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Re: Newbie question solving lin sys

From: John W. Eaton
Subject: Re: Newbie question solving lin sys
Date: Fri, 27 Dec 2002 19:58:34 -0600

On 27-Dec-2002, John B. Thoo <address@hidden> wrote:

| That works to give me a polygonal curve that connects the 12 points.  
| That's good, but how would I plot the polynomial with more grid points 
| (smoother)?  It works reasonably when I do "format long" and then copy & 
| paste the coefficients to plot
| octave:139> gplot [0:11] "22 + 129.316052352359*x - 
| 330.008191451222*x**2 + 356.743251568416*x**3 - 211.743557625161*x**4 + 
| 77.5116682813202*x**5 - 18.4455900615644*x**6 + 2.90919225949428*x**7 - 
| 0.301909718420775*x**8 + 0.0198178458639753*x**9 - 
| 0.000745701048375786*x**10 + 0.0000122504808365439*x**11"
| but there must be an easier way. :-)
| Once again, thanks very much for all your help.

In Octave, a polynomial is represented by its coefficients (arranged
in descending order).  For example, a vector C of length N:

     p(x) = C(1) x^N + ... + C(N) x + C(N+1).

Try something like

  A = [...];
  T = [...];
  c = flipud (A\T);  ## note that you ended up with coefficients in
                     ## the opposite order from what Octave's poly
                     ## functions want...
  x = 0:0.1:11;
  y = polyval (c, x)
  plot (x, y);

Makes a nice plot for me given your original A and T.


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