Science:Math Exam Resources/Courses/MATH101/April 2013/Question 10 (b)

MATH101 April 2013

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Question 10 (b)

Full-Solution Problems. In questions 4–12, justify your answers and show all your work. If a box is provided, write your final answer there. Unless otherwise indicated, simplification of numerical answers is required in these questions. The power series representation above is used to approximate $\displaystyle \int _{0}^{1/4}{\frac {1}{1+x^{3}}}\ dx$. How many terms are required to guarantee that the resulting approximation is within $\displaystyle 10^{-5}$ of the exact value? Justify your answer.

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
Use part (a) to find a series representation of the definite integral.

Hint 2
What theorem of series do we have at our disposal that allows us to estimate remainders?

Hint 3
Try to Alternating Series Remainder Theorem.

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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 part (a), $\displaystyle {\begin{aligned}\int _{0}^{1/4}{\frac {1}{1+x^{3}}}\ dx&=\left.\left[\left(\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{3n+1}}x^{3n+1}\right)+C\right]\right|_{0}^{1/4}\\&=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{3n+1}}\left({\frac {1}{4}}\right)^{3n+1}\end{aligned}}$ This is an alternating sum with $\displaystyle b_{n}={\frac {1}{3n+1}}\left({\frac {1}{4}}\right)^{3n+1}.$ Clearly $\displaystyle b_{n+1}\leq b_{n}$ and $\displaystyle \lim _{n\to \infty }b_{n}=0$. Then by Alternating Series Remainder Theorem, the error in the approximation is bounded as follows: $\displaystyle |{\text{error}}|\leq b_{n+1}$ Then we are guaranteed the error is less than $\displaystyle 10^{-5}$ if $\displaystyle b_{n+1}\leq 10^{-5}$, that is, we must solve $\displaystyle {\frac {1}{3(n+1)+1}}\left({\frac {1}{4}}\right)^{3(n+1)+1}\leq 10^{-5}={\frac {1}{10^{5}}}.$ For $\displaystyle n+1=1$, $\displaystyle {\frac {1}{3+1}}\left({\frac {1}{4}}\right)^{3+1}={\frac {1}{4^{5}}}>{\frac {1}{10^{5}}}$ so we do not have enough terms yet. For $\displaystyle n+1=2$, $\displaystyle {\frac {1}{3(2)+1}}\left({\frac {1}{4}}\right)^{3(2)+1}\leq {\frac {1}{7}}{\frac {1}{16000}}={\frac {1}{112000}}<{\frac {1}{10^{5}}}$ where we have used $\displaystyle 4^{7}=2^{14}=2^{10}\cdot 16=1024\cdot 16\geq 1000\cdot 16=16000$ to approximate. Thus, $\displaystyle n+1=2$ is sufficient. Since the series starts at 0, we need 2 terms (the $\displaystyle n=0$ and $\displaystyle n=1$ term, specifically).

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