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?

## 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 – Ronak Agarwal · 2 years, 1 month ago

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– Mvs Saketh · 2 years, 1 month ago

Thankyou , that was awesome :)Log in to reply

@Mvs Saketh – Ronak Agarwal · 2 years, 1 month ago

Are you learning statistical mechanics.Log in to reply

@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 ? – Mvs Saketh · 2 years, 1 month ago

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– Ronak Agarwal · 2 years, 1 month ago

Even I am wondering about this, this should be \(\frac{1}{e}\) but then the identity remains unproven.Log in to reply

– Mvs Saketh · 2 years, 1 month ago

but it is definitely e because function must attain maxima at m=0, and also be positive on either side of m=0, and hence e^(-x^2) is only suitableLog in to reply

– Mvs Saketh · 2 years, 1 month ago

indeed , and thanks for helping :)Log in to reply

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 – Shashwat Shukla · 2 years, 1 month ago

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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 – Mvs Saketh · 2 years, 1 month ago

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– Shashwat Shukla · 2 years, 1 month ago

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...Log in to reply

– Mvs Saketh · 2 years, 1 month ago

Yes, indeed , i was studying the second law of thermodynamics, and the mathematical meaning of entropyLog in to reply

– Shashwat Shukla · 2 years, 1 month ago

Cool :)Log in to reply

please help @Ronak Agarwal @Shashwat Shukla @Deepanshu Gupta @Pratik Shastri and any one else who can – Mvs Saketh · 2 years, 1 month ago

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