User:Sintel/Dual Weil-Euclidean norm - Revision history

Sintel: /* Relation to other metrics */ -stray subscript

2025-06-06T12:32:22Z

Relation to other metrics: -stray subscript

← Older revision		Revision as of 12:32, 6 June 2025
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	</math>		</math>

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

	:<math>		:<math>

Sintel at 16:33, 26 April 2025

2025-04-26T16:33:21Z

← Older revision		Revision as of 16:33, 26 April 2025
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	:<math>		:<math>
	~~\begin{gather}~~		\left\langle \alpha, \beta \right\rangle^{\ast} = \alpha G^{-1} \beta^{\mathsf T}
	\left\langle \alpha, \beta \right\rangle^{\ast} = \alpha G^{-1} \beta^{\mathsf T} \\		</math>

			:<math>
	\|\|\alpha\|\|^{\ast} = \sqrt{\left\langle \alpha,\alpha \right\rangle^{\ast}} = \sqrt{\alpha G^{-1} \alpha^{\mathsf T}}		\|\|\alpha\|\|^{\ast} = \sqrt{\left\langle \alpha,\alpha \right\rangle^{\ast}} = \sqrt{\alpha G^{-1} \alpha^{\mathsf T}}
	~~\end{gather}~~
	</math>		</math>

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	<ref>Miller, K. S. (1981). On the Inverse of the Sum of Matrices. Mathematics Magazine, 54(2), 67–72. https://doi.org/10.2307/2690437</ref>		<ref>Miller, K. S. (1981). On the Inverse of the Sum of Matrices. Mathematics Magazine, 54(2), 67–72. https://doi.org/10.2307/2690437</ref>

	: If <math>A</math> and <math>A+B</math> are invertible, and <math>B</math> has rank 1, then let <math>g = \text{tr}(BA^{-1})</math>. Then <math>g \neq -1</math> and		::If <math>A</math> and <math>A+B</math> are invertible, and <math>B</math> has rank 1, then let <math>g = \text{tr}(BA^{-1})</math>. Then <math>g \neq -1</math> and

	: <math>(A+B)^{−1}=A^{-1} − \frac{1}{1+g}A^{-1}BA^{-1}</math>		::<math>(A+B)^{−1}=A^{-1} − \frac{1}{1+g}A^{-1}BA^{-1}</math>

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

			:<math> l = W^{-2}j </math>

	:<math>		:<math>
	~~\begin{gather}~~
	~~l = W^{-2}j \\~~
	G^{-1} = W^{-2} - \frac{1}{1+g} l^{\mathsf T}l		G^{-1} = W^{-2} - \frac{1}{1+g} l^{\mathsf T}l
	~~\end{gather}~~
	</math>		</math>

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	:<math>		:<math>
	~~\begin{gather}~~		j^{\mathsf T}j \circ W^{-2} = I_n
	j^{\mathsf T}j \circ W^{-2} = I_n \\		</math>

			:<math>
	g = \text{tr}(BA^{-1}) = \text{tr}(j^{\mathsf T}j W^{-2}) = n		g = \text{tr}(BA^{-1}) = \text{tr}(j^{\mathsf T}j W^{-2}) = n
	~~\end{gather}~~
	</math>		</math>

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	\hline		\hline
	k\cdot j		k\cdot j
	\end{bmatrix} \\		\end{bmatrix}
			</math>

			:<math>
	G(k) = X_k^{\mathsf T} X_k = W^2 + k^2j^{\mathsf T}j		G(k) = X_k^{\mathsf T} X_k = W^2 + k^2j^{\mathsf T}j
	</math>		</math>
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	:<math>		:<math>
	\begin{~~gather~~}		\begin{aligned}
	nk^2E^2 = 1\\		nk^2E^2 &= 1\\
	E = \sqrt{\frac{1}{nk^2}}\\		E &= \sqrt{\frac{1}{nk^2}}\\
	k = \sqrt{\frac{1}{nE^2}}		k &= \sqrt{\frac{1}{nE^2}}
	\end{~~gather~~}		\end{aligned}
	</math>		</math>
	== References ==		== References ==

Sintel: more fixes

2025-04-26T16:20:15Z

more fixes

← Older revision		Revision as of 16:20, 26 April 2025
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	:<math>		:<math>
	W = \begin{bmatrix}		W =
	\log_2 2 & 0 & \cdots & 0 \\		\begin{bmatrix}
	0 & \log_2 3 & \cdots & 0\\		\log_2 2 & 0 & \cdots & 0 \\
	\vdots & \vdots & \ddots & \vdots \\		0 & \log_2 3 & \cdots & 0\\
	0 & 0 & \cdots & \log_2 p		\vdots & \vdots & \ddots & \vdots \\
	\end{bmatrix}		0 & 0 & \cdots & \log_2 p
			\end{bmatrix}
	</math>		</math>

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	:<math>		:<math>
	X = \begin{bmatrix}		X =
	W \\		\begin{bmatrix}
	\hline		W \\
	j		\hline
	\end{bmatrix}		j
			\end{bmatrix}
	</math>		</math>

