The Riemann zeta function and tuning: Difference between revisions

Line 48:

If we take exponentials of both sides, then

<math>\displaystyle \exp(F_s(x)) = \~~left|~~\zeta(s + 2 \pi i x/\ln 2)~~\right|~~</math>

<math>\displaystyle \exp(F_s(x)) = \abs{\zeta(s + 2 \pi i x/\ln 2)}</math>

so that we see that the absolute value of the zeta function serves to measure the relative error of an equal division.

Line 118:

<math> \displaystyle

\~~left|~~ \zeta(s) ~~\right|~~^2 = \left[\sum_n n^{-(\sigma+it)}\right] \cdot \left[\sum_d d^{-(\sigma-it)}\right]</math>

\abs{ \zeta(s) }^2 = \left[\sum_n n^{-(\sigma+it)}\right] \cdot \left[\sum_d d^{-(\sigma-it)}\right]</math>

where d is a new variable used internally in the second summation.

Line 125:

<math> \displaystyle

\~~left|~~ \zeta(s) ~~\right|~~^2 = \sum_{n,d} \left[n^{-(\sigma+it)} \cdot d^{-(\sigma-it)}\right] = \sum_{n,d} \frac{\left({\tfrac{n}{d}}\right)^{-it}}{(nd)^{\sigma}}</math>

\abs{ \zeta(s) }^2 = \sum_{n,d} \left[n^{-(\sigma+it)} \cdot d^{-(\sigma-it)}\right] = \sum_{n,d} \frac{\left({\tfrac{n}{d}}\right)^{-it}}{(nd)^{\sigma}}</math>

<span style="line-height: 1.5;">Now let's do a bit of algebra with the exponential function, and use Euler's identity:</span>

<math> \displaystyle

\~~left|~~ \zeta(s) ~~\right|~~^2 = \sum_{n,d} \frac{e^{-it \ln\left({\tfrac{n}{d}}\right)}}{(nd)^{\sigma}}

\abs{ \zeta(s) }^2 = \sum_{n,d} \frac{e^{-it \ln\left({\tfrac{n}{d}}\right)}}{(nd)^{\sigma}}

= \sum_{n,d} \frac{\cos\left(-t \ln\left({\tfrac{n}{d}}\right)\right) + i\sin\left(-t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}}

= \sum_{n,d} \frac{\cos\left(t \ln\left({\tfrac{n}{d}}\right)\right) - i\sin\left(t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}}</math>

Line 139:

<math> \displaystyle

\~~left|~~ \zeta(s) ~~\right|~~^2 = \sum_{n=d} \left[ \frac{\cos\left(t \ln\left({\tfrac{n}{d}}\right)\right) - i\sin\left(t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}} \right] +

\abs{ \zeta(s) }^2 = \sum_{n=d} \left[ \frac{\cos\left(t \ln\left({\tfrac{n}{d}}\right)\right) - i\sin\left(t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}} \right] +

\sum_{n>d} \left[ \frac{\cos\left(t \ln\left({\tfrac{n}{d}}\right)\right) - i\sin\left(t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}} \right] +

\sum_{n< d} \left[ \frac{\cos\left(t \ln\left({\tfrac{n}{d}}\right)\right) - i\sin\left(t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}} \right]</math>

Line 148:

<math> \displaystyle

\~~left|~~ \zeta(s) ~~\right|~~^2 = \sum_{n=d} \left[ \frac{\cos\left( t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}} \right] +

\abs{ \zeta(s) }^2 = \sum_{n=d} \left[ \frac{\cos\left( t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}} \right] +

\sum_{n>d} \left[ \frac{\cos\left(t \ln\left({\tfrac{n}{d}}\right)\right) - i\sin\left(t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}} \right] +

\sum_{n< d} \left[ \frac{\cos\left(t \ln\left({\tfrac{n}{d}}\right)\right) - i\sin\left(t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}} \right]</math>

Line 169:

<math> \displaystyle

\~~left|~~ \zeta(s) ~~\right|~~^2 = \sum_{n=d} \left[ \frac{\cos\left(t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}} \right] +

\abs{ \zeta(s) }^2 = \sum_{n=d} \left[ \frac{\cos\left(t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}} \right] +

\sum_{n>d} \left[ \frac{\cos\left(t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}} \right] +

