User:Arseniiv/Isodifferential subdivision

This page is a stub. You can help the Xenharmonic Wiki by expanding it.

Isodifferential or linear subdivision^{[idiosyncratic term]} of an interval [math]\displaystyle{ s }[/math] into [math]\displaystyle{ d }[/math] parts is an isodifferential chord of [math]\displaystyle{ d+1 }[/math] notes, spanning [math]\displaystyle{ s }[/math] —that is, a chord [math]\displaystyle{ a_0 : a_1 : a_2 : \ldots : a_{d-1} : a_d }[/math] with [math]\displaystyle{ \frac{a_d}{a_0} = s }[/math], where frequency differences between consecutive notes are the same: [math]\displaystyle{ a_1 - a_0 = a_2 - a_1 = \ldots = a_d - a_{d-1} }[/math]. If the interval [math]\displaystyle{ s }[/math] is rational, this (then isoharmonic) chord can be represented with integer [math]\displaystyle{ a_k }[/math] values.

Isodifferential subdivision of [math]\displaystyle{ s }[/math] into two intervals [math]\displaystyle{ a, h }[/math] is already quite notable: [math]\displaystyle{ a }[/math] is the arithmetic mean of [math]\displaystyle{ s }[/math] and 1, the unison —this is just by definition chosen above,— and [math]\displaystyle{ h }[/math] is the harmonic mean of [math]\displaystyle{ s }[/math] and 1.

Dividing a superparticular interval in this way gives two superparticular intervals, which gives rise to "the Archytas's pyramid", a binary tree

            1:2
          /     \
      2:3         3:4
     /   \       /   \
   4:5   5:6   6:7   7:8
   /  \  /  \  /  \  /  \
  8: 9:10:11:12:13:14:15:16

which generates every superparticular interval by linearly dividing an octave into 2, 4, 8 parts and so on.

One can also successively divide a tritave into 3 parts, generating a ternary tree with every interval [math]\displaystyle{ \tfrac{n+2}n }[/math] that isn't also a superparticular (so, with [math]\displaystyle{ n }[/math] odd).

Properties

1. Subdividing an interval into [math]\displaystyle{ d }[/math] parts and then each of those into [math]\displaystyle{ d' }[/math] parts is the same as doing it in the opposite order or dividing into [math]\displaystyle{ d d' }[/math] parts at once.

This is obvious from looking at isoharmonic chords like 5:7:9:11:13:15:17: such a chord is a linear subdivision of both 5:11:17 into three and 5:9:13:17 into two, which are divisions of 5:17 themselves. Dividing irrational intervals works exactly the same, which can be visualized by dividing segments on the frequency line [math]\displaystyle{ \mathbb R_{>0} }[/math] into equal parts, segments themselves representing intervals.

2. The first part is a weighted arithmetic mean of 1 and [math]\displaystyle{ s }[/math] with weights [math]\displaystyle{ d - 1 }[/math] and 1.

Example: in 3:5:7:9, 5/3 is 2 parts 1/1 and 1 part 3/1: [math]\displaystyle{ \frac{2\cdot\mathbf1 + 1\cdot\mathbf3}{2 + 1} = \frac53 }[/math].

3. The last part is a weighted harmonic mean of 1 and [math]\displaystyle{ s }[/math] with weights [math]\displaystyle{ d - 1 }[/math] and 1.

Example: in 3:5:7:9, 7/9 is 2 parts 1/1 and 1 part 1/3 (NB harmonic mean): [math]\displaystyle{ \frac{2\cdot\mathbf1 + 1\cdot\mathbf{1/3}}{2 + 1} = \frac{7/3}3 = \frac79 }[/math].

4. Other parts can be realized as weighted harmonic and arithmetic means taken one after another with correct weights to get a suffix of a prefix (or vice versa).

Example: in 3:5:7:9, we can first select the prefix 3:(5):7 as AM with weights 1 and 2: [math]\displaystyle{ \frac{1\cdot\mathbf1 + 2\cdot\mathbf3}{1 + 2} = \frac73 }[/math], and then take the last part of 3:5:7 as HM with weights 1 and 1 (the common one): [math]\displaystyle{ \frac{\mathbf1 + \mathbf{3/7}}2 = \frac{10/7}2 = \frac57 }[/math]. We can do it in reverse order: first focus on 5:(7):9 taking HM with weights 1 and 2, then take AM with weights 1 and 1 to get the first part.

Formula

We obtain the formula for [math]\displaystyle{ k }[/math]-th part [math]\displaystyle{ a_k / a_{k-1} }[/math].

For simplicity, choose [math]\displaystyle{ a_0 = 1, a_d = s }[/math] and thus get [math]\displaystyle{ s - 1 = d\Delta }[/math] for [math]\displaystyle{ \Delta }[/math] the difference between consecutive [math]\displaystyle{ a_k }[/math] here, which gives [math]\displaystyle{ \Delta = \tfrac{s - 1}d }[/math] and [math]\displaystyle{ a_k = 1 + k\Delta = \tfrac{d + k(s - 1)}d }[/math]. So the part is [math]\displaystyle{ \frac{a_k}{a_{k-1}} = \frac{d + k(s - 1)}{d + (k - 1)(s - 1)} }[/math].

If [math]\displaystyle{ s }[/math] is rational [math]\displaystyle{ \tfrac{m + D}m }[/math], an intuitive approach is as follows:

Let [math]\displaystyle{ L = \operatorname{lcm}(D, d) }[/math]. If [math]\displaystyle{ D }[/math] isn't already divisible by [math]\displaystyle{ d }[/math], multiply the fraction with [math]\displaystyle{ a = L / D }[/math] to get [math]\displaystyle{ \tfrac{a m + L}{a m} }[/math]. Then integer [math]\displaystyle{ \Delta = L / d }[/math] and we get the [math]\displaystyle{ k }[/math]-th part being [math]\displaystyle{ \frac{a m + k \Delta}{a m + (k-1) \Delta} }[/math] or, writing everything out, [math]\displaystyle{ \frac{m d + k D}{m d + (k-1) D} }[/math] (if you work mentally, this one can have unnecessary cancellations after working with larger numbers).