Limit generator for 5-limit meantone

This turns out to be 1/4-comma meantone; the sequence of fifths falls
into a regular pattern of three pure fifts followed by a 40/27.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:

> This turns out to be 1/4-comma meantone; the sequence of fifths falls
> into a regular pattern of three pure fifts followed by a 40/27.

I should add for this and the miracle example that the limit octave is
2. This is not always the case; Paul may or may not like to know that
for Pajara, for instance, it is distinctly flat. Carl I hope is taking
note of this, as it seems to be along the lines he has been agitating for.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<gwsmith@s...>
> wrote:
>
> > This turns out to be 1/4-comma meantone; the sequence of fifths
falls
> > into a regular pattern of three pure fifts followed by a 40/27.
>
> I should add for this and the miracle example that the limit octave
is
> 2. This is not always the case; Paul may or may not like to know
that
> for Pajara, for instance, it is distinctly flat. Carl I hope is
taking
> note of this, as it seems to be along the lines he has been
agitating for.

can you explain, step by step, all the details of this calculation?

--- In tuning-math@yahoogroups.com, "wallyesterpaulrus"
<wallyesterpaulrus@y...> wrote:
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
> wrote:
> > --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
> <gwsmith@s...>
> > wrote:
> >
> > > This turns out to be 1/4-comma meantone; the sequence of fifths
> falls
> > > into a regular pattern of three pure fifts followed by a 40/27.
> >
> > I should add for this and the miracle example that the limit octave
> is
> > 2. This is not always the case; Paul may or may not like to know
> that
> > for Pajara, for instance, it is distinctly flat. Carl I hope is
> taking
> > note of this, as it seems to be along the lines he has been
> agitating for.
>
> can you explain, step by step, all the details of this calculation?

Which one? In general, you show something is going to be the best
possible choice given the commas you have--for instance 2^n is often
the most reduced for any n, but for some temperaments, such as pajara
with its 64/63 in the mix, it isn't (64 reduces to 63.)

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "wallyesterpaulrus"
> <wallyesterpaulrus@y...> wrote:
> > --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<gwsmith@s...>
> > wrote:
> > > --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
> > <gwsmith@s...>
> > > wrote:
> > >
> > > > This turns out to be 1/4-comma meantone; the sequence of
fifths
> > falls
> > > > into a regular pattern of three pure fifts followed by a
40/27.
> > >
> > > I should add for this and the miracle example that the limit
octave
> > is
> > > 2. This is not always the case; Paul may or may not like to
know
> > that
> > > for Pajara, for instance, it is distinctly flat. Carl I hope is
> > taking
> > > note of this, as it seems to be along the lines he has been
> > agitating for.
> >
> > can you explain, step by step, all the details of this
calculation?
>
> Which one? In general, you show something is going to be the best
> possible choice given the commas you have--for instance 2^n is often
> the most reduced for any n, but for some temperaments, such as
pajara
> with its 64/63 in the mix, it isn't (64 reduces to 63.)

right, show all the details of *this* calculation.

--- In tuning-math@yahoogroups.com, "wallyesterpaulrus"
<wallyesterpaulrus@y...> wrote:

> > Which one? In general, you show something is going to be the best
> > possible choice given the commas you have--for instance 2^n is often
> > the most reduced for any n, but for some temperaments, such as
> pajara
> > with its 64/63 in the mix, it isn't (64 reduces to 63.)
>
> right, show all the details of *this* calculation.

I'm confused. Obviously, 63 is less than 64, so 2^6 reduces to 63.
What is there to show? On the other hand, for 5-limit meantone, if we
have 5^n then multiplying it by 80/81 obviously increases its height,
whereas attempting to lower the power of 5 by multiplying by 81/80
gives 81*5^(n-1) / 16, with height 2^4 * 3^4 * 5^(n-1) > 5^n. Hence,
5^n is always reduced, and so 5 must be exactly generated by g^4 where
g is the generator, so that g = 5^(1/4). If we look at the TM
reduction of 1, 3/2, (3/2)^2, (3/2)^3, (3/2)^4 we get 1, 3/2, 9/4,
10/3, 5, ...giving ratios of 3/2, 3/2, 40/27, 3/2, ...

Unfortunately it seems to get a lot more complicated in the 7-limit
and beyond.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:

> Unfortunately it seems to get a lot more complicated in the 7-limit
> and beyond.

"A lot" is exaggerated; I find these can be done. The 7-limit meantone
case is easy--exactly the same argument applies to 126/125 as to
81/80, and the limit generator is 1/4-comma meantone again. Miracle is
more difficult, but you can show that (27/5)^n cannot be further
reduced by either 225/224 or 1029/1024; since 27/5 is represented by
25 generator steps, the limit secor is (27/5)^(1/25), or 116.782 cents.

Now the question is how to turn it into an algorithm.

>This turns out to be 1/4-comma meantone; the sequence of fifths falls
>into a regular pattern of three pure fifts followed by a 40/27.

It's not clear to me why big n should be special. Just because it
causes convergence? Why should I optimize my generator for whatever
a chain of 500 generators approximates, over what a chain of 5 of
them approximates?

>I should add for this and the miracle example that the limit octave is
>2. This is not always the case; Paul may or may not like to know that
>for Pajara, for instance, it is distinctly flat. Carl I hope is taking
>note of this, as it seems to be along the lines he has been agitating
>for.

Indeed. -C.

i have no idea what you're talking about. :(

--- In tuning-math@yahoogroups.com, "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@yahoogroups.com, "wallyesterpaulrus"
> <wallyesterpaulrus@y...> wrote:
>
> > > Which one? In general, you show something is going to be the
best
> > > possible choice given the commas you have--for instance 2^n is
often
> > > the most reduced for any n, but for some temperaments, such as
> > pajara
> > > with its 64/63 in the mix, it isn't (64 reduces to 63.)
> >
> > right, show all the details of *this* calculation.
>
> I'm confused. Obviously, 63 is less than 64, so 2^6 reduces to 63.
> What is there to show? On the other hand, for 5-limit meantone, if
we
> have 5^n then multiplying it by 80/81 obviously increases its
height,
> whereas attempting to lower the power of 5 by multiplying by 81/80
> gives 81*5^(n-1) / 16, with height 2^4 * 3^4 * 5^(n-1) > 5^n.
Hence,
> 5^n is always reduced, and so 5 must be exactly generated by g^4
where
> g is the generator, so that g = 5^(1/4). If we look at the TM
> reduction of 1, 3/2, (3/2)^2, (3/2)^3, (3/2)^4 we get 1, 3/2, 9/4,
> 10/3, 5, ...giving ratios of 3/2, 3/2, 40/27, 3/2, ...
>
> Unfortunately it seems to get a lot more complicated in the 7-limit
> and beyond.