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

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

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Solution 1

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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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