ichbin
09-15-2013, 01:23 AM
Section 6.5 describes Miller's method for computing Bessel functions of integer order in the lower transition region ($\sqrt{n} < x < n$). As the text says, this gives rise to the question of "at how large an n we need to start the downward recurrence". The text says that there is a heuristic argument involving the limiting forms in various regions by which "you should be able to convince yourself" that the answer is $m = n + C \sqrt{n}$.
First, I have never been able to produce such a heuristic argument. If someone can sketch such an argument, I would love to see it.
Second, the literature in this area does not seem to agree that such a simple formula for a satisfactory m exists. Olver and Skoone ["Note on backward recurrence algorithms", Math. Comp. 26 (1972) 941] sketch a method of first running the recurrence upward to find the point at which a sufficiently large differential factor between J_m and J_n is reached, then running Miller's algorithm down from that m. The DLMF [http://dlmf.nist.gov/10.74] endorses this method.
I would love for NR's suggested formula to work, so please convince me that it does. I don't find it reassuring that the NR code doubles the value that of m produced by the suggested formula: not $m = n + 2 C \sqrt{n}$ but $m = 2(n + 2 C \sqrt{n})$!
First, I have never been able to produce such a heuristic argument. If someone can sketch such an argument, I would love to see it.
Second, the literature in this area does not seem to agree that such a simple formula for a satisfactory m exists. Olver and Skoone ["Note on backward recurrence algorithms", Math. Comp. 26 (1972) 941] sketch a method of first running the recurrence upward to find the point at which a sufficiently large differential factor between J_m and J_n is reached, then running Miller's algorithm down from that m. The DLMF [http://dlmf.nist.gov/10.74] endorses this method.
I would love for NR's suggested formula to work, so please convince me that it does. I don't find it reassuring that the NR code doubles the value that of m produced by the suggested formula: not $m = n + 2 C \sqrt{n}$ but $m = 2(n + 2 C \sqrt{n})$!