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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.

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