Update.

[thirdparty/glibc.git] / 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.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~