S-expression: Difference between revisions

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== Quick rules of S-expressions ==
== Quick rules of S-expressions ==
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 when they are [[tempering out|tempered out]]. The linked sections provide deeper information into each comma family.
[[File:S(expression).svg|thumb|Here is a summarized infographic of what many of these S-expression types do to a harmonic series segment when tempered out.]]
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.


* [[#Sk (square-particulars)|Square-particulars]]: '''S''k''''', superparticular fractions of the form {{sfrac|''k''<sup>2</sup>|''k''<sup>2</sup> − 1}}. <br>Tempering out S''k'' equates {{sfrac|''k'' + 1|''k''}} with {{sfrac|''k''|''k'' − 1}} and splits {{sfrac|''k'' + 1|''k'' − 1}} in two.
* [[#Sk (square-particulars)|Square-particulars]]: '''S''k''''', superparticular fractions of the form {{sfrac|''k''<sup>2</sup>|''k''<sup>2</sup> − 1}}. <br>Tempering out S''k'' equates {{sfrac|''k'' + 1|''k''}} with {{sfrac|''k''|''k'' − 1}} and splits {{sfrac|''k'' + 1|''k'' − 1}} in two.
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* [[#Sk/S(k + 1) (ultraparticulars)|Ultraparticulars]]: {{nowrap|'''S''k''/S(''k'' + 1)'''}}. Tempering this out splits {{sfrac|''k'' + 2|''k'' − 1}} into {{pars|{{sfrac|''k'' + 1|''k''}}}}<sup>3</sup>.
* [[#Sk/S(k + 1) (ultraparticulars)|Ultraparticulars]]: {{nowrap|'''S''k''/S(''k'' + 1)'''}}. Tempering this out splits {{sfrac|''k'' + 2|''k'' − 1}} into {{pars|{{sfrac|''k'' + 1|''k''}}}}<sup>3</sup>.
* [[#Sk/S(k + 2) (semiparticulars)|Semiparticulars]]: {{nowrap|'''S''k''/S(''k'' + 2)'''}}. Tempering this out splits {{sfrac|''k'' + 3|''k'' − 1}} into {{pars|{{sfrac|''k'' + 2|''k''}}}}<sup>2</sup>.
* [[#Sk/S(k + 2) (semiparticulars)|Semiparticulars]]: {{nowrap|'''S''k''/S(''k'' + 2)'''}}. Tempering this out splits {{sfrac|''k'' + 3|''k'' − 1}} into {{pars|{{sfrac|''k'' + 2|''k''}}}}<sup>2</sup>.
* [[#Ck and Cpk (cube-particulars)|Cube-particulars]]: '''C''k''''' and '''Cp''k''''', superparticular fractions of the form {{sfrac|''k''<sup>3</sup>|''k''<sup>3</sup> − 1}} and {{sfrac|''k''<sup>3</sup> + 1|''k''<sup>3</sup>}}, respectively.  
* [[#Ck and Cpk (cube-particulars)|Cube-particulars]]: '''C''k''''' and '''Cp''k''''', superparticular fractions of the form {{sfrac|''k''<sup>3</sup>|''k''<sup>3</sup> − 1}} and {{sfrac|''k''<sup>3</sup> + 1|''k''<sup>3</sup>}}, respectively.


== S''k'' (square-particulars) ==
== S''k'' (square-particulars) ==
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From this, it can be deduced that {{nowrap| (6/5)<sup>3</sup> ~ [[7/4]] }}, because one of the 6/5's can be lowered by S6 to 7/6 and another of the 6/5's can be raised by S5 to 5/4. Then because we have tempered S5 and S6 together, we have 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.
From this, it can be deduced that {{nowrap| (6/5)<sup>3</sup> ~ [[7/4]] }}, because one of the 6/5's can be lowered by S6 to 7/6 and another of the 6/5's can be raised by S5 to 5/4. Then because we have tempered S5 and S6 together, we have 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.


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:
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:


$$ \frac{k+2}{k+1} \leftarrow S(k+1)~Sk \rightarrow \frac{k+1}{k} \leftarrow S(k+1)~Sk \rightarrow \frac{k}{k-1} $$
$$ \frac{k+2}{k+1} \leftarrow S(k+1)~Sk \rightarrow \frac{k+1}{k} \leftarrow S(k+1)~Sk \rightarrow \frac{k}{k-1} $$
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$$ \frac{L/M}{M/s} = \frac{ \left(\frac{k}{k-2}\right)/\left(\frac{k+1}{k-1}\right) }{ \left(\frac{k+1}{k-1}\right)/\left(\frac{k+2}{k}\right) } = \frac{Ls}{M^2} = \frac{\frac{k+2}{k-2}}{\left(\frac{k+1}{k-1}\right)^2} $$
$$ \frac{L/M}{M/s} = \frac{ \left(\frac{k}{k-2}\right)/\left(\frac{k+1}{k-1}\right) }{ \left(\frac{k+1}{k-1}\right)/\left(\frac{k+2}{k}\right) } = \frac{Ls}{M^2} = \frac{\frac{k+2}{k-2}}{\left(\frac{k+1}{k-1}\right)^2} $$


… 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.  
… 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.  


