Home > Combinatorics and Probability, Group Theory, Linear Algebra, Markov Chains, Mathematics > On Patchix by Patchix Products – Tying Up Loose Ends - (RWLA,MCT,GT,AM Part V)

## On Patchix by Patchix Products – Tying Up Loose Ends - (RWLA,MCT,GT,AM Part V)

In this post I want to "tie up a few lose ends."  For example, in my last post I stated that the patchix pattern

$\begin{array}{ccc} p_1(x,y) & = & 1 - cos(2 \pi x) cos(2 \pi y) \\ p_2(x,y) & = & 1 + \frac{cos(2 \pi x) cos(2 \pi y)}{2} \\ p_3(x,y) & = & 1 - \frac{cos(2 \pi x) cos(2 \pi y)}{4} \\ p_2(x,y) & = & 1 + \frac{cos(2 \pi x) cos(2 \pi y)}{8} \\ \vdots \\ p_t(x,y) & = & 1 - \frac{cos(2 \pi x) cos(2 \pi y)}{(-2)^{t-1}} \end{array}$

for $t \in \mathbb{Z^+}$, but I didn't prove it.  It's simple to do by induction: by the inductive hypothesis,

$p_1(x,y) = 1 - cos(2 \pi x) cos(2 \pi y) = 1 - \frac{cos(2 \pi x) cos(2 \pi t)}{(-2)^{1-1}}$

By the inductive step, assume

$p_k(x,y) = 1 - \frac{cos(2 \pi x) cos(2 \pi y)}{(-2)^{k-1}}$

Then,

$\begin{array}{ccc} p_{k+1}(x,t) & = & \int_0^1 p_1(1-y,t) \cdot p_k(x,y) dy \\ & = & \int_0^1 \left( 1 - cos(2 \pi (1-y))cos(2 \pi t) \right) \cdot \left( 1 - \frac{cos(2 \pi x) cos(2 \pi y)}{(-2)^{k-1}} \right) dy \end{array}$

Now, if one dislikes shortcuts one can expand the product and integrate term by term to one's heart's content.  The "shorter" version is to relate the story: notice the product of 1 with itself is 1, and such will integrate to 1 in the unit interval.  So we save it.  The integrals $\int_0^1 cos(2 \pi y) dy$ and $\int_0^1 cos(2 \pi - 2\pi y) dy$ both evaluate to zero, so we are left only with the task of evaluating the crossterm:

$\begin{array}{ccc} && \int_0^1 cos(2 \pi (1-y))cos(2 \pi t) \cdot \frac{cos(2 \pi x) cos(2 \pi y)}{(-2)^{k-1}} dy \\ & = & \frac{cos(2 \pi t) cos (2 \pi x)}{(-2)^{k-1}} \int_0^1 cos(2 \pi - 2 \pi y) cos(2 \pi y) dy \\ & = & \frac{cos(2 \pi t) cos (2 \pi x)}{(-2)^{k-1}} \int_0^1 cos^2(2 \pi y) dy \\ & = & \frac{cos(2 \pi t) cos (2 \pi x)}{(-2)^{k-1}} \cdot \frac{1}{2} \\ & = & -\frac{cos(2 \pi t) cos (2 \pi x)}{(-2)^{k}} \end{array}$

Let's not forget the 1 we had saved, so:

$p_{k+1}(x,t) = 1 - \frac{cos(2 \pi x) cos(2 \pi t)}{(-2)^{k}} \rightsquigarrow 1 - \frac{cos(2 \pi x) cos(2 \pi y)}{(-2)^{k}} = p_{k+1}(x,y)$

as we wanted to show.

So finally notice that, of course, if we take the limit as $t$ approaches infinity, the patch evolution tendency is to become 1, the uniform distribution:

$\lim_{t \rightarrow \infty} p_t(x,y) = 1 = u(x,y)$

From here on out, I want to set up the operative framework of patchixes, in analogy with discrete matrices.  I want to show that in general, patchix products are non-commutative.  This is easily done by counterexample:

We want to show that $p(x,y) \star q(x,y) \neq q(x,y) \star p(x,y)$. So suppose the patchixes $p(x,y) = x$ and $q(x,y) = y$. Then

$p(x,y) \star q(x,y) = \int_0^1 p(1-y,t) \cdot q(x,y) dy = \int_0^1 (1-y) y dy = \int_0^1 y - y^2 dy = \frac{1}{6}$

and

$q(x,y) \star p(x,y) = \int_0^1 q(1-y,t) \cdot p(x,y) dy = \int_0^1 (t \cdot x) dy = t \cdot x \rightsquigarrow x \cdot y$

are clearly not-equal.  It would be great to say that, because patchixes are non-commutative, patches are too, but we don't know that patches as a whole subset of patchixes commute, so let's disprove it.  Now suppose the patches $p(x,y) = x + \frac{1}{2}$ and $q(x,y) = 1 + xy - \frac{y}{2}$.  Then

