exp.mx raw

   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  //
  11  //	Exp(+Inf) = +Inf
  12  //	Exp(NaN) = NaN
  13  //
  14  // Very large values overflow to 0 or +Inf.
  15  // Very small values underflow to 1.
  16  func Exp(x float64) float64 {
  17  	if haveArchExp {
  18  		return archExp(x)
  19  	}
  20  	return exp(x)
  21  }
  22  
  23  // The original C code, the long comment, and the constants
  24  // below are from FreeBSD's /usr/src/lib/msun/src/e_exp.c
  25  // and came with this notice. The go code is a simplified
  26  // version of the original C.
  27  //
  28  // ====================================================
  29  // Copyright (C) 2004 by Sun Microsystems, Inc. All rights reserved.
  30  //
  31  // Permission to use, copy, modify, and distribute this
  32  // software is freely granted, provided that this notice
  33  // is preserved.
  34  // ====================================================
  35  //
  36  //
  37  // exp(x)
  38  // Returns the exponential of x.
  39  //
  40  // Method
  41  //   1. Argument reduction:
  42  //      Reduce x to an r so that |r| <= 0.5*ln2 ~ 0.34658.
  43  //      Given x, find r and integer k such that
  44  //
  45  //               x = k*ln2 + r,  |r| <= 0.5*ln2.
  46  //
  47  //      Here r will be represented as r = hi-lo for better
  48  //      accuracy.
  49  //
  50  //   2. Approximation of exp(r) by a special rational function on
  51  //      the interval [0,0.34658]:
  52  //      Write
  53  //          R(r**2) = r*(exp(r)+1)/(exp(r)-1) = 2 + r*r/6 - r**4/360 + ...
  54  //      We use a special Remez algorithm on [0,0.34658] to generate
  55  //      a polynomial of degree 5 to approximate R. The maximum error
  56  //      of this polynomial approximation is bounded by 2**-59. In
  57  //      other words,
  58  //          R(z) ~ 2.0 + P1*z + P2*z**2 + P3*z**3 + P4*z**4 + P5*z**5
  59  //      (where z=r*r, and the values of P1 to P5 are listed below)
  60  //      and
  61  //          |                  5          |     -59
  62  //          | 2.0+P1*z+...+P5*z   -  R(z) | <= 2
  63  //          |                             |
  64  //      The computation of exp(r) thus becomes
  65  //                             2*r
  66  //              exp(r) = 1 + -------
  67  //                            R - r
  68  //                                 r*R1(r)
  69  //                     = 1 + r + ----------- (for better accuracy)
  70  //                                2 - R1(r)
  71  //      where
  72  //                               2       4             10
  73  //              R1(r) = r - (P1*r  + P2*r  + ... + P5*r   ).
  74  //
  75  //   3. Scale back to obtain exp(x):
  76  //      From step 1, we have
  77  //         exp(x) = 2**k * exp(r)
  78  //
  79  // Special cases:
  80  //      exp(INF) is INF, exp(NaN) is NaN;
  81  //      exp(-INF) is 0, and
  82  //      for finite argument, only exp(0)=1 is exact.
  83  //
  84  // Accuracy:
  85  //      according to an error analysis, the error is always less than
  86  //      1 ulp (unit in the last place).
  87  //
  88  // Misc. info.
  89  //      For IEEE double
  90  //          if x >  7.09782712893383973096e+02 then exp(x) overflow
  91  //          if x < -7.45133219101941108420e+02 then exp(x) underflow
  92  //
  93  // Constants:
  94  // The hexadecimal values are the intended ones for the following
  95  // constants. The decimal values may be used, provided that the
  96  // compiler will convert from decimal to binary accurately enough
  97  // to produce the hexadecimal values shown.
