S-expression: Difference between revisions

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== Sk/S(k + 1) (ultraparticulars) ==

== {{nowrap|S''k''/S(''k'' + 1)}} (ultraparticulars) ==

=== Motivational example ===

Often it is desirable to make consecutive [[superparticular]] intervals equidistant. This has a number of nice consequences, many of which not explained ~~here — see~~ the motivation section for each infinite family of commas defined on this page.

Often it is desirable to make consecutive [[superparticular]] intervals equidistant. This has a number of nice consequences, many of which not explained here—see the motivation section for each infinite family of commas defined on this page.

For example, if you want 6/5 equidistant from 5/4 and 7/6, you must equate ([[5/4]]~~)/(~~[[6/5]]) = [[25/24]] = S5 with ([[6/5]]~~)/(~~[[7/6]]) = [[36/35]] = S6, hence tempering S5/S6 = ~~([[~~25/24~~]])/([[~~36/35~~]])~~ = [[875/864]], but it's actually often not necessary to know the specific numbers, often familiarizing yourself with and understanding the "S''k''" notation will give you a lot of insight, as we'll see.

For example, if you want 6/5 equidistant from 5/4 and 7/6, you must equate {{nowrap|{{sfrac|[[5/4]]|[[6/5]]}} {{=}} [[25/24]]}} {{nowrap|{{=}} S5}} with {{nowrap|{{sfrac|[[6/5]]|[[7/6]]}} {{=}} [[36/35]]}} {{nowrap|{{=}} S6}}, hence tempering {{nowrap|{{sfrac|S5|S6}} {{=}} {{sfrac|25/24|36/35}}}} {{nowrap|{{=}} [[875/864]]}}, but it's actually often not necessary to know the specific numbers, often familiarizing yourself with and understanding the "S''k''" notation will give you a lot of insight, as we'll see.

Back to our example: we know that S5 ~ S6 (because we're tempering S5/S6); from this we can deduce that the intervals must be arranged like this: 7/6 <— S5~S6 —> 6/5 <— S5~S6 —> 5/4.

Back to our example: we know that {{nowrap|S5 ~ S6}} (because we're tempering S5/S6); from this we can deduce that the intervals must be arranged like this: {{nowrap|7/6 ← S5~S6 → 6/5 ← S5~S6 → 5/4}}.

From this you can deduce that ([[6/5]])<sup>3</sup> = [[7/4]], because you can lower one of the 6/5's to [[7/6]] (lowering it by S6) and raise another of the 6/5's to [[5/4]] (raising it by S5). Then because we've tempered S5 and S6 together, we've lowered and raised by the same amount, so the result of 7/6 * 6/5 * 5/4 = 7/4 must be the same as the result of 6/5 * 6/5 * 6/5 in this temperament.

From this you can deduce that {{nowrap|([[6/5]])<sup>3</sup> → [[7/4]]}}, because you can lower one of the 6/5's to [[7/6]] (lowering it by S6) and raise another of the 6/5's to [[5/4]] (raising it by S5). Then because we've tempered S5 and S6 together, we've lowered and raised by the same amount, so the result of {{nowrap|7/6 * 6/5 * 5/4 {{=}} 7/4}} must be the same as the result of {{nowrap|6/5 * 6/5 * 6/5}} in this temperament.

Familiarize yourself with the structure of this argument, as [[S-expression/Advanced results#Mathematical derivations|it generalizes to arbitrary S''k'']]; the algebraic proof is tedious, but the intuition is the same:

~~(''~~k''+2~~)/(''~~k''+1~~) <—~~ S(''k''+1)~~~S''k'' —> (''~~k''+1~~)/''~~k~~'' <—~~ S(''k''+1)~~~S''k'' —> ''~~k~~''/(''~~k''-1)

<math>\displaystyle\frac{k+2}{k+1} \leftarrow S(k+1)~Sk \rightarrow \frac{k+1}{k} \leftarrow S(k+1)~Sk \rightarrow \frac{k}{k-1}</math>

...implies that three (''k''+1)/''k''~~'s is equal to a (~~''k''+2~~)/(~~''k''-1) iff we temper S''k''/S(''k''+1). {{qed}}

...implies that three {{sfrac|''k'' + 1|''k''}} give {{sfrac|''k'' + 2|''k'' − 1}} iff we temper {{sfrac|S''k''|S(''k'' + 1)}}. {{qed}}

