Question

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

$\pi -\frac{{\pi }^3}{3!}+\frac{{\pi }^5}{5!}-\frac{{\pi }^7}{7!}+\cdots$

Short answer

The values of the series$\pi -\frac{{\pi }^3}{3!}+\frac{{\pi }^5}{5!}-\frac{{\pi }^7}{7!}+\cdots$is$\pi -\frac{{\pi }^3}{3!}+\frac{{\pi }^5}{5!}-\frac{{\pi }^7}{7!}+\cdots =0$

Step-by-step solution

Given information

The series is$\pi -\frac{{\pi }^3}{3!}+\frac{{\pi }^5}{5!}-\frac{{\pi }^7}{7!}+\cdots$

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

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

$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots$

The series$\pi -\frac{{\pi }^3}{3!}+\frac{{\pi }^5}{5!}-\frac{{\pi }^7}{7!}+\cdots$is the Maclaurin series for$\sin x$at$x=\pi$

Find the values of the series. 

$x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots =\sin x$

Therefore,

$$\begin{gathered} \pi -\frac{{\pi }^3}{3!}+\frac{{\pi }^5}{5!}-\frac{{\pi }^7}{7!}+\cdots =\sin \pi \\ \pi -\frac{{\pi }^3}{3!}+\frac{{\pi }^5}{5!}-\frac{{\pi }^7}{7!}+\cdots =0 \end{gathered}$$