Pepini

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4 years, 97 days

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These are questions asked by Pepini

Hi! Do you know how to compute this point of intersection? I thought that would be points given from

solve(f(t)=g(t),t)

but it's not true. I don't want to calculate this numerically because I think it should be seen from formula but ofc I can be wrong.

I'll give the code below:

restart;
with(plots);
f := t -> exp(t) + exp(-t^2 - 2*t);
g := t -> exp(-t^2) + exp(-t^2 - 2*t);
display(plot([f(t), g(t), t = -3 .. 3], discont = true, color = red), pointplot([[2, 2], [exp(-1) + exp(1), exp(-1) + exp(1)]], color = blue, symbol = solidcircle), view = [0 .. 6, 0 .. 4], plot([2, t, t = 0 .. 4]));

 

 


hi! how can I write it correct?

this is the parametric curve in cartesian coordinates:

spacecurve([2*t^3 - 2*t, 4*t^2, t^3 + t], t = -1 .. 1, color = black, numpoints = 1000, orientation = [90, 90, 90]);

and this should be the same curve but in spherical coordinates:

spacecurve([sqrt(5)*(t^3 + t), arccos(1/sqrt(5)), arccot((t^2 - 1)/(2*t))], t = -1 .. 1, coords = spherical, color = black);

I rewrite that parametrization using formula from this page 

https://en.wikipedia.org/wiki/Spherical_coordinate_system

why doesn't it work?

hi, do you know how can we smooth this ellipse and make it more clear?

thickness, grid, numpoints don't help

restart;
with(plots);
f := (x, y) -> x^2 + y^2 - 12*x + 16*y;
display(plot3d(f(x, y), x = -9 .. 9, y = -9 .. 9), pointplot3d([[6, -8, f(6, -8)]], color = red, symbol = solidcircle, symbolsize = 18), view = [-4.2 .. 8.2, -8.2 .. 4.2, -100 .. 100], plot3d([cos(t), sin(t), 1 - 12*cos(t) + 15*sin(t)], t = 0 .. 2*Pi, orientation = [-15, 68, 5]));
 

hi! is there any kind of formula for coloring functions by HUE like this

contourplot(abs(sin(x) + y^2 + y + x*I), x = -6 .. 6, y = -5 .. 5, contours = [seq(1/2^n, n = 1 .. 3), seq(2^n, n = 1 .. 5)], numpoints = 200000, thickness = 1.5, color = black);

I wonder about situation when we don't have function for example f(z)=z but f(x,y)=x+I*y and we want to use domain coloring for that. Thanks in advance.

hi I want to color points like phase portraits of complex functions, so I did something like this:

with(plots);
with(Statistics);
with(LinearAlgebra);
data := Matrix(<Sample(Uniform(-2, 2), [10, 2]) | RandomVector(10, generator = rand(0 .. 3))>);
pointplot(data[() .. (), 1], data[() .. (), 2], symbolsize = 20, symbol = solidbox, colorscheme = ["valuesplit", data[() .. (), 3], [0 = "Red", 1 = "Blue", 2 = "Green", 3 = "Purple"]]);

f := (x, y) -> x^2 - y^2 + 2*I*x*y;
LR := <seq(Re(f(data[1 .. 10, 1 .. 2][i, 1], data[1 .. 10, 1 .. 2][i, 2])), i = 1 .. 10)>;
LI := <seq(Im(f(data[1 .. 10, 1 .. 2][i, 1], data[1 .. 10, 1 .. 2][i, 2])), i = 1 .. 10)>;
pointplot(LR(), LI(), symbolsize = 20, symbol = solidbox, colorscheme = ["valuesplit", data[() .. (), 3], [0 = "Red", 1 = "Blue", 2 = "Green", 3 = "Purple"]]);

there should color points like this in the input (domain coloring of the f(z)=z^2 ):

and output space:

for example these two stripes of pink points on the input space tell us that all those points get mapped
somewhere to the pink direction, the lower right of the output space, thanks in advance

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