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/* @(#)e_jn.c 1.4 95/01/18 */
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/*
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* ====================================================
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* Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
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*
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* Developed at SunSoft, a Sun Microsystems, Inc. business.
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* Permission to use, copy, modify, and distribute this
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* software is freely granted, provided that this notice
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* is preserved.
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* ====================================================
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*/
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/*
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* __ieee754_jn(n, x), __ieee754_yn(n, x)
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* floating point Bessel's function of the 1st and 2nd kind
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* of order n
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*
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* Special cases:
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* y0(0)=y1(0)=yn(n,0) = -inf with division by zero signal;
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* y0(-ve)=y1(-ve)=yn(n,-ve) are NaN with invalid signal.
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* Note 2. About jn(n,x), yn(n,x)
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* For n=0, j0(x) is called,
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* for n=1, j1(x) is called,
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* for n<x, forward recursion us used starting
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* from values of j0(x) and j1(x).
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* for n>x, a continued fraction approximation to
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* j(n,x)/j(n-1,x) is evaluated and then backward
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* recursion is used starting from a supposed value
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* for j(n,x). The resulting value of j(0,x) is
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* compared with the actual value to correct the
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* supposed value of j(n,x).
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*
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* yn(n,x) is similar in all respects, except
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* that forward recursion is used for all
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* values of n>1.
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*
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*/
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#include "fdlibm.h"
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#ifdef __STDC__
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static const double
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#else
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static double
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#endif
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invsqrtpi= 5.64189583547756279280e-01, /* 0x3FE20DD7, 0x50429B6D */
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two = 2.00000000000000000000e+00, /* 0x40000000, 0x00000000 */
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one = 1.00000000000000000000e+00; /* 0x3FF00000, 0x00000000 */
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static double zero = 0.00000000000000000000e+00;
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#ifdef __STDC__
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double __ieee754_jn(int n, double x)
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#else
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double __ieee754_jn(n,x)
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int n; double x;
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#endif
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{
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int i,hx,ix,lx, sgn;
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double a, b, temp, di;
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double z, w;
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/* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
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* Thus, J(-n,x) = J(n,-x)
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*/
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hx = __HI(x);
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ix = 0x7fffffff&hx;
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lx = __LO(x);
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/* if J(n,NaN) is NaN */
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if((ix|((unsigned)(lx|-lx))>>31)>0x7ff00000) return x+x;
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if(n<0){
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n = -n;
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x = -x;
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hx ^= 0x80000000;
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}
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if(n==0) return(__ieee754_j0(x));
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if(n==1) return(__ieee754_j1(x));
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sgn = (n&1)&(hx>>31); /* even n -- 0, odd n -- sign(x) */
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x = fabs(x);
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if((ix|lx)==0||ix>=0x7ff00000) /* if x is 0 or inf */
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b = zero;
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else if((double)n<=x) {
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/* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
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if(ix>=0x52D00000) { /* x > 2**302 */
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/* (x >> n**2)
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* Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
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* Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
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* Let s=sin(x), c=cos(x),
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* xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
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*
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* n sin(xn)*sqt2 cos(xn)*sqt2
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* ----------------------------------
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* 0 s-c c+s
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* 1 -s-c -c+s
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* 2 -s+c -c-s
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* 3 s+c c-s
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*/
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switch(n&3) {
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case 0: temp = cos(x)+sin(x); break;
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case 1: temp = -cos(x)+sin(x); break;
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case 2: temp = -cos(x)-sin(x); break;
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case 3: temp = cos(x)-sin(x); break;
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}
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b = invsqrtpi*temp/sqrt(x);
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} else {
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a = __ieee754_j0(x);
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b = __ieee754_j1(x);
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for(i=1;i<n;i++){
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temp = b;
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b = b*((double)(i+i)/x) - a; /* avoid underflow */
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a = temp;
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}
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}
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} else {
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if(ix<0x3e100000) { /* x < 2**-29 */
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/* x is tiny, return the first Taylor expansion of J(n,x)
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* J(n,x) = 1/n!*(x/2)^n - ...
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*/
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if(n>33) /* underflow */
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b = zero;
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else {
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temp = x*0.5; b = temp;
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for (a=one,i=2;i<=n;i++) {
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a *= (double)i; /* a = n! */
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b *= temp; /* b = (x/2)^n */
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}
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b = b/a;
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}
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} else {
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/* use backward recurrence */
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/* x x^2 x^2
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* J(n,x)/J(n-1,x) = ---- ------ ------ .....
