# Power function fitting with my data

22 views (last 30 days)

Show older comments

Ahmed Abdalazeez
on 30 Jun 2018

Commented: Ahmed Abdalazeez
on 2 Jul 2018

I have a question regarding the power function fitting.... I have attached my two pictures. They may help you to understand what I want to do.

1. I am looking for the best fitting with my data (blue curve).

using a power function respect to x. I tried one, but it's not as good as I want.

2. For linear fitting with (blue 'o') respect to x.

Thank you so much in advance.

Best regards,

Ahmed

##### 12 Comments

Image Analyst
on 2 Jul 2018

### Accepted Answer

John D'Errico
on 1 Jul 2018

Edited: John D'Errico
on 1 Jul 2018

Since you have not posted your data, it is difficult to help you very much. But...

You can use polyfit to fit the linear curve. However, I see that it is not really a straight line. And you seem to have connected only two points as you drew the line. But just use polyfit to fit a straight line. A quadratic polynomial would be a much better fit however.

Next, you described the first curve as a power function. But if you chose to use the WRONG power function, it will fit terribly poorly. In fact, the curve as it has been drawn does not look like a true power law curve. I'd guess it to be a hyperbolic curve, just looking at the shape of the tails. They look to be asymptotically linear in both tails, and there simply is no way a power fit will be linear in both tails. Anyway, if you think that a pure power fit, thus something like

a*x^b

will fit that first curve, you are simply wrong. I might consider a model of the more general form

a + b*(x-c)^d

Because both a constant offset in x and y will be necessary. It still will not have the correct shape in the upper tail from what I see.

Really, the model I would try fitting to that curve would be something like this:

y = a - b/(x-c)

It won't fit that well, because that curve has a vertical asymptote (so vertical slope on the right end) at x==c. But it would be a start. So then you would next consider a more general hyperbolic curve. I'd suggest the model, if and when you post some data.

So there are several reasons that may explain why you got a poor fit.

So what model were you trying to fit there exactly? Why do you think that curve follows a power law (in any form)?

AND POST YOUR DATA, if you want serious help.

Edit: just for kicks, with no data at all, here is a curve that is starting to look much like your curve.

yhyp = @(x,a,b,x0,y0) y0 + (a*b*(x-x0 + sqrt(- a^2 + b^2 + (x-x0).^2)))/(b^2 - a^2);

ezplot(@(x) yhyp(x,.001,10,265,0.028),[220 290])

It is an arc of a hyperbola, with linear asymptotes as x goes in both directions. The curve fitting toolbox would fit it pretty easily, although I would suggest providing starting values similar to those I used.

##### 2 Comments

John D'Errico
on 2 Jul 2018

Edited: Image Analyst
on 2 Jul 2018

Now that you have provided data for the curves y(x), though I note that you still called the variables LL and eta_norm. Sigh. You are asking for trouble in your code by doing things like that.

Using the hyperbolic model I suggested in my answer...

ft = fittype('y0 + (a*b*(x-x0 + sqrt(- a^2 + b^2 + (x-x0).^2)))/(b^2 - a^2)','independent','x','coeff',{'a','b','x0','y0'})

ft =

General model:

ft(a,b,x0,y0,x) = y0 + (a*b*(x-x0 + sqrt(- a^2 + b^2 + (x-x0).^2)))/(b^2 - a^2)

mdl = fit(LL,eta_norm,ft,'start',[.01,10,950,0.028])

mdl =

General model:

mdl(x) = y0 + (a*b*(x-x0 + sqrt(- a^2 + b^2 + (x-x0).^2)))/(b^2 - a^2)

Coefficients (with 95% confidence bounds):

a = 0.002316 (0.002088, 0.002545)

b = 25.34 (23.75, 26.92)

x0 = 945.8 (943.8, 947.9)

y0 = 0.02786 (0.0278, 0.02792)

plot(LL,eta_norm,'o')

hold on

plot(mdl)

I get a very reasonable fit. The horizontal asymptote is at y0 = 0.02786, which seems reasonable from the plot.

You can also compute the linear asymptote on the right side simply enough.

