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Math Help - Simpson's Rule

  1. #16
    MHF Contributor chisigma's Avatar
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    Milton Abramowitz & Irene Stegun, Handbook of Mathematical Function, National Bureau of Standards, June 1964, pag.27...

    \int \sqrt{x^{2} \pm a^{2}}\cdot dx = \frac{x}{2} \cdot \sqrt {x^{2} \pm a^{2}} \pm \frac{a^{2}}{2}\cdot \ln |x + \sqrt {x^{2} \pm a^{2}}| + c (1)

    Applying (1) we find...

    \int_{0}^{2\pi} \sqrt{1+x^{2}}\cdot dx = 21.256294148209\dots (2)

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    \chi \sigma
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  2. #17
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    Last edited by harold; March 5th 2010 at 01:42 AM.
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  3. #18
    MHF Contributor chisigma's Avatar
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    The error in of the approximate integration is given by...

    \epsilon \sim - \frac{(b-a)^{5}}{180\cdot n^{4}}\cdot f^{(4)} (\xi) (1)

    ... where a \le \xi \le b. From (1) it is evident that it depends strongly from n. The problem in applying (1) is that You have to compute the derivative of fourth order of f(*) and find the value of \xi \in [a,b] that maximizes |f^{(4)} (*)| in that interval... not a very comfortable job ! ...

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  4. #19
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    Last edited by harold; March 5th 2010 at 01:42 AM.
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  5. #20
    Grand Panjandrum
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    Quote Originally Posted by harold View Post
    I'm sorry if I mistated it but I already have the formula for the error bound. I'm having trouble actually doing the computation. That's where I need the help. I'm not even sure the value to choose for c. I've even been trying to carry out the trapezoidal rule but with this integral it's tough.
    I don't know if you are allowed to do this, but the attachment shows the Maxima caculation of the fourth derivative:



    So we see that |f^{(4)}(x)|\le 3 on [0, 2\pi]

    CB
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  6. #21
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    Last edited by harold; March 5th 2010 at 01:44 AM.
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  7. #22
    Grand Panjandrum
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    Quote Originally Posted by harold View Post
    Hi CB,
    Your derivative is of course right. But when choosing a "c" value to substitute in for x, dont you choose 1? In which I get .7954951286. Now to use the error formula from Simpson's Rule which is: -\frac{h^5}{90}f^{iv}(c). I remember giving the error formula with something like b - a over 180 but that was the composite rule--I think we're using the regular rule (not composite) in which the error is the one above. When I compute everything, I get 2.704857428 but this can't be correct, the error should be small. (in the formula above,  h=x_2 - x_1 = x_1 - x_0 and the integral is from x_2 to x_0.

    Also CB, when it says "Determine the number of subintervals n required to approximate the integral to within 10^{-6} by using the composite trapezoidal rule," I can find it using math software myself but I'm stuck on the 10^{-6} part...does that mean six decimal places must be correct or six digits or..?
    You use the value which maximises |f^{(4)}| , which in this case is 0 see post #9 of this thread.

    CB
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  8. #23
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    Last edited by harold; March 5th 2010 at 01:44 AM.
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  9. #24
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    Quote Originally Posted by harold View Post
    If we choose 0 then we obtain an even bigger error but this isn't good right? Computing the error using -\frac{h^5}{90}f^{iv}(c), you obtain 10.20065616 since  h = \pi but the error is supposed to be small so why is this happening? The problem I'm working on doesnt say to use the composite simpson rule so error formula above has to be right, right?
    How are h and n defined?

    CB
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  10. #25
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    Last edited by harold; March 5th 2010 at 01:44 AM.
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  11. #26
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    Last edited by harold; March 5th 2010 at 01:44 AM.
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