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Hello I found two equations that lead to constant e. I wonder if they are known. I think especially first one is most likely known but I couldn't find, it is hard to search google with all these symbols, and please forgive my bad formatting.

  • I found this one a while ago while trying to re-find gaussian distribution from pascal triangle.

$$\lim_{n \to \infty} \frac{ (n + \sqrt{n})! \cdot (n - \sqrt{n})!} { (n!)^2} = e $$

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This one is recursive and infinite. After trying to write its formula, i thought it is best for me to write in steps. first term is

$$2* ( 1 - \frac{\left( n-1\right) }{2 \cdot n} ) + $$

now it is recursive

$$\frac{\left( n-1\right) }{2 \cdot n} \cdot [ 3* (1 - \frac{\left(n-2 \right)}{ 3\cdot n} ) + \frac{\left(n-2 \right) }{3 \cdot n} \cdot [ 4* (1 - \frac{\left(n-3 \right) }{4 \cdot n} ) + \frac{\left(n-3 \right) }{4 \cdot n}\cdot [ 5* (1 - \frac{\left(n-4 \right) }{5 \cdot n} ) .. $$

and it goes on.

Now that I am not mathematician, far from it but i want to be, an amateur one that plays with numbers. I want to know not with formula but with pure intuition, why a variable with probability $e^{-1}$ has highest randomness by itself; I still dont know. But while searching internet, i read a numeric simulation of e on wikipedia page and broke it apart to understand, this second formula came out. So if i didnt make an error, here it is.

Edit: I edited second function, there was an error, but now more simple it seems, please look again. Thanks

xcvbnm
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    It wouldn't surprise me if the first one could be deduced without too much effort from Stirling's approximation, at least for $n$ square. – Arthur Mar 05 '15 at 23:12
  • @Arthur Stirling's approximation is really an approximation to $\Gamma$ function: it works equally well for integer and non-integer arguments. –  Mar 11 '15 at 00:03
  • @Woodface But the factorial as in the question does not work for non-integer inputs, so $n$ a square is a necessary condition for the question as stated to make sense. – Arthur Mar 11 '15 at 00:30
  • Hello, I looked at wolfram alpha, for non integer numbers, it still gives answer e, also i noticed i made an error while calculating second equation and now i updated it. Can you look again, maybe it is analog to a known equation, maybe not but it is easier to know now. thanks – xcvbnm Mar 11 '15 at 13:54

1 Answers1

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Not an answer, but a quick heuristic for the first one:

$$ \frac{(n+m)!(n-m)!}{{n!}^2}=\frac{(n+1)(n+2)\cdots (n+m)}{(n-m+1)(n-m+2)\cdots n}=\prod_{k=1}^m \frac{n+k}{n-m+k}$$

For large $n$, and assuming $m\ll n$, or $m=o(n)$, this can be approximated by the middle term

$$ \left(\frac{n+m/2}{n-m/2}\right)^m\approx \left(1+\frac{m}{n}\right)^m\approx e^{m^2/n}$$

leonbloy
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  • i still dont know much, but i think here m is about being sqrt n, is about approximating (1+1/n) ^n more directly. there are sqrt n times fractions, which are near 1. in official e formula, all fractions are perfectly uniform and very very close to 1. but here closeness to 1 is broken when we get distant to n. we multiply with slightly bigger numbers. but then we cut off multiplication in fewer steps. somehow sqrt n is the number that satisfies this trade off – xcvbnm Mar 06 '15 at 00:36