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Message: 8375 - Contents - Hide Contents

Date: Wed, 19 Nov 2003 18:43:28

Subject: Re: "does not work in the 11-limit"

From: George D. Secor

--- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote:
>> No, because it's not actually "written", >
> Are you entering notes from a keyboard? > > -C.
No, I have to mouse back and forth around the screen, clicking on a note duration in one place (if it needs to be changed from what was set for the previous note) and then clicking on the staff in the appropriate place to draw the note. This part would go much faster if Cakewalk allowed me to use the keyboard to change the note durations (with the left hand; only about a half-dozen different keys would be needed) while I inserted the notes with the mouse (with the right hand) -- MONZ, THIS IS AN IDEA YOU SHOULD USE IN YOUR NEW SOFTWARE!!! (It would also be nice to be able to set sharps/flats and microtonal accidentals to be applied to the next note with the keyboard.) Then, if the note isn't something that occurs in the key signature (or if I chose not to have anything in the key signature), then I have to right-click on the note and set a chromatic alteration with the mouse. And if it isn't spelled the way I wanted it (e.g., if I wanted G# instead of Ab), then I have to have to fix that with a keyboard entry. In case you're wondering why I'm finicky about the spelling (since I'm not intending to print out any music): this enables me to use more than 12 pitch names in an octave, so when it comes time to insert the pitch bends, I can easily see that G# and Ab are different tones in my tuning and that I should bend them by different amounts. Once I've done all of the notes in a section, then I have to insert the pitch bends manually in event view (which is mostly done with the keyboard), also making sure that each note has the intended midi channel selected. As you can see, this is all very time-consuming, so I'm looking forward to Monz's software to make all of this a whole lot easier. --George
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Message: 8376 - Contents - Hide Contents

Date: Wed, 19 Nov 2003 23:28:34

Subject: Re: "does not work in the 11-limit"

From: Gene Ward Smith

--- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote:

> I too like to keep one voice per MIDI channel. However, Scala's > seq format provides a higher-level music-description language. It > has "tracks". Right now, you have to code seq files by hand (if > someone were to come up with a 'Cakewalk' for seq files...).
Well...I normally have a score which can be in a wide variety of formats in Maple, allowing latitude in compositional practice, and then compute something and write formatted output to a file which becomes a seq file once the right headers are added.
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Message: 8377 - Contents - Hide Contents

Date: Wed, 19 Nov 2003 09:07:17

Subject: Re: Vals?

From: Dave Keenan

--- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...>
wrote:
> --- In tuning-math@xxxxxxxxxxx.xxxx "Dave Keenan" <d.keenan@b...> > wrote:
>> I like the prefix tilde for complement since it supports >> De-Morgan-like intuitions from Boolean algebra. >
> Postfix seems more natural to me; that's normally how these things > are done.
Well, in ASCII * is of course most commonly infix for multiplication. As postfix I'm used to it being the complex-conjugate operator which doesn't seem as analogous as prefix ~ for the logical complement.
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Message: 8378 - Contents - Hide Contents

Date: Wed, 19 Nov 2003 18:44:49

Subject: Re: "does not work in the 11-limit" (was:: Vals?)

