1 /* origin: FreeBSD /usr/src/lib/msun/src/e_jn.c */
3 * ====================================================
4 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
6 * Developed at SunSoft, a Sun Microsystems, Inc. business.
7 * Permission to use, copy, modify, and distribute this
8 * software is freely granted, provided that this notice
10 * ====================================================
14 * floating point Bessel's function of the 1st and 2nd kind
18 * y0(0)=y1(0)=yn(n,0) = -inf with division by zero signal;
19 * y0(-ve)=y1(-ve)=yn(n,-ve) are NaN with invalid signal.
20 * Note 2. About jn(n,x), yn(n,x)
21 * For n=0, j0(x) is called,
22 * for n=1, j1(x) is called,
23 * for n<x, forward recursion us used starting
24 * from values of j0(x) and j1(x).
25 * for n>x, a continued fraction approximation to
26 * j(n,x)/j(n-1,x) is evaluated and then backward
27 * recursion is used starting from a supposed value
28 * for j(n,x). The resulting value of j(0,x) is
29 * compared with the actual value to correct the
30 * supposed value of j(n,x).
32 * yn(n,x) is similar in all respects, except
33 * that forward recursion is used for all
40 static const double invsqrtpi = 5.64189583547756279280e-01; /* 0x3FE20DD7, 0x50429B6D */
42 double jn(int n, double x)
44 int32_t i,hx,ix,lx,sgn;
45 double a, b, temp, di;
48 /* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
49 * Thus, J(-n,x) = J(n,-x)
51 EXTRACT_WORDS(hx, lx, x);
53 /* if J(n,NaN) is NaN */
54 if ((ix|((uint32_t)(lx|-lx))>>31) > 0x7ff00000)
61 if (n == 0) return j0(x);
62 if (n == 1) return j1(x);
64 sgn = (n&1)&(hx>>31); /* even n -- 0, odd n -- sign(x) */
66 if ((ix|lx) == 0 || ix >= 0x7ff00000) /* if x is 0 or inf */
68 else if ((double)n <= x) {
69 /* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
70 if (ix >= 0x52D00000) { /* x > 2**302 */
72 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
73 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
74 * Let s=sin(x), c=cos(x),
75 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
77 * n sin(xn)*sqt2 cos(xn)*sqt2
78 * ----------------------------------
85 case 0: temp = cos(x)+sin(x); break;
86 case 1: temp = -cos(x)+sin(x); break;
87 case 2: temp = -cos(x)-sin(x); break;
88 case 3: temp = cos(x)-sin(x); break;
90 b = invsqrtpi*temp/sqrt(x);
96 b = b*((double)(i+i)/x) - a; /* avoid underflow */
101 if (ix < 0x3e100000) { /* x < 2**-29 */
102 /* x is tiny, return the first Taylor expansion of J(n,x)
103 * J(n,x) = 1/n!*(x/2)^n - ...
105 if (n > 33) /* underflow */
110 for (a=1.0,i=2; i<=n; i++) {
111 a *= (double)i; /* a = n! */
112 b *= temp; /* b = (x/2)^n */
117 /* use backward recurrence */
119 * J(n,x)/J(n-1,x) = ---- ------ ------ .....
120 * 2n - 2(n+1) - 2(n+2)
123 * (for large x) = ---- ------ ------ .....
125 * -- - ------ - ------ -
128 * Let w = 2n/x and h=2/x, then the above quotient
129 * is equal to the continued fraction:
131 * = -----------------------
133 * w - -----------------
138 * To determine how many terms needed, let
139 * Q(0) = w, Q(1) = w(w+h) - 1,
140 * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
141 * When Q(k) > 1e4 good for single
142 * When Q(k) > 1e9 good for double
143 * When Q(k) > 1e17 good for quadruple
150 w = (n+n)/(double)x; h = 2.0/(double)x;
163 for (t=0.0, i = 2*(n+k); i>=m; i -= 2)
167 /* estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
168 * Hence, if n*(log(2n/x)) > ...
169 * single 8.8722839355e+01
170 * double 7.09782712893383973096e+02
171 * long double 1.1356523406294143949491931077970765006170e+04
172 * then recurrent value may overflow and the result is
173 * likely underflow to zero
177 tmp = tmp*log(fabs(v*tmp));
178 if (tmp < 7.09782712893383973096e+02) {
179 for (i=n-1,di=(double)(i+i); i>0; i--) {
187 for (i=n-1,di=(double)(i+i); i>0; i--) {
193 /* scale b to avoid spurious overflow */
203 if (fabs(z) >= fabs(w))
209 if (sgn==1) return -b;
215 double yn(int n, double x)
221 EXTRACT_WORDS(hx, lx, x);
222 ix = 0x7fffffff & hx;
223 /* if Y(n,NaN) is NaN */
224 if ((ix|((uint32_t)(lx|-lx))>>31) > 0x7ff00000)
233 sign = 1 - ((n&1)<<1);
239 if (ix == 0x7ff00000)
241 if (ix >= 0x52D00000) { /* x > 2**302 */
243 * Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
244 * Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
245 * Let s=sin(x), c=cos(x),
246 * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
248 * n sin(xn)*sqt2 cos(xn)*sqt2
249 * ----------------------------------
256 case 0: temp = sin(x)-cos(x); break;
257 case 1: temp = -sin(x)-cos(x); break;
258 case 2: temp = -sin(x)+cos(x); break;
259 case 3: temp = sin(x)+cos(x); break;
261 b = invsqrtpi*temp/sqrt(x);
266 /* quit if b is -inf */
267 GET_HIGH_WORD(high, b);
268 for (i=1; i<n && high!=0xfff00000; i++){
270 b = ((double)(i+i)/x)*b - a;
271 GET_HIGH_WORD(high, b);
275 if (sign > 0) return b;