Why is the saddlepoint approximation called a "saddlepoint" approximation?

Question

I am just doing some research on the saddlepoint approximation and before I delved to deep into actually applying the approximation method, I was just curious as to why it is in fact called a saddlepoint approximation. I know that a saddlepoint in calculus is a point that can be a maximum or a minimum depending on the direction that you are looking at the graph, but I wasn't sure how this had anything to do with the technique. Thanks!

Answer 1

As you probably know, the saddlepoint method (also known as or stationary phase method or method of steepest descent) used for computing integrals of the form

$$ I = \int_{\gamma} f\left(z\right) e^{\lambda g\left(z\right)}dz, \qquad z \in \mathbb C, \qquad 0 \ll \lambda \in \mathbb R%, \qquad \gamma \quad \text{ is a contour }. $$

over a contour $\gamma$ in the complex plane, where both $f$ and $g$ are analytic functions. It is essentially the same as Laplace's method but modified for taking integrals over complex plane.

The idea of the method is to deform contour $\gamma$ in a way that it passes through a saddle point $z_0$ of function $g(z)$, so that at this point function $g$ has a maximum if you move along direction of steepest descend. Recall that analytic functions do not have minima or maxima, but only saddle-points in the domain of analyticity, and that the value of a contour integral of an analytical function does not change if you deform contour continuously. Since real constant $\lambda\gg 0$ is supposed to be large, it determines range of magnitude of integrand in the vicinity of saddle-point $z_0$. Thus, deforming $C$ in a way so that it passes through the $z_0$ in the direction of steepest descent we can ensure that the integrand will be decreasing in magnitude nearly as quickly as possible.

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To summarize, the reason the it is called "saddle point method" is because finding saddle points is the crucial step in the algorithm:

1. Write integral in the form $\displaystyle\qquad I\left(\lambda\right) = \int_{\gamma}f\left(z\right)e^{\lambda g\left(z\right)}dz$.

2. Since the exponent determines behavior of $\,I(\lambda)$ as $\lambda\to\infty,\,$ investigate $g(z)$ in a following way:

   • find saddle point(s) of $g \iff$ find all points where $g'(z)=0$ (since $g$ is analytic)
   • construct paths of steepest descent

3. Deform $\gamma$ to match path of steepest descent.

4. Obtain asymptotical approximation of the integral $I(\lambda)$ using Laplace's Method.