\sum_{n< d} \left[ \frac{\cos\left(t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}} \right]</math>

Line 176:

<math> \displaystyle

\~~left|~~ \zeta(s) ~~\right|~~^2 = \sum_{n,d} \frac{\cos\left(t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}}</math>

\abs{ \zeta(s) }^2 = \sum_{n,d} \frac{\cos\left(t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}}</math>

Finally, by making the mysterious substitution {{nowrap|''t'' {{=}} {{frac|2π|ln(2)}} ''x''}}, the musical implications of the above will start to reveal themselves:

<math> \displaystyle

\~~left|~~ \zeta(s) ~~\right|~~^2 = \sum_{n,d} \frac{\cos\left(2\pi x \log_2\left(\tfrac{n}{d}\right)\right)}{(nd)^{\sigma}}</math>

\abs{ \zeta(s) }^2 = \sum_{n,d} \frac{\cos\left(2\pi x \log_2\left(\tfrac{n}{d}\right)\right)}{(nd)^{\sigma}}</math>

Let's take a breather and see what we've got.

Line 225:

<math> \displaystyle

\~~left|~~ \zeta(s) ~~\right|~~^2 = \sum_{n,d} \frac{\cos\left(t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}}</math>

\abs{ \zeta(s) }^2 = \sum_{n,d} \frac{\cos\left(t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}}</math>

Note that since there's no restriction that n and d be coprime, the "rationals" we're using here don't have to be reduced. So this shows that zeta yields an error metric over all unreduced rationals, but leaves open the question of how reduced rationals are handled. It turns out that the same function also measures the error of reduced rationals, scaled only by a rolloff-dependent constant factor across all EDOs.

Line 234:

<math> \displaystyle

\~~left|~~ \zeta(s) ~~\right|~~^2 = \sum_{n',d',c} \frac{\cos\left(t \ln\left({\tfrac{cn'}{cd'}}\right)\right)}{(cn' \cdot cd')^{\sigma}}</math>

\abs{ \zeta(s) }^2 = \sum_{n',d',c} \frac{\cos\left(t \ln\left({\tfrac{cn'}{cd'}}\right)\right)}{(cn' \cdot cd')^{\sigma}}</math>

Now, the common factor ''c''/''c'' cancels out inside the log in the numerator. However, in the denominator, we get an extra factor of ''c''<sup>2</sup> to contend with. This yields

<math> \displaystyle

\~~left|~~ \zeta(s) ~~\right|~~^2 = \sum_{n',d',c} \frac{\cos\left(t \ln\left({\tfrac{n'}{d'}}\right)\right)}{(c^2 \cdot n'd')^{\sigma}}

\abs{ \zeta(s) }^2 = \sum_{n',d',c} \frac{\cos\left(t \ln\left({\tfrac{n'}{d'}}\right)\right)}{(c^2 \cdot n'd')^{\sigma}}

= \sum_{n',d',c} \left[ \frac{1}{c^{2\sigma}} \cdot \frac{\cos\left(t \ln\left({\tfrac{n'}{d'}}\right)\right)}{(n'd')^{\sigma}} \right]</math>

Line 245:

<math> \displaystyle

\~~left|~~ \zeta(s) ~~\right|~~^2 = \left[ \sum_c \frac{1}{c^{2\sigma}} \right] \cdot \left[ \sum_{n',d'} \frac{\cos\left(t \ln\left({\tfrac{n'}{d'}}\right)\right)}{(n'd')^{\sigma}} \right]</math>

\abs{ \zeta(s) }^2 = \left[ \sum_c \frac{1}{c^{2\sigma}} \right] \cdot \left[ \sum_{n',d'} \frac{\cos\left(t \ln\left({\tfrac{n'}{d'}}\right)\right)}{(n'd')^{\sigma}} \right]</math>

Finally, we note that on the left summation we simply have another zeta series, yielding

<math> \displaystyle

\~~left|~~ \zeta(s) ~~\right|~~^2 = \zeta(2\sigma) \cdot \left[ \sum_{n',d'} \frac{\cos\left(t \ln\left({\tfrac{n'}{d'}}\right)\right)}{(n'd')^{\sigma}} \right]</math>

\abs{ \zeta(s) }^2 = \zeta(2\sigma) \cdot \left[ \sum_{n',d'} \frac{\cos\left(t \ln\left({\tfrac{n'}{d'}}\right)\right)}{(n'd')^{\sigma}} \right]</math>

