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Topic: Zeroes of n! (Read 1074 times) |
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ThudnBlunder
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Zeroes of n!
« on: Aug 27th, 2010, 8:26am » |
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Nay, not with how many naughts doth n! endeth.
Rather, as n   what fraction of the digits of n! are these end zeroes?
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| « Last Edit: Nov 30th, 2010, 6:19am by ThudnBlunder » |
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towr
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Re: Zeroes of n!
« Reply #1 on: Aug 27th, 2010, 2:24pm » |
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There are about n/4 zeroes,
n! is about sqrt((2n+1/3)pi) nn e-n
So, divide the two and take the limit and you're left with nothing.
Another way to put it, the number of trailing zeroes grow linearly, the number of digits grows exponentially, hence the fraction of trailing zeroes goes to zero.
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ThudnBlunder
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Re: Zeroes of n!
« Reply #2 on: Aug 27th, 2010, 4:13pm » |
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on Aug 27th, 2010, 2:24pm, towr wrote:
n! is about sqrt((2n+1/3)pi) nn e-n
So, divide the two... |
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No harm in taking logs first...
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ThudnBlunder
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Re: Zeroes of n!
« Reply #3 on: Aug 29th, 2010, 1:20pm » |
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on Aug 27th, 2010, 2:24pm, towr wrote:
...the number of digits grows exponentially.... |
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NOT...
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towr
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Re: Zeroes of n!
« Reply #4 on: Aug 29th, 2010, 1:38pm » |
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In order of the size of the input they do.
[edit]Well, okay, it doesn't really help to use different measures in both terms..
So, rather let's just take Stirling's approximation, and go with [n/4] / [n ln(n)-n] -> 0 as n -> inf.
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| « Last Edit: Aug 29th, 2010, 1:46pm by towr » |
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