Also note that in the above, (''k'' + 1)/(''k'' - 1) is the mediant of the adjacent two intervals, meaning that division of an interval into two via tempering out a semiparticular is in some sense 'optimal' relative to the complexity. This also means that if ''k'' is a multiple of 2, this corresponds to a natural way to split the square superparticular S(''k''/2) into two parts. For example, if ''k'' = 10, then we have (10 + 2)/10, (10 + 1)/(10 - 1), 10/(10 - 2) as equidistant, which simplified is 6/5, 11/9, 5/4, with 11/9 being the mediant of 6/5 and 5/4, and therefore the corresponding superparticular S5 = (5/4)/(6/5) is split into two parts which are tempered together: (5/4)/(11/9) = 45/44 and (11/9)/(6/5) = 55/54. The semiparticular is therefore S(10 - 1)/S(10 + 1) = S9/S11 = 243/242 = (45/44)/(55/54) = ((10 + 2)/(10 - 2))/((10 + 1)/(10 - 1))<sup>2</sup>.
Also note that in the above, (''k'' + 1)/(''k'' - 1) is the mediant of the adjacent two intervals, meaning that division of an interval into two via tempering out a semiparticular is in some sense 'optimal' relative to the complexity. This also means that if ''k'' is a multiple of 2, this corresponds to a natural way to split the square superparticular S(''k''/2) into two parts. For example, if ''k'' = 10, then we have (10 + 2)/10, (10 + 1)/(10 - 1), 10/(10 - 2) as equidistant, which simplified is 6/5, 11/9, 5/4, with 11/9 being the mediant of 6/5 and 5/4, and therefore the corresponding superparticular S5 = (5/4)/(6/5) is split into two parts which are tempered together: (5/4)/(11/9) = 45/44 and (11/9)/(6/5) = 55/54. The semiparticular is therefore S(10 - 1)/S(10 + 1) = S9/S11 = 243/242 = (45/44)/(55/54) = ((10 + 2)/(10 - 2))/((10 + 1)/(10 - 1))<sup>2</sup>.
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{{Note| While a lot of these have pages, not all of them do, although that does not mean they should not. A noticeable streak of commas currently without pages correspond to when dividing a superparticular interval implicates intervals from a higher [[prime limit]], as a surprising amount of 23-limit semiparticulars shown here already have pages. }}
{{Note| While a lot of these have pages, not all of them do, although that does not mean they should not. A noticeable streak of commas currently without pages correspond to when dividing a superparticular interval implicates intervals from a higher [[prime limit]], as a surprising amount of 23-limit semiparticulars shown here already have pages. }}


== S''k''<sup>2</sup>⋅S(''k'' + 1) and S(''k'' − 1)⋅S''k''<sup>2</sup> (lopsided commas) ==
== {{nowrap|S''k''/S(''k'' + 3)}} (three-particulars) ==
=== Significance ===
 
# Tempering out any two consecutive square-particulars, S''k'' and S(''k'' + 1), implies tempering out the two associated lopsided commas as well as the associated triangle-particular and ultraparticular, so the lopsided commas represent the general form of the highest-damage relations/consequences of doing so.
=== Significance ===
# If a comma (such as the diaschisma, [[2048/2025]]), admits an expression as a lopsided comma, it means that one is likely missing out on tempering opportunities by not also tempering out the square-particulars composing it (such as [[256/255|S16]] and [[289/288|S17]] in the case of the diaschisma), often involving expanding the subgroup and adding a number of new equivalence relations (as previously explained) while simultaneously making the temperament more efficient and more precise.
# It it surprising that this form corresponds to yet another infinite family to do with spacing of obviously-related intervals. It thusly also helps further motivate expressing comma lists in terms of chains of square-particulars that are equated, either via {{nowrap| S''x'' {{=}} S''y'' {{=}} S''z'' {{=}} ... }} or via {{nowrap| { S''x''/''y''/''z''/... }, with ''x'' < ''y'' < ''z'' }}.
# It is surprising that there are fairly simple general equivalence relations for these S-expressions, essentially being "free" to read off of an S-expression-based comma list, once you know the general form.
# Relatedly to the previous point, they are used in the construction of ''pentaparticulars'', which take the form S''k''/S(''k'' + 3) * ( S(''k'' + 1)/S(''k'' + 2) )<sup>2</sup>, which are equal to how (''k'' + 2)/(''k'' + 1) is approximately a fifth of (''k'' + 4)/(''k'' - 1), and are intuitively obvious from the idea of making superparticulars equidistant via ultraparticulars.
 
# More trivially, they are implied in various simple S-expression-based comma lists, like [[58edo]]'s [[17-limit]] as describable by {S6/S7, S8/9/10/11/13, S12, S14, S16, S17}, where S8/9/10/11/13 is a shorthand for [[64/63|S8]] = [[81/80|S9]] = [[100/99|S10]] = [[121/120|S11]] = [[169/168|S13]], wherein we have {{nowrap| three-particulars }} {{nowrap| { S8/S11 {{=}} ( (10/7)/(11/8) )/( (11/8)/(4/3) ), S10/S13 {{=}} ( (4/3)/(13/10) )/( (13/10)/(14/11) ), S14/S17 {{=}} ( (16/13)/(17/14) )/( (17/14)/(6/5) ) }. }}
=== Derivation of equivalence relation ===
# Three-particulars currently have no ''simple'' known splitting property due to the three intervals made equidistant not obviously composing to some other interval of significance, so seem to instead be about making exact spacings that feel approximately equidistant. For example, in 7:8:9:10:11:12, the closest is to notice the smallest (7:10) and largest (9:12) interval multiply to an interval which is 11/10 more than 7:12, and hence in general splits an interval (''k'' + 4)/(''k'' - 1) * (''k'' + 2)/(''k'' + 1) into two parts of (''k'' + 3)/''k''. Because the multiplication by a superparticular interval is a rather peculiar requirement, we leave it here as a note.
Using the clarity of [[#Using S-factorizations to understand the significance of S-expressions|S-factorizations]], we can show the interval relations implicated by these two new "lopsided" forms, which will make clear the reason for their name:
 
