\dfrac{x}{3} \right) +\left( -\dfrac{x}{3}\right) ^{2}+\left( -\dfrac{x}{3}\right) ^{3}+\cdots\right] =\dfrac{1}{3}\sum\limits_{k\,=0}^{\infty }(-1) ^{k}\left( \dfrac{x}{3}\right) ^{k}\nonumber \\ &=&\sum\limits_{k\,=0}^{\infty }\dfrac{ (-1) ^{k}x^{k}}{3^{k+1}} \end{eqnarray*}$$

(c) $F(x) =\dfrac{x^{2}}{1-x}=x^{2}\left( \dfrac{1}{ 1-x}\right)$. Now for all numbers in the interval $(-1,\,1)$, $$\dfrac{1}{1-x}=1+x+x^{2}+\cdots+x^{n}+\cdots$$

So for any number $x$ in the interval of convergence, $-1< x < 1$, we have $$F(x) =x^{2}\left( 1+x+x^{2}+\cdots+x^{n}+\cdots\right) =x^{2}+x^{3}+x^{4}+\cdots+x^{n+2}+\cdots\,=\sum\limits_{k\,=2}^{\infty }x^{k}$$