// The following is adapted from fdlibm (http://www.netlib.org/fdlibm). // // ==================================================== // Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved. // // Developed at SunSoft, a Sun Microsystems, Inc. business. // Permission to use, copy, modify, and distribute this // software is freely granted, provided that this notice // is preserved. // ==================================================== // // The original source code covered by the above license above has been // modified significantly by Google Inc. // Copyright 2016 the V8 project authors. All rights reserved. // Copyright 2020 The Chromium Authors #include "third_party/fdlibm/ieee754.h" #include <cmath> #include <limits> #include "base/bit_cast.h" #include "base/compiler_specific.h" #include "build/build_config.h" #include "third_party/fdlibm/overflowing-math.h" namespace fdlibm { namespace { /* * The original fdlibm code used statements like: * n0 = ((*(int*)&one)>>29)^1; * index of high word * * ix0 = *(n0+(int*)&x); * high word of x * * ix1 = *((1-n0)+(int*)&x); * low word of x * * to dig two 32 bit words out of the 64 bit IEEE floating point * value. That is non-ANSI, and, moreover, the gcc instruction * scheduler gets it wrong. We instead use the following macros. * Unlike the original code, we determine the endianness at compile * time, not at run time; I don't see much benefit to selecting * endianness at run time. */ /* Get two 32 bit ints from a double. */ #define EXTRACT_WORDS … /* Get the more significant 32 bit int from a double. */ #define GET_HIGH_WORD … /* Get the less significant 32 bit int from a double. */ #define GET_LOW_WORD … /* Set a double from two 32 bit ints. */ #define INSERT_WORDS … /* Set the more significant 32 bits of a double from an int. */ #define SET_HIGH_WORD … /* Set the less significant 32 bits of a double from an int. */ #define SET_LOW_WORD … [[nodiscard]] int32_t __ieee754_rem_pio2(double x, double* y); [[nodiscard]] double __kernel_cos(double x, double y); [[nodiscard]] int __kernel_rem_pio2(double* x, double* y, int e0, int nx, int prec, const int32_t* ipio2); [[nodiscard]] double __kernel_sin(double x, double y, int iy); /* __ieee754_rem_pio2(x,y) * * return the remainder of x rem pi/2 in y[0]+y[1] * use __kernel_rem_pio2() */ int32_t __ieee754_rem_pio2(double x, double *y) { … } /* __kernel_cos( x, y ) * kernel cos function on [-pi/4, pi/4], pi/4 ~ 0.785398164 * Input x is assumed to be bounded by ~pi/4 in magnitude. * Input y is the tail of x. * * Algorithm * 1. Since cos(-x) = cos(x), we need only to consider positive x. * 2. if x < 2^-27 (hx<0x3E400000 0), return 1 with inexact if x!=0. * 3. cos(x) is approximated by a polynomial of degree 14 on * [0,pi/4] * 4 14 * cos(x) ~ 1 - x*x/2 + C1*x + ... + C6*x * where the remez error is * * | 2 4 6 8 10 12 14 | -58 * |cos(x)-(1-.5*x +C1*x +C2*x +C3*x +C4*x +C5*x +C6*x )| <= 2 * | | * * 4 6 8 10 12 14 * 4. let r = C1*x +C2*x +C3*x +C4*x +C5*x +C6*x , then * cos(x) = 1 - x*x/2 + r * since cos(x+y) ~ cos(x) - sin(x)*y * ~ cos(x) - x*y, * a correction term is necessary in cos(x) and hence * cos(x+y) = 1 - (x*x/2 - (r - x*y)) * For better accuracy when x > 0.3, let qx = |x|/4 with * the last 32 bits mask off, and if x > 0.78125, let qx = 0.28125. * Then * cos(x+y) = (1-qx) - ((x*x/2-qx) - (r-x*y)). * Note that 1-qx and (x*x/2-qx) is EXACT here, and the * magnitude of the latter is at least a quarter of x*x/2, * thus, reducing the rounding error in the subtraction. */ ALWAYS_INLINE double __kernel_cos(double x, double y) { … } /* __kernel_rem_pio2(x,y,e0,nx,prec,ipio2) * double x[],y[]; int e0,nx,prec; int ipio2[]; * * __kernel_rem_pio2 return the last three digits of N with * y = x - N*pi/2 * so that |y| < pi/2. * * The method is to compute the integer (mod 8) and fraction parts of * (2/pi)*x without doing the full multiplication. In general we * skip the