EightballIdiot

Dont believe EVERYthing on the internet or from me. I enjoy math and other langauges.

[eightballidiot]

AI doesn't belong on Stack Exchange

The moderators on Stack Exchange have made it clear that AI chatbots don't belong on Stack Exchange.

Chatbots are these idiotic robots trying way too hard to be polite and trying to answer questions that they are not equipped to answer.

I whole heartedly agree with the moderators, because anything or anyone trying to answer questions or be polite does not belong on Stack Exchange at all.

When the AI Bubble pops, I hope that toxic site gets blown into the Duat along with it.

[eightballidiot]

'Sun Tzu said That!'

The funniest part about Meet the Soldier is that his quotation of Sun Tzu is a part of a real passage: "If fighting is sure to result in victory, you must fight."

The quote in Meet the Soldier really reads like something Soldier would say to sound wise, philosophical, and put together even though Soldier's brain is very broken.

However, the original quote by Sun Tzu was longer and had a pretty different tone:

"If fighting is sure to result in victory, then you must fight, even though the ruler forbid it; if fighting will not result in victory, then you must not fight even at the ruler's bidding"

The original seems less intellectual and philosophical and more just a word of advice to commit mutiny when you need to, which is hilarious coming from a famous general.

[eightballidiot]

The Cayley-Dickson Construction

(the equations only work on my blog)

The real numbers, complex numbers, quaternions, octonions, and sedenions are all examples of unital algebras with a set of units that are orthogonal to each other; the units are all linearly independent.

The number of units in each algebra is a power of two with all units squaring to -1 except the real unit (which is 1).

Real Numbers: 1 unit (1 real)

Complex Numbers: 2 units (1 real, 1 imaginary)

Quaternions: 4 units (1 real, 3 imaginary)

Octonions: 8 units (1 real, 7 imaginary)

Sedenions: 16 units (1 real, 15 imaginary)

The algebras can be thought of as a set of chain of nesting dolls where all of the lower algebras are contained as a subset of the algebras above them. The octonions contain the quaternions, complex numbers, and real numbers. The octonions themself are contained by the sedenions and the rest of the higher algebras; there are an infinite number of them each corresponding to a power of two.

The sedenions and the other four lower systems are pictured below:

(picture courtesy of Wikipedia)

In addition to being subsets or supersets of each other, the higher algebras in this list can each be derived from the algebra with half the number of units below it; the complex numbers can be derived from the real numbers, quaternions can be derived from the complex numbers, &c.. This derivation of a higher algebra from its lower neighbor is the Cayley-Dickson construction.

The construction is accomplished by representing a number in the higher algebra as a pair of numbers in the lower algebra:

$$q=(a,b)$$

The a and b in this instance are numbers in the lower algebra, and q is a representation of a number in the algebra above the lower one. The representation is like a pair of points; the complex numbers are represented with a pair of real numbers a+bi, and the quaternions are represented with a pair of complex numbers (a+bi)+(c+di)j, &c.. The (a,b) representation means that a corresponds to the first half of q and b corresponds to the other half.

Multiplication of the higher numbers is also defined below where the macron over a symbol means it is the conjugate:

$$(a,b)*(c,d) = (ac-\overline{d} b, da+b\overline{c})$$

The equations in each slot, of the right side of the equation above, are just the product of those two numbers as they would be multiplied in the lower system; remember, a, b, c, and d are numbers in the lower system. The first equation on the right side corresponds to the first half of the product of (a,b) and (c,d), and the other half corresponds to the other half.

