Hermite integral formula

1Complex Analysis¶

We first recall some results from complex variables.

The necessary and sufficient condition is given by

2Laurent expansion and residue¶

If $f(z)$ is not analytic at a single point $z_0$, then we can use a Laurent series. Let $f(z)$ be analytic in the region

and let $z_0$ be an isolated singular point of $f(z)$. The Laurent expansion of $f(z)$ in $D$ is given by

with

where $C$ is a simple closed contour lying in $D$. The coefficient $C_{-1}$ is called the residue of $f(z)$ at $z_0$. When $n=-1$, we have

3Lagrange interpolation¶

Let $x_0,x_1,\ldots,x_n$ be a set of $n+1$ distinct interpolation points. Define

The Lagrange polynomials are given by

and the interpolant of degree $n$ is

Let $x$ be any point which is distinct from the interpolation points. Let $\Gamma_j$ be a contour in the complex plane that encloses $x_j$ but none of the other points, nor the point $x$. Using the Cauchy integral formula, we see that

Hence

Now let $\Gamma'$ be a curve which encloses all of the points $\{x_j\}$ but not the point $x$ and let $f$ be analytic on and interior to $\Gamma'$. Then we can combine the $\Gamma_j$ integrals to get an expression for the interpolant

by using the same type of argument we used above.

Now suppose we enlarge the contour of integration to a new contour Γ that encloses $x$ as well as $\{x_j\}$, and we assume $f$ is analytic on and inside Γ. Then

where we used the Cauchy integral formula to get $-f(x)$.

4Effect of point distribution $\{x_j\}$¶

On a fixed contour Γ, the quantities $f(t)$ and $t-x$ in the error formula are independent of $\{x_j\}$. The error depends on the ratio

If Γ is far away from $\{x_j\}$, then for each $t \in \Gamma$,

this ratio will shrink exponentially as $n \to \infty$, and if this happens, we may conclude that $p(x)$ converges exponentially to $f(x)$ as $n \to \infty$. The crucial condition is that it must be possible to continue $f$ analytically far out to Γ.