simultaneously, fitting two functions with two databases

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I have two data sets y1(x00) and y2(x00) in the x00 range, as:
x00=[0,1,1.5,2,2.5,3,3.5,4,4.5,5]*2*pi/3;
y1=[0,1.082,2.27,3.91,5.93,8.24,10.69,13.02,14.83,15.55];
y2=[36.39,36.16,35.81,35.24,34.40,33.27,31.88,30.39,29.11,29.00];
I need fitting both functions f1=@(c1,c21,c22,c3,x) and f2=@(c1,c21,c22,c3,x), with these data in ranges x00.
Note that [c1,..,c4] are required constant coefficients and x is the varibale in range of the x00. I need to obtain the [c1,..,c4] parameters.
In my search, I found nonlinear least squre methodes with matlab, but please take into account that I am new to Matlab and can only curve fit very basic data points.
x00=[0,1,1.5,2,2.5,3,3.5,4,4.5,5]*2*pi/3;
y1=[0,1.082,2.27,3.91,5.93,8.24,10.69,13.02,14.83,15.55];
y2=[36.39,36.16,35.81,35.24,34.40,33.27,31.88,30.39,29.11,29.00];
f1=@(c1,c21,c22,c3,x) sqrt( 2.0*c21*cos(1.5*x) - 3.0*c3 - 2.0*c21 - 2.0*c22 - 3.0*c1 + 2.0*c22*cos(1.5*x) - 1.0*exp(-x*1.5i)*(c1*c3 + 2.0*c1^2*exp(x*4.5i) + 2.0*c1^2*exp(x*1.5i) + 5.0*c1^2*exp(x*3.0i) + 5.0*c3^2*exp(x*3.0i) + 2.0*c3^2*exp(x*6.0i) + 6.0*c21^2*exp(x*3.0i) + 6.0*c22^2*exp(x*3.0i) + 2.0*c3^2 - 8.0*c21^2*exp(x*3.0i)*cos(1.5*x) + 2.0*c21^2*exp(x*3.0i)*cos(3.0*x) - 8.0*c22^2*exp(x*3.0i)*cos(1.5*x) + 2.0*c22^2*exp(x*3.0i)*cos(3.0*x) + 6.0*c1*c3*exp(x*4.5i) + 6.0*c1*c3*exp(x*1.5i) + 4.0*c1*c3*exp(x*3.0i) + c1*c3*exp(x*6.0i) - 12.0*c21*c22*exp(x*3.0i) + 16.0*c21*c22*exp(x*3.0i)*cos(1.5*x) - 4.0*c21*c22*exp(x*3.0i)*cos(3.0*x))^(1/2));
f2=@(c1,c21,c22,c3,x) sqrt( 2.0*c21*cos(1.5*x) - 3.0*c3 - 2.0*c21 - 2.0*c22 - 3.0*c1 + 2.0*c22*cos(1.5*x) + exp(-x*1.5i)*(c1*c3 + 2.0*c1^2*exp(x*4.5i) + 2.0*c1^2*exp(x*1.5i) + 5.0*c1^2*exp(x*3.0i) + 5.0*c3^2*exp(x*3.0i) + 2.0*c3^2*exp(x*6.0i) + 6.0*c21^2*exp(x*3.0i) + 6.0*c22^2*exp(x*3.0i) + 2.0*c3^2 - 8.0*c21^2*exp(x*3.0i)*cos(1.5*x) + 2.0*c21^2*exp(x*3.0i)*cos(3.0*x) - 8.0*c22^2*exp(x*3.0i)*cos(1.5*x) + 2.0*c22^2*exp(x*3.0i)*cos(3.0*x) + 6.0*c1*c3*exp(x*4.5i) + 6.0*c1*c3*exp(x*1.5i) + 4.0*c1*c3*exp(x*3.0i) + c1*c3*exp(x*6.0i) - 12.0*c21*c22*exp(x*3.0i) + 16.0*c21*c22*exp(x*3.0i)*cos(1.5*x) - 4.0*c21*c22*exp(x*3.0i)*cos(3.0*x))^(1/2));
Any support will help me, thanks.
Please assist.
  3 Comments
Arad
Arad on 14 Aug 2022
Edited: Arad on 14 Aug 2022
Thank you for your attentions.
  1. The f1 and f2 are roots of the 2-th equation and have same components in their terms. They have form as: F1=sqrt[A+sqrt[B]] and F2=sqrt[A-sqrt[B]].
  2. Although, they have complex terms in their boady, but they are real functions.
  3. You can neglicated the sqrt, and I change the database after it.
  4. the c1,c21,c22,c3 are real parameters.
A22= 2*c22*(2*cos((3*x)/2) - 2) - 3*c3 - 3*c1;
A21= 2*c21*(2*cos((3*x)/2) - 2) - 3*c3 - 3*c1;
B1=c1*(exp(x*1i) + 2*exp(-(x*1i)/2))
B3=c3*(2*exp(x*1i) + exp(-x*2i))
f1=sqrt((A11+A22)/2 - (abs(B1)^2 + abs(B3)^2 + B1*conj(B3) + B3*conj(B1) + ((A11-A22)/2)^2)^(1/2))
f2=sqrt((A11+A22)/2 + (abs(B1)^2 + abs(B3)^2 + B1*conj(B3) + B3*conj(B1) + ((A11-A22)/2)^2)^(1/2))
If needed, I calculate the functions analytically and write their real form which is based on the Cos(x)???

