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Message: 8300 - Contents - Hide Contents Date: Sat, 15 Nov 2003 03:56:25 Subject: Re: Vals? From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote:> Wait... is this true: 'For a scale with card k, if there is > no standard val with n=k that consistently maps the scale, the > scale is not a Constant Structure.'Sorry, but no. You are far too hung up on standard vals!

Message: 8301 - Contents - Hide Contents Date: Sat, 15 Nov 2003 04:00:20 Subject: Re: Vals? From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> wrote:> right, but i still want to understand it, since it was in my > relativity textbooks . . .It's more complicated in relativity. There you have tangent spaces and cotangent spaces *at every point*, which have to connect together, plus you have a non-positive inner product which changes from point to point. We've got it easy and should enjoy ourseves.

Message: 8302 - Contents - Hide Contents Date: Sat, 15 Nov 2003 04:04:04 Subject: Re: Vals? From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "Dave Keenan" <d.keenan@b...> wrote:> Drop the word "standard". There's absolutely no relationship between > "standard" vals and Constant Structure. Sorry if anything I said > mislead you in that direction.I may have started it by remarking that all the vals we get from complete p-limit chords turn out to be standard. Which is true, but of small importance; it simply serves as a way to keep easily in mind what they are.

Message: 8303 - Contents - Hide Contents Date: Sat, 15 Nov 2003 04:06:58 Subject: Re: Vals? From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "Dave Keenan" <d.keenan@b...> wrote:> So [-4 4 -1> (x) [-11 4 2> = <12 19 28]Or if you like, |-4 4 -1> ^ |-11 4 2> = <12 19 28|. Welcome to the wonderful world of wedge products.

Message: 8304 - Contents - Hide Contents Date: Sat, 15 Nov 2003 04:12:49 Subject: Re: Vals? From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "Dave Keenan" <d.keenan@b...> wrote:> How does that generalise to other than the 5-limit? i.e. vectors with > other than 3 components?In the 7-limit, the wedge product of two monzos is a 6D wedge product vector, (which is the two intervals are commas gives us on reduction a wedgie for a temperament) wedging it with a monzo again gives us a val. The wedge product of two vals (I'm assuming things are set up the way I define them) gives us, once again, a 6D wedge product vector, (which if the two vals are et vals gives us on reduction a wedgie for a temperament) wedging it with a val again gives us a monzo. This has to be done carefully in terms of basis elements to make the equivalencies work.

Message: 8305 - Contents - Hide Contents Date: Sat, 15 Nov 2003 04:14:44 Subject: Re: Vals? From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "Dave Keenan" <d.keenan@b...> wrote:> Me too, since I want it to be generalisable to matrices, and it seems > Gene doesn't care about that.It's not a standard notation; however we might find it useful if it doesn't confuse people.

Message: 8306 - Contents - Hide Contents Date: Sat, 15 Nov 2003 04:30:48 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: >>> But if I've made a serious mistake I really need to know: >> What do the integers [the val's coefficients] represent in tuning >> terms? What are they counting? >> They can be counting different things; steps of an equal temperament, > number of octaves, number of generators,But these are all temperament "generators" in its most general sense. As I said.> exponent of a particular > prime numberNow I'm really confused. I thought these were called monzos, not vals.

Message: 8307 - Contents - Hide Contents Date: Sat, 15 Nov 2003 05:24:36 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 agree it's awkward. Carl objected so vehemently to EDO and I > wanted>> to reserve ET for the most general term (including EDOs ED3s cETs). >> Perhaps this would be a misuse of ET. Do we have some other term for >> the most general category of 1D temperaments, i.e. any single >> generator temperament whether or not it is an integer fraction of > any>> ratio? I guess "1D-temperament" will do. >> Not 1D. These are 0-dimensional temperaments, I'm afraid.Yes. You're right. Of course I was coming from the mathematical standpoint. Either way, we have descending dimensions planar temperaments, linear temperaments, <then what?> temperaments?

Message: 8308 - Contents - Hide Contents Date: Sat, 15 Nov 2003 05:29:14 Subject: Re: Vals? From: Dave Keenan --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote:> --- In tuning-math@xxxxxxxxxxx.xxxx "Paul Erlich" <perlich@a...> > wrote: >>> not sure how gene would do this notationally, probably i did >> something terrible, but without it i could not have made those >> charts . . . >> It's a special case of the wedge product, and I'd notate it that way.So the wedge product is a generalisation of the 3D Cartesian product or cross product. Awesome! There are really some light-bulbs coming on in my head today. :-) Thanks Gene.

