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Q-34E

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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 .34. f(x)=$\sqrt{x,}{}{{x}}_{{0}}{=}{1}$

The required expression is$1+\frac{1}{2}\left(X-1\right)+\sum _{n-2}^{\infty }\frac{{\left(-1\right)}^{n+1}\left(2n-3\right)!}{{2}^{2n-2}\left(n-2\right)!\left(n\right)!}{\left(x-1\right)}^{n}$

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,$f\left(x-{x}_{0}\right)=f\left({x}_{0}\right)+f\text{'}\left({x}_{0}\right).\left(x-{x}_{0}\right)+f"\left({x}_{0}\right)-\frac{\left(x-{x}_{0}\right)}{2!}+f\text{'}\text{'}\text{'}\left({x}_{0}\right)-\frac{{\left(x-{x}_{0}\right)}^{3}}{3!}+....$

## Step 2: Derivatives of function at x0

We have to calculate the Taylor series expansion for,$f\left(x\right)=\sqrt{x}$ at x0 =1 .

Calculating the derivatives of function at x0 .

f (x) =$\sqrt{x}$then f (x0) =1

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

f''(x) = $\frac{-1}{4}{x}^{-3/2}$then f''(x0) =$\frac{-1}{4}$

f'''(x) = $\frac{3}{8}{x}^{-5/2}$ then f'''(x0) = $\frac{3}{8}$

f''''(x) = $\frac{-15}{16}{x}^{-7/2}$ then f''''(x0) = $\frac{-15}{16}$

## Step 3: Substitute the derivatives in Taylor series

Substituting the above derivatives in Taylor series expansion for the function at x0=1, then,

$\sqrt{x}=1+\frac{1}{2}\left(x-1\right)-\frac{1}{4}\frac{{\left(x-1\right)}^{2}}{2!}+\frac{3}{8}\frac{{\left(x-1\right)}^{3}}{3!}-\frac{15}{16}\frac{{\left(x-1\right)}^{4}}{4!}+....$

= $1+\frac{1}{2}\left(x-1\right)+\sum _{n-2}^{\infty }\frac{{\left(-1\right)}^{n+1}}{{2}^{n}}\frac{\left(2n-3\right)!}{{2}^{n}{2}^{n-2}\left(n-2\right)!\left(n\right)!}{\left(x-1\right)}^{n}$

= role="math" localid="1664103079471" $1+\frac{1}{2}\left(x-1\right)+\sum _{n-2}^{\infty }\frac{{\left(-1\right)}^{n+1}\left(2n-3\right)!}{{2}^{2n-2}\left(n-2\right)!\left(n\right)!}{\left(x-1\right)}^{n}$

Hence, the required expression is $1+\frac{1}{2}\left(x-1\right)+\sum _{n-2}^{\infty }\frac{{\left(-1\right)}^{n+1}\left(2n-3\right)!}{{2}^{2n-2}\left(n-2\right)!\left(n\right)!}{\left(x-1\right)}^{n}$