Square superparticular
Sk (square-particulars)
A square superparticular, or square-particular for short, is a superparticular interval whose numerator is a square number, which is to say, a superparticular of the form
[math]\Large \frac {k^2}{k^2 - 1} = \frac {k/(k - 1)}{(k + 1)/k}[/math]
which is square-(super)particular k for a given integer k > 1. A suggested shorthand for this interval is Sk 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 temperaments. Note that this means S2 = 4/3 is the first musically meaningful square-particular, as S1 = 1/0.
Also note that we use the notation Skp to mean (Sk)p rather than S(kp) for convenience in the practical analysis of regular temperaments using S-expressions.
Significance/motivation
Square-particulars are important structurally because they are the intervals between consecutive superparticular intervals while simultaneously being superparticular themselves, which means that whether and how they are tempered tells us information about how well a temperament can represent the harmonic series up to the (k + 1)th harmonic, as well as the potential representational sacrifices that must be made from that point onward. In other words, understanding the mappings of Sk in a given temperament is equivalent to understanding the spacing of consecutive superparticular intervals, and thereby to understanding the way it represents (or tries to represent) the harmonic series.
Table of square-particulars
Below is a table of 23-limit square-particulars:
Alternatives to tempering square-particulars
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 Uk = Sk/S(k + 1), we get ultraparticulars*. Ultraparticulars have a secondary (and mathematically equivalent) consequence: Because (k + 2)/(k + 1) and k/(k - 1) are equidistant from (k + 1)/k (because of tempering Sk/S(k + 1)), this means that another expression for Sk/S(k + 1) is the following:
[math]\large {\rm S}k / {\rm S} (k + 1) = \frac{(k + 2) / (k - 1)}{((k + 1)/k)^3}[/math]
This means you can read the k and k + 1 from the S-expression of an ultraparticular as being the interval involved in the cubing equivalence (abbreviated to "cube relation" in the table of ultraparticulars).
* 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 Uk = Sk/S(k - 1) and Uk = 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.
Furthermore, defining another sequence of commas with formula Sk/S(k + 2) leads to semiparticulars which inform many natural ways in which one might want to halve intervals with other intervals, and with their own more structural consequences, talked about there. These also arise from tempering consecutive ultraparticulars.
S-expressions
An S-expression is any product, or ratio of products, of square superparticulars Sk.
Sk*S(k + 1) (triangle-particulars)
Significance
1. Every triangle-particular is superparticular, so these are efficient commas. (See also the #Short proof of the superparticularity of triangle-particulars.)
2. Often each individual triangle-particular, taken as a comma, implies other useful equivalences not necessarily corresponding to the general form, speaking of which...
3. Every triangle-particular is the difference between two nearly-adjacent superparticular intervals (k + 2)/(k + 1) and k/(k - 1).
4. Tempering any two consecutive square-particulars Sk and S(k + 1) implies tempering a triangle-particular, so these are common commas. (See also: lopsided commas.)
5. If we temper Sk * S(k + 1) but not Sk or S(k + 1), then one or more intervals of k/(k - 1), (k + 1)/k and (k + 2)/(k + 1) must be mapped inconsistently, because:
- if (k + 1)/k is mapped above (k + 2)/(k + 1) ~ k/(k-1) we have (k + 1)/k > k/(k - 1) and if it is mapped below we have (k + 1)/k < (k + 2)/(k + 1).
- (Generalisations of this and their implications for consistency are discussed in #Sk*S(k + 1)*...*S(k + n - 1) (1/n-square-particulars).)
Meaning
Notice that if we equate (k + 2)/(k + 1) with k/(k - 1) (by tempering out their difference), then multiply both sides by (k + 1)/k, we have:
(k + 2)/(k + 1) * (k + 1)/k = (k + 1)/k * k/(k - 1)
...which simplifies to: (k + 2)/k = (k + 1)/(k - 1).
This means that if we temper: [math] {\rm S}k \cdot {\rm S}(k+1) \large = \frac{k/(k-1)}{(k+1)/k} \cdot \frac{(k+1)/k}{(k+2)/(k+1)} = \frac{k/(k-1)}{(k+2)/(k+1)}[/math]
...then this equivalence is achieved. Note that there is little to no reason to not also temper Sk and S(k + 1) individually unless other considerations seem to force your hand.
Short proof of the superparticularity of triangle-particulars
Sk*S(k + 1) = ( k/(k - 1) )/( (k + 2)/(k + 1) ) = ( k(k + 1) )/( (k - 1)(k + 2) ) = (k2 + k)/(k2 + k - 2)
Then notice that k2 + 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 numbers! (Hence the alternative name "triangle-particular".)
Table of triangle-particulars
For completeness, all the intervals of this form are included, because of their structural importance for JI, and for the possibility of (in)consistency of mappings when tempered for the above reason.
| S-expression | Interval relation | Comma |
|---|---|---|
| S2*S3 | (3/1)/(2/1) | 3/2 |
| S3*S4 | (3/2)/(5/4) | 6/5 |
| S4*S5 | (4/3)/(6/5) | 10/9 |
| S5*S6 | (5/4)/(7/6) | 15/14 |
| S6*S7 | (6/5)/(8/7) | 21/20 |
| S7*S8 = S4/S6 | (7/6)(9/8) | 28/27 |
| S8*S9 = S6 | (8/7)/(10/9) | 36/35 |
| S9*S10 | (9/8)/(11/10) | 45/44 |
| S10*S11 | (10/9)/(12/11) | 55/54 |
| S11*S12 | (11/10)/(13/12) | 66/65 |
| S12*S13 | (12/11)/(14/13) | 78/77 |
| S13*S14 | (13/12)/(15/14) | 91/90 |
| S14*S15 | (14/13)/(16/15) | 105/104 |
| S15*S16 | (15/14)/(17/16) | 120/119 |
| S16*S17 | (16/15)/(18/17) | 136/135 |
| S17*S18 | (17/16)/(19/18) | 153/152 |
| S18*S19 | (18/17)/(20/19) | 171/170 |
| S19*S20 | (19/18)/(21/20) | 190/189 |
| S20*S21 | (20/19)/(22/21) | 210/209 |
| S21*S22 | (21/20)/(23/22) | 231/230 |
| S22*S23 | (22/21)/(24/23) | 253/252 |
| S23*S24 | (23/22)/(25/24) | 276/275 |
| S24*S25 | (24/23)/(26/25) | 300/299 |
| S25*S26 = S10/S12 | (25/24)/(27/26) | 325/324 |
| S26*S27 | (26/25)/(28/27) | 351/350 |
Also included are some higher-up 23-limit 1/2-square-particulars (as many of the prior intervals were quite large):
| S-expression | Interval relation | Comma |
|---|---|---|
| S33*S34 | (33/32)/(35/34) | 561/560 |
| S34*S35 | (34/33)/(36/35) | 595/594 |
| S49*S50 = S35 | (49/48)/(51/50) | 1225/1224 |
| S50*S51 | (50/49)/(52/51) | 1275/1274 |
| S55*S56 | (55/54)/(57/56) | 1540/1539 |
| S64*S65 | (64/63)/(66/65) | 2080/2079 |
| S76*S77 | (76/75)/(78/77) | 2926/2925 |
| S169*S170 | (169/168)/(171/170) | 14365/14364 |
| S208*S209 | (208/207)/(210/209) | 21736/21735 |
(Note: after 75, 76, 77, 78, streaks of four consecutive harmonics in the 23-limit become very sparse. The last few streaks are deeply related to the consistency and structure of 311edo, as 311edo can be described as the unique 23-limit temperament that tempers all triangle-particulars from 595/594 up to 21736/21735. It also tempers all the square-particulars composing those triangle-particulars with the exception of S169 and S170. It also maps the corresponding intervals of the 77-odd-limit consistently. 170/169 is the only place where the logic seems to 'break' as it is mapped to 2 steps instead of 3 meaning the mapping of that superparticular is inconsistent.)
