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Size of x!

  1. Jan 27, 2009 #1
    is there a way to determine (does not need to be exact) the number of digits in x! ?

    sorry if this is kind of pointless
     
  2. jcsd
  3. Jan 27, 2009 #2

    mathman

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    For large n, use Stirling's formula:

    n! ~ (2[pi]n)1/2 (n/e)n
     
  4. Jan 27, 2009 #3

    CRGreathouse

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    O(x log x).

    To be more precise, x log x - x log e + O(log x), where the base of the logarithms is the base you want to express the number.
     
  5. Jan 27, 2009 #4

    Gib Z

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    Don't forget to then apply

    [tex]D = 1 + \lfloor \log_{10} \lfloor n! \rfloor \rfloor [/tex]

    after that for the number of digits.
     
  6. Jan 28, 2009 #5

    disregardthat

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    If [tex]n![/tex] is hard to calculate, the series

    [tex]D = 1 + \lfloor \sum^n_{i=1} \log_{10}i \rfloor[/tex]

    should be easier to calculate - for integer [tex]n[/tex].
     
  7. Jan 28, 2009 #6

    Gib Z

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    Depends how large n is. If it's quite large, calculating logs for each term up to it may not be easier than using Stirlings Formula which simplifies into something that doesn't look that bad after a log anyway. But that definitely is a good idea =]
     
  8. Jan 28, 2009 #7
    Anyone know of a way to get an answer just to a certain precision in mathematica?
    I do no mean truncating the answer, but more the way a person would, so for example,
    1/"insert incredibly long hard to calculate mess here" as zero, or instead of getting the inside of a natural log to hundreds of decimal points, just get it to 10 or so.
    is there a way to do that?
     
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