Science:Math Exam Resources/Courses/MATH101/April 2016/Question 11 (b)

MATH101 April 2016

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q2 (a)  •  Q2 (b)  •  Q2 (c)  •  Q3 (a)  •  Q3 (b)  •  Q4 (a)  •  Q4 (b)  •  Q5 (a)  •  Q5 (b)  •  Q6 (a)  •  Q6 (b)  •  Q7  •  Q8 (a)  •  Q8 (b)  •  Q9 (a)  •  Q9 (b)  •  Q10 (a)  •  Q10 (b)  •  Q11 (a)  •  Q11 (b)  •  

   Other MATH101 Exams

  •  April 2011  •  April 2010  •  April 2009  •  April 2008  •  April 2007  •  April 2005  •  April 2006  •  April 2012  •  April 2013  •  April 2014  •  April 2015  •  April 2016  •  April 2018  •  April 2017  •  

[hide] Question 11 (b)
Both parts of this question concern the series $\displaystyle S=\sum _{n=1}^{\infty }(-1)^{n-1}24n^{2}e^{-n^{3}}$. (b) Suppose that you approximate the series $S$ by its fifth partial sum $S_{5}$. Give an upper bound for the error resulting from this approximation. Explain why your error bound is valid 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.

[show] Hint 1
Recall that Alternating series estimation theorem: If $s=\sum _{n=1}^{\infty }(-1)^{n-1}b_{n}$ is the sum of an alternating series that satisfies: $b_{n+1}\leq b_{n}$ and $\lim _{n\to \infty }b_{n}=0,$ then $|R_{n}|=|s-s_{n}|\leq b_{n+1}$, where $s_{n}=\sum _{k=1}^{n}(-1)^{k-1}b_{k}$.

[show] Hint 2

(Alternative way) Since $|S-S_{5}|\leq \sum _{n=6}^{\infty }24n^{2}e^{-n^{3}}$, try to estimate the right hand side, using the Remainder estimate for the integral test: Suppose $f(k)=a_{k}$ where $f$ is a continuous, positive, decreasing function for $x\geq n$ and $s=\sum _{k=1}^{\infty }a_{k}$ is convergent. Then, $\int _{n+1}^{\infty }f(x)dx\leq s-s_{n}\leq \int _{n}^{\infty }f(x)dx$ where $s_{n}=\sum _{k=1}^{n}a_{k}$. Compare this error with the one obtained in Hint 1.

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.

[show] 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. In part (a), to show the convergence of the series we checked that $b_{n}=24n^{2}e^{-n^{3}}$ is decreasing and positive. The limit of this sequence is equal to $0$, because by applying l'Hopital Rule: $\lim _{n\rightarrow \infty }b_{n}=\lim _{n\rightarrow \infty }{\frac {24n^{2}}{e^{n^{3}}}}=\lim _{n\rightarrow \infty }{\frac {48n}{3n^{2}e^{n^{3}}}}=\lim _{n\rightarrow \infty }{\frac {48}{3ne^{n^{3}}}}=0$ Thus by the Alternating Series Test, this series is convergent. For a convergent Alternating Series, as mentioned in hint (1), we would like to approximate the sum $S$ with the sum of the first five terms $S_{5}$ as the following $S_{5}={\frac {24}{e}}-{\frac {24\cdot 2^{2}}{e^{8}}}+{\frac {24\cdot 3^{2}}{e^{27}}}-{\frac {24\cdot 4^{2}}{e^{64}}}+{\frac {24\cdot 5^{2}}{e^{125}}}$ By the error formula for Alternating Series, the upper bound for the error of approximation is the immediate term of the sequence $b_{n}$ after $S_{5}$ which is $b_{6}={\frac {24\times 36}{e^{6^{3}}}}.$ This can be justified by the following argument: Let's write the error term in the expanded form: $R_{5}=S-S_{5}=-{\frac {24\cdot 6^{2}}{e^{6^{3}}}}+{\frac {24\cdot 7^{2}}{e^{7^{3}}}}-{\frac {24\cdot 8^{2}}{e^{8^{3}}}}+{\frac {24\cdot 9^{2}}{e^{9^{3}}}}-{\frac {24\cdot 10^{2}}{e^{10^{3}}}}+...$ By playing with brackets we can find the bound for the error. First: $R_{5}=S-S_{5}={\Big (}-{\frac {24\cdot 6^{2}}{e^{6^{3}}}}+{\frac {24\cdot 7^{2}}{e^{7^{3}}}}{\Big )}+{\Big (}-{\frac {24\cdot 8^{2}}{e^{8^{3}}}}+{\frac {24\cdot 9^{2}}{e^{9^{3}}}}{\Big )}-{\frac {24\cdot 10^{2}}{e^{10^{3}}}}+...$ due to the decreasing terms of the sequence, the sum of the terms in each of the ( ) will be a negative number, so $R_{5}=S-S_{5}<0$ This time, $R_{5}=S-S_{5}=-{\frac {24\cdot 6^{2}}{e^{6^{3}}}}+{\Big (}{\frac {24\cdot 7^{2}}{e^{7^{3}}}}-{\frac {24\cdot 8^{2}}{e^{8^{3}}}}{\Big )}+{\Big (}{\frac {24\cdot 9^{2}}{e^{9^{3}}}}-{\frac {24\cdot 10^{2}}{e^{10^{3}}}}{\Big )}+...$ Again because of decreasing terms, the sum of the terms in each ( ) is positive $R_{5}=S-S_{5}=-{\frac {24\cdot 6^{2}}{e^{6^{3}}}}+$ (positive terms) $>-{\frac {24\cdot 6^{2}}{e^{6^{3}}}}$ so $-{\frac {24\cdot 6^{2}}{e^{6^{3}}}}<S-S_{5}<0$ , which implies $\color {blue}|S-S_{5}|<{\frac {24\cdot 6^{2}}{e^{6^{3}}}}.$

[show] 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. An alternative method is to use the Integral test to approximate the error. First we note that $|S-S_{5}|=|\sum _{n=6}^{\infty }(-1)^{n-1}24n^{2}e^{-n^{3}}|\leq \sum _{n=6}^{\infty }|(-1)^{n-1}24n^{2}e^{-n^{3}}|=\sum _{n=6}^{\infty }24n^{2}e^{-n^{3}}.$ so it is sufficient to find an upper bound for $\sum _{n=6}^{\infty }24n^{2}e^{-n^{3}}.$ Now if $a_{n}=24n^{2}e^{-n^{3}}$, then we showed that $a_{n}$ is positive and decreasing, so we can apply the integral test to estimate $\sum _{n=6}^{\infty }24n^{2}e^{-n^{3}}$ with $\int _{5}^{\infty }24x^{2}e^{-x^{3}}\,dx$. This integral represents the area under the graph of $f(x)=24x^{2}e^{-x^{3}}$ But $\sum _{n=6}^{\infty }24n^{2}e^{-n^{3}}$ represents the sum of the area of rectangles starting at $x=5$ with width $1$, which is in fact the Reimann sum of $f(x)$ with right endpoints on each subinterval $[5,6],[6,7],...$, so it underestimates the actual area under the graph, therefore; $\sum _{n=6}^{\infty }24n^{2}e^{-n^{3}}\leq \int _{5}^{\infty }24x^{2}e^{-x^{3}}\,dx=8e^{-5^{3}}.$ $\color {blue}|S-S_{5}|\leq 8e^{-5^{3}}.$