Q 26 - Calculus

Question

Use the Maclaurin series for $\sin x$,cosx, and  $e^x$to find the values of the following series.
$\pi -\frac{{\pi }^2}{2!}+\frac{{\pi }^3}{3!}-\frac{{\pi }^4}{4!}+\frac{{\pi }^5}{5!}-\cdots$

Step-by-Step Solution

Answer

The values of the series $\pi -\frac{{\pi }^2}{2!}+\frac{{\pi }^3}{3!}-\frac{{\pi }^4}{4!}+\frac{{\pi }^5}{5!}-\cdots$is $\pi -\frac{{\pi }^2}{2!}+\frac{{\pi }^3}{3!}-\frac{{\pi }^4}{4!}+\frac{{\pi }^5}{5!}-\cdots =1-e^{-\pi }$

Step 1: Given information

The series is $\pi -\frac{{\pi }^2}{2!}+\frac{{\pi }^3}{3!}-\frac{{\pi }^4}{4!}+\frac{{\pi }^5}{5!}-\cdots$

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

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

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

So, $$\begin{gathered} e^{-x}=1-x+\frac{x^2}{2!}-\frac{x^3}{3!}+\frac{x^4}{4!}-\cdots \\ -e^{-x}=-1+x-\frac{x^2}{2!}+\frac{x^3}{3!}-\frac{x^4}{4!}+\cdots \end{gathered}$$

Or,

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

The series $\pi -\frac{{\pi }^2}{2!}+\frac{{\pi }^3}{3!}-\frac{{\pi }^4}{4!}+\frac{{\pi }^5}{5!}-\cdots$is the Maclaurin series for $1-e^{-x}$at $x=\pi$

Step 3: Find the values of the series.

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

Therefore,

$$\begin{gathered} \pi -\frac{{\pi }^2}{2!}+\frac{{\pi }^3}{3!}-\frac{{\pi }^4}{4!}+\frac{{\pi }^5}{5!}-\cdots =1-e^{-\pi } \\ \pi -\frac{{\pi }^2}{2!}+\frac{{\pi }^3}{3!}-\frac{{\pi }^4}{4!}+\frac{{\pi }^5}{5!}-\cdots =1-e^{-\pi } \end{gathered}$$