Is there a simple method to find or estimate how large ⌊lgn!⌋
is ? I'd like to find (or estimate) how much digits 2017!2017 has, or how much is big that number .
I tried for some little number and general form of n!n=¯ak...a4a3a2a1k=? so
I took logarithm of n!n↦nlog(n!).
It is easy when n is little number , but when go for a large number ...what we can do ?
My question can be translate as nn∑i=1log(i)=?
Answer
Stirling's approximation says that
n!≈√2πn(ne)n
Taking logarithms on both sides should be easy enough.
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