How to set linspace to infinity

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verzhen Ligai
verzhen Ligai on 8 May 2016
Commented: verzhen Ligai on 8 May 2016
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.

Accepted Answer

jgg
jgg on 8 May 2016
Edited: jgg on 8 May 2016
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.
  4 Comments
Star Strider
Star Strider on 8 May 2016
Give logspace a go. It might at least help you describe your function at extreme values.
verzhen Ligai
verzhen Ligai on 8 May 2016
Thank you for your kind answer.

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