Tenney–Euclidean metrics: Difference between revisions

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[-0.1440 1.8086 3.0624]

If we want to find the temperamental seminorm T(250/243) of 250/243, we convert it into a monzo as {{monzo| 1 -5 3 }}. Now we may multiply '''P''' by this on the left, obtaining the zero vector. Taking the dot product of the zero vector {{monzo| 1 -5 3 }} gives zero, and taking the square root of zero we get zero, the temperametal seminorm T(250/243) of 250/243. All of this is telling us that 250/243 is a comma of this temperament, which is 5-limit [[Porcupine family|porcupine]].

If we want to find the temperamental seminorm T (250/243) of 250/243, we convert it into a monzo as {{monzo| 1 -5 3 }}. Now we may multiply '''P''' by this on the left, obtaining the zero vector. Taking the dot product of the zero vector {{monzo| 1 -5 3 }} gives zero, and taking the square root of zero we get zero, the temperametal seminorm T (250/243) of 250/243. All of this is telling us that 250/243 is a comma of this temperament, which is 5-limit [[Porcupine family|porcupine]].

Similarly, starting from the monzo {{monzo| -1 1 0 }} for 3/2, we may multiply this by '''P''', obtaining {{val| -0.8793 0.9957 1.9526 }}, and taking the dot product of this with {{monzo| -1 1 0 }} gives 1.875 with square root 1.3693, which is T(3/2).

Similarly, starting from the monzo {{monzo| -1 1 0 }} for 3/2, we may multiply this by '''P''', obtaining {{val| -0.8793 0.9957 1.9526 }}, and taking the dot product of this with {{monzo| -1 1 0 }} gives 1.875 with square root 1.3693, which is T (3/2).

We can, however, map the monzos to elements of a rank ''r'' abelian group (where ''r'' is the rank of the temperament) which abstractly represents the elements of the temperament without regard to tuning, the [[abstract regular temperament]]. If b is a monzo, this mapping is given by ~~bAT~~. Hence we have {{monzo| 1 -5 3 }}~~AT~~ maps to [0 0] for the interval associated to 250/243, and {{monzo| -1 1 0 }}~~AT~~ maps to [9 13] for the interval assciated to 3/2. This is the number of steps needed to get to 3/2 in 15et and 22et respectively. We now may obtain a matrix defining the temperamental norm on this abstract temperament by ''P'' = (AW2AT)-1, which is approximately

We can, however, map the monzos to elements of a rank ''r'' abelian group (where ''r'' is the rank of the temperament) which abstractly represents the elements of the temperament without regard to tuning, the [[abstract regular temperament]]. If b is a monzo, this mapping is given by Ab. Hence we have A{{monzo| 1 -5 3 }} maps to {{monzo| 0 0 }} for the interval associated to 250/243, and A{{monzo| -1 1 0 }} maps to {{monzo| 9 13 }} for the interval assciated to 3/2. This is the number of steps needed to get to 3/2 in 15et and 22et respectively. We now may obtain a matrix defining the temperamental norm on this abstract temperament by ''P'' = (AW2AT)-1, which is approximately

[175.3265 -120.0291]

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[-120.0291 82.1730]

Using this, we find the temperamental norm of [9 13] to be sqrt ([9 13]''P''[9 13]T) ~ sqrt (1.875) ~ 1.3693, identical to the temperamental seminorm of 3/2. Note however that while '''P''' does not depend on the choice of basis vals for the temperament, ''P'' does; if we choose [{{val| 1 2 3 }}, {{val| 0 -3 -5 }}] for our basis instead, then 3/2 is represented by [1 -3] and ''P'' changes coordinates to produce the same final result of temperamental complexity.

Using this, we find the temperamental norm of {{monzo| 9 13 }} to be sqrt ([9 13]''P''[9 13]T) ~ sqrt (1.875) ~ 1.3693, identical to the temperamental seminorm of 3/2. Note however that while '''P''' does not depend on the choice of basis vals for the temperament, ''P'' does; if we choose [{{val| 1 2 3 }}, {{val| 0 -3 -5 }}] for our basis instead, then 3/2 is represented by {{monzo| 1 -3 }} and ''P'' changes coordinates to produce the same final result of temperamental complexity.

