On Convergent Sequences

November 13th, 2016 No comments

So here's a question that I've been preoccupied with in the last couple of days.  Suppose I have a convergent (at the sum) sequence, if you like a geometric sequence, say  1/2, 1/4, 1/8, ... = 1.  If one were to multiply the elements of the sequence by its corresponding index, would this still be a convergent sequence? In this case, I'm concerned with 1/2 * 1, 1/4 * 2, 1/8 * 3, ... and its convergence (at the sum).  I wish to explore whether there are particular circumstances in which this is indeed the case.

I'll leave the question open ended for a bit.  I have solved this via derivative considerations, I think (I do not mean it as a pun... I think I have solved this by considering derivatives of functions).

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 Riding the Wave

May 30th, 2016 No comments

So here it is, the pinnacle of my research effort thus far. I'll start by the definitions:

Definition 1. Let f(x,y) and g(x,y) be surfaces so that f,g \colon [0,1] \times [0,1] \to \mathbb{R}. The star operator \star \colon [0,1]^2 \times [0,1]^2 \to [0,1]^2 takes two surfaces and creates another in the following way:

 \left( f(x,y), g(x,y) \right) \rightsquigarrow \left( f(1-y, z), g(x,y) \right) \rightsquigarrow h(x,z) \rightsquigarrow h(x,y)

with the central transformation being defined by \diamond \colon [0,1]^2 \times [0,1]^2 \to [0,1]^2

 f(1-y,z) \diamond g(x,y) = \int_{0}^{1} f(1-y,z) g(x,y) \, dy = h(x,z)

and the last transformation that takes h(x,z) \rightsquigarrow h(x,y) we will call j \colon [0,1]^2 \to [0,1]^2. Thus

 \boxed{f(x,y) \star g(x,y) = j \left( f(1-y, z) \diamond g(x,y) \right) = j \left( \int_0^1 f(1-y,z) g(x,y) \, dy \right)}

Definition 2. Define a continuous, bounded surface p(x,y), with p \colon [ 0, 1 ] \times [ 0, 1 ] \to \mathbb{R}^+ \cup \{0\} ,  and let  \int_0^1 p(x,y) \, dx = 1 be true regardless of the value of  y .    In other words, integrating such surface with respect to x yields the uniform  probability distribution u(y), u \colon [0,1] \to \{1\} . We will call this a strict Pasquali patch, and such is intimately related to probability notions.  With p \colon [ 0, 1 ] \times [ 0, 1 ] \to \mathbb{R}, we have a more general definition for a Pasquali patch.

Construction 1. Let p(x,y) = \sum_{i = 1}^n f_i(x) \cdot g_i(y) = \mathbf{f}(x) \cdot \mathbf{g}(y), a function which consists of a finite sum of pairs of functions of x and y. In the spirit of conciseness, we omit the transpose symbology, thusly understanding the first vector listed in the dot product as a row vector and the second vector as a column vector.  Then p(x,y) is a Pasquali patch provided

g_n(y) = \frac{1- \sum_{i = 1}^{n-1} g_i(y) F_i}{F_n} = \frac{1 - \mathbf{g}_{n-1}(y) \cdot \mathbf{F}_{n-1}}{F_n}

and F_n \neq 0. Thus, we may choose n-1 arbitrary functions of x,  n-1 arbitrary functions of y, an nth function of x so that F_n \neq 0, and

p(x,y) = \sum_{i = 1}^{n-1} f_i(x) \cdot g_i(y) + f_n(x) \cdot \frac{1 - \sum_{i = 1}^{n-1} g_i(y) F_i}{F_n} = \mathbf{f}_{n-1}(x) \cdot \mathbf{g}_{n-1}(y) + f_n(x) \cdot \frac{1 - \mathbf{g}_{n-1}(y) \cdot \mathbf{F}_{n-1}}{F_n}

 We may write the normalized version as:

 \boxed{ p(x,y) = \left( \mathbf{f}_{n-1}(x) - f_n^*(x) \cdot \mathbf{F}_{n-1} \right) \cdot \mathbf{g}_{n-1}(y) + f_n^*(x)}

and again observe that the unit contribution to the integral of the Pasquali patch is provided by f_n^*(x), so that F_n^* = 1.