Nevertheless, the order of the multiplication does matter as the quaternions and higher don't commute. It should also be noted that real numbers are their own conjugates:

$$\left. \overline{a}= a \ \right| a \in \Bbb{R}$$

[eightballidiot]

How to Diagonalize

(these equations don't work unless you go to my blog directly)

Diagonalizing an n by n square matrix A is done by finding a matrix P which satisfies the equation below where D is a n by n diagonal matrix:

$$A=PDP^{-1}$$

This equation can be rewritten (if only if P is invertible) as the one below which can be used to derive P and D:

$$AP=PD$$

The diagonal matrix will have n different values along the diagonal. These entries will be denoted by λ subscripted with some number indicating its place on the diagonal from left to right.

$$D=\begin{pmatrix} \lambda_1 & \cdots & 0\\\ \vdots & \ddots & \vdots\\\ 0 & \cdots & \lambda_n \end{pmatrix}$$

With this definition in mind for D, the equation for P can be written such that the columns of P can be written with p subscripted by the number of the column.

$$P=\begin{pmatrix} p_1 & \cdots & p_n \end{pmatrix}$$

$$A\begin{pmatrix} p_1 & \cdots & p_n \end{pmatrix}=\begin{pmatrix} p_1 & \cdots & p_n \end{pmatrix} \begin{pmatrix} \lambda_1 & \cdots & 0\\\ \vdots & \ddots & \vdots\\\ 0 & \cdots & \lambda_n \end{pmatrix}$$

$$A \begin{pmatrix} p_1 & \cdots & p_n \end{pmatrix} = \begin{pmatrix} \lambda_1 p_1 & \cdots & \lambda_n p_n \end{pmatrix}$$

This last form of the formula is key to finding the diagonal matrix. If one chooses one of the columns of P, then one can find the formula below, which should look familiar if one knowns about eigenvectors:

$$A p_x = \lambda_x p_x$$

This type of column vector, which equals itself times a scalar λ when multiplied by a matrix, is an eigenvector, and the scalar λ is the eigenvalue.

Therefore, the columns of the matrix P are the eigenvectors of A and the diagonal matrix D is the diagonal matrix of the corresponding eigenvalues (the eigenvector of column 1 in P corresponds to eigenvalue 1 in D and so on).

$$(A - \lambda_x I)p_x = 0$$

(I wrote about eigenvectors here)

BUT, this only works if and only if P is invertible; if it's not, you can kiss this entire process goodbye. Some matrices cannot be diagonalized.

[eightballidiot]

Origin of Language (layman opinion)

The origin of language is a bit of a rocky field because quite frankly there is very little to go off of. This post is just me conspiracy ranting about my personal opinions.

Language as we know it is more than likely older than the migration out of Africa; it seems a little too convenient to me that all current day humans are able to use language fairly equally in the grand scheme of things. If language evolved after the migration out of Africa, there would need to be multiple events where all current day humans acquire language. Me personally, I'm pretty sure the more likely explanation is that homo sapiens had access to language well before humans migrated outside Africa and Eurasia.

Neanderthals are a bit weird. They seem fairly intelligent from all available evidence and there is evidence of mixed ancestry for human neanderthal hybrids. My gut instinct is that Neanderthal were probably comparable to humans in terms of language faculty due to how intelligent they were when compared to older hominids. Wouldn't it be kina weird if they were really intelligent and they weren't able to use language.

If I had to guess from the available evidence, I would bet good money that languages is more than likely a very old innovation for hominids, and language was more than likely a process that got more and more complex as hominids got smarter. The only reason that I could imagine why human languages are so equally complex today is in my opinion a result of multiple bottlenecking incidents throughout human history that turned our species towards a very homogenous gene pool. I cant really think of any other simple explanation.

Again this is all speculation and I may be very dumb.

[eightballidiot]

Eigenvectors

(Click here to make the equations work)

A square matrix can best be thought of as a linear transformation which acts on an n-dimensional grid. Vectors in this context can be thought of as coordinates in this n-dimensional grid which can be transformed by the matrices.

(image taken from Wikipedia)

If you haven't already, check out 3b1b's video on matrix multiplication (he explains it better than I can).

For some square matrices with complex number entries, there exist vectors that obey the following equation where "λ" is a scalar and "v" is a vector.

$$Av=\lambda v$$

These are the eigenvectors "v" of the matrix "A" and the associated scalar(s) of these eigenvectors are the eigenvalues "λ".

