Data fitting in between following variables x and b for quadratic equation
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clear all
clc
a=[0.450 0.486 0.546 0.589 0.633 0.656];
b=[1.5435 1.536 1.5273 1.523 1.5197 1.5179];
x=1./a.^2;
3 Commenti
Mathieu NOE
il 20 Gen 2023
it's unclear what the input and output are
usually a and b would be parameters / coefficients, x and y the input / output arrays
MOHD UWAIS
il 20 Gen 2023
Torsten
il 20 Gen 2023
So you want to determine p0,p1,p2 such that (in the least-squares sense)
p0 + p1*(1/0.45)^2 + p2*(1/0.45)^4 = 1.5435
p0 + p1*(1/0.486)^2 + p2*(1/0.486)^4 = 1.536
p0 + p1*(1/0.546)^2 + p2*(1/0.546)^4 = 1.5273
...
p0 + p1*(1/0.656)^2 + p2*(1/0.656)^4 = 1.5179
?
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