Patrick Welche <prlw1@cam.ac.uk>

[netbsd-mini2440.git] / lib / libm / noieee_src / n_jn.c

/*      $NetBSD: n_jn.c,v 1.5 2002/06/15 00:10:17 matt Exp $    */
/*-
 * Copyright (c) 1992, 1993
 *      The Regents of the University of California.  All rights reserved.
 *
 * Redistribution and use in source and binary forms, with or without
 * modification, are permitted provided that the following conditions
 * are met:
 * 1. Redistributions of source code must retain the above copyright
 *    notice, this list of conditions and the following disclaimer.
 * 2. Redistributions in binary form must reproduce the above copyright
 *    notice, this list of conditions and the following disclaimer in the
 *    documentation and/or other materials provided with the distribution.
 * 3. Neither the name of the University nor the names of its contributors
 *    may be used to endorse or promote products derived from this software
 *    without specific prior written permission.
 *
 * THIS SOFTWARE IS PROVIDED BY THE REGENTS AND CONTRIBUTORS ``AS IS'' AND
 * ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT LIMITED TO, THE
 * IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR A PARTICULAR PURPOSE
 * ARE DISCLAIMED.  IN NO EVENT SHALL THE REGENTS OR CONTRIBUTORS BE LIABLE
 * FOR ANY DIRECT, INDIRECT, INCIDENTAL, SPECIAL, EXEMPLARY, OR CONSEQUENTIAL
 * DAMAGES (INCLUDING, BUT NOT LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS
 * OR SERVICES; LOSS OF USE, DATA, OR PROFITS; OR BUSINESS INTERRUPTION)
 * HOWEVER CAUSED AND ON ANY THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT
 * LIABILITY, OR TORT (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY
 * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
 * SUCH DAMAGE.
 */

#ifndef lint
#if 0
static char sccsid[] = "@(#)jn.c        8.2 (Berkeley) 11/30/93";
#endif
#endif /* not lint */

/*
 * 16 December 1992
 * Minor modifications by Peter McIlroy to adapt non-IEEE architecture.
 */

/*
 * ====================================================
 * Copyright (C) 1992 by Sun Microsystems, Inc.
 *
 * Developed at SunPro, a Sun Microsystems, Inc. business.
 * Permission to use, copy, modify, and distribute this
 * software is freely granted, provided that this notice
 * is preserved.
 * ====================================================
 *
 * ******************* WARNING ********************
 * This is an alpha version of SunPro's FDLIBM (Freely
 * Distributable Math Library) for IEEE double precision
 * arithmetic. FDLIBM is a basic math library written
 * in C that runs on machines that conform to IEEE
 * Standard 754/854. This alpha version is distributed
 * for testing purpose. Those who use this software
 * should report any bugs to
 *
 *              fdlibm-comments@sunpro.eng.sun.com
 *
 * -- K.C. Ng, Oct 12, 1992
 * ************************************************
 */

/*
 * jn(int n, double x), yn(int n, double x)
 * floating point Bessel's function of the 1st and 2nd kind
 * of order n
 *
 * Special cases:
 *      y0(0)=y1(0)=yn(n,0) = -inf with division by zero signal;
 *      y0(-ve)=y1(-ve)=yn(n,-ve) are NaN with invalid signal.
 * Note 2. About jn(n,x), yn(n,x)
 *      For n=0, j0(x) is called,
 *      for n=1, j1(x) is called,
 *      for n<x, forward recursion us used starting
 *      from values of j0(x) and j1(x).
 *      for n>x, a continued fraction approximation to
 *      j(n,x)/j(n-1,x) is evaluated and then backward
 *      recursion is used starting from a supposed value
 *      for j(n,x). The resulting value of j(0,x) is
 *      compared with the actual value to correct the
 *      supposed value of j(n,x).
 *
 *      yn(n,x) is similar in all respects, except
 *      that forward recursion is used for all
 *      values of n>1.
 *
 */

#include "mathimpl.h"
#include <float.h>
#include <errno.h>

#if defined(__vax__) || defined(tahoe)
#define _IEEE   0
#else
#define _IEEE   1
#define infnan(x) (0.0)
#endif

static const double
invsqrtpi= 5.641895835477562869480794515607725858441e-0001,
two  = 2.0,
zero = 0.0,
one  = 1.0;

double
jn(int n, double x)
{
        int i, sgn;
        double a, b, temp;
        double z, w;