Thus, for large x, we have the limiting behavior:

sqrt(- a^2 + b^2 + (x-x0).^2) --> (x-x0)

So that model reduces, for large x, to:

y0 + 2*a*b*(x-x0)/(b^2 - a^2)

Thus the slope of the linear asymptote is 2*a*b/(b^2-a^2)

In fact, as I predicted, a hyperbolic fit does do quite nicely here, because your data really seems to approach linearity in each wing.

Note that a power function does not fit this data terribly well. In fact, a power curve fit will be more difficult to do, because of numerical issues.

### More Answers (1)

Image Analyst
on 1 Jul 2018

Since you're not providing us with data, I had to make code to create some. It's at the top of this script. But it seems to do a pretty good job.

% Uses fitnlm() to fit a non-linear model (an power law curve) through noisy data.

% Requires the Statistics and Machine Learning Toolbox, which is where fitnlm() is contained.

% Initialization steps.

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 = 20;

% Create the X coordinates: 30 points from 0.01 to 20, inclusive.

X = linspace(230, 290, 50);

% Define function that the X values obey.

a = .000005;

b = 2 % Arbitrary sample values I picked.

c = .2

d = 225

e = 0.028

Y = a * b .^ (c*( X - d)) + e; % Get a vector. No noise in this Y yet.

% Add noise to Y.

Y = Y + 0.001 * randn(1, length(Y));

% Now we have noisy training data that we can send to fitnlm().

% Plot the noisy initial data.

plot(X, Y, 'b*', 'LineWidth', 2, 'MarkerSize', 8);

grid on;

% Convert X and Y into a table, which is the form fitnlm() likes the input data to be in.

tbl = table(X', Y');

% Define the model as Y = a * (x .^ b) + c

% Note how this "x" of modelfun is related to big X and big Y.

% x((:, 1) is actually X and x(:, 2) is actually Y - the first and second columns of the table.

modelfun = @(b,x) b(1) * b(2) .^ (b(3)*(x(:, 1) - b(4))) + b(5);

beta0 = [.0001, 2, .2, 225, .028]; % Guess values to start with. Just make your best guess.

% Now the next line is where the actual model computation is done.

mdl = fitnlm(tbl, modelfun, beta0);

% Now the model creation is done and the coefficients have been determined.

% YAY!!!!

% Extract the coefficient values from the the model object.

% The actual coefficients are in the "Estimate" column of the "Coefficients" table that's part of the mode.

coefficients = mdl.Coefficients{:, 'Estimate'}

% Create smoothed/regressed data using the model:

yFitted = coefficients(1) * coefficients(2) .^ (coefficients(3)*(X - coefficients(4))) + coefficients(5);

% Now we're done and we can plot the smooth model as a red line going through the noisy blue markers.

hold on;

plot(X, yFitted, 'r-', 'LineWidth', 2);

grid on;

title('Power Law Regression with fitnlm()', 'FontSize', fontSize);

xlabel('X', 'FontSize', fontSize);

ylabel('Y', 'FontSize', fontSize);

legendHandle = legend('Noisy Y', 'Fitted Y', 'Location', 'north');

legendHandle.FontSize = 25;

message = sprintf('Coefficients for Y = a * b .^ (c*(X ^ d)) + e:\n a = %8.5f\n b = %8.5f\n c = %8.5f\n d = %8.5f\n e = %8.5f',...

coefficients(1), coefficients(2), coefficients(3), coefficients(5), coefficients(5));

text(235, .055, message, 'FontSize', 23, 'Color', 'r', 'FontWeight', 'bold', 'Interpreter', 'none');

% Set up figure properties:

% Enlarge figure to full screen.

set(gcf, 'Units', 'Normalized', 'OuterPosition', [0, 0.04, 1, 0.96]);

% Get rid of tool bar and pulldown menus that are along top of figure.

% set(gcf, 'Toolbar', 'none', 'Menu', 'none');

% Give a name to the title bar.

set(gcf, 'Name', 'Demo by ImageAnalyst', 'NumberTitle', 'Off')

##### 0 Comments

### See Also

### Categories

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!