| |
| /* @(#)e_jn.c 1.4 95/01/18 */ |
| /* |
| * ==================================================== |
| * 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. |
| * ==================================================== |
| */ |
| |
| /* |
| * __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)=ieee_y1(0)=ieee_yn(n,0) = -inf with division by zero signal; |
| * y0(-ve)=ieee_y1(-ve)=ieee_yn(n,-ve) are NaN with invalid signal. |
| * Note 2. About ieee_jn(n,x), ieee_yn(n,x) |
| * For n=0, ieee_j0(x) is called, |
| * for n=1, ieee_j1(x) is called, |
| * for n<x, forward recursion us used starting |
| * from values of ieee_j0(x) and ieee_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 "fdlibm.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 */ |
| |
| static double zero = 0.00000000000000000000e+00; |
| |
| #ifdef __STDC__ |
| double __ieee754_jn(int n, double x) |
| #else |
| double __ieee754_jn(n,x) |
| int n; double x; |
| #endif |
| { |
| int 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) |
| */ |
| hx = __HI(x); |
| ix = 0x7fffffff&hx; |
| lx = __LO(x); |
| /* if J(n,NaN) is NaN */ |
| if((ix|((unsigned)(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 = ieee_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) = ieee_cos(x-(2n+1)*pi/4)*ieee_sqrt(2/x*pi) |
| * Yn(x) = ieee_sin(x-(2n+1)*pi/4)*ieee_sqrt(2/x*pi) |
| * Let s=ieee_sin(x), c=ieee_cos(x), |
| * xn=x-(2n+1)*pi/4, sqt2 = ieee_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 = ieee_cos(x)+ieee_sin(x); break; |
| case 1: temp = -ieee_cos(x)+ieee_sin(x); break; |
| case 2: temp = -ieee_cos(x)-ieee_sin(x); break; |
| case 3: temp = ieee_cos(x)-ieee_sin(x); break; |
| } |
| b = invsqrtpi*temp/ieee_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; int 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 ieee_log((2/x)^n*n!) = n*ieee_log(2/x)+n*ln(n) |
| * Hence, if n*(ieee_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(ieee_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 |
| { |
| int i,hx,ix,lx; |
| int sign; |
| double a, b, temp; |
| |
| hx = __HI(x); |
| ix = 0x7fffffff&hx; |
| lx = __LO(x); |
| /* if Y(n,NaN) is NaN */ |
| if((ix|((unsigned)(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) = ieee_cos(x-(2n+1)*pi/4)*ieee_sqrt(2/x*pi) |
| * Yn(x) = ieee_sin(x-(2n+1)*pi/4)*ieee_sqrt(2/x*pi) |
| * Let s=ieee_sin(x), c=ieee_cos(x), |
| * xn=x-(2n+1)*pi/4, sqt2 = ieee_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 = ieee_sin(x)-ieee_cos(x); break; |
| case 1: temp = -ieee_sin(x)-ieee_cos(x); break; |
| case 2: temp = -ieee_sin(x)+ieee_cos(x); break; |
| case 3: temp = ieee_sin(x)+ieee_cos(x); break; |
| } |
| b = invsqrtpi*temp/ieee_sqrt(x); |
| } else { |
| a = __ieee754_y0(x); |
| b = __ieee754_y1(x); |
| /* quit if b is -inf */ |
| for(i=1;i<n&&(__HI(b) != 0xfff00000);i++){ |
| temp = b; |
| b = ((double)(i+i)/x)*b - a; |
| a = temp; |
| } |
| } |
| if(sign>0) return b; else return -b; |
| } |
