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      1 
      2 /* @(#)e_jn.c 1.4 95/01/18 */
      3 /*
      4  * ====================================================
      5  * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
      6  *
      7  * Developed at SunSoft, a Sun Microsystems, Inc. business.
      8  * Permission to use, copy, modify, and distribute this
      9  * software is freely granted, provided that this notice
     10  * is preserved.
     11  * ====================================================
     12  */
     13 
     14 #ifndef lint
     15 static char rcsid[] = "$FreeBSD: src/lib/msun/src/e_jn.c,v 1.9 2005/02/04 18:26:06 das Exp $";
     16 #endif
     17 
     18 /*
     19  * __ieee754_jn(n, x), __ieee754_yn(n, x)
     20  * floating point Bessel's function of the 1st and 2nd kind
     21  * of order n
     22  *
     23  * Special cases:
     24  *	y0(0)=y1(0)=yn(n,0) = -inf with division by zero signal;
     25  *	y0(-ve)=y1(-ve)=yn(n,-ve) are NaN with invalid signal.
     26  * Note 2. About jn(n,x), yn(n,x)
     27  *	For n=0, j0(x) is called,
     28  *	for n=1, j1(x) is called,
     29  *	for n<x, forward recursion us used starting
     30  *	from values of j0(x) and j1(x).
     31  *	for n>x, a continued fraction approximation to
     32  *	j(n,x)/j(n-1,x) is evaluated and then backward
     33  *	recursion is used starting from a supposed value
     34  *	for j(n,x). The resulting value of j(0,x) is
     35  *	compared with the actual value to correct the
     36  *	supposed value of j(n,x).
     37  *
     38  *	yn(n,x) is similar in all respects, except
     39  *	that forward recursion is used for all
     40  *	values of n>1.
     41  *
     42  */
     43 
     44 #include "math.h"
     45 #include "math_private.h"
     46 
     47 static const double
     48 invsqrtpi=  5.64189583547756279280e-01, /* 0x3FE20DD7, 0x50429B6D */
     49 two   =  2.00000000000000000000e+00, /* 0x40000000, 0x00000000 */
     50 one   =  1.00000000000000000000e+00; /* 0x3FF00000, 0x00000000 */
     51 
     52 static const double zero  =  0.00000000000000000000e+00;
     53 
     54 double
     55 __ieee754_jn(int n, double x)
     56 {
     57 	int32_t i,hx,ix,lx, sgn;
     58 	double a, b, temp, di;
     59 	double z, w;
     60 
     61     /* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
     62      * Thus, J(-n,x) = J(n,-x)
     63      */
     64 	EXTRACT_WORDS(hx,lx,x);
     65 	ix = 0x7fffffff&hx;
     66     /* if J(n,NaN) is NaN */
     67 	if((ix|((u_int32_t)(lx|-lx))>>31)>0x7ff00000) return x+x;
     68 	if(n<0){
     69 		n = -n;
     70 		x = -x;
     71 		hx ^= 0x80000000;
     72 	}
     73 	if(n==0) return(__ieee754_j0(x));
     74 	if(n==1) return(__ieee754_j1(x));
     75 	sgn = (n&1)&(hx>>31);	/* even n -- 0, odd n -- sign(x) */
     76 	x = fabs(x);
     77 	if((ix|lx)==0||ix>=0x7ff00000) 	/* if x is 0 or inf */
     78 	    b = zero;
     79 	else if((double)n<=x) {
     80 		/* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
     81 	    if(ix>=0x52D00000) { /* x > 2**302 */
     82     /* (x >> n**2)
     83      *	    Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
     84      *	    Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
     85      *	    Let s=sin(x), c=cos(x),
     86      *		xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
     87      *
     88      *		   n	sin(xn)*sqt2	cos(xn)*sqt2
     89      *		----------------------------------
     90      *		   0	 s-c		 c+s
     91      *		   1	-s-c 		-c+s
     92      *		   2	-s+c		-c-s
     93      *		   3	 s+c		 c-s
     94      */
     95 		switch(n&3) {
     96 		    case 0: temp =  cos(x)+sin(x); break;
     97 		    case 1: temp = -cos(x)+sin(x); break;
     98 		    case 2: temp = -cos(x)-sin(x); break;
     99 		    case 3: temp =  cos(x)-sin(x); break;
    100 		}
    101 		b = invsqrtpi*temp/sqrt(x);
    102 	    } else {
    103 	        a = __ieee754_j0(x);
    104 	        b = __ieee754_j1(x);
    105 	        for(i=1;i<n;i++){
    106 		    temp = b;
    107 		    b = b*((double)(i+i)/x) - a; /* avoid underflow */
    108 		    a = temp;
    109 	        }
    110 	    }
    111 	} else {
    112 	    if(ix<0x3e100000) {	/* x < 2**-29 */
    113     /* x is tiny, return the first Taylor expansion of J(n,x)
    114      * J(n,x) = 1/n!*(x/2)^n  - ...
    115      */
    116 		if(n>33)	/* underflow */
    117 		    b = zero;
    118 		else {
    119 		    temp = x*0.5; b = temp;
    120 		    for (a=one,i=2;i<=n;i++) {
    121 			a *= (double)i;		/* a = n! */
    122 			b *= temp;		/* b = (x/2)^n */
    123 		    }
    124 		    b = b/a;
    125 		}
    126 	    } else {
    127 		/* use backward recurrence */
    128 		/* 			x      x^2      x^2
    129 		 *  J(n,x)/J(n-1,x) =  ----   ------   ------   .....
