Patent val/Properties: Difference between revisions

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

This page shows some properties of [[patent val]]s as well as [[generalized patent val]]s (GPVs), although currently it is all about the latter.

== To tell if a val is a GPV ==

Suppose we have a ''p''-limit val v, to tell if it 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

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>

~~for ''~~n''. 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>

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

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

~~\mathrm {~~N} = \bigcap_{i = 1}^{\pi (p)} ~~\mathrm {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 ~~not~~ empty; 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 ==

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 {{w|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 vk and its next GPV vk + 1 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''''k'' and its next GPV {{nowrap|''V''''k'' + 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 {{val| 12 19 28 }} is a GPV, then the next GPV is one of {{val| 13 19 28 }}, {{val| 12 20 28 }}, or {{val| 12 19 29 }}.

This also holds for any rationally independent subgroups, such as 2.3.7 and 2.9.7. It does not hold, however, for rationally dependent subgroups, such as 2.3.9.7, where at certain points of edo number ''N'', both the ~~mapping~~ for 3 and 9 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 get an increment.

{{Proof

| title=Proof

| contents=By definition, the ''p''-limit GPV of ''n''-edo is {{nowrap|''V''(''n'') {{=}} round(''n'' log2(''Q''))}}, where ''Q'' is the prime basis {{val| 2 3 5 … ''p'' }}.

To prove the sorting property, we need to prove

# For any prime ''q''''i'' in ''Q'', there is a point of ''n'' at which ''v''''i'' gets an increment; and

# For any distinct primes ''q''''i'', ''q''''j'' in ''Q'', there is ''not'' a point of ''n'' at which both ''v''''i'' and ''v''''j'' get an increment;

where an increment of ''f''(''x'') at ''x''0 is defined as lim {{nowrap|''x'' → ''x''0+ ''f''(''x'')}} {{=}} {{nowrap|lim ''x'' → ''x''0− ''f''(''x'') + 1}}.

<nowiki/>#1 holds immediately following the definition of the round function, and the point is {{nowrap|''n'' {{=}} (''v''''i'' + 1/2)/log2(''q''''i'')}}.

To prove #2, let us assume there exists such an ''n''. By the definition of the round function, an increment of {{nowrap|''y'' {{=}} round(''x'')}} occurs only if {{nowrap|2''x'' ∈ '''Z'''}}. Thus, for any distinct primes {{nowrap|''q''''i'', ''q''''j'' ∈ ''Q''}}, {{nowrap|2''n'' log2(''q''''i'') ∈ '''Z'''}}, and {{nowrap|2''n'' log2(''q''''j'') ∈ '''Z'''}}. If that is the case, then their quotient {{nowrap|(2''n'' log2(''q''''i''))/(2''n'' log2(''q''''j'')) {{=}} log''q''''j''(''q''''i'') ∈ '''Q'''}}, which contradicts {{w|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.

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

# Enter a GPV. Set {{nowrap|''i'' {{=}} 1}}.

# Copy the input and increase its ''i''-th entry by 1.

# Test if it is a GPV.

# If it is, return it and go back to step 1; otherwise, increase ''i'' by 1 and go back to step 2.

# Enter a GPV. Set ''i'' = 1.

To roll backwards:

# Copy the input and ~~increment~~ its ''i''-th entry by 1.

# Enter a GPV. Set {{nowrap|''i'' {{=}} 1}}.

# Copy the input and decrease its ''i''-th entry by 1.

# Test if it is a GPV.

# If it is, return it and back to step 1; otherwise, ~~increment~~ ''i'' by 1 and back to step 2.

# If it is, return it and go back to step 1; otherwise, increase ''i'' by 1 and go back to step 2.

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>

~~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~~.