$\begin{array}{ccc} p(x,y) \star q(x,y) & = & \int_0^1 p(1-y,t) \cdot q(x,y) dy \\ & = & \int_0^1 \left( \frac{3}{2} - y \right) \cdot \left( 1 + xy - \frac{y}{2} \right) dy \\ & = & \frac{5x}{12} + \frac{19}{24} \end{array}$

where

$\begin{array}{ccc} q(x,y) \star p(x,y) & = & \int_0^1 q(1-y,t) \cdot p(x,y) dy \\ & = & \int_0^1 q(1-y,t) \cdot p(x) dy \\ & = & p(x) \int_0^1 q(1-y,t) dy \\ & = & p(x) \cdot u(t) = p(x) \\ & = & x + \frac{1}{2} \end{array}$

By refraining from calculating this last bit explicitly, we have (serendipitously) proved that any patch by a patch that is solely a function of $x$ returns the last patch, a result which reminds us of the analogous distribution by patch result I have shown in my previous post (a distribution on [0,1] times a patch that is solely a function of $x$ returns the patch, that viewed from the point of view of functions is a distribution on [0,1]).  A quick note: the integral $\int_0^1 q(1-y,t) dy$ is the unit distribution because $\int_0^1 q(x,y) dx = u(y)$ and $x \rightsquigarrow (1-y)$ and $dx \rightsquigarrow -dy$.

The end result of these observations is that patches are also, in general, non-commutative.

Next, I want to show that patchixes in general are associative.  This is a bit tricky because of the "after integral" transformations we have to do, but it is doable if we keep careful track of our accounting.  We want to show that $[p(x,y) \star q(x,y)] \star r(x,y) = p(x,y) \star [q(x,y) \star r(x,y)]$.  Let's begin with the left hand side.

$\begin{array}{ccc} [p(x,y) \star q(x,y)] \star r(x,y) & \rightsquigarrow & [p(x,w) \star q(x,w)] \star r(x,y) \\ & = & \left( \int_0^1 p(1-w, y) \cdot q(x, w) dw \right) \star r(x, y) \\ & = & \int_0^1 \left( \int_0^1 p(1-w, t) \cdot q(1-y, w) dw \right) \cdot r(x, y) dy \\ & = & \int_0^1 \int_0^1 p(1-w, t) \cdot q(1-y, w) \cdot r(x, y) dw dy \\ & = & s(x,t) \rightsquigarrow s(x,y) \end{array}$

Now the right hand side

$\begin{array}{ccc} p(x,y) \star [q(x,y) \star r(x,y)] & \rightsquigarrow & p(x,w) \star \left( \int_0^1 q(1-y, w) \cdot r(x,y) dy \right) \\ & = & \int_0^1 p(1-w, t) \cdot \left( \int_0^1 q(1-y, w) \cdot r(x,y) dy \right ) dw \\ & = & \int_0^1 \int_0^1 p(1-w,t) \cdot q(1-y, w) \cdot r(x,y) dy dw \\ & = & s(x,t) \rightsquigarrow s(x,y) \end{array}$

The two sides are equal when we can apply the Fubini theorem to exchange the order of integration.

Of course, patches, being a subset of patchixes, inherit associativity.

Defining a patchix left and right identity is extremely difficult, in the sense that, if we take a hint from discrete matrices, we'd be looking at a very special function on the $xy$ plane, so that $i(1-y,y) = i(x,1-x) = 1$ and $0$ everywhere else.  Because there is no "pretty" way to define this as a function of $x$ and $y$ both, showing that when we multiply a patchix by this function on either the right or the left requires elaborate explication. Unless we take it as axiomatic high ground, postulating the existence of an identity function $i(x,y)$ so that $i(x,y) \star p(x,y) = p(x,y) = p(x,y) \star i(x,y)$ to make the framework work, there is no easy way out.  Let's give it a shot then.

Left identity:

$i(x,y) \star p(x,y) = \int_0^1 i(1-y,t) \cdot p(x,y) dy$

Now $i(1-y,t) = 1$ only for values where $t = y$, as we've defined it, otherwise the integral is zero and there is nothing to solve.  So then we've got

$\int_0^1 i(1-t,t) \cdot p(x,t) dy = \int_0^1 (1) \cdot p(x,t) dy = p(x,t) \rightsquigarrow p(x,y)$

which is essentially the argument I make for the zero patch power in my informal paper on continuous Markov transition matrices or patches (however, there's a problem with this definition on patches, more of this below).  There's the question of why we didn't force the change of $dy \rightsquigarrow dt$, and this is because the only way to obtain a function of both $x$ and $t$ is to force the patchix to the $x t$ plane and let the integral be taken in the $x y$ plane.  If this argument is unsatisfactory, consider this one:  at $t = 0 = y$ the patchix takes the values $p(x, 0)$ which is a function of $x$ alone.  Thus,

