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.

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