Chapter 10.2, Problem 25E

(a)

To determine

The first three non-zero terms.

Answer to Problem 25E

The value is $e^{x/2}=1+\frac{x}{2}+\frac{x^2}{8}+\mathrm{.....}+\frac{x^n}{2^n.n!}+\mathrm{.....}$

Explanation of Solution

Given information:

The general term of the Taylor series generated by

  $e^{x/2}atx=0$

Formula used:

  $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\mathrm{.....}+\frac{x^n}{n!}+\mathrm{.....}$

Calculation:

Maclaurin series of a function $e^x$ is given by

  $e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\mathrm{.....}+\frac{x^n}{n!}+\mathrm{.....}$

Substitute $\frac{x}{2}$ for x in the Maclaurin series for $e^x$

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

  $e^{x/2}=1+\frac{x}{2}+\frac{x^2}{8}+\mathrm{.....}+\frac{x^n}{2^n.n!}+\mathrm{.....}$

Conclusion:

The value is $e^{x/2}=1+\frac{x}{2}+\frac{x^2}{8}+\mathrm{.....}+\frac{x^n}{2^n.n!}+\mathrm{.....}$

(b)

To determine

The first three non-zero terms using the power series.

Answer to Problem 25E

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

Explanation of Solution

Given information:

The general term of a power series to represent

  $g\left(x\right)=\frac{e^x-1}{x}$

Formula used:

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

Calculation:

Given,

  $g\left(x\right)=\frac{e^x-1}{x}$

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

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

This can also be written as

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

Conclusion:

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

(c)

To determine

The value of function g.

Answer to Problem 25E

The value is ${\displaystyle \sum _{n=1}^{\infty }\frac{n}{\left(n+1\right)!}=1}$

Explanation of Solution

Given information:

The given function is

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

Formula used:

Differentiation method is used.

Calculation:

  $g\left(x\right)=\frac{e^x-1}{x}$

Differentiating $g\left(x\right)$ with respect to x

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

Putting $x=1$

  $g\text{'}\left(1\right)=\mathrm{1.........}\left(1\right)$

Since,

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

Differentiating both the sides with respect to x,

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

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

Putting $x=1$

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

From equation (1) and equation (2)

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

Therefore,

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

Conclusion:

The value is ${\displaystyle \sum _{n=1}^{\infty }\frac{n}{\left(n+1\right)!}=1}$