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Math Help - How to numericall compute a series with a big factorial in it

  1. #1
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    How to numericall compute a series with a big factorial in it

    Hi,

    Imagine I got the following series:

    f(\alpha)=(n!/\alpha!)\sum_{i=0}^n(-1)^{(i-\alpha)}u^i/[(i-\alpha)!(n-i)!]

    In which u \in[0,1] (and u \in REAL)
    While \alpha and i and n are INTEGERS

    The (-1)^something causes an alternating sign, in this way the sum doesn't explode, but seems to end up with a real number between 0 and 1,
    adding something big, substracting something even bigger and so on...

    (In case the factorial of a negative number is needed, for example in (i- \alpha)! this is set to 1.

    The problem is that I need to compute the result of this series for large n (say 10000) and my computer breaks down on n! (of couse) for large n.

    Is there an easier way to get to a result without computer overflow?

    Thanks,

    Gerrit
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  2. #2
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    Re: How to numericall compute a series with a big factorial in it

    Consider:
    1) Stirling's Formula has applications to #2 and #3
    2) Gamma Functions instead of factorials
    3) Combinations: Excluding the weird negative factorials, what you have there is "i choose alpha" times "n choose i"
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  3. #3
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    Re: How to numericall compute a series with a big factorial in it

    Thanks a lot! I'll look at it tomorrow. I forgot to tell that \alpha \leq n

    I don't see how the negative factorial can be avoided, as in "i choose \alpha" you will also get (i- \alpha)<0

    I looked the Gamma function up, but what's the advantage of using it? It's just the same as a factorial, so it will result in huge numbers as well, won't it?

    The stirling formula looks promising (neither knew that one), I'll have a closer look at that.

    Are you sure that adding and substracting huge numbers can be avoided by use of one of these functions? The term u^i seems to destroy the possibility to divide huge numbers between each other to small ones before adding them in the sum, as u^i is different for every i term.

    BTW1: the formula came from: (n choose \alpha) times ([n- \alpha] choose [i- \alpha])
    BTW2: What do you mean by #2 and #3?
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  4. #4
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    Re: How to numericall compute a series with a big factorial in it

    Compute one term in terms of the preceding one.
    Suppose you have a series with general term \frac{x^{n}}{n!}, then if you have the value of, say, the 100th term, (n=99), the next term would be calculated by dividing that value by 100 and then multiplying by x.
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  5. #5
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    Re: How to numericall compute a series with a big factorial in it

    Thanks, but that doesn't make you get rid of the big numbers, does it?

    Imagine that n=4 and \alpha=0, then you get:

    +1*[4!/(0!0!4!)]u^0-1*[4!/(0!1!3!)]u^1+1*[4!/(0!2!2!)]u^2-1*[4!/(0!3!1!)]u^3+1*[4!/(0!4!0!)]u^4 =

    = u^0 -4u^1 +6u^2 -4u^3 + u^4

    Which you would compute as: u^0*(1-u*[4-u*{6-u*(4-u)}])

    In that way, for huge n, you will still have the huge integers adding and substracting to eachother, or not?

    Doesn't this series converge to something?
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