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Convergence with number of sample points--theory

What is the minimum $M$ required for a given accuracy in (G.2)? Writing $f(s)$ as a complex Fourier series on the periodic interval $L$,

\begin{displaymath}
f(s)\; = \; \sum_{n=-\infty}^{\infty} \tilde{f}_n e^{i k_n s} ,
\hspace{1in} \mbox{where} \; k_n = \frac{2 \pi n}{L} .
\end{displaymath} (G.4)

it is clear that the desired integral $F$ is just $L \tilde{f}_0$. The approximation (G.2) instead gives the sum $L (\tilde{f}_0 + \tilde{f}_{\pm M} + \tilde{f}_{\pm 2M} + \cdots )$. The notation $\tilde{f}_{\pm n}$ implies the sum of $\tilde{f}_{n}$ and $\tilde{f}_{-n}$ both with arbitrary phase factors. This is an example of the Nyquist sampling theorem[161]: $f$ is completely represented up to spatial frequencies $\pm k_{M/2}$, but beyond this the spectrum is folded back into this range. In particular the frequencies $k_{jM}$ for integer $j \neq 0$ get folded back to zero-frequency, corrupting the estimate of $F$. Thus one should choose sample spacing $L/M < 2\pi/k_{{\mbox{\tiny max}}}$, where $k_{{\mbox{\tiny max}}}$ is the bandwidth of $f(s)$. That is, $f(s)$ is assumed to have negligible components beyond wavenumber $k_{{\mbox{\tiny max}}}$.

The functions $g(s)$ and $h(s)$ are wavefunctions (Helmholtz equation solutions existing in the plane), measured along a closed curve, which I first take as smooth and of large curvature radius. So to a first approximation $g(s)$ and $h(s)$ are bandwidth-limited to $k_{{\mbox{\tiny max}}} = k$, where the free-space wavenumber is $k \equiv 2\pi/\lambda_{{\mbox{\tiny B}}}$. Therefore $\tilde{f} = \tilde{g} \ast \tilde{h}$ is bandwidth-limited to $2k$. This would suggest that just over 2 samples per free-space wavelength are needed to achieve high accuracy, and that convergence is exponential beyond this.

Two facts heavily modify this simple semiclassical picture: 1) there exist evanescent wave components in eigenstates or basis functions (for instance, evanescent basis functions can oscillate at up to $4k$, see Section 6.3.3), and 2) the boundary may not be infinitely differentiable (required for exponential convergence), rather corners (and `kinks') may introduce discontinuities into the function $f(s)$ which give power-law Fourier tails (and power-law convergence).


next up previous
Next: Convergence in practice Up: Appendix G: Numerical evaluation Previous: Appendix G: Numerical evaluation
Alex Barnett 2001-10-03