User:Sintel/Dual Weil-Euclidean norm: Difference between revisions

On some <math>p</math>-limit subgroup with <math>n</math> primes, define the <math>n \times n</math> Tenney weighting matrix <math>W</math>:

W = \begin{bmatrix}

\log_2 2 & 0 & \cdots & 0 \\

0 & 0 & \cdots & \log_2 p

\end{bmatrix}

And the row vector <math>j</math> containing the log-primes (aka the [[JIP]]): <math>j = \begin{bmatrix}

Then the block matrix <math>X</math> obtained from these:

X = \begin{bmatrix}

W \\

j

\end{bmatrix}

defines an inner product, with positive definite <math>G =  X^{\mathsf T} X </math>:

\left\langle x,y \right\rangle = x^{\mathsf T} X^{\mathsf T} X y = x^{\mathsf T} G y

and an induced norm <math>||x|| = \sqrt{\left\langle x,x \right\rangle}</math>, which is the [[Weil_Norms,_Tenney-Weil_Norms,_and_TWp_Interval_and_Tuning_Space#Weil-Euclidean_Norm|Weil-Euclidean norm]].

, which gives the dual norm:

\left\langle \alpha, \beta \right\rangle^{\ast} = \alpha G^{-1} \beta^{\mathsf T} \\

||\alpha||^{\ast} = \sqrt{\left\langle \alpha,\alpha \right\rangle^{\ast}} = \sqrt{\alpha G^{-1} \alpha^{\mathsf T}}

The goal is now to find an expression for <math>G^{-1}</math>.

First, note that:

G =  X^{\mathsf T} X = W^2 + j^{\mathsf T}j

Since the outer product <math>j^{\mathsf T}j</math> is rank-1 we can use a theorem on the inverse of matrix sums which states:

Now identifying <math>A = W^2</math> and <math>B = j^{\mathsf T}j</math>. We can see that

G^{-1} = (W^2 + j^{\mathsf T}j)^{-1} = W^{-2} - \frac{1}{1+g} W^{-2}j^{\mathsf T}jW^{-2}

Now let <math>l = \begin{bmatrix}

\end{bmatrix} </math>, then

l = W^{-2}j \\

G^{-1} = W^{-2} - \frac{1}{1+g} l^{\mathsf T}l

Now we only need to find <math>g</math>. The trace of a matrix product is equal to the sum of the elements of their Hadamard (elementwise) product. Since

j^{\mathsf T}j \circ W^{-2} = I_n \\

g = \text{tr}(BA^{-1}) = \text{tr}(j^{\mathsf T}j W^{-2}) = n

Which leads to the final expression:

G^{-1} = W^{-2} - \frac{1}{n+1} l^{\mathsf T}l

== Relation to other metrics ==

[[Graham Breed]] gives the following formula (adapted for the notation introduced here):<ref>See formula (16) in section 3.1 "Cross-Weighted Metrics" In Breed, G. (2008). RMS-Based Error and Complexity Measures Involving Composite Intervals http://x31eq.com/composite.pdf</ref>

\begin{aligned}

G_a(\lambda) &= W^{-2} - \lambda \frac{W^{-2}j^{\mathsf T}jW^{-2}}{jW^{-2}j^{\mathsf T}} \\

&= W^{-2} - \lambda \frac{l^{\mathsf T}l}{n}

\end{aligned}

So this is equivalent to <math>G^{-1}</math> when we pick <math>\lambda = \frac{n}{n+1}</math>.

His [[Cangwu badness|parametric badness]] is given:<ref>Breed, G. (2016). http://x31eq.com/badness.pdf</ref>

\begin{aligned}

G_b(E) &= \frac{W^{-2}}{jW^{-2}j^{\mathsf T}} (1+E^2) -  \frac{W^{-2}j^{\mathsf T}jW^{-2}}{(jW^{-2}j^{\mathsf T})^2} \\

&= \frac{W^{-2}}{n} (1+E^2) -  \frac{l^{\mathsf T}l}{n^2}

\end{aligned}

Since the metric is equivalent up to scaling, we multiply by <math>\frac{n}{1+E^2_k}</math> to obtain:

G^{\prime}_b(E) = W^{-2} - \frac{1}{n(1+E^2)}l^{\mathsf T}l

Again, this is equivalent to <math>G^{-1}</math>, when we pick <math>E = \sqrt{\frac{1}{n}}</math>

For some parameter <math>k > 0</math>, set:

X_k = \begin{bmatrix}

W \\

\end{bmatrix}\\

G(k) =  X_k^{\mathsf T} X_k = W^2 + k^2j^{\mathsf T}j

Going through the same derivation, we find:

G^{-1}(k) = W^{-2} - \frac{k^2}{nk^2+1} l^{\mathsf T}l

Which leads to a simple relation to <math>E</math>:

nk^2E^2 = 1\\

E = \sqrt{\frac{1}{nk^2}}\\

k = \sqrt{\frac{1}{nE^2}}

== References ==