Percent of Common Tones

I have recently come up with a parallel to Benedetti/Tenney Height, which I believe will be useful in comparing dyads of harmonic timbres (and therefore well suited for Harmonic Entropy seeding). It counts the percent of common partials shared by two tones. Here is the method:

Suppose we have a ratio A:B. The lowest note has an infinite amount of partials, and the highest note has an infinite amount as well. We therefore have 2*infinity partials altogether.

Assuming A and B are in lowest terms (and are therefore coprime), every "Bth" partial of the upper tone will be a common tone, as well as every "Ath" partial of the lower tone. As ratios, these are infinity/A partials and infinity/B partials.

Now to find the percent of common partials, we divide the shared partials by the total partials. This is (infinity/A+infinity/B)/(2*infinity), which simplifies to (A+B)/(2*A+B).

To use this as a measurement tool for complexity requires that we simply take the inverse. This insures that complex ratios are higher in height, which agrees with other institutions such as Benedetti and Tenney Height. This gives us the final form of 2AB/(A+B), which we can reduce to A*B/(A+B) for ease of use.

Side Note: Mike has proposed a similar but weighted version of this, which is 1/A^2+1/B^2. It is partly based on the idea that a harmonic timbre with 1/N^2 roll-off reflects an ideal timbre of sorts (it sounds a lot more natural than 1/N roll-off).

-Ryan

On Mon, Nov 7, 2011 at 2:28 AM, Ryan Avella <domeofatonement@yahoo.com> wrote:
>
> Side Note: Mike has proposed a similar but weighted version of this, which is 1/A^2+1/B^2. It is partly based on the idea that a harmonic timbre with 1/N^2 roll-off reflects an ideal timbre of sorts (it sounds a lot more natural than 1/N roll-off).

This should be inverted. AB/(A+B) is the same thing as 1/(1/A + 1/B).
Mine is the same thing as 1/(1/A^2 + 1/B^2) = A^2B^2/(A^2+B^2).

I remember reading a study a while ago about how the optimum exponent
was subjective and somewhere between 1 and 2. But for now, might as
well just go with 2 to keep it simple.

-Mike

--- In tuning-math@yahoogroups.com, "Ryan Avella" <domeofatonement@...> wrote:
> Suppose we have a ratio A:B. The lowest note has an infinite amount of partials, and the highest note has an infinite amount as well. We therefore have 2*infinity partials altogether.
>
> Assuming A and B are in lowest terms (and are therefore coprime), every "Bth" partial of the upper tone will be a common tone, as well as every "Ath" partial of the lower tone. As ratios, these are infinity/A partials and infinity/B partials.
>
> Now to find the percent of common partials, we divide the shared partials by the total partials. This is (infinity/A+infinity/B)/(2*infinity), which simplifies to (A+B)/(2*A+B).

First of all, I assume (A+B)/(2*A+B) is a typo for (A+B)/(2*A*B). Now that we have that out of the way...

This is completely meaningless because infinity/infinity is undefined. By changing the details of the calculation you can get a completely different number.

For example, if the two pitches are A Hz and B Hz, I know that every partial is an integer number of Hz. I'm going to try to calculate the same thing as you (fraction of total partials that are shared) using a different method. Out of "infinity" possibilities, we have

Partials of A: infinity/A
Partials of B: infinity/B
Shared partials: infinity/(A*B)
Total partials, counting duplicates twice: infinity/A + infinity/B
Total partials, counting each only once: inf/A + inf/B - inf/(A*B)

Fraction of partials that are shared, with duplicate counting:
infinity/(A*B) / (infinity/A + infinity/B) = 1/(A+B)
(or 2/(A+B) is the shared partials also count twice)

Fraction of partials that are shared, with non-duplicate counting:
infinity/(A*B) / (infinity/A + infinity/B - infinity/(A*B)) = 1/(A+B-1)

Neither of these answers is equivalent to your answer! Does this mean one of us made a mistake in calculation? No, it simply means that this kind of "arithmetic with infinity" is always meaningless, and we're actually calculating the values of two different limits (as in calculus limits).

You need to give a more precise definition of what "percent of common partials" means in order for any formula to be "correct" and others "incorrect". Percentages are easy to define when all the sets involved are finite, but you have to be very careful to define things properly when infinite sets are involved.

See http://en.wikipedia.org/wiki/Riemann_rearrangement_theorem , http://en.wikipedia.org/wiki/User:Tlogmer/Monty_Hell_problem , http://en.wikipedia.org/wiki/Hilbert%27s_paradox_of_the_Grand_Hotel , etc.

Keenan