Message: 8309 - Contents - Hide Contents Date: Sun, 16 Nov 2003 18:21:39 Subject: Re: Vals? From: monz hi Graham, --- In tuning-math@xxxxxxxxxxx.xxxx Graham Breed <graham@m...> wrote:> monz wrote: >>> unfortunately, i have no clue what "bivectors" and >> "trivectors" are. >> A bivector will be the wedge product of two vectors, > and a trivector the wedge product of three vectors.OK, thanks. i knew that they referred to two and three of something. now if only i understood what a "wedge product" is ... i understand it intuitively, but not mathematically.> A bi-val must be the wedge product of two vals. > If I can get it the right way round this time > > <12 19 28] ^ <22 35 51] = [-11 4 2> > > you can see that this particular bi-vector is the same > as a monzo. But it only works in the 5-limit. A more > familiar equation should be > > <12 19 28] ^ <19 30 44] = [-4 4 -1> > > so the intersection of 12-equal and 19-equal is a > linear temperament in which the syntonic comma vanishes. > In general terms, you can write that > > <h12] ^ <h19] = <<meantone]] > > where <<meantone]] is a bi-val defining some kind of meantone.thanks so much for this! it's *much* clearer now!> > <snip> > > so the general equation for linear temperaments can be written > > <<lintemp]] ^ [octave> = +/- <mapping] > > And, if you're really daring, you can even get rid of the ^ > > <<lintemp]][octave> = +/- <mapping] > > > I don't know if anybody's following this, but I think > the multi-bra notation makes it clearer than anything > we've had before.yes, i totally agree. i couldn't follow the part of your post which i snipped, or that last bit quoted above, but the new notation *is* helping a lot. using plain terms like "meantone" and "meantone7" also helps. -monz

Message: 8310 - Contents - Hide Contents Date: Sun, 16 Nov 2003 23:22:36 Subject: Re: Vals? From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx Graham Breed <graham@m...> wrote:> As the dimensions don't always make it unique, it may be worth > specifying whether to take the complement of the left or right hand side > in an operation involving mixed bras and kets.And probably coercing to> the right hand side would make more sense, because I think the brakets > usually give the right results then when the results are scalars.My approach has been coercion; I routinely convert multimonzos into multivals. In my practice multivals are only complimented over to multimonzos when the result is simply a monzo, ie a 1-monzo. Of course if you have a multival and want to wedge it with a monzo, it has to be converted first to the complimentary multimonzo, and then (if you are playing the coercion game) converted back again. This can also be looked at as a matter of having two kinds of product. My own software just calls it "up" vs "down", a matter of the dimension of the corresponding temperament increasing (wedge with a val) or decreasing (wedge with a monzo.)> I don't know if anybody's following this, but I think the multi-bra > notation makes it clearer than anything we've had before.I think you and I are probably the only ones without headaches, but I do agree. We seem to have passed rapidly from disappointing incomprehension to far more understanding of this stuff than I've learned to expect, and I think the notation thing is a big help. I'm going to revise my web pages, and I think revised dictionary entries are in order.

Message: 8311 - Contents - Hide Contents Date: Sun, 16 Nov 2003 23:50:03 Subject: Re: Vals? From: Dave Keenan --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote:> --- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote: >>>> But yeah. What do others think? Square brackets or vertical >>> bars (pipes)? >>>> I don't care, but I think we should standardize. >> I'd also suggest that, unless we specify otherwise, a 3D val or monzo > is 5-limit, a 4D val or monzo 7-limit, etc. and that the basis for > the n-multimonzos is ordered so that they correspond to the (n-pi(p))- > multivals. That, at least, is how I wrote my software.Yes. I totally agree. It is unfortunate that John Browne didn't do it that way. See chapter 2 page 18 Grassmann Algebra Book * [with cont.] (Wayb.) Although he does mention complementary bases. I assume your "pi()" function is the prime index function where pi(2) = 1, pi(3) = 2, pi(5) = 3, etc. So does this mean that since a 5-limit monzo has coefficients [e2 e3 e5> then a 5-limit bimonzo has coefficients ordered [[e35 e52 e23>> where the subscripts indicate the primes referred-to in the basis? And the same with maps and bimaps (vals and bivals). And since a 7-limit monzo has coefficients [e2 e3 e5 e7> then a 7-limit trimonzo will have coefficients ordered [[[e357 e572 e723 e235>>>. Is this how your software does it too Graham? But how do you order the coefficents of a 7-limit bimonzo or bimap (bival) so it's its own complement???