$\int_0^1 i(1,0) \cdot p(x,0) dy = p(x,0) \int_0^1 (1) dy = p(x,0)$

if we do this for all $t \in [0,1]$, we are certainly left with $p(x,t)$.  We may raise the objection that, if we create a mental picture of the situation, at $t = 0$, $i$ takes a value of 1 only at $y = 0$, so that, on the $x y$ plane, all values of $p(x, y)$ are zeroed except those at $y = 0$.  Thinking about it this way creates the difficulty of the integral with respect to $y$: it evaluates to zero (there is no "area" in the $x y$ plane anymore, only a filament or fiber at $y=0$), and we would be left with the zero patchix.  There is no way to resolve this except two ways: to send the patchix $p(x,y)$ to $p(x,t)$ before we take the integral in the $x,y$ plane, and then toss the integral out the window (or take it on the uniform distribution), or, to think of the filament $p(x,0) = p_0(x)$ as $p_0(x) \times [0,1] = p_0(x,y)$ and then integrate in the $x y$ plane to obtain $p_0(x) \rightsquigarrow p(x,0)$ and do this for all $t$ to get $p(x,t)$.  Hence yes, the difficulty of defining the identity function on "surface" matrices (because it is not smooth like they are and because it is defined piece-wise).

Right identity:

$p(x,y) \star i(x,y) = \int_0^1 p(1-y,t) \cdot i(x,y) dy$

Here we remind ourselves that $i(x,1-x) = 1$ and zero otherwise, so that we can make the substitution

$\int_0^1 p(x,t) \cdot i(x,1-x) dy = \int_0^1 p(x,t) \cdot (1) dy = p(x,t) \rightsquigarrow p(x,y)$

We of course have issues: it may seem redundant to send $x \rightsquigarrow 1-y \rightsquigarrow x$, sending $x$ back to itself, but again this is the only way to remain consistent and get back the original function.  Again there's an issue of why we didn't send the integral $dy \rightsquigarrow -dx$, but this has to remain in the $x y$ plane for the mechanics to work.  Other objections are likewise not easily resolved; but the argument would work out algebraically if we concede on a few things: otherwise we cannot but shrug at the fact that it is, indeed, a little bit of hocus pocus, and we return to our suggestion to postulate the identity function as an axiom. Perhaps maybe these issues can be resolved or elucidated a little later, I don't lose hope.

Defining inverse patchixes will also present a great difficulty, particularly because they have to produce the identity function when we "patchix multiply" two mutually inverse patchixes  together.  I was thinking that we could perhaps determine whether a particular patchix has one, by extending Sarrus's rule (for determinants) to be continuous, which would involve, I'm sure, multiple integrations.  This will be a topic of further investigation for me. The cool thing is, if we can elucidate how this "continuous version" of the determinant works, many different results from Linear Algebra could follow.  I am also trying to figure out how two inverse patchixes would look like, and if I can produce an example (at all), virtually from thin air.  If I can, then perhaps we're on our way to constructing patchix groups of all flavors.

Unfortunately, patches can't inherit the identity as we've defined it: the integral with respect to $x$ of $i(x,y)$ is zero for all $y$.  Thus $i(x,y)$ is not a patch.

This problem makes us want to think of the uniform distribution $u(x,y)$ as another possible candidate for the identity for patchixes all, and it might just work if we agree that, when we don't have a function of $t$ or of $x$ after doing the setup-transformations for the integral, we send whatever function remains there before taking the integral.

Left identity:

$u(x,y) \star p(x,y) = \int_0^1 u(1-y,t) \cdot p(x,y) dy \rightsquigarrow \int_0^1 (1) \cdot p(x,t) dy = p(x,t) \rightsquigarrow p(x,y)$

Right identity:

$p(x,y) \star u(x,y) = \int_0^1 p(1-y,t) \cdot u(x,y) dy \rightsquigarrow \int_0^1 p(x,t) \cdot (1) dy = p(x,t) \rightsquigarrow p(x,y)$

This has several happy consequences: we avoid dealing with a piece-wise defined function $i(x,y)$ which is zero everywhere except on $y = 1-x$, the uniform distribution is smooth, we can now more easily define inverses (by finding multiplicative inverse functions, more on this below), and, specifically regarding patches, $\int_0^1 u(x,y) dx = u(y) = 1$ so the uniform distribution is indeed a patch.

In my mental picture, the "patchix product" of the uniform distribution with a patchix (and vice versa) doesn't "add up" (pun intended), but the algebraic trickery would seem to be the same even when using the alternative $i(x,y)$.  So.  At this point I sort of have to convince myself into accepting this for now.