Marchetto update

I accepted Microsoft Excel's rounding of the large fractions
too readily and too unthinkingly in my earlier posting on
the new interpretation I have of Marchetto's monochord
measurements, so it contains a few minor errors.

I wrote:

> With this method of division, Marchetto got a diesis at
> the end which was much larger than all the rest (92 cents!
> - the others are 22, 37, 30, and 24 cents), and ratios
> with some very high prime factors: his 'enharmonic semitone'
> is 182/173 [= 2^1 * 7^1 * 13^1 * 173^-1] and his 'chromatic
> semitone' is 577/541 [= 541^-1 * 577^1].

The cents measurements are all correct, but the actual
ratios for these two semitones are:

enharmonic
= 531441 / 505160
= 2^-3 * 3^12 * 5^-1 * 73^-1 * 173^-1

chromatic
= 43046721 / 40360904
= 2^-3 * 3^16 * 5045113^-1

So the enharmonic still contains 173 as a factor,
but the chromatic has the huge prime factor 5045113!

The formula for calculating the monochord string-length
goes like this:

('p' with a number is Marchetto's 1/9th 'part' of a whole tone,
'1' is the whole string = lowest note, and 8/9 is the monochord
bridge marking off a 'whole tone' above the lowest note)

p1 = 8/9 + (8/9 * (1 - 8/9))

p2 = 8/9 + (8/9 * (p1 - 8/9))

p3 = 8/9 + (8/9 * (p2 - 8/9))

etc.

I'm a big fan of elegance in mathematical formulae,
and I'm pretty sure that this can be simplified,
but I don't know how. I'd appreciate the explanation.

Also, because the power of 3 increases by 2 for each
part, the numbers get absurdly large. After doing
all the calculations on paper with the regular numbers
to make sure I had the correct ratios, I tried
prime-factoring the whole process, and I follow most
of it, but I don't understand how to get the new
primes that result in the numerators of the answers
from the prime-factor calculations.

For example, the last step in the formula for p1 would be:

( (2^3 * 3^2) / (3^4) ) + ( (2^3) / (3^4) )

but the answer is ( (2^4) * (5^1) ) / (3^4)

Where does that 5 come from?

I can see it in the regular equation:
n = numerator
d = denominator
p = part
x = ordinal number of the part

for n(p(x)), it's the sum of d(p(x-1)) + 2^(3*(x-1))

and d(p(x-1)) = 3^(x*2+2)

Please help me simplify this!

-monz

Joseph L. Monzo....................monz@juno.com
http://www.ixpres.com/interval/monzo/homepage.html
|"...I had broken thru the lattice barrier..."|
| - Erv Wilson |
--------------------------------------------------

___________________________________________________________________
You don't need to buy Internet access to use free Internet e-mail.
Get completely free e-mail from Juno at http://www.juno.com/getjuno.html
or call Juno at (800) 654-JUNO [654-5866]