User:Frostburn/Theory From First Principles: Difference between revisions

Line 66:

</math>

However rational approximations to the <math>\mathsf{JIP}</math> do belong in our space. Equal temperaments can be represented by such approximations called [[~~Val~~|vals]] which we define as

However rational approximations to the <math>\mathsf{JIP}</math> do belong in our space. Equal temperaments can be represented by such approximations called [[Patent_val#Generalized_patent_val|generalized patent vals]] which we define as

<math>\mathrm{~~val~~}(n; a, b, \ldots, z) := n \overrightarrow{a}^{-1} + \lfloor n \log_a(b) + \frac{1}{2}\rfloor \overrightarrow{b}^{-1} + \ldots + \lfloor n \log_a(z) + \frac{1}{2}\rfloor \overrightarrow{z}^{-1}</math> ,

<math>\mathrm{gpv}(n; a, b, \ldots, z) := n \overrightarrow{a}^{-1} + \lfloor n \log_a(b) + \frac{1}{2}\rfloor \overrightarrow{b}^{-1} + \ldots + \lfloor n \log_a(z) + \frac{1}{2}\rfloor \overrightarrow{z}^{-1}</math> ,

which belong to a class of [[Val|vals]] complementary to [[Monzo|monzos]].

Usually the basis is obvious from context e.g. <math>a = 2, b = 3, c = 5</math>. In these cases we use a left-facing arrow e.g.

<math>\overleftarrow{12} := \mathrm{~~val~~}(12; 2, 3, 5) = 12 e^2 + 19 e^3 + 28 e^5 =: \langle 12, 19, 28 \rbrack</math>

<math>\overleftarrow{12} := \mathrm{gpv}(12; 2, 3, 5) = 12 e^2 + 19 e^3 + 28 e^5 =: \langle 12, 19, 28 \rbrack</math>

We can use these new objects to calculate how many steps of 12edo a tempered interval spans e.g.

Line 84:

Line 86:

The geometric inverses are mainly relevant for subgroup temperaments. Consider [[The_Archipelago#Subgroup_temperaments|Barbados]]:

<math>\overleftarrow{5} := \mathrm{~~val~~}(5; 2, 3, 13/5) = 5 \cdot \overrightarrow{2}^{-1} + 8 \cdot \overrightarrow{3}^{-1} + 7 \cdot \overrightarrow{13/5}^{-1} = 5 e^2 + 8 e^3 - \frac{7}{2}e^5 + \frac{7}{2}e^{13}</math>

<math>\overleftarrow{5} := \mathrm{gpv}(5; 2, 3, 13/5) = 5 \cdot \overrightarrow{2}^{-1} + 8 \cdot \overrightarrow{3}^{-1} + 7 \cdot \overrightarrow{13/5}^{-1} = 5 e^2 + 8 e^3 - \frac{7}{2}e^5 + \frac{7}{2}e^{13}</math>

We can verify that the comma 676/675 indeed vanishes using this val:

@@ Line 66: / Line 66: @@
 </math>
-However rational approximations to the <math>\mathsf{JIP}</math> do belong in our space. Equal temperaments can be represented by such approximations called [[Val|vals]] which we define as
+However rational approximations to the <math>\mathsf{JIP}</math> do belong in our space. Equal temperaments can be represented by such approximations called [[Patent_val#Generalized_patent_val|generalized patent vals]] which we define as
-<math>\mathrm{val}(n; a, b, \ldots, z) := n \overrightarrow{a}^{-1} + \lfloor n \log_a(b) + \frac{1}{2}\rfloor \overrightarrow{b}^{-1} + \ldots + \lfloor n \log_a(z) + \frac{1}{2}\rfloor \overrightarrow{z}^{-1}</math> ,
+<math>\mathrm{gpv}(n; a, b, \ldots, z) := n \overrightarrow{a}^{-1} + \lfloor n \log_a(b) + \frac{1}{2}\rfloor \overrightarrow{b}^{-1} + \ldots + \lfloor n \log_a(z) + \frac{1}{2}\rfloor \overrightarrow{z}^{-1}</math> ,
+which belong to a class of [[Val|vals]] complementary to [[Monzo|monzos]].
 Usually the basis is obvious from context e.g. <math>a = 2, b = 3, c = 5</math>. In these cases we use a left-facing arrow e.g.
-<math>\overleftarrow{12} := \mathrm{val}(12; 2, 3, 5) = 12 e^2 + 19 e^3 + 28 e^5 =: \langle 12, 19, 28 \rbrack</math>
+<math>\overleftarrow{12} := \mathrm{gpv}(12; 2, 3, 5) = 12 e^2 + 19 e^3 + 28 e^5 =: \langle 12, 19, 28 \rbrack</math>
 We can use these new objects to calculate how many steps of 12edo a tempered interval spans e.g.
@@ Line 84: / Line 86: @@
 The geometric inverses are mainly relevant for subgroup temperaments. Consider [[The_Archipelago#Subgroup_temperaments|Barbados]]:
-<math>\overleftarrow{5} := \mathrm{val}(5; 2, 3, 13/5) = 5 \cdot \overrightarrow{2}^{-1} + 8 \cdot \overrightarrow{3}^{-1} + 7 \cdot \overrightarrow{13/5}^{-1} = 5 e^2 + 8 e^3 - \frac{7}{2}e^5 + \frac{7}{2}e^{13}</math>
+<math>\overleftarrow{5} := \mathrm{gpv}(5; 2, 3, 13/5) = 5 \cdot \overrightarrow{2}^{-1} + 8 \cdot \overrightarrow{3}^{-1} + 7 \cdot \overrightarrow{13/5}^{-1} = 5 e^2 + 8 e^3 - \frac{7}{2}e^5 + \frac{7}{2}e^{13}</math>
 We can verify that the comma 676/675 indeed vanishes using this val: