Q 27 - Calculus

Question

Use the Maclaurin series for $\sin x$,

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

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

Step-by-Step Solution

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 1: Given information

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

Step 2: Find the Maclaurin series for the function f ( x ) = sin x  

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 $x=\frac{\pi }{2}$

Step 3: Find the values of the series

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

Therefore,

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