 
=== Derivation ===
S''k''<sup>2</sup>⋅S(''k'' + 1) = [''k'' - 1, ''k'', ''k'' + 1, ''k'' + 2]^(2[-1, 2, -1, 0] + [0, -1, 2, -1] = [-2, 4, -2, 0] + [0, -1, 2, -1] = [-2, 3, 0, -1]) implies:
For a general introduction to method we use here, see [[#Using S-factorizations to understand the significance of S-expressions|S-factorizations]], as I will not re-explain the method here (though I find it quite straightforward).
 
 
S''k''<sup>2</sup>⋅S(''k'' + 1) = (''k''/(''k'' - 1))<sup>2</sup> / ((''k'' + 2)/''k'') through [-2, 3, 0, -1] = [-2, 2, 0, 0] - [0, -1, 0, 1].
We start with:
 
<pre>
S(''k'' - 1)⋅S''k''<sup>2</sup> = [''k'' - 2, ''k'' - 1, ''k'', ''k'' + 1]^([-1, 2, -1, 0] + 2[0, -1, 2, -1] = [-1, 2, -1, 0] + [0, -2, 4, -2] = [-1, 0, 3, -2]) implies:
Sk = [k-1, k, k+1]^[-1, 2, -1]
 
</pre>
S(''k'' - 1)⋅S''k''<sup>2</sup> = (''k''/(''k'' - 2)) / ((''k'' + 1)/''k'')<sup>2</sup> through [-1, 0, 3, -2] = [-1, 0, 1, 0] - [0, 0, -2, 2].
From which we can examine
 
<pre>
=== Tables ===
Sk / S(k+3) = [k-1, k, k+1]^[-1, 2, -1] / [k+2, k+3, k+4]^[-1, 2, -1]
Below are two tables of [[43-limit]] lopsided commas. First, the "top heavy" lopsided commas, where the squared interval is in the numerator, then the "bottom heavy" lopsided commas, where the squared interval is in the denominator. These tables are so big because these commas are quite large, so the more interesting commas appear later. For this reason and for completeness, the tables show up to until a little past the largest known lopsided commas that have their own page: the [[olympia]] and the [[phaotic comma]].
= [k-1, k, k+1]^[-1, 2, -1] * [k+2, k+3, k+4]^[1, -2, 1]
 
= [k-1, k, k+1, k+2, k+3, k+4]^[-1, 2, -1, 1, -2, 1]
==== Top-heavy lopsided commas ====
</pre>
{| class="wikitable center-all
It is initially not obvious that this corresponds to anything interesting. However, as discovered by [[Lériendil]] (though not by this method), tempering out the comma makes (''k'' + 2)/(''k'' - 1), (''k'' + 3)/''k'', (''k'' + 4)/(''k'' + 1) equidistant, which is not arcane to understand either; we have three intervals of the form (''a'' + 3)/''a'' for adjacent values of ''a'' made equidistant, and the placement of ''a'' is obvious w.r.t the S-expression, being "the only place it could be" (in the center). Thus, we will show that the S-factorization derived from making these equidistant is equivalent:
|-
<pre>
! S-expression
let F(k + 1) = ( (k + 2)/(k-1) )/( (k + 3)/k ) = [k-1, k, k+1, k+2, k+3]^[-1, 1, 0, 1, -1]
! Square relation
 
! Ratio
let F(k + 2) = ( (k + 3)/k )/( (k + 4)/(k + 1) ) = [k, k+1, k+2, k+3, k+4]^[-1, 1, 0, 1, -1]
|-
 
| S2<sup>2</sup>⋅S3 = ([[3/2]])⋅([[4/3]])
then F(k + 1)/F(k + 2) = [k-1, k, k+1, k+2, k+3, k+4]^([-1, 1, 0, 1, -1, 0] - [0, -1, 1, 0, 1, -1])
| ([[2/1]])<sup>2</sup>/([[2/1]])
= [k-1, k, k+1, k+2, k+3, k+4]^( [-1, 1, 0, 1, -1, 0]
| [[2/1]]
                                + [0, 1, -1, 0, -1, 1] )
|-
= [k-1, k, k+1, k+2, k+3, k+4]^[ -1, 2, -1, 1, -2, 1 ]
| S3<sup>2</sup>⋅S4 = ([[6/5]])⋅([[9/8]])
</pre>
| ([[3/2]])<sup>2</sup>/([[5/3]])
...as expected. In fact, if you look at the expressions in the table, it becomes obvious why we have the central element squared (via a 2 and -2 in the S-factorization), as it follows from the general algebraic form of {{nowrap| ''n''<sup>2</sup> / (''n''<sup>2</sup> - 1) {{=}} ( ''n''/(''n'' - 1) )/( (''n'' + 1)/''n'' ) }}.
| [[27/20]]
 
