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

MATH101 April 2008

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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.

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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

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