--- old/src/share/native/java/lang/fdlibm/src/e_jn.c 2011-08-04 10:54:16.000000000 -0700 +++ /dev/null 2011-08-03 01:04:26.146397172 -0700 @@ -1,285 +0,0 @@ - -/* - * Copyright (c) 1998, 2001, Oracle and/or its affiliates. All rights reserved. - * DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER. - * - * This code is free software; you can redistribute it and/or modify it - * under the terms of the GNU General Public License version 2 only, as - * published by the Free Software Foundation. Oracle designates this - * particular file as subject to the "Classpath" exception as provided - * by Oracle in the LICENSE file that accompanied this code. - * - * This code is distributed in the hope that it will be useful, but WITHOUT - * ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or - * FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License - * version 2 for more details (a copy is included in the LICENSE file that - * accompanied this code). - * - * You should have received a copy of the GNU General Public License version - * 2 along with this work; if not, write to the Free Software Foundation, - * Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA. - * - * Please contact Oracle, 500 Oracle Parkway, Redwood Shores, CA 94065 USA - * or visit www.oracle.com if you need additional information or have any - * questions. - */ - -/* - * __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 nx, 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 "fdlibm.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 */ - -static double zero = 0.00000000000000000000e+00; - -#ifdef __STDC__ - double __ieee754_jn(int n, double x) -#else - double __ieee754_jn(n,x) - int n; double x; -#endif -{ - int i,hx,ix,lx, sgn; - double a, b, temp = 0, 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) - */ - hx = __HI(x); - ix = 0x7fffffff&hx; - lx = __LO(x); - /* if J(n,NaN) is NaN */ - if((ix|((unsigned)(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;i33) /* 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; int 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 -{ - int i,hx,ix,lx; - int sign; - double a, b, temp = 0; - - hx = __HI(x); - ix = 0x7fffffff&hx; - lx = __LO(x); - /* if Y(n,NaN) is NaN */ - if((ix|((unsigned)(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 { - a = __ieee754_y0(x); - b = __ieee754_y1(x); - /* quit if b is -inf */ - for(i=1;i0) return b; else return -b; -}