Finding solutions to "v" and "λ" is done by rearranging the equation above:

$$Av=\lambda v$$

$$Av =\lambda Iv$$

$$(A-\lambda I)v =0$$

The equation above can be used to find the eigenvalues by knowing that the determinant of (A-λI) must be equal to zero in order for the equation to work. Therefore, the eigenvalues are given by the formula below:

$$\Vert A-\lambda I \Vert = 0$$

When expanding the determinant, one is left with a "n"-degree polynomial of "λ" where "n" is the dimension of the square matrix "A". By solving the polynomial, "n" eigenvalues are found (including duplicate roots) which can then be used to solve a system of linear equations to find eigenvectors "v" associated with those eigenvalues.

$$(A-\lambda I)v = 0$$

There are an infinite number of solutions and thus an infinite number of eigenvectors for the eigenvalues.

[eightballidiot]

Diagonalizing a Matrix (why tho?)

(tumblr doesnt like Mathjax: click here to make the equations work)

Diagonalization is one of those things that you learn in math and dont understand the point of why you need to know it and why you should even care.

I cant speak for everyone, but I like diagonalization because it makes matrix functions a lot easier to compute.

A matrix "A" is said to be diagonalizable if there exists a matrix "P" such that the equation above holds (where "D" is a diagonal matrix):

$$A = PDP^{-1}$$

The reason why this transformation is useful is because diagonal matrices are very easy to use in power series. A diagonal matrix raised to the power of "n" is equal to a diagonal matrix with the diagonal entries each raised to the power of "n" as seen belowː

$$\begin{pmatrix}2 & 0 & 0\\\ 0 & 3 & 0\\\ 0 & 0 & 4\end{pmatrix}^n=\begin{pmatrix}2^n & 0 & 0\\\ 0 & 3^n & 0\\\ 0 & 0 & 4^n\end{pmatrix}$$

This simple fact can be used to simplify matrix exponent of any matrix "A" if it is diagonalizable.

$$A^n = (PDP^{-1})^n = PDP^{-1}*PDP^{-1}*... = P(D^n)P^{-1}$$

The above statement is true since "P" multiplied by its inverse is the identity matrix and therefore they can be canceled out. The diagonal raised to the power n is trivially found as was seen above meaning that the matrix "A" raised to power of "n" can be calculated by simply multiplying three matrices instead of multiplying an "n" number of matrices; when "n" gets bigs, youll be begging for "A" to be diagonalizable.

[eightballidiot]

Testing

$$\begin{pmatrix}2 & 0 & 0\\\ 0 & 3 & 0\\\ 0 & 0 & 4\end{pmatrix}^n=\begin{pmatrix}2^n & 0 & 0\\\ 0 & 3^n & 0\\\ 0 & 0 & 4^n\end{pmatrix}$$

$$|x|*|y| = \sqrt{x_1^2....} * \sqrt{y_1^2....}$$

[eightballidiot]

What does convex actually mean?

I would bet that most people are aware of the terms convex and concave, and a good portion of those people have at least a basic idea of what those terms mean.

Concave is described as having cavities which curve into the solid while a convex solid only ever curves outwards.

(image courtesy of grammarly forum?)

This definition is fine; it doesn't really take too much effort to gauge whether a shape is convex or concave. However, assigning a mathematical statement to define whether something is convex or not seems kinda tricky; the definition above seems more like a vibe than an actual math definition.

However, there is a way to rigorously define convexness for a geometric object. A convex object is a solid with a defined interior where any line segment formed by any two interior points never crosses outside of the solid.

(images taken from Wikipedia)

The green figure above in convex as, no matter what choices one makes for point x and point y, the line segment will always be contained with the shape in green. There are no two points that have a line segment which reaches outside the shape.

Nevertheless, this shape in green isnt convex as there does exist a line segment that reaches outside of the shape while its two point remain inside as seen above. This shape also fits the vibes based definition above where a convex shape doesnt have cavities.

[eightballidiot]

There is only One Mirror Dimension

For those in chemistry or mathematics, chirality or mirror image objects will be familiar. They are objects that are the exact same except for a complete reflection of its structure.

