Fitting equation in matlab
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I have an equation of the form y= a0+a1log(x)+ a2log(1/x)
I want to use polyfit but I don't know how to fix the degree of the polynomial in this case. Can anyone help me please?
采纳的回答
Since log(1/x)=-log(x) your equation model has redundant terms. It is equivalent to
y = a0+(a1-a2)*log(x)
= A+B*log(x)
where a0 has been relabeled as A and B has replaced a1-a2.
You could fit A and B, I suppose, by doing
AB=polyfit(log(x),y,1);
A=AB(2);
B=AB(1);
18 个评论
sisay
2013-8-16
Matt J
2013-8-16
If you agreed with me when I told you that your equation has redundant terms, then you shouldn't be interested in a1 and a2 anymore. You should be interested only in B.
sisay
2013-8-16
There is no unique solution for a1 and a2. Once you have B, any pair of a1 and a2 satisyfing
B=a1-a2
will produce the same fit. That's what happens when you try to fit an equation with redundant terms.
sisay
2013-8-16
sisay
2013-8-16
It doesn't change anything. After re-arranging the equation, you will have
y = a0+(a1-a2)log(x)+(a2-a1)*log(22)
= A+B*log(x)
which is the same form as before with the change of variables
A=a0+(a2-a1)*log(22)
B=a2-a1
You could also rewrite as
y=A+B*(log(x/22)
and do
AB=polyfit(log(x/22),y,1)
Image Analyst
2013-8-16
Why not just say a2=0? Since it's only the sum that has to be a certain value (B), then just say a2=0, and a1=B and be done with it.
sisay
2013-8-16
sisay
2013-8-16
Walter Roberson
2013-8-16
If a2 cannot be 0, can it be 1?
Image Analyst
2013-8-16
But what you're saying doesn't make sense. See my code below where I set a2 = 0 and it does the fit just fine. If you think it doesn't then say why. If you want some different value of a2, then pick one and I'll adjust the a1 and it will fit just fine again.
Image Analyst
2013-8-16
sisay, you don't have enough parameters to pin down both a1 and a2. That's what everyone is trying to tell you. You know the laws of logarithms don't you? So you can see that a1 and a2 are always go together in a pair "a1-a2" - there is no way to specify each, all you can do is to specify the difference "a1-a2". Why did my code below not convince you of that?
I have no idea what "that finally used ss check ups" means. But when Matt asked you if there were some other constraints that you weren't telling us, you said no. So in that case, all answerers are in agreement that a2 can be 0 or 1 or anything you want it to be because a1 will just adjust so that the difference is what it needs to be. Why can't you follow our arguments/explanations? Did you read my comments in my code where I proved that?
Walter Roberson
2013-8-16
Your equation is underdetermined. There are an infinite number of (a1, a2) pairs that will work in your equation; everything along the ray starting from a2 = epsilon (epsilon being positive and arbitrarily close to zero) and up, with only the difference calculatable.
Unless, that is, you have additional information that can be used to constrain the two values.
sisay
2013-8-16
更多回答（2 个）
Walter Roberson
2013-8-16
0 个投票
Any fixed degree that you use will result in a polynomial that tends to be infinitely wrong as x tends to infinity.
Image Analyst
2013-8-16
sisay, try this:
clc; % Clear the command window.
close all; % Close all figures (except those of imtool.)
clear; % Erase all existing variables. Or clearvars if you want.
workspace; % Make sure the workspace panel is showing.
format long g;
format compact;
fontSize = 24;
% Construct x.
x = linspace(.01, 40, 50);
a0 = 1;
a1 = 2;
a2 = 3;
% Create the perfect equation.
y = a0 + a1 * log(x)+ a2 * log(1./x);
subplot(3,1,1);
plot(x, y, 'b.-');
title('Noise-free signal', 'FontSize', fontSize);
% Enlarge figure to full screen.
set(gcf, 'units','normalized','outerposition',[0 0 1 1]);
% Add some noise to make a noisy signal that we will fit.
yNoisy = y + 1.5 * rand(1, length(y));
subplot(3,1,2);
plot(x, yNoisy, 'b.-');
title('Noisy signal', 'FontSize', fontSize);
% Now get the fit
% y = a0 + a1 * log(x) - a2 * log(x)
% y = a0 + (a1 - a2) * log(x)
% y = (a1 - a2) * log(x) + a0
% Let newX = log(x), and (a1-a2) = coeffs(1), then
% y = coefficients(1) * newX + coefficients(2)
% so now we can use polyfit to fit a line.
newX = log(x);
coefficients = polyfit(newX, yNoisy, 1);
% Now get the fitted values
a0 = coefficients(2);
a1 = coefficients(1);
a2 = 0; % Might as well be 0 as any other value.
yFitted = a0 + a1 * log(x)+ a2 * log(1./x);
% and plot them
subplot(3,1,3);
plot(x, yNoisy, 'b.');
hold on;
plot(x, yFitted, 'r-', 'LineWidth', 3);
title('Fitted signal', 'FontSize', fontSize);
legend('Noisy data', 'Fitted signal');
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