part of the product that are known to be a huge integer ( * more accurately, = 0 mod 8 ). Thus the number of operations are * independent of the exponent of the input. * * (2/pi) is represented by an array of 24-bit integers in ipio2[]. * * Input parameters: * x[] The input value (must be positive) is broken into nx * pieces of 24-bit integers in double precision format. * x[i] will be the i-th 24 bit of x. The scaled exponent * of x[0] is given in input parameter e0 (i.e., x[0]*2^e0 * match x's up to 24 bits. * * Example of breaking a double positive z into x[0]+x[1]+x[2]: * e0 = ilogb(z)-23 * z = scalbn(z,-e0) * for i = 0,1,2 * x[i] = floor(z) * z = (z-x[i])*2**24 * * * y[] output result in an array of double precision numbers. * The dimension of y[] is: * 24-bit precision 1 * 53-bit precision 2 * 64-bit precision 2 * 113-bit precision 3 * The actual value is the sum of them. Thus for 113-bit * precison, one may have to do something like: * * long double t,w,r_head, r_tail; * t = (long double)y[2] + (long double)y[1]; * w = (long double)y[0]; * r_head = t+w; * r_tail = w - (r_head - t); * * e0 The exponent of x[0] * * nx dimension of x[] * * prec an integer indicating the precision: * 0 24 bits (single) * 1 53 bits (double) * 2 64 bits (extended) * 3 113 bits (quad) * * ipio2[] * integer array, contains the (24*i)-th to (24*i+23)-th * bit of 2/pi after binary point. The corresponding * floating value is * * ipio2[i] * 2^(-24(i+1)). * * External function: * double scalbn(), floor(); * * * Here is the description of some local variables: * * jk jk+1 is the initial number of terms of ipio2[] needed * in the computation. The recommended value is 2,3,4, * 6 for single, double, extended,and quad. * * jz local integer variable indicating the number of * terms of ipio2[] used. * * jx nx - 1 * * jv index for pointing to the suitable ipio2[] for the * computation. In general, we want * ( 2^e0*x[0] * ipio2[jv-1]*2^(-24jv) )/8 * is an integer. Thus * e0-3-24*jv >= 0 or (e0-3)/24 >= jv * Hence jv = max(0,(e0-3)/24). * * jp jp+1 is the number of terms in PIo2[] needed, jp = jk. * * q[] double array with integral value, representing the * 24-bits chunk of the product of x and 2/pi. * * q0 the corresponding exponent of q[0]. Note that the * exponent for q[i] would be q0-24*i. * * PIo2[] double precision array, obtained by cutting pi/2 * into 24 bits chunks. * * f[] ipio2[] in floating point * * iq[] integer array by breaking up q[] in 24-bits chunk. * * fq[] final product of x*(2/pi) in fq[0],..,fq[jk] * * ih integer. If >0 it indicates q[] is >= 0.5, hence * it also indicates the *sign* of the result. * */ int __kernel_rem_pio2(double *x, double *y, int e0, int nx, int prec, const int32_t *ipio2) { … } /* __kernel_sin( x, y, iy) * kernel sin function on [-pi/4, pi/4], pi/4 ~ 0.7854 * Input x is assumed to be bounded by ~pi/4 in magnitude. * Input y is the tail of x. * Input iy indicates whether y is 0. (if iy=0, y assume to be 0). * * Algorithm * 1. Since sin(-x) = -sin(x), we need only to consider positive x. * 2. if x < 2^-27 (hx<0x3E400000 0), return x with inexact if x!=0. * 3. sin(x) is approximated by a polynomial of degree 13 on * [0,pi/4] * 3 13 * sin(x) ~ x + S1*x + ... + S6*x * where * * |sin(x) 2 4 6 8 10 12 | -58 * |----- - (1+S1*x +S2*x +S3*x +S4*x +S5*x +S6*x )| <= 2 * | x | * * 4. sin(x+y) = sin(x) + sin'(x')*y * ~ sin(x) + (1-x*x/2)*y * For better accuracy, let * 3 2 2 2 2 * r = x *(S2+x *(S3+x *(S4+x *(S5+x *S6)))) * then 3 2 * sin(x) = x + (S1*x + (x *(r-y/2)+y)) */ ALWAYS_INLINE double __kernel_sin(double x, double y, int iy) { … } /* __kernel_tan( x, y, k ) * kernel tan function on [-pi/4, pi/4], pi/4 ~ 0.7854 * Input x is assumed to be bounded by ~pi/4 in magnitude. * Input y is the tail of x. * Input k indicates whether tan (if k=1) or * -1/tan (if k= -1) is returned. * * Algorithm * 1. Since tan(-x) = -tan(x), we need only to consider positive x. * 2. if x < 2^-28 (hx<0x3E300000 0), return x with inexact if x!