Assuming that \(f(x)=e^{x}\) can be represented by a power series in \(x\), find its Maclaurin series.
Solution To express a function \(f\) as a Maclaurin series, we begin by evaluating \(f\) and its derivatives at \(0\). \[ \begin{array}{rl@{\qquad}rll} f(x) & =e^{x} & f(0) &=1 \\ f^{\prime} (x)& =e^{x} & f^{\prime} (0) & =1 \\ f^{\prime \prime} (x) & =e^{x} & f^{\prime \prime} (0) &=1 \\ & \vdots & & \vdots \end{array} \]
Then we use the definition of a Maclaurin series. \[ \begin{eqnarray} f(x)=\sum\limits_{k\,=\,0}^{\infty }\frac{f^{(k) }(0) }{k!}x^{k} \underset{\underset{\color{#0066A7}{{f^{(k) }(0) =1 }}} {\color{#0066A7}{{{\uparrow}}}} }{=} 1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\cdots +\frac{ x^{n}}{n!}+\cdots =\sum_{k\,=\,0}^{\infty }\frac{x^{k}}{k!}\tag{3} \end{eqnarray} \]