# Help limits

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

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
6 years, 3 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 $\displaystyle L ={ 2 }^{ N }{ e }^{ { -m }^{ 2 }/2N }$

- 6 years, 3 months ago

Thankyou , that was awesome :)

- 6 years, 3 months ago

Are you learning statistical mechanics. @Mvs Saketh

- 6 years, 3 months ago

indeed , and thanks for helping :)

- 6 years, 3 months ago

@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 ?

- 6 years, 3 months ago

Even I am wondering about this, this should be $\frac{1}{e}$ but then the identity remains unproven.

- 6 years, 3 months 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 suitable

- 6 years, 3 months ago

- 6 years, 3 months ago

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

- 6 years, 3 months ago

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

- 6 years, 3 months 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...

- 6 years, 3 months ago

Yes, indeed , i was studying the second law of thermodynamics, and the mathematical meaning of entropy

- 6 years, 3 months ago

Cool :)

- 6 years, 3 months ago