# Graphs



## ben1996123 (Jan 6, 2011)

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))))))


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## bluecloe45 (Jan 6, 2011)

I totally get it.


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## buelercuber (Jan 6, 2011)

lolben


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## ben1996123 (Jan 6, 2011)

buelercuber said:


> lolben



LOLOLOLOLOLOLOLOL


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## steph1389 (Jan 6, 2011)

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


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## emolover (Jan 6, 2011)

_Hmm, I... wait... what the **** is this ****._

Ohhhh. Now I get it.


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## aronpm (Jan 7, 2011)

Sup


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## maggot (Jan 7, 2011)

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.


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## qqwref (Jan 7, 2011)

maggot said:


> 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.


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## maggot (Jan 7, 2011)

qqwref said:


> 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 ^^;


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## Kirjava (Jan 7, 2011)

buelercuber said:


> lolben


 
Grow up.

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


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## Rinfiyks (Jan 7, 2011)

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.:


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## ben1996123 (Jan 7, 2011)

Kirjava said:


> 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...



Rinfiyks said:


> 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...


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## Athefre (Jan 7, 2011)

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


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## Rinfiyks (Jan 7, 2011)

ben1996123 said:


> 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


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