Patent val/Properties: Difference between revisions

(2 intermediate revisions by 2 users not shown)

Line 7:

For every prime ''q'' in the ''p''-limit, solve for ''n'':

<math> \displaystyle

<center><math> \displaystyle

\operatorname {round} (n \log_2 q) = v_{\pi (q)} </math>

\operatorname {round} (n \log_2 q) = v_{\pi (q)} </math></center>

The solution is

Note that <math>\pi (n)</math> is the {{W|prime counting function}}; that is, the number of prime numbers less than or equal to <math>n</math>. The solution is

<math> \displaystyle

<center><math> \displaystyle

\frac {v_{\pi (q)} - 1/2}{\log_2 (q)} < n < \frac {v_{\pi (q)} + 1/2}{\log_2 (q)} </math>

\frac {v_{\pi (q)} - 1/2}{\log_2 (q)} \le n < \frac {v_{\pi (q)} + 1/2}{\log_2 (q)} </math></center>

Denote the solution sets as ''N''1, ''N''2, …, ''N''π(''p''). Find their {{w|intersection (set theory)|intersection}} ''N'', that is,

<math> \displaystyle

<center><math> \displaystyle

N = \bigcap_{i = 1}^{\pi (p)} N_i </math>

N = \bigcap_{i = 1}^{\pi (p)} N_i </math></center>

Then ''V'' is a GPV of every edo in ''N'' if ''N'' is non-empty ~~and not a singleton (i.e. a single point)~~; otherwise it is not a GPV.

Then ''V'' is a GPV of every edo in ''N'' if ''N'' is non-empty; otherwise it is not a GPV.

== Cardinality ==

Line 63:

Notice that the all-zero val is a GPV, so you can always enter it for the first one. It is guaranteed that you can iterate through all GPVs in this method. In practice it is recommended to start with the last entry and going backwards in step 2 for better performance, because the largest prime is most likely to increment.

== A more efficient method ==

We consider the values of <math>n</math> for which the mapping of a prime increments by a step. The just tuning of a prime <math>p</math> is <math>n\log_2(p)</math> steps of <math>n</math>-edo, and as <math>n</math> increases, the mapping of prime <math>p</math> will increase by a step when <math>n\log_2(p)</math> is a half-integer. In particular, if prime <math>p</math> is mapped to <math>m_p</math> steps of <math>n</math>-edo, the smallest value of <math>n</math> where prime <math>p</math> is instead mapped to <math>m_p + 1</math> steps occurs when

<center><math>n\log_2 (p) = m_p + \dfrac {1}{2},</math></center>

and solving for <math>n</math>, we get

<center><math>n = \dfrac {m_p + \tfrac{1}{2}}{\log_2(p)}.</math></center>

If we start from any GPV <math>⟨m_2 \: m_3 \: m_5 \: \dots ]</math>, we find the list

<center><math>\left[ \dfrac {m_2 + \tfrac{1}{2}}{\log_2(2)} \:

\dfrac {m_3 + \tfrac{1}{2}}{\log_2(3)} \:

\dots \right],</math></center>

and the next GPV occurs when <math>n</math> reaches the smallest number in that list, say the one corresponding to prime <math>p</math>. Then the next GPV increments the mapping of prime <math>p</math> by one step, while keeping the mappings of all other primes the same. We also add <math>\tfrac {1}{\log_2(p)}</math> to the entry of the list corresponding to prime <math>p</math>. We repeat the steps of finding the smallest entry in the list, incrementing the mapping of the corresponding prime by one step, and adding <math>\tfrac {1}{\log_2(p)}</math> to the corresponding entry in the list to repeatedly find the next GPV.

[[Category:Regular temperament theory]]

[[Category:Math]]

[[Category:Val]]

@@ Line 7: / Line 7: @@
 For every prime ''q'' in the ''p''-limit, solve for ''n'':
-<math> \displaystyle
+<center><math> \displaystyle
-\operatorname {round} (n \log_2 q) = v_{\pi (q)} </math>
+\operatorname {round} (n \log_2 q) = v_{\pi (q)} </math></center>
-The solution is
+Note that <math>\pi (n)</math> is the {{W|prime counting function}}; that is, the number of prime numbers less than or equal to <math>n</math>. The solution is
-<math> \displaystyle
+<center><math> \displaystyle
-\frac {v_{\pi (q)} - 1/2}{\log_2 (q)} < n < \frac {v_{\pi (q)} + 1/2}{\log_2 (q)} </math>
+\frac {v_{\pi (q)} - 1/2}{\log_2 (q)} \le n < \frac {v_{\pi (q)} + 1/2}{\log_2 (q)} </math></center>
 Denote the solution sets as ''N''<sub>1</sub>, ''N''<sub>2</sub>, …, ''N''<sub>π(''p'')</sub>. Find their {{w|intersection (set theory)|intersection}} ''N'', that is,
-<math> \displaystyle
+<center><math> \displaystyle
-N = \bigcap_{i = 1}^{\pi (p)} N_i </math>
+N = \bigcap_{i = 1}^{\pi (p)} N_i </math></center>
-Then ''V'' is a GPV of every edo in ''N'' if ''N'' is non-empty and not a singleton (i.e. a single point); otherwise it is not a GPV.
+Then ''V'' is a GPV of every edo in ''N'' if ''N'' is non-empty; otherwise it is not a GPV.
 == Cardinality ==
@@ Line 63: / Line 63: @@
 Notice that the all-zero val is a GPV, so you can always enter it for the first one. It is guaranteed that you can iterate through all GPVs in this method. In practice it is recommended to start with the last entry and going backwards in step 2 for better performance, because the largest prime is most likely to increment.
+== A more efficient method ==
+We consider the values of <math>n</math> for which the mapping of a prime increments by a step. The just tuning of a prime <math>p</math> is <math>n\log_2(p)</math> steps of <math>n</math>-edo, and as <math>n</math> increases, the mapping of prime <math>p</math> will increase by a step when <math>n\log_2(p)</math> is a half-integer. In particular, if prime <math>p</math> is mapped to <math>m_p</math> steps of <math>n</math>-edo, the smallest value of <math>n</math> where prime <math>p</math> is instead mapped to <math>m_p + 1</math> steps occurs when
+<center><math>n\log_2 (p) = m_p + \dfrac {1}{2},</math></center>
+and solving for <math>n</math>, we get
+<center><math>n = \dfrac {m_p + \tfrac{1}{2}}{\log_2(p)}.</math></center>
+If we start from any GPV <math>⟨m_2 \: m_3 \: m_5 \: \dots ]</math>, we find the list
+<center><math>\left[ \dfrac {m_2 + \tfrac{1}{2}}{\log_2(2)} \:
+\dfrac {m_3 + \tfrac{1}{2}}{\log_2(3)} \:
+\dots \right],</math></center>
+and the next GPV occurs when <math>n</math> reaches the smallest number in that list, say the one corresponding to prime <math>p</math>. Then the next GPV increments the mapping of prime <math>p</math> by one step, while keeping the mappings of all other primes the same. We also add <math>\tfrac {1}{\log_2(p)}</math> to the entry of the list corresponding to prime <math>p</math>. We repeat the steps of finding the smallest entry in the list, incrementing the mapping of the corresponding prime by one step, and adding <math>\tfrac {1}{\log_2(p)}</math> to the corresponding entry in the list to repeatedly find the next GPV.
 [[Category:Regular temperament theory]]
 [[Category:Math]]
 [[Category:Val]]