Home | History | Annotate | Download | only in math
      1 // Copyright 2009 The Go Authors. All rights reserved.
      2 // Use of this source code is governed by a BSD-style
      3 // license that can be found in the LICENSE file.
      4 
      5 package math
      6 
      7 // Exp returns e**x, the base-e exponential of x.
      8 //
      9 // Special cases are:
     10 //	Exp(+Inf) = +Inf
     11 //	Exp(NaN) = NaN
     12 // Very large values overflow to 0 or +Inf.
     13 // Very small values underflow to 1.
     14 func Exp(x float64) float64
     15 
     16 // The original C code, the long comment, and the constants
     17 // below are from FreeBSD's /usr/src/lib/msun/src/e_exp.c
     18 // and came with this notice. The go code is a simplified
     19 // version of the original C.
     20 //
     21 // ====================================================
     22 // Copyright (C) 2004 by Sun Microsystems, Inc. All rights reserved.
     23 //
     24 // Permission to use, copy, modify, and distribute this
     25 // software is freely granted, provided that this notice
     26 // is preserved.
     27 // ====================================================
     28 //
     29 //
     30 // exp(x)
     31 // Returns the exponential of x.
     32 //
     33 // Method
     34 //   1. Argument reduction:
     35 //      Reduce x to an r so that |r| <= 0.5*ln2 ~ 0.34658.
     36 //      Given x, find r and integer k such that
     37 //
     38 //               x = k*ln2 + r,  |r| <= 0.5*ln2.
     39 //
     40 //      Here r will be represented as r = hi-lo for better
     41 //      accuracy.
     42 //
     43 //   2. Approximation of exp(r) by a special rational function on
     44 //      the interval [0,0.34658]:
     45 //      Write
     46 //          R(r**2) = r*(exp(r)+1)/(exp(r)-1) = 2 + r*r/6 - r**4/360 + ...
     47 //      We use a special Remez algorithm on [0,0.34658] to generate
     48 //      a polynomial of degree 5 to approximate R. The maximum error
     49 //      of this polynomial approximation is bounded by 2**-59. In
     50 //      other words,
     51 //          R(z) ~ 2.0 + P1*z + P2*z**2 + P3*z**3 + P4*z**4 + P5*z**5
     52 //      (where z=r*r, and the values of P1 to P5 are listed below)
     53 //      and
     54 //          |                  5          |     -59
     55 //          | 2.0+P1*z+...+P5*z   -  R(z) | <= 2
     56 //          |                             |
     57 //      The computation of exp(r) thus becomes
     58 //                             2*r
     59 //              exp(r) = 1 + -------
     60 //                            R - r
     61 //                                 r*R1(r)
     62 //                     = 1 + r + ----------- (for better accuracy)
     63 //                                2 - R1(r)
     64 //      where
     65 //                               2       4             10
     66 //              R1(r) = r - (P1*r  + P2*r  + ... + P5*r   ).
     67 //
     68 //   3. Scale back to obtain exp(x):
     69 //      From step 1, we have
     70 //         exp(x) = 2**k * exp(r)
     71 //
     72 // Special cases:
     73 //      exp(INF) is INF, exp(NaN) is NaN;
     74 //      exp(-INF) is 0, and
     75 //      for finite argument, only exp(0)=1 is exact.
     76 //
     77 // Accuracy:
     78 //      according to an error analysis, the error is always less than
     79 //      1 ulp (unit in the last place).
     80 //
     81 // Misc. info.
     82 //      For IEEE double
     83 //          if x >  7.09782712893383973096e+02 then exp(x) overflow
     84 //          if x < -7.45133219101941108420e+02 then exp(x) underflow
     85 //
     86 // Constants:
     87 // The hexadecimal values are the intended ones for the following
     88 // constants. The decimal values may be used, provided that the
     89 // compiler will convert from decimal to binary accurately enough
     90 // to produce the hexadecimal values shown.
