A limit converging to exp(-x²)

[Feynman's fan]

Here is the problem at hand:

[itex]
\lim\limits_{n\to\infty} \frac{\sqrt{n}}{2^{2n}} \binom {2n} {n+\lfloor x\sqrt{n} \rfloor} = \frac{1}{\sqrt{\pi}} e^{-x^2}
[/itex]

I've tried to evaluate the limit using the Stirling's formula, many things canceled out yet I got stuck.

Thanks a lot for any help or hints.

ADDED:

Here is what I did:

I was trying to squeeze the limit by using the Stirling's formula, which is:
[itex]\sqrt{2\pi}n^{n+\frac{1}{2}}e^{-n}<n!<\sqrt{2\pi}n^{n+\frac{1}{2}}e^{-n+\frac{1}{12(n-1)}}[/itex].

So I started with the RHS of the squeezing:

For simplicity: let [itex]\lfloor x\sqrt{n} \rfloor:=p[/itex].

Then

[itex]
\begin{align}
\frac{\sqrt{n}}{2^{2n}} \binom {2n} {n+p} &= \frac{\sqrt{n}(2n)!}{2^{2n}(n+p)!(n-p)!}
\\ &< \frac{\sqrt{n}\sqrt{2\pi}(2n)^{(2n)+\frac{1}{2}}e^{-(2n)+\frac{1}{12((2n)-1)}}}{2^{2n}\sqrt{2\pi}(n+p)^{(n+p)+\frac{1}{2}}e^{-(n+p)}\sqrt{2\pi}(n-p)^{(n-p)+\frac{1}{2}}e^{-(n-p)}}
\\ &< \frac{n^{2n+1}e^{\frac{1}{12(2n-1)}}}{\sqrt{\pi}(n+p)^{n+p+\frac{1}{2}}(n-p)^{n-p+\frac{1}{2}}}
\\ &< ?
\end{align}
[/itex]

 

[mfb]

Please use the homework template and show your work - the approach with Stirling's formula looks good, so it is impossible to tell where you did a mistake or where you need help.

 

[Feynman's fan]

Thank you for looking into it!

Here is what I did.

I was trying to squeeze the limit by using the Stirling's formula, which is:
[itex]\sqrt{2\pi}n^{n+\frac{1}{2}}e^{-n}<n!<\sqrt{2\pi}n^{n+\frac{1}{2}}e^{-n+\frac{1}{12(n-1)}}[/itex].

So I started with the RHS of the squeezing:

For simplicity: let [itex]\lfloor x\sqrt{n} \rfloor:=p[/itex].

Then

[itex]
\begin{align}
\frac{\sqrt{n}}{2^{2n}} \binom {2n} {n+p} &= \frac{\sqrt{n}(2n)!}{2^{2n}(n+p)!(n-p)!}
\\ &< \frac{\sqrt{n}\sqrt{2\pi}(2n)^{(2n)+\frac{1}{2}}e^{-(2n)+\frac{1}{12((2n)-1)}}}{2^{2n}\sqrt{2\pi}(n+p)^{(n+p)+\frac{1}{2}}e^{-(n+p)}\sqrt{2\pi}(n-p)^{(n-p)+\frac{1}{2}}e^{-(n-p)}}
\\ &< \frac{n^{2n+1}e^{\frac{1}{12(2n-1)}}}{\sqrt{\pi}(n+p)^{n+p+\frac{1}{2}}(n-p)^{n-p+\frac{1}{2}}}
\\ &< ?
\end{align}
[/itex]

 

[Ray Vickson]

Here is the problem at hand:

[itex]
\lim\limits_{n\to\infty} \frac{\sqrt{n}}{2^{2n}} \binom {2n} {n+\lfloor x\sqrt{n} \rfloor} = \frac{1}{\sqrt{\pi}} e^{-x^2}
[/itex]

I've tried to evaluate the limit using the Stirling's formula, many things canceled out yet I got stuck.

Thanks a lot for any help or hints.

ADDED:

Here is what I did:

I was trying to squeeze the limit by using the Stirling's formula, which is:
[itex]\sqrt{2\pi}n^{n+\frac{1}{2}}e^{-n}<n!<\sqrt{2\pi}n^{n+\frac{1}{2}}e^{-n+\frac{1}{12(n-1)}}[/itex].

So I started with the RHS of the squeezing:

For simplicity: let [itex]\lfloor x\sqrt{n} \rfloor:=p[/itex].

