# How to set linspace to infinity

18 visualizaciones (últimos 30 días)
verzhen Ligai el 8 de Mayo de 2016
Comentada: verzhen Ligai el 8 de Mayo de 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.
##### 0 comentariosMostrar -2 comentarios más antiguosOcultar -2 comentarios más antiguos

Iniciar sesión para comentar.

jgg el 8 de Mayo de 2016
Editada: jgg el 8 de Mayo de 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 comentariosMostrar 2 comentarios más antiguosOcultar 2 comentarios más antiguos
Star Strider el 8 de Mayo de 2016
verzhen Ligai el 8 de Mayo de 2016

Iniciar sesión para comentar.

### Categorías

Más información sobre Line Plots en Help Center y File Exchange.

### Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by