S-expression: Difference between revisions

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== Sk/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 [[#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)~~.)

For differences between square-particulars of the form S(''k'' - 1)/S(''k'' + 1) the resulting comma is either [[superparticular]] or [[#Glossary|odd-particular]]. (This terminology also suggests [[#Glossary|throdd-particular]] and [[#Glossary|quodd-particular]] as generalizations.)

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'' (=~~small~~) and ''k''/(''k'' - 2) (=~~large~~) are equidistant from (''k'' + 1)/(''k'' - 1) (=~~medium~~) because to make them equidistant we need to temper:

Tempering S(''k'' - 1)/S(''k'' + 1) 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'' (=s) and ''k''/(''k'' - 2) (=L) are equidistant from (''k'' + 1)/(''k'' - 1) (=M) because to make them equidistant we need to temper:

~~( large/medium )~~/~~( medium~~/~~small )~~

<math>

\LARGE \frac{L/M}{M/s} = \frac{ \left(\frac{k}{k-2}\right)/\left(\frac{k+1}{k-1}\right) }{ \left(\frac{k+1}{k-1}\right)/\left(\frac{k+2}{k}\right) } = \frac{Ls}{m^2} = \frac{\frac{k+2}{k-2}}{\left(\frac{k+1}{k-1}\right)^2}

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

</math>

= ~~large * small / medium2~~

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

@@ Line 889: / Line 889: @@
 == Sk/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 [[#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).)
+For differences between square-particulars of the form S(''k'' - 1)/S(''k'' + 1) the resulting comma is either [[superparticular]] or [[#Glossary|odd-particular]]. (This terminology also suggests [[#Glossary|throdd-particular]] and [[#Glossary|quodd-particular]] as generalizations.)
-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'' (=small) and ''k''/(''k'' - 2) (=large) are equidistant from (''k'' + 1)/(''k'' - 1) (=medium) because to make them equidistant we need to temper:
+Tempering S(''k'' - 1)/S(''k'' + 1) 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'' (=s) and ''k''/(''k'' - 2) (=L) are equidistant from (''k'' + 1)/(''k'' - 1) (=M) because to make them equidistant we need to temper:
-( large/medium )/( medium/small )
+<math>
+\LARGE \frac{L/M}{M/s} = \frac{ \left(\frac{k}{k-2}\right)/\left(\frac{k+1}{k-1}\right) }{ \left(\frac{k+1}{k-1}\right)/\left(\frac{k+2}{k}\right) } = \frac{Ls}{m^2} = \frac{\frac{k+2}{k-2}}{\left(\frac{k+1}{k-1}\right)^2}
-= ( (''k''/(''k''-2))/((''k''+1)/(''k''-1)) )/( ((''k''+1)/(''k''-1))/((''k''+2)/''k'')  )
+</math>
-= large * small / medium<sup>2</sup>
-= (''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: