I have a curve that looks like an exponentiel function, I would like to fit this curve with this equation :
The goal is to find the value of A, T and d which will minimize V with my initial curve.
I did a function that is able to do it but it takes 10 seconds to run.
3 loops that test all the values that I want to and at the end of the 3 loops I calculate the RMSD (root mean square deviation) between my 2 curves and I put the result in a vector min_RMSN, at the end I check the minimum value of min_RMSD and it's done...
But this is not the best way for sure.
Thank you for your help, ideas :)
Matlab has a built in fminsearch function that does pretty much exactly what you want. You define a function handle that takes the RMSE of your data vs. the function fit, pass in your initial guess for A, T and d, and get a result:
x0 = [A0, T0, d0]
fn = #(x) sum((x(1) * (1 - exp(-x[2] / (t - x[3]))) - y).^2)
V = fminsearch(#fn, x0)
Here t is the x-data for the curve you have, y are the corresponding y-values and, A0, T0, d0 are the initial guesses for your parameters. fn computes the suquare of the RMSE between your ideal curve and y. No need to take the square root since you minimizing the square will also minimize the RMSE itself, and computing square roots takes time.
Related
I have a Matlab function G(x,y,z). At each given (x,y,z), G(x,y,z) is a scalar. x=(x1,x2,...,xK) is a Kx1 vector.
Let us fix y,z at some given values. I would like your help to understand how to compute the derivative of G with respect to xk evaluated at a certain x.
For example, suppose K=3
function f= G(x1,x2,x3,y,z)
f=3*x1*sin(z)*cos(y)+3*x2*sin(z)*cos(y)+3*x3*sin(z)*cos(y);
end
How do I compute the derivative of G(x1,x2,x3,4,3) wrto x2 and then evaluate it at x=(1,2,6)?
You're looking for the partial derivative of dG/dx2
So the first thing would be getting rid of your fixed variables
G2 = #(x2) G(1,x2,6,4,3);
The numerical derivatives are finite differences, you need to choose an step h for your finite difference, and an appropriate method
The simplest one is
(G2(x2+h)-G2(x2))/h
You can make h as small as your numeric precision allows you to. At the limit h -> 0 the finite difference is the partial derivative
The task is to create a cone hat in Matlab by creating a developable surface with numerical methods. There are 3 parts of which I have done 2. My question is regarding part 3 where I need to calculate the least rectangular paper surface that can contain the hat. And I need to calculate the material waste of the paper.
YOU CAN MAYBE SKIP THE LONG BACKGROUND AND GO TO LAST PARAGRAPH
BACKGROUND:
The cone hat can be created with a skewed cone with its tip located at (a; 0; b) and with a circle-formed base.
x = Rcos u,
y = Rsin u
z = 0
0<_ u >_2pi
with
known values for R, a and b
epsilon and eta ('n') is the curves x- and y- values when the parameter u goes from 0 to 2pi and alpha is the angle of inclination for the curve at point (epsilon, eta). Starting values at A:
u=0, alhpa=0, epsilon=0, eta=0.
Curve stops at B where the parameter u has reached 2pi.
1.
I plotted the curve by using Runge-Kutta-4 and showed that the tip is located at P = (0, sqrt(b^2 + (R-alpha)^2))
2.
I showed that by using smaller intervals in RK4 I still get quite good accuracy but the problem then is that the curve isn't smooth. Therefore I used Hermite-Interpolation of epsilon and eta as functions of u in every interval to get a better curve.
3.
Ok so now I need to calculate the least rectangular paper surface that can contain the hat and the size of the material waste of the paper. If the end angle alpha(2pi) in the template is pi or pi/2 the material waste will be less. I now get values for R & alpha (R=7.8 and alpha=5.5) and my task is to calculate which height, b the cone hat is going to get with the construction-criteria alpha(2pi)=pi (and then for alpha(2pi)=pi/2 for another sized hat).
So I took the first equation above (the expression containing b) and rewrote it like an integral:
TO THE QUESTION
What I understand is that I need to solve this integral in matlab and then choose b so that alpha(2pi)-pi=0 (using the given criteria above).
The values for R and alpha is given and t is defined as an interval earlier (in part 1 where I did the RK4). So when the integral is solved I get f(b) = 0 which I should be able to solve with for example the secant method? But I'm not able to solve the integral with the matlab function 'integral'.. cause I don't have the value of b of course, that is what I am looking for. So how am I going to go about this? Is there a function in matlab which can be used?
