Thought of another way to do this...
One-liner with size 21, but fails because of trivial roundoff error :(
This isnt a particularly difficult problem
you can call this function powerpoly
function ppower = powerpoly(p,n)
ppower = p;
i = 1
while i < n
ppower = conv(ppower,p);
i = i + 1;
This is a numerical algorithm and is not exact, but very accurate. I'm failing the test because the output for p=1:5; N=3 is off by this amount...
-3.4972e-14 8.8818e-15 3.5527e-14 0.0000e+00 0.0000e+00 -2.8422e-14
Columns 7 through 12:
0.0000e+00 0.0000e+00 0.0000e+00 0.0000e+00 0.0000e+00 -5.6843e-14
COMPLETELY RIDICULOUS!!!!. Using isequal() is a poor choice for evaluating numerical algorithms.
Here is my algorithm that "failed" the isequal test by -5e-14 on a few values. I thought I would actually try to write a somewhat fast algorithm instead of just a for loop calling conv() repeatedly and reallocating memory each time.
function q = polypow(p,N)
Given you comments above, you might be interested in solving my convolution series at http://www.mathworks.com/matlabcentral/cody/?term=Fast+1-D+Convolution
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