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	:<math>		:<math>
	\left\langle \alpha, \beta \right\rangle^{\ast} = \alpha G^{-1} \beta^{\mathsf T} \\		\begin{gather}
	\|\|\alpha\|\|^{\ast} = \sqrt{\left\langle \alpha,\alpha \right\rangle^{\ast}} = \sqrt{\alpha G^{-1} \alpha^{\mathsf T}}		\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}}
			\end{gather}
	</math>		</math>

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	:<math>		:<math>
	G = X^{\mathsf T} X = W^2 + j^{\mathsf T}j		G = X^{\mathsf T} X = W^2 + j^{\mathsf T}j
	</math>		</math>

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	:<math>		:<math>
	G^{-1} = (W^2 + j^{\mathsf T}j)^{-1} = W^{-2} - \frac{1}{1+g} W^{-2}j^{\mathsf T}jW^{-2}		G^{-1} = (W^2 + j^{\mathsf T}j)^{-1} = W^{-2} - \frac{1}{1+g} W^{-2}j^{\mathsf T}jW^{-2}
	</math>		</math>

	Now let <math>l = \begin{bmatrix}		Now let <math>
	\frac{1}{\log_2 2} & \frac{1}{\log_2 3} & \cdots & \frac{1}{\log_2 p} \\		l = \begin{bmatrix}
	\end{bmatrix} </math>, then		\frac{1}{\log_2 2} & \frac{1}{\log_2 3} & \cdots & \frac{1}{\log_2 p} \\
			\end{bmatrix}
			</math>, then

	:<math>		:<math>
	l = W^{-2}j \\		\begin{gather}
	G^{-1} = W^{-2} - \frac{1}{1+g} l^{\mathsf T}l		l = W^{-2}j \\
			G^{-1} = W^{-2} - \frac{1}{1+g} l^{\mathsf T}l
			\end{gather}
	</math>		</math>

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	:<math>		:<math>
	j^{\mathsf T}j \circ W^{-2} = I_n \\		\begin{gather}
	g = \text{tr}(BA^{-1}) = \text{tr}(j^{\mathsf T}j W^{-2}) = n		j^{\mathsf T}j \circ W^{-2} = I_n \\
			g = \text{tr}(BA^{-1}) = \text{tr}(j^{\mathsf T}j W^{-2}) = n
			\end{gather}
	</math>		</math>

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	:<math>		:<math>
	G^{-1} = W^{-2} - \frac{1}{n+1} l^{\mathsf T}l		G^{-1} = W^{-2} - \frac{1}{n+1} l^{\mathsf T}l
	</math>		</math>

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	:<math>		:<math>
	\begin{aligned}		\begin{aligned}
	G_a(\lambda) &= W^{-2} - \lambda \frac{W^{-2}j^{\mathsf T}jW^{-2}}{jW^{-2}j^{\mathsf T}} \\		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}		&= W^{-2} - \lambda \frac{l^{\mathsf T}l}{n}
	\end{aligned}		\end{aligned}
	</math>		</math>
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	:<math>		:<math>
	\begin{aligned}		\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} \\		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}		&= \frac{W^{-2}}{n} (1+E^2) - \frac{l^{\mathsf T}l}{n^2}
	\end{aligned}		\end{aligned}
	</math>		</math>
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	:<math>		:<math>
	G^{\prime}_b(E) = W^{-2} - \frac{1}{n(1+E^2)}l^{\mathsf T}l		G^{\prime}_b(E) = W^{-2} - \frac{1}{n(1+E^2)}l^{\mathsf T}l
	</math>		</math>

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	:<math>		:<math>
	X_k = \begin{bmatrix}		X_k =
	W \\		\begin{bmatrix}
	\hline		W \\
	k\cdot j		\hline
	\end{bmatrix}\\		k\cdot j
	G(k) = X_k^{\mathsf T} X_k = W^2 + k^2j^{\mathsf T}j		\end{bmatrix} \\
			G(k) = X_k^{\mathsf T} X_k = W^2 + k^2j^{\mathsf T}j
	</math>		</math>

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	:<math>		:<math>
	G^{-1}(k) = W^{-2} - \frac{k^2}{nk^2+1} l^{\mathsf T}l		G^{-1}(k) = W^{-2} - \frac{k^2}{nk^2+1} l^{\mathsf T}l
	</math>		</math>

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	:<math>		:<math>
	\begin{gather}		\begin{gather}
	nk^2E^2 = 1\\		nk^2E^2 = 1\\
	E = \sqrt{\frac{1}{nk^2}}\\		E = \sqrt{\frac{1}{nk^2}}\\
	k = \sqrt{\frac{1}{nE^2}}		k = \sqrt{\frac{1}{nE^2}}
	\end{gather}		\end{gather}
	</math>		</math>
	== References ==		== References ==

Sintel: math formatting

2025-04-26T16:15:05Z

math formatting

Show changes

Sintel: simplify fraction

2022-04-17T01:45:20Z

simplify fraction

← Older revision		Revision as of 01:45, 17 April 2022
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	$$		$$

