Q. 53 - 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.

$53. \ln \left(x\right),3$

Step-by-Step Solution

Answer

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

Step 1. Given data

We have the function $f\left(x\right)=\ln \left(x\right)$

Step 2. Table of the taylor series

Any function $f$ with a derivative of order $n$, the taylor series at $x=3$ is given by,

$\begin{array}{rl}{P}_n\left(x\right)=& f\left(3\right)+f^{\mathrm{\prime }}\left(3\right)(x-3)+\frac{f^{\prime \prime }\left(3\right)}{2!}(x-3{)}^2+\frac{f^{\prime \prime \mathrm{\prime }}\left(3\right)}{3!}(x-3{)}^3+\frac{f^{\prime \prime \mathrm{\prime }}\left(3\right)}{4!}(x-3{)}^4+\dots \end{array}$

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 $f\left(x\right)=\ln \left(x\right)$ at $x=3$.

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

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

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

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