S(k - 1)*Sk*S(k + 1) (1/3-square-particulars)
This section concerns commas of the form S(k - 1) * Sk * S(k + 1) = ( (k-1)/(k-2) )/( (k+2)/(k+1) ) which therefore do not (directly) involve the kth harmonic.
Significance
1. Two thirds of all 1/3-square-particulars are superparticular and the other third are throdd-particular, so these are efficient commas. (See also the #Proof of simplification of 1/3-square-particulars.)
2. They are often implied in a variety of ways by combinations of other commas discussed on this page.
3. Their omission of direct relation to the kth harmonic make them theoretically interesting and potentially useful. (The other type of comma on this page that does this is semiparticulars.)
4. Square-particulars, 1/2-square-particulars (a.k.a. triangle-particulars) and 1/3-square-particulars are part of a more general sequence with interesting properties: 1/n-square-particulars.
Proof of simplification of 1/3-square-particulars
This section concerns commas of the form S(k - 1) * Sk * S(k + 1) = ( (k-1)/(k-2) )/( (k+2)/(k+1) ) which therefore do not (directly) involve the kth harmonic. We can check their general algebraic expression for any potential simplifications:
S(k-1) * Sk * S(k+1) = ( (k-1)/(k-2) )/( k/(k-1) ) * ( k/(k-1) )/( (k+1)/k ) * ( (k+1)/k )/( (k+2)/(k+1) ) = ( (k-1)/(k-2) )/( (k+2)/(k+1) ) = (k-1)(k+1)/((k-2)(k+2)) = (k^2 - 1)/(k^2 - 4) if k=3n+1 then: S(k-1) * Sk * S(k+1) = (9n^2 + 6n)/(9n^2 + 6n - 3) = (3n^2 + 2n)/(3n^2 + 2n - 1) if k=3n+2 then: S(k-1) * Sk * S(k+1) = (9n^2 + 12n + 3)/(9n^2 + 12n) = (3n^2 + 4n + 1)/(3n^2 + 4n) if k=3n then: S(k-1) * Sk * S(k+1) = (9n^2 - 1)/(9n^2 - 4)
In other words, what this shows is all 1/3-square-particulars of the form S(k - 1) * Sk * S(k + 1) are superparticular iff k is throdd (not a multiple of 3), and all 1/3-square-particulars of the form S(3k - 1) * S(3k) * S(3k + 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).
Table of 1/3-square-particulars
Below is a table of such commas in the 41-prime-limited 199-odd-limit:
| S-expression | Interval relation | Comma |
|---|---|---|
| S2*S3*S4 | (2/1)/(5/4) | 8/5 |
| S3*S4*S5 | (3/2)/(6/5) | 5/4 |
| S4*S5*S6 | (4/3)/(7/6) | 8/7 |
| S5*S6*S7 | (5/4)/(8/7) | 35/32 |
| S6*S7*S8 | (6/5)/(9/8) | 16/15 |
| S7*S8*S9 | (7/6)/(10/9) | 21/20 |
| S8*S9*S10 | (8/7)/(11/10) | 80/77 |
| S9*S10*S11 | (9/8)/(12/11) | 33/32 |
| S10*S11*S12 | (10/9)/(13/12) | 40/39 |
| S11*S12*S13 | (11/10)/(14/13) | 143/140 |
| S12*S13*S14 | (12/11)/(15/14) | 56/55 |
| S13*S14*S15 | (13/12)/(16/15) | 65/64 |
| S14*S15*S16 | (14/13)/(17/16) | 224/221 |
| S15*S16*S17 | (15/14)/(18/17) | 85/84 |
| S16*S17*S18 | (16/15)/(19/18) | 96/95 |
| S17*S18*S19 | (17/16)/(20/19) | 323/320 |
| S18*S19*S20 | (18/17)/(21/20) | 120/119 |
| S19*S20*S21 | (19/18)/(22/21) | 133/132 |
| S20*S21*S22 | (20/19)/(23/22) | 440/437 |
| S21*S22*S23 | (21/20)/(24/23) | 161/160 |
| S22*S23*S24 | (22/21)/(25/24) | 176/175 |
| S23*S24*S25 | (23/22)/(26/25) | 575/572 |
| S24*S25*S26 | (24/23)/(27/26) | 208/207 |
| S25*S26*S27 | (25/24)/(28/27) | 225/224 |
| S26*S27*S28 | (26/25)/(29/28) | 728/725 |
| S27*S28*S29 | (27/26)/(30/29) | 261/260 |
| S28*S29*S30 | (28/27)/(31/30) | 280/279 |
| S29*S30*S31 | (29/28)/(32/31) | 899/896 |
| S30*S31*S32 | (30/29)/(33/32) | 320/319 |
| S31*S32*S33 | (31/30)/(34/33) | 341/340 |
| S32*S33*S34 | (32/31)/(35/34) | 1088/1085 |
| S33*S34*S35 | (33/32)/(36/35) | 385/384 |
| S34*S35*S36 | (34/33)/(37/36) | 408/407 |
| S35*S36*S37 | (35/34)/(38/37) | 1295/1292 |
| S36*S37*S38 | (36/35)/(39/38) | 456/455 |
| S37*S38*S39 | (37/36)/(40/39) | 481/480 |
| S38*S39*S40 | (38/37)/(41/40) | 1520/1517 |
| S39*S40*S41 | (39/38)/(42/41) | 533/532 |
| S42*S43*S44 | (42/41)/(45/44) | 616/615 |
| S46*S47*S48 | (46/45)/(49/48) | 736/735 |
| S49*S50*S51 | (49/48)/(52/51) | 833/832 |
| S52*S53*S54 | (52/51)/(55/54) | 936/935 |
| S55*S56*S57 | (55/54)/(58/57) | 1045/1044 |
| S63*S64*S65 | (63/62)/(66/65) | 1365/1364 |
| S66*S67*S68 | (66/65)/(69/68) | 1496/1495 |
| S75*S76*S77 | (75/74)/(78/77) | 1925/1924 |
| S78*S79*S80 | (78/77)/(81/80) | 2080/2079 |
| S82*S83*S84 | (82/81)/(85/84) | 2296/2295 |
| S85*S86*S87 | (85/84)/(88/87) | 2465/2464 |
| S88*S89*S90 | (88/87)/(91/90) | 2640/2639 |
| S93*S94*S95 | (93/92)/(96/95) | 2945/2944 |
| S96*S97*S98 | (96/95)/(99/98) | 3136/3135 |
| S112*S113*S114 | (112/111)/(115/114) | 4256/4255 |
| S117*S118*S119 | (117/116)/(120/119) | 4641/4640 |
| S121*S122*S123 | (121/120)/(124/123) | 4961/4960 |
| S133*S134*S135 | (133/132)/(136/135) | 5985/5984 |
| S145*S146*S147 | (145/144)/(148/147) | 7105/7104 |
| S153*S154*S155 | (153/152)/(156/155) | 7905/7904 |
| S162*S163*S164 | (162/161)/(165/164) | 8856/8855 |
| S187*S188*S189 | (187/186)/(190/189) | 11781/11780 |
Sk*S(k + 1)*...*S(k + n - 1) (1/n-square-particulars)
Motivation
