BOP tuning: Difference between revisions

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First, we ~~define~~ a ~~"naive" scaled L1 norm on monzos~~, where ~~each prime is weighted p^s. We will write this naive L1 norm on a monzo as~~ <math>~~\text{naive}_s(m)~~</math>.

First, we note that for any tuning map, call it <math>T</math>, there is an associated vector called a '''signed error map'''. This is given by <math>E = T-J</math>, where <math>J</math> is the JIP. If we assume that we are in a prime-limit, then the signed error map will contain the signed error on each prime. We assume that all errors are unweighted so far.

For ~~an arbitrary monzo~~ <math>~~m = |a\, b\, c\~~, ...\~~rangle~~</math>~~, we have:~~

For any such error map, we can then divide the unweighted error by the desired weighting on each prime to get the weighted error map. This can be viewed as a change of coordinates to a "weighted" coordinate system. Assuming our tuning maps are row vectors, this can be expressed as a matrix right-multiplication by some diagonal weighting matrix <math>W</math>, where the diagonal entries are the desired weights on each prime. We will then denote the weighted error by <math>F = E\cdot W</math>.

~~<math>\text{naive}_s(m) = 2^s|a| + 3^s|b| + 5^s|c| +~~ ...</~~math>~~

For any such error map, we can then evaluate the maximum ''unsigned'' weighted error on the primes. This quantity is equal to the maximum of the absolute values of the coefficients of the signed weighted error map, or equivalently the [https://en.wikipedia.org/wiki/Norm_(mathematics)#Maximum_norm_(special_case_of:_infinity_norm,_uniform_norm,_or_supremum_norm) L-inf] norm of the signed error map in the weighted coordinate system.

The Linf norm has the property of being the [https://en.wikipedia.org/wiki/Dual_norm dual norm] of the L1 norm on the dual space, which is the space of monzos. This means that it has the following property:

~~In comparison, the true weighting we want on monzos is:~~

<math>\|F\|_\infty = \sup_M \frac{|<F|M>|}{\|M\|_1}</math>

where the supremum is taken over all monzos <math>M</math> in the ''inverse-weighted'' coordinate system on the dual space. That is, if <math>m</math> were a monzo in unweighted coordinates, its representation in this space would be <math>M=W^{-1}\cdot m</math>, where we are left-multiplying by the inverse of the weighting matrix from before.

where ~~<math>\|...\|_{T1}</math>~~ is the ~~[[Tenney_Height]]~~.

This tells us that the max weighted error on the primes has the property of also being the max weighted error on ''all'' monzos in the prime-limit, and that this weighted error will always be obtained at a prime, where the weighting is given by the dual L1 norm.

Note that the true weighting is not a norm on monzos, but rather the exponential of a norm: the Tenney height. Our naive weighting agrees with the true one on the primes (and only the primes), and differs for all composite rationals.

Now, if our weighting matrix is the usual <math>1/log(p)</math> Tenney-weighting matrix, then the above is equivalent to [[Paul Erlich]]'s theorem that minimizing the max Tenney-weighted error on the primes minimizes the max Tenney-weighted error on all intervals. However, if we instead change the weighting matrix to <math>1/p^s</math> instead, then our Linf norm will be dual to a different, somewhat unusual weighted L1 norm on monzos: the one where the weighting on the primes is given by <math>p^s</math>, and the weighting for an arbitrary monzo <math>m = |a\, b\, c\, ...\rangle</math> is given by

~~Now, suppose~~ <math>~~T</math> is some tuning map in weighted coordinates, giving the tuning on each prime, where the weighting is now given by the <math>1/p~~^s~~</math> on the primes rather than the usual <math>1/\log~~(p)~~</math>. Then we can define the associated "weighted error map" <math>E~~=~~T-J</math>, where <math>J</math> is the JIP. This map gives the <math>1/p~~^s</math>~~-weighted error of each prime.~~

<math>\text{sopfr}^s(m) = 2^s|a| + 3^s|b| + 5^s|c| + ...</math>

where the notation on the left will be explained shortly.

