Logo Search packages:      
Sourcecode: glibc version File versions

e_jn.c

/* @(#)e_jn.c 5.1 93/09/24 */
/*
 * ====================================================
 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
 *
 * 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.
 * ====================================================
 */

#if defined(LIBM_SCCS) && !defined(lint)
static char rcsid[] = "$NetBSD: e_jn.c,v 1.9 1995/05/10 20:45:34 jtc Exp $";
#endif

/*
 * __ieee754_jn(n, x), __ieee754_yn(n, 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 "math.h"
#include "math_private.h"

#ifdef __STDC__
static const double
#else
static double
#endif
invsqrtpi=  5.64189583547756279280e-01, /* 0x3FE20DD7, 0x50429B6D */
two   =  2.00000000000000000000e+00, /* 0x40000000, 0x00000000 */
one   =  1.00000000000000000000e+00; /* 0x3FF00000, 0x00000000 */

#ifdef __STDC__
static const double zero  =  0.00000000000000000000e+00;
#else
static double zero  =  0.00000000000000000000e+00;
#endif

#ifdef __STDC__
      double __ieee754_jn(int n, double x)
#else
      double __ieee754_jn(n,x)
      int n; double x;
#endif
{
      int32_t i,hx,ix,lx, sgn;
      double a, b, temp, di;
      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)
     */
      EXTRACT_WORDS(hx,lx,x);
      ix = 0x7fffffff&hx;
    /* if J(n,NaN) is NaN */
      if((ix|((u_int32_t)(lx|-lx))>>31)>0x7ff00000) return x+x;
      if(n<0){
            n = -n;
            x = -x;
            hx ^= 0x80000000;
      }
      if(n==0) return(__ieee754_j0(x));
      if(n==1) return(__ieee754_j1(x));
      sgn = (n&1)&(hx>>31);   /* even n -- 0, odd n -- sign(x) */
      x = fabs(x);
      if((ix|lx)==0||ix>=0x7ff00000)      /* 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(ix>=0x52D00000) { /* 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
     */
            double s;
            double c;
            __sincos (x, &s, &c);
            switch(n&3) {
                case 0: temp =  c + s; break;
                case 1: temp = -c + s; break;
                case 2: temp = -c - s; break;
                case 3: temp =  c - s; break;
            }
            b = invsqrtpi*temp/__ieee754_sqrt(x);
          } else {
              a = __ieee754_j0(x);
              b = __ieee754_j1(x);
              for(i=1;i<n;i++){
                temp = b;
                b = b*((double)(i+i)/x) - a; /* avoid underflow */
                a = temp;
              }
          }
      } else {
          if(ix<0x3e100000) { /* 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; int32_t 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 is
             *  likely underflow to zero
             */
            tmp = n;
            v = two/x;
            tmp = tmp*__ieee754_log(fabs(v*tmp));
            if(tmp<7.09782712893383973096e+02) {
                for(i=n-1,di=(double)(i+i);i>0;i--){
                    temp = b;
                  b *= di;
                  b  = b/x - a;
                    a = temp;
                  di -= two;
                }
            } else {
                for(i=n-1,di=(double)(i+i);i>0;i--){
                    temp = b;
                  b *= di;
                  b  = b/x - a;
                    a = temp;
                  di -= two;
                /* scale b to avoid spurious overflow */
                  if(b>1e100) {
                      a /= b;
                      t /= b;
                      b  = one;
                  }
                }
            }
            b = (t*__ieee754_j0(x)/b);
          }
      }
      if(sgn==1) return -b; else return b;
}

#ifdef __STDC__
      double __ieee754_yn(int n, double x)
#else
      double __ieee754_yn(n,x)
      int n; double x;
#endif
{
      int32_t i,hx,ix,lx;
      int32_t sign;
      double a, b, temp;

      EXTRACT_WORDS(hx,lx,x);
      ix = 0x7fffffff&hx;
    /* if Y(n,NaN) is NaN */
      if((ix|((u_int32_t)(lx|-lx))>>31)>0x7ff00000) return x+x;
      if((ix|lx)==0) return -one/zero;
      if(hx<0) return zero/zero;
      sign = 1;
      if(n<0){
            n = -n;
            sign = 1 - ((n&1)<<1);
      }
      if(n==0) return(__ieee754_y0(x));
      if(n==1) return(sign*__ieee754_y1(x));
      if(ix==0x7ff00000) return zero;
      if(ix>=0x52D00000) { /* 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
     */
            double c;
            double s;
            __sincos (x, &s, &c);
            switch(n&3) {
                case 0: temp =  s - c; break;
                case 1: temp = -s - c; break;
                case 2: temp = -s + c; break;
                case 3: temp =  s + c; break;
            }
            b = invsqrtpi*temp/__ieee754_sqrt(x);
      } else {
          u_int32_t high;
          a = __ieee754_y0(x);
          b = __ieee754_y1(x);
      /* quit if b is -inf */
          GET_HIGH_WORD(high,b);
          for(i=1;i<n&&high!=0xfff00000;i++){
            temp = b;
            b = ((double)(i+i)/x)*b - a;
            GET_HIGH_WORD(high,b);
            a = temp;
          }
      }
      if(sign>0) return b; else return -b;
}

Generated by  Doxygen 1.6.0   Back to index