From: George D. Secor

--- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> 
wrote:
> --- In tuning-math@xxxxxxxxxxx.xxxx "George D. Secor" <gdsecor@y...> wrote:
>> --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote:
>>> --- In tuning-math@xxxxxxxxxxx.xxxx "George D. Secor" <gdsecor@y...> >>> wrote: >>> ... >>> i realized, since i made my original post, that the "dominant >>> pentatonic" is not CS in 12-equal. perhaps that's one source of my >>> difficulty? >>
>> I wouldn't think so. >
> Well, I'm interested in investigating further . . .
Okay. Once you're satisfied that we've adequately pinned down the concept of functional scale disorientation (or whatever it is we end up calling it -- the word "disorientation" suggests subjectivity), then we can take another look at this.
>>>> 2) But if there are two intervals in a scale that are *not >>>> functionally different* (such as the two 2:3s or 3:4s in our 11-limit >>>> hexatonic otonality), >>>
>>> why aren't they functionally different? because we don't have a well- >>> defined sense of hexatonic musical function, while we know all too >>> much about the history and theory of the diatonic scale? i don't >>> think that the "happen to" above can be defined in any precise or >>> perceptually relevant sense -- though it would be nice . . . >>
>> As I see it, interval function is independent of the number of tones >> in the scale, but instead has to do with the (just) *ratio* that is >> either directly expressed (in JI) or implied (in a temperament) by >> that interval. So two tempered intervals that (in a given context) >> are implying the same just interval are functionally the same, even >> if they are not exactly the same size (such as in a well- >> temperament). But two tempered intervals that (by context) imply >> different just intervals are functionally different, even if they are >> exactly the same size in a particular tuning. >
> What just interval does the 12-equal augmented second imply? And how > is this implication effected, exactly?
The diatonic system and traditional harmony do not exist only in 12- ET, so I would first want to answer this question for the whole range of tunings in which diatonic music could be performed. The 6th and 7th members of a harmonic minor scale could be given ratios of 8/5 and 15/8 (as 3rds of the subdominant and dominant triads), so the interval between these tones would be 64:75. (In the case of an extended meantone temperament, if you wanted to go to a 7- prime-limit and call this 6:7, I would see no problem with that. The only requirement is that, since the two tones are related by a different number of positions in a chain of fifths than is the minor 3rd, the augmented 2nd must be identified by a different ratio than 5:6.) To answer the question for 12-ET: that's a special case in which 5:6 and 64:75 are conflated, but even in 12-ET an augmented 2nd in the context of traditional (diatonic) harmony still functions as a 64:75 (a dissonance), not a (consonant) 5:6.
>> In the context of a diatonic scale the tones are all assumed to be in >> a chain of fifths. If one member of that chain is taken to represent >> 1/1, then each of the other members can be assigned at least one >> (rational) ratio that is unique to that member. An augmented 4th and >> diminished 5th (or a minor 3rd and augmented 2nd, etc.) will >> therefore be considered to be serving different harmonic functions, >> since they represent different ratios. >
> I'd like to see this made more explicit.
In a C major scale the diminished 5th B-F has the upper note of the interval -6 positions in the chain of 5ths, relative to the lower note. F to B (an augmented 4th) is +6 positions. If F is taken as 4/3 and B as 15/8, then a diminished 5th represents the ratio 45:64 and the augmented 4th represents 32:45. Thus their harmonic functions differ.
>> In an 8:9:10:11:12:14:16 scale there is no question that the two 2:3s >> (or the two 3:4s) are for all intents and purposes identical (since >> this is JI), >
> What if I tuned a harmonic minor scale in JI with a 6:5 augmented > second?
You wouldn't have consonant (5-limit) thirds on both the subdominant and dominant triads, so I think that your scale would depart from conventional thinking about what constitutes a harmonic minor scale. Let's develop this further. Suppose that you kept the 8/5 and raised the 15/8 to 48/25 so that you have a dissonant dominant triad (a very interesting possibility, since a higher leading tone would be desirable from a melodic perspective). You would also have a consonant minor triad Ab-B-Eb that makes a very nice resolution to G-C-E. Problem is, the spelling of this minor triad should be Ab-Cb-Eb, so something's wrong here: the tones don't *function* properly (i.e., consistently) in a heptatonic scale, because there are two tones competing to be some sort of "C". Now suppose that you keep the 15/8 but lower the 8/5 to 25/16 so that you have a dissonant subdominant triad. Now let's make the tonic triad major, so the 3rd degree of the scale is 5/4. You would then have a consonant major triad E-Ab-B (that would also resolve nicely to E-G-C). Again there's a problem with the spelling, since Ab should be G#, so again the tones don't *function* properly in a heptatonic scale, since now you have two tones that are some sort of "G". I'm not saying that you can't compose music using intervals this way, only that these tonal relationships are not consistent with a heptatonic scale.
>> so on a *harmonic* level they are functionally >> equivalent. But since these pairs of intervals subtend different >> steps in the scale, the potential for _functional scale >> disorientation_ (if you don't like the term, then please suggest >> something else) exists. >
> I'm hoping we can make this precise. Right now it seems fuzzy, with > meaning adapted differently to fit this fact and that. Please help me > remove the ambiguity.
Are I making any progress with my examples?
> Perhaps we are talking about epimorphic vs. non-epimorphic scales? If > so, realizing this could be a breakthrough. At least we could have a > precise (and very relevant to the material on this list) mathematical > characterization of what makes a scale have or not have "functional > scale disorientation" to you. That could be very helpful. Gene, would > you chime in?
I looked up this term in Monz's dictionary but gave up trying to figure it out when I saw "val" in the definition. I'll wait for Gene's comments before I try again. --George
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Message: 8379 - Contents - Hide Contents

Date: Wed, 19 Nov 2003 19:15:15

Subject: Re: "does not work in the 11-limit" (was:: Vals?)

