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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=== Significance ===
=== Significance ===
# Tempering out any two consecutive square-particulars S''k'' and S({{nowrap|''k'' + 1}}) will naturally imply tempering out the ultraparticular between them, {{sfrac|S''k''|S(''k'' + 1)}}, meaning they are very common implicit commas.
# Tempering out any two consecutive square-particulars S''k'' and S({{nowrap|''k'' + 1}}) will naturally imply tempering out the ultraparticular between them, {{sfrac|S''k''|S(''k'' + 1)}}, meaning they are very common implicit commas.
# 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.{{clarify}}
# 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''<sup>2</sup>⋅S(''k'' + 1)}} or {{nowrap|S''k''⋅S(''k'' + 1)<sup>2</sup>}} 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{{clarify}} 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, S16<sup>2</sup>⋅S17, S16⋅S17<sup>2</sup>} }}, and any of the two commas in the latter set imply all the other commas too.
# 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''<sup>2</sup>⋅S(''k'' + 1)}} or {{nowrap|S''k''⋅S(''k'' + 1)<sup>2</sup>}} 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, S16<sup>2</sup>⋅S17, S16⋅S17<sup>2</sup>} }}, and any of the two commas in the latter set imply all the other commas too.


=== Table of ultraparticulars ===
=== Table of ultraparticulars ===
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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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== 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 ===
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$$
$$


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 ===
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| [[176/175]]
| [[176/175]]
| S8/S10, S22⋅S23⋅S24
| S8/S10, S22⋅S23⋅S24
|-
|[[225/224]]
|S15, S25*S26*S27
|-
|-
| [[243/242]]
| [[243/242]]
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| [[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]]
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| [[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
|}
|}