libgo/go/math/exp.go

   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
  15 //extern exp
  16 func libc_exp(float64) float64
  17
  18 func Exp(x float64) float64 {
  19         return libc_exp(x)
  20 }
  21
  22 // The original C code, the long comment, and the constants
  23 // below are from FreeBSD's /usr/src/lib/msun/src/e_exp.c
  24 // and came with this notice.  The go code is a simplified
  25 // version of the original C.
  26 //
  27 // ====================================================
  28 // Copyright (C) 2004 by Sun Microsystems, Inc. All rights reserved.
  29 //
  30 // Permission to use, copy, modify, and distribute this
  31 // software is freely granted, provided that this notice
  32 // is preserved.
  33 // ====================================================
  34 //
  35 //
  36 // exp(x)
  37 // Returns the exponential of x.
  38 //
  39 // Method
  40 //   1. Argument reduction:
  41 //      Reduce x to an r so that |r| <= 0.5*ln2 ~ 0.34658.
  42 //      Given x, find r and integer k such that
  43 //
  44 //               x = k*ln2 + r,  |r| <= 0.5*ln2.
  45 //
  46 //      Here r will be represented as r = hi-lo for better
  47 //      accuracy.
  48 //
  49 //   2. Approximation of exp(r) by a special rational function on
  50 //      the interval [0,0.34658]:
  51 //      Write
  52 //          R(r**2) = r*(exp(r)+1)/(exp(r)-1) = 2 + r*r/6 - r**4/360 + ...
  53 //      We use a special Remes algorithm on [0,0.34658] to generate
  54 //      a polynomial of degree 5 to approximate R. The maximum error
  55 //      of this polynomial approximation is bounded by 2**-59. In
  56 //      other words,
  57 //          R(z) ~ 2.0 + P1*z + P2*z**2 + P3*z**3 + P4*z**4 + P5*z**5
  58 //      (where z=r*r, and the values of P1 to P5 are listed below)
  59 //      and
  60 //          |                  5          |     -59
  61 //          | 2.0+P1*z+...+P5*z   -  R(z) | <= 2
  62 //          |                             |
  63 //      The computation of exp(r) thus becomes
  64 //                             2*r
  65 //              exp(r) = 1 + -------
  66 //                            R - r
  67 //                                 r*R1(r)
  68 //                     = 1 + r + ----------- (for better accuracy)
  69 //                                2 - R1(r)
  70 //      where
  71 //                               2       4             10
  72 //              R1(r) = r - (P1*r  + P2*r  + ... + P5*r   ).
  73 //
  74 //   3. Scale back to obtain exp(x):
  75 //      From step 1, we have
  76 //         exp(x) = 2**k * exp(r)
  77 //
  78 // Special cases:
  79 //      exp(INF) is INF, exp(NaN) is NaN;
  80 //      exp(-INF) is 0, and
  81 //      for finite argument, only exp(0)=1 is exact.
  82 //
  83 // Accuracy:
  84 //      according to an error analysis, the error is always less than
  85 //      1 ulp (unit in the last place).
  86 //
  87 // Misc. info.
  88 //      For IEEE double
  89 //          if x >  7.09782712893383973096e+02 then exp(x) overflow
  90 //          if x < -7.45133219101941108420e+02 then exp(x) underflow
  91 //
  92 // Constants:
  93 // The hexadecimal values are the intended ones for the following
  94 // constants. The decimal values may be used, provided that the
  95 // compiler will convert from decimal to binary accurately enough
  96 // to produce the hexadecimal values shown.
  97
  98 func exp(x float64) float64 {
  99         const (
 100                 Ln2Hi = 6.93147180369123816490e-01
 101                 Ln2Lo = 1.90821492927058770002e-10
 102                 Log2e = 1.44269504088896338700e+00
 103
 104                 Overflow  = 7.09782712893383973096e+02
 105                 Underflow = -7.45133219101941108420e+02
 106                 NearZero  = 1.0 / (1 << 28) // 2**-28
 107         )
 108
 109         // special cases
 110         switch {
 111         case IsNaN(x) || IsInf(x, 1):
 112                 return x
 113         case IsInf(x, -1):
 114                 return 0
 115         case x > Overflow:
 116                 return Inf(1)
 117         case x < Underflow:
 118                 return 0
 119         case -NearZero < x && x < NearZero:
 120                 return 1 + x
 121         }
 122
 123         // reduce; computed as r = hi - lo for extra precision.
 124         var k int
 125         switch {
 126         case x < 0:
 127                 k = int(Log2e*x - 0.5)
 128         case x > 0:
 129                 k = int(Log2e*x + 0.5)
 130         }
 131         hi := x - float64(k)*Ln2Hi
 132         lo := float64(k) * Ln2Lo
 133
 134         // compute
 135         return expmulti(hi, lo, k)
 136 }
 137
 138 // Exp2 returns 2**x, the base-2 exponential of x.
 139 //
 140 // Special cases are the same as Exp.
 141 func Exp2(x float64) float64 {
 142         return exp2(x)
 143 }
 144
 145 func exp2(x float64) float64 {
 146         const (
 147                 Ln2Hi = 6.93147180369123816490e-01
 148                 Ln2Lo = 1.90821492927058770002e-10
 149
 150                 Overflow  = 1.0239999999999999e+03
 151                 Underflow = -1.0740e+03
 152         )
 153
 154         // special cases
 155         switch {
 156         case IsNaN(x) || IsInf(x, 1):
 157                 return x
 158         case IsInf(x, -1):
 159                 return 0
 160         case x > Overflow:
 161                 return Inf(1)
 162         case x < Underflow:
 163                 return 0
 164         }
 165
 166         // argument reduction; x = r×lg(e) + k with |r| ≤ ln(2)/2.
 167         // computed as r = hi - lo for extra precision.
 168         var k int
 169         switch {
 170         case x > 0:
 171                 k = int(x + 0.5)
 172         case x < 0:
 173                 k = int(x - 0.5)
 174         }
 175         t := x - float64(k)
 176         hi := t * Ln2Hi
 177         lo := -t * Ln2Lo
 178
 179         // compute
 180         return expmulti(hi, lo, k)
 181 }
 182
 183 // exp1 returns e**r × 2**k where r = hi - lo and |r| ≤ ln(2)/2.
 184 func expmulti(hi, lo float64, k int) float64 {
 185         const (
 186                 P1 = 1.66666666666666019037e-01  /* 0x3FC55555; 0x5555553E */
 187                 P2 = -2.77777777770155933842e-03 /* 0xBF66C16C; 0x16BEBD93 */
 188                 P3 = 6.61375632143793436117e-05  /* 0x3F11566A; 0xAF25DE2C */
 189                 P4 = -1.65339022054652515390e-06 /* 0xBEBBBD41; 0xC5D26BF1 */
 190                 P5 = 4.13813679705723846039e-08  /* 0x3E663769; 0x72BEA4D0 */
 191         )
 192
 193         r := hi - lo
 194         t := r * r
 195         c := r - t*(P1+t*(P2+t*(P3+t*(P4+t*P5))))
 196         y := 1 - ((lo - (r*c)/(2-c)) - hi)
 197         // TODO(rsc): make sure Ldexp can handle boundary k
 198         return Ldexp(y, k)
 199 }