~~We then look at~~ the ~~quantity~~

Now, we note that this unusually weighted L1 norm does not provide the weighting we want on all rationals, which is <math>(nd)^s</math>. Rather, when <math>s=1</math>, this is called the [[Wilson Height|Wilson norm]], and is equivalent to the [http://mathworld.wolfram.com/SumofPrimeFactors.html sum of prime factors with repetition] of the ratio <math>n/d</math> in question, often written <math>\text{sopfr}(n/d)</math>. For arbitrary <math>s\geq 1</math>, this is the sum of the <math>s</math>'th power of the prime factors with repetition of the ratio, which we will denote <math>\text{sopfr}^s(n/d)</math>. However, we note that this strange weighting ''does'' equal the weighting we want on the primes, where we have <math>\text{sopfr}^s(n/d) = (n/d)^s</math> - it only diverges on the composite numbers.

~~<math>\sup \frac{|\langle E|m \rangle|}{\text{naive}_s(m)}</math>~~

~~where <math>\langle E|m \rangle</math>~~ is the ~~signed~~ error ~~of the monzo m~~ according to the ~~error map E~~. ~~This expression represents~~ the ~~maximum "naive"~~-weighted error on all rationals.

We now want to show that minimizing the max weighted-error according to this strangely weighted L1 norm is equivalent to minimizing the max weighted-error according to the true weighting that we want. To do this, we consider what happens to the evaluation of max-weighted error on all rationals, for any given tuning, if we keep the tuning as is but simply change the weighting on the rationals to the thing we want. To prove our result, we need only show that

~~Since the naive norm is simply the L1 norm on the weighted monzos, we know via the definition of the [https://en.wikipedia~~.~~org/wiki/Dual_norm dual norm] that this is equal~~ the ~~Linf norm of~~ the ~~inverse-weighted error map. Since the Linf norm is just the max of the abs of the error map,~~ we ~~know that~~ the max ~~naive-~~weighted error on all rationals ~~will be equal to the max naive-weighted error on the primes~~.

1. Changing the weighting scheme to the thing we want does not make the max weighted error on all rationals any better

2. Likewise, it doesn't make it any worse

The first part is easy: for any tuning map, we know that the max "strange L1"-weighted error is equal to that of one of the primes, and we know that the primes agree in weighting on both metrics. Since we know that the weighting on the worst prime doesn't change, we know that after we switch weightings, the new max error can be no better than the thing we had.

~~We now want~~ to show that ~~this same property holds for the~~ ''~~true~~'' ~~weighting~~ we want ~~on monzos, not the naive one~~. ~~That is~~, we want to show:

To show the second part, all we need to show is that all rationals are weighted more '''strongly''' with our strange L1 metric than with the true metric we want. In particular, we want to show the following:

<math>~~\sup \frac{|\langle E|m \rangle|}{~~\text{~~true~~}_s(m)~~} =~~ \~~sup \frac{|\langle E|m \rangle|}{\text{naive}_s~~(m)}</math>

<math>\text{sopfr}^s(n/d) \leq (n/d)^s</math>

~~To see~~ this, we ~~need only to show that~~

where the left hand side is our strange metric and the right hand side is the true metric. Since we are ''dividing'' by this weighting, for any rational, the weighted error for the same amount of mistuning will be larger given the weighting scheme on the left than the weighting scheme on the right. This means that we have

~~# The max weighted error gets no better.~~

<math>\frac{\text{err}_{n/d}}{\text{sopfr}^s(n/d)} \geq \frac{\text{err}_{n/d}}{(n/d)^s}</math>

~~# The max weighted error gets no worse.~~

To show this is fairly easy. Remember our definition of the strange L1 metric:

~~To show that the weighted error gets no better, we need only refer to our previous result that for any given tuning map the max "naive"-weighted error is obtained on~~ a ~~prime~~. ~~Since the naive and true weightings are identical for primes, we know the weighted error for each prime will be identical for both metrics~~. ~~Since the supremum is obtained on a prime, we know that the "true"-weighted error cannot be any better than the naive supremum~~.

