Chapter 10, Problem 71RE

a.

To determine

To find: the first three terms and the general term for the Taylor series for f at x=1.

Answer to Problem 71RE

1^st term = $\frac{1}{2}$ , 2^nd term = $\frac{-1}{4}$ , 3^rd term = $\frac{1}{8}$

General Term

  $f(x)={\displaystyle \sum _{n=1}^{\infty }\frac{{(-1)}^{n+1}{(x-1)}^{n-1}}{2^n}}$

Explanation of Solution

Given information: given function f is

  $f(x)=\frac{1}{x+1}$

Concept Used:- Taylor series for f(x) at $x=a$ is

  $f(x)=f(a)+f\text{'}(a)\frac{(x-a)}{1!}+f\text{'}\text{'}(a)\frac{{(x-a)}^2}{2!}+\mathrm{.....}$ ------ (1)

Binomial Expansion of ${\left(1+x\right)}^n$ is

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

Calculation:

  $\begin{array}{l}f(x)=\frac{1}{x+1}=\frac{1}{x-1+1+1}=\frac{1}{(x-1)+2}\\ f(x)=\frac{1}{2\left(1+\frac{(x-1)}{2}\right)}\\ f(x)=\frac{1}{2}{\left(1+\frac{(x-1)}{2}\right)}^{-1}\\ f(x)=\frac{1}{2}\left(1-\frac{(x-1)}{2}+{\left(\frac{(x-1)}{2}\right)}^2-{\left(\frac{(x-1)}{2}\right)}^3+\mathrm{.....}\right)\\ f(x)=\frac{1}{2}\left(1-\frac{(x-1)}{2}+\frac{{(x-1)}^2}{4}-{\frac{(x-1)}{8}}^3+\mathrm{.....}\right)\\ f(x)=\left(\frac{1}{2}-\frac{(x-1)}{4}+\frac{{(x-1)}^2}{8}-{\frac{(x-1)}{16}}^3+\mathrm{.....}\right)---(2)\\ f(x)={\displaystyle \sum _{n=1}^{\infty }\frac{{(-1)}^{n+1}{(x-1)}^{n-1}}{2^n}}x=a\end{array}$

On comparing with equation (a), we get

1^st term = $\frac{1}{2}$ , 2^nd term = $\frac{-1}{4}$ , 3^rd term = $\frac{1}{8}$

General Term

  $f(x)={\displaystyle \sum _{n=1}^{\infty }\frac{{(-1)}^{n+1}{(x-1)}^{n-1}}{2^n}}$

b.

To determine

To find: the radius of convergence.

Answer to Problem 71RE

  $R=2$

Explanation of Solution

Given information: by sub part (a)

  $f(x)={\displaystyle \sum _{n=1}^{\infty }\frac{{(-1)}^{n+1}{(x-1)}^{n-1}}{2^n}}$

Concept used:-

  $\frac{1}{R}={\lim }_{n\to \infty }\sup \left|{\left(a_n\right)}^{\frac{1}{n}}\right|$

Calculation:-

  $\begin{array}{l}\frac{1}{R}={\lim }_{n\to \infty }\sup \left|{\left(a_n\right)}^{\frac{1}{n}}\right|\\ Here,a_n=\frac{{(-1)}^{n+1}}{2^n}\\ \frac{1}{R}=\frac{1}{2}\\ R=2\end{array}$

c.

To determine

To find: third order polynomial for f(x) at $x=1$ and use it to approximate $x=0.5$ .

Answer to Problem 71RE

  $f(x)=\left(\frac{1}{2}-\frac{(x-1)}{4}+\frac{{(x-1)}^2}{8}-{\frac{(x-1)}{16}}^3\right)$

  $f(0.5)=0.695$

Explanation of Solution

Given information:

By subpart (a) and equation (1), third order Taylor polynomial of f(x) at $x=a$ is

  $f(x)=f(a)+f\text{'}(a)\frac{(x-a)}{1!}+f\text{'}\text{'}(a)\frac{{(x-a)}^2}{2!}+f\text{'}\text{'}\text{'}(a)\frac{{(x-a)}^3}{3!}$

Calculation:

Now, by equation (2), we have third order Taylor polynomial of f(x) at $x=1$ is

  $f(x)=\left(\frac{1}{2}-\frac{(x-1)}{4}+\frac{{(x-1)}^2}{8}-{\frac{(x-1)}{16}}^3\right)$ ---- (3)

Now, approximating $x=0.5$ using above Taylor polynomial that is, in equation (3) put $x=0.5=\frac{1}{2}$

  $\begin{array}{l}f(x)=\left(\frac{1}{2}-\frac{(x-1)}{4}+\frac{{(x-1)}^2}{8}-{\frac{(x-1)}{16}}^3\right)---(2)\\ f(0.5)=f\left(\frac{1}{2}\right)=\left(\frac{1}{2}-\frac{\left(\frac{1}{2}-1\right)}{4}+\frac{{\left(\frac{1}{2}-1\right)}^2}{8}-{\frac{\left(\frac{1}{2}-1\right)}{16}}^3\right)\\ f(0.5)=\left(\frac{1}{2}-\frac{\left(\frac{-1}{2}\right)}{4}+\frac{{\left(\frac{-1}{2}\right)}^2}{8}-{\frac{\left(-\frac{1}{2}\right)}{16}}^3\right)\\ f(0.5)=\left(\frac{1}{2}+\frac{1}{8}+\frac{1}{32}+\frac{1}{128}\right)\\ f(0.5)=\frac{1}{2}\left(1+\frac{1}{4}+\frac{1}{8}+\frac{1}{64}\right)\\ f(0.5)=\frac{1}{2}\left(\frac{64+16+8+1}{64}\right)=0.6953125\\ f(0.5)\approx 0.695\end{array}$