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/* origin: FreeBSD /usr/src/lib/msun/src/e_jn.c */
/*
* ====================================================
* Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
*
* Developed at SunSoft, a Sun Microsystems, Inc. business.
* Permission to use, copy, modify, and distribute this
* software is freely granted, provided that this notice
* is preserved.
* ====================================================
*/
/*
* jn(n, x), 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 is 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 "libm.h"
static const double invsqrtpi = 5.64189583547756279280e-01; /* 0x3FE20DD7, 0x50429B6D */
double jn(int n, double x)
{
uint32_t ix, lx;
int nm1, i, sign;
double a, b, temp;
EXTRACT_WORDS(ix, lx, x);
sign = ix>>31;
ix &= 0x7fffffff;
if ((ix | (lx|-lx)>>31) > 0x7ff00000) /* nan */
return x;
/* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
* Thus, J(-n,x) = J(n,-x)
*/
/* nm1 = |n|-1 is used instead of |n| to handle n==INT_MIN */
if (n == 0)
return j0(x);
if (n < 0) {
nm1 = -(n+1);
x = -x;
sign ^= 1;
} else
nm1 = n-1;
if (nm1 == 0)
return j1(x);
sign &= n; /* even n: 0, odd n: signbit(x) */
x = fabs(x);
if ((ix|lx) == 0 || ix == 0x7ff00000) /* if x is 0 or inf */
b = 0.0;
else if (nm1 < x) {
/* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
if (ix >= 0x52d00000) { /* x > 2**302 */
/* (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
*/
switch(nm1&3) {
case 0: temp = -cos(x)+sin(x); break;
case 1: temp = -cos(x)-sin(x); break;
case 2: temp = cos(x)-sin(x); break;
default:
case 3: temp = cos(x)+sin(x); break;
}
b = invsqrtpi*temp/sqrt(x);
} else {
a = j0(x);
b = j1(x);
for (i=0; i<nm1; ) {
i++;
temp = b;
b = b*(2.0*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 (nm1 > 32) /* underflow */
b = 0.0;
else {
temp = x*0.5;
b = temp;
a = 1.0;
for (i=2; i<=nm1+1; i++) {
a *= (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 */
double t,q0,q1,w,h,z,tmp,nf;
int k;
nf = nm1 + 1.0;
w = 2*nf/x;
h = 2/x;
z = w+h;
q0 = w;
q1 = w*z - 1.0;
k = 1;
while (q1 < 1.0e9) {
k += 1;
z += h;
tmp = z*q1 - q0;
q0 = q1;
q1 = tmp;
}
for (t=0.0, i=k; i>=0; i--)
t = 1/(2*(i+nf)/x - t);
a = t;
b = 1.0;
/* 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 = nf*log(fabs(w));
if (tmp < 7.09782712893383973096e+02) {
for (i=nm1; i>0; i--) {
temp = b;
b = b*(2.0*i)/x - a;
a = temp;
}
} else {
for (i=nm1; i>0; i--) {
temp = b;
b = b*(2.0*i)/x - a;
a = temp;
/* scale b to avoid spurious overflow */
if (b > 0x1p500) {
a /= b;
t /= b;
b = 1.0;
}
}
}
z = j0(x);
w = j1(x);
if (fabs(z) >= fabs(w))
b = t*z/b;
else
b = t*w/a;
}
}
return sign ? -b : b;
}
double yn(int n, double x)
{
uint32_t ix, lx, ib;
int nm1, sign, i;
double a, b, temp;
EXTRACT_WORDS(ix, lx, x);
sign = ix>>31;
ix &= 0x7fffffff;
if ((ix | (lx|-lx)>>31) > 0x7ff00000) /* nan */
return x;
if (sign && (ix|lx)!=0) /* x < 0 */
return 0/0.0;
if (ix == 0x7ff00000)
return 0.0;
if (n == 0)
return y0(x);
if (n < 0) {
nm1 = -(n+1);
sign = n&1;
} else {
nm1 = n-1;
sign = 0;
}
if (nm1 == 0)
return sign ? -y1(x) : y1(x);
if (ix >= 0x52d00000) { /* x > 2**302 */
/* (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
*/
switch(nm1&3) {
case 0: temp = -sin(x)-cos(x); break;
case 1: temp = -sin(x)+cos(x); break;
case 2: temp = sin(x)+cos(x); break;
default:
case 3: temp = sin(x)-cos(x); break;
}
b = invsqrtpi*temp/sqrt(x);
} else {
a = y0(x);
b = y1(x);
/* quit if b is -inf */
GET_HIGH_WORD(ib, b);
for (i=0; i<nm1 && ib!=0xfff00000; ){
i++;
temp = b;
b = (2.0*i/x)*b - a;
GET_HIGH_WORD(ib, b);
a = temp;
}
}
return sign ? -b : b;
}
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