glibc/sysdeps/ieee754/ldbl-96/e_jnl.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 long double contributed by
Stephen L. Moshier <moshier@na-net.ornl.gov> */
/*
* __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"
#ifdef __STDC__
static const long double
#else
static long double
#endif
invsqrtpi = 5.64189583547756286948079e-1L, two = 2.0e0L, one = 1.0e0L;
#ifdef __STDC__
static const long double zero = 0.0L;
#else
static long double zero = 0.0L;
#endif
#ifdef __STDC__
long double
__ieee754_jnl (int n, long double x)
#else
long double
__ieee754_jnl (n, x)
int n;
long double x;
#endif
{
u_int32_t se, i0, i1;
int32_t i, ix, sgn;
long double a, b, temp, di;
long double z, w;
/* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
* Thus, J(-n,x) = J(n,-x)
*/
GET_LDOUBLE_WORDS (se, i0, i1, x);
ix = se & 0x7fff;
/* if J(n,NaN) is NaN */
if ((ix == 0x7fff) && ((i0 & 0x7fffffff) != 0))
return x + x;
if (n < 0)
{
n = -n;
x = -x;
se ^= 0x8000;
}
if (n == 0)
return (__ieee754_j0l (x));
if (n == 1)
return (__ieee754_j1l (x));
sgn = (n & 1) & (se >> 15); /* even n -- 0, odd n -- sign(x) */
x = fabsl (x);
if ((ix | i0 | i1) == 0 || ix >= 0x7fff) /* 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 >= 0x412D)
{ /* x > 2**302 */
/* ??? This might be a futile gesture.
If x exceeds X_TLOSS anyway, the wrapper function
will set the result to zero. */
/* (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 < 0x3fde)
{ /* x < 2**-33 */
/* 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 *= (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 < 1.0e11L)
{
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;
}
}
}
b = (t * __ieee754_j0l (x) / b);
}
}
if (sgn == 1)
return -b;
else
return b;
}
#ifdef __STDC__
long double
__ieee754_ynl (int n, long double x)
#else
long double
__ieee754_ynl (n, x)
int n;
long double x;
#endif
{
u_int32_t se, i0, i1;
int32_t i, ix;
int32_t sign;
long double a, b, temp;
GET_LDOUBLE_WORDS (se, i0, i1, x);
ix = se & 0x7fff;
/* if Y(n,NaN) is NaN */
if ((ix == 0x7fff) && ((i0 & 0x7fffffff) != 0))
return x + x;
if ((ix | i0 | i1) == 0)
return -one / zero;
if (se & 0x8000)
return zero / zero;
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 == 0x7fff)
return zero;
if (ix >= 0x412D)
{ /* 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 */
GET_LDOUBLE_WORDS (se, i0, i1, b);
for (i = 1; i < n && se != 0xffff; i++)
{
temp = b;
b = ((long double) (i + i) / x) * b - a;
GET_LDOUBLE_WORDS (se, i0, i1, b);
a = temp;
}
}
if (sign > 0)
return b;
else
return -b;
}