Patent val/Properties: Difference between revisions

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link back to GPV page
Adjacent GPVs property -> sorting property, and improve and databox the proof
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\mathrm {N} = \bigcap_{i = 1}^{\pi (p)} \mathrm {N}_i </math>
\mathrm {N} = \bigcap_{i = 1}^{\pi (p)} \mathrm {N}_i </math>


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


== Cardinality ==
== Cardinality ==
Given a finite prime limit, the set of all GPVs are [[Wikipedia: Countably infinite|countably infinite]].  
Given a finite prime limit, the set of all GPVs are [[Wikipedia: Countably infinite|countably infinite]].  


== Adjacent GPVs property ==
== Sorting property ==
Given a finite prime limit, the set of all GPVs can be ordered in this way such that all but one entry in the GPV V<sub>''k''</sub> and its next GPV V<sub>''k'' + 1</sub> are the same, and for the different entry, the latter increments the former by 1.  
Given a finite prime limit, the set of all GPVs can be ordered in this way such that all but one entry in the GPV V<sub>''k''</sub> and its next GPV V<sub>''k'' + 1</sub> are the same, and for the different entry, the latter increments the former by 1.  


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This also holds for any rationally independent subgroups, such as 2.3.7 and 2.9.7, and even ''some'' rationally dependent subgroups, such as 2.3.9.7. It does not hold, however, for other rationally dependent subgroups, such as 2.3.27.7, where at certain points of edo number ''n'', both the mappings for 3 and 27 increment.
This also holds for any rationally independent subgroups, such as 2.3.7 and 2.9.7, and even ''some'' rationally dependent subgroups, such as 2.3.9.7. It does not hold, however, for other rationally dependent subgroups, such as 2.3.27.7, where at certain points of edo number ''n'', both the mappings for 3 and 27 increment.


=== Proof ===
{{Databox|Proof|
By definition, the ''p''-limit GPV of ''n''-edo is V (''n'') = round (''n'' log<sub>2</sub> (Q)), where Q is the prime basis {{val| 2, 3, 5, , ''p'' }}.  
By definition, the ''p''-limit GPV of ''n''-edo is V (''n'') &#61; round (''n'' log<sub>2</sub> (Q)), where Q is the prime basis {{val| 2 3 5 … ''p'' }}.  


The adjacent GPVs property is equivalent to
The sorting property is equivalent to
# for any prime ''q''<sub>''i''</sub> in Q, there is a point of ''n'' to cause ''v''<sub>''i''</sub> to increment to ''v''<sub>''i''</sub> + 1; and
# for any prime ''q''<sub>''i''</sub> in Q, there is a point of ''n'' to cause ''v''<sub>''i''</sub> to increment to ''v''<sub>''i''</sub> + 1; and
# for any distinct primes ''q''<sub>''i''</sub>, ''q''<sub>''j''</sub> in Q, there is ''not'' a point of ''n'' to cause both ''v''<sub>''i''</sub> and ''v''<sub>''j''</sub> to increment to ''v''<sub>''i''</sub> + 1 and ''v''<sub>''j''</sub> + 1, respectively.
# for any distinct primes ''q''<sub>''i''</sub>, ''q''<sub>''j''</sub> in Q, there is ''not'' a point of ''n'' to cause both ''v''<sub>''i''</sub> and ''v''<sub>''j''</sub> to increment to ''v''<sub>''i''</sub> + 1 and ''v''<sub>''j''</sub> + 1, respectively;


<nowiki>#1</nowiki> holds because the point is ''n'' = (''v''<sub>''i''</sub> + 1/2)/log<sub>2</sub> (''q''<sub>''i''</sub>).  
where an increment of ''f'' (''x'') at ''x''<sub>0</sub> is defined as lim ''x''''x''<sub>0</sub><sup>+</sup> ''f'' (''x'') &#61; lim ''x''→ ''x''<sub>0</sub><sup>-</sup> ''f'' (''x'') + 1.  


To prove <nowiki>#2</nowiki>, let us assume there exists such an ''n''. By the property of the round function, an increment of ''y'' = round (''x'') occurs only if 2''x'' ∈ '''Z'''. Thus, for any distinct primes ''q''<sub>''i''</sub>, ''q''<sub>''j''</sub> in Q, 2''n'' log<sub>2</sub> (''q''<sub>''i''</sub>) ∈ '''Z''', and 2''n'' log<sub>2</sub> (''q''<sub>''j''</sub>) ∈ '''Z'''. If that is the case, then their quotient (2''n'' log<sub>2</sub> (''q''<sub>''i''</sub>))/(2''n'' log<sub>2</sub> (''q''<sub>''j''</sub>)) = log<sub>''q''<sub>''j''</sub></sub> (''q''<sub>''i''</sub>) ∈ '''Q''', which contradicts [[Wikipedia: Gelfond–Schneider theorem|Gelfond–Schneider theorem]]. Therefore, the hypothesis is false, and such an ''n'' does not exist.
<nowiki>#1</nowiki> holds immediately following the definition of the round function, and the point is ''n'' &#61; (''v''<sub>''i''</sub> + 1/2)/log<sub>2</sub> (''q''<sub>''i''</sub>).
 
