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.

${\cos }^{2}x$(Hint: Use the identity ${\cos }^{2}x=\frac{1}{2}(1+cos2x)$)

Short answer

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

Step-by-step solution

Step 1. Given Information

Consider the function$f\left(x\right)={\cos }^{2}x$

Calculation

The formula for ${\cos }^{2}x$is as follows,

${\cos }^{2}x=\frac{1}{2}(1+cos2x)$

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

Substituting $2x$with $x$in the above equation we get role="math" localid="1649871853799" $\cos 2x=\sum _{k=0}^{\infty }\frac{{(-1)}^k}{\left(2k\right)!}{\left(2x\right)}^{2k}$

Now we add $1$to the above series and divide by 2 to get the Maclaurin series for ${\cos }^{2}x$

Thus, role="math" localid="1649872484707" $\begin{array}{rcl}\frac{1+\cos 2x}{2}& =& \frac{1+\sum _{k=0}^{\infty }\frac{{(-1)}^k}{\left(2k\right)!}{\left(2x\right)}^{2k}}{2}\\ & =& \frac{1}{2}+\frac{1}{2}\sum _{k=0}^{\infty }{(-1)}^k\frac{22kx2k}{\left(2k\right)!}=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\sum _{k=1}^{\infty }{(-1)}^k\frac{2^{2k}{x}^{2k}}{\left(2k\right)!}\\ & =& 1+\sum _{k=1}^{\infty }{(-1)}^k\frac{2^{2k-1}{x}^{2k}}{\left(2k\right)!}\end{array}$

Simplification

The interval of the convergence for the Maclaurin series of the given funtion is $\mathrm{ℝ}$, since the interval for the convergence of $\cos x$is $\mathrm{ℝ}$. Therefore, the required answer isrole="math" localid="1649872900734" ${\cos }^{2}x=1+\sum _{k=1}^{\infty }{(-1)}^k\frac{2^{2k-1}{x}^{2k}}{\left(2k\right)!}$