| 1 | #include "Python.h"
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| 2 |
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| 3 | #ifdef X87_DOUBLE_ROUNDING
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| 4 | /* On x86 platforms using an x87 FPU, this function is called from the
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| 5 | Py_FORCE_DOUBLE macro (defined in pymath.h) to force a floating-point
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| 6 | number out of an 80-bit x87 FPU register and into a 64-bit memory location,
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| 7 | thus rounding from extended precision to double precision. */
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| 8 | double _Py_force_double(double x)
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| 9 | {
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| 10 | volatile double y;
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| 11 | y = x;
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| 12 | return y;
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| 13 | }
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| 14 | #endif
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| 15 |
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| 16 | #ifndef HAVE_HYPOT
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| 17 | double hypot(double x, double y)
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| 18 | {
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| 19 | double yx;
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| 20 |
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| 21 | x = fabs(x);
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| 22 | y = fabs(y);
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| 23 | if (x < y) {
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| 24 | double temp = x;
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| 25 | x = y;
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| 26 | y = temp;
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| 27 | }
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| 28 | if (x == 0.)
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| 29 | return 0.;
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| 30 | else {
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| 31 | yx = y/x;
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| 32 | return x*sqrt(1.+yx*yx);
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| 33 | }
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| 34 | }
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| 35 | #endif /* HAVE_HYPOT */
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| 36 |
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| 37 | #ifndef HAVE_COPYSIGN
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| 38 | double
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| 39 | copysign(double x, double y)
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| 40 | {
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| 41 | /* use atan2 to distinguish -0. from 0. */
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| 42 | if (y > 0. || (y == 0. && atan2(y, -1.) > 0.)) {
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| 43 | return fabs(x);
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| 44 | } else {
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| 45 | return -fabs(x);
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| 46 | }
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| 47 | }
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| 48 | #endif /* HAVE_COPYSIGN */
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| 49 |
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| 50 | #ifndef HAVE_LOG1P
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| 51 | #include <float.h>
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| 52 |
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| 53 | double
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| 54 | log1p(double x)
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| 55 | {
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| 56 | /* For x small, we use the following approach. Let y be the nearest
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| 57 | float to 1+x, then
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| 58 |
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| 59 | 1+x = y * (1 - (y-1-x)/y)
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| 60 |
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| 61 | so log(1+x) = log(y) + log(1-(y-1-x)/y). Since (y-1-x)/y is tiny,
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| 62 | the second term is well approximated by (y-1-x)/y. If abs(x) >=
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| 63 | DBL_EPSILON/2 or the rounding-mode is some form of round-to-nearest
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| 64 | then y-1-x will be exactly representable, and is computed exactly
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| 65 | by (y-1)-x.
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| 66 |
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| 67 | If abs(x) < DBL_EPSILON/2 and the rounding mode is not known to be
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| 68 | round-to-nearest then this method is slightly dangerous: 1+x could
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| 69 | be rounded up to 1+DBL_EPSILON instead of down to 1, and in that
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| 70 | case y-1-x will not be exactly representable any more and the
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| 71 | result can be off by many ulps. But this is easily fixed: for a
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| 72 | floating-point number |x| < DBL_EPSILON/2., the closest
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| 73 | floating-point number to log(1+x) is exactly x.
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| 74 | */
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| 75 |
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| 76 | double y;
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| 77 | if (fabs(x) < DBL_EPSILON/2.) {
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| 78 | return x;
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| 79 | } else if (-0.5 <= x && x <= 1.) {
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| 80 | /* WARNING: it's possible than an overeager compiler
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| 81 | will incorrectly optimize the following two lines
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| 82 | to the equivalent of "return log(1.+x)". If this
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| 83 | happens, then results from log1p will be inaccurate
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| 84 | for small x. */
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| 85 | y = 1.+x;
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| 86 | return log(y)-((y-1.)-x)/y;
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| 87 | } else {
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| 88 | /* NaNs and infinities should end up here */
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| 89 | return log(1.+x);
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| 90 | }
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| 91 | }
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| 92 | #endif /* HAVE_LOG1P */
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| 93 |
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| 94 | /*
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| 95 | * ====================================================
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| 96 | * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
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| 97 | *
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| 98 | * Developed at SunPro, a Sun Microsystems, Inc. business.
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| 99 | * Permission to use, copy, modify, and distribute this
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| 100 | * software is freely granted, provided that this notice
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| 101 | * is preserved.