=0. * 3. tan(x) is approximated by a odd polynomial of degree 27 on * [0,0.67434] * 3 27 * tan(x) ~ x + T1*x + ... + T13*x * where * * |tan(x) 2 4 26 | -59.2 * |----- - (1+T1*x +T2*x +.... +T13*x )| <= 2 * | x | * * Note: tan(x+y) = tan(x) + tan'(x)*y * ~ tan(x) + (1+x*x)*y * Therefore, for better accuracy in computing tan(x+y), let * 3 2 2 2 2 * r = x *(T2+x *(T3+x *(...+x *(T12+x *T13)))) * then * 3 2 * tan(x+y) = x + (T1*x + (x *(r+y)+y)) * * 4. For x in [0.67434,pi/4], let y = pi/4 - x, then * tan(x) = tan(pi/4-y) = (1-tan(y))/(1+tan(y)) * = 1 - 2*(tan(y) - (tan(y)^2)/(1+tan(y))) */ double __kernel_tan(double x, double y, int iy) { … } } // namespace /* acos(x) * Method : * acos(x) = pi/2 - asin(x) * acos(-x) = pi/2 + asin(x) * For |x|<=0.5 * acos(x) = pi/2 - (x + x*x^2*R(x^2)) (see asin.c) * For x>0.5 * acos(x) = pi/2 - (pi/2 - 2asin(sqrt((1-x)/2))) * = 2asin(sqrt((1-x)/2)) * = 2s + 2s*z*R(z) ...z=(1-x)/2, s=sqrt(z) * = 2f + (2c + 2s*z*R(z)) * where f=hi part of s, and c = (z-f*f)/(s+f) is the correction term * for f so that f+c ~ sqrt(z). * For x<-0.5 * acos(x) = pi - 2asin(sqrt((1-|x|)/2)) * = pi - 0.5*(s+s*z*R(z)), where z=(1-|x|)/2,s=sqrt(z) * * Special cases: * if x is NaN, return x itself; * if |x|>1, return NaN with invalid signal. * * Function needed: sqrt */ double acos(double x) { … } /* acosh(x) * Method : * Based on * acosh(x) = log [ x + sqrt(x*x-1) ] * we have * acosh(x) := log(x)+ln2, if x is large; else * acosh(x) := log(2x-1/(sqrt(x*x-1)+x)) if x>2; else * acosh(x) := log1p(t+sqrt(2.0*t+t*t)); where t=x-1. * * Special cases: * acosh(x) is NaN with signal if x<1. * acosh(NaN) is NaN without signal. */ double acosh(double x) { … } /* asin(x) * Method : * Since asin(x) = x + x^3/6 + x^5*3/40 + x^7*15/336 + ... * we approximate asin(x) on [0,0.5] by * asin(x) = x + x*x^2*R(x^2) * where * R(x^2) is a rational approximation of (asin(x)-x)/x^3 * and its remez error is bounded by * |(asin(x)-x)/x^3 - R(x^2)| < 2^(-58.75) * * For x in [0.5,1] * asin(x) = pi/2-2*asin(sqrt((1-x)/2)) * Let y = (1-x), z = y/2, s := sqrt(z), and pio2_hi+pio2_lo=pi/2; * then for x>0.98 * asin(x) = pi/2 - 2*(s+s*z*R(z)) * = pio2_hi - (2*(s+s*z*R(z)) - pio2_lo) * For x<=0.98, let pio4_hi = pio2_hi/2, then * f = hi part of s; * c = sqrt(z) - f = (z-f*f)/(s+f) ...f+c=sqrt(z) * and * asin(x) = pi/2 - 2*(s+s*z*R(z)) * = pio4_hi+(pio4-2s)-(2s*z*R(z)-pio2_lo) * = pio4_hi+(pio4-2f)-(2s*z*R(z)-(pio2_lo+2c)) * * Special cases: * if x is NaN, return x itself; * if |x|>1, return NaN with invalid signal. */ double asin(double x) { … } /* asinh(x) * Method : * Based on * asinh(x) = sign(x) * log [ |x| + sqrt(x*x+1) ] * we have * asinh(x) := x if 1+x*x=1, * := sign(x)*(log(x)+ln2)) for large |x|, else * := sign(x)*log(2|x|+1/(|x|+sqrt(x*x+1))) if|x|>2, else * := sign(x)*log1p(|x| + x^2/(1 + sqrt(1+x^2))) */ double asinh(double x) { … } /* atan(x) * Method * 1. Reduce x to positive by atan(x) = -atan(-x). * 2. According to the integer k=4t+0.25 chopped, t=x, the argument * is further reduced to one of the following intervals and the * arctangent of t is evaluated by the corresponding formula: * * [0,7/16] atan(x) = t-t^3*(a1+t^2*(a2+...(a10+t^2*a11)...) * [7/16,11/16] atan(x) = atan(1/2) + atan( (t-0.5)/(1+t/2) ) * [11/16.19/16] atan(x) = atan( 1 ) + atan( (t-1)/(1+t) ) * [19/16,39/16] atan(x) = atan(3/2) + atan( (t-1.5)/(1+1.5t) ) * [39/16,INF] atan(x) = atan(INF) + atan( -1/t ) * * Constants: * The hexadecimal values are the intended ones for the following * constants. The decimal values may be used, provided that the * compiler will convert from decimal to binary accurately enough * to produce the hexadecimal values shown. */ double atan(double x) { … } /* atan2(y,x) * Method : * 1. Reduce y to positive by atan2(y,x)=-atan2(-y,x). * 2. Reduce x to positive by (if x and y are