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38ae768d88
After the last addition to the math test suite PPC routines haven't been adjusted so far.
165 lines
4.5 KiB
C
165 lines
4.5 KiB
C
/* expm1l.c
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*
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* Exponential function, minus 1
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* 128-bit long double precision
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*
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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 one.
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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 -79,+MAXLOG 100,000 1.7e-34 4.5e-35
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*
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*/
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/* Copyright 2001 by Stephen L. Moshier
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This library is free software; you can redistribute it and/or
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modify it under the terms of the GNU Lesser General Public
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License as published by the Free Software Foundation; either
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version 2.1 of the License, or (at your option) any later version.
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This library is distributed in the hope that it will be useful,
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but WITHOUT ANY WARRANTY; without even the implied warranty of
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MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
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Lesser General Public License for more details.
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You should have received a copy of the GNU Lesser General Public
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License along with this library; if not, write to the Free Software
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Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA */
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#include <errno.h>
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#include "math.h"
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#include "math_private.h"
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#include <math_ldbl_opt.h>
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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 = 8.1e-36 */
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static const long double
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P0 = 2.943520915569954073888921213330863757240E8L,
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P1 = -5.722847283900608941516165725053359168840E7L,
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P2 = 8.944630806357575461578107295909719817253E6L,
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P3 = -7.212432713558031519943281748462837065308E5L,
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P4 = 4.578962475841642634225390068461943438441E4L,
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P5 = -1.716772506388927649032068540558788106762E3L,
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P6 = 4.401308817383362136048032038528753151144E1L,
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P7 = -4.888737542888633647784737721812546636240E-1L,
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Q0 = 1.766112549341972444333352727998584753865E9L,
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Q1 = -7.848989743695296475743081255027098295771E8L,
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Q2 = 1.615869009634292424463780387327037251069E8L,
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Q3 = -2.019684072836541751428967854947019415698E7L,
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Q4 = 1.682912729190313538934190635536631941751E6L,
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Q5 = -9.615511549171441430850103489315371768998E4L,
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Q6 = 3.697714952261803935521187272204485251835E3L,
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Q7 = -8.802340681794263968892934703309274564037E1L,
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/* Q8 = 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 (2^16384 * (1 - 2^-113)) */
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maxlog = 1.1356523406294143949491931077970764891253E4L,
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/* ln 2^-114 */
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minarg = -7.9018778583833765273564461846232128760607E1L, big = 2e307L;
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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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int32_t ix, sign;
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ieee854_long_double_shape_type u;
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int k;
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/* Detect infinity and NaN. */
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u.value = x;
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ix = u.parts32.w0;
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sign = ix & 0x80000000;
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ix &= 0x7fffffff;
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if (ix >= 0x7ff00000)
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{
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/* Infinity. */
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if (((ix & 0xfffff) | u.parts32.w1 | (u.parts32.w2&0x7fffffff) | u.parts32.w3) == 0)
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{
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if (sign)
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return -1.0L;
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else
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return x;
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}
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/* NaN. No invalid exception. */
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return x;
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}
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/* expm1(+- 0) = +- 0. */
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if ((ix == 0) && (u.parts32.w1 | (u.parts32.w2&0x7fffffff) | u.parts32.w3) == 0)
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return x;
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/* Overflow. */
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if (x > maxlog)
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{
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__set_errno (ERANGE);
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return (big * big);
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}
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/* Minimum value. */
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if (x < minarg)
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return (4.0/big - 1.0L);
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/* Express x = ln 2 (k + remainder), remainder not exceeding 1/2. */
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xx = C1 + C2; /* ln 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 = (((((((P7 * x
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+ P6) * x
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+ P5) * x + P4) * x + P3) * x + P2) * x + P1) * x + P0) * x;
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qx = (((((((x
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+ Q7) * x
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+ Q6) * x + Q5) * x + Q4) * x + Q3) * x + Q2) * x + Q1) * x + 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
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= 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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libm_hidden_def (__expm1l)
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long_double_symbol (libm, __expm1l, expm1l);
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