From d20d4ac2e0fd671899042a8c3bbaf3b69c93f2c8 Mon Sep 17 00:00:00 2001 From: Joseph Myers Date: Tue, 15 May 2012 10:24:47 +0000 Subject: [PATCH] Remove README.libm. --- ChangeLog | 4 + README.libm | 856 ---------------------------------------------------- 2 files changed, 4 insertions(+), 856 deletions(-) delete mode 100644 README.libm diff --git a/ChangeLog b/ChangeLog index 0d145922a7..51288eb675 100644 --- a/ChangeLog +++ b/ChangeLog @@ -1,3 +1,7 @@ +2012-05-15 Joseph Myers + + * README.libm: Remove file. + 2012-05-14 H.J. Lu * sysdeps/x86_64/start.S: Simulate popping 4-byte argument diff --git a/README.libm b/README.libm deleted file mode 100644 index f058cf846c..0000000000 --- a/README.libm +++ /dev/null @@ -1,856 +0,0 @@ -The following functions for the `long double' versions of the libm -function have to be written: - -e_acosl.c -e_asinl.c -e_atan2l.c -e_expl.c -e_fmodl.c -e_hypotl.c -e_j0l.c -e_j1l.c -e_jnl.c -e_lgammal_r.c -e_logl.c -e_log10l.c -e_powl.c -e_rem_pio2l.c -e_sinhl.c -e_sqrtl.c - -k_cosl.c -k_rem_pio2l.c -k_sinl.c -k_tanl.c - -s_atanl.c -s_erfl.c -s_expm1l.c -s_log1pl.c - -~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ - Methods -~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ -arcsin -~~~~~~ - * 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)) -~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ -arccos -~~~~~~ - * 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) -~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ -atan2 -~~~~~ - * 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, -~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ -atan -~~~~ - * 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 ) -~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ -exp -~~~ - * 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 Reme 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) -~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ -hypot -~~~~~ - * If (assume round-to-nearest) z=x*x+y*y - * has error less than sqrt(2)/2 ulp, than - * sqrt(z) has error less than 1 ulp (exercise). - * - * So, compute sqrt(x*x+y*y) with some care as - * follows to get the error below 1 ulp: - * - * Assume x>y>0; - * (if possible, set rounding to round-to-nearest) - * 1. if x > 2y use - * x1*x1+(y*y+(x2*(x+x1))) for x*x+y*y - * where x1 = x with lower 32 bits cleared, x2 = x-x1; else - * 2. if x <= 2y use - * t1*y1+((x-y)*(x-y)+(t1*y2+t2*y)) - * where t1 = 2x with lower 32 bits cleared, t2 = 2x-t1, - * y1= y with lower 32 bits chopped, y2 = y-y1. - * - * NOTE: scaling may be necessary if some argument is too - * large or too tiny -~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ -j0/y0 -~~~~~ - * Method -- j0(x): - * 1. For tiny x, we use j0(x) = 1 - x^2/4 + x^4/64 - ... - * 2. Reduce x to |x| since j0(x)=j0(-x), and - * for x in (0,2) - * j0(x) = 1-z/4+ z^2*R0/S0, where z = x*x; - * (precision: |j0-1+z/4-z^2R0/S0 |<2**-63.67 ) - * for x in (2,inf) - * j0(x) = sqrt(2/(pi*x))*(p0(x)*cos(x0)-q0(x)*sin(x0)) - * where x0 = x-pi/4. It is better to compute sin(x0),cos(x0) - * as follow: - * cos(x0) = cos(x)cos(pi/4)+sin(x)sin(pi/4) - * = 1/sqrt(2) * (cos(x) + sin(x)) - * sin(x0) = sin(x)cos(pi/4)-cos(x)sin(pi/4) - * = 1/sqrt(2) * (sin(x) - cos(x)) - * (To avoid cancellation, use - * sin(x) +- cos(x) = -cos(2x)/(sin(x) -+ cos(x)) - * to compute the worse one.) - * - * Method -- y0(x): - * 1. For x<2. - * Since - * y0(x) = 2/pi*(j0(x)*(ln(x/2)+Euler) + x^2/4 - ...) - * therefore y0(x)-2/pi*j0(x)*ln(x) is an even function. - * We use the following function to approximate y0, - * y0(x) = U(z)/V(z) + (2/pi)*(j0(x)*ln(x)), z= x^2 - * where - * U(z) = u00 + u01*z + ... + u06*z^6 - * V(z) = 1 + v01*z + ... + v04*z^4 - * with absolute approximation error bounded by 2**-72. - * Note: For tiny x, U/V = u0 and