? Basins of attraction

Given x and r, we might define a ?-function basin of attraction for x
as all y such that |?(x)-?(y)| < r. Since I think the inverse ?
function is not too difficult to compute, this could be computed by
means of that.

The ? basin is bigger around more portenous intervals, such as 3/2.
One could look for tunings that put every approximation for a set of
intervals into its basin, on the grounds that in this way the
intervals would be disambiguated. This favors higher limit intervals
over lower limit ones; the higher limit intervals are presumed to be
delicate. Or, it might be interesting to pick an r and take everything
which isn't already in a basin as a scale element, going up by
denominator size.

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

We can define a scale box_n by the set {box(1+i*2^(-n)} for
0 < i <= 1. Then the even scale degrees of box_{n+1} are all the scale
degrees of box_n, and the odd scale degrees are the mediants of the
even scale degrees, so that odd degrees are more complex than even
degrees, and degrees divisible by four less complex yet, and so forth.
Box_n can also be descibed in terms of the Stern-Brocot tree
restricted to 1 < q <= 2 for some level of the tree. Box4 is in the
Scala archive already as mediant16.

These box scales, if taken to a high enough level, might be used to
define the basins of attraction. For instance, box29 gives about 10
cents on either side of 3/2, its zone being 85/57 < q < 86/57. The
next interval over is box(3/2+2^(-28)), which is 83/55, and it gets
the much smaller zone 86/57 < q < 163/108, a total basin of attraction
less than a cent wide. However, both basins satisfy |?(N) - ?(q)| <
2^(-29), where N is 3/2 and 83/55 respectively.

Here is a list of the size in cents of the basin of attraction for the
11-limit diamond, using 2^(-29):

1 115.458378
12/11 1.345774
11/10 1.536061
10/9 1.789992
9/8 2.137942
8/7 2.631694
7/6 3.365555
6/5 4.526801
11/9 1.427534
5/4 6.533664
14/11 .917726
9/7 2.155300
4/3 10.492938
11/8 1.543030
7/5 3.733159
10/7 1.939886
16/11 .837305
3/2 20.248582
14/9 1.121777
11/7 1.763549
8/5 3.266538
18/11 .713824
5/3 8.394063
12/7 1.616480
7/4 4.666758
16/9 .981431
9/5 3.017789
20/11 .669726
11/6 2.141670

Here is a listing by decreasing basin size:

1 115.458378
3/2 20.248582
4/3 10.492938
5/3 8.394063
5/4 6.533664
7/4 4.666758
6/5 4.526801
7/5 3.733159
7/6 3.365555
8/5 3.266538
9/5 3.017789
8/7 2.631694
9/7 2.155300
11/6 2.141670
9/8 2.137942
10/7 1.939886
10/9 1.789992
11/7 1.763549
12/7 1.616480
11/8 1.543030
11/10 1.536061
11/9 1.427534
12/11 1.345774
14/9 1.121777
16/9 .981431
14/11 .917726
16/11 .837305
18/11 .713824
20/11 .669726

>Here is a list of the size in cents of the basin of attraction for the
>11-limit diamond, using 2^(-29):
>
>1 115.458378
>12/11 1.345774
>11/10 1.536061
>10/9 1.789992
>9/8 2.137942
>8/7 2.631694
>7/6 3.365555
>6/5 4.526801
>11/9 1.427534
>5/4 6.533664
>14/11 .917726
>9/7 2.155300
>4/3 10.492938
>11/8 1.543030
>7/5 3.733159
>10/7 1.939886
>16/11 .837305
>3/2 20.248582
>14/9 1.121777
>11/7 1.763549
>8/5 3.266538
>18/11 .713824
>5/3 8.394063
>12/7 1.616480
>7/4 4.666758
>16/9 .981431
>9/5 3.017789
>20/11 .669726
>11/6 2.141670

Is this proportional to tenneyheight or log(tenneyheight)?

