A formula of the product of the first n integers?

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SUMMARY

The discussion centers on the absence of a direct formula for the product of the first n integers, commonly represented as n!. Participants confirm that while there is a formula for the sum of the first n integers, n(n+1)/2, no equivalent exists for the product. Instead, they reference Stirling's approximation for n! and Gosper's reformulation as useful alternatives. Additionally, the integral form of the factorial function, n! = Γ(n+1) = ∫₀^∞ e^(-t) t^n dt, is mentioned as a mathematical representation of the factorial.

PREREQUISITES
  • Understanding of factorial notation (n!)
  • Familiarity with Stirling's approximation
  • Knowledge of the Gamma function (Γ)
  • Basic calculus concepts, particularly integrals
NEXT STEPS
  • Research Stirling's approximation and its applications in combinatorics
  • Explore Gosper's reformulation of Stirling's formula
  • Study the properties and applications of the Gamma function
  • Learn about integral calculus, focusing on improper integrals
USEFUL FOR

Mathematicians, students studying combinatorics, and anyone interested in advanced mathematical concepts related to factorials and approximations.

quasar987
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I'm sure it exists, and it'd help me to have it. Thx!
 
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n!

......
 
No no, like there's n(n+1)/2 for the sum of the first n integers. Is there an equivalent formula for the product?
 
quasar987 said:
No no, like there's n(n+1)/2 for the sum of the first n integers. Is there an equivalent formula for the product?

I want to say no since if there was we would probably use that instead of n! everywhere, but there's stirling's approximation to n!.
 
Last edited:
You're probably right.
 
Stirling's formula gives a good approximation. I prefer to use Gosper's reformulation of same.
 
Well, there's always the integral form of the factorial function; although I doubt it's what you're looking for:

n! = \Gamma(n+1) = \int_0^{\infty} e^{-t} t^n dt
 
it's n(n+1)/2 if you start at 1 and n(n-1)/2 if you start at zero if I recall...
 
Robokapp said:
it's n(n+1)/2 if you start at 1 and n(n-1)/2 if you start at zero if I recall...

I think you mean n and n-1, respectively, if you're talking about sums. Obviously adding in 0 has no effect.
 

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