S-expression: Difference between revisions

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...as expected. In fact, if you look at the expressions in the table, it becomes obvious why we have the central element squared (via a 2 and -2 in the S-factorization), as it follows from the general algebraic form of {{nowrap| ''n''<sup>2</sup> / (''n''<sup>2</sup> - 1) {{=}} ( ''n''/(''n'' - 1) )/( (''n'' + 1)/''n'' ) }}.

'''(It should be noted:''' The same derivation above can be observed as not depending on the number of 0's in the S-factorization of the expressions of {{nowrap| <code>F(k + 1)</code> and <code>F(k + 2)</code> }}, and that therefore we have S''a''/S''b'' being an equidistance relation of intervals, where the generalized versions of the terms {{nowrap| <code>F(k + 1)</code> and <code>F(k + 2)</code> }} correspond to the two differences of intervals of the form (''n'' + ''k'')/''n'', where ''k'' = 3 corresponds to one zero in the monzo and zero in {{nowrap| <code>[ -1, 2, -1, 1, -2, 1 ]</code> }}, and where the terms {{nowrap| <code>F(k + 1)</code> and <code>F(k + 2)</code> }} therefore correspond to the two differences used to make three intervals of the form (''n'' + ''k'')/''n'' equidistant. "Which three intervals" is thus answered as those of 3 consecutive/adjacent values of ''n''. '''Therefore:''' What makes three-particulars special in this more general S-comma family is that the largest and smallest interval of the three almost compose to something simple, up to a superparticular difference, s.t the corresponding splitting relation can often simplify for cases of interest, as discussed in [[S-expression#Significance_7|#Significance]]'''.)'''

'''(An important corollary that should be noted:''' The same derivation above can be observed as not depending on the number of 0's in the S-factorization of the expressions of {{nowrap| <code>F(k + 1)</code> and <code>F(k + 2)</code> }}, and that therefore we have S''a''/S''b'' being an equidistance relation of intervals, where the generalized versions of the terms {{nowrap| <code>F(k + 1)</code> and <code>F(k + 2)</code> }} correspond to the two differences of intervals of the form (''n'' + ''k'')/''n'', where ''k'' = 3 corresponds to one zero in the monzo and zero in {{nowrap| <code>[ -1, 2, -1, 1, -2, 1 ]</code> }}, and where the terms {{nowrap| <code>F(k + 1)</code> and <code>F(k + 2)</code> }} therefore correspond to the two differences used to make three intervals of the form (''n'' + ''k'')/''n'' equidistant. "Which three intervals" is thus answered as those of 3 consecutive/adjacent values of ''n''. '''Therefore:''' What makes three-particulars special in this more general S-comma family is that the largest and smallest interval of the three almost compose to something simple, up to a superparticular difference, s.t the corresponding splitting relation can often simplify for cases of interest, as discussed in [[S-expression#Significance_7|#Significance]]'''.)'''

=== Table of three-particulars ===

@@ Line 2,185: / Line 2,185: @@
 ...as expected. In fact, if you look at the expressions in the table, it becomes obvious why we have the central element squared (via a 2 and -2 in the S-factorization), as it follows from the general algebraic form of {{nowrap| ''n''<sup>2</sup> / (''n''<sup>2</sup> - 1) {{=}} ( ''n''/(''n'' - 1) )/( (''n'' + 1)/''n'' ) }}.
-'''(It should be noted:''' The same derivation above can be observed as not depending on the number of 0's in the S-factorization of the expressions of {{nowrap| <code>F(k + 1)</code> and <code>F(k + 2)</code> }}, and that therefore we have S''a''/S''b'' being an equidistance relation of intervals, where the generalized versions of the terms {{nowrap| <code>F(k + 1)</code> and <code>F(k + 2)</code> }} correspond to the two differences of intervals of the form (''n'' + ''k'')/''n'', where ''k'' = 3 corresponds to one zero in the monzo and zero in {{nowrap| <code>[ -1, 2, -1, 1, -2, 1 ]</code> }}, and where the terms {{nowrap| <code>F(k + 1)</code> and <code>F(k + 2)</code> }} therefore correspond to the two differences used to make three intervals of the form (''n'' + ''k'')/''n'' equidistant. "Which three intervals" is thus answered as those of 3 consecutive/adjacent values of ''n''. '''Therefore:''' What makes three-particulars special in this more general S-comma family is that the largest and smallest interval of the three almost compose to something simple, up to a superparticular difference, s.t the corresponding splitting relation can often simplify for cases of interest, as discussed in [[S-expression#Significance_7|#Significance]]'''.)'''
+'''(An important corollary that should be noted:''' The same derivation above can be observed as not depending on the number of 0's in the S-factorization of the expressions of {{nowrap| <code>F(k + 1)</code> and <code>F(k + 2)</code> }}, and that therefore we have S''a''/S''b'' being an equidistance relation of intervals, where the generalized versions of the terms {{nowrap| <code>F(k + 1)</code> and <code>F(k + 2)</code> }} correspond to the two differences of intervals of the form (''n'' + ''k'')/''n'', where ''k'' = 3 corresponds to one zero in the monzo and zero in {{nowrap| <code>[ -1, 2, -1, 1, -2, 1 ]</code> }}, and where the terms {{nowrap| <code>F(k + 1)</code> and <code>F(k + 2)</code> }} therefore correspond to the two differences used to make three intervals of the form (''n'' + ''k'')/''n'' equidistant. "Which three intervals" is thus answered as those of 3 consecutive/adjacent values of ''n''. '''Therefore:''' What makes three-particulars special in this more general S-comma family is that the largest and smallest interval of the three almost compose to something simple, up to a superparticular difference, s.t the corresponding splitting relation can often simplify for cases of interest, as discussed in [[S-expression#Significance_7|#Significance]]'''.)'''
 === Table of three-particulars ===