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

Step-by-step solution

Step 1. Given Information

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

Step 2

We know that the Maclaurin series for the function $g\left(x\right)=e^x$is $e^x=\sum _{k-0}^{\infty }\frac{1}{k!}{x}^k$

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 result for the function role="math" localid="1649869343697" $f\left(x\right)=\frac{e^x+e^{-x}}{2}$we add the series for $e^x$and $e^{-x}$and then divide by $2$

Thus,

$\frac{e^x+e^{-x}}{2}=1+\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+...$implies that,

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