glibc/sysdeps/ieee754/ldbl-128/s_expm1l.c
Ulrich Drepper 7f3394bdf3 Fix errno for boundary conditions in 128-bit long double.
Similar to the changes which went already in for the other formats,
follow POSIX rules for errno.
2009-05-29 12:00:22 -07:00

166 lines
4.4 KiB
C

/* expm1l.c
*
* Exponential function, minus 1
* 128-bit long double precision
*
*
*
* SYNOPSIS:
*
* long double x, y, expm1l();
*
* y = expm1l( x );
*
*
*
* DESCRIPTION:
*
* Returns e (2.71828...) raised to the x power, minus one.
*
* Range reduction is accomplished by separating the argument
* into an integer k and fraction f such that
*
* x k f
* e = 2 e.
*
* An expansion x + .5 x^2 + x^3 R(x) approximates exp(f) - 1
* in the basic range [-0.5 ln 2, 0.5 ln 2].
*
*
* ACCURACY:
*
* Relative error:
* arithmetic domain # trials peak rms
* IEEE -79,+MAXLOG 100,000 1.7e-34 4.5e-35
*
*/
/* Copyright 2001 by Stephen L. Moshier
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, write to the Free Software
Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA */
#include <errno.h>
#include "math.h"
#include "math_private.h"
/* exp(x) - 1 = x + 0.5 x^2 + x^3 P(x)/Q(x)
-.5 ln 2 < x < .5 ln 2
Theoretical peak relative error = 8.1e-36 */
static const long double
P0 = 2.943520915569954073888921213330863757240E8L,
P1 = -5.722847283900608941516165725053359168840E7L,
P2 = 8.944630806357575461578107295909719817253E6L,
P3 = -7.212432713558031519943281748462837065308E5L,
P4 = 4.578962475841642634225390068461943438441E4L,
P5 = -1.716772506388927649032068540558788106762E3L,
P6 = 4.401308817383362136048032038528753151144E1L,
P7 = -4.888737542888633647784737721812546636240E-1L,
Q0 = 1.766112549341972444333352727998584753865E9L,
Q1 = -7.848989743695296475743081255027098295771E8L,
Q2 = 1.615869009634292424463780387327037251069E8L,
Q3 = -2.019684072836541751428967854947019415698E7L,
Q4 = 1.682912729190313538934190635536631941751E6L,
Q5 = -9.615511549171441430850103489315371768998E4L,
Q6 = 3.697714952261803935521187272204485251835E3L,
Q7 = -8.802340681794263968892934703309274564037E1L,
/* Q8 = 1.000000000000000000000000000000000000000E0 */
/* C1 + C2 = ln 2 */
C1 = 6.93145751953125E-1L,
C2 = 1.428606820309417232121458176568075500134E-6L,
/* ln (2^16384 * (1 - 2^-113)) */
maxlog = 1.1356523406294143949491931077970764891253E4L,
/* ln 2^-114 */
minarg = -7.9018778583833765273564461846232128760607E1L, big = 2e4932L;
long double
__expm1l (long double x)
{
long double px, qx, xx;
int32_t ix, sign;
ieee854_long_double_shape_type u;
int k;
/* Detect infinity and NaN. */
u.value = x;
ix = u.parts32.w0;
sign = ix & 0x80000000;
ix &= 0x7fffffff;
if (ix >= 0x7fff0000)
{
/* Infinity. */
if (((ix & 0xffff) | u.parts32.w1 | u.parts32.w2 | u.parts32.w3) == 0)
{
if (sign)
return -1.0L;
else
return x;
}
/* NaN. No invalid exception. */
return x;
}
/* expm1(+- 0) = +- 0. */
if ((ix == 0) && (u.parts32.w1 | u.parts32.w2 | u.parts32.w3) == 0)
return x;
/* Overflow. */
if (x > maxlog)
{
__set_errno (ERANGE);
return (big * big);
}
/* Minimum value. */
if (x < minarg)
return (4.0/big - 1.0L);
/* Express x = ln 2 (k + remainder), remainder not exceeding 1/2. */
xx = C1 + C2; /* ln 2. */
px = __floorl (0.5 + x / xx);
k = px;
/* remainder times ln 2 */
x -= px * C1;
x -= px * C2;
/* Approximate exp(remainder ln 2). */
px = (((((((P7 * x
+ P6) * x
+ P5) * x + P4) * x + P3) * x + P2) * x + P1) * x + P0) * x;
qx = (((((((x
+ Q7) * x
+ Q6) * x + Q5) * x + Q4) * x + Q3) * x + Q2) * x + Q1) * x + Q0;
xx = x * x;
qx = x + (0.5 * xx + xx * px / qx);
/* exp(x) = exp(k ln 2) exp(remainder ln 2) = 2^k exp(remainder ln 2).
We have qx = exp(remainder ln 2) - 1, so
exp(x) - 1 = 2^k (qx + 1) - 1
= 2^k qx + 2^k - 1. */
px = __ldexpl (1.0L, k);
x = px * qx + (px - 1.0);
return x;
}
libm_hidden_def (__expm1l)
weak_alias (__expm1l, expm1l)