Question

Use the Maclaurin series for $\sin x$,

$\cos x$, and $e^x$to find the values of the following series.

$1-\frac{{\pi }^2}{2^2·2!}+\frac{{\pi }^4}{2^4·4!}-\frac{{\pi }^6}{2^6·6!}+\cdots$

Short answer

The values of the series$1-\frac{{\pi }^2}{2^2·2!}+\frac{{\pi }^4}{2^4·4!}-\frac{{\pi }^6}{2^6·6!}+\cdots$is$1-\frac{{\pi }^2}{2^2·2!}+\frac{{\pi }^4}{2^4·4!}-\frac{{\pi }^6}{2^6·6!}+\cdots =0$

Step-by-step solution

Given information

The series is$1-\frac{{\pi }^2}{2^2·2!}+\frac{{\pi }^4}{2^4·4!}-\frac{{\pi }^6}{2^6·6!}+\cdots$

Find the Maclaurin series for the function f(x)=sinx 

The Maclaurin series for the function $f\left(x\right)=\sin x$is

$\cos x=1-\frac{x^2}{2!}+\frac{x^3}{3!}-\frac{x^4}{4!}+\cdots$

The series $1-\frac{{\pi }^2}{2^2·2!}+\frac{{\pi }^4}{2^4·4!}-\frac{{\pi }^6}{2^6·6!}+\cdots$is the Maclaurin series for

$\cos x$at$x=\frac{\pi }{2}$

Find the values of the series

$1-\frac{x^2}{2!}+\frac{x^3}{3!}-\frac{x^4}{4!}+\cdots =\cos x$

Therefore,

$$\begin{gathered} 1-\frac{{\pi }^2}{2^2·2!}+\frac{{\pi }^4}{2^4·4!}-\frac{{\pi }^6}{2^6·6!}+\cdots =\cos \frac{\pi }{2} \\ 1-\frac{{\pi }^2}{2^2·2!}+\frac{{\pi }^4}{2^4·4!}-\frac{{\pi }^6}{2^6·6!}+\cdots =0 \end{gathered}$$