mat =xlsread('example2.xlsx','Sheet1','A2:C32');
density = mat(:,1);
Z_MD = mat(:,3);
eta = mat(:,2);
eta_c = 1/1.55;
Your model is:
Z(eta) = 1 + 3*eta/(eta_c - eta) + sum(k*A_k*(eta/eta_c)^k)
where the sum operates over k = 1:4. The goal being a minimax fit. Since the first two terms in that model are not variable at all, we will just subtract them for the fit. But at first, plot EVERYTHING. Always plot your data. Look at it. Think about it.
The result being a simple power type function. It might seem things will work at least reasonably well, as polynomials like that sort of stuff. So plot more. Think more. Look at what we see. First, remember that your function has a singularity in it. The singularity lives at eta==eta_c.
eta_c
eta_c =
0.64516
plot(eta,Z_MD - (1 + 3*eta./(eta_c - eta)),'ro')
So this is the function you wish to fit with a polynomial.
The first problem is, it does not seem to pass throughzero at eta==0. It is less than zero down there.
Z(eta) - 1 - 3*eta/(eta_c - eta) = sum(k*A_k*(eta/eta_c)^k)
You should observe that the right hand side is identically zero when eta == 0. The left hand side (what I just plotted)is not so. So this will create some amount of lack of fit down there. But worse, what happens when eta approaches 0.64516? The left hand side has a singularity.
Does the right hand side? NO. There can be NO singularities in a polynomial. And the right hand side is purely a polynomial, and only a 4th degree one at that.
Do you really want to go through with this? Sigh.
First, ets see what happens when we just use a simple linear least squares. Backslash can handle that well enough.
A2 = [(eta./eta_c).^(1:4).*(1:4)]\(Z_MD - (1 + 3*eta./(eta_c - eta)))
A2 =
5.302
-20.991
33.224
-16.441
So A2 is the 2-norm solution. Given those coefficients, we can look at the fit.
Zpred = [(eta./eta_c).^(1:4).*(1:4)]*A2;
plot(eta,Z_MD - (1 + 3*eta./(eta_c - eta)),'ro')
hold on
grid on
plot(eta,Zpred,'b-')
Sigh. Double sigh. Well, it does look reasonably close to a minimax fit as it is. A 2-norm fit like this will often not be unreasonable. Lets look at the residuals.
plot(eta,Z_MD - (1 + 3*eta./(eta_c - eta)) - Zpred,'ro')
grid on
yline(0);
So indeed pretty close to a minimax fit, as I said. It only deviates by a bit. Is it really worth the effort, for a fit that does not fit that well in the first place?
I'd suggest that looking for a true minimax fit here is a bit extreme, given the significant problems in the fit. Do you really need it? Regardless, fminimax is much easier to use than you seem to be making it appear. Two lines suffice, although it would always be a good idea to use the 2-norm solution as a start point. So three lines.
mmfun = @(A) abs(Z_MD - 1 - 3*eta./(eta_c - eta) - [(eta./eta_c).^(1:4).*(1:4)]*A);
[Amm,~,maxres,exitflag] = fminimax(mmfun,A2)
Local minimum possible. Constraints satisfied.
fminimax stopped because the size of the current search direction is less than
twice the default value of the step size tolerance and constraints are
satisfied to within the default value of the constraint tolerance.
<stopping criteria details>
Amm =
5.7492
-23.245
36.892
-18.375
maxres =
0.15832
exitflag =
4
So a little different from A2. But honestly, hardly worth the effort. Note that I was carful to use abs in mmfun, since fminimax minimizes the maxima, but not the absolute max. From the help:
fminimax finds a minimax solution of a function of several variables.
fminimax attempts to solve the following problem:
min (max {FUN(X} ) where FUN and X can be vectors or matrices.
So you do need an abs in there.
Zpredmm = [(eta./eta_c).^(1:4).*(1:4)]*Amm;
plot(eta,Z_MD - (1 + 3*eta./(eta_c - eta)) - Zpredmm,'ro')
yline(0);
grid on
From a different perspective, can you honestly see the difference below, between the green and blue curves? Is that difference significant?
plot(eta,Z_MD,'ro',eta,(1 + 3*eta./(eta_c - eta)) + Zpredmm,'b-',eta,(1 + 3*eta./(eta_c - eta)) + Zpred,'g-')
grid on
Whatever floats your boat. Personally, I think focusing on a minimax approximation here is somewhat misguided, because there are essential flaws in that approximation that far exceed the tiny differences.
Best Answer