which is true but which I cannot prove. Taylor-expanding the log and resumming in , , doesn’t seem to help.

Consider a {\it one-dimensional random walk} of a point $\cal Q$
that starts at $x=0$ and at each step jumps from the point it
occupies to either right or
left adjacent integer.
The probability of finding point $\cal P$ in
location $x=K$ after $N$ steps from its initial position is
given by the binomial distribution
$$
{P}(K,N)=\begin{cases} \dfrac{N !} {2^N\; \left(\frac{N+ K
}{2}!\right) \left(\frac{N- K }{2}!\right) }, \text{ if
}\vert K\vert \le N; \; N-K \text{ is
even},\\
0, \text{ otherwise}.\end{cases} \eqno{\rm (3)}
$$

Let us now assume that
$$ N=2n \text{ is even } \eqno{\rm (4a)}$$
and $\varepsilon$ satisfies
$$
0<\varepsilon<\dfrac 5{30}. \eqno{\rm (4b)}
$$
Formula (3) can be rewritten as
$$
\hskip-0.2cm {P}(-2k,2n)= {P}(2k,2n)=\begin{cases} \dfrac{(2n) !}
{2^{2n}\; \left(n!\right)^2 }\cdot
\dfrac{\left(n!\right)^2 } { (n+ k)!
(n- k)! }=\\
\hskip0.5cm \dfrac{(2n) !} {2^{2n}\; \left(n!\right)^2 }
\prod\limits_{j=1}^{k} \dfrac{\;\;n-\ (k-j) \;\;}{n+ { j} }, \text{ if } 0\le k\le n,\\
0, \text{ otherwise}.\end{cases} {\rm (5)}
$$
Formula (5) implies that sufficiently large $ n$
$$
1-\sum\limits_{\vert
k\vert < n^{0.5+\varepsilon} } P\big(2k, 2n \big) =\sum\limits_{\vert
k\vert \ge n^{0.5+\varepsilon} } P\big(2k, 2n \big)\le 2^{-
n^{2\varepsilon}} \eqno{\rm (6)}
$$
and for $\vert k\vert \le n^{0.5+\varepsilon}$
$$
\frac 1{\sqrt{ n}} e^{-\frac{k^2}{n} -8
n^{-0.5+3\varepsilon} } \le\frac {2^{2n}\; \left(n!\right)^2 }{(2n)
!\sqrt{ n}}
\;\; P(2k,2n)
\le \frac 1{\sqrt{ n}} e^{-\frac{k^2}{n} +8
n^{-0.5+3\varepsilon} }.\eqno{\rm (7)}
$$
The proof of formulas (6,7) is given in the Appendix.

Formula (7) implies
$$
\bigg[ \frac 1{\sqrt{ n}}\sum\limits_{ \vert k \vert \le
n^{0.5+\varepsilon} } e^{-\frac{k^2}{n}}\bigg] e^{ -8
n^{-0.5+3\varepsilon} } \le\frac {2^{2n}\; \left(n!\right)^2 }{(2n)
!\sqrt{ n}}
{ \sum\limits_{\vert k \vert \le n^{0.5+\varepsilon}}P(2k,2n)}
\le \bigg[
\frac 1{\sqrt{ n}}\sum\limits_{ \vert k \vert \le
n^{0.5+\varepsilon} } e^{-\frac{k^2}{n}}\bigg] e^{ 8
n^{-0.5+3\varepsilon} }.
$$
\color{black} Taking now limit as $n\to +\infty$ and using
$\lim\limits_{n\to +\infty}e^{ 8 n^{-0.5+3\varepsilon}
}=\lim\limits_{n\to +\infty}e^{- 8n^{-0.5+3\varepsilon} }=1$,
$\lim\limits_{n\to +\infty} \sum\limits_{\vert k \vert \le
n^{0.5+\varepsilon}}P(2k,2n)=1$ due to (6), $\lim\limits_{N\to
+\infty}\dfrac 1{\sqrt{n}}\sum\limits_{{\vert k \vert \le
n^{0.5+\varepsilon},}
\atop{k \text{ is even}}}
e^{-\frac{k^2}{n}}=\int\limits_{-\infty}^{+\infty}e^{-t^2}dt=\sqrt{\pi}$
we obtain (1).

