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On Naturally Arising Differential Equations

October 18th, 2016 No comments

So if you have been following the argument a bit, it turns out that

 p(x,y,t) = \alpha^{t-1} \mathbf{P}_x(x) \cdot \mathbf{P}_y(y) + f_n^*(x)

is the starting at time t = 1 transition probability propagator of a probability distribution, say c_0(x) at t=0, in the interval x = 0 to 1.  A question that I tried to answer was how zeros are propagated via the propagator or at the probability distribution, which lead to theorems that I dubbed "Shadow Casting" because, under that context, it turns out that a zero, if found on the propagator, remained in place until infinity, and via the propagator it appears on the probability distribution we want to propagate as well (therefore casting a "shadow").  I hadn't thought of the following approach until recently, and I haven't worked it out completely, but it connects to the theory of Ordinary Differential Equations which seems interesting to me. Here's the argument:

Suppose we focus on p(x,y,1) for the time being, and wish to find the zeros on the transition probability surface.  Thus we seek p(x,y,1) = 0 and suppose y(x) is an implicit function of x. We have

 p(x,y,1) = 0 = \mathbf{P}_x(x) \cdot \mathbf{P}_y(y(x)) + f_n^*(x)

 Now let \mathbf{P}_y(y) is a collection derived from y(x), so that, for example,

 \mathbf{P}_y(y(x)) = \left[ \begin{array}{c} y(x) \\ y^{\prime}(x) \\ \vdots \\ y^{n-1}(x) \end{array} \right]

and I think we have successfully created a link to ODEs.  To find the zeros on the surface (and other time surfaces of the propagator) we stick in the correct \alpha and solve, using the familiar methods (solve the homogeneous equation and the particular solution via sin-cos-exponential solutions, variation of parameters, power series, etc.).

I'm working out the specificities, for example including the constraints we know on f_n^*(x) or \mathbf{P}_x(x).  Perhaps this approach will help us broaden the spectrum of differential equations we can solve, by linking via Shadow Casting.

It may seem basic, but I think there is some untapped power here.

Additionally, I have been working on clarifying some thoughts on polynomials that converge in area in the interval from 0 to 1, but all those details tend to be a bit drab and I keep having trouble focusing.  Nevertheless, there is a lot of clarity that I have been able to include, and it is now in the newest draft of "Compendium".  By the way, I renamed it. It is now "Compendium of Claims and Proofs on How Probability Distributions Transform".  There's still soooo much more to do.

Here it is! part-i-v28

On Proving Eigenvalue = 1 for Particular Surfaces

March 21st, 2016 No comments

So, as you have read here, I've been saying for quite a few years now that (bounded) smooth surfaces on \left[ 0, 1 \right]^2 \to \mathbb{R} have certain invariants when viewed from a particular perspective (eigenvalues, eigenfunctions).  There are particular surfaces which I dubbed Pasquali patches (for lack of a better word) and a particular construction, namely

 p(x,y) = f_1(x) g_1(y) + f_2(x) \frac{1-g_1(y) F_1}{F_2}

with F_2 = \int_0^1 f_2(x) \, dx \neq 0 which are very special in that they have lots of properties which are interesting and closely tied to probability theory.  I have now proven for this particular construction that it possesses two very specific eigenvalues given a particular operator "star" \star... one of which is \lambda_1 = 1, regardless of function choices for f_1(x), g_1(y) and almost arbitrary choice of f_2(x), which, by requirement needs F_2 \neq 0.  This mimics well known probability mathematics (except in the surface realm) and operator theory/linear algebra. I think of this as a very proud accomplishment.

Thusly, I have revamped the relevant sections in Compendium full of new and juicy recharacterizations in order to be able to do just this... particularly Section 2, the definition of Pasquali patches and Section 11.6, Relevant Generalizations as Applied to Pasquali Patches (where I have included such a proof).  Section 2 is now a lot tighter than it used to be, and I'm trying really hard to go over everything to close all the loopholes I've left so that Compendium isn't just notes but an actual... Volume of Mathematics or Book or something.

I am very close to a full eigenvalue theory which I will apply to a generalization of the Pasquali patch formula above, giving a multiplicity of calculable eigenvalues for, not just Pasquali patches, but any sufficiently well behaved surface on \left[ 0, 1 \right]^2 \to \mathbb{R} .

Part I v26

More On Convergent-in-Area Polynomials in the Interval [0,1]

October 9th, 2014 No comments

So remember last time I wrote about the rich algebraic structure of polynomials that have convergent area in the interval [0,1]? Turns out there is a lot that can be done, even more than I imagined.  I have already begun to sketch the properties out and have a long ways to go still, but at least I have defined operations on these polynomials that I think can be very useful.

Here's my newest version of Compendium where I have added these things.

Part I v23

On convergent-in-area polynomials in the interval [0,1]

August 13th, 2014 No comments

Take any finite or infinite polynomial which converges in area in the interval between [0,1].  We can define equivalence classes on such space by mapping them back to distinct probability distributions, such that each equivalence class with all its elements is a semigroup under a particular operation.  We can extend each semigroup into a group if we incorporate an identity element and inverses.  Now we've created fun to last for a lifetime!  I've included these ideas in version 20 of Compendium.

Part I v20

There is more to come, as we spawn mathematical objects of weirdness that can be related back or extend probability theory.

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

October 17th, 2010 No comments

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.