A limit converging to exp(-x²)

In summary: I think we should leave the proof, because it is not a summary.In summary, the problem at hand involves evaluating the limit \lim\limits_{n\to\infty} \frac{\sqrt{n}}{2^{2n}} \binom {2n} {n+\lfloor x\sqrt{n} \rfloor} and using Stirling's formula to manipulate the expression. Various approaches have been attempted, such as using the squeeze theorem and simplifying the greatest-integer function, but the desired result of e^{-x^2} has not been achieved. Further exploration is needed.
  • #1
Feynman's fan
14
0
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]
 
Last edited:
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  • #2
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.
 
  • #3
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]
 
  • #4
Feynman's fan said:
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).
 
  • #5
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:
$$p\geq x\sqrt{n}-1$$
 
  • #6
Feynman's fan said:
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.
 
  • #7
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].


mfb said:
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]?


Ray Vickson said:
I think use of the squeeze theorem is unnecessary; the simple form of Stirling is enough
Could you please elaborate on this?
 
  • #8
Feynman's fan said:
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.
 
  • #9
Ray Vickson said:
[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?
 
Last edited:
  • #10
Feynman's fan said:
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.
 
  • #11
Ray Vickson said:
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!)
 

FAQ: A limit converging to exp(-x²)

What is a limit converging to exp(-x²)?

A limit converging to exp(-x²) is an expression that describes the behavior of a mathematical function as the input values approach negative infinity. In this case, the function is exp(-x²), which represents the exponential of a negative squared value.

What is the significance of exp(-x²) in mathematics?

Exp(-x²) is a commonly used function in mathematics, particularly in the field of probability and statistics. It appears in the normal distribution function, which is used to model many real-world phenomena.

How is a limit converging to exp(-x²) calculated?

A limit converging to exp(-x²) is calculated by evaluating the function for increasingly large negative values of x. As x approaches negative infinity, the value of exp(-x²) approaches 0, so the limit is equal to 0.

What is the graphical representation of a limit converging to exp(-x²)?

A limit converging to exp(-x²) can be represented graphically as a curve that approaches the x-axis as x approaches negative infinity. The curve will never actually touch the x-axis, but it will get closer and closer to it as x becomes more negative.

What is the practical application of a limit converging to exp(-x²)?

A limit converging to exp(-x²) is used to model phenomena that follow a normal distribution, such as heights or IQ scores. It can also be used in calculations involving probabilities, such as finding the area under a normal curve.

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