If instead we want the OETES, we may remove the first row of [{{val| 1 2 3 }}, {{val| 0 -3 -5 }}], leaving just [{{val| 0 -3 -5 }}]. If we now call this 1×3 matrix A, then (AW2AT)-1 is a 1×1 matrix; in effect a scalar, with value [{{val| 0.1215588 }}]. Multiplying a monzo b ~~times~~ A~~T~~ gives a 1×1 matrix ~~bAT~~ whose value is the number of generator steps of porcupine (of size a tempered 10/9) it takes to get to the octave class to which b belongs. Performing the multiplication and taking the square root, we conclude the OE complexity is simply proportional to this number of steps.

If instead we want the OETES, we may remove the first row of [{{val| 1 2 3 }}, {{val| 0 -3 -5 }}], leaving just [{{val| 0 -3 -5 }}]. If we now call this 1×3 matrix A, then ''P'' = (AW2AT)-1 is a 1×1 matrix; in effect a scalar, with value [{{val| 0.1215588 }}]. Multiplying a monzo b by A on the left gives a 1×1 matrix Ab whose value is the number of generator steps of porcupine (of size a tempered 10/9) it takes to get to the octave class to which b belongs. Performing the multiplication and taking the square root, we conclude the OE complexity is simply proportional to this number of generator steps.

For a more substantial example we need to consider at least a rank three temperament, so let us turn to 7-limit marvel, the 7-limit temperament tempering out 225/224. The 2×4 matrix of monzos whose first row represents 2 and whose second row 225/224 is [{{monzo| 1 0 0 0 }}, {{monzo| -5 2 2 -1 }}]. If we denote log2 of the odd primes by p3, p5, p7 etc, then the monzo weighting of this matrix is M = [{{monzo| 1 0 0 0 }}, {{monzo| -5 2p3 2p5 -p7 }}], and P = I - M+M = [{{monzo| 1 0 0 0 }}, {{monzo| 0 4(p5)2+(p7)2 -4(p3)(p5) 2(p3)(p7) }}/H, {{monzo| 0 -4(p3)(p5) 4(p3)2+(p7)2 2(p5)(p7) }}/H, {{monzo| 0 2(p3)(p7) 2(p5)(p7) 4((p3)2+(p5)2) }}/H], where H = 4(p3)2+4(p5)2+(p7)2. On the other hand, we may start from the normal val list for the temperament, which is [{{val| 1 0 0 -5 }}, {{val| 0 1 0 2 }}, {{val| 0 0 1 2 }}]. Removing the first row gives [{{val| 0 1 0 2 }}, {{val| 0 0 1 2 }}], and val weighting this gives C = [{{val| 0 1/p3 0 2/p7 }}, {{val| 0 0 1/p5 2/p7 }}]. Then P = C+C is precisely the same matrix we obtained before.

For a more substantial example we need to consider at least a rank three temperament, so let us turn to 7-limit marvel, the 7-limit temperament tempering out 225/224. The 2×4 matrix of monzos whose first row represents 2 and whose second row 225/224 is [{{monzo| 1 0 0 0 }}, {{monzo| -5 2 2 -1 }}]. If we denote log2 of the odd primes by p3, p5, p7 etc, then the monzo weighting of this matrix is M = [{{monzo| 1 0 0 0 }}, {{monzo| -5 2p3 2p5 -p7 }}], and P = I - MM+ = [{{monzo| 1 0 0 0 }}, {{monzo| 0 4(p5)2+(p7)2 -4(p3)(p5) 2(p3)(p7) }}/H, {{monzo| 0 -4(p3)(p5) 4(p3)2+(p7)2 2(p5)(p7) }}/H, {{monzo| 0 2(p3)(p7) 2(p5)(p7) 4((p3)2+(p5)2) }}/H], where H = 4(p3)2+4(p5)2+(p7)2. On the other hand, we may start from the normal val list for the temperament, which is [{{val| 1 0 0 -5 }}, {{val| 0 1 0 2 }}, {{val| 0 0 1 2 }}]. Removing the first row gives [{{val| 0 1 0 2 }}, {{val| 0 0 1 2 }}], and val weighting this gives C = [{{val| 0 1/p3 0 2/p7 }}, {{val| 0 0 1/p5 2/p7 }}]. Then P = C+C is precisely the same matrix we obtained before.

Octaves are now projected to the origin as well as commas. We can as before form the quotient space with respect to the seminorm, and obtain a normed space in which octave-equivalent interval classes of the intervals of the temperament are the lattice points. This seminorm applied to monzos gives the OE complexity.

