how to plot level curves f(x,y)=yln(x)+xln(y) through point (1,1) in matlab
MATLAB: How to plot level curves f(x,y)=yln(x)+xln(y) through point (1,1)
contourMATLAB
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Show us code to prove it, but, from the symptom described you wrote something like:
x=-7:7;y=yourfunction(x);plot(y)which would display what you described (and have done precisely what you told MATLAB to do). With no x passed to it, plot doesn't know anything at all about the fact you defined x somewhere else; all it can do is use what it was given.
plot(x,y)is the correct syntax.
HINT: Read the documentation before crying wolf! It'll be faster most of the time than waiting for somebody here to see and answer. :)
What did you do wrong? It looks like Jon may be correct, in that you did not use the newly estimated set of parameters from the fit.
xy = [0.0 6.0815 5.330 4.2845 2.8960 1.3175 -0.1690 -0.47105 0.23120 1.68135 3.55150 5.32165 6.51180 6.86195 6.29210 4.92225 3.04240 1.19255 0270 -0.09285 0.75300 2.21315 3.89330 5.29345 6.09360 6.08];x = xy(:,1);y = xy(:,2);FIRST, PLOT YOUR DATA. LOOK AT IT. Does this simple trig model make sense?
plot(x,y,'ro')
And, yes, the result does seem to be a very simple sinusoidally varying curve.
However, there is no need to use lsqcurvefit to fit this, since it is a linear model in the parameters. Using a nonlinear estimation tool to fit a linear model is like using a Mack truck to carry a pea to Boston. A bit of over kill. But, yes, you could use lsqcurvefit. I won't bother to do so here.
A = [ones(size(x)),sind(x),sind(2*x),cosd(x),cosd(2*x)];coefs = A\yplot(x,y,'ro')hold onfun = @(u,coefs) coefs(1) + coefs(2)*sind(u) + coefs(3)*sind(2*u) + coefs(4)*cosd(u) + coefs(5)*cosd(2*u);plot(x,fun(x,coefs),'b-')hold off
While there is a little lack of fit, it is not terrible. Look at the residual errors.
plot(x,y - fun(x,coefs),'o-')
Lack of fit is usually seen as patterning in the residuals. Here we see exactly that. So your model is probably not the correct model for this data, merely a good approximation to that data. By way of comparison, how much of the information in your original signal has been explained by the model?
[var(y),var(y - fun(x,coefs))]As a percentage of the total variance...
100*(1 - var(y - fun(x,coefs))/var(y))So around 99% of the total variance in y has been explained by your model. So pretty good, but not perfect.
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