Consistency: Difference between revisions

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'''Theorem:''' A (1/(2''d'') steps of ''n'' ED''k'') threshold can be interpreted as allowing stacking ''d'' copies of a chord C, including the original chord, via dyads that occur in the chord, so that the resulting chord will always satisfy condition #2 of chord consistency.

Proof: Consider a dyad D = {''x'', ''y''} on two notes ''x'' and ''y'' that occur in the resulting chord C' = C1 ∪ C2 ∪ … ∪ C''d'' in the equal temperament, where the Ci are copies of the (approximations of) chord C. We may assume that the notes ''x'' and ''y'' belong in two different copies of C, C''i'' and C''i'' + ''m'', where 1 ≤ ''i'' ≤ ''i'' + ''m'' ≤ ''d''. Suppose C''i'', C''i'' + 1, …, C''i'' + ''m'' are separated by dyads D1, D2, …, D''m'' that occur in C. Let ''x' '' be the interval ''x'' + D1 + D2 + … + D''m'', the counterpart of ''x'' in C''i'' + ''m''. Since ''m'' ≤ ''d'' - 1, by consistency to span ''d'', each dyad D''j'' have relative error 1/(2''d'') since D''i'' occurs in C. The relative error on D1 + D2 + … + D''m'' compared to its just counterpart is < (''d'' - 1)/(2''d''), and again by assumption of consistency to span ''d'', the dyad ''y'' - ''x' '' has error 1/(2''d''). Hence the total relative error on D is strictly less than 1/2. Since D is arbitrary, we have proved condition #2 of chord consistency. QED.

Proof: Consider a dyad D = {''x'', ''y''} consisting of two notes ''x'' and ''y'' that occur in the resulting chord C' = C1 ∪ C2 ∪ … ∪ C''d'' in the equal temperament, where the Ci are copies of the (approximations of) chord C. We may assume that the notes ''x'' and ''y'' belong in two different copies of C, C''i'' and C''i'' + ''m'', where 1 ≤ ''i'' ≤ ''i'' + ''m'' ≤ ''d''. Suppose C''i'', C''i'' + 1, …, C''i'' + ''m'' are separated by dyads D1, D2, …, D''m'' that occur in C. Let ''x' '' be the interval ''x'' + D1 + D2 + … + D''m'', the counterpart of ''x'' in C''i'' + ''m''. Since ''m'' ≤ ''d'' - 1, by consistency to span ''d'', each dyad D''j'' have relative error 1/(2''d'') since D''i'' occurs in C. The relative error on D1 + D2 + … + D''m'' compared to its just counterpart is < (''d'' - 1)/(2''d''), and again by assumption of consistency to span ''d'', the dyad ''y'' - ''x' '' has error 1/(2''d''). Hence the total relative error on D is strictly less than 1/2. Since D is arbitrary, we have proved condition #2 of chord consistency. QED.

Question: Will the resulting chord always satisfy condition #1 as well?

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 '''Theorem:''' A (1/(2''d'') steps of ''n'' ED''k'') threshold can be interpreted as allowing stacking ''d'' copies of a chord C, including the original chord, via dyads that occur in the chord, so that the resulting chord will always satisfy condition #2 of chord consistency.
-Proof: Consider a dyad D = {''x'', ''y''} on two notes ''x'' and ''y'' that occur in the resulting chord C' = C<sub>1</sub> ∪ C<sub>2</sub> ∪ … ∪ C<sub>''d''</sub> in the equal temperament, where the C<sub>i</sub> are copies of the (approximations of) chord C. We may assume that the notes ''x'' and ''y'' belong in two different copies of C, C<sub>''i''</sub> and C<sub>''i'' + ''m''</sub>, where 1 ≤ ''i'' ≤ ''i'' + ''m'' ≤ ''d''. Suppose C<sub>''i''</sub>, C<sub>''i'' + 1</sub>, …, C<sub>''i'' + ''m''</sub> are separated by dyads D<sub>1</sub>, D<sub>2</sub>, …, D<sub>''m''</sub> that occur in C. Let ''x' '' be the interval ''x'' + D<sub>1</sub> + D<sub>2</sub> + … + D<sub>''m''</sub>, the counterpart of ''x'' in C<sub>''i'' + ''m''</sub>. Since ''m'' ≤ ''d'' - 1, by consistency to span ''d'', each dyad D<sub>''j''</sub> have relative error 1/(2''d'') since D<sub>''i''</sub> occurs in C. The relative error on D<sub>1</sub> + D<sub>2</sub> + … + D<sub>''m''</sub> compared to its just counterpart is < (''d'' - 1)/(2''d''), and again by assumption of consistency to span ''d'', the dyad ''y'' - ''x' '' has error 1/(2''d''). Hence the total relative error on D is strictly less than 1/2. Since D is arbitrary, we have proved condition #2 of chord consistency. QED.
+Proof: Consider a dyad D = {''x'', ''y''} consisting of two notes ''x'' and ''y'' that occur in the resulting chord C' = C<sub>1</sub> ∪ C<sub>2</sub> ∪ … ∪ C<sub>''d''</sub> in the equal temperament, where the C<sub>i</sub> are copies of the (approximations of) chord C. We may assume that the notes ''x'' and ''y'' belong in two different copies of C, C<sub>''i''</sub> and C<sub>''i'' + ''m''</sub>, where 1 ≤ ''i'' ≤ ''i'' + ''m'' ≤ ''d''. Suppose C<sub>''i''</sub>, C<sub>''i'' + 1</sub>, …, C<sub>''i'' + ''m''</sub> are separated by dyads D<sub>1</sub>, D<sub>2</sub>, …, D<sub>''m''</sub> that occur in C. Let ''x' '' be the interval ''x'' + D<sub>1</sub> + D<sub>2</sub> + … + D<sub>''m''</sub>, the counterpart of ''x'' in C<sub>''i'' + ''m''</sub>. Since ''m'' ≤ ''d'' - 1, by consistency to span ''d'', each dyad D<sub>''j''</sub> have relative error 1/(2''d'') since D<sub>''i''</sub> occurs in C. The relative error on D<sub>1</sub> + D<sub>2</sub> + … + D<sub>''m''</sub> compared to its just counterpart is < (''d'' - 1)/(2''d''), and again by assumption of consistency to span ''d'', the dyad ''y'' - ''x' '' has error 1/(2''d''). Hence the total relative error on D is strictly less than 1/2. Since D is arbitrary, we have proved condition #2 of chord consistency. QED.
 Question: Will the resulting chord always satisfy condition #1 as well?