unexceptional): * ARG (x+iy) = arctan(y/x) ... if x > 0, * ARG (x+iy) = pi - arctan[y/(-x)] ... if x < 0, * * Special cases: * * ATAN2((anything), NaN ) is NaN; * ATAN2(NAN , (anything) ) is NaN; * ATAN2(+-0, +(anything but NaN)) is +-0 ; * ATAN2(+-0, -(anything but NaN)) is +-pi ; * ATAN2(+-(anything but 0 and NaN), 0) is +-pi/2; * ATAN2(+-(anything but INF and NaN), +INF) is +-0 ; * ATAN2(+-(anything but INF and NaN), -INF) is +-pi; * ATAN2(+-INF,+INF ) is +-pi/4 ; * ATAN2(+-INF,-INF ) is +-3pi/4; * ATAN2(+-INF, (anything but,0,NaN, and INF)) is +-pi/2; * * Constants: * The hexadecimal values are the intended ones for the following * constants. The decimal values may be used, provided that the * compiler will convert from decimal to binary accurately enough * to produce the hexadecimal values shown. */ double atan2(double y, double x) { … } /* cos(x) * Return cosine function of x. * * kernel function: * __kernel_sin ... sine function on [-pi/4,pi/4] * __kernel_cos ... cosine function on [-pi/4,pi/4] * __ieee754_rem_pio2 ... argument reduction routine * * Method. * Let S,C and T denote the sin, cos and tan respectively on * [-PI/4, +PI/4]. Reduce the argument x to y1+y2 = x-k*pi/2 * in [-pi/4 , +pi/4], and let n = k mod 4. * We have * * n sin(x) cos(x) tan(x) * ---------------------------------------------------------- * 0 S C T * 1 C -S -1/T * 2 -S -C T * 3 -C S -1/T * ---------------------------------------------------------- * * Special cases: * Let trig be any of sin, cos, or tan. * trig(+-INF) is NaN, with signals; * trig(NaN) is that NaN; * * Accuracy: * TRIG(x) returns trig(x) nearly rounded */ double cos(double x) { … } /* exp(x) * Returns the exponential of x. * * Method * 1. Argument reduction: * Reduce x to an r so that |r| <= 0.5*ln2 ~ 0.34658. * Given x, find r and integer k such that * * x = k*ln2 + r, |r| <= 0.5*ln2. * * Here r will be represented as r = hi-lo for better * accuracy. * * 2. Approximation of exp(r) by a special rational function on * the interval [0,0.34658]: * Write * R(r**2) = r*(exp(r)+1)/(exp(r)-1) = 2 + r*r/6 - r**4/360 + ... * We use a special Remes algorithm on [0,0.34658] to generate * a polynomial of degree 5 to approximate R. The maximum error * of this polynomial approximation is bounded by 2**-59. In * other words, * R(z) ~ 2.0 + P1*z + P2*z**2 + P3*z**3 + P4*z**4 + P5*z**5 * (where z=r*r, and the values of P1 to P5 are listed below) * and * | 5 | -59 * | 2.0+P1*z+...+P5*z - R(z) | <= 2 * | | * The computation of exp(r) thus becomes * 2*r * exp(r) = 1 + ------- * R - r * r*R1(r) * = 1 + r + ----------- (for better accuracy) * 2 - R1(r) * where * 2 4 10 * R1(r) = r - (P1*r + P2*r + ... + P5*r ). * * 3. Scale back to obtain exp(x): * From step 1, we have * exp(x) = 2^k * exp(r) * * Special cases: * exp(INF) is INF, exp(NaN) is NaN; * exp(-INF) is 0, and * for finite argument, only exp(0)=1 is exact. * * Accuracy: * according to an error analysis, the error is always less than * 1 ulp (unit in the last place). * * Misc. info. * For IEEE double * if x > 7.09782712893383973096e+02 then exp(x) overflow * if x < -7.45133219101941108420e+02 then exp(x) underflow * * Constants: * The hexadecimal values are the intended ones for the following * constants. The decimal values may be used, provided that the * compiler will convert from decimal to binary accurately enough * to produce the hexadecimal values shown. */ double exp(double x) { … } /* * Method : * 1.Reduced x to positive by atanh(-x) = -atanh(x) * 2.For x>=0.5 * 1 2x x * atanh(x) = --- * log(1 + -------) = 0.5 * log1p(2 * --------) * 2 1 - x 1 - x * * For x<0.5 * atanh(x) = 0.5*log1p(2x+2x*x/(1-x)) * * Special cases: * atanh(x) is NaN if |x| > 1 with signal; * atanh(NaN) is that NaN with no signal; * atanh(+-1) is +-INF with signal. * */ double atanh(double x) { … } /* log(x) * Return the logrithm of x * * Method : * 1. Argument Reduction: find k and f such that * x = 2^k * (1+f), * where sqrt(2)/2 < 1+f < sqrt(2) . * * 2. Approximation of log(1+f). * Let s = f/(2+f) ; based on log(1+f) = log(1+s) - log(1-s) * = 2s + 2/3 s**3 + 2/5 s**5 + ....., * = 2s + s*R * We use a special Reme algorithm on [0,0.1716] to generate * a polynomial of degree 14 to approximate R The maximum error * of this polynomial approximation is bounded by 2**-58.45. In * other words, * 2 4 6 8 10 12 14 * R(z) ~ Lg1*s +Lg2*s +Lg3*s +Lg4*s +Lg5*s +Lg6*s +Lg7*s * (the values of Lg1 to Lg7 are listed in the program) * and * | 2 14 | -58.45 * | Lg1*s +...