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 1: / Line 1: @@
-This page shows some properties of the generalized patent val (GPV).
+{{Breadcrumb}}
+This page shows some properties of [[patent val]]s as well as [[generalized patent val]]s (GPVs), although currently it is all about the latter.
 == To tell if a val is a GPV ==
-Suppose we have a ''p''-limit val v, to tell if it 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
+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>
-for ''n''. 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>
-Let N<sub>1</sub>, N<sub>2</sub>, …, N<sub>π (''p'')</sub> denote the solution sets. Find
+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
-\mathrm {N} = \bigcap_{i = 1}^{\pi (p)} \mathrm {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 not empty; 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 ==
-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 {{w|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 {{nowrap|''V''<sub>''k'' + 1</sub>}} 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 {{val| 12 19 28 }} is a GPV, then the next GPV is one of {{val| 13 19 28 }}, {{val| 12 20 28 }}, or {{val| 12 19 29 }}.
-This also holds for any rationally independent subgroups, such as 2.3.7 and 2.9.7. It does not hold, however, for rationally dependent subgroups, such as 2.3.9.7, where at certain points of edo number ''N'', both the mapping for 3 and 9 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 get an increment.
+{{Proof
+| title=Proof
+| contents=By definition, the ''p''-limit GPV of ''n''-edo is {{nowrap|''V''(''n'') {{=}} round(''n'' log<sub>2</sub>(''Q''))}}, where ''Q'' is the prime basis {{val| 2 3 5 … ''p'' }}.
+To prove the sorting property, we need to prove
+# For any prime ''q''<sub>''i''</sub> in ''Q'', there is a point of ''n'' at which ''v''<sub>''i''</sub> gets an increment; and
+# For any distinct primes ''q''<sub>''i''</sub>, ''q''<sub>''j''</sub> in ''Q'', there is ''not'' a point of ''n'' at which both ''v''<sub>''i''</sub> and ''v''<sub>''j''</sub> get an increment;
+where an increment of ''f''(''x'') at ''x''<sub>0</sub> is defined as lim {{nowrap|''x'' → ''x''<sub>0</sub><sup>+</sup> ''f''(''x'')}} {{=}} {{nowrap|lim ''x'' → ''x''<sub>0</sub><sup>&minus;</sup> ''f''(''x'') + 1}}.
+<nowiki/>#1 holds immediately following the definition of the round function, and the point is {{nowrap|''n'' {{=}} (''v''<sub>''i''</sub> + 1/2)/log<sub>2</sub>(''q''<sub>''i''</sub>)}}.
+To prove #2, let us assume there exists such an ''n''. By the definition of the round function, an increment of {{nowrap|''y'' {{=}} round(''x'')}} occurs only if {{nowrap|2''x'' ∈ '''Z'''}}. Thus, for any distinct primes {{nowrap|''q''<sub>''i''</sub>, ''q''<sub>''j''</sub> ∈ ''Q''}}, {{nowrap|2''n'' log<sub>2</sub>(''q''<sub>''i''</sub>) ∈ '''Z'''}}, and {{nowrap|2''n'' log<sub>2</sub>(''q''<sub>''j''</sub>) ∈ '''Z'''}}. If that is the case, then their quotient {{nowrap|(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 {{w|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.
+Given a finite prime limit, the above properties offer a way to iterate through all GPVs. To roll forwards:
+# Enter a GPV. Set {{nowrap|''i'' {{=}} 1}}.
+# Copy the input and increase its ''i''-th entry by 1.
+# Test if it is a GPV.
+# If it is, return it and go back to step 1; otherwise, increase ''i'' by 1 and go back to step 2.
-# Enter a GPV. Set ''i'' = 1.
+To roll backwards:
-# Copy the input and increment its ''i''-th entry by 1.
+# Enter a GPV. Set {{nowrap|''i'' {{=}} 1}}.
+# Copy the input and decrease its ''i''-th entry by 1.
 # Test if it is a GPV.
-# If it is, return it and back to step 1; otherwise, increment ''i'' by 1 and back to step 2.
+# If it is, return it and go back to step 1; otherwise, increase ''i'' by 1 and go back to step 2.
+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>
-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.
+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]]