How did Chu Tsai-yü calculate 12-tone equal temperament in the 16th century?

[2^Oscar]

Hey guys,

I've been writing an essay on 12-tone equal temperament and I've reached a point where I would like to show one of the early methods of calculating it.

I have searched on the internet but I have been unable to find his method... I can see how it can be done using continued fractions and logarithms but I'm unsure if they had the ability to calculate logarithms decimally in the 16^th century? I would really appreciate it if someone could enlighten me :)


Thanks in advance,
Oscar

 

[arildno]

1. "but I'm unsure if they had the ability to calculate logarithms decimally in the 16th century? "
Nope.
Logarithms were introduced in the early 17th century by Napier.

2. "I've been writing an essay on 12-tone equal temperament "

i have no idea what this is, but I am intrigued:
I thought 12-tones (something?) was invented by Schönberg in the early 20th century?
But that is not what you are referring to?

 

[2^Oscar]

Schoenberg created 12 tone music - which is to say music which uses all 12 notes equally without a tonal centre - it is a tonal structure (or anti-structure if you will) and are how notes are played in relationship with each other. 12 tone temperaments are basically musical scales with an octave divided into 12 notes, and are ways of actually assigning certain frequencies to each note. Pythagoras is credited with the earliest system by using ratios from the harmonic series to 'stack' frequencies on top of each other and thus form a chromatic scale (within the confines of 2:1 ratio i.e. an octave).

The problem with this is that notes don't have equal 'gaps' between them they are irregular, and when modulating to other keys or playing with other instruments it becomes difficult. 12-tone equal temperament is essentially a way of tuning the chromatic scale with logarithmically equal gaps between the notes (the actual ratio being 1:2^1/12. (It is how modern instruemnts are tuned - interestingly it also means that in terms of creating pure harmony everything nowadays is out of tune!)

Inability to calculate 2^1/12 easily meant that equal temperament could not be tuned easily for many years after it was first theorized by Vincenzo Gallilei.

Chu Tsai-yü was the first person to calculate it and I was curious as to how he achieved this (I figured it would add some colour to my essay)

Here's some useful links if you're interested (I find this whole topic one of the most fascinating parts of mathematics):

http://www.skytopia.com/project/scale.html"
http://www.davesabine.com/Music/Articles/PythagorasMathematicalTheoruminMusic/tabid/169/Default.aspx"
http://en.wikipedia.org/wiki/Equal_temperament"
http://en.wikipedia.org/wiki/Serialism"

Hope this clarifies some points - thank you for your interest,
Oscar

 

[arildno]

So, if I understand you correctly, the mathematical issue involved here boils down to calculating 2^(1/12)?

In that case, you may set up an iterative scheme with what is known as Newton's method:
$$x_{n}=x_{n-1}-\frac{x_{n-1}^{12}-2}{12x_{n-1}^{11}}, n=1,2...$$
Setting, say $x_{0}=1$,
we get that the next appoximate value is (by setting n=1):
$$x_{1}=x_{0}-\frac{x_{0}^{12}-2}{12x_{0}^{11}}=1+\frac{1}{12}=\frac{13}{12}$$

This first-order approximation, 13/12, is only 2.2% off the correct value; maybe that is what they used?

If you want a (vastly) better estimate, you might calculate $x_{2}$ in a similar manner.

 

[2^Oscar]

Oh of course I hadn't thought of Newton-Raphson.

Chu Tsai-yü expressed it to 9 decimal places so it was very much an approximation... but I am unsure as to whether this was the method he used as he calculated it at the end of the 16th century before Newton was even alive.

Mersenne was the first western mathematician to calculate it some years later, so he may possibly have used this method.

Thank you for the response, it is appreciated greatly,
Oscar

 

[arildno]

I forgot to mention that the Newton-Rapsohn method, in the special case of estimating square roots, was invented by the Babylonians 3000 years ago.

