/* * Copyright (c) 1998, 2003, Oracle and/or its affiliates. All rights reserved. * DO NOT ALTER OR REMOVE COPYRIGHT NOTICES OR THIS FILE HEADER. * * This code is free software; you can redistribute it and/or modify it * under the terms of the GNU General Public License version 2 only, as * published by the Free Software Foundation. Oracle designates this * particular file as subject to the "Classpath" exception as provided * by Oracle in the LICENSE file that accompanied this code. * * This code 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 General Public License * version 2 for more details (a copy is included in the LICENSE file that * accompanied this code). * * You should have received a copy of the GNU General Public License version * 2 along with this work; if not, write to the Free Software Foundation, * Inc., 51 Franklin St, Fifth Floor, Boston, MA 02110-1301 USA. * * Please contact Oracle, 500 Oracle Parkway, Redwood Shores, CA 94065 USA * or visit www.oracle.com if you need additional information or have any * questions.
*/
/* double log1p(double x) * * Method : * 1. Argument Reduction: find k and f such that * 1+x = 2^k * (1+f), * where sqrt(2)/2 < 1+f < sqrt(2) . * * Note. If k=0, then f=x is exact. However, if k!=0, then f * may not be representable exactly. In that case, a correction * term is need. Let u=1+x rounded. Let c = (1+x)-u, then * log(1+x) - log(u) ~ c/u. Thus, we proceed to compute log(u), * and add back the correction term c/u. * (Note: when x > 2**53, one can simply return log(x)) * * 2. Approximation of log1p(f). * Let s = f/(2+f) ; based on log(1+f) = log(1+s) - log(1-s) * = 2s + 2/3 s**3 + 2/5 s**5 + ....., * = 2s + s*R * We use a special Reme algorithm on [0,0.1716] to generate * a polynomial of degree 14 to approximate R The maximum error * of this polynomial approximation is bounded by 2**-58.45. In * other words, * 2 4 6 8 10 12 14 * R(z) ~ Lp1*s +Lp2*s +Lp3*s +Lp4*s +Lp5*s +Lp6*s +Lp7*s * (the values of Lp1 to Lp7 are listed in the program) * and * | 2 14 | -58.45 * | Lp1*s +...+Lp7*s - R(z) | <= 2 * | | * Note that 2s = f - s*f = f - hfsq + s*hfsq, where hfsq = f*f/2. * In order to guarantee error in log below 1ulp, we compute log * by * log1p(f) = f - (hfsq - s*(hfsq+R)). * * 3. Finally, log1p(x) = k*ln2 + log1p(f). * = k*ln2_hi+(f-(hfsq-(s*(hfsq+R)+k*ln2_lo))) * Here ln2 is split into two floating point number: * ln2_hi + ln2_lo, * where n*ln2_hi is always exact for |n| < 2000. * * Special cases: * log1p(x) is NaN with signal if x < -1 (including -INF) ; * log1p(+INF) is +INF; log1p(-1) is -INF with signal; * log1p(NaN) is that NaN with no signal. * * Accuracy: * according to an error analysis, the error is always less than * 1 ulp (unit in the last place). * * Constants: * The hexadecimal values are the intended ones for the following * constants. The decimal values may be used, provided that the * compiler will convert from decimal to binary accurately enough * to produce the hexadecimal values shown. * * Note: Assuming log() return accurate answer, the following * algorithm can be used to compute log1p(x) to within a few ULP: * * u = 1+x; * if(u==1.0) return x ; else * return log(u)*(x/(u-1.0)); * * See HP-15C Advanced Functions Handbook, p.193.
*/
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