i/test3

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    /*
       TEST3.I

       Pure scalar math problem for timing Yorick.

       $Id: test3.i,v 1.1 1993/08/27 18:50:06 munro Exp $
     */
    /*    Copyright (c) 1994.  The Regents of the University of California.
                        All rights reserved.  */

    func test3 (npass)
    /* DOCUMENT test3
             or test3, npass
         Computes the ratio r which solves 1 + r^2 + r^3 +...+ r^n = s,
         given n and s.  If NPASS is given, the calculation is repeated
         that many times (actually the equation is solved many times for

         each pass).  The worker routine invgeom can actually be
         vectorized; the vector version is gseries_r in series.i.  */
    {

      if (is_void(npass) || npass<=0) npass= 1;

      extern s, n, answer;
      nsums= numberof(answer);
      r= 0.0*answer;

      now= split= array(0.0, 3);
      timer, now;
      m= npass;
      while (m--) {

        for (i=1 ; i<=nsums ; i++) r(i)= invgeom(s(i), n);
      }
      timer, now, split;

      timer_print, "Time per pass", split/npass, "Total time", split;

      if (anyof(abs(r-answer)>1.e-9*answer))

        write, "***WARNING*** values returned by invgeom are not correct";
    }

    #if 0
    CPU seconds per pass:
                IDL    Yorick     Basis       DAP
    HP730        -      0.11       5.08      5.00
    Solbourne  0.31     0.21       11.8      13.7
        (varies by ~10%)

         forum[10] yorick

          Yorick ready.  For help type 'help'

         > #include "/home/miggle/munro/Yorick/include/test3.i"
         > test3,1
    		    Timing Category     CPU sec  System sec    Wall sec
    		      Time per pass       0.190       0.000       0.230
    			 Total time       0.190       0.000       0.230
         > test3,51
    		    Timing Category     CPU sec  System sec    Wall sec
    		      Time per pass       0.206       0.000       0.207
    			 Total time      10.520       0.000      10.560

         forum[6] time idl

         IDL> .rnew /home/miggle/munro/Yorick/include/test3
         % Compiled module: INVGEOM.
         % Compiled module: TEST3.
         % Compiled module: SPAN.
         % Compiled module: $MAIN$.
         IDL> test3,1
         IDL> exit
         0.5u 0.3s 0:08 11% 0+928k 0+2io 0pf+0w
         forum[7] time idl

         IDL> .rnew /home/miggle/munro/Yorick/include/test3
         % Compiled module: INVGEOM.
         % Compiled module: TEST3.
         % Compiled module: SPAN.
         % Compiled module: $MAIN$.
         IDL> test3,51
         IDL> exit
         16.0u 0.4s 0:41 39% 0+1400k 0+0io 0pf+0w

         forum[13] basis
         Initializing Basis System
         Basis 7.0
         Initializing EZCURVE/NCAR Graphics
         Ezcurve/NCAR 2.0
         Basis> echo=0

         Basis> read /home/miggle/munro/Yorick/include/test3.bas

         End of input file /home/miggle/munro/Yorick/include/test3.bas
         Resuming input from TERMINAL
         Basis> test3(1)

    	CPU (sec)   SYS (sec)
    	   11.817       0.017

         forum[14] dap
         DAP> echo=0

         DAP> read /home/miggle/munro/Yorick/include/test3.bas
         DAP> test3(1)

    	CPU (sec)   SYS (sec)
    	   13.700       0.033
    #endif

    /* invgeom(s, n)

         returns the ratio r of the finite geometric series, given the sum s:
            1 + r + r^2 + r^3 + ... + r^n = s     */

    func invgeom (s, n)
    {
      nn= n+1;
      if (nn<2) return s-1.0;

      /* compute an approximate result which has exact values and
         derivatives at s==1, s==n, and s->infinity */

      /* different approximations apply for s>nn and s<nn */
      if (s>nn) {
        pow= 1.0/n;
        npow= nn^pow - 1.0;
        n2r= 1.0/(nn-2.0);
        A= (2.0-nn*npow)*n2r;
        B= (2.0*npow-nn*pow)*nn*n2r;
        r= s^pow - pow + A*(nn/s)^pow + B/s;
      } else {
        sn= (s-1.0)/n;
        n2r= 1.0/(nn*nn);
        r= 1.0 - 1.0/s + n2r*sn*sn*(nn+1.0 - sn);
      }

      /* Polish the approximation using Newton-Raphson iterations.  */
      for (;;) {
        rr= r-1.0;
        rn= r^(nn-1);
        delta= rr*s - (r*rn-1.0);

        if (abs(delta)<=1.e-9*abs(rr*s)) break;
        r+= delta/(nn*rn-s);
      }
      /* try to get it to machine precision */
      if (delta) r+= delta/(nn*rn-s);

      return r;
    }

    /* Set up a bunch of values to compute.  */

    answer= span(0.0, 2.0, 200);  /* number must be even to avoid 1.0000 */
    n= 100;
    s= (answer^(n+1)-1.0)/(answer-1.0);