<math>\text{sopfr}^s(m) = 2^s|a| + 3^s|b| + 5^s|c| + ...</math>

In comparison, the true weighting we want on monzos is:

~~To show the weighted error also gets no worse, we can show that for any monzo <math>m</math>, we have~~

<math>\text{~~naive~~}_s(m) \~~leq~~ \~~text~~{~~true~~}~~_s(m)~~</math>

~~Due to this result, we know that while the unweighted error stays the same, the weighting can only increase as we move from the "naive" to the "true" metric, so the weighted error can only go down.~~

~~To see~~ that ~~this~~ inequality holds~~, note that we can rewrite it~~ for ~~some monzo <math>|a\, b\, c\, ... \rangle</math> as~~

So we want to show that the following inequality holds for all monzos:

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# For any particular prime <math>p</math> with integer coefficient <math>q</math>, <math>p^s|q| \leq p^{s|q|}</math>

# Since the nonzero terms on the left side are strictly positive, their sum must be less than their product, which is in turn less than the product of the associated "true" terms given above.

As a result, we have shown that for any particular tuning map, the max weighted error according to our strange L1 weighting will be the max weighted error according to the true metric we want: it gets no better and gets no worse. Since both are given by the Linf norm on error maps -- aka the max-abs <math>1/p^s</math> weighted error on the primes -- finding the tuning map in a particular temperament that minimizes this quantity is equivalent to finding the tuning map that minimizes the weighted error on all rationals according to ''both'' metrics.

~~Since we have shown the supremum does no better and no worse, we now know that the supremums are the same.~~ As a result, we have our main theorem:

As a result, we have our main theorem:

'''Theorem:''' to minimize the <math>1/(nd)^s</math> weighted error on all rationals, one needs only minimize the <math>1/p^s</math> weighted error on the primes.

