teaching machines

Twelve Steps

Music theory is a daunting subject, built out of centuries of mathematical analysis and aristocratic pride. Sometimes I think I’d be better off reinventing it than trying to learn it at this late stage of its maturity. Here’s how I might get started.

One day I’d be absently fooling around with a bit of wire, which I’d stretch between two nails. I’d absently pluck the wire and hear it make an intriguing sound. I’d pluck it over and over again until I grew tired, and then I’d move on to something else.

Later on I’d recreate the experiment using another bit of wire, and I’d notice that the sound was similar to but different from my original wire. I’d observe that my second wire was of a different length than my first. Intrigued by this difference, I’d start to experiment. Soon I’d have a series of wires of different lengths that I could pluck. I’d pluck them individually in succession, but I’d also pluck them simultaneously. Some multi-plucks would bring discomfort to my ears, but others would please me.

I don’t have such an instrument to play for you, but I can simulate one. This one has ten wires, with the shortest having length 1 and the longest having length 2:

Much to my surprise, I would discover that a wire of a certain length would “cooperate” with a wire twice its length. I’d find the same to be true of a wire four times as long and half as long. All of them would fit together nicely, as shown in this simulation:

I’d name the region between cooperating wire lengths a doubling. I’d make a device out of several doublings that I would use to entertain friends and pass the time. I wouldn’t be able to have very many doublings, though, before the device became unwieldy and the sound too shrill. I’d want more qualities of sounds. The only way to do this, I’d reason, is to add back in wires with lengths in between each doubling. But I’d need to be careful to only play wires that sounded pleasing together.

Suppose I inserted just one extra wire. What length would I make this new wire? Probably I’m make it the average of the base wire and its double. All told, the wire lengths would be 1, 1.5, and 2. If I named the shortest wire wire 0 and the longest wire 2, I’d document my instrument using the following table of wire numbers and lengths:

Wire Length
0 1
1 1.5
2 2

I’d name this arrangement a 2-partition.

To create a 3-partition, I’d insert another wire, but redistribute the wires evenly, resulting in these lengths:

Wire Length
0 1
1 1.3
2 1.6
3 2

In a 4-partition, I’d have these lengths:

Wire Length
0 1
1 1.25
2 1.5
3 1.75
4 2

And so on and so forth. In general, if I placed n wires in each doubling to make an n-partition, I’d measure each wire according to this formula:

$$\textit{length of wire}_i = 1 + \frac{i}{n}$$

Suppose I settled on a making 12-partition instrument. Visually, a plot of my wire lengths would look like this, with the wires on the x-axis and their lengths on the y-axis:

Eventually, I would try adding more wires after the double: wire 13, wire 14, and so on. Following my length formula, my wires would have these lengths:

Wire Length
0 1
1 1.083
2 1.16
3 1.25
4 1.3
5 1.416
6 1.5
7 1.583
8 1.6
9 1.75
10 1.83
11 1.916
12 2
13 2.083
14 2.16
15 2.25
16 2.3
17 2.416
18 2.5
19 2.583
20 2.6
21 2.75
22 2.83
23 2.916
24 3

In scanning this table, I would eventually find a pernicious flaw in my design. I’d find that going from wire 0 of length 1 to wire 12 of length 2 would produce the doubling that I liked so much. But going up another 12 wires to wire 24 would produce a 3. “That’s not a doubling,” I’d say to myself. Nor is going another 12-partition from wire 1 to wire 13. In fact, there would only be the one doubling in my entire instrument. Given the pleasure I found in doublings and my instrument’s failure to provide them, I would throw my instrument down, smashing it into tiny pieces.

After taking some time cool off, I would vent about my failure with some trusted friends, and they would think through the problem with me. In particular, one might suggest trying a different partitioning scheme. Perhaps instead of dividing the doubling into equally-sized chunks, in which each neighboring pair of wires has the same additive difference, the friend would wonder what would happen if I made each pair of wires have the same multiplicative ratio.

I’d wonder what ratio I should use. A ratio of 1 would make every string have the same length. A ratio of 2 would make every string a doubling. I’d conclude that I needed a ratio x somewhere between 1 and 2. It would take me a while to figure out exactly what ratio to use. But in my notebook I’d mark down the following observations about the wire lengths and their relationships if their ratio was held constant. Each wire would be some fixed proportion x of the previous wire:

Wire Length
0 1
1 $\textrm{length}_0 \cdot x$
2 $\textrm{length}_1 \cdot x$
3 $\textrm{length}_2 \cdot x$
4 $\textrm{length}_3 \cdot x$
i $\textrm{length}_{i – 1} \cdot x$

After considering this table, I’d find that I could simplify the lengths by substituting in the previous wires’ lengths:

Wire Length Simplified Length
0 1 1
1 $\textrm{length}_0 \cdot x$ $1 \cdot x = x$
2 $\textrm{length}_1 \cdot x$ $x \cdot x = x^2$
3 $\textrm{length}_2 \cdot x$ $x^2 \cdot x = x^3$
4 $\textrm{length}_3 \cdot x$ $x^3 \cdot x = x^4$
i $\textrm{length}_{i – 1} \cdot x$ $x^{i – 1} \cdot x = x^i$

From this table, I would know that wire 12 would have length $x^{12}$. But I would also excitedly recall that I wanted wire 12 to have length 2. With two measures of the same string, I would figure out what x should be:

$$\begin{array}{rcl}x^{12} &=& 2 \\\sqrt[12]{x^{12}} &=& \sqrt[12]{2} \\x &=& \sqrt[12]{2} = 2^\frac{1}{12}\end{array}$$

Plugging this magic value of x into my table, I’d find the following exact lengths of the wires:

Wire Length
0 $x^0 = (2^\frac{1}{12})^0 = 1$
1 $x^1 = (2^\frac{1}{12})^1 = 1.059463$
2 $x^2 = (2^\frac{1}{12})^2 = 1.122462$
3 $x^3 = (2^\frac{1}{12})^3 = 1.189207$
4 $x^4 = (2^\frac{1}{12})^4 = 1.259921$
i $x^i = (2^\frac{1}{12})^i = 2^\frac{i}{12}$
12 $x^{12} = (2^\frac{1}{12})^{12} = 2$
13 $x^{13} = (2^\frac{1}{12})^{13} = 2.118926$
24 $x^{24} = (2^\frac{1}{12})^{24} = 4$

To my delight, I would find doublings all over—in fact, every 12-wire jump would yield a doubling!

Visually, a plot of my wire lengths for a ratio-based 12-partition would look like the blue line, with the old scheme plotted in green:

This table and plot show the lengths for a 12-partition. For a more general n-partition under this new ratio-based partitioning scheme, I’d measure each wire i according to this formula:

$$\textit{length of wire}_i = 2^\frac{i}{n}$$

I would then rebuild my instrument to span several doublings, with each wire being $2^\frac{1}{12}$ times the length of the preceding wire:

I would still find combinations of wires that didn’t sound pleasing, but I’d also find a number of combinations that did sound pleasing. Furthermore, the combinations that sounded good together in one doubling would sound just as good if I shifted them to a different doubling. And I’d find doublings between each pair of 12 notes.

Thus would I become the father of music theory. But really, I’d just think of myself as a musician trying to please my own ears.

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