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Math Help - Find maximum

  1. #1
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    Find maximum

    I would like to find the n that maximizes this discrete function

    d(n) = \frac{R^{n+1}}{\sin(\theta)}\left[R\sin(\theta(n+1)) - \sin(\theta(n+2))\right]

    where n is 0,1,2,....
    0<R<1 and  0<\theta < \pi

    Is that even possible?
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  2. #2
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    Re: Find maximum

    Hey niaren.

    Hint: Try treating n as a continuous variable and using calculus to get the maximum points and look at answers that are integer values around those points.

    With regards to the constraints on R and theta, consider using multi-variable optimization with those constraints.

    If you get a small number of turning points, then it will be easy to find the values of N that maximizes the function.
    Thanks from niaren
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  3. #3
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    Re: Find maximum

    I will write t instead of n, just to make clear that t is continuous.

    d'(t) = R^{t+1}\log(R) \left[R\sin(\theta(t+1)) + \sin(\theta(t+2))\right] + R^{t+1}\left[R\cos(\theta(t+1))\theta + \cos(\theta(t+2))\theta\right] \\ = R^{t+1}\left[\log(R) \left(R\sin(\theta(t+1)) + \sin(\theta(t+2))\right) +\left(R\cos(\theta(t+1))\theta + \cos(\theta(t+2))\theta\right)\right]

    Now setting equal to zero gives

    0 = R^{t+1}\left[\log(R) \left(R\sin(\theta(t+1)) + \sin(\theta(t+2))\right) +\left(R\cos(\theta(t+1))\theta + \cos(\theta(t+2))\theta\right)\right] \\ 0 = \log(R) \left(R\sin(\theta(t+1)) + \sin(\theta(t+2))\right) +\left(R\cos(\theta(t+1))\theta + \cos(\theta(t+2))\theta\right) \\ 0 = \log(R)R\sin(\theta(t+1)) + \log(R)\sin(\theta(t+2)) + R\cos(\theta(t+1))\theta + \cos(\theta(t+2))\theta

    Is it possible to isolate t? It seems hard but my PC says it is possible, it also gives a result but I haven't found out yet how to do it manually. Any hints?
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  4. #4
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    Re: Find maximum

    I'm not sure but you might want to look at the various trig identities to see if you can simply the RHS in terms of a single trig representation.
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  5. #5
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    Re: Find maximum

    Thanks chiro.
    I think I've got it now. I worked backwards and figured out the trick. I found the trig identity to use and the underlying idea. The idea is to take the trig terms with argument \theta(t+2) and factor them out in terms of \theta(t+1). Then the terms involving \theta(t+1) can be boiled down to a tangent. The remaining terms does not depend on t. Very nice .

    Actually, this max value I have been working on finding an expression for is a special case of a more general problem which I am now more prepared to tackle (a max overshoot value in some digital filters).
    Thanks for your feedback.

    Although not relevant anymore, I though I just mention that the expression for the derivative I posted is not correct. It has some sign issues. I thought I just mention it. Anyway thanks.
    Last edited by niaren; May 24th 2013 at 03:54 AM.
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