|
|
| Line 1,622: |
Line 1,622: |
| From this you can deduce that ([[6/5]])<sup>3</sup> = [[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. | | From this you can deduce that ([[6/5]])<sup>3</sup> = [[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 [[Square superparticular#Mathematical derivation|it generalizes to arbitrary S''k'']]; the algebraic proof is tedious, but the intuition is the same: | | Familiarize yourself with the structure of this argument, as [[S-expression/Advanced results#Mathematical derivations|it generalizes to arbitrary S''k'']]; the algebraic proof is tedious, but the intuition is the same: |
|
| |
|
| (''k''+2)/(''k''+1) <— S(''k''+1)~S''k'' —> (''k''+1)/''k'' <— S(''k''+1)~S''k'' —> ''k''/(''k''-1) | | (''k''+2)/(''k''+1) <— S(''k''+1)~S''k'' —> (''k''+1)/''k'' <— S(''k''+1)~S''k'' —> ''k''/(''k''-1) |
| Line 1,847: |
Line 1,847: |
| </math> | | </math> |
|
| |
|
| ...and notice that the latter expression is the one we've [[#Mathematical derivation|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: | | ...and notice that the latter expression is the one we've [[S-expression/Advanced results#Mathematical derivations|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) | | (''k''+2)/''k'', (''k''+1)/(''k''-1), ''k''/(''k''-2) |
|
| |
|
| Line 2,041: |
Line 2,041: |
|
| |
|
| (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.) | | (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 * ...
| |
|
| |
| <nowiki>*</nowiki> except [[lopsided comma]]s 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 S''k'' as [''k''-1, ''k'', ''k''+1]^[-1, 2, -1] then if we write something like S''k''/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.
| |
|
| |
| <pre>
| |
| Sk = [k-1, k, k+1]^[-1, 2, -1]
| |
| </pre>
| |
| <pre>
| |
| 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]
| |
| </pre>
| |
| <pre>
| |
| 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
| |
| </pre>
| |
| <pre>
| |
| 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
| |
| </pre>
| |
| <pre>
| |
| 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
| |
| </pre>
| |
|
| |
| This technique will be called "'''S-factorizations'''", as it is uses a certain format for expressing factorization (analogous to [[monzo]]s) 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|equivalent S-expressions]].
| |
|
| |
| The generalisation of this method using commutative group theory is discussed in [[square superparticular#Abstraction|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:
| |
| <pre>
| |
| 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] )
| |
| </pre>
| |
| 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 S''k''<sup>''p''</sup> to mean (S''k'')<sup>''p''</sup> rather than S(''k''<sup>''p''</sup>) for convenience in the practical analysis of [[regular temperament]]s using [[S-expression]]s.
| |
|
| |
| 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-particular]]s, 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 O''k'' = (''k'' / (''k'' - 2))/((''k'' + 2) / ''k'') for odd ''k'' as relevant to [[no-twos subgroup temperaments]].
| |
|
| |
|
| == Sk<sup>2</sup> * S(k + 1) and S(k - 1) * Sk<sup>2</sup> (lopsided commas) == | | == Sk<sup>2</sup> * S(k + 1) and S(k - 1) * Sk<sup>2</sup> (lopsided commas) == |
| Line 2,130: |
Line 2,052: |
|
| |
|
| === Derivation of equivalence relation === | | === 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: | | Using the clarity of [[S-expression/Advanced results#Using S-factorizations to understand the significance of S-expressions|S-factorizations]], we can show the interval relations implicated by these two new "lopsided" forms, which will make clear the reason for their name: |
|
| |
|
| S''k''<sup>2</sup> * 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: | | S''k''<sup>2</sup> * 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: |
| Line 2,683: |
Line 2,605: |
| </math> | | </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. | | 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, as 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 == | | == Glossary == |
| Line 2,744: |
Line 2,666: |
| ; Indirect S-comma | | ; 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! | | : 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'') = ''a''<sup>2</sup> - ''b''<sup>2</sup> 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 S''k''/S(''k'' + 1) = ((''k'' + 2)/(''k'' - 1)) / ((''k'' + 1)/''k'')<sup>3</sup>:
| |
|
| |
| <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 S''k''/S(''k'' + 2) = ((''k'' + 3)/(''k'' - 1)) / ((''k'' + 2)/''k'')<sup>2</sup>:
| |
|
| |
| <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 S''k''/S(''k'' + 2) is superparticular for all but the case of S(4''n'' - 1)/S(4''n'' + 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]''': S''k''/S(''k'' + 2) = (k<sup>3</sup> + 3k<sup>2</sup>)/(k<sup>3</sup> + 3k<sup>2</sup> - 4)
| |
|
| |
| : Note that when ''k'' = 2''n'' 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 (2''n'')<sup>2</sup> = 4''n''<sup>2</sup>, therefore S''k''/S(''k'' + 2) is superparticular when ''k'' is even.
