I have a vector than can be plotted and I would like to compute its integral. I don't mean the total area below, but how it evolves over the domain of integration. Basically, its "indefinite" integral. Is this possible? maybe via interpolation? Since I am working with Chebyshev differentiation matrices, do you know if there is an equivalent for integration?
Thank you
You probably want cumtrapz:
x = linspace(0,2*pi,1000); % example x axis values
y = sin(x); % example function
I = cumtrapz(x, y); % compute its integral
plot(x, y, x, I) % plot them
grid on % grid
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
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
I have never used MATLAB before, so I am very lost. For my calculus class, we were tasked with finding a certain function and then using MATLAB to graph it. Finding the function was no problem. However, trying to graph it has me pulling my hair out. The function is z(x,y)= xy(x+y)(2x+y)(3x+y)(x-2y)(x-3y)(x-4y). Any help or advice is GREATLY appreciated.
You can define a anonymous function handle.
% define function
% .* denotes element wise multiplication
f = #(x,y) x.*y.*(x+y).*(2*x+y).*(3*x+y).*(x-2*y).*(x-3*y).*(x-4*y);
% define range and resolution for x and y
x = -20:0.5:20;
y = -20:0.5:20;
% create meshgrid for 3d plotting
[X, Y] = meshgrid(x,y);
% calculate z values (for meshgrid)
z = f(X, Y);
% plot the function
figure()
surf(x,y,z)
To explain further, since you want to calculate the z value for x and y pairs, you should use a element wise multiplication .*.
Then you have to create a meshgrid for the x and y values, to have all the possible x and y pairs in the two new matrices X and Y. Providing these to your function will calculate the corresponding z value for all these pairs. You can use these for plotting, e.g. surf.
I'm wondering if anyone has any insight into why these two plot commands produce domains that are orders of magnitude different?
syms t
x1Axis = 0:.01:10;
fun1(t) = sin(t)
plot(sin(x1Axis))
hold on
y = sin(x1Axis)
plot(x1Axis, y)
fun1(t) is plotted "symbolically" while y is evaluated and plotted "discreetly". Should I be using a different plot method in the case of the first function?
No, you are not plotting the symbolic function correctly.
In your code, the instruction plot(sin(x1Axis)) is not a symbolic plot, but a numeric plot of the sine versus the index of each value.
From the plot documentation page:
plot(Y) creates a 2-D line plot of the data in Y versus the index of
each value.
If Y is a vector, then the x-axis scale ranges from 1 to length(Y).
To plot the symbolic function use fplot.
The following example will allow you to see that both the symbolic and numeric plots are the same:
xmin = 0;
xmax = 10;
% Symbolic Plot.
syms t
fun1(t) = sin(t);
fplot(fun1, [xmin xmax], '-r');
hold on;
% Numeric Plot.
x = xmin:.01:xmax;
y = sin(x);
plot(x, y, '--g');
% Add legend.
legend('Symbolic Plot', 'Numeric Plot', 'Location', 'north');
This is the result:
if the function F is available it is easy to draw surf plot i.e.
x=1:0.1:4;
y=1:0.1:4;
[X,Y]=meshgrid(x,y);
Z=sin(X).^2+cos(Y).^2;
surf(X,Y,Z);
view(2) ;
in my case I calculated F function using least square:
for example I have x and y vectors
x = [0 9.8312 77.1256 117.9810 99.9979];
y = [0 2754.5 4043.3 5376.3 5050.4];
the linear function of these two vector is define by
F= [1149.73 , 37.63];
therefore the estimation is equal to
z= [ones(5,1) x']*F';
which is
z = [1149.73 1519.67 4051.96 5589.35 4912.65];
and if it is plotted
plot(x,y,'b.');
hold on;plot(x,y,'b-');
hold on; plot(x,z,'-r');
The linear z ( red line) is showing correctly. Now I want to plot it for all possible combination of x and y using grid and I need to have a mesh for all inputs
[X,Y] = meshgrid(x,y);
but how to make the Z matrix to show the intensity plot of function Z? The Z suppose to have high intensity close to z value and less value far from it. I should suppose to get something like this
Thanks
P.S: the F is calculated using pinv (least square).
You have to interpolate the scattered data to plot it on grid. Here is a simple example for your x, y and z vectors
xi=linspace(min(x),max(x),100)
yi=linspace(min(y),max(y),100)
[XI YI]=meshgrid(xi,yi);
ZI = griddata(x,y,z,XI,YI);
contourf(XI,YI,ZI)