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

	== Generalized norm ==		== Generalized norm ==

Sintel: all done now I think

2022-04-17T01:42:31Z

all done now I think

← Older revision		Revision as of 01:42, 17 April 2022
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	== Generalized norm ==		== Generalized norm ==
	For some parameter <math>k</math>, set:		For some parameter <math>k > 0</math>, set:

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

	~~And the~~ relation to ~~parametric badness is~~ <math>E = \sqrt{\frac{n+1}{nk^2} ~~- 1~~}~~</math> or <math>~~k = \sqrt{\frac{n+1}{~~n}(1+E~~^2)}~~</math>~~		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 ==		== References ==

Sintel at 01:23, 17 April 2022

2022-04-17T01:23:25Z

← Older revision		Revision as of 01:23, 17 April 2022
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	$$		$$
	\begin{aligned}		\begin{aligned}
	G_b(~~E_k~~) &= \frac{W^{-2}}{jW^{-2}j^{\mathsf T}} (1+E^~~2_k~~) - \frac{W^{-2}j^{\mathsf T}jW^{-2}}{(jW^{-2}j^{\mathsf T})^2} \\		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_k~~) - \frac{l^{\mathsf T}l}{n^2}		&= \frac{W^{-2}}{n} (1+E^2) - \frac{l^{\mathsf T}l}{n^2}
	\end{aligned}		\end{aligned}
	$$		$$
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	$$		$$
	G^{\prime}_b(~~E_k~~) = W^{-2} - \frac{1}{n(1+E^~~2_k~~)}l^{\mathsf T}l		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_k~~ = \sqrt{\frac{n+1}{n} - 1}</math>		Again, this is equivalent to <math>G^{-1}</math>, when we pick <math>E = \sqrt{\frac{n+1}{n} - 1}</math>
	==== ~~References~~ ====
			== Generalized norm ==
			For some parameter <math>k</math>, set:

			$$
			X_k = \begin{bmatrix}
			W \\
			\hline
			k\cdot j
			\end{bmatrix}\\
			G(k) = X_k^{\mathsf T} X_k = W^2 + k^2j^{\mathsf T}j
			$$

			Going through the same derivation, we find:

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

			And the relation to parametric badness is <math>E = \sqrt{\frac{n+1}{nk^2} - 1}</math> or <math>k = \sqrt{\frac{n+1}{n}(1+E^2)}</math>
			== References ==

Sintel: parametric badness

2022-04-17T00:30:03Z

parametric badness

← Older revision		Revision as of 00:30, 17 April 2022
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	$$		$$
	G^{-1} = W^{-2} - \frac{1}{1+n} l^{\mathsf T}l		G^{-1} = W^{-2} - \frac{1}{n+1} l^{\mathsf T}l
	$$		$$

	== ~~Cross-Weighted Metric~~ ==		== 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>		[[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>
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	$$		$$
	\begin{aligned}		\begin{aligned}
	G'(\lambda) &= W^{-2} - \lambda \frac{W^{-2}j^{\mathsf T}jW^{-2}}{jW^{-2}j^{\mathsf T}} \\		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}		&= W^{-2} - \lambda \frac{l^{\mathsf T}l}{n}
	\end{aligned}		\end{aligned}
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	So this is equivalent to <math>G^{-1}</math> when we pick <math>\lambda = \frac{n}{n+1}</math>.		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>

			$$
			\begin{aligned}
			G_b(E_k) &= \frac{W^{-2}}{jW^{-2}j^{\mathsf T}} (1+E^2_k) - \frac{W^{-2}j^{\mathsf T}jW^{-2}}{(jW^{-2}j^{\mathsf T})^2} \\
			&= \frac{W^{-2}}{n} (1+E^2_k) - \frac{l^{\mathsf T}l}{n^2}
			\end{aligned}
			$$

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

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

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

Sintel at 23:49, 16 April 2022

2022-04-16T23:49:15Z

@@ Line 1: / Line 1: @@
 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>:
@@ Line 84: / Line 86: @@
 $$
-<hr>
+== Cross-Weighted Metric ==

Sintel at 22:46, 16 April 2022

2022-04-16T22:46:06Z

← Older revision		Revision as of 22:46, 16 April 2022
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	The inner product on the dual space can then be derived by simply inverting <math>G</math>		The inner product on the dual space can then be derived by simply inverting <math>G</math>
	<ref>Taking the map <math>\Gamma: V\to V^{\ast}: x \mapsto \left\langle \cdot ,x \right\rangle</math>, we require <math>\left\langle \Gamma(x),\Gamma(y) \right\rangle^{\ast} = \left\langle x,y \right\rangle</math>.</ref>		<ref>Taking the natural map to the dual space <math>\Gamma: V\to V^{\ast}: x \mapsto \left\langle x, \cdot \right\rangle</math>, we require <math>\left\langle \Gamma(x),\Gamma(y) \right\rangle^{\ast} = \left\langle x,y \right\rangle</math>.</ref>
	, which gives the dual norm:		, which gives the dual norm: