How to set linspace to infinity

Good day, everyone!

This is my code for the 3D plot of the equation Z= (X^2+3*Y^2)*exp(-X^2-Y^2). In my code, I set my linspace to (-2,2) as an example. But I was wondering if there is a way to set the linspace for x, y to (0, inf). If I have tried writing that way, no change occurred. (I assume it's an error). Following is my code. x=linspace(-2,2); y=linspace(-2,2); [X,Y] = meshgrid(x,y); Z=(X.^2+3*Y.^2)*exp(-X.^2-Y.^2); meshz(X,Y,Z)

Thank you.

Answer 1

[jgg]

No, this isn't possible because linspace generates a uniformly spaced vector over the two endpoints. Such a vector on (0,Inf) would have an infinite number of entries and would not be practical. (For example, if would take an infinite amount of memory).

A good alternative would be to identify a limit where the behaviour of your function is "close" to the limiting behavior you want to view then using that point instead. Since your function is exponential, something like:

 x = linspace(0,10); y = linspace(0,10);

is probably sufficient.

[verzhen Ligai]

Thank you for your answer. It seems like I wasn't so clear on the function of linspace. I tried the linspace values that you have suggested, I didn't get the results that I was expecting. Is there any other way to generate graphs of functions without using linspace? The mathematical questions that I am trying to plot right now already come with answers. And in those sample models, there aren't endpoints. I am a bit confused... Thank you very much.

[Image Analyst]

You'll have to be more specific. Otherwise, just plot them out to some big number and see where the graphs level off to some steady state or asymptote, if they do.

[Star Strider]

Give logspace a go. It might at least help you describe your function at extreme values.

[verzhen Ligai]

Thank you for your kind answer.