From: Paul Erlich

--- In tuning-math@xxxxxxxxxxx.xxxx "George D. Secor" <gdsecor@y...> 
wrote:

> To answer the question for 12-ET: that's a special case in which 5:6 > and 64:75 are conflated, but even in 12-ET an augmented 2nd in the > context of traditional (diatonic) harmony still functions as a 64:75 > (a dissonance), not a (consonant) 5:6.
I'm not sure i buy the idea that it 'functions as a 64:75' -- at least not if you're excluding simultaneous 'functioning' as 108:125 and 1024:1215 . . .
>> Perhaps we are talking about epimorphic vs. non-epimorphic scales? > If
>> so, realizing this could be a breakthrough. At least we could have > a
>> precise (and very relevant to the material on this list) > mathematical
>> characterization of what makes a scale have or not have "functional >> scale disorientation" to you. That could be very helpful. Gene, > would
>> you chime in? >
> I looked up this term in Monz's dictionary but gave up trying to > figure it out when I saw "val" in the definition.
Uh-oh -- luckily Dave Keenan, at least, has recently cleared his hurdle. You may want to look at his most recent posts here, where he was trying to come up with a friendlier term for what 'val' means. Although his attempts weren't entirely satisfactory, they should get the relevant meaning of the term across to you.
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Message: 8380 - Contents - Hide Contents

Date: Wed, 19 Nov 2003 19:29:53

Subject: Re: "does not work in the 11-limit" (was:: Vals?)

From: monz

--- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> 
wrote:
> --- In tuning-math@xxxxxxxxxxx.xxxx "George D. Secor" <gdsecor@y...> > wrote: >
>>> Perhaps we are talking about epimorphic vs. >>> non-epimorphic scales? If so, realizing this >>> could be a breakthrough. At least we could have >>> a precise (and very relevant to the material on >>> this list) mathematical characterization of what >>> makes a scale have or not have "functional scale >>> disorientation" to you. That could be very helpful. >>> Gene, would you chime in? >>
>> I looked up this term in Monz's dictionary which? "epimorphic"? >> but gave up trying to figure it out when I saw "val" >> in the definition. >
> Uh-oh -- luckily Dave Keenan, at least, has recently > cleared his hurdle. You may want to look at his most > recent posts here, where he was trying to come up with > a friendlier term for what 'val' means. Although his > attempts weren't entirely satisfactory, they should get > the relevant meaning of the term across to you.
i'd really like to amend, alter, or replace the definition i have of "val". i suppose Gene's definition should stand, but a hefty amendment of the kind of stuff Dave wrote would help others to understand, and me too. anyone willing to provide something that i can just paste in? -monz
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Message: 8381 - Contents - Hide Contents

Date: Wed, 19 Nov 2003 19:31:23

Subject: Re: Vals?

From: Gene Ward Smith

--- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> 
wrote:

> what was your original basis choice, and what do the patterns of > signs for duals look like under it?
I'd suggest we forget about that. Alphabetical is the most usual approach, and we are already using it. Moreover, it does allow us to use the formula Sum indicies + m(m+1)/2 to determine the sign of the compliment.
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Message: 8382 - Contents - Hide Contents

Date: Wed, 19 Nov 2003 19:34:30

Subject: Re: "does not work in the 11-limit"

From: Gene Ward Smith

--- In tuning-math@xxxxxxxxxxx.xxxx "George D. Secor" <gdsecor@y...> 
wrote:

> As you can see, this is all very time-consuming, so I'm looking > forward to Monz's software to make all of this a whole lot easier.
There's always the possibility of simply creating a Scala seq file directly.
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Message: 8383 - Contents - Hide Contents

Date: Wed, 19 Nov 2003 01:10:04

Subject: Re: Vals?

From: Paul Erlich

i don't think my answer was right.

rather, i think it has something to do with the fact that the 
complement of a vector (a,b) in the plane is a line parallel to the 
vector (-b,a) or (b,-a), while the complement of a vector (a,b,c) in 
3D space is a plane whose *normal* is simply (a,b,c) or -(a,b,c) . . .

--- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> 
wrote:
> --- In tuning-math@xxxxxxxxxxx.xxxx "Dave Keenan" <d.keenan@b...> > wrote: >
>> E.g. With Pauls's example of the syntonic comma and diaschisma and >> 12-ET, wedging the two comma monzos gives >> >> [-4 4 -1> ^ [-11 4 2> = [[28 19 12>> a bimonzo >> >> whereas their cross product gives >> >> [-4 4 -1> (x) [-11 4 2> = <12 19 28] a map >> >> and one is the complement of the other >> >> ~[[28 19 12>> = <12 19 28] >> >> So why no problems with minus signs in 3D? >
> maybe this is why: > > Tensor -- from MathWorld * [with cont.] > > "While the distinction between covariant and contravariant indices > must be made for general tensors, the two are equivalent for tensors > in three-dimensional Euclidean space, and such tensors are known as > Cartesian tensors."
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Message: 8384 - Contents - Hide Contents

Date: Wed, 19 Nov 2003 19:53:08

Subject: Re: "does not work in the 11-limit" (was:: Vals?)