In fact, I think my posting was chopped!!! So, instead, please click here.

http://www.ma.hw.ac.uk/~takis/commentonstirling.pdf

(If someone could tell me how to post properly, I’d appreciate!)

Sorry for the mess above!

David Speyer’s method, starting from $n! = \int_0^\infty x^n e^{-x} dx$,
resembles very much Laplace’s method.
Also, as he observed, the connection to the CLT is apparent.
But there is one thing in David’s method,
which is not very natural.
Namely (following Laplace) it is better to make the
change of variable
\[
t=x/n
\]
first, instead of $t=x-n$. Then the gamma integral gives
\[
n! = n^{n+1} \int_0^\infty (t e^{-t})^n dt
\]
so that
\[
\frac{n!}{n^n e^{-n} \sqrt{n}} =
\sqrt{n} e^n \int_0^\infty e^{-n g(t)} dt,
\]
where
\[
g(t) = t - \log t
\]
is a function with minimum at $t=1$,
and $g(1)=1$, $g’(1)=0$, $g”(1)=1$.
Following Laplace, the action is played by the second derivative of $g$,
so fixing and $\epsilon > 0$ pick $0< \delta < 1$ so that
\[
|g''(t) - 1| < \epsilon \mbox{ if } |t-1| 0$, both bounds converge to zero.
This means that
\[
\frac{n!}{n^n e^{-n} \sqrt{n}} =
o(1) + \sqrt{n} e^n \int_{1-\delta}^{1+\delta}
e^{-n g(t)} dt.
\]
Now write
\[
g(t) = 1 + \frac{1}{2} g''(\tau)(t-1)^2
\]
for some $\tau=\tau(t)$ lying between $1$ and $t$, and therefore
$|g”(\tau) -1| 0$, (iv) the infimum of $g(x)$ over any compact
set not containing $x_0$ is strictly larger than $g(x_0)$,
(v) $f$ is continuous around $x_0$ with $f(x_0) \neq 0$.
Then
\[
I_n \sim
f(x_0) e^{-(n+\frac{1}{2}) g(x_0)} \sqrt{2\pi/n g''(x_0)}.
\]
The proof is along the same lines as in the special case above.

Regarding the connection to the CLT:
Let $X_0, X_1, X_2$, … be a sequence of independent random variables,
all exponential with mean 1:
\[
P(X_j > t) = e^{-t}.
\]
Then $S_n = X_0+ X_1 + \cdots + X_n$ has density
\[
\frac{1}{n!} x^n e^{-x}.
\]
So the density of $(S_n-n)/\sqrt{n}$ is
\[
f_n(x) = \frac{1}{n!} (n+\sqrt{n}~x)^n e^{-n-\sqrt{n}~x} \sqrt{n}.
\]
A local version of the CLT tells us that, for all $x$ (in fact, uniformly),
\[
\lim_{n \to \infty} f_n(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}
\]
Setting $x=0$ we obtain Stirling’s approximation.
Of course, the heavy-duty machinery invoked is not simpler
than the proof given above.

And here is an exact statement of the LOCAL CLT
(see, e.g. Fristedt and Gray, a Modern Approach to Probability Theory,
Thm 15.9, p.284):

Let $S_n = X_1+ … + X_n$ be sum of independent
and identically distributed random variables, where $X_1$ has
square integrable density, mean $\mu$, and finite variance $\sigma^2$.
Let $f_n$ be the density of $(S_n-n\mu)/\sqrt{n\sigma^2}$.
Then $f_n(x)$ converges (uniformly over all $x$)
to the standard Gaussian density.

Neil

My point is, you get the same thing with (1-x^2)^n or exp(-x*x/2) when raised to the n – the normal curve.

So to me, it seems like a product version of the central limit theorem – we keep reapplying by multiplying yet again by something roughly a normal distribution, and converges (nicely!) to the real normal distribution.

Turning the above sentence into math .. ie. stating why in the limit the repeated product converges to a scaled normal distribution… I havent done that yet – it might be obvious to others reading this blog …

Great, stimulating discussion/exposion, thanks!
[apologies for lack of latex]

Gil Kalai told me about your demystification of Stirling’s formula. Indeed, this easy result deserves to be better explained and understood, and I recall as an undergraduate I was quite frustrated at not being given a proper explanation for it. Then, about 10 years ago I discovered a new derivation of the constant sqrt(2 pi) which I believe to be slightly simpler than the one you presented, and conceptually better motivated, since it relates to the probabilistic viewpoint involving the central limit theorem. The difference with the approach you and a few of the readers mention is that I look at the cumulative probabilities of the symmetric binomial distribution instead of at just the “local” probabilities. By computing the sum of the first half of the binomial coefficients in a given row in two ways (first, using the obvious symmetry, and second, using a simple integration formula that converges to the integral of the Gaussian distribution), one gets the constant immediately.

