Yahoo Tuning Groups Ultimate Backup tuning truncating the range of a given ratio

I have an interesting idea that somebody here might find worthwhile to
explore more. Using the Stern-Brocott tree you can delineate the range
of Just Intonated ratios by way Equal Temperaments:

For example, if we seed two trees with fractions of an octave at 0/2
and 3/3, and 0/3 and 4/4 we create an upper limit and a lower limit
respectively for each position in the tree as if it were seeded for JI
with 1/1 and 2/1

It would be interesting to set up some diagram (say another view of
the same idea expressed by Helmholtz and Partch and Erlich et al) were
the pitch continuum is given something akin to a "bounds of general
recognize-ability". I think you'll find the borders will soon meet and
blend and the range of independent ratios will be far smaller than i
would've initially suspected. As an example of this note that even an
frequency ratio as "simple" as 6/5 will be seen to blur as its sharp
border will be difficult to distinguish from the flat border of an
11/9--according to this relatively simple seeding (there are many ways
to seed the tree in this fashion), these borders cross and slightly blur:

6/5 = 4/16 (lower limit) 3/11 (upper limit)
11/9 = 8/29 (lower limit) 6/20 (upper limit)

any ideas/ input?

http://www.myspace.com/danstearns

🔗Carl Lumma <clumma@yahoo.com>

🔗Graham Breed <gbreed@gmail.com>

🔗daniel_anthony_stearns <daniel_anthony_stearns@yahoo.com>

Hi Carl and Graham .The second paragraph showed how to seed the tree
with ET fractions of an octave: "if we seed two trees with fractions
of an octave at 0/2 and 3/3, and 0/3 and 4/4 we create an upper limit
and a lower limit respectively for each position in the tree as if it
were seeded for JI with 1/1 and 2/1" . So this seeding would give a
scale tree in ET fractions of an octave as:

0/5 7/7
7/12
7/17 14/19
7/22 14/29 21/31 21/26
(etc)

So ,in case this isn't clear, here's the lower limit:

0/3 4/4
4/7
4/10 8/11
4/13 8/17 12/18 12/15
(etc)

And here's the upper limit:

0/2 3/3
3/5
3/7 6/8
3/9 6/12 9/13 9/11
(etc)

What's interesting here to me, is that this gives some pretty clear
results in terms of creating a "bounds of general recognize-ability",
and the results line up pretty interestingly with similar efforts,
though the method is quite different.

http://www.myspace.com/danstearns

--- In tuning@yahoogroups.com, Graham Breed <gbreed@...> wrote:
>
> daniel_anthony_stearns wrote:
> > I have an interesting idea that somebody here might find worthwhile to
> > explore more. Using the Stern-Brocott tree you can delineate the range
> > of Just Intonated ratios by way Equal Temperaments:
>
> That'd be the scale tree of ratios then. What do equal temperaments
> have to do with it? You don't mention any in your post.
>
>
> Graham
>

🔗Gene Ward Smith <genewardsmith@coolgoose.com>

🔗daniel_anthony_stearns <daniel_anthony_stearns@yahoo.com>

🔗Gene Ward Smith <genewardsmith@coolgoose.com>

🔗daniel_anthony_stearns <daniel_anthony_stearns@yahoo.com>

well i think it's important to see the 5 and 7 bounds in the context
more than their mean, but let's say it's two ways to say the same
thing; because I'm not interested in that and the distinction between
what you're saying and what i'm saying doesn't matter (other than if
it somehow makes it easier to understand). But what interests me is
the results, any ideas/observations etc?