From: Gene Ward Smith

--- In tuning-math@xxxxxxxxxxx.xxxx "monz" <monz@a...> wrote:

> i'd really like to amend, alter, or replace the definition > i have of "val". i suppose Gene's definition should stand, > but a hefty amendment of the kind of stuff Dave wrote > would help others to understand, and me too.
No, that needs to be rewritten now that we've agreed on a notation.
> anyone willing to provide something that i can just paste in?
We could try something like this: A monzo is a ket vector of exponents of a positive rational number in a certain prime limit p; if q = 2^e2 3^e3 ... p^ep, then the corresponding monzo is [e2 e3 ... ep>. A val, in the same prime limit, is a bra vector of integers. For prime limit p, both the monzo and the val have dimension pi(p), meaning the number of primes up to p. The inner product of a val and monzo therefore defines a mapping from p-limit positive rational numbers to integers; if v = <v2 v3 ... vp] is a p-limit val and e = [e2 e3 ... ep> is a p-limit monzo, then <v|e> = <v2 v3 ... vp |e2 e3 ... ep> = v2e2 + v3e3 + ... + vpep is the homomorphic mapping v(e) defined by v. The following may be too mathematical for your dictionary, but is from my web site: Intervals and Vals For p an odd prime, the intervals of the p-limit Np may be taken as the set of all frequency ratios which are positive rational numbers whose factorization involves only primes less than or equal to p. If q is such a ratio, it may be written in factored form as q = 2^e2 3^e3 ... p^ep where e2, e3, ... ep are integer exponents. We may write this in factored form as a ket vector of the exponents, or monzo: [e2 e3 ... ep> The p-limit rational numbers Np form an abelian group, or Z-module, under multiplication, so that it acts on itself as a transformation group of a musical space; this becomes an additive group using vector addition when written additively as a monzo. Np is a free abelian group of rank pi(p), where pi(p) is the number of primes less than or equal to p. The rank is the dimension of the vector space in which Np written additively can be embedded as a lattice; saying it is free means this embedding can be done, since there are no torsion elements, meaning there are no positive rational numbers q (called roots of unity) other than 1 itself, with the property that for some positive power n, q^n = 1. Given the p-limit group Np of intervals, there is a non-canonically isomorphic dual group Vp of vals. A val is a homomorphism of Np to the integers Z. Just as an interval may be regarded as a Z-linear combination of basis elements representing the prime numbers, a val may be regarded as a Z-linear combination of a dual basis, consisting of the p-adic valuations. For a given prime p, the corresponding p-adic valuation vp gives the p-exponent of an interval q, so for instance v2(5/4) = -2, v3(5/4) = 0, v5(5/4) = 1. If intervals are written as ket vectors, or monzos, vals are denoted by the corresponding bra vector. The 5-limit 12-et val, for instance, would be written <12 19 28].
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Message: 8385 - Contents - Hide Contents

Date: Wed, 19 Nov 2003 02:29:26

Subject: Re: Vals?

From: Dave Keenan

--- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote:
> i don't think my answer was right. > > rather, i think it has something to do with the fact that the > complement of a vector (a,b) in the plane is a line parallel to the > vector (-b,a) or (b,-a), while the complement of a vector (a,b,c) in > 3D space is a plane whose *normal* is simply (a,b,c) or -(a,b,c) . .
But did I actually get it right when I wrote [-4 4 -1> ^ [-11 4 2> = [[28 19 12>> a bimonzo Could a minus sign or two be required in the bimonzo, which then disappear when you take the complement? The basis for the monzos is lg(2) lg(3) lg(5) where lg() is a logarithm function of arbitrary base. According to John Browne, the above bimonzo is correct if its basis is lg(2)^lg(3) lg(5)^lg(2) lg(3)^lg(5) (indices 12 31 23) where ^ is the wedge-product operator, not exponentiation. = [[ (-4*4)-(4*-11) (-1*-11)-(-4*2) (4*2)-(-1*4) >> = [[28 19 12>> But if the basis is instead lg(2)^lg(3) lg(2)^lg(5) lg(3)^lg(5) (indices 12 13 23) (just swapped the order of lg(2) and lg(5) in the middle one) then the bimonzo is [[28 -19 12>> And it starts to look like the general complement (for any grade and dimension) should not only reverse the order of coefficients, but negate every second one. But I'm really not sure. Gene, help!
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Message: 8386 - Contents - Hide Contents

Date: Wed, 19 Nov 2003 19:54:04

Subject: Re: Vals?

