From c4e44e97f8562254d9da898f6ed7e6cb4d8a3ce4 Mon Sep 17 00:00:00 2001 From: Eric Andersen Date: Sun, 6 Mar 2005 07:11:53 +0000 Subject: Trim off whitespace --- libm/e_jn.c | 42 +++++++++++++++++++++--------------------- 1 file changed, 21 insertions(+), 21 deletions(-) (limited to 'libm/e_jn.c') diff --git a/libm/e_jn.c b/libm/e_jn.c index 870824cf8..857c4a3f5 100644 --- a/libm/e_jn.c +++ b/libm/e_jn.c @@ -5,7 +5,7 @@ * * 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. * ==================================================== */ @@ -18,7 +18,7 @@ static char rcsid[] = "$NetBSD: e_jn.c,v 1.9 1995/05/10 20:45:34 jtc Exp $"; * __ieee754_jn(n, x), __ieee754_yn(n, x) * floating point Bessel's function of the 1st and 2nd kind * of order n - * + * * Special cases: * y0(0)=y1(0)=yn(n,0) = -inf with division by zero signal; * y0(-ve)=y1(-ve)=yn(n,-ve) are NaN with invalid signal. @@ -37,7 +37,7 @@ static char rcsid[] = "$NetBSD: e_jn.c,v 1.9 1995/05/10 20:45:34 jtc Exp $"; * yn(n,x) is similar in all respects, except * that forward recursion is used for all * values of n>1. - * + * */ #include "math.h" @@ -76,7 +76,7 @@ static double zero = 0.00000000000000000000e+00; ix = 0x7fffffff&hx; /* if J(n,NaN) is NaN */ if((ix|((u_int32_t)(lx|-lx))>>31)>0x7ff00000) return x+x; - if(n<0){ + if(n<0){ n = -n; x = -x; hx ^= 0x80000000; @@ -87,13 +87,13 @@ static double zero = 0.00000000000000000000e+00; x = fabs(x); if((ix|lx)==0||ix>=0x7ff00000) /* if x is 0 or inf */ b = zero; - else if((double)n<=x) { + 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) + /* (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 @@ -110,7 +110,7 @@ static double zero = 0.00000000000000000000e+00; case 3: temp = cos(x)-sin(x); break; } b = invsqrtpi*temp/sqrt(x); - } else { + } else { a = __ieee754_j0(x); b = __ieee754_j1(x); for(i=1;i33) /* underflow */ @@ -136,14 +136,14 @@ static double zero = 0.00000000000000000000e+00; } } 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 @@ -159,9 +159,9 @@ static double zero = 0.00000000000000000000e+00; * 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; @@ -183,7 +183,7 @@ static double zero = 0.00000000000000000000e+00; * single 8.8722839355e+01 * double 7.09782712893383973096e+02 * long double 1.1356523406294143949491931077970765006170e+04 - * then recurrent value may overflow and the result is + * then recurrent value may overflow and the result is * likely underflow to zero */ tmp = n; @@ -219,9 +219,9 @@ static double zero = 0.00000000000000000000e+00; } #ifdef __STDC__ - double __ieee754_yn(int n, double x) + double __ieee754_yn(int n, double x) #else - double __ieee754_yn(n,x) + double __ieee754_yn(n,x) int n; double x; #endif { @@ -244,10 +244,10 @@ static double zero = 0.00000000000000000000e+00; if(n==1) return(sign*__ieee754_y1(x)); if(ix==0x7ff00000) return zero; if(ix>=0x52D00000) { /* 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 @@ -270,7 +270,7 @@ static double zero = 0.00000000000000000000e+00; b = __ieee754_y1(x); /* quit if b is -inf */ GET_HIGH_WORD(high,b); - for(i=1;i