S-expression: Difference between revisions

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== Sk*S(k + 1) (triangle-particulars) ==

=== Significance ===

1. Every triangle-particular is superparticular, so these are efficient commas. (See also the [[#short proof of the superparticularity of triangle-particulars]].)

If we ~~examine~~ (''k'' + 1)/''k'' ~~then~~ we ~~can notice~~ that if we equate (''k'' + 2)/(''k'' + 1) with ''k''/(''k'' - 1), we have:

2. Often each individual triangle-particular, taken as a comma, implies other useful equivalences not necessarily corresponding to the general form, speaking of which...

3. Every triangle-particular is the difference between two nearly-adjacent superparticular intervals (''k'' + 2)/(''k'' + 1) and ''k''/(''k'' - 1).

4. Tempering any two consecutive square-particulars S''k'' and S(''k'' + 1) implies tempering a triangle-particular, so these are common commas. (See also: [[lopsided comma]]s.)

5. If we temper S''k'' * S(''k'' + 1) but not S''k'' or S(''k'' + 1), then one or more intervals of ''k''/(''k'' - 1), (''k'' + 1)/''k'' and (''k'' + 2)/(''k'' + 1) ''must'' be mapped inconsistently, because:

: if (''k'' + 1)/''k'' is mapped above (''k'' + 2)/(''k'' + 1) ~ k/(k-1) we have (''k'' + 1)/''k'' > ''k''/(''k'' - 1) and if it is mapped below we have (''k'' + 1)/''k'' < (''k'' + 2)/(''k'' + 1).

: (Generalisations of this and their implications for consistency are discussed in [[#Sk*S(k + 1)*...*S(k + n - 1) (1/n-square-particulars)]].)

=== Meaning ===

Notice that if we equate (''k'' + 2)/(''k'' + 1) with ''k''/(''k'' - 1) (by [[tempering out]] their difference), then multiply both sides by (''k'' + 1)/''k'', we have:

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

~~Which is~~ to ~~say that if we temper S''k''*S~~(''k'' + 1) ~~= (''k''~~/(''k'' ~~- 1))/((''k'' + 1)/''k'') * ((''k'' + 1)/''k'')/((''k'' + 2)/~~(''k'' + 1)~~) = (''k''~~/(''k'' - 1))/((''k'' + 2)/(''k'' + 1)) then this equivalence is achieved. Note that there is little to no reason to not also temper S''k'' and S(''k''+1) individually unless other considerations seem to force your hand. Another reason commas of this form are of note is they are always [[superparticular]].

...which simplifies to: (''k'' + 2)/''k'' = (''k'' + 1)/(''k'' - 1).

~~It is also an interesting consequence~~ that if we temper S~~''k''*S(''k'' + 1) but not S''~~k~~'' or~~ S(''k'' + 1)~~, then one or more intervals of ''~~k''/(''k'' - 1), (''k'' + 1)/''k~~'' and~~ (''k~~'' + 2)/(''k'' + 1) ''must'' be mapped inconsistently, because if (''k''~~ + 1)/''k~~'' is mapped above~~ (''k'' + 2)/(~~''k'' + 1) ~ k/(k-1) we have (''~~k'' + 1)~~/''~~k~~'' > ''k''~~/(~~''k'' - 1) and if it is mapped below we have (''~~k'' + 1)~~/''k'' <~~ (''k'' + 2)/(''k'' + 1)~~. (Generalisations of this and their implications for consistency are discussed in [[#Sk*S(k + 1)*...*S(k + n - 1) (1~~/~~n-square-particulars)]].)~~

This means that if we temper: <math> {\rm S}k \cdot {\rm S}(k+1) \large = \frac{k/(k-1)}{(k+1)/k} \cdot \frac{(k+1)/k}{(k+2)/(k+1)} = \frac{k/(k+1)}{(k+2)/(k+1)}</math>

~~A short proof of the superparticularity of~~ S''k''*S(''k'' + 1) ~~is as follows:~~

...then this equivalence is achieved. Note that there is little to no reason to not also temper S''k'' and S(''k'' + 1) individually unless other considerations seem to force your hand.