>Here is a listing by decreasing basin size:
>
>1 115.458378
>3/2 20.248582
>4/3 10.492938
>5/3 8.394063
>5/4 6.533664
>7/4 4.666758
>6/5 4.526801
>7/5 3.733159
>7/6 3.365555
>8/5 3.266538
>9/5 3.017789
>8/7 2.631694
>9/7 2.155300
>11/6 2.141670
>9/8 2.137942
>10/7 1.939886
>10/9 1.789992
>11/7 1.763549
>12/7 1.616480
>11/8 1.543030
>11/10 1.536061
>11/9 1.427534
>12/11 1.345774
>14/9 1.121777
>16/9 .981431
>14/11 .917726
>16/11 .837305
>18/11 .713824
>20/11 .669726

What about intervals > an octave?

I don't know why you think ? is useful. If it makes the higher-limit
intervals more delicate, that seems wrong, since if anything the
lower-limit ones feel mistuning more. That's the reasoning behind
Tenney weighting, anyway, which is backed up by harmonic entropy.
Speaking of which, why don't you just start calculating harmonic
entropy for chords? It's definitely the most elegant model of
discordance I know of, and if it's well-behaved enough to use for
optimizations it should definitely replace cents error.

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@...> wrote:

> I don't know why you think ? is useful. If it makes the higher-limit
> intervals more delicate, that seems wrong, since if anything the
> lower-limit ones feel mistuning more. That's the reasoning behind
> Tenney weighting, anyway, which is backed up by harmonic entropy.

Does harmonic entropy really say that mistuning of fifths is more
delicate than mistuning of thirds, and if so, how? Anyway, it doesn't
say a thing about mistuning per se. I could, of course, go in exactly
the opposite direction and look at |box(N)-box(q)| instead.

> Speaking of which, why don't you just start calculating harmonic
> entropy for chords?

Because calculating harmonic entropy looks like a huge, horrible nasty
computational headache, of course.

>> I don't know why you think ? is useful. If it makes the higher-limit
>> intervals more delicate, that seems wrong, since if anything the
>> lower-limit ones feel mistuning more. That's the reasoning behind
>> Tenney weighting, anyway, which is backed up by harmonic entropy.
>
>Does harmonic entropy really say that mistuning of fifths is more
>delicate than mistuning of thirds, and if so, how?

I dunno about that. But for just intervals, Tenney height agrees
with harmonic entropy. There are deeper minima around simpler
ratios, but I don't know if they're any wider. If they really are
the same width, that would mean a given amount of mistuning causes
a greater increase in entropy for simple ratios.

When the mistuning knocks you out of one basin and into another is
less clear. Harmonic entropy is continuous, so you just get a
value -- it doesn't tell you what approximates what (that I know of).

There is this thing called Keenan justness, which is the 2nd
derivative of h. entropy...
/tuning-math/files/Erlich/keenan.jpg

>> Speaking of which, why don't you just start calculating harmonic
>> entropy for chords?
>
>Because calculating harmonic entropy looks like a huge, horrible nasty
>computational headache, of course.

Hm.

-Carl

Carl Lumma wrote:
>>>I don't know why you think ? is useful. If it makes the higher-limit
>>>intervals more delicate, that seems wrong, since if anything the
>>>lower-limit ones feel mistuning more. That's the reasoning behind
>>>Tenney weighting, anyway, which is backed up by harmonic entropy.
>>
>>Does harmonic entropy really say that mistuning of fifths is more
>>delicate than mistuning of thirds, and if so, how?
> > I dunno about that. But for just intervals, Tenney height agrees
> with harmonic entropy. There are deeper minima around simpler
> ratios, but I don't know if they're any wider. If they really are
> the same width, that would mean a given amount of mistuning causes
> a greater increase in entropy for simple ratios.

Yes, I think that's the idea.

