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57
copysign(double x, double y)
59
/* use atan2 to distinguish -0. from 0. */
60
if (y > 0. || (y == 0. && atan2(y, -1.) > 0.)) {
59
/* use atan2 to distinguish -0. from 0. */
60
if (y > 0. || (y == 0. && atan2(y, -1.) > 0.)) {
66
66
#endif /* HAVE_COPYSIGN */
87
/* For x small, we use the following approach. Let y be the nearest
90
1+x = y * (1 - (y-1-x)/y)
92
so log(1+x) = log(y) + log(1-(y-1-x)/y). Since (y-1-x)/y is tiny,
93
the second term is well approximated by (y-1-x)/y. If abs(x) >=
94
DBL_EPSILON/2 or the rounding-mode is some form of round-to-nearest
95
then y-1-x will be exactly representable, and is computed exactly
98
If abs(x) < DBL_EPSILON/2 and the rounding mode is not known to be
99
round-to-nearest then this method is slightly dangerous: 1+x could
100
be rounded up to 1+DBL_EPSILON instead of down to 1, and in that
101
case y-1-x will not be exactly representable any more and the
102
result can be off by many ulps. But this is easily fixed: for a
103
floating-point number |x| < DBL_EPSILON/2., the closest
104
floating-point number to log(1+x) is exactly x.
108
if (fabs(x) < DBL_EPSILON/2.) {
110
} else if (-0.5 <= x && x <= 1.) {
111
/* WARNING: it's possible than an overeager compiler
112
will incorrectly optimize the following two lines
113
to the equivalent of "return log(1.+x)". If this
114
happens, then results from log1p will be inaccurate
117
return log(y)-((y-1.)-x)/y;
119
/* NaNs and infinities should end up here */
87
/* For x small, we use the following approach. Let y be the nearest
90
1+x = y * (1 - (y-1-x)/y)
92
so log(1+x) = log(y) + log(1-(y-1-x)/y). Since (y-1-x)/y is tiny,
93
the second term is well approximated by (y-1-x)/y. If abs(x) >=
94
DBL_EPSILON/2 or the rounding-mode is some form of round-to-nearest
95
then y-1-x will be exactly representable, and is computed exactly
98
If abs(x) < DBL_EPSILON/2 and the rounding mode is not known to be
99
round-to-nearest then this method is slightly dangerous: 1+x could
100
be rounded up to 1+DBL_EPSILON instead of down to 1, and in that
101
case y-1-x will not be exactly representable any more and the
102
result can be off by many ulps. But this is easily fixed: for a
103
floating-point number |x| < DBL_EPSILON/2., the closest
104
floating-point number to log(1+x) is exactly x.
108
if (fabs(x) < DBL_EPSILON/2.) {
110
} else if (-0.5 <= x && x <= 1.) {
111
/* WARNING: it's possible than an overeager compiler
112
will incorrectly optimize the following two lines
113
to the equivalent of "return log(1.+x)". If this
114
happens, then results from log1p will be inaccurate
117
return log(y)-((y-1.)-x)/y;
119
/* NaNs and infinities should end up here */
123
123
#endif /* HAVE_LOG1P */
129
129
* Developed at SunPro, a Sun Microsystems, Inc. business.