From: Paul Erlich

--- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> 
wrote:
> --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> > wrote: >
>> what was your original basis choice, and what do the patterns of >> signs for duals look like under it? >
> I'd suggest we forget about that. Alphabetical is the most usual > approach, and we are already using it.
OK -- but it's interesting to note that the cross product immediately gives you the quantity of interest in 3D, regardless of indexing conventions. The GABLE tutorial claims that cross products are useless and should be dispensed with since geometric algebra has better ways of solving all the problems that the cross product is used for. I don't know . . .
> Moreover, it does allow us to > use the formula Sum indicies + m(m+1)/2 to determine the sign of the > compliment.
my highest compliments, but it's spelled complement.
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Message: 8387 - Contents - Hide Contents

Date: Wed, 19 Nov 2003 02:41:53

Subject: Re: Vals?

From: Paul Erlich

--- In tuning-math@xxxxxxxxxxx.xxxx "Dave Keenan" <d.keenan@b...> 
wrote:
> --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote:
>> i don't think my answer was right. >> >> rather, i think it has something to do with the fact that the >> complement of a vector (a,b) in the plane is a line parallel to the >> vector (-b,a) or (b,-a), while the complement of a vector (a,b,c) in >> 3D space is a plane whose *normal* is simply (a,b,c) or -
(a,b,c) . .
> > But did I actually get it right when I wrote > > [-4 4 -1> ^ [-11 4 2> = [[28 19 12>> a bimonzo
GABLE gives 28*e2^e3 + 12*e3^e5 + 19*e5^e2, where "e" is the unit vector.
> Could a minus sign or two be required in the bimonzo, which then > disappear when you take the complement? > > The basis for the monzos is > lg(2) lg(3) lg(5) > where lg() is a logarithm function of arbitrary base. > > According to John Browne, the above bimonzo is correct if its basis is > > lg(2)^lg(3) lg(5)^lg(2) lg(3)^lg(5) (indices 12 31 23) > > where ^ is the wedge-product operator, not exponentiation. > > = [[ (-4*4)-(4*-11) (-1*-11)-(-4*2) (4*2)-(-1*4) >> > = [[28 19 12>> > > But if the basis is instead > > lg(2)^lg(3) lg(2)^lg(5) lg(3)^lg(5) (indices 12 13 23) > > (just swapped the order of lg(2) and lg(5) in the middle one) > then the bimonzo is > [[28 -19 12>>
right, but if you keep the (directed) angle between the two vectors in each basis bivector the same, you don't get this behavior in 3D (since you use e5^e2 and not e2^e5 in your basis) -- but you *do* get it in 2D.
> And it starts to look like the general complement (for any grade and > dimension) should not only reverse the order of coefficients, but > negate every second one.
what could be special about every second one? think about this purely geometrically, so the order of the primes loses its significance . . .
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Message: 8388 - Contents - Hide Contents

Date: Wed, 19 Nov 2003 20:09:31

Subject: Re: "does not work in the 11-limit" (was:: Vals?)

From: Paul Erlich

--- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> 
wrote:
> --- In tuning-math@xxxxxxxxxxx.xxxx "monz" <monz@a...> wrote: >
>> i'd really like to amend, alter, or replace the definition >> i have of "val". i suppose Gene's definition should stand, >> but a hefty amendment of the kind of stuff Dave wrote >> would help others to understand, and me too. >
> No, that needs to be rewritten now that we've agreed on a notation. >
>> anyone willing to provide something that i can just paste in? >
> We could try something like this: > > A monzo is a ket vector of exponents of a positive rational number in > a certain prime limit p; if q = 2^e2 3^e3 ... p^ep, then the > corresponding monzo is [e2 e3 ... ep>. A val, in the same prime limit, > is a bra vector of integers. For prime limit p, both the monzo and > the val have dimension pi(p), meaning the number of primes up to p. > The inner product of a val and monzo therefore defines a mapping from > p-limit positive rational numbers to integers; if v = <v2 v3 ... vp] > is a p-limit val and e = [e2 e3 ... ep> is a p-limit monzo, then > > <v|e> = <v2 v3 ... vp |e2 e3 ... ep> = v2e2 + v3e3 + ... + vpep > > is the homomorphic mapping v(e) defined by v.
this is the definition for what? no offense, but regardless, it needs to be *greatly* expanded upon to be useful to 99.9% of its likely audience. i don't have time right now as i'm chatting with monz's business partner and then have to leave . . .
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Message: 8389 - Contents - Hide Contents

Date: Wed, 19 Nov 2003 03:27:28

Subject: Re: Vals?