--- In tuning@yahoogroups.com, "Gene Ward Smith" <genewardsmith@...>
wrote:
>
> --- In tuning@yahoogroups.com, "daniel_anthony_stearns"
> <daniel_anthony_stearns@> wrote:
> >
> > well you have to set the boundaries, and in this sence it's important
> > how you "seed" the tree
>
> The boundries are set by the root of the tree, for example 7/12.
> Nothing further is needed. If you go up, 7/12 lies between 4/7 and
> 3/5, and this seems to be the same as your seeding. It's not necessary.
>
> > --- In tuning@yahoogroups.com, "Gene Ward Smith" <genewardsmith@>
> > wrote:
> > >
> > > --- In tuning@yahoogroups.com, "daniel_anthony_stearns"
> > > <daniel_anthony_stearns@> wrote:
> > >
> > > I think it would be clearer if you didn't talk of "seeding", and
> > > looked at it as simply the Stern-Brocot tree depending from 7/12.
> > >
> >
>

🔗Kraig Grady <kraiggrady@anaphoria.com>

🔗daniel_anthony_stearns <daniel_anthony_stearns@yahoo.com>

🔗Gene Ward Smith <genewardsmith@coolgoose.com>

🔗daniel_anthony_stearns <daniel_anthony_stearns@yahoo.com>

This method is not about generators, but rather interval span, and
uniqueness so this method has more in common with say Partch or
Erlich than it does Yasser or Wilson. Here's a quick and dirty example
that might perhaps give a better idea of what i'm getting at here. It
uses the 5 and 7 tree seeding, as it's the simplest JI to ET (or vice
versa) conversion :

1108 19/10 1117
1097 17/9 1108
1084 15/8 1096
1067 13/7 1080
1043 11/6 1059
1011 9/5 1029
988 16/9 1008
960 7/4 982
923 12/7 947
873 5/3 900
828 13/8 857
800 8/5 831
768 11/7 800
686 3/2 720
600 10/7 635
565 7/5 600
533 11/8 568
480 4/3 514
417 9/7 450
369 5/4 400
331 11/9 360
300 6/5 327
274 13/11 300
253 7/6 277
218 8/7 240
192 9/8 212
171 10/9 189
155 11/10 171

The cents in the left theoretically represent the "flat" threshold and
the cents in the right-hand column the "sharp" threshold. I eliminated
any instances where the overlap was greater than a few cents, or
overlap represents a continuum; as it is at the 1/1 and the 2/1. So
according to this idea we have pretty wide pockets, or gaps where
distinguishing ratios from one another is quite fuzzy. What's
interesting to me about this method is its pantheistic simplicity and
its explicitness; giving an exact flat and a sharp border (even if we
ultimately interpret these loosely), and so handily joining frequency
ratios and fractions of an octave.

http://www.myspace.com/danstearns

--- In tuning@yahoogroups.com, "Gene Ward Smith" <genewardsmith@...>
wrote:
>
> --- In tuning@yahoogroups.com, "daniel_anthony_stearns"
> <daniel_anthony_stearns@> wrote:
>
> But what interests me is
> > the results, any ideas/observations etc?
>
> An observation one might make is that picking on a "border" generator
> like 7/12 is rather different than picking on something like 18/31,
> which is right in the middle of a temperament range. From 18/31, you get
> 29/50 < 18/31 < 25/43, and continuing to fill that in gives a whole
> range of meantones. Similarly, looking at what depends from 31/53
> gives you a whole range of schismatics.
>

🔗Carl Lumma <clumma@yahoo.com>

🔗Ozan Yarman <ozanyarman@ozanyarman.com>

Thresholds? Why that concept applies to maqam perdes... Notice that I
remained within the boundaries specified with 79/80 MOS 159-tET.

----- Original Message -----
From: "daniel_anthony_stearns" <daniel_anthony_stearns@yahoo.com>
To: <tuning@yahoogroups.com>
Sent: 04 Kasï¿½m 2006 Cumartesi 0:27
Subject: [tuning] Re: truncating the range of a given ratio

This method is not about generators, but rather interval span, and
uniquenessï¿½ so this method has more in common with say Partch or
Erlich than it does Yasser or Wilson. Here's a quick and dirty example
that might perhaps give a better idea of what i'm getting at here. It
uses the 5 and 7 tree seeding, as it's the simplest JI to ET (or vice
versa) conversion :

http://www.myspace.com/danstearns

🔗Robert walker <robertwalker@robertinventor.com>

Hi Dan,

There's a notation I find useful in this situation where one wants to talk about fractions of an octave considered logarithmically, and ratios understood normally, at the same time.

