Size of x!

  • Thread starter soandos
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  • #1
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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
 

Answers and Replies

  • #2
mathman
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For large n, use Stirling's formula:

n! ~ (2[pi]n)1/2 (n/e)n
 
  • #3
CRGreathouse
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is there a way to determine (does not need to be exact) the number of digits in x! ?
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.
 
  • #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.
 
  • #5
disregardthat
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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.
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].
 
  • #6
Gib Z
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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].
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 =]
 
  • #7
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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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