+Lg7*s - R(z) | <= 2 * | | * Note that 2s = f - s*f = f - hfsq + s*hfsq, where hfsq = f*f/2. * In order to guarantee error in log below 1ulp, we compute log * by * log(1+f) = f - s*(f - R) (if f is not too large) * log(1+f) = f - (hfsq - s*(hfsq+R)). (better accuracy) * * 3. Finally, log(x) = k*ln2 + log(1+f). * = k*ln2_hi+(f-(hfsq-(s*(hfsq+R)+k*ln2_lo))) * Here ln2 is split into two floating point number: * ln2_hi + ln2_lo, * where n*ln2_hi is always exact for |n| < 2000. * * Special cases: * log(x) is NaN with signal if x < 0 (including -INF) ; * log(+INF) is +INF; log(0) is -INF with signal; * log(NaN) is that NaN with no signal. * * Accuracy: * according to an error analysis, the error is always less than * 1 ulp (unit in the last place). * * Constants: * The hexadecimal values are the intended ones for the following * constants. The decimal values may be used, provided that the * compiler will convert from decimal to binary accurately enough * to produce the hexadecimal values shown. */ double log(double x) { … } /* double log1p(double x) * * Method : * 1. Argument Reduction: find k and f such that * 1+x = 2^k * (1+f), * where sqrt(2)/2 < 1+f < sqrt(2) . * * Note. If k=0, then f=x is exact. However, if k!=0, then f * may not be representable exactly. In that case, a correction * term is need. Let u=1+x rounded. Let c = (1+x)-u, then * log(1+x) - log(u) ~ c/u. Thus, we proceed to compute log(u), * and add back the correction term c/u. * (Note: when x > 2**53, one can simply return log(x)) * * 2. Approximation of log1p(f). * Let s = f/(2+f) ; based on log(1+f) = log(1+s) - log(1-s) * = 2s + 2/3 s**3 + 2/5 s**5 + ....., * = 2s + s*R * We use a special Reme algorithm on [0,0.1716] to generate * a polynomial of degree 14 to approximate R The maximum error * of this polynomial approximation is bounded by 2**-58.45. In * other words, * 2 4 6 8 10 12 14 * R(z) ~ Lp1*s +Lp2*s +Lp3*s +Lp4*s +Lp5*s +Lp6*s +Lp7*s * (the values of Lp1 to Lp7 are listed in the program) * and * | 2 14 | -58.45 * | Lp1*s +...+Lp7*s - R(z) | <= 2 * | | * Note that 2s = f - s*f = f - hfsq + s*hfsq, where hfsq = f*f/2. * In order to guarantee error in log below 1ulp, we compute log * by * log1p(f) = f - (hfsq - s*(hfsq+R)). * * 3. Finally, log1p(x) = k*ln2 + log1p(f). * = k*ln2_hi+(f-(hfsq-(s*(hfsq+R)+k*ln2_lo))) * Here ln2 is split into two floating point number: * ln2_hi + ln2_lo, * where n*ln2_hi is always exact for |n| < 2000. * * Special cases: * log1p(x) is NaN with signal if x < -1 (including -INF) ; * log1p(+INF) is +INF; log1p(-1) is -INF with signal; * log1p(NaN) is that NaN with no signal. * * Accuracy: * according to an error analysis, the error is always less than * 1 ulp (unit in the last place). * * Constants: * The hexadecimal values are the intended ones for the following * constants. The decimal values may be used, provided that the * compiler will convert from decimal to binary accurately enough * to produce the hexadecimal values shown. * * Note: Assuming log() return accurate answer, the following * algorithm can be used to compute log1p(x) to within a few ULP: * * u = 1+x; * if(u==1.0) return x ; else * return log(u)*(x/(u-1.0)); * * See HP-15C Advanced Functions Handbook, p.193. */ double log1p(double x) { … } /* * k_log1p(f): * Return log(1+f) - f for 1+f in ~[sqrt(2)/2, sqrt(2)]. * * The