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[/] [openrisc/] [trunk/] [gnu-old/] [newlib-1.17.0/] [newlib/] [libm/] [math/] [e_jn.c] - Blame information for rev 816

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Line No. Rev Author Line
1 148 jeremybenn
 
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/* @(#)e_jn.c 5.1 93/09/24 */
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/*
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 * ====================================================
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 * Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
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 *
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 * Developed at SunPro, a Sun Microsystems, Inc. business.
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 * Permission to use, copy, modify, and distribute this
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 * software is freely granted, provided that this notice
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 * is preserved.
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 * ====================================================
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 */
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/*
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 * __ieee754_jn(n, x), __ieee754_yn(n, x)
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 * floating point Bessel's function of the 1st and 2nd kind
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 * of order n
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 *
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 * Special cases:
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 *      y0(0)=y1(0)=yn(n,0) = -inf with division by zero signal;
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 *      y0(-ve)=y1(-ve)=yn(n,-ve) are NaN with invalid signal.
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 * Note 2. About jn(n,x), yn(n,x)
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 *      For n=0, j0(x) is called,
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 *      for n=1, j1(x) is called,
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 *      for n<x, forward recursion us used starting
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 *      from values of j0(x) and j1(x).
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 *      for n>x, a continued fraction approximation to
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 *      j(n,x)/j(n-1,x) is evaluated and then backward
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 *      recursion is used starting from a supposed value
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 *      for j(n,x). The resulting value of j(0,x) is
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 *      compared with the actual value to correct the
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 *      supposed value of j(n,x).
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 *
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 *      yn(n,x) is similar in all respects, except
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 *      that forward recursion is used for all
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 *      values of n>1.
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 *
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 */
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#include "fdlibm.h"
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#ifndef _DOUBLE_IS_32BITS
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#ifdef __STDC__
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static const double
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#else
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static double
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#endif
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invsqrtpi=  5.64189583547756279280e-01, /* 0x3FE20DD7, 0x50429B6D */
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two   =  2.00000000000000000000e+00, /* 0x40000000, 0x00000000 */
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one   =  1.00000000000000000000e+00; /* 0x3FF00000, 0x00000000 */
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#ifdef __STDC__
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static const double zero  =  0.00000000000000000000e+00;
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#else
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static double zero  =  0.00000000000000000000e+00;
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#endif
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#ifdef __STDC__
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        double __ieee754_jn(int n, double x)
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#else
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        double __ieee754_jn(n,x)
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        int n; double x;
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#endif
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{
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        __int32_t i,hx,ix,lx, sgn;
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        double a, b, temp, di;
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        double z, w;
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    /* J(-n,x) = (-1)^n * J(n, x), J(n, -x) = (-1)^n * J(n, x)
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     * Thus, J(-n,x) = J(n,-x)
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     */
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        EXTRACT_WORDS(hx,lx,x);
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        ix = 0x7fffffff&hx;
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    /* if J(n,NaN) is NaN */
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        if((ix|((__uint32_t)(lx|-lx))>>31)>0x7ff00000) return x+x;
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        if(n<0){
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                n = -n;
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                x = -x;
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                hx ^= 0x80000000;
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        }
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        if(n==0) return(__ieee754_j0(x));
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        if(n==1) return(__ieee754_j1(x));
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        sgn = (n&1)&(hx>>31);   /* even n -- 0, odd n -- sign(x) */
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        x = fabs(x);
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        if((ix|lx)==0||ix>=0x7ff00000)   /* if x is 0 or inf */
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            b = zero;
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        else if((double)n<=x) {
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                /* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
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            if(ix>=0x52D00000) { /* x > 2**302 */
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    /* (x >> n**2)
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     *      Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
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     *      Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
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     *      Let s=sin(x), c=cos(x),
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     *          xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
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     *
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     *             n    sin(xn)*sqt2    cos(xn)*sqt2
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     *          ----------------------------------
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     *             0     s-c             c+s
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     *             1    -s-c            -c+s
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     *             2    -s+c            -c-s
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     *             3     s+c             c-s
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     */
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                switch(n&3) {
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                    case 0: temp =  cos(x)+sin(x); break;
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                    case 1: temp = -cos(x)+sin(x); break;
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                    case 2: temp = -cos(x)-sin(x); break;
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                    case 3: temp =  cos(x)-sin(x); break;
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                }
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                b = invsqrtpi*temp/__ieee754_sqrt(x);
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            } else {
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                a = __ieee754_j0(x);
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                b = __ieee754_j1(x);
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                for(i=1;i<n;i++){
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                    temp = b;
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                    b = b*((double)(i+i)/x) - a; /* avoid underflow */
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                    a = temp;
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                }
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            }
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        } else {
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            if(ix<0x3e100000) { /* x < 2**-29 */
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    /* x is tiny, return the first Taylor expansion of J(n,x)
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     * J(n,x) = 1/n!*(x/2)^n  - ...
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     */
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                if(n>33)        /* underflow */
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                    b = zero;
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                else {
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                    temp = x*0.5; b = temp;
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                    for (a=one,i=2;i<=n;i++) {
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                        a *= (double)i;         /* a = n! */
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                        b *= temp;              /* b = (x/2)^n */
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                    }
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                    b = b/a;
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                }
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            } else {
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                /* use backward recurrence */
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                /*                      x      x^2      x^2
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                 *  J(n,x)/J(n-1,x) =  ----   ------   ------   .....
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                 *                      2n  - 2(n+1) - 2(n+2)
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                 *
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                 *                      1      1        1
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                 *  (for large x)   =  ----  ------   ------   .....
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                 *                      2n   2(n+1)   2(n+2)
