S-expression: Difference between revisions

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== Sk/S(k + 2) (semiparticulars) ==

~~For differences between square-particulars~~ of ~~the form S''k''~~/S(~~''k'' +~~ 2)~~, the resulting comma~~ is ~~either~~ [[~~superparticular~~]] or [[~~#Glossary|odd-particular~~]].

=== Motivational examples ===

If we want to halve one JI interval into two of another JI interval, there is a powerful and elegant pattern for doing so:

* [[4/3]] is approximately half of [[9/5]]

* [[9/7]] is approximately half of [[5/3]](=10/6)

* [[5/4]] is approximately half of [[11/7]]

* [[11/9]] is approximately half of [[3/2]](=12/8)

* [[6/5]] is approximately half of [[13/9]]

* [[13/11]] is approximately half of [[7/5]](14/10)

* [[7/6]] is approximately half of [[15/11]]

* [[15/13]] is approximately half of [[4/3]](=16/12)

* [[8/7]] is approximately half of [[17/13]]

* [[17/15]] is approximately half of [[9/7]](=18/14)

* [[9/8]] is approximately half of [[19/15]]

* [[19/17]] is approximately half of [[5/4]](=20/16)

Do you see the pattern?

~~(This terminology also suggests [[#Glossary|throdd-particular]] and~~ [[#Glossary|quodd-particular]] ~~as generalizations~~.)

The pattern is this: take some arbitrary [[#Glossary|quodd-particular]] (''k'' + 4)/''k''; observe that we can split it into (''k'' + 4)/(''k'' + 2) * (''k'' + 2)/''k''.

~~In the below, we use S~~(''k'' ~~- 1~~)/S(''k'' + 1) ~~for symmetry around~~ ''k'' ~~to make the math visually simpler~~, ~~but keep in mind~~ it'~~s equivalent to using an offset~~ ''k''.

Now observe that (''k'' + 2)/''k'' > (''k'' + 3)/(''k'' + 1) > (''k'' + 4)/(''k'' + 2); in fact, it can be shown fairly easily that (''k'' + 3)/(''k'' + 1) is the [[mediant]] of (''k'' + 4)/(''k'' + 2) and (''k'' + 2)/''k''.

~~Also keep in mind~~ that ''k'' ~~- ''a''~~ (~~for positive~~ ''a'') ~~is smaller than ''k'', so that~~ ''k''/(''k'' ~~- ''a'') >~~ (''k'' + ~~''a''~~)/''k'' ~~(because the former appears earlier in the harmonic series and is thus larger~~)~~; this is an important and useful intuition to learn~~.

It turns out that making this mediant — (''k'' + 3)/(''k'' + 1) — equal to half of (''k'' + 4)/''k'' is equivalent to tempering S(''k'' + 1)/S(''k'' + 3).

=== Significance ===

1. For differences between square-particulars of the form S''k''/S(''k'' + 2), the resulting comma is either [[superparticular]] or [[#Glossary|odd-particular]], so these are efficient commas. (This terminology also suggests [[#Glossary|throdd-particular]] and [[#Glossary|quodd-particular]] as generalizations.)

2. Tempering any two consecutive [[ultraparticular]]s implies tempering a semiparticular, so from two adjacent "thirding" equivalences you get a "halving" equivalence for free!

3. Tempering any two nearly-consecutive square-particulars (S''k'' and S(''k'' + 2)) implies tempering a semiparticular; this is generally much more ideal than tempering two consecutive S''k'' because it is a lot lower damage (see [[lopsided comma]]s for (relatively) large commas implied by this higher-damage strategy).

4. On top of the halving equivalence, there is a number of subtler structural implications, [[discussed below, that may be desirable to the temperament designer.

=== Meaning ===

: '''Reader notes:''' In the below, we use S(''k'' - 1)/S(''k'' + 1) for symmetry around ''k'' to make the math visually simpler, but keep in mind it's equivalent to using an offset ''k''.

: '''Also:''' keep in mind that ''k'' - ''a'' (for positive ''a'') is smaller than ''k'', so that ''k''/(''k'' - ''a'') > (''k'' + ''a'')/''k'' (because the former appears earlier in the harmonic series & is thus larger); this is an important and useful intuition to learn.

