Science:Math Exam Resources/Courses/MATH101/April 2014/Question 08/Solution 1

By the Taylor series formula (taking $a=2$),

${\begin{aligned}f(x)=\sum _{n=0}^{\infty }c_{n}(x-2)^{n}\end{aligned}}$

where

${\begin{aligned}c_{0}=f(2)\quad {\text{and}}\quad c_{n}={\frac {f^{(n)}(2)}{n!}}.\end{aligned}}$

First, $c_{0}=f(2)=\ln 2$.

Next, if we take a couple derivatives,

${\begin{aligned}f^{(1)}(x)={\frac {1}{x}},\quad f^{(2)}(x)=-{\frac {1}{x^{2}}},\quad f^{(3)}(x)={\frac {2}{x^{3}}},\quad f^{(4)}(x)=-{\frac {6}{x^{4}}}\end{aligned}}$

we can find the following patterns: the sign keeps changing, the power of $x$ in denominator of $f^{(n)}(x)$ is $n$, and the number in numerator of $f^{(n)}(x)$ is $(n-1)!$. Thus, we can get a general formula for $f^{(n)}(x)$:

${\begin{aligned}f^{(n)}(x)={\frac {(-1)^{n+1}(n-1)!}{x^{n}}},\end{aligned}}$

which implies

${\begin{aligned}f^{(n)}(2)={\frac {(-1)^{n+1}(n-1)!}{2^{n}}}\end{aligned}}$

and

${\begin{aligned}c_{n}={\frac {(-1)^{n+1}(n-1)!}{2^{n}n!}}={\frac {(-1)^{n+1}}{n2^{n}}}.\end{aligned}}$

Thus, we have the Taylor series:

${\begin{aligned}\ln x=\ln 2+\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n2^{n}}}(x-2)^{n}.\end{aligned}}$

For the interval of convergence, we will use the ratio test. First, compute

${\begin{aligned}\lim _{n\to \infty }\left|{\dfrac {{\frac {(-1)^{n+2}}{(n+1)2^{n+1}}}(x-2)^{n+1}}{{\frac {(-1)^{n+1}}{n2^{n}}}(x-2)^{n}}}\right|=\lim _{n\to \infty }\left|{\frac {(-1)^{n+2}(x-2)^{n+1}}{(n+1)2^{n+1}}}{\frac {n2^{n}}{(-1)^{n+1}(x-2)^{n}}}\right|=\lim _{n\to \infty }{\frac {n}{2(n+1)}}\cdots |x-2|={\frac {|x-2|}{2}}.\end{aligned}}$

The series converges when this ratio is smaller than 1 and so for $x$ in $|x-2|<2$, the series converges and for $x$ in $|x-2|>2$ the series diverges. We still have to check the endpoints, that is when $|x-2|=2$. This occurs at $x=0$ and $x=4$. At $x=0$, the series becomes,

$\ln x=\ln 2+\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n2^{n}}}(-2)^{n}=\ln 2+\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}(-1)^{n}}{n}}=\ln 2+\sum _{n=1}^{\infty }{\frac {(-1)}{n}}.$

We know that this series diverges and therefore the Taylor series does not converge at $x=0$. Note that we expect this to happen because $\lim _{x\to 0}\ln(x)=-\infty$. At $x=4$, the series becomes,

$\ln x=\ln 2+\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n2^{n}}}(2)^{n}=\ln 2+\sum _{n=1}^{\infty }{\frac {(-1)^{n+1}}{n}}.$

This is an alternating series with decreasing terms that tend to zero and therefore by the alternating series test, converges. Once again, this makes sense since $\ln(4)$ has a finite value. Combining everything together, the interval of convergence for the series is $x\in (0,4]$.