Yahoo Tuning Groups Ultimate Backup tuning-math formula for meantone implications?

Hello all,

Please take a look at
/tuning-math/files/monz/formula1-6-cmt.gif

This is an x-y plot of the numeric relationship between the
pitches of 1/6-comma meantone their their acoustically closest
implied 5-limit JI ratios, as illustrated on my lattice at
http://www.ixpres.com/interval/monzo/meantone/lattices/lattices.htm

I suppose calculus is need to derive this numerically,
since some values of x have two values for y and z, yes?

In cases where there are two values for both y and z
(i.e., -3 and +3 on my graph, and -3-6x and 3+6x if
it were to be continued), the lower z goes with the
top y and vice versa.

If someone is interested in this, I would also appreciate formulas
for the other meantone systems on my webpage, as well as a
generalized formula if one can be derived.

love / peace / harmony ...

-monz
http://www.monz.org
"All roads lead to n^0"

_________________________________________________________
Do You Yahoo!?
Get your free @yahoo.com address at http://mail.yahoo.com

🔗unidala <JGill99@imajis.com>

--- In tuning-math@y..., "monz" <joemonz@y...> wrote:
> Hello all,
>
>
> Please take a look at
> /tuning-math/files/monz/formula1-6-
cmt.gif
>
> This is an x-y plot of the numeric relationship between the
> pitches of 1/6-comma meantone their their acoustically closest
> implied 5-limit JI ratios, as illustrated on my lattice at
> http://www.ixpres.com/interval/monzo/meantone/lattices/lattices.htm
>
>
> I suppose calculus is need to derive this numerically,
> since some values of x have two values for y and z, yes?

J Gill: If (to characterize what you are doing) there is a necessity
for the (independent variable) domain (in X) to result in *two*
values of the (dependent variables) ranges (in Y and in Z), then it
is not describable as a "function" (where there cannot be a "one to
two" correspondence between the independent variable (X) and either
of the independent variables (Y and Z).

>
> In cases where there are two values for both y and z
> (i.e., -3 and +3 on my graph, and -3-6x and 3+6x if
> it were to be continued), the lower z goes with the
> top y and vice versa.

JG: Despite this, there are more than one values (in Y and Z) for a
single point in (X), right? If so, the above holds true (despite
the "solver" approach described below!).

>
> If someone is interested in this, I would also appreciate formulas
> for the other meantone systems on my webpage, as well as a
> generalized formula if one can be derived.

JG: It seems like you already know what you've got (in your diagram).
Over the three (of what must be independently approached) sections of
the X axis (to qualify as being a "function"), the Z value is a
constant, and the Y value is a linear function of X. So you would
have:

Z + (SLOPE of dY/dX) * (X) [in log domain, log range].

If you want to play with "solving for Y" (or "solving for Z"):

*IF* one places a '=' sign in your diagram's identity (as opposed to
the '~=' sign, and algebraically re-arranges it, the form is:

(LOG[base 3]of(5^(3*Z))) - (LOG[base 3]of(5^(X/2))) + 3Y - X = 0

Since LOG[base B] = (LOG[base A]of(U)) / (LOG[base a]of(B))
the above is restated (where LN = natural logarithm base e) as:

(LN of(5^(3*Z)))/LN of(3) - (LN of(5^(X/2)))/LN of(3) + 3Y - X = 0

Note that the value of Z (over "piece-wise" sections of the domain of
the "independent variable" in X, IF these "sections" can be ISOLATED)
is a CONSTANT value (over each of such "piece-wise" ranges). In that
sense, it is a solution in ONE variable [*one* X value corresponds to
*one* value for each value of Y as the independent variable, or the
converse case (*one* Y value corresponding to *one* value in X).

Solving for Y (with Z held as a constant value between X2 and X1):

Y = (((LN of (5^(X/2)) - LN of (5^(3*Z))) / (LN of(3)) + X) / 3

Solving for Z [where you insert Y as = Z + (dY/dX)*X in each of the
three equations for the three (isolated sections) falling between
Xmin and Xmax) will require using a "solver" program (since the
independent variable Z to be determined appears in *TWO* places in
this restated identity (Mathematica should have such capabilities):

Z = (LN of (LN of (5^(X/2)-(3*Y + X)/(LN of 3))) / (3 * (LN of 5)))

I think that I got these algebraic restatements right (double-check
me on that... :)

Regards, J Gill

🔗unidala <JGill99@imajis.com>

--- In tuning-math@y..., "unidala" <JGill99@i...> wrote:

