Any other ways to program tan(x) between [-3*pi/2 to 3*pi/2])?

Hello!
I write down the code for the tan(x) between [-3*pi/2 to 3*pi/2]), but I want to reduce it, or doing it in another way. Any ideas?
x1=-(3*pi/2)+0.01:0.01:-(pi/2)-0.01;
plot(x1,tan(x1),'r','linewidth',2.5)
hold on
x2=-(pi/2)+0.01:0.01:(pi/2)-0.01;
plot(x2,tan(x2),'r','linewidth',2.5)
x3=(pi/2)+0.01:0.01:(3*pi/2)-0.01;
plot(x3,tan(x3),'r','linewidth',2.5)
hold off

 Respuesta aceptada

x = -3*pi/2 + .1 : 0.1 : 3*pi/2 - 0.1;
y = tan(x);
plot(x, tan(x), 'r','linewidth',2);
grid on;

5 comentarios

Thanks but I know that! The problem is tan(x) doesn't converge to [-3*pi/2 -pi/2 pi/2 3*pi/2]...That was my reason to program it like that!
Image Analyst
Image Analyst el 13 de Jul. de 2013
Editada: Image Analyst el 14 de Jul. de 2013
It works. I don't know what you want to do. Do you just want to display it but not have the axes go from -inf to +inf? If so, just use ylim() to set it to whatever you want. MATLAB does handle infinity so you just need to define exactly what you want to plot.
x = -3*pi/2 : 0.1 : 3*pi/2;
y = tan(x);
plot(x, y, 'r','linewidth',2);
ylim([-10, 10]);
grid on;
Now it is correct, What about if I want to remove the vertical lines??
Image Analyst
Image Analyst el 14 de Jul. de 2013
Editada: Image Analyst el 14 de Jul. de 2013
You can cover them up with black lines using the line() function, or you can set values more than 10 or 1000 or whatever to nan:
x = -3*pi/2 : 0.01 : 3*pi/2;
y = tan(x);
maxYtoPlot = 10;
minYtoPlot = -10;
y(y > maxYtoPlot) = nan;
y(y < minYtoPlot) = nan;
plot(x, y, 'r','linewidth',2);
ylim([minYtoPlot, maxYtoPlot])
grid on;
Good JOB!

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Más respuestas (1)

x=-3*pi/2+0.01:0.01:-(pi/2)-0.01
y=tan(x)
y=[y nan y nan y]
x=linspace(x(1),pi/2-0.01,numel(y))
plot(x,y)

3 comentarios

I am sorry to say it is not correct because the tan(x) must pass through [-pi 0 and pi] which yours doesn't!!!
Look at x, you will see that x takes values near 0, if you want to be more closer to 0, you have to choose a very small sample time
Thanks, I need to try it to see!

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