@@ Line 12: / Line 12: @@
-First, we define a "naive" scaled L1 norm on monzos, where each prime is weighted p^s. We will write this naive L1 norm on a monzo as <math>\text{naive}_s(m)</math>.
+First, we note that for any tuning map, call it <math>T</math>, there is an associated vector called a '''signed error map'''. This is given by <math>E = T-J</math>, where <math>J</math> is the JIP. If we assume that we are in a prime-limit, then the signed error map will contain the signed error on each prime. We assume that all errors are unweighted so far.
-For an arbitrary monzo <math>m = |a\, b\, c\, ...\rangle</math>, we have:
+For any such error map, we can then divide the unweighted error by the desired weighting on each prime to get the weighted error map. This can be viewed as a change of coordinates to a "weighted" coordinate system. Assuming our tuning maps are row vectors, this can be expressed as a matrix right-multiplication by some diagonal weighting matrix <math>W</math>, where the diagonal entries are the desired weights on each prime. We will then denote the weighted error by <math>F = E\cdot W</math>.
-<math>\text{naive}_s(m) = 2^s|a| + 3^s|b| + 5^s|c| + ...</math>
+For any such error map, we can then evaluate the maximum ''unsigned'' weighted error on the primes. This quantity is equal to the maximum of the absolute values of the coefficients of the signed weighted error map, or equivalently the [https://en.wikipedia.org/wiki/Norm_(mathematics)#Maximum_norm_(special_case_of:_infinity_norm,_uniform_norm,_or_supremum_norm) L-inf] norm of the signed error map in the weighted coordinate system.
+The Linf norm has the property of being the [https://en.wikipedia.org/wiki/Dual_norm dual norm] of the L1 norm on the dual space, which is the space of monzos. This means that it has the following property:
-In comparison, the true weighting we want on monzos is:
+<math>\|F\|_\infty = \sup_M \frac{|<F|M>|}{\|M\|_1}</math>
-<math>\text{true}_s(m) = 2^{s|a|} \cdot 3^{s|b|} \cdot 5^{s|c|} \cdot ... = 2^{s\|m\|_{T1}}</math>
+where the supremum is taken over all monzos <math>M</math> in the ''inverse-weighted'' coordinate system on the dual space. That is, if <math>m</math> were a monzo in unweighted coordinates, its representation in this space would be <math>M=W^{-1}\cdot m</math>, where we are left-multiplying by the inverse of the weighting matrix from before.
-where <math>\|...\|_{T1}</math> is the [[Tenney_Height]].
+This tells us that the max weighted error on the primes has the property of also being the max weighted error on ''all'' monzos in the prime-limit, and that this weighted error will always be obtained at a prime, where the weighting is given by the dual L1 norm.
-Note that the true weighting is not a norm on monzos, but rather the exponential of a norm: the Tenney height. Our naive weighting agrees with the true one on the primes (and only the primes), and differs for all composite rationals.
+Now, if our weighting matrix is the usual <math>1/log(p)</math> Tenney-weighting matrix, then the above is equivalent to [[Paul Erlich]]'s theorem that minimizing the max Tenney-weighted error on the primes minimizes the max Tenney-weighted error on all intervals. However, if we instead change the weighting matrix to <math>1/p^s</math> instead, then our Linf norm will be dual to a different, somewhat unusual weighted L1 norm on monzos: the one where the weighting on the primes is given by <math>p^s</math>, and the weighting for an arbitrary monzo <math>m = |a\, b\, c\, ...\rangle</math> is given by
-Now, suppose <math>T</math> is some tuning map in weighted coordinates, giving the tuning on each prime, where the weighting is now given by the <math>1/p^s</math> on the primes rather than the usual <math>1/\log(p)</math>. Then we can define the associated "weighted error map" <math>E=T-J</math>, where <math>J</math> is the JIP. This map gives the <math>1/p^s</math>-weighted error of each prime.
+<math>\text{sopfr}^s(m) = 2^s|a| + 3^s|b| + 5^s|c| + ...</math>
+where the notation on the left will be explained shortly.
-We then look at the quantity
+Now, we note that this unusually weighted L1 norm does not provide the weighting we want on all rationals, which is <math>(nd)^s</math>. Rather, when <math>s=1</math>, this is called the [[Wilson Height|Wilson norm]], and is equivalent to the [http://mathworld.wolfram.com/SumofPrimeFactors.html sum of prime factors with repetition] of the ratio <math>n/d</math> in question, often written <math>\text{sopfr}(n/d)</math>. For arbitrary <math>s\geq 1</math>, this is the sum of the <math>s</math>'th power of the prime factors with repetition of the ratio, which we will denote <math>\text{sopfr}^s(n/d)</math>. However, we note that this strange weighting ''does'' equal the weighting we want on the primes, where we have <math>\text{sopfr}^s(n/d) = (n/d)^s</math> - it only diverges on the composite numbers.
-<math>\sup \frac{|\langle E|m \rangle|}{\text{naive}_s(m)}</math>
-where <math>\langle E|m \rangle</math> is the signed error of the monzo m according to the error map E. This expression represents the maximum "naive"-weighted error on all rationals.
