Q. 54 - Calculus

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.

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

Step-by-Step Solution

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