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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>
and are incorporated herein by permission of the author. The author
reserves the right to distribute this material elsewhere under different
copying permissions. These modifications are distributed here under
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, see
<http://www.gnu.org/licenses/>. */
/*
* __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.
*
*/
#include "math.h"
#include "math_private.h"
static const long double
invsqrtpi = 5.6418958354775628694807945156077258584405E-1L,
two = 2.0e0L,
one = 1.0e0L,
zero = 0.0L;
long double
__ieee754_jnl (int n, long double x)
{
u_int32_t se;
int32_t i, ix, sgn;
long double a, b, temp, di;
long double z, w;
ieee854_long_double_shape_type 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.parts32.w0;
ix = se & 0x7fffffff;
/* if J(n,NaN) is NaN */
if (ix >= 0x7ff00000)
{
if ((u.parts32.w0 & 0xfffff) | u.parts32.w1
| (u.parts32.w2 & 0x7fffffff) | u.parts32.w3)
return x + x;
}
if (n < 0)
{
n = -n;
x = -x;
se ^= 0x80000000;
}
if (n == 0)
return (__ieee754_j0l (x));
if (n == 1)
return (__ieee754_j1l (x));
sgn = (n & 1) & (se >> 31); /* even n -- 0, odd n -- sign(x) */
x = fabsl (x);
if (x == 0.0L || ix >= 0x7ff00000) /* if x is 0 or inf */
b = zero;
else if ((long double) n <= x)
{
/* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
if (ix >= 0x52d00000)
{ /* 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
*/
long double s;
long double c;
__sincosl (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 / __ieee754_sqrtl (x);
}
else
{
a = __ieee754_j0l (x);
b = __ieee754_j1l (x);
for (i = 1; i < n; i++)
{
temp = b;
b = b * ((long double) (i + i) / x) - a; /* avoid underflow */
a = temp;
}
}
}
else
{
if (ix < 0x3e100000)
{ /* x < 2**-29 */
/* x is tiny, return the first Taylor expansion of J(n,x)
* J(n,x) = 1/n!*(x/2)^n - ...
*/
if (n >= 33) /* underflow, result < 10^-300 */
b = zero;
else
{
temp = x * 0.5;
b = temp;
for (a = one, i = 2; i <= n; i++)
{
a *= (long double) 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 */
long double t, v;
long double q0, q1, h, tmp;
int32_t k, m;
w = (n + n) / (long double) x;
h = 2.0L / (long double) x;
q0 = w;
z = w + h;
q1 = w * z - 1.0L;
k = 1;
while (q1 < 1.0e17L)
{
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
* long double 1.1356523406294143949491931077970765006170e+04
* then recurrent value may overflow and the result is
* likely underflow to zero
*/
tmp = n;
v = two / x;
tmp = tmp * __ieee754_logl (fabsl (v * tmp));
if (tmp < 1.1356523406294143949491931077970765006170e+04L)
{
for (i = n - 1, di = (long double) (i + i); i > 0; i--)
{
temp = b;
b *= di;
b = b / x - a;
a = temp;
di -= two;
}
}
else
{
for (i = n - 1, di = (long double) (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 > 1e100L)
{
a /= b;
t /= b;
b = one;
}
}
}
/* 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 = __ieee754_j0l (x);
w = __ieee754_j1l (x);
if (fabsl (z) >= fabsl (w))
b = (t * z / b);
else
b = (t * w / a);
}
}
if (sgn == 1)
return -b;
else
return b;
}
strong_alias (__ieee754_jnl, __jnl_finite)
long double
__ieee754_ynl (int n, long double x)
{
u_int32_t se;
int32_t i, ix;
int32_t sign;
long double a, b, temp;
ieee854_long_double_shape_type u;
u.value = x;
se = u.parts32.w0;
ix = se & 0x7fffffff;
/* if Y(n,NaN) is NaN */
if (ix >= 0x7ff00000)
{
if ((u.parts32.w0 & 0xfffff) | u.parts32.w1
| (u.parts32.w2 & 0x7fffffff) | u.parts32.w3)
return x + x;
}
if (x <= 0.0L)
{
if (x == 0.0L)
return -HUGE_VALL + x;
if (se & 0x80000000)
return zero / (zero * x);
}
sign = 1;
if (n < 0)
{
n = -n;
sign = 1 - ((n & 1) << 1);
}
if (n == 0)
return (__ieee754_y0l (x));
if (n == 1)
return (sign * __ieee754_y1l (x));
if (ix >= 0x7ff00000)
return zero;
if (ix >= 0x52D00000)
{ /* 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
*/
long double s;
long double c;
__sincosl (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 / __ieee754_sqrtl (x);
}
else
{
a = __ieee754_y0l (x);
b = __ieee754_y1l (x);
/* quit if b is -inf */
u.value = b;
se = u.parts32.w0 & 0xfff00000;
for (i = 1; i < n && se != 0xfff00000; i++)
{
temp = b;
b = ((long double) (i + i) / x) * b - a;
u.value = b;
se = u.parts32.w0 & 0xfff00000;
a = temp;
}
}
if (sign > 0)
return b;
else
return -b;
}
strong_alias (__ieee754_ynl, __ynl_finite)
|
