github.com/twelsh-aw/go/src@v0.0.0-20230516233729-a56fe86a7c81/math/exp.go (about)

     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  }