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jeremybenn |
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/* @(#)e_exp.c 5.1 93/09/24 */
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/*
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* ====================================================
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* Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
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*
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* Developed at SunPro, a Sun Microsystems, Inc. business.
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* Permission to use, copy, modify, and distribute this
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* software is freely granted, provided that this notice
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* is preserved.
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* ====================================================
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*/
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/* __ieee754_exp(x)
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* Returns the exponential of x.
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*
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* Method
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* 1. Argument reduction:
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* Reduce x to an r so that |r| <= 0.5*ln2 ~ 0.34658.
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* Given x, find r and integer k such that
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*
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* x = k*ln2 + r, |r| <= 0.5*ln2.
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*
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* Here r will be represented as r = hi-lo for better
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* accuracy.
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*
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* 2. Approximation of exp(r) by a special rational function on
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* the interval [0,0.34658]:
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* Write
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* R(r**2) = r*(exp(r)+1)/(exp(r)-1) = 2 + r*r/6 - r**4/360 + ...
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* We use a special Reme algorithm on [0,0.34658] to generate
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* a polynomial of degree 5 to approximate R. The maximum error
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* of this polynomial approximation is bounded by 2**-59. In
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* other words,
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* R(z) ~ 2.0 + P1*z + P2*z**2 + P3*z**3 + P4*z**4 + P5*z**5
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* (where z=r*r, and the values of P1 to P5 are listed below)
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* and
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* | 5 | -59
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* | 2.0+P1*z+...+P5*z - R(z) | <= 2
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* | |
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* The computation of exp(r) thus becomes
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* 2*r
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* exp(r) = 1 + -------
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* R - r
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* r*R1(r)
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* = 1 + r + ----------- (for better accuracy)
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* 2 - R1(r)
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* where
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* 2 4 10
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* R1(r) = r - (P1*r + P2*r + ... + P5*r ).
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*
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* 3. Scale back to obtain exp(x):
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* From step 1, we have
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* exp(x) = 2^k * exp(r)
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*
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* Special cases:
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* exp(INF) is INF, exp(NaN) is NaN;
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* exp(-INF) is 0, and
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* for finite argument, only exp(0)=1 is exact.
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*
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* Accuracy:
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* according to an error analysis, the error is always less than
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* 1 ulp (unit in the last place).
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*
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* Misc. info.
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* For IEEE double
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* if x > 7.09782712893383973096e+02 then exp(x) overflow
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* if x < -7.45133219101941108420e+02 then exp(x) underflow
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*
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* Constants:
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* The hexadecimal values are the intended ones for the following
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* constants. The decimal values may be used, provided that the
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* compiler will convert from decimal to binary accurately enough
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* to produce the hexadecimal values shown.
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*/
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#include "fdlibm.h"
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#ifndef _DOUBLE_IS_32BITS
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#ifdef __STDC__
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static const double
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#else
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static double
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#endif
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one = 1.0,
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halF[2] = {0.5,-0.5,},
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huge = 1.0e+300,
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twom1000= 9.33263618503218878990e-302, /* 2**-1000=0x01700000,0*/
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o_threshold= 7.09782712893383973096e+02, /* 0x40862E42, 0xFEFA39EF */
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u_threshold= -7.45133219101941108420e+02, /* 0xc0874910, 0xD52D3051 */
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ln2HI[2] ={ 6.93147180369123816490e-01, /* 0x3fe62e42, 0xfee00000 */
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-6.93147180369123816490e-01,},/* 0xbfe62e42, 0xfee00000 */
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ln2LO[2] ={ 1.90821492927058770002e-10, /* 0x3dea39ef, 0x35793c76 */
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-1.90821492927058770002e-10,},/* 0xbdea39ef, 0x35793c76 */
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invln2 = 1.44269504088896338700e+00, /* 0x3ff71547, 0x652b82fe */
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P1 = 1.66666666666666019037e-01, /* 0x3FC55555, 0x5555553E */
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P2 = -2.77777777770155933842e-03, /* 0xBF66C16C, 0x16BEBD93 */
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P3 = 6.61375632143793436117e-05, /* 0x3F11566A, 0xAF25DE2C */
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P4 = -1.65339022054652515390e-06, /* 0xBEBBBD41, 0xC5D26BF1 */
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P5 = 4.13813679705723846039e-08; /* 0x3E663769, 0x72BEA4D0 */
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#ifdef __STDC__
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double __ieee754_exp(double x) /* default IEEE double exp */
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#else
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double __ieee754_exp(x) /* default IEEE double exp */
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double x;
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#endif
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{
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double y,hi,lo,c,t;
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__int32_t k,xsb;
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__uint32_t hx;
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GET_HIGH_WORD(hx,x);
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xsb = (hx>>31)&1; /* sign bit of x */
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hx &= 0x7fffffff; /* high word of |x| */
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/* filter out non-finite argument */
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if(hx >= 0x40862E42) { /* if |x|>=709.78... */
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if(hx>=0x7ff00000) {
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__uint32_t lx;
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GET_LOW_WORD(lx,x);
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if(((hx&0xfffff)|lx)!=0)
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return x+x; /* NaN */
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else return (xsb==0)? x:0.0; /* exp(+-inf)={inf,0} */
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}
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if(x > o_threshold) return huge*huge; /* overflow */
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if(x < u_threshold) return twom1000*twom1000; /* underflow */
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}
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/* argument reduction */
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if(hx > 0x3fd62e42) { /* if |x| > 0.5 ln2 */
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if(hx < 0x3FF0A2B2) { /* and |x| < 1.5 ln2 */
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hi = x-ln2HI[xsb]; lo=ln2LO[xsb]; k = 1-xsb-xsb;
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} else {
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k = invln2*x+halF[xsb];
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t = k;
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hi = x - t*ln2HI[0]; /* t*ln2HI is exact here */
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lo = t*ln2LO[0];
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}
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x = hi - lo;
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}
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else if(hx < 0x3e300000) { /* when |x|<2**-28 */
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if(huge+x>one) return one+x;/* trigger inexact */
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}
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else k = 0;
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/* x is now in primary range */
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t = x*x;
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c = x - t*(P1+t*(P2+t*(P3+t*(P4+t*P5))));
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if(k==0) return one-((x*c)/(c-2.0)-x);
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else y = one-((lo-(x*c)/(2.0-c))-hi);
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if(k >= -1021) {
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__uint32_t hy;
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GET_HIGH_WORD(hy,y);
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SET_HIGH_WORD(y,hy+(k<<20)); /* add k to y's exponent */
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return y;
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} else {
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__uint32_t hy;
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GET_HIGH_WORD(hy,y);
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SET_HIGH_WORD(y,hy+((k+1000)<<20)); /* add k to y's exponent */
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return y*twom1000;
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}
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}
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#endif /* defined(_DOUBLE_IS_32BITS) */
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