Chapter 9.2, Problem 24E

(a)

To determine

The value of $f\text{'}\left(0\right)$ and $f^{\left(10\right)}\left(0\right)$ of the function $f\left(x\right)$ .

Answer to Problem 24E

The value of $f\text{'}\left(0\right)$ for the function $f\left(x\right)$ is $f\text{'}\left(0\right)=\frac{1}{2}$ .

The value of $f^{\left(10\right)}\left(0\right)$ for the function $f\left(x\right)$ is $f^{\left(10\right)}\left(0\right)=\frac{1}{11}$ .

Explanation of Solution

Given:

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

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

Differentiate the Maclaurin series for $f\left(x\right)$ with respect $x$ .

  $\begin{array}{c}\frac{d}{dx}f\left(x\right)=\frac{d}{dx}\left(1+\frac{x}{2!}+\frac{x^2}{3!}+\frac{x^3}{4!}+\mathrm{..................}+\frac{x^n}{\left(n+1\right)!}+\mathrm{...........}\right)\\ f\text{'}\left(x\right)=\frac{1}{2!}+\frac{2x}{3!}+\frac{3x^2}{4!}+\mathrm{.................}+\frac{n{x}^{n-1}}{\left(n+1\right)!}+\mathrm{.........}\\ f\text{'}\left(0\right)=\frac{1}{2}\end{array}$

Differentiate the Maclaurin series for $f\left(x\right)$ with respect $x$ ten times.

  $\begin{array}{l}f\left(x\right)=1+\frac{x}{2!}+\frac{x^2}{3!}+\frac{x^3}{4!}+\mathrm{..................}+\frac{x^n}{\left(n+1\right)!}+\mathrm{...........}\\ f^{\left(10\right)}\left(x\right)=\frac{10!}{11!}+\frac{11!}{12!}x+\frac{12!}{13!\times 2!}+\mathrm{............}\\ f^{\left(10\right)}\left(0\right)=\frac{1}{11}\end{array}$

Conclusion:

The value of $f\text{'}\left(0\right)$ for the function $f\left(x\right)$ is $f\text{'}\left(0\right)=\frac{1}{2}$ .

The value of $f^{\left(10\right)}\left(0\right)$ for the function $f\left(x\right)$ is $f^{\left(10\right)}\left(0\right)=\frac{1}{11}$ .

(b)

To determine

The first four nonzero terms and general terms of Maclaurin series for $g\left(x\right)$ .

Answer to Problem 24E

The first four non zero terms are

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

General term of function is

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

Explanation of Solution

Given:

The given function is

  $g\left(x\right)=xf\left(x\right)$

The general term of the new function is:

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

First four non zero terms are:

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

Conclusion:

The first four nonzero terms are

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

General term of function is

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

(c)

To determine

The familiar function of $g\left(x\right)$ without using the series.

Answer to Problem 24E

The familiar function of $g\left(x\right)$ without using the series is $g\left(x\right)=e^x-1$ .

Explanation of Solution

Given:

The given function is

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

The Maclaurin series is given by:

  $F\left(x\right)=F\left(0\right)+F\text{'}\left(0\right)x+\frac{F\text{'}\text{'}\left(0\right)}{2!}{x}^2+\frac{F\text{'}\text{'}\text{'}\left(0\right)}{3!}{x}^3+\mathrm{.....................}$

On comparing coefficient of different power of the Maclaurin series with the given function:

  $\begin{array}{l}F\left(0\right)=0\\ F\text{'}\left(0\right)=1\\ \frac{F\text{'}\text{'}\left(0\right)}{2!}=\frac{1}{2!}\Rightarrow F\text{'}\text{'}\left(0\right)=1\\ \frac{F\text{'}\text{'}\left(0\right)}{3!}=\frac{1}{3!}\Rightarrow F\text{'}\text{'}\text{'}\left(0\right)=1\end{array}$

As all the derivative of function has same value at $x=a=0$ except the first value so the familiar function for this is

  $g\left(x\right)=e^x-1$

Hence, The familiar function of $g\left(x\right)$ without using the series is $g\left(x\right)=e^x-1$ .