S-expression: Difference between revisions

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which is square-(super)particular ''k'' for a given integer ''k > 1''. A suggested shorthand for this interval is '''Sk''' for the ''k''-th square superparticular, where the ''S'' stands for "(Shorthand for) Second-order/Square Superparticular". This will be used later in this article. Note that this means S2 = [[4/3]] is the first musically meaningful square-particular, as S1 = 1/0.

which is square-(super)particular ''k'' for a given integer ''k > 1''. A suggested shorthand for this interval is '''S''k''''' for the ''k''-th square superparticular, where the ''S'' stands for "(Shorthand for) Second-order/Square Superparticular". This will be used later in this article. Note that this means S2 = [[4/3]] is the first musically meaningful square-particular, as S1 = 1/0.

Square-particulars are important structurally because they are the intervals between consecutive superparticular intervals while simultaneously being superparticular themselves, which means that whether and how they are tempered tells us information about how well a temperament can represent the harmonic series up to the (''n'' + 1)th harmonic, as well as the potential representational sacrifices that must be made from that point onward.

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== S''k''*S(''k'' + 1) (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:

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

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

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

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]].

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~~2<~~/~~sup> + k~~ - 2)

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)]].)

Then notice that k2 + 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 commas ~~(or intervals)~~ are [[triangular number]]s!

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

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!

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.

@@ Line 4: / Line 4: @@
 <math>\frac {k^2}{k^2 - 1} = \frac {k/(k - 1)}{(k + 1)/k}</math>
-which is square-(super)particular ''k'' for a given integer ''k > 1''. A suggested shorthand for this interval is '''Sk''' for the ''k''-th square superparticular, where the ''S'' stands for "(Shorthand for) Second-order/Square Superparticular". This will be used later in this article. Note that this means S2 = [[4/3]] is the first musically meaningful square-particular, as S1 = 1/0.
+which is square-(super)particular ''k'' for a given integer ''k > 1''. A suggested shorthand for this interval is '''S''k''''' for the ''k''-th square superparticular, where the ''S'' stands for "(Shorthand for) Second-order/Square Superparticular". This will be used later in this article. Note that this means S2 = [[4/3]] is the first musically meaningful square-particular, as S1 = 1/0.
 Square-particulars are important structurally because they are the intervals between consecutive superparticular intervals while simultaneously being superparticular themselves, which means that whether and how they are tempered tells us information about how well a temperament can represent the harmonic series up to the (''n'' + 1)th harmonic, as well as the potential representational sacrifices that must be made from that point onward.
@@ Line 20: / Line 20: @@
 == S''k''*S(''k'' + 1) (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:
+If we examine (''k'' + 1)/''k'' then we can notice that if we equate (''k'' + 2)/(''k'' + 1) with ''k''/(''k'' - 1), we have:
-(k+2)/(k+1) * (k+1)/k = (k+1)/k * k/(k-1)
+(''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]]. 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)]].) A short proof of the superparticularity of S''k''*S(''k'' + 1) is as follows:
+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]].
-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)
+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)]].)
-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 commas (or intervals) are [[triangular number]]s!
+A short proof of the superparticularity of S''k''*S(''k'' + 1) is as follows:
+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!
 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.