I have three vectors, one of X locations, another of Y locations and the third is a f(x, y). I want to find the algebraic expression interpolation polynomial (using matlab) since I will later on use the result in an optimization problem in AMPL. As far as I know, there are not any functions that return the interpolation polynomial.
I have tried https://la.mathworks.com/help/matlab/ref/griddedinterpolant.html, but this function only gives the interpolated values at certain points.
I have also tried https://la.mathworks.com/help/matlab/ref/triscatteredinterp.html as sugested in Functional form of 2D interpolation in Matlab, but the output isn't the coefficents of the polynomial. I cannot see it, it seems to be locked inside of a weird variable.
This is a small program that I have done to test what I am doing:
close all
clear
clc
[X,Y] = ndgrid(1:10,1:10);
V = X.^2 + 3*(Y).^2;
F = griddedInterpolant(X,Y,V,'cubic');
[Xq,Yq] = ndgrid(1:0.5:10,1:0.5:10);
Vq = F(Xq,Yq);
mesh(Xq,Yq,Vq)
figure
mesh(X, Y, V)
I want an output that instead of returning the value at grid points returns whatever it has used to calculate said values. I am aware that it can be done in mathematica with https://reference.wolfram.com/language/ref/InterpolatingPolynomial.html, so I find weird that matlab can't.
You can use fit if you have the curve fitting toolbox.
If it's not the case you can use a simple regression, if I take your example:
% The example data
[X,Y] = ndgrid(1:10,1:10);
V = X.^2 + 3*(Y).^2;
% The size of X
s = size(X(:),1);
% Let's suppose that you want to fit a polynome of degree 2.
% Create all the possible combination for a polynome of degree 2
% cst x y x^2 y^2 x*y
A = [ones(s,1), X(:), Y(:), X(:).^2, Y(:).^2, X(:).*Y(:)]
% Then using mldivide
p = A\V(:)
% We obtain:
p =
0 % cst
0 % x
0 % y
1 % x^2
3 % y^2
0 % x*y
Related
I have this hard coded version which fits data to a curve for linear, quadratic and cubic polynomials:
for some data x and a function y
M=[x.^0 x.^1];
L=[x.^0 x.^1 x.^2];
linear = (M'*M)\(M'*y);
plot(x, linear(1)+linear(2)*x, ';linear;r');
deg2 = (L'*L)\(L'*y);
plot(x, deg2(1)+deg2(2)*x+deg2(3)*(x.*x), ';quadratic plot;b');
I am wondering how can I turn this into a for loop to plot curves for degree n polynomials? The part I'm stuck on is the plotting part, how would I be able to translate the increase in the number of coefficients in to the for loop?
what I have:
for i = 1:5 % say we're trying to plot curves up to degree 5 polynomials...
curr=x.^(0:i);
degI = (curr'*curr)\(curr'*y);
plot(x, ???) % what goes in here<-
end
If it is only the plotting, you can use the polyval function to evaluate polynomials of desired grade by supplying a vector of coefficients
% For example, some random coefficients for a 5th order polynomial
% degI = (curr'*curr)\(curr'*y) % Your case
degi = [3.2755 0.8131 0.5950 2.4918 4.7987 1.5464]; % for 5th order polynomial
x = linspace(-2, 2, 10000);
hold on
% Using polyval to loop over the grade of the polynomials
for i = 1:length(degI)
plot(x, polyval(degI(1:i), x))
end
gives the all polynomials in one plot
I believe this should answer your question exactly. You just have to be careful with the matrix dimensions.
for i = 1:5
curr=x.^(0:i);
degI = (curr'*curr)\(curr'*y);
plot(x, x.^(0:i)*degI)
end
Interpolate the Runge function of Example 10.6 at Chebyshev points for n from 10 to 170
in increments of 10. Calculate the maximum interpolation error on the uniform evaluation
mesh x = -1:.001:1 and plot the error vs. polynomial degree as in Figure 10.8 using
semilogy. Observe spectral accuracy.
The runge function is given by: f(x) = 1 / (1 + 25x^2)
My code so far:
x = -1:0.001:1;
n = 170;
i = 10:10:170;
cx = cos(((2*i + 1)/(2*(n+1)))*pi); %chebyshev pts
y = 1 ./ (1 + 25*x.^2); %true fct
%chebyshev polynomial, don't know how to construct using matlab
yc = polyval(c, x); %graph of approx polynomial fct
plot(x, yc);
mErr = (1 / ((2.^n).*(n+1)!))*%n+1 derivative of f evaluated at max x in [-1,1], not sure how to do this
%plotting stuff
I know very little matlab, so I am struggling on creating the interpolating polynomial. I did some google work, but I was confused with the current functions as I didn't find one that just simply took in points and the polynomial to be interpolated. I am also a bit confused in this case of whether I should be doing i = 0:1:n and n=10:10:170 or if n is fixed here. Any help is appreciated, thank you
Since you know very little about MATLAB, I will try explain everything step by step:
First, to visualize the Runge function, you can type:
f = #(x) 1./(1+25*x.^2); % Runge function
% plot Runge function over [-1,1];
x = -1:1e-3:1;
y = f(x);
figure;
plot(x,y); title('Runge function)'); xlabel('x');ylabel('y');
The #(x) part of the code is a function handle, a very useful feature of MATLAB. Notice the function is properly vecotrized, so it can receive as an argument a variable or an array. The plot function is straightforward.
