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Math Help - Deriving the equation

  1. #1
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    Deriving the equation

    Hi all!

    I have the following general equation which is: s(s+t)=((1-t)s(x)^\alpha + ts(x+1)^\alpha)^\frac{1}{\alpha}. where s is survival function

    It is told to me that when I substitute \alpha =0, I will be able to obtain

    log(s(x+t))=(1-t)log (s(x))+(t) log (s(x+1)).

    But how do I do that since when \alpha=0, 1/\alpha goes to infinity??
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  2. #2
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    Quote Originally Posted by noob mathematician View Post
    Hi all!

    I have the following general equation which is: s(s+t)=((1-t)s(x)^\alpha + ts(x+1)^\alpha)^\frac{1}{\alpha}. where s is survival function

    It is told to me that when I substitute \alpha =0, I will be able to obtain

    log(s(x+t))=(1-t)log (s(x))+(t) log (s(x+1)).

    But how do I do that since when \alpha=0, 1/\alpha goes to infinity??
    Since direct substitution leads to the indeterminant form 1^{\infty} you would be expected to take the limit \alpha \to 0.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Since direct substitution leads to the indeterminant form 1^{\infty} you would be expected to take the limit \alpha \to 0.
    Thanks for the hint.
    But I still finding it hard to take the limit to 0. How do I go about doing it? Substitute \alpha with a very small number? 0.000001 etc
    Or do I differentiate the equation?
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  4. #4
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    Quote Originally Posted by noob mathematician View Post
    Thanks for the hint.
    But I still finding it hard to take the limit to 0. How do I go about doing it? Substitute \alpha with a very small number? 0.000001 etc Mr F says: NO. That's not how you take a limit.
    Or do I differentiate the equation?
    The technique can be found here: Indeterminate form - Wikipedia, the free encyclopedia
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  5. #5
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    Oh great! L'H˘pital's rule.. Thanks so much
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  6. #6
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    take the logarithm first
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