1/n-square-particulars continue the pattern (1/2-square-particulars, 1/3-square-particulars, ...) to a comma/interval whose S-expression is can be written in the form of a product of n consecutive square-particulars (including Sk but not including S(k + n)) and which can therefore be written as the ratio between the two superparticulars k/(k - 1) and (k + n)/(k + n - 1).
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 EG 10/9 and 11/10 are considered as having 1 distance between them, corresponding to (1/1-)square-particulars (in this case S10).
These commas are important in a few ways: 1. As a generalization of important special cases n=0, n=1 and n=2, (which are almost all superparticular; the only case where they aren't is that 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.
2. Because of being the ratio of two superparticular intervals, in higher-complexity cases they often correspond to small commas between large commas which we don't want to temper, for example (81/80)/(91/90) = S81 * S82 * ... * S90 = 729/728 = S27. They also often simplify in cases like these; note that a suggested shorthand is S81..90 for S81 * S82 * ... * S90 and thus more generally Sa..b for Sa * S(a + 1) * ... * Sb.
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, 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.
5. They're relevant to understanding how much damage is present in a temperament's harmonic series representation, because they show how many superparticular intervals are either not distinguished or worse mapped inconsistently, bringing us finally to...
6. They're relevant to understanding limitations of consistency (or more precisely, monotonicity) of any given temperament, as the next section will discuss.
Significance/implications for consistency
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 Sk but not including S(k + n)) and which can therefore be written as the ratio between the two superparticulars k/(k - 1) and (k + n)/(k + n - 1) have implications for the consistency of the (k + n)-odd-limit when tempered. Specifically:
If a temperament tempers a 1/n-square-particular of the form Sk*S(k+1)*...*S(k+n-1), it must temper all of the n square-particulars that compose it, which is to say it must also temper all of Sk, S(k+1), ..., S(k+n-1). If it does not, it is necessarily inconsistent (more formally & weakly, not monotonic) in the (k + n)-odd-limit(*). A proof is as follows:
Consider the following sequence of superparticular intervals, all of which in the (k + n)-odd-limit:
(k + n)/(k + n - 1), (k + n - 1)/(k + n - 2), ..., (k + 1)/k, k/(k - 1)
Because of tempering Sk*S(k+1)*...*S(k+n-1), we require that (k + n)/(k + n - 1) = k/(k - 1) consistently. Therefore, if any superparticular x/(x - 1) imbetween (meaning k + n > x > k) is not tempered to the same tempered interval, it must be mapped to a different tempered interval. But this means that one of the following must be true:
mapping((k + n)/(k + n - 1)) > mapping(x/(x - 1))
mapping(k/(k - 1)) < mapping(x/(x - 1))
Therefore any superparticular interval x/(x - 1) between the extrema must be mapped to the same interval as those extrema in order for a consistent tuning in the (k + n)-odd-limit to even potentially be possible. Another way of phrasing this conclusion is that tempering Sk*S(k+1)*...*S(k+n-1) but not all of the constituent square-particulars limits the possible odd-limit consistency of a temperament to the (k - 1)-odd-limit.
(* Note that this statement is a slight inaccuracy, because technically the tuning of the higher rank temperament corresponding to the lower rank temperament that tempers all of these commas is the unique and only (continuum of) tuning(s) for which this statement is false, but it's reasonable to simplify this technicality as this (continuum of) tuning(s) corresponds exactly and uniquely to tempering all the square-particulars we said were not tempered.)
Sk/S(k + 1) (ultraparticulars)
Motivational example
Often it is desirable to make consecutive superparticular intervals equidistant. This has a number of nice consequences, many of which not explained here — see the motivation section for each infinite family of commas defined on this page.
For example, if you want 6/5 equidistant from 5/4 and 7/6, you must equate (5/4)/(6/5) = 25/24 = S5 with (6/5)/(7/6) = 36/35 = S6, hence tempering S5/S6 = (25/24)/(36/35) = 875/864, but it's actually often not necessary to know the specific numbers, often familiarizing yourself with and understanding the "Sk" notation will give you a lot of insight, as we'll see.
Back to our example: we know that S5 ~ S6 (because we're tempering S5/S6); from this we can deduce that the intervals must be arranged like this: 7/6 <— S5~S6 —> 6/5 <— S5~S6 —> 5/4.
From this you can deduce that (6/5)3 = 7/4, because you can lower one of the 6/5's to 7/6 (lowering it by S6) and raise another of the 6/5's to 5/4 (raising it by S5). Then because we've tempered S5 and S6 together, we've lowered and raised by the same amount, so the result of 7/6 * 6/5 * 5/4 = 7/4 must be the same as the result of 6/5 * 6/5 * 6/5 in this temperament.
