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Re: Complete rank 2 temperament surveys

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  • Gene Ward Smith
    ... Here s another thought. Suppose we start with an R2 wedgie, and divide through each coefficient by the corresponding product of logs base 2,
    Message 1 of 10 , Jan 20, 2006
      --- In tuning-math@yahoogroups.com, Graham Breed <gbreed@g...> wrote:
      >
      > I'm making progress on proving that all pairs of a set of equal
      > temperaments can find all R2 temperaments with certain properties.

      Here's another thought.

      Suppose we start with an R2 wedgie, and divide through each
      coefficient by the corresponding product of logs base 2,
      1/log2(q)*log2(r) for the q-r coefficient. Then we get a weighted
      wedgie W = <<w1 w2 .., wn||, and C = max(|wi|) is a complexity measure.

      If v is a val belonging to the temperament, let w be the corresponding
      weighted val, dividing through by 1/log2(q) for the prime q term. Then
      the wedge product W^w = zero trival. Therefore, if u = w/N, where the
      val is an N-et val, we also have W^u = zero trival.

      If j = <1 1 1 ... 1| is the JI val, we have that the maximum of the
      absolute values of the coefficients of W^j, call it E, is an error
      measurement for W. We also have that the maximum of the absolute
      values of j-u, call it e, is an error measurement for u, and hence for
      the original val w.

      Now

      E = ||W^j|| = ||W^(j-u)|| <= 3 ||W|| ||j-u|| = 3Ce

      Here here "3" appears because there are three added products in the
      computation of the trival.
    • Graham Breed
      ... So u s the weighted tuning map. There isn t anything to stop the octave being optimized, is there? ... Yes, j-u is the weighted errors of the primes so
      Message 2 of 10 , Jan 21, 2006
        Gene Ward Smith wrote:

        > Here's another thought.
        >
        > Suppose we start with an R2 wedgie, and divide through each
        > coefficient by the corresponding product of logs base 2,
        > 1/log2(q)*log2(r) for the q-r coefficient. Then we get a weighted
        > wedgie W = <<w1 w2 .., wn||, and C = max(|wi|) is a complexity measure.
        >
        > If v is a val belonging to the temperament, let w be the corresponding
        > weighted val, dividing through by 1/log2(q) for the prime q term. Then
        > the wedge product W^w = zero trival. Therefore, if u = w/N, where the
        > val is an N-et val, we also have W^u = zero trival.

        So u's the weighted tuning map. There isn't anything to stop the octave
        being optimized, is there?

        > If j = <1 1 1 ... 1| is the JI val, we have that the maximum of the
        > absolute values of the coefficients of W^j, call it E, is an error
        > measurement for W. We also have that the maximum of the absolute
        > values of j-u, call it e, is an error measurement for u, and hence for
        > the original val w.

        Yes, j-u is the weighted errors of the primes so any function of that is
        a weighted prime error.

        > Now
        >
        > E = ||W^j|| = ||W^(j-u)|| <= 3 ||W|| ||j-u|| = 3Ce
        >
        > Here here "3" appears because there are three added products in the
        > computation of the trival.

        Does the modulus have to be a sum-abs or would an RMS work as well? It
        looks bigger than my limit. One thing you aren't doing is selecting the
        good equal temperaments. This mapping for 32-equal:

        (32, 50, 75, 90, 109)

        is valid in 11-limit miracle, but it doesn't get past my error
        threshold. I'm not interested in the poor ETs. Only the ones that
        converge on the optimal generator -- 10, 31, 41 and 72. It happens that
        11 and 21 work but I don't think they have to. You're being much more
        liberal.

        It might be a better threshold for very small equal temperaments though,
        if I can reconcile the units. It's nice and simple!

        And really, where does that 3 come from?


        Graham


        p.s. In my original message, there's an equation

        Dopt = sqrt(CmaxEmax)

        which should be

        Dopt = sqrt(Cmax/Emax)

        Sorry about that. The example calculation was correct.
      • Gene Ward Smith
        ... octave ... I don t see how that helps in connection with finding suitable vals. For instance, if you have a tuning map M =
        Message 3 of 10 , Jan 21, 2006
          --- In tuning-math@yahoogroups.com, Graham Breed <gbreed@g...> wrote:

          > So u's the weighted tuning map. There isn't anything to stop the
          octave
          > being optimized, is there?

