/* origin: FreeBSD /usr/src/lib/msun/src/e_jn.c */ /* * ==================================================== * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved. * * Developed at SunSoft, a Sun Microsystems, Inc. business. * Permission to use, copy, modify, and distribute this * software is freely granted, provided that this notice * is preserved. * ==================================================== */ /* * jn(n, x), 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 is 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 "libm.h" static const double invsqrtpi = 5.64189583547756279280e-01; /* 0x3FE20DD7, 0x50429B6D */ double jn(int n, double x) { uint32_t ix, lx; int nm1, i, sign; double a, b, temp; EXTRACT_WORDS(ix, lx, x); sign = ix>>31; ix &= 0x7fffffff; if ((ix | (lx|-lx)>>31) > 0x7ff00000) /* nan */ return x; /* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x) * Thus, J(-n,x) = J(n,-x) */ /* nm1 = |n|-1 is used instead of |n| to handle n==INT_MIN */ if (n == 0) return j0(x); if (n < 0) { nm1 = -(n+1); x = -x; sign ^= 1; } else nm1 = n-1; if (nm1 == 0) return j1(x); sign &= n; /* even n: 0, odd n: signbit(x) */ x = fabs(x); if ((ix|lx) == 0 || ix == 0x7ff00000) /* if x is 0 or inf */ b = 0.0; else if (nm1 < 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(nm1&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; default: case 3: temp = cos(x)+sin(x); break; } b = invsqrtpi*temp/sqrt(x); } else { a = j0(x); b = j1(x); for (i=0; i<nm1; ) { i++; temp = b; b = b*(2.0*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 (nm1 > 32) /* underflow */ b = 0.0; else { temp = x*0.5; b = temp; a = 1.0; for (i=2; i<=nm1+1; 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,q0,q1,w,h,z,tmp,nf; int k; nf = nm1 + 1.0; w = 2*nf/x; h = 2/x; z = w+h; q0 = w; q1 = w*z - 1.0; k = 1; while (q1 < 1.0e9) { k += 1; z += h; tmp = z*q1 - q0; q0 = q1; q1 = tmp; } for (t=0.0, i=k; i>=0; i--) t = 1/(2*(i+nf)/x - t); a = t; b = 1.0; /* 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 = nf*log(fabs(w)); if (tmp < 7.09782712893383973096e+02) { for (i=nm1; i>0; i--) { temp = b; b = b*(2.0*i)/x - a; a = temp; } } else { for (i=nm1; i>0; i--) { temp = b; b = b*(2.0*i)/x - a; a = temp; /* scale b to avoid spurious overflow */ if (b > 0x1p500) { a /= b; t /= b; b = 1.0; } } } z = j0(x); w = j1(x); if (fabs(z) >= fabs(w)) b = t*z/b; else b = t*w/a; } } return sign ? -b : b; } double yn(int n, double x) { uint32_t ix, lx, ib; int nm1, sign, i; double a, b, temp; EXTRACT_WORDS(ix, lx, x); sign = ix>>31; ix &= 0x7fffffff; if ((ix | (lx|-lx)>>31) > 0x7ff00000) /* nan */ return x; if (sign && (ix|lx)!=0) /* x < 0 */ return 0/0.0; if (ix == 0x7ff00000) return 0.0; if (n == 0) return y0(x); if (n < 0) { nm1 = -(n+1); sign = n&1; } else { nm1 = n-1; sign = 0; } if (nm1 == 0) return sign ? -y1(x) : y1(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(nm1&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; default: case 3: temp = sin(x)-cos(x); break; } b = invsqrtpi*temp/sqrt(x); } else { a = y0(x); b = y1(x); /* quit if b is -inf */ GET_HIGH_WORD(ib, b); for (i=0; i<nm1 && ib!=0xfff00000; ){ i++; temp = b; b = (2.0*i/x)*b - a; GET_HIGH_WORD(ib, b); a = temp; } } return sign ? -b : b; }