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

=== Short proof of the superparticularity of triangle-particulars ===

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

Then notice that ''k''2 + ''k'' is always a multiple of 2, therefore the above always simplifies to a superparticular. Half of this superparticular is halfway between the corresponding square-particulars, and because of its composition it could therefore be reasoned that it'd likely be half as accurate as tempering either of the square-particulars individually, so these are "1/2-square-particulars" in a sense, and half of a square is a triangle, which is not a coincidence here because the numerators of all of these superparticular intervals/commas are [[triangular number]]s!

Then notice that ''k''2 + ''k'' is always a multiple of 2, therefore the above always simplifies to a superparticular. Half of this superparticular is halfway between the corresponding square-particulars, and because of its composition it could therefore be reasoned that it'd likely be half as accurate as tempering either of the square-particulars individually, so these are "1/2-square-particulars" in a sense, and half of a square is a triangle, which is not a coincidence here because the numerators of all of these superparticular intervals/commas are [[triangular number]]s! (Hence the alternative name "[[triangle-particular]]".)

For completeness, all the ~~commas~~ of this form are included, because ~~these "commas" (intervals rather) have~~ structural importance for JI, and for the possibility of consistency of mappings for the above reason.

=== Table of triangle-particulars ===

For completeness, all the intervals of this form are included, because of their structural importance for JI, and for the possibility of (in)consistency of mappings when tempered for the above reason.

