Graphs

[ben1996123]

Saw this, and decided to find some others. using MathGV (Direct download)

Green one is R = sin(theta^x)+pi
Brownish one is R = sin(cos(tan(sec(csc(cot(theta))))))

 

[bluecloe45]

I totally get it.

 

[buelercuber]

lolben

 

[ben1996123]

lolben

LOLOLOLOLOLOLOLOL

 

[steph1389]

Green one follows chaos quite well. Didn't realise it did that.

 

[emolover]

Hmm, I... wait... what the **** is this ****.

Ohhhh. Now I get it.

 

[aronpm]

Sup

 

[maggot]

i find the green fascinating. since the wavelength is approaching 0, you would figure the graph would show a solid bar after a while, however it looks like there is discontinuities.

 

[qqwref]

i find the green fascinating. since the frequency is approaching 0, you would figure the graph would show a solid bar after a while, however it looks like there is discontinuities.

From experience I would say that is almost certainly an issue with the graphing program and/or graphics card, rather than a characteristic of the function itself.

 

[maggot]

From experience I would say that is almost certainly an issue with the graphing program and/or graphics card, rather than a characteristic of the function itself.

im sure, wish i could zoom ^^;

 

[Kirjava]

lolben

Grow up.

EDIT; Also x^2 + ((5y/4) - sqrt(|x|))^2 = 1

 

[Rinfiyks]

What I find more fun is sketching a shape of a graph on paper/mspaint and trying to find an equation that fits it.
E.g.:

 

[ben1996123]

Grow up.

EDIT; Also x^2 + ((5y/4) - sqrt(|x|))^2 = 1

What program are you using? Both that I have come up with like 5 errors...

What I find more fun is sketching a shape of a graph on paper/mspaint and trying to find an equation that fits it.
E.g.:

Thought about that before, but no idea how you would do it...

 

[Athefre]

The green one is obviously a slinky being played with by a careless child. The brown one is a CD.

 

[Rinfiyks]

Thought about that before, but no idea how you would do it...

Spoiler

Well the symmetry about x axis makes me think it will be y^2 = f(x)
f(0) = 0
f(1) = 1
f(x tends to -inf) = inf
f(>1) = negative (because it doesn't exist on the graph, so it must be imaginary/complex because of the y^2)
After a lot of guessing you might assume f(x) contains 2 parts
f(x) = p(x) + q(x)
At 0 and 1, p(x) = -q(x)
at x>1, p(x) + q(x) < 0
as x tends to -inf, p(x) + q(x) tend to inf
After more guessing, you find
y^2 = x^2 - x^3