From: Dave Keenan

--- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote:
>> According to John Browne, the above bimonzo is correct if its basis > is >>
>> lg(2)^lg(3) lg(5)^lg(2) lg(3)^lg(5) (indices 12 31 23) >> >> where ^ is the wedge-product operator, not exponentiation. >> >> = [[ (-4*4)-(4*-11) (-1*-11)-(-4*2) (4*2)-(-1*4) >> >> = [[28 19 12>> >> >> But if the basis is instead >> >> lg(2)^lg(3) lg(2)^lg(5) lg(3)^lg(5) (indices 12 13 23) >> >> (just swapped the order of lg(2) and lg(5) in the middle one) >> then the bimonzo is >> [[28 -19 12>> >
> right, but if you keep the (directed) angle between the two vectors > in each basis bivector the same, you don't get this behavior in 3D > (since you use e5^e2 and not e2^e5 in your basis) -- but you *do* get > it in 2D. >
>> And it starts to look like the general complement (for any grade and >> dimension) should not only reverse the order of coefficients, but >> negate every second one. >
> what could be special about every second one? think about this purely > geometrically, so the order of the primes loses its significance . . .
True, but we have to agree on _some_ standard ordering of the coefficients in a multivector of any grade and dimension, and in addition to that, we have to agree on the ordering of the grade-1 basis components making up higher-grade basis components. It should be something we can easily remember for any grade and dimension. Lexicographic ("alphabetical") ordering (in both of the above cases), is something that's easy to remember. It's what Browne uses in his Mathematica package. And it seems like it might give rise to a uniform complementation rule of "negate every second one and reverse the order". Clearly this rule is unavoidable in the 3-limit (2D) case. There's no choice in the matter there. I think I was wrong before when I said it sounded like Gene's ordering was different to John Browne's. It also sounds like they might both be the same as Graham's. But I'm not sure. You needn't worry about this giving you unfamiliar results in 3D. Remember that while the cross-product of two vectors gives you another vector at right angles to both of them, the wedge-product of two vectors does not. Instead it gives you a bivector representing the plane (or "planar direction") containing those two vectors. If you want the vector normal to that plane you have to take the complement of that bivector. So it doesn't matter how we shuffle the indices in the basis of the bivector representing the plane, because the definition of the complement operation will change accordingly, so the normal always comes out the way you would expect. A (x) B = ~(A ^ B) At least I think that's right. Sigh.
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Message: 8390 - Contents - Hide Contents

Date: Wed, 19 Nov 2003 19:58:30

Subject: Re: "does not work in the 11-limit" (was:: Vals?)

From: monz

whew!  thanks, Gene!

... i appreciate this, but ... um ... really also need a
musician-friendly version along the lines of what Dave
was doing.  i'm hoping that he, paul, or Graham can help
with that.

but i will update the definitions with yours.



-monz


--- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> 
wrote:
> --- In tuning-math@xxxxxxxxxxx.xxxx "monz" <monz@a...> wrote: >
>> i'd really like to amend, alter, or replace the definition >> i have of "val". i suppose Gene's definition should stand, >> but a hefty amendment of the kind of stuff Dave wrote >> would help others to understand, and me too. >
> No, that needs to be rewritten now that we've agreed on a notation. >
>> anyone willing to provide something that i can just paste in? >
> We could try something like this: > > A monzo is a ket vector of exponents of a positive rational number in > a certain prime limit p; if q = 2^e2 3^e3 ... p^ep, then the > corresponding monzo is [e2 e3 ... ep>. A val, in the same prime limit, > is a bra vector of integers. For prime limit p, both the monzo and > the val have dimension pi(p), meaning the number of primes up to p. > The inner product of a val and monzo therefore defines a mapping from > p-limit positive rational numbers to integers; if v = <v2 v3 ... vp] > is a p-limit val and e = [e2 e3 ... ep> is a p-limit monzo, then > > <v|e> = <v2 v3 ... vp |e2 e3 ... ep> = v2e2 + v3e3 + ... + vpep > > is the homomorphic mapping v(e) defined by v. > > > The following may be too mathematical for your dictionary, but is > from my web site: > > Intervals and Vals > > For p an odd prime, the intervals of the p-limit Np may be taken as > the set of all frequency ratios which are positive rational numbers > whose factorization involves only primes less than or equal to p. If > q is such a ratio, it may be written in factored form as > > q = 2^e2 3^e3 ... p^ep > > where e2, e3, ... ep are integer exponents. We may write this in > factored form as a ket vector of the exponents, or monzo: > > [e2 e3 ... ep> > > The p-limit rational numbers Np form an abelian group, or Z-module, > under multiplication, so that it acts on itself as a transformation > group of a musical space; this becomes an additive group using vector > addition when written additively as a monzo. > > Np is a free abelian group of rank pi(p), where pi(p) is the number > of primes less than or equal to p. The rank is the dimension of the > vector space in which Np written additively can be embedded as a > lattice; saying it is free means this embedding can be done, since > there are no torsion elements, meaning there are no positive rational > numbers q (called roots of unity) other than 1 itself, with the > property that for some positive power n, q^n = 1. > > Given the p-limit group Np of intervals, there is a non-canonically > isomorphic dual group Vp of vals. A val is a homomorphism of Np to > the integers Z. Just as an interval may be regarded as a Z-linear > combination of basis elements representing the prime numbers, a val > may be regarded as a Z-linear combination of a dual basis, consisting > of the p-adic valuations. For a given prime p, the corresponding > p-adic valuation vp gives the p-exponent of an interval q, so for > instance v2(5/4) = -2, v3(5/4) = 0, v5(5/4) = 1. If intervals are > written as ket vectors, or monzos, vals are denoted by the > corresponding bra vector. The 5-limit 12-et val, for instance, would > be written <12 19 28].
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Message: 8391 - Contents - Hide Contents

Date: Wed, 19 Nov 2003 04:18:08

Subject: Re: "does not work in the 11-limit" (was:: Vals?)

