I frigging love math

Yarrun XIX

I love math so much

Calica

exercise: find the degree of the function given only the colors in the boundary

Odradek

Klinotaxis

Calica said:

exercise: find the degree of the function given only the colors in the boundary

i'm gonna say... bachelor's

glennmagusharvey

what is this plot?

Bee

Calica said:

exercise: find the degree of the function given only the colors in the boundary

I count 4 rainbows and 2 double rainbows, so one would assume 8 even without peeking.

Calica

I said only the boundary so you'd just count the eight rainbows :p

Calica

I also said that because that's kind of the proof of the fundamental theorem of algebra from a paper i recall with graphs almost exactly like that one

Calica

by which i mean literally that graph

Yarrun XIX

A discussion about higher-level math notation was had. A paper was linked. And I found joy where I had previously found very little.

Yarrun XIX

glennmagusharvey said:

what is this plot?

Details are in the paper Calica linked, but,

Okay, so, let's say you have a function that has a complex input and output. You want to graph that sucker without using 4-D graphs, because nobody needs that. So, you map all relevant points to a color wheel using polar coordinates. The degree of the point determines the color, with red being 0 and cyan being pi and so forth, and the distance from the origin determines the brightness, with dark = low and light = high. And that color is used for the output of that function

So the position of each point is the 'z' value, mapped onto the real and imaginary axes, and f(z) is the color. You following me?

glennmagusharvey

Yarrun XIX said:

glennmagusharvey said:

what is this plot?

Details are in the paper Calica linked, but,

Okay, so, let's say you have a function that has a complex input and output. You want to graph that sucker without using 4-D graphs, because nobody needs that. So, you map all relevant points to a color wheel using polar coordinates. The degree of the point determines the color, with red being 0 and cyan being pi and so forth, and the distance from the origin determines the brightness, with dark = low and light = high. And that color is used for the output of that function

So the position of each point is the 'z' value, mapped onto the real and imaginary axes, and f(z) is the color. You following me?

Ahh, I see.

vtk

I'm up an hour past bedtime because I was focused on an art project involving both projective and hyperbolic geometry, plus quaternions.

Yarrun XIX

Ah jeez, I was in a depressive funk when they taught quaternions

Calica

vtk said:

I'm up an hour past bedtime because I was focused on an art project involving both projective and hyperbolic geometry, plus quaternions.

What's the art project if you don't mind me asking

vtk

Six-coloring the real protective plane, significantly over-thought. What else would it be? :-P