To understand the Runge phenomenon, consider a linearly spaced vector of [-1,1] of 10 elements and use these points to obtain the interpolating (Lagrange) polynomial. You get the following:
% 10 linearly spaced points
xc = linspace(-1,1,10);
yc = f(xc);
p = polyfit(xc,yc,9); % gives the coefficients of the polynomial of degree 10
hold on; plot(xc,yc,'o',x,polyval(p,x));
The polyfit function does a polynomial curve fitting - it obtains the coefficients of the interpolating polynomial, given the poins x,y and the degree of the polynomial n. You can easily evaluate the polynomial at other points with the polyval function.
Obseve that, close to the end domains, you get an oscilatting polynomial and the interpolation is not a good approximation of the function. As a matter of fact, you can plot the absolute error, comparing the value of the function f(x) and the interpolating polynomial p(x):
plot(x,abs(y-polyval(p,x))); xlabel('x');ylabel('|f(x)-p(x)|');title('Error');
This error can be reduced if, instead of using a linearly space vector, you use other points to do the interpolation. A good choice is to use the Chebyshev nodes, which should reduce the error. As a matter of fact, notice that:
% find 10 Chebyshev nodes and mark them on the plot
n = 10;
k = 1:10; % iterator
xc = cos((2*k-1)/2/n*pi); % Chebyshev nodes
yc = f(xc); % function evaluated at Chebyshev nodes
hold on;
plot(xc,yc,'o')
% find polynomial to interpolate data using the Chebyshev nodes
p = polyfit(xc,yc,n-1); % gives the coefficients of the polynomial of degree 10
plot(x,polyval(p,x),'--'); % plot polynomial
legend('Runge function','Chebyshev nodes','interpolating polynomial','location','best')
Notice how the error is reduced close to the end domains. You don't get now that high oscillatory behaviour of the interpolating polynomial. If you plot the error, you will observe:
plot(x,abs(y-polyval(p,x))); xlabel('x');ylabel('|f(x)-p(x)|');title('Error');
If, now, you change the number of Chebyshev nodes, you will get an even better approximation. A little modification on the code lets you run it again for different numbers of nodes. You can store the maximum error and plot it as a function of the number of nodes:
n=1:20; % number of nodes
% pre-allocation for speed
e_ln = zeros(1,length(n)); % error for the linearly spaced interpolation
e_cn = zeros(1,length(n)); % error for the chebyshev nodes interpolation
for ii=1:length(n)
% linearly spaced vector
x_ln = linspace(-1,1,n(ii)); y_ln = f(x_ln);
p_ln = polyfit(x_ln,y_ln,n(ii)-1);
e_ln(ii) = max( abs( y-polyval(p_ln,x) ) );
% Chebyshev nodes
k = 1:n(ii); x_cn = cos((2*k-1)/2/n(ii)*pi); y_cn = f(x_cn);
p_cn = polyfit(x_cn,y_cn,n(ii)-1);
e_cn(ii) = max( abs( y-polyval(p_cn,x) ) );
end
figure
plot(n,e_ln,n,e_cn);
xlabel('no of points'); ylabel('maximum absolute error');
legend('linearly space','chebyshev nodes','location','best')
[X,Y] = meshgrid(-8:.5:8);
R = sqrt(X.^2 + Y.^2) + eps;
Z = sin(R)./R;
scatter3(X,Y,Z)
Error using scatter3 (line 64)
X, Y and Z must be vectors of the same length.
Matlab R2018b windows x64
As shown in the documentation, X, Y, Z must be vectors. (When you enter an article on mathworks from Googling, say, "matlab scatter3", you will first see the syntax for the function. Blue text means hyperlink. All the inputs are linked to the bottom of the page where their exact typing is defined.)
The reason is (probably) as follows.
As stated in the documentation, scatter3 puts circles (or other symbols of your choice if you modify the graphic object) on 3D coordinates of your choice. The coordinates are the ith element of X, Y, Z respectively. For example, the x-coordinate of the 10th point you wish to plot in 3D is X(10).
Thus it is not natural to input matrices into scatter3. If you know X(i), Y(i), Z(i) are indeed the coordinates you want to plot for all i, even though your X, Y, Z are not vectors for some reason, you need to reshape X, Y, Z.
In order to reshape, you can simply do scatter3(X(:), Y(:), Z(:)) which tells Matlab to read your arrays as a vectors. (You should look up in what order this is done. But it is in the intuitive way.) Or you can use reshape. Chances are: reshape is faster for large data set. But ofc (:) is more convenient.