> When the mistuning knocks you out of one basin and into another is
> less clear. Harmonic entropy is continuous, so you just get a
> value -- it doesn't tell you what approximates what (that I know of).

As long as it's differentiable, it's clear. You move from one basin to another when the gradient becomes zero. That's what "basin" means after all.

For scoring temperaments it probably doesn't matter. You can set an average entropy of intervals, weighted by their complexity. There are problems with optimization because you can leave a basin. It might be interesting for somebody to look into this (and it'd only need a one-dimensional harmonic entropy function).

> There is this thing called Keenan justness, which is the 2nd
> derivative of h. entropy...
> /tuning-math/files/Erlich/keenan.jpg
> > >>>Speaking of which, why don't you just start calculating harmonic
>>>entropy for chords? >>
>>Because calculating harmonic entropy looks like a huge, horrible nasty
>>computational headache, of course.
> > > Hm.

Does anybody have the algorithm then? Is the problem speed or memory, and what's the algorithmic complexity?

Graham

>> When the mistuning knocks you out of one basin and into another is
>> less clear. Harmonic entropy is continuous, so you just get a
>> value -- it doesn't tell you what approximates what (that I know of).
>
>As long as it's differentiable, it's clear. You move from one basin to
>another when the gradient becomes zero. That's what "basin" means after
>all.

Eyeballing the graphs and looking at printouts of the values, it
doesn't seem that clear to me. Some minima may not even have
JI analogs.

>For scoring temperaments it probably doesn't matter. You can set an
>average entropy of intervals, weighted by their complexity.

How would complexity be relevant to harmonic entropy?

>> There is this thing called Keenan justness, which is the 2nd
>> derivative of h. entropy...
>> /tuning-math/files/Erlich/keenan.jpg
>>
>>>>Speaking of which, why don't you just start calculating harmonic
>>>>entropy for chords?
>>>
>>>Because calculating harmonic entropy looks like a huge, horrible nasty
>>>computational headache, of course.
>>
>> Hm.
>
>Does anybody have the algorithm then? Is the problem speed or memory,
>and what's the algorithmic complexity?

Last I heard, Paul knew what to do, but was waiting for computer
time.

-Carl

Carl Lumma wrote:
>>>When the mistuning knocks you out of one basin and into another is
>>>less clear. Harmonic entropy is continuous, so you just get a
>>>value -- it doesn't tell you what approximates what (that I know of).
>>
>>As long as it's differentiable, it's clear. You move from one basin to >>another when the gradient becomes zero. That's what "basin" means after >>all.
> > Eyeballing the graphs and looking at printouts of the values, it
> doesn't seem that clear to me. Some minima may not even have
> JI analogs.

If they don't have JI analogs, then they're basins without JI analogs. What does the ear care? It may be that the function isn't differentiable, or has too many small basins, in which case you'll need to filter it.

>>For scoring temperaments it probably doesn't matter. You can set an >>average entropy of intervals, weighted by their complexity.
> > How would complexity be relevant to harmonic entropy?

It's relevant to temperaments. You want simple intervals from the temperament to be good consonances. You can set the weighting so that the dissonance of arbitrarily complex intervals doesn't matter.

> > >>>There is this thing called Keenan justness, which is the 2nd
>>>derivative of h. entropy...
>>>/tuning-math/files/Erlich/keenan.jpg
>>>
>>>
>>>>>Speaking of which, why don't you just start calculating harmonic
>>>>>entropy for chords? >>>>
>>>>Because calculating harmonic entropy looks like a huge, horrible nasty
>>>>computational headache, of course.
>>>
>>>Hm.
>>
>>Does anybody have the algorithm then? Is the problem speed or memory, >>and what's the algorithmic complexity?
> > Last I heard, Paul knew what to do, but was waiting for computer
> time.

I expect there are ways of reducing the complexity, but I can't speculate without knowing the algorithm, or the problems.