    /* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
     * Thus, J(-n,x) = J(n,-x)
     */
    /* if J(n,NaN) is NaN */
        if (_IEEE && isnan(x)) return x+x;
        if (n<0){
                n = -n;
                x = -x;
        }
        if (n==0) return(j0(x));
        if (n==1) return(j1(x));
        sgn = (n&1)&(x < zero);         /* even n -- 0, odd n -- sign(x) */
        x = fabs(x);
        if (x == 0 || !finite (x))      /* if x is 0 or inf */
            b = zero;
        else if ((double) n <= x) {
                        /* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
            if (_IEEE && x >= 8.148143905337944345e+090) {
                                        /* x >= 2**302 */
    /* (x >> n**2)
     *      Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
     *      Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
     *      Let s=sin(x), c=cos(x),
     *          xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
     *
     *             n    sin(xn)*sqt2    cos(xn)*sqt2
     *          ----------------------------------
     *             0     s-c             c+s
     *             1    -s-c            -c+s
     *             2    -s+c            -c-s
     *             3     s+c             c-s
     */
                switch(n&3) {
                    case 0: temp =  cos(x)+sin(x); break;
                    case 1: temp = -cos(x)+sin(x); break;
                    case 2: temp = -cos(x)-sin(x); break;
                    case 3: temp =  cos(x)-sin(x); break;
                }
                b = invsqrtpi*temp/sqrt(x);
            } else {
                a = j0(x);
                b = j1(x);
                for(i=1;i<n;i++){
                    temp = b;
                    b = b*((double)(i+i)/x) - a; /* avoid underflow */
                    a = temp;
                }
            }
        } else {
            if (x < 1.86264514923095703125e-009) { /* x < 2**-29 */
    /* x is tiny, return the first Taylor expansion of J(n,x)
     * J(n,x) = 1/n!*(x/2)^n  - ...
     */
                if (n > 33)     /* underflow */
                    b = zero;
                else {
                    temp = x*0.5; b = temp;
                    for (a=one,i=2;i<=n;i++) {
                        a *= (double)i;         /* a = n! */
                        b *= temp;              /* b = (x/2)^n */
                    }
                    b = b/a;
                }
            } else {
                /* use backward recurrence */
                /*                      x      x^2      x^2
                 *  J(n,x)/J(n-1,x) =  ----   ------   ------   .....
                 *                      2n  - 2(n+1) - 2(n+2)
                 *
                 *                      1      1        1
                 *  (for large x)   =  ----  ------   ------   .....
                 *                      2n   2(n+1)   2(n+2)
                 *                      -- - ------ - ------ -
                 *                       x     x         x
                 *
                 * Let w = 2n/x and h=2/x, then the above quotient
                 * is equal to the continued fraction:
                 *                  1
                 *      = -----------------------
                 *                     1
                 *         w - -----------------
                 *                        1
                 *              w+h - ---------
                 *                     w+2h - ...
                 *
                 * To determine how many terms needed, let
                 * Q(0) = w, Q(1) = w(w+h) - 1,
                 * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
                 * When Q(k) > 1e4      good for single
                 * When Q(k) > 1e9      good for double
                 * When Q(k) > 1e17     good for quadruple
                 */
            /* determine k */
                double t,v;
                double q0,q1,h,tmp; int k,m;
                w  = (n+n)/(double)x; h = 2.0/(double)x;
                q0 = w;  z = w+h; q1 = w*z - 1.0; k=1;
                while (q1<1.0e9) {
                        k += 1; z += h;
                        tmp = z*q1 - q0;
                        q0 = q1;
                        q1 = tmp;
                }
                m = n+n;
                for(t=zero, i = 2*(n+k); i>=m; i -= 2) t = one/(i/x-t);
                a = t;
                b = one;
                /*  estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
                 *  Hence, if n*(log(2n/x)) > ...
                 *  single 8.8722839355e+01
                 *  double 7.09782712893383973096e+02
                 *  long double 1.1356523406294143949491931077970765006170e+04
                 *  then recurrent value may overflow and the result will
                 *  likely underflow to zero
                 */
                tmp = n;
                v = two/x;
                tmp = tmp*log(fabs(v*tmp));
                for (i=n-1;i>0;i--){
                        temp = b;
                        b = ((i+i)/x)*b - a;
                        a = temp;
                    /* scale b to avoid spurious overflow */
#                       if defined(__vax__) || defined(tahoe)
#                               define BMAX 1e13
#                       else
#                               define BMAX 1e100
#                       endif /* defined(__vax__) || defined(tahoe) */
                        if (b > BMAX) {
                                a /= b;
                                t /= b;
                                b = one;
                        }
                }
                b = (t*j0(x)/b);
            }
        }
        return ((sgn == 1) ? -b : b);
}

double
yn(int n, double x)
{
        int i, sign;
        double a, b, temp;

    /* Y(n,NaN), Y(n, x < 0) is NaN */
        if (x <= 0 || (_IEEE && x != x))
                if (_IEEE && x < 0) return zero/zero;
                else if (x < 0)     return (infnan(EDOM));
                else if (_IEEE)     return -one/zero;
                else                return(infnan(-ERANGE));
        else if (!finite(x)) return(0);
        sign = 1;
        if (n<0){
                n = -n;
                sign = 1 - ((n&1)<<2);
        }
        if (n == 0) return(y0(x));
        if (n == 1) return(sign*y1(x));
        if(_IEEE && x >= 8.148143905337944345e+090) { /* x > 2**302 */
    /* (x >> n**2)
     *      Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
     *      Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
     *      Let s=sin(x), c=cos(x),
     *          xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
     *
     *             n    sin(xn)*sqt2    cos(xn)*sqt2
     *          ----------------------------------
     *             0     s-c             c+s
     *             1    -s-c            -c+s
     *             2    -s+c            -c-s
     *             3     s+c             c-s
     */
                switch (n&3) {
                    case 0: temp =  sin(x)-cos(x); break;
                    case 1: temp = -sin(x)-cos(x); break;
                    case 2: temp = -sin(x)+cos(x); break;
                    case 3: temp =  sin(x)+cos(x); break;
                }
                b = invsqrtpi*temp/sqrt(x);
        } else {
            a = y0(x);
            b = y1(x);
        /* quit if b is -inf */
            for (i = 1; i < n && !finite(b); i++){
                temp = b;
                b = ((double)(i+i)/x)*b - a;
                a = temp;
            }
        }
        if (!_IEEE && !finite(b))
                return (infnan(-sign * ERANGE));
        return ((sign > 0) ? b : -b);
}