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

MATH101 April 2014

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Question 08
Long Problem. Show your work. No credit will be given for the answer without the correct accompanying work. Find the Taylor series for $\displaystyle f(x)=\ln(x)$ centered at a = 2. Find the interval of convergence for this series.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
What is the Taylor series formula for some function $f$ at $x=a$?

Hint 2
Find the first few derivative, general form for nth derivative. Keep track of contributions towards constant and the sign.

Hint 3
(Interval of convergence) What test do we need to use for convergence? Don't forget to check the endpoints!

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 $x\in (0,4]$.

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