From: Gene Ward Smith

--- In tuning-math@xxxxxxxxxxx.xxxx "George D. Secor" <gdsecor@y...> 
wrote:

Sure, 
> I would say that it's not necessary or even important for a chord to > be a constant structure.
It'sn certainly very important to me; I've used the idea extensively.
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Message: 8392 - Contents - Hide Contents

Date: Wed, 19 Nov 2003 20:32:22

Subject: Re: "does not work in the 11-limit"

From: George D. Secor

--- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> 
wrote:
> --- In tuning-math@xxxxxxxxxxx.xxxx "George D. Secor" <gdsecor@y...> wrote: >
>> As you can see, this is all very time-consuming, so I'm looking >> forward to Monz's software to make all of this a whole lot easier. >
> There's always the possibility of simply creating a Scala seq file > directly.
I haven't figured out how to do that yet. (Yes, I did see the recent postings about that on the main list.) Anyway, I haven't had much incentive to figure it out, because I don't use midi channels (in Cakewalk) the same way Scala does -- I prefer to keep a single melodic line in a single track and channel so I can copy and paste something from one track to another, including the pitch-bend events. I then change the channel for each note in the new track to another number (which goes fairly quickly in Cakewalk, once I figured out how to do it). Another reason for assigning channels this way is that I experienced that having two different patches assigned to the same channel results tends to corrupt the quality of the sound. --George
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Message: 8393 - Contents - Hide Contents

Date: Wed, 19 Nov 2003 04:22:51

Subject: Re: Vals?

From: Gene Ward Smith

--- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> 
wrote:

> so it's pretty much a matter of convention which ones you consider > covariant and which ones you consider contravariant, but you're ok as > long as you keep the two categories straight? a little math wouldn't > hurt :)
The idea is that contravariant vectors are ordinary, garden variety vectors, and linear mappings of vectors are covariant vectors. What confuses the issue is that mappings of the mappings are canonically isomorphic to the vectors you started with.
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Message: 8394 - Contents - Hide Contents

Date: Wed, 19 Nov 2003 21:10:25

Subject: Re: "does not work in the 11-limit" (was:: Vals?)

From: George D. Secor

--- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> 
wrote:
> --- In tuning-math@xxxxxxxxxxx.xxxx "George D. Secor" <gdsecor@y...> > wrote: >
>> To answer the question for 12-ET: that's a special case in which 5:6 >> and 64:75 are conflated, but even in 12-ET an augmented 2nd in the >> context of traditional (diatonic) harmony still functions as a 64:75 >> (a dissonance), not a (consonant) 5:6. >
> I'm not sure i buy the idea that it 'functions as a 64:75' -- at > least not if you're excluding simultaneous 'functioning' as 108:125 > and 1024:1215 . . .
But yes, I would also include *both* 108:125 and 1024:1215 as ratios (or *roles*, as Dave and I call them when a Sagittal symbol is used to represent multiple ratios) that would be covered by the interval function of an augmented 2nd in traditional diatonic harmony, just as I would include both 8:9 and 9:10 as roles in which the interval of a major 2nd functions. My only requirement is that a single ratio *not* be represented by multiple vectors (or in this case multiple scalars) in the scale construct. I mentioned vectors in the previous paragraph, because in the case of the 11-limit hexatonic otonal scale that we've been using as our other example, the tones occur in a 4-dimensional structure. And there will be only one ratio (or role) for each tone, since the scale is JI.
>>> Perhaps we are talking about epimorphic vs. non-epimorphic scales? If >>> so, realizing this could be a breakthrough. At least we could have a >>> precise (and very relevant to the material on this list) mathematical >>> characterization of what makes a scale have or not have "functional >>> scale disorientation" to you. That could be very helpful. Gene, would >>> you chime in? >>
>> I looked up this term in Monz's dictionary but gave up trying to >> figure it out when I saw "val" in the definition. >
> Uh-oh -- luckily Dave Keenan, at least, has recently cleared his > hurdle. You may want to look at his most recent posts here, where he > was trying to come up with a friendlier term for what 'val' means. > Although his attempts weren't entirely satisfactory, they should get > the relevant meaning of the term across to you.
I was just reacting to this like Gene did when he saw Dave's "komma". Okay, I studied the definitions of "epimorphic" and "val" in Monz's dictionary for a couple of minutes, and I think I got somewhere with this. Gene, did you intend the term "epimorphic" to apply to temperaments as well as rational scales? If so, then is it possible that "qn" *could* be an irrational number? If the answer to the second question is yes, then I don't think that Gene and I are dealing with the same thing, because I require that the "qn" for an irrational scale (or tempered) interval be a rational ratio (or set of ratios) which that interval is intended to approximate or represent. We could test this out by seeing whether Gene considers the various scale examples given here: Yahoo groups: /tuning-math/message/7827 * [with cont.] to be epimorphic: 1) If each scale is in 1/4-comma meantone temperament, and 2) If each scale is in 12-ET. Different answers for 1) and 2) would indicate that we're *not* dealing with the same thing. --George
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Message: 8395 - Contents - Hide Contents

