Literature:Elements of Harmony: Difference between revisions
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**Continued fractions | **Continued fractions | ||
**Calculus? | **Calculus? | ||
*Book 2 discusses basic acoustics using the number theory proved in Book 1 | *Book 2 discusses basic acoustics using the number theory proved in Book 1 (don't mention frequencies) | ||
**harmonic series | **harmonic series | ||
**intervals as string length ratios (given equal thickness and tension); these can be written as tuples/monzos | **intervals as string length ratios (given equal thickness and tension); these can be written as tuples/monzos |
Revision as of 10:30, 9 June 2017
Elements of Harmony (placeholder name) is a collection of Netagin-language textbooks covering elementary number theory, acoustics, and just intonation music theory.
Contents
- Book 1 discusses mathematical results:
- Basically the ones in Euclid's Elements plus...
- Continued fractions
- Calculus?
- Book 2 discusses basic acoustics using the number theory proved in Book 1 (don't mention frequencies)
- harmonic series
- intervals as string length ratios (given equal thickness and tension); these can be written as tuples/monzos
- motivates logs; gives computation of logs
- Book 3 discusses harmonic properties of various scales.
- odd- and prime-limit
- chord voicings
- otonal and utonal chords
Full text (Netagin)
Full text (English)
Sketch
Motivation for log: Assume strings 1 and 2 have equal tension and thickness, and have lengths $L_1$ and $L_2$ respectively. Find a function $\log$ that, given the ratio between $L_1$ and $L_2$, measures the corresponding subjective difference in pitch.
$\log(\frac{L_1}/{L_2})$ gives the subjective intervallic change when moving from string 1 to string 2.
Thus: if $r_1$ and $r_2$ are string length ratios, then we want
$\log(1) = 0;$ $\log(a) > 0$ when $a > 1;$ $\log(r_1 r_2) = \log(r_1) + \log(r_2).$
The area $A(t)$ under the hyperbola $y = \frac{1}/{x}$ from $x = 1$ to $x = t$ satisfies the desired properties: $A(1) = 0; A(tu) = A(t) + A(u)$.
Ts-T derives that the desired function is well-approximated by Taylor polynomials: namely, $log(x+1) = A(x+1) ≈ \sum_{n=1}^{N} (-1)^n \frac{x^n}/{n}$, when $|x| < 1$. (Need calculus?)