What is the best estimation of the factorial function?

Question

I have to calculate factorials of an arbitrarily large integers mod another arbitrarily large integer. i have considered Stirling's Approximation and Ramanujan’s factorial approximation. is it possible to get better estimates?

If you are using modular arithmetic, analytic estimates might not be the most useful thing for you, depending on what exactly your problem is. Instead things like https://en.wikipedia.org/wiki/Wilson%27s_theorem might help. — Alex J Best, Nov 22 '18 at 11:53
There's the gamma function but I don't think it will help you much in contexts of modular arithmetic... — Prasun Biswas, Nov 22 '18 at 11:57
lets say, for the time being i just need to estimate the factorial. what will the nearest approximation be? — Malik Hamza Murtaza, Nov 22 '18 at 12:05
Have a look at https://math.stackexchange.com/questions/676952/is-ramanujans-approximation-for-the-factorial-optimal-or-can-it-be-tweaked-a/677163#677163. For $n=20$, what I wrote gives $2432902008176665737$ to be compared to the exact $2432902008176640000$ — Claude Leibovici, Nov 22 '18 at 12:32
Yes that's exactly what got me started on this path... that's a great answer. — Malik Hamza Murtaza, Nov 22 '18 at 12:39
I don't know if this could be of interest since the coefficients start to be huge. Using the same kind of expression with degree $2$ in numerator and degree $3$ in denominator, for $n=20$, I get $2432902008176639896$. If you want, I could post the numbers in an answer. — Claude Leibovici, Nov 22 '18 at 13:05
if i am not wrong totally, it is my understanding that as the factorial grows bigger and bigger, there are a certain number of trailing zeroes, if the percentage error can somehow be reduced to magnitude, wont we be able to just ignore the error? — Malik Hamza Murtaza, Nov 22 '18 at 13:15
Also how did you come up with these numbers? could you please explain? — Malik Hamza Murtaza, Nov 22 '18 at 13:27

Claude Leibovici · Accepted Answer · 2018-11-23T05:40:13.973

Writing $$n!\approx\sqrt{\pi}\left(\frac{n}{e}\right)^n\root\LARGE{6}\of{8n^3+4n^2+n+\frac 1 {30}\color{red}{-}x(n)}$$ with $$x(n)=\frac{a_0+a_1n+a_2n^2}{b_0+b_1n+b_2n^2+b_3n^3}$$ the coefficients are $$\left( \begin{array}{cc} a_0 & 12521740840824081\\ a_1 & 132077016740516320 \\ a_2 & 261892615461486240 \\ b_0 &3339455095907419720 \\ b_1 & 7902477164268212400 \\ b_2 & 5812898776788230400 \end{array} \right)$$ A few values $$\left( \begin{array}{ccc} n & \text{approximation} & \text{exact} \\ 10 & 3628800 & 3628800 \\ 15 & 1307674368000 & 1307674368000 \\ 20 & 2432902008176639896 & 2432902008176640000 \\ 25 & 15511210043330985910414618 & 15511210043330985984000000 \\ 30 & 265252859812191058429178640362769 & 265252859812191058636308480000000 \end{array} \right)$$

Edit

Using ratios of polyomials (as done in this answer and the previous one) leads to incredibly huge coefficients. Thinking more about it, I thought that is would be better to just write $$x(n)=\sum_{k=1}^m \frac{a_k}{n^k}$$ and, when required, transform this expansion to the desired $[p,p+1]$ Padé approximant. The coefficients so obtained are listed below $$\left( \begin{array}{cc} k & a_k \\ 1 & \frac{11}{240} \\ 2 & -\frac{79}{3360} \\ 3 & -\frac{3539}{201600} \\ 4 & \frac{9511}{403200} \\ 5 & \frac{10051}{716800} \\ 6 & -\frac{233934691}{6386688000} \\ 7 & -\frac{3595113569}{178827264000} \\ 8 & \frac{403527851669}{4649508864000} \\ 9 & \frac{25622861661869}{557941063680000} \\ 10 & -\frac{30016604936501}{101443829760000} \\ 11 & -\frac{685661227463561}{4463528509440000} \\ 12 & \frac{109896661164737049961}{79673983893504000000} \end{array} \right)$$ For $n=25$, this would lead to $\color{blue}{1551121004333098598400000}5$.

For $n=30$ , this would lead to $\color{blue}{26525285981219105863630848}5359781$.

This seems to be significantly better.

score 0 · Answer 2 · answered Nov 23 '18 at 07:06

Stirlings approximation is good ... especially if in the expression for $\ln(n!)$ you include the series in odd powers of 1/n. Are you aware of that series - Stirlings aporoximation is all-too-often given without it. The coefficients of it are not too compliated - Bernoulli numbers multiplied by simple factors. Ah yes! maybe it is troublesome for very large n then because of that - the way the Bernoulli numbers 'turn round'.

This might answer your question though.

What is the best estimation of the factorial function?

2 Answers2