It's
7//12 for seven twelfths of an octave. So 7//12 = 700 cents.

Then if one proceeds to construct your trees:

(this needs to be seen with fixed width font - see Show Message Option > Use fixed width font in yahoougroups).

Upper limits tree:
0//2 3//3
3//5
3//7 3//8
3//9 6//12 6//13 6//11
3//11 6//16 9//19 9//17 9//18 9//21 9//19 9//14
...
generated Stern Brocott fashion
eg. 6//12 from the fractions above to left
and right 3//7 and 3//5 by 3+3 = 6 and 7+5 = 12

Lower limits tree:
0//3 4//4
4//7
4//10 8//11
4//13 8//17 12//18 12//15
4//16 8//23 12//27 12//25 16//25 20//29 20//26 16//19

In cents:

Upper limits:
0 1200
720
514 450
400 600 554 655
327 450 568 635 600 514 568 771

Lower limits:
0 1200
686
480 872
369 565 800 960
300 417 533 576 768 827 923 1011
....

Then the ratios these bracket should be
1/1 2/1
3/2
4/3 5/3
5/4 7/5 8/5 7/4
6/5 9/7 11/8 10/7 11/7 13/8 12/7 9/5
....

So one would assume,
4//7 < 3/2 < 3//5
which indeed they do bracket quite nicely:
in cents:
686 < 702 < 720

similarly
4//13 < 5/4 < 3//9
in cents:
369 < 386 < 400

and the first of your examples:
4//16 < 6/5 < 3//11
in cents:
300 < 316 < 327

So anyway it seems this raises a number of interesting questions, not least, why this bracketting occurs in the first place - and does it continue all the way down the tree to infinity?

That's something a mathematician should enjoy proving :-). Sometime when I have time I may give it a go, but maybe Gene can see why it is true. Or do you have a proof of it yourself?

Then it is interesting as you say to contemplate whether it has any connection with consonance limits arrived at through more elaborate methods, at any rate it is an easily computed upper / lower bound for the region to regard as a particular ratio, which one could imagine could be useful in a number of contexts, and the simpler ratios have larger areas around them, so it is generally the right kind of a shape.

Eventually one must get overlaps because every ratio occurs in the tree, so for instance, ratios arbitrarily close to 3/2. But it seems it goes a fair way before you get to them.

Though you say there are many possible seedings that work. What would be other examples? Do they produce similar "consonance" limits or are some much narrower and some much wider? And do they all bracket the ratios nicely like this?

Thanks,

Robert

🔗Carl Lumma <clumma@yahoo.com>

🔗Gene Ward Smith <genewardsmith@coolgoose.com>

🔗Robert walker <robertwalker@robertinventor.com>

Hi Gene,

What you do is to consider the three trees together and look at the
same position in each tree.

So if you look at the position of the 6/5 in the ratios tree - then look
in the same position in the lower limit tree, you will see a 4//16
If you look at the same position in the upper limit
tree you see a 3//11

Observe that 4//16 < 6/5 < 3//11

There I hope the construction of the upper and lower limit trees is
clear - that they are constructed in the same way that you construct
the Stern Brocot tree but instead of using ordinary ratios you use
fractional octaves - e.g. 7//12 means
7/12 of 1200 cents, 4//16 means 4/16 of 1200 cents and so on.

So what I was asking is - does this pattern continue all the
way down the tree, or if I were to continue it for a few more
rows, would I eventually find a counter example.

Also what choices of numbers to seed the ET tree give this
correlation, how do you find upper and lower bound trees
that work, and so on. Just meant as some outline questions
to get one started thinking about what Dan has found out, and
what it means. I wish I had a bit of time to spend a few days
thinking about it and exploring it but this is a time when I
a have a lot on in other areas.

Robert

🔗Gene Ward Smith <genewardsmith@coolgoose.com>

--- In tuning@yahoogroups.com, "Robert walker" <robertwalker@...> wrote:
>
> Hi Gene,
>
> What you do is to consider the three trees together and look at the
> same position in each tree.

