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glibc  2.9
e_jn.c
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00001 /* @(#)e_jn.c 5.1 93/09/24 */
00002 /*
00003  * ====================================================
00004  * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
00005  *
00006  * Developed at SunPro, a Sun Microsystems, Inc. business.
00007  * Permission to use, copy, modify, and distribute this
00008  * software is freely granted, provided that this notice
00009  * is preserved.
00010  * ====================================================
00011  */
00012 
00013 #if defined(LIBM_SCCS) && !defined(lint)
00014 static char rcsid[] = "$NetBSD: e_jn.c,v 1.9 1995/05/10 20:45:34 jtc Exp $";
00015 #endif
00016 
00017 /*
00018  * __ieee754_jn(n, x), __ieee754_yn(n, x)
00019  * floating point Bessel's function of the 1st and 2nd kind
00020  * of order n
00021  *
00022  * Special cases:
00023  *     y0(0)=y1(0)=yn(n,0) = -inf with overflow signal;
00024  *     y0(-ve)=y1(-ve)=yn(n,-ve) are NaN with invalid signal.
00025  * Note 2. About jn(n,x), yn(n,x)
00026  *     For n=0, j0(x) is called,
00027  *     for n=1, j1(x) is called,
00028  *     for n<x, forward recursion us used starting
00029  *     from values of j0(x) and j1(x).
00030  *     for n>x, a continued fraction approximation to
00031  *     j(n,x)/j(n-1,x) is evaluated and then backward
00032  *     recursion is used starting from a supposed value
00033  *     for j(n,x). The resulting value of j(0,x) is
00034  *     compared with the actual value to correct the
00035  *     supposed value of j(n,x).
00036  *
00037  *     yn(n,x) is similar in all respects, except
00038  *     that forward recursion is used for all
00039  *     values of n>1.
00040  *
00041  */
00042 
00043 #include "math.h"
00044 #include "math_private.h"
00045 
00046 #ifdef __STDC__
00047 static const double
00048 #else
00049 static double
00050 #endif
00051 invsqrtpi=  5.64189583547756279280e-01, /* 0x3FE20DD7, 0x50429B6D */
00052 two   =  2.00000000000000000000e+00, /* 0x40000000, 0x00000000 */
00053 one   =  1.00000000000000000000e+00; /* 0x3FF00000, 0x00000000 */
00054 
00055 #ifdef __STDC__
00056 static const double zero  =  0.00000000000000000000e+00;
00057 #else
00058 static double zero  =  0.00000000000000000000e+00;
00059 #endif
00060 
00061 #ifdef __STDC__
00062        double __ieee754_jn(int n, double x)
00063 #else
00064        double __ieee754_jn(n,x)
00065        int n; double x;
00066 #endif
00067 {
00068        int32_t i,hx,ix,lx, sgn;
00069        double a, b, temp, di;
00070        double z, w;
00071 
00072     /* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
00073      * Thus, J(-n,x) = J(n,-x)
00074      */
00075        EXTRACT_WORDS(hx,lx,x);
00076        ix = 0x7fffffff&hx;
00077     /* if J(n,NaN) is NaN */
00078        if((ix|((u_int32_t)(lx|-lx))>>31)>0x7ff00000) return x+x;
00079        if(n<0){
00080               n = -n;
00081               x = -x;
00082               hx ^= 0x80000000;
00083        }
00084        if(n==0) return(__ieee754_j0(x));
00085        if(n==1) return(__ieee754_j1(x));
00086        sgn = (n&1)&(hx>>31);       /* even n -- 0, odd n -- sign(x) */
00087        x = fabs(x);
00088        if((ix|lx)==0||ix>=0x7ff00000)     /* if x is 0 or inf */
00089            b = zero;
00090        else if((double)n<=x) {
00091               /* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
00092            if(ix>=0x52D00000) { /* x > 2**302 */
00093     /* (x >> n**2)
00094      *     Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
00095      *     Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
00096      *     Let s=sin(x), c=cos(x),
00097      *        xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
00098      *
00099      *           n   sin(xn)*sqt2  cos(xn)*sqt2
00100      *        ----------------------------------
00101      *           0    s-c           c+s
00102      *           1   -s-c          -c+s
00103      *           2   -s+c          -c-s
00104      *           3    s+c           c-s
00105      */
00106               double s;
00107               double c;
00108               __sincos (x, &s, &c);
00109               switch(n&3) {
00110                   case 0: temp =  c + s; break;
00111                   case 1: temp = -c + s; break;
00112                   case 2: temp = -c - s; break;
00113                   case 3: temp =  c - s; break;
00114               }
00115               b = invsqrtpi*temp/__ieee754_sqrt(x);
00116            } else {
00117                a = __ieee754_j0(x);
00118                b = __ieee754_j1(x);
00119                for(i=1;i<n;i++){
00120                   temp = b;
00121                   b = b*((double)(i+i)/x) - a; /* avoid underflow */
00122                   a = temp;
00123                }
00124            }
00125        } else {
00126            if(ix<0x3e100000) {     /* x < 2**-29 */
00127     /* x is tiny, return the first Taylor expansion of J(n,x)
00128      * J(n,x) = 1/n!*(x/2)^n  - ...
00129      */
00130               if(n>33)      /* underflow */
00131                   b = zero;
00132               else {
00133                   temp = x*0.5; b = temp;
00134                   for (a=one,i=2;i<=n;i++) {
00135                      a *= (double)i;             /* a = n! */
00136                      b *= temp;           /* b = (x/2)^n */
00137                   }
00138                   b = b/a;
00139               }
00140            } else {
00141               /* use backward recurrence */
00142               /*                   x      x^2      x^2
00143                *  J(n,x)/J(n-1,x) =  ----   ------   ------   .....
