diff options
Diffstat (limited to 'sysdeps/ieee754')
| -rw-r--r-- | sysdeps/ieee754/ldbl-96/e_jnl.c | 368 |
1 files changed, 368 insertions, 0 deletions
diff --git a/sysdeps/ieee754/ldbl-96/e_jnl.c b/sysdeps/ieee754/ldbl-96/e_jnl.c new file mode 100644 index 0000000000..6b69a4588f --- /dev/null +++ b/sysdeps/ieee754/ldbl-96/e_jnl.c @@ -0,0 +1,368 @@ +/* + * ==================================================== + * 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 long double contributed by + Stephen L. Moshier <moshier@na-net.ornl.gov> */ + +/* + * __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 long double +#else +static long double +#endif + invsqrtpi = 5.64189583547756286948079e-1L, two = 2.0e0L, one = 1.0e0L; + +#ifdef __STDC__ +static const long double zero = 0.0L; +#else +static long double zero = 0.0L; +#endif + +#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, i0, i1; + int32_t i, ix, sgn; + long double a, b, temp, di; + long 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) + */ + + GET_LDOUBLE_WORDS (se, i0, i1, x); + ix = se & 0x7fff; + + /* if J(n,NaN) is NaN */ + if ((ix == 0x7fff) && ((i0 & 0x7fffffff) != 0)) + return x + x; + if (n < 0) + { + n = -n; + x = -x; + se ^= 0x8000; + } + if (n == 0) + return (__ieee754_j0l (x)); + if (n == 1) + return (__ieee754_j1l (x)); + sgn = (n & 1) & (se >> 15); /* even n -- 0, odd n -- sign(x) */ + x = fabsl (x); + if ((ix | i0 | i1) == 0 || ix >= 0x7fff) /* 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 >= 0x412D) + { /* x > 2**302 */ + + /* ??? This might be a futile gesture. + If x exceeds X_TLOSS anyway, the wrapper function + will set the result to zero. */ + + /* (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 < 0x3fde) + { /* x < 2**-33 */ + /* 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 < 1.0e11L) + { + 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, i0, i1; + int32_t i, ix; + int32_t sign; + long double a, b, temp; + + + GET_LDOUBLE_WORDS (se, i0, i1, x); + ix = se & 0x7fff; + /* if Y(n,NaN) is NaN */ + if ((ix == 0x7fff) && ((i0 & 0x7fffffff) != 0)) + return x + x; + if ((ix | i0 | i1) == 0) + return -one / zero; + if (se & 0x8000) + 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 == 0x7fff) + return zero; + if (ix >= 0x412D) + { /* 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 */ + GET_LDOUBLE_WORDS (se, i0, i1, b); + for (i = 1; i < n && se != 0xffff; i++) + { + temp = b; + b = ((long double) (i + i) / x) * b - a; + GET_LDOUBLE_WORDS (se, i0, i1, b); + a = temp; + } + } + if (sign > 0) + return b; + else + return -b; +} |