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Stirling formula Azimuth

Laplace method This entry was posted on Sunday, October 3rd, 2021 at 3:05 pm and is filed under You can follow any responses to this entry through the RSS You can leave a response, or trackback from…

Stirling’s formula says

on  sim means that the ratio of the two quantities goes to 1 how n  to  infty.

Where does this formula come from? In particular, how the name works 2  pi participate in it? To understand these things, I think a non-rigorous argument that can be made rigorous is more useful than a rigorous test with all the i points and t crossed. I think it’s important to keep the argument short. So let me do this. The punchline will be that the 2  pi comes from this formula:

 displaystyle { int _ {-  infty} ^  infty e ^ {- x ^ 2/2} , dx =  sqrt {2  pi}}

And that, I hope you know, comes from squaring the two sides and turning the left side into an integral integral that you can do in polar coordinates, taking out a factor of 2  pi because it only depends on what you are integrating r, no  theta.

Okay, here it goes. Let’s start with

 displaystyle { int_0 ^  infty x ^ ne ^ {- x} , dx = n!  }

It can be easily shown by repeated integration by parts.

Then we do this:

 begin {array} {ccl} n!  & = &  displaystyle { int_0 ^  infty x ^ ne ^ {- x} , dx}   & = &  displaystyle { int_0 ^  infty e ^ {n  ln x -x} , dx}   & = &  displaystyle {n  int_0 ^  infty e ^ {n  ln (ny) -ny} , dy}    end {array}

In the first step, we are writing x ^ n how e ^ {n  ln x}. In the second we change the variables: x = n i.

Then we can use  ln (ny) =  ln n +  ln y to ruin things:

 displaystyle {n!  = ne ^ {n  ln n}  int_0 ^  infty e ^ {n  ln y -ny} , dy}

All the hard work will be done proving this:

 displaystyle { int_0 ^  infty e ^ {n  ln y -ny} , dy  sim  sqrt { frac {2  pi} {n}} ;  e ^ {- n}}

With that in mind, we get it

 displaystyle {n!  = ne ^ {n  ln n}  sqrt { frac {2  pi} {n}} ;  e ^ {- n}}

and simplifying we obtain the formulas of Stirling:

 displaystyle {n!  =  sqrt {2  pi n}  left ( frac {n} {e}  right) ^ n}

Laplace method

How do we achieve this?

 displaystyle { int_0 ^  infty e ^ {n  ln y -ny} , dy  sim  sqrt { frac {2  pi} {n}} ;  e ^ {- n}} ;  ?

Let’s write it like

 displaystyle { int_0 ^  infty e ^ {- n (y -  ln y)} , dy  sim  sqrt { frac {2  pi} {n}} ;  e ^ {- n}}

The trick is to watch it as n becomes large, the integral will be dominated by the point at which y -  ln y is as small as possible. Then we can approximate the integral for a Gaussian with a maximum point at this time!

Realize that

 Displaystyle { frac {d} {dy} (i -  ln i) = 1 - i ^ {- 1}}

 Displaystyle { frac {d ^ 2} {dy ^ 2} (i -  ln i) = i ^ {- 2}}

thus the function y -  ln y has a critical point a i = 1 and its second derivative is 1 there, so it’s a local minimum. In fact, this point is a minimum throughout the range (0,  infty).

Then we use this:

Laplace method. Suppose f  two points [a,b]  to  mathbb {R} has a unique minimum at some point x_0  in (a, b) i f 0.” class=”latex”/> Then

 displaystyle { int_a ^ be ^ {- nf (x)} dx  sim  sqrt { frac {2  pi} {nf '' (x_0)}} ;  e ^ {- nf (x_0)}}

how n  to  infty.

This is demonstrated by approximating the integral by a Gaussian. Applying it to our case at hand, we succeed

 displaystyle { int_a ^ be ^ {- n (y -  ln y)} dy  sim  sqrt { frac {2  pi} {nf '' (y_0)} ;  e - {nf (y_0)}

on f (i) = i -  ln i, i_0 = 1, $ f (y_0) = 1 $ i $ f ”(y_0) = 1. $ So we get

 displaystyle { int_a ^ be ^ {- n (y -  ln y)} dy  sim  sqrt { frac {2  pi} {n} ;  e ^ {- n}}

and then leaving it a  a 0, b  a +  infty we get what we want.

Therefore, from this point of view — and there are others — the key to Stirling’s formula is Laplace’s method of approximating an integral as

 displaystyle { int_a ^ be ^ {- nf (x)} dx}

with a Gaussian integral as n  to  infty. In the end, the crucial calculation is

 displaystyle { int _ {-  infty} ^  infty e ^ {- x ^ 2/2} , dx =  sqrt {2  pi}}

You can see the full proof of Laplace’s method here:

• Wikipedia, Laplace method.

Physicists who have made quantum field theory will know that when the thrust is shown it is mostly Gaussian integrals: the limit n  to  infty what we’re seeing here is like a “classic limit” where  hbar  a 0. Therefore, they will know this idea.

Here there should be a deeper morality, about how n! it is related to some Gaussian process, but I do not know, although I know how binomial coefficients approximate a Gaussian distribution. Know some deeper explanation, perhaps in terms of probability theory and combinatorics, of why n! ends up being described asymptotically by an integral of a Gaussian?

This entry was posted on Sunday, October 3rd, 2021 at 3:05 pm and is filed under Mathematics. You can follow any responses to this entry through the RSS 2.0 feed. You can leave a response, or trackback from your own site.

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