lx = __LO(x);
lx = __LO(u);
	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 = fd_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 =  fd_cos(x)+fd_sin(x); break;		    case 1: temp = -fd_cos(x)+fd_sin(x); break;		    case 2: temp = -fd_cos(x)-fd_sin(x); break;		    case 3: temp =  fd_cos(x)-fd_sin(x); break;		}		b = invsqrtpi*temp/fd_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(fd_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;}