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                 *                      -- - ------ - ------ -
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                 *                       x     x         x
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                 *
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                 * Let w = 2n/x and h=2/x, then the above quotient
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                 * is equal to the continued fraction:
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                 *                  1
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                 *      = -----------------------
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                 *                     1
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                 *         w - -----------------
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                 *                        1
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                 *              w+h - ---------
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                 *                     w+2h - ...
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                 *
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                 * To determine how many terms needed, let
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                 * Q(0) = w, Q(1) = w(w+h) - 1,
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                 * Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
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                 * When Q(k) > 1e4      good for single
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                 * When Q(k) > 1e9      good for double
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                 * When Q(k) > 1e17     good for quadruple
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                 */
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            /* determine k */
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                double t,v;
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                double q0,q1,h,tmp; __int32_t k,m;
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                w  = (n+n)/(double)x; h = 2.0/(double)x;
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                q0 = w;  z = w+h; q1 = w*z - 1.0; k=1;
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                while(q1<1.0e9) {
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                        k += 1; z += h;
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                        tmp = z*q1 - q0;
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                        q0 = q1;
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                        q1 = tmp;
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                }
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                m = n+n;
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                for(t=zero, i = 2*(n+k); i>=m; i -= 2) t = one/(i/x-t);
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                a = t;
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                b = one;
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                /*  estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
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                 *  Hence, if n*(log(2n/x)) > ...
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                 *  single 8.8722839355e+01
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                 *  double 7.09782712893383973096e+02
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                 *  long double 1.1356523406294143949491931077970765006170e+04
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                 *  then recurrent value may overflow and the result is
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                 *  likely underflow to zero
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                 */
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                tmp = n;
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                v = two/x;
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                tmp = tmp*__ieee754_log(fabs(v*tmp));
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                if(tmp<7.09782712893383973096e+02) {
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                    for(i=n-1,di=(double)(i+i);i>0;i--){
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                        temp = b;
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                        b *= di;
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                        b  = b/x - a;
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                        a = temp;
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                        di -= two;
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                    }
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                } else {
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                    for(i=n-1,di=(double)(i+i);i>0;i--){
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                        temp = b;
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                        b *= di;
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                        b  = b/x - a;
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                        a = temp;
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                        di -= two;
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                    /* scale b to avoid spurious overflow */
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                        if(b>1e100) {
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                            a /= b;
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                            t /= b;
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                            b  = one;
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                        }
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                    }
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                }
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                b = (t*__ieee754_j0(x)/b);
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            }
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        }
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        if(sgn==1) return -b; else return b;
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}
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#ifdef __STDC__
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        double __ieee754_yn(int n, double x)
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#else
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        double __ieee754_yn(n,x)
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        int n; double x;
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#endif
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{
226
        __int32_t i,hx,ix,lx;
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        __int32_t sign;
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        double a, b, temp;
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        EXTRACT_WORDS(hx,lx,x);
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        ix = 0x7fffffff&hx;
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    /* if Y(n,NaN) is NaN */
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        if((ix|((__uint32_t)(lx|-lx))>>31)>0x7ff00000) return x+x;
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        if((ix|lx)==0) return -one/zero;
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        if(hx<0) return zero/zero;
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        sign = 1;
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        if(n<0){
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                n = -n;
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                sign = 1 - ((n&1)<<1);
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        }
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        if(n==0) return(__ieee754_y0(x));
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        if(n==1) return(sign*__ieee754_y1(x));
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        if(ix==0x7ff00000) return zero;
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        if(ix>=0x52D00000) { /* x > 2**302 */
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    /* (x >> n**2)
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     *      Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
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     *      Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
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     *      Let s=sin(x), c=cos(x),
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     *          xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
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     *
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     *             n    sin(xn)*sqt2    cos(xn)*sqt2
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     *          ----------------------------------
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     *             0     s-c             c+s
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     *             1    -s-c            -c+s
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     *             2    -s+c            -c-s
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     *             3     s+c             c-s
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     */
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                switch(n&3) {
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                    case 0: temp =  sin(x)-cos(x); break;
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                    case 1: temp = -sin(x)-cos(x); break;
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                    case 2: temp = -sin(x)+cos(x); break;
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                    case 3: temp =  sin(x)+cos(x); break;
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                }
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                b = invsqrtpi*temp/__ieee754_sqrt(x);
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        } else {
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            __uint32_t high;
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            a = __ieee754_y0(x);
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            b = __ieee754_y1(x);
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        /* quit if b is -inf */
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            GET_HIGH_WORD(high,b);
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            for(i=1;i<n&&high!=0xfff00000;i++){
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                temp = b;
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                b = ((double)(i+i)/x)*b - a;
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                GET_HIGH_WORD(high,b);
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                a = temp;
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            }
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        }
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        if(sign>0) return b; else return -b;
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}
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#endif /* defined(_DOUBLE_IS_32BITS) */

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