  98  
  99  func exp(x float64) float64 {
 100  	const (
 101  		Ln2Hi = 6.93147180369123816490e-01
 102  		Ln2Lo = 1.90821492927058770002e-10
 103  		Log2e = 1.44269504088896338700e+00
 104  
 105  		Overflow  = 7.09782712893383973096e+02
 106  		Underflow = -7.45133219101941108420e+02
 107  		NearZero  = 1.0 / (1 << 28) // 2**-28
 108  	)
 109  
 110  	// special cases
 111  	switch {
 112  	case IsNaN(x) || IsInf(x, 1):
 113  		return x
 114  	case IsInf(x, -1):
 115  		return 0
 116  	case x > Overflow:
 117  		return Inf(1)
 118  	case x < Underflow:
 119  		return 0
 120  	case -NearZero < x && x < NearZero:
 121  		return 1 + x
 122  	}
 123  
 124  	// reduce; computed as r = hi - lo for extra precision.
 125  	var k int
 126  	switch {
 127  	case x < 0:
 128  		k = int(Log2e*x - 0.5)
 129  	case x > 0:
 130  		k = int(Log2e*x + 0.5)
 131  	}
 132  	hi := x - float64(k)*Ln2Hi
 133  	lo := float64(k) * Ln2Lo
 134  
 135  	// compute
 136  	return expmulti(hi, lo, k)
 137  }
 138  
 139  // Exp2 returns 2**x, the base-2 exponential of x.
 140  //
 141  // Special cases are the same as [Exp].
 142  func Exp2(x float64) float64 {
 143  	if haveArchExp2 {
 144  		return archExp2(x)
 145  	}
 146  	return exp2(x)
 147  }
 148  
 149  func exp2(x float64) float64 {
 150  	const (
 151  		Ln2Hi = 6.93147180369123816490e-01
 152  		Ln2Lo = 1.90821492927058770002e-10
 153  
 154  		Overflow  = 1.0239999999999999e+03
 155  		Underflow = -1.0740e+03
 156  	)
 157  
 158  	// special cases
 159  	switch {
 160  	case IsNaN(x) || IsInf(x, 1):
 161  		return x
 162  	case IsInf(x, -1):
 163  		return 0
 164  	case x > Overflow:
 165  		return Inf(1)
 166  	case x < Underflow:
 167  		return 0
 168  	}
 169  
 170  	// argument reduction; x = r×lg(e) + k with |r| ≤ ln(2)/2.
 171  	// computed as r = hi - lo for extra precision.
 172  	var k int
 173  	switch {
 174  	case x > 0:
 175  		k = int(x + 0.5)
 176  	case x < 0:
 177  		k = int(x - 0.5)
 178  	}
 179  	t := x - float64(k)
 180  	hi := t * Ln2Hi
 181  	lo := -t * Ln2Lo
 182  
 183  	// compute
 184  	return expmulti(hi, lo, k)
 185  }
 186  
 187  // exp1 returns e**r × 2**k where r = hi - lo and |r| ≤ ln(2)/2.
 188  func expmulti(hi, lo float64, k int) float64 {
 189  	const (
 190  		P1 = 1.66666666666666657415e-01  /* 0x3FC55555; 0x55555555 */
 191  		P2 = -2.77777777770155933842e-03 /* 0xBF66C16C; 0x16BEBD93 */
 192  		P3 = 6.61375632143793436117e-05  /* 0x3F11566A; 0xAF25DE2C */
 193  		P4 = -1.65339022054652515390e-06 /* 0xBEBBBD41; 0xC5D26BF1 */
 194  		P5 = 4.13813679705723846039e-08  /* 0x3E663769; 0x72BEA4D0 */
 195  	)
 196  
 197  	r := hi - lo
 198  	t := r * r
 199  	c := r - t*(P1+t*(P2+t*(P3+t*(P4+t*P5))))
 200  	y := 1 - ((lo - (r*c)/(2-c)) - hi)
 201  	// TODO(rsc): make sure Ldexp can handle boundary k
 202  	return Ldexp(y, k)
 203  }
 204