|-
'''(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]]'''.)'''
| S4<sup>2</sup>⋅S5 = ([[10/9]])⋅([[16/15]])
 
| ([[4/3]])<sup>2</sup>/([[3/2]])
=== Table of three-particulars ===
| [[32/27]]
Beyond the obvious range, 43-limit three-particulars become very sparse, hence S286/S289 is the first to break this pattern.
|-
 
| S5<sup>2</sup>⋅S6 = ([[15/14]])⋅([[25/24]])
However, as it concerns intervals so small they can be considered commas, we do not extend the table further, as it's equal to {{sfrac| ([[96/95]])/([[289/286]]) | ([[289/286]])/([[290/287]]) }}.
| ([[5/4]])<sup>2</sup>/([[7/5]])
 
| [[125/112]]
(The 43-limit is picked due to 47/32 being close to 3/2, and 43/32 being close to 4/3, as well as to avoid spam.)
|-
{| class="wikitable center-all
| S6<sup>2</sup>⋅S7 = ([[21/20]])⋅([[36/35]])
|-
| ([[6/5]])<sup>2</sup>/([[4/3]])
! S-expression
| [[27/25]]
! Square relation
! Ratio
! Subgroup
|-
| S2/S5
| ( ([[4/1]])/([[5/2]]) )/( ([[5/2]])/([[2/1]]) )
| [[32/25]]
| 2.5
|-
| S3/S6
| ( ([[5/2]])/([[2/1]]) )/( ([[2/1]])/([[7/4]]) )
| [[35/32]]
| 2.5.7
|-
| S4/S7
| ( ([[2/1]])/([[7/4]]) )/( ([[7/4]])/([[8/5]]) )
| [[256/245]]
| 2.5.7
|-
| S5/S8
| ( ([[7/4]])/([[8/5]]) )/( ([[8/5]])/([[3/2]]) )
| [[525/512]]
| 2.3.5.7
|-
| S6/S9
| ( ([[8/5]])/([[3/2]]) )/( ([[3/2]])/([[10/7]]) )
| [[64/63]]
| 2.3.7
|-
| S7/S10
| ( ([[3/2]])/([[10/7]]) )/( ([[10/7]])/([[11/8]]) )
| [[1617/1600]]
| 2.3.5.7.11
|-
| S8/S11
| ( ([[10/7]])/([[11/8]]) )/( ([[11/8]])/([[4/3]]) )
| [[2560/2541]]
| 2.3.5.7.11
|-
| S9/S12
| ( ([[11/8]])/([[4/3]]) )/( ([[4/3]])/([[13/10]]) )
| [[1287/1280]]
| 2.3.5.11.13
|-
| S10/S13
| ( ([[4/3]])/([[13/10]]) )/( ([[13/10]])/([[14/11]]) )
| [[5600/5577]]
| 2.3.5.7.11.13
|-
| S11/S14
| ( ([[13/10]])/([[14/11]]) )/( ([[14/11]])/([[5/4]]) )
| [[1573/1568]]
| 2.7.11.13
|-
| S12/S15
| ( ([[14/11]])/([[5/4]]) )/( ([[5/4]])/([[16/13]]) )
| [[3584/3575]]
| 2.5.7.11.13
|-
| S13/S16
| ( ([[5/4]])/([[16/13]]) )/( ([[16/13]])/([[17/14]]) )
| [[14365/14336]]
| 2.5.7.13.17
|-
| S14/S17
| ( ([[16/13]])/([[17/14]]) )/( ([[17/14]])/([[6/5]]) )
| [[18816/18785]]
| 2.3.5.7.13.17
|-
| S15/S18
| ( ([[17/14]])/([[6/5]]) )/( ([[6/5]])/([[19/16]]) )
| [[8075/8064]]
| 2.3.5.7.17.19
|-
| S16/S19
| ( ([[6/5]])/([[19/16]]) )/( ([[19/16]])/([[20/17]]) )
| [[6144/6137]]
| 2.3.17.19
|-
| S17/S20
| ( ([[19/16]])/([[20/17]]) )/( ([[20/17]])/([[7/6]]) )
| [[38437/38400]]
| 2.3.5.7.17.19
|-
| S18/S21
| ( ([[20/17]])/([[7/6]]) )/( ([[7/6]])/([[22/19]]) )
| [[15840/15827]]
| 2.3.5.7.11.17.19
|-
| S19/S22
| ( ([[7/6]])/([[22/19]]) )/( ([[22/19]])/([[23/20]]) )
| [[58121/58080]]
| 2.3.5.7.11.19.23
|-
| S20/S23
| ( ([[22/19]])/([[23/20]]) )/( ([[23/20]])/([[8/7]]) )
| [[70400/70357]]
| 2.5.7.11.19.23
|-
| S21/S24
| ( ([[23/20]])/([[8/7]]) )/( ([[8/7]])/([[25/22]]) )
| [[5635/5632]]
| 2.5.7.11.23
|-
| S22/S25
| ( ([[8/7]])/([[25/22]]) )/( ([[25/22]])/([[26/23]]) )
| [[100672/100625]]
| 2.5.7.11.13.23
|-
| S23/S26
| ( ([[25/22]])/([[26/23]]) )/( ([[26/23]])/([[9/8]]) )
| [[119025/118976]]
| 2.3.5.11.13.23
|-
| S24/S27
| ( ([[26/23]])/([[9/8]]) )/( ([[9/8]])/([[28/25]]) )
| [[46592/46575]]
| 2.3.5.7.13.23
|-
| S25/S28
| ( ([[9/8]])/([[28/25]]) )/( ([[28/25]])/([[29/26]]) )
| [[163125/163072]]
| 2.3.5.7.13.29
|-
| S26/S29
| ( ([[28/25]])/([[29/26]]) )/( ([[29/26]])/([[10/9]]) )
| [[37856/37845]]
| 2.3.5.7.13.29
|-
| S27/S30
| ( ([[29/26]])/([[10/9]]) )/( ([[10/9]])/([[31/28]]) )
| [[72819/72800]]
| 2.3.5.7.13.29.31
|-
| S28/S31
| ( ([[10/9]])/([[31/28]]) )/( ([[31/28]])/([[32/29]]) )
| [[250880/250821]]
| 2.3.5.7.29.31
|-
| S29/S32
| ( ([[31/28]])/([[32/29]]) )/( ([[32/29]])/([[11/10]]) )
| [[286781/286720]]
| 2.5.7.11.29.31
|-
| S30/S33
| ( ([[32/29]])/([[11/10]]) )/( ([[11/10]])/([[34/31]]) )
| [[108800/108779]]
| 2.5.11.17.29.31
|-
| S31/S34
| ( ([[11/10]])/([[34/31]]) )/( ([[34/31]])/([[35/32]]) )
| [[73997/73984]]
| 2.7.11.17.31
|-
| S32/S35
| ( ([[34/31]])/([[35/32]]) )/( ([[35/32]])/([[12/11]]) )
| [[417792/417725]]
| 2.3.5.7.11.17.31
|-
| S33/S36
| ( ([[35/32]])/([[12/11]]) )/( ([[12/11]])/([[37/34]]) )
| [[156695/156672]]
| 2.3.5.7.11.17.37
|-
| S34/S37
| ( ([[12/11]])/([[37/34]]) )/( ([[37/34]])/([[38/35]]) )
| [[527136/527065]]
| 2.3.5.7.11.17.19.37
|-
| S35/S38
| ( ([[37/34]])/([[38/35]]) )/( ([[38/35]])/([[13/12]]) )
| [[589225/589152]]
| 2.3.5.7.13.17.19.37
|-
| S36/S39
| ( ([[38/35]])/([[13/12]]) )/( ([[13/12]])/([[40/37]]) )
| [[43776/43771]]
| 2.3.7.13.19.37
|-
| S37/S40
| ( ([[13/12]])/([[40/37]]) )/( ([[40/37]])/([[41/38]]) )
| [[729677/729600]]
| 2.3.5.13.19.37.41
|-
| S38/S41
| ( ([[40/37]])/([[41/38]]) )/( ([[41/38]])/([[14/13]]) )
| [[808640/808561]]
| 2.5.7.13.19.37.41
|-
| S39/S42
| ( ([[41/38]])/([[14/13]]) )/( ([[14/13]])/([[43/40]]) )
| [[297947/297920]]
| 2.5.7.13.19.41.43
|-
| S40/S43
| ( ([[14/13]])/([[43/40]]) )/( ([[43/40]])/([[44/41]]) )
| [[985600/985517]]
| 2.5.7.11.13.41.43
|-
| S41/S44
| ( ([[43/40]])/([[44/41]]) )/( ([[44/41]])/([[15/14]]) )
| [[216849/216832]]
| 2.3.7.11.41.43
|-
| S42/S45
| ( ([[44/41]])/([[15/14]]) )/( ([[15/14]])/([[46/43]]) )
| [[396704/396675]]
| 2.3.5.7.11.23.41.43
|-
| S286/S289
| ( ([[96/95]])/([[289/286]]) )/( ([[289/286]])/([[290/287]]) )
| [[455440128/455440013]]
| 2.3.7.11.13.17.19.29.41
|}
 
