How to fit a quadratic equation to the experimental data by manipulating quadratic coefficients?
15 views (last 30 days)
Show older comments
I have been trying to fit a quadratic equation to my data where I can manipulate the quadratic coefficients to get the closest fit to my data.
Here is my data,
y=[4.98279822987897e-07; 4.083e-07; 2.913e-07; 2.525e-07; 2.293e-07; 2.08788064633478e-07; 1.8974e-07; 1.8209e-07; 1.75186e-07;1.76257830379737e-07; 1.85224500066765e-07; 2.04176252688011e-07; 2.32655118657041e-07;2.71795522568243e-07; 3.22436473806901e-07;3.83535479445256e-07;];
x=[5.5; 5.9; 6.9; 7.8; 8.5; 9.4; 11.2; 12.3; 13.9; 15.5; 17.2; 18.9; 20.5; 21.9; 23.1; 23.98;];
Tried using polyfit but I need to be able to manipulate the quadratic coefficients or are there any other fits to use other than polyfit? Also tried using a weighted linear regression but I could not figure it out. Thanks!
0 Comments
Answers (2)
James Phillips
on 9 Nov 2019
I got a good fit using the 4-parameter Nelder yield density equation:
y = (a + x) / (b + c * (a + x) + d * (a + x) * (a + x))
with parameters:
a = 1.7313390100742435E+01
b = -1.1993161122097411E+09
c = 8.2891507074461997E+07
d = -1.2404871966288472E+06
yielding RMSE = 3.772E-09 and R-squared = 0.9983
0 Comments
Alex Sha
on 4 Nov 2019
If you don't care the type of fitting function, try the following one:
y = p1+p2*x+p3*x^3+p4/x+p5/x^2
Sum of Squared Residual: 3.237636036555E-16
Correlation Coef. (R): 0.998819161526896
R-Square: 0.997639717433292
Adjusted R-Square: 0.997276597038414
Determination Coef. (DC): 0.997639717433292
Chi-Square: 5.89888356402621E-10
F-Statistic: 1162.36473616939
Parameter Best Estimate
---------- -------------
p1 2.73712499948838E-6
p2 -1.11638197940077E-7
p3 8.10419259268286E-11
p4 -2.22763441216914E-5
p5 7.28628219858989E-5
0 Comments
See Also
Categories
Find more on Fit Postprocessing in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!