You can use the differential equation for alpha and test different values for b until the condition alpha(2pi)=pi is met. For example:
b0=1 %initial seed
b=fsolve(#find_b,b0) %use the function fsolve or any of your choice
The function to be solved is:
function[obj]=find_b(b)
alpha0=0 %initual valur for alpha at u=0
uspan=[0 2*pi] %range for u
%Use the internal ode solver or any of your choice
[u,alpha] = ode45(#(u,alpha) integrate_alpha(u,alpha,b), uspan, alpha0);
alpha_final=alpha(end) %Get the last value for alpha (at u=2*pi)
obj=alpha_final-pi %Function to be solved
end
And the integration can be done like this:
function[dalpha]=integrate_alpha(u,alpha,b)
a=1; %you can use the right value here
R=1; %you can use the right value here
dalpha=(R-a*cos(u))/sqrt(b^2+(R-a*cos(u))^2);
end
I have the following equation:
I want to do a exponential curve fitting using MATLAB for the above equation, where y = f(u,a). y is my output while (u,a) are my inputs. I want to find the coefficients A,B for a set of provided data.
I know how to do this for simple polynomials by defining states. As an example, if states= (ones(size(u)), u u.^2), this will give me L+Mu+Nu^2, with L, M and N being regression coefficients.
However, this is not the case for the above equation. How could I do this in MATLAB?
Building on what #eigenchris said, simply take the natural logarithm (log in MATLAB) of both sides of the equation. If we do this, we would in fact be linearizing the equation in log space. In other words, given your original equation:
We get:
However, this isn't exactly polynomial regression. This is more of a least squares fitting of your points. Specifically, what you would do is given a set of y and set pair of (u,a) points, you would build a system of equations and solve for this system via least squares. In other words, given the set y = (y_0, y_1, y_2,...y_N), and (u,a) = ((u_0, a_0), (u_1, a_1), ..., (u_N, a_N)), where N is the number of points that you have, you would build your system of equations like so:
This can be written in matrix form:
To solve for A and B, you simply need to find the least-squares solution. You can see that it's in the form of:
Y = AX
To solve for X, we use what is called the pseudoinverse. As such:
X = A^{*} * Y
A^{*} is the pseudoinverse. This can eloquently be done in MATLAB using the \ or mldivide operator. All you have to do is build a vector of y values with the log taken, as well as building the matrix of u and a values. Therefore, if your points (u,a) are stored in U and A respectively, as well as the values of y stored in Y, you would simply do this:
x = [u.^2 a.^3] \ log(y);
x(1) will contain the coefficient for A, while x(2) will contain the coefficient for B. As A. Donda has noted in his answer (which I embarrassingly forgot about), the values of A and B are obtained assuming that the errors with respect to the exact curve you are trying to fit to are normally (Gaussian) distributed with a constant variance. The errors also need to be additive. If this is not the case, then your parameters achieved may not represent the best fit possible.
See this Wikipedia page for more details on what assumptions least-squares fitting takes:
http://en.wikipedia.org/wiki/Least_squares#Least_squares.2C_regression_analysis_and_statistics
One approach is to use a linear regression of log(y) with respect to u² and a³:
Assuming that u, a, and y are column vectors of the same length:
AB = [u .^ 2, a .^ 3] \ log(y)
After this, AB(1) is the fit value for A and AB(2) is the fit value for B. The computation uses Matlab's mldivide operator; an alternative would be to use the pseudo-inverse.
The fit values found this way are Maximum Likelihood estimates of the parameters under the assumption that deviations from the exact equation are constant-variance normally distributed errors additive to A u² + B a³. If the actual source of deviations differs from this, these estimates may not be optimal.
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I'm going to write a program where the input is a data set of 2D points and the output is the regression coefficients of the line of best fit by minimizing the minimum MSE error.
I have some sample points that I would like to process:
X Y
1.00 1.00
2.00 2.00
3.00 1.30
4.00 3.75
5.00 2.25
How would I do this in MATLAB?
Specifically, I need to get the following formula:
y = A + Bx + e
A is the intercept and B is the slope while e is the residual error per point.
Judging from the link you provided, and my understanding of your problem, you want to calculate the line of best fit for a set of data points. You also want to do this from first principles. This will require some basic Calculus as well as some linear algebra for solving a 2 x 2 system of equations. If you recall from linear regression theory, we wish to find the best slope m and intercept b such that for a set of points ([x_1,y_1], [x_2,y_2], ..., [x_n,y_n]) (that is, we have n data points), we want to minimize the sum of squared residuals between this line and the data points.
In other words, we wish to minimize the cost function F(m,b,x,y):
m and b are our slope and intercept for this best fit line, while x and y are a vector of x and y co-ordinates that form our data set.
This function is convex, so there is an optimal minimum that we can determine. The minimum can be determined by finding the derivative with respect to each parameter, and setting these equal to 0. We then solve for m and b. The intuition behind this is that we are simultaneously finding m and b such that the cost function is jointly minimized by these two parameters. In other words:
OK, so let's find the first quantity :
We can drop the factor 2 from the derivative as the other side of the equation is equal to 0, and we can also do some distribution of terms by multiplying the -x_i term throughout:
Next, let's tackle the next parameter :
We can again drop the factor of 2 and distribute the -1 throughout the expression:
Knowing that is simply n, we can simplify the above to:
Now, we need to simultaneously solve for m and b with the above two equations. This will jointly minimize the cost function which finds the best line of fit for our data points.