== {{nowrap|(S''k''/S(''k'' + 1))<sup>2</sup> * S(''k'' - 1)/S(''k'' + 2)}} (pentaparticulars) ==
 
=== Significance ===
# Pentaparticulars represent the obvious way of splitting an interval (''k'' + 3)/(''k'' - 2) into five parts of (''k'' + 1)/''k'', so represent a generalization of ultraparticulars by observing the harmonic series chord {{nowrap|''k''-2 : ''k''-1 : ''k'' : ''k''+1 : ''k''+2 : ''k''+3.}}
# Interestingly, while three-particulars are about equidistance but (currently) have no (clean) known splitting property (due to the three intervals made equidistant not obviously composing to some other significant interval), pentaparticulars are in a sense symmetric to this and related algebraically, by instead being specifically about splitting.
# They are obviously implied by any system that equates S''k'' with S(''k'' + 1) and S(''k'' - 1) with S(''k'' + 2) simultaneously, so are not as uncommon as might be guessed, but tend to be accurate equivalences, and 5 is a rather specific number of parts to divide an interval into, even if obviously the most natural for an interval whose numerator is 5 more than the denominator (up to simplification).
 
=== Derivation ===
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 I will not re-explain the method here. However, we will use a previous result from three-particulars as to immediately say that:
<pre>
Sk / S(k+3) = [k-1, k, k+1, k+2, k+3, k+4]^[-1, 2, -1, 1, -2, 1]
</pre>
and also relevant to our problem are
<pre>
S(k+1) / S(k+2) = [k, k+1, k+2]^[-1, 2, -1] / [k+1, k+2, k+3]^[-1, 2, -1]
= [k, k+1, k+2, k+3]^( [-1, 2, -1, 0] - [0, -1, 2, -1] )
= [k, k+1, k+2, k+3]^[-1, 3, -3, 1]
pentaparticular((k+2)/(k+1)) = [k-1, k, k+1, k+2, k+3, k+4]^[-1, 0, -5, 5, 0, 1]
</pre>
Then, dividing the pentaparticular by S''k''/S(''k'' + 3) yields:
<pre>
[k-1, k, k+1, k+2, k+3, k+4]^( [-1, 0, 5, -5, 0, 1]
                              - [-1, 2, -1, 1, -2, 1] )
= [k-1, k, k+1, k+2, k+3, k+4]^[0, -2, 6, -6, 2, 0]
= ( [k, k+1, k+2, k+3]^[-1, 3, -3, 1] )^2 = ( S(k+1) / S(k+2) )^2
</pre>
 
Thus proving the general form, which is for any given superparticular, the corresponding pentaparticular is equal to the square of the ultraparticular times the correspondingly-centered three-particular.
 
=== Table of pentaparticulars ===
Up to what is currently the largest pentaparticular on the wiki.
{| class="wikitable center-all
|-
! S-expression
! Relation
! Ratio
! Subgroup
|-
| ([[135/128|S3/S4]])<sup>2</sup> * [[32/25|S2/S5]]
| ([[6/1]])/([[4/3]])<sup>5</sup>
| [[729/512]]
| 2.3
|-
| ([[128/125|S4/S5]])<sup>2</sup> * [[35/32|S3/S6]]
| ([[7/2]])/([[5/4]])<sup>5</sup>
| [[3584/3125]]
| 2.5.7
|-
| ([[875/864|S5/S6]])<sup>2</sup> * [[256/245|S4/S7]]
| ([[8/3]])/([[6/5]])<sup>5</sup>
| [[3125/2916]]
| 2.3.5
|-
| ([[1728/1715|S6/S7]])<sup>2</sup> * [[525/512|S5/S8]]
| ([[9/4]])/([[7/6]])<sup>5</sup>
| [[17496/16807]]
| 2.3.7
|-
| ([[1029/1024|S7/S8]])<sup>2</sup> * [[64/63|S6/S9]]
| ([[2/1]])/([[8/7]])<sup>5</sup>
| [[16807/16384]]
| 2.7
|-
| ([[5120/5103|S8/S9]])<sup>2</sup> * [[1617/1600|S7/S10]]
| ([[11/6]])/([[9/8]])<sup>5</sup>
| [[180224/177147]]
| 2.3.11
|-
| ([[8019/8000|S9/S10]])<sup>2</sup> * [[2560/2541|S8/S11]]
| ([[12/7]])/([[10/9]])<sup>5</sup>
| [[177147/175000]]
| 2.3.5.7
|-
