Question

In Exercises 49–56 find the Taylor series for the specified function and the given value of $x_0$. Note: These are the same functions and values as in Exercises 41–48.

$52.\sqrt{x},1$

Short answer

The Taylor series for the function $f\left(x\right)=\sqrt{x}$at$x=1$is$P_n\left(x\right)=1+\frac{1}{2}(x-1)+\sum _{k=2}^{\mathrm{\infty }}(-1{)}^{k+1}\frac{1\cdot 3\cdot 5\cdots (2k-3)}{2^{k}k!}(x-1{)}^k$

Step-by-step solution

Step 1. Given data

We have the function$f\left(x\right)=\sqrt{x}$.

Step 2. Table of the taylor series 

Any function $f$with a derivative of order $n$, the taylor series at $x=1$ is given by,

$P_n\left(x\right)=f\left(1\right)+f^{\mathrm{\prime }}\left(1\right)(x-1)+\frac{f^{\prime \prime }\left(1\right)}{2!}(x-1{)}^2+\frac{f^{\prime \prime \mathrm{\prime }}\left(1\right)}{3!}(x-1{)}^3+\frac{f^{\prime \prime \prime \prime }\left(1\right)}{4!}(x-1{)}^4+\dots$

we can write the general of the Taylor series of the function $f$is,

$P_n\left(x\right)=\sum _{k=0}^{\mathrm{\infty }}\frac{f^k\left(x_0\right)}{k!}{(x-x_0)}^n$

So, let us first construct the table of the Taylor series for the function $f\left(x\right)=\sqrt{x}$at $x=1$

n	$f^n\left(x\right)$	$f^n\left(0\right)$	$\frac{f^n\left(0\right)}{n!}$
0	$\sqrt{x}$	1	1
1	$\frac{1}{2\sqrt{x}}$	$\frac{1}{2}$	$\frac{1}{2}$
2	$-\frac{1}{2^2}(x{)}^{-\frac{3}{2}}$	$-\frac{1}{2^2}$	$\frac{-\frac{1}{2^2}}{2!}$
3	$\frac{1\cdot 3}{2^3}(x{)}^{-\frac{5}{2}}$	$\frac{1\cdot 3}{2^3}$	$\frac{\frac{1\cdot 3}{2^3}}{3!}$
4	$-\frac{1\cdot 3\cdot 5}{2^4}(x{)}^{-\frac{7}{2}}$	$-\frac{1\cdot 3\cdot 5}{2^4}$	$\frac{{\displaystyle -\frac{1\cdot 3\cdot 5}{2^4}}}{4!}$
. . .	. . .	. . .	. . .
k	$(-1{)}^{k+1}\frac{1\cdot 3\cdot 5\cdots (2k-3)}{2^4}(1-x{)}^{-\left(\frac{2k-1}{2}\right)}$	$(-1{)}^{k+1}\frac{1\cdot 3\cdot 5\cdots (2k-3)}{2^4}$	$(-1{)}^{k+1}\frac{{\displaystyle \frac{1\cdot 3\cdot 5\cdots (2k-3)}{2^4}}}{k!}$

Step 3. Taylor series for the f(x)=x

The Taylor series for the function $f\left(x\right)=\sqrt{x}$at $x=1$ is$P_n\left(x\right)=\begin{array}{rl}1& +\frac{1}{2}\cdot (x-1)+\frac{-\frac{1}{2^2}}{2!}(x-1{)}^2+\frac{\frac{1\cdot 3}{2^3}}{3!}(x-1{)}^3+\frac{-\frac{1\cdot 3\cdot 5}{2^4}}{4!}(x-1{)}^4+....\dots +\frac{(-1{)}^{k+1}\frac{1\cdot 3\cdot 5\cdots (2k-3)}{2^k}}{k!}(x-1{)}^k\end{array}$

Or we can write this as$P_n\left(x\right)=1+\frac{1}{2}(x-1)+\sum _{k=2}^{\mathrm{\infty }}(-1{)}^{k+1}\frac{1\cdot 3\cdot 5\cdots (2k-3)}{2^{k}k!}(x-1{)}^k$