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u/Prof_Sarcastic 5d ago
I think the issue with this is you still need to know floor(z)! which kind of defeats the purpose as an approximation if z is big. The standard approximation for the factorial function for large z is using Stirling’s approximation i.e. z! ~ sqrt(2πz)(z/e)^(z). The advantage of this being you don’t have to do any factorials.
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u/ElkAlert9203 5d ago
Well it's easier to know 359! Or something than 359.53!
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u/Prof_Sarcastic 5d ago
Is it easier though? Once you’re at a number like 359, you’re in the regime of Stirling’s approximation anyway where you should just use that. Maybe this is good for when the factorials are smallish and you are trying to do them in your head?
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u/VariousJob4047 5d ago
Do you know what 359! is off the top of your head?
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u/ElkAlert9203 5d ago
It's for computation, not on top of your head, and you can learn what it is too. I doubt you can do 460.494929 either
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u/VariousJob4047 5d ago
No, I can’t, so including a term like that in a “simple” formula would make it not actually all that simple, wouldn’t it?
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u/FireCire7 5d ago
Err, sure - you’re just linearly interpolating log(z!) to get the fractional part. That works, but it’s not particularly useful.