diff options
Diffstat (limited to 'lib/libm/common_source/jn.c')
| -rw-r--r-- | lib/libm/common_source/jn.c | 44 |
1 files changed, 22 insertions, 22 deletions
diff --git a/lib/libm/common_source/jn.c b/lib/libm/common_source/jn.c index 85a5401..28d9687 100644 --- a/lib/libm/common_source/jn.c +++ b/lib/libm/common_source/jn.c @@ -46,18 +46,18 @@ static char sccsid[] = "@(#)jn.c 8.2 (Berkeley) 11/30/93"; * * Developed at SunPro, a Sun Microsystems, Inc. business. * Permission to use, copy, modify, and distribute this - * software is freely granted, provided that this notice + * software is freely granted, provided that this notice * is preserved. * ==================================================== * * ******************* WARNING ******************** * This is an alpha version of SunPro's FDLIBM (Freely - * Distributable Math Library) for IEEE double precision + * Distributable Math Library) for IEEE double precision * arithmetic. FDLIBM is a basic math library written - * in C that runs on machines that conform to IEEE - * Standard 754/854. This alpha version is distributed - * for testing purpose. Those who use this software - * should report any bugs to + * in C that runs on machines that conform to IEEE + * Standard 754/854. This alpha version is distributed + * for testing purpose. Those who use this software + * should report any bugs to * * fdlibm-comments@sunpro.eng.sun.com * @@ -69,7 +69,7 @@ static char sccsid[] = "@(#)jn.c 8.2 (Berkeley) 11/30/93"; * jn(int n, double x), yn(int n, double 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. @@ -88,7 +88,7 @@ static char sccsid[] = "@(#)jn.c 8.2 (Berkeley) 11/30/93"; * yn(n,x) is similar in all respects, except * that forward recursion is used for all * values of n>1. - * + * */ #include <math.h> @@ -120,7 +120,7 @@ double jn(n,x) */ /* if J(n,NaN) is NaN */ if (_IEEE && isnan(x)) return x+x; - if (n<0){ + if (n<0){ n = -n; x = -x; } @@ -134,10 +134,10 @@ double jn(n,x) /* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */ if (_IEEE && x >= 8.148143905337944345e+090) { /* x >= 2**302 */ - /* (x >> n**2) + /* (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), + * Let s=sin(x), c=cos(x), * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then * * n sin(xn)*sqt2 cos(xn)*sqt2 @@ -154,7 +154,7 @@ double jn(n,x) case 3: temp = cos(x)-sin(x); break; } b = invsqrtpi*temp/sqrt(x); - } else { + } else { a = j0(x); b = j1(x); for(i=1;i<n;i++){ @@ -165,7 +165,7 @@ double jn(n,x) } } else { if (x < 1.86264514923095703125e-009) { /* x < 2**-29 */ - /* x is tiny, return the first Taylor expansion of J(n,x) + /* x is tiny, return the first Taylor expansion of J(n,x) * J(n,x) = 1/n!*(x/2)^n - ... */ if (n > 33) /* underflow */ @@ -180,14 +180,14 @@ double jn(n,x) } } else { /* use backward recurrence */ - /* x x^2 x^2 + /* x x^2 x^2 * J(n,x)/J(n-1,x) = ---- ------ ------ ..... * 2n - 2(n+1) - 2(n+2) * - * 1 1 1 + * 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 @@ -203,9 +203,9 @@ double jn(n,x) * 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 + * 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; @@ -254,7 +254,7 @@ double jn(n,x) } return ((sgn == 1) ? -b : b); } -double yn(n,x) +double yn(n,x) int n; double x; { int i, sign; @@ -275,10 +275,10 @@ double yn(n,x) if (n == 0) return(y0(x)); if (n == 1) return(sign*y1(x)); if(_IEEE && x >= 8.148143905337944345e+090) { /* x > 2**302 */ - /* (x >> n**2) + /* (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), + * Let s=sin(x), c=cos(x), * xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then * * n sin(xn)*sqt2 cos(xn)*sqt2 |
