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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?

Note by Mvs Saketh
1 year, 11 months ago

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Well 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 · 1 year, 11 months ago

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@Ronak Agarwal Thankyou , that was awesome :) Mvs Saketh · 1 year, 11 months ago

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@Mvs Saketh Are you learning statistical mechanics. @Mvs Saketh Ronak Agarwal · 1 year, 11 months ago

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@Ronak Agarwal @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 · 1 year, 11 months ago

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@Mvs Saketh Even I am wondering about this, this should be \(\frac{1}{e}\) but then the identity remains unproven. Ronak Agarwal · 1 year, 11 months ago

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@Ronak Agarwal 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 suitable Mvs Saketh · 1 year, 11 months ago

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@Ronak Agarwal indeed , and thanks for helping :) Mvs Saketh · 1 year, 11 months ago

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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 Shashwat Shukla · 1 year, 11 months ago

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@Shashwat Shukla 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 Mvs Saketh · 1 year, 11 months ago

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@Mvs Saketh 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... Shashwat Shukla · 1 year, 11 months ago

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@Shashwat Shukla Yes, indeed , i was studying the second law of thermodynamics, and the mathematical meaning of entropy Mvs Saketh · 1 year, 11 months ago

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@Mvs Saketh Cool :) Shashwat Shukla · 1 year, 11 months ago

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please help @Ronak Agarwal @Shashwat Shukla @Deepanshu Gupta @Pratik Shastri and any one else who can Mvs Saketh · 1 year, 11 months ago

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