Q. 53 - 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.

$\sin (-5x^2)$

Step-by-Step Solution

Answer

The interval of convergence for the series $\begin{array}{rcl}& & \sum _{k=0}^{\infty }\frac{{(-1)}^{k+2k+1} 5^{2k+1}}{(2k+1)!}{x}^{4k+2}\end{array}$ .

Step 1. Given Information

The function $f\left(x\right)=\sin (-5x^2)$

The objective is to find the series of intervals of convergence  .

Step 2. Calculation

So, to get the Maclaurin series for the function $f\left(x\right)=\sin (-5x^2)$, we replace $x$ by $-5x^2$ in the Maclaurin series of the function $\sin x$

  $\begin{array}{rcl}\sin \left(-5x^2\right)& =& \sum _{k=0}^{\infty }\frac{{(-1)}^k}{(2k+1)!}{(-5x^2)}^{2k+1}\\ & =& \sum _{k=0}^{\infty }\frac{{(-1)}^k{(-1)}^{2k+1}{5}^{2k+1}}{(2k+1)!}{x}^{4k+2}\\ & =& \sum _{k=0}^{\infty }\frac{{(-1)}^{k+2k+1} 5^{2k+1}}{(2k+1)!}{x}^{4k+2}\end{array}$

Therefore, the required series is $\begin{array}{rcl}& & \sum _{k=0}^{\infty }\frac{{(-1)}^{k+2k+1} 5^{2k+1}}{(2k+1)!}{x}^{4k+2}\end{array}$.