This makes no sense until you define the trees.

> So if you look at the position of the 6/5 in the ratios tree - then look
> in the same position in the lower limit tree, you will see a 4//16
> If you look at the same position in the upper limit
> tree you see a 3//11

Definitions?

> Observe that 4//16 < 6/5 < 3//11

1/4 and 3/11 are both semiconvergents to log2(6/5), but so are a lot
of things.

> There I hope the construction of the upper and lower limit trees is
> clear - that they are constructed in the same way that you construct
> the Stern Brocot tree but instead of using ordinary ratios you use
> fractional octaves - e.g. 7//12 means
> 7/12 of 1200 cents, 4//16 means 4/16 of 1200 cents and so on.

It's exactly the same. Both simply *are* the Stern-Brocot tree. What,
then, are you saying?

> So what I was asking is - does this pattern continue all the
> way down the tree, or if I were to continue it for a few more
> rows, would I eventually find a counter example.

WHAT pattern??

> Also what choices of numbers to seed the ET tree give this
> correlation, how do you find upper and lower bound trees
> that work, and so on.

What's an ET tree, and how do you seed it? Do you mean a ratio that
all branches depend from?

We could do this in tuning-math if that would be better.

🔗Robert walker <robertwalker@robertinventor.com>

Hi Gene,

I was referring back to the previous post where the trees
are defined.

By an et tree I just meant one that is based on
fractional octaves so e.g. 3//5 in the tree
means 3/5 of 1200 cents (or 3 steps of 5-et).

Take Dan's example of the 0//2 and 3//3 tree.
So just like the Stern Brocot tree,except that you start with
0//2 and 3//3 at the top.

Those are the seeds - the 0//2 etc.

The Stern Brocot tree here is seeded
with 1/1 and 2/1.

The reason the seed ratios for the tree are significant is because
f the correlation between the three trees.

All three trees are constructed like the Stern Brocot tree, the only
difference is in the interpretation of the numbers as ratios or
as fractional octaves.

http://www.cut-the-knot.org/blue/Stern.shtml

So - here is the Stern Brocot tree (3 rows) interpreted as ratios:
1/1 2/1
3/2
4/3 5/3

Upper limits tree, interpreted as fractional octaves so I use the 0//2 notation:
0//2 3//3
3//5
3//7 3//8

Lower limits tree:
0//3 4//4
4//7
4//10 8//11

Then look in the same position in all three trees. For instance 4/3 is the first ratio in the
third row of the ratios tree.

In the same position in the upper limits tree we find 3//7
In the same position in the lower limits tree we find 4//10

Then the result, conjecture, or whatever it is is that
4//10 < 4/3 < 3//7.
and similarly for the other entries in the three trees
So for instance, 8//11 < 5/3 < 3//8 etc.

For these few rows anyway, and for a few more rows as far as I explored it,
the lower and upper bounds nicely bracket the ratio with
a roughly equal sized gap to either side.

I hope this is clear now. One could set it all out using recurrence relations etc,
but hopefully this is enough now to make it clear how it works. Do ask again if
not - it is a simple idea and if it isn't coming over clearly, perhaps I have somehow
made it seem to complicated.

These sorts of things are tricky to explain because of the similarity to something else
which the reader anticipates that you are trying to say.

- for instance we know that the ratios further down a single tree bracket the ratios further
up which seemed to be where the confusion arose. In that case of course what
one uses to seed the tree is irrelevant. But here, because we are looking at a correlation
of three different trees, then the seeding of the et trees _is_ significant because the
correlation depends on it.

If you look back to Dan's original post again, you will find that it all seems clear
with hindsight and you wonder why you didn't get the message when you first
read it, these sorts of things seem to work like that. You kind of skip over the
details that don't seem to make sense because one has an expectation that
the author is saying something more ordinary in a clumsy way,when in fact
he is saying something rather more novel and saying it precisely.

