blob: ae318ac36b7d9d18acafd13e5e4ae4cf396e4345 [file] [log] [blame]
 /* * ==================================================== * 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. * ==================================================== */ /* Modifications for 128-bit long double are Copyright (C) 2001 Stephen L. Moshier and are incorporated herein by permission of the author. The author reserves the right to distribute this material elsewhere under different copying permissions. These modifications are distributed here under the following terms: This library is free software; you can redistribute it and/or modify it under the terms of the GNU Lesser General Public License as published by the Free Software Foundation; either version 2.1 of the License, or (at your option) any later version. This library 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 Lesser General Public License for more details. You should have received a copy of the GNU Lesser General Public License along with this library; if not, see . */ /* * __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 "quadmath-imp.h" static const __float128 invsqrtpi = 5.6418958354775628694807945156077258584405E-1Q, two = 2, one = 1, zero = 0; __float128 jnq (int n, __float128 x) { uint32_t se; int32_t i, ix, sgn; __float128 a, b, temp, di, ret; __float128 z, w; ieee854_float128 u; /* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x) * Thus, J(-n,x) = J(n,-x) */ u.value = x; se = u.words32.w0; ix = se & 0x7fffffff; /* if J(n,NaN) is NaN */ if (ix >= 0x7fff0000) { if ((u.words32.w0 & 0xffff) | u.words32.w1 | u.words32.w2 | u.words32.w3) return x + x; } if (n < 0) { n = -n; x = -x; se ^= 0x80000000; } if (n == 0) return (j0q (x)); if (n == 1) return (j1q (x)); sgn = (n & 1) & (se >> 31); /* even n -- 0, odd n -- sign(x) */ x = fabsq (x); { SET_RESTORE_ROUNDF128 (FE_TONEAREST); if (x == 0 || ix >= 0x7fff0000) /* if x is 0 or inf */ return sgn == 1 ? -zero : zero; else if ((__float128) n <= x) { /* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */ if (ix >= 0x412D0000) { /* x > 2**302 */ /* ??? Could use an expansion for large x here. */ /* (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 */ __float128 s; __float128 c; sincosq (x, &s, &c); switch (n & 3) { case 0: temp = c + s; break; case 1: temp = -c + s; break; case 2: temp = -c - s; break; case 3: temp = c - s; break; } b = invsqrtpi * temp / sqrtq (x); } else { a = j0q (x); b = j1q (x); for (i = 1; i < n; i++) { temp = b; b = b * ((__float128) (i + i) / x) - a; /* avoid underflow */ a = temp; } } } else { if (ix < 0x3fc60000) { /* x < 2**-57 */ /* x is tiny, return the first Taylor expansion of J(n,x) * J(n,x) = 1/n!*(x/2)^n - ... */ if (n >= 400) /* underflow, result < 10^-4952 */ b = zero; else { temp = x * 0.5; b = temp; for (a = one, i = 2; i <= n; i++) { a *= (__float128) 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 */ __float128 t, v; __float128 q0, q1, h, tmp; int32_t k, m; w = (n + n) / (__float128) x; h = 2 / (__float128) x; q0 = w; z = w + h; q1 = w * z - 1; k = 1; while (q1 < 1.0e17Q) { 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 * logq (fabsq (v * tmp)); if (tmp < 1.1356523406294143949491931077970765006170e+04Q) { for (i = n - 1, di = (__float128) (i + i); i > 0; i--) { temp = b; b *= di; b = b / x - a; a = temp; di -= two; } } else { for (i = n - 1, di = (__float128) (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 > 1e100Q) { a /= b; t /= b; b = one; } } } /* j0() and j1() suffer enormous loss of precision at and * near zero; however, we know that their zero points never * coincide, so just choose the one further away from zero. */ z = j0q (x); w = j1q (x); if (fabsq (z) >= fabsq (w)) b = (t * z / b); else b = (t * w / a); } } if (sgn == 1) ret = -b; else ret = b; } if (ret == 0) { ret = copysignq (FLT128_MIN, ret) * FLT128_MIN; errno = ERANGE; } else math_check_force_underflow (ret); return ret; } __float128 ynq (int n, __float128 x) { uint32_t se; int32_t i, ix; int32_t sign; __float128 a, b, temp, ret; ieee854_float128 u; u.value = x; se = u.words32.w0; ix = se & 0x7fffffff; /* if Y(n,NaN) is NaN */ if (ix >= 0x7fff0000) { if ((u.words32.w0 & 0xffff) | u.words32.w1 | u.words32.w2 | u.words32.w3) return x + x; } if (x <= 0) { if (x == 0) return ((n < 0 && (n & 1) != 0) ? 1 : -1) / 0.0Q; if (se & 0x80000000) return zero / (zero * x); } sign = 1; if (n < 0) { n = -n; sign = 1 - ((n & 1) << 1); } if (n == 0) return (y0q (x)); { SET_RESTORE_ROUNDF128 (FE_TONEAREST); if (n == 1) { ret = sign * y1q (x); goto out; } if (ix >= 0x7fff0000) return zero; if (ix >= 0x412D0000) { /* x > 2**302 */ /* ??? See comment above on the possible futility of this. */ /* (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 */ __float128 s; __float128 c; sincosq (x, &s, &c); switch (n & 3) { case 0: temp = s - c; break; case 1: temp = -s - c; break; case 2: temp = -s + c; break; case 3: temp = s + c; break; } b = invsqrtpi * temp / sqrtq (x); } else { a = y0q (x); b = y1q (x); /* quit if b is -inf */ u.value = b; se = u.words32.w0 & 0xffff0000; for (i = 1; i < n && se != 0xffff0000; i++) { temp = b; b = ((__float128) (i + i) / x) * b - a; u.value = b; se = u.words32.w0 & 0xffff0000; a = temp; } } /* If B is +-Inf, set up errno accordingly. */ if (! finiteq (b)) errno = ERANGE; if (sign > 0) ret = b; else ret = -b; } out: if (isinfq (ret)) ret = copysignq (FLT128_MAX, ret) * FLT128_MAX; return ret; }