Date: Wed, 19 Nov 2003 04:26:35

Subject: Re: Vals?

From: Gene Ward Smith

--- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> 
wrote:

> if linear temperaments are 2-dimesional as you always stress, why > would these be 0-dimensional and not 1-dimensional?
Don't blame me--you are the one who insisted linear temperaments were to be called linear, not planar. If they are linear--ie 1D, then what are really 1D temperaments (ets) now have to be called 0D. for example,
> 88cET has a single generator of 88 cents . . . seems 1 dimensional to > me!
Of course, but I was mugged for saying this sort of thing in the first place. If you have octave equivalence, you can reduce mod octaves, and get cyclic groups, which is about as 0D a thing as this business will afford you.
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Message: 8396 - Contents - Hide Contents

Date: Wed, 19 Nov 2003 21:22:41

Subject: Re: Vals?

From: Dave Keenan

--- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote:
> --- In tuning-math@xxxxxxxxxxx.xxxx "Dave Keenan" <d.keenan@b...> > wrote:
>> OK. With lexicographic ordering of the indices, it isn't as simple > as
>> negating every second coefficient. There's sometimes a hiccup in the >> middle. It's explained in Section 5.4 of >> >> > Index of /homes/browne/grassmannalgebra/book/b... * [with cont.] (Wayb.) > TheComplement.pdf >
> The page cannot be displayed
It does seem to be offline at the moment. I'll keep trying occasionally and let you know if I get thru again.
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Message: 8397 - Contents - Hide Contents

Date: Wed, 19 Nov 2003 04:33:05

Subject: Re: Vals?

From: Gene Ward Smith

--- In tuning-math@xxxxxxxxxxx.xxxx "Dave Keenan" <d.keenan@b...> 
wrote:

> I'm only concerned with whether (our extension of) the notation makes > the Grassman Algebra clearer for us. And it certainly seems to be > doing so.
I would also suggest extending the inner product notation to multivals with corresponding multimonzos <<u|v]] <<<u|v]]] and so forth. This is one step towards defining conjugacy.
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Message: 8398 - Contents - Hide Contents

Date: Wed, 19 Nov 2003 21:36:23

Subject: Re: Vals?

From: Dave Keenan

--- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote:
> --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> > wrote:
>> --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> >> wrote: >>
>>> what was your original basis choice, and what do the patterns of >>> signs for duals look like under it? >>
>> I'd suggest we forget about that. Alphabetical is the most usual >> approach, and we are already using it. >
> OK -- but it's interesting to note that the cross product immediately > gives you the quantity of interest in 3D, regardless of indexing > conventions.
Paul. You must have missed where I explained that the cross-product stays the same no matter what the indexing conventions, because the wedge-product and the complement change in "complementary" ways when you change the indexing and A(x)B = ~(A^B). Gene:
>> Moreover, it does allow us to >> use the formula Sum indicies + m(m+1)/2 to determine the sign of >> the compliment.
"m" here is the grade of the object, i.e. the number of nested brakets. A more intuitive (for me) alternative to m(m+1)/2 is Ceiling(m/2). If the sum of the indices plus this quantity is even then you negate it when complementing.
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Message: 8399 - Contents - Hide Contents

Date: Wed, 19 Nov 2003 04:35:43

Subject: Re: Vals?

From: Gene Ward Smith

--- In tuning-math@xxxxxxxxxxx.xxxx "Dave Keenan" <d.keenan@b...> 
wrote:

> There must be a convenient way of dealing with these. Does it actually > matter if you use non-consecutive primes, as long as you do it > consistently throughout the calculations. Isn't it really just the > _dimension_ of the multi-vectors that must be fixed for any given set > of calculations?
The dimensions only need to be fixed if you want to define the compliment.
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