> From: genewardsmith <genewardsmith@juno.com>

> To: <tuning-math@yahoogroups.com>

> Sent: Sunday, February 03, 2002 10:04 PM

> Subject: [tuning-math] Re: Gene's PB formula, generalized

>

>

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

>

> > i don't really know what to call them, so i'll just make this

> > do: {hv, hw, hx, hy, hz}. it's the top row of numbers in the

> > adjoint (or is it a unimodular inverse?) of the kernel.

>

> I would call those hv(2), hw(2), hx(2), hy(2) and hz(2),

> where the h's are vals, which you can equate to column vectors.

thanks for stepping in with that, Gene. the feeling was nagging

away at me that some reference to the prime-factors ought to be

included in there somewhere, and now i see where.

so how's this? ...

for a set of i rational unison-vectors {u1/v1, u2/v2,... ui/vi},

where {hx, hy, ...hq} is the top row of the unimodular adjoint

of the kernel matrix of the unison-vectors, for any non-zero

I can define a scale by calculating for 0 <= n < d :

step[n] = (u1/v1)^round(hx(2)*n/d) * (u2/v2)^round(hy(2)*n/d)

* ... (ui/vi)^round(hi(2)*n/d) .

-monz

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

> so how's this? ...

>

>

> for a set of i rational unison-vectors {u1/v1, u2/v2,... ui/vi},

> where {hx, hy, ...hq} is the top row of the unimodular adjoint

> of the kernel matrix of the unison-vectors, for any non-zero

> I can define a scale by calculating for 0 <= n < d :

>

> step[n] = (u1/v1)^round(hx(2)*n/d) * (u2/v2)^round(hy(2)*n/d)

> * ... (ui/vi)^round(hi(2)*n/d) .

they can't all be unison vectors!

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

> for a set of i rational unison-vectors {u1/v1, u2/v2,... ui/vi},

> where {hx, hy, ...hq} is the top row of the unimodular adjoint

> of the kernel matrix of the unison-vectors,

This should be h1, h2, ... hi to correspond to your unison vectors.

Also, u1/v1 is a step vector, and the matrix therefore is not just the kernel, but a set of generators for the kernel plus a step vector.

"Unimodular adjoint" should just be "matrix inverse", and you should note that since u1/v1 is a step vector, the matrix is unimodular, and hence is invertible to an integral matrix.

> From: genewardsmith <genewardsmith@juno.com>

> To: <tuning-math@yahoogroups.com>

> Sent: Sunday, February 03, 2002 10:18 PM

> Subject: [tuning-math] Re: Gene's PB formula, generalized

>

>

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

>

> > for a set of i rational unison-vectors {u1/v1, u2/v2,... ui/vi},

> > where {hx, hy, ...hq} is the top row of the unimodular adjoint

> > of the kernel matrix of the unison-vectors,

>

> This should be h1, h2, ... hi to correspond to your unison vectors.

absolutely.

> Also, u1/v1 is a step vector, and the matrix therefore is not just

> the kernel, but a set of generators for the kernel plus a step vector.

right, got it.

> "Unimodular adjoint" should just be "matrix inverse", and you should

> note that since u1/v1 is a step vector, the matrix is unimodular,

> and hence is invertible to an integral matrix.

ok, thanks. how's this? :

where M is the matrix composed of a set of i rational

vectors {u1/v1, u2/v2,... ui/vi} in which u1/v1 is a

step-vector and {u2/v2 ... ui/vi} are commatic unison-vector

generators of the kernel, and where {h1, h2, ...hi} is the

top row of M^-1,

for any non-zero a scale can be defined by calculating

for 0 <= n < d :

step[n] = (u1/v1)^round(h1(2)*n/d) * (u2/v2)^round(h2(2)*n/d)

* ... (ui/vi)^round(hi(2)*n/d) .

(i think we should describe what n and d are.)

-monz

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

> where M is the matrix composed of a set of i rational

> vectors {u1/v1, u2/v2,... ui/vi} in which u1/v1 is a

> step-vector and {u2/v2 ... ui/vi} are commatic unison-vector

not necessarily commatic. if we view this as an application of a

smithian notation, these may be chromatic unison vectors.