It's a stupid question but these two line of code do the same thing wthere dt1 is a constant. Is it right?
prova= sum((out1.Y{i}(:,3:end)).^2)*dt1;
h=1;
for i=3:end;
y1=out.Y(:,i);
prova(h)=sum(y1.^2)*dt1;
h=h+1;
end;

1 commentaire

Jan
Jan le 22 Fév 2014
Replace for i=3:end by for i = 3:size(out.Y, 2) .

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Walter Roberson
Walter Roberson le 22 Fév 2014

0 votes

Yes, they appear to be, at least to the limits of round-off error.

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