> I think that I got these algebraic restatements right (double-check
> me on that... :)

JG: I just "checked-myself", and the identity for Z was incorrect:

CORRECTED:

Z = (LN of (5^(X/2)-(3*Y + X)/(LN of 3))) / (3 * (LN of 5)))

>
>
>
> Regards, J Gill

🔗unidala <JGill99@imajis.com>

--- In tuning-math@y..., "unidala" <JGill99@i...> wrote:

J Gill: I'll get it right yet...(2nd revision):

CORRECTED (again):

Solving for Z [where you insert Y as = 2 ^ (Z + (dY/dX)* X) in each
of the three equations for the three (isolated sections) falling
between Xmin and Xmax, where dY/dX above is the log-log slope and Z
and X are in *octaves*] will require using a "solver" program
(since the "independent" variable Z to be determined appears in *TWO*
places in this restated identity, thus requiring an "iterative" solver
(Mathematica should have such capabilities):

Z = (LN of (5^(X/2)-(3*Y + X)/(LN of 3))) / (3 * (LN of 5)))

Regards, J Gill

🔗monz <joemonz@yahoo.com>

Hi J,

Thanks for the explanation and corrections.

- monz

> From: unidala <JGill99@imajis.com>
> To: <tuning-math@yahoogroups.com>
> Sent: Sunday, December 16, 2001 7:28 AM
> Subject: [tuning-math] Re: formula for meantone implications?
>
>
> J Gill: If (to characterize what you are doing) there is a necessity
> for the (independent variable) domain (in X) to result in *two*
> values of the (dependent variables) ranges (in Y and in Z), then it
> is not describable as a "function" (where there cannot be a "one to
> two" correspondence between the independent variable (X) and either
> of the independent variables (Y and Z).
>
> <etc. -- snip>

_________________________________________________________
Do You Yahoo!?
Get your free @yahoo.com address at http://mail.yahoo.com

🔗monz <joemonz@yahoo.com>

🔗unidala <JGill99@imajis.com>

--- In tuning-math@y..., "monz" <joemonz@y...> wrote:
>
> > From: monz <joemonz@y...>
> > To: <tuning-math@y...>
> > Sent: Sunday, December 16, 2001 1:31 PM
> > Subject: Re: [tuning-math] Re: formula for meantone implications?
> >
> >
> > Hi J,
> >
> > Thanks for the explanation and corrections.
> >
> >
> > - monz
>
>
> It seems to work OK, but I'm still confused. What I'm looking
> for is a way to mathematically define the implied ratios, with
> the requirement that when the meantone pitch is *exactly* midway
> between two ratios, both ratios must be given as answers.

J Gill: Monz, it sounds like you want to build a machine than can "think" (like people do)! I guess if you can define a set of JI ratios (which you like, or which meet some "man-made" criteria for the numerical size of the numerator/denominator involved, etc.), you could write a program to "decide" which of those ratios your meantone pitch value is "closest" to [by some pre-determined measure such as RMS error in deviation from a function such as 2^(pitch/reference)].

But those are all things which *you* can do ... Making the choice (among a predetermined set of JI ratio possibilities) automatically, would be (it seems to me) only as interesting as the members of the set which you (the human) essentially pre-determine! The rest would just be automata... ["regression analysis" of RMS error from a particular "function" in the log (octave) domain]. The problem is, that "function" would probably be a particular discrete set of JI intervals chosen by a human (which introduces an imposed limitations on the choices which the "machine" would then be biased towards selecting). Only as satisfying as the perceived "quality" of the set of possible JI ratios which... a human essentially pre-determines. IF *you* can decide on such an ultimate "gamut", it would seem do-able...

What you need (perhaps) is a sort of ... "nerd in a box" :)

But "the box" can't "change it's mind" in the wonderful ways that humans do...

J Gill

🔗monz <joemonz@yahoo.com>

Hi J,

> From: unidala <JGill99@imajis.com>
> To: <tuning-math@yahoogroups.com>
> Sent: Sunday, December 16, 2001 7:56 PM
> Subject: [tuning-math] Re: formula for meantone implications?
>
>
> J Gill: Monz, it sounds like you want to build a machine
> than can "think" (like people do)! I guess if you can
> define a set of JI ratios (which you like, or which meet
> some "man-made" criteria for the numerical size of the
> numerator/denominator involved, etc.), you could write
> a program to "decide" which of those ratios your meantone
> pitch value is "closest" to [by some pre-determined measure
> such as RMS error in deviation from a function such as
> 2^(pitch/reference)].

