Asked by sisay
on 16 Aug 2013

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?

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Answer by Matt J
on 16 Aug 2013

Edited by Matt J
on 16 Aug 2013

Accepted answer

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);

Show 15 older comments

Image Analyst
on 16 Aug 2013

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
on 16 Aug 2013

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
on 16 Aug 2013

Thank you guys for your helping me

Answer by Walter Roberson
on 16 Aug 2013

*Any* fixed degree that you use will result in a polynomial that tends to be infinitely wrong as x tends to infinity.

sisay
on 16 Aug 2013

Is there anyway of fitting this type of equation then? I find it difficult.

Answer by Image Analyst
on 16 Aug 2013

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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