following describes the overall strategy for computing * logarithms in base e. The argument reduction and adding the final * term of the polynomial are done by the caller for increased accuracy * when different bases are used. * * Method : * 1. Argument Reduction: find k and f such that * x = 2^k * (1+f), * where sqrt(2)/2 < 1+f < sqrt(2) . * * 2. Approximation of log(1+f). * Let s = f/(2+f) ; based on log(1+f) = log(1+s) - log(1-s) * = 2s + 2/3 s**3 + 2/5 s**5 + ....., * = 2s + s*R * We use a special Reme algorithm on [0,0.1716] to generate * a polynomial of degree 14 to approximate R The maximum error * of this polynomial approximation is bounded by 2**-58.45. In * other words, * 2 4 6 8 10 12 14 * R(z) ~ Lg1*s +Lg2*s +Lg3*s +Lg4*s +Lg5*s +Lg6*s +Lg7*s * (the values of Lg1 to Lg7 are listed in the program) * and * | 2 14 | -58.45 * | Lg1*s +...+Lg7*s - R(z) | <= 2 * | | * Note that 2s = f - s*f = f - hfsq + s*hfsq, where hfsq = f*f/2. * In order to guarantee error in log below 1ulp, we compute log * by * log(1+f) = f - s*(f - R) (if f is not too large) * log(1+f) = f - (hfsq - s*(hfsq+R)). (better accuracy) * * 3. Finally, log(x) = k*ln2 + log(1+f). * = k*ln2_hi+(f-(hfsq-(s*(hfsq+R)+k*ln2_lo))) * Here ln2 is split into two floating point number: * ln2_hi + ln2_lo, * where n*ln2_hi is always exact for |n| < 2000. * * Special cases: * log(x) is NaN with signal if x < 0 (including -INF) ; * log(+INF) is +INF; log(0) is -INF with signal; * log(NaN) is that NaN with no signal. * * Accuracy: * according to an error analysis, the error is always less than * 1 ulp (unit in the last place). * * Constants: * The hexadecimal values are the intended ones for the following * constants. The decimal values may be used, provided that the * compiler will convert from decimal to binary accurately enough * to produce the hexadecimal values shown. */ static const double Lg1 = …, /* 3FE55555 55555593 */ Lg2 = …, /* 3FD99999 9997FA04 */ Lg3 = …, /* 3FD24924 94229359 */ Lg4 = …, /* 3FCC71C5 1D8E78AF */ Lg5 = …, /* 3FC74664 96CB03DE */ Lg6 = …, /* 3FC39A09 D078C69F */ Lg7 = …; /* 3FC2F112 DF3E5244 */ /* * We always inline k_log1p(), since doing so produces a * substantial performance improvement (~40% on amd64). */ static inline double k_log1p(double f) { … } /* * Return the base 2 logarithm of x. See e_log.c and k_log.h for most * comments. * * This reduces x to {k, 1+f} exactly as in e_log.c, then calls the kernel, * then does the combining and scaling steps * log2(x) = (f - 0.5*f*f + k_log1p(f)) / ln2 + k * in not-quite-routine extra precision. */ double log2(double x) { … } /* * Return the base 10 logarithm of x * * Method : * Let log10_2hi = leading 40 bits of log10(2) and * log10_2lo = log10(2) - log10_2hi, * ivln10 = 1/log(10) rounded. * Then * n = ilogb(x), * if(n<0) n = n+1; * x = scalbn(x,-n); * log10(x) := n*log10_2hi + (n*log10_2lo + ivln10*log(x)) * * Note 1: * To guarantee log10(10**n)=n, where 10**n is normal, the rounding * mode must set to Round-to-Nearest. * Note 2: * [1/log(10)] rounded to 53 bits has error .198 ulps; * log10 is monotonic at all binary break points. * * Special cases: * log10(x) is NaN if x < 0; * log10(+INF) is +INF; log10(0) is -INF; * log10(NaN) is that NaN; * log10(10**N) = N for N=0,1,...,22. */ double log10(double x) { … } /* expm1(x) * Returns exp(x)-1, the exponential of x minus 1. * * Method * 1. Argument reduction: * Given x, find r and integer k such that * * x = k*ln2 + r, |r| <= 0.5*ln2 ~ 0.34658 * * Here a correction term c will be computed to compensate * the error in r when rounded to a floating-point number. * * 2. Approximating expm1(r) by a special rational function on * the interval [0,0.34658]: * Since * r*(exp(r)+1)/(exp(r)-1) = 2+ r^2/6 - r^4/360 + ... * we define R1(r*r) by * r*(exp(r)+1)/(exp(r)-1) = 2+ r^2/6 * R1(r*r) * That is, * R1(r**2) = 6/r *((exp(r)+1)/(exp(r)-1) - 2/r) * = 6/r * ( 1 + 2.0*(1/(exp(r)-1) - 1/r)) * = 1 - r^2/60 + r^4/2520 - r^6/100800 + ... * We use a special Reme algorithm on [0,0.347] to generate * a polynomial of degree 5 in r*r to approximate R1. The * maximum error of this polynomial approximation is bounded * by 2**-61. In other words, * R1(z) ~ 1.0 + Q1*z + Q2*z**2 + Q3*z**3 + Q4*z**4 + Q5*z**5 * where Q1 = -1.6666666666666567384E-2, * Q2 = 3.9682539681370365873E-4, * Q3 = -9.9206344733435987357E-6, * Q4 = 2.5051361420808517002E-7, * Q5 = -6.2843505682382617102E-9; * z = r*r, * with error bounded by * | 5 | -61 * | 1.0+Q1*z+...