          I don't see how that helps in connection with finding suitable vals.
          For instance, if you have a tuning map M = <1 m3 m5 ... mp| belonging
          to a temperament, and then round off, you get an infinite number of
          vals belonging to the temperament appearing at a fixed density. If you
          did that for a TOP tuning map, it wouldn't work.

          > > If j = <1 1 1 ... 1| is the JI val, we have that the maximum of the
          > > absolute values of the coefficients of W^j, call it E, is an error
          > > measurement for W. We also have that the maximum of the absolute
          > > values of j-u, call it e, is an error measurement for u, and hence for
          > > the original val w.
          >
          > Yes, j-u is the weighted errors of the primes so any function of
          that is
          > a weighted prime error.

          And minimums for it give TOP and NOT tunings.

          > > Now
          > >
          > > E = ||W^j|| = ||W^(j-u)|| <= 3 ||W|| ||j-u|| = 3Ce
          > >
          > > Here here "3" appears because there are three added products in the
          > > computation of the trival.
          >
          > Does the modulus have to be a sum-abs or would an RMS work as well?

          I didn't use either one; I meant max-abs.

          > And really, where does that 3 come from?

          You take terms, one of which comes from the weighted wedgie, the other
          from the weighted val, and multiply them. Sums of three of such terms
          make up the coefficients of the trival. The 3, in other words, results
          from this being a trival.
        • Graham Breed
          ... I can t follow this at all. Round off to what? You mean like your standard vals? So what difference does it make starting with TOP? And where did we
          Message 4 of 10 , Jan 22, 2006
            Gene Ward Smith wrote:
            > --- In tuning-math@yahoogroups.com, Graham Breed <gbreed@g...> wrote:
            >
            >>So u's the weighted tuning map. There isn't anything to stop the
            > octave
            >>being optimized, is there?
            >
            > I don't see how that helps in connection with finding suitable vals.
            > For instance, if you have a tuning map M = <1 m3 m5 ... mp| belonging
            > to a temperament, and then round off, you get an infinite number of
            > vals belonging to the temperament appearing at a fixed density. If you
            > did that for a TOP tuning map, it wouldn't work.

            I can't follow this at all. Round off to what? You mean like your
            standard vals? So what difference does it make starting with TOP? And
            where did we get this tuning map from in the first place?

            >>Does the modulus have to be a sum-abs or would an RMS work as well?
            >
            > I didn't use either one; I meant max-abs.

            Then does the modulus have to be a max-abs, or would an RMS work as well?


            Graham
          • Gene Ward Smith
            ... Exactly. ... You can start with TOP, actually. It would work, except when you multiply by N, you might not get a division into N parts. And ... Anywhere,
            Message 5 of 10 , Jan 23, 2006
              --- In tuning-math@yahoogroups.com, Graham Breed <gbreed@g...> wrote:

              > > For instance, if you have a tuning map M = <1 m3 m5 ... mp| belonging
              > > to a temperament, and then round off, you get an infinite number of
              > > vals belonging to the temperament appearing at a fixed density. If you
              > > did that for a TOP tuning map, it wouldn't work.
              >
              > I can't follow this at all. Round off to what? You mean like your
              > standard vals?

              Exactly.

              > So what difference does it make starting with TOP?

              You can start with TOP, actually. It would work, except when you
              multiply by N, you might not get a division into N parts.

              And
              > where did we get this tuning map from in the first place?

              Anywhere, so long as the values are irrational and it maps all the
              commas to zero. However, I don't know how to pull them out of the air
              without already knowing a temperament.

              > >>Does the modulus have to be a sum-abs or would an RMS work as well?
              > >
              > > I didn't use either one; I meant max-abs.
              >
              > Then does the modulus have to be a max-abs, or would an RMS work as
              well?

              Max-abs was how I got my formula.
            • Graham Breed
              ... For either approximate-TOP or NOT, that puts a limit of E
              Message 6 of 10 , Jan 24, 2006
                Gene Ward Smith wrote:
                >>I can't follow this at all. Round off to what? You mean like your
                >>standard vals?
                >
                > Exactly.

                For either approximate-TOP or NOT, that puts a limit of

                E < k/2d + Emax

                where E is the error in the ET, d is the number of notes, Emax is the
                maximum error in the R2 temperaments you're looking for (using the same
                error measure as for ETs, not your vaguely correlated wedgie-error) and
                k is some constant deriving from the prime intervals that I haven't
                worked out.

                That may well be a practical limit for small ETs, but how do you show
                that such ETs (that round off correctly, and still suppport the R2
                temperament that's being rounded) will exist?

                I can't see that it has anything to do with the rest of your proof, either.

