| 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 n<x, forward recursion us used starting |
| * from values of j0(x) and j1(x). |
| * for n>x, 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<0 |
| * = 2.0 - tiny (if x <= -6) |
| * erf(x) = sign(x)*(1.0 - erfc(x)) if x < 6, else |
| * erf(x) = sign(x)*(1.0 - tiny) |
| * where |
| * R2(z) = degree 6 poly in z, (z=1/x^2) |
| * S2(z) = degree 7 poly in z |
| * |
| * Note1: |
| * To compute exp(-x*x-0.5625+R/S), let s be a single |
| * precision number and s := x; then |
| * -x*x = -s*s + (s-x)*(s+x) |
| * exp(-x*x-0.5626+R/S) = |
| * exp(-s*s-0.5625)*exp((s-x)*(s+x)+R/S); |
| * Note2: |
| * Here 4 and 5 make use of the asymptotic series |
| * exp(-x*x) |
| * erfc(x) ~ ---------- * ( 1 + Poly(1/x^2) ) |
| * x*sqrt(pi) |
| * We use rational approximation to approximate |
| * g(s)=f(1/x^2) = log(erfc(x)*x) - x*x + 0.5625 |
| * Here is the error bound for R1/S1 and R2/S2 |
| * |R1/S1 - f(x)| < 2**(-62.57) |
| * |R2/S2 - f(x)| < 2**(-61.52) |
| * |
| * 5. For inf > 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. |
| ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ |
