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Math Help - Series and continuity

  1. #1
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    Series and continuity

    I have two formula;


    f(x) = \frac{a^x}{\displaystyle{\sum_{i=0}^{d-1} s^i}}

    g(x) = \frac{\displaystyle{\sum_{i=1}^{x-1} a^i}}{\displaystyle{\sum_{i=0}^{x-1} a^i}}

    With 'a' being a constant, 1>a>0.

    I was looking for the equivalent continuous function, that is, functions that would be exactly the same as f(x) and g(x) but would be continuous on [0, +inf

    I know that these functions;

    f(x) = \displaystyle\frac{a^x(a-1)}{a^x-1}

    g(x) = \displaystyle\frac{a^x-a}{a^x-1}

    ...work fine, I'm pretty they're exactly what I'm looking for, but I have no idea how to get (with some rigor) from the discrete form to the continuous form
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  2. #2
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    Thus sums do not make sense if x is not a positive integer.
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  3. #3
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    I don't understand how your equation helps related my discrete and continuous equation ?
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  4. #4
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    As TPH pointed out, the sums make sense only if x is an integer.
    To elaborate, the symbol \sum_{i=1}^{x-1} means to add all the terms in the summation with i= 1,2,3... upto x-1. This summation implicitly implies x-1 \in \mathbb{Z} and thus x is an integer.

    The first function can be continuous, but the second one is clearly discrete!
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  5. #5
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    But;

    g(x) = \displaystyle\frac{a^x-a}{a^x-1}

    It works, for every single point, it works ! So, at worst, it's a very good approximation.

    And there is a mistake, the first equation should read;

    <br />
f(x) = \frac{a^x}{\displaystyle{\sum_{i=0}^{x-1} a^i}}<br />

    I'm not looking for the exact same function, I'm looking for a continuous function that would be exactly the same as the discrete one on for integers. The two equations I found do this, but it was done through induction, how can I get the continuous forms from the discrete equations.
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  6. #6
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by AnnM View Post
    But;

    g(x) = \displaystyle\frac{a^x-a}{a^x-1}

    It works, for every single point, it works ! So, at worst, it's a very good approximation.
    however, the point there is domain of g.
    since x is in the upper limit of your summation, it should be a positive interger like what they told. in fact, it should be x \in \mathbb{Z}^+ such that x > 1.

    EDIT: now i get your point.. that g works fine..
    Last edited by kalagota; July 10th 2008 at 10:19 PM.
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  7. #7
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by AnnM View Post

    And there is a mistake, the first equation should read;

    <br />
f(x) = \frac{a^x}{\displaystyle{\sum_{i=0}^{x-1} a^i}}<br />

    I'm not looking for the exact same function, I'm looking for a continuous function that would be exactly the same as the discrete one on for integers. The two equations I found do this, but it was done through induction, how can I get the continuous forms from the discrete equations.
    this gives a clearer view..

    i think, this will work.. f(x) = \frac{a^{x+1}-a^x}{a^x - 1} = \frac{a^x(a-1)}{a^x - 1}
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  8. #8
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    I found how I did it in the first place, by definition;

     \sum_{i=0}^{x-1} a^i = \frac{a^x-1}{a-1}

    So,

     \sum_{i=1}^{x-1} a^i = \sum_{i=0}^{x-1} a^i -1 = \frac{a^x-1}{a-1} - 1 = \frac{a^x-1-a+1}{a-1}  = \frac{a^x-a}{a-1}

     f(x) = a^x\times \left(\frac{a^x-1}{a-1}\right)^{-1}

     f(x) = a^x\times \frac{a-1}{a^x-1}

     f(x) = \frac{a^x(a-1)}{a^x-1}

    For g(x);

     g(x) = \left(\frac{a^x-a}{a-1}\right) \times \left(\frac{a^x-1}{a-1}\right)^{-1}

     g(x) = \left(\frac{a^x-a}{a-1}\right) \times \left(\frac{a-1}{a^x-1}\right)

     g(x) = \frac{a^x-a}{a^x-1}
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  9. #9
    Senior Member nikhil's Avatar
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    Hi!AnnM
    hmmm..Though its very strange that to me this question seem to be easy if I have understood what you actually meant(talking about concept)(generally it doesn't happens).
    Let me first talk about f(x).
    First you have given a sigma notation.which make it a discreate function that is you can not put x=.7 as x should be a positive integer.but then you described it using geometric prog..(I hope this conversion was not your main question)this changed the behaviour of f(x) that is now x=.7 is permissible .but still f(x) is not continuous.so now find the points where f(x) is discontinuous.now at the point of discontinuity take the limit
    and express that point separatly.This will make the function continuous for sure.if no limit exist then it will be impossible to find an continuous function.you have already given a domain i.e[0,infinity).so wo don't have to bother about negative values of x.so for example first find the limit of f(x) when x approaching to 0 (positive).I guess you will get +infinity so define it separatly.f(x) will remain the same where it is defined.Do same with g(x).
    Above procedure (taking limit and defining) is used tn make a function continuous.
    Hope this helped you.
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