As for the first-order approximation, the n-th root of "a" (x0=1) has the pleasingly simple form:
$$x_{1}=1+\frac{a-1}{n}$$
Many cultures may have fiddled out this one prior to Newton!

Possibly, a similar iterative scheme was used by Chu.

 

[2^Oscar]

Ok thank you very much for your help!

I had no idea Newton-Raphson went that far back, I find it fascinating to think that work that old could be applied to what is a relatively modern concept :)

Again, thank you very much for your help - I'd been stuck on this one for hours! This kind of mathematics really fascinates me - I've heard that many universities go into it in even greater depth which I am very much looking forward to :)


Many appreciations,
Oscar

 

[arildno]

Thank YOU for providing many interesting links I am sure I, and many others here at PF, will enjoy reading!

 

[HallsofIvy]

Ok thank you very much for your help!

I had no idea Newton-Raphson went that far back, I find it fascinating to think that work that old could be applied to what is a relatively modern concept :)

Well, they didn't call it "Newton-Raphson" then!

And, in fact, you don't need even the basic idea of "Newton-Raphson" (approximating the function by its tangent line). The equation, $x^2= a$ can be rewritten $x= a/x$ so if your first approximation, x₀ is less than the correct square root of a, just dividing a by x₀ gives $x_1= a/x_0$ and x₁ must be larger than the correct square root. Similarly, if x₀ is larger than the correct square root, x₁ will be smaller: the correct square root of a is the "geometric average" of x₀ and x₁ so it is always between them.

We don't know exactly where between, so use the "midpoint", (x₀+ x₁)/2 as your next "approximation. In other words use the arithmetic average of x₀ and x₁ rather than the geometric average because it is easier to calculate. That also lies between the two numbers and, with any luck, will be a better approximation than either x₀ or x₁. Of course, with x₁= a/x₀, x₀+ x₁= x₀+ a/x₀ so (x₀+ x₁)/2= x₀/2+ a/(2x₀)= x₀- x₀/2+ a/(2x₀)= x₀- x₀²/(2x₀)+ a/(2x₀)= x₀- (x₀^2- a)/(2x₀).

Using "Newton-Raphson" to solve x²= a, we would use f(x)= x²- a so that f'(x)= 2x and the formula x- f(x)/f'(x) happens to be exactly what we got above. But we certainly don't need any knowledge of calculus to get that formula.

Again, thank you very much for your help - I'd been stuck on this one for hours! This kind of mathematics really fascinates me - I've heard that many universities go into it in even greater depth which I am very much looking forward to :)


Many appreciations,
Oscar

 

[jambaugh]

One can of course go about this process mechanically. Build a long 12 stringed harp with a sliding fret. Make a guess at the fret position (and use a gauge to keep it straight and clamp it down). Tune each sucessive string to the previous one as it is fingered at the fret. When done see if you have an octive. If high or low adjust the fret accordingly and repeat. I would imagine in a day you could get your 12 tone score (and 12th root of 1/2 from the fret position) to the limits of your ability to distinguish tones by ear.

A second fret at the octive (half-way) would allow the use of beat frequencies to get it really percise. Also a wooden "finger" to avoid any added tension from fingering the fret.

And of course one could speed the process up by working out the square root and cube root of 1/2 positions first.

(note roots of 1/2 for wavelength = roots of 2 for frequency)

 

[2^Oscar]

Well, they didn't call it "Newton-Raphson" then!

And, in fact, you don't need even the basic idea of "Newton-Raphson" (approximating the function by its tangent line). The equation, $x^2= a$ can be rewritten $x= a/x$ so if your first approximation, x₀ is less than the correct square root of a, just dividing a by x₀ gives $x_1= a/x_0$ and x₁ must be larger than the correct square root. Similarly, if x₀ is larger than the correct square root, x₁ will be smaller: the correct square root of a is the "geometric average" of x₀ and x₁ so it is always between them.

