S-expression: Difference between revisions

== Quick rules of S-expressions ==

== S''k'' (square-particulars) ==

which is square-(super)particular ''k'' for a given integer {{nowrap|''k'' > 1}}. A suggested shorthand for this interval is '''S''k''''' for the ''k''-th square superparticular, where the ''S'' stands for "(Shorthand for) Second-order/Square Superparticular". This will be used later in this article as the notation will prove powerful in understanding the commas and implied tempered structures of [[regular temperament]]s. Note that this means {{nowrap|S2 {{=}} [[4/3]]}} is the first musically meaningful square-particular, as {{nowrap|S1 {{=}} 1/0}}.

It is common to temper square superparticulars, equating two adjacent superparticulars at some point in the harmonic series, but for higher accuracy or structural reasons it can be more beneficial to instead temper differences between consecutive square superparticulars so that the corresponding consecutive superparticulars are tempered to have equal spacing between them. If we define a sequence of commas {{nowrap|U''k'' {{=}} {{sfrac|S''k''|S(''k'' + 1)}}}}, we get [[#Sk/S(k + 1) (ultraparticulars)|ultraparticulars]]<ref group="note">In analogy with the "super-", "ultra-" progression and because these would be differences between adjacent differences between adjacent superparticulars, which means a higher order of "particular", and as we will see, no longer [[superparticular]], hence the need for a new name. Also note that the choice of indexing is rather arbitrary and up to debate, as U''k'' = S(''k'' - 1)/S''k'' and U''k'' = S(''k'' + 1)/S(''k'' + 2) also make sense. Therefore it is advised to use the S-expression to refer to an ultraparticular unambiguously, or the comma itself.</ref>. Ultraparticulars have a secondary (and mathematically equivalent) consequence: Because {{sfrac|''k'' + 2|''k'' + 1}} and {{sfrac|''k''|''k'' − 1}} are equidistant from {{sfrac|''k'' + 1|''k''}} (because of tempering {{sfrac|S''k''|S(''k'' + 1)}}), this means that another expression for {{sfrac|S''k''|S(''k'' + 1)}} is the following:

This means you can read the ''k'' and {{nowrap|''k'' + 1}} from the S-expression of an ultraparticular as being the interval involved in the cubing equivalence (abbreviated to "cube relation" in [[#Sk/S(k + 1) (ultraparticulars)|the table of ultraparticulars]]).

1. Every triangle-particular is superparticular, so these are efficient commas. (See also the [[#Short proof of the superparticularity of triangle-particulars]].)

3. Every triangle-particular is the difference between two nearly-adjacent superparticular intervals {{nowrap|''k'' + 2|''k'' + 1}} and {{nowrap|''k''|''k'' − 1}}.

5. If we temper {{nowrap|S''k'' * S(''k'' + 1)}} but not S''k'' or {{nowrap|S(''k'' + 1)}}, then one or more intervals of {{sfrac|''k''|''k'' − 1}}, {{sfrac|''k'' + 1|''k''}}, and {{sfrac|''k'' + 2|''k'' + 1}} ''must'' be mapped inconsistently, because:

: If {{nowrap|''k'' + 1|''k''}} is mapped above {{nowrap|{{sfrac|''k'' + 2|''k'' + 1}} ~ {{sfrac|''k''|''k'' − 1}} we have {{nowrap|{{sfrac|''k'' + 1|''k''}} &gt; {{sfrac|''k''|''k'' − 1}}}} and if it is mapped below we have {{nowrap|{{sfrac|''k'' + 1|''k''}} &lt; {{sfrac|''k'' + 2|''k'' + 1}}}}.

=== Meaning ===

Notice that if we equate {{sfrac|''k'' + 2|''k'' + 1}} with {{sfrac|''k''|''k'' − 1}} (by [[tempering out]] their difference), then multiply both sides by {{sfrac|''k'' + 1|''k''}}, we have:

which simplifies to:

=== Short proof of the superparticularity of triangle-particulars ===

Then notice that {{nowrap|''k''² + ''k''}} is always a multiple of 2, therefore the above always simplifies to a superparticular. Half of this superparticular is halfway between the corresponding square-particulars, and because of its composition it could therefore be reasoned that it'd likely be half as accurate as tempering either of the square-particulars individually, so these are "1/2-square-particulars" in a sense, and half of a square is a triangle, which is not a coincidence here because the numerators of all of these superparticular intervals/commas are [[triangular number]]s! (Hence the alternative name "[[triangle-particular]]".)