Not at all! It's much simpler than that. I'm just looking
for an elegant mathematical formula to explain what I'm showing
on my lattices.

The only measure I'm using is simple closeness in pitch-height.
The only reason it gets complicated and requires two solutions
sometimes is because some meantone pitches are exactly midway
between the two closest implied ratios.

-monz

_________________________________________________________
Do You Yahoo!?
Get your free @yahoo.com address at http://mail.yahoo.com

🔗paulerlich <paul@stretch-music.com>

🔗unidala <JGill99@imajis.com>

--- In tuning-math@y..., "monz" <joemonz@y...> wrote:
> Hi J,
>
>
> > From: unidala <JGill99@i...>
> > To: <tuning-math@y...>
> > Sent: Sunday, December 16, 2001 7:56 PM
> > Subject: [tuning-math] Re: formula for meantone implications?
> >
> >
> > J Gill: Monz, it sounds like you want to build a machine
> > than can "think" (like people do)! I guess if you can
> > define a set of JI ratios (which you like, or which meet
> > some "man-made" criteria for the numerical size of the
> > numerator/denominator involved, etc.), you could write
> > a program to "decide" which of those ratios your meantone
> > pitch value is "closest" to [by some pre-determined measure
> > such as RMS error in deviation from a function such as
> > 2^(pitch/reference)].
>
>
> Not at all! It's much simpler than that.

JG: So, you *do not* have a predetermined finite set of JI scale pitch-ratios in mind with which to compare with your mean-tone scale pitches? Your response (above) *could* indicate that you do *not* want to limit the possible JI scale-ratios to a finite set???

> I'm just looking
> for an elegant mathematical formula to explain what I'm showing
> on my lattices.

JG: In message #1924 (corrected and revised in mesaages #1925 and #1926) I algebraically re-arranged the "mathematical formula" which your diagram showed, in order that you would have the identities necessary in order to solve for Y or Z as a function of X:

Solving for Y (with Z held as a constant value between X2 and X1):

Y = (((LN of (5^(X/2)) - LN of (5^(3*Z))) / (LN of(3)) + X) / 3

Z = (LN of (5^(X/2)-(3*Y + X)/(LN of 3))) / (3 * (LN of 5)))

The (combined) form for Z is as follows:

Z=(LN of(5^(X/2)-(3*(2^(Z+(dY/dX)*X))+ X)/(LNof 3)))/(3*(LN of 5)))

I guess I'm still confused as to what you want you algorithm to do...

> The only measure I'm using is simple closeness

JG: of your mean-tone pitches to *what* (finite or infinite) set of your intended JI pitch-ratios to be compared to your mean-tone pitches?

> in pitch-height.

> The only reason it gets complicated and requires two solutions
> sometimes is because some meantone pitches are exactly midway
> between the two closest implied ratios.

JG: That "special case" could be dealt with, it seems.

🔗unidala <JGill99@imajis.com>

Joe,

The more that I stared at my "correction" in message #1926, the more I realized that I (erroneously) added something (2^) to the solution for Z as a function of X. With the understanding that the dependent variables Y and Z (as well as the independent variable X), being in the *exponents" of your (nicely done) diagram's algebraic identity [which was (3^(X/3))*(5^(X/6)) ~= (3^Y)*(5^Z)] represent quantaties in the "logarithmic domain" (and not the "linear domain"), the correction below should work well for the such determinations of Y and Z as functions of X as (it would seem) you have requested.