+Q5*z - R1(z) | <= 2 * | | * * expm1(r) = exp(r)-1 is then computed by the following * specific way which minimize the accumulation rounding error: * 2 3 * r r [ 3 - (R1 + R1*r/2) ] * expm1(r) = r + --- + --- * [--------------------] * 2 2 [ 6 - r*(3 - R1*r/2) ] * * To compensate the error in the argument reduction, we use * expm1(r+c) = expm1(r) + c + expm1(r)*c * ~ expm1(r) + c + r*c * Thus c+r*c will be added in as the correction terms for * expm1(r+c). Now rearrange the term to avoid optimization * screw up: * ( 2 2 ) * ({ ( r [ R1 - (3 - R1*r/2) ] ) } r ) * expm1(r+c)~r - ({r*(--- * [--------------------]-c)-c} - --- ) * ({ ( 2 [ 6 - r*(3 - R1*r/2) ] ) } 2 ) * ( ) * * = r - E * 3. Scale back to obtain expm1(x): * From step 1, we have * expm1(x) = either 2^k*[expm1(r)+1] - 1 * = or 2^k*[expm1(r) + (1-2^-k)] * 4. Implementation notes: * (A). To save one multiplication, we scale the coefficient Qi * to Qi*2^i, and replace z by (x^2)/2. * (B). To achieve maximum accuracy, we compute expm1(x) by * (i) if x < -56*ln2, return -1.0, (raise inexact if x!=inf) * (ii) if k=0, return r-E * (iii) if k=-1, return 0.5*(r-E)-0.5 * (iv) if k=1 if r < -0.25, return 2*((r+0.5)- E) * else return 1.0+2.0*(r-E); * (v) if (k<-2||k>56) return 2^k(1-(E-r)) - 1 (or exp(x)-1) * (vi) if k <= 20, return 2^k((1-2^-k)-(E-r)), else * (vii) return 2^k(1-((E+2^-k)-r)) * * Special cases: * expm1(INF) is INF, expm1(NaN) is NaN; * expm1(-INF) is -1, and * for finite argument, only expm1(0)=0 is exact. * * Accuracy: * according to an error analysis, the error is always less than * 1 ulp (unit in the last place). * * Misc. info. * For IEEE double * if x > 7.09782712893383973096e+02 then expm1(x) overflow * * Constants: * The hexadecimal values are the intended ones for the following * constants. The decimal values may be used, provided that the * compiler will convert from decimal to binary accurately enough * to produce the hexadecimal values shown. */ double expm1(double x) { … } double cbrt(double x) { … } /* sin(x) * Return sine function of x. * * kernel function: * __kernel_sin ... sine function on [-pi/4,pi/4] * __kernel_cos ... cose function on [-pi/4,pi/4] * __ieee754_rem_pio2 ... argument reduction routine * * Method. * Let S,C and T denote the sin, cos and tan respectively on * [-PI/4, +PI/4]. Reduce the argument x to y1+y2 = x-k*pi/2 * in [-pi/4 , +pi/4], and let n = k mod 4. * We have * * n sin(x) cos(x) tan(x) * ---------------------------------------------------------- * 0 S C T * 1 C -S -1/T * 2 -S -C T * 3 -C S -1/T * ---------------------------------------------------------- * * Special cases: * Let trig be any of sin, cos, or tan. * trig(+-INF) is NaN, with signals; * trig(NaN) is that NaN; * * Accuracy: * TRIG(x) returns trig(x) nearly rounded */ double sin(double x) { … } /* tan(x) * Return tangent function of x. * * kernel function: * __kernel_tan ... tangent function on [-pi/4,pi/4] * __ieee754_rem_pio2 ... argument reduction routine * * Method. * Let S,C and T denote the sin, cos and tan respectively on * [-PI/4, +PI/4]. Reduce the argument x to y1+y2 = x-k*pi/2 * in [-pi/4 , +pi/4], and let n = k mod 4. * We have * * n sin(x) cos(x) tan(x) * ---------------------------------------------------------- * 0 S C T * 1 C -S -1/T * 2 -S -C T * 3 -C S -1/T * ---------------------------------------------------------- * * Special