RE: [tuning] Re: TD 861 -- Wilson's footprints on plateau! (for K raig Graidy)

David Keenan wrote,

>A concise statement of the significance of these noble numbers in
>the logarithmic-melodic case eluded me (and apparently Paul Erlich) until
>Dan Stearns provided it.

Did you find my concise statement on this (in response to your request for
one) to be incorrect?

>Margo, In your excellent articles on Keenan Pepper's neo-Gothic tuning you
>make use of noble mediants for both purposes (melodic and harmonic). I
>think the reader would benefit from a clear distinction between the two.

Yes, and Kraig has often pointed out these two different applications of
these same underlying mathematics, and so should continue to do so to avoid
confusion.

>Anyway, from the perspective of Keenan P's original post and all that
>followed, maybe it would be nice to have two terms that relate to
>mediants (logarithmically or linearly) in a sequential (i.e., low to
>high) and reversed (high to low) Fibonacci expansion. The first is
>obviously a "Golden" ______ (whatever; generator, mediant, etc.). The
>second, if not a "Silver" ______ (whatever; generator, mediant, etc.),
>then what?

So far, you've only given one example of each of these. As far as I can
tell, they both produce MOS scales with a 1:phi step-size ratio, so they are
both equally "Golden" or "Noble". It is unclear how loose you're being with
the term "Fibonacci". There is a so-called Lucas sequence that begins 1, 3,
4, 7, 11, 18, . . . which one of Keenan P.'s generators generated . . .
anyway, give some examples . . .

>Despite its friction with the standard definition, I think in the
>generalized sense that "Silver" (whatever) works here. Any better
>ideas?

Why not preserve the standard definition so that [(2-sqrt(2))*1200] can be
called a Silver generator?

>In the logarithmic sense, a generalized "Golden generator" would
>always give a uniquely optimized strictly proper scale, and a
>generalized "Silver generator" would always give a uniquely optimized
>improper scale.

I must have missed something. Can you show with examples? Do any of your
"Silver" scales show up on Wilson's horagrams?

This is really interesting -- I'd like to understand and to help -- why
don't we go with "Bronze" rather than "Silver" for now until we clear this
up . . .

Paul H. Erlich wrote,

> So far, you've only given one example of each of these. As far as I
can tell, they both produce MOS scales with a 1:phi step-size ratio,
so they are both equally "Golden" or "Noble".

The idea being that a generator is culled from two terms, say a/b,
c/d, and that a Fibonacci expansion of those terms taken a/b, c/d, ...
and c/d, a/b, ... results in two different generators. One, where the
two terms are in an a/b, c/d, ... low to high L/s = Phi order, I'd
term "Golden". The other, where the two terms are taken in an c/d,
a/b, ... reversed L/s = Phi+1 order, I'd suggest calling something
else: I like "Silver", but perhaps (as you suggest) "Bronze" or
something is a better fit...

> It is unclear how loose you're being with the term "Fibonacci".
There is a so-called Lucas sequence that begins 1, 3, 4, 7, 11, 18, .
.

Hmm, I'm aware of a three term "tribonacci" series and a related
constant, but an abuse of the term Fibonacci series never really
occurred to me. I guess I see any two term expansion converging on Phi
as "Fibonacci"... perhaps that's too loose?

The standard issue example is 3/5 and 4/7, where the sequential
Fibonacci's "Golden generator" expansion would be Kornerup's, and the
reversed Fibonacci's "Silver generator" expansion would be Pepper's.
But this can be generalized to any two terms logarithmically or
linearly.

As another logarithmic example I'll use your 8s2L decatonic. "P" =
1200/2, and the Golden a/b, c/d Fibonacci expansion is 1/1, 4/5, ...
which gives the following 8s2L "Golden" standard pentachordals:

0 107 214 386 493 600 707 880 986 1093 1200
0 107 280 386 493 600 773 880 986 1093 1200
0 173 280 386 493 666 773 880 986 1093 1200
0 107 214 320 493 600 707 814 920 1027 1200
0 107 214 386 493 600 707 814 920 1093 1200
0 107 280 386 493 600 707 814 986 1093 1200
0 173 280 386 493 600 707 880 986 1093 1200
0 107 214 320 427 534 707 814 920 1027 1200
0 107 214 320 427 600 707 814 920 1093 1200
0 107 214 320 493 600 707 814 986 1093 1200

The Silver c/d, a/b Fibonacci expansion is 4/5, 1/1, ... which gives
the following 8s2L "Silver" standard pentachordals:

