Mathaematics.

The rants of a phisycs-mathematics student. Now with \LaTeX extensions.

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Just some pictures of the Venus transit, (and, in a more general way, the mini-event we made at school for it)

We had better ways of seeing it there, sadly, I couldn''t get any screenshots from the computer connected to one of the telescopes

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So I opened a submissions box.

For my exam time absence~.

(Photos, videos, concepts, or interesting problems are well appreciated)

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By Abstruse Goose

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The Double Pendulum Orchestrated, by NorlamFile is an awesome sonic representation of a conservative double pendulum oscilation system, one of the most beautiful physical and mathematical models

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On space-filling curves

Formal definition.

Let \(\phi :[0,1]\to T\) (where \( T\) is a topological space) be a continous function (i.e. \(\phi ^{-1}\) of any open set \( O\) of \( T\) is an open set of the unit interval) from the unit interval to a topological space. Such map is said to be an space-filling curve if \(\phi\) is onto (i.e. if \phi passes through every element of \( T\) ).

Note

Generally, space filling curves are represented on spaces homeomorphic to \( E^{n}\) euclidian spaces. So to say, the plane, or any geometric solid, since these are the easiest ways to visualize a space filling curve. The most popular example, perhaps, is the two-dimensional Hilbert curve.

Images

A Hilbert curve from the unit interval to the unit cube

Peano Curve from the unit interval to the unit cube.

Flow Snake sapce filling curve.

[page-avenue-deactivated20150628]

Antithesis

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Julia set. by page-avenue.

[dvdp]

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The Lorenz attractor

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the Julia set.

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Sorry guys.

Last three school weeks. Final projects, exams, and kind-of complicated topics.

When I end my Group Theory, advanced linear algebra courses and the overwhelming amounts of multivariable calc and mechanics I'll be back. In the meantime, I'll ony be posting pretty images.

[galoisphrenic]

Evariste Galois’s manuscript

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Rant on education in mathematics

So, this is something that constantly comes into my mind.

During elementary school, one is thaught arithmethics, the base for every other areas of mathematics, since all of them rely, in any amount, on arithmetics. One is taught the basic operations with natural numbers, and, eventually, the set of real numbers.

At least in my country, middle school is where one starts with basic algebra, I mean, solving linear equations, coloquially called "finding x". This is still no problem, since showing how equations work is pretty self-explanatory.

The messy part on education in mathematics starts when one reaches high school, that is the part when students ask "Why does that happen?" when they are thaught trigonometric functions, identities, limits, derivatives, integrals, et cetera. However, it is not precissely easy to answer those questions. One cannot explain the linearity of calculus operators to their students if they have no idea of linear algebra concepts such as linear transformation. Whan cannot explain the reason of integrals to a group of students if they have no idea of mathematical analysis. Then, what is the point?

I personally believe that, in mathemathics, theory is first. This means, high school should first teach formal logic and set theory (instead of having a recap in middle school topics), and notions on geometry and abstract algebra, at least, defining functions, binary operations, notions on group theory and analysis, since one can construct every topic in mathematics by knowing that, and then, reasons for concepts seen later would be known. And then, math wouldn't be seen as "the hardest subject in school" (well, perhaps at the beginning).

Yet, again, it's just me drabbling and ranting on a topic that will never be concluded.

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Group theory: The conjugacy problem.

In absract algebra, mainly group theory, given a group \( G\) aconjugateof an element \( x\in G\)  is another element \( y\in G\) of the form \( y=zxz'\) with \( z\in G\) (and obviously \( z'\) , since as a group all element must have its inverse included.

Theconjugacy problemis a decision problem, which consists in determining if an element \( y\in G\) can be represented as a conjugate of \( x\in G\) .

An easy example is the group \(\left( Z,+\right)\) where \( Z\) is the set of integer numbers.

¿Is it possible to express \( y=z+x-z\) for \( y\neq x\) and \( z\in Z\) ?

For example, for:

\( 3=z+1-z\)

¿Is there a value of \( z\) for which this equation is valid?

[isomorphismes]

In 20th-century abstract mathematics, one builds up ideas and properties—not assuming anything except what one is told. You think 2+3=5? Well in my space that I just made up, e₂⊕e₃ = e₁, and “5” doesn’t even exist!

Concepts are added in incrementally,

∥ A ∥ means the “size” of A. ...

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Vacation

I'll post something when I'm back home or when I'm not drunk.

Which will be... pretty much at the same time.

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The Rubik's Cube as a permutation group.

In abstract algebra, apermutation groupis a group \( G\) that consists in all possible permutations of a set. So to say, the elements of a permutation group are the possible permutations of the elements of a different, \( H\) group. Then we get a group \(\left( G,+\right)\) where \( +\) is the composition of permutations. Also, each permutation \(\phi_{i}\left(H\right)\) . We can assume that permutations are bijective functions, since as a group under composition of permutations, an inverse permutation that yields the identity permutation when composed with one specific permutation exists within \( G\) .

But how is the Rubik's 3x3 cube a permutation group?

Let's talk about the cube itself, disregarding the colors from now. The cube has six fixed piecesand 20 permutable pieces, therefore, every piece itself can be placed in \( 20!\) different positions. Hence, the Rubik's Cube pieces form an order 20! permutation group. But why would we even care on a rubiks cube without distinction of sides. Even a rock would solve one of those.

The cube WITH face distinction is another permutation group, over a greater set \(\bar{H}\) , this set is not formed by the pieces, but by the stickers on the pieces. The mathematics yielding this are slightly more complex, but the number of possible permutations without disassembling the cube is \( 8!\times 3^{7}\times 12!/2\times 2^{11}\). So we have a permutation group on this order. The fact scrambling and solvingn the cube shows the existence of inverse permutations, let's say that the solved cube is the identity permutation, the resut of that scramble is \(\psi\left( \bar{H}\right)\) and, the permutation that solves the cube would be \(\psi^{-1}\left( \bar{H}\right)\). But \(\psi^{-1}\) can be expressed as the composition of multiple, elemental permutations, which would be, face rotations. Which means that\(\psi^{-1}\left(\bar{H}\right) =\sum_{i=1}^{12}\alpha_{j}\varphi_{i} (\bar{H} )\) where \(\varphi_{i} (\bar{H} )\) are the single step face rotations and \(\alpha\in \{ 1,2,3\}\) since one-face rotations form cyclic subgroups of our permutation group \( G\) and the fourth rotation is equal to the identity rotation which is, not rotating the face at all. The single face rotations that \(\varphi_{i}\) represent are the steps in the solution of the cube.

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Yes, this is pretty much something I will need to show my abstract Algebra I groub by the end of the semester.

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Formal definition of a vector space.

Let \( V\) be a nonempty set, whose elements are called "vectors" , let \( K\) be a field, whose elements are called "scalars", and define two operations:

\( +:V\times V\to V\)

\( ·:K\times V\to V \)

\( V\) is called a vector space over a field \( K\) if:

\(\left( V, +\right)\) is an abelian group.

Let \(\alpha ,\beta\in K, v,w\in V\) then:

\(\alpha ·\left( v+w\right) =\alpha ·v+\alpha ·w\)

\(\left(\alpha +\beta\right) ·v=\alpha ·v+\beta ·v\)

\(\left(\alpha\beta\right) ·v=\alpha\left(\beta ·v\right)\)

\(\dot{e} ·v=v\) where \(\dot{e}\) is the multiplicative identity on the field \( K\)