The wedgie: Difference between revisions
Wikispaces>genewardsmith **Imported revision 289936115 - Original comment: ** |
Wikispaces>genewardsmith **Imported revision 289970701 - Original comment: ** |
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<h2>IMPORTED REVISION FROM WIKISPACES</h2> | <h2>IMPORTED REVISION FROM WIKISPACES</h2> | ||
This is an imported revision from Wikispaces. The revision metadata is included below for reference:<br> | This is an imported revision from Wikispaces. The revision metadata is included below for reference:<br> | ||
: This revision was by author [[User:genewardsmith|genewardsmith]] and made on <tt>2012-01- | : This revision was by author [[User:genewardsmith|genewardsmith]] and made on <tt>2012-01-06 03:21:38 UTC</tt>.<br> | ||
: The original revision id was <tt> | : The original revision id was <tt>289970701</tt>.<br> | ||
: The revision comment was: <tt></tt><br> | : The revision comment was: <tt></tt><br> | ||
The revision contents are below, presented both in the original Wikispaces Wikitext format, and in HTML exactly as Wikispaces rendered it.<br> | The revision contents are below, presented both in the original Wikispaces Wikitext format, and in HTML exactly as Wikispaces rendered it.<br> | ||
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[[math]] | [[math]] | ||
For this we can conclude that d, e and f satisfy |d - q3b + q5a| ≤ | For this we can conclude that d, e and f satisfy |d - q3b + q5a| ≤ 2E√q3√q5, |e - q3c + q7a| ≤ 2E√q3√q7 and |f - q5c + q7d| ≤ 2E√q5q7. This has an interesting interpretation: since <1 q3 q5 q7|∧<0 a b c| = <<a b c q3b-q5a q3c-q7a q5c-q7b||, if E ≤ 1/(4√q5√q7), then the full wedgie can be recovered from the octave equivalent (OE) portion of the wedgie simply by wedging it with <1 q3 q5 q7| and rounding to the nearest integer. This is not a very serious constraint to place on relative error; it seems unlikely anyone would be interested in a temperament which did not fall well under this low standard. Hence we may compile lists of reasonable temperaments by presuming "reasonable" requires this bound to be met, searching through triples <<a b c ...|| (note that if all of these are zero, 2 is being tempered out) up to some complexity bound, wedging with <1 q3 q5 q7| and rounding, then checking if the GCD is one and the Pfaffian af-be+cd is zero. Then we may toss everthing which does not meet the bound on relative error; however, for a reasonable list we will want a tighter bound. | ||
If C = ||W|| is the TE complexity, then the formula for the [[Tenney-Euclidean metrics#Logflat TE badness|logflat badness]] B in the 7-limit rank-two case is particularly simple: B = CE. If complexity is bounded by, for example, 20 (which allows for some quite complex temperaments) then since E ≤ 1/( | If C = ||W|| is the TE complexity, then the formula for the [[Tenney-Euclidean metrics#Logflat TE badness|logflat badness]] B in the 7-limit rank-two case is particularly simple: B = CE. If complexity is bounded by, for example, 20 (which allows for some quite complex temperaments) then since E ≤ 1/(4√q5√q7), B ≤ 20/(4√q5√q7) = 240.250. This is an absurdly high badness figure; while simply bounding complexity will lead to a finite list, the list would be enormous. An alternative is also to bound badness; for instance, we might produce a list of 7-limit rank-two temperaments with complexity less than 20 and a more reasonable badness limit, such as 0.05 or 0.06. | ||
Essentially the same situation obtains for rank two temperaments in higher limits. The rule then is that if E ≤ 1/(C(n, 3)√lb(q)√lb(p)) then wedging <1 lb(3) lb(5) ... lb(p)| with the val consisting of 0 followed by the first n-1 coefficients of the wedgie and rounding will give the wedgie. Here p and q are the largest and second largest primes in the prime limit, lb(x) is log base two, and C(n, 3) is n choose three, n(n-1)(n-2)/6. | |||
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+(\frac{f}{q_5q_7}-\frac{e}{q_3q_7}+\frac{d}{q_3q_5})^2 = 4 E^2</script><!-- ws:end:WikiTextMathRule:0 --><br /> | +(\frac{f}{q_5q_7}-\frac{e}{q_3q_7}+\frac{d}{q_3q_5})^2 = 4 E^2</script><!-- ws:end:WikiTextMathRule:0 --><br /> | ||
<br /> | <br /> | ||
For this we can conclude that d, e and f satisfy |d - q3b + q5a| ≤ | For this we can conclude that d, e and f satisfy |d - q3b + q5a| ≤ 2E√q3√q5, |e - q3c + q7a| ≤ 2E√q3√q7 and |f - q5c + q7d| ≤ 2E√q5q7. This has an interesting interpretation: since &lt;1 q3 q5 q7|∧&lt;0 a b c| = &lt;&lt;a b c q3b-q5a q3c-q7a q5c-q7b||, if E ≤ 1/(4√q5√q7), then the full wedgie can be recovered from the octave equivalent (OE) portion of the wedgie simply by wedging it with &lt;1 q3 q5 q7| and rounding to the nearest integer. This is not a very serious constraint to place on relative error; it seems unlikely anyone would be interested in a temperament which did not fall well under this low standard. Hence we may compile lists of reasonable temperaments by presuming &quot;reasonable&quot; requires this bound to be met, searching through triples &lt;&lt;a b c ...|| (note that if all of these are zero, 2 is being tempered out) up to some complexity bound, wedging with &lt;1 q3 q5 q7| and rounding, then checking if the GCD is one and the Pfaffian af-be+cd is zero. Then we may toss everthing which does not meet the bound on relative error; however, for a reasonable list we will want a tighter bound. <br /> | ||
<br /> | |||
If C = ||W|| is the TE complexity, then the formula for the <a class="wiki_link" href="/Tenney-Euclidean%20metrics#Logflat TE badness">logflat badness</a> B in the 7-limit rank-two case is particularly simple: B = CE. If complexity is bounded by, for example, 20 (which allows for some quite complex temperaments) then since E ≤ 1/(4√q5√q7), B ≤ 20/(4√q5√q7) = 240.250. This is an absurdly high badness figure; while simply bounding complexity will lead to a finite list, the list would be enormous. An alternative is also to bound badness; for instance, we might produce a list of 7-limit rank-two temperaments with complexity less than 20 and a more reasonable badness limit, such as 0.05 or 0.06.<br /> | |||
<br /> | <br /> | ||
Essentially the same situation obtains for rank two temperaments in higher limits. The rule then is that if E ≤ 1/(C(n, 3)√lb(q)√lb(p)) then wedging &lt;1 lb(3) lb(5) ... lb(p)| with the val consisting of 0 followed by the first n-1 coefficients of the wedgie and rounding will give the wedgie. Here p and q are the largest and second largest primes in the prime limit, lb(x) is log base two, and C(n, 3) is n choose three, n(n-1)(n-2)/6.</body></html></pre></div> | |||