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Math Help - Simpson approximation method

  1. #1
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    Simpson approximation method

    Use the Simpson method to estimate \displaystyle\int_{0}^{1}\cos(x^2)dx with an approximation error less than 0.001.

    Well, I have a problem. Actually I'm looking for a bound for the error of approximate method integration by using Simpson's method.

    I have to bring \displaystyle\int_{0}^{1}\cos(x^2)dx with an error less than 0.001.

    I started looking for the fourth order derivative, and got:

    f^4(x)=48x^2\sin(x^2)-12\cos(x^2)+16x^4\cos(x^2)

    Now, I have to find a bound K for this derivative in the interval [0,1]. What I did do was watch for if they had maximum and minimum on the interval, then calculate the derivative of order five:
    f^5(x)=120x\sin(x^2)+160x^3\cos(x^2)-32x^5\sin(x^2)

    From here I did was ask:

    If f^5(x)\in{[0,1]}\Rightarrow{f^5(x)=0\Longleftrightarrow{x=0}}

    So here I know is that zero is a maximum or minimum. Then I wanted to look for on the concavity of the curve, and I thought the easiest thing would be to look at the sixth derivative, to know how it would behave the fourth derivative of the original function.

    f^6(x)=720x^2\cos(x^2)-480x^4\sin(x^2)-64x^6\cos(x^2)+120\sin(x^2)

    The problem is that when evaluated

    f^6(0)=120\sin(0^2)=0

    So, I get zero the derivative sixth, and I do not say whether it is concave upwards or downwards is concave.

    What should I do? there may be a less cumbersome to work this, if so I would know.

    Greetings.
    Last edited by Ulysses; August 8th 2010 at 07:25 PM.
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  2. #2
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    I get ...

    f^4(x) = 48x^2 \sin(x^2) + (16x^4-12)\cos(x^2)
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  3. #3
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    Yes, sorry. I've corrected some typos. Anyway, I've solved it. It was trivial, I could use any point on the interval to see the concavity of the curve, I didn't even needed to use the 6th derivative, as it only has an inflection point on zero for the given interval.

    Thanks.
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