# Math 262A: Lecture 4

Let’s begin by going over the computation from the end of Lecture 3, where we were trying to compute the coefficients $m_d(N)$ in the expansion

$e^{\sum_{d=1}^\infty c_d(N) \frac{z^d}{d!}} = 1 + \sum\limits_{d=1}^\infty m_d(N) \frac{z^d}{d!}$

with

$c_d(N) = \delta_{d \geq 3} N(-1)^{d-1} (d-1)!.$

From the Exponential Formula, we get

$m_d(N) = \sum\limits_{\pi \in \mathrm{P}(d)} \prod\limits_{B \in \pi} c_{|B|}(N),$

so for each partition $\pi$ of $\{1,\dots,d\}$ the corresponding term in the sum $m_d(N)$ is

$\prod\limits_{B \in \pi} \delta_{|B| \geq 3} N(-1)^{|B|-1} (|B|-1)! = N^{b(\pi)} (-1)^{d-b(\pi)} \prod\limits_{B \in \pi} \delta_{|B| \geq 3} (|B|-1)!,$

where $b(\pi)$ is the number of blocks in $\pi.$ The product

$\prod\limits_{B \in \pi} \delta_{|B| \geq 3} (|B|-1)!$

is counting the number of ways to build a cyclic permutation of length at least $3$ on each block of $\pi,$ so we conclude that

$m_d(N) = \sum\limits_{k=1}^d (-1)^{d-k} N^k t_k(d)$

with $t_k(d)$ the number of permutations in the symmetric group $\mathrm{S}(d)$ made up of exactly $k$ cycles, each of length at least $3.$ We can write this a bit more precisely as

$m_d(N) = \sum\limits_{k=1}^{\lfloor \frac{d}{3} \rfloor} (-1)^{d-k} N^k t_k(d).$

It remains to solve Problem 2 from Lecture 3, which was to evaluate the integral

$\int\limits_{-\varepsilon}^{+\varepsilon} x^d e^{-\frac{N}{2}x^2} \mathrm{d}x.$

Unlike Problem 1, Problem 2 cannot be solved exactly. In the case $d=0,$ this was discussed in Lecture 2: we can approximate

$\int\limits_{-\varepsilon}^{+\varepsilon} e^{-\frac{N}{2}x^2} \mathrm{d}x = \int\limits_{-\infty}^{+\infty} e^{-\frac{N}{2}x^2} \mathrm{d}x + E_0(\varepsilon,N)$

with $E_0(\varepsilon,N)$ an exponentially small errror, and then evaluate the Gaussian integral

$\int\limits_{-\infty}^{+\infty} e^{-\frac{N}{2}x^2} \mathrm{d}x = \sqrt{\frac{2\pi}{N}}.$

For any fixed $d \in \mathbb{N}$ we have an analogous approximation

$\int\limits_{-\varepsilon}^{+\varepsilon} x^d e^{-\frac{N}{2}x^2} \mathrm{d}x = \int\limits_{-\infty}^{+\infty}x^d e^{-\frac{N}{2}x^2} \mathrm{d}x +E_d(\varepsilon,N),$

with $E_d(\varepsilon,N)$ an error term that remains to be quantified precisely but is clearly rapidly decaying given the exponentially small tails of the Gaussian density. We now evaluate the total integral

$\int\limits_{-\infty}^{+\infty}x^d e^{-\frac{N}{2}x^2} \mathrm{d}x.$

In fact, we will evaluate all such integrals simultaneously by looking instead at the integral

$L(z) = \int\limits_{-\infty}^{+\infty} e^{-\frac{N}{2}x^2+xz} \mathrm{d}x,$

which is the Laplace transform of the Gaussian density. On one hand, observe that the integral converges to define an entire function of $z \in \mathbb{C}$ whose derivatives at $z=0$ can be computed by differentiating under the integral sign, and are exactly the integrals we’re trying to evaluate:

$\bigg[\frac{\mathrm{d}}{\mathrm{d}z}\bigg]^d L(z) \big{|}_{z=0}=\int\limits_{-\infty}^{+\infty}x^d e^{-\frac{N}{2}x^2} \mathrm{d}x.$

(Differentiation under the integral sign is our first contact with the ghost of Richard Feynman, a congenial specter that will continue to haunt us). On the other hand, we can compute $L(z)$ exactly by completing the square: since

$-\frac{N}{2}x^2+xz = -\frac{N}{2}\left( x^2 -\frac{2}{N}xz + \frac{1}{N^2}z^2\right) + \frac{1}{2N}z^2 = -\frac{N}{2}\left( x-\frac{1}{N}z\right)^2 + \frac{1}{2N}z^2,$

we obtain

$L(z) = e^{\frac{1}{2N}z^2} \int\limits_{-\infty}^{+\infty} e^{-\frac{N}{2}\left(x-\frac{1}{N}z\right)^2} \mathrm{d}x = e^{\frac{1}{2N}z^2} \int\limits_{-\infty}^{+\infty} e^{-\frac{N}{2}x^2} \mathrm{d}x,$

by translation invariance of Lebesgue measure, and hence

$L(z) = \sqrt{\frac{2\pi}{N}} e^{\frac{1}{2N}z^2}.$

This is an important feature of the Gaussian density: up to minor details, it is a fixed point of the Laplace transform. In particular, this makes the Maclaurin series of $L(z)$ is easy to compute directly:

