BOP tuning: Difference between revisions

Line 47:

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

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. That is, the above would entail we have

where the left hand side is our strange metric and the right hand side is the true metric. This would would entail we have

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

~~Where that <math>\text{err}_{n/d}</math> denotes unweighted~~ error (~~which~~ is the same on both sides, ~~as we are not changing the tuning, just the weighting~~).

where the left hand side is the strangely-weighted error, and the right hand side is the true-weighted error. (Note the numerator is the same on both sides, representing unweighted error.)

If we can show the above, this means we have shown that after re-weighting, the weighted error on all rationals goes ''down'' -- except at the primes, where the error (and in particular the worst error) remains the same. This would give us our desired result.

The proof is fairly ~~easy. Remember~~ our definition of the strange L1 metric:

The proof is fairly simple: remember our definition of the strange L1 metric:

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

@@ Line 47: / Line 47: @@
 <math>\text{sopfr}^s(n/d) \leq (n/d)^s</math>
-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. That is, the above would entail we have
+where the left hand side is our strange metric and the right hand side is the true metric. This would would entail we have
 <math>\frac{\text{err}_{n/d}}{\text{sopfr}^s(n/d)} \geq \frac{\text{err}_{n/d}}{(n/d)^s}</math>
-Where that <math>\text{err}_{n/d}</math> denotes unweighted error (which is the same on both sides, as we are not changing the tuning, just the weighting).
+where the left hand side is the strangely-weighted error, and the right hand side is the true-weighted error. (Note the numerator is the same on both sides, representing unweighted error.)
 If we can show the above, this means we have shown that after re-weighting, the weighted error on all rationals goes ''down'' -- except at the primes, where the error (and in particular the worst error) remains the same. This would give us our desired result.
-The proof is fairly easy. Remember our definition of the strange L1 metric:
+The proof is fairly simple: remember our definition of the strange L1 metric:
 <math>\text{sopfr}^s(m) = 2^s|a| + 3^s|b| + 5^s|c| + ...</math>