If we start from a normal val list and remove the first val, the remaining vals map to the octave classes of the notes of the temperament. If we call this reduced list of vals R, then the inner product on note classes in this basis is defined by the symmetric matrix S = (RW2RT)-1. In the case of marvel, we obtain S = [[(p3)2(4(p5)2+(p7)2) -4(p3)2(p5)2], [-4(p3)2(p5)2 (p5)2(4(p3)2+(p7)2)]]/H. If k = ~~[a b]~~ is a note class of marvel in the coordinates defined by the truncated val list R, which in this case has a basis corresponding to tempered 3 and 5, then sqrt (~~kSk~~T) gives the OE complexity of the note class.

If we start from a normal val list and remove the first val, the remaining vals map to the octave classes of the notes of the temperament. If we call this reduced list of vals R, then the inner product on note classes in this basis is defined by the symmetric matrix S = (RW2RT)-1. In the case of marvel, we obtain S = [[(p3)2(4(p5)2+(p7)2) -4(p3)2(p5)2], [-4(p3)2(p5)2 (p5)2(4(p3)2+(p7)2)]]/H. If k = {{monzo| ''k''1 ''k''2 }} is a note class of marvel in the coordinates defined by the truncated val list R, which in this case has a basis corresponding to tempered 3 and 5, then sqrt (kTSk) gives the OE complexity of the note class.

[[Category:Math]]

@@ Line 33: / Line 33: @@
 [-0.1440 1.8086 3.0624]
-If we want to find the temperamental seminorm T(250/243) of 250/243, we convert it into a monzo as {{monzo| 1 -5 3 }}. Now we may multiply '''P''' by this on the left, obtaining the zero vector. Taking the dot product of the zero vector {{monzo| 1 -5 3 }} gives zero, and taking the square root of zero we get zero, the temperametal seminorm T(250/243) of 250/243. All of this is telling us that 250/243 is a comma of this temperament, which is 5-limit [[Porcupine family|porcupine]].
+If we want to find the temperamental seminorm T (250/243) of 250/243, we convert it into a monzo as {{monzo| 1 -5 3 }}. Now we may multiply '''P''' by this on the left, obtaining the zero vector. Taking the dot product of the zero vector {{monzo| 1 -5 3 }} gives zero, and taking the square root of zero we get zero, the temperametal seminorm T (250/243) of 250/243. All of this is telling us that 250/243 is a comma of this temperament, which is 5-limit [[Porcupine family|porcupine]].
-Similarly, starting from the monzo {{monzo| -1 1 0 }} for 3/2, we may multiply this by '''P''', obtaining {{val| -0.8793 0.9957 1.9526 }}, and taking the dot product of this with {{monzo| -1 1 0 }} gives 1.875 with square root 1.3693, which is T(3/2).
+Similarly, starting from the monzo {{monzo| -1 1 0 }} for 3/2, we may multiply this by '''P''', obtaining {{val| -0.8793 0.9957 1.9526 }}, and taking the dot product of this with {{monzo| -1 1 0 }} gives 1.875 with square root 1.3693, which is T (3/2).
-We can, however, map the monzos to elements of a rank ''r'' abelian group (where ''r'' is the rank of the temperament) which abstractly represents the elements of the temperament without regard to tuning, the [[abstract regular temperament]]. If b is a monzo, this mapping is given by bA<sup>T</sup>. Hence we have {{monzo| 1 -5 3 }}A<sup>T</sup> maps to [0 0] for the interval associated to 250/243, and {{monzo| -1 1 0 }}A<sup>T</sup> maps to [9 13] for the interval assciated to 3/2. This is the number of steps needed to get to 3/2 in 15et and 22et respectively. We now may obtain a matrix defining the temperamental norm on this abstract temperament by ''P'' = (AW<sup>2</sup>A<sup>T</sup>)<sup>-1</sup>, which is approximately
+We can, however, map the monzos to elements of a rank ''r'' abelian group (where ''r'' is the rank of the temperament) which abstractly represents the elements of the temperament without regard to tuning, the [[abstract regular temperament]]. If b is a monzo, this mapping is given by Ab. Hence we have A{{monzo| 1 -5 3 }} maps to {{monzo| 0 0 }} for the interval associated to 250/243, and A{{monzo| -1 1 0 }} maps to {{monzo| 9 13 }} for the interval assciated to 3/2. This is the number of steps needed to get to 3/2 in 15et and 22et respectively. We now may obtain a matrix defining the temperamental norm on this abstract temperament by ''P'' = (AW<sup>2</sup>A<sup>T</sup>)<sup>-1</sup>, which is approximately
 [175.3265 -120.0291]
@@ Line 43: / Line 43: @@
 [-120.0291 82.1730]
-Using this, we find the temperamental norm of [9 13] to be sqrt ([9 13]''P''[9 13]<sup>T</sup>) ~ sqrt (1.875) ~ 1.3693, identical to the temperamental seminorm of 3/2. Note however that while '''P''' does not depend on the choice of basis vals for the temperament, ''P'' does; if we choose [{{val| 1 2 3 }}, {{val| 0 -3 -5 }}] for our basis instead, then 3/2 is represented by [1 -3] and ''P'' changes coordinates to produce the same final result of temperamental complexity.
+Using this, we find the temperamental norm of {{monzo| 9 13 }} to be sqrt ([9 13]''P''[9 13]<sup>T</sup>) ~ sqrt (1.875) ~ 1.3693, identical to the temperamental seminorm of 3/2. Note however that while '''P''' does not depend on the choice of basis vals for the temperament, ''P'' does; if we choose [{{val| 1 2 3 }}, {{val| 0 -3 -5 }}] for our basis instead, then 3/2 is represented by {{monzo| 1 -3 }} and ''P'' changes coordinates to produce the same final result of temperamental complexity.
-If instead we want the OETES, we may remove the first row of [{{val| 1 2 3 }}, {{val| 0 -3 -5 }}], leaving just [{{val| 0 -3 -5 }}]. If we now call this 1×3 matrix A, then (AW<sup>2</sup>A<sup>T</sup>)<sup>-1</sup> is a 1×1 matrix; in effect a scalar, with value [{{val| 0.1215588 }}]. Multiplying a monzo b times A<sup>T</sup> gives a 1×1 matrix bA<sup>T</sup> whose value is the number of generator steps of porcupine (of size a tempered 10/9) it takes to get to the octave class to which b belongs. Performing the multiplication and taking the square root, we conclude the OE complexity is simply proportional to this number of steps.
