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Archive for January 13th, 2014

## More On Collections of Powers (via Starring) of Functions

These proofs really belong to the Compendium, but I have no access to the tex file at the moment, so I am using this post to jot down ideas.

If we accept powering of functions via the defintions of the star product, then the following will make sense. Â On the other hand if we view this from the standpoint of simple matrices (and replace the star product by matrix multiplication), the following claims may also make sense.

Definition 1.Â  For a countably infinite collection $\mathbb{F} = \{F^1, F^2, \ldots, F^j, \ldots \}_{j \in \mathbb{Z}^+}$, the "generator" isÂ $F^1$ since all other elements in the collection are powers of such. Â (As an aside, we may contemplate finite collections in the event of powering periodicity or nilpotency, but not presently).

Claim 1. Take the countably infinite collectionsÂ $\mathbb{F}$ andÂ $\mathbb{G}$ with generators $F^1$ and $G^1$ respectively. Â If $F^1 = G^1$, then $\mathbb{F} = \mathbb{G}$.

Proof. Â By hypothesis we have thatÂ $F^1 = G^1$, and let us assume thatÂ $F^k = G^k$. Then:

Â Thus we have shown that Â $\mathbb{F} \subset \mathbb{G}$ and $\mathbb{G} \subset \mathbb{F}$ and so naturally we concludeÂ $\mathbb{F} = \mathbb{G}$.

Claim 2.Â Take the countably infinite collectionsÂ $\mathbb{F}$ andÂ $\mathbb{G}$ with generators $F^1$ and $G^1$ respectively. Â If $F^1 = G^n$ for some $n \in \mathbb{Z}^+$, then $\mathbb{F} \subset \mathbb{G}$.

Proof.Â By hypothesisÂ $F^1 = G^n$ for some positive integer $n$. Â Also,

using Claim 1.10 of the Compendium. Â Next suppose thatÂ $F^k = G^{kn}$. Â Then:Â

Â Thus $\mathbb{F} \subset \mathbb{G}$ since all elements of the collection $\mathbb{F}$ are contained in $\mathbb{G}$.

Definition 2.Â Â For a countably infinite collectionÂ $\mathbb{F} = \{F^1, F^2, \ldots, F^j, \ldots \}_{j \in \mathbb{Z}^+}$, we call the subset $\mathbb{F}_{n} \subset \mathbb{F}$ the "subsequent power subset at $n$", and it is composed of all those powers greater than or equal to $n \in \mathbb{Z}^+$.

Claim 3. Take the countably infinite collectionsÂ $\mathbb{F}$ andÂ $\mathbb{G}$ with generators $F^1$ and $G^1$ respectively. Â If $F^m = G^n$ for some $m, n \in \mathbb{Z}^+$ and $m (since we haven't defined "roots" it must also be true that $\frac{n}{m} \in \mathbb{Z}^+$), then $\mathbb{F}_m \subset \mathbb{G}_n \subset \mathbb{G}$. (At this point we don't really care about containment of powers less than $n$, since we haven't established a mechanism to obtain them by "unpowering" or "taking roots").

Proof. Â By hypothesisÂ $F^m = G^n$. Â In particular, we have the statement $F^{m+1} = G^{n + p}$ for some integer $p \in \mathbb{Z}^+$. Â (Advancing one element in the collection $\mathbb{F}$ starting from $m$, advances us $p$ units in the collection $\mathbb{G}$ starting from $n$.) Â Taking this statement to the logical consequence, $F^m \star F^1 = G^n \star G^p \Rightarrow F^1 = G^p$. Â (Notice that since we have not defined negative powers of functions, $p$ is necessarily a positive integer).

Next let us assume thatÂ $F^{m+k} = G^{n + kp}$. Â Then:Â

It can be seen now thatÂ Â $\mathbb{F}_m \subset \mathbb{G}_n \subset \mathbb{G}$.

