Chapter 10, Problem 22RE

To determine

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

Answer to Problem 22RE

Function is $f\left(x\right)={\tan }^{-1}x$ , point is $x=\frac{1}{\sqrt{3}}$ and sum is $S=\frac{\pi }{6}$

Explanation of Solution

Given:

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

  $S=\frac{1}{\sqrt{3}}-\frac{1}{9\sqrt{3}}+\frac{1}{45\sqrt{3}}-\mathrm{...}+{\left(-1\right)}^{n-1}\frac{1}{\left(2n-1\right){\left(\sqrt{3}\right)}^{2n-1}}+\mathrm{...}$

Calculation:

As, the Maclaurin series for the function $f\left(x\right)={\tan }^{-1}x$ is

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

Now, at $x=\frac{1}{\sqrt{3}}$

  $f\left(\frac{1}{\sqrt{3}}\right)={\tan }^{-1}\frac{1}{\sqrt{3}}=\frac{1}{\sqrt{3}}-\frac{1}{9\sqrt{3}}+\frac{1}{45\sqrt{3}}-\mathrm{...}+{\left(-1\right)}^{n-1}\frac{1}{\left(2n-1\right){\left(\sqrt{3}\right)}^{2n-1}}+\mathrm{...}$

This series is similar to the given series.

So, the function is

  $f\left(x\right)={\tan }^{-1}x$ and $x=\frac{1}{\sqrt{3}}$

And the sum of the given series is,

  $\begin{array}{l}f\left(\frac{1}{\sqrt{3}}\right)={\tan }^{-1}\frac{1}{\sqrt{3}}\\ \Rightarrow S=\frac{\pi }{6}\end{array}$

Therefore, function is $f\left(x\right)={\tan }^{-1}x$ , point is $x=\frac{1}{\sqrt{3}}$ and sum is $S=\frac{\pi }{6}$