cases: * Let trig be any of sin, cos, or tan. * trig(+-INF) is NaN, with signals; * trig(NaN) is that NaN; * * Accuracy: * TRIG(x) returns trig(x) nearly rounded */ double tan(double x) { … } /* * ES6 draft 09-27-13, section 20.2.2.12. * Math.cosh * Method : * mathematically cosh(x) if defined to be (exp(x)+exp(-x))/2 * 1. Replace x by |x| (cosh(x) = cosh(-x)). * 2. * [ exp(x) - 1 ]^2 * 0 <= x <= ln2/2 : cosh(x) := 1 + ------------------- * 2*exp(x) * * exp(x) + 1/exp(x) * ln2/2 <= x <= 22 : cosh(x) := ------------------- * 2 * 22 <= x <= lnovft : cosh(x) := exp(x)/2 * lnovft <= x <= ln2ovft: cosh(x) := exp(x/2)/2 * exp(x/2) * ln2ovft < x : cosh(x) := huge*huge (overflow) * * Special cases: * cosh(x) is |x| if x is +INF, -INF, or NaN. * only cosh(0)=1 is exact for finite x. */ double cosh(double x) { … } /* * ES2019 Draft 2019-01-02 12.6.4 * Math.pow & Exponentiation Operator * * Return X raised to the Yth power * * Method: * Let x = 2 * (1+f) * 1. Compute and return log2(x) in two pieces: * log2(x) = w1 + w2, * where w1 has 53-24 = 29 bit trailing zeros. * 2. Perform y*log2(x) = n+y' by simulating muti-precision * arithmetic, where |y'|<=0.5. * 3. Return x**y = 2**n*exp(y'*log2) * * Special cases: * 1. (anything) ** 0 is 1 * 2. (anything) ** 1 is itself * 3. (anything) ** NAN is NAN * 4. NAN ** (anything except 0) is NAN * 5. +-(|x| > 1) ** +INF is +INF * 6. +-(|x| > 1) ** -INF is +0 * 7. +-(|x| < 1) ** +INF is +0 * 8. +-(|x| < 1) ** -INF is +INF * 9. +-1 ** +-INF is NAN * 10. +0 ** (+anything except 0, NAN) is +0 * 11. -0 ** (+anything except 0, NAN, odd integer) is +0 * 12. +0 ** (-anything except 0, NAN) is +INF * 13. -0 ** (-anything except 0, NAN, odd integer) is +INF * 14. -0 ** (odd integer) = -( +0 ** (odd integer) ) * 15. +INF ** (+anything except 0,NAN) is +INF * 16. +INF ** (-anything except 0,NAN) is +0 * 17. -INF ** (anything) = -0 ** (-anything) * 18. (-anything) ** (integer) is (-1)**(integer)*(+anything**integer) * 19. (-anything except 0 and inf) ** (non-integer) is NAN * * Accuracy: * pow(x,y) returns x**y nearly rounded. In particular, * pow(integer, integer) always returns the correct integer provided it is * representable. * * Constants: * The hexadecimal values are the intended ones for the following * constants. The decimal values may be used, provided that the * compiler will convert from decimal to binary accurately enough * to produce the hexadecimal values shown. */ double pow(double x, double y) { … } /* * ES6 draft 09-27-13, section 20.2.2.30. * Math.sinh * Method : * mathematically sinh(x) if defined to be (exp(x)-exp(-x))/2 * 1. Replace x by |x| (sinh(-x) = -sinh(x)). * 2. * E + E/(E+1) * 0 <= x <= 22 : sinh(x) := --------------, E=expm1(x) * 2 * * 22 <= x <= lnovft : sinh(x) := exp(x)/2 * lnovft <= x <= ln2ovft: sinh(x) := exp(x/2)/2 * exp(x/2) * ln2ovft < x : sinh(x) := x*shuge (overflow) * * Special cases: * sinh(x) is |x| if x is +Infinity, -Infinity, or NaN. * only sinh(0)=0 is exact for finite x. */ double sinh(double x) { … } /* Tanh(x) * Return the Hyperbolic Tangent of x * * Method : * x -x * e - e * 0. tanh(x) is defined to be ----------- * x -x * e + e * 1. reduce x to non-negative by tanh(-x) = -tanh(x). * 2. 0 <= x < 2**-28 : tanh(x) := x with inexact if x != 0 * -t * 2**-28 <= x < 1 : tanh(x) := -----; t = expm1(-2x) * t + 2 * 2 * 1 <= x < 22 : tanh(x) := 1 - -----; t = expm1(2x) * t + 2 * 22 <= x <= INF : tanh(x) := 1. * * Special cases: * tanh(NaN) is NaN; * only tanh(0)=0 is exact for finite argument. */ double tanh(double x) { … } float powf(float x, float y) { … } float expf(float x) { … } float log10f(float x) { … } float sinf(double x) { … } float asinf(double x) { … } #undef EXTRACT_WORDS #undef GET_HIGH_WORD #undef GET_LOW_WORD #undef INSERT_WORDS #undef SET_HIGH_WORD #undef SET_LOW_WORD } // namespace fdlibm
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