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https://git.planet-casio.com/Lephenixnoir/OpenLibm.git
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138 lines
3.5 KiB
C
138 lines
3.5 KiB
C
/* $OpenBSD: s_expm1l.c,v 1.2 2011/07/20 21:02:51 martynas Exp $ */
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/*
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* Copyright (c) 2008 Stephen L. Moshier <steve@moshier.net>
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*
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* Permission to use, copy, modify, and distribute this software for any
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* purpose with or without fee is hereby granted, provided that the above
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* copyright notice and this permission notice appear in all copies.
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*
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* THE SOFTWARE IS PROVIDED "AS IS" AND THE AUTHOR DISCLAIMS ALL WARRANTIES
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* WITH REGARD TO THIS SOFTWARE INCLUDING ALL IMPLIED WARRANTIES OF
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* MERCHANTABILITY AND FITNESS. IN NO EVENT SHALL THE AUTHOR BE LIABLE FOR
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* ANY SPECIAL, DIRECT, INDIRECT, OR CONSEQUENTIAL DAMAGES OR ANY DAMAGES
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* WHATSOEVER RESULTING FROM LOSS OF USE, DATA OR PROFITS, WHETHER IN AN
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* ACTION OF CONTRACT, NEGLIGENCE OR OTHER TORTIOUS ACTION, ARISING OUT OF
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* OR IN CONNECTION WITH THE USE OR PERFORMANCE OF THIS SOFTWARE.
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*/
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/* expm1l.c
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*
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* Exponential function, minus 1
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* Long double precision
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*
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*
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* SYNOPSIS:
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*
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* long double x, y, expm1l();
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*
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* y = expm1l( x );
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*
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*
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*
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* DESCRIPTION:
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*
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* Returns e (2.71828...) raised to the x power, minus 1.
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*
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* Range reduction is accomplished by separating the argument
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* into an integer k and fraction f such that
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*
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* x k f
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* e = 2 e.
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*
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* An expansion x + .5 x^2 + x^3 R(x) approximates exp(f) - 1
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* in the basic range [-0.5 ln 2, 0.5 ln 2].
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*
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*
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* ACCURACY:
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*
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* Relative error:
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* arithmetic domain # trials peak rms
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* IEEE -45,+MAXLOG 200,000 1.2e-19 2.5e-20
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*
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* ERROR MESSAGES:
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*
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* message condition value returned
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* expm1l overflow x > MAXLOG MAXNUM
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*
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*/
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#include <openlibm.h>
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static const long double MAXLOGL = 1.1356523406294143949492E4L;
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/* exp(x) - 1 = x + 0.5 x^2 + x^3 P(x)/Q(x)
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-.5 ln 2 < x < .5 ln 2
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Theoretical peak relative error = 3.4e-22 */
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static const long double
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P0 = -1.586135578666346600772998894928250240826E4L,
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P1 = 2.642771505685952966904660652518429479531E3L,
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P2 = -3.423199068835684263987132888286791620673E2L,
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P3 = 1.800826371455042224581246202420972737840E1L,
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P4 = -5.238523121205561042771939008061958820811E-1L,
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Q0 = -9.516813471998079611319047060563358064497E4L,
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Q1 = 3.964866271411091674556850458227710004570E4L,
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Q2 = -7.207678383830091850230366618190187434796E3L,
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Q3 = 7.206038318724600171970199625081491823079E2L,
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Q4 = -4.002027679107076077238836622982900945173E1L,
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/* Q5 = 1.000000000000000000000000000000000000000E0 */
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/* C1 + C2 = ln 2 */
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C1 = 6.93145751953125E-1L,
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C2 = 1.428606820309417232121458176568075500134E-6L,
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/* ln 2^-65 */
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minarg = -4.5054566736396445112120088E1L;
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static const long double huge = 0x1p10000L;
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long double
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expm1l(long double x)
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{
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long double px, qx, xx;
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int k;
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/* Overflow. */
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if (x > MAXLOGL)
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return (huge*huge); /* overflow */
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if (x == 0.0)
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return x;
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/* Minimum value. */
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if (x < minarg)
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return -1.0L;
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xx = C1 + C2;
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/* Express x = ln 2 (k + remainder), remainder not exceeding 1/2. */
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px = floorl (0.5 + x / xx);
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k = px;
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/* remainder times ln 2 */
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x -= px * C1;
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x -= px * C2;
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/* Approximate exp(remainder ln 2). */
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px = (((( P4 * x
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+ P3) * x
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+ P2) * x
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+ P1) * x
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+ P0) * x;
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qx = (((( x
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+ Q4) * x
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+ Q3) * x
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+ Q2) * x
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+ Q1) * x
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+ Q0;
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xx = x * x;
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qx = x + (0.5 * xx + xx * px / qx);
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/* exp(x) = exp(k ln 2) exp(remainder ln 2) = 2^k exp(remainder ln 2).
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We have qx = exp(remainder ln 2) - 1, so
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exp(x) - 1 = 2^k (qx + 1) - 1 = 2^k qx + 2^k - 1. */
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px = ldexpl(1.0L, k);
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x = px * qx + (px - 1.0);
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return x;
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}
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