123 lines
2.5 KiB
C
123 lines
2.5 KiB
C
/*
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* (c) copyright 1988 by the Vrije Universiteit, Amsterdam, The Netherlands.
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* See the copyright notice in the ACK home directory, in the file "Copyright".
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*
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* Author: Ceriel J.H. Jacobs
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*/
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/* $Header$ */
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#include <math.h>
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#include <errno.h>
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extern int errno;
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double
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yn(n, x)
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double x;
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{
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/* Use y0, y1, and the recurrence relation
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y(n+1,x) = 2*n*y(n,x)/x - y(n-1, x).
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According to Hart & Cheney, this is stable for all
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x, n.
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Also use: y(-n,x) = (-1)^n * y(n, x)
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*/
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int negative = 0;
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extern double y0(), y1();
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double yn1, yn2;
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register int i;
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if (x <= 0) {
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errno = EDOM;
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return -HUGE;
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}
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if (n < 0) {
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n = -n;
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negative = (n % 2);
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}
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if (n == 0) return y0(x);
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if (n == 1) return y1(x);
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yn2 = y0(x);
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yn1 = y1(x);
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for (i = 1; i < n; i++) {
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double tmp = yn1;
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yn1 = (i*2)*yn1/x - yn2;
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yn2 = tmp;
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}
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if (negative) return -yn1;
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return yn1;
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}
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double
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jn(n, x)
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double x;
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{
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/* Unfortunately, according to Hart & Cheney, the recurrence
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j(n+1,x) = 2*n*j(n,x)/x - j(n-1,x) is unstable for
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increasing n, except when x > n.
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However, j(n,x)/j(n-1,x) = 2/(2*n-x*x/(2*(n+1)-x*x/( ....
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(a continued fraction).
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We can use this to determine KJn and KJn-1, where K is a
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normalization constant not yet known. This enables us
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to determine KJn-2, ...., KJ1, KJ0. Now we can use the
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J0 or J1 approximation to determine K.
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Use: j(-n, x) = (-1)^n * j(n, x)
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j(n, -x) = (-1)^n * j(n, x)
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*/
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extern double j0(), j1();
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if (n < 0) {
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n = -n;
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x = -x;
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}
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if (n == 0) return j0(x);
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if (n == 1) return j1(x);
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if (x > n) {
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/* in this case, the recurrence relation is stable for
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increasing n, so we use that.
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*/
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double jn2 = j0(x), jn1 = j1(x);
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register int i;
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for (i = 1; i < n; i++) {
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double tmp = jn1;
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jn1 = (2*i)*jn1/x - jn2;
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jn2 = tmp;
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}
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return jn1;
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}
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{
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/* we first compute j(n,x)/j(n-1,x) */
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register int i;
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double quotient = 0.0;
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double xsqr = x*x;
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double jn1, jn2;
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for (i = 20; /* ??? how many do we need ??? */
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i > 0; i--) {
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quotient = xsqr/(2*(i+n) - quotient);
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}
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quotient = x / (2*n - quotient);
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jn1 = quotient;
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jn2 = 1.0;
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for (i = n-1; i > 0; i--) {
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/* recurrence relation is stable for decreasing n
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*/
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double tmp = jn2;
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jn2 = (2*i)*jn2/x - jn1;
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jn1 = tmp;
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}
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/* So, now we have K*Jn = quotient and K*J0 = jn2.
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Now it is easy; compute real j0, this gives K = jn2/j0,
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and this then gives Jn = quotient/K = j0 * quotient / jn2.
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*/
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return j0(x)*quotient/jn2;
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}
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}
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