     91 
     92 func exp(x float64) float64 {
     93 	const (
     94 		Ln2Hi = 6.93147180369123816490e-01
     95 		Ln2Lo = 1.90821492927058770002e-10
     96 		Log2e = 1.44269504088896338700e+00
     97 
     98 		Overflow  = 7.09782712893383973096e+02
     99 		Underflow = -7.45133219101941108420e+02
    100 		NearZero  = 1.0 / (1 << 28) // 2**-28
    101 	)
    102 
    103 	// special cases
    104 	switch {
    105 	case IsNaN(x) || IsInf(x, 1):
    106 		return x
    107 	case IsInf(x, -1):
    108 		return 0
    109 	case x > Overflow:
    110 		return Inf(1)
    111 	case x < Underflow:
    112 		return 0
    113 	case -NearZero < x && x < NearZero:
    114 		return 1 + x
    115 	}
    116 
    117 	// reduce; computed as r = hi - lo for extra precision.
    118 	var k int
    119 	switch {
    120 	case x < 0:
    121 		k = int(Log2e*x - 0.5)
    122 	case x > 0:
    123 		k = int(Log2e*x + 0.5)
    124 	}
    125 	hi := x - float64(k)*Ln2Hi
    126 	lo := float64(k) * Ln2Lo
    127 
    128 	// compute
    129 	return expmulti(hi, lo, k)
    130 }
    131 
    132 // Exp2 returns 2**x, the base-2 exponential of x.
    133 //
    134 // Special cases are the same as Exp.
    135 func Exp2(x float64) float64
    136 
    137 func exp2(x float64) float64 {
    138 	const (
    139 		Ln2Hi = 6.93147180369123816490e-01
    140 		Ln2Lo = 1.90821492927058770002e-10
    141 
    142 		Overflow  = 1.0239999999999999e+03
    143 		Underflow = -1.0740e+03
    144 	)
    145 
    146 	// special cases
    147 	switch {
    148 	case IsNaN(x) || IsInf(x, 1):
    149 		return x
    150 	case IsInf(x, -1):
    151 		return 0
    152 	case x > Overflow:
    153 		return Inf(1)
    154 	case x < Underflow:
    155 		return 0
    156 	}
    157 
    158 	// argument reduction; x = rlg(e) + k with |r|  ln(2)/2.
    159 	// computed as r = hi - lo for extra precision.
    160 	var k int
    161 	switch {
    162 	case x > 0:
    163 		k = int(x + 0.5)
    164 	case x < 0:
    165 		k = int(x - 0.5)
    166 	}
    167 	t := x - float64(k)
    168 	hi := t * Ln2Hi
    169 	lo := -t * Ln2Lo
    170 
    171 	// compute
    172 	return expmulti(hi, lo, k)
    173 }
    174 
    175 // exp1 returns e**r  2**k where r = hi - lo and |r|  ln(2)/2.
    176 func expmulti(hi, lo float64, k int) float64 {
    177 	const (
    178 		P1 = 1.66666666666666657415e-01  /* 0x3FC55555; 0x55555555 */
    179 		P2 = -2.77777777770155933842e-03 /* 0xBF66C16C; 0x16BEBD93 */
    180 		P3 = 6.61375632143793436117e-05  /* 0x3F11566A; 0xAF25DE2C */
    181 		P4 = -1.65339022054652515390e-06 /* 0xBEBBBD41; 0xC5D26BF1 */
    182 		P5 = 4.13813679705723846039e-08  /* 0x3E663769; 0x72BEA4D0 */
    183 	)
    184 
    185 	r := hi - lo
    186 	t := r * r
    187 	c := r - t*(P1+t*(P2+t*(P3+t*(P4+t*P5))))
    188 	y := 1 - ((lo - (r*c)/(2-c)) - hi)
    189 	// TODO(rsc): make sure Ldexp can handle boundary k
    190 	return Ldexp(y, k)
    191 }
    192