S-expression: Difference between revisions

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== S''k''/S(''k'' + 2) (semiparticulars) ==

For differences between square-particulars of the form S(''k'' + 1)/S(''k'' + 3) the resulting comma is either [[superparticular]] or "odd-particular~~", meaning an interval of the form (2''n'' + 1)/(2''n'' - 1) for some positive integer ''n''~~. (This terminology also suggests "throdd-particular" for intervals of the form (3''n'' + 1)/(3''n'' - 2) and (3''n'' + 2)/(3''n'' - 1) and maybe "quodd-particular" (sounding like "quad-particular") for (4''n'' + 3)/(4''n'' - 1) and (4''n'' + 1)/(4''n'' - 3).) Furthermore, S(''k'' + 1)/S(''k'' + 3), when tempered, implies that (''k'' + 4)/''k'' is divisible exactly into two halves of (''k'' + 3)/(''k'' + 1) which is equated with ((''k'' + 1)/''k'')((''k'' + 4)/(''k'' + 3)). It is for this reason that the indexing choice of S(''k'' + 1)/S(''k'' + 3) was chosen, as two (''k'' + 3)/(''k'' + 1)'s make a (''k'' + 4)/''k''. This form of comma does not yet have an official name, but a proposed name is "semiparticular", because most of the time it is superparticular but less often it is odd-particular, and because when tempered they all cause an interval to be divided into two equal parts where each part is a tempered version of a superparticular or odd-particular, and the interval being divided in half is sometimes quodd-particular, sometimes odd-particular and sometimes superparticular.

For differences between square-particulars of the form S(''k'' + 1)/S(''k'' + 3) the resulting comma is either [[superparticular]] or [[#Glossary|odd-particular]]. (This terminology also suggests "throdd-particular" for intervals of the form (3''n'' + 1)/(3''n'' - 2) and (3''n'' + 2)/(3''n'' - 1) and maybe "quodd-particular" (sounding like "quad-particular") for (4''n'' + 3)/(4''n'' - 1) and (4''n'' + 1)/(4''n'' - 3).)

~~Specifically~~, S''k''/S(''k'' + 2) is superparticular ~~when~~ ''k'' is not a ~~multiple~~ of 4, and odd-particular ~~otherwise~~. ~~See [[#Mathematical derivation]] for details on~~ this ~~and other facts stated here.~~

Tempering S(''k'' - 1)/S(''k'' + 3) implies that (''k'' + 2)/(''k'' - 2) is divisible exactly into two halves of (''k'' + 1)/(''k'' - 1). It also implies that the intervals (''k'' + 2)/''k'' and ''k''/(''k'' - 2) are equidistant from (''k'' + 1)/(''k'' - 1) because to make them equidistant we need to temper:

( (''k''/(''k''-2))/((''k''+1)/(''k''-1)) )/( ((''k''+1)/(''k''-1))/((''k''+2)/''k'') )

= (''k''+2)/''k'' * ''k''/(''k''-2) / ((''k''+1)/(''k''-1))2 = ((''k''+2)/(''k''-2))/((''k''+1)/(''k''-1))2

...and notice that the latter expression is the one we've [[#Mathematical derivation|shown is equal to S(''k''-1)/S(''k''+1)]]. 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)

Also note that in the above, (''k''+1)/(''k''-1) is the [[mediant]] of the adjacent two intervals, meaning that division of an interval into two via tempering a semiparticular is in some sense 'optimal' relative to the complexity. This also means that if ''k'' is a multiple of 2, this corresponds to a natural way to split the square superparticular S(''k''/2) into two parts. For example, if ''k'' = 10 then we have (10+2)/10, (10+1)/(10-1), 10/(10-2) as equidistant, which simplified is 6/5, 11/9, 5/4, with 11/9 being the mediant of 6/5 and 5/4, and therefore the corresponding superparticular S5 = (5/4)/(6/5) is split into two parts which are tempered together: (5/4)/(11/9) = 45/44 and (11/9)/(6/5) = 55/54. The semiparticular is therefore S(10-1)/S(10+1) = S9/S11 = 243/242 = (45/44)/(55/54) = ((10+2)/(10-2))/((10+1)/(10-1))2.

This form of comma does not yet have an official name, but a proposed name is "semiparticular", because most of the time it is superparticular but less often it is odd-particular, and because when tempered they all cause an interval to be divided into two equal parts where each part is a (tempered version of a) superparticular or odd-particular, and the interval being divided in half is sometimes quodd-particular, sometimes odd-particular and sometimes superparticular. Specifically:

To find out what a superparticular (''a''+1)/''a'' is approximately half of, temper the semiparticular S(2''a'')/S(2''a''+2) and you can observe that (2''a''+3)/(2''a''-1) is the interval it is approximately half of.

To find out what an odd-particular (2''a''+1)/(2''a''-1) is approximately half of, temper the semiparticular S(2''a''-1)/S(2''a''+1) and you can observe that (2''a''+2)/(2''a''-2) = (''a''+1)/(''a''-1), a superparticular or odd-particular, is the interval it is approximately half of.

To find out what splits a superparticular (''a''+1)/''a'' in half, temper the semiparticular S(4''a''+1)/S(4''a''+3) and you can observe that (4''a''+3)/(4''a''+1), an odd-particular, is the interval that is approximately half of it.

To find out what splits an odd-particular (2''a''+1)/(2''a''-1) in half, temper the semiparticular S(4''a''-2)/S(4''a''+2) and you can observe that (4''a''-1)/(4''a''+1), an odd-particular, is the interval that is approximately half of it.

