Question

$\mathit{f}\mathbf{\left(}\mathit{x}\mathbf{\right)}\mathbf{=}\mathbf{\surd }\mathit{x}\mathbf{,}\mathit{a}\mathbf{=}\mathbf{25}\mathbf{.}$

Short answer

The first few terms of Taylor’s series expansion are:

$\sqrt{x}=5+\left(\frac{x-25}{10}\right)-\frac{(x-25{)}^2}{1000}+\frac{(x-25{)}^3}{50000}+\dots$

Step-by-step solution

Given Information

The given function.

$f\left(x\right)=\surd x$

Definition of Taylor’s series

The Taylor’s series is a power series that yields the expansion of a function in the vicinity of a point if the function is continuous, all of its derivatives exist, and the series converges.

Find the Taylor’s series for x

Rewrite $\sqrt{x}$,

$$\begin{gathered} \sqrt{x}=\sqrt{25+(x-25)} \\ \sqrt{x}=5\sqrt{1+\frac{(x-25)}{25}} \end{gathered}$$

Use the Maclaurin series of $\sqrt{1+x}$and replace $x$by $\frac{x-25}{25}$,

localid="1657421695758" $$\begin{gathered} \sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16}-\frac{5x^4}{128}+\dots \\ \sqrt{1+\frac{(x-25)}{25}}=1+\frac{\left(\frac{x-25}{25}\right)}{2}-\frac{{\left(\frac{x-25}{25}\right)}^2}{8}+\frac{{\left(\frac{x-25}{25}\right)}^3}{16}+\dots \\ \sqrt{x}=5\times \sqrt{1+\frac{(x-25)}{25}} \end{gathered}$$

$=5+\left(\frac{x-25}{10}\right)-\frac{(x-25{)}^2}{1000}+\frac{(x-25{)}^3}{50000}+\dots$

Hence, first few terms of Taylor’s series expansion about $a=25$are:

$\sqrt{x}=5+\left(\frac{x-25}{10}\right)-\frac{(x-25{)}^2}{1000}+\frac{(x-25{)}^3}{50000}+\dots$