Message: 8312 - Contents - Hide Contents Date: Sun, 16 Nov 2003 00:07:15 Subject: Re: Vals? From: Dave Keenan --- In tuning-math@xxxxxxxxxxx.xxxx Graham Breed <graham@m...> wrote:>> If you only temper out two [7-limit] commas, you get a linear temperament. >> |-4 4 -1> ^ |2 2 -1 -1> = 7-limit meantone.I'd have added the zero to the end to make it [-1 4 -1 0> just to keep things clear.> I don't know how to write linear temperaments as bras, ...Here's an idea. Use << ... ]] for a linear temperament (bivector), <<< ... ]]] for a planar temperament (trivector) etc. So we have a Pascal's triangle of covariant-multivector (multival) types like this (where the number indicates how many coefficients there are). ,vals 1 , ' ,bivals 1-limit 1 <1] , ' ,trivals 2-limit 1 <2] <<1]] , ' 3-limit 1 <3] <<3]] <<<1]]] 5-limit 1 <4] <<6]] <<<4]]] <<<<1]]]] 7-limit 1 <5] <<10]] <<<10]]] <<<<5]]]]<<<<<1]]]]] 11-limit etc. And I would have thought we'd have this for the contravariant multivectors (multimonzos). ,monzos 1 , ' ,bimonzos 1-limit 1 [1> , ' ,trimonzos 2-limit 1 [2> [[1>> , ' 3-limit 1 [3> [[3>> [[[1>>> 5-limit 1 [4> [[6>> [[[4>>> [[[[1>>>> 7-limit 1 [5> [[10>> [[[10>>> [[[[5>>>>[[[[[1>>>>> 11-limit etc. However it seems that, at least in our application, we have 1 1-limit 1 1 2-limit 1 <2] 1 3-limit 1 <3] [3> 1 5-limit 1 <4] <<6]] [4> 1 7-limit 1 <5] <<10]] [[10>> [5> 1 11-limit where those on the midline could go either way. Why is that? I thought they had incomensurable bases. One lot has units of "log-prime" and the other "per-log-prime". e.g. Why is a 5-limit bi-val a monzo?

Message: 8313 - Contents - Hide Contents Date: Sun, 16 Nov 2003 00:09:33 Subject: Re: Vals? From: Dave Keenan --- In tuning-math@xxxxxxxxxxx.xxxx Graham Breed <graham@m...> wrote:> Dave Keenan wrote: >>> So the wedge product is a generalisation of the 3D Cartesian product >> or cross product. Awesome! There are really some light-bulbs coming on >> in my head today. :-) Thanks Gene. >> No, it's the other way around. The 3-D dot and cross products are > special cases of Grassman algebra, which came first.I wasn't implying any history. If B is a special case of A then A is a generalisation of B. But thanks for the history.

Message: 8314 - Contents - Hide Contents Date: Sun, 16 Nov 2003 00:18:09 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: >>>> exponent of a particular >>> prime number >>>> Now I'm really confused. I thought these were called monzos, not > vals. >> The mapping which for any rational number gives the p-exponent of > that rational number is called a padic valuation, and is the basis > for the vals in the same way that prime numbers are the basis for > monzos.So what I said was correct. Putting it another way, more in engineering terminology that pure math, the "units" for the coefficients of a monzo are the logs of the respective primes, and the "units" for the coefficients of the vals (in our application) or prime-mappings are "generators per log prime" so that the dot product of a val and a monzo has units of simply "generators".

Message: 8315 - Contents - Hide Contents Date: Sun, 16 Nov 2003 00:20:45 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: >>> So the wedge product is a generalisation of the 3D Cartesian product >> or cross product. Awesome! There are really some light-bulbs coming > on>> in my head today. :-) Thanks Gene. >> You've got it. It also defines the determinant, come to that.Right. I reread John Browne's Introduction to Grassman Algebra and it's making a lot more sense this time.

