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.

$x\cos \left(x^2\right)$

Short answer

The answer is$x\cos \left(x^2\right)=\begin{array}{rcl}& & x\end{array}\begin{array}{rcl}& & \sum _{k=0}^{\infty }\end{array}\begin{array}{rcl}& & \frac{{(-1)}^k}{\left(2k\right)!}\end{array}\begin{array}{rcl}& & x\end{array}\begin{array}{rcl}& & {}^{4k+1}\end{array}$

Step-by-step solution

Step 1. Given Information

Consider the function$x\cos \left(x^2\right)$

Step 2 

We know that the Maclaurin series for the function $g\left(x\right)=\cos \left(x\right)$is $\cos x=\sum _{k=0}^{\infty }\frac{{(-1)}^k}{\left(2k\right)!}{x}^{2k}$

So, to find the Maclaurin series for the function $f\left(x\right)=x\cos \left(x^2\right)$, we replace $x$by $x^2$and then multiply by $\cos x$

Therefore,$\begin{array}{rcl}x\cos \left(x^2\right)& =& x\sum _{k=0}^{\infty }\frac{{(-1)}^k}{\left(2k\right)!}{\left(x^2\right)}^{2k}=x\sum _{k=0}^{\infty }\frac{{(-1)}^k}{\left(2k\right)!}{x}^{4k}\end{array}$implies that,role="math" localid="1649865748409" $x\cos \left(x^2\right)=\begin{array}{rcl}& & x\end{array}\begin{array}{rcl}& & \sum _{k=0}^{\infty }\end{array}\begin{array}{rcl}& & \frac{{(-1)}^k}{\left(2k\right)!}\end{array}\begin{array}{rcl}& & x\end{array}\begin{array}{rcl}& & {}^{4k+1}\end{array}$