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e_jn.c (7109B)


      1 /* @(#)e_jn.c 5.1 93/09/24 */
      2 /*
      3  * ====================================================
      4  * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
      5  *
      6  * Developed at SunPro, a Sun Microsystems, Inc. business.
      7  * Permission to use, copy, modify, and distribute this
      8  * software is freely granted, provided that this notice
      9  * is preserved.
     10  * ====================================================
     11  */
     12 
     13 #ifndef lint
     14 static char rcsid[] = "$FreeBSD: src/lib/msun/src/e_jn.c,v 1.8 2002/05/28 18:15:04 alfred Exp $";
     15 #endif
     16 
     17 /*
     18  * __ieee754_jn(n, x), __ieee754_yn(n, x)
     19  * floating point Bessel's function of the 1st and 2nd kind
     20  * of order n
     21  *
     22  * Special cases:
     23  *	y0(0)=y1(0)=yn(n,0) = -inf with division by zero signal;
     24  *	y0(-ve)=y1(-ve)=yn(n,-ve) are NaN with invalid signal.
     25  * Note 2. About jn(n,x), yn(n,x)
     26  *	For n=0, j0(x) is called,
     27  *	for n=1, j1(x) is called,
     28  *	for n<x, forward recursion us used starting
     29  *	from values of j0(x) and j1(x).
     30  *	for n>x, a continued fraction approximation to
     31  *	j(n,x)/j(n-1,x) is evaluated and then backward
     32  *	recursion is used starting from a supposed value
     33  *	for j(n,x). The resulting value of j(0,x) is
     34  *	compared with the actual value to correct the
     35  *	supposed value of j(n,x).
     36  *
     37  *	yn(n,x) is similar in all respects, except
     38  *	that forward recursion is used for all
     39  *	values of n>1.
     40  *
     41  */
     42 
     43 #include "math.h"
     44 #include "math_private.h"
     45 
     46 static const double
     47 invsqrtpi=  5.64189583547756279280e-01, /* 0x3FE20DD7, 0x50429B6D */
     48 two   =  2.00000000000000000000e+00, /* 0x40000000, 0x00000000 */
     49 one   =  1.00000000000000000000e+00; /* 0x3FF00000, 0x00000000 */
     50 
     51 static const double zero  =  0.00000000000000000000e+00;
     52 
     53 double
     54 __ieee754_jn(int n, double x)
     55 {
     56 	int32_t i,hx,ix,lx, sgn;
     57 	double a, b, temp, di;
     58 	double z, w;
     59 
     60     /* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
     61      * Thus, J(-n,x) = J(n,-x)
     62      */
     63 	EXTRACT_WORDS(hx,lx,x);
     64 	ix = 0x7fffffff&hx;
     65     /* if J(n,NaN) is NaN */
     66 	if((ix|((u_int32_t)(lx|-lx))>>31)>0x7ff00000) return x+x;
     67 	if(n<0){
     68 		n = -n;
     69 		x = -x;
     70 		hx ^= 0x80000000;
     71 	}
     72 	if(n==0) return(__ieee754_j0(x));
     73 	if(n==1) return(__ieee754_j1(x));
     74 	sgn = (n&1)&(hx>>31);	/* even n -- 0, odd n -- sign(x) */
     75 	x = fabs(x);
     76 	if((ix|lx)==0||ix>=0x7ff00000) 	/* if x is 0 or inf */
     77 	    b = zero;
     78 	else if((double)n<=x) {
     79 		/* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
     80 	    if(ix>=0x52D00000) { /* x > 2**302 */
     81     /* (x >> n**2)
     82      *	    Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
     83      *	    Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
     84      *	    Let s=sin(x), c=cos(x),
     85      *		xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
     86      *
     87      *		   n	sin(xn)*sqt2	cos(xn)*sqt2
     88      *		----------------------------------
     89      *		   0	 s-c		 c+s
     90      *		   1	-s-c 		-c+s
     91      *		   2	-s+c		-c-s
     92      *		   3	 s+c		 c-s
     93      */
     94 		switch(n&3) {
     95 		    case 0: temp =  cos(x)+sin(x); break;
     96 		    case 1: temp = -cos(x)+sin(x); break;
     97 		    case 2: temp = -cos(x)-sin(x); break;
     98 		    case 3: temp =  cos(x)-sin(x); break;
     99 		}
    100 		b = invsqrtpi*temp/sqrt(x);
    101 	    } else {
    102 	        a = __ieee754_j0(x);
    103 	        b = __ieee754_j1(x);
    104 	        for(i=1;i<n;i++){
    105 		    temp = b;
    106 		    b = b*((double)(i+i)/x) - a; /* avoid underflow */
    107 		    a = temp;
    108 	        }
    109 	    }
    110 	} else {
    111 	    if(ix<0x3e100000) {	/* x < 2**-29 */
    112     /* x is tiny, return the first Taylor expansion of J(n,x)
    113      * J(n,x) = 1/n!*(x/2)^n  - ...
    114      */
    115 		if(n>33)	/* underflow */
    116 		    b = zero;
    117 		else {
    118 		    temp = x*0.5; b = temp;
    119 		    for (a=one,i=2;i<=n;i++) {
    120 			a *= (double)i;		/* a = n! */
    121 			b *= temp;		/* b = (x/2)^n */
    122 		    }
    123 		    b = b/a;
    124 		}
    125 	    } else {
    126 		/* use backward recurrence */
    127 		/* 			x      x^2      x^2
    128 		 *  J(n,x)/J(n-1,x) =  ----   ------   ------   .....
    129 		 *			2n  - 2(n+1) - 2(n+2)
    130 		 *
    131 		 * 			1      1        1
    132 		 *  (for large x)   =  ----  ------   ------   .....
    133 		 *			2n   2(n+1)   2(n+2)
    134 		 *			-- - ------ - ------ -
    135 		 *			 x     x         x
    136 		 *
    137 		 * Let w = 2n/x and h=2/x, then the above quotient
    138 		 * is equal to the continued fraction:
    139 		 *		    1
    140 		 *	= -----------------------
    141 		 *		       1
    142 		 *	   w - -----------------
    143 		 *			  1
    144 		 * 	        w+h - ---------
    145 		 *		       w+2h - ...
    146 		 *
    147 		 * To determine how many terms needed, let
    148 		 * Q(0) = w, Q(1) = w(w+h) - 1,
    149 		 * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
    150 		 * When Q(k) > 1e4	good for single
    151 		 * When Q(k) > 1e9	good for double
    152 		 * When Q(k) > 1e17	good for quadruple
    153 		 */
    154 	    /* determine k */
    155 		double t,v;
    156 		double q0,q1,h,tmp; int32_t k,m;
    157 		w  = (n+n)/(double)x; h = 2.0/(double)x;
    158 		q0 = w;  z = w+h; q1 = w*z - 1.0; k=1;
    159 		while(q1<1.0e9) {
    160 			k += 1; z += h;
    161 			tmp = z*q1 - q0;
    162 			q0 = q1;
    163 			q1 = tmp;
    164 		}
    165 		m = n+n;
    166 		for(t=zero, i = 2*(n+k); i>=m; i -= 2) t = one/(i/x-t);
    167 		a = t;
    168 		b = one;
    169 		/*  estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
    170 		 *  Hence, if n*(log(2n/x)) > ...
    171 		 *  single 8.8722839355e+01
    172 		 *  double 7.09782712893383973096e+02
    173 		 *  long double 1.1356523406294143949491931077970765006170e+04
    174 		 *  then recurrent value may overflow and the result is
    175 		 *  likely underflow to zero
    176 		 */
    177 		tmp = n;
    178 		v = two/x;
    179 		tmp = tmp*__ieee754_log(fabs(v*tmp));
    180 		if(tmp<7.09782712893383973096e+02) {
    181 	    	    for(i=n-1,di=(double)(i+i);i>0;i--){
    182 		        temp = b;
    183 			b *= di;
    184 			b  = b/x - a;
    185 		        a = temp;
    186 			di -= two;
    187 	     	    }
    188 		} else {
    189 	    	    for(i=n-1,di=(double)(i+i);i>0;i--){
    190 		        temp = b;
    191 			b *= di;
    192 			b  = b/x - a;
    193 		        a = temp;
    194 			di -= two;
    195 		    /* scale b to avoid spurious overflow */
    196 			if(b>1e100) {
    197 			    a /= b;
    198 			    t /= b;
    199 			    b  = one;
    200 			}
    201 	     	    }
    202 		}
    203 	    	b = (t*__ieee754_j0(x)/b);
    204 	    }
    205 	}
    206 	if(sgn==1) return -b; else return b;
    207 }
    208 
    209 double
    210 __ieee754_yn(int n, double x)
    211 {
    212 	int32_t i,hx,ix,lx;
    213 	int32_t sign;
    214 	double a, b, temp;
    215 
    216 	EXTRACT_WORDS(hx,lx,x);
    217 	ix = 0x7fffffff&hx;
    218     /* if Y(n,NaN) is NaN */
    219 	if((ix|((u_int32_t)(lx|-lx))>>31)>0x7ff00000) return x+x;
    220 	if((ix|lx)==0) return -one/zero;
    221 	if(hx<0) return zero/zero;
    222 	sign = 1;
    223 	if(n<0){
    224 		n = -n;
    225 		sign = 1 - ((n&1)<<1);
    226 	}
    227 	if(n==0) return(__ieee754_y0(x));
    228 	if(n==1) return(sign*__ieee754_y1(x));
    229 	if(ix==0x7ff00000) return zero;
    230 	if(ix>=0x52D00000) { /* x > 2**302 */
    231     /* (x >> n**2)
    232      *	    Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
    233      *	    Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
    234      *	    Let s=sin(x), c=cos(x),
    235      *		xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
    236      *
    237      *		   n	sin(xn)*sqt2	cos(xn)*sqt2
    238      *		----------------------------------
    239      *		   0	 s-c		 c+s
    240      *		   1	-s-c 		-c+s
    241      *		   2	-s+c		-c-s
    242      *		   3	 s+c		 c-s
    243      */
    244 		switch(n&3) {
    245 		    case 0: temp =  sin(x)-cos(x); break;
    246 		    case 1: temp = -sin(x)-cos(x); break;
    247 		    case 2: temp = -sin(x)+cos(x); break;
    248 		    case 3: temp =  sin(x)+cos(x); break;
    249 		}
    250 		b = invsqrtpi*temp/sqrt(x);
    251 	} else {
    252 	    u_int32_t high;
    253 	    a = __ieee754_y0(x);
    254 	    b = __ieee754_y1(x);
    255 	/* quit if b is -inf */
    256 	    GET_HIGH_WORD(high,b);
    257 	    for(i=1;i<n&&high!=0xfff00000;i++){
    258 		temp = b;
    259 		b = ((double)(i+i)/x)*b - a;
    260 		GET_HIGH_WORD(high,b);
    261 		a = temp;
    262 	    }
    263 	}
    264 	if(sign>0) return b; else return -b;
    265 }