1 /* origin: FreeBSD /usr/src/lib/msun/src/e_jnf.c */
3 * Conversion to float by Ian Lance Taylor, Cygnus Support, ian@cygnus.com.
6 * ====================================================
7 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
9 * Developed at SunPro, a Sun Microsystems, Inc. business.
10 * Permission to use, copy, modify, and distribute this
11 * software is freely granted, provided that this notice
13 * ====================================================
19 two = 2.0000000000e+00, /* 0x40000000 */
20 one = 1.0000000000e+00; /* 0x3F800000 */
22 static const float zero = 0.0000000000e+00;
24 float jnf(int n, float x)
30 /* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
31 * Thus, J(-n,x) = J(n,-x)
33 GET_FLOAT_WORD(hx, x);
35 /* if J(n,NaN) is NaN */
43 if (n == 0) return j0f(x);
44 if (n == 1) return j1f(x);
46 sgn = (n&1)&(hx>>31); /* even n -- 0, odd n -- sign(x) */
48 if (ix == 0 || ix >= 0x7f800000) /* if x is 0 or inf */
50 else if((float)n <= x) {
51 /* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
56 b = b*((float)(i+i)/x) - a; /* avoid underflow */
60 if (ix < 0x30800000) { /* x < 2**-29 */
61 /* x is tiny, return the first Taylor expansion of J(n,x)
62 * J(n,x) = 1/n!*(x/2)^n - ...
64 if (n > 33) /* underflow */
69 for (a=one,i=2; i<=n; i++) {
70 a *= (float)i; /* a = n! */
71 b *= temp; /* b = (x/2)^n */
76 /* use backward recurrence */
78 * J(n,x)/J(n-1,x) = ---- ------ ------ .....
79 * 2n - 2(n+1) - 2(n+2)
82 * (for large x) = ---- ------ ------ .....
84 * -- - ------ - ------ -
87 * Let w = 2n/x and h=2/x, then the above quotient
88 * is equal to the continued fraction:
90 * = -----------------------
92 * w - -----------------
97 * To determine how many terms needed, let
98 * Q(0) = w, Q(1) = w(w+h) - 1,
99 * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
100 * When Q(k) > 1e4 good for single
101 * When Q(k) > 1e9 good for double
102 * When Q(k) > 1e17 good for quadruple
115 while (q1 < 1.0e9f) {
123 for (t=zero, i = 2*(n+k); i>=m; i -= 2)
127 /* estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
128 * Hence, if n*(log(2n/x)) > ...
129 * single 8.8722839355e+01
130 * double 7.09782712893383973096e+02
131 * long double 1.1356523406294143949491931077970765006170e+04
132 * then recurrent value may overflow and the result is
133 * likely underflow to zero
137 tmp = tmp*logf(fabsf(v*tmp));
138 if (tmp < 88.721679688f) {
139 for (i=n-1,di=(float)(i+i); i>0; i--) {
147 for (i=n-1,di=(float)(i+i); i>0; i--){
153 /* scale b to avoid spurious overflow */
163 if (fabsf(z) >= fabsf(w))
169 if (sgn == 1) return -b;
173 float ynf(int n, float x)
179 GET_FLOAT_WORD(hx, x);
180 ix = 0x7fffffff & hx;
181 /* if Y(n,NaN) is NaN */
191 sign = 1 - ((n&1)<<1);
197 if (ix == 0x7f800000)
202 /* quit if b is -inf */
203 GET_FLOAT_WORD(ib,b);
204 for (i = 1; i < n && ib != 0xff800000; i++){
206 b = ((float)(i+i)/x)*b - a;
207 GET_FLOAT_WORD(ib, b);