The following should work:
[X,Y] = meshgrid(-8:.5:8);
R = sqrt(X.^2 + Y.^2) + eps;
Z = sin(R)./R;
X = X(:);
Y = Y(:);
Z = Z(:);
scatter3(X,Y,Z)
scatter3 needs vectors, not matrices as far as I can see here
this is my result:
If you want to use meshgrid without reshaping the matrices you have to use plot3 and the 'o' symbol. So you can get a similar result with:
plot3(X,Y,Z,'o')
EDIT:
A question that arose in association with this post was, which of the following methods is more efficient in terms of computation speed: The function reshape(X,[],1), suggested by me, or the simpler colon version X(:), suggested by #Argyll.
After timing the reshape function versus the : method, I have to admit that the latter is more efficient.
I added my results and the code I used to time both functions:
sizes = linspace(100,10000,100);
time_reshape = [];
time_col = [];
for i=1:length(sizes)
X = rand(sizes(i)); % Create random squared matrix
r = #() ResFcn(X);
c = #() ColFcn(X);
time_reshape = [time_reshape timeit(r)/1000] % Take average of 1000 measurements
time_col = [time_col timeit(c)/1000] % Take average of 1000 measurements
end
figure()
hold on
grid on
plot(sizes(2:end), time_col(2:end))
plot(sizes(2:end), time_reshape(2:end))
legend("Colon","Reshape","Location","northwest")
title("Comparison: Reshape vs. Colon Method")
xlabel("Length of squared matrix")
ylabel("Average execution time [s]")
hold off
function res = ResFcn(X)
for i = 1:1000 % Repeat 1000 times
res = reshape(X,[],1);
end
end
function res = ColFcn(X)
for i = 1:1000 % Repeat 1000 times
res = X(:);
end
end
I have 2 vectors x and y to which I want to fit a polynomial as y = f(x) in MATLAB.
I could have used polyfit. However, I want to fit only selective power terms of the polynomial. For example, y = f(x) = a*x^3 + b*x + c. Notice that I don't have the x^2 term in there. Is there any built-in function in MATLAB to achieve this?
I am not sure if simply ignoring the coefficient that MATLAB gives for x^2 is same as fitting the polynomial without x^2 term.
If you don't have the curve fitting tool box (see #thewaywewalk's comment), or anyway, it is easy to use mldivide:
x=rand(10,1); % your x data
y=5*x.^3+2*x+7+rand(10,1)*.01; % some y data with noise
[x.^3 x ones(size(x))]\y % least squares solve to y = a*x^3 + b*x + c
gives
ans =
4.9799
2.0211
6.9980
Note that "simply ignoring the coefficient that MATLAB gives for x^2" is definitely not the "same as fitting the polynomial without x^2 term".
It sounds like you have the fitting toolbox and want to just remove a possible coefficient. If that's the case, here's a way to do it.
%What is the degree of the polynomial (cubic)
polyDegree = 3;
%What powers do you want to skip (x^2 and x)
skipPowers = [2, 1];
%This sets up the options
opts = fitoptions( 'Method', 'LinearLeastSquares' );
%All coefficients of degrees not specified between x^n and x^0 can have any value
opts.Lower = -inf(1, polyDegree + 1);
opts.Upper = inf(1, polyDegree + 1);
%The coefficients you want to skip have a range from 0 to 0.
opts.Lower(polyDegree + 1 - skipPowers) = 0;
opts.Upper(polyDegree + 1 - skipPowers) = 0;
%Do the fit using the specified polynomial degree.
[fitresult, gof] = fit( x, y, ['poly', num2str(polyDegree)] , opts );
In recent Matlab versions you can do it easily with a GUI by selecting the APPS tab and then the Curve Fitting Tool.
For example, I define:
x = 1:10;
y = x + randn(1,10);
I then select those variables as X data, Y data in the tool and I choose a custom equation defined as a*x^3 + b*x + c. The results are:
I've some x.mat and y.mat.
And I would like to find the polynomial equation from this.
I've tried
p = polyfit(x,y,3);
with y2 = p(1)*x.^3 + p(2)*x.^2 + p(3)*x
but my y2 is not equal to the original y . What is wrong ?
Thanks you
As radarhead wrote in his comment, you forgot the coefficient of zero degree (p(4) here).
Assuming x and y are vectors of same length n, polyfit(x,y,n-1) will return a vector containing the coefficients of the interpolating polynomial (of degree n-1) in descending order.
Then, the value of the interpolating polynomial at a point z will be given by:
p(1)*z^3 + p(2)*z^2 + p(3)*z + p(4)
Don't forget p(4)! As Bas suggested, you can use the polyval function to easily compute the value of a polynomial at a a given point:
polyval(p,z);
To illustrate this, see the code below, which generates 4 data points, plots those points and the polynomial interpolating them:
n = 4;
x = sort(rand(n,1));
y = rand(n,1);
p = polyfit(x,y,n-1);
figure
hold on
plot(x,y,'bo');
xx=linspace(x(1),x(end),100);
plot(xx,polyval(p,xx),'r');
hold off