S-expression: Difference between revisions

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# Relatedly to the previous point, they are used in the construction of ''pentaparticulars'', which take the form S''k''/S(''k'' + 3) * ( S(''k'' + 1)/S(''k'' + 2) )2, which are equal to how (''k'' + 2)/(''k'' + 1) is approximately a fifth of (''k'' + 4)/(''k'' - 1), and are intuitively obvious from the idea of making superparticulars equidistant via ultraparticulars.

# More trivially, they are implied in various simple S-expression-based comma lists, like [[58edo]]'s [[17-limit]] as describable by {S6/S7, S8/9/10/11/13, S12, S14, S16, S17}, where S8/9/10/11/13 is a shorthand for [[64/63|S8]] = [[81/80|S9]] = [[100/99|S10]] = [[121/120|S11]] = [[169/168|S13]], wherein we have {{nowrap| three-particulars }} {{nowrap| { S8/S11 {{=}} ( (10/7)/(11/8) )/( (11/8)/(4/3) ), S10/S13 {{=}} ( (4/3)/(13/10) )/( (13/10)/(14/11) ), S14/S17 {{=}} ( (16/13)/(17/14) )/( (17/14)/(6/5) ) }. }}

# Three-particulars currently have no ''simple'' known splitting property due to the three intervals made equidistant not obviously composing to some other interval of significance, so seem to instead be about making exact spacings that feel approximately equidistant. For example, in 7:8:9:10:11:12, the closest is to notice the smallest (7:10) and largest (9:12) interval multiply to an interval which is 11/10 more than 7:12, and hence in general splits an interval (''k'' + 4)/(''k'' - 1) * (''k'' + 2)/(''k'' + 1) into two parts of (''k'' + 3)/''k''. Because the multiplication by a superparticular interval is a rather peculiar requirement, we leave it here as a note.

=== Derivation ===

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From which we can examine

<pre>

Sk * S(k+3) = [k-1, k, k+1]^[-1, 2, -1] / [k+2, k+3, k+4]^[-1, 2, -1]

Sk / S(k+3) = [k-1, k, k+1]^[-1, 2, -1] / [k+2, k+3, k+4]^[-1, 2, -1]

= [k-1, k, k+1]^[-1, 2, -1] * [k+2, k+3, k+4]^[1, -2, 1]

= [k-1, k, k+1, k+2, k+3, k+4]^[-1, 2, -1, 1, -2, 1]

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| 2.3.7.11.13.17.19.29.41

|}

== {{nowrap|(S''k''/S(''k'' + 1))2 * S(''k'' - 1)/S(''k'' + 2)}} (pentaparticulars) ==

=== Significance ===

# Pentaparticulars represent the obvious way of splitting an interval (''k'' + 3)/(''k'' - 2) into five parts of (''k'' + 1)/''k'', so represent a generalization of ultraparticulars by observing the harmonic series chord {{nowrap|''k''-2:''k''-1:''k'':''k''+1:''k''+2:''k''+3}}.

# Interestingly, while three-particulars are about equidistance but (currently) have no (clean) known splitting property (due to the three intervals made equidistant not obviously composing to some other significant interval), pentaparticulars are in a sense symmetric to this and related algebraically, by instead being specifically about splitting.

=== Derivation ===

As before, for a general introduction to the method we use here, see [[#Using S-factorizations to understand the significance of S-expressions|S-factorizations]], as I will not re-explain the method here. However, we will use a previous result from three-particulars as to immediately say that:

<pre>

Sk / S(k+3) = [k-1, k, k+1, k+2, k+3, k+4]^[-1, 2, -1, 1, -2, 1]

</pre>

and also relevant to our problem are

<pre>

S(k+1) / S(k+2) = [k, k+1, k+2]^[-1, 2, -1] / [k+1, k+2, k+3]^[-1, 2, -1]

= [k, k+1, k+2, k+3]^( [-1, 2, -1, 0] - [0, -1, 2, -1] )

= [k, k+1, k+2, k+3]^[-1, 3, -3, 1]

pentaparticular((k+2)/(k+1)) = [k-1, k, k+1, k+2, k+3, k+4]^[-1, 0, -5, 5, 0, 1]

</pre>

Then, dividing the pentaparticular by S''k''/S(''k'' + 3) yields:

<pre>

[k-1, k, k+1, k+2, k+3, k+4]^( [-1, 0, 5, -5, 0, 1]

- [-1, 2, -1, 1, -2, 1] )

= [k-1, k, k+1, k+2, k+3, k+4]^[0, -2, 6, -6, 2, 0]

= ( [k, k+1, k+2, k+3]^[-1, 3, -3, 1] )^2 = ( S(k+1) / S(k+2) )^2

</pre>

Thus proving the general form, which is for any given superparticular, the corresponding pentaparticular is equal to the square of the ultraparticular times the correspondingly-centered three-particular.

=== Table of pentaparticulars ===

Here is a table of 43-limit pentaparticulars.