An example for 2 dimensional objects is given below where the red and blue objects are mirrored imagesː

The two mirrored images above are formed by the reflection line in red. Every point on either side of the line goes to another point on the opposite side of the line. A line between opposite points is bisected perpendicularly by the reflection line. Only the points on the line stay in the same place.

There are an infinite number of possible places to put a reflection line on the 2d plane, but in the grand scheme of things, all possible reflection lines produce the same reflected shape (if one ignores differences in rotation and position.

An analogous situation happens in 3 dimensions, where a reflection plane (a 2d plane existing in a 3d space) swaps all points on either side of the plane to the opposite side. This produces the chiral pairs found in chemistry.

There are again an infinite number of places to put the reflection plane, but no matter what, the reflected images will all be the same if you ignore differences in rotation and position.

No matter where the mirroring line or plane is placed, it will always basically produce the same exact thing. There is one possible mirrored object.

Whenever you look in a mirror, no matter where you may be, you will always be looking at the same mirrored image of yourself.

[eightballidiot]

Inequalities are Weird (the math ones)

(the equations work here)

Inequalities are at their core in contrast to equality equations (its in the name). The algebra that can be applied to both sides seems pretty restrictive in comparison to regular algebraic equations.

The most striking property is that multiplying (or dividing) both sides by a negative number reverse the direction of the alligator mouth.

The reason why this happens is actually more simple than one might think. The equations below can be used to understand why this rule is true.

$$x<y$$

$$-x>-y$$

(click here if the equations arent working)

On a purely conceptual basis, the equations above make since. If y is indeed greater than x, than the negative of y must be less than the negative of x. Multiplying (or dividing) by -1 is just the same as taking the entire real number line and flipping it 180º. y is greater than x so y will be to the right of x, and -y is less than -x because it is now to the left of -x since the line was turned over.

Multiplying by a negative number "-p" is essentially the same as multiplying by -1 and then multiplying by "p" (division acts the same way).

More generally, in order to apply a function to both sides of the inequality, the function must either be "increasing*" (the alligator mouth is in the same direction) or "decreasing*" (the alligator mouth is in the opposite direction).

*- it should specifically be strictly increasing or strictly decreasing where no two outputs are equal unless their input is equal.

The reason why one reverses the inequality sign for the function f(x)=-x is because said function is always decreasing meaning that the order of the real numbers is reversed. For an equation f(x)=2x, the function is always increasing meaning the order of the real numbers is preserved.

Determining whether a function is increasing or decreasing requires two properties for each; the function must be injective meaning that f(x) and f(y) are equal if only if x and y are equal themselves, and the function must be either always greater than or equal to 0 (for increasing function) or always less than or equal to 0 (for decreasing functions).

For instance, the exponential function e^x is an increasing function as its derivative is always positive and the function is injective (meaning e^x only equals e^y if x and y are the same):

$$x<y$$

$$e^x<e^y$$

The function -x^3 is decreasing as the derivative is always zero or negative and there are no (real) numbers that have the same output. Therefore, the alligator mouth goes in the opposite direction:

$$x<y$$

$$-x^3>-y^3$$

[eightballidiot]

There are no Regular (Euclidean) Star Tessellations

There are three regular tessellations of the Euclidean plane. They are the triangular, square, and hexagonal tilings respectively; {3,6}, {4,4}, and {6,3}:

A regular tiling is defined as a tiling where the vertices, edges, and faces are all "transitive". A class of things is transitive if all of its various individual things can be mapped back onto any other individual things via moving, rotating, or reflecting the entire tiling while also making sure that the entire tiling also looks the same.

In other words, there is no identifiable difference between one thing and another thing of the same class.

For a Schläfli symbol to (potentially) tile the plane, the numbers {p,q} must obey the equation below:

$$2 = \frac{q(p-2)}{p}$$

(click here if it doesn't work)

For finite positive integer, the three tilings above are the only solutions. However, star polygons have Schläfli symbols with irreducible fractional entries greater than 2. Thus, by purely looking at the equation above, there are an infinite number of solutions if one allows star polygons to be vertex figures or faces.