If {{nowrap|''k'' {{=}} 3''n'' + 1}} then:

if {{nowrap|''k'' {{=}} 3''n'' + 2}} then:

if {{nowrap|''k'' {{=}} 3''n''}} then:

In other words, what this shows is all {{frac|1|3}}-square-particulars of the form {{frac|S(''k'' − 1) * S''k'' * S(''k'' + 1)}} are superparticular iff ''k'' is throdd (not a multiple of 3), and all {{frac|1|3}}-square-particulars of the form {{nowrap|S(3''k'' − 1) * S(3''k'') * S(3''k'' + 1)}} are throdd-particular with the numerator and denominator always being one less than a multiple of 3 (which is to say, commas of this form are throdd-particular iff ''k'' is threven and superparticular iff ''k'' is throdd).

|}

=== Motivation ===

In other words, each and every S-expression of a comma as a 1/''n''-square-particular corresponds exactly to expressing it as the ratio between two [[superparticular]] intervals, with ''n'' distance between them, where, for example, 10/9 and 11/10 are considered as having 1 distance between them, corresponding to (1/1-)square-particulars (in this case [[100/99|S10]]).

1. As a generalization of important special cases {{nowrap|''n'' {{=}} 0|''n'' {{=}} 1}}, and {{nowrap|''n'' {{=}} 2}}, (which are almost all superparticular; the only case where they aren't is that {{nowrap|''n'' {{=}} 3}} (1/3-square-particulars) are throdd-particular one third of the time, so this suggests these are efficient commas. A cursory look will show that many 1/n-square-particulars for small n are superparticular, and many more are the next best things (odd-particular, throdd-particular, quodd-particular, etc.) so this confirms them being a family of efficient commas.

3. They often correspond to "nontrivial" equivalences that need to be dug up which are not obvious from their expression as a ratio of two superparticular intervals, for example, [[385/384|S33*S34*S35]], suggesting they are a goldmine for valuable tempering opportunities.  

4. Their expressions naturally make them implied by tempering consecutive square-particulars, so if you notice them present and that the individual square-particulars aren't tempered, if you want to extend your temperament and/or reduce its rank (tempering it down) and/or hope to make your temperament more efficient, you can try tempering the untempered square-particulars that a tempered 1/''n''-square-particular is composed of (although this is not always possible). There is also good theoretical motivation for wanting to do this, as the next section will discuss.

6. They're relevant to understanding limitations of consistency (or more precisely, monotonicity) of any given temperament, as the next section will discuss.

1/n-square-particulars, which is to say, commas which can be written in the form of a product of ''n'' consecutive square-particulars (including S''k'' but not including {{nowrap|S(''k'' + ''n'')}}) and which can therefore be written as the ratio between the two superparticulars {{sfrac|''k''|''k'' − 1}} and {{sfrac|''k'' + ''n''|''k'' + ''n'' − 1}} have implications for the [[consistency]] of the ({{nowrap|''k'' + ''n''}})-[[odd-limit]] when tempered. Specifically:

Consider the following sequence of superparticular intervals, all of which in the ({{nowrap|''k'' + ''n''}})-odd-limit:

<math>\displaystyle\begin{align}

\end{align}</math>

{| class="wikitable center-all\

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:

=== Significance ===

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:

(''k''+2)/''k'', (''k''+1)/(''k''-1), ''k''/(''k''-2)

It suffices to show every superparticular number including 2/1 has an expression using square-particulars:

</math>

== Glossary ==

; 1/''n''-square-particular

: A comma which is the product of ''n'' consecutive square-particulars and which can therefore be expressed as the ratio between two superparticulars.

: Replacing/substituting ''a'' with ''k'' and ''b'' with ''k'' + ''n'' - 1 gives us an equivalent expression that includes the number of square-particulars ''n'':

: For ''b'' = ''a'' + 1 these can also be called triangle-particulars, in which case they are always superparticular.

; Odd-particular