* Re: A patch for libm-ieee754
1998-02-22 13:03 ` Toon Moene
1998-02-22 13:33 ` Ulrich Drepper
@ 1998-02-22 13:42 ` Andreas Jaeger
1 sibling, 0 replies; 9+ messages in thread
From: Andreas Jaeger @ 1998-02-22 13:42 UTC (permalink / raw)
To: Toon Moene; +Cc: egcs, GNU C Library
>>>>> Toon Moene writes:
Toon> Andreas wrote:
>>>>> H J Lu writes:
>> Hi,
>> It turns out those libm-ieee754 bugs are not in egcs. Here is
Toon> the patch
>> for glibc 2.1 to fix a few libm-ieee754 bugs. Ulrich, could you please
>> take a look?
>> I've added all your math patches (and removed
>> isless/islessequal i486 macros) and it's really an
>> improvement
Toon> Andreas, would it be useful to mobilise some people to help do the
Toon> math related stuff in glibc-2.x ?
It would be better to ask Ulrich Drepper for details since he should
know the situation better than I do. He might add some comments to
the following personal view of the issue.
Toon> I might be able to stirr some interest in the group that did
Toon> `enquire.c' originally (CWI - het Centrum voor Wiskunde en
Toon> Informatica in Amsterdam).
Toon> If you think that would be useful, could you outline the current
Toon> problem areas in glibc ?
The problems I know of are:
- The libm-ieee `gamma' function needs to be reimplemented. At the
moment the function just calls exp(lgamma) :-(.
- A number of functions need to be implemented as `long double'
versions, I'm appending below the document README.libm which gives a
description.
- The libm-ieee `scalb' function gives wrong results for
non-integral second parameters (Please note that scalb is defined as
scalb (double x, double n)).
- We've got a test suite that just tests some single values (only 2152
at the moment :-), e.g. behaviour for NaN, Inf, 0, some defined
properties (e.g. sin (0), exp(1)) and some random values. The
current libm passes all tests. I hope that we caught most of the
problems - but better tests might be added. For some functions
(exp, exp2, sin, cos) the taylor polynomials are calculated with GNU
MP for a range of number and checked against the functions in libm.
It might be a good idea to add also such tests for other functions.
- Implement optimized complex math functions in assembler. Uli has
implemented most of these functions in C and states:
> Beside this most of the complex math functions which are new in
> ISO C 9X should be improved. Writing some of them in assembler is
> useful to exploit the parallelism which often is available.
If you want to start you might take the current glibc 2.1 snapshot as
basis since it contains the full ISO C9X set of math functions and
also more tests than glibc 2.0.x.
Andreas
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
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.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
--
Andreas Jaeger aj@arthur.rhein-neckar.de jaeger@informatik.uni-kl.de
for pgp-key finger ajaeger@alma.student.uni-kl.de
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