+We now want to show that minimizing the max weighted-error according to this strangely weighted L1 norm is equivalent to minimizing the max weighted-error according to the true weighting that we want. To do this, we consider what happens to the evaluation of max-weighted error on all rationals, for any given tuning, if we keep the tuning as is but simply change the weighting on the rationals to the thing we want. To prove our result, we need only show that
-Since the naive norm is simply the L1 norm on the weighted monzos, we know via the definition of the [https://en.wikipedia.org/wiki/Dual_norm dual norm] that this is equal the Linf norm of the inverse-weighted error map. Since the Linf norm is just the max of the abs of the error map, we know that the max naive-weighted error on all rationals will be equal to the max naive-weighted error on the primes.
+. Changing the weighting scheme to the thing we want does not make the max weighted error on all rationals any better
+. Likewise, it doesn't make it any worse
+The first part is easy: for any tuning map, we know that the max "strange L1"-weighted error is equal to that of one of the primes, and we know that the primes agree in weighting on both metrics. Since we know that the weighting on the worst prime doesn't change, we know that after we switch weightings, the new max error can be no better than the thing we had.
-We now want to show that this same property holds for the ''true'' weighting we want on monzos, not the naive one. That is, we want to show:
+To show the second part, all we need to show is that all rationals are weighted more '''strongly''' with our strange L1 metric than with the true metric we want. In particular, we want to show the following:
-<math>\sup \frac{|\langle E|m \rangle|}{\text{true}_s(m)} = \sup \frac{|\langle E|m \rangle|}{\text{naive}_s(m)}</math>
+<math>\text{sopfr}^s(n/d) \leq (n/d)^s</math>
-To see this, we need only to show that
+where the left hand side is our strange metric and the right hand side is the true metric. Since we are ''dividing'' by this weighting, for any rational, the weighted error for the same amount of mistuning will be larger given the weighting scheme on the left than the weighting scheme on the right. This means that we have
-# The max weighted error gets no better.
+<math>\frac{\text{err}_{n/d}}{\text{sopfr}^s(n/d)} \geq \frac{\text{err}_{n/d}}{(n/d)^s}</math>
-# The max weighted error gets no worse.
+To show this is fairly easy. Remember our definition of the strange L1 metric:
-To show that the weighted error gets no better, we need only refer to our previous result that for any given tuning map the max "naive"-weighted error is obtained on a prime. Since the naive and true weightings are identical for primes, we know the weighted error for each prime will be identical for both metrics. Since the supremum is obtained on a prime, we know that the "true"-weighted error cannot be any better than the naive supremum.
+<math>\text{sopfr}^s(m) = 2^s|a| + 3^s|b| + 5^s|c| + ...</math>
+In comparison, the true weighting we want on monzos is:
-To show the weighted error also gets no worse, we can show that for any monzo <math>m</math>, we have
+<math>\text{true}^s(m) = 2^{s|a|} \cdot 3^{s|b|} \cdot 5^{s|c|} \cdot ...</math>
-<math>\text{naive}_s(m) \leq \text{true}_s(m)</math>
-Due to this result, we know that while the unweighted error stays the same, the weighting can only increase as we move from the "naive" to the "true" metric, so the weighted error can only go down.
-To see that this inequality holds, note that we can rewrite it for some monzo <math>|a\, b\, c\, ... \rangle</math> as
+So we want to show that the following inequality holds for all monzos:
 <math>2^s|a| + 3^s|b| + 5^s|c| + ... \leq 2^{s|a|} \cdot 3^{s|b|} \cdot 5^{s|c|} \cdot ...</math>
@@ Line 68: / Line 69: @@
 # For any particular prime <math>p</math> with integer coefficient <math>q</math>, <math>p^s|q| \leq p^{s|q|}</math>
 # Since the nonzero terms on the left side are strictly positive, their sum must be less than their product, which is in turn less than the product of the associated "true" terms given above.
+As a result, we have shown that for any particular tuning map, the max weighted error according to our strange L1 weighting will be the max weighted error according to the true metric we want: it gets no better and gets no worse. Since both are given by the Linf norm on error maps -- aka the max-abs <math>1/p^s</math> weighted error on the primes -- finding the tuning map in a particular temperament that minimizes this quantity is equivalent to finding the tuning map that minimizes the weighted error on all rationals according to ''both'' metrics.
-Since we have shown the supremum does no better and no worse, we now know that the supremums are the same. As a result, we have our main theorem:
+As a result, we have our main theorem:
 '''Theorem:''' to minimize the <math>1/(nd)^s</math> weighted error on all rationals, one needs only minimize the <math>1/p^s</math> weighted error on the primes.