Question

Question: In Problems 29–34, determine the Taylor series about the point x₀for the given functions and values of x₀.

34. f(x)=$\sqrt{x,}{x}_0=1$

Short answer

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

Step-by-step solution

Taylor series

For a function ^f(x)the Taylor series expansion about a point x₀ is given by,$f(x-x_0)=f\left(x_0\right)+f\text{'}\left(x_0\right).\left(x-x_0\right)+f"\left(x_0\right)-\frac{(x-x_0)}{2!}+f\text{'}\text{'}\text{'}\left(x_0\right)-\frac{{\left(x-x_0\right)}^3}{3!}+....$

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) =$\frac{1}{2}{x}^{-1/2}$thenf'(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) =$\frac{-15}{16}$

Substitute the derivatives in Taylor series

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

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

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

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

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