Graham

>>>>When the mistuning knocks you out of one basin and into another is
>>>>less clear. Harmonic entropy is continuous, so you just get a
>>>>value -- it doesn't tell you what approximates what (that I know of).
>>>
>>>As long as it's differentiable, it's clear. You move from one basin to
>>>another when the gradient becomes zero. That's what "basin" means after
>>>all.
>>
>> Eyeballing the graphs and looking at printouts of the values, it
>> doesn't seem that clear to me. Some minima may not even have
>> JI analogs.
>
>If they don't have JI analogs, then they're basins without JI analogs.
>What does the ear care?

The ear doesn't, but you have to know what to ask for when you
search for a temperament.

>It may be that the function isn't
>differentiable, or has too many small basins, in which case you'll need
>to filter it.

It's the latter, I think.

>>>For scoring temperaments it probably doesn't matter. You can set an
>>>average entropy of intervals, weighted by their complexity.
>>
>> How would complexity be relevant to harmonic entropy?
>
>It's relevant to temperaments. You want simple intervals from the
>temperament to be good consonances.

Oh, I thought you meant harmonic complexity!

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <ekin@...> wrote:

> I dunno about that. But for just intervals, Tenney height agrees
> with harmonic entropy.

Which means what, exactly? Harmonic entropy is a parametrized family
of badness functions, not a single badness function.

>> I dunno about that. But for just intervals, Tenney height agrees
>> with harmonic entropy.
>
>Which means what, exactly? Harmonic entropy is a parametrized family
>of badness functions, not a single badness function.

The only parameter for dyadic entropy is s. Typical values of s
lead to rankings for just intervals that agree with Tenney height.

There are actually two other parameters -- the 'limit' of the series
used to carve up the audio spectrum, and the type of series used
(Farey, Mann, etc.) . As the limit goes to infinity, the entropy
curve converges on a single shape, so that takes care of that.
Meanwhile, the entropy values for just intervals are proportionate
to log(n*d) whether a Farey, Mann, or Tenney series is used to begin
with. Is what I remember.

-Carl

--- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<genewardsmith@...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
> <genewardsmith@> wrote:
>
> We can define a scale box_n by the set {box(1+i*2^(-n)} for
> 0 < i <= 1. Then the even scale degrees of box_{n+1} are all the
scale
> degrees of box_n, and the odd scale degrees are the mediants of the
> even scale degrees, so that odd degrees are more complex than even
> degrees, and degrees divisible by four less complex yet, and so
forth.
> Box_n can also be descibed in terms of the Stern-Brocot tree
> restricted to 1 < q <= 2 for some level of the tree. Box4 is in the
> Scala archive already as mediant16.
>
> These box scales, if taken to a high enough level, might be used to
> define the basins of attraction. For instance, box29 gives about 10
> cents on either side of 3/2, its zone being 85/57 < q < 86/57. The
> next interval over is box(3/2+2^(-28)), which is 83/55, and it gets
> the much smaller zone 86/57 < q < 163/108, a total basin of
attraction
> less than a cent wide. However, both basins satisfy |?(N) - ?(q)| <
> 2^(-29), where N is 3/2 and 83/55 respectively.
>
> Here is a list of the size in cents of the basin of attraction for
the
> 11-limit diamond, using 2^(-29):

I know you've done "the box thing" before, could you please define
the use of 2^(negative power)? I've reviewed both Farey sequences
and the Stern-Brocot tree. Thanks!

