/* @(#)e_jn.c 1.4 95/01/18 */
/*
* ====================================================
* 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.
* ====================================================
*/
/*
* __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 "fdlibm.h"
#ifdef __STDC__
static const double
#else
static double
#endif
invsqrtpi= 5.64189583547756279280e-01, /* 0x3FE20DD7, 0x50429B6D */
two = 2.00000000000000000000e+00, /* 0x40000000, 0x00000000 */
one = 1.00000000000000000000e+00; /* 0x3FF00000, 0x00000000 */
static double zero = 0.00000000000000000000e+00;
#ifdef __STDC__
double __ieee754_jn(int n, double x)
#else
double __ieee754_jn(n,x)
int n; double x;
#endif
{
int i,hx,ix,lx, sgn;
double a, b, temp, di;
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)
*/
hx = __HI(x);
ix = 0x7fffffff&hx;
lx = __LO(x);
/* if J(n,NaN) is NaN */
if((ix|((unsigned)(lx|-lx))>>31)>0x7ff00000) return x+x;
if(n<0){
n = -n;
x = -x;
hx ^= 0x80000000;
}
if(n==0) return(__ieee754_j0(x));
if(n==1) return(__ieee754_j1(x));
sgn = (n&1)&(hx>>31); /* even n -- 0, odd n -- sign(x) */
x = fabs(x);
if((ix|lx)==0||ix>=0x7ff00000) /* if x is 0 or inf */
b = zero;
else if((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 */
/* (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(n&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;
case 3: temp = cos(x)-sin(x); break;
}
b = invsqrtpi*temp/sqrt(x);
} else {
a = __ieee754_j0(x);
b = __ieee754_j1(x);
for(i=1;i<n;i++){
temp = b;
b = b*((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 */
b = zero;
else {
temp = x*0.5; b = temp;
for (a=one,i=2;i<=n;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,v;
double q0,q1,h,tmp; int k,m;
w = (n+n)/(double)x; h = 2.0/(double)x;
q0 = w; z = w+h; q1 = w*z - 1.0; k=1;
while(q1<1.0e9) {
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_log(fabs(v*tmp));
if(tmp<7.09782712893383973096e+02) {
for(i=n-1,di=(double)(i+i);i>0;i--){
temp = b;
b *= di;
b = b/x - a;
a = temp;
di -= two;
}
} else {
for(i=n-1,di=(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>1e100) {
a /= b;
t /= b;
b = one;
}
}
}
b = (t*__ieee754_j0(x)/b);
}
}
if(sgn==1) return -b; else return b;
}
#ifdef __STDC__
double __ieee754_yn(int n, double x)
#else
double __ieee754_yn(n,x)
int n; double x;
#endif
{
int i,hx,ix,lx;
int sign;
double a, b, temp;
hx = __HI(x);
ix = 0x7fffffff&hx;
lx = __LO(x);
/* if Y(n,NaN) is NaN */
if((ix|((unsigned)(lx|-lx))>>31)>0x7ff00000) return x+x;
if((ix|lx)==0) return -one/zero;
if(hx<0) return zero/zero;
sign = 1;
if(n<0){
n = -n;
sign = 1 - ((n&1)<<1);
}
if(n==0) return(__ieee754_y0(x));
if(n==1) return(sign*__ieee754_y1(x));
if(ix==0x7ff00000) return zero;
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(n&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;
case 3: temp = sin(x)+cos(x); break;
}
b = invsqrtpi*temp/sqrt(x);
} else {
a = __ieee754_y0(x);
b = __ieee754_y1(x);
/* quit if b is -inf */
for(i=1;i<n&&(__HI(b) != 0xfff00000);i++){
temp = b;
b = ((double)(i+i)/x)*b - a;
a = temp;
}
}
if(sign>0) return b; else return -b;
}
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