Claim 1. Pasquali patches constructed as by Construction 1 are closed.

Proof. To make the proof clear, let us relabel the normalized version of Construction 1 as

 p(x,y) = \overbrace{\left( \mathbf{f}_{n-1}(x) - f_n^*(x) \cdot \mathbf{F}_{n-1} \right)}^{\mathbf{P}_x(x)} \cdot \overbrace{\mathbf{g}_{n-1}(y)}^{\mathbf{P}_y(y)} + f_n^*(x)

with  \int_0^1 \mathbf{P}_x(x) \, dx = 0 so as to manipulate the equation more simply, and

q(x,y) = \mathbf{Q}_x(x) \cdot \mathbf{Q}_y(y) + h_n^*(x)

with F_n^* = 1 and H_n^* = 1. Then

 \begin{array}{ccc} p(x,y) \star q(x,y) & = & j \left( \int_0^1 \left( \mathbf{P}_x(1-y) \cdot \mathbf{P}_y(t) + f_n^*(1-y) \right) \cdot \left( \mathbf{Q}_x(x) \cdot \mathbf{Q}_y(y) + h_n^*(x) \right) \, dy \right) \\ & = & \left[ \alpha \cdot \mathbf{Q}_x(x) + 0 \cdot \{\beta \cdot h_n^*(x)\} \right] \cdot \mathbf{P}_y(y) + \gamma \cdot \mathbf{Q}_x(x) \cdot \mathbf{1} + h_n^*(x) \end{array}

with \alpha = \int_0^1 \mathbf{P}_x(1-y) \cdot \mathbf{Q}_y(y) \, dy, \beta = \int_0^1 \mathbf{P}_x(1-y) \cdot \mathbf{1} \, dy = 0, and \gamma = \int_0^1 f_n^*(1-y) \cdot \mathbf{Q}_y(y) \, dy . We're not too concerned of the form of the resultant star product as much as its structure. Observe

 r(x,y) = \overbrace{ \alpha \cdot \mathbf{Q}_x(x)}^{\mathbf{R}_x^a(x)} \cdot \overbrace{\mathbf{P}_y(y)}^{\mathbf{R}_y^a(y)} + \overbrace{\gamma \cdot \mathbf{Q}_x(x)}^{\mathbf{R}^b_x(x)} \cdot \overbrace{\mathbf{1}}^{\mathbf{R}_y^b(y)} + h_n^*(x)

can be folded back into function vectors \mathbf{R}_x(x) and \mathbf{R}_y(y). Thus the structure of Construction 1 functions is preserved when we multiply one by another, showing closure. Of course the property of Construction 1 being Pasquali patches means r(x,y) is closed under that property, and so is a Pasquali patch also, as can be seen when we integrate across x:

 \int_0^1 r(x,y) \, dx = \int_0^1 0 \cdot \{ \alpha \cdot \mathbf{Q}_x(x) \cdot \mathbf{P}_y(y) \} + 0 \cdot \{\gamma \cdot \mathbf{Q}_x(x) \cdot \mathbf{1} \} + h_n^*(x) \, dx = 1

and the unit contribution is given by h_n^*(x). \qed

Claim 2. Pasquali patches constructed as by Construction 1 have powers:

 p_n(x,y) = \alpha^{n-1} \cdot \mathbf{P}_x(x) \cdot \mathbf{P}_y(y) + \sum_{i=0}^{n-2} \alpha^{i} \cdot \gamma \cdot \mathbf{P}_x(x) \cdot \mathbf{1} + f_n^*(x)

with

 \alpha = \int_0^1 \mathbf{P}_x(1-y) \cdot \mathbf{P}_y(y) \, dy = \mathbf{P}_x(x) \star \mathbf{P}_y(y) = \mathrm{str} \left[ \mathbf{P}_x(x) \cdot \mathbf{P}_y(y) \right]

and

 \gamma = \int_0^1 f_n^*(1-y) \cdot \mathbf{P}_y(y) \, dy = f_n^*(x) \star \mathbf{P}_y(y)