--- In tuning-math@y..., "unidala" <JGill99@i...> wrote:
> --- In tuning-math@y..., "monz" <joemonz@y...> wrote:
> > Hi J,
> >
> >
> > > From: unidala <JGill99@i...>
> > > To: <tuning-math@y...>
> > > Sent: Sunday, December 16, 2001 7:56 PM
> > > Subject: [tuning-math] Re: formula for meantone implications?
> > >
> > >
> > > J Gill: Monz, it sounds like you want to build a machine
> > > than can "think" (like people do)! I guess if you can
> > > define a set of JI ratios (which you like, or which meet
> > > some "man-made" criteria for the numerical size of the
> > > numerator/denominator involved, etc.), you could write
> > > a program to "decide" which of those ratios your meantone
> > > pitch value is "closest" to [by some pre-determined measure
> > > such as RMS error in deviation from a function such as
> > > 2^(pitch/reference)].
> >
> >
> > Not at all! It's much simpler than that.
>
> JG: So, you *do not* have a predetermined finite set of JI scale pitch-ratios in mind with which to compare with your mean-tone scale pitches? Your response (above) *could* indicate that you do *not* want to limit the possible JI scale-ratios to a finite set???
>
> > I'm just looking
> > for an elegant mathematical formula to explain what I'm showing
> > on my lattices.
>
> JG: In message #1924 (corrected and revised in mesaages #1925 and #1926) I algebraically re-arranged the "mathematical formula" which your diagram showed, in order that you would have the identities necessary in order to solve for Y or Z as a function of X:
>
> Solving for Y (with Z held as a constant value between X2 and X1):
>
> Y = (((LN of (5^(X/2)) - LN of (5^(3*Z))) / (LN of(3)) + X) / 3
>
> Solving for Z [WHERE YOU INSERT Y AS = (Z + (dY/dX)* X) in each
> of the three equations for the three (isolated sections) falling
> between Xmin and Xmax, where dY/dX above is the log-log slope and Z
> and X are in *octaves*] will require using a "solver" program
> (since the "independent" variable Z to be determined appears in >*TWO*
> places in this restated identity, thus requiring an "iterative" >solver (OR PERHAPS, LINEAR ALGEBRA, WHICH IS NOT MY STRONG POINT)
> (Mathematica should have such capabilities):
>
> Z = (LN of (5^(X/2)-(3*Y + X)/(LN of 3))) / (3 * (LN of 5)))
>
>
> The (combined, AND CORRECTED) form for Z is as follows:
>
> Z=(LN of(5^(X/2)-(3*(Z+(dY/dX)*X)+ X)/(LN of 3)))/(3*(LN of 5)))
>
> I guess I'm still confused as to what you want you algorithm to do...
>
>
> > The only measure I'm using is simple closeness
>
> JG: of your mean-tone pitches to *what* (finite or infinite) set of your intended JI pitch-ratios to be compared to your mean-tone pitches?
>
> > in pitch-height.
>
> > The only reason it gets complicated and requires two solutions
> > sometimes is because some meantone pitches are exactly midway
> > between the two closest implied ratios.
>
> JG: That "special case" could be dealt with, it seems.

Sincerely, J Gill :)

🔗joemonz <joemonz@yahoo.com>

--- In tuning-math@y..., "unidala" <JGill99@i...> wrote:
/tuning-math/message/1955

> Joe,
>
> The more that I stared at my "correction" in message #1926,
> the more I realized that I (erroneously) added something (2^)
> to the solution for Z as a function of X. With the understanding
> that the dependent variables Y and Z (as well as the independent
> variable X), being in the *exponents" of your (nicely done)
> diagram's algebraic identity [which was (3^(X/3))*(5^(X/6))
> ~= (3^Y)*(5^Z)] represent quantaties in the "logarithmic domain"
> (and not the "linear domain"), the correction below should work
> well for the such determinations of Y and Z as functions of X as
> (it would seem) you have requested.
>
> > Z=(LN of(5^(X/2)-(3*(Z+(dY/dX)*X)+ X)/(LN of 3)))/(3*(LN of 5)))

There's one parenthesis too many at the end.
Can you go over this again and remove it? Thanks.
I'm running your formulas thru Excel to see what the produce.

-monz

🔗unidala <JGill99@imajis.com>

--- In tuning-math@y..., "unidala" <JGill99@i...> wrote:
> --- In tuning-math@y..., "joemonz" <joemonz@y...> wrote:
>
> > There's one parenthesis too many at the end.
>
> > -monz

J Gill: Joe, I'm afraid that it is "worse than that". My original algebraic "manipulations" done "on-the-fly" (sloppy work) led to a log-antilog "mess". I'll bet (if you did try to plug it in to the Excel solver) that "FUBAR" is all that resulted... :(

The solutions for Y(X) and Z(X) are *much* simpler than I managed to make them (where LN stands for the natural logarithm, and the independent variable X and the dependent variables Y and Z are understood to be numerical values in the log domain):

Y = (LN OF ((3^(X/3))*(5^(X/6))/(5^(Z)))/(LN of (3)))

AND, IN A VERY SIMILAR FORM AS ABOVE,

Z = (LN of ((3^(X/3))*(5^(X/6))/(3^(Y)))/(LN of (3)))

There are some extra parenthesis added (for clarity) which should not matter to Excel (which may remove them automatically, anyway).

From your original identity (where the ~= is set to =), as one increases the value of X, test for Z as well as Y (Z AND Y) having the (JI) integer exponents which you have specified in your program that you are "looking for" (at least I *think* that's what you are trying to accomplish with this...).

Regards, J Gill :)