Tempering S(''k'' - 1)/S(''k'' + 1) implies that (''k'' + 2)/(''k'' - 2) is divisible exactly into two halves of (''k'' + 1)/(''k'' - 1). It also implies that the intervals (''k'' + 2)/''k'' (=s) and ''k''/(''k'' - 2) (=L) are equidistant from (''k'' + 1)/(''k'' - 1) (=M) because to make them equidistant we need to temper:

@@ Line 889: / Line 889: @@
 == Sk/S(k + 2) (semiparticulars) ==
-For differences between square-particulars of the form S''k''/S(''k'' + 2), the resulting comma is either [[superparticular]] or [[#Glossary|odd-particular]].
+=== Motivational examples ===
+If we want to halve one JI interval into two of another JI interval, there is a powerful and elegant pattern for doing so:
+* [[4/3]] is approximately half of [[9/5]]
+* [[9/7]] is approximately half of [[5/3]](=10/6)
+* [[5/4]] is approximately half of [[11/7]]
+* [[11/9]] is approximately half of [[3/2]](=12/8)
+* [[6/5]] is approximately half of [[13/9]]
+* [[13/11]] is approximately half of [[7/5]](14/10)
+* [[7/6]] is approximately half of [[15/11]]
+* [[15/13]] is approximately half of [[4/3]](=16/12)
+* [[8/7]] is approximately half of [[17/13]]
+* [[17/15]] is approximately half of [[9/7]](=18/14)
+* [[9/8]] is approximately half of [[19/15]]
+* [[19/17]] is approximately half of [[5/4]](=20/16)
+Do you see the pattern?
-(This terminology also suggests [[#Glossary|throdd-particular]] and [[#Glossary|quodd-particular]] as generalizations.)
+The pattern is this: take some arbitrary [[#Glossary|quodd-particular]] (''k'' + 4)/''k''; observe that we can split it into (''k'' + 4)/(''k'' + 2) * (''k'' + 2)/''k''.
-In the below, we use S(''k'' - 1)/S(''k'' + 1) for symmetry around ''k'' to make the math visually simpler, but keep in mind it's equivalent to using an offset ''k''.
+Now observe that (''k'' + 2)/''k'' > (''k'' + 3)/(''k'' + 1) > (''k'' + 4)/(''k'' + 2); in fact, it can be shown fairly easily that (''k'' + 3)/(''k'' + 1) is the [[mediant]] of (''k'' + 4)/(''k'' + 2) and (''k'' + 2)/''k''.
-Also keep in mind that ''k'' - ''a'' (for positive ''a'') is smaller than ''k'', so that ''k''/(''k'' - ''a'') > (''k'' + ''a'')/''k'' (because the former appears earlier in the harmonic series and is thus larger); this is an important and useful intuition to learn.
+It turns out that making this mediant — (''k'' + 3)/(''k'' + 1) — equal to half of (''k'' + 4)/''k'' is equivalent to tempering S(''k'' + 1)/S(''k'' + 3).
 === Significance ===
+. For differences between square-particulars of the form S''k''/S(''k'' + 2), the resulting comma is either [[superparticular]] or [[#Glossary|odd-particular]], so these are efficient commas. (This terminology also suggests [[#Glossary|throdd-particular]] and [[#Glossary|quodd-particular]] as generalizations.)
+. Tempering any two consecutive [[ultraparticular]]s implies tempering a semiparticular, so from two adjacent "thirding" equivalences you get a "halving" equivalence for free!
+. Tempering any two nearly-consecutive square-particulars (S''k'' and S(''k'' + 2)) implies tempering a semiparticular; this is generally much more ideal than tempering two consecutive S''k'' because it is a lot lower damage (see [[lopsided comma]]s for (relatively) large commas implied by this higher-damage strategy).
+. On top of the halving equivalence, there is a number of subtler structural implications, [[discussed below, that may be desirable to the temperament designer.
+=== Meaning ===
+: '''Reader notes:''' In the below, we use S(''k'' - 1)/S(''k'' + 1) for symmetry around ''k'' to make the math visually simpler, but keep in mind it's equivalent to using an offset ''k''.
+: '''Also:''' keep in mind that ''k'' - ''a'' (for positive ''a'') is smaller than ''k'', so that ''k''/(''k'' - ''a'') > (''k'' + ''a'')/''k'' (because the former appears earlier in the harmonic series & is thus larger); this is an important and useful intuition to learn.
 Tempering S(''k'' - 1)/S(''k'' + 1) implies that (''k'' + 2)/(''k'' - 2) is divisible exactly into two halves of (''k'' + 1)/(''k'' - 1). It also implies that the intervals (''k'' + 2)/''k'' (=s) and ''k''/(''k'' - 2) (=L) are equidistant from (''k'' + 1)/(''k'' - 1) (=M) because to make them equidistant we need to temper: