polynomial evaluation for a simple curve.

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Paul Rogers
Paul Rogers am 30 Jun. 2020
Beantwortet: Paul Rogers am 1 Jul. 2020
After I got the coefficent from the command:
p1=polyfit(x,y,4);
the p1 vector is:
p1 =
-35.4628 39.9257 -14.8278 1.6022 1.1031
the x vector is:
x=[0:0.001:0.4000]
Now, how do I write like this:
I don't need the polyval command.

Akzeptierte Antwort

Paul Rogers
Paul Rogers am 1 Jul. 2020
formula = ((x.^4)*-35.4628) + ((x.^3)*39.9257) + ((x.^2)*-14.8278) + ((x.^1)*1.6022) + ((x.^0)*1.1031);

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