>
> 1 115.458378
> 12/11 1.345774
> 11/10 1.536061
> 10/9 1.789992
> 9/8 2.137942
> 8/7 2.631694
> 7/6 3.365555
> 6/5 4.526801
> 11/9 1.427534
> 5/4 6.533664
> 14/11 .917726
> 9/7 2.155300
> 4/3 10.492938
> 11/8 1.543030
> 7/5 3.733159
> 10/7 1.939886
> 16/11 .837305
> 3/2 20.248582
> 14/9 1.121777
> 11/7 1.763549
> 8/5 3.266538
> 18/11 .713824
> 5/3 8.394063
> 12/7 1.616480
> 7/4 4.666758
> 16/9 .981431
> 9/5 3.017789
> 20/11 .669726
> 11/6 2.141670
>
> Here is a listing by decreasing basin size:
>
> 1 115.458378
> 3/2 20.248582
> 4/3 10.492938
> 5/3 8.394063
> 5/4 6.533664
> 7/4 4.666758
> 6/5 4.526801
> 7/5 3.733159
> 7/6 3.365555
> 8/5 3.266538
> 9/5 3.017789
> 8/7 2.631694
> 9/7 2.155300
> 11/6 2.141670
> 9/8 2.137942
> 10/7 1.939886
> 10/9 1.789992
> 11/7 1.763549
> 12/7 1.616480
> 11/8 1.543030
> 11/10 1.536061
> 11/9 1.427534
> 12/11 1.345774
> 14/9 1.121777
> 16/9 .981431
> 14/11 .917726
> 16/11 .837305
> 18/11 .713824
> 20/11 .669726
>

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@...> wrote:

> > These box scales, if taken to a high enough level, might be used to
> > define the basins of attraction. For instance, box29 gives about 10
> > cents on either side of 3/2, its zone being 85/57 < q < 86/57. The
> > next interval over is box(3/2+2^(-28)), which is 83/55, and it gets
> > the much smaller zone 86/57 < q < 163/108, a total basin of
> attraction
> > less than a cent wide. However, both basins satisfy |?(N) - ?(q)| <
> > 2^(-29), where N is 3/2 and 83/55 respectively.
> >
> > Here is a list of the size in cents of the basin of attraction for
> the
> > 11-limit diamond, using 2^(-29):
>
> I know you've done "the box thing" before, could you please define
> the use of 2^(negative power)? I've reviewed both Farey sequences
> and the Stern-Brocot tree. Thanks!

2^(-29) means that this is the interval between sucessive dyadic
rationals that I'm taking "box" of. One can also use it to measure the
"depth" of something on the Stern-Brocot tree: if we take 18/31, then
?(18/31) = 75/128, and 18/31 is what might be called a "level seven"
fraction, in that it is of the form Odd/2^7. We can list everything in
a certain range of some level or less using the box function; for
instance, going from 37/64 = 296/512 to 19/32 = 304/512 we get, upon
taking the box, 11/19, 40/69, 29/50, 47/81, 18/31, 43/74, 25/43,
32/55, 7/12. These are the level nine, or less, meantone fifths. This
approach is the Stern-Brocot alternative to looking at the nth Farey
sequence.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<genewardsmith@...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@> wrote:
>
> > > These box scales, if taken to a high enough level, might be
used to
> > > define the basins of attraction. For instance, box29 gives
about 10
> > > cents on either side of 3/2, its zone being 85/57 < q < 86/57.
The
> > > next interval over is box(3/2+2^(-28)), which is 83/55, and it
gets
> > > the much smaller zone 86/57 < q < 163/108, a total basin of
> > attraction
> > > less than a cent wide. However, both basins satisfy |?(N) - ?(q)
| <
> > > 2^(-29), where N is 3/2 and 83/55 respectively.
> > >
> > > Here is a list of the size in cents of the basin of attraction
for
> > the
> > > 11-limit diamond, using 2^(-29):
> >
> > I know you've done "the box thing" before, could you please define
> > the use of 2^(negative power)? I've reviewed both Farey sequences
> > and the Stern-Brocot tree. Thanks!
>
> 2^(-29) means that this is the interval between sucessive dyadic
> rationals that I'm taking "box" of. One can also use it to measure
the
> "depth" of something on the Stern-Brocot tree: if we take 18/31,
then
> ?(18/31) = 75/128, and 18/31 is what might be called a "level seven"
> fraction, in that it is of the form Odd/2^7. We can list everything
in
> a certain range of some level or less using the box function; for
> instance, going from 37/64 = 296/512 to 19/32 = 304/512 we get, upon
> taking the box, 11/19, 40/69, 29/50, 47/81, 18/31, 43/74, 25/43,
> 32/55, 7/12. These are the level nine, or less, meantone fifths.
This
> approach is the Stern-Brocot alternative to looking at the nth Farey
> sequence.
>

