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Message: 8377 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.
Message: 8379 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.
Message: 8380 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
Message: 8383 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 * > > "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."
Message: 8385 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!
Message: 8386 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.
Message: 8387 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 . . .
Message: 8388 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 . . .
Message: 8389 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.
Message: 8390 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].
Message: 8396 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/bookpdf * > 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.
Message: 8398 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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