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/* @(#)e_exp.c 5.1 93/09/24 */
/*
* ====================================================
* Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
*
* Developed at SunPro, a Sun Microsystems, Inc. business.
* 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 Reme 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"
#ifndef _DOUBLE_IS_32BITS
#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;
__int32_t k,xsb;
__uint32_t hx;
GET_HIGH_WORD(hx,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) {
__uint32_t lx;
GET_LOW_WORD(lx,x);
if(((hx&0xfffff)|lx)!=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 = 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) {
__uint32_t hy;
GET_HIGH_WORD(hy,y);
SET_HIGH_WORD(y,hy+(k<<20)); /* add k to y's exponent */
return y;
} else {
__uint32_t hy;
GET_HIGH_WORD(hy,y);
SET_HIGH_WORD(y,hy+((k+1000)<<20)); /* add k to y's exponent */
return y*twom1000;
}
}
#endif /* defined(_DOUBLE_IS_32BITS) */
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