Chapter 10, Problem 19RE

To determine

To find: The function $f\left(x\right)$ , value of x and the sum of the given series.

Answer to Problem 19RE

Function is $f\left(x\right)=\sin x$ , point is $x=\pi$ and sum is $S=0$

Explanation of Solution

Given:

The given Maclaurin series for the function $f\left(x\right)$ at a particular point is

  $S=\pi -\frac{{\pi }^3}{3!}+\frac{{\pi }^5}{5!}-\mathrm{...}+{\left(-1\right)}^n\frac{{\pi }^{2n=1}}{\left(2n+1\right)!}+\mathrm{...}$

Calculation:

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

  $f\left(x\right)=\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\mathrm{....}+{\left(-1\right)}^n\frac{x^{2n+1}}{\left(2n+1\right)!}+\mathrm{...}$

Now, at $x=\pi$

  $f\left(\pi \right)=\sin \pi =\pi -\frac{{\pi }^3}{3!}+\frac{{\pi }^5}{5!}-\mathrm{....}+{\left(-1\right)}^n\frac{{\pi }^{2n+1}}{\left(2n+1\right)!}+\mathrm{...}$

This series is similar to the given series.

So, the function is

  $f\left(x\right)=\sin x$ and $x=\pi$

And the sum of the given series is,

  $\begin{array}{l}f\left(\pi \right)=\sin \pi \\ \Rightarrow S=0\end{array}$

Therefore, function is $f\left(x\right)=\sin x$ , point is $x=\pi$ and sum is $S=0$