Sometimes there are undefined terms and that is why one misses
what the author is syaing - but in the case of Dan's message, everything that
one needed to know was actually there in the original post, and it is just a matter
of somehow overcoming the reader's natural expectations that one is saying
something else from what one is actually saying. So it needs, like more emphasis,
repetition, and so on until the other person gets what it is about, this is normal
and I don't think there is anything one can do about it except be patient and keep trying.
It took me a day or two mulling it over from time to time before I finally got what he was saying.

So then that's why the next question for Dan is - how does one arrive at such seeds
and what happens if one tries other seeds.

For mathematical proof I suppose one question is, does this pattern continue
all the way down the tree to infinity, if so why, and if not, is there any choice of seed for
which it does, or what is the maximum distance one can go down the tree in this
way while preserving the bracketting of the ratios like this.

It is intriguing, looking at just the few rows I mapped out because the limits do bracket the
ratios very nicely with the ratio right near the middle of the range between the two
limits.

Anyway I hope the idea is clear now but if not do ask again. I'm not sure if it needs to
go to tuning math - complicated mathematical notation would perhaps obscure it unless
one needed it of course e.g. to construct a proof.

Thanks,

Robert

🔗Gene Ward Smith <genewardsmith@coolgoose.com>

Shouldn't that be 6//8 = 3//4?

> Lower limits tree:
> 0//3 4//4
> 4//7
> 4//10 8//11
>
> Then look in the same position in all three trees. For instance 4/3
is the first ratio in the
> third row of the ratios tree.

OK, I get it now. If 1 < t < 2, then let a = 4 (t-1)/(t+2) and
b = 3 (t-1)/(t+1). Then a < log2(t) < b; a is the lower limit, and b
is the upper limit.

🔗daniel_anthony_stearns <daniel_anthony_stearns@yahoo.com>

Thanks Robert. You can distort or fine tune the Just Intonation
relationships with different seeding, but I use this seeding because
it's the simplest way to create a tree where the smallest fractions of
an octave correlate to the frequency ratio seeding of 1/1 and 2/1--and
of course it's also important to note that the smaller the numbers in
the fractions the larger the "span" between them. This is very
important ,and the only seeding that works this way, if you want to
interpret the results as unique interval spans that frame the pitch
continuum with something akin to "bounds of general
recognize-ability", and in this sense I think it's quite interesting.
(And as I've said a few times, I'm also attracted to its simplicity
and pantheistic overtones.)

Anyway, I have no proof, but my instincts tell me the relationships
extend into infinity just as the 1/1 and 2/1 seeding would--but as you
might expect, the factions of an octave representing unique space are
quite finite!

