The Math Thread
Re: The Math Thread
φ+1 = φ^2
φ-1 = 1/φ
φ-1 = 1/φ
Re: The Math Thread
Wait what.
TT: I guess one could use those words to describe it.
TT: If armed with a predilection for the inapt.
TT: If armed with a predilection for the inapt.
Re: The Math Thread
You can check it with a calculator 

- The Kakama
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Re: The Math Thread
Ok, to restart the thread, here's something cool:
As you guys perhaps know, the complex numbers (real+imaginary) form a plane. If we warp the plane adding an extra point, infinity, to close it, we get a sphere, called the Riemann sphere (infinity IS an number there!
)
Now, the five Platonic solids (tetrahedron, cube, dodecahedron, etc.) can be embedded inside a sphere, with their vertices touching the surface. So I thought, are there any equations whose solutions are exactly the complex number coordinates of the vertices in the Riemann sphere?
I spent some time looking for them, and finally I found that the answer is: yes! There are three different ones, here they are with their plots:
-Tetrahedron:
-Octahedron:
-Icosahedron:
To obtain the cube and dodecahedron ones, you just flip the fractions' numerator and denominator in the last two equations. Cool or what? 
EDIT: if anyone is interested, I used the plugin "conformal" in the Gimp to make the graphs. You can try them yourself, or any other function you like
As you guys perhaps know, the complex numbers (real+imaginary) form a plane. If we warp the plane adding an extra point, infinity, to close it, we get a sphere, called the Riemann sphere (infinity IS an number there!

Now, the five Platonic solids (tetrahedron, cube, dodecahedron, etc.) can be embedded inside a sphere, with their vertices touching the surface. So I thought, are there any equations whose solutions are exactly the complex number coordinates of the vertices in the Riemann sphere?
I spent some time looking for them, and finally I found that the answer is: yes! There are three different ones, here they are with their plots:
-Tetrahedron:

EDIT: if anyone is interested, I used the plugin "conformal" in the Gimp to make the graphs. You can try them yourself, or any other function you like

- The Kakama
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Re: The Math Thread
Wow. What's with the colours and lines though?
e^(pi(i))+1 =0
e^(pi(i))+1 =0
Is this my final form?
Re: The Math Thread
The colours represent the phase (the "sign") of the complex number. For example, positive numbers are red, negative numbers are light blue, pure imaginary numbers are green and violet, etc. Two shadowed lines next to each other means that the modulus ("size") of the numbers in one of them is double the modulus in the other. It's better explained here:
https://en.wikipedia.org/wiki/Domain_co ... ex_numbers
About the non-shadowed grey lines, I think they're just a square grid, to see how the complex plane deforms when you apply the function.
We have to do that weird color trick to properly draw complex functions, because we need a plane to graph real functions (1D+1D=2D), so to graph complex functions without color we would need 2D+2D=4 dimensions.
it also has a very cool geometric interpretation.
https://en.wikipedia.org/wiki/Domain_co ... ex_numbers
About the non-shadowed grey lines, I think they're just a square grid, to see how the complex plane deforms when you apply the function.
We have to do that weird color trick to properly draw complex functions, because we need a plane to graph real functions (1D+1D=2D), so to graph complex functions without color we would need 2D+2D=4 dimensions.
Good choice of equatione^(pi(i))+1 =0

- ENIHCAMBUS
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Re: The Math Thread
I was thinking, if you use a component in an equation that leads up to infinite, and then graph it, you end up having a fractal?
Is this right?
Is this right?
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Re: The Math Thread
It's not really about equations, but more about transformations.
Fractals have the property of self-similarity, which means that if you apply a certain transformation to a fractal it stays the same. Things who are unchanged by a certain transformation are said to be a fixed point of that transformation. Written it looks like this:
Transf(Fixpoint) = Fixpoint
Now, if you start with any object, and apply the transformation a lot of times, if you have luck, when you apply it again it won't make much of a difference:
Transf(Transf(Transf(Transf(Transf(Transf(Transf(Transf(Transf(Transf(Object)))))))))) = Transf^10(Object)
won't look very different from
Transf(Transf(Transf(Transf(Transf(Transf(Transf(Transf(Transf(Transf(Object))))))))))=Transf^11(Object)
In the limit, when the number of transformations you apply is infinity, since ∞ = ∞+1 when you apply it one more time it makes
Transf(Transf^∞(Object)) = Transf^(∞+1)(Object) = Transf^∞(Object)
So Transf^∞(Object) is a fixpoint, because the transformation hasn't changed it! So in a sense, the component that is infinite when you are working with fractals is the number of transformations you have applied. Not every fixed point is a fractal, but all fractals are fixed points.
For example, if the transformation is the sprout of two small triangles at each angle of a curve, Transf^∞(Triangle) = Koch Snowflake.
-------------------------------------------------------------------------------------------------------------------------------------
When a function/equation returns infinity, what you have is (usually) a pole. Poles and infinities are not much mysterious, everything becomes much clearer when you study the Riemann sphere. There you see that a pole is something like a reverse zero/root of a function. If you graph functions with pole singularities of degree N, they look exactly like zeros of degree N, with the color wheel inverted.
Fractals have the property of self-similarity, which means that if you apply a certain transformation to a fractal it stays the same. Things who are unchanged by a certain transformation are said to be a fixed point of that transformation. Written it looks like this:
Transf(Fixpoint) = Fixpoint
Now, if you start with any object, and apply the transformation a lot of times, if you have luck, when you apply it again it won't make much of a difference:
Transf(Transf(Transf(Transf(Transf(Transf(Transf(Transf(Transf(Transf(Object)))))))))) = Transf^10(Object)
won't look very different from
Transf(Transf(Transf(Transf(Transf(Transf(Transf(Transf(Transf(Transf(Object))))))))))=Transf^11(Object)
In the limit, when the number of transformations you apply is infinity, since ∞ = ∞+1 when you apply it one more time it makes
Transf(Transf^∞(Object)) = Transf^(∞+1)(Object) = Transf^∞(Object)
So Transf^∞(Object) is a fixpoint, because the transformation hasn't changed it! So in a sense, the component that is infinite when you are working with fractals is the number of transformations you have applied. Not every fixed point is a fractal, but all fractals are fixed points.
For example, if the transformation is the sprout of two small triangles at each angle of a curve, Transf^∞(Triangle) = Koch Snowflake.
-------------------------------------------------------------------------------------------------------------------------------------
When a function/equation returns infinity, what you have is (usually) a pole. Poles and infinities are not much mysterious, everything becomes much clearer when you study the Riemann sphere. There you see that a pole is something like a reverse zero/root of a function. If you graph functions with pole singularities of degree N, they look exactly like zeros of degree N, with the color wheel inverted.
- ENIHCAMBUS
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Re: The Math Thread
Interesting, I just wanted to know if there was a correlation between the 2 types of diagrams. 

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