S-expression: Difference between revisions
m fix |
m add three-particulars (W.I.P) |
||
| Line 2,150: | Line 2,150: | ||
{{Note| While a lot of these have pages, not all of them do, although that does not mean they should not. 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| While a lot of these have pages, not all of them do, although that does not mean they should not. 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. }} | ||
== {{nowrap|S''k''/S(''k'' + 3)}} (three-particulars) == | |||
=== Significance === | |||
# It it surprising that this form corresponds to yet another infinite family to do with spacing of obviously-related intervals. It thusly also helps further motivate expressing comma lists in terms of chains of square-particulars that are equated, either via {{nowrap| S''x'' {{=}} S''y'' {{=}} S''z'' {{=}} ... }} or via {{nowrap| { S''x''/''y''/''z''/... }, with ''x'' < ''y'' < ''z'' }}. | |||
# 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. | |||
=== Derivation === | |||
For a general introduction to 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 (though I find it quite straightforward). | |||
We start with: | |||
<pre> | |||
Sk = [k-1, k, k+1]^[-1, 2, -1] | |||
</pre> | |||
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] | |||
= [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] | |||
</pre> | |||
It is initially not obvious that this corresponds to anything interesting. However, as discovered by [[Lériendil]] (though not by this method), tempering out the comma makes (''k'' + 2)/(''k'' - 1), (''k'' + 3)/''k'', (''k'' + 4)/(''k'' + 1) equidistant, which is not arcane to understand either; we have three intervals of the form (''a'' + 3)/''a'' for adjacent values of ''a'' made equidistant, and the placement of ''a'' is obvious w.r.t the S-expression, being "the only place it could be" (in the center). Thus, we will show that the S-factorization derived from making these equidistant is equivalent: | |||
<pre> | |||
let F(k + 1) = ( (k + 2)/(k-1) )/( (k + 3)/k ) = [k-1, k, k+1, k+2, k+3]^[-1, 1, 0, 1, -1] | |||
let F(k + 2) = ( (k + 3)/k )/( (k + 4)/(k + 1) ) = [k, k+1, k+2, k+3, k+4]^[-1, 1, 0, 1, -1] | |||
then F(k + 1)/F(k + 2) = [k-1, k, k+1, k+2, k+3, k+4]^([-1, 1, 0, 1, -1, 0] - [0, -1, 1, 0, 1, -1]) | |||
= [k-1, k, k+1, k+2, k+3, k+4]^( [-1, 1, 0, 1, -1, 0] | |||
+ [0, 1, -1, 0, -1, 1] ) | |||
= [k-1, k, k+1, k+2, k+3, k+4]^[ -1, 2, -1, 1, -2, 1 ] | |||
</pre> | |||
...as expected. | |||
=== Table of three-particulars === | |||
W.I.P. | |||
{| class="wikitable center-all | |||
|- | |||
! S-expression | |||
! Square relation | |||
! Ratio | |||
|- | |||
| S2<sup>2</sup>⋅S3 = ([[3/2]])⋅([[4/3]]) | |||
| ([[2/1]])<sup>2</sup>/([[2/1]]) | |||
| [[2/1]] | |||
|} | |||
== S''k''<sup>2</sup>⋅S(''k'' + 1) and S(''k'' − 1)⋅S''k''<sup>2</sup> (lopsided commas) == | == S''k''<sup>2</sup>⋅S(''k'' + 1) and S(''k'' − 1)⋅S''k''<sup>2</sup> (lopsided commas) == | ||