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Math Help - simple calculus question

  1. #1
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    simple calculus question

    Hi guys,

    I have a simple calculus question, (so simple it's actually embarrassing)
    So I'm reading this article that says:
     \mbox{if we have: } p(D_i,D_i^* \mid \Psi) = \frac{1}{4w_i} sech^2(\frac{D_i^* - D_i}{2w_i})
    \mbox{then: }\sum\limits_{i=1}^{N}log \frac{\int_{d_i}^{\infty} p(D_i, D_i^* \mid \Psi) dD^* }{\int_{0}^{d_i} p(D_i, D_i^* \mid \Psi) dD^* } = \sum\limits_{i=1}^{N}w_i (d_i - D_i)
    But I can't get that result!!!
    Can anyone show me the intermediate steps please?
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  2. #2
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    Re: simple calculus question

    I assume that you are integrating with respect to D_i^*. You mostly need the following identities

    \int p(D_i,D_i^* | \Psi ) dD_i^* = \frac{1}{2}\tanh\left(\frac{D_i^*-D_i}{2w_i}\right)
    \tanh(x) = \frac{e^x-e^{-x}}{e^x+e^{-x}}

    Then you find the limit as D_i^*\to \infty. It is 1/2 or -1/2 depending on whether w_i>0 or not. Suppose that w_i>0. Then

    \int_{d_i}^{\infty} p(D_i,D_i^* | \Psi ) dD_i^* = \frac{1}{2}\left(1-\tanh\left(\frac{d_i-D_i}{2w_i}\right)\right)

    The integral in the denominator is

    \int_{0}^{d_i} p(D_i,D_i^* | \Psi ) dD_i^* = \frac{1}{2}\left(\tanh\left(\frac{d_i-D_i}{2w_i}\right) - \tanh\left(\frac{-D_i}{2w_i}\right)\right)

    Converting to trigonometric form, the top integral becomes

    \int_{d_i}^{\infty} p(D_i,D_i^* | \Psi ) dD_i^* = \frac{2 e^{D_i/w_i}}{e^{d_i/w_i}+e^{D_i/w_i}}

    and the bottom integral becomes

    \int_{0}^{d_i} p(D_i,D_i^* | \Psi ) dD_i^* = \frac{2 \left(e^{d_i/w_i}-1\right) e^{D_i/w_i}}{\left(e^{D_i/w_i}+1\right)\left(e^{d_i/w_i}+e^{D_i/w_i}\right)}

    The fraction of these simplifies to

    \frac{\int_{d_i}^{\infty} p(D_i,D_i^* | \Psi ) dD_i^*}{\int_{0}^{d_i} p(D_i,D_i^* | \Psi ) dD_i^*} = \frac{e^{D_i/w_i}+1}{e^{d_i/w_i}-1}

    The whole sum is now equal to

    \sum_{i=1}^N \log \left(\frac{e^{D_i/w_i}+1}{e^{d_i/w_i}-1}\right)

    I'm not sure how one can proceed from here.
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  3. #3
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    Re: simple calculus question

    Thank you Vlasev,

    I am getting quite similar results to yours, which is obviously different the one I expected!
    My original reference is the appendix section of this paper:
    http://carlos-hernandez.org/papers/hernandez_cvpr07.pdf

    The authors are quite well known in their field and the paper was published in a very high ranked conference, so I will be very surprised if their calculation is wrong.
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  4. #4
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    Re: simple calculus question

    Even thought they are well-known and the conference is highly-ranked, it's not guarantee that there won't be any mistakes. I'd say that there will be more mistakes in places where a derivation should be "obvious". The referees will probably not check. As for this, I don't know whether it's wrong or not.

    Edit: I think their identity might well be wrong.
    Last edited by Vlasev; August 16th 2012 at 10:28 PM.
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