We don't know exactly where between, so use the "midpoint", (x₀+ x₁)/2 as your next "approximation. In other words use the arithmetic average of x₀ and x₁ rather than the geometric average because it is easier to calculate. That also lies between the two numbers and, with any luck, will be a better approximation than either x₀ or x₁. Of course, with x₁= a/x₀, x₀+ x₁= x₀+ a/x₀ so (x₀+ x₁)/2= x₀/2+ a/(2x₀)= x₀- x₀/2+ a/(2x₀)= x₀- x₀²/(2x₀)+ a/(2x₀)= x₀- (x₀^2- a)/(2x₀).

Using "Newton-Raphson" to solve x²= a, we would use f(x)= x²- a so that f'(x)= 2x and the formula x- f(x)/f'(x) happens to be exactly what we got above. But we certainly don't need any knowledge of calculus to get that formula.

I had a look at ways of computing roots and is the above method the Babylonian method? If so its a very elegant way of calculating it!

One can of course go about this process mechanically.

The problem with this is it could be inaccurate, especially if done by ear. A semi-tone in equal temperament is merely a 100 cent interval and only a few cents discrepancy between, say two harpsichords, could sound truly awful. Furthermore no one had really heard equal temperament around the baroque era when they were trying to achieve it, and so it could be with all the many well temperaments around the time people wouldn't recognise the 100cent wide semi-tone as an equally tempered semi-tone...

To anyone else who is interested in the differences between the temperaments and all the different tunings the following link is to a video of a Bach prelude and fugue (quite a famous one) being played in just intonation (a tuning consisting of integer ratios) - I played this to some of my friends and they either thought it sounded pure and clean or they thought it simply sounded out of tune. You can really hear the draw backs of this kind of tuning when the harmony gets more intense in the fugue.

http://www.youtube.com/watch?v=72ukYpfpgDI"

Thank you for the replies,
Oscar

 

[arildno]

The formula given through Halls' derivation is exactly the one given by Newton-Raphson.

I didn't mean to say that Babylonians did calculus, and Halls has made a strong case for how they MIGHT have reached that formula.

I don't think anyone knows how the Babylonians actually reached the formula (and, in my view, most Babylonians didn't know that, either)

 

[2^Oscar]

Ohh I see thank you for clarifying (my own stupidity )

Oscar

 

[CRGreathouse]

The problem with this is it could be inaccurate, especially if done by ear. A semi-tone in equal temperament is merely a 100 cent interval and only a few cents discrepancy between, say two harpsichords, could sound truly awful. Furthermore no one had really heard equal temperament around the baroque era when they were trying to achieve it, and so it could be with all the many well temperaments around the time people wouldn't recognise the 100cent wide semi-tone as an equally tempered semi-tone...

That's not a hard problem to solve, even if you can't distinguish a difference as small as a cent. Play the 'same' note on the two and listen to the interference pattern. The closer the actual notes produced, the longer its period. This should let you get down to a fraction of a cent.

Some back-of-the-envelope calculations suggest that a 1-cent difference around middle C should give rise to interference of period roughly 2 seconds. Closer and closer notes give rise to (approximately) proportionately longer period interference.

 

[jambaugh]

The problem with this is it could be inaccurate, especially if done by ear.

I disagree. The method I describe would involve tuning the next string to the same frequency as the previous. Using the beat frequency one can get this quite accurate say a quarter cycle per second. Even better if one is patient.

Again the longer the total length the higher precision you can position the fret as a percentage of the whole length. Say a meter total length and fret position to within a quarter mm.

Though the 12 stages does involve cumulative error but it should be unbiased error so high vs. low errors would partially cancel. Again this can be mitigated by working first the 4th and 3rd root problems.

I'd wager you could get the frequencies down to about two tenths of a percent error. The only real drawback I see is the tedious repetition involved in cycling through the 12 tunings after each fret adjustment... that and getting good strings which hold their tune through the process.