Stirling's Approximation for a factorial raised to a power

In summary, the conversation discusses using log identities and Stirling's Approximation to find an approximation for ##[(\alpha - 1)!]^2##. However, the proposed solution is incorrect and instead, the correct approach is to first use Stirling's Approximation to approximate ##(\alpha - 1)!## and then square that result.
  • #1
rmiller70015
110
1
Homework Statement
Find Stirling's Approximation for ##[(\alpha - 1)!]^2##
Relevant Equations
For large N: ##log(N!) \approx Nlog(N)-N##
Using log identities:
##log((\alpha - 1)!^2) = 2(log(\alpha - 1)!)##
Then apply Stirling's Approximation
##(2[(\alpha - 1)log(\alpha - 1) - (\alpha - 1)##
## = 2(\alpha -1)log(\alpha -1) - 2\alpha+2##

Is this correct? I can't find a way to check this computationally.
 
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  • #2
rmiller70015 said:
Homework Statement:: Find Stirling's Approximation for ##[(\alpha - 1)!]^2##
Relevant Equations:: For large N: ##log(N!) \approx Nlog(N)-N##

Using log identities:
##log((\alpha - 1)!^2) = 2(log(\alpha - 1)!)##
Then apply Stirling's Approximation
##(2[(\alpha - 1)log(\alpha - 1) - (\alpha - 1)##
## = 2(\alpha -1)log(\alpha -1) - 2\alpha+2##

Is this correct? I can't find a way to check this computationally.
I don't think it's correct, and I get something different. If you want to approximate ##[(\alpha - 1)!]^2##, first use Stirling's to approximate ##(\alpha - 1)!##, and then square that result.
 
Last edited:
  • #3
Thanks, that was bugging me.
 
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