Q. 54 - Calculus

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.

$54. \sqrt[3]{x},1$

Step-by-Step Solution

Answer

The Taylor series for the function $f\left(x\right)=\sqrt[3]{x}$ at $x=1$ is $P_n\left(x\right)=1+\frac{1}{3}(x-1)+\sum _{k=2}^{\mathrm{\infty }}(-1{)}^{k+1}\frac{5\cdot 8\cdot 11\cdots (2k-3)}{3^{k}k!}(x-1{)}^k$

Step 1. Given data

We have the function $f\left(x\right)=\sqrt[3]{x}$

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,

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

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 $f\left(x\right)=\sqrt[3]{x}$ at $x=1$

n	$f^n\left(x\right)$	$f^n\left(0\right)$	$\frac{f^n\left(0\right)}{n!}$
0	$\sqrt[3]{x}$	1	1
1	$\frac{1}{3}{x}^{\frac{-2}{3}}$	$\frac{1}{3}$	$\frac{1}{3}$
2	$-\frac{2}{3^2}(x{)}^{-\frac{5}{3}}$	$-\frac{2}{3^2}$	$\frac{-\frac{2}{3^2}}{2!}$
3	$\frac{2\cdot 5}{3^3}(x{)}^{-\frac{8}{3}}$	$\frac{2\cdot 5}{3^3}$	$\frac{\frac{2\cdot 5}{3^3}}{3!}$
4	$-\frac{2\cdot 5\cdot 8}{3^4}(x{)}^{-\frac{11}{3}}$	$-\frac{2\cdot 5\cdot 8}{3^4}$	$\frac{-\frac{2\cdot 5\cdot 8}{3^4}}{4!}$
. . .	. . .	. . .	. . .
k	$(-1{)}^{k+1}\frac{2[5\cdot 8\cdots (2k-3\left)\right]}{3^k}(x{)}^{-\left(\frac{3k-1}{2}\right)}$	$(-1{)}^{k+1}\frac{2[5\cdot 8\cdots (2k-3\left)\right]}{3^k}$	$(-1{)}^{k+1}\frac{\frac{2[5\cdot 8\cdots (2k-3\left)\right]}{3^k}}{k!}$

Step 3. Taylor series for the f ( x ) = x 3

The Taylor series for the function $f\left(x\right)=\sqrt[3]{x}$ at $x=1$ is

$P_n\left(x\right)=\begin{array}{rl}1& +\frac{1}{3}\cdot (x-1)+\frac{-\frac{2}{3^2}}{2!}(x-1{)}^2+\frac{\frac{2\cdot 5}{3^3}}{3!}(x-1{)}^3+\frac{-\frac{2\cdot 5\cdot 8}{3^4}}{4!}(x-1{)}^4+\dots +\frac{(-1{)}^{k+1}\frac{5\cdot 8\cdot 11\cdots (2k-3)}{3^k}}{k!}(x-1{)}^k\end{array}$

Or we can write this as $P_n\left(x\right)=1+\frac{1}{3}(x-1)+\sum _{k=2}^{\mathrm{\infty }}(-1{)}^{k+1}\frac{5\cdot 8\cdot 11\cdots (2k-3)}{3^{k}k!}(x-1{)}^k$