January 24, 2022

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

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 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?

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