+If instead we want the OETES, we may remove the first row of [{{val| 1 2 3 }}, {{val| 0 -3 -5 }}], leaving just [{{val| 0 -3 -5 }}]. If we now call this 1×3 matrix A, then ''P'' = (AW<sup>2</sup>A<sup>T</sup>)<sup>-1</sup> is a 1×1 matrix; in effect a scalar, with value [{{val| 0.1215588 }}]. Multiplying a monzo b by A on the left gives a 1×1 matrix Ab whose value is the number of generator steps of porcupine (of size a tempered 10/9) it takes to get to the octave class to which b belongs. Performing the multiplication and taking the square root, we conclude the OE complexity is simply proportional to this number of generator steps.
-For a more substantial example we need to consider at least a rank three temperament, so let us turn to 7-limit marvel, the 7-limit temperament tempering out 225/224. The 2×4 matrix of monzos whose first row represents 2 and whose second row 225/224 is [{{monzo| 1 0 0 0 }}, {{monzo| -5 2 2 -1 }}]. If we denote log<sub>2</sub> of the odd primes by p3, p5, p7 etc, then the monzo weighting of this matrix is M = [{{monzo| 1 0 0 0 }}, {{monzo| -5 2p3 2p5 -p7 }}], and P = I - M<sup>+</sup>M = [{{monzo| 1 0 0 0 }}, {{monzo| 0 4(p5)<sup>2</sup>+(p7)<sup>2</sup> -4(p3)(p5) 2(p3)(p7) }}/H, {{monzo| 0 -4(p3)(p5) 4(p3)<sup>2</sup>+(p7)<sup>2</sup> 2(p5)(p7) }}/H, {{monzo| 0 2(p3)(p7) 2(p5)(p7) 4((p3)<sup>2</sup>+(p5)<sup>2</sup>) }}/H], where H = 4(p3)<sup>2</sup>+4(p5)<sup>2</sup>+(p7)<sup>2</sup>. On the other hand, we may start from the normal val list for the temperament, which is [{{val| 1 0 0 -5 }}, {{val| 0 1 0 2 }}, {{val| 0 0 1 2 }}]. Removing the first row gives [{{val| 0 1 0 2 }}, {{val| 0 0 1 2 }}], and val weighting this gives C = [{{val| 0 1/p3 0 2/p7 }}, {{val| 0 0 1/p5 2/p7 }}]. Then P = C<sup>+</sup>C is precisely the same matrix we obtained before.
+For a more substantial example we need to consider at least a rank three temperament, so let us turn to 7-limit marvel, the 7-limit temperament tempering out 225/224. The 2×4 matrix of monzos whose first row represents 2 and whose second row 225/224 is [{{monzo| 1 0 0 0 }}, {{monzo| -5 2 2 -1 }}]. If we denote log<sub>2</sub> of the odd primes by p3, p5, p7 etc, then the monzo weighting of this matrix is M = [{{monzo| 1 0 0 0 }}, {{monzo| -5 2p3 2p5 -p7 }}], and P = I - MM<sup>+</sup> = [{{monzo| 1 0 0 0 }}, {{monzo| 0 4(p5)<sup>2</sup>+(p7)<sup>2</sup> -4(p3)(p5) 2(p3)(p7) }}/H, {{monzo| 0 -4(p3)(p5) 4(p3)<sup>2</sup>+(p7)<sup>2</sup> 2(p5)(p7) }}/H, {{monzo| 0 2(p3)(p7) 2(p5)(p7) 4((p3)<sup>2</sup>+(p5)<sup>2</sup>) }}/H], where H = 4(p3)<sup>2</sup>+4(p5)<sup>2</sup>+(p7)<sup>2</sup>. On the other hand, we may start from the normal val list for the temperament, which is [{{val| 1 0 0 -5 }}, {{val| 0 1 0 2 }}, {{val| 0 0 1 2 }}]. Removing the first row gives [{{val| 0 1 0 2 }}, {{val| 0 0 1 2 }}], and val weighting this gives C = [{{val| 0 1/p3 0 2/p7 }}, {{val| 0 0 1/p5 2/p7 }}]. Then P = C<sup>+</sup>C is precisely the same matrix we obtained before.
 Octaves are now projected to the origin as well as commas. We can as before form the quotient space with respect to the seminorm, and obtain a normed space in which octave-equivalent interval classes of the intervals of the temperament are the lattice points. This seminorm applied to monzos gives the OE complexity.
-If we start from a normal val list and remove the first val, the remaining vals map to the octave classes of the notes of the temperament. If we call this reduced list of vals R, then the inner product on note classes in this basis is defined by the symmetric matrix S = (RW<sup>2</sup>R<sup>T</sup>)<sup>-1</sup>. In the case of marvel, we obtain S = [[(p3)<sup>2</sup>(4(p5)<sup>2</sup>+(p7)<sup>2</sup>) -4(p3)<sup>2</sup>(p5)<sup>2</sup>], [-4(p3)<sup>2</sup>(p5)<sup>2</sup> (p5)<sup>2</sup>(4(p3)<sup>2</sup>+(p7)<sup>2</sup>)]]/H. If k = [a b] is a note class of marvel in the coordinates defined by the truncated val list R, which in this case has a basis corresponding to tempered 3 and 5, then sqrt (kSk<sup>T</sup>) gives the OE complexity of the note class.
+If we start from a normal val list and remove the first val, the remaining vals map to the octave classes of the notes of the temperament. If we call this reduced list of vals R, then the inner product on note classes in this basis is defined by the symmetric matrix S = (RW<sup>2</sup>R<sup>T</sup>)<sup>-1</sup>. In the case of marvel, we obtain S = [[(p3)<sup>2</sup>(4(p5)<sup>2</sup>+(p7)<sup>2</sup>) -4(p3)<sup>2</sup>(p5)<sup>2</sup>], [-4(p3)<sup>2</sup>(p5)<sup>2</sup> (p5)<sup>2</sup>(4(p3)<sup>2</sup>+(p7)<sup>2</sup>)]]/H. If k = {{monzo| ''k''<sub>1</sub> ''k''<sub>2</sub> }} is a note class of marvel in the coordinates defined by the truncated val list R, which in this case has a basis corresponding to tempered 3 and 5, then sqrt (k<sup>T</sup>Sk) gives the OE complexity of the note class.
 [[Category:Math]]