Totally rework the math library, this time based on the MacOs X

math library (which is itself based on the math lib from FreeBSD).
 -Erik

1 files changed, 281 insertions, 0 deletions

diff --git a/libm/e_jn.c b/libm/e_jn.c
new file mode 100644
index 000000000..27a8a1969
--- /dev/null
+++ b/libm/e_jn.c
@@ -0,0 +1,281 @@
+/* @(#)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
+     */
+		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 = __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
+     */
+		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 {
+	    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;
+}