=== Significance ===

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-== Sk/S(k + 1) (ultraparticulars) ==
+== {{nowrap|S''k''/S(''k'' + 1)}} (ultraparticulars) ==
 === Motivational example ===
-Often it is desirable to make consecutive [[superparticular]] intervals equidistant. This has a number of nice consequences, many of which not explained here — see the motivation section for each infinite family of commas defined on this page.
+Often it is desirable to make consecutive [[superparticular]] intervals equidistant. This has a number of nice consequences, many of which not explained here—see the motivation section for each infinite family of commas defined on this page.
-For example, if you want 6/5 equidistant from 5/4 and 7/6, you must equate ([[5/4]])/([[6/5]]) = [[25/24]] = S5 with ([[6/5]])/([[7/6]]) = [[36/35]] = S6, hence tempering S5/S6 = ([[25/24]])/([[36/35]]) = [[875/864]], but it's actually often not necessary to know the specific numbers, often familiarizing yourself with and understanding the "S''k''" notation will give you a lot of insight, as we'll see.
+For example, if you want 6/5 equidistant from 5/4 and 7/6, you must equate {{nowrap|{{sfrac|[[5/4]]|[[6/5]]}} {{=}} [[25/24]]}} {{nowrap|{{=}} S5}} with {{nowrap|{{sfrac|[[6/5]]|[[7/6]]}} {{=}} [[36/35]]}} {{nowrap|{{=}} S6}}, hence tempering {{nowrap|{{sfrac|S5|S6}} {{=}} {{sfrac|25/24|36/35}}}} {{nowrap|{{=}} [[875/864]]}}, but it's actually often not necessary to know the specific numbers, often familiarizing yourself with and understanding the "S''k''" notation will give you a lot of insight, as we'll see.
-Back to our example: we know that S5 ~ S6 (because we're tempering S5/S6); from this we can deduce that the intervals must be arranged like this: 7/6 <— S5~S6 —> 6/5 <— S5~S6 —> 5/4.
+Back to our example: we know that {{nowrap|S5 ~ S6}} (because we're tempering S5/S6); from this we can deduce that the intervals must be arranged like this: {{nowrap|7/6 &larr; S5~S6 &rarr; 6/5 &larr; S5~S6 &rarr; 5/4}}.
-From this you can deduce that ([[6/5]])<sup>3</sup> = [[7/4]], because you can lower one of the 6/5's to [[7/6]] (lowering it by S6) and raise another of the 6/5's to [[5/4]] (raising it by S5). Then because we've tempered S5 and S6 together, we've lowered and raised by the same amount, so the result of 7/6 * 6/5 * 5/4 = 7/4 must be the same as the result of 6/5 * 6/5 * 6/5 in this temperament.
+From this you can deduce that {{nowrap|([[6/5]])<sup>3</sup> &rarr; [[7/4]]}}, because you can lower one of the 6/5's to [[7/6]] (lowering it by S6) and raise another of the 6/5's to [[5/4]] (raising it by S5). Then because we've tempered S5 and S6 together, we've lowered and raised by the same amount, so the result of {{nowrap|7/6 * 6/5 * 5/4 {{=}} 7/4}} must be the same as the result of {{nowrap|6/5 * 6/5 * 6/5}} in this temperament.
 Familiarize yourself with the structure of this argument, as [[S-expression/Advanced results#Mathematical derivations|it generalizes to arbitrary S''k'']]; the algebraic proof is tedious, but the intuition is the same:
-(''k''+2)/(''k''+1) <— S(''k''+1)~S''k'' —> (''k''+1)/''k'' <— S(''k''+1)~S''k'' —> ''k''/(''k''-1)
+<math>\displaystyle\frac{k+2}{k+1} \leftarrow S(k+1)~Sk \rightarrow \frac{k+1}{k} \leftarrow S(k+1)~Sk \rightarrow \frac{k}{k-1}</math>
-...implies that three (''k''+1)/''k'''s is equal to a (''k''+2)/(''k''-1) iff we temper S''k''/S(''k''+1). {{qed}}
+...implies that three {{sfrac|''k'' + 1|''k''}} give {{sfrac|''k'' + 2|''k'' − 1}} iff we temper {{sfrac|S''k''|S(''k'' + 1)}}.&nbsp;{{qed}}
 === Significance ===