Q 25 - Calculus

Question

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

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

Step-by-Step Solution

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 1: Given information

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

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 

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

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

Step 3: Find the values of the series.

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

Therefore,

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