MATH101 April 2008
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Question 05 (b)
It can be shown that the 4th derivative of
has absolute value at most 60 on the interval [0,1]. Using this bound, find the smallest positive integer n you can such that the Simpson's Rule approximation for has error less or equal to 0.001. You may use the fact that if
on the interval , then the error in using to approximate
has absolute value less than or equal to
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!
Justify to yourself that it suffices to find an n so that
Don't forget that n has to be even! (By the constraints of Simpson's rule)
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The question tells us that we can choose in our solution. Plugging this in, it suffices to find an n so that
Cross multiplying yields
dividing both sides by 3 gives
To find the smallest value for n, we start enumerating the first few even integers to the power of 4.
so we see that the first even integer n so that is completing the proof.
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