130
130
* Permission to use, copy, modify, and distribute this
131
* software is freely granted, provided that this notice
131
* software is freely granted, provided that this notice
133
133
* ====================================================
144
* asinh(x) = sign(x) * log [ |x| + sqrt(x*x+1) ]
146
* asinh(x) := x if 1+x*x=1,
147
* := sign(x)*(log(x)+ln2)) for large |x|, else
148
* := sign(x)*log(2|x|+1/(|x|+sqrt(x*x+1))) if|x|>2, else
149
* := sign(x)*log1p(|x| + x^2/(1 + sqrt(1+x^2)))
144
* asinh(x) = sign(x) * log [ |x| + sqrt(x*x+1) ]
146
* asinh(x) := x if 1+x*x=1,
147
* := sign(x)*(log(x)+ln2)) for large |x|, else
148
* := sign(x)*log(2|x|+1/(|x|+sqrt(x*x+1))) if|x|>2, else
149
* := sign(x)*log1p(|x| + x^2/(1 + sqrt(1+x^2)))
152
152
#ifndef HAVE_ASINH
157
double absx = fabs(x);
159
if (Py_IS_NAN(x) || Py_IS_INFINITY(x)) {
162
if (absx < two_pow_m28) { /* |x| < 2**-28 */
163
return x; /* return x inexact except 0 */
165
if (absx > two_pow_p28) { /* |x| > 2**28 */
168
else if (absx > 2.0) { /* 2 < |x| < 2**28 */
169
w = log(2.0*absx + 1.0 / (sqrt(x*x + 1.0) + absx));
171
else { /* 2**-28 <= |x| < 2= */
173
w = log1p(absx + t / (1.0 + sqrt(1.0 + t)));
175
return copysign(w, x);
157
double absx = fabs(x);
159
if (Py_IS_NAN(x) || Py_IS_INFINITY(x)) {
162
if (absx < two_pow_m28) { /* |x| < 2**-28 */
163
return x; /* return x inexact except 0 */
165
if (absx > two_pow_p28) { /* |x| > 2**28 */
168
else if (absx > 2.0) { /* 2 < |x| < 2**28 */
169
w = log(2.0*absx + 1.0 / (sqrt(x*x + 1.0) + absx));
171
else { /* 2**-28 <= |x| < 2= */
173
w = log1p(absx + t / (1.0 + sqrt(1.0 + t)));
175
return copysign(w, x);
178
178
#endif /* HAVE_ASINH */
183
* acosh(x) = log [ x + sqrt(x*x-1) ]
183
* acosh(x) = log [ x + sqrt(x*x-1) ]
185
* acosh(x) := log(x)+ln2, if x is large; else
186
* acosh(x) := log(2x-1/(sqrt(x*x-1)+x)) if x>2; else
187
* acosh(x) := log1p(t+sqrt(2.0*t+t*t)); where t=x-1.
185
* acosh(x) := log(x)+ln2, if x is large; else
186
* acosh(x) := log(2x-1/(sqrt(x*x-1)+x)) if x>2; else
187
* acosh(x) := log1p(t+sqrt(2.0*t+t*t)); where t=x-1.
190
190
* acosh(x) is NaN with signal if x<1.
201
if (x < 1.) { /* x < 1; return a signaling NaN */
201
if (x < 1.) { /* x < 1; return a signaling NaN */
209
else if (x >= two_pow_p28) { /* x > 2**28 */
210
if (Py_IS_INFINITY(x)) {
213
return log(x)+ln2; /* acosh(huge)=log(2x) */
217
return 0.0; /* acosh(1) = 0 */
219
else if (x > 2.) { /* 2 < x < 2**28 */
221
return log(2.0*x - 1.0 / (x + sqrt(t - 1.0)));
223
else { /* 1 < x <= 2 */
225
return log1p(t + sqrt(2.0*t + t*t));
209
else if (x >= two_pow_p28) { /* x > 2**28 */
210
if (Py_IS_INFINITY(x)) {
213
return log(x)+ln2; /* acosh(huge)=log(2x) */
217
return 0.0; /* acosh(1) = 0 */
219
else if (x > 2.) { /* 2 < x < 2**28 */
221
return log(2.0*x - 1.0 / (x + sqrt(t - 1.0)));
223
else { /* 1 < x <= 2 */
225
return log1p(t + sqrt(2.0*t + t*t));
228
228
#endif /* HAVE_ACOSH */
258
if (absx >= 1.) { /* |x| >= 1 */
258
if (absx >= 1.) { /* |x| >= 1 */
266
if (absx < two_pow_m28) { /* |x| < 2**-28 */
269
if (absx < 0.5) { /* |x| < 0.5 */
271
t = 0.5 * log1p(t + t*absx / (1.0 - absx));
273
else { /* 0.5 <= |x| <= 1.0 */
274
t = 0.5 * log1p((absx + absx) / (1.0 - absx));
276
return copysign(t, x);
266
if (absx < two_pow_m28) { /* |x| < 2**-28 */
269
if (absx < 0.5) { /* |x| < 0.5 */
271
t = 0.5 * log1p(t + t*absx / (1.0 - absx));
273
else { /* 0.5 <= |x| <= 1.0 */
274
t = 0.5 * log1p((absx + absx) / (1.0 - absx));
276
return copysign(t, x);
278
278
#endif /* HAVE_ATANH */