Familiarize yourself with the structure of this argument, as it generalizes to arbitrary Sk; the algebraic proof is tedious, but the intuition is the same:
(k+2)/(k+1) <— S(k+1)~Sk —> (k+1)/k <— S(k+1)~Sk —> k/(k-1)
...implies that three (k+1)/k's is equal to a (k+2)/(k-1) iff we temper Sk/S(k+1). [math]\square[/math]
Significance
1. Tempering any two consecutive square-particulars Sk and S(k+1) will naturally imply tempering the ultraparticular between them (Sk/S(k+1)), meaning they are very common implicit commas.
2. Tempering any two consecutive ultraparticulars will imply tempering the semiparticular which is their sum/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 Sk notation to achieve that compactness.
3. Tempering the ultraparticular Sk/S(k+1) along with either the corresponding 1/2-square-particular Sk * S(k+1) or one of the two corresponding lopsided commas Sk2 * S(k+1) or Sk * S(k+1)2 implies tempering both of Sk and S(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 Sk and S(k+1). This only gets better if you temper a third consecutive square-particular. This is an abundance of "at a glance" essential tempering information that is fully general so only needs to be learned once, and is the motivation of the use of S-expressions. (For example, {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!)
Table of ultraparticulars
The above table is a list of all 23-limit ultraparticulars corresponding to Sk with k < 77, plus ultraparticulars corresponding to dividing a superparticular interval into three equal parts up to 17/16 (or up to 19/18 but excluding 18/17 because of it requiring a large prime, 71), plus S27/S28 so that we have all ultraparticulars up to S28/S29 listed rather than up to S26/S27.
This table has been expanded following every ultraparticular from S2/S3 to S16/S17 having its own page, with S12/S13 and S13/S14 expected to have their own pages soon. Note that ultraparticulars are, in general, extremely precise commas so that usually one wouldn't consider tempering them directly rather than through tempering the square-particulars Sk which they are composed of. As an example of this, notice that S10/S11 is the largest ultraparticular categorised as an unnoticeable comma, which means not unnoticeable in the absolute sense but rather in the sense of being smaller than the melodic just-noticeable difference, despite only dividing a superparticular as simple and unremarkable as 4/3. For this reason, a cents column has been included to aid an appreciation of their precision. The cent value of a semiparticular is roughly double that of any of the two ultraparticulars it is composed of; this becomes more true the higher you go.
Note also from this table how the shorthand becomes increasingly convenient higher up the series, where (preferably consistent) temperaments that temper out the ultraparticular but neither of the superparticulars which it is a difference between are of increasing precision. Note also how every three superparticulars the interval divided into three equal parts simplifies to a superparticular. This happens for S(3k + 1)/S(3k+ 2) for a positive integer k, because then the superparticular can be expressed as:
[math]\frac{(3k + 3)/3k}{((3k + 2)(3k + 1))^3} = \frac{(k + 1)/k}{((3k + 2)(3k + 1))^3}[/math]
Also note that if you temper multiple adjacent ultraparticulars, you sometimes are not required to use those ultraparticulars in the comma list as description of (the bulk of) the tempering may be possible through semiparticulars, discussed next.
Sk/S(k + 2) (semiparticulars)
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?
The pattern is this: take some arbitrary quodd-particular (k + 4)/k; observe that we can split it into (k + 4)/(k + 2) * (k + 2)/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.
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 Sk/S(k + 2), the resulting comma is either superparticular or odd-particular, so these are efficient commas. (This terminology also suggests throdd-particular and quodd-particular as generalizations.)
2. Tempering any two consecutive ultraparticulars 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 (Sk and S(k + 2)) implies tempering a semiparticular; this is generally much more ideal than tempering two consecutive Sk because it is a lot lower damage (see lopsided commas 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:
[math] \LARGE \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} [/math]
...and notice that the latter expression is the one we've shown is equal to S(k-1)/S(k+1) (up to an offset k). In other words, you could interpret that a reason that tempering S(k-1)/S(k+1) results in (k+1)/(k-1) being half of (k+2)/(k-2) is because it makes the following three intervals equidistant: (k+2)/k, (k+1)/(k-1), k/(k-2)
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 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))2.
This form of comma has been named "semiparticular", because most of the time it is superparticular but less often it is odd-particular, and because when tempered out they all cause an interval to be divided into two equal parts where each part is a (tempered version of a) superparticular or odd-particular, and the interval being divided in half is sometimes quodd-particular, sometimes odd-particular and sometimes superparticular. Specifically:
- To find out what a superparticular (a+1)/a is approximately half of, temper the semiparticular S(2a)/S(2a+2) and you can observe that (2a+3)/(2a-1) is the interval it is approximately half of.
- To find out what an odd-particular (2a+1)/(2a-1) is approximately half of, temper the semiparticular S(2a-1)/S(2a+1) and you can observe that (2a+2)/(2a-2) = (a+1)/(a-1), a superparticular or odd-particular, is the interval it is approximately half of.
- To find out what splits a superparticular (a+1)/a in half, temper the semiparticular S(4a+1)/S(4a+3) and you can observe that (4a+3)/(4a+1), an odd-particular, is the interval that is approximately half of it.
- To find out what splits an odd-particular (2a+1)/(2a-1) in half, temper the semiparticular S(4a-2)/S(4a+2) and you can observe that (4a-1)/(4a+1), an odd-particular, is the interval that is approximately half of it.
Also, the interval in the denominator of an expression of a semiparticular of the form (a/b)/(c/d)2 is significant in that it has a special relationship: specifically, consider tempering (a/b)/(c/d)2 so therefore the interval c/d is equal to the interval (a/b)/(c/d). This is significant because it allows the intuitive replacement of two consecutive superparticulars (whose product is a superparticular or odd-particular) with the two superparticulars directly adjacent to them.
For example, as 9/8 = 18/17 * 17/16 we can replace 18/17 with 19/18 and 17/16 with 16/15 by tempering S16/S18 = (19/15)/(9/8)2 because we can multiply 9/8 by the tempered comma (19/15)/(9/8)2 to get (19/15)/(9/8) = (19/18)(16/15) (because 9/8 = 18/16), or as 13/11 = 13/12 * 12/11 we can replace 13/12 with 14/13 and 12/11 with 11/10 by tempering S11/S13 = (7/5)/(13/11)2 because we can multiply 13/11 by the tempered comma (7/5)/(13/11)2 to get (7/5)/(13/11) = (14/13)(11/10) (because 7/5 = 14/10). Note we have to replace both intervals simultaneously as this is lower error, and note that if we want to be able to replace them individually we must pick the higher error route of tempering S16 and S18 or S11 and S13 individually (for which tempering the semiparticular is then an implied consequence). (The broader lesson is that you can rewrite exact JI equivalences with the commas you are tempering to find new interesting consequences of those commas.)
