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derive an equation to represent n! Instead of multiplying from 1 to n. One straight equation not involving factorial like the way we represent summation of r as n(n+1)/2, where r varies from 1 to n. A similar type of expression iam looking for

Mohd atif
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    $\sum_{i=1}^n (n-1)!$. There, I only multiplied from $1$ to $n-1$... No, seriously, what are you looking for, where are your own approaches and ideas? – Dirk Apr 27 '17 at 13:12
  • $n!=\Gamma (n+1)$? You really need to give us some idea of what you are looking for. – lulu Apr 27 '17 at 13:13
  • One single straight equation not involving any factorial – Mohd atif Apr 27 '17 at 13:14
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    So, what's wrong with $n!=\Gamma (n+1)=\int_0^{\infty}t^{n}e^{-t}dt$? – lulu Apr 27 '17 at 13:15
  • If you are looking for some easy formula to plug into your calculator, you're not going to find one (but your calculator might have a factorial function). Stirling's approximation (https://en.wikipedia.org/wiki/Stirling%27s_approximation) is a formula to approximate $n!$ when $n$ is large: $n! \approx \sqrt{2\pi n} \left(\frac{n}{e}\right)^n$. But if you want $n!$ exactly, you just have to multiply $1 \cdot 2 \cdot 3 \cdots n$, or get a computer to do it for you. – kccu Apr 27 '17 at 13:19
  • More simpler , solution should contain powers of n and no integration – Mohd atif Apr 27 '17 at 13:21
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    @Mohdatif There is no such formula. – kccu Apr 27 '17 at 13:21
  • An exact solution would be more appreciated – Mohd atif Apr 27 '17 at 13:22
  • There is and u can derive using Euler mclaurins formula – Mohd atif Apr 27 '17 at 13:23
  • @Mohdatif That gives you Stirling's approximation, which I posted above. – kccu Apr 27 '17 at 13:26
  • Yeah I know that's why I asked for exact equation not approximation – Mohd atif Apr 27 '17 at 13:30
  • $n! =(1-t)^{n+1} \frac{d^n}{dt^n}(1-t)^{-1}$ – achille hui Apr 27 '17 at 13:31
  • No differentiation, Not that equation – Mohd atif Apr 27 '17 at 13:34
  • @kccu not approximate equation , need an exact equation – Mohd atif Apr 27 '17 at 13:40
  • @Mohidatif The Euler-Maclaurin formula does not lead to an exact equation, unless you allow infinite sums. Stirling's approximation does not lead to an exact equation, unless you allow infinite sums. – Ben Grossmann Apr 27 '17 at 14:11
  • even taylor series doesnt lead exact solution unless you allow infinite terms. but we always can try to come closer to the solution – Mohd atif Apr 27 '17 at 17:33
  • here is your solution n! = √2π e^( (n+0.5) logn - n + 1 ⁄ (12n ) - 1 ⁄ (360 n^3 )+1 ⁄ (1260 n^5) - 1 ⁄ (1680 n^7 ) ) sorry for not formatting – Mohd atif Apr 27 '17 at 18:03
  • @Mohdatif If you know this formula why are you asking us for it? – kccu Apr 27 '17 at 18:46

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There is no formula for $n!$ involving the usual arithmetic operations because $n!$ grows much faster than any rational function.

lhf
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  • See also https://math.stackexchange.com/questions/98481/is-a-factorial-an-algebraic-function-and-an-elementary-function. – lhf Apr 27 '17 at 13:32
  • There is as the kccu said above Stirling equation gives approximately, but Euler mclaurins equation gives you exact equation – Mohd atif Apr 27 '17 at 13:32