Continuous matrices

This will be a naive discussion about functions roughly of the form $A: \mathbb{R}^2\to\mathbb{R}$ but allowing delta behaviour which I only know of informally so far. $A$ is a continuous matrix. Addition $A+B$ is done componentwise and multiplication $AB$ is carried out by integrating
$$(AB)(x,y)=\int_{-\infty}^{+\infty}A(x,z)B(z,y)dz.$$

The plot operator $P$, defined through $P_f(x,y)=\delta(x-f(y))$, creates a continuous matrix. One interesting property, when combined with the usual associativity and distributivity of matrix operations, is that
$$ P_fP_g=\int_{-\infty}^{+\infty}\delta(x-f(z))\delta(z-g(y))dz=\delta(x-f(g(y)))=P_{f\circ g}.$$ This property makes it possible to interpret an arbitrary continuous matrix as a superposition of graphs $A = \sum \alpha_iP_{f_i}$ of single variable functions $f_i$, and to interpret multiplication of two matrices $A,B$ as a weaving together of all possible compositions: $AB = \sum\alpha_i\beta_jP_{f_i\circ g_j}.$ But all this is just conceptual speculation.


Here is a new light on fourier transforms: $F(x,y)=e^{-ixy}$. The inversion formula $FFf(x)=\tau f(-x)$ can be written as $F^2=\tau\delta(x+y)$ and verified using the definition of matrix multiplication.


In simple cases of signal analysis a linear operator $T$ transforming the signal is characterized through its impulse response $h=T\delta$ in the fashion $Tx(t)=\int_{-\infty}^{+\infty}h(t-u)x(u)du$, i.e. through the convolution $Tx=h*x$. The $h*$ part can be written as a matrix. Indeed, for any functions $x,y$,
$$ x*y(t) = \int_{-\infty}^{+\infty}x(t-u)y(u)du = C_xy(t) $$ where $C_f(x,y)=f(x-y)$ in general. In this case we have $Tx=h*x=C_hx$ so that $T=C_h$ in the same sense that a finite-dimensional linear transformation is equal to its transformation matrix.


This reminds me of the familiar way to define complex multiplication:
$$\left(\begin{matrix}a\\b\end{matrix}\right)\left(\begin{matrix}c\\d\end{matrix}\right)=\left(\begin{matrix}a&-b\\b&a\end{matrix}\right)\left(\begin{matrix}c\\d\end{matrix}\right).$$ Maybe it's more common to use matrices in both steads but we see that with this formulation $zw=M_zw$ where $M_z$ is chosen in the obvious way. The similarity between $C$ and $M$ is that both are obtained by repeating data across diagonals.

Another similarity: In the same way that matrix models verify the associativity $M_uM_vM_w$ of complex number multiplication $uvw$, they verify the associativity $C_fC_gC_h$ of convolutions $f*g*h$. And the same holds for the distributive property.

The square root algorithm

I was taught that to get the square root of a number \(a\) you iterate \(x\mapsto \frac{x+\frac ax}2\) until you're happy with the result. More generally, to scale a vector \(v\) to have quadrance \(a\) you iterate \(v\mapsto \frac{1+\frac a{v\cdot v}}2v\). That's a direct analog. With negative quadrances you get chaotic behavior.

Tetrahedra on a hyperbolic parabola

Instructed by a youtube video I folded a hyperbolic parabola. It looked like a regular tetrahedron so I wondered if there is any regular tetrahedron on the rational surface \(z=x^2-y^2\).

It should be centered at the origin and have the following vertices:\[[a,b,a^2-b^2], [-a,-b,a^2-b^2], [-b, a, -a^2+b^2], [b, -a, -a^2+b^2].\]We want all quadrances to be the same. By symmetry, only one equation is needed:\[4a^2+4b^2=(a-b)^2+(a+b)^2+4(a^2-b^2)^2\]or equivalently\[a^2+b^2=2(a^2-b^2)^2.\]