Interesting. I wonder why they all turn out to be meantone fifths?
I take it, that by interval, you mean arithmetic, like (304-
296)/512=8/512=1/64? I think I have the Stern-Brocot tree figured
out, but could you tell me if/how levels in the tree correspond to
your dyadic levels...Thanks
(304-296)/512?

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@...> wrote:

> Interesting. I wonder why they all turn out to be meantone fifths?

Simply because that's where I was looking.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<genewardsmith@...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@> wrote:
>
> > Interesting. I wonder why they all turn out to be meantone fifths?
>
> Simply because that's where I was looking.

Hmm. Kind of circular! Maybe I'm dumb, but where were you looking?

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
> <genewardsmith@> wrote:
> >
> > --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > <paul_hjelmstad@> wrote:
> >
> > > Interesting. I wonder why they all turn out to be meantone fifths?
> >
> > Simply because that's where I was looking.
>
> Hmm. Kind of circular! Maybe I'm dumb, but where were you looking?

There are 512 level 9 intervals 1 <= q < 2, and I chose to look at
the range associated to meantone is all it is.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<genewardsmith@...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@> wrote:
> >
> > --- In tuning-math@yahoogroups.com, "Gene Ward Smith"
> > <genewardsmith@> wrote:
> > >
> > > --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> > > <paul_hjelmstad@> wrote:
> > >
> > > > Interesting. I wonder why they all turn out to be meantone
fifths?
> > >
> > > Simply because that's where I was looking.
> >
> > Hmm. Kind of circular! Maybe I'm dumb, but where were you looking?
>
> There are 512 level 9 intervals 1 <= q < 2, and I chose to look at
> the range associated to meantone is all it is.

I see. (Gee, I miss "Up Thread" - Why did Yahoo have to get rid of
it?) Let me figure this out. Since meantone goes from 0-comma to 1/4
comma, which would be 3/2 down to 5^(1/4), is this the range you are
looking at, or is the range wider, going down to the gruesome 1-comma
meantone? Of course, there are the negative-comma meantones, too
right? To take in fifths > 3/2...
>

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@...> wrote:

> I see. (Gee, I miss "Up Thread" - Why did Yahoo have to get rid of
> it?) Let me figure this out. Since meantone goes from 0-comma to 1/4
> comma, which would be 3/2 down to 5^(1/4), is this the range you are
> looking at, or is the range wider, going down to the gruesome 1-comma
> meantone? Of course, there are the negative-comma meantones, too
> right? To take in fifths > 3/2...

I would consider anything between 1/3 comma and 12-et to be a meantone
tuning, so I looked between 12 and 19. As for how far down the flat
end extends, I guess I'd cut it off at 26, so you could toss 26 and 45
in there.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<genewardsmith@...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@> wrote:
>
> > I see. (Gee, I miss "Up Thread" - Why did Yahoo have to get rid
of
> > it?) Let me figure this out. Since meantone goes from 0-comma to
1/4
> > comma, which would be 3/2 down to 5^(1/4), is this the range you
are
> > looking at, or is the range wider, going down to the gruesome 1-
comma
> > meantone? Of course, there are the negative-comma meantones, too
> > right? To take in fifths > 3/2...
>
> I would consider anything between 1/3 comma and 12-et to be a
meantone
> tuning, so I looked between 12 and 19. As for how far down the flat
> end extends, I guess I'd cut it off at 26, so you could toss 26 and
45
> in there.
>
I see. I think I've almost got it - the box thing that is - I am
going to find your earlier post(s) on it, which I learned (and
subsequently forgot). Stern-Brocot also ties into Polya's methods,
as I am sure you know - I found that on EIS - Basin of attraction -
sounds reminiscent of Chaos theory - but as usual, I'm rambling.

