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Math Help - upper bounds numerical integration error

  1. #1
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    upper bounds numerical integration error

    okay so the integral is \int_{0}^{3} \sqrt{x+1}dx and using midpoint approximation so I use

     \frac{(b-a)^3k_{2}}{24n^2}, where\,n=10\,and\,k_{2} = max_{[a,b]} |f"(x)|

    f"(x) = -\frac{1}{4}(x+1)^{-3/2} found k_{2} = \frac{1}{32}

    I am getting \frac{9}{25600} book says 9/3200

    also, latex sure is being a pain on here lately
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  2. #2
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    Re: upper bounds numerical integration error

    The question then goes on to ask for the upper bounds from trapezoid and simpsons rule but the book has different answers then I am getting. Did I do the first one correctly?
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    Re: upper bounds numerical integration error

    now the question asks to find the number n of subintervals that will make the error less than 5*10^{-4}

    just typing out the midpoint because the book is giving not the same answer I am giving

    |E_{m}| \le 5*10^{-4}

    \frac{3^3*1/32}{24n^2} \le 5*10^{-4} \approx 8.3852

    I am getting n= 9 book is saying n = 24
    Last edited by Jonroberts74; February 17th 2014 at 05:08 PM.
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  4. #4
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    Re: upper bounds numerical integration error

    I'd like to help you out Jon but w/o a reference to what you're doing I'm not able to figure out what you need figured out.

    Is there some online reference that uses the same notation and method?
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  5. #5
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    Re: upper bounds numerical integration error

    k_{2}=1/4 isn't it ? (Max at x=0.)
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    Re: upper bounds numerical integration error

    Quote Originally Posted by romsek View Post
    I'd like to help you out Jon but w/o a reference to what you're doing I'm not able to figure out what you need figured out.

    Is there some online reference that uses the same notation and method?
    http://math.cmu.edu/~mittal/Recitation_notes.pdf
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    Re: upper bounds numerical integration error

    Quote Originally Posted by BobP View Post
    k_{2}=1/4 isn't it ? (Max at x=0.)
    the second derivative of the given function is \frac{-1}{4}(x+1)^{-3/2} and has a maximum on [0,3] of -1/32 of which you take the absolute value of

    still with latex issues
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    Re: upper bounds numerical integration error

    re your first post on this thread
    BobP is correct. The max of the absolute value of the 2nd derivative is 1/4 at x=0 which leads to the bound of 9/3200

    your 2nd problem is wrong for the same reason. The max value is 1/4 not 1/32
    Last edited by romsek; February 18th 2014 at 01:53 PM.
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    Re: upper bounds numerical integration error

    To reiterate what romsek and BobP are saying:

    You are taking k_2 = \left| \max_{[a,b]} f''(x) \right|, which is the absolute value of the maximum of the second derivative on the given interval. BobP and romsek are saying, take the absolute value function first: |f''(x)| = \left|-\dfrac{1}{4}(x+1)^{-3/2} \right| = \dfrac{1}{4}(x+1)^{-3/2}. Then, find the maximum. Hence, k_2 = \max_{[0,3]} \dfrac{1}{4}(x+1)^{-3/2} = \dfrac{1}{4}.
    Thanks from Jonroberts74
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    Re: upper bounds numerical integration error

    Quote Originally Posted by SlipEternal View Post
    To reiterate what romsek and BobP are saying:

    You are taking $$k_2 = \left| \max_{[a,b]} f''(x) \right|$$, which is the absolute value of the maximum of the second derivative on the given interval. BobP and romsek are saying, take the absolute value function first: $$|f''(x)| = \left|-\dfrac{1}{4}(x+1)^{-3/2} \right| = \dfrac{1}{4}(x+1)^{-3/2}$$. Then, find the maximum. Hence, $$k_2 = \max_{[0,3]} \dfrac{1}{4}(x+1)^{-3/2} = \dfrac{1}{4}$$.
    new latex
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  11. #11
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    Re: upper bounds numerical integration error

    I see how I was messing up, taking the absolute of the answer not of the function. thanks
    Last edited by Jonroberts74; February 18th 2014 at 03:57 PM.
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