Akselai at 11:39, 28 January 2024

2024-01-28T11:39:09Z

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	In general, the more elements ''S'' contains, the smaller the comma after a fixed number of steps. The rate of diminishing of RES commas is an ~~active~~ unsolved problem ~~(at least [[user:Akselai]] is working on it)~~.		In general, the more elements ''S'' contains, the smaller the comma after a fixed number of steps. The monzos coordinates appear to grow at the rate of factorial (or ''x''<sup>''x''</sup>). However, the rate of diminishing of RES commas is an unsolved problem. Nevertheless, we have an absolute upper bound of 2<sup>-''n''</sup>''a''<sub>0</sub>, because the distance from an interval to two equal steps cannot exceed half the distance between the two equal steps.


	= Miscellaneous =		= Miscellaneous =
	It is possible for an RES comma to be 1/1. Consider a modified version of the above example. We have ''a'' = 5/4, but this time ''S'' = {[-120 1 51⟩, 3/2, 2/1}. We get ''a''<sub>1</sub> = 128/125 and ''a''<sub>2</sub> = [-120 1 51⟩ again. But in the case of ''a''<sub>3</sub>, since ''a''<sub>2</sub> is already in the set of intervals, no other interval in the set is going to be closer than a power of ''a''<sub>2</sub>, so ''a''<sub>3</sub> = ''a''<sub>2</sub>/''a''<sub>2</sub> = 1/1. Since all the powers of 1/1 are 1/1, the closest interval is just the smallest interval in the set, so ''a''<sub>4</sub> = [-120 1 51⟩, ''a''<sub>5</sub> = 1/1, and so on.		It is possible for an RES comma to be 1/1. Consider a modified version of the above example. We have ''a'' = 5/4, but this time ''S'' = {[-120 1 51⟩, 3/2, 2/1}. We get ''a''<sub>1</sub> = 128/125 and ''a''<sub>2</sub> = [-120 1 51⟩ again. But in the case of ''a''<sub>3</sub>, since ''a''<sub>2</sub> is already in the set of intervals, no other interval in the set is going to be closer than a power of ''a''<sub>2</sub>, so ''a''<sub>3</sub> = ''a''<sub>2</sub>/''a''<sub>2</sub> = 1/1. Since all the powers of 1/1 are 1/1, the closest interval is just the smallest interval in the set, so ''a''<sub>4</sub> = [-120 1 51⟩, ''a''<sub>5</sub> = 1/1, and so on.

	~~= Code =~~
	~~See [[user:Akselai]]'s SandBox page.~~

Akselai: Created page with "'''Recursive equal step (RES) commas''' are a sequence of commas {''a''₁, ''a''₂, ...} generated by an initial interval ''a''₀ and a set o..."

2024-01-21T15:16:43Z

Created page with "'''Recursive equal step (RES) commas''' are a sequence of commas {''a''1, ''a''2, ...} generated by an initial interval ''a''0 and a set o..."

New page

'''Recursive equal step (RES) commas''' are a sequence of [[comma]]s {''a''1, ''a''2, ...} generated by an initial interval ''a''0 and a set of intervals {''b''1, ''b''2, ..., ''b''''k''}. They are constructed by repeatedly comparing stacks of an interval with each of the intervals, and finding the smallest difference between them.

= Mathematical Definition =
By convention, all intervals are in fraction form (instead of monzo form). Suppose ''S'' is a finite set of intervals {''b''k} and ''a'' is an interval (generically ''a''''m'' ∉ ''S'' for all ''m'' ∈ '''Z'''). Then for all ''i'' > 0,

''a''0 = ''a''

''a''''i''+1 = min''m'' ∈ '''Z''', 1 ≤ j ≤ k ‖''a''''i''''m''/''b''k‖

where ‖''x''‖ = exp|log(''x'')| denotes the multiplicative absolute value, which turns an interval into an ascending interval.

We can write ''a''n as RESn(''a''; ''S'').

= Worked out example =
We start with ''a'' = 5/4, ''S'' = {3/2, 2/1}. The closest power of ''a'' to 3/2 is (5/4)2, and the closest power of ''a'' to 2/1 is (5/4)3. Since (2/1)/(5/4)3 < (5/4)2/(3/2), we have ''a''1 = (2/1)/(5/4)3 = [[128/125]], the augmented comma.

With the value of ''a''1, we can find ''a''2. The closest power of ''a''1 to 3/2 is (128/125)17, and the closest power of ''a''1 to 2/1 is (128/125)29. Since (128/125)17/(3/2) < (2/1)/(128/125)29, we have ''a''2 = (3/2)/(128/125)17 = [-120 1 51⟩.

We can repeat this process to get ''a''3, ''a''4 and so forth.

The complexity of the comma after 2 steps is already very large. However, [-120 1 51⟩ is not small compared to other commas with a similar complexity; it is larger than 3.9 cents, which is roughly the size of [24 -21 4⟩, the [[vulture comma]].

= More examples =
{| class="wikitable"
|+''a'' = 5/4, ''S'' = {3/2, 2/1}
|-
! n
! RES''n''(''a''; ''S'')
! Size (cents)
|-
| 0
| 5/4
| 386.31
|-
| 1
| 128/125
| 41.059
|-
| 2
| [-120 1 51⟩
| 3.9544
|-
| 3
| [36361 -303 -15453⟩
| 1.8144
|-
| 4
| [14071708 -117262 -5980311⟩
| 0.21180
|}

{| class="wikitable"
|+''a'' = 7/5, ''S'' = {6/5, 3/2, 7/4, 2/1}
|-
! n
! RES''n''(''a''; ''S'')
! Size (cents)
|-
| 0
| 7/5
| 582.51
|-
| 1
| 50/49
| 34.976
|-
| 2
| [-8 1 -19 18⟩
| 0.86075
|-
| 3
| [11153 -1394 26486 -25092⟩
| 0.10909
|}

In general, the more elements ''S'' contains, the smaller the comma after a fixed number of steps. The rate of diminishing of RES commas is an active unsolved problem (at least [[user:Akselai]] is working on it).

= Miscellaneous =
It is possible for an RES comma to be 1/1. Consider a modified version of the above example. We have ''a'' = 5/4, but this time ''S'' = {[-120 1 51⟩, 3/2, 2/1}. We get ''a''1 = 128/125 and ''a''2 = [-120 1 51⟩ again. But in the case of ''a''3, since ''a''2 is already in the set of intervals, no other interval in the set is going to be closer than a power of ''a''2, so ''a''3 = ''a''2/''a''2 = 1/1. Since all the powers of 1/1 are 1/1, the closest interval is just the smallest interval in the set, so ''a''4 = [-120 1 51⟩, ''a''5 = 1/1, and so on.

= Code =
See [[user:Akselai]]'s SandBox page.

User:Akselai/RES comma - Revision history

Akselai at 11:39, 28 January 2024

Akselai: Created page with "'''Recursive equal step (RES) commas''' are a sequence of commas {''a''1, ''a''2, ...} generated by an initial interval ''a''0 and a set o..."

Akselai: Created page with "'''Recursive equal step (RES) commas''' are a sequence of commas {''a''₁, ''a''₂, ...} generated by an initial interval ''a''₀ and a set o..."