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Accepted Answer

Torsten
Torsten on 14 Aug 2022
I think you will have to work on the representation of f1 and f2 without the imaginary unit:
c10 = 3.0;
c210 = 1.0;
c220 = -1.0;
c30 = 0.5;
C0 = [c10,c210,c220,c30];
x = [0,1,1.5,2,2.5,3,3.5,4,4.5,5]*2*pi/3;
y1 = [0,1.082,2.27,3.91,5.93,8.24,10.69,13.02,14.83,15.55];
y2 = [36.39,36.16,35.81,35.24,34.40,33.27,31.88,30.39,29.11,29.00];
f1 =@(c1,c21,c22,c3) real(2.0*c21*cos(1.5*x) - 3.0*c3 - 2.0*c21 - 2.0*c22 - 3.0*c1 + 2.0*c22*cos(1.5*x) - exp(-x*1.5i).*(c1*c3 + 2.0*c1^2*exp(x*4.5i) + 2.0*c1^2*exp(x*1.5i) + 5.0*c1^2*exp(x*3.0i) + 5.0*c3^2*exp(x*3.0i) + 2.0*c3^2*exp(x*6.0i) + 6.0*c21^2*exp(x*3.0i) + 6.0*c22^2*exp(x*3.0i) + 2.0*c3^2 - 8.0*c21^2*exp(x*3.0i).*cos(1.5*x) + 2.0*c21^2*exp(x*3.0i).*cos(3.0*x) - 8.0*c22^2*exp(x*3.0i).*cos(1.5*x) + 2.0*c22^2*exp(x*3.0i).*cos(3.0*x) + 6.0*c1*c3*exp(x*4.5i) + 6.0*c1*c3*exp(x*1.5i) + 4.0*c1*c3*exp(x*3.0i) + c1*c3*exp(x*6.0i) - 12.0*c21*c22*exp(x*3.0i) + 16.0*c21*c22*exp(x*3.0i).*cos(1.5*x) - 4.0*c21*c22*exp(x*3.0i).*cos(3.0*x))).^(1/2) - y1.^2;
f2 =@(c1,c21,c22,c3) real(2.0*c21*cos(1.5*x) - 3.0*c3 - 2.0*c21 - 2.0*c22 - 3.0*c1 + 2.0*c22*cos(1.5*x) + exp(-x*1.5i).*(c1*c3 + 2.0*c1^2*exp(x*4.5i) + 2.0*c1^2*exp(x*1.5i) + 5.0*c1^2*exp(x*3.0i) + 5.0*c3^2*exp(x*3.0i) + 2.0*c3^2*exp(x*6.0i) + 6.0*c21^2*exp(x*3.0i) + 6.0*c22^2*exp(x*3.0i) + 2.0*c3^2 - 8.0*c21^2*exp(x*3.0i).*cos(1.5*x) + 2.0*c21^2*exp(x*3.0i).*cos(3.0*x) - 8.0*c22^2*exp(x*3.0i).*cos(1.5*x) + 2.0*c22^2*exp(x*3.0i).*cos(3.0*x) + 6.0*c1*c3*exp(x*4.5i) + 6.0*c1*c3*exp(x*1.5i) + 4.0*c1*c3*exp(x*3.0i) + c1*c3*exp(x*6.0i) - 12.0*c21*c22*exp(x*3.0i) + 16.0*c21*c22*exp(x*3.0i).*cos(1.5*x) - 4.0*c21*c22*exp(x*3.0i).*cos(3.0*x))).^(1/2) - y2.^2;
F = @(c)[f1(real(c(1)),real(c(2)),real(c(3)),real(c(4))),f2(real(c(1)),real(c(2)),real(c(3)),real(c(4)))];
C = lsqnonlin(F,C0)
Local minimum possible. lsqnonlin stopped because the size of the current step is less than the value of the step size tolerance.
C =
61.9452 +11.9296i 1.4621 -43.5196i -1.8836 +33.6286i 56.3354 -80.1942i
norm(F(C))
ans = 3.3212e+03
C = real(C)
C = 1×4
61.9452 1.4621 -1.8836 56.3354
  2 Comments
Torsten
Torsten on 15 Aug 2022
At least this gives you some hints on how to set up your problem. The fitting results itself are bad - so I think you will have to further modify f1 and f2.

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