User:IssaRice/Logical induction notation: Difference between revisions

Revision as of 02:15, 3 August 2018

This is in user space because it's not really about machine learning.

Term	Notation	Type	Definition	Notes
${\mathcal {F}}$ -combination	$A$	${\mathcal {S}}\cup \{0,1\}\to {\mathcal {F}}_{n}$		Function application of an ${\mathcal {F}}$ -combination uses square brackets instead of parentheses. Why? As far as I can tell, this is because each coefficient is in ${\mathcal {F}}$ so is itself a function. This means we have two senses of "application": we can pick out the specific coefficient we want (square brackets), or we can apply each coefficient to return something (parentheses).
Holdings from $T$ against ${\overline {\mathbb {P} }}$ (a $\mathbb {Q}$ -combination)	$T({\overline {\mathbb {P} }})$	${\mathcal {S}}\cup \{0,1\}\to \mathbb {Q}$

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 ! Term !! Notation !! Type !! Definition !! Notes
 |-
-| <math>\mathcal F</math>-combination || <math>A</math> || <math>\mathcal S \cup \{0,1\} \to \mathcal F_n</math> || || Function application of an <math>\mathcal F</math>-combination uses square brackets instead of parentheses.
+| <math>\mathcal F</math>-combination || <math>A</math> || <math>\mathcal S \cup \{0,1\} \to \mathcal F_n</math> || || Function application of an <math>\mathcal F</math>-combination uses square brackets instead of parentheses. Why? As far as I can tell, this is because each coefficient is in <math>\mathcal F</math> so is itself a function. This means we have two senses of "application": we can pick out the specific coefficient we want (square brackets), or we can apply each coefficient to return something (parentheses).
 |-
 | Holdings from <math>T</math> against <math>\overline{\mathbb P}</math> (a <math>\mathbb Q</math>-combination)|| <math>T(\overline{\mathbb P})</math> || <math>\mathcal S \cup \{0,1\} \to \mathbb Q</math> || ||