Chapter 9.2, Problem 25E

(a)

To determine

To find: The first four nonzero terms and general terms of Taylor series for $e^{x/2}atx=0$ .

Answer to Problem 25E

The first four nonzero terms are

  $f\left(x\right)=1+\frac{x}{2}+\frac{x^2}{8}+\frac{x^3}{48}$

General term of function is

  $f\left(x\right)=1+\frac{x}{2}+\frac{x^2}{8}+\frac{x^3}{48}+\frac{x^4}{384}+\mathrm{.............}$

Explanation of Solution

Given:

The given function is $f\left(x\right)=e^{x/2}$ .

The Taylor series for $f\left(x\right)$ is

  $\begin{array}{c}f\left(x\right)={\displaystyle \sum _{n=0}^{\infty }\frac{f^{\left(n\right)}\left(a\right)}{n!}}{\left(x-a\right)}^n\\ =f\left(a\right)+f\text{'}\left(a\right)\left(x-a\right)+\frac{f\text{'}\text{'}\left(a\right)}{2!}{\left(x-a\right)}^2+\frac{f\text{'}\text{'}\text{'}\left(a\right)}{3!}{\left(x-a\right)}^3+\mathrm{............}\end{array}$

The value of function at $x=a=0$ .

  $\begin{array}{c}f\left(x\right)=e^{x/2}\\ f\left(0\right)=e^{0/2}\\ =1\end{array}$

Differentiate the function with respect to $x$

  $\begin{array}{c}\frac{d}{dx}f\left(x\right)=\frac{d}{dx}\left(e^{x/2}\right)\\ f\text{'}\left(x\right)=\frac{\left(e^{x/2}\right)}{2}\mathrm{.........}\left(1\right)\\ f\text{'}\left(0\right)=\frac{1}{2}\end{array}$

Differentiate the equation $2$ with respect $x$ five times.

  $\begin{array}{l}\frac{d}{dx}{f}^{\text{'}}\left(x\right)=\frac{d}{dx}\left(\frac{e^{x/2}}{2}\right)\\ f^{\text{'}\text{'}}\left(x\right)=\frac{e^{x/2}}{4}\Rightarrow f^{\text{'}\text{'}}\left(0\right)=\frac{1}{4}\\ f^{\text{'}\text{'}\text{'}}\left(x\right)=\frac{e^{x/2}}{8}\Rightarrow f^{\text{'}\text{'}\text{'}}\left(0\right)=\frac{1}{8}\\ f^{\left(4\right)}\left(x\right)=\frac{e^{x/2}}{16}\Rightarrow f^{\left(4\right)}\left(0\right)=\frac{1}{16}\\ f^{\left(5\right)}\left(x\right)=\frac{e^{x/2}}{32}\Rightarrow f^{\left(5\right)}\left(0\right)=\frac{1}{32}\end{array}$

The first four non zero terms of function are:

  $\begin{array}{c}f\left(x\right)=f\left(0\right)+f^{\text{'}}\left(0\right)\left(x\right)+\frac{f^{\text{'}\text{'}}\left(0\right)}{2!}{\left(x\right)}^2+\frac{f^{\text{'}\text{'}\text{'}}\left(a\right)}{3!}{\left(x\right)}^3+\mathrm{............}\\ =1+\frac{x}{2}+\frac{1}{4}\times \frac{x^2}{2!}+\frac{1}{8}\times \frac{x^3}{3!}\\ =1+\frac{x}{2}+\frac{x^2}{8}+\frac{x^3}{48}\end{array}$

The general term of Taylor series of function

  $\begin{array}{c}f\left(x\right)=f\left(0\right)+f^{\text{'}}\left(0\right)\left(x\right)+\frac{f^{\text{'}\text{'}}\left(0\right)}{2!}{\left(x\right)}^2+\frac{f^{\text{'}\text{'}\text{'}}\left(a\right)}{3!}{\left(x\right)}^3+\mathrm{............}\\ =1+\frac{x}{2}+\frac{1}{4}\times \frac{x^2}{2!}+\frac{1}{8}\times \frac{x^3}{3!}+\frac{1}{16}\times \frac{x^4}{4!}+\mathrm{..........}\\ =1+\frac{x}{2}+\frac{x^2}{8}+\frac{x^3}{48}+\frac{x^4}{384}+\mathrm{.............}\end{array}$

Conclusion:

The first four nonzero terms are

  $f\left(x\right)=1+\frac{x}{2}+\frac{x^2}{8}+\frac{x^3}{48}$

General term of function is

  $f\left(x\right)=1+\frac{x}{2}+\frac{x^2}{8}+\frac{x^3}{48}+\frac{x^4}{384}+\mathrm{.............}$

(b)

To determine

To find: The first four nonzero terms and general terms of Taylor series for $g\left(x\right)=\frac{e^x-1}{x}atx=0$ .

Answer to Problem 25E

The first four non zero terms are

  $g\left(x\right)=1+\frac{x}{2!}+\frac{x^2}{3!}+\frac{x^3}{4!}$

General term of function is

  $g\left(x\right)=1+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\mathrm{..................}+\frac{x^{n+1}}{\left(n+1\right)!}+\mathrm{...........}$

Explanation of Solution

Given:

The given function is

  $g\left(x\right)=\frac{e^x-1}{x}atx=0$

The Taylor series for $f\left(x\right)$ is

  $\begin{array}{c}f\left(x\right)={\displaystyle \sum _{n=0}^{\infty }\frac{f^{\left(n\right)}\left(a\right)}{n!}}{\left(x-a\right)}^n\\ =f\left(a\right)+f\text{'}\left(a\right)\left(x-a\right)+\frac{f\text{'}\text{'}\left(a\right)}{2!}{\left(x-a\right)}^2+\frac{f\text{'}\text{'}\text{'}\left(a\right)}{3!}{\left(x-a\right)}^3+\mathrm{............}\end{array}$

The Taylor series expansion of $e^{x}atx=0$

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

Therefore,

  $\begin{array}{c}{e}^x-1=x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}\\ \frac{e^x-1}{x}=\frac{1}{x}\left(x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}\right)\\ =1+\frac{x}{2!}+\frac{x^2}{3!}+\frac{x^3}{4!}\end{array}$

The general term for the function is

  $\begin{array}{c}\frac{e^x-1}{x}=1+\frac{x}{2!}+\frac{x^2}{3!}+\frac{x^3}{4!}+\mathrm{........}\\ =1+\frac{x}{2!}+\frac{x^2}{3!}+\frac{x^3}{4!}+\mathrm{........}\frac{x^n}{\left(n+1\right)!}\end{array}$

Conclusion:

The first four nonzero terms are

  $g\left(x\right)=1+\frac{x}{2!}+\frac{x^2}{3!}+\frac{x^3}{4!}$

General term of function is

  $g\left(x\right)=1+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\mathrm{..................}+\frac{x^{n+1}}{\left(n+1\right)!}+\mathrm{...........}$

(c)

To determine

To find: The value of $g\text{'}\left(1\right)$ for the function $g\left(x\right)=\frac{e^x-1}{x}$ and prove that ${\displaystyle \sum _{n=1}^{\infty }\frac{n}{\left(n+1\right)!}=1}$ .

Answer to Problem 25E

The value of $g\text{'}\left(1\right)$ for the function $g\left(x\right)=\frac{e^x-1}{x}$ is $g\text{'}\left(1\right)=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\mathrm{.......}\frac{n}{\left(n+1\right)!}$ .

It is proved that $g\text{'}\left(1\right)={\displaystyle \sum _{n=1}^{\infty }\frac{n}{\left(n+1\right)!}}=1$ .

Explanation of Solution

Given:

The general term for the function is

  $\begin{array}{c}\frac{e^x-1}{x}=1+\frac{x}{2!}+\frac{x^2}{3!}+\frac{x^3}{4!}+\mathrm{........}\\ =1+\frac{x}{2!}+\frac{x^2}{3!}+\frac{x^3}{4!}+\mathrm{........}\frac{x^n}{\left(n+1\right)!}\end{array}$

On differentiating the function.

  $\begin{array}{l}\frac{d}{dx}\left(\frac{e^x-1}{x}\right)=\frac{d}{dx}\left(1+\frac{x}{2!}+\frac{x^2}{3!}+\frac{x^3}{4!}+\mathrm{........}\frac{x^n}{\left(n+1\right)!}\right)\\ g\text{'}\left(x\right)=\frac{1}{2!}+\frac{2x}{3!}+\frac{3x^2}{4!}+\mathrm{.......}\frac{n{x}^{n-1}}{\left(n+1\right)!}\\ g\text{'}\left(1\right)=\frac{1}{2!}+\frac{2\times 1}{3!}+\frac{3\times 1^2}{4!}+\mathrm{.......}\frac{n\times 1^{n-1}}{\left(n+1\right)!}\\ g\text{'}\left(1\right)=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\mathrm{.......}\frac{n}{\left(n+1\right)!}\end{array}$

The general term of $g\text{'}\left(1\right)$ for the function $g\left(x\right)=\frac{e^x-1}{x}$ is

  $\begin{array}{c}g\text{'}\left(1\right)=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\mathrm{.......}\frac{n}{\left(n+1\right)!}\\ ={\displaystyle \sum _{n=1}^{\infty }\frac{n}{\left(n+1\right)!}}\mathrm{................}\left(1\right)\end{array}$

On differentiating the function $g\left(x\right)=\frac{e^x-1}{x}$

  $\begin{array}{c}\frac{d}{dx}\left(g\left(x\right)\right)=\frac{d}{dx}\left(\frac{e^x-1}{x}\right)\\ g\text{'}\left(x\right)=\frac{e^x\left(x\right)-1\left(e^x-1\right)}{x^2}\\ g\text{'}\left(1\right)=\frac{e^1\left(1\right)-1\left(e^1-1\right)}{1^2}\\ g\text{'}\left(1\right)=1\mathrm{...............}\left(2\right)\end{array}$

From $\left(1\right)$ and $\left(2\right)$

  $g\text{'}\left(1\right)={\displaystyle \sum _{n=1}^{\infty }\frac{n}{\left(n+1\right)!}}=1$

Conclusion:

The value of $g\text{'}\left(1\right)$ for the function $g\left(x\right)=\frac{e^x-1}{x}$ is $g\text{'}\left(1\right)=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\mathrm{.......}\frac{n}{\left(n+1\right)!}$ .

It is proved that $g\text{'}\left(1\right)={\displaystyle \sum _{n=1}^{\infty }\frac{n}{\left(n+1\right)!}}=1$ .