Gene wrote:

> 2^(-29) means that this is the interval between sucessive dyadic
> rationals that I'm taking "box" of. One can also use it to measure
the
> "depth" of something on the Stern-Brocot tree: if we take 18/31,
then
> ?(18/31) = 75/128, and 18/31 is what might be called a "level seven"
> fraction, in that it is of the form Odd/2^7. We can list everything
in
> a certain range of some level or less using the box function; for
> instance, going from 37/64 = 296/512 to 19/32 = 304/512 we get, upon
> taking the box, 11/19, 40/69, 29/50, 47/81, 18/31, 43/74, 25/43,
> 32/55, 7/12. These are the level nine, or less, meantone fifths.
This
> approach is the Stern-Brocot alternative to looking at the nth Farey
> sequence.

Could you please give me the continued fraction notation
for box(3/2+2^(-28)). I get (1;0,1,26,1) which gives 55/27, not
83/55. Oops.

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@...> wrote:
>
> Gene wrote:
>
> > 2^(-29) means that this is the interval between sucessive dyadic
> > rationals that I'm taking "box" of. One can also use it to
measure
> the
> > "depth" of something on the Stern-Brocot tree: if we take 18/31,
> then
> > ?(18/31) = 75/128, and 18/31 is what might be called a "level
seven"
> > fraction, in that it is of the form Odd/2^7. We can list
everything
> in
> > a certain range of some level or less using the box function; for
> > instance, going from 37/64 = 296/512 to 19/32 = 304/512 we get,
upon
> > taking the box, 11/19, 40/69, 29/50, 47/81, 18/31, 43/74, 25/43,
> > 32/55, 7/12. These are the level nine, or less, meantone fifths.
> This
> > approach is the Stern-Brocot alternative to looking at the nth
Farey
> > sequence.
>
> Could you please give me the continued fraction notation
> for box(3/2+2^(-28)). I get (1;0,1,26,1) which gives 55/27, not
> 83/55. Oops.

Found it - its (1;1,1,26,1). I am having trouble with definition
of the Conway box function on Wikipedia. Definition conflicts
with definition of the question mark function given previously.

(The binary string further up includes 0 to the left of the decimal
point, to obtain (0;2,1,2,1,2,1...) for the continued fraction
expansion...)

Also, in this case, where 3/2 + 2^-28 expands to 1.10..1 in binary,
you seem to have to assume 0.10..1 to get 1,1,26,1 and then put
the 1 in there for n(sub-0): (1;1,1,26,1). I'll test some others.

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@...> wrote:

> Found it - its (1;1,1,26,1). I am having trouble with definition
> of the Conway box function on Wikipedia. Definition conflicts
> with definition of the question mark function given previously.

Conflicts how?

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@...> wrote:

> Could you please give me the continued fraction notation
> for box(3/2+2^(-28)). I get (1;0,1,26,1) which gives 55/27, not
> 83/55. Oops.

I'm not sure what you are asking; the continued fraction for
83/55 is (1;1,1,27) though. You can also write that (1;1,1,26,1).

--- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<genewardsmith@...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@> wrote:
>
> > Found it - its (1;1,1,26,1). I am having trouble with definition
> > of the Conway box function on Wikipedia. Definition conflicts
> > with definition of the question mark function given previously.
>
> Conflicts how?

The binary string further up includes 0 to the left of the decimal
point, to obtain (0;2,1,2,1,2,1...) for the continued fraction
expansion...In the Conway box function, they make no such provision,
only considers digits to the right of the decimal point, unless I
am misreading it.