| ([[4000/3993|S10/S11]])<sup>2</sup> * [[1287/1280|S9/S12]]
| ([[13/8]])/([[11/10]])<sup>5</sup>
| [[162500/161051]]
| 2.5.11.13
|-
| ([[17303/17280|S11/S12]])<sup>2</sup> * [[5600/5577|S10/S13]]
| ([[14/9]])/([[12/11]])<sup>5</sup>
| [[1127357/1119744]]
| 2.3.7.11
|-
| ([[24192/24167|S12/S13]])<sup>2</sup> * [[1573/1568|S11/S14]]
| ([[3/2]])/([[13/12]])<sup>5</sup>
| [[373248/371293]]
| 2.3.13
|-
| ([[10985/10976|S13/S14]])<sup>2</sup> * [[3584/3575|S12/S15]]
| ([[16/11]])/([[14/13]])<sup>5</sup>
| [[371293/369754]]
| 2.7.11.13
|-
| ([[43904/43875|S14/S15]])<sup>2</sup> * [[14365/14336|S13/S16]]
| ([[17/12]])/([[15/14]])<sup>5</sup>
| [[2285752/2278125]]
| 2.3.5.7.17
|-
| ([[57375/57344|S15/S16]])<sup>2</sup> * [[18816/18785|S14/S17]]
| ([[18/13]])/([[16/15]])<sup>5</sup>
| [[6834375/6815744]]
| 2.3.5.13
|-
| ([[24576/24565|S16/S17]])<sup>2</sup> * [[8075/8064|S15/S18]]
| ([[19/14]])/([[17/16]])<sup>5</sup>
| [[9961472/9938999]]
| 2.7.17.19
|-
| ([[93347/93312|S17/S18]])<sup>2</sup> * [[6144/6137|S16/S19]]
| ([[4/3]])/([[18/17]])<sup>5</sup>
| [[1419857/1417176]]
| 2.3.17
|-
| ([[116640/116603|S18/S19]])<sup>2</sup> * [[38437/38400|S17/S20]]
| ([[21/16]])/([[19/18]])<sup>5</sup>
| [[2480058/2476099]]
| 2.3.7.19
|-
| ([[48013/48000|S19/S20]])<sup>2</sup> * [[15840/15827|S18/S21]]
| ([[22/17]])/([[20/19]])<sup>5</sup>
| [[27237089/27200000]]
| 2.5.11.17.19
|-
| ([[176000/175959|S20/S21]])<sup>2</sup> * [[58121/58080|S19/S22]]
| ([[23/18]])/([[21/20]])<sup>5</sup>
| [[36800000/36756909]]
| 2.3.5.7.23
|-
| ([[213003/212960|S21/S22]])<sup>2</sup> * [[70400/70357|S20/S23]]
| ([[24/19]])/([[22/21]])<sup>5</sup>
| [[12252303/12239876]]
| 2.3.7.11.19
|-
| ([[85184/85169|S22/S23]])<sup>2</sup> * [[5635/5632|S21/S24]]
| ([[5/4]])/([[23/22]])<sup>5</sup>
| [[6442040/6436343]]
| 2.5.11.23
|-
| ([[304175/304128|S23/S24]])<sup>2</sup> * [[100672/100625|S22/S25]]
| ([[26/21]])/([[24/23]])<sup>5</sup>
| [[83672459/83607552]]
| 2.3.7.13.23
|-
| ([[359424/359375|S24/S25]])<sup>2</sup> * [[119025/118976|S23/S26]]
| ([[27/22]])/([[25/24]])<sup>5</sup>
| [[107495424/107421875]]
| 2.3.5.11
|-
| ([[140625/140608|S25/S26]])<sup>2</sup> * [[46592/46575|S24/S27]]
| ([[28/23]])/([[26/25]])<sup>5</sup>
| [[68359375/68317912]]
| 2.5.7.13.23
|-
| ([[492128/492075|S26/S27]])<sup>2</sup> * [[163125/163072|S25/S28]]
| ([[29/24]])/([[27/26]])<sup>5</sup>
| [[43069988/43046721]]
| 2.3.13.29
|-
| ([[570807/570752|S27/S28]])<sup>2</sup> * [[37856/37845|S26/S29]]
| ([[6/5]])/([[28/27]])<sup>5</sup>
| [[43046721/43025920]]
| 2.3.5.7
|-
| ([[219520/219501|S28/S29]])<sup>2</sup> * [[72819/72800|S27/S30]]
| ([[31/26]])/([[29/28]])<sup>5</sup>
| [[266760704/266644937]]
| 2.7.13.29.31
|-
| ([[756059/756000|S29/S30]])<sup>2</sup> * [[250880/250821|S28/S31]]
| ([[32/27]])/([[30/29]])<sup>5</sup>
| [[20511149/20503125]]
| 3.5.29
|-
| ([[864000/863939|S30/S31]])<sup>2</sup> * [[286781/286720|S29/S32]]
| ([[33/28]])/([[31/30]])<sup>5</sup>
| [[200475000/200404057]]
| 2.3.5.7.11.31
|-
| ([[327701/327680|S31/S32]])<sup>2</sup> * [[108800/108779|S30/S33]]
| ([[34/29]])/([[32/31]])<sup>5</sup>
| [[486695567/486539264]]
| 2.17.29.31
|-
| ([[1114112/1114047|S32/S33]])<sup>2</sup> * [[73997/73984|S31/S34]]
| ([[7/6]])/([[33/32]])<sup>5</sup>
| [[117440512/117406179]]
| 2.3.7.11
|}
 