Message: 8316 - Contents - Hide Contents Date: Sun, 16 Nov 2003 00:42:06 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: >>> The square brackets alook more enclosing, and can't be mistaken for > a one. >> I'm happy either way. Paul? What does a physics major think?Hey I majored in Physics too (as well as Comp Sci). :-) I learned the Dirac notation in second year Quantum Physics, from the Feynman Lectures Volume III. But that _was_ a long time ago. When I just looked at that book again now. You hardly ever see < .. | or |.. > alone, but usually < ... | ... >. Also, you hardly ever see actual numbers inside, usually variables, so the issue of confusion of | with 1 or it's lack of visual enclosingness (is that a word?) don't arise. But yeah. What do others think? Square brackets or vertical bars (pipes)?

Message: 8317 - Contents - Hide Contents Date: Sun, 16 Nov 2003 01:07:02 Subject: Re: Integrating the Riemann-Siegel Zeta function and ets From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote:> If we look at successively larger values, we get 2, 5, 7, 12, 19, 31, > 41, 53, 72 ..., and this makes a lot of sense to me.This continues 72, 130, 171 ...> If we take the values over one by decades, we get: > > 1-10: 2, 3, 5, 7, 10 > 11-20: 12, 15, 17, 19 > 21-30: 22, 24, 26, 27, 29 > 31-40: 31, 34, 36 > 41-50: 41, 43, 46, 50 > 51-50: 53, 58, 60 > 61-70: 63, 65, 68 > 71-80: 72, 77, 80 > 81-90: 84, 87, 89 > 91-100: 94, 96, 99 > > It seems the density may be falling off slowly.This pattern also continues. The density is falling off, but very slowly. However the really significant ets are marked by major spikes, which grow ever more major. 101-110: 103, 106, 109 111-120: 111, 113, 118 121-130: 121, 125, 130 131-140: 135, 137, 140 141-150: 144, 145, 149 151-160: 152, 159 161-170: 161, 164, 167 171-180: 171 181-190: 183, 190 191-200: 193, 198 Here is what we have for 101-200; over 1.3 we have 103, 111, 118, 121, 130, 140, 152, 159, 171, 183, 190, 193 and 198. [102.7999443, 103.0793314] 1.340775 [105.8454247, 106.0892716] 1.181299 [108.8567520, 109.1034903] 1.025274 [110.9293379, 111.2010820] 1.394739 [112.8424527, 113.1050591] 1.261675 [117.8183072, 118.0990707] 1.544280 [120.9526475, 121.2217233] 1.316426 [124.8240486, 125.0682415] 1.239477 [129.8557778, 130.1420437] 1.634018 [134.9134275, 135.1413271] 1.054602 [136.8838124, 137.1336386] 1.209806 [139.8355968, 140.1167252] 1.548424 [143.7742322, 144.0090406] 1.101443 [144.9633882, 145.2016246] 1.161664 [148.8281106, 149.0720246] 1.280257 [151.9272068, 152.2036245] 1.593855 [158.8160194, 159.0739639] 1.436994 [160.8867930, 161.1399165] 1.272837 [163.9511536, 164.1657206] 1.046830 [166.9758931, 167.2096100] 1.086573 [170.8680796, 171.1331178] 1.652856 [182.8591559, 183.1297521] 1.643410 [189.8015621, 190.0604329] 1.520966 [192.8977322, 193.1469215] 1.453742 [197.9298229, 198.1820821] 1.464293

Message: 8318 - Contents - Hide Contents Date: Sun, 16 Nov 2003 03:29:40 Subject: Re: Vals? From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx "Dave Keenan" <d.keenan@b...> wrote:> Here's an idea. Use << ... ]] for a linear temperament (bivector), <<< > ... ]]] for a planar temperament (trivector) etc.That's a thought. I haven't seen this notation ever used, but it is logical.> And I would have thought we'd have this for the contravariant > multivectors (multimonzos).If we fix a prime limit, we have duality between multivals and multimonzos; in the 7-limit, a bival and a bimonzo can be identified. In the 11-limit, bivals and 4-monzos, and bimonzos and 4-vals, can be identified, as can trivals with trimonzos. This involves changing the basis of the n-monzos to make them numerically correspond to the (pi(p)-n)-vals. My approach to all this has been to swap the n-monzo for the corresponding (pi(p)-n)-val, and use that, but this does require we fix a prime limit.> e.g. Why is a 5-limit bi-val a monzo?The above duality. It isn't, except if you make the identification.

