Retro68/gcc/libquadmath/math/jnq.c

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/*
* ====================================================
* Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
*
* Developed at SunPro, a Sun Microsystems, Inc. business.
* Permission to use, copy, modify, and distribute this
* software is freely granted, provided that this notice
* is preserved.
* ====================================================
*/
/* Modifications for 128-bit long double are
Copyright (C) 2001 Stephen L. Moshier <moshier@na-net.ornl.gov>
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and are incorporated herein by permission of the author. The author
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reserves the right to distribute this material elsewhere under different
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copying permissions. These modifications are distributed here under
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the following terms:
This library is free software; you can redistribute it and/or
modify it under the terms of the GNU Lesser General Public
License as published by the Free Software Foundation; either
version 2.1 of the License, or (at your option) any later version.
This library is distributed in the hope that it will be useful,
but WITHOUT ANY WARRANTY; without even the implied warranty of
MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU
Lesser General Public License for more details.
You should have received a copy of the GNU Lesser General Public
License along with this library; if not, write to the Free Software
Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA */
/*
* __ieee754_jn(n, x), __ieee754_yn(n, x)
* floating point Bessel's function of the 1st and 2nd kind
* of order n
*
* Special cases:
* y0(0)=y1(0)=yn(n,0) = -inf with division by zero signal;
* y0(-ve)=y1(-ve)=yn(n,-ve) are NaN with invalid signal.
* 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.
*
*/
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#include <errno.h>
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#include "quadmath-imp.h"
static const __float128
invsqrtpi = 5.6418958354775628694807945156077258584405E-1Q,
two = 2.0e0Q,
one = 1.0e0Q,
zero = 0.0Q;
__float128
jnq (int n, __float128 x)
{
uint32_t se;
int32_t i, ix, sgn;
__float128 a, b, temp, di;
__float128 z, w;
ieee854_float128 u;
/* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
* Thus, J(-n,x) = J(n,-x)
*/
u.value = x;
se = u.words32.w0;
ix = se & 0x7fffffff;
/* if J(n,NaN) is NaN */
if (ix >= 0x7fff0000)
{
if ((u.words32.w0 & 0xffff) | u.words32.w1 | u.words32.w2 | u.words32.w3)
return x + x;
}
if (n < 0)
{
n = -n;
x = -x;
se ^= 0x80000000;
}
if (n == 0)
return (j0q (x));
if (n == 1)
return (j1q (x));
sgn = (n & 1) & (se >> 31); /* even n -- 0, odd n -- sign(x) */
x = fabsq (x);
if (x == 0.0Q || ix >= 0x7fff0000) /* if x is 0 or inf */
b = zero;
else if ((__float128) n <= x)
{
/* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
if (ix >= 0x412D0000)
{ /* x > 2**302 */
/* ??? Could use an expansion for large x here. */
/* (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
*/
__float128 s;
__float128 c;
sincosq (x, &s, &c);
switch (n & 3)
{
case 0:
temp = c + s;
break;
case 1:
temp = -c + s;
break;
case 2:
temp = -c - s;
break;
case 3:
temp = c - s;
break;
}
b = invsqrtpi * temp / sqrtq (x);
}
else
{
a = j0q (x);
b = j1q (x);
for (i = 1; i < n; i++)
{
temp = b;
b = b * ((__float128) (i + i) / x) - a; /* avoid underflow */
a = temp;
}
}
}
else
{
if (ix < 0x3fc60000)
{ /* x < 2**-57 */
/* x is tiny, return the first Taylor expansion of J(n,x)
* J(n,x) = 1/n!*(x/2)^n - ...
*/
if (n >= 400) /* underflow, result < 10^-4952 */
b = zero;
else
{
temp = x * 0.5;
b = temp;
for (a = one, i = 2; i <= n; i++)
{
a *= (__float128) i; /* a = n! */
b *= temp; /* b = (x/2)^n */
}
b = b / a;
}
}
else
{
/* 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
*/
/* determine k */
__float128 t, v;
__float128 q0, q1, h, tmp;
int32_t k, m;
w = (n + n) / (__float128) x;
h = 2.0Q / (__float128) x;
q0 = w;
z = w + h;
q1 = w * z - 1.0Q;
k = 1;
while (q1 < 1.0e17Q)
{
k += 1;
z += h;
tmp = z * q1 - q0;
q0 = q1;
q1 = tmp;
}
m = n + n;
for (t = zero, i = 2 * (n + k); i >= m; i -= 2)
t = one / (i / x - t);
a = t;
b = one;
/* 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
* __float128 1.1356523406294143949491931077970765006170e+04
* then recurrent value may overflow and the result is
* likely underflow to zero
*/
tmp = n;
v = two / x;
tmp = tmp * logq (fabsq (v * tmp));
if (tmp < 1.1356523406294143949491931077970765006170e+04Q)
{
for (i = n - 1, di = (__float128) (i + i); i > 0; i--)
{
temp = b;
b *= di;
b = b / x - a;
a = temp;
di -= two;
}
}
else
{
for (i = n - 1, di = (__float128) (i + i); i > 0; i--)
{
temp = b;
b *= di;
b = b / x - a;
a = temp;
di -= two;
/* scale b to avoid spurious overflow */
if (b > 1e100Q)
{
a /= b;
t /= b;
b = one;
}
}
}
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/* j0() and j1() suffer enormous loss of precision at and
* near zero; however, we know that their zero points never
* coincide, so just choose the one further away from zero.
*/
z = j0q (x);
w = j1q (x);
if (fabsq (z) >= fabsq (w))
b = (t * z / b);
else
b = (t * w / a);
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}
}
if (sgn == 1)
return -b;
else
return b;
}
__float128
ynq (int n, __float128 x)
{
uint32_t se;
int32_t i, ix;
int32_t sign;
__float128 a, b, temp;
ieee854_float128 u;
u.value = x;
se = u.words32.w0;
ix = se & 0x7fffffff;
/* if Y(n,NaN) is NaN */
if (ix >= 0x7fff0000)
{
if ((u.words32.w0 & 0xffff) | u.words32.w1 | u.words32.w2 | u.words32.w3)
return x + x;
}
if (x <= 0.0Q)
{
if (x == 0.0Q)
return -HUGE_VALQ + x;
if (se & 0x80000000)
return zero / (zero * x);
}
sign = 1;
if (n < 0)
{
n = -n;
sign = 1 - ((n & 1) << 1);
}
if (n == 0)
return (y0q (x));
if (n == 1)
return (sign * y1q (x));
if (ix >= 0x7fff0000)
return zero;
if (ix >= 0x412D0000)
{ /* x > 2**302 */
/* ??? See comment above on the possible futility of this. */
/* (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
*/
__float128 s;
__float128 c;
sincosq (x, &s, &c);
switch (n & 3)
{
case 0:
temp = s - c;
break;
case 1:
temp = -s - c;
break;
case 2:
temp = -s + c;
break;
case 3:
temp = s + c;
break;
}
b = invsqrtpi * temp / sqrtq (x);
}
else
{
a = y0q (x);
b = y1q (x);
/* quit if b is -inf */
u.value = b;
se = u.words32.w0 & 0xffff0000;
for (i = 1; i < n && se != 0xffff0000; i++)
{
temp = b;
b = ((__float128) (i + i) / x) * b - a;
u.value = b;
se = u.words32.w0 & 0xffff0000;
a = temp;
}
}
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/* If B is +-Inf, set up errno accordingly. */
if (! finiteq (b))
errno = ERANGE;
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if (sign > 0)
return b;
else
return -b;
}