== S''k''<sup>2</sup>⋅S(''k'' + 1) and S(''k'' − 1)⋅S''k''<sup>2</sup> (lopsided commas) ==
=== Significance ===
# Tempering out any two consecutive square-particulars, S''k'' and S(''k'' + 1), implies tempering out the two associated lopsided commas as well as the associated triangle-particular and ultraparticular, so the lopsided commas represent the general form of the highest-damage relations/consequences of doing so.
# If a comma (such as the diaschisma, [[2048/2025]]), admits an expression as a lopsided comma, it means that one is likely missing out on tempering opportunities by not also tempering out the square-particulars composing it (such as [[256/255|S16]] and [[289/288|S17]] in the case of the diaschisma), often involving expanding the subgroup and adding a number of new equivalence relations (as previously explained) while simultaneously making the temperament more efficient and more precise.
# It is surprising that there are fairly simple general equivalence relations for these S-expressions, essentially being "free" to read off of an S-expression-based comma list, once you know the general form.
 
=== Derivation of equivalence relation ===
Using the clarity of [[#Using S-factorizations to understand the significance of S-expressions|S-factorizations]], we can show the interval relations implicated by these two new "lopsided" forms, which will make clear the reason for their name:
 
S''k''<sup>2</sup>⋅S(''k'' + 1) = [''k'' - 1, ''k'', ''k'' + 1, ''k'' + 2]^(2[-1, 2, -1, 0] + [0, -1, 2, -1] = [-2, 4, -2, 0] + [0, -1, 2, -1] = [-2, 3, 0, -1]) implies:
 
S''k''<sup>2</sup>⋅S(''k'' + 1) = (''k''/(''k'' - 1))<sup>2</sup> / ((''k'' + 2)/''k'') through [-2, 3, 0, -1] = [-2, 2, 0, 0] - [0, -1, 0, 1].
 
