Can someone please prove that
limN→∞(2N+m)!(2N−m)!N!=2Ne−2Nm2
I tried to use strilings approximation (simple form) and arrived here
N→∞lim(2N+m)!(2N−m)!N!≃eNNN(N+m)(N+m)/2(N−m)(N−m)/2e(N+m)/2+(N−m)/22N+m)/2+(N−m)/2=2N(N+m)(N+m)/2(N−m)(N−m)/2NN
but how to go beyond?
#Combinatorics
#Calculus
#Limits
#Statisticalmechanics
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Well it's easy from here rewrite it as :
L=n→∞lim(1+Nm)N/2(1−Nm)N/2(N−mN+m)m/22N
⇒n→∞lim(1−N2m2)N/2(N−mN+m)m/22N
⇒L=n→∞lim(N−mN+m)m/22N((1−N2m2)N2/m2)−m2/2N
Now n→∞lim((1−N2m2)N2/m2=e and
n→∞lim(N−mN+m)m/2=1
Resulting in L=2Ne−m2/2N
@Mvs Saketh
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Thankyou , that was awesome :)
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Are you learning statistical mechanics. @Mvs Saketh
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@Ronak Agarwal - sorry but one small doubt again, there is a minus sign there,
i mean (1+x)^(1/x) = e so here it should be (1-x)^(1/x) =1/e right, why did you write e ?
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e1 but then the identity remains unproven.
Even I am wondering about this, this should beLog in to reply
please help @Ronak Agarwal @Shashwat Shukla @Deepanshu Gupta @Pratik Shastri and any one else who can
You're missing a 2πn in the denominator.
It can be shown (using the central limit theorem) that :
(kn)=2nπ2ne−2n(k−2n)2 (As n→∞)
Here, setting k=2n±m yields the desired result.
Mvs Saketh
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yes thats why i said that i have used the simpler form of stirlings theorem, without involving the pi terms,
The real expression is the one you said but i was studying just the introductory parts of the statistical limits and there the simpler form without the pi part was given, yours is indeed the more accurate one
and yes thanks for giving the general limit for nCr, that is helpful, thanks
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You're welcome :)...Are you studying Boltzmann's law for entropy by any chance? This expression looks like one for the multiplicity of a system...
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