Question

In Problems 11-16 use an appropriate series in (2) to find the Maclaurin series of the given function. Write your answer in summation notation.

${\mathit{e}}^{\mathbf{-}\mathbf{x}\mathbf{/}\mathbf{2}}$

Short answer

$\sum _{n=0}^{\infty }\frac{{(-1)}^n{x}^n}{n!2^n}$

Step-by-step solution

Definition

Maclaurin seriesare a type of series expansion in which all terms are nonnegative integer powers of the variable.

Find Maclaurin series

We have Maclaurin series of $e^x$.

$$\begin{gathered} e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots \\ =\sum _{n=0}^{\infty }\frac{1}{n!}{x}^n \end{gathered}$$

Now replace$x$ as$-\frac{x}{2}$ we get

$$\begin{gathered} e^x=1+\frac{\left(\frac{-x}{2}\right)}{1!}+\frac{{\left(\frac{-x}{2}\right)}^2}{2!}+\frac{{\left(\frac{-x}{2}\right)}^3}{3!}+\dots \\ =\sum _{n=0}^{\infty }\frac{1}{n!}{\left(\frac{-x}{2}\right)}^n \\ =\sum _{n=0}^{\infty }\frac{{(-1)}^n{x}^n}{n!2^n} \end{gathered}$$

Thus the Maclaurin series of$e^{-x/2}$ is $\sum _{n=0}^{\infty }\frac{{(-1)}^n{x}^n}{n!2^n}$.