Here follows a table of semiparticulars. Perhaps many of the patterns will become clearer if you examine this table:

{| class="wikitable center-all

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 == S''k''/S(''k'' + 2) (semiparticulars) ==
-For differences between square-particulars of the form S(''k'' + 1)/S(''k'' + 3) the resulting comma is either [[superparticular]] or "odd-particular", meaning an interval of the form (2''n'' + 1)/(2''n'' - 1) for some positive integer ''n''. (This terminology also suggests "throdd-particular" for intervals of the form (3''n'' + 1)/(3''n'' - 2) and (3''n'' + 2)/(3''n'' - 1) and maybe "quodd-particular" (sounding like "quad-particular") for (4''n'' + 3)/(4''n'' - 1) and (4''n'' + 1)/(4''n'' - 3).) Furthermore, S(''k'' + 1)/S(''k'' + 3), when tempered, implies that (''k'' + 4)/''k'' is divisible exactly into two halves of (''k'' + 3)/(''k'' + 1) which is equated with ((''k'' + 1)/''k'')((''k'' + 4)/(''k'' + 3)). It is for this reason that the indexing choice of S(''k'' + 1)/S(''k'' + 3) was chosen, as two (''k'' + 3)/(''k'' + 1)'s make a (''k'' + 4)/''k''. This form of comma does not yet have an official name, but a proposed name is "semiparticular", because most of the time it is superparticular but less often it is odd-particular, and because when tempered they all cause an interval to be divided into two equal parts where each part is a tempered version of a superparticular or odd-particular, and the interval being divided in half is sometimes quodd-particular, sometimes odd-particular and sometimes superparticular.
+For differences between square-particulars of the form S(''k'' + 1)/S(''k'' + 3) the resulting comma is either [[superparticular]] or [[#Glossary|odd-particular]]. (This terminology also suggests "throdd-particular" for intervals of the form (3''n'' + 1)/(3''n'' - 2) and (3''n'' + 2)/(3''n'' - 1) and maybe "quodd-particular" (sounding like "quad-particular") for (4''n'' + 3)/(4''n'' - 1) and (4''n'' + 1)/(4''n'' - 3).)
-Specifically, S''k''/S(''k'' + 2) is superparticular when ''k'' is not a multiple of 4, and odd-particular otherwise. See [[#Mathematical derivation]] for details on this and other facts stated here.
+Tempering S(''k'' - 1)/S(''k'' + 3) implies that (''k'' + 2)/(''k'' - 2) is divisible exactly into two halves of (''k'' + 1)/(''k'' - 1). It also implies that the intervals (''k'' + 2)/''k'' and ''k''/(''k'' - 2) are equidistant from (''k'' + 1)/(''k'' - 1) because to make them equidistant we need to temper:
+( (''k''/(''k''-2))/((''k''+1)/(''k''-1)) )/( ((''k''+1)/(''k''-1))/((''k''+2)/''k'')  )
+= (''k''+2)/''k'' * ''k''/(''k''-2) / ((''k''+1)/(''k''-1))<sup>2</sup> = ((''k''+2)/(''k''-2))/((''k''+1)/(''k''-1))<sup>2</sup>
+...and notice that the latter expression is the one we've [[#Mathematical derivation|shown is equal to S(''k''-1)/S(''k''+1)]]. 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)
+Also note that in the above, (''k''+1)/(''k''-1) is the [[mediant]] of the adjacent two intervals, meaning that division of an interval into two via tempering a semiparticular is in some sense 'optimal' relative to the complexity. This also means that if ''k'' is a multiple of 2, this corresponds to a natural way to split the square superparticular S(''k''/2) into two parts. For example, if ''k'' = 10 then we have (10+2)/10, (10+1)/(10-1), 10/(10-2) as equidistant, which simplified is 6/5, 11/9, 5/4, with 11/9 being the mediant of 6/5 and 5/4, and therefore the corresponding superparticular S5 = (5/4)/(6/5) is split into two parts which are tempered together: (5/4)/(11/9) = 45/44 and (11/9)/(6/5) = 55/54. The semiparticular is therefore S(10-1)/S(10+1) = S9/S11 = 243/242 = (45/44)/(55/54) = ((10+2)/(10-2))/((10+1)/(10-1))<sup>2</sup>.
+This form of comma does not yet have an official name, but a proposed name is "semiparticular", because most of the time it is superparticular but less often it is odd-particular, and because when tempered they all cause an interval to be divided into two equal parts where each part is a (tempered version of a) superparticular or odd-particular, and the interval being divided in half is sometimes quodd-particular, sometimes odd-particular and sometimes superparticular. Specifically:
+To find out what a superparticular (''a''+1)/''a'' is approximately half of, temper the semiparticular S(2''a'')/S(2''a''+2) and you can observe that (2''a''+3)/(2''a''-1) is the interval it is approximately half of.
+To find out what an odd-particular (2''a''+1)/(2''a''-1) is approximately half of, temper the semiparticular S(2''a''-1)/S(2''a''+1) and you can observe that (2''a''+2)/(2''a''-2) = (''a''+1)/(''a''-1), a superparticular or odd-particular, is the interval it is approximately half of.
+To find out what splits a superparticular (''a''+1)/''a'' in half, temper the semiparticular S(4''a''+1)/S(4''a''+3) and you can observe that (4''a''+3)/(4''a''+1), an odd-particular, is the interval that is approximately half of it.
+To find out what splits an odd-particular (2''a''+1)/(2''a''-1) in half, temper the semiparticular S(4''a''-2)/S(4''a''+2) and you can observe that (4''a''-1)/(4''a''+1), an odd-particular, is the interval that is approximately half of it.
+Here follows a table of semiparticulars. Perhaps many of the patterns will become clearer if you examine this table:
 {| class="wikitable center-all

S-expression: Difference between revisions

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