Analyzing a Function Defined by a Power Series

A function \(f\) is defined by the power series \(f(x)=\sum\limits_{k\,=\,0}^{\infty }x^{k}\).

1. Find the domain of \(f\).
2. Evaluate \(f\left(\dfrac{1}{2}\right)\) and \(f\left(-\dfrac{1}{3}\right)\).
3. Find \(f\).

Solution (a) \(\sum\limits_{k\,=\,0}^{\infty }x^{k}\) is a power series centered at \(0\) with \(a_{k}=1\). Then \[ f(x) =1+x+x^{2}+x^{3}+x^{4}+ \cdots \]

The domain of \(f\) equals the interval of convergence of the power series. Since the series \(\sum\limits_{k\,=\,0}^{\infty }x^{k}\) is a geometric series, it converges for \(\vert x\vert < 1\). The radius of convergence is \(1\), and the interval of convergence is \((-1,1)\). The domain of \(f\) is the open interval \((-1,1) \).

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(b) The numbers \(\dfrac{1}{2}\) and \(-\dfrac{1}{3}\) are in the interval \((-1,1)\), so they are in the domain of \(f\). Then \(f\left( \dfrac{1}{2}\right)\) is a geometric series with \(r=\dfrac{1}{2}\), \(a=1\), and \[ f\left(\dfrac{1}{2}\right) =1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right) ^{2}+\left(\dfrac{1}{2}\right) ^{3}+\cdots \,= \frac{a}{1-r} = \dfrac{1}{1-\dfrac{1}{2}}=2 \]

Similarly, \[ f\left( -\dfrac{1}{3}\right) =1-\dfrac{1}{3}+\left( -\dfrac{1}{3}\right) ^{2}+\left( -\dfrac{1}{3}\right) ^{3}+\cdots \,=\dfrac{1}{1+\dfrac{1}{3}}=\dfrac{3}{4} \]

(c) Since \(f\) is defined by a geometric series, we can find \(f\) by summing the series. \[ \begin{eqnarray*} &&f(x) =\sum\limits_{k\,=\,0}^{\infty }x^{k} = 1 + x + x^{2} + \cdots + x^{n} + \cdots \underset{\underset{\color{#00ADEE}{a=1; r = x }}{\color{#00ADEE}{\uparrow}}}{=}\dfrac{1}{1-x}\quad -1 < x < 1\\ \end{eqnarray*} \]