Table of semiparticulars
Here follows a table of 23-limit semiparticulars corresponding to square-particulars Sk for k < 96, plus all semiparticulars up to S33/S35 = S99, an exceptional 11-limit comma, plus all semiparticulars dividing superparticular intervals up to 13/12 (corresponding to the 17-limit semiparticular S49/S51) for completeness. This table also shows all semiparticulars corresponding to splitting an odd-particular in two up to 17/15 (although a common strategy is to temper the square-particular that is the difference between the two superparticular intervals the odd-particular is composed of instead). The bound k < 96 was chosen as it corresponds to another remarkable semiparticular S78/S80 = S351. Perhaps many of the patterns will become clearer if you examine this table:
(Note that while a lot of these have pages, not all of them do, although that doesn't mean they shouldn't. 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.)
Using S-factorizations to understand the significance of S-expressions
This section deals with the forms of the main infinite comma families listed on this page* as expressed in terms of nearby harmonics in the harmonic series and as related to square-particulars; note that this uses a mathematical notation of [a, b, c, ...]^[x, y, z, ...] to denote a^x * b^y * c^z * ...
* except lopsided commas which are dealt with in their respective subsection of the page (which is conveniently just under this section, partly as it demonstrates a less easy application of this technique).
If instead of working through things algebraically we look at square-particulars as describing a relationship between adjacent harmonics, we can use this to understand why certain simplifications and equivalences exist in a way that is equivalent to the sometimes harder-to-understand usual algebraic form:
If we describe Sk as [k-1, k, k+1]^[-1, 2, -1] then if we write something like Sk/S(k + 2) (semiparticulars) in this form we get:
[k-1, k, k+1, k+2, k+3]^([-1, 2, -1, 0, 0] - [0, 0, -1, 2, -1] = [-1, 2, 0, -2, 1]) from which we can clearly see that we have two (k+2)/k's making up a (k+3)/(k-1). An exercise to the reader is to go through the other forms discussed on this page to derive similar expressions.
Sk = [k-1, k, k+1]^[-1, 2, -1]
Sk * S(k+1) = [k-1, k, k+1, k+2]^[-1, 1, 1, -1] = [k-1, k, k+1(, k+2)]^[-1, 2, -1(, 0)] * [(k-1,) k, k+1, k+2]^[(0,) -1, 2, -1]
S(k-1) * Sk * S(k+1) = [k-2, k-1, k, k+1, k+2]^[-1, 1, 0, 1, -1] = ( (k-1)/(k-2) )( k/(k-1) ) * ( k/(k-1) )/( (k+1)/k ) * ( (k+1)/k )/( (k+2)/(k+1) ) = ( (k-1)/(k-2) )/( (k+2)/(k+1) ) = ( (k-1)(k+1) )/( (k-2)(k+2) ) k-2 k-1 k k+1 k+2 -1 2 -1 0 0 0 -1 2 -1 0 0 0 -1 2 -1 ======================== -1 1 0 1 -1
Sk / S(k+1) = [k-1, k, k+1, k+2]^[-1, 3, -3, 1] = [k-1, k, k+1]^[-1, 2, -1] * [k, k+1, k+2]^[1, -2, 1] = (k+2)/(k-1) * ( k/(k+1) )^3 = (k+2)/(k-1) / ((k+1)/k)^3
S(k-1) / S(k+1) = [k-2, k-1, k, k+1, k+2]^[-1, 2, 0, -2, 1] = [k-2, k-1, k]^[-1, 2, -1] * [k, k+1, k+2]^[ 1, -2, 1] = [k-2, k-1, k]^[-1, 2, -1] / [k, k+1, k+2]^[-1, 2, -1] = (k+2)/(k-2) * ((k-1)/(k+1))^2 = (k+2)/(k-2) / ((k+1)/(k-1))^2 k-2 k-1 k k+1 k+2 -1 2 -1 0 0 0 0 1 -2 1 ======================== -1 2 0 -2 1
This technique will be called "S-factorizations", as it is uses a certain format for expressing factorization (analogous to monzos) that is uniquely suited for interpreting the relationships described by S-expressions.
Note that the redundancy in these factorizations (in the sense that there are generators that are not linearly independent of the others) is a property that reflects the reality of equivalent S-expressions.
The generalisation of this method using commutative group theory is discussed in the abstraction section of this page.
Using S-factorizations to show a useful equivalence/redundancy of S-expressions
Absent of restrictions on the form that an S-expression may take, there is no unique S-expression for any given rational number. This is in fact a huge advantage, because it allows one to understand the landscape of commas in a way that sees interconnectedness of subgroups and corresponding tempering opportunities. But then what S-expressions are equivalent, other than mathematical one-offs? The most important general rule can be derived quite simply using S-factorizations:
The general S-expression equivalence
Consider:
Sk = [k-1, k, k+1]^[-1, 2, -1] versus what it is claimed to be equivalent to:
S(2k-1) * S(2k) * S(2k) * S(2k+1)
= [2k-2, 2k-1, 2k, 2k+1, 2k+2]^(
[-1, 2, -1]
+ [-2, 4, -2]
+ [-1, 2, -1]
= [-1, 0, 2, 0, -1] )
From here we can observe that the exponents are on even integers and that the factors of 2 involved cancel (we divide by 2 once for 2k-2 and 2k+2 having -1 as the power and we multiply by 2 twice for 2k having 2 as the power). Therefore the expressions are algebraically equivalent, which leads to the surprising fact that the following equivalence is true for all real and complex k:
[math] \large {\rm S}k = \large {\rm S}(2k-1) \cdot \large {\rm S}(2k)^2 \cdot \large {\rm S}(2k+1) [/math]
...where we use the notation Skp to mean (Sk)p rather than S(kp) for convenience in the practical analysis of regular temperaments using S-expressions.
For tuning theory only integer k > 1 is of relevance. Technically, rational k other than 1 correspond to rational commas too; the most relevant case for tuning theory is that half-integer k work as an alternative notation for odd-particulars, though for intuitively understanding the notation, the method described in #Abstraction may be recommendable as having (in a mathematical sense) exact analogues for every infinite family of commas defined in terms of an analogue of an S-expression, for which the most musically fruitful example is Ok = (k / (k - 2))/((k + 2) / k) for odd k as relevant to no-twos subgroup temperaments.
