3 ms·
A decent approximation to n! is sqrt(2pi n) * (n/e)^n. A worse approximation is that n! is about (n/e)^n. So, if b(n) is the base in b^n=n!, then b(n) is about
by tylerneylon 3y ago
A decent approximation to n! is sqrt(2pi n) * (n/e)^n. A worse approximation is that n! is about (n/e)^n.
So, if b(n) is the base in b^n=n!, then b(n) is about (n/e). [I see this as the key observation of the article.]
Keeping in mind that (n/e)^n is a bit smaller than n!, a more accurate value for b(n) is a little higher than n/e.
So it makes sense that 25! ~= 10^25 because 25/e is a bit below 10, and the adjusted value for b(n) ought to be a little higher than 25/e; 10 will do in this case.
(And 270! ~= 100^270, etc.)
- bo1024 3y agoYes! I'm surprised he didn't mention that the factor is e.
- svat 3y agoStirling's approximation carried out to more terms gives: b = n/e + ln(2πn)/2e + (3(ln(2πn))^2 + 2)/(24en) + o(1/n) which becomes a good approximation for large n: https://colab.research.google.com/drive/17cC8YUqKjEBLsNJkJ4IxA_LKDdg9Hflv https://colab.research.google.com/drive/17cC8YUqKjEBLsNJkJ4I...