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| 102 | * ====================================================
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| 103 | */
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| 104 |
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| 105 | static const double ln2 = 6.93147180559945286227E-01;
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| 106 | static const double two_pow_m28 = 3.7252902984619141E-09; /* 2**-28 */
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| 107 | static const double two_pow_p28 = 268435456.0; /* 2**28 */
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| 108 | static const double zero = 0.0;
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| 109 |
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| 110 | /* asinh(x)
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| 111 | * Method :
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| 112 | * Based on
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| 113 | * asinh(x) = sign(x) * log [ |x| + sqrt(x*x+1) ]
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| 114 | * we have
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| 115 | * asinh(x) := x if 1+x*x=1,
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| 116 | * := sign(x)*(log(x)+ln2)) for large |x|, else
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| 117 | * := sign(x)*log(2|x|+1/(|x|+sqrt(x*x+1))) if|x|>2, else
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| 118 | * := sign(x)*log1p(|x| + x^2/(1 + sqrt(1+x^2)))
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| 119 | */
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| 120 |
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| 121 | #ifndef HAVE_ASINH
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| 122 | double
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| 123 | asinh(double x)
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| 124 | {
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| 125 | double w;
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| 126 | double absx = fabs(x);
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| 127 |
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| 128 | if (Py_IS_NAN(x) || Py_IS_INFINITY(x)) {
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| 129 | return x+x;
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| 130 | }
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| 131 | if (absx < two_pow_m28) { /* |x| < 2**-28 */
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| 132 | return x; /* return x inexact except 0 */
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| 133 | }
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| 134 | if (absx > two_pow_p28) { /* |x| > 2**28 */
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| 135 | w = log(absx)+ln2;
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| 136 | }
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| 137 | else if (absx > 2.0) { /* 2 < |x| < 2**28 */
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| 138 | w = log(2.0*absx + 1.0 / (sqrt(x*x + 1.0) + absx));
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| 139 | }
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| 140 | else { /* 2**-28 <= |x| < 2= */
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| 141 | double t = x*x;
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| 142 | w = log1p(absx + t / (1.0 + sqrt(1.0 + t)));
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| 143 | }
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| 144 | return copysign(w, x);
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| 145 |
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| 146 | }
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| 147 | #endif /* HAVE_ASINH */
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| 148 |
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| 149 | /* acosh(x)
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| 150 | * Method :
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| 151 | * Based on
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| 152 | * acosh(x) = log [ x + sqrt(x*x-1) ]
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| 153 | * we have
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| 154 | * acosh(x) := log(x)+ln2, if x is large; else
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| 155 | * acosh(x) := log(2x-1/(sqrt(x*x-1)+x)) if x>2; else
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| 156 | * acosh(x) := log1p(t+sqrt(2.0*t+t*t)); where t=x-1.
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| 157 | *
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| 158 | * Special cases:
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| 159 | * acosh(x) is NaN with signal if x<1.
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| 160 | * acosh(NaN) is NaN without signal.
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| 161 | */
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| 162 |
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| 163 | #ifndef HAVE_ACOSH
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| 164 | double
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| 165 | acosh(double x)
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| 166 | {
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| 167 | if (Py_IS_NAN(x)) {
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| 168 | return x+x;
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| 169 | }
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| 170 | if (x < 1.) { /* x < 1; return a signaling NaN */
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| 171 | errno = EDOM;
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| 172 | #ifdef Py_NAN
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| 173 | return Py_NAN;
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| 174 | #else
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| 175 | return (x-x)/(x-x);
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| 176 | #endif
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| 177 | }
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| 178 | else if (x >= two_pow_p28) { /* x > 2**28 */
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| 179 | if (Py_IS_INFINITY(x)) {
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| 180 | return x+x;
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| 181 | } else {
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| 182 | return log(x)+ln2; /* acosh(huge)=log(2x) */
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| 183 | }
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| 184 | }
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| 185 | else if (x == 1.) {
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| 186 | return 0.0; /* acosh(1) = 0 */
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| 187 | }
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| 188 | else if (x > 2.) { /* 2 < x < 2**28 */
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| 189 | double t = x*x;
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| 190 | return log(2.0*x - 1.0 / (x + sqrt(t - 1.0)));
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| 191 | }
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| 192 | else { /* 1 < x <= 2 */
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| 193 | double t = x - 1.0;
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| 194 | return log1p(t + sqrt(2.0*t + t*t));
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| 195 | }
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| 196 | }
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| 197 | #endif /* HAVE_ACOSH */
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| 198 |
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| 199 | /* atanh(x)
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| 200 | * Method :
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| 201 | * 1.Reduced x to positive by atanh(-x) = -atanh(x)
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| 202 | * 2.For x>=0.5
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| 203 | * 1 2x x
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| 204 | * atanh(x) = --- * log(1 + -------) = 0.5 * log1p(2 * --------)
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| 205 | * 2 1 - x 1 - x
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| 206 | *
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| 207 | * For x<0.5
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| 208 | * atanh(x) = 0.5*log1p(2x+2x*x/(1-x))
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| 209 | *
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| 210 | * Special cases:
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| 211 | * atanh(x) is NaN if |x| >= 1 with signal;
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| 212 | * atanh(NaN) is that NaN with no signal;
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| 213 | *
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| 214 | */
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| 215 |
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| 216 | #ifndef HAVE_ATANH
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| 217 | double
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| 218 | atanh(double x)
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| 219 | {
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| 220 | double absx;
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| 221 | double t;
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| 222 |
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| 223 | if (Py_IS_NAN(x)) {
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| 224 | return x+x;
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| 225 | }
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| 226 | absx = fabs(x);
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| 227 | if (absx >= 1.) { /* |x| >= 1 */
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| 228 | errno = EDOM;
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| 229 | #ifdef Py_NAN
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| 230 | return Py_NAN;
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| 231 | #else
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| 232 | return x/zero;
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| 233 | #endif
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| 234 | }
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| 235 | if (absx < two_pow_m28) { /* |x| < 2**-28 */
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| 236 | return x;
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| 237 | }
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| 238 | if (absx < 0.5) { /* |x| < 0.5 */
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| 239 | t = absx+absx;
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| 240 | t = 0.5 * log1p(t + t*absx / (1.0 - absx));
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| 241 | }
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| 242 | else { /* 0.5 <= |x| <= 1.0 */
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| 243 | t = 0.5 * log1p((absx + absx) / (1.0 - absx));
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| 244 | }
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| 245 | return copysign(t, x);
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| 246 | }
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| 247 | #endif /* HAVE_ATANH */
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