Q. 59 - 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.

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

Step-by-Step Solution

Answer

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

Step 1. Given Information

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

Step 2 : 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 $\cos 2x=\sum _{k=0}^{\infty }\frac{{(-1)}^k}{\left(2k\right)!}{\left(2x\right)}^{2k}$

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

Thus, $\begin{array}{rcl}\frac{1+\cos 2x}{2}& =& \frac{1+\sum _{k=0}^{\infty }\frac{{(-1)}^k}{\left(2k\right)!}{\left(2x\right)}^{2k}}{2}\\ & =& \frac{1}{2}+\frac{1}{2}\sum _{k=0}^{\infty }{(-1)}^k\frac{22k x2k}{\left(2k\right)!}=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}\sum _{k=1}^{\infty }{(-1)}^k\frac{2^{2k} x^{2k}}{\left(2k\right)!}\\ & =& 1+\sum _{k=1}^{\infty }{(-1)}^k\frac{2^{2k-1} x^{2k}}{\left(2k\right)!}\end{array}$

Step 3: 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 is  ${\cos }^{2}x=1+\sum _{k=1}^{\infty }{(-1)}^k\frac{2^{2k-1}{x}^{2k}}{\left(2k\right)!}$