S-expression: Difference between revisions

Line 2,181:

=== Derivation of equivalence relation ===

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:

Using the clarity of [[#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''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:

Line 2,665:

| [[256000/255879]]

|}

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

This section deals with the forms of the infinite comma families as expressed in terms of nearby harmonics in the harmonic series and as related to square-superparticulars; 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 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. (For example, through cancellation it's easy to prove that 1/n-square-particulars (the product of n consecutive square-(super)particulars) are equal to the ratio of the two superparticular intervals on the ends.)

<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 [[S-expression/Advanced_results#Abstraction]], though the ideas are very simple for anyone with simple mathematical training willing to learn the very basics needed.

=== 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''''p'' to mean (S''k'')''p'' rather than S(''k''''p'') 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]].

== Equivalent S-expressions ==

@@ Line 2,181: / Line 2,181: @@
 === Derivation of equivalence relation ===
-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:
+Using the clarity of [[#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:
@@ Line 2,665: / Line 2,665: @@
 | [[256000/255879]]
 |}
+== Using S-factorizations to understand the significance of S-expressions ==
+This section deals with the forms of the infinite comma families as expressed in terms of nearby harmonics in the harmonic series and as related to square-superparticulars; 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 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. (For example, through cancellation it's easy to prove that 1/n-square-particulars (the product of n consecutive square-(super)particulars) are equal to the ratio of the two superparticular intervals on the ends.)
+<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
+   -1    2   -1    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    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 [[S-expression/Advanced_results#Abstraction]], though the ideas are very simple for anyone with simple mathematical training willing to learn the very basics needed.
+=== 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]].
 == Equivalent S-expressions ==