>

--- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<genewardsmith@...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@> wrote:
>
> > Could you please give me the continued fraction notation
> > for box(3/2+2^(-28)). I get (1;0,1,26,1) which gives 55/27, not
> > 83/55. Oops.
>
> I'm not sure what you are asking; the continued fraction for
> 83/55 is (1;1,1,27) though. You can also write that (1;1,1,26,1).

Right, I finally found (1;1,1,26,1) which works. I had to assume
(0.1000000000000000000000000001) though, to get 1,1,26,1; this is
confusing to me, and then put 1=(floor(x)) back to get(1;1,1,26,1)
Does one always assume an initial 0 to calculate the n1-nn part of
the continued fraction?

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@...> wrote:
>
> The binary string further up includes 0 to the left of the decimal
> point, to obtain (0;2,1,2,1,2,1...) for the continued fraction
> expansion...In the Conway box function, they make no such provision,
> only considers digits to the right of the decimal point, unless I
> am misreading it.

To find box(x), first take y = x-floor(x), then box(x) = floor(x)+box(y).

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@...> wrote:

> Does one always assume an initial 0 to calculate the n1-nn part of
> the continued fraction?

I don't know what you mean, but if you read the Wikipedia definition
of box, you start out looking at the base 2 expansion to the right of
the decimal point, and add floor(x) afterwards.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<genewardsmith@...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@> wrote:
> >
> > The binary string further up includes 0 to the left of the decimal
> > point, to obtain (0;2,1,2,1,2,1...) for the continued fraction
> > expansion...In the Conway box function, they make no such provision,
> > only considers digits to the right of the decimal point, unless I
> > am misreading it.
>
> To find box(x), first take y = x-floor(x), then box(x) = floor(x)+box
(y).
>
Thanks. That's what I came up with, but its nice to have a formula

--- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<genewardsmith@...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@> wrote:
>
> > Does one always assume an initial 0 to calculate the n1-nn part of
> > the continued fraction?
>
> I don't know what you mean, but if you read the Wikipedia definition
> of box, you start out looking at the base 2 expansion to the right of
> the decimal point, and add floor(x) afterwards.

Okay - my point was, that in calculating a continued fraction, like
(0;2,1,2,1,2,1..) you have to base it on the binary string 0011001100
which is based on 0.01100110, that's all I meant. "I just want it to
work" as my customers always say! (And it does...)
>

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@...> wrote:

> Okay - my point was, that in calculating a continued fraction, like
> (0;2,1,2,1,2,1..) you have to base it on the binary string 0011001100
> which is based on 0.01100110, that's all I meant. "I just want it to
> work" as my customers always say! (And it does...)

No, that should be .001001001.

--- In tuning-math@yahoogroups.com, "Gene Ward Smith"
<genewardsmith@...> wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
> <paul_hjelmstad@> wrote:
>
> > Okay - my point was, that in calculating a continued fraction,
like
> > (0;2,1,2,1,2,1..) you have to base it on the binary string
0011001100
> > which is based on 0.01100110, that's all I meant. "I just want it
to
> > work" as my customers always say! (And it does...)
>
> No, that should be .001001001.

But ---

Wikipedia:

One obvious way to interpret such a string is to place a binary point
after the first 0 and read the string as a binary expansion: thus,
for instance, the string 001001001001001001001001... represents the
binary number 0.010010010010..., or 2/7

Sorry to be contrary, but this seems true for box(75/128)-> 18/31 as
well:

0.1001011 -> (0;1,1,2,1,1,2) (I verified this is the continued
fraction using the subtract-and-invert method). This is all new to me
so maybe I'm missing some subtlety in this...

--- In tuning-math@yahoogroups.com, "Paul G Hjelmstad"
<paul_hjelmstad@...> wrote:
>
> 0.1001011 -> (0;1,1,2,1,1,2) (I verified this is the continued
> fraction using the subtract-and-invert method). This is all new to me
> so maybe I'm missing some subtlety in this...
>

I'll check and see if Wikipedia needs fixing.