Sk2 * S(k + 1) and S(k - 1) * Sk2 (lopsided commas)
Significance
1. Tempering any two consecutive square-particulars, Sk and S(k + 1), implies tempering 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.
2. 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 the square-particulars composing it (such as S16 and 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.
3. 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 S-factorizations, we can show the interval relations implicated by these two new "lopsided" forms, which will make clear the reason for their name:
Sk2 * 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:
Sk2 * S(k+1) = (k/(k-1))2 / ((k+2)/k) through [-2, 3, 0, -1] = [-2, 2, 0, 0] - [0, -1, 0, 1].
S(k-1) * Sk2 = [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) * Sk2 = (k/(k-2)) / ((k+1)/k)2 through [-1, 0, 3, -2] = [-1, 0, 1, 0] - [0, 0, -2, 2].
Tables
Below is 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
Bottom-heavy lopsided commas
Equivalent S-expressions
Significance and meaning
All S-expressions have other equivalent S-expressions, however when the equivalence makes one comma a member of two of the infinite families discussed on this page, or otherwise makes it equal to a product or ratio between two such commas, this basically always means exceptional and nontrivial ("deep") tempering opportunities, usually leading to multiple of the most elegant and efficient temperaments that we know of depending how you temper further. 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 as you allow higher n, but are still quite rare for small n.
Incomplete table
Here is an incomplete list of examples (feel free to expand with any equivalences you find that you think are valuable):
| Comma | S-expressions |
|---|---|
| 81/80 | S6/S8 = S9 |
| 176/175 | S8/S10 = S22*S23*S24 |
| 243/242 | S9/S11 = S15/(S22/S24 = S55) |
| 325/324 | S10/S12 = S25*S26 |
| 676/675 | S13/S15 = S26 |
| 1225/1224 | S35 = S49*S50 |
| 3025/3024 | S22/S24 = S55 = S25/S27 * S99 |
| 2601/2600 | S17/(S25*S26) = S51 |
| 9801/9800 | S99 = S33/S35 |
| 123201/123200 | S351 = S78/S80 |
Note: Where a comma written in the form a/b is used in an S-expression, this means to replace that comma with any equivalent S-expression. This is done in the case of 3025/3024 as there are many S-expressions for it so restating them each time it appears seems inconvenient.
A proof that every positive rational number (and thus every JI interval) can be written as an S-expression follows.
It suffices to show every superparticular number including 2/1 has an expression using square-particulars:
[math] \small 2/1 = S_2 \cdot S_2 \cdot S_3\ ,\\ \small 3/2 = S_2 \cdot S_3\ ,\\ \small 4/3 = S_2\ ,\\ \normalsize \frac{a/(a-1)}{(b+1)/b} = \prod_{k=a}^b \left( S_k = \frac{k/(k-1)}{(k+1)/k} \right) \\ \ \ \ = \frac{a/(a-1)}{(a+1)/a} \cdot \frac{(a+1)/a}{(a+2)/(a+1)} \cdot \frac{(a+2)/(a+1)}{(a+3)/(a+2)} \cdot\ ...\ \cdot \frac{b/(b-1)}{(b+1)/b} = \frac{a/(a-1)}{(b+1)/b} \\ \implies \frac{a/(a-1)}{(b+1)/b} = S_a \cdot S_{a+1} \cdot S_{a+2} \cdot\ ...\ \cdot S_b \\ \implies \frac{S_2 \cdot S_2 \cdot S_3}{\prod_{a=2}^k S_a} = 2\cdot\left( \frac{2/(2-1)}{(k+1)/k} \right)^{-1} = 2\cdot\left( \frac{(k+1)/k}{2} \right) = (k+1)/k [/math]
From here it should not be hard to see how to make any positive rational number. For 11/6, for example, we can do (11/10)(10/9)(9/8)...(2/1) = 11 and then divide that by (6/5)(5/4)(4/3)(3/2)(2/1), meaning 11/6 = (11/10)(10/9)(9/8)(8/7)(7/6) because of the cancellations, then each of those superparticulars we replace with the corresponding S-expression to get the final S-expression. This final S-expression is likely to be far from the most efficient or interesting expression; the redundancy in S-expressions is a strength and feature; it tells us that there are more than the trivial connections between commas and intervals and that S-expressions can be wielded as a mathematical tool/language to investigate and identify them.
Glossary
- Superparticular
- The interval/comma between two consecutive harmonics. See superparticular.
- These are of the form (k + 1)/k.
- Square-particular
- A superparticular interval/comma whose numerator is a square number. A shorthand (nick)name for square superparticular.
- These are of the form k2/(k2 - 1) = Sk.
- Triangle-particular
- A superparticular interval/comma whose numerator is a triangular number. A shorthand (nick)name for triangular superparticular. An alternative name for 1/2-square-particular.
- These are of the form (k2 + k)/(k2 + k - 2). (This always simplifies to a superparticular.)
- 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.
- These are of the form Sa*S(a+1)*...*Sb = (a/(a - 1))/((b + 1)/b) = ab/((a - 1)(b + 1)).
- Replacing/substituting a with k and b with k + n - 1 gives us an equivalent expression that includes the number of square-particulars n:
- Sk*S(k+1)*...*S(k+n-1) = (k/(k - 1))/((k + n)/(k + n - 1)) = k(k + n - 1)/((k - 1)(k + n))
- For b = a + 1 these can also be called triangle-particulars, in which case they are always superparticular.
- These have implications for whether consistency in the (n+k)=(b+1)-odd-limit is potentially possible in a given temperament; see the section on 1/n-square-particulars.
- Odd-particular
- An interval/comma between two consecutive odd harmonics. The odd analogue of superparticular.
- These are of the form (2k + 1)/(2k - 1).
- Throdd-particular
- An interval/comma between two harmonics 3 apart which is not superparticular.
- These are of the form (3k + 1)/(3k - 2) or (3k + 2)/(3k - 1).
- Quodd-particular
- An interval/comma between two harmonics 4 apart which is not superparticular or odd-particular.
- These are of the form (4k + 1)/(4k - 3) and (4k + 3)/(4k - 1).
- Ultraparticular
- An interval/comma which is the ratio of two consecutive square-particulars.
- These are of the form Sk/S(k + 1).
- Semiparticular
- A superparticular or odd-particular interval/comma which is the ratio between two adjacent-to-adjacent square-particulars, which is to say:
- These are of the form Sk/S(k + 2).
