Monday Math 54 « Twisted One 151’s Weblog

Monday Math 54

By twistedone151

Let us consider an integral of the form $I(\lambda)=\int_a^be^{-\frac{f(x)}{\lambda}}\,dx$ , with parameter λ>0, and where f(x) has a single local minimum in the interval (a,b). The method of steepest decent, also known as saddle-point integration, is a useful method for approximating such an integral in the small λ limit. Namely, if the local minimum occurs at x=x₀, then we can expand f(x) in the Taylor series about this point; as it is a local minimum, f‘(x₀)=0 and f”(x₀)≥0. Here, we assume that this second derivative is nonzero. Thus, our Taylor series is
$f(x)=f(x_0)+\frac{1}{2}f''(x_0)(x-x_0)^2+\cdots$ .
Now, we note that as λ→0⁺, the term in the exponent becomes ever more negative, and the integrand approaches zero; the local minimum at x₀ is the “slowest” to approach, and thus the function near that point comes to dominate the rest of the integral; and so
$I(\lambda)\approx\int_a^be^{-\frac{f(x_0)}{\lambda}-\frac{f''(x_0)}{2\lambda}(x-x_0)^2}\,dx$
and
$\begin{eqnarray}\int_a^be^{-\frac{f(x_0)}{\lambda}-\frac{f''(x_0)}{2\lambda}(x-x_0)^2}\,dx&=&\int_a^be^{-\frac{f(x_0)}{\lambda}}e^{-\frac{f''(x_0)}{2\lambda}(x-x_0)^2}\,dx\\&=&e^{-\frac{f(x_0)}{\lambda}}\int_a^be^{-\frac{f''(x_0)}{2\lambda}(x-x_0)^2}\,dx\end{eqnarray}$
Now, we see that the last integral is part of a gaussian integral; the peak of our gaussian is in the region of integration, and as λ→0⁺, the gaussian’s width goes to zero as well, so that the tails become negligible in the limit, and
$\int_a^be^{-\frac{f''(x_0)}{2\lambda}(x-x_0)^2}\,dx\approx\int_{-\infty}^{\infty}e^{-\frac{f''(x_0)}{2\lambda}(x-x_0)^2}\,dx$ .
Now, recall that the integral of the gaussian over all real numbers is given by
$\int_{-\infty}^{\infty}e^{-ax^2}\,dx=\sqrt{\frac{\pi}{a}}$ .
Thus $\int_{-\infty}^{\infty}e^{-\frac{f''(x_0)}{2\lambda}(x-x_0)^2}\,dx=\sqrt{\frac{2\pi\lambda}{f''(x_0)}}$ ,
and so
$I(\lambda)\approx{e^{-\frac{f(x_0)}{\lambda}}}\sqrt{\frac{2\pi\lambda}{f''(x_0)}}$ .

Let us perform an example: $n!=\operatorname{\Gamma}(n+1)=\int_0^{\infty}x^{n}e^{-x}\,dx$ . This does not fit the form as it is, but can be transformed to do so. First, let us make the substitution $x=nu\to\,dx=n\,du$ . Then we have:
$n!=n^{n+1}\int_0^{\infty}u^{n}e^{-nu}\,du$ . Second, we note that for positive u, $u^n=e^{\ln(u^n)}=e^{n\ln{u}}$ .
Thus, we have
$n!=n^{n+1}\int_0^{\infty}e^{-n(u-\ln{u})}\,du$ ,
and the integral above fits our form with $\lambda=\frac{1}{n}$ and $f(u)=u-\ln{u}$ ; thus our approximation will be for n→∞. Now, $f'(u)=1-\frac{1}{u}$ , which is zero for u=1. $f''(u)=\frac{1}{u^2}$ , so $f''(1)=1$ , and $f(1)=1-\ln{1}=1$ . Thus, our first saddle-point approximation says
$\int_0^{\infty}e^{-n(u-\ln{u})}\,du\approx{e}^{-\frac{f(1)}{\lambda}}\sqrt{\frac{2\pi\lambda}{f''(1)}}=e^{-n}\sqrt{\frac{2\pi}{n}}$
and so
$n!\approx{n}^{n+1}e^{-n}\sqrt{\frac{2\pi}{n}}=\sqrt{2\pi{n}}n^ne^{-n}$
which you may recognize as Stirling’s approximation (see here).

Tags: Approximation, Factorial, Gamma Function, Integration, Math, Method of Steepest Descent, Monday Math, Saddle-Point Approximation, Saddle-Point Integration, Stirling's Approximation

This entry was posted on January 12, 2009 at 1:29 am and is filed under Math/Science. 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.

One Response to “Monday Math 54”

GrowthHormoneRelease.com Says:
January 15, 2009 at 12:45 pm | Reply
I’m Confused :s

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Monday Math 54

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