Can someone please prove that

\(\lim _{ N \to \infty }{ \frac { N! }{ \left( \frac { N+m }{ 2 } \right) !\left( \frac { N-m }{ 2 } \right) ! } \quad } =\quad { 2 }^{ N }{ e }^{ -\frac { { m }^{ 2 } }{ 2N } }\)

I tried to use strilings approximation (simple form) and arrived here

\( \displaystyle\lim _{ N \to \infty }{ \frac { N! }{ \left( \frac { N+m }{ 2 } \right) !\left( \frac { N-m }{ 2 } \right) ! } \quad } \simeq \quad \frac { { N }^{ N } }{ { e }^{ N } } \frac { { e }^{ (N+m)/2\quad +\quad (N-m)/2 }{ 2 }^{ N+m)/2\quad +\quad (N-m)/2 } }{ { (N+m) }^{ (N+m)/2 }\left( N-m \right) ^{ (N-m)/2 } } ={ 2 }^{ N }\frac { { N }^{ N } }{ { (N+m) }^{ (N+m)/2 }\left( N-m \right) ^{ (N-m)/2 } } \)

but how to go beyond?

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## Comments

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TopNewestWell it's easy from here rewrite it as :

\[ \displaystyle L = \lim _{ n\rightarrow \infty }{\dfrac { { 2 }^{ N } }{ { (1+\frac { m }{ N } ) }^{ N/2 }{ (1-\frac { m }{ N } ) }^{ N/2 }{ (\frac { N+m }{ N-m } ) }^{ m/2 } }} \]

\[\displaystyle \Rightarrow \lim _{ n\rightarrow \infty }{ \frac { { 2 }^{ N } }{ { (1-\frac { { m }^{ 2 } }{ { N }^{ 2 } } ) }^{ N/2 }{ (\frac { N+m }{ N-m } ) }^{ m/2 } } } \]

\[\displaystyle \Rightarrow L = \lim _{ n\rightarrow \infty }{ \frac { { 2 }^{ N }{ { ((1-\frac { { m }^{ 2 } }{ { N }^{ 2 } } ) }^{ { N }^{ 2 }/{ m }^{ 2 } }) }^{ { -m }^{ 2 }/2N } }{ { (\frac { N+m }{ N-m } ) }^{ m/2 } } } \]

Now \[\displaystyle \lim _{ n\rightarrow \infty }{ { ((1-\frac { { m }^{ 2 } }{ { N }^{ 2 } } ) }^{ { N }^{ 2 }/{ m }^{ 2 } } } =e\] and

\[\displaystyle \lim _{ n\rightarrow \infty }{ { (\frac { N+m }{ N-m } ) }^{ m/2 } } =1\]

Resulting in \[\displaystyle L ={ 2 }^{ N }{ e }^{ { -m }^{ 2 }/2N }\]

@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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please help @Ronak Agarwal @Shashwat Shukla @Deepanshu Gupta @Pratik Shastri and any one else who can

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You're missing a \(\sqrt{\frac{\pi n}{2}}\) in the denominator.

It can be shown (using the central limit theorem) that :

\[\binom{n}{k}=\frac{2^n}{\sqrt{\frac{n\pi}{2}}}e^{-\frac{(k-\frac{n}{2})^2}{\frac{n}{2}}}\] (As \(n\rightarrow \infty\))

Here, setting \(k=\frac{n \pm m}{2}\) 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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