/* @(#)e_exp.c 1.6 04/04/22 */ /* * ==================================================== * Copyright (C) 2004 by Sun Microsystems, Inc. All rights reserved. * * Permission to use, copy, modify, and distribute this * software is freely granted, provided that this notice * is preserved. * ==================================================== */ /* __ieee754_exp(x) * Returns the exponential of x. * * Method * 1. Argument reduction: * Reduce x to an r so that |r| <= 0.5*ln2 ~ 0.34658. * Given x, find r and integer k such that * * x = k*ln2 + r, |r| <= 0.5*ln2. * * Here r will be represented as r = hi-lo for better * accuracy. * * 2. Approximation of exp(r) by a special rational function on * the interval [0,0.34658]: * Write * R(r**2) = r*(exp(r)+1)/(exp(r)-1) = 2 + r*r/6 - r**4/360 + ... * We use a special Remes algorithm on [0,0.34658] to generate * a polynomial of degree 5 to approximate R. The maximum error * of this polynomial approximation is bounded by 2**-59. In * other words, * R(z) ~ 2.0 + P1*z + P2*z**2 + P3*z**3 + P4*z**4 + P5*z**5 * (where z=r*r, and the values of P1 to P5 are listed below) * and * | 5 | -59 * | 2.0+P1*z+...+P5*z - R(z) | <= 2 * | | * The computation of exp(r) thus becomes * 2*r * exp(r) = 1 + ------- * R - r * r*R1(r) * = 1 + r + ----------- (for better accuracy) * 2 - R1(r) * where * 2 4 10 * R1(r) = r - (P1*r + P2*r + ... + P5*r ). * * 3. Scale back to obtain exp(x): * From step 1, we have * exp(x) = 2^k * exp(r) * * Special cases: * exp(INF) is INF, exp(NaN) is NaN; * exp(-INF) is 0, and * for finite argument, only exp(0)=1 is exact. * * Accuracy: * according to an error analysis, the error is always less than * 1 ulp (unit in the last place). * * Misc. info. * For IEEE double * if x > 7.09782712893383973096e+02 then exp(x) overflow * if x < -7.45133219101941108420e+02 then exp(x) underflow * * 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. */ #include "fdlibm.h" #ifdef __STDC__ static const double #else static double #endif one = 1.0, halF[2] = {0.5,-0.5,}, huge = 1.0e+300, twom1000= 9.33263618503218878990e-302, /* 2**-1000=0x01700000,0*/ o_threshold= 7.09782712893383973096e+02, /* 0x40862E42, 0xFEFA39EF */ u_threshold= -7.45133219101941108420e+02, /* 0xc0874910, 0xD52D3051 */ ln2HI[2] ={ 6.93147180369123816490e-01, /* 0x3fe62e42, 0xfee00000 */ -6.93147180369123816490e-01,},/* 0xbfe62e42, 0xfee00000 */ ln2LO[2] ={ 1.90821492927058770002e-10, /* 0x3dea39ef, 0x35793c76 */ -1.90821492927058770002e-10,},/* 0xbdea39ef, 0x35793c76 */ invln2 = 1.44269504088896338700e+00, /* 0x3ff71547, 0x652b82fe */ P1 = 1.66666666666666019037e-01, /* 0x3FC55555, 0x5555553E */ P2 = -2.77777777770155933842e-03, /* 0xBF66C16C, 0x16BEBD93 */ P3 = 6.61375632143793436117e-05, /* 0x3F11566A, 0xAF25DE2C */ P4 = -1.65339022054652515390e-06, /* 0xBEBBBD41, 0xC5D26BF1 */ P5 = 4.13813679705723846039e-08; /* 0x3E663769, 0x72BEA4D0 */ #ifdef __STDC__ double __ieee754_exp(double x) /* default IEEE double exp */ #else double __ieee754_exp(x) /* default IEEE double exp */ double x; #endif { double y,hi,lo,c,t; int k,xsb; unsigned hx; hx = __HI(x); /* high word of x */ xsb = (hx>>31)&1; /* sign bit of x */ hx &= 0x7fffffff; /* high word of |x| */ /* filter out non-finite argument */ if(hx >= 0x40862E42) { /* if |x|>=709.78... */ if(hx>=0x7ff00000) { if(((hx&0xfffff)|__LO(x))!=0) return x+x; /* NaN */ else return (xsb==0)? x:0.0; /* exp(+-inf)={inf,0} */ } if(x > o_threshold) return huge*huge; /* overflow */ if(x < u_threshold) return twom1000*twom1000; /* underflow */ } /* argument reduction */ if(hx > 0x3fd62e42) { /* if |x| > 0.5 ln2 */ if(hx < 0x3FF0A2B2) { /* and |x| < 1.5 ln2 */ hi = x-ln2HI[xsb]; lo=ln2LO[xsb]; k = 1-xsb-xsb; } else { k = (int)(invln2*x+halF[xsb]); t = k; hi = x - t*ln2HI[0]; /* t*ln2HI is exact here */ lo = t*ln2LO[0]; } x = hi - lo; } else if(hx < 0x3e300000) { /* when |x|<2**-28 */ if(huge+x>one) return one+x;/* trigger inexact */ } else k = 0; /* x is now in primary range */ t = x*x; c = x - t*(P1+t*(P2+t*(P3+t*(P4+t*P5)))); if(k==0) return one-((x*c)/(c-2.0)-x); else y = one-((lo-(x*c)/(2.0-c))-hi); if(k >= -1021) { __HI(y) += (k<<20); /* add k to y's exponent */ return y; } else { __HI(y) += ((k+1000)<<20);/* add k to y's exponent */ return y*twom1000; } }