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- /* @(#)e_jn.c 5.1 93/09/24 */
- /*
- * ====================================================
- * 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.
- * ====================================================
- */
- #if defined(LIBM_SCCS) && !defined(lint)
- static char rcsid[] = "$NetBSD: e_jn.c,v 1.9 1995/05/10 20:45:34 jtc Exp $";
- #endif
- /*
- * __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 double
- #else
- static double
- #endif
- invsqrtpi= 5.64189583547756279280e-01, /* 0x3FE20DD7, 0x50429B6D */
- two = 2.00000000000000000000e+00, /* 0x40000000, 0x00000000 */
- one = 1.00000000000000000000e+00; /* 0x3FF00000, 0x00000000 */
- #ifdef __STDC__
- static const double zero = 0.00000000000000000000e+00;
- #else
- static double zero = 0.00000000000000000000e+00;
- #endif
- #ifdef __STDC__
- double __ieee754_jn(int n, double x)
- #else
- double __ieee754_jn(n,x)
- int n; double x;
- #endif
- {
- int32_t 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)
- */
- EXTRACT_WORDS(hx,lx,x);
- ix = 0x7fffffff&hx;
- /* if J(n,NaN) is NaN */
- if((ix|((u_int32_t)(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; int32_t 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
- {
- int32_t i,hx,ix,lx;
- int32_t sign;
- double a, b, temp;
- EXTRACT_WORDS(hx,lx,x);
- ix = 0x7fffffff&hx;
- /* if Y(n,NaN) is NaN */
- if((ix|((u_int32_t)(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 {
- u_int32_t high;
- a = __ieee754_y0(x);
- b = __ieee754_y1(x);
- /* quit if b is -inf */
- GET_HIGH_WORD(high,b);
- for(i=1;i<n&&high!=0xfff00000;i++){
- temp = b;
- b = ((double)(i+i)/x)*b - a;
- GET_HIGH_WORD(high,b);
- a = temp;
- }
- }
- if(sign>0) return b; else return -b;
- }
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