0 91 181 419 509 600 691 928 1019 1109 1200
0 91 328 419 509 600 837 928 1019 1109 1200
0 237 328 419 509 747 837 928 1019 1109 1200
0 91 181 272 509 600 691 781 872 963 1200
0 91 181 419 509 600 691 781 872 1109 1200
0 91 328 419 509 600 691 781 1019 1109 1200
0 237 328 419 509 600 691 928 1019 1109 1200
0 91 181 272 363 453 691 781 872 963 1200
0 91 181 272 363 600 691 781 872 1109 1200
0 91 181 272 509 600 691 781 1019 1109 1200

As a linear, or acoustic example, I'll use the 4:5 and 7:9. The Golden
a/b, c/d Fibonacci expansion is 4:5, 7:9, ... which gives a "Golden
mediant" of ~422�. The Silver c/d, a/b Fibonacci expansion is 7:9,
4:5, ... which gives a "Silver mediant" ~412�.

> Why not preserve the standard definition so that [(2-sqrt(2))*1200]
can be called a Silver generator?

Well I think your right, but it just sounds so good the other way!

> Do any of your "Silver" scales show up on Wilson's horagrams?

Well, as I've said a couple of times in previous posts, I tend to see
these "Silver" scales as points on the Phi paths, of which there are
many indeed. But as the L/s = Phi+1 expansion is a "natural" result of
reversing the initial two Fibonacci terms, I think it also warrants a
special recognition of sorts. (Which I think is also in tune with
Keenan P's original post.)

> why don't we go with "Bronze" rather than "Silver" for now until we
clear this up . . .

Maybe your right. (How about a true generalized Silver generator
though?)

--Dan Stearns

Dan -- I'm confused. If you start with 1/1 and 4/5, the sequence of
generators will be

1/1 4/5 5/6 9/11 14/17 . . .

And the limit of this sequence is a "Golden" right?

If you start with 4/5 and 1/1, the sequence of generators will be

4/5 1/1 5/6 6/7 11/13 17/20 . . .

And the limit of this sequence is a "Bronze" generator right?

But, if you start with 1/1 and 5/6, you get the sequence

1/1 5/6 6/7 11/13 17/12 . . .

So is that "Golden" or "Bronze"?

Whoops . . . "17/12" should read "17/20", and of course the limit of that
last sequence is what I'm asking about, since it should be the same as the
limit of the previous sequence and yet it appears than Dan is proposing a
different classification for it, though I'm probably misunderstanding him .
. .

Paul H. Erlich wrote,

> But, if you start with 1/1 and 5/6, you get . . . is that "Golden"
or "Bronze"?

"Bronze", as it converges on the same ~509� generator. But I'm really
only thinking of these Fibonacci expansions in relation to the two
terms that are derived from the initial Ls scale index.

--Dan Stearns

I wrote,

> But, if you start with 1/1 and 5/6, you get . . . is that "Golden"
>or "Bronze"?

>"Bronze", as it converges on the same ~509¢ generator.

So I'm not understanding what makes the 1/1, 4/5, . . . series "Golden"
while the 1/1, 5/6, . . . series is "Bronze".

>But I'm really
>only thinking of these Fibonacci expansions in relation to the two
>terms that are derived from the initial Ls scale index.

Can you expand on that please?

Paul H. Erlich wrote,

> So I'm not understanding what makes the 1/1, 4/5, . . . series
"Golden" while the 1/1, 5/6, . . . series is "Bronze".

The two resulting generators, here those are ~493� and ~509�. Again,
depending on your perspective, the "Bronze" generator could be seen as
nothing more than a ("Golden") stop along the Phi trail... though I've
tried to explain why I think it is perhaps deserving of more.

> Can you expand on that please?

Well the 1/1 and the 4/5 are acquire their meaning here because they
are a reduction of the 2L8s index.

--Dan Stearns

Dan, I would like to see a clearer explanation of all this, but it seems
that all "Bronze" generators are also "Golden" generators, and vice versa .
. . in either case, the set of generators is simple the set of noble
fractions of the octave . . . the difference is in the way they are used.
Right? So how are they used? Can you make reference in your explanation, if
possible, to the Wilson horagrams, which exist for many of the simpler noble
fractions of the octave.

BTW, the Hahn/Breed "Silver" generator is _not_ one of the above . . .

Paul H. Erlich wrote,

> Dan, I would like to see a clearer explanation of all this,

I'm sure it'll all make more sense coming from someone else, as I'm
afraid that we've just come to a logjam of our different ways of
explaining things or something... so rather than dragging it all out
I'll just wait for someone else to jump on in.

> BTW, the Hahn/Breed "Silver" generator is _not_ one of the above . .
.

