Edit File: README.libm
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. ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Back to File Manager