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/** AUTHOR* Catherine Loader, catherine@research.bell-labs.com.* October 23, 2000.** Merge in to R:* Copyright (C) 2000-2021 The R Core Team** This program is free software; you can redistribute it and/or modify* it under the terms of the GNU General Public License as published by* the Free Software Foundation; either version 2 of the License, or* (at your option) any later version.** This program 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 for more details.** You should have received a copy of the GNU General Public License* along with this program; if not, a copy is available at* https://www.R-project.org/Licenses/*** DESCRIPTION* Evaluates the "deviance part"* bd0(x,M) := M * D0(x/M) = M*[ x/M * log(x/M) + 1 - (x/M) ] =* = x * log(x/M) + M - x* where M = E[X] = n*p (or = lambda), for x, M > 0** in a manner that should be stable (with small relative error)* for all x and M=np. In particular for x/np close to 1, direct* evaluation fails, and evaluation is based on the Taylor series* of log((1+v)/(1-v)) with v = (x-M)/(x+M) = (x-np)/(x+np).** Martyn Plummer had the nice idea to use log1p() and Martin Maechler* emphasized the extra need to control cancellation.** MP: t := (x-M)/M ( <==> 1+t = x/M ==>** bd0 = M*[ x/M * log(x/M) + 1 - (x/M) ] = M*[ (1+t)*log1p(t) + 1 - (1+t) ]* = M*[ (1+t)*log1p(t) - t ] =: M * p1log1pm(t) =: M * p1l1(t)* MM: The above is very nice, as the "simple" p1l1() function would be useful* to have available in a fast numerical stable way more generally.*/#include "nmath.h"double attribute_hidden bd0(double x, double np){if(!R_FINITE(x) || !R_FINITE(np) || np == 0.0) ML_WARN_return_NAN;if (fabs(x-np) < 0.1*(x+np)) {doublev = (x-np)/(x+np), // might underflow to 0s = (x-np)*v;if(fabs(s) < DBL_MIN) return s;double ej = 2*x*v;v *= v; // "v = v^2"for (int j = 1; j < 1000; j++) { /* Taylor series; 1000: no infinite loopas |v| < .1, v^2000 is "zero" */ej *= v;// = 2 x v^(2j+1)double s_ = s;s += ej/((j<<1)+1);if (s == s_) { /* last term was effectively 0 */#ifdef DEBUG_bd0REprintf("bd0(%g, %g): T.series w/ %d terms -> bd0=%g\n", x, np, j, s);#endifreturn s;}}MATHLIB_WARNING4("bd0(%g, %g): T.series failed to converge in 1000 it.; s=%g, ej/(2j+1)=%g\n",x, np, s, ej/((2*1000)+1));}/* else: | x - np | is not too small */return(x*log(x/np)+np-x);}//- NOTA BENE (R Bugzilla PR#15628) : Morten Welinder proposed ebd0() -- in file ./ebd0.c as more accurate --------