/* * Copyright (c) 1985 Regents of the University of California. * * Use and reproduction of this software are granted in accordance with * the terms and conditions specified in the Berkeley Software License * Agreement (in particular, this entails acknowledgement of the programs' * source, and inclusion of this notice) with the additional understanding * that all recipients should regard themselves as participants in an * ongoing research project and hence should feel obligated to report * their experiences (good or bad) with these elementary function codes, * using "sendbug 4bsd-bugs@BERKELEY", to the authors. */ #ifndef lint static char sccsid[] = "@(#)log1p.c 1.3 (Berkeley) 8/21/85"; #endif not lint /* LOG1P(x) * RETURN THE LOGARITHM OF 1+x * DOUBLE PRECISION (VAX D FORMAT 56 bits, IEEE DOUBLE 53 BITS) * CODED IN C BY K.C. NG, 1/19/85; * REVISED BY K.C. NG on 2/6/85, 3/7/85, 3/24/85, 4/16/85. * * Required system supported functions: * scalb(x,n) * copysign(x,y) * logb(x) * finite(x) * * Required kernel function: * log__L(z) * * Method : * 1. Argument Reduction: find k and f such that * 1+x = 2^k * (1+f), * where sqrt(2)/2 < 1+f < sqrt(2) . * * 2. 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 + ....., * log(1+f) is computed by * * log(1+f) = 2s + s*log__L(s*s) * where * log__L(z) = z*(L1 + z*(L2 + z*(... (L6 + z*L7)...))) * * See log__L() for the values of the coefficients. * * 3. Finally, log(1+x) = k*ln2 + log(1+f). * * Remarks 1. In step 3 n*ln2 will be stored in two floating point numbers * n*ln2hi + n*ln2lo, where ln2hi is chosen such that the last * 20 bits (for VAX D format), or the last 21 bits ( for IEEE * double) is 0. This ensures n*ln2hi is exactly representable. * 2. In step 1, f may not be representable. A correction term c * for f is computed. It follows that the correction term for * f - t (the leading term of log(1+f) in step 2) is c-c*x. We * add this correction term to n*ln2lo to attenuate the error. * * * Special cases: * log1p(x) is NaN with signal if x < -1; log1p(NaN) is NaN with no signal; * log1p(INF) is +INF; log1p(-1) is -INF with signal; * only log1p(0)=0 is exact for finite argument. * * Accuracy: * log1p(x) returns the exact log(1+x) nearly rounded. In a test run * with 1,536,000 random arguments on a VAX, the maximum observed * error was .846 ulps (units 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. */ #ifdef VAX /* VAX D format */ #include /* double static */ /* ln2hi = 6.9314718055829871446E-1 , Hex 2^ 0 * .B17217F7D00000 */ /* ln2lo = 1.6465949582897081279E-12 , Hex 2^-39 * .E7BCD5E4F1D9CC */ /* sqrt2 = 1.4142135623730950622E0 ; Hex 2^ 1 * .B504F333F9DE65 */ static long ln2hix[] = { 0x72174031, 0x0000f7d0}; static long ln2lox[] = { 0xbcd52ce7, 0xd9cce4f1}; static long sqrt2x[] = { 0x04f340b5, 0xde6533f9}; #define ln2hi (*(double*)ln2hix) #define ln2lo (*(double*)ln2lox) #define sqrt2 (*(double*)sqrt2x) #else /* IEEE double */ double static ln2hi = 6.9314718036912381649E-1 , /*Hex 2^ -1 * 1.62E42FEE00000 */ ln2lo = 1.9082149292705877000E-10 , /*Hex 2^-33 * 1.A39EF35793C76 */ sqrt2 = 1.4142135623730951455E0 ; /*Hex 2^ 0 * 1.6A09E667F3BCD */ #endif double log1p(x) double x; { static double zero=0.0, negone= -1.0, one=1.0, half=1.0/2.0, small=1.0E-20; /* 1+small == 1 */ double logb(),copysign(),scalb(),log__L(),z,s,t,c; int k,finite(); #ifndef VAX if(x!=x) return(x); /* x is NaN */ #endif if(finite(x)) { if( x > negone ) { /* argument reduction */ if(copysign(x,one)= sqrt2 ) { k += 1 ; z *= half; t *= half; } t += negone; x = z + t; c = (t-x)+z ; /* correction term for x */ /* compute log(1+x) */ s = x/(2+x); t = x*x*half; c += (k*ln2lo-c*x); z = c+s*(t+log__L(s*s)); x += (z - t) ; return(k*ln2hi+x); } /* end of if (x > negone) */ else { #ifdef VAX extern double infnan(); if ( x == negone ) return (infnan(-ERANGE)); /* -INF */ else return (infnan(EDOM)); /* NaN */ #else /* IEEE double */ /* x = -1, return -INF with signal */ if ( x == negone ) return( negone/zero ); /* negative argument for log, return NaN with signal */ else return ( zero / zero ); #endif } } /* end of if (finite(x)) */ /* log(-INF) is NaN */ else if(x<0) return(zero/zero); /* log(+INF) is INF */ else return(x); }