| Current Path : /compat/linux/proc/68247/root/compat/linux/proc/68247/root/compat/linux/usr/share/doc/glibc-2.9/ |
FreeBSD hs32.drive.ne.jp 9.1-RELEASE FreeBSD 9.1-RELEASE #1: Wed Jan 14 12:18:08 JST 2015 root@hs32.drive.ne.jp:/sys/amd64/compile/hs32 amd64 |
| Current File : //compat/linux/proc/68247/root/compat/linux/proc/68247/root/compat/linux/usr/share/doc/glibc-2.9/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.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~