Proof by Induction. First, using the formula observe

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

and the second power

 p_2(x,y) = \alpha \cdot \mathbf{P}_x(x) \cdot \mathbf{P}_y(y) + \gamma \cdot \mathbf{P}_x(x) + f_n(x)

which is exactly what we expect from the definition of Construction 1 and Claim 1. Next, let us assume that the formula works and

 p_k(x,y) = \alpha^{k-1} \cdot \mathbf{P}_x(x) \cdot \mathbf{P}_y(y) + \sum_{i=0}^{k-2} \alpha^i \cdot \gamma \cdot \mathbf{P}_x(x) \cdot \mathbf{1} + f_n^*(x)

Let us examine p_{k+1}(x,y) = p_k(x,y) \star p(x,y)

 p_{k+1}(x,y) = j \left( \int_0^1 \left( \alpha^{k-1} \cdot \mathbf{P}_x(1-y) \cdot \mathbf{P}_y(t) + \sum_{i=0}^{k-2} \alpha^i \cdot \gamma \cdot \mathbf{P}_x(1-y) \cdot \mathbf{1} + f_n^*(1-y) \right) \cdot \left( \mathbf{P}_x(x) \cdot \mathbf{P}_y(y) + f_n^*(x) \right) \, dy \right)

term by term upon dotting. The first dot the first term is:

 \alpha^{k-1} \cdot \mathbf{P}_x(x) \cdot \mathbf{P}_y(y) \cdot \underbrace{\int_0^1 \mathbf{P}_x(1-y) \cdot \mathbf{P}_y(y) \, dy}_\alpha = \alpha^k \cdot \mathbf{P}_x(x) \cdot \mathbf{P}_y(y)

The first dot the last term is:

 \alpha^{k-1} \cdot \mathbf{P}_x(x) \cdot \mathbf{P}_y(y) \cdot f_n^*(x) \cdot \underbrace{\int_0^1 \mathbf{P}_x(1-y) \cdot \mathbf{1} \, dy}_\beta = 0 \cdot \{ \alpha^{k-1} \cdot \mathbf{P}_x(x) \cdot \mathbf{P}_y(y) \cdot f_n^*(x) \cdot 0 \}

The last dot the first term is:

 \mathbf{P}_x(x) \cdot \int_0^1 f_n^*(1-y) \cdot \mathbf{P}_y(y) \, dy = \gamma \cdot \mathbf{P}_x(x) \cdot \mathbf{1}

The last dot the last term is:

 f_n^*(x) \cdot \int_0^1 f_n^*(1-y) \, dy = f_n^*(x)

The middle dot the first term is:

 \sum_{i=0}^{k-2} \alpha^i \cdot \gamma \cdot \mathbf{P}_x(x) \cdot \underbrace{\int_0^1 \mathbf{P}_x(1-y) \cdot \mathbf{P}_y(y) \, dy}_\alpha = \sum_{i=0}^{k-2} \alpha^{i+1} \cdot C \cdot \mathbf{P}_x(x) \cdot \mathbf{1}

Finally, the middle dot the last term vanishes:

\sum_{i=0}^{k-2} \alpha^{i} \cdot \gamma \cdot f_n^*(x) \cdot \underbrace{\int_0^1 \mathbf{P}_x(1-y) \cdot \mathbf{1} \, dy}_\beta = 0 \cdot \{ \sum_{i=0}^{k-2} \alpha^{i} \cdot \gamma \cdot f_n^*(x) \cdot 0 \}

Putting all this information together we get:

\begin{array}{ccc} p_{k+1}(x,y) & = & \alpha^k \cdot \mathbf{P}_x(x) \cdot \mathbf{P}_y(y) + \gamma \cdot \mathbf{P}_x(x) \cdot \mathbf{1} + \sum_{i=0}^{k-2} \alpha^{i+1} \cdot \gamma \cdot \mathbf{P}_x(x) \cdot \mathbf{1} + f_n^*(x) \\ & = & \alpha^k \cdot \mathbf{P}_x(x) \cdot \mathbf{P}_y(y) + \left( \sum_{i=0}^{k-2} \alpha^{i+1} + 1 \right) \cdot \gamma \cdot \mathbf{P}_x(x) \cdot \mathbf{1} + f_n^*(x) \\ & = & \alpha^k \cdot \mathbf{P}_x(x) \cdot \mathbf{P}_y(y) + \sum_{i=0}^{k-1} \alpha^{i} \cdot \gamma \cdot \mathbf{P}_x(x) \cdot \mathbf{1} + f_n^*(x) \end{array}


\qed

Claim 3. It follows that

 p_\infty(x) = \frac{\gamma}{1-\alpha} \cdot \mathbf{P}_x(x) \cdot \mathbf{1} + f_n^*(x)

and  \lvert \alpha \rvert < 1, \gamma bounded, both conditions necessary and sufficient to establish that such a limiting surface indeed exists (convergence criterion). Furthermore, we check that this is indeed a Pasquali patch.

Proof. To reach a steady state limit,

 p_\infty(x) = \lim_{n \to \infty} p_n(x,y) = \lim_{n \to \infty} \left[ \alpha^{n-1} \cdot \mathbf{P}_x(x) \cdot \mathbf{P}_y(y) + \sum_{i=0}^{n-2} \alpha^{i} \cdot \gamma \cdot \mathbf{P}_x(x) \cdot \mathbf{1} + f_n^*(x) \right]

Next, the steady state limit must be solely functions of x, so the functions of y must vanish at the limit. Thus, it follows that  \lvert \alpha \rvert < 1. We have now established bounds on \alpha, which happen to be exactly the radius of convergence of the geometric series:

 \sum_{i=0}^\infty \alpha^i = \frac{1}{1-\alpha}

and

p_\infty(x) = 0 \cdot \{ \lim_{n \to \infty} \left[ \alpha^{n-1} \cdot \mathbf{P}_x(x) \cdot \mathbf{P}_y(y) \right] \} + \lim_{n \to \infty} \left[ \sum_{i=0}^{n-2} \alpha^{i} \cdot \gamma \cdot \mathbf{P}_x(x) \cdot \mathbf{1} \right] + f_n^*(x)

gives the desired result:

p_\infty(x) = \frac{\gamma}{1-\alpha} \cdot \mathbf{P}_x(x) \cdot \mathbf{1} + f_n^*(x)

As a check, we integrate across x to corroborate the definition of Pasquali patch:

 \int_0^1 p_\infty(x) \, dx = 0 \cdot \{ \frac{\gamma}{1-\alpha} \cdot \int_0^1 \mathbf{P}_x(x) \cdot \mathbf{1} \, dx \} + \int_0^1 f_n^*(x) \, dx = 1

Claim 4. 

 p_\infty(x) = \frac{\gamma}{1-\alpha} \cdot \mathbf{P}_x(x) \cdot \mathbf{1} + f_n^*(x)

is the eigenfunction corresponding to eigenvalue \lambda = 1 of all Construction 1 functions, through each power independently.

Proof. An eigenfunction  e(x) has the property

e(x) \star h(x,y) = \lambda e(x)

where the eigenfunction's   corresponding eigenvalue is \lambda.  The claim is more ambitious, and we will show that  p_\infty(x) \star p_n(x,y) = 1 \cdot p_\infty(x) for any n \in \mathbb{Z}^+.  The left-hand side is

 j \left( \int_0^1 \left( \frac{\gamma}{1-\alpha} \cdot \mathbf{P}_x(1-y) \cdot \mathbf{1} + f_n^*(1-y) \right) \cdot \left( \alpha^{n-1} \cdot \mathbf{P}_x(x) \cdot \mathbf{P}_y(y) + \sum_{i=0}^{n-2} \alpha^{i} \cdot \gamma \cdot \mathbf{P}_x(x) \cdot \mathbf{1} + f_n^*(x) \right) \, dy \right)

Observe the first term dotted with the middle and last term produce \beta which annihilates the results, so that the only relevant term is the first dot the first:

 \frac{\gamma}{1-\alpha} \cdot \alpha^{n-1} \cdot \mathbf{P}_x(x) \cdot \underbrace{\int_0^1 \mathbf{P}_x(1-y) \cdot \mathbf{P}_y(y) \, dy}_\alpha = \frac{\gamma}{1-\alpha} \cdot \alpha^{n} \cdot \mathbf{P}_x(x) \cdot \mathbf{1}