Corollary 1. Â Take the countably infinite collectionsÂ $\mathbb{F}$ andÂ $\mathbb{G}$.Â Â $\mathbb{F}_m = \mathbb{G}_n \subset \mathbb{G}$ if and only if $m = n$ and $p = 1$.

Proof.

$\Rightarrow$ SinceÂ $\mathbb{F}_m = \mathbb{G}_n$ there is no element of either collection that does not match one in the other collection. Â Thus the step between elements must be 1, and $p = 1$. Â Furthermore the elements in a collection can be put in bijective correspondence (by definition of equality of sets) with the elements in the other collection, and the indices must exactly coincide by the ordering being the usual order on the positive integers (we order by power). Â Thus $m=n$.

$\Leftarrow$ Since $m = n$ we have that $F^m = G^n \Rightarrow F^m = G^m$ for all $m \in \mathbb{Z}^+$, and also $F^{m+1} = G^{n + p} \Rightarrow F^{m+1} = G^{m + 1}$Â since $p = 1$. Â So now assume that this works for the $k$th element $k>m+1$, so that $F^{m+k} = G^{m+k} \Rightarrow F^k = G^k$. Â Then we have that:

and we are done.

The following claims and proofs are applicable to Pasquali patches as they are to Â Markov matrices, where we shift from a function point-of-view to a matricial point-of-view using in one the star product and in the other normal matrix multiplication.

Claim 4. Take the countably infinite collections of Pasquali patch self-powers $\mathbb{P} = \{ P^1, P^2, \ldots, P^j, \ldots \}_{j \in \mathbb{Z}^+}$ andÂ $\mathbb{Q} = \{ Q^1, Q^2, \ldots, Q^k, \ldots \}_{k \in \mathbb{Z}^+}$ with stationary limiting $P^\infty$ and $Q^\infty$ respectively. If $\mathbb{P}$ and $\mathbb{Q}$ coincide at ($m, n$) so that $P^m = Q^n$ ($\frac{n}{m} \in \mathbb{Z}^+$) then $P^\infty = Q^\infty$.

Proof.

The infinite collections coincide at $(m,n)$ for $m, n, \frac{n}{m} \in\mathbb{Z}^+$, thus we have that $P^m = Q^n$. Â By Claim 3 we have thatÂ $\mathbb{P}_m \subset \mathbb{Q}_n$. Â Now the limit $\lim_{n \to \infty} Q_n = Q^\infty$ by hypothesis of the existence of the stationary patch, and sinceÂ $\mathbb{P}_m$ is a subset of $\mathbb{Q}_n$, it follows that$\lim_{m \to \infty} P_m = Q^\infty$. Â We also know thatÂ $\lim_{m \to \infty} P_m = P^\infty$ however (again by hypothesis of the existence of the stationary patch), so it must therefore be true thatÂ Â $P^\infty = Q^\infty$.

The reverse claim is not (always) true:

False Claim 1. Â IfÂ $P^\infty = Q^\infty$, thenÂ $\mathbb{P}$ and $\mathbb{Q}$ coincide at ($m, n$) so that $P^m = Q^n$ ($\frac{n}{m} \in \mathbb{Z}^+$).

Disproof. Â Take the collection of Pasquali patches $\mathbb{P} = \{P^1 = 1, P^2 = 1, \ldots, P^j = 1, \ldots\}_{j \in \mathbb{Z}^+}$ with $P^\infty = 1$, and any collection of Pasquali patches $\mathbb{Q} = \{Q^1, Q^2, \ldots, Q^k, \ldots\}_{k \in \mathbb{Z}^+}$ so that the generator $Q^1 = q(x,y)$ is an explicit function of $y$ and $Q^\infty = 1$, as in an example in the Compendium. Â By a Claim in the Compendium, all elements in the collectionÂ $\mathbb{Q}$ are explicit functions of $y$ (even as $Q^\infty$ is not). Â Clearly, no $P \in \mathbb{P}$ is equal to an element $Q \in \mathbb{Q}$ (all $P$ are constant where all $Q$ vary with $y$), yet they have the same stationary patch.