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]\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. Note that this means S2 = 4/3 is the first musically meaningful square-particular, as S1 = 1/0.
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.
Below is a table of 23-limit 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]{\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.
Sk*S(k + 1) (triangle-particulars)
If we examine (k + 1)/k then we can notice that if we equate (k + 2)/(k + 1) with k/(k - 1), we have:
(k + 2)/(k + 1) * (k + 1)/k = (k + 1)/k * k/(k - 1)
Which is to say that if we temper Sk*S(k + 1) = (k/(k - 1))/((k + 1)/k) * ((k + 1)/k)/((k + 2)/(k + 1)) = (k/(k - 1))/((k + 2)/(k + 1)) 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. Another reason commas of this form are of note is they are always superparticular.
It is also an interesting consequence that 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).)
A short proof of the superparticularity of Sk*S(k + 1) is as follows:
Sk*S(k + 1) = (k/(k - 1))/((k + 2)/(k + 1)) = (k(k + 1))/((k - 1)(k + 2)) = (k^{2} + k)/(k^{2} + k - 2)
Then notice that k^{2} + k is always a multiple of 2, therefore the above always simplifies to a superparticular. Half of this superparticular is halfway between the corresponding square-particulars, and because of its composition it could therefore be reasoned that it'd likely be half as accurate as tempering either of the square-particulars individually, so these are "1/2-square-particulars" in a sense, and half of a square is a triangle, which is not a coincidence here because the numerators of all of these superparticular intervals/commas are triangular numbers!
For completeness, all the commas of this form are included, because these "commas" (intervals rather) have structural importance for JI, and for the possibility of consistency of mappings 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. 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).
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 |
Note how there
Sk*S(k + 1)*...*S(k + n - 1) (1/n-square-particulars)
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 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.
Sk/S(k + 1) (ultraparticulars)
Note that tempering any two consecutive square-particulars will naturally imply tempering the ultraparticular between them (meaning they are very common implicit commas), and that 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.
S-expression | Cube Relation | Comma |
---|---|---|
S2/S3 = (4/3)/(9/8) | (4/1) / (3/2)^{3} | 32/27 |
S3/S4 = (9/8)/(16/15) | (5/2) / (4/3)^{3} | 135/128 |
S4/S5 = (16/15)/(25/24) | (2/1) / (5/4)^{3} | 128/125 |
S5/S6 = (25/24)/(36/35) | (7/4) / (6/5)^{3} | 875/864 |
S6/S7 = (36/35)/(49/48) | (8/5) / (7/6)^{3} | 1728/1715 |
S7/S8 = (49/48)/(64/63) | (3/2) / (8/7)^{3} | 1029/1024 |
S8/S9 = (64/63)/(81/80) | (10/7) / (9/8)^{3} | 5120/5103 |
S9/S10 = (81/80)/(100/99) | (11/8) / (10/9)^{3} | 8019/8000 |
S10/S11 = (100/99)/(121/120) | (4/3) / (11/10)^{3} | 4000/3993 |
S11/S12 = (121/120)/(144/143) | (13/10) /(12/11)^{3} | 17303/17280 |
S12/S13 = (144/143)/(169/168) | (14/11) / (13/12)^{3} | 24192/24167 |
S13/S14 = (169/168)/(196/195) | (5/4) / (14/13)^{3} | 10985/10976 |
S14/S15 = (196/195)/(225/224) | (16/13) / (15/14)^{3} | 43904/43875 |
S15/S16 = (225/224)/(256/255) | (17/14) / (16/15)^{3} | 57375/57344 |
S16/S17 = (256/255)/(289/288) | (6/5) / (17/16)^{3} | 24576/24565 |
(Note that while a lot of these have pages, not all of them do, although that doesn't mean they shouldn't.)
Note 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)
For differences between square-particulars of the form S(k + 1)/S(k + 3) the resulting comma is either superparticular or odd-particular. (This terminology also suggests "throdd-particular" for intervals of the form (3n + 1)/(3n - 2) and (3n + 2)/(3n - 1) and maybe "quodd-particular" (sounding like "quad-particular") for (4n + 3)/(4n - 1) and (4n + 1)/(4n - 3).)
Tempering S(k - 1)/S(k + 3) 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 (=small) and k/(k - 2) (=large) are equidistant from (k + 1)/(k - 1) (=medium) because to make them equidistant we need to temper:
( large/medium )/( medium/small )
= ( (k/(k-2))/((k+1)/(k-1)) )/( ((k+1)/(k-1))/((k+2)/k) )
= large * small / medium^{2}
= (k+2)/k * k/(k-2) / ((k+1)/(k-1))^{2} = ((k+2)/(k-2))/((k+1)/(k-1))^{2}
...and notice that the latter expression is the one we've shown is equal to S(k-1)/S(k+1). 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 does not yet have an official name, but a proposed name is "semiparticular", because most of the time it is superparticular but less often it is odd-particular, and because when tempered 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 the two superparticulars composing a superparticular or oddparticular 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.)
