S-expression: Difference between revisions

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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).)

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'' + 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:

( large/medium )/( medium/small )

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 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).)
-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'' + 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:
 ( large/medium )/( medium/small )