S-expression
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
k2/(k2 - 1) = (k/(k-1)) / ((k+1)/k)
which is square-(super)particular k for a given integer k > 1. A suggested shorthand for this interval is Sk for the kth 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 (n+1)th harmonic, as well as the potential representational sacrifices that must be made from that point onward.
It is common to temper square superparticulars, equating two adjacent superparticulars at some point in the harmonic series, but for higher accuracy 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 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:
Sk/S(k+1) = (k+2)/(k-1) / ((k+1)/k)3
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.
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. An arithmetic of rather interesting commas whose elements are square-particulars 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:
((3k+3)/(3k))/((3k+2)/(3k+1))3 = (k+1)/k / ((3k+2)/(3k+1))3
Also note that if you temper multiple adjacent ultraparticulars, you sometimes aren't 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", meaning an interval of the form (2n+1)/(2n-1) for some positive integer n. (This terminology also suggests "throdd-particular" for intervals of the form (3n+2)/(3n) and (3n+1)/(3n-1) and maybe "quodd-particular" (sounding like "quad-particular") for (4n+3)/(4n-1) and (4n+1)/(4n-3).) Furthermore, S(k+1)/S(k+3), when tempered, implies that (k+4)/k is divisible exactly into two halves of (k+3)/(k+1) which is equated with (k+1)/k * (k+4)/(k+3). It is for this reason that the indexing choice of S(k+1)/S(k+3) was chosen, as two (k+3)/(k+1)'s make a (k+4)/k. 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, Sk/S(k+2) is superparticular when k is not a multiple of 4, and odd-particular otherwise. See derivation for details on this and other facts stated here.
(Note that while a lot of these have pages, not all of them do, although that doesn't mean they shouldn't.)
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.
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) 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-1) 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.
Metaparticulars: A suggested name for the general class of commas describable by S-expressions.
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:
Sk/S(k+1) = ( k2/(k2 - 1) )/( (k+1)2/((k+1)2 - 1) )
- = k2/(k+1)/(k-1) * (k2 + 2k)/(k+1)2
- = k2/(k+1)3/(k-1) * (k + 2)k
- = k3/(k+1)3 * (k + 2)/(k - 1) = ((k+2)/(k-1)) / ((k+1)/k)3
For semiparticulars, we want to show that Sk/S(k+2) = ((k+3)/(k-1)) / ((k+2)/k)2:
Sk/S(k+2) = ( k2/(k2 - 1) )/( (k+2)2/((k+2)2 - 1) )
- = k2/(k+1)/(k-1) * (k+3)(k+1)/(k+2)2
- = k2/(k-1) * (k+3)/(k+2)2
- = (k+3)/(k-1) * k2 / (k+2)2 = ((k+3)/(k-1)) / ((k+2)/k)2
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:
Sk/S(k+2) = (k+3)/(k-1) * k2 / (k+2)2
- = (k3 + 3k2)/( (k-1)(k2 + 4k + 4) )
- = (k3 + 3k2)/( k3 + 4k2 + 4k - k2 - 4k - 4 )
- = (k3 + 3k2)/( k3 + 3k2 - 4 )
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:
- S(4n+1)/S(4n+3) = ( (4n+1)3 + 3(4n+1)2 )/( (4n+1)3 + 3(4n+1)2 - 4 )
- = ( 4m+1 + 3(4a+1) )/( 4m+1 + 3(4a+1) - 4)
- = ( 4m + 4(3a) + 4 )/( 4m + 4(3a) ) = ( m + 3a + 1 )/( m + 3a )
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:
- S(4n-1)/S(4n+1) = ( 4m-1 + 3(4a+1) )/( 4m-1 + 3(4a+1) - 4 )
- = ( 4m-1 + 4(3a) + 3 )/( 4m-1 + 4(3a) + 3 - 4 )
- = ( 4m + 4(3a) + 2 )/( 4m + 4(3a) - 2 )
- = ( 2m + 2(3a) + 1 )/( 2m + 2(3a) - 1 )
- ...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).