Suggested languages for you:

Americas

Europe

Q- 31 E

Expert-verified
Found in: Page 1

### Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

# Question: In Problems 29–34, determine the Taylor series about the point X0 for the given functions and values of X0.31. $f\left(x\right)=\frac{1+x}{1-x}.$x0 = 0 ,

The required expression is $1+\sum _{n-1}^{\infty }2{\left(x\right)}^{n}.$

See the step by step solution

## Step 1: Taylor series

For a function $f\left(x\right)$ the Taylor series expansion about a point ${x}_{0}$ is given by,$f\left(x-{x}_{0}\right)=f\left({x}_{0}\right)+f\text{'}\left({x}_{0\right)}.\left(x-{x}_{0}\right)+f\text{'}\text{'}\left({x}_{0}\right).\frac{{\left(x-{x}_{0}\right)}^{2}}{2!}+f\text{'}\text{'}\text{'}\left({x}_{0}\right)\frac{{\left(x-{x}_{0}\right)}^{3}}{3!}+....$

## Step 2: Derivatives of function at x0

We have to calculate the Taylor series expansion for, f(x) = $\frac{1+x}{1-x}$ at x0=0.

The function f(x) can be further simplified for easier calculations,

$\frac{1+x}{1-x}=\frac{-\left(1+x\right)}{x-1}\phantom{\rule{0ex}{0ex}}=\frac{-\left(2+x-1\right)}{x-1}\phantom{\rule{0ex}{0ex}}=\frac{-2}{x-1}-\frac{x-1}{x-1}\phantom{\rule{0ex}{0ex}}=\frac{2}{1-x}-1$

Calculating the derivatives of function at x0 .

$f\left(x\right)=\frac{2}{1-x}-1$then $f\left({x}_{0}\right)=1$

$f\text{'}\left(x\right)=\frac{2}{{\left(1-x\right)}^{2}}$ then$f\text{'}\left({x}_{0}\right)=2$

$f\text{'}\text{'}\left({x}_{}\right)=\frac{4}{{\left(1-x\right)}^{2}}$then $f\text{'}\text{'}\left({x}_{0}\right)=4$

$f\text{'}\text{'}\text{'}\left(x\right)=\frac{12}{{\left(1-x\right)}^{4}}$ then $f\text{'}\text{'}\text{'}\left({x}_{0}\right)=12$

$f\text{'}\text{'}\text{'}\text{'}\left({x}_{0}\right)=\frac{48}{{\left(1-x\right)}^{5}}$then $f\text{'}\text{'}\text{'}\text{'}\left({x}_{0}\right)=48$

## Step 3: Substitute the derivatives in Taylor series

Substituting the above derivatives in Taylor series expansion for the function at x0=0, then,

$\frac{1+x}{1-x}=1-2.\left(x-0\right)+4\frac{{\left(x-0\right)}^{2}}{2!}-12.\frac{{\left(x-0\right)}^{3}}{3!}+48.\frac{{\left(x-0\right)}^{4}}{4!}+....$

= $1+2x+2{x}^{2}+2{x}^{3}+2{x}^{4}+....$

= $1+\sum _{n-1}^{\infty }2{\left(x\right)}^{n}$

Hence, the required expression is

## Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.