http://www.myspace.com/danstearns

--- In tuning@yahoogroups.com, "Robert walker" <robertwalker@...> wrote:
>
> Hi Gene,
>
> I was referring back to the previous post where the trees
> are defined.
>
> By an et tree I just meant one that is based on
> fractional octaves so e.g. 3//5 in the tree
> means 3/5 of 1200 cents (or 3 steps of 5-et).
>
> Take Dan's example of the 0//2 and 3//3 tree.
> So just like the Stern Brocot tree,except that you start with
> 0//2 and 3//3 at the top.
>
> Those are the seeds - the 0//2 etc.
>
> The Stern Brocot tree here is seeded
> with 1/1 and 2/1.
>
> The reason the seed ratios for the tree are significant is because
> f the correlation between the three trees.
>
> All three trees are constructed like the Stern Brocot tree, the only
> difference is in the interpretation of the numbers as ratios or
> as fractional octaves.
>
> http://www.cut-the-knot.org/blue/Stern.shtml
>
> So - here is the Stern Brocot tree (3 rows) interpreted as ratios:
> 1/1 2/1
> 3/2
> 4/3 5/3
>
>
> Upper limits tree, interpreted as fractional octaves so I use the
0//2 notation:
> 0//2 3//3
> 3//5
> 3//7 3//8
>
>
> Lower limits tree:
> 0//3 4//4
> 4//7
> 4//10 8//11
>
> Then look in the same position in all three trees. For instance 4/3
is the first ratio in the
> third row of the ratios tree.
>
> In the same position in the upper limits tree we find 3//7
> In the same position in the lower limits tree we find 4//10
>
> Then the result, conjecture, or whatever it is is that
> 4//10 < 4/3 < 3//7.
> and similarly for the other entries in the three trees
> So for instance, 8//11 < 5/3 < 3//8 etc.
>
> For these few rows anyway, and for a few more rows as far as I
explored it,
> the lower and upper bounds nicely bracket the ratio with
> a roughly equal sized gap to either side.
>
> I hope this is clear now. One could set it all out using recurrence
relations etc,
> but hopefully this is enough now to make it clear how it works. Do
ask again if
> not - it is a simple idea and if it isn't coming over clearly,
perhaps I have somehow
> made it seem to complicated.
>
> These sorts of things are tricky to explain because of the
similarity to something else
> which the reader anticipates that you are trying to say.
>
> - for instance we know that the ratios further down a single tree
bracket the ratios further
> up which seemed to be where the confusion arose. In that case of
course what
> one uses to seed the tree is irrelevant. But here, because we are
looking at a correlation
> of three different trees, then the seeding of the et trees _is_
significant because the
> correlation depends on it.
>
> If you look back to Dan's original post again, you will find that it
all seems clear
> with hindsight and you wonder why you didn't get the message when
you first
> read it, these sorts of things seem to work like that. You kind of
skip over the
> details that don't seem to make sense because one has an expectation
that
> the author is saying something more ordinary in a clumsy way,when
in fact
> he is saying something rather more novel and saying it precisely.
>
> Sometimes there are undefined terms and that is why one misses
> what the author is syaing - but in the case of Dan's message,
everything that
> one needed to know was actually there in the original post, and it
is just a matter
> of somehow overcoming the reader's natural expectations that one is
saying
> something else from what one is actually saying. So it needs, like
more emphasis,
> repetition, and so on until the other person gets what it is about,
this is normal
> and I don't think there is anything one can do about it except be
patient and keep trying.
> It took me a day or two mulling it over from time to time before I
finally got what he was saying.
>
> So then that's why the next question for Dan is - how does one
arrive at such seeds
> and what happens if one tries other seeds.
>
> For mathematical proof I suppose one question is, does this pattern
continue
> all the way down the tree to infinity, if so why, and if not, is
there any choice of seed for
> which it does, or what is the maximum distance one can go down the
tree in this
> way while preserving the bracketting of the ratios like this.
>
> It is intriguing, looking at just the few rows I mapped out because
the limits do bracket the
> ratios very nicely with the ratio right near the middle of the range
between the two
> limits.
>
> Anyway I hope the idea is clear now but if not do ask again. I'm not
sure if it needs to
> go to tuning math - complicated mathematical notation would perhaps
obscure it unless
> one needed it of course e.g. to construct a proof.
>
> Thanks,
>
> Robert
>

🔗Gene Ward Smith <genewardsmith@coolgoose.com>

🔗daniel_anthony_stearns <daniel_anthony_stearns@yahoo.com>

--- In tuning@yahoogroups.com, "Gene Ward Smith" <genewardsmith@...>
wrote:
>
> > Upper limits tree, interpreted as fractional octaves so I use the
> 0//2 notation:
> > 0//2 3//3
> > 3//5
> > 3//7 3//8
>
> Shouldn't that be 6//8 = 3//4?

just thought i'd mention that it should be 6//8, but not written as
3//4 while constructing the tree. This is the same with seeding the
tree for some non-octave periodicity, say 7/4, where you'd use 4/4 and
7/4 as your seeding.

> OK, I get it now. If 1 < t < 2, then let a = 4 (t-1)/(t+2) and
> b = 3 (t-1)/(t+1). Then a < log2(t) < b; a is the lower limit, and b
> is the upper limit.