 The second term dot the first produces:

 \alpha^{n-1} \cdot \mathbf{P}_x(x) \cdot \underbrace{\int_0^1 f_n^*(1-y) \cdot \mathbf{P}_y(y) \, dy}_\gamma = \alpha^{n-1} \cdot \gamma \cdot \mathbf{P}_x(x) \cdot \mathbf{1}

 The second term and the second:

 \sum_{i=0}^{n-1} \alpha^i \cdot \gamma \cdot \mathbf{P}_x(x) \cdot \mathbf{1} \cdot \int_0^1 f_n^*(1-y) \, dy = \sum_{i=0}^{n-1} \alpha^i \cdot \gamma \cdot \mathbf{P}_x(x) \cdot \mathbf{1}

and the second by the last term gives

 \int_0^1 f^*_n(1-y) \cdot f^*_n(x)\, dy = f_n^*(x)

Factoring gives

 \left( \frac{\alpha^n}{1-\alpha} + \alpha^{n-1} + \sum_{i=0}^{n-2} \alpha^i \right) \cdot \gamma \cdot \mathbf{P}_x(x) \cdot \mathbf{1} + f_n^*(x) = \left( \frac{\alpha^n}{1-\alpha} + \sum_{i=0}^{n-1} \alpha^i \right) \cdot \gamma \cdot \mathbf{P}_x(x) \cdot \mathbf{1} + f_n^*(x)

The parenthetical part of this last formulation is equivalent to

\begin{array}{ccc} \frac{\alpha^n}{1-\alpha} + \frac{1-\alpha}{1-\alpha} \cdot \sum_{i=0}^{n-1} \alpha^i & = & \frac{1}{1-\alpha} \left( \alpha^n + \sum_{i=0}^{n-1} \alpha^i - \sum_{i=0}^{n-1} \alpha^{i+1} \right) \\ & = & \frac{1}{1-\alpha} \left( \sum_{i=0}^n \alpha^i - \sum_{i=1}^n \alpha^i \right) \\ & = & \frac{1}{1-\alpha} \end{array}

within the bounds already established for \alpha, and the result of the star product is

\frac{\gamma}{1-\alpha} \cdot \mathbf{P}_x(x) \cdot \mathbf{1} + f_n^*(x) = p_\infty(x)

as we wanted to show. \qed

Claim 5. 

 e(x) = A \cdot \mathbf{P}_x(x)

A is a constant, is the eigenfunction corresponding to eigenvalue \lambda = \alpha of

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

Proof. The eigenfunction equation is suggestive of what we must do to prove the claim:  e(x) \star p(x,y) = \lambda e(x). We must show that, starring the eigenfunction with p(x,y), we obtain \alpha times the eigenfunction.

Thus:

\begin{array}{ccc} e(x) \star p(x,y) & = & A \cdot \mathbf{P}_x(x) \star \left( \mathbf{P}_x(x) \cdot \mathbf{P}_y(y) + f_n^*(x) \right) \\ & = & A \cdot \mathbf{P}_x(x) \star \left( \mathbf{P}_x(x) \cdot \mathbf{P}_y(y) \right) + 0 \cdot \{ A \cdot \mathbf{P}_x(x) \star f_n^*(x) \} \\ & = & A \cdot \mathbf{P}_x(x) \cdot \underbrace{\mathbf{P}_x(x) \star \mathbf{P}_y(y)}_\alpha \\ & = & \alpha \cdot A \cdot \mathbf{P}_x(x) \\ & = & \alpha \cdot e(x)\end{array}


\qed

Part I v27

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

On Shadow Casting

February 26th, 2016 2 comments

Well, time to shake things up a bit! I have decided to change a few things on my blog, including the theme.

The latex plugin broke with the new WP auto-installation features and I haven't been writing on here much, but I've kept going on new Theorems whenever I can.  One of my latest successes, of which I am very proud of, is Shadow Casting.  It is in this new version of my Compendium of Claims and Proofs.  I will be trying to fix how the equations display in the meantime...

Part I v25