Science:Math Exam Resources/Courses/MATH101/April 2008/Question 05 (a)

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MATH101 April 2008

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q1 (f)  •  Q1 (g)  •  Q2 (a)  •  Q2 (b)  •  Q2 (c)  •  Q2 (d)  •  Q3 (a)  •  Q3 (b)  •  Q3 (c)  •  Q3 (d)  •  Q4 (a)  •  Q4 (b)  •  Q5 (a)  •  Q5 (b)  •  Q6  •  Q7 (a)  •  Q7 (b)  •  

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Question 05 (a)
Let $I=\int _{0}^{1}\cos(x^{2})\,dx$ Write down the first three nonzero terms obtained by using Maclaurin series to estimate $I$, and explain why the error in using this estimate is less than 0.001.

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
The Maclaurin series for $\cos(x)$ is $\cos(x)=1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-...=\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{2n}}{(2n)!}}$

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Solution

As suggested in the hint, we know that $\cos(y)=1-{\frac {y^{2}}{2!}}+{\frac {y^{4}}{4!}}-{\frac {y^{6}}{6!}}...=\sum _{n=0}^{\infty }{\frac {(-1)^{n}y^{2n}}{(2n)!}}$ Plugging in $y=x^{2}$ gives $\cos(x^{2})=1-{\frac {x^{4}}{2}}+{\frac {x^{8}}{24}}-{\frac {x^{12}}{720}}+...=\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{4n}}{(2n)!}}$ Now, integrating gives $\int \cos(x^{2})=C+x-{\frac {x^{5}}{(2)(5)}}+{\frac {x^{9}}{(24)(9)}}+{\frac {x^{13}}{(720)(13)}}+...=\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{4n+1}}{(4n+1)(2n)!}}$ Now, using just the first three terms and integrating from 0 to 1 gives $\int _{0}^{1}\cos(x^{2})=1-{\frac {1}{10}}+{\frac {1}{216}}$ This expansion is correct to within $0.001$ which we can show by using the alternating series estimation theorem. We know that $b_{n}={\frac {1}{(4n+1)(2n)!}}$ is decreasing. The terms also limit to 0 and clearly this is an alternating series. Hence, the error above using three terms is no worse than what the fourth term tells us it is. Hence, the error is bounded above by $Error\leq {\frac {1}{(720)(13)}}<{\frac {1}{(100)(10)}}={\frac {1}{1000}}=0.001$ and this completes the question.

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