Here follows a table of semiparticulars. Perhaps many of the patterns will become clearer if you examine this table:
S-expression | Square Relation | Comma |
---|---|---|
S2/S4 = (4/3)/(16/15) | (5/1) / (2/1)^{2} | 5/4 |
S3/S5 = (9/8)/(25/24) | (3/1) / (5/3)^{2} | 27/25 |
S4/S6 = (16/15)/(36/35) | (7/3) / (3/2)^{2} | 28/27 |
S5/S7 = (25/24)/(49/48) | (2/1) / (7/5)^{2} | 50/49 |
S6/S8 = (36/35)/(64/63) | (9/5) / (4/3)^{2} | 81/80 |
S7/S9 = (49/48)/(81/80) | (5/3) / (9/7)^{2} | 245/243 |
S8/S10 = (64/63)/(100/99) | (11/7) / (5/4)^{2} | 176/175 |
S9/S11 = (81/80)/(121/120) | (3/2) / (11/9)^{2} | 243/242 |
S10/S12 = (100/99)/(144/143) | (13/9) / (6/5)^{2} | 325/324 |
S11/S13 = (121/120)/(169/168) | (7/5) / (13/11)^{2} | 847/845 |
S12/S14 = (144/143)/(196/195) | (15/11) / (7/6)^{2} | 540/539 |
S13/S15 = (169/168)/(225/224) | (4/3) / (15/13)^{2} | 676/675 |
S14/S16 = (196/195)/(256/255) | (17/13) / (8/7)^{2} | 833/832 |
S15/S17 = (225/224)/(289/288) | (9/7) / (17/15)^{2} | 2025/2023 |
S16/S18 = (256/255)/(324/323) | (19/15) / (9/8)^{2} | 1216/1215 |
S17/S19 = (289/288)/(361/360) | (5/4) / (19/17)^{2} | 1445/1444 |
S18/S20 = (324/323)/(400/399) | (21/17) / (10/9)^{2} | 1701/1700 |
S19/S21 = (361/360)/(441/440) | (11/9) / (21/19)^{2} | 3971/3969 |
S20/S22 = (400/399)/(484/483) | (23/19) / (11/10)^{2} | 2300/2299 |
S21/S23 = (441/440)/(529/528) | (6/5) / (23/21)^{2} | 2646/2645 |
S22/S24 = (484/483)/(576/575) | (25/21) / (12/11)^{2} | 3025/3024 |
S23/S25 = (529/528)/(625/624) | (13/11) / (25/23)^{2} | 6877/6875 |
S24/S26 = (576/575)/(676/675) | (27/23) / (13/12)^{2} | 3888/3887 |
S25/S27 = (625/624)/(729/728) | (7/6) / (27/25)^{2} | 4375/4374 |
(Note that while a lot of these have pages, not all of them do, although that doesn't mean they shouldn't.)
Sk^{2} * S(k+1) and S(k-1) * Sk^{2} (lopsided commas)
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. For now however, we want to focus on showing the equivalences implied by these two new "lopsided" forms:
Sk^{2} * 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:
Sk^{2} * 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) * Sk^{2} = [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) * Sk^{2} = (k/(k+1))^{2} / ((k-2)/k) through [-1, 0, 3, -2] = [0, 0, 2, -2] - [1, 0, -1, 0].
This technique will be called "S-monzos", as it is a form analogous to monzos that is uniquely suited for interpreting the relationships described by S-expressions.
A computer-generated table of both of these forms of commas is currently pending creation and will be added soon.
Using S-monzos to understand the significance of S-expressions
This section deals with the forms of the main 5 infinite comma families 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 * ...
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
Equivalent S-expressions
Some S-expressions have other equivalent S-expressions. This is easiest to observe in semiparticulars. Here is an incomplete list of examples (feel free to expand):
Comma | S-expressions |
---|---|
81/80 | S6/S8 = S9 |
243/242 | S9/S11 = S15/(3025/3024) |
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 |
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.
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 k^{2}/(k^{2} - 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 (k^{2} + k)/(k^{2} + 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-monzo
- An expression that takes a list of consecutive integer harmonics including the kth harmonic and raises them to integer powers, analogous to a monzo but uniquely suited to analysing S-expressions.
- For example: Sk = [k-1, k, k+1]^[-1, 2, -1] because Sk = (k-1)^{-1}k^{2}(k+1)^{-1}.
- Metaparticulars
- A potential name for square-particulars, ultraparticulars and semiparticulars. This list may expand if more interesting simple infinite sequences of commas are found.
Mathematical derivation
(Note for readers: Familiarity with basic algebra is advised, especially the rules (a + b)(a - b) = a^{2} - b^{2} 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) = (k^{3} + 3k^{2})/(k^{3} + 3k^{2} - 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} = 4n^{2}, 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.