                >>So what difference does it make starting with TOP?
                >
                > You can start with TOP, actually. It would work, except when you
                > multiply by N, you might not get a division into N parts.

                You can always divide through by the octave stretch, get the ET
                approximation, and stretch the octave again. What's so special about
                the octave, anyway?


                As for your limit, I get the max-abs of the weighted wedgie for 11-limit
                miracle as 1.469 where the weighting matches the interval sizes. The
                maximum of the weighted wedgie wedged with ones is 0.007. That gives
                3Ce as 0.032. The actual TOP error for the version of 1-equal mapped <1
                1 3 3 2] is 0.382 (NOT is 0.422), and it belongs to miracle temperament.
                So something's not working.


                Graham
              • Gene Ward Smith
                ... 11-limit ... I get totally different numbers. The max-abs of the weighted wedgie I find to be C = 7.345, and the max of weighted ^ ones I get to be E =
                Message 7 of 10 , Jan 24, 2006
                  --- In tuning-math@yahoogroups.com, Graham Breed <gbreed@g...> wrote:

                  > As for your limit, I get the max-abs of the weighted wedgie for
                  11-limit
                  > miracle as 1.469 where the weighting matches the interval sizes. The
                  > maximum of the weighted wedgie wedged with ones is 0.007.

                  I get totally different numbers. The max-abs of the weighted wedgie I
                  find to be C = 7.345, and the max of "weighted ^ ones" I get to be
                  E = 0.0048. For such high error it doesn't make sense to use the val
                  you gave, and in fact the best bound is found with TOP. The weighted
                  error there (maximum TOP error in octave terms) is e = 0.0005258. Then
                  E = 0.0048 < 3Ce = 0.0116.
                • Graham Breed
                  ... Yes, 7.345 is correct, sorry. Ah, you have the wedgie complexity on the left. So E = E/3C as a constraint on the TOP error of the equal
                  Message 8 of 10 , Jan 25, 2006
                    Gene Ward Smith wrote:
                    > --- In tuning-math@yahoogroups.com, Graham Breed <gbreed@g...> wrote:
                    >
                    >
                    >>As for your limit, I get the max-abs of the weighted wedgie for
                    >
                    > 11-limit
                    >
                    >>miracle as 1.469 where the weighting matches the interval sizes. The
                    >>maximum of the weighted wedgie wedged with ones is 0.007.
                    >
                    >
                    > I get totally different numbers. The max-abs of the weighted wedgie I
                    > find to be C = 7.345, and the max of "weighted ^ ones" I get to be
                    > E = 0.0048. For such high error it doesn't make sense to use the val
                    > you gave, and in fact the best bound is found with TOP. The weighted
                    > error there (maximum TOP error in octave terms) is e = 0.0005258. Then
                    > E = 0.0048 < 3Ce = 0.0116.

                    Yes, 7.345 is correct, sorry.

                    Ah, you have the wedgie complexity on the left. So

                    E <= 3Ce

                    becomes

                    e >= E/3C

                    as a constraint on the TOP error of the equal temperaments. Yes, that's
                    probably going to work, but what's the point of it? It means we have to
                    look for equal temperaments with an error at least as high as something
                    so low it couldn't be in the right class anyway??? Well, I suppose we
                    have a wedgie badness measure of some value but it doesn't have anything
                    to do with the original question.

                    For that wedgie badness calculation, W is

                    <<3.786 -3.015 -0.712 4.336 -6.793 -4.495 0.547 2.301 7.345 5.045]]

                    and W^j is

                    <<<0.0071 0.0032 -0.0033 -0.0012 -0.0056 -0.0030 0.0028 0.0048 0.0034
                    0.0014]]]

                    so there is a 0.0048 but it isn't the maximum.


                    Graham
                  • wallyesterpaulrus
                    Thanks for this analysis, Graham. I don t have time to sit down and work through it all right now; hopefully some free time will come down the pike my way. But
                    Message 9 of 10 , Feb 10, 2006
                      Thanks for this analysis, Graham. I don't have time to sit down and
                      work through it all right now; hopefully some free time will come down
                      the pike my way. But it doesn't look complete anyway so I don't feel
                      too bad about moving on . . .


                      --- In tuning-math@yahoogroups.com, Graham Breed <gbreed@...> wrote:
                      >
                      > I'm making progress on proving that all pairs of a set of equal
                      > temperaments can find all R2 temperaments with certain properties. I
                      > don't know enough number theory to actually make it a proof.

                      [snip]
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