Our approach is to try to solve for one of the variables, say \(a\) in terms of \(b\). We will plug in different rational \(b\) and filter out those which give rational \(a\). A major insight is that filtering can be done by parametrization, or a change of variables if you wish. Let's start with completing the square:\[\left(a^2-\frac{4b^2+1}4\right)^2=b^2+\frac1{16}.\]Denote the left-hand side by \(c\). We can find points on the hyperbola \(c^2-b^2=\frac1{16}\) by first parametrizing the unit circle and noticing that\[c^2-b^2=\frac1{16}\iff\left(\frac bc\right)^2+\left(\frac1{4c}\right)^2=1.\]The parametrization becomes (we choose \(2t\) rather than \(1-t^2\) in the denominator)\[(b,c)=\left(\frac{1-t^2}{8t},\frac{1+t^2}{8t}\right).\]and thus the completed square can be written (we arbitrarily choose a plus sign when taking roots)\begin{align*}a^2&=\frac{\dfrac{(1-t^2)^2}{16t^2}+1}4+\frac{1+t^2}{8t}=\frac{1+14t^2+t^4+8t+8t^3}{64t^2}\\&=\frac{(1+t)^4+4t(1+t)^2}{64t^2}=\frac{(1+t)^2}{64t^2}((1+t)^2+4t).\end{align*}We now turn to the equation\[d^2=(1+t)^2+4t=(t+3)^2-8=e^2-8.\]In this case let's try a different substitution:\begin{align*}(d,e)&=(u+v,u-v)\\uv&=-2\\(d,e)&=\left(u-\frac2u,u+\frac2u\right)=\left(\frac{u^2-2}u,\frac{u^2+2}u\right)\\t&=e-3=\frac{u^2-3u+2}u=\frac{(u-1)(u-2)}u.\end{align*}Finally,\begin{align*}(a,b)&=\left(\frac{1+t}{8t}d,\frac{1-t^2}{8t}\right)=\left(\frac{\frac{u^2-2u+2}u\cdot\frac{u^2-2}u}{8\frac{(u-1)(u-2)}u},\frac{-\frac{u^2-2u+2}u\cdot\frac{u^2-4u+2}u}{8\frac{(u-1)(u-2)}u}\right)\\&=\frac{u^2-2u+2}{8u(u-1)(u-2)}(u^2-2, -u^2+4u-2).\end{align*}In particular, one possible tetrahedron, obtained when \(u=-1\), has vertices\[[a,b,a^2-b^2]=\frac1{48}[5, 35, -25],\text{ etc}.\]

More music

Yesterday I was determined to write a song, and so I did.

 

Theorem classification

I claim (as an experiment) that there are three kinds of theorems "If A then B". Let \(|A|\) be the number of arrangements of objects that fit the description, i.e. interpret \(A\) as a set of tuples.

1. \(|A| = \aleph_0\) – these give exact answers
2. \(|A| = c\) – these give approximate results, usually to arbitrary precision
3. \(|A| > \aleph_0\text{ but }|A|\not=c\) – these are even deeper set-theoretical fantasies

If \(|A|\in\mathbb{N}\text{ or }|A| = 0\) then the theorem doesn't deserve to be called a theorem because the use of the implication can be avoided by enumerating all statements \((a_i\in A,b_i\in B)_{i=1}^n\).

A subtle issue is that my use of sets is not justified. For example, according to present set theory, there is no set of all sets, so my classification cannot handle statements like "All sets have a power set". I count all such theorems into category 3 because they are set-theoretic.

How many functions are there?

Claim
There are more real functions than there are real numbers.

Explanation
\(\{f:\mathbb{R}\to\mathbb{R}\} \geq \{f:\mathbb{R}\to\{0, 1\}\} = \mathcal{P}(\mathbb{R}) > \mathbb{R}\)

Claim
There are just as many continuous functions as there are real numbers.

