## On Eigen(patch(ix))values, II - (RWLA,MCT,GT,AM Part IX)

So remember my little conjecture from last time, that the number of patch(ix) (kernel) eigenvalues would depend on the number of x terms that composed it? Â I started working it out by writing all expressions and trying to substitute them and I got sums of sums of sums and it became nightmarish, and since math is supposed to be elegant, I opted for a different track. Â A little **proposition** did result, but I'm not sure yet if it means what I want it to mean. Haha.

If you recall, last time we figured that

and

Let's rename the sums by indexing over the subscripts, so that

Renaming therefore the constants we get:

and

Last time we substituted one equation into the other to figure additional restrictions on . Â A faster way to do this is to Â notice:

and

If we multiply these two expressions we get

Finally, dividing out both we arrive at the quadratic expression on of before:

Now. Â Let's posit that, instead of we have . Â Then by all the same arguments we should have an expression of that is the same, and an expression of that is:

with the similar implication that

An would give the implication

If we are to multiply all similar expressions, we get

or

In other words, we want to make a pairwise argument to obtain the product of the -expressions and a polynomial in . Â Next I'd like to show the **proposition**:

and for this I want to begin with a combinatorial argument. Â On the left hand side, the number of pairwise comparisons we can make depends on the number of factors of the polynomial (or, the highest degree of the polynomial). Â That is to say, we can make pairwise comparisons, or comparisons. Â Now, I don't know whether anyone has ever noticed this, but this last simplified part looks exceptionally like Gauss's sum of consecutive integers (pyramidal series), so in other words, this last part is in effect which I find very cool, because we have just shown, quite accidentally, the equivalence:

The way I actually figured this out is by noticing that, in our pairwise comparisons, say for the 3rd-degree-polynomial-in- case, by writing the pairwise comparisons first of the products, then of the (in other words, ordering logically all products), there were 2 of the first and 1 of the second (and none of the ). Â If we do the same for the 4th-degree, there are 3 of the , 2 of theÂ , and 1 of theÂ , with none of theÂ . Â In other words, the pair-products could be written as the sum of the cardinality of the groupings: .

Now Gauss's sum of integers formula is already known to work in the general case (just use an inductive proof, e.g.), so the substitution of it into the binomial equivalence needs no further elaboration: it generalizes automatically for all .

So if we are to multiply all pairwise comparisons, notice there will be products of each -factor: there are products belonging to theÂ grouping (because this first grouping has entries, from the Gauss formula equivalence), there are products belonging to theÂ PLUS the one already counted in the grouping, for a total of, again, . Â The th grouping has products listed for itself PLUS one for each of the previous groupings, for a total of , and the th grouping has products listed for itself PLUS one for each of the previous groupings, for a total of . Â We are left in effect with:

The right hand side of each pairwise comparison was nothing more than the simple product on the cross indexes of , so it's not difficult to argue then that, if we multiply such pairs, we get . Â Â We then take the th root on both sides of the equation.

Since the case follows the same basic structure of the argument, we are done with proving our proposition.

What I want this proposition to mean may be very different than what it actually is, I'm hopeful nevertheless but I agree that it requires a bit of further investigation. Â As I hinted before, I would like that

with, for example, represent the constraint on the eigen(patch(ix))values of or, if not that, maybe , which brings into question the superposition of functions and their effect on the eigenvalues. Â I may be wildly speculating, but hey! Â I don't really know better! Â I'll do a few experiments and see what shows.