Scale step iterations

Suppose you have a scale with step sizes proportional to 1 and phi,
and you multiply by phi. You now have scales with steps of size phi
and phi^2, or in other words phi and phi+1, which means a
transformation of s small steps and L large steps by s->L, L->s+L;
which can be expressed in terms of the matrix

[1 1]
[0 1]

The inverse of this unimodular matrix is

[0 1]
[1 -1]

This tells us to go to smaller step sizes, we send L->s, s->L-s. The
characteristic polynomial of the first matrix is x^2-x-1, with root
phi, and of the second matrix is x^2+x-1, with root 1/phi.

Similarly, suppose we have steps A of size 1, B of size t+1/t and C
of size t^2, where t^3-t^2-t-1=0. We can also write this as
[1,t^2-1,t], and multiplying by t gives [t,t^2+1,t^2], which is to say
[C,B+2A,B+A], so that the transformation matrix is

[0 0 1]
[2 1 0]
[1 1 0]

The characteristic polynomial of this is x^3-x^2-x-1, and its inverse
is

[0 1 -1]
[0 -1 2]
[1 0 0]

with characteristic polynomial x^3+x^2+x-1, with root 1/t. It tells
us that to go to smaller step sizes, we send A->B-C, B->2C-B, C->A.
It would appear we can do this sort of thing in a lot of different
ways.

--- In tuning-math@y..., genewardsmith@j... wrote:

> Similarly, suppose we have steps A of size 1, B of size t+1/t and C
> of size t^2, where t^3-t^2-t-1=0.

That doesn't agree with what Dan said. He said that C/A = t, not t^2.
I think maybe you just made a typo, Gene?

--- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:
> --- In tuning-math@y..., genewardsmith@j... wrote:

> > Similarly, suppose we have steps A of size 1, B of size t+1/t and
C
> > of size t^2, where t^3-t^2-t-1=0.

> That doesn't agree with what Dan said. He said that C/A = t, not
t^2.
> I think maybe you just made a typo, Gene?

Do I ever post anything without a mistake in it somewhere, I wonder?

Dan set C/A = (A+C)/B = t, so that C = t A. If we set A to 1, C is t,
and (1+t)/B = t, so B = (1+t)/t = 1+1/t = t^2-t. Then
t * [1, 1+1/t, t] = [t, t+1, t^2] = [C, A+C, B+C], leading to a
matrix of

[0 0 1]
[1 0 1]
[0 1 1]

with characteristic polynomial x^3-x^2-x-1 and inverse matrix

[-1 1 0]
[-1 0 1]
[ 1 0 0]

so that to go to smaller steps we transform by A->B-A, B->C-A, and
C->A.

--- In tuning-math@y..., genewardsmith@j... wrote:

> Dan set C/A = (A+C)/B = t, so that C = t A. If we set A to 1, C is
t,

Ah . . . much better.

> and (1+t)/B = t, so B = (1+t)/t = 1+1/t = t^2-t. Then
> t * [1, 1+1/t, t] = [t, t+1, t^2] = [C, A+C, B+C], leading to a
> matrix of
>
> [0 0 1]
> [1 0 1]
> [0 1 1]

So the constructing rule is

small(old) = Large(new)
medium(old) = small(new) + Large(new)
Large(old) = medium(new) + Large(new)

Right?

I'd think other, similar constructing rules could also make some
sense as ways of obtaining three-step-size scales? For the two-step-
size case, we discussed the Silver Mean case, with ratio sqrt(2)-1
(IIRC).

--- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:

> I'd think other, similar constructing rules could also make some
> sense as ways of obtaining three-step-size scales? For the two-step-
> size case, we discussed the Silver Mean case, with ratio sqrt(2)-1
> (IIRC).

That's what I was trying to do with my example. The polynomial (and
associated number field) x^3-x^2-x-1 has discriminant -44. We have
only two smaller discriminants in absolute value, -31 and -23. They
are associated to the smallest and second smallest Pisot numbers (by
a theorem of Siegel); x^3-x-1 of disciminant -23 gives us the
smallest Pisot number, and x^3-x^2-1 of discriminant -31 the second
smallest. These seem like the two most obvious characteristic
polynomials to use to look for alternative schemes. They suggest we
could for instance try the following replacement schemes: A->b, B->c,
C->a+b and A->b, B->c, C->a+c.

--- In tuning-math@y..., genewardsmith@j... wrote:

> That's what I was trying to do with my example. The polynomial (and
> associated number field) x^3-x^2-x-1 has discriminant -44. We have
> only two smaller discriminants in absolute value, -31 and -23. They
> are associated to the smallest and second smallest Pisot numbers
(by
> a theorem of Siegel); x^3-x-1 of disciminant -23 gives us the
> smallest Pisot number, and x^3-x^2-1 of discriminant -31 the second
> smallest. These seem like the two most obvious characteristic
> polynomials to use to look for alternative schemes. They suggest we
> could for instance try the following replacement schemes: A->b, B-
>c,
> C->a+b and A->b, B->c, C->a+c.

Cool! What does the discriminant tell us?

--- In tuning-math@y..., "Paul Erlich" <paul@s...> wrote:

> Cool! What does the discriminant tell us?

Lots of things, but for our purposes the most significant seems to be
that these are the smallest cubic fields in some sense. Just as
the Golden Field Q(phi) has the smallest real quadratic field
discriminant (at 5) and Q(sqrt(2)) the second smallest (at 8) and so
supply the least complicated examples, the cubic fields of
discriminants -23, -31 and -44 would seem to be the right place to
start. The smallest totally real cubic field has discriminant 49, but
this does not give us a Pisot number and that might be important. One
polynomial for it has roots 2*cos(2^i pi/7) for i from 1 to 3, and is
x^3+x^2-2*x-1.