Question

Find the Maclaurin series for the functions in Exercises 51–60

by substituting into a known Maclaurin series. Also, give the

interval of convergence for the series.

$\frac{e^x-e^{-x}}{2}$

Short answer

The answer is$\frac{e^x-e^{-x}}{2}=\sum _{k=0}^{\infty }\frac{1}{(2k+1)!}{x}^{2k+1}$

Step-by-step solution

Step 1. Given Information

Consider the function$\frac{e^x-e^{-x}}{2}$

Find the interval of convergence for the series.

We know that the Maclaurin series for the func$e^x=\sum _{k=0}^{\infty }\frac{1}{k!}{x}^k$

So, the series for $h\left(x\right)=e^{-x}$can be found by substituting $x$by $-x$

That is, $e^{-x}=\sum _{k=0}^{\infty }\frac{1}{k!}{(-x)}^k$

Finally, to find the series for the function $f\left(x\right)=\frac{e^x-e^{-x}}{2}$we substract the series for $e^{-x}$from $e^x$and divide the result by 2

Thus, role="math" localid="1649870849460" $\begin{array}{rcl}{e}^x-e^{-x}& =& \sum _{k=0}^{\infty }\frac{1}{k!}{x}^k-\sum _{k=0}^{\infty }\frac{1}{k!}{(-x)}^k\\ & =& \sum _{k=0}^{\infty }\frac{1}{k!}[x^k-(-x^k\left)\right]\\ & =& 2x+\frac{2x^3}{3!}+\frac{2x^5}{5!}+\frac{2x^7}{7!}+...\end{array}$

Then dividing by $2$we get

$\frac{e^x-e^{-x}}{2}=x+\frac{x^3}{3!}+\frac{x^5}{5!}+\frac{x^7}{7!}+...$

implies that,role="math" localid="1649871145815" $\frac{e^x-e^{-x}}{2}=\sum _{k=0}^{\infty }\frac{1}{(2k+1)!}{x}^{2k+1}$