To prove <nowiki>#2</nowiki>, let us assume there exists such an ''n''. By the definition of the round function, an increment of ''y'' &#61; round (''x'') occurs only if 2''x'' ∈ '''Z'''. Thus, for any distinct primes ''q''<sub>''i''</sub>, ''q''<sub>''j''</sub> in Q, 2''n'' log<sub>2</sub> (''q''<sub>''i''</sub>) ∈ '''Z''', and 2''n'' log<sub>2</sub> (''q''<sub>''j''</sub>) ∈ '''Z'''. If that is the case, then their quotient (2''n'' log<sub>2</sub> (''q''<sub>''i''</sub>))/(2''n'' log<sub>2</sub> (''q''<sub>''j''</sub>)) &#61; log<sub>''q''<sub>''j''</sub></sub> (''q''<sub>''i''</sub>) ∈ '''Q''', which contradicts [[Wikipedia: Gelfond–Schneider theorem|Gelfond–Schneider theorem]]. Therefore, the hypothesis is false, and such an ''n'' does not exist.
}}


== Application ==
== Application ==

Revision as of 17:59, 1 January 2022

This page shows some properties of the generalized patent val (GPV).

To tell if a val is a GPV

Suppose we have a p-limit val V, to tell if it is a GPV:

For every prime q in the p-limit, solve

[math]\displaystyle{ \displaystyle \operatorname {round} (n \log_2 q) = v_{\pi (q)} }[/math]

for n. The solution is

[math]\displaystyle{ \displaystyle \frac {v_{\pi (q)} - 1/2}{\log_2 (q)} \lt n \lt \frac {v_{\pi (q)} + 1/2}{\log_2 (q)} }[/math]

Let N1, N2, …, Nπ (p) denote the solution sets. Find

[math]\displaystyle{ \displaystyle \mathrm {N} = \bigcap_{i = 1}^{\pi (p)} \mathrm {N}_i }[/math]

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.

Cardinality

Given a finite prime limit, the set of all GPVs are countably infinite.

Sorting property

Given a finite prime limit, the set of all GPVs can be ordered in this way such that all but one entry in the GPV Vk and its next GPV Vk + 1 are the same, and for the different entry, the latter increments the former by 1.

This property states that, for example, if it is known that 12 19 28] is a GPV, then the next GPV is one of 13 19 28], 12 20 28], or 12 19 29].

This also holds for any rationally independent subgroups, such as 2.3.7 and 2.9.7, and even some rationally dependent subgroups, such as 2.3.9.7. It does not hold, however, for other rationally dependent subgroups, such as 2.3.27.7, where at certain points of edo number n, both the mappings for 3 and 27 increment.

Proof

By definition, the p-limit GPV of n-edo is V (n) = round (n log2 (Q)), where Q is the prime basis 2 3 5 … p].

The sorting property is equivalent to

  1. for any prime qi in Q, there is a point of n to cause vi to increment to vi + 1; and
  2. for any distinct primes qi, qj in Q, there is not a point of n to cause both vi and vj to increment to vi + 1 and vj + 1, respectively;

where an increment of f (x) at x0 is defined as lim xx0+ f (x) = lim xx0- f (x) + 1.

#1 holds immediately following the definition of the round function, and the point is n = (vi + 1/2)/log2 (qi).

To prove #2, let us assume there exists such an n. By the definition of the round function, an increment of y = round (x) occurs only if 2xZ. Thus, for any distinct primes qi, qj in Q, 2n log2 (qi) ∈ Z, and 2n log2 (qj) ∈ Z. If that is the case, then their quotient (2n log2 (qi))/(2n log2 (qj)) = logqj (qi) ∈ Q, which contradicts Gelfond–Schneider theorem. Therefore, the hypothesis is false, and such an n does not exist.

Application

Given a finite prime limit, the above properties offer a way to iterate through all GPVs.

  1. Enter a GPV. Set i = 1.
  2. Copy the input and increment its i-th entry by 1.
  3. Test if it is a GPV.
  4. If it is, return it and back to step 1; otherwise, increment i by 1 and back to step 2.

Notice that the all-zero val is a GPV, 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 suggested starting with the last entry and going backwards for better performance, because the largest prime is most likely to increment.