Explanation
\(\{f:\mathbb{R}\to\mathbb{R}, f\text{ continuous}\}\\\qquad\begin{cases}\geq\{f:\{0\}\to\mathbb{R}\} = \mathbb{R}\\\leq\{f:\mathbb{N}\times\mathbb{N}\to\mathbb{R}\}=\mathbb{R}^{\mathbb{N}\cdot\mathbb{N}}=2^{\mathbb{N}\cdot\mathbb{N}\cdot\mathbb{N}}=2^\mathbb{N}=\mathbb{R}\end{cases}\)

The first step on the second line is justified by the fact that for every continuous function, there is a sequence of parallelogram functions converging pointwise to it. By parallelogram function, I mean a function whose value is specified at evenly distributed points and interpolated linearly in between. For example, a parallelogram function approximating \(\sin(x)\) is \[f(x) = \sin(\lfloor x\rfloor) + (x-\lfloor x\rfloor)\sin(\lfloor x\rfloor+1).\]
To narrow the number of parallelogram functions down to \(\{f:\mathbb{N}\times\mathbb{N}\to\mathbb{R}\}\), require that the value at \(0\) be specified and that the distance between specified points for the \(n\)th function in the sequence be \(1/n\).

Central motion excludes the center

Central motion is the motion of a particle under the influence of a central force. I have known for some year that the particle cannot not pass through the origin. Mimicking a proof in my multivariable calculus book, I'll formalize the fact.

Theorem
If \(r:(a, b)\to\mathbb{R}^3\) is twice differentiable, \(r(t)\times r''(t) = \vec{0}\) for all \(t\) and \(r(t_0)\times r'(t_0) \not=\vec{0}\) for some \(t_0\), then \(r(t)\) is never \(\vec{0}\).

Proof
\[\frac{d}{dt}(r(t)\times r'(t)) = r'(t)\times r'(t) + r(t)\times r''(t) = \vec{0}\]Thus \(r(t)\times r'(t)=c\) for some \(c\). Because \(r(t_0)\times r'(t_0) \not=\vec{0}\), \(c\not=\vec{0}\). If \(r(t_1)\) were \(\vec{0}\) for some \(t_1\) then \(c\) would be \(\vec{0}\) so there is no such \(t_1\).

I wonder how to generalize this to higher dimensions.

Continuity at infinity

Usually \(f(x)\to L\text{ as }x\to \infty\) is defined in an epsilon N fashion. With a projective point of view, infinity becomes a point among the usual numbers, and the definitions of convergence at finite and infinite points can be unified. What strikes me is that this means that if f(x) converges as x goes to infinity, then f can be continuously extended at infinity.

Cauchy–Schwarz

Finally I understand the Cauchy–Schwarz inequality! No induction needed.

Theorem
\(a, b\in \mathrm{Rat}^n\)
\(n\in\mathrm{Nat}\)
                     
\((a\bullet b)^2 \leq \mathop{\rm Q}(a)\mathop{\rm Q}(b)\)

Proof
Recall that by definition
\[a\bullet b=\sum_{i=1}^n a_ib_i\]\[\mathop{\rm Q}(a)=\sum_{i=1}^n a_i^2\]\[\mathop{\rm Q}(b)=\sum_{i=1}^n b_i^2.\]
The claim, reformulated as
\[\left(\sum_{i=1}^n a_ib_i\right)^2 \leq \left(\sum_{i=1}^n a_i^2\right)\left(\sum_{i=1}^n b_i^2\right),\]
is equivalent to (by distributing terms)
\[\left(\sum_{i=1}^n a_i^2b_i^2\right)+\sum_{1\leq i<j\leq n}2a_ib_ia_jb_j \leq \left(\sum_{i=1}^n a_i^2b_i^2\right)+\sum_{1\leq i<j\leq n}a_i^2b_j^2+a_j^2b_i^2\]
and (by collecting terms)
\[0 \leq \sum_{1\leq i<j\leq n}a_i^2b_j^2+a_j^2b_i^2-2a_ib_ia_jb_j = \sum_{1\leq i<j\leq n}(a_ib_j-a_jb_i)^2.\]
The last statement is true because squares are positive and sums of positive numbers are positive.

A spiral

Let \(\mathbb{O}\) be the countable set of all infinite rational number sequences which are ultimately but not identically zero, and let \(H\) contain those elements in \(\mathbb{O}\) whose last nonzero element is positive. With respect to \(\mathbb{O}\), define \(H^\complement\). Then \(f(x) = -x\) is a bijection from \(H\) to \(H^\complement\).

Also, \(H\) looks like a spiral. To see that, require the last nonzero element to be the first, then the second, and then the third one, and watch this sequence of subsets twist into a new dimension at each step.

That's really something! A spiral whose reflection in the origin is its complement.