Hi How can i Solve the complex equation below. j = √−1 2z4 + (1 − j)z3 + (1 + 3j)z2 + (2 − 6j)z + (6 − 2j) = 0 i want to Visulize the location of the roots in the complex plane with the command ComplexListPlot[ ]
Good question. For Plot3D, we're plotting f(x,y), two independent variables. We can have three independent variables with ContourPlot3D to ploy f(x,y,z) = constant. I hope that helps!
HI Selamo, with an implicitly define function, you'll want to use ContourPlot3D. So try this: ContourPlot3D[ Sin[x]*Cos[y] + Sin[y]*Cos[z] + Sin[z]*Cos[x] == 0, {x, -2 Pi, 2 Pi}, {y, -2 Pi, 2 Pi}, {z, -2 Pi, 2 Pi}]. That's a lot of computations, so your computer may have to think about it for a little while.
Hi John, how do I plot a wave equation u(x,t) = sech^2((x/3)-(t/3))? I have tried with Plot3D[Sech^(2)((x/4)-(t/3)), {x, -4, 4}, {t,0, .5}, AxesLabel-> {"x", "t","u"}] but just a box came out without the graph.
Hi Farisa, I believe you have a couple of syntax problems. Remember to use hard brackets for inputs, and you'll need to move your exponent. Try this: Plot3D[(Sech[(x/4)-(t/3)])^2, {x, -4, 4}, {t,0, .5}, AxesLabel-> {"x", "t","u"}] .
Why do you call ContourPlot3D[a * b * c - c^2, {a, -2,2},{b, -2,2},{c, -2,2} ] a 4-Dimensional Object? Is it because it has got 3 independent variables? Sorry for ignorance.
Hi John, do you mean plotting a set of coordinates? If so, use ListPlot3D[{x1,y1,z1},...{xk,yk,zk}]. If you want to plot f(x,y,z) in 3-space, remember that f(x,y,z) is a 4-dimensional object in general... so we'll have to use ContourPlot3D[f[x,y,z]==k, {x,...},{y,...},{z,...}]. Here f(x,y,z)= k is an implicitly defined 3-dimensional function. I hope this helps! Thanks for the feedback.
This basics series helped me a lot to prepare for my online particle one night before ,,,,so thank you so much
I'm happy to hear that!
Hi How can i Solve the complex equation below. j = √−1
2z4 + (1 − j)z3 + (1 + 3j)z2 + (2 − 6j)z + (6 − 2j) = 0
i want to Visulize the location of the roots in the complex plane with the command ComplexListPlot[ ]
How many independent variables are required to plot 3d command in Mathematica
Good question. For Plot3D, we're plotting f(x,y), two independent variables. We can have three independent variables with ContourPlot3D to ploy f(x,y,z) = constant. I hope that helps!
Good morning, how can i plot this equation of a sherical gyroid sinx*cosy + siny*cosz + sinz*cosx = 0
HI Selamo, with an implicitly define function, you'll want to use ContourPlot3D. So try this:
ContourPlot3D[
Sin[x]*Cos[y] + Sin[y]*Cos[z] + Sin[z]*Cos[x] == 0, {x, -2 Pi,
2 Pi}, {y, -2 Pi, 2 Pi}, {z, -2 Pi, 2 Pi}]. That's a lot of computations, so your computer may have to think about it for a little while.
Hi John, how do I plot a wave equation u(x,t) = sech^2((x/3)-(t/3))?
I have tried with Plot3D[Sech^(2)((x/4)-(t/3)), {x, -4, 4}, {t,0, .5}, AxesLabel-> {"x", "t","u"}] but just a box came out without the graph.
Hi Farisa, I believe you have a couple of syntax problems. Remember to use hard brackets for inputs, and you'll need to move your exponent. Try this: Plot3D[(Sech[(x/4)-(t/3)])^2, {x, -4, 4}, {t,0, .5}, AxesLabel-> {"x", "t","u"}] .
Why do you call ContourPlot3D[a * b * c - c^2, {a, -2,2},{b, -2,2},{c, -2,2} ] a 4-Dimensional Object? Is it because it has got 3 independent variables? Sorry for ignorance.
That's exactly right. We have three independent variables and one dependent one. That's why Plot3D won't work in this situation.
John, how do you plot 3D functions of 3 Variables (x= [1,4,5,....]; y= [3,7,8....]; z= [6,9,10....] ;)
Hi John, do you mean plotting a set of coordinates? If so, use ListPlot3D[{x1,y1,z1},...{xk,yk,zk}]. If you want to plot f(x,y,z) in 3-space, remember that f(x,y,z) is a 4-dimensional object in general... so we'll have to use ContourPlot3D[f[x,y,z]==k, {x,...},{y,...},{z,...}]. Here f(x,y,z)= k is an implicitly defined 3-dimensional function. I hope this helps! Thanks for the feedback.
In mathematica 12 a lot of things from here dont apply.
Oh no! I did not realize that. What in particular does not apply?