Help limits

Can someone please prove that

limNN!(N+m2)!(Nm2)!=2Nem22N\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

limNN!(N+m2)!(Nm2)!NNeNe(N+m)/2+(Nm)/22N+m)/2+(Nm)/2(N+m)(N+m)/2(Nm)(Nm)/2=2NNN(N+m)(N+m)/2(Nm)(Nm)/2 \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
4 years, 8 months ago

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1 vote

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Well it's easy from here rewrite it as :

L=limn2N(1+mN)N/2(1mN)N/2(N+mNm)m/2 \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 } }}

limn2N(1m2N2)N/2(N+mNm)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 } } }

L=limn2N((1m2N2)N2/m2)m2/2N(N+mNm)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 limn((1m2N2)N2/m2=e\displaystyle \lim _{ n\rightarrow \infty }{ { ((1-\frac { { m }^{ 2 } }{ { N }^{ 2 } } ) }^{ { N }^{ 2 }/{ m }^{ 2 } } } =e and

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

Resulting in L=2Nem2/2N\displaystyle L ={ 2 }^{ N }{ e }^{ { -m }^{ 2 }/2N }

@Mvs Saketh

Ronak Agarwal - 4 years, 8 months ago

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Thankyou , that was awesome :)

Mvs Saketh - 4 years, 8 months ago

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Are you learning statistical mechanics. @Mvs Saketh

Ronak Agarwal - 4 years, 8 months ago

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@Ronak Agarwal indeed , and thanks for helping :)

Mvs Saketh - 4 years, 8 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 - 4 years, 7 months ago

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

Ronak Agarwal - 4 years, 7 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 - 4 years, 7 months ago

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

Mvs Saketh - 4 years, 8 months ago

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You're missing a πn2\sqrt{\frac{\pi n}{2}} in the denominator.

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

(nk)=2nnπ2e(kn2)2n2\binom{n}{k}=\frac{2^n}{\sqrt{\frac{n\pi}{2}}}e^{-\frac{(k-\frac{n}{2})^2}{\frac{n}{2}}} (As nn\rightarrow \infty)

Here, setting k=n±m2k=\frac{n \pm m}{2} yields the desired result.

Mvs Saketh

Shashwat Shukla - 4 years, 8 months ago

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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

Mvs Saketh - 4 years, 8 months ago

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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...

Shashwat Shukla - 4 years, 8 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 - 4 years, 8 months ago

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@Mvs Saketh Cool :)

Shashwat Shukla - 4 years, 8 months ago

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