MATLAB: Haw can I plot these two equations

3d plotsMATLABmatlab functionplotting

I want to plot these two equation in the same image if it is possible. Thank yous a lot

exp(x)*(sin(y)-cos(y)-(exp(x)*sin(2*y)))==0

exp(x)*(-sin(y)-cos(y)+(2*exp(x)*cos(2*y))) ==0

Best Answer

Probably the easiest way is to use the contour function:

f1 = @(x,y) exp(x).*(sin(y)-cos(y)-(exp(x).*sin(2*y)));
f2 = @(x,y) exp(x).*(-sin(y)-cos(y)+(2*exp(x).*cos(2*y)));
[X,Y] = ndgrid(-10:0.1:10);
figure
contour(X, Y, f1(X,Y), [0 0], 'r')
hold on
contour(X, Y, f2(X,Y), [0 0], 'g')
hold off
grid on

Experiment to get the result you want.

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MATLAB: How to fit an exponential curve

Try this:

filename1 = 'x2.mat';
D1 = load(filename1);
x = D1.x2;
filename2 = 'y2.mat';
D2 = load(filename2);
y = D2.y2;
fcn = @(b,x) b(1).*exp(b(2).*x) + b(3).*exp(b(4).*x);
[B,fval] = fminsearch(@(b) norm(y - fcn(b,x)), ones(4,1));
figure
plot(x, y, 'p')
hold on
plot(x, fcn(B,x), '-')
hold off
grid

The fitted parameters are:

B =
          1.13024777245481
         -2.75090020576997
         -2.09865110252493
         -5.48051415288241

How can I fit an exponential curve - 2018 11 14.png

EDIT — Added plot figure.

MATLAB: How to find local and global minima of a 2D non continuous function

Thanks for your answer !

I understand what you mean by how I should define my numbers. I guess the best thing to do right now is to stick with floating point numbers to solve fonc. Using your method, I found the global minima (0, 1/15), which is relevant with the experiment I did. However, in order to fund the zeros of eqn, I am trying to use fzero function, with a lot of differents starting points located within (-0.05; 0.25) (interval of interest here) and I don't find the same results as yours.

Best Answer

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