Doing some re-arranging, we can isolate m and b on one side of the equations and the rest on the other sides:
As you can see, we can formulate this into a 2 x 2 system of equations to solve for m and b. Specifically, let's re-arrange the two equations above so that it's in matrix form:
With regards to above, we can decompose the problem by solving a linear system: Ax = b. All you have to do is solve for x, which is x = A^{-1}*b. To find the inverse of a 2 x 2 system, given the matrix:
The inverse is simply:
Therefore, by substituting our quantities into the above equation, we solve for m and b in matrix form, and it simplifies to this:
Carrying out this multiplication and solving for m and b individually, this gives:
As such, to find the best slope and intercept to best fit your data, you need to calculate m and b using the above equations.
Given your data specified in the link in your comments, we can do this quite easily:
%// Define points
X = 1:5;
Y = [1 2 1.3 3.75 2.25];
%// Get total number of points
n = numel(X);
% // Define relevant quantities for finding quantities
sumxi = sum(X);
sumyi = sum(Y);
sumxiyi = sum(X.*Y);
sumxi2 = sum(X.^2);
sumyi2 = sum(Y.^2);
%// Determine slope and intercept
m = (sumxi * sumyi - n*sumxiyi) / (sumxi^2 - n*sumxi2);
b = (sumxiyi * sumxi - sumyi * sumxi2) / (sumxi^2 - n*sumxi2);
%// Display them
disp([m b])
... and we get:
0.4250 0.7850
Therefore, the line of best fit that minimizes the error is:
y = 0.4250*x + 0.7850
However, if you want to use built-in MATLAB tools, you can use polyfit (credit goes to Luis Mendo for providing the hint). polyfit determines the line (or nth order polynomial curve rather...) of best fit by linear regression by minimizing the sum of squared errors between the best fit line and your data points. How you call the function is so:
coeff = polyfit(x,y,order);
x and y are the x and y points of your data while order determines the order of the line of best fit you want. As an example, order=1 means that the line is linear, order=2 means that the line is quadratic and so on. Essentially, polyfit fits a polynomial of order order given your data points. Given your problem, order=1. As such, given the data in the link, you would simply do:
X = 1:5;
Y = [1 2 1.3 3.75 2.25];
coeff = polyfit(X,Y,1)
coeff =
0.4250 0.7850
The way coeff works is that these are the coefficients of the regression line, starting from the highest order in decreasing value. As such, the above coeff variable means that the regression line was fitted as:
y = 0.4250*x + 0.7850
The first coefficient is the slope while the second coefficient is the intercept. You'll also see that this matches up with the link you provided.
If you want a visual representation, here's a plot of the data points as well as the regression line that best fits these points:
plot(X, Y, 'r.', X, polyval(coeff, X));
Here's the plot:
polyval takes an array of coefficients (usually produced by polyfit), and you provide a set of x co-ordinates and it calculates what the y values are given the values of x. Essentially, you are evaluating what the points are along the best fit line.
Edit - Extending to higher orders
If you want to extend so that you're finding the best fit for any nth order polynomial, I won't go into the details, but it boils down to constructing the following linear system. Given the relationship for the ith point between (x_i, y_i):
You would construct the following linear system:
Basically, you would create a vector of points y, and you would construct a matrix X such that each column denotes taking your vector of points x and applying a power operation to each column. Specifically, the first column is the zero-th power, the first column is the first power, the second column is the second power and so on. You would do this up until m, which is the order polynomial you want. The vector of e would be the residual error for each point in your set.
Specifically, the formulation of the problem can be written in matrix form as:
Once you construct this matrix, you would find the parameters by least-squares by calculating the pseudo-inverse. How the pseudo-inverse is derived, you can read it up on the Wikipedia article I linked to, but this is the basis for minimizing a system by least-squares. The pseudo-inverse is the backbone behind least-squares minimization. Specifically:
(X^{T}*X)^{-1}*X^{T} is the pseudo-inverse. X itself is a very popular matrix, which is known as the Vandermonde matrix and MATLAB has a command called vander to help you compute that matrix. A small note is that vander in MATLAB is returned in reverse order. The powers decrease from m-1 down to 0. If you want to have this reversed, you'd need to call fliplr on that output matrix. Also, you will need to append one more column at the end of it, which is the vector with all of its elements raised to the mth power.
I won't go into how you'd repeat your example for anything higher order than linear. I'm going to leave that to you as a learning exercise, but simply construct the vector y, the matrix X with vander, then find the parameters by applying the pseudo-inverse of X with the above to solve for your parameters.