Message: 8319 - Contents - Hide Contents Date: Sun, 16 Nov 2003 03:39:01 Subject: Re: Vals? From: Gene Ward Smith --- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote:>> But yeah. What do others think? Square brackets or vertical >> bars (pipes)? >> I don't care, but I think we should standardize.I'd also suggest that, unless we specify otherwise, a 3D val or monzo is 5-limit, a 4D val or monzo 7-limit, etc. and that the basis for the n-multimonzos is ordered so that they correspond to the (n-pi(p))- multivals. That, at least, is how I wrote my software.

Message: 8320 - Contents - Hide Contents Date: Sun, 16 Nov 2003 09:38:07 Subject: Re: Vals? From: monz --- In tuning-math@xxxxxxxxxxx.xxxx Graham Breed <graham@m...> wrote:> monz wrote: >>> i sure wish i knew what the hell this was all about. >> especially since my name is being used as a term all thru it. >> >> you guys (Gene, paul, Dave) lost me on this long ago. >> but it sure seems interesting. >> We've established that the wedge product of two monzos > corresponds to the temperament in which they vanish. So, > with Gene's notation, a comma > and diaschisma give 12-equal. > > |-4 4 -1> ^ |-11 4 2> = <12 19 28| > > Well, on top of that, you can temper out 36:35, or |2 2 -1 -1> > > |-4 4 -1> ^ |-11 4 2> ^ |2 2 -1 -1> = <12 19 28 34| > > Which, to check with my Python module: >>>>> from temper import Wedgie as Monzo >>>> syntonic = Monzo((-4,4,-1)) >>>> diaschisma = Monzo((-11,4,2)) >>>> septimal = Monzo((2,2,-1,-1)) >>>> (syntonic^diaschisma^septimal).complement().flatten()> (12, 19, 28, 34) > > If you only temper out two commas, you get a linear temperament. > > |-4 4 -1> ^ |2 2 -1 -1> = 7-limit meantone. > > I don't know how to write linear temperaments as bras, but > there are some things you can show. For example, an > octave equivalent mapping is like tempering out the octave. >>>>> octave = Monzo((1,0,0)) >>>> (syntonic^septimal^octave).complement().flatten()> (0, 1, 4, -2) > > which means > > |-4 4 -1> ^ |2 2 -1 -1> ^ |1 0 0> = <0 1 4 -2| > > and (1 4 -2) is the octave-equivalent mapping for this > particular version of meantone, where C-Bb approximates 4:7, > rather than C-A#. For the more accurate one, you can do > > |-4 4 -1> ^ |1 2 -3 1> ^ |1 0 0> = <0 1 4 10|thanks, Graham ... this helps a lot! -monz

Message: 8321 - Contents - Hide Contents Date: Sun, 16 Nov 2003 09:40:42 Subject: Re: Vals? From: monz --- In tuning-math@xxxxxxxxxxx.xxxx Carl Lumma <ekin@l...> wrote:>> [Dave Keenan:] >> But yeah. What do others think? Square brackets or vertical >> bars (pipes)? >> I don't care, but I think we should standardize. > > -Carli think the best solution is to use square brackets for the bra and ket when written separately, and to use the pipe when they're put together. bra <...] ket [...> braket <...|...> didn't someone just suggest that yesterday? -monz

Message: 8322 - Contents - Hide Contents Date: Sun, 16 Nov 2003 09:43:56 Subject: Re: Vals? From: monz --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote:> --- In tuning-math@xxxxxxxxxxx.xxxx "Dave Keenan" <d.keenan@b...> > wrote: >>> Here's an idea. Use << ... ]] for a linear temperament >> (bivector), <<< ... ]]] for a planar temperament (trivector) >> etc. >> That's a thought. I haven't seen this notation ever used, > but it is logical.i'll second the notational idea, because i can see the logic in it. unfortunately, i have no clue what "bivectors" and "trivectors" are.>> And I would have thought we'd have this for the contravariant >> multivectors (multimonzos). >> If we fix a prime limit, we have duality between multivals > and multimonzos; in the 7-limit, a bival and a bimonzo can > be identified. > In the 11-limit, bivals and 4-monzos, and bimonzos and 4-vals, > can be identified, as can trivals with trimonzos. This > involves changing the basis of the n-monzos to make them > numerically correspond to the (pi(p)-n)-vals. My approach > to all this has been to swap the n-monzo for the corresponding > (pi(p)-n)-val, and use that, but this does require we fix > a prime limit. >>> e.g. Why is a 5-limit bi-val a monzo? >> The above duality. It isn't, except if you make the > identification.oh dear ... Graham, i was beginning to understand this stuff after your last post, but now i'm hopelessly lost again. help! -monz

