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Q-32E

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Found in: Page 435

### 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 X0.32. f(x)=ln(1+x), x0 =0

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

See the step by step solution

## Step 1: Taylor series

For a function f(x) the Taylor series expansion about a point x0 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)= ln (1 + x) at.

Calculating the derivatives of function at x0 ,

f (x) =ln (1+x) then f (x0)=0

f'(x) = $\frac{1}{1+x}$then f '(x0) =1

f''(x) = $\frac{-1}{{\left(1+x\right)}^{2}}$ then f''(x0) = -1

f'''(x) = $\frac{2}{{\left(1+x\right)}^{3}}$then f'''(x0) = 2

f''''(x) = $\frac{-6}{{\left(1+x\right)}^{4}}$ then f''''(x0) = -6

## Step 3: Substitute the derivatives in Taylor series

Substituting the above derivatives in Taylor series expansion for the function at ${x}_{0}=0$, then,

$\mathrm{ln}\left(1+x\right)=0+1.\left(x-0\right)-1.\frac{\left(x-{0}^{2}}{2!}+2\frac{{\left(x-0\right)}^{3}}{3!}-6.\frac{{\left(x-0\right)}^{4}}{4!}+....$

= $x-\frac{{x}^{2}}{2}+\frac{{x}^{3}}{3}-\frac{{x}^{4}}{4}+...$

= role="math" localid="1664200514125" $\sum _{n-1}^{\infty }\frac{{\left(-1\right)}^{n+1}}{n}.{\left(x\right)}^{n}$

Hence, the required expression is $\sum _{n-1}^{\infty }\frac{{\left(-1\right)}^{n+1}}{n}.{\left(x\right)}^{n}$