Chapter 9.2, Problem 27E

(a)

To determine

The first four non zero terms of Taylor series for $f\left(x\right)=\sqrt{1+x}atx=0$ .

Answer to Problem 27E

The first four non zero terms of Taylor series for $f\left(x\right)=\sqrt{1+x}atx=0$ is $f\left(x\right)=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}$ .

Explanation of Solution

Given:

The given function is $f\left(x\right)=\sqrt{1+x}$ .

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 $x=a=0$ .

  $\begin{array}{c}f\left(x\right)=\sqrt{1+x}\\ f\left(0\right)=\sqrt{1+0}\\ =1\end{array}$

Differentiate the function with respect to $x$

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

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

  $\begin{array}{l}\frac{d}{dx}{f}^{\text{'}}\left(x\right)=\frac{d}{dx}\left(\frac{1}{2\sqrt{1+x}}\right)\\ f^{\text{'}\text{'}}\left(x\right)=\frac{-1}{4{\left(1+x\right)}^{3/2}}\Rightarrow f^{\text{'}\text{'}}\left(0\right)=-\frac{1}{4}\\ f^{\text{'}\text{'}\text{'}}\left(x\right)=-\frac{-3}{8{\left(1+x\right)}^{5/2}}\Rightarrow f^{\text{'}\text{'}\text{'}}\left(0\right)=\frac{3}{8}\\ f^{\left(4\right)}\left(x\right)=\frac{3\times \left(-5\right)}{16{\left(1+x\right)}^{7/2}}\Rightarrow f^{\left(4\right)}\left(0\right)=\frac{-15}{16}\\ f^{\left(5\right)}\left(x\right)=\frac{-15\times \left(-7\right)}{32{\left(1+x\right)}^{9/2}}\Rightarrow f^{\left(5\right)}\left(0\right)=\frac{105}{32}\end{array}$

The first four non zero terms of function are:

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

(b)

To determine

The first four non zero terms of Taylor series for $g\left(x\right)=\sqrt{1+x^2}atx=0$ .

Answer to Problem 27E

The first four non zero terms of Taylor series for $g\left(x\right)=\sqrt{1+x^2}atx=0$ is $g\left(x\right)=1+\frac{x^2}{2}-\frac{x^4}{8}-\frac{x^6}{240}$ .

Explanation of Solution

Given:

The given function is $g\left(x\right)=\sqrt{1+x^2}$ .

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 $x=a=0$ .

  $\begin{array}{c}g\left(x\right)=\sqrt{1+x^2}\\ g\left(0\right)=\sqrt{1+0}\\ =1\end{array}$

Differentiate the function with respect to $x$

  $\begin{array}{c}\frac{d}{dx}g\left(x\right)=\frac{d}{dx}\left(\sqrt{1+x^2}\right)\\ g\text{'}\left(x\right)=\frac{1}{2\left(\sqrt{1+x^2}\right)}\times 2x\mathrm{.........}\left(1\right)\\ g\text{'}\left(0\right)=0\end{array}$

Differentiate the equation $\left(1\right)$ with respect $x$ five times.

  $\begin{array}{l}\frac{d}{dx}{g}^{\text{'}}\left(x\right)=\frac{d}{dx}\left(\frac{x}{\sqrt{1+x^2}}\right)\\ g^{\text{'}\text{'}}\left(x\right)=\frac{2\left(1+x^2\right)-x\left(2x\right)}{2{\left(1+x^2\right)}^{3/2}}=\frac{1}{{\left(1+x^2\right)}^{3/2}}\Rightarrow g^{\text{'}\text{'}}\left(0\right)=1\\ g^{\text{'}\text{'}\text{'}}\left(x\right)=\frac{-3}{2{\left(1+x^2\right)}^{5/2}}\left(2x\right)=\frac{-3x}{{\left(1+x^2\right)}^{5/2}}\Rightarrow g^{\text{'}\text{'}\text{'}}\left(0\right)=0\\ g^{\left(4\right)}\left(x\right)=\frac{-3}{{\left(1+x^2\right)}^{5/2}}+\frac{-3x}{{\left(1+x^2\right)}^{7/2}}\left(\frac{-5}{2}\right)\left(2x\right)=\frac{-3+12x^2}{{\left(1+x^2\right)}^{7/2}}\Rightarrow g^{\left(4\right)}\left(0\right)=-3\\ g^{\left(5\right)}\left(x\right)=\frac{24x}{{\left(1+x^2\right)}^{7/2}}+\frac{\left(-3+12x^2\right)}{{\left(1+x^2\right)}^{9/2}}\left(\frac{9}{2}\right)\left(2x\right)=\frac{3x\left(-1+44x^2\right)}{{\left(1+x^2\right)}^{9/2}}\Rightarrow g^{\left(5\right)}\left(0\right)=0\\ g^{\left(6\right)}\left(0\right)=-3\end{array}$

The first four non zero terms of function are:

  $\begin{array}{c}g\left(x\right)=g\left(0\right)+g^{\text{'}}\left(0\right)\left(x\right)+\frac{g^{\text{'}\text{'}}\left(0\right)}{2!}{\left(x\right)}^2+\frac{g^{\text{'}\text{'}\text{'}}\left(0\right)}{3!}{\left(x\right)}^3+\frac{g^{\left(4\right)}\left(0\right)}{4!}{\left(x\right)}^4+\frac{g^{{}^{\left(5\right)}}\left(0\right)}{5!}{\left(x\right)}^5+\frac{g^{{}^{\left(6\right)}}\left(0\right)}{6!}{\left(x\right)}^6\mathrm{............}\\ =1+\frac{x^2}{2!}-3\frac{x^4}{4!}-3\frac{x^6}{6!}\\ =1+\frac{x^2}{2}-\frac{x^4}{8}-\frac{x^6}{240}\end{array}$

(c)

To determine

The first four nonzero value for the function $h$ .

Answer to Problem 27E

The first four nonzero value for the function $h$ is $h\left(x\right)=5+x-\frac{x^3}{3}+\frac{x^5}{5}$ .

Explanation of Solution

Given:

The differentiate form of function is

  $h\text{'}\left(x\right)=\sqrt{1+x^2}$

Initial value of function is $h\left(0\right)=5$

On integrating the given function.

  $\begin{array}{c}{\displaystyle \int h\text{'}\left(x\right)}={\displaystyle \int \sqrt{1+x^2}}dx\\ h\left(x\right)={\tan }^{-1}x+C\\ h\left(0\right)=5={\tan }^{-1}0+C\\ C=5\\ hence\\ h\left(x\right)={\tan }^{-1}x+5\end{array}$

Taylor expansion of the function $h\left(x\right)$ is

  $\begin{array}{c}h\left(x\right)={\tan }^{-1}x+5\\ =x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\mathrm{..........}+5\\ =5+x-\frac{x^3}{3}+\frac{x^5}{5}\end{array}$

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 $\left(2\right)$

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