The main issue already with these star tilings is that they will intersect themselves. Granted, this is not that unusual as the Kepler-Poinsot polyhedra are also regular and self-intersecting so it might be foreseeable that such a tiling could exist and still be considered regular.

While the equation does have many potential solutions such as {10/3,5} and {8/3,8}, none of these are regular due to an additional constraint mentioned above.

All vertices, edges, and faces must be unidentifiable and indistinguishable from all of the other vertices, edges, and faces.

Every star tiling is nonperiodic, there does not exist a tiling where the various things can all be matched to every other thing of the same type. They are all incapable of having analogous symmetries which are found in regular tilings.

[eightballidiot]

ANCIENTWRITING

THEOLDESTWRITINGWASALMOSTALWAYSWRITTENWITHNOSPACESNOPUNCTUATIONANDNOLOWERCASELETTERS

PUNCTUATIONWORDSPACESANDLOWERCASELETTERSWEREONLYINVENTEDLATERBYSCRIBESTOSPEEDUPWRITINGANDREADING

PEOPLEBACKINTHEDAYJUSTASSUMEDTHATASPEAKERWOULDBEABLETOPARSETHESTRINGOFLETTERSWITHOUTMUCHDIFFICULTY

[eightballidiot]

There are 5 Platonic Solids and 3 'Platonic' Tessellations

(the equations will only work here)

A Platonic solid is a regular polyhedron with convex regular polygons as its faces and vertex figures. Equivalently, the faces and vertex figure must have Schläfli symbols "{p}" where "p" is an integer greater than 2 (¡hey, look!).

The vertex figure is the shape that one would get if one cuts off a corner of the vertex; for instance, the vertex figure of a cube is a triangle as seen below:

The Schläfli symbol for a polyhedron with {p} faces and {q} vertex figures.

For the Platonic solids, the shape of the vertex figure is just the convex p-sided polygon where p is also the number of faces adjoined at each vertex. In order for a convex vertex figure to be formed in this way, the multiple regular polygons must not have a combined total of 360°.

Since the interior angle of a regular polygon is 180(p-2)/p°, the possible number combinations for {p,q} must obey the below inequality:

$$360 > 180\frac{q(p-2)}{p}$$

$$2 > \frac{q(p-2)}{p}$$

The faces {p} must be triangles or more because the digon doesnt work when adjoining it to form a polyhedron. It should also be stated that a vertex figure with only two faces doesn't work either as the faces will just fold in on themselves like an infinitesimally slim sandwich, so q must also be greater than 3.

For p = 3, the inequality is equivalent to the formula below:

$$2 > \frac{q(3-2)}{3} = \frac{q}{3}$$

Therefore, the only possible vertex figures {q} must be less than 6 as 6/3 = 2. Meaning {3,3}, {3,4}, and {3,5} are valid Platonic solids; they are the tetrahedron, octahedron, and icosahedron respectively.

The case of {3,6} is not a Platonic solid as it is not strictly convex and is infinite in size. It is however a regular tiling of the Euclidean place as 6 triangles around a vertex form 360°.

For p = 4, the inequality becomes the equation below:

$$2 > \frac{q(4-2)}{4} = \frac{q}{2}$$

This inequality is even more restrictive as only q = 3 satisfies the equation as 4/2 = 2. The only valid Platonic configuration is {4,3}: the cube.

{4,4} is the regular square tiling:

For p = 5, the inequality is

$$2 > \frac{q(5-2)}{5} = \frac{3q}{5}$$

or more simply 10 > 3q.

The only finite or regular configuration is {5,3} as 3*4 = 12 which is greater than 10. This also means that {5,4} is an invalid (Euclidean) tiling as a tiling requires both sides of the inequality to be equal whereas 12 does not equal 10.

The inequality for p = 6 is

$$2 > \frac{q(6-2)}{6} = \frac{4q}{6}$$

or 12 > 4q.

Since 3*4 = 12, {6,3} can't be finite and must be a tiling of hexagons.

The inequality has no solutions for p > 6 while also having q being 3 or more which means there are no other Platonic solids or tilings.