00144                *                   2n  - 2(n+1) - 2(n+2)
00145                *
00146                *                   1      1        1
00147                *  (for large x)   =  ----  ------   ------   .....
00148                *                   2n   2(n+1)   2(n+2)
00149                *                   -- - ------ - ------ -
00150                *                    x     x         x
00151                *
00152                * Let w = 2n/x and h=2/x, then the above quotient
00153                * is equal to the continued fraction:
00154                *                1
00155                *     = -----------------------
00156                *                   1
00157                *        w - -----------------
00158                *                     1
00159                *             w+h - ---------
00160                *                   w+2h - ...
00161                *
00162                * To determine how many terms needed, let
00163                * Q(0) = w, Q(1) = w(w+h) - 1,
00164                * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
00165                * When Q(k) > 1e4   good for single
00166                * When Q(k) > 1e9   good for double
00167                * When Q(k) > 1e17  good for quadruple
00168                */
00169            /* determine k */
00170               double t,v;
00171               double q0,q1,h,tmp; int32_t k,m;
00172               w  = (n+n)/(double)x; h = 2.0/(double)x;
00173               q0 = w;  z = w+h; q1 = w*z - 1.0; k=1;
00174               while(q1<1.0e9) {
00175                      k += 1; z += h;
00176                      tmp = z*q1 - q0;
00177                      q0 = q1;
00178                      q1 = tmp;
00179               }
00180               m = n+n;
00181               for(t=zero, i = 2*(n+k); i>=m; i -= 2) t = one/(i/x-t);
00182               a = t;
00183               b = one;
00184               /*  estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
00185                *  Hence, if n*(log(2n/x)) > ...
00186                *  single 8.8722839355e+01
00187                *  double 7.09782712893383973096e+02
00188                *  long double 1.1356523406294143949491931077970765006170e+04
00189                *  then recurrent value may overflow and the result is
00190                *  likely underflow to zero
00191                */
00192               tmp = n;
00193               v = two/x;
00194               tmp = tmp*__ieee754_log(fabs(v*tmp));
00195               if(tmp<7.09782712893383973096e+02) {
00196                   for(i=n-1,di=(double)(i+i);i>0;i--){
00197                       temp = b;
00198                      b *= di;
00199                      b  = b/x - a;
00200                       a = temp;
00201                      di -= two;
00202                   }
00203               } else {
00204                   for(i=n-1,di=(double)(i+i);i>0;i--){
00205                       temp = b;
00206                      b *= di;
00207                      b  = b/x - a;
00208                       a = temp;
00209                      di -= two;
00210                   /* scale b to avoid spurious overflow */
00211                      if(b>1e100) {
00212                          a /= b;
00213                          t /= b;
00214                          b  = one;
00215                      }
00216                   }
00217               }
00218               b = (t*__ieee754_j0(x)/b);
00219            }
00220        }
00221        if(sgn==1) return -b; else return b;
00222 }
00223 
00224 #ifdef __STDC__
00225        double __ieee754_yn(int n, double x)
00226 #else
00227        double __ieee754_yn(n,x)
00228        int n; double x;
00229 #endif
00230 {
00231        int32_t i,hx,ix,lx;
00232        int32_t sign;
00233        double a, b, temp;
00234 
00235        EXTRACT_WORDS(hx,lx,x);
00236        ix = 0x7fffffff&hx;
00237     /* if Y(n,NaN) is NaN */
00238        if((ix|((u_int32_t)(lx|-lx))>>31)>0x7ff00000) return x+x;
00239        if((ix|lx)==0) return -HUGE_VAL+x; /* -inf and overflow exception.  */;
00240        if(hx<0) return zero/(zero*x);
00241        sign = 1;
00242        if(n<0){
00243               n = -n;
00244               sign = 1 - ((n&1)<<1);
00245        }
00246        if(n==0) return(__ieee754_y0(x));
00247        if(n==1) return(sign*__ieee754_y1(x));
00248        if(ix==0x7ff00000) return zero;
00249        if(ix>=0x52D00000) { /* x > 2**302 */
00250     /* (x >> n**2)
00251      *     Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
00252      *     Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
00253      *     Let s=sin(x), c=cos(x),
00254      *        xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
00255      *
00256      *           n   sin(xn)*sqt2  cos(xn)*sqt2
00257      *        ----------------------------------
00258      *           0    s-c           c+s
00259      *           1   -s-c          -c+s
00260      *           2   -s+c          -c-s
00261      *           3    s+c           c-s
00262      */
00263               double c;
00264               double s;
00265               __sincos (x, &s, &c);
00266               switch(n&3) {
00267                   case 0: temp =  s - c; break;
00268                   case 1: temp = -s - c; break;
00269                   case 2: temp = -s + c; break;
00270                   case 3: temp =  s + c; break;
00271               }
00272               b = invsqrtpi*temp/__ieee754_sqrt(x);
00273        } else {
00274            u_int32_t high;
00275            a = __ieee754_y0(x);
00276            b = __ieee754_y1(x);
00277        /* quit if b is -inf */
00278            GET_HIGH_WORD(high,b);
00279            for(i=1;i<n&&high!=0xfff00000;i++){
00280               temp = b;
00281               b = ((double)(i+i)/x)*b - a;
00282               GET_HIGH_WORD(high,b);
00283               a = temp;
00284            }
00285        }
00286        if(sign>0) return b; else return -b;
00287 }