j0(x)~1, hence - * y0(tiny) = u0 + (2/pi)*ln(tiny), (choose tiny<2**-27) - * 2. For x>=2. - * y0(x) = sqrt(2/(pi*x))*(p0(x)*cos(x0)+q0(x)*sin(x0)) - * where x0 = x-pi/4. It is better to compute sin(x0),cos(x0) - * by the method mentioned above. -~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ -j1/y1 -~~~~~ - * Method -- j1(x): - * 1. For tiny x, we use j1(x) = x/2 - x^3/16 + x^5/384 - ... - * 2. Reduce x to |x| since j1(x)=-j1(-x), and - * for x in (0,2) - * j1(x) = x/2 + x*z*R0/S0, where z = x*x; - * (precision: |j1/x - 1/2 - R0/S0 |<2**-61.51 ) - * for x in (2,inf) - * j1(x) = sqrt(2/(pi*x))*(p1(x)*cos(x1)-q1(x)*sin(x1)) - * y1(x) = sqrt(2/(pi*x))*(p1(x)*sin(x1)+q1(x)*cos(x1)) - * where x1 = x-3*pi/4. It is better to compute sin(x1),cos(x1) - * as follow: - * cos(x1) = cos(x)cos(3pi/4)+sin(x)sin(3pi/4) - * = 1/sqrt(2) * (sin(x) - cos(x)) - * sin(x1) = sin(x)cos(3pi/4)-cos(x)sin(3pi/4) - * = -1/sqrt(2) * (sin(x) + cos(x)) - * (To avoid cancellation, use - * sin(x) +- cos(x) = -cos(2x)/(sin(x) -+ cos(x)) - * to compute the worse one.) - * - * Method -- y1(x): - * 1. screen out x<=0 cases: y1(0)=-inf, y1(x<0)=NaN - * 2. For x<2. - * Since - * y1(x) = 2/pi*(j1(x)*(ln(x/2)+Euler)-1/x-x/2+5/64*x^3-...) - * therefore y1(x)-2/pi*j1(x)*ln(x)-1/x is an odd function. - * We use the following function to approximate y1, - * y1(x) = x*U(z)/V(z) + (2/pi)*(j1(x)*ln(x)-1/x), z= x^2 - * where for x in [0,2] (abs err less than 2**-65.89) - * U(z) = U0[0] + U0[1]*z + ... + U0[4]*z^4 - * V(z) = 1 + v0[0]*z + ... + v0[4]*z^5 - * Note: For tiny x, 1/x dominate y1 and hence - * y1(tiny) = -2/pi/tiny, (choose tiny<2**-54) - * 3. For x>=2. - * y1(x) = sqrt(2/(pi*x))*(p1(x)*sin(x1)+q1(x)*cos(x1)) - * where x1 = x-3*pi/4. It is better to compute sin(x1),cos(x1) - * by method mentioned above. -~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ -jn/yn -~~~~~ - * Note 2. About jn(n,x), yn(n,x) - * For n=0, j0(x) is called, - * for n=1, j1(x) is called, - * for nx, a continued fraction approximation to - * j(n,x)/j(n-1,x) is evaluated and then backward - * recursion is used starting from a supposed value - * for j(n,x). The resulting value of j(0,x) is - * compared with the actual value to correct the - * supposed value of j(n,x). - * - * yn(n,x) is similar in all respects, except - * that forward recursion is used for all - * values of n>1. - -jn: - /* (x >> n**2) - * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi) - * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi) - * Let s=sin(x), c=cos(x), - * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then - * - * n sin(xn)*sqt2 cos(xn)*sqt2 - * ---------------------------------- - * 0 s-c c+s - * 1 -s-c -c+s - * 2 -s+c -c-s - * 3 s+c c-s -... - /* x is tiny, return the first Taylor expansion of J(n,x) - * J(n,x) = 1/n!*(x/2)^n - ... -... - /* use backward recurrence */ - /* x x^2 x^2 - * J(n,x)/J(n-1,x) = ---- ------ ------ ..... - * 2n - 2(n+1) - 2(n+2) - * - * 1 1 1 - * (for large x) = ---- ------ ------ ..... - * 2n 2(n+1) 2(n+2) - * -- - ------ - ------ - - * x x x - * - * Let w = 2n/x and h=2/x, then the above quotient - * is equal to the continued fraction: - * 1 - * = ----------------------- - * 1 - * w - ----------------- - * 1 - * w+h - --------- - * w+2h - ... - * - * To determine how many terms needed, let - * Q(0) = w, Q(1) = w(w+h) - 1, - * Q(k) = (w+k*h)*Q(k-1) - Q(k-2), - * When Q(k) > 1e4 good for single - * When Q(k) > 1e9 good for double - * When Q(k) > 1e17 good for quadruple - -... - /* estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n) - * Hence, if n*(log(2n/x)) > ... - * single 8.8722839355e+01 - * double 7.09782712893383973096e+02 - * long double 1.1356523406294143949491931077970765006170e+04 - * then recurrent value may overflow and the result is - * likely underflow to zero - -yn: - /* (x >> n**2) - * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi) - * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi) - * Let s=sin(x), c=cos(x), - * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then - * - * n sin(xn)*sqt2 cos(xn)*sqt2 - * ---------------------------------- - * 0 s-c c+s - * 1 -s-c -c+s - * 2 -s+c -c-s - * 3 s+c c-s -~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ -lgamma -~~~~~~ - * Method: - * 1. Argument Reduction for 0 < x <= 8 - * Since gamma(1+s)=s*gamma(s), for x in [0,8], we may - * reduce x to a number in [1.5,2.5] by - * lgamma(1+s) = log(s) + lgamma(s) - * for example, - * lgamma(7.3) = log(6.3) + lgamma(6.3) - * = log(6.3*5.3) + lgamma(5.3) - * = log(6.3*5.3*4.3*3.3*2.3) + lgamma(2.3) - * 2. Polynomial approximation of lgamma around its - * minimun ymin=1.461632144968362245 to maintain monotonicity. - * On [ymin-0.23, ymin+0.27] (i.e., [1.23164,1.73163]), use - * Let z = x-ymin; - * lgamma(x) = -1.214862905358496078218 + z^2*poly(z) - * where - * poly(z) is a 14 degree polynomial. - * 2. Rational approximation in the primary interval [2,3] - * We use the following approximation: - * s = x-2.0; - * lgamma(x) = 0.5*s + s*P(s)/Q(s) - * with accuracy - * |P/Q - (lgamma(x)-0.5s)| < 2**-61.71 - * Our algorithms are based on the following observation - * - * zeta(2)-1 2 zeta(3)-1 3 - * lgamma(2+s) = s*(1-Euler) + --------- * s - --------- * s + ... - * 2 3 - * - * where Euler = 0.5771... is the Euler constant, which is very - * close to 0.5. - * - * 3. For x>=8, we have - * lgamma(x)~(x-0.5)log(x)-x+0.5*log(2pi)+1/(12x)-1/(360x**3)+.... - * (better formula: - * lgamma(x)~(x-0.5)*(log(x)-1)-.5*(log(2pi)-1) + ...) - * Let z = 1/x, then we approximation - * f(z) = lgamma(x) - (x-0.5)(log(x)-1) - * by - * 3 5 11 - * w = w0 + w1*z + w2*z + w3*z + ... + w6*z - * where - * |w - f(z)| < 2**-58.74 - * - * 4. For negative x, since (G is gamma function) - * -x*G(-x)*G(x) = pi/sin(pi*x), - * we have - * G(x) = pi/(sin(pi*x)*(-x)*G(-x)) - * since G(-x) is positive, sign(G(x)) = sign(sin(pi*x)) for x<0 - * Hence, for x<0, signgam = sign(sin(pi*x)) and - * lgamma(x) = log(|Gamma(x)|) - * = log(pi/(|x*sin(pi*x)|)) - lgamma(-x); - * Note: one should avoid compute pi*(-x) directly in the - * computation of sin(pi*(-x)). -~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ -log -~~~ - * 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. -~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ -log10 -~~~~~ - * 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. -~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ -pow -~~~ - * 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 -~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ -rem_pio2 return the remainder of x rem pi/2 in y[0]+y[1] -~~~~~~~~ -This is one of the basic functions which is written with highest accuracy -in mind. -~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ -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) -~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ -sqrt -~~~~ - * Method: - * Bit by bit method using integer arithmetic. (Slow, but portable) - * 1. Normalization - * Scale x to y in [1,4) with even powers of 2: - * find an integer k such that 1 <= (y=x*2^(-2k)) < 4, then - * sqrt(x) = 2^k * sqrt(y) - * 2. Bit by bit computation - * Let q = sqrt(y) truncated to i bit after binary point (q = 1), - * i 0 - * i+1 2 - * s = 2*q , and y = 2 * ( y - q ). (1) - * i i i i - * - * To compute q from q , one checks whether - * i+1 i - * - * -(i+1) 2 - * (q + 2 ) <= y. (2) - * i - * -(i+1) - * If (2) is false, then q = q ; otherwise q = q + 2 . - * i+1 i i+1 i - * - * With some algebric manipulation, it is not difficult to see - * that (2) is equivalent to - * -(i+1) - * s + 2 <= y (3) - * i i - * - * The advantage