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Math Help - Proof for Continuous function

  1. #1
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    Proof for Continuous function

    first, we have a function: y_k with variables of x_m , (m=1,2,...,M)
    y_k = \frac{\sum_{m=1}^M x_m \alpha_{m,k}}{\sum_{m=1}^M x_m \beta_{m,k} +1}
    we know that the  x_m \geq 0 , \alpha_{m,k}, \beta_{m,k} can be considered as constants.
    Now, based on y_k we have a function of z that to the maximize the minimum of y_k ,  (k=1,2,...,K) by optimally allocating the x_m as
    z = \max_{x_m} \min_{k=1.\cdot,K} (y_k).,
    where \sum_{m=1}^M x_m = x,
    Actually, we can alternatively express the above function of z as
     z=f(x),
    where the x now becomes a variable which is the sum of the x_m (m=1,2,...,M).
    then, we want to prove  z=f(x), is continuous to x and the function f(') returns the maiximized minimum value of y_k with an input variable of x .

    for the z=f(x), I have prove it is a strictly monotonically increasing function to x, but I failed to prove the continuity of it.

    Thanks
    Last edited by clover616; March 30th 2010 at 07:40 PM.
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by clover616 View Post
    y_k = \frac{\sum_{m=1}^M x_m \alpha_{m,k}}{\sum_{m=1}^M x_m \beta_{m,k} +1}, \sum_{m=1}^M x_m = x, x_m \geq 0, <br />
z = \max_{x_m} \min_{k=1.\cdot,K} (y_k).<br />
    \alpha, \beta are constants


    if we define a function of z=f(x), prove f(.) is continuous to x.

    Thanks
    I barely understand what the functions are. Could you maybe give a little background?
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    I have modified the contents. thank!
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by clover616 View Post
    I have modified the contents. thank!
    Please forgive my ignorance but I am still having trouble understanding. Here is the gist of what I've gotten as of now:

    We have a fixed set of constants (the \alpha,\betas) and a function y:\mathbb{R}^M\to\mathbb{R} by a function (I understand the function itself). Clearly the order of the tuple is very relevant since for example \frac{1}{100}\cdot 10+\frac{1}{10}\cdot 1\leqslant \frac{1}{10}\cdot 1+\frac{1}{100}\cdot 10.

    Now here is where I get lost. So you would agree that the above implies that we could maximize our function based on the order of our xs. But, this maximization depends on the fixed \alpha,\betas and so really it seems that z:\mathbb{R}^M\to\mathbb{R}^M where (\alpha_1,\beta_1,\cdots)\mapsto (x_1,\cdots,x_M) where x_1,\cdots,x_M are the values which maximize the function for the values of alpha and beta specified.

    So, how can we fix the alpha and betas? Is this function even well defined?
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    Quote Originally Posted by Drexel28 View Post
    Please forgive my ignorance but I am still having trouble understanding. Here is the gist of what I've gotten as of now:

    We have a fixed set of constants (the \alpha,\betas) and a function y:\mathbb{R}^M\to\mathbb{R} by a function (I understand the function itself). Clearly the order of the tuple is very relevant since for example \frac{1}{100}\cdot 10+\frac{1}{10}\cdot 1\leqslant \frac{1}{10}\cdot 1+\frac{1}{100}\cdot 10.

    Now here is where I get lost. So you would agree that the above implies that we could maximize our function based on the order of our xs. But, this maximization depends on the fixed \alpha,\betas and so really it seems that z:\mathbb{R}^M\to\mathbb{R}^M where (\alpha_1,\beta_1,\cdots)\mapsto (x_1,\cdots,x_M) where x_1,\cdots,x_M are the values which maximize the function for the values of alpha and beta specified.

    So, how can we fix the alpha and betas? Is this function even well defined?
    sorry about my ambiguous definition of \alpha_{m,k},\beta_{m,k}.
    actually, in the function of y_k, the \alpha_{m,k},\beta_{m,k} are predetermined in my model regardness of the x_m or we can say \alpha_{m,k},\beta_{m,k} are independent to the x_m.

    please have a look of my proof for the monotonicity of the funtion, then may be you can know about the problem.

    In my problem , I need to focus on the continuity between z and x, instead of that between z and x_m and there is no closed form solution for x_m, so this make the problem difficult.
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    Last edited by clover616; March 30th 2010 at 09:39 PM.
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by clover616 View Post
    sorry about my ambiguous definition of \alpha_{m,k},\beta_{m,k}.
    actually, in the function of y_k, the \alpha_{m,k},\beta_{m,k} are predetermined in my model regardness of the x_m or we can say \alpha_{m,k},\beta_{m,k} are independent to the x_m.

    please have a look of my proof for the monotonicity of the funtion, then may be you can know about the problem.

    In my problem , I need to focus on the continuity between z and x, instead of that between z and x_m and there is no closed form solution for x_m, so this make the problem difficult.
    I really don't want to leave you hanging so I'll be straight with you. This problem seems pretty involved and I personally don't have the time to devote it the time it deserves.. I'll give it one more go a little later but don't hold your breath.
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    Quote Originally Posted by Drexel28 View Post
    I really don't want to leave you hanging so I'll be straight with you. This problem seems pretty involved and I personally don't have the time to devote it the time it deserves.. I'll give it one more go a little later but don't hold your breath.

    Thank you very much for your kind assistance and help!
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by clover616 View Post
    Thank you very much for your kind assistance and help!
    I'm sorry to say problems like this (those which take time to understand...especially because there HAS to be some kind of background to this problem) usually go unanswered on forums. People just don't have the time and aren't willing to put the effort into them. I wish you the best of luck with this problem and if you have any smaller bits of the problem that you would like help on (or another problem for that matter) please don't hesitate to re-post. Most of your question will be answered.
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