Good luck!
My project require me to use Matlab to create a symbolic equation with square wave inside.
I tried to write it like this but to no avail:
syms t;
a=square(t);
Input arguments must be 'double'.
What can i do to solve this problem? Thanks in advance for the helps offered.
here are a couple of general options using floor and sign functions:
f=#(A,T,x0,x) A*sign(sin((2*pi*(x-x0))/T));
f=#(A,T,x0,x) A*(-1).^(floor(2*(x-x0)/T));
So for example using the floor function:
syms x
sqr=2*floor(x)-floor(2*x)+1;
ezplot(sqr, [-2, 2])
Here is something to get you started. Recall that we can express a square wave as a Fourier Series expansion. I won't bother you with the details, but you can represent any periodic function as a summation of cosines and sines (à la #RTL). Without going into the derivation, this is the closed-form equation for a square wave of frequency f, with a peak-to-peak amplitude of 2 (i.e. it goes from -1 to 1). Recall that the frequency is the amount of cycles per seconds. Therefore, f = 1 means that we repeat our square wave every second.
Basically, what you have to do is code up the first line of the equation... but how in the world would you do that? Welcome to the world of the Symbolic Math Toolbox. What we will need to do before hand is declare what our frequency is. Let's assume f = 1 for now. With the Symbolic Math Toolbox, you can define what are considered as mathematics variables within MATLAB. After, MATLAB has a whole suite of tools that you can use to evaluate functions that rely on these variables. A good example would be if you want to use this to define a closed-form solution of a function f(x). You can then use diff to differentiate and see what the derivative is. Try it yourself:
syms x;
f = x^4;
df = diff(f);
syms denotes that you are declaring anything coming after the statement to be a mathematical variable. In this case, x is just that. df should now give you 4x^3. Cool eh? In any case, let's get back to our problem at hand. We see that there are in fact two variables in the periodic square function that need to be defined: t and k. Once we do this, we need to create our function that is inside the summation first. We can do this by:
syms t k;
f = 1; %//Define frequency here
funcSum = (sin(2*pi*(2*k - 1)*f*t) / (2*k - 1));
That settles that problem... now how do we encapsulate this into an infinite sum!? The sum command in MATLAB assumes that we have a finite array to sum over. If you want to symbolically sum over a function, we must use the symsum function. We usually call it like this:
funcOut = symsum(func, v, start, finish);
func is the function we wish to sum over. v is the summation variable that we wish to use to index in the sum. In our case, that's k. start is the beginning of the sum, which is 1 in our case, and finish is where we wish to finish up our summation. In our case, that's infinity, and so MATLAB has a special keyword called Inf to denote that. Therefore:
xsquare = (4/pi) * symsum(funcSum, k, 1, Inf);
xquare now contains your representation of a square wave defined in terms of the Symbolic Math Toolbox. Now, if you want to plot your square wave and see if we have this right. We can do the following. Let's go between -3 <= t <= 3. As such, you would do something like this:
tVector = -3 : 0.01 : 3; %// Choose a step size of 0.01
yout = subs(xsquare, t, tVector);
You will notice though that there will be some values that are NaN. The reason why is because right at a multiple of the period (T = 1, 2, 3, ...), the behaviour is undefined as the derivative right at these points is undefined. As such, we can fill this in using either 1 or -1. Let's just choose 1 for now. Also, because the Fourier Series is generally a complex-valued function, and the square-wave is purely real, the output of this function will actually give you a complex-valued vector. As such, simply chop off the complex parts to get the real parts only:
yout = real(double(yout)); %// To cast back to double.
yout(isnan(yout)) = 1;
plot(tVector, yout);
You'll get something like:
You could also do this the ezplot way by doing: ezplot(xsquare). However, you'll see that at the points where the wave repeats itself, we get NaN values and so there is a disconnect between the high peak and low peak.
Note:
Natan's solution is much more elegant. I was still writing this post by the time he put something up. Either way, I wanted to give a more signal processing perspective to how to do this. Go Fourier!
A Fourier series for the square wave of unit amplitude is:
alpha + 2/Pi*sum(sin( n * Pi*alpha)/n*cos(n*theta),n=1..infinity)
Here is a handy trick:
cos(n*theta) = Re( exp( I * n * theta))
and
1/n*exp(I*n*theta) = I*anti-derivative(exp(I*n*theta),theta)
Put it all together: pull the anti-derivative ( or integral ) operator out of the sum, and you get a geometric series. Then integrate and finally take the real part.
Result:
squarewave=
alpha+ 1/Pi*Re(I*ln((1-exp(I*(theta+Pi*alpha)))/(1-exp(I*(theta-Pi*alpha)))))
I tried it in MAPLE and it works great! (probably not very practical though)