Science:Math Exam Resources/Courses/MATH103/April 2013/Question 10

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MATH103 April 2013

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Question 10
Taylor Series: Find the first five terms of the Taylor series centred at 0 for $\displaystyle F(x)=\int _{0}^{x}\ln(1+s^{2})\,ds$ i.e. the polynomial $\displaystyle T_{4}(x)$ with $\displaystyle x^{4}$ the highest power of x.

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
First notice that the question is asking for the first 5 terms NOT the first 5 nonzero terms. Thus, it's possible to grind this out simply by taking derivatives.

Hint 2
The Taylor Series Formula about 0 is given by $\displaystyle F(x)=\sum _{n=0}^{\infty }{\frac {F^{(n)}(0)x^{n}}{n!}}$

Hint 3
When taking derivatives, don't forget that your first function is the function defined by the integral and that you'll have to use the chain rule.

Hint 4
(Alternate Solution) If you want, you can start with the Taylor expansion of $\displaystyle \ln(1+x)$ and go from there.

Hint 5
The Taylor expansion is given by $\displaystyle \ln(1+x)=\sum _{n=1}^{\infty }(-1)^{n+1}{\frac {x^{n}}{n}}$.

Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution 1

We mechanically plug in the derivatives to the Taylor series formula given by $\displaystyle F(x)=\sum _{n=0}^{\infty }{\frac {F^{(n)}(0)x^{n}}{n!}}$ where $\displaystyle F(x)=\int _{0}^{x}\ln(1+s^{2})\,ds$. Now, we have $\displaystyle F(0)=\int _{0}^{0}\ln(1+s^{2})\,ds=0$. Using the FTC, we get $\displaystyle F'(0)=\left.{\frac {d}{dx}}\right|_{x=0}\int _{0}^{x}\ln(1+s^{2})\,ds=\ln(1+x^{2}){\big |}_{x=0}=\ln(1)=0$. The next derivative is (via the chain rule) $\displaystyle F''(0)=\left.{\frac {d}{dx}}\right|_{x=0}\ln(1+x^{2})=\left.{\frac {2x}{1+x^{2}}}\right|_{x=0}=0$. The next derivative is (via the quotient rule) $\displaystyle F'''(0)=\left.{\frac {d}{dx}}\right|_{x=0}{\frac {2x}{1+x^{2}}}=\left.{\frac {2(1+x^{2})-(2x)^{2}}{(1+x^{2})^{2}}}\right|_{x=0}=2$. The last derivative is ${\begin{aligned}F''''(0)&=\left.{\frac {d}{dx}}\right|_{x=0}{\frac {2(1+x^{2})-(2x)^{2}}{(1+x^{2})^{2}}}\\&=\left.{\frac {d}{dx}}\right|_{x=0}{\frac {2-2x^{2}}{(1+x^{2})^{2}}}\\&=\left.{\frac {-4x(1+x^{2})^{2}-2(1-x^{2})\cdot 2(1+x^{2})(2x)}{(1+x^{2})^{4}}}\right|_{x=0}\\&=0\end{aligned}}$. Thus, the fourth Taylor polynomial is $\displaystyle T_{4}(x)={\frac {x^{3}}{3}}$ as required.

Solution 2

Starting with $\displaystyle \ln(1+x)=\sum _{n=1}^{\infty }(-1)^{n+1}{\frac {x^{n}}{n}}$ We see that $\displaystyle \ln(1+s^{2})=\sum _{n=1}^{\infty }(-1)^{n+1}{\frac {s^{2n}}{n}}$. Integrating yields $\displaystyle \int _{0}^{x}\ln(1+s^{2})=\sum _{n=1}^{\infty }(-1)^{n+1}\int _{0}^{x}{\frac {s^{2n}}{n}}=\sum _{n=1}^{\infty }(-1)^{n+1}{\frac {x^{2n+1}}{n(2n+1)}}$ Plugging in the first few values yields $\displaystyle \sum _{n=1}^{\infty }(-1)^{n+1}{\frac {x^{2n+1}}{n(2n+1)}}={\frac {x^{3}}{3}}-{\frac {x^{5}}{10}}+...$ Since the question is seeking $\displaystyle T_{4}(x)$, we have that $\displaystyle T_{4}(x)={\frac {x^{3}}{3}}$ as required.

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