Right, I realize that. My question was is there a *generalized* way to
create true "Silver" generators (something I hadn't given much thought
to at all).

--Dan Stearns

Dan wrote,

>I'm sure it'll all make more sense coming from someone else, as I'm
>afraid that we've just come to a logjam of our different ways of
>explaining things or something... so rather than dragging it all out
>I'll just wait for someone else to jump on in.

Would you at least try?

>My question was is there a *generalized* way to
>create true "Silver" generators (something I hadn't given much thought
>to at all).

Well, the first one (continued fraction all 1s) is "THE" Golden, Phi
generator itself, coming from the convergence of ratios of successive
members of "THE" Fibonacci series: 1, 1, 2, 3, 5, 8, 13, 21 . . ., generated
by the rule
S(n+1) = S(n) + S(n-1).

The second one (continued fraction all 2s) is the Breed/Hahn one and comes
from the convergence of ratios of successive members of this series: 1, 2,
5, 12, 29, 70 . . ., and that series can be generated by the rule
S(n+1) = 2*S(n) + S(n-1).

The third one (continued fraction all 3s) comes from the series 1, 3, 10,
33, 109, 360 . . ., rule is
S(n+1) = 3*S(n) + S(n-1).

I think you can see the pattern . . .

Does that answer your question?

Paul H. Erlich wrote,

> Would you at least try?

Well of course I have been, so that's not really the problem (lack of
trying that is)... but past a certain point it just becomes an issue
of diminishing returns, and probably a good bit of an annoyance to all
those non participants who also field the rapid-fire barrages of
increasingly unproductive back and forth.

> Does that answer your question?

Not really. I'm thinking of something else, but I haven't even begun
to look at it so it very well might be nothing.

--Dan Stearns

I wrote,

>> Does that answer your question?

>Not really. I'm thinking of something else,

You mean you'd like me to frame the answer in a different way, or apply it
more directly, or do you mean your mind is on a completely different subject
right now?

>but I haven't even begun
>to look at it so it very well might be nothing.

You mean it very well might be OK?

Paul H. Erlich wrote,

> You mean you'd like me to frame the answer in a different way, or
apply it more directly,

What I'm thinking is that "Silver" generator Graham gave would be the
same as weighting the fractions with Pythagoras' constant right? If
so, wouldn't this simple weighting scheme create a true generalized
"Silver" generator with other adjacent fractions as well?

--you know who

Dan wrote,

>What I'm thinking is that "Silver" generator Graham gave would be the
>same as weighting the fractions with Pythagoras' constant right?

Unfortuntately, no (I have no idea what Pythagoras' constant is but I tried
sqrt(2), sqrt(2)-1, etc . . . they didn't work). I'll try to go home and
derive the proper formula tonight.

>If
>so, wouldn't this simple weighting scheme create a true generalized
>"Silver" generator with other adjacent fractions as well?

I'm afraid that wouldn't correspond to the meaning of "Silver" either but it
should give us a general formula for continued fractions that end in all 2's
(like the noble/"golden" ones end in all 1's). I think are free to come up
with a name for these particular irrationals, the "next layer" after the
noble numbers.

"Silver" means the continued fraction expansion consists only of an infinite
number of occurences of a single integer.

Paul H. Erlich wrote,

> Unfortuntately, no

Try X = P/(N*D), where "P" = 1200, "N" = (7+sqrt(2)*5) and "D" =
(4+sqrt(2)*3). This should give X = [(2-sqrt(2))*1200]. Right?

> I'm afraid that wouldn't correspond to the meaning of "Silver"
either but it should give us a general formula for continued fractions
that end in all 2's (like the noble/"golden" ones end in all 1's).

Yes, that's what I had in mind; barely!

--Dan Stearns

Dan wrote,

>Try X = P/(N*D), where "P" = 1200, "N" = (7+sqrt(2)*5) and "D" =
>(4+sqrt(2)*3). This should give X = [(2-sqrt(2))*1200]. Right?

If you mean X = P/(N/D), then yes. But 3/5 and 7/4 are not part of the
sequence
1/2, 2/5, 5/12, 12/29, 29/70 . . .
So where are they coming from?

I figured I would go home and derive a formula that would use the numbers in
the sequence, like we saw for the phi case. But now you've thrown in 7, 4,
and 3 and I'm all confused.

Dave Keenan, any help here?

Paul H. Erlich wrote,

> If you mean X = P/(N/D), then yes.

Hmm. I think I got it right actually... In any event, it's not very
pretty, and I'm sure I could do a lot better when I get some time to
actually look at it all.