- S-expression
- An expression using the Sk shorthand notation corresponding strictly to multiplying and dividing only (arbitrary) square-particulars. S-expressions include singular square superparticulars and expressions for other superparticulars in terms of square superparticulars.
- S-factorization
- An expression that takes a list of consecutive integer harmonics including the kth harmonic and raises them to integer powers, similar to a smonzo but uniquely suited to analysing S-expressions.
- For example: Sk = [k-1, k, k+1]^[-1, 2, -1] because Sk = (k-1)-1k2(k+1)-1.
- S-comma
- Any comma within one of the infinite families of commas discussed here, excluding 1/n-square-particulars for n>5 (so square-particulars, triangle-particulars, 1/3-square-particulars, 1/4-square-particulars and 1/5-square-particulars are included, but not anything beyond; this bound is used for exclusion (rather than n>3) to allow the utility of 1/5-square-particulars in avoiding twin primes by equating superparticular intervals on either side of the twin primes).
- Indirect S-comma
- Any comma that is the product or ratio of two S-commas. These appear frequently as S-expressions for commas that are more challenging/nontrivial to represent from the perspective of S-expressions, for example the schisma admits at least three such representations!
Mathematical derivation
(Note for readers: Familiarity with basic algebra is advised, especially the rules (a + b)(a - b) = a2 - b2 and 1/(a/b) = b/a. Often multiple steps are combined into a single step for brevity, so follow carefully.)
For ultraparticulars, we want to show that Sk/S(k + 1) = ((k + 2)/(k - 1)) / ((k + 1)/k)3:
[math] \begin {align} {\rm S}k/{\rm S}(k + 1) &= \frac {k^2/(k^2 - 1)}{(k + 1)^2 / ((k + 1)^2 - 1)} \\ &= \frac {k^2}{(k + 1)(k - 1)} \cdot \frac{k^2 + 2k}{(k + 1)^2} \\ &= \frac {k^2}{(k + 1)^3 (k - 1)} \cdot (k + 2)k \\ &= \frac {k^3}{(k + 1)^3} \cdot \frac{(k + 2)}{(k - 1)} \\ &= \frac {(k + 2)/(k - 1)}{((k + 1)/k)^3} \end {align} [/math]
For semiparticulars, we want to show that Sk/S(k + 2) = ((k + 3)/(k - 1)) / ((k + 2)/k)2:
[math] \begin {align} {\rm S}k/{\rm S}(k + 2) &= \frac {k^2/(k^2 - 1)}{(k + 2)^2 / ((k + 2)^2 - 1)} \\ &= \frac {k^2}{(k + 1)(k - 1)} \cdot \frac{(k + 3)(k + 1)}{(k + 2)^2} \\ &= \frac {k^2}{k - 1} \cdot \frac {k + 3}{(k + 2)^2} \\ &= \frac {k + 3}{k - 1} \cdot \frac{k^2}{(k + 2)^2} \\ &= \frac {(k + 3)/(k - 1)}{((k + 2)/k)^2} \end {align} [/math]
For semiparticulars, we also want to show that Sk/S(k + 2) is superparticular for all but the case of S(4n - 1)/S(4n + 1) which is odd-particular:
[math] \begin {align} {\rm S}k/{\rm S}(k + 2) &= \frac {k + 3}{k - 1} \cdot \frac{k^2}{(k + 2)^2} \\ &= \frac {k^3 + 3k^2}{(k - 1)(k^2 + 4k + 4)} \\ &= \frac {k^3 + 3k^2}{k^3 + 4k^2 + 4k - k^2 - 4k - 4} \\ &= \frac {k^3 + 3k^2}{k^3 + 3k^2 - 4} \end {align} [/math]
This result will be useful, so we will refer to it as [Eq. 1]: Sk/S(k + 2) = (k3 + 3k2)/(k3 + 3k2 - 4)
- Note that when k = 2n in [Eq. 1], everything in the numerator and denominator is divisible by 4 because the only instances of k have it raised to a power of 2 or greater meaning there will be a factor of (2n)2 = 4n2, therefore Sk/S(k + 2) is superparticular when k is even.
- When k = 4n + 1 in [Eq. 1], we have to do some extra work to show the result is superparticular:
- (4n + 1)3 = (4n)3 + 3·(4n)2 + 3·4n + 1 is of the form 4m + 1.
- (4n + 1)2 = (4n)2 + 2·4n + 1 is also of the form 4m + 1.
- Therefore we can replace their occurrences in [Eq. 1] with 4m + 1 and 4a + 1 respectively, without having to worry about what m and a are (as we only need to know that m and a are positive integers). Therefore to show S(4n + 1)/S(4n + 3) is superparticular, we set k = 4n + 1 in [Eq. 1] and then do the replacements and simplify to a superparticular:
- [math] \begin {align} {\rm S}(4n + 1)/{\rm S}(4n + 3) &= \frac {(4n + 1)^3 + 3(4n + 1)^2}{(4n + 1)^3 + 3(4n + 1)^2 - 4} \\ &= \frac {(4m + 1) + 3(4a + 1)}{(4m + 1) + 3(4a + 1) - 4} \\ &= \frac {4m + 4(3a) + 4}{4m + 4(3a)} \\ &= \frac {m + 3a + 1}{m + 3a} \end {align} [/math]
- Then for the final case we want to show that S(4n - 1)/S(4n + 1) is odd-particular by setting k = 4n - 1 in [Eq. 1]:
- S(4n - 1)/S(4n + 1) = ( (4n - 1)3 + 3(4n - 1)2 )/( (4n - 1)3 + 3(4n - 1)2 - 4 )
- As before we make replacements:
- (4n - 1)3 = (4n)3 - 3·(4n)2 + 3·4n - 1 is of the form 4m - 1 so will be replaced with such.
- (4n + 1)2 = (4n)2 - 2·4n + 1 is of the form 4a + 1 so will be replaced with such.
- Therefore:
- [math] \begin {align} {\rm S}(4n - 1)/{\rm S}(4n + 1) &= \frac {(4n - 1)^3 + 3(4n - 1)^2}{(4n - 1)^3 + 3(4n - 1)^2 - 4} \\ &= \frac {(4m - 1) + 3(4a + 1)}{(4m - 1) + 3(4a + 1) - 4} \\ &= \frac {4m + 4(3a) + 2}{4m + 4(3a) - 2} \\ &= \frac {2m + 2(3a) + 1}{2m + 2(3a) - 1} \end {align} [/math]
- … which is of the form (2x + 1)/(2x - 1) meaning it is odd-particular.
In conclusion: Sk/S(k + 2) is superparticular for k ≠ 3 (mod 4) and is odd-particular when k = 3 (mod 4).
