/* * ==================================================== * 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 contributed by Stephen L. Moshier */ /* * __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 "math.h" #include "math_private.h" #ifdef __STDC__ static const long double #else static long double #endif invsqrtpi = 5.6418958354775628694807945156077258584405E-1L, two = 2.0e0L, one = 1.0e0L, zero = 0.0L; #ifdef __STDC__ long double __ieee754_jnl (int n, long double x) #else long double __ieee754_jnl (n, x) int n; long double x; #endif { u_int32_t se; int32_t i, ix, sgn; long double a, b, temp, di; long double z, w; ieee854_long_double_shape_type 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.parts32.w0; ix = se & 0x7fffffff; /* if J(n,NaN) is NaN */ if (ix >= 0x7fff0000) { if ((u.parts32.w0 & 0xffff) | u.parts32.w1 | u.parts32.w2 | u.parts32.w3) return x + x; } if (n < 0) { n = -n; x = -x; se ^= 0x80000000; } if (n == 0) return (__ieee754_j0l (x)); if (n == 1) return (__ieee754_j1l (x)); sgn = (n & 1) & (se >> 31); /* even n -- 0, odd n -- sign(x) */ x = fabsl (x); if (x == 0.0L || ix >= 0x7fff0000) /* if x is 0 or inf */ b = zero; else if ((long double) 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 */ long double s; long double c; __sincosl (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 / __ieee754_sqrtl (x); } else { a = __ieee754_j0l (x); b = __ieee754_j1l (x); for (i = 1; i < n; i++) { temp = b; b = b * ((long double) (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 *= (long 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 */ long double t, v; long double q0, q1, h, tmp; int32_t k, m; w = (n + n) / (long double) x; h = 2.0L / (long double) x; q0 = w; z = w + h; q1 = w * z - 1.0L; k = 1; while (q1 < 1.0e17L) { 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_logl (fabsl (v * tmp)); if (tmp < 1.1356523406294143949491931077970765006170e+04L) { for (i = n - 1, di = (long double) (i + i); i > 0; i--) { temp = b; b *= di; b = b / x - a; a = temp; di -= two; } } else { for (i = n - 1, di = (long 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 > 1e100L) { a /= b; t /= b; b = one; } } } b = (t * __ieee754_j0l (x) / b); } } if (sgn == 1) return -b; else return b; } #ifdef __STDC__ long double __ieee754_ynl (int n, long double x) #else long double __ieee754_ynl (n, x) int n; long double x; #endif { u_int32_t se; int32_t i, ix; int32_t sign; long double a, b, temp; ieee854_long_double_shape_type u; u.value = x; se = u.parts32.w0; ix = se & 0x7fffffff; /* if Y(n,NaN) is NaN */ if (ix >= 0x7fff0000) { if ((u.parts32.w0 & 0xffff) | u.parts32.w1 | u.parts32.w2 | u.parts32.w3) return x + x; } if (x <= 0.0L) { if (x == 0.0L) return -one / zero; if (se & 0x80000000) return zero / zero; } sign = 1; if (n < 0) { n = -n; sign = 1 - ((n & 1) << 1); } if (n == 0) return (__ieee754_y0l (x)); if (n == 1) return (sign * __ieee754_y1l (x)); 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 */ long double s; long double c; __sincosl (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 / __ieee754_sqrtl (x); } else { a = __ieee754_y0l (x); b = __ieee754_y1l (x); /* quit if b is -inf */ u.value = b; se = u.parts32.w0 & 0xffff0000; for (i = 1; i < n && se != 0xffff0000; i++) { temp = b; b = ((long double) (i + i) / x) * b - a; u.value = b; se = u.parts32.w0 & 0xffff0000; a = temp; } } if (sign > 0) return b; else return -b; }