--Dan Stearns

I wrote,

>> If you mean X = P/(N/D), then yes.

Dan wrote,

>Hmm. I think I got it right actually...

Dan, you wrote P/(N*D), which equals 10.346, while you want X to equal
702.94, right?

Paul H. Erlich wrote,

> Dan, you wrote P/(N*D), which equals 10.346, while you want X to
equal 702.94, right?

Sorry, getting tiered. That should have read P/((7+S*5))*(4+S*3),
where "P" = 1200 and "S" = sqrt(2).

--Dan Stearns

Dave Keenan wrote,

>Sorry Paul, this _was_ concise. I just failed to unpack from it the reason
>why it was useful for L/s to be close to Phi, rather than merely proper or
>strictly proper. i.e. the fact that this gives maximum distinction between
>interval classes.

Well, if you're going to create an infinite number of intervals, then in a
sense this is true. But for a single MOS scale along the path, is there any
way you can define "distinction between interval classes" such that L/s =
Phi implies "maximum distinction between interval classes"?

David Keenan wrote,

>Kraig, what does Wilson claim is special about logarithmic noble
>generators? What is the musical significance of the ratio of large to small
>step sizes (in cents) being phi rather than say 1.5 or 1.7?

>What does Kornerup claim for his golden meantone?

To me (and to some extent, to Wilson and Kornerup), the significance is that
the noble generators have _infinite_ potential for new MOS scales with the
golden step-size ratio, by virtue of the most familiar property of the
phi-division: the ratio of the smaller to the larger is the same as the
ratio of the larger to the whole. So in each expansion of the scale, the
large intervals are subdivided into two intervals: an L, which is the same
as the last scale's small intervals, joining them in the new "L" class; and
an s, which is new and hence occurs as many times in the new scale as the
old large interval occured in the old scale. In other words, where n is a
"counting" function:

n(s(new)) = n(L(old))
n(L(new)) = n(L(old)) + n(S(old))

Of course, this was Yasser's "Theory of Evolution", but it was Kornerup who
realized that only a noble generator would obey this theory.

Dan Stearns wrote,

>But it would seem to me that one must ask the same types of questions
>of the linear use of Golden or Nobel mediants: is ~833¢ really any
>better point of complexity between 1:1 and 1:2 (etc., etc.) than any
>other; what's the "proof"?

Once again, the proof requires one to go infinitely high, this time in the
overtone series, and shows that you're always farther from having coinciding
harmonics (or from implying a fundamental) that you would be with any other
interval.

In reality, though, your hearing resolution will limit you quite early in
this process, which is why, although the _local_ maxima of harmonic entropy
do tend to occur at noble ratios, the _global_ maximum is at a very flat
minor second.

Paul H. Erlich wrote,

> (continued fraction all 1s) . . . generated by the rule S(n+1) =
S(n) + S(n-1).

> (continued fraction all 2s) . . . generated by the rule S(n+1) =
2*S(n) + S(n-1).

> (continued fraction all 3s) . . . rule is S(n+1) = 3*S(n) + S(n-1).

> I think you can see the pattern

I've had a chance to look at this a bit now, and I think this might be
yet another interesting way to incrementally express the range of a
given generator by walking a weighted series across two fractions (or
a given Ls indexes borders).

Take 1/2, 3/5 for example... If you derive a series of constants by
walking up the continued fraction ladder by 1s and create a series of
generators as say P/((2+Y*5))*(1+Y*3), where "P" = periodicity and "Y"
= a continued fraction constant, you'd have (I'll somewhat arbitrarily
stop where there would be consecutive roundings of the same
generator):

X = 696.2145
X = 702.9437
X = 707.0368
X = 709.6464
X = 711.4172
X = 712.6855
X = 713.6340

Now if you take the next the mediant of 1/2 and 3/5 and use the last
two terms as P/((5+Y*7))*(3+Y*4) and carry out the same process you'd
have (I'll again stop where there would be consecutive roundings of
the same generator):

X = 696.2145
X = 693.5423
X = 691.8107
X = 690.6614
X = 689.8603
X = 689.2756

So, another interesting way to walk strictly proper and improper
generators up and down the paths that converge towards their proper
borders.

--Dan Stearns

Joseph wrote,

>I did have a question, though concerning the diagram. The "golden
>mean" is linked to pentagonal symmetry in a diagram which contains,
>within it, a hexany figure...

No, sir -- the hexany has hexagonal (actually, octahedral or cubic), not
pentagonal symmetry.

>So, could somebody explain to me, in layman's terms, whether the
>hexany CPS process is linked to the golden mean?? It rather looks
>like it might be...

I'm afraid it's not.