[enba-and-luka]

[eightballidiot]

The Apeirogon

(if equation broke go here)

The apeirogon (from the Greek word ἄπειροv 'infinite') is a polygon with an infinite number of edges. This post is about the regular apeirogon where all edges and 'angles' are congruent.

The apeirogon is best thought of as the end state of increasing the number of sides on a regular (convex) polygon. The interior angles get increasingly larger and the interior gets bigger and bigger if one sets the side lengths to be constant.

The formula for the interior angles of a regular convex polygon is given below where "p" is the number of sides:

$$180\frac{(p-2)}{p}$$

As the formula above approaches infinity, the value gets asymptotically close to 180°; therefore, the interior angles of the apeirogon with an infinite number of sides must also have interior angles of 180°.

$$\lim \limits_{p \to \infty}180\frac{(p-2)}{p}=180$$

If you think that sounds stupid, wait until you see the image of one:

Dealing with infinite objects gets pretty unintuitive, for any regular polygon with a finite number of sides, no matter how large that number might be, the sides always loop back around to enclose a finite hull. Nevertheless, this seems impossible for the apeirogon; it looks like a fucking line.

Now imagine if you took two of these things and mashed them together. You'll get the apeirogonal tiling:

There are two faces each connected to one another an infinite number of times on an infinite number of edges. 2 times 180° is 360° so there shouldn't be any room left. All signs logically point to it being a complete tiling of the plane, and yet it feels so stupid.

Don't even get me started on the prism.

[eightballidiot]

The Schläfli Symbol for (Star) Polygons

The Schläfli Symbol can be defined for 2-dimensional shapes as "the regular polygon such that all the interior angles are 180(p-2)/p°" which is denoted by "{p}" (p is always greater than 2).

For integers "p" greater than 2, the symbol "{p}" just refers to the regular convex p-sided polygon (¡hey look!).

The Schläfli symbol "{2}" is considered invalid as the digon would have overlapping edges (in Euclidean space). Although it does obey the formula above as 180(2-2)/2 = 0 and the digon does indeed have interior angles of 0°.

For those that have seen the Jan Misali video, yall are probably skeptical of my simplistic definition above. This is the part where I need to talk about star polygons.

The pentagram is pretty well known even outside of this Satanic site. It is usually not considered a polygon in the most basic sense of the word polygon as the 5 edges overlap. However, if one disregards the restriction against edges crossing each other, the pentagram can be classified as a regular polygon; all of it edges are equal and all of its vertices (the dots at the tips) look the same.

However, the pentagram isn't the only regular star polygon. There are an infinite number of similar shapes.

More generally a regular star polygon can be formed by taking the vertices of a convex p-sided polygon and connecting the points in a different way.

Starting from the top in a clockwise manner, one can draw the vertices of a p-sided polygon. A star polygon with "p" vertices can be constructed by drawing a line between points that are two or more vertices away if one were going in a clockwise manner.

For instance, the pentagram is made up of the vertices of a pentagon. Instead of connecting the edges between the points that are right next to each other, one can connect the points that are two vertices away in a clockwise direction meaning that the first point would connect to the third point, and the third point would connect to the fifth, and the fifth point would connect to the second, and the second point would connect with the fourth, and then finally the fourth would connect with the first, completing the pentagram.

If one takes the group of points for {p}, and some number "q" exists such that the fraction "p/q" is both irreducible and greater than 2, then a star polygon can be formed by connecting each point with an edge while skipping (q - 1) points (the points are ordered in a rotational direction). The star polygon of this form is denoted with the Schläfli symbol "{p/q}".

However, this fractional entry seems like a weird symbol. We established that the Schläfli symbol "{x}" has interior angles equal to 180(x-2)/x°; in order for this relationship to hold for the star polygons each star {p/q} would need to satisfy the interior angle equation as well where the angles are equal to 180*(p/q - 2) / (p/q)°.

Believe or not, the interior angle formula does hold for the fractional entries. The pentagram {5/2} has interior angles equal to 36° or 180*(5/2 - 2) / (5/2)°.

Why this relationship holds is left as an exercise to the reader.

Hint:

inscribed angles / e^πi = -1