/* * CDDL HEADER START * * The contents of this file are subject to the terms of the * Common Development and Distribution License (the "License"). * You may not use this file except in compliance with the License. * * You can obtain a copy of the license at usr/src/OPENSOLARIS.LICENSE * or http://www.opensolaris.org/os/licensing. * See the License for the specific language governing permissions * and limitations under the License. * * When distributing Covered Code, include this CDDL HEADER in each * file and include the License file at usr/src/OPENSOLARIS.LICENSE. * If applicable, add the following below this CDDL HEADER, with the * fields enclosed by brackets "[]" replaced with your own identifying * information: Portions Copyright [yyyy] [name of copyright owner] * * CDDL HEADER END */ /* * Copyright 2011 Nexenta Systems, Inc. All rights reserved. */ /* * Copyright 2006 Sun Microsystems, Inc. All rights reserved. * Use is subject to license terms. */ #if defined(ELFOBJ) #pragma weak jnl = __jnl #pragma weak ynl = __ynl #endif /* * floating point Bessel's function of the 1st and 2nd kind * of order n: jn(n,x),yn(n,x); * * 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 "libm.h" #include "longdouble.h" #include /* LDBL_MAX */ #define GENERIC long double static const GENERIC invsqrtpi = 5.641895835477562869480794515607725858441e-0001L, two = 2.0L, zero = 0.0L, one = 1.0L; GENERIC jnl(n, x) int n; GENERIC x; { int i, sgn; GENERIC a, b, temp = 0, 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) */ if (n < 0) { n = -n; x = -x; } if (n == 0) return (j0l(x)); if (n == 1) return (j1l(x)); if (x != x) return x+x; if ((n&1) == 0) sgn = 0; /* even n */ else sgn = signbitl(x); /* old n */ x = fabsl(x); if (x == zero || !finitel(x)) b = zero; else if ((GENERIC)n <= x) { /* * Safe to use * J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */ if (x > 1.0e91L) { /* * 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 = cosl(x)+sinl(x); break; case 1: temp = -cosl(x)+sinl(x); break; case 2: temp = -cosl(x)-sinl(x); break; case 3: temp = cosl(x)-sinl(x); break; } b = invsqrtpi*temp/sqrtl(x); } else { a = j0l(x); b = j1l(x); for (i = 1; i < n; i++) { temp = b; b = b*((GENERIC)(i+i)/x) - a; /* avoid underflow */ a = temp; } } } else { if (x < 1e-17L) { /* use J(n,x) = 1/n!*(x/2)^n */ b = powl(0.5L*x, (GENERIC) n); if (b != zero) { for (a = one, i = 1; i <= n; i++) a *= (GENERIC)i; 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 quaduple */ /* determin k */ GENERIC 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 < 1.0e17) { 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*logl(fabsl(v*tmp)); if (tmp < 1.1356523406294143949491931077970765e+04L) { for (i = n-1; i > 0; i--) { temp = b; b = ((i+i)/x)*b - a; a = temp; } } else { for (i = n-1; i > 0; i--) { temp = b; b = ((i+i)/x)*b - a; a = temp; if (b > 1e1000L) { a /= b; t /= b; b = 1.0; } } } b = (t*j0l(x)/b); } } if (sgn == 1) return -b; else return b; } GENERIC ynl(n, x) int n; GENERIC x; { int i; int sign; GENERIC a, b, temp = 0; if (x != x) return x+x; if (x <= zero) { if (x == zero) return -one/zero; else return zero/zero; } sign = 1; if (n < 0) { n = -n; if ((n&1) == 1) sign = -1; } if (n == 0) return (y0l(x)); if (n == 1) return (sign*y1l(x)); if (!finitel(x)) return zero; if (x > 1.0e91L) { /* * 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 = sinl(x)-cosl(x); break; case 1: temp = -sinl(x)-cosl(x); break; case 2: temp = -sinl(x)+cosl(x); break; case 3: temp = sinl(x)+cosl(x); break; } b = invsqrtpi*temp/sqrtl(x); } else { a = y0l(x); b = y1l(x); /* * fix 1262058 and take care of non-default rounding */ for (i = 1; i < n; i++) { temp = b; b *= (GENERIC) (i + i) / x; if (b <= -LDBL_MAX) break; b -= a; a = temp; } } if (sign > 0) return b; else return -b; }