    130 		 *			2n  - 2(n+1) - 2(n+2)
    131 		 *
    132 		 * 			1      1        1
    133 		 *  (for large x)   =  ----  ------   ------   .....
    134 		 *			2n   2(n+1)   2(n+2)
    135 		 *			-- - ------ - ------ -
    136 		 *			 x     x         x
    137 		 *
    138 		 * Let w = 2n/x and h=2/x, then the above quotient
    139 		 * is equal to the continued fraction:
    140 		 *		    1
    141 		 *	= -----------------------
    142 		 *		       1
    143 		 *	   w - -----------------
    144 		 *			  1
    145 		 * 	        w+h - ---------
    146 		 *		       w+2h - ...
    147 		 *
    148 		 * To determine how many terms needed, let
    149 		 * Q(0) = w, Q(1) = w(w+h) - 1,
    150 		 * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
    151 		 * When Q(k) > 1e4	good for single
    152 		 * When Q(k) > 1e9	good for double
    153 		 * When Q(k) > 1e17	good for quadruple
    154 		 */
    155 	    /* determine k */
    156 		double t,v;
    157 		double q0,q1,h,tmp; int32_t k,m;
    158 		w  = (n+n)/(double)x; h = 2.0/(double)x;
    159 		q0 = w;  z = w+h; q1 = w*z - 1.0; k=1;
    160 		while(q1<1.0e9) {
    161 			k += 1; z += h;
    162 			tmp = z*q1 - q0;
    163 			q0 = q1;
    164 			q1 = tmp;
    165 		}
    166 		m = n+n;
    167 		for(t=zero, i = 2*(n+k); i>=m; i -= 2) t = one/(i/x-t);
    168 		a = t;
    169 		b = one;
    170 		/*  estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
    171 		 *  Hence, if n*(log(2n/x)) > ...
    172 		 *  single 8.8722839355e+01
    173 		 *  double 7.09782712893383973096e+02
    174 		 *  long double 1.1356523406294143949491931077970765006170e+04
    175 		 *  then recurrent value may overflow and the result is
    176 		 *  likely underflow to zero
    177 		 */
    178 		tmp = n;
    179 		v = two/x;
    180 		tmp = tmp*__ieee754_log(fabs(v*tmp));
    181 		if(tmp<7.09782712893383973096e+02) {
    182 	    	    for(i=n-1,di=(double)(i+i);i>0;i--){
    183 		        temp = b;
    184 			b *= di;
    185 			b  = b/x - a;
    186 		        a = temp;
    187 			di -= two;
    188 	     	    }
    189 		} else {
    190 	    	    for(i=n-1,di=(double)(i+i);i>0;i--){
    191 		        temp = b;
    192 			b *= di;
    193 			b  = b/x - a;
    194 		        a = temp;
    195 			di -= two;
    196 		    /* scale b to avoid spurious overflow */
    197 			if(b>1e100) {
    198 			    a /= b;
    199 			    t /= b;
    200 			    b  = one;
    201 			}
    202 	     	    }
    203 		}
    204 	    	b = (t*__ieee754_j0(x)/b);
    205 	    }
    206 	}
    207 	if(sgn==1) return -b; else return b;
    208 }
    209 
    210 double
    211 __ieee754_yn(int n, double x)
    212 {
    213 	int32_t i,hx,ix,lx;
    214 	int32_t sign;
    215 	double a, b, temp;
    216 
    217 	EXTRACT_WORDS(hx,lx,x);
    218 	ix = 0x7fffffff&hx;
    219     /* if Y(n,NaN) is NaN */
    220 	if((ix|((u_int32_t)(lx|-lx))>>31)>0x7ff00000) return x+x;
    221 	if((ix|lx)==0) return -one/zero;
    222 	if(hx<0) return zero/zero;
    223 	sign = 1;
    224 	if(n<0){
    225 		n = -n;
    226 		sign = 1 - ((n&1)<<1);
    227 	}
    228 	if(n==0) return(__ieee754_y0(x));
    229 	if(n==1) return(sign*__ieee754_y1(x));
    230 	if(ix==0x7ff00000) return zero;
    231 	if(ix>=0x52D00000) { /* x > 2**302 */
    232     /* (x >> n**2)
    233      *	    Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
    234      *	    Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
    235      *	    Let s=sin(x), c=cos(x),
    236      *		xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
    237      *
    238      *		   n	sin(xn)*sqt2	cos(xn)*sqt2
    239      *		----------------------------------
    240      *		   0	 s-c		 c+s
    241      *		   1	-s-c 		-c+s
    242      *		   2	-s+c		-c-s
    243      *		   3	 s+c		 c-s
    244      */
    245 		switch(n&3) {
    246 		    case 0: temp =  sin(x)-cos(x); break;
    247 		    case 1: temp = -sin(x)-cos(x); break;
    248 		    case 2: temp = -sin(x)+cos(x); break;
    249 		    case 3: temp =  sin(x)+cos(x); break;
    250 		}
    251 		b = invsqrtpi*temp/sqrt(x);
    252 	} else {
    253 	    u_int32_t high;
    254 	    a = __ieee754_y0(x);
    255 	    b = __ieee754_y1(x);
    256 	/* quit if b is -inf */
    257 	    GET_HIGH_WORD(high,b);
    258 	    for(i=1;i<n&&high!=0xfff00000;i++){
    259 		temp = b;
    260 		b = ((double)(i+i)/x)*b - a;
    261 		GET_HIGH_WORD(high,b);
    262 		a = temp;
    263 	    }
    264 	}
    265 	if(sign>0) return b; else return -b;
    266 }
    267