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@@ Line 205: / Line 205: @@
 == Sk*S(k + 1) (triangle-particulars) ==
+=== Significance ===
+. Every triangle-particular is superparticular, so these are efficient commas. (See also the [[#short proof of the superparticularity of triangle-particulars]].)
-If we examine (''k'' + 1)/''k'' then we can notice that if we equate (''k'' + 2)/(''k'' + 1) with ''k''/(''k'' - 1), we have:
+. Often each individual triangle-particular, taken as a comma, implies other useful equivalences not necessarily corresponding to the general form, speaking of which...
+. Every triangle-particular is the difference between two nearly-adjacent superparticular intervals (''k'' + 2)/(''k'' + 1) and ''k''/(''k'' - 1).
+. Tempering any two consecutive square-particulars S''k'' and S(''k'' + 1) implies tempering a triangle-particular, so these are common commas. (See also: [[lopsided comma]]s.)
+. If we temper S''k'' * S(''k'' + 1) but not S''k'' or S(''k'' + 1), then one or more intervals of ''k''/(''k'' - 1), (''k'' + 1)/''k'' and (''k'' + 2)/(''k'' + 1) ''must'' be mapped inconsistently, because:
+: if (''k'' + 1)/''k'' is mapped above (''k'' + 2)/(''k'' + 1) ~ k/(k-1) we have (''k'' + 1)/''k'' > ''k''/(''k'' - 1) and if it is mapped below we have (''k'' + 1)/''k'' < (''k'' + 2)/(''k'' + 1).
+: (Generalisations of this and their implications for consistency are discussed in [[#Sk*S(k + 1)*...*S(k + n - 1) (1/n-square-particulars)]].)
+=== Meaning ===
+Notice that if we equate (''k'' + 2)/(''k'' + 1) with ''k''/(''k'' - 1) (by [[tempering out]] their difference), then multiply both sides by (''k'' + 1)/''k'', we have:
 (''k'' + 2)/(''k'' + 1) * (''k'' + 1)/''k'' = (''k'' + 1)/''k'' * ''k''/(''k'' - 1)
-Which is to say that if we temper S''k''*S(''k'' + 1) = (''k''/(''k'' - 1))/((''k'' + 1)/''k'') * ((''k'' + 1)/''k'')/((''k'' + 2)/(''k'' + 1)) = (''k''/(''k'' - 1))/((''k'' + 2)/(''k'' + 1)) then this equivalence is achieved. Note that there is little to no reason to not also temper S''k'' and S(''k''+1) individually unless other considerations seem to force your hand. Another reason commas of this form are of note is they are always [[superparticular]].
+...which simplifies to: (''k'' + 2)/''k'' = (''k'' + 1)/(''k'' - 1).
-It is also an interesting consequence that if we temper S''k''*S(''k'' + 1) but not S''k'' or S(''k'' + 1), then one or more intervals of ''k''/(''k'' - 1), (''k'' + 1)/''k'' and (''k'' + 2)/(''k'' + 1) ''must'' be mapped inconsistently, because if (''k'' + 1)/''k'' is mapped above (''k'' + 2)/(''k'' + 1) ~ k/(k-1) we have (''k'' + 1)/''k'' > ''k''/(''k'' - 1) and if it is mapped below we have (''k'' + 1)/''k'' < (''k'' + 2)/(''k'' + 1). (Generalisations of this and their implications for consistency are discussed in [[#Sk*S(k + 1)*...*S(k + n - 1) (1/n-square-particulars)]].)
+This means that if we temper: <math> {\rm S}k \cdot {\rm S}(k+1) \large = \frac{k/(k-1)}{(k+1)/k} \cdot \frac{(k+1)/k}{(k+2)/(k+1)} = \frac{k/(k+1)}{(k+2)/(k+1)}</math>
-A short proof of the superparticularity of S''k''*S(''k'' + 1) is as follows:
+...then this equivalence is achieved. Note that there is little to no reason to not also temper S''k'' and S(''k'' + 1) individually unless other considerations seem to force your hand.
-S''k''*S(''k'' + 1) = (''k''/(''k'' - 1))/((''k'' + 2)/(''k'' + 1)) = (''k''(''k'' + 1))/((''k'' - 1)(''k'' + 2)) = (''k''<sup>2</sup> + ''k'')/(''k''<sup>2</sup> + ''k'' - 2)
+=== Short proof of the superparticularity of triangle-particulars ===
+S''k''*S(''k'' + 1) = ( ''k''/(''k'' - 1) )/( (''k'' + 2)/(''k'' + 1) ) = ( ''k''(''k'' + 1) )/( (''k'' - 1)(''k'' + 2) ) = (''k''<sup>2</sup> + ''k'')/(''k''<sup>2</sup> + ''k'' - 2)
-Then notice that ''k''<sup>2</sup> + ''k'' is always a multiple of 2, therefore the above always simplifies to a superparticular. Half of this superparticular is halfway between the corresponding square-particulars, and because of its composition it could therefore be reasoned that it'd likely be half as accurate as tempering either of the square-particulars individually, so these are "1/2-square-particulars" in a sense, and half of a square is a triangle, which is not a coincidence here because the numerators of all of these superparticular intervals/commas are [[triangular number]]s!
+Then notice that ''k''<sup>2</sup> + ''k'' is always a multiple of 2, therefore the above always simplifies to a superparticular. Half of this superparticular is halfway between the corresponding square-particulars, and because of its composition it could therefore be reasoned that it'd likely be half as accurate as tempering either of the square-particulars individually, so these are "1/2-square-particulars" in a sense, and half of a square is a triangle, which is not a coincidence here because the numerators of all of these superparticular intervals/commas are [[triangular number]]s! (Hence the alternative name "[[triangle-particular]]".)
-For completeness, all the commas of this form are included, because these "commas" (intervals rather) have structural importance for JI, and for the possibility of consistency of mappings for the above reason.
+=== Table of triangle-particulars ===
+For completeness, all the intervals of this form are included, because of their structural importance for JI, and for the possibility of (in)consistency of mappings when tempered for the above reason.
 {| class="wikitable center-all