| |
|
| |
| : When ''k'' = 4''n'' + 1 in [Eq. 1], we have to do some extra work to show the result is superparticular:
| |
|
| |
| : (4''n'' + 1)<sup>3</sup> = (4''n'')<sup>3</sup> + 3·(4''n'')<sup>2</sup> + 3·4''n'' + 1 is of the form 4''m'' + 1.
| |
|
| |
| : (4''n'' + 1)<sup>2</sup> = (4''n'')<sup>2</sup> + 2·4''n'' + 1 is also of the form 4''m'' + 1.
| |
|
| |
| : Therefore we can replace their occurrences in [Eq. 1] with 4''m'' + 1 and 4''a'' + 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(4''n'' + 1)/S(4''n'' + 3) is superparticular, we set k = 4''n'' + 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(4''n'' - 1)/S(4''n'' + 1) is odd-particular by setting k = 4''n'' - 1 in [Eq. 1]:
| |
|
| |
| : S(4''n'' - 1)/S(4''n'' + 1) = ( (4''n'' - 1)<sup>3</sup> + 3(4''n'' - 1)<sup>2</sup> )/( (4''n'' - 1)<sup>3</sup> + 3(4''n'' - 1)<sup>2</sup> - 4 )
| |
|
| |
| : As before we make replacements:
| |
|
| |
| : (4''n'' - 1)<sup>3</sup> = (4''n'')<sup>3</sup> - 3·(4''n'')<sup>2</sup> + 3·4''n'' - 1 is of the form 4''m'' - 1 so will be replaced with such.
| |
|
| |
| : (4''n'' + 1)<sup>2</sup> = (4''n'')<sup>2</sup> - 2·4''n'' + 1 is of the form 4''a'' + 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 (2''x'' + 1)/(2''x'' - 1) meaning it is odd-particular.
| |
|
| |
| In conclusion: S''k''/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))<sup>2</sup> meaning the ''k''th 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 [[Wikipedia:Abelian group|commutative group]] with generators h<sub>i</sub>, ..., h<sub>k</sub>, ..., h<sub>j</sub> (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<sub>k+1</sub> h<sub>k</sub><sup>-1</sup> = h<sub>k+1</sub> / h<sub>k</sub> (we'll use multiplicative notation) meaning that (the analogue of) S''k'' 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<sub>+</sub>) 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<sub>k</sub>, 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 Sa''k'', 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 Sa''k''<sup>p</sup> 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 [[exotemperament]]s can shine as representing a deep, coarse logic.
| |
|
| |
| === h<sub>k</sub> = k ===
| |
| The trivial example, equal to normal S-expressions and S-factorizations, before abstraction.
| |
|
| |
| === h<sub>k</sub> = 2k + 1 ===
| |
| An analogy of S-expressions and S-factorizations for [[EDT]]s 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 So''k'', 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>
| |
|
| |
| === h<sub>k</sub> = (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 Ss''k'', if this is not unambiguous, with the letter "s" standing for "superparticular" or "super" generally.
| |
|
| |
| We will see that this implies Ss''k'' is the difference between two adjacent S''k'' (an ultraparticular), implying Ss''k'' * 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 Ss''a''/Ss''b'' has utility.
| |
|
| |
|
| [[Category:Elementary math]] | | [[Category:Elementary math]] |