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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.


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Accepted Answer

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.


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Image Analyst
Image Analyst on 8 May 2016
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
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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