| |
| /* |
| * 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. |
| */ |
| |
| #include "fdlibm.h" |
| |
| #ifdef __STDC__ |
| static const double |
| #else |
| static double |
| #endif |
| ln2_hi = 6.93147180369123816490e-01, /* 3fe62e42 fee00000 */ |
| ln2_lo = 1.90821492927058770002e-10, /* 3dea39ef 35793c76 */ |
| two54 = 1.80143985094819840000e+16, /* 43500000 00000000 */ |
| Lp1 = 6.666666666666735130e-01, /* 3FE55555 55555593 */ |
| Lp2 = 3.999999999940941908e-01, /* 3FD99999 9997FA04 */ |
| Lp3 = 2.857142874366239149e-01, /* 3FD24924 94229359 */ |
| Lp4 = 2.222219843214978396e-01, /* 3FCC71C5 1D8E78AF */ |
| Lp5 = 1.818357216161805012e-01, /* 3FC74664 96CB03DE */ |
| Lp6 = 1.531383769920937332e-01, /* 3FC39A09 D078C69F */ |
| Lp7 = 1.479819860511658591e-01; /* 3FC2F112 DF3E5244 */ |
| |
| static double zero = 0.0; |
| |
| #ifdef __STDC__ |
| double log1p(double x) |
| #else |
| double log1p(x) |
| double x; |
| #endif |
| { |
| double hfsq,f=0,c=0,s,z,R,u; |
| int k,hx,hu=0,ax; |
| |
| hx = __HI(x); /* high word of x */ |
| ax = hx&0x7fffffff; |
| |
| k = 1; |
| if (hx < 0x3FDA827A) { /* x < 0.41422 */ |
| if(ax>=0x3ff00000) { /* x <= -1.0 */ |
| /* |
| * Added redundant test against hx to work around VC++ |
| * code generation problem. |
| */ |
| if(x==-1.0 && (hx==0xbff00000)) /* log1p(-1)=-inf */ |
| return -two54/zero; |
| else |
| return (x-x)/(x-x); /* log1p(x<-1)=NaN */ |
| } |
| if(ax<0x3e200000) { /* |x| < 2**-29 */ |
| if(two54+x>zero /* raise inexact */ |
| &&ax<0x3c900000) /* |x| < 2**-54 */ |
| return x; |
| else |
| return x - x*x*0.5; |
| } |
| if(hx>0||hx<=((int)0xbfd2bec3)) { |
| k=0;f=x;hu=1;} /* -0.2929<x<0.41422 */ |
| } |
| if (hx >= 0x7ff00000) return x+x; |
| if(k!=0) { |
| if(hx<0x43400000) { |
| u = 1.0+x; |
| hu = __HI(u); /* high word of u */ |
| k = (hu>>20)-1023; |
| c = (k>0)? 1.0-(u-x):x-(u-1.0);/* correction term */ |
| c /= u; |
| } else { |
| u = x; |
| hu = __HI(u); /* high word of u */ |
| k = (hu>>20)-1023; |
| c = 0; |
| } |
| hu &= 0x000fffff; |
| if(hu<0x6a09e) { |
| __HI(u) = hu|0x3ff00000; /* normalize u */ |
| } else { |
| k += 1; |
| __HI(u) = hu|0x3fe00000; /* normalize u/2 */ |
| hu = (0x00100000-hu)>>2; |
| } |
| f = u-1.0; |
| } |
| hfsq=0.5*f*f; |
| if(hu==0) { /* |f| < 2**-20 */ |
| if(f==zero) { if(k==0) return zero; |
| else {c += k*ln2_lo; return k*ln2_hi+c;}} |
| R = hfsq*(1.0-0.66666666666666666*f); |
| if(k==0) return f-R; else |
| return k*ln2_hi-((R-(k*ln2_lo+c))-f); |
| } |
| s = f/(2.0+f); |
| z = s*s; |
| R = z*(Lp1+z*(Lp2+z*(Lp3+z*(Lp4+z*(Lp5+z*(Lp6+z*Lp7)))))); |
| if(k==0) return f-(hfsq-s*(hfsq+R)); else |
| return k*ln2_hi-((hfsq-(s*(hfsq+R)+(k*ln2_lo+c)))-f); |
| } |