S-expression/Advanced results

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Using S-factorizations to understand the significance of S-expressions

This section deals with the forms of the main infinite comma families listed on the main S-expression 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 * ...

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]\displaystyle{ \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 Sk^p to mean (Sk)^p rather than S(k^p) 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.

Mathematical derivations

(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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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]\displaystyle{ \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 h_i, ..., h_k, ..., h_j (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 h_k+1 h_k^-1 = h_k+1 / h_k (we'll use multiplicative notation) meaning that (the analogue of) Sk is:

[math]\displaystyle{ \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]\displaystyle{ \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 h_k, 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 Sak^p 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.

h_k = k

The trivial example, equal to normal S-expressions and S-factorizations, before abstraction.

h_k = 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]\displaystyle{ \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]

h_k = (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]\displaystyle{ \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.