TODO

== S''k''2⋅S(''k'' + 1) and S(''k'' − 1)⋅S''k''2 (lopsided commas) ==

@@ Line 2,157: / Line 2,157: @@
 # Relatedly to the previous point, they are used in the construction of ''pentaparticulars'', which take the form S''k''/S(''k'' + 3) * ( S(''k'' + 1)/S(''k'' + 2) )<sup>2</sup>, which are equal to how (''k'' + 2)/(''k'' + 1) is approximately a fifth of (''k'' + 4)/(''k'' - 1), and are intuitively obvious from the idea of making superparticulars equidistant via ultraparticulars.
 # More trivially, they are implied in various simple S-expression-based comma lists, like [[58edo]]'s [[17-limit]] as describable by {S6/S7, S8/9/10/11/13, S12, S14, S16, S17}, where S8/9/10/11/13 is a shorthand for [[64/63|S8]] = [[81/80|S9]] = [[100/99|S10]] = [[121/120|S11]] = [[169/168|S13]], wherein we have {{nowrap| three-particulars }} {{nowrap| { S8/S11 {{=}} ( (10/7)/(11/8) )/( (11/8)/(4/3) ), S10/S13 {{=}} ( (4/3)/(13/10) )/( (13/10)/(14/11) ), S14/S17 {{=}} ( (16/13)/(17/14) )/( (17/14)/(6/5) ) }. }}
+# Three-particulars currently have no ''simple'' known splitting property due to the three intervals made equidistant not obviously composing to some other interval of significance, so seem to instead be about making exact spacings that feel approximately equidistant. For example, in 7:8:9:10:11:12, the closest is to notice the smallest (7:10) and largest (9:12) interval multiply to an interval which is 11/10 more than 7:12, and hence in general splits an interval (''k'' + 4)/(''k'' - 1) * (''k'' + 2)/(''k'' + 1) into two parts of (''k'' + 3)/''k''. Because the multiplication by a superparticular interval is a rather peculiar requirement, we leave it here as a note.
 === Derivation ===
@@ Line 2,167: / Line 2,168: @@
 From which we can examine
 <pre>
-Sk * S(k+3) = [k-1, k, k+1]^[-1, 2, -1] / [k+2, k+3, k+4]^[-1, 2, -1]
+Sk / S(k+3) = [k-1, k, k+1]^[-1, 2, -1] / [k+2, k+3, k+4]^[-1, 2, -1]
   = [k-1, k, k+1]^[-1, 2, -1] * [k+2, k+3, k+4]^[1, -2, 1]
   = [k-1, k, k+1, k+2, k+3, k+4]^[-1, 2, -1, 1, -2, 1]
@@ Line 2,407: / Line 2,408: @@
 | 2.3.7.11.13.17.19.29.41
 |}
+== {{nowrap|(S''k''/S(''k'' + 1))<sup>2</sup> * S(''k'' - 1)/S(''k'' + 2)}} (pentaparticulars) ==
+=== Significance ===
+# Pentaparticulars represent the obvious way of splitting an interval (''k'' + 3)/(''k'' - 2) into five parts of (''k'' + 1)/''k'', so represent a generalization of ultraparticulars by observing the harmonic series chord {{nowrap|''k''-2:''k''-1:''k'':''k''+1:''k''+2:''k''+3}}.
+# Interestingly, while three-particulars are about equidistance but (currently) have no (clean) known splitting property (due to the three intervals made equidistant not obviously composing to some other significant interval), pentaparticulars are in a sense symmetric to this and related algebraically, by instead being specifically about splitting.
+=== Derivation ===
+As before, for a general introduction to the method we use here, see [[#Using S-factorizations to understand the significance of S-expressions|S-factorizations]], as I will not re-explain the method here. However, we will use a previous result from three-particulars as to immediately say that:
+<pre>
+Sk / S(k+3) = [k-1, k, k+1, k+2, k+3, k+4]^[-1, 2, -1, 1, -2, 1]
+</pre>
+and also relevant to our problem are
+<pre>
+S(k+1) / S(k+2) = [k, k+1, k+2]^[-1, 2, -1] / [k+1, k+2, k+3]^[-1, 2, -1]
+ = [k, k+1, k+2, k+3]^( [-1, 2, -1, 0] - [0, -1, 2, -1] )
+ = [k, k+1, k+2, k+3]^[-1, 3, -3, 1]
+pentaparticular((k+2)/(k+1)) = [k-1, k, k+1, k+2, k+3, k+4]^[-1, 0, -5, 5, 0, 1]
+</pre>
+Then, dividing the pentaparticular by S''k''/S(''k'' + 3) yields:
+<pre>
+ [k-1, k, k+1, k+2, k+3, k+4]^( [-1, 0, 5, -5, 0, 1]
+                              - [-1, 2, -1, 1, -2, 1] )
+ = [k-1, k, k+1, k+2, k+3, k+4]^[0, -2, 6, -6, 2, 0]
+ = ( [k, k+1, k+2, k+3]^[-1, 3, -3, 1] )^2 = ( S(k+1) / S(k+2) )^2
+</pre>
+Thus proving the general form, which is for any given superparticular, the corresponding pentaparticular is equal to the square of the ultraparticular times the correspondingly-centered three-particular.
+=== Table of pentaparticulars ===
+Here is a table of 43-limit pentaparticulars.
+TODO
 == S''k''<sup>2</sup>⋅S(''k'' + 1) and S(''k'' − 1)⋅S''k''<sup>2</sup> (lopsided commas) ==