Message: 8323 - Contents - Hide Contents Date: Sun, 16 Nov 2003 09:46:49 Subject: Re: Vals? From: monz hi Gene, --- In tuning-math@xxxxxxxxxxx.xxxx "Gene Ward Smith" <gwsmith@s...> wrote:> I'd also suggest that, unless we specify otherwise, a 3D val > or monzo is 5-limit, a 4D val or monzo 7-limit, etc. and that > the basis for the n-multimonzos is ordered so that they > correspond to the (n-pi(p))-multivals. That, at least, is > how I wrote my software.i'll second these suggestions. but even tho i understand your last sentence, what's (n-pi(p)) ? -monz

Message: 8324 - Contents - Hide Contents Date: Sun, 16 Nov 2003 11:23:06 Subject: Re: Vals? From: Graham Breed monz wrote:> unfortunately, i have no clue what "bivectors" and "trivectors" > are.A bivector will be the wedge product of two vectors, and a trivector the wedge product of three vectors.>>> e.g. Why is a 5-limit bi-val a monzo? >>>> The above duality. It isn't, except if you make the >> identification. >> oh dear ... Graham, i was beginning to understand this stuff > after your last post, but now i'm hopelessly lost again. help!A bi-val must be the wedge product of two vals. If I can get it the right way round this time <12 19 28] ^ <22 35 51] = [-11 4 2> you can see that this particular bi-vector is the same as a monzo. But it only works in the 5-limit. A more familiar equation should be <12 19 28] ^ <19 30 44] = [-4 4 -1> so the intersection of 12-equal and 19-equal is a linear temperament in which the syntonic comma vanishes. In general terms, you can write that <h12] ^ <h19] = <<meantone]] where <<meantone]] is a bi-val defining some kind of meantone. I've previously shown the usual 7-limit case [-4 4 -1 0> ^ [1 2 -3 1> = <<meantone7]] so combining the two equations gives <12 19 28 34] ^ <19 30 44 53] = [-4 4 -1 0> ^ [1 2 -3 1> where a bi-val happens to equal a bi-monzo. Here's the Python to check with:>>> import temper >>> h12 = temper.BestET(12, temper.limit7) >>> h19 = temper.BestET(19, temper.limit7) >>> syntonic = temper.Wedgie((-4, 4, -1)) >>> septimal = temper.Wedgie((1, 2, -3, 1)) >>> h12^h19 == ~(syntonic^septimal) 1 >>> h12[12, 19, 28, 34]>>> h19[19, 30, 44, 53] understanding the bi-val <<meantone]] itself is difficult, but you should always be able to get the octave-equivalent mapping by taking the wedge product with the octave <<meantone]] ^ [1,0...> = +/- <0, 1, 4, ...] Here's how you calculate it:>>> octave = temper.Wedgie((1,)) >>> (octave^(h12^h19).complement()).complement().flatten()(0, -1, -4, -10) There are two ways of doing the left hand side. Firstly, you could take the complement of the octave, which will be a tri-val. <<meantone]] ^ <<<octave_complement]]] = ???? in fact, this is illegal, because you can't have a 5-val in 4-space. So instead, we have to take the complement of <<meantone]] which is a bi-monzo [[meantone_complement>> ^ [1,> = [[[mapping_complement>>> In this space, the complement of a tri-monzo is a val, so we can say that [[[mapping_complement>>> = <mapping] = -<0 1 4 10] As the dimensions don't always make it unique, it may be worth specifying whether to take the complement of the left or right hand side in an operation involving mixed bras and kets. And probably coercing to the right hand side would make more sense, because I think the brakets usually give the right results then when the results are scalars. That is <et|interval> = steps to interval or <primes|interval> = size of interval always give positive numbers for an ascending interval. The calculation is as follows: Take the complement of [interval> to get a multi-val Multiply this by <primes] to get an even multi-er val Take the complement of this to get a scalar. so the general equation for linear temperaments can be written <<lintemp]] ^ [octave> = +/- <mapping] And, if you're really daring, you can even get rid of the ^ <<lintemp]][octave> = +/- <mapping] I don't know if anybody's following this, but I think the multi-bra notation makes it clearer than anything we've had before. Graham

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