of (3) is that s and y can be computed by - * i i - * the following recurrence formula: - * if (3) is false - * - * s = s , y = y ; (4) - * i+1 i i+1 i - * - * otherwise, - * -i -(i+1) - * s = s + 2 , y = y - s - 2 (5) - * i+1 i i+1 i i - * - * One may easily use induction to prove (4) and (5). - * Note. Since the left hand side of (3) contain only i+2 bits, - * it does not necessary to do a full (53-bit) comparison - * in (3). - * 3. Final rounding - * After generating the 53 bits result, we compute one more bit. - * Together with the remainder, we can decide whether the - * result is exact, bigger than 1/2ulp, or less than 1/2ulp - * (it will never equal to 1/2ulp). - * The rounding mode can be detected by checking whether - * huge + tiny is equal to huge, and whether huge - tiny is - * equal to huge for some floating point number "huge" and "tiny". -~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ -cos -~~~ - * 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. -~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ -sin -~~~ - * 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)) -~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ -tan -~~~ - * 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))) -~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ -atan -~~~~ - * 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 ) -~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ -erf -~~~ - * x - * 2 |\ - * erf(x) = --------- | exp(-t*t)dt - * sqrt(pi) \| - * 0 - * - * erfc(x) = 1-erf(x) - * Note that - * erf(-x) = -erf(x) - * erfc(-x) = 2 - erfc(x) - * - * Method: - * 1. For |x| in [0, 0.84375] - * erf(x) = x + x*R(x^2) - * erfc(x) = 1 - erf(x) if x in [-.84375,0.25] - * = 0.5 + ((0.5-x)-x*R) if x in [0.25,0.84375] - * where R = P/Q where P is an odd poly of degree 8 and - * Q is an odd poly of degree 10. - * -57.90 - * | R - (erf(x)-x)/x | <= 2 - * - * - * Remark. The formula is derived by noting - * erf(x) = (2/sqrt(pi))*(x - x^3/3 + x^5/10 - x^7/42 + ....) - * and that - * 2/sqrt(pi) = 1.128379167095512573896158903121545171688 - * is close to one. The interval is chosen because the fix - * point of erf(x) is near 0.6174 (i.e., erf(x)=x when x is - * near 0.6174), and by some experiment, 0.84375 is chosen to - * guarantee the error is less than one ulp for erf. - * - * 2. For |x| in [0.84375,1.25], let s = |x| - 1, and - * c = 0.84506291151 rounded to single (24 bits) - * erf(x) = sign(x) * (c + P1(s)/Q1(s)) - * erfc(x) = (1-c) - P1(s)/Q1(s) if x > 0 - * 1+(c+P1(s)/Q1(s)) if x < 0 - * |P1/Q1 - (erf(|x|)-c)| <= 2**-59.06 - * Remark: here we use the taylor series expansion at x=1. - * erf(1+s) = erf(1) + s*Poly(s) - * = 0.845.. + P1(s)/Q1(s) - * That is, we use rational approximation to approximate - * erf(1+s) - (c = (single)0.84506291151) - * Note that |P1/Q1|< 0.078 for x in [0.84375,1.25] - * where - * P1(s) = degree 6 poly in s - * Q1(s) = degree 6 poly in s - * - * 3. For x in [1.25,1/0.35(~2.857143)], - * erfc(x) = (1/x)*exp(-x*x-0.5625+R1/S1) - * erf(x) = 1 - erfc(x) - * where - * R1(z) = degree 7 poly in z, (z=1/x^2) - * S1(z) = degree 8 poly in z - * - * 4. For x in [1/0.35,28] - * erfc(x) = (1/x)*exp(-x*x-0.5625+R2/S2) if x > 0 - * = 2.0 - (1/x)*exp(-x*x-0.5625+R2/S2) if -6 x >= 28 - * erf(x) = sign(x) *(1 - tiny) (raise inexact) - * erfc(x) = tiny*tiny (raise underflow) if x > 0 - * = 2 - tiny if x<0 -~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ -expm1 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; - * (where z=r*r, and the values of Q1 to Q5 are listed below) - * 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. -~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ -log1p -~~~~~ - * 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. -~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~