S(''k'' - 1)⋅S''k''<sup>2</sup> = [''k'' - 2, ''k'' - 1, ''k'', ''k'' + 1]^([-1, 2, -1, 0] + 2[0, -1, 2, -1] = [-1, 2, -1, 0] + [0, -2, 4, -2] = [-1, 0, 3, -2]) implies:
 
S(''k'' - 1)⋅S''k''<sup>2</sup> = (''k''/(''k'' - 2)) / ((''k'' + 1)/''k'')<sup>2</sup> through [-1, 0, 3, -2] = [-1, 0, 1, 0] - [0, 0, -2, 2].
 
=== Tables ===
Below are two tables of [[43-limit]] lopsided commas. First, the "top heavy" lopsided commas, where the squared interval is in the numerator, then the "bottom heavy" lopsided commas, where the squared interval is in the denominator. These tables are so big because these commas are quite large, so the more interesting commas appear later. For this reason and for completeness, the tables show up to until a little past the largest known lopsided commas that have their own page: the [[olympia]] and the [[phaotic comma]].
 
==== Top-heavy lopsided commas ====
{| class="wikitable center-all
|-
! S-expression
! Square relation
! Ratio
|-
| S2<sup>2</sup>⋅S3 = ([[3/2]])⋅([[4/3]])
| ([[2/1]])<sup>2</sup>/([[2/1]])
| [[2/1]]
|-
| S3<sup>2</sup>⋅S4 = ([[6/5]])⋅([[9/8]])
| ([[3/2]])<sup>2</sup>/([[5/3]])
| [[27/20]]
|-
| S4<sup>2</sup>⋅S5 = ([[10/9]])⋅([[16/15]])
| ([[4/3]])<sup>2</sup>/([[3/2]])
| [[32/27]]
|-
| S5<sup>2</sup>⋅S6 = ([[15/14]])⋅([[25/24]])
| ([[5/4]])<sup>2</sup>/([[7/5]])
| [[125/112]]
|-
| S6<sup>2</sup>⋅S7 = ([[21/20]])⋅([[36/35]])
| ([[6/5]])<sup>2</sup>/([[4/3]])
| [[27/25]]
|-
|-
| S7<sup>2</sup>⋅S8 = ([[28/27]])⋅([[49/48]])
| S7<sup>2</sup>⋅S8 = ([[28/27]])⋅([[49/48]])
Line 2,876: Line 3,326:


== Equivalent S-expressions ==
== Equivalent S-expressions ==
All S-expressions have other equivalent S-expressions; however, when the equivalence makes one comma a member of two of the infinite families discussed above, or otherwise makes it equal to a product or ratio between two such commas, 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''.
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''.


=== A useful general rule ===
=== A useful general rule ===
Line 2,885: Line 3,335:
$$
$$


This is important to note because using this simple rule we can derive an infinite amount of trivially and obviously equivalent S-expressions. See [[S-expression/Advanced results]] for mathematical details.
This is important to note because using this simple rule we can derive an infinite amount of trivial equivalent S-expressions. See [[S-expression/Advanced results]] for mathematical details.


=== Examples ===
=== Examples ===
Line 2,909: Line 3,359:
| [[176/175]]
| [[176/175]]
| S8/S10, S22⋅S23⋅S24
| S8/S10, S22⋅S23⋅S24
|-
|[[225/224]]
|S15, S25*S26*S27
|-
|-
| [[243/242]]
| [[243/242]]
Line 2,921: Line 3,374:
| [[676/675]]
| [[676/675]]
| S26, S13/S15
| S26, S13/S15
|-
|[[1216/1215]]
|S16/S18, S64*S65*S76*S77
|-
|-
| [[1225/1224]]
| [[1225/1224]]
| S35, S49⋅S50
| S35, S49⋅S50
|-
|[[2080/2079]]
|S64⋅S65, S78⋅S79⋅S80
|-
|-
| [[2601/2600]]
| [[2601/2600]]
Line 2,930: Line 3,389:
| [[3025/3024]]
| [[3025/3024]]
| S55, S22/S24, (S25/S27)⋅S99
| S55, S22/S24, (S25/S27)⋅S99
|-
|[[3136/3135]]
|S56, S96*S97*S98
|-
|[[4000/3993]]
|S10/S11, (S12/S14)/S99, S25*(S64*S65)/S55
|-
|[[4375/4374]]
|S25/S27, S55/S99
|-
|[[6656/6655]]
|(S64*S65)/S55, S64*S351/S99
|-
|-
| [[9801/9800]]
| [[9801/9800]]
| S99, S33/S35
| S99, S33/S35
|-
|[[10985/10976]]
|S13/S14, (S64*S65^2)*S99
|-
|-
| [[25921/25920]]
| [[25921/25920]]
| S161, S46/S48
| S161, S46/S48
|-
|[[32805/32768]]
|S15/(S8/S9), S19/(S16/S18)^2
|-
|[[43904/43875]]
|S14/S15, S49*S64, S56*S76*S77
|-
|-
| <small>[[123201/123200]]</small>
| <small>[[123201/123200]]</small>
| S351, S78/S80
| S351, S78/S80, S25/S27/S65
|}
|}