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* 2n - 2(n+1) - 2(n+2)
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*
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* 1 1 1
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* (for large x) = ---- ------ ------ .....
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* 2n 2(n+1) 2(n+2)
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* -- - ------ - ------ -
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* x x x
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*
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* Let w = 2n/x and h=2/x, then the above quotient
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* is equal to the continued fraction:
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* 1
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* = -----------------------
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* 1
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* w - -----------------
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* 1
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* w+h - ---------
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* w+2h - ...
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*
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* To determine how many terms needed, let
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* Q(0) = w, Q(1) = w(w+h) - 1,
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* Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
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* When Q(k) > 1e4 good for single
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* When Q(k) > 1e9 good for double
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* When Q(k) > 1e17 good for quadruple
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*/
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/* determine k */
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double t,v;
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double q0,q1,h,tmp; int k,m;
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w = (n+n)/(double)x; h = 2.0/(double)x;
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q0 = w; z = w+h; q1 = w*z - 1.0; k=1;
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while(q1<1.0e9) {
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k += 1; z += h;
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tmp = z*q1 - q0;
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q0 = q1;
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q1 = tmp;
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}
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m = n+n;
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for(t=zero, i = 2*(n+k); i>=m; i -= 2) t = one/(i/x-t);
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a = t;
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b = one;
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/* estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
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* Hence, if n*(log(2n/x)) > ...
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* single 8.8722839355e+01
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* double 7.09782712893383973096e+02
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* long double 1.1356523406294143949491931077970765006170e+04
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* then recurrent value may overflow and the result is
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* likely underflow to zero
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*/
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tmp = n;
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v = two/x;
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tmp = tmp*__ieee754_log(fabs(v*tmp));
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if(tmp<7.09782712893383973096e+02) {
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for(i=n-1,di=(double)(i+i);i>0;i--){
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temp = b;
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b *= di;
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b = b/x - a;
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a = temp;
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di -= two;
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}
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} else {
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for(i=n-1,di=(double)(i+i);i>0;i--){
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temp = b;
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b *= di;
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b = b/x - a;
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a = temp;
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di -= two;
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/* scale b to avoid spurious overflow */
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if(b>1e100) {
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a /= b;
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t /= b;
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b = one;
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}
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}
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}
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b = (t*__ieee754_j0(x)/b);
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}
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}
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if(sgn==1) return -b; else return b;
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}
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#ifdef __STDC__
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double __ieee754_yn(int n, double x)
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#else
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double __ieee754_yn(n,x)
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int n; double x;
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#endif
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{
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int i,hx,ix,lx;
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int sign;
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double a, b, temp;
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hx = __HI(x);
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ix = 0x7fffffff&hx;
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lx = __LO(x);
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/* if Y(n,NaN) is NaN */
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if((ix|((unsigned)(lx|-lx))>>31)>0x7ff00000) return x+x;
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if((ix|lx)==0) return -one/zero;
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if(hx<0) return zero/zero;
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sign = 1;
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if(n<0){
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n = -n;
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sign = 1 - ((n&1)<<1);
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}
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if(n==0) return(__ieee754_y0(x));
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if(n==1) return(sign*__ieee754_y1(x));
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if(ix==0x7ff00000) return zero;
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if(ix>=0x52D00000) { /* x > 2**302 */
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/* (x >> n**2)
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* Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
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* Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
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* Let s=sin(x), c=cos(x),
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* xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
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*
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* n sin(xn)*sqt2 cos(xn)*sqt2
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* ----------------------------------
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* 0 s-c c+s
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* 1 -s-c -c+s
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* 2 -s+c -c-s
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* 3 s+c c-s
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*/
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switch(n&3) {
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case 0: temp = sin(x)-cos(x); break;
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case 1: temp = -sin(x)-cos(x); break;
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case 2: temp = -sin(x)+cos(x); break;
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case 3: temp = sin(x)+cos(x); break;
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}
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b = invsqrtpi*temp/sqrt(x);
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} else {
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a = __ieee754_y0(x);
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b = __ieee754_y1(x);
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/* quit if b is -inf */
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for(i=1;i<n&&(__HI(b) != 0xfff00000);i++){
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temp = b;
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b = ((double)(i+i)/x)*b - a;
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a = temp;
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}
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}
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if(sign>0) return b; else return -b;
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}
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