Chapter 9, Problem 20RE

To determine

To find: The sum of the series $1-\frac{{\pi }^2}{9\cdot 2!}+\frac{{\pi }^3}{81\cdot 3!}-\mathrm{...}+{\left(-1\right)}^n\frac{{\pi }^n}{3^{2n}\cdot \left(2n\right)!}+\mathrm{...}$ and function and at what point.

Answer to Problem 20RE

Sum: $\frac{1}{2}$

Function: $f\left(x\right)=\cos x$

Point: $x=\frac{\pi }{3}$

Explanation of Solution

Given information:

  $1-\frac{{\pi }^2}{9\cdot 2!}+\frac{{\pi }^3}{81\cdot 3!}-\mathrm{...}+{\left(-1\right)}^n\frac{{\pi }^n}{3^{2n}\cdot \left(2n\right)!}+\mathrm{...}$

Concept Used:

Maclaurin Series of $f\left(x\right)={\displaystyle \sum _{n=0}^{\infty }\frac{f^n\left(0\right)}{n!}{x}^n}$

  ${\displaystyle \sum _{n=0}^{\infty }\frac{f^n\left(0\right)}{n!}{x}^n}=f\left(0\right)+\frac{f\text{'}\left(0\right)}{1!}x+\frac{f\text{'}\text{'}\left(0\right)}{2!}{x}^2+\mathrm{...}$

Calculation:

  $1-\frac{{\pi }^2}{9\cdot 2!}+\frac{{\pi }^3}{81\cdot 3!}-\mathrm{...}+{\left(-1\right)}^n\frac{{\pi }^n}{3^{2n}\cdot \left(2n\right)!}+\mathrm{...}$

First rewrite the series as summation form.

So, the summation form of the series is ${\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n\frac{{\pi }^n}{3^{2n}\cdot \left(2n\right)!}}$ .

Now compare the summation with Maclaurin series summation and yields,

  $\begin{array}{c}\cos x={\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n\frac{x^{2n}}{\left(2n\right)!}}\\ ={\displaystyle \sum _{n=0}^{\infty }{\left(-1\right)}^n\frac{{\left(\frac{\pi }{3}\right)}^{2n}}{\left(2n\right)!}}\end{array}$

So, the point is at $x=\frac{\pi }{3}$ .

The Maclaurin series of $\cos x$ is same.

So, the function is $f\left(x\right)=\cos x$ .

Substitute $x=\frac{\pi }{3}$ into the function and the sum of the series is $\frac{1}{2}$ .