My proof appeared in the American Math. Monthly 107 (2000), 556–557, and can be found on my web site. I wonder if you and others will think that my approach is indeed simpler than integrating powers of cosine – I find it hard to be objective about this, but of course both approaches are quite easy. By the way, I apologize for the pompous title of my paper (so pompous that I’m embarrassed to write it here) – in my defense, I was 21 when I wrote it…

In particular, on pp. 354-5 of his EXCURSIONS, Professor Young notes that summing the values of an analytic function f at the first n non-negative integers boils down to finding another function F related to it by

f(x) = F(x + 1) – F(x) = F’(x) + (1/2!)F”(x) + (1/3!)F”’(x) + … ,

since f(0) + f(1) + … + f(n-1) is then just F(n) – F(0). And now, to Euler, the striking formal resemblance between

f(x) = F’(x) + (1/2!)F”(x) + (1/3!)F”’(x) + …

and

y = x + ax^2 + bx^3 + …

naturally suggested solving for F in much the same way as one solves for x in terms of y in a power series of this sort. One simply writes

x = y + ry^2 + sy^3 + … ,

with r, s, … undetermined constants, and substitutes this form into the power series. Then one fixes r, s, … successively by requiring the coefficients of terms of like degree to match on both sides of the equation.

Of course there’s more to it than this in the end, but it’s hard to think of a more straightforward argument for putting

F’(x) = f(x) + rf’(x) + sf”(x) + … ,

and substituting back. And not only does this astonishingly simple idea actually work, it exposes the essence of the Euler-Maclaurin summation formula.

Chapter 6 (pp 109-126) is an excellent, motivated, wonderfully illustrated expository account of the path traveled largely by Gowers to Stirling’s formula. The Central Limit Theorem is discussed in the previous chapter, and so the estimate of the middle binomial coefficient 2n choose n is available. His approach to Central Limit Theorem is also excellent, starting from coin-tossing, and leading to the standard improper double integral exploited for the normal distribution. Explicit use of the big-Oh notation is avoided.

Ball also mentions that historically this approach is backwards; the Central Limit Theorem was first established using Stirling-like ideas. Also, Stirling used the Wallis product formula.

tsm [at] usna.edu

I was wondering if you’d mind if I used your blog article on Stirling’s formula for the next edition of “Eureka” (the journal of the Archimedeans undergrad. maths society here at Cambridge). It would most likely reach quite a broad audience of first years next year if you would consent. It might be worth appending David Speyer’s remarks also.

Of course if there is something else you’d prefer I’m open to suggestions.

Contact me at archim-eureka-editor@srcf.ucam.org if you are interested.

[Many thanks for the comment, and I like the explanation for the appearance of the binomial
coefficient even if it doesn't get the whole way to answering the question. I've corrected the bounds
in the integrals and fixed your LaTeX bug now.]

Also (though in some sense this ties in with Lior’s post above) one non-magic reason to get out of these integrals is that it is the constant coefficient in — a fairly natural trick to try — so putting and rescaling one gets

I can’t see how to get round using (the proof of) Wallis’ formula, though.

Thanks.

The formula that doesn’t parse is

.

I dropped the coefficient 1/2 in the Taylor series
, and in several of the following formulas.

[Many thanks for the comment, and I've fixed the bugs now.]

Here is a different approach where the constant comes out relatively naturally. The first step is less motivated than in your argument but, after that, everything feels natural to me.

Start with the integral formula for the Gamma function:

Now, a quick computation with derivatives shows that is optimized at x=n, so lets make a change of variables to put x=n at the origin. Set , so

The formula is screaming to be replaced by . Just making this replacement gives us far too much error, so keep one more term.

so the integrand is

Intuitively, we are done at this point, because . Being careful, we have the problem that the error term will overwhelm the terms we care about when u is too large. So the careful way to do this is to chop our integrand at for some between 1/2 and 2/3, and use some straightforward bounds to show that those tails don’t contribute.