Talk:The Riemann zeta function and tuning: Difference between revisions
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:: This has the behaviour that it "rewards" for being in-tune and "punishes" for being out-of-tune, and it looks only at harmonics that are prime powers, with a simplicity weighting. | :: This has the behaviour that it "rewards" for being in-tune and "punishes" for being out-of-tune, and it looks only at harmonics that are prime powers, with a simplicity weighting. | ||
:: It should hopefully be obvious from this expression that what it is measuring is in terms of steps of ''x'' equal temperament (because "(near-)perfectly in tune" means the cosine in the numerator evaluates to (nearly) 1 and "(near-)perfectly out of tune" means it evaluates to (nearly) -1), which we can confirm by noticing that the unaltered zeta graph has its record peaks grow very slowly/subtly, which should intuitively make sense: if you are measuring by cosine, for every interval that's in there's gonna be an interval that's out; you can't beat this, you can only try to prioritise the intervals that are lower complexity, and even that only works to | :: It should hopefully be obvious from this expression that what it is measuring is in terms of steps of ''x'' equal temperament (because "(near-)perfectly in tune" means the cosine in the numerator evaluates to (nearly) 1 and "(near-)perfectly out of tune" means it evaluates to (nearly) -1), which we can confirm by noticing that the unaltered zeta graph has its record peaks grow very slowly/subtly, which should intuitively make sense: if you are measuring by cosine, for every interval that's in there's gonna be an interval that's out; you can't beat this, you can only try to prioritise the intervals that are lower complexity, and even that only works to a slight extent (as is evident from the flatness of the graph). | ||
:: If we divide by ''n'', then we would get a very strange graph that as far as I can tell doesn't make any sense, because the "valleys in error" (which here, because of the sign flip, are peaks) would become smaller and smaller, so how would you even tell what a record is? (Because the result would cross the zero line many times, suggesting those tunings are "perfectly in tune". So if you apply this reasoning correctly by using something based on 1 - cos(2pi x), then you get something fairly different.) | :: If we divide by ''n'', then we would get a very strange graph that as far as I can tell doesn't make any sense, because the "valleys in error" (which here, because of the sign flip, are peaks) would become smaller and smaller, so how would you even tell what a record is? (Because the result would cross the zero line many times, suggesting those tunings are "perfectly in tune". So if you apply this reasoning correctly by using something based on 1 - cos(2pi x), then you get something fairly different.) | ||