Q. 60 - 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.

${\sin }^{2}x$(Hint: use the identity ${\sin }^{2}x = \frac{1}{2}(1-\cos 2x)$)

Step-by-Step Solution

Answer

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

Step 1. Given Information

We consider the function $f\left(x\right)={\sin }^{2}x$

Step 2 : Simplification

The formula for ${\sin }^{2}x$ is as follows ${\sin }^{2}x=\frac{1}{2}(1-\cos 2x)$

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

Substituting $2x$with $x$ in the above equation we get $\cos 2x=\sum _{k=0}^{\infty }\frac{{(-1)}^k}{\left(2k\right)!}{\left(2x\right)}^{2k}$

Step 3 : Calculation

We now evaluate the value of $\frac{1-\cos 2x}{2}$ using the above series

Thus, $\begin{array}{rcl}\frac{1-\cos 2x}{2}& =& \frac{1-{\displaystyle \sum _{k=0}^{\infty }}{\displaystyle \frac{{(-1)}^k}{\left(2k\right)!}}{\left(2k\right)}^{2k}}{2}\\ & =& \frac{1}{2}-\frac{1}{2}\sum _{k=0}^{\infty }{(-1)}^k\frac{22kx2k}{\left(2k\right)!}=\frac{1}{2}-\frac{1}{2}-\frac{1}{2}\sum _{k=1}^{\infty }{(-1)}^k\frac{2^{2k}{x}^{2k}}{\left(2k\right)!}\end{array}$

The interval of convergence of the Maclaurin series of the given function is $\mathrm{ℝ}$, as the interval of convergence for the Maclaurin series of $\cos x$ is $\mathrm{ℝ}$

Therefore the required Maclaurin series for the given function is ${\sin }^{2}x=\sum _{k=1}^{\infty }{(-1)}^{k+1}\frac{22k-1x2k}{\left(2k\right)!}$