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/* $OpenBSD: s_expm1l.c,v 1.3 2016/09/12 19:47:03 guenther Exp $ */
/*
* Copyright (c) 2008 Stephen L. Moshier <steve@moshier.net>
*
* Permission to use, copy, modify, and distribute this software for any
* purpose with or without fee is hereby granted, provided that the above
* copyright notice and this permission notice appear in all copies.
*
* THE SOFTWARE IS PROVIDED "AS IS" AND THE AUTHOR DISCLAIMS ALL WARRANTIES
* WITH REGARD TO THIS SOFTWARE INCLUDING ALL IMPLIED WARRANTIES OF
* MERCHANTABILITY AND FITNESS. IN NO EVENT SHALL THE AUTHOR BE LIABLE FOR
* ANY SPECIAL, DIRECT, INDIRECT, OR CONSEQUENTIAL DAMAGES OR ANY DAMAGES
* WHATSOEVER RESULTING FROM LOSS OF USE, DATA OR PROFITS, WHETHER IN AN
* ACTION OF CONTRACT, NEGLIGENCE OR OTHER TORTIOUS ACTION, ARISING OUT OF
* OR IN CONNECTION WITH THE USE OR PERFORMANCE OF THIS SOFTWARE.
*/
/* expm1l.c
*
* Exponential function, minus 1
* Long double precision
*
*
* SYNOPSIS:
*
* long double x, y, expm1l();
*
* y = expm1l( x );
*
*
*
* DESCRIPTION:
*
* Returns e (2.71828...) raised to the x power, minus 1.
*
* 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 -45,+MAXLOG 200,000 1.2e-19 2.5e-20
*
* ERROR MESSAGES:
*
* message condition value returned
* expm1l overflow x > MAXLOG MAXNUM
*
*/
#include <math.h>
static const long double MAXLOGL = 1.1356523406294143949492E4L;
/* 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 = 3.4e-22 */
static const long double
P0 = -1.586135578666346600772998894928250240826E4L,
P1 = 2.642771505685952966904660652518429479531E3L,
P2 = -3.423199068835684263987132888286791620673E2L,
P3 = 1.800826371455042224581246202420972737840E1L,
P4 = -5.238523121205561042771939008061958820811E-1L,
Q0 = -9.516813471998079611319047060563358064497E4L,
Q1 = 3.964866271411091674556850458227710004570E4L,
Q2 = -7.207678383830091850230366618190187434796E3L,
Q3 = 7.206038318724600171970199625081491823079E2L,
Q4 = -4.002027679107076077238836622982900945173E1L,
/* Q5 = 1.000000000000000000000000000000000000000E0 */
/* C1 + C2 = ln 2 */
C1 = 6.93145751953125E-1L,
C2 = 1.428606820309417232121458176568075500134E-6L,
/* ln 2^-65 */
minarg = -4.5054566736396445112120088E1L;
static const long double huge = 0x1p10000L;
long double
expm1l(long double x)
{
long double px, qx, xx;
int k;
/* Overflow. */
if (x > MAXLOGL)
return (huge*huge); /* overflow */
if (x == 0.0)
return x;
/* Minimum value. */
if (x < minarg)
return -1.0L;
xx = C1 + C2;
/* Express x = ln 2 (k + remainder), remainder not exceeding 1/2. */
px = floorl (0.5 + x / xx);
k = px;
/* remainder times ln 2 */
x -= px * C1;
x -= px * C2;
/* Approximate exp(remainder ln 2). */
px = (((( P4 * x
+ P3) * x
+ P2) * x
+ P1) * x
+ P0) * x;
qx = (((( 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;
}
DEF_STD(expm1l);
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