Why 12 notes in an octave?

(unfinished)

– why build scales from octaves and 5ths?
simple ratios of frequencies
– simple ratios, intervals, Pythagoras..

simplest ratios are x2 (octave) and x3 (oct + 5th)
*** why these sound good – the hear etc.. no beats.
*** ok, what is the sound of some other ratios that one might think may sound good?
pi, phi, root 2, e, 4/3 etc.

– simple ratios are primes, and so.. there’s always a pyth gap.
– going up in 5ths (3/2 times the freq)..
continued fractions
– have to approximate, adjust, fudge.. how best to do it? with continued fractions. This is a very useful technique, used, among other things, for finding best approximations of ratios with simple fractions, for example \frac{22}{7} as an approximation for \displaystyle\pi – or \frac{355}{113}, known in ancient China, and accurate to within XXXXX – or finding good calendar systems, reconciling the incommensurable periods of the day, month and year.

We have to find a particular number of octaves, call it x, that’s very close to a particular number of 5ths, call it y. i.e.
x number of octaves \approx y number of 5ths.
Since going up by octaves is the same as multiplying the frequency by 2 repeatedly, up 1 8ve is x2, up two is x2x2 or 2^2, up 3 octaves is x2x2x2 or 2^3, we can represent the frequency/pitch by 2^x. Similarly the number of fifths we go up is y lots of x3/2×3/2×3/2×3/2.. or (\frac{3}{2})^y

    \begin{align*} 2^x&=\left(\dfrac{3}{2}\right)^y\\ \log_2 2^x&=log_2\left(\dfrac{3}{2}\right)^y\qquad\qquad\text{(Take the logarithm base 2 of both sides)}\\ x\log_2\,2&=y\log_2\,\frac{3}{2}\qquad\qquad\text{(since }\log a^b=b\log a)\\ \frac{x}{y}\times1&=\log_2\,\frac{3}{2}\qquad\qquad\quad\text{(since }\log_2 2=1)\\ &=\log_2 3-\log_2 2\qquad\qquad\text{(since }\log\frac{a}{b}=\log a-\log b)\\ &=\log_2 3-1\\ \frac{x}{y}&=\frac{\log3}{\log2}-1\qquad\qquad\text{(since }\log_a b=\frac{\log b}{\log a})\\ &=1.58496250072115\ldots-1\qquad\qquad\text{(using a calculator - or previously, logarithm tables)}\\ &=0.5849625\ldots\\ \end{align*}

which is very close to \dfrac{585}{1000} – the difference is only 0.0000374992788\ldots – less than 4 parts in 100,000.
Which I believe is much more accuracy than is needed, and won’t make any difference to the result.

    \[\frac{585}{1000}=\frac{117}{200}\]

The next step is to turn \frac{117}{200} into a continued fraction. It’s not so useful to know that 117 octaves is close to 200 fifths, unless you want an octave with 200 notes in it!

    \begin{align*} \frac{117}{200}\quad&=\quad\cfrac{1}{\frac{200}{117}}\quad=\quad\cfrac{1}{1+\frac{83}{117}} \quad=\quad\cfrac{1}{1+\cfrac{1}{\frac{117}{83}}} \quad = \quad \cfrac{1}{1+\cfrac{1}{1+\frac{34}{83}}}\\ &=\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{2+\frac{15}{34}}}} \quad = \quad \cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{2+\frac{4}{15}}}}}\\ &=\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{3+\frac{3}{4}}}}}} \quad = \quad \cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{2+\cfrac{1}{3+\cfrac{1}{1+\frac{1}{3}}}}}}}\\ \end{align*}

This is the same process as Euclid’s algorithm, and can be illustrated in a 200×117 rectangle:
euclid 220 over 117
The successively closer approximations, convergents,are then gotten by using successively using more and of the continued fraction numbers.
The fractions that are produced, \frac{a}{b}, will represent our sought-for quantities x and y, of numbers of octaves, and numbers of fifths, that will be ‘assumed’ to be equal for our tuning.

convergent 1 = 1.
convergent 2 = 1+1/1=2
So we start with the \quad \cfrac{1}{1+\cfrac{1}{1+\frac{34}{83}}}\quad one, ignore the \frac{34}{83} part, and convert the rest back to a normal fraction: \quad \displaystyle\frac{1}{1+\frac{1}{1}}=\frac{1}{2}
So in this case, we have \, x=1 \, and \,y=2\, – a scale with two ‘fifths’ in an octave (very flat ones!). Converting the other fractions we get:

    \begin{align*} \cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{2+\frac{15}{34}}}}\, &\to \,\cfrac{1}{1+\cfrac{1}{1+\frac{1}{2+}}} \,=\,\cfrac{1}{1+\cfrac{1}{1+\frac{1}{2}}}\\ &=\,\cfrac{1}{1+\cfrac{1}{\frac{3}{2}}}\,=\,\cfrac{1}{1+\frac{2}{3}} \,=\, \cfrac{1}{\frac{5}{3}} \,=\,\frac{3}{5} \end{align*}

convergents to 200x117
to be continued…

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