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Thread: gamma-function

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    gamma-function




    gamma-function


    g(t) = ymax * t^(alpha) * exp( alpha * (1-t))

    With known ymax and alpha, what is the maximum slope of g(t) (with t>0)?

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    Re: gamma-function

    g(t) = y_{\max}t^{\alpha}\exp\{\alpha(1-t)\}

    To compute the slope, you differentiate the function with respect to t:

    g'(t) = y_{\max}\alpha t^{\alpha-1}\exp\{\alpha(1-t)\} -y_{\max}\alpha t^{\alpha}\exp\{\alpha(1-t)\}

    =  y_{\max}\alpha t^{\alpha-1}\exp\{\alpha(1-t)\}(1 - t)

    To find the maximum slope, we differentiate again:

    g''(t) = y_{\max}\alpha(\alpha - 1)t^{\alpha-2}\exp\{\alpha(1-t)\}(1 - t)- y_{\max}\alpha^2 t^{\alpha-1}\exp\{\alpha(1-t)\}(1 - t) - y_{\max}\alpha t^{\alpha-1}\exp\{\alpha(1-t)\}

    = y_{\max}\alpha t^{\alpha-2}\exp\{\alpha(1-t)\}[(\alpha - 1)(1-t) - \alpha t(1-t) - t ]

    = y_{\max}\alpha t^{\alpha-2}\exp\{\alpha(1-t)\}(\alpha t^2 - 2\alpha t + \alpha - 1)

    Assume that y_{\max} > 0, \alpha > 0 such that y_{\max}\alpha t^{\alpha-2}\exp\{\alpha(1-t)\} > 0

    In such case the sign of g''(t) solely depends on the sign of \alpha t^2 - 2\alpha t + \alpha - 1

    By quadratic formula, we have

    \alpha t^2 - 2\alpha t + \alpha - 1 = \alpha \left(t - 1 + \frac {1} {\sqrt{\alpha}}\right)\left(t - 1 - \frac {1} {\sqrt{\alpha}}\right)

    \left\{\begin{matrix} > 0 & \text{if} & t < 1 - \frac {1} {\sqrt{\alpha}} \\ = 0 & \text{if} & t = 1 - \frac {1} {\sqrt{\alpha}} \\  < 0 & \text{if} & 1 - \frac {1} {\sqrt{\alpha}} < t < 1 + \frac {1} {\sqrt{\alpha}}\\ = 0 & \text{if} & t = 1 + \frac {1} {\sqrt{\alpha}} \\ > 0 & \text{if} & t > 1 + \frac {1} {\sqrt{\alpha}}\end{matrix}\right.

    So t = 1 - \frac {1} {\sqrt{\alpha}} is a local maximum. And you just need to compare this with the boundary t = +\infty

    But as you have the constraint t > 0, so when 0 < \alpha < 1 you compare the endpoints t = 0 and t = +\infty instead.

    It is easy to see that \lim_{t\to+\infty} g'(t) = 0, therefore we conclude that the maximum slope is at t = 1 - \frac {1} {\sqrt{\alpha}} when \alpha \geq 1

    Otherwise when 0 < \alpha < 1, we see that \lim_{t\to 0^+} g'(t) = +\infty which means that the slope is unbounded, or more strictly speaking the maximum point/maximum slope does not exist.

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    Re: gamma-function

    It’s so perfect. I do appreciate your kindness.

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    Re: gamma-function

    What value does the maximum slope have?

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    Re: gamma-function

    The maximum slope is just g'\left(1 - \frac {1} {\sqrt{\alpha}}\right) when \alpha \geq 1

    and does not exist when 0 < \alpha < 1 (+\infty)

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    Re: gamma-function

    What about highest value (that can be possible) such as 0.5, 1, 1.5, 3, 10, ..... something like that

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    Re: gamma-function

    Not sure what do you want. You just have to plug in all the values \alpha, y_{\max} you have.

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    Re: gamma-function



    highest value = 1
    Last edited by pongtep; 05-27-2013 at 08:18 AM.

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    Re: gamma-function

    Ok let say you have the difficulty to plug-in the formula and I do it for you:

    g'\left(1 - \frac {1} {\sqrt{\alpha}}\right) = y_{\max} \sqrt{\alpha} \left(1 - \frac {1} {\sqrt{\alpha}}\right)^{\alpha - 1}\exp\left\{\sqrt{\alpha}\right\}

    Now you can further put the (exact) values of y_{\max}, \alpha to obtain the numerical value which you desperately want.

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    Re: gamma-function

    It is highest result (numerical value) that it can calculate from this formula. I not sure, perhaps it is infinity.
    Last edited by pongtep; 05-27-2013 at 08:57 AM.

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    Re: gamma-function

    Let say you treat y_{\max} as a variable instead of a constant and you want to maximize g' over y_{\max} \in (0, +\infty). Then you can see g' increase without bound as y_{\max} increase without bound. It is the same for \alpha

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    Re: gamma-function

    Yes, and g' will increase without bound, is it correct?

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    Re: gamma-function

    Yes you see the functions g, g', g'' etc all have a y_{\max} factor in front. So they must increase as y_{\max} increase (when they are positive).

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    Re: gamma-function

    Thank you so much. If i have further questions would you mind if I asked you again?

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    Re: gamma-function


    Please Example if

    ymax = 3
    alpha = 2

    What is maximum slope?

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