S-expression: Difference between revisions

Line 368:

(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, and maps the corresponding intervals consistently.)

== S(''k'' - 1)*S''k''*S(''k'' + 1) (1/3-square-particulars) ==

This section concerns commas of the form S(''k'' - 1) * S''k'' * 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:

<pre>

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)

</pre>

In other words, what this shows is all 1/3-square-particulars of the form S(''k'' - 1) * S''k'' * S(''k'' + 1) are superparticular iff ''k'' is throdd (not a multiple of 3), and all 1/3-square-particulars of the form S(3''k'' - 1) * S(3''k'') * S(3''k'' + 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 99-odd-limit:

{| class="wikitable center-all

|-

! S-expression

! Interval relation

! Comma

|-

| S2*S3*S4

| ([[2/1]])/([[5/4]])

| [[8/5]]

|}

== S''k''*S(''k'' + 1)*...*S(''k'' + ''n'' - 1) (1/n-square-particulars) ==

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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.

== Redundant Integer-Contiguous Harmonic monzos (RICH monzos) ==

This section deals with the forms of these 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 * ...

<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>

== Glossary ==

@@ Line 368: / Line 368: @@
 (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, and maps the corresponding intervals consistently.)
+== S(''k'' - 1)*S''k''*S(''k'' + 1) (1/3-square-particulars) ==
+This section concerns commas of the form S(''k'' - 1) * S''k'' * 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:
+<pre>
+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)
+</pre>
+In other words, what this shows is all 1/3-square-particulars of the form S(''k'' - 1) * S''k'' * S(''k'' + 1) are superparticular iff ''k'' is throdd (not a multiple of 3), and all 1/3-square-particulars of the form S(3''k'' - 1) * S(3''k'') * S(3''k'' + 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 99-odd-limit:
+{| class="wikitable center-all
+|-
+! S-expression
+! Interval relation
+! Comma
+|-
+| S2*S3*S4
+| ([[2/1]])/([[5/4]])
+| [[8/5]]
+|}
 == S''k''*S(''k'' + 1)*...*S(''k'' + ''n'' - 1) (1/n-square-particulars) ==
@@ Line 645: / Line 672: @@
 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.
+== Redundant Integer-Contiguous Harmonic monzos (RICH monzos) ==
+This section deals with the forms of these 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 * ...
+<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>
 == Glossary ==