I have z * ( cos(1.57 * (a^2+b^2+1)^(1/2)) / sin(1.57 * (a^2+b^2+1)^(1/2)) ), which project the coordinates to the infinite on a circle. But this is only a fancy zoom, straight lines are not projected to circles.
"It's important to note that When rendering with these infinitely-tiled images, the mapping that is rendered is actually the inverse of the function given. This is because it is prohibitively expensive to compute images of the function itself. (In fact, we are rendering a kind of "pull-back".) Specifically, when rendering a pixel in the image, the location of that pixel in the complex plane is passed to your function, which produces a complex value. That transformed value specifies a location in the original image. This procedure is fast, but plots the inverse of the function given. So, if you want to plot e.g. log(z), then you should instead put in e^z."
I should probably try to find a way to make it clearer what all can be done, or make it more obvious where the documentation is ("t" is mentioned there.)
Keyframe|10 years ago
edit: a nicer rotozoom: http://brandonpelfrey.github.io/complex-function-plot/?expre...
BenoitP|10 years ago
I have spend the last hour trying to display the Poincaré disk[1], but to no avail.
I have z * ( cos(1.57 * (a^2+b^2+1)^(1/2)) / sin(1.57 * (a^2+b^2+1)^(1/2)) ), which project the coordinates to the infinite on a circle. But this is only a fancy zoom, straight lines are not projected to circles.
[1]https://en.wikipedia.org/wiki/Poincar%C3%A9_disk_model
marcosscriven|10 years ago
"It's important to note that When rendering with these infinitely-tiled images, the mapping that is rendered is actually the inverse of the function given. This is because it is prohibitively expensive to compute images of the function itself. (In fact, we are rendering a kind of "pull-back".) Specifically, when rendering a pixel in the image, the location of that pixel in the complex plane is passed to your function, which produces a complex value. That transformed value specifies a location in the original image. This procedure is fast, but plots the inverse of the function given. So, if you want to plot e.g. log(z), then you should instead put in e^z."
BenoitP|10 years ago
((i-(z)/10000)/(i+(z)/10000))^(50000)
which is sort of like the Poincaré half plane
arcatek|10 years ago
Ex: t*z
brandonpelfrey|10 years ago
VikingCoder|10 years ago
I love thinking about this, as it goes negative.