On Fri, 09 Aug 2002 22:06:40 -0000 "wallyesterpaulrus"

<perlich@aya.yale.edu> writes:

> --- In tuning-math@y..., "nasnas_100" <nasnas_100@y...> wrote:

> > hi i am new student i have a problem

> > find the root of f(x)=x^3+30x-30

> > thanks

> what does this have to do with tuning?

Nothing; it looks like a homework problem. However, for your

consideration I present the following.

I can represent, and so study, a scale by the polynomial whose roots are

the scale elements. To do this right, I want the octave to be represented

by a prime number; that is, I want a map to primes h such that h(2)=p, p

a prime. In that way I have no zero divisors in the ring mod p, or in

other words I am in a field.

Suppose I want to study Blackjack, which is a scale in Miracle. I can't

do it mod 72, since that is composite, and 31 and 41 are a little small

and may give me extraneous relationships. The 103-et would probably be

fine, but instead I choose the map [4447, 7039, 10317, 12477], using the

prime 4447 to represent 2. This gives me the rms optimal values for

Miracle tuning, with a secor of 432/4447. I now take the polynomial with

roots 432i where i ranges from -10 to 10, which I can reduce mod 4447

without loss of information:

x^21-61*x^19-1724*x^17-1045*x^15-28*x^13+1971*x^11-1724*x^9-1326*x^7+114

*x^5-846*x^3+1260*x

I also define a polynomial whose roots correspond to the twelve 7-limit

consonant intervals, obtaining

x^12-1314*x^10+1560*x^8-1735*x^6-1921*x^4+1244*x^2+1202

If I take the resultant of the first polynomial with x-n substituted for

x with the second polynomial and factor mod 4447,

I get

(n+1601)^5*(n-1550)^3*(n+254)*(n+1169)^5*(n+432)^10*(n+1855)^7*(n+559)

^6*(n+686)*(n+2033)^5*(n-1169)^5*(n-254)*(n+991)^7*(n-1855)^7*(n-610)*(n-

1982)^4*(n-686)*(n-991)^7*(n+305)^6*(n-737)^6*(n-432)^10*(n+1423)^7*(n+15

50)

^3*n^10*(n-1296)^10*(n-864)^10*(n-1423)^7*(n-1601)^5*(n-305)^6*(n+864)^10

*(n

+610)*(n-1042)*(n-127)^6*(n+1042)*(n-1118)^2*(n-1728)^9*(n+1296)^10*(n+11

18)

^2*(n+2160)^8*(n-559)^6*(n+127)^6*(n+1728)^9*(n-2160)^8*(n-178)*(n+1982)^

4*(

n+178)*(n+737)^6*(n-2033)^5

This gives me all the places mod 4447 where we have consonant interval

relationships to Blackjack--including those in Blackjack. The

multiplicities give the number of consonances. The steps of Blackjack,

centered at the unison, are

[-2160, -1855, -1728, -1423, -1296, -991, -864, -559, -432, -127,

0, 127, 432, 559, 864, 991, 1296, 1423, 1728, 1855, 2160]

and the corresponding multiplicities are

[8, 7, 9, 7, 10, 7, 10, 6, 10, 6, 10, 6, 10, 6, 10, 7, 10, 7, 9, 7, 8]

If we add this up, we get 170, which is twice the number of 7-limit

consonances for Blackjack--twice since we count them twice, once for each

scale step.