User:Akselai/On the infinite division of the octave

Abstract

A construction of ∞edo by vals is given, such that its structure is compatible with the regular temperament theory of finite edos.

Introduction

Equal divisions of the octave (edos) are, historically, a trick to deal with the (countably) infinite pitches in just intonation (JI), arguably the basis of almost all music and hearing. It reduces the infinite to the finite (after octave equivalence), and multiplication to addition. In light of the additive structure of edos, people have constructed larger and larger edos to approximate just intonation more and more accurately. A relatively famous example is 11358058edo.

A natural extension of this is called ∞edo, the infinite division of the octave. We already know from the definition of edos, that for all integers n, k>1, that nedo is a subset of (kn)edo, and is in fact a subgroup. So we also suppose that ∞edo contains finite edos. I put this in bold because this is a key assumption in our investigation of ∞edo.

This construction is evidently problematic. The first and most obvious problem is that the step sizes are not well defined. What does it mean by 1 step of ∞edo? We have, by definition, 1 step of nedo equal to [math]\displaystyle{ 1 \backslash n = 2^{1/n} }[/math], so naturally [math]\displaystyle{ 1 \backslash \infty = \lim_{n \rightarrow \infty} 2^{1/n} = 1 }[/math]. Thus every step of ∞edo is the unison. We are not going anywhere by moving a finite amount of scale steps, and an infinite amount of scale steps (e.g. to get to the octave) is even more absurd since infinity is not a quantity.

Another problem with this is structure. Suppose we divide the octave into countably infinite many steps, whatever that may mean. We can label each interval with a positive integer, according to its appearance in the sequence 1edo, 2edo, 3edo, ... This has the advantage that finite scale steps no longer "pile up infinitesimally near the unison" as we have seen above. But now our labels don't make sense algebraically, i.e. the stacking of the intervals corresponding to 3 and 4 is not the one corresponding with 7.

In fact, by restricting to the countably infinite, there is also a mismatch of cardinality of this construction, if the goal is to (I paraphrase) recreate all harmonics [and intervals] perfectly, since the continuum is uncountable.

So is there even a way to see ∞edo by a formal construction other than by a facetious meme in xenharmonic circles? I say the answer is yes.

Akselai's construction of ∞edo

Remember our key assumption: we suppose that ∞edo contains finite edos, in a natural way compatible with the embedding of a smaller edo into a larger edo. Every edo has a mapping (called a val) to a subgroup of JI, specified with a (co)vector with finitely many coordinates. For example 12edo has the val 2.3.5 <12 19 28] because it maps 12 steps to the harmonic 2, 19 steps to 3, and 28 steps to 5.

We also recall the concept of an edo extension. The better known construction is in the case of 12edo to 24edo, where the mapping of the 11th harmonic is adjoined to our tone system. In this system, we give the val 2.3.5 <24 38 56] to 24edo, since the amount of scale steps is doubled, and also the val 2.3.5.11 <24 38 56 83] to accomodate the 11th harmonic.

The extension of an edo is not entirely representative by the behaviour of the larger edo alone, i.e. the val is not always patent. For example, the 2.3.5 val in 60edo is <60 95 139], which is not the same thing by multiplying each entry in the 12edo val by 5 (it would be <60 95 140], the mapping of 5 is due to the inflection of the syntonic comma). Thus, the information of the smaller edos are actually important.

To illustrate another examples, here is a tower of edo extensions with length 3:

2.3.5 <19 30 44] ⊆ 2.3.5.7 <57 90 132 160] ⊆ 2.3.5.7.11 <285 450 660 800 986]

By extending this tower to the infinity prime limit, we obtain a strictly ascending chain of edo mappings

[math]\displaystyle{ I_1 \langle a_1 \ a_2 \ \cdots \ a_n] \subseteq I_2 \langle m_1a_1 \ m_1a_2 \ \cdots \ m_1a_n \ a_{n+1}] \subseteq I_3 \langle m_1m_2a_1 \ m_1m_2a_2 \ \cdots \ m_1m_2a_n \ m_2a_{n+1} \ a_{n+2}] \subseteq \cdots }[/math]

where [math]\displaystyle{ I_1 \subseteq I_2 \subseteq I_3 \subseteq \cdots }[/math] as JI subgroups.

Thus, we obtain an ∞edo JI mapping by means of the chain of inclusions, and there are [math]\displaystyle{ |\mathbb{Z}^\mathbb{Z}| }[/math] such mappings.

Operations

Given a mapping of ∞edo, the intervals of ∞edo can be specified by that of a JI interval α, and some nedo, defined as the least edo with its associated subgroup containing α. The actual number of scale steps in nedo can be inferred from the val chain. The good news: there is now a natural algebraic structure on ∞edo with respect to JI intervals! Suppose we have two scale steps of ∞edo, (α, m) and (β, n) (with m ≤ n), and we want to stack them. Suppose we have also calculated the scale steps in their respective edos as s and t. Then the result is simply (αβ, n), and it is readily verified that the number of scale steps of this interval is (n/m)s + t.

[More operations at your request.]

Properties

∞edo, by this construction is a flexible object. Some have defined ∞edo as simply the union of all edos, which is actually supported by this construction. At the h-th height of the chain with medo, we only need to adjoin (mh)edo to obtain the (h+1)-th height, and we would have encompassed all integer factors along the tower and hence all edos. (Though, the intervals of an arbitrary subset edo do not follow a val mapping.)

On the other hand, ∞edo can also be built from, say 5ⁿedos. Then it would not contain 2edo, among other edos that are not powers of 5.

Implementation

∞edo is readily implemented by calculators by the above definitions and operations. The only downside is that an infinite stream of JI basis intervals and another infinite stream of edos are to be read for the algorithms to work. However, the calculations are guaranteed to be finitary.

[I'll make a program here if I've got the time.]