Alternatively stated: S(k - 1)/S(k + 1) is superparticular for k ≠ 0 (mod 4) and is odd-particular when k = 0 (mod 4). This alternative statement highlights an interesting fact that the four harmonics related by tempering S(k - 1)/S(k + 1) are (k - 2):(k - 1):(k + 1):(k + 2) through tempering ((k+2)/(k-2)) / ((k+1)/(k-1))2 meaning the kth harmonic is the only one not included and therefore a semiparticular is odd-particular if the excluded "harmonic in the middle" (around which the two on each side are symmetric in terms of placement) is a multiple of 4 and is superparticular otherwise.
Abstraction
The maths
Let H be a commutative group with generators hi, ..., hk, ..., hj (such that i ≤ k ≤ j).
These generators are a series indexed by the integers that are analogous to a portion of the harmonic series, but "analogous" is extremely abstract here, because:
The fact that they are indexed by a range of integers is the only analogy that is guaranteed to hold, but as it turns out, is sufficient for defining analogies of superparticulars and thus S-expressions.
Thus: (the analogue of) a superparticular is of the form hk+1 hk-1 = hk+1 / hk (we'll use multiplicative notation) meaning that (the analogue of) Sk is:
[math] \begin {align} {\rm S}(k) = \frac{h_k^2}{h_{k-1} h_{k+1}} = (h_k / h_{k-1})/(h_{k+1} / h_k) = h_k h_{k-1}^{-1} (h_{k+1} h_k^{-1})^{-1} = h_k^2 h_{k-1}^{-1} h_{k+1}^{-1} \end {align} [/math]
Then (the analogues of) S-factorizations correspond to the exponents of the generators, such that:
[math] \begin {align} {\rm S}(k) =\ .. h_{k-3}^0 h_{k-2}^0 h_{k-1}^{-1} h_k^2 h_{k+1}^{-1} h_{k+2}^0 h_{k+3}^0 ..\ = [..,\ h_{k-3},\ h_{k-2},\ h_{k-1},\ h_k,\ h_{k+1},\ h_{k+2},\ h_{k+3},\ ..]^{\Large[..,\ 0,\ 0, -1,\ 2, -1,\ 0,\ 0,\ ..]} \end {align} [/math]
This completes the analogy. What this means is:
Every infinite comma family defined in terms of an S-expression will have an infinite number of analogues, because of the maths of S-factorizations continuing to work as expected.
The meanings of these analogues are up to us to interpret, however. This brings us to applications, which we will examine next, with a focus on the musical ones:
Applications
Unless otherwise specified, we will let k be in the positive integers (Z+) and we will let the group operation be the multiplication of rationals, but this abstraction is much more powerful than that.
If you are working with a certain expression for hk, it is suggested to use Sa where "a" is a letter that abbreviates the meaning of what you are using.
Letter suggestions are provided below for expressions suspected to be theoretically useful/interesting in the direction of designing desirable temperaments.
If we use the letter "a" for some analogy Sak, then because of the guarantees of the analogy, we will always be able to speak of a-square-particulars, a-ultraparticulars, a-semiparticulars, a-1/n-square-particulars and a-lopsided-commas, and have it make abstract sense.
To emphasize: this is because all comma families expressed in terms of expressions involving H's group operation applied to elements Sakp for k, p in Z will have analogues if k is a valid index for Sa.
Finally, while we usually speak of temperaments, note that S-expressions, as a tool for aiding RTT, have a wide variety of fruitful applications, exactly because RTT already itself has a wide variety of fruitful applications not explicitly involving tempering, especially in the design of scales where the constant structure property is desirable, where exotemperaments can shine as representing a deep, coarse logic.
hk = k
The trivial example, equal to normal S-expressions and S-factorizations, before abstraction.
hk = 2k + 1
An analogy of S-expressions and S-factorizations for EDTs which can be used as a corresponding RTT tool for when we only want to focus on the arithmetic of odds, using 3/1 as the new equave.
A suggestion is to use the notation Sok, if this is not unambiguous, with the letter "o" standing for "odd". Thus:
[math] \begin {align} {\rm So}(k) = h_k^2 h_{k-1}^{-1} h_{k+1}^{-1} = \frac{ (2k+1)^2 }{ (2k-1)(2k+3) } = \frac{ 4k^2 + 4k + 1 }{ 4k^2 + 4k - 3 } \end {align} [/math]
hk = (k + 1)/k
An analogy of S-expressions and S-factorizations aiming at deeply faithful modelling of the harmonic series through modelling distances between superparticulars accurately.
A suggestion is to use the notation Ssk, if this is not unambiguous, with the letter "s" standing for "superparticular" or "super" generally.
We will see that this implies Ssk is the difference between two adjacent Sk (an ultraparticular), implying Ssk * Ss(k - 1) is a semiparticular. Thus:
[math] \begin {align} {\rm Ss}(k) = h_k^2 h_{k-1}^{-1} h_{k+1}^{-1} = \large \frac{ \frac{(k+1)}{k} \cdot \frac{(k+1)}{k} }{ \frac{k}{k-1}\cdot\frac{k+2}{k+1} } \normalsize = \frac{ {\rm S}(k+1) }{ {\rm S}(k) } \implies {\rm Ss}(k){\rm Ss}(k-1) = \frac{ {\rm S}(k+1) }{ {\rm S}(k) } \cdot \frac{ {\rm S}(k) }{ {\rm S}(k-1) } = \frac{ {\rm S}(k+1) }{ {\rm S}(k-1) } \end {align} [/math]
This implies that s-ultraparticulars, s-semiparticulars, etc. are now about ratios between ultraparticulars, and that s-1/n-square-superparticulars are about ratios between square-particulars.
While the utility of the latter should be clear to those familiar with using square-particulars for analysis, it is not clear how useful a concept a ratio between ultraparticulars is, thus we should investigate why we might want such:
When we equate square-particulars, we lose accuracy the larger streak of them we equate, so eventually we are forced to break the chain.
When this happens, we have an ultraparticular that should, in any accurate temperament, be mapped positively, but as a result, we will have multiple such ultraparticulars.
Therefore, it is of interest to describe when these ultraparticulars are equated.
Now consider that as temperaments get more accurate, the streaks of equated square-particulars will either get higher/smaller or shorter, thus in the limit, we do not want to equate them.
Rather, we want to equate the ultraparticulars between them in order that we have a smooth growing of distances.
While the question of where it is most appropriate and accurate to equate ultraparticulars is beyond the scope of this section, it nonetheless shows that the 2D comma family Ssa/Ssb has utility.