S-expression: Difference between revisions

As S-expressions are deployed widely on the wiki and in the broader xen community, below is a list of what the most common S-expression categories imply is equated when they are [[tempering out|tempered out]]. The linked sections provide deeper information into each comma family.

* [[#Ck and Cpk (cube-particulars)|Cube-particulars]]: '''C''k''''' and '''Cp''k''''', superparticular fractions of the form {{sfrac|''k''³|''k''³ − 1}} and {{sfrac|''k''³ + 1|''k''³}}, respectively.

$$ \frac {k²}{k² - 1} = \frac {k/(k - 1)}{(k + 1)/k} $$

<nowiki>$$ S(k) \cdot S(k + 1) = \frac{\frac{k}{k - 1}}{\frac{k + 2}{k + 1}} = \frac{k(k + 1)}{(k - 1)(k + 2)} = \frac{k² + k}{k² + k - 2} $$</nowiki>

The reader is encouraged to familiarize themself with the structure of this argument, as it generalizes to arbitrary S''k'' (→ [[S-expression/Advanced_results#Mathematical_derivations]]); the algebraic proof is tedious, but the intuition is the same:

# Tempering out any two consecutive ultraparticulars will imply tempering out the [[#Sk/S(k + 2) (semiparticulars)|semiparticular]], which is their product. A rather-interesting arithmetic of square-particular and related commas exists. This arithmetic can be described compactly with S-expressions, which is to say, expressions composed of square superparticulars multiplied and divided together, using the S''k'' notation to achieve that compactness.

# Tempering out the ultraparticular S''k''/S({{nowrap|''k'' + 1}}) along with either the corresponding 1/2-square-particular {{nowrap|S''k''⋅S(''k'' + 1)}} or one of the two corresponding lopsided commas {{nowrap|S''k''²⋅S(''k'' + 1)}} or {{nowrap|S''k''⋅S(''k'' + 1)²}} implies tempering both of S''k'' and S({{nowrap|''k'' + 1}}) individually, and vice versa, so that there is a total of ''five'' equivalences—corresponding to ''five'' infinite families of commas—for every such S''k'' and {{nowrap|S(''k'' + 1)}}. This only gets better if you temper out a third consecutive square-particular. This is an abundance of "at-a-glance" essential tempering information that is fully general, so it only needs to be learned once, and is another motivation of the use of S-expressions. For example, {{nowrap|{S16, S17} → {S16⋅S17, S16/S17, S16²⋅S17, S16⋅S17²} }}, and any of the two commas in the latter set imply all the other commas too.

<nowiki>$$ \frac{L/M}{M/s} = \frac{ \left(\frac{k}{k-2}\right)/\left(\frac{k+1}{k-1}\right) }{ \left(\frac{k+1}{k-1}\right)/\left(\frac{k+2}{k}\right) } = \frac{Ls}{M²} = \frac{\frac{k+2}{k-2}}{\left(\frac{k+1}{k-1}\right)²} $$</nowiki>

… and notice that the latter expression is the one we have shown is equal to S(''k'' - 1)/S(''k'' + 1) up to an offset ''k'' (→ [[S-expression/Advanced results #Mathematical derivations]]). In other words, that tempering out S(''k'' - 1)/S(''k'' + 1) results in (''k'' + 1)/(''k'' - 1) being half of (''k'' + 2)/(''k'' - 2) is an implication that it makes (''k'' + 2)/''k'', (''k'' + 1)/(''k'' - 1), and ''k''/(''k'' - 2) equidistant.  

For a general introduction to method we use here, see [[#Using S-factorizations to understand the significance of S-expressions|S-factorizations]], as we will not re-explain the method here.

'''(An important corollary that should be noted:''' The same derivation above can be observed as not depending on the number of 0's in the S-factorization of the expressions of {{nowrap| <code>F(k + 1)</code> and <code>F(k + 2)</code> }}, and that therefore we have S''a''/S''b'' being an equidistance relation of intervals, where the generalized versions of the terms {{nowrap| <code>F(k + 1)</code> and <code>F(k + 2)</code> }} correspond to the two differences of intervals of the form (''n'' + ''k'')/''n'', where ''k'' = 3 corresponds to one zero in the monzo and zero in {{nowrap| <code>[ -1, 2, -1, 1, -2, 1 ]</code> }}, and where the terms {{nowrap| <code>F(k + 1)</code> and <code>F(k + 2)</code> }} therefore correspond to the two differences used to make three intervals of the form (''n'' + ''k'')/''n'' equidistant. "Which three intervals" is thus answered as those of 3 consecutive/adjacent values of ''n''. '''Therefore:''' What makes three-particulars special in this more general S-comma family is that the largest and smallest interval of the three almost compose to something simple, up to a superparticular difference, s.t the corresponding splitting relation can often simplify for cases of interest, as discussed in [[S-expression#Significance_7|#Significance]]'''.)'''

As before, for a general introduction to the method we use here, see [[#Using S-factorizations to understand the significance of S-expressions|S-factorizations]], as we will not re-explain the method here. However, we will use a previous result from three-particulars as to immediately say that:

This family of superparticular interval is of the form {{nowrap|{{sfrac|''k''³|''k''³ − 1}} {{=}} C''k''}} and {{nowrap|{{sfrac|''k''³ + 1|''k''³}} {{=}} Cp''k''}} (for ''cube-particular complement''). Both C''k'' and Cp''k'' are notable because ''k''³ + 1 and ''k''³ − 1 are always composite for ''k'' ≥ 2, unlike with square-particulars, where ''k''² + 1 can be prime. The term ''S-expression'' arguably does not apply here, but they are noted on this page nonetheless because they appear as an extension of [[ultraparticular]]s, as {{nowrap| S''k''/S(''k'' + 1) {{=}} C''k''⋅Cp(''k'' + 1) }}, therefore every temperament tempering out an ultraparticular can potentially be extended in a non-obvious way to tempering out a pair of cube-particulars.

All S-expressions have other equivalent S-expressions; however, a comma might appear in more than one family of S-commas or arise as a specific product of them. This often means nontrivial, deep tempering opportunities, usually leading to multiple of the most elegant and efficient temperaments that we know of depending on how the tempering is further realized. Generally we exclude 1/''n''-square-particulars, only noting up to 1/3-square-particulars, because equivalent 1/''n''-square-particular expressions become very common for higher ''n'', but are still quite rare for small ''n''.

{\rm S}k = {\rm S}(2k - 1) \cdot {\rm S}(2k)² \cdot {\rm S}(2k + 1)

This is important to note because using this simple rule we can derive an infinite amount of trivial equivalent S-expressions. See [[#The general S-expression equivalence]] for mathematical details.

! rowspan="2" | Comma<br>(by size)

! colspan="4" | S-expressions

| S7⋅S8

|S4/S6

| S6

|S8⋅S9

| S8

|S6/S9

|S4/(S6⋅S7)

|(S4⋅S5⋅S6)/S3

| S9

|S6/S8

| S8/S10

|S22⋅S23⋅S24

| S9/S11

|S15/S55

|S15/(S22/S24)

|S25⋅S26

|[[352/351]]*

|S12/(S9/S11)

| S12/S14

|(S9⋅S10)/S7

|(S6/S7)/(S8/S10)

|S26

|S13/S14

| S35

|S49⋅S50

| S51

|S17/(S25⋅S26)

| S55

|S22/S24

|(S25/S27)⋅S99

| S99

|S33/S35

| S161

|S46/S48

| S351

|S78/S80