Then

[itex]
\begin{align}
\frac{\sqrt{n}}{2^{2n}} \binom {2n} {n+p} &= \frac{\sqrt{n}(2n)!}{2^{2n}(n+p)!(n-p)!}
\\ &< \frac{\sqrt{n}\sqrt{2\pi}(2n)^{(2n)+\frac{1}{2}}e^{-(2n)+\frac{1}{12((2n)-1)}}}{2^{2n}\sqrt{2\pi}(n+p)^{(n+p)+\frac{1}{2}}e^{-(n+p)}\sqrt{2\pi}(n-p)^{(n-p)+\frac{1}{2}}e^{-(n-p)}}
\\ &< \frac{n^{2n+1}e^{\frac{1}{12(2n-1)}}}{\sqrt{\pi}(n+p)^{n+p+\frac{1}{2}}(n-p)^{n-p+\frac{1}{2}}}
\\ &< ?
\end{align}
[/itex]

A slightly better upper bound on n! is
[tex] \sqrt{2 \pi} n^{n + \frac{1}{2}} e^{-n + \frac{1}{12n}} [/tex]
For a derivation, see Feller, Introduction to Probability Theory, Vol. 1, Wiley (1968).

 

[mfb]

You can simplify your denominator if you split the products into two factors each (with ^(n+1/2) and ^+-p).
p^2 is an integer so you can replace it back, for remaining factors of p this could be useful:
$$p\geq x\sqrt{n}-1$$

 

[Ray Vickson]

Here is the problem at hand:

[itex]
\lim\limits_{n\to\infty} \frac{\sqrt{n}}{2^{2n}} \binom {2n} {n+\lfloor x\sqrt{n} \rfloor} = \frac{1}{\sqrt{\pi}} e^{-x^2}
[/itex]

I've tried to evaluate the limit using the Stirling's formula, many things canceled out yet I got stuck.

Thanks a lot for any help or hints.

ADDED:

Here is what I did:

I was trying to squeeze the limit by using the Stirling's formula, which is:
[itex]\sqrt{2\pi}n^{n+\frac{1}{2}}e^{-n}<n!<\sqrt{2\pi}n^{n+\frac{1}{2}}e^{-n+\frac{1}{12(n-1)}}[/itex].

So I started with the RHS of the squeezing:

For simplicity: let [itex]\lfloor x\sqrt{n} \rfloor:=p[/itex].

Then

[itex]
\begin{align}
\frac{\sqrt{n}}{2^{2n}} \binom {2n} {n+p} &= \frac{\sqrt{n}(2n)!}{2^{2n}(n+p)!(n-p)!}
\\ &< \frac{\sqrt{n}\sqrt{2\pi}(2n)^{(2n)+\frac{1}{2}}e^{-(2n)+\frac{1}{12((2n)-1)}}}{2^{2n}\sqrt{2\pi}(n+p)^{(n+p)+\frac{1}{2}}e^{-(n+p)}\sqrt{2\pi}(n-p)^{(n-p)+\frac{1}{2}}e^{-(n-p)}}
\\ &< \frac{n^{2n+1}e^{\frac{1}{12(2n-1)}}}{\sqrt{\pi}(n+p)^{n+p+\frac{1}{2}}(n-p)^{n-p+\frac{1}{2}}}
\\ &< ?
\end{align}
[/itex]

I think use of the squeeze theorem is unnecessary; the simple form of Stirling is enough, because it is an asymptotic form, meaning that
[tex] \lim_{k \to \infty} \frac{k!}{\text{St}(k)} = 1, \text{ where }
\text{St}(k) = \sqrt{2 \pi} k^{k + (1/2)} e^{-k}[/tex]
Also, if I were doing it I would start by eliminating the greatest-integer function and just use ##x \sqrt{n}## in Stirling's formula. After that I would use ##x \sqrt{n}-1 \leq \lfloor x \sqrt{n} \rfloor \leq x \sqrt{n}+1## to show that the limit would not change.

 

[Feynman's fan]

I've tried the suggested things yet unfortunately I still cannot arrive at [itex]e^{-x^2}[/itex] which supposedly should follow from something like [itex](1+\frac{-x^2}{n})^n[/itex].

p^2 is an integer so you can replace it back

But how do I proceed with [itex]\lfloor x\sqrt{n} \rfloor^2[/itex]?

I think use of the squeeze theorem is unnecessary; the simple form of Stirling is enough

Could you please elaborate on this?

 

[Ray Vickson]

I've tried the suggested things yet unfortunately I still cannot arrive at [itex]e^{-x^2}[/itex] which supposedly should follow from something like [itex](1+\frac{-x^2}{n})^n[/itex].



But how do I proceed with [itex]\lfloor x\sqrt{n} \rfloor^2[/itex]?



Could you please elaborate on this?

Well, ## (1+u/n)^n \to e^u## if ##u## does not depend on ##n##, so you get the limit you want!

We can drop the floor function because
[tex]
\lim_{n \to \infty}
\frac{\Gamma(n+x\sqrt{n}+p)\Gamma(n-x\sqrt{n}-p)}
{\Gamma(n+x \sqrt{n}) \Gamma(n-x\sqrt{n})} = 1,[/tex]
as you can, and should, verify.

 

[Feynman's fan]

[tex]\lim_{n \to \infty}
\frac{\Gamma(n+x\sqrt{n}+p)\Gamma(n-x\sqrt{n}-p)}
{\Gamma(n+x \sqrt{n}) \Gamma(n-x\sqrt{n})} = 1,[/tex]

This approach sounds very interesting! Yet how did you arrive at the limit?

I thought that if we want to drop [itex]\lfloor · \rfloor[/itex] at the very beginning, we need to proceed as follows:

\begin{align}
\lim_{n \to \infty} \frac{\sqrt{n}}{2^{2n}} \binom {2n} {n+\lfloor x\sqrt{n} \rfloor} &= \lim\frac{\sqrt{n}}{2^{2n}}\frac{(2n)!}{(n+\lfloor x\sqrt{n} \rfloor)!(n-\lfloor x\sqrt{n} \rfloor)!}
\\ &= \lim\frac{\sqrt{n}}{2^{2n}} \frac{(2n)!}{ \Gamma (n+\lfloor x\sqrt{n} \rfloor+1)\ \ \Gamma(n-\lfloor x\sqrt{n} \rfloor+1)}
\\ &= \lim\frac{\sqrt{n}}{2^{2n}} \frac{(2n)!}{ \Gamma (n+ x\sqrt{n}+1)\ \ \Gamma(n- x\sqrt{n} +1)}
\end{align}
And the last step is okay because $$\lim_{n\to\infty}\frac{\Gamma (n+\lfloor x\sqrt{n} \rfloor+1)\ \ \Gamma(n-\lfloor x\sqrt{n} \rfloor+1)}{\Gamma (n+ x\sqrt{n}+1)\ \ \Gamma(n- x\sqrt{n} +1)}=1 $$
What am I doing wrong here?

 

[Ray Vickson]

This approach sounds very interesting! Yet how did you arrive at the limit?

I thought that if we want to drop [itex]\lfloor · \rfloor[/itex] at the very beginning, we need to proceed as follows:

\begin{align}
\lim_{n \to \infty} \frac{\sqrt{n}}{2^{2n}} \binom {2n} {n+\lfloor x\sqrt{n} \rfloor} &= \lim\frac{\sqrt{n}}{2^{2n}}\frac{(2n)!}{(n+\lfloor x\sqrt{n} \rfloor)!(n-\lfloor x\sqrt{n} \rfloor)!}
\\ &= \lim\frac{\sqrt{n}}{2^{2n}} \frac{(2n)!}{ \Gamma (n+\lfloor x\sqrt{n} \rfloor+1)\ \ \Gamma(n-\lfloor x\sqrt{n} \rfloor+1)}
\\ &= \lim\frac{\sqrt{n}}{2^{2n}} \frac{(2n)!}{ \Gamma (n+ x\sqrt{n}+1)\ \ \Gamma(n- x\sqrt{n} +1)}
\end{align}
And the last step is okay because $$\lim_{n\to\infty}\frac{\Gamma (n+\lfloor x\sqrt{n} \rfloor+1)\ \ \Gamma(n-\lfloor x\sqrt{n} \rfloor+1)}{\Gamma (n+ x\sqrt{n}+1)\ \ \Gamma(n- x\sqrt{n} +1)}=1 $$
What am I doing wrong here?

You are not doing anything wrong here; you just are not nearly finished. You need to use Stirling's formula on all the Gamma functions and on the factorials, then simplify. It is messy and will take time.

 

[Feynman's fan]

You are not doing anything wrong here; you just are not nearly finished. You need to use Stirling's formula on all the Gamma functions and on the factorials, then simplify. It is messy and will take time.

Thanks a lot for pointing out the direction!

By the way, can we use the result from above, which is [tex]\lim\limits_{n\to\infty} \frac{\sqrt{n}}{2^{2n}} \binom {2n} {n+\lfloor x\sqrt{n} \rfloor} = \frac{1}{\sqrt{\pi}} e^{-x^2}[/tex]as step functions to show that [tex]\int_{-\infty}^{\infty}e^{-x^2}=\sqrt{\pi}[/tex]?

It seems to fit quite well.

(I know there are other conventional ways to show it but this floor function makes it exactly to a sequence of step functions!)