$L(z) = \sqrt{\frac{2\pi}{N}}\sum\limits_{k=0}^\infty \frac{(2k)!}{N^k 2^k k!} \frac{z^{2k}}{(2k)!}.$

Taking a closer look at the nonzero coefficients of this series, we observe that

$\frac{(2k)!}{2^k k!} = \frac{(2k)(2k-1)(2(k-1))(2k-3) \dots 1}{2^k k!} = (2k-1)(2k-3) \dots 1,$

the product of all odd numbers below $2k.$ Thus in terms of the double factorial $n!!,$ which is the product of all numbers below $n$ of the same parity (and should not be confused with the much larger number $(n!)!$),\$ we have

$L(z) = \sqrt{\frac{2\pi}{N}}\sum\limits_{k=0}^\infty N^{-k} (2k-1)!!\frac{z^{2k}}{(2k)!}.$

We thus obtain the integral formula

$\int\limits_{-\infty}^{+\infty}x^d e^{-\frac{N}{2}x^2} \mathrm{d}x = \sqrt{\frac{2\pi}{N}}\begin{cases} 0, \text{ if } d \text{ odd } \\ N^{-\frac{d}{2}} (d-1)!!, \text{ if } d \text{ even.} \end{cases}$

Exercise 1: Show that for $d$ even, $(d-1)!!$ is the number of permutations in $\mathrm{S}(d)$ all of whose cycles have length exactly $2,$ i.e. the number of fixed point free involutions.

We have finally reached the point where we can legitimately write down Stirling’s formula. Here’s a step-by-step reconstruction of the argument we’ve been building.

Step 1: Obtain the integral representation

$N! = \frac{N^{N+1}}{e^N}Z_N(S),$

where $S(w) = w- \log(1+w)$ and

$Z_N(S) =\int\limits_{-1}^{+\infty} e^{-N(w-\log(1+w)} \mathrm{d}w.$

Step 2: Demonstrate that the integral $Z_N(S)$ localizes in an $\varepsilon$-neighborhood of the minimizer of $S:$

$Z_N(S) = Z_{N,\varepsilon}(S),$

where

$Z_{N,\varepsilon}(S) = \int_{-\varepsilon}^{+\varepsilon}e^{-NS(w)}\mathrm{d}w + O(e^{-c(\varepsilon)}N).$

Step 3: Write the integrand of $Z_{N,\varepsilon}(S)$ in the form

$e^{-NS(w)} = e^{-\frac{N}{2}w^2} e^{\sum_{d=3}^\infty c_d(N) \frac{w^d}{d!}}.$

Step 4: Expand the non-Gaussian part of the integrand,

$e^{\sum_{d=3}^\infty c_d(N) \frac{w^d}{d!}} = 1 + \sum\limits_{d=1}^\infty m_d(N) \frac{w^d}{d!},$

with the coefficients $m_d(N)$ being the “disconnected” version of the coefficients $c_d(N).$

Step 5: Integrate term-by-term:

$Z_{N,\varepsilon}(S) = \int_{-\varepsilon}^{+\varepsilon}e^{-\frac{N}{2}w^2}\mathrm{d}w + \sum\limits_{d=1}^\infty m_d(N) \int_{-\varepsilon}^{+\varepsilon}w^de^{-\frac{N}{2}w^2}\mathrm{d}w.$

Step 6: Replace each truncated Gaussian integral with a total Gaussian integral at the cost of an exponentially small error $E_{d,\varepsilon}(N),$ and evaluate:

$m_d(N) \int_{-\varepsilon}^{+\varepsilon}w^de^{-\frac{N}{2}w^2}\mathrm{d}w = [d \text{ even}] m_d(N) N^{-\frac{d}{2}} \left(\frac{d}{2}-1\right)!! + E_{d,\varepsilon}(N).$

The bracket in Step 6 is an Iverson bracket.

Problem 1: Carefully executing the above steps, show that we have obtained Stirling’s approximation: as $N \to \infty,$ we have

$N! = \sqrt{2\pi}\frac{N^{N+\frac{1}{2}}}{e^N}\left(1 + O(\frac{1}{N}) \right).$

Problem 2: Develop a good bookkeeping system allowing the computation of subleading terms in this approximation.

Problem 3: Go through Steps 1 to 6, consider what features of $S$ were used, and examine to what extent these features were needed.

These are the problems which we will focus on in the next lectures. In particular, Problem 2 leads to the concept of Feynman diagrams. You can start thinking about these problems now, and see what you can see.