Ok, now Blackwood understood this in terms of the fifth as a generator
that creates a 7 note scale with 5 large steps and 2 small steps, and
Wilson the same only in a much more generalized and thorough way.What
i'm saying here is that if one follows the process down the whole tree
and allows their observation to notice that it's also creating an
upper and a lower limit in terms of the intervals that make up the
periodicity (the octave here). So,what i was perhaps hoping for here
was that someone who knows their way around excel or some other such
program might make up a chart/diagram where these borders would be
clearer, as i think the results are quite interesting given this
interpretation.

http://www.myspace.com/danstearns

🔗Robert Walker <robertwalker@robertinventor.com>

Hi Gene,

> OK, I get it now. If 1 < t < 2, then let a = 4 (t-1)/(t+2) and
> b = 3 (t-1)/(t+1). Then a < log2(t) < b; a is the lower limit, and b
> is the upper limit.

Congratulations on proving it so quickly. You may need to fill in some
of the steps for me.

BTW something that might puzzle a non mathematician for a moment or
two - by the t there Gene means any arbitrary ratio in the tree, e.g.
the 6/5 etc. So the lower bound for instance would be 4(t-1)/(t+2)
becomes in the case of 3/2 then its 4 (3/2-1) / (3/2 + 2) i.e. 4*
(1/2) / 7/2) i.e. 4//7

Then the a < log2(t) < b means that the a and b in the et trees bracket
the log of the ratio (e.g. putting it all in cents for more
familiarity, 3/2 is 702 cents, and can also be written as log2(3/2)
*1200, so we want a*1200 < log2(3/2)*1200 (say) < b *1200 or what
amounts to the same thing and is mathematically simpler, a < log2(t) <
b)

So, anyway, Gene, with the a < log2(t) < b, can you explain how that is
derived? I wondered if it somehow used one of the series derivations
for log, but couldn't see it right away if it does. Maybe it is
something else altogether and I'm following a red herring?

With the derivations of the formulae for a and b in terms of t, I
assume it is proved inductively through the way the tree is
constructed, and could try deriving it, don't expect it would take me
long to do, just check it is preserved after addition of the ratios -
unless there is another shortcut way to see it?

Thanks,

Robert

🔗Robert Walker <robertwalker@robertinventor.com>

Hi Dan,

Thanks, that's clear now.

Well one possibility I assume would be to start with some
more complicated ratio e.g. if 0//2 and 3//3 works, maybe
also 0/2000 and 3001/3002 will work or something, I mean
not necessarily those particular examples but you get the idea.
There - well without actually trying it out, but I think they
would be likely to again create large ranges for each
interval just like the similar simple ratios
- while also giving scope for fine tuning. At least
it is the sort of thing that can work.

But simpler seems best in absence of any reason to choose
something else. Just mentioning the possibility in case it
is interesting.

Yes of course as far as heard consonances one isn't going
very far down the tree. However it is interesting to find
out as a matter of pure maths, why it works. I suppose
pure maths spin off research type of thing.

Gene has sketched a proof. But he may need to fill in some
details before I can follow it.

Thanks,

Robert

🔗Gene Ward Smith <genewardsmith@coolgoose.com>

🔗Robert Walker <robertwalker@robertinventor.com>

🔗Gene Ward Smith <genewardsmith@coolgoose.com>

🔗Robert Walker <robertwalker@robertinventor.com>

Hi Gene,

Thanks, that's neat. I mean - the idea to look at the general shape
of the graph of the difference between two curves as a way to
establish an inequality between them. Here, observing that the graph
of the difference starts at zero at t = 0, and ends at zero at t = 1,
and in between has a single maximum and no minimum so given that the
curve is smooth, the only way you can "join the dots" as it were to
achieve such a shape must give a graph which is always positive, and
a positive difference establishes that first of the original curves
is always less than the other.

I can see that the details get a little messy but the approach is a
neat one. I mean it is all straightforward calculus as you say, just
application of the standard rules of differentiation once you have
that idea. It is the idea of using it to prove an inequality that is
a bit novel to me. If I knew the technique at any time then I've
forgotten about it. It is one of those things that seems kind of
obvious only in hindsight once you've seen it. I'm not sure if I'd
have ever thought of applying calculus to this problem or not for a
fair while at least.

Thanks again,

Robert