<math> \displaystyle

\frac{\~~left|~~ \zeta(s) ~~\right|~~^2}{\zeta(2\sigma)} = \sum_{n',d'} \frac{\cos\left(t \ln\left({\tfrac{n'}{d'}}\right)\right)}{(n'd')^{\sigma}}</math>

\frac{\abs{ \zeta(s) }^2}{\zeta(2\sigma)} = \sum_{n',d'} \frac{\cos\left(t \ln\left({\tfrac{n'}{d'}}\right)\right)}{(n'd')^{\sigma}}</math>

Now, since we're fixing σ and letting ''t'' vary, the left zeta term is constant for all EDOs. This demonstrates that the zeta function also measures cosine error over all the reduced rationals, up to a constant factor. QED.

Line 260:

So far we have shown the following:

* Error on prime powers: <math>\log \~~left|~~\zeta(\sigma + it)~~\right|~~</math>

* Error on prime powers: <math>\log \abs{\zeta(\sigma + it)}</math>

* Error on unreduced rationals: <math>\~~left|~~\zeta(\sigma+it)~~\right|~~^2</math>

* Error on unreduced rationals: <math>\abs{\zeta(\sigma+it)}^2</math>

* Error on reduced rationals: <math>\frac{\~~left|~~\zeta(\sigma+it)~~\right|~~^2}{\zeta(2\sigma)}</math>

* Error on reduced rationals: <math>\frac{\abs{\zeta(\sigma+it)}^2}{\zeta(2\sigma)}</math>

Since the second is a simple monotonic transformation of the first, we can see that the same function basically measures both the relative error on just the prime powers, and also on all unreduced rationals, at least in the sense that EDOs will be ranked identically by both measures. The third function is really just the second function divided by a constant, since we only really care about letting <math>t</math> vary—we instead typically set <math>\sigma</math> to some value which represents the weighting "rolloff" on rationals. So, all three of these functions will rank EDOs identically.

Line 270:

It turns out that using the same principles of derivation above, we can also derive another expression, this time for the relative error on only the harmonics—i.e. those intervals of the form <math>1/1, 2/1, 3/1, ... n/1, ...</math>. This was studied in a paper by Peter Buch called [[:File:Zetamusic5.pdf|"Favored cardinalities of scales"]]. The expression is:

Error on harmonics only: <math>\~~left|~~\textbf{Re}\left[\zeta(\sigma + it)\right]~~\right|~~</math>

Error on harmonics only: <math>\abs{\textbf{Re}\left[\zeta(\sigma + it)\right]}</math>

Note that, although the last four expressions were all monotonic transformations of one another, this one is not—this is the 'real part' of the zeta function, whereas the others were all some simple monotonic function of the 'absolute value' of the zeta function. The results, however, are very similar—in particular, the peaks are approximately to one another, shifted by only a small amount (at least for reasonably-sized EDOs up to a few hundred).

Line 277:

The expression

<math>\displaystyle{\~~left|~~\zeta\left(\frac{1}{2} + it\right)~~\right|~~^2 \cdot \overline {\phi(t)}}</math>

<math>\displaystyle{\abs{\zeta\left(\frac{1}{2} + it\right)}^2 \cdot \overline {\phi(t)}}</math>

is, up to a flip in sign, the Fourier transform of the unnormalized Harmonic Shannon Entropy for {{nowrap|''N'' {{=}} ∞}}, where φ(''t'') is the characteristic function (aka Fourier transform) of the spreading distribution and {{overline|φ(''t'')}} denotes complex conjugation.

Line 630:

<math>

\displaystyle\~~left|~~1 - p^{\frac{1}{2} - it}~~\right|~~ = \sqrt{1 + \frac{1}{p} - \frac{2 \cos(t \ln p)}{\sqrt{p}}}

\displaystyle\abs{1 - p^{\frac{1}{2} - it}} = \sqrt{1 + \frac{1}{p} - \frac{2 \cos(t \ln p)}{\sqrt{p}}}

</math>

@@ Line 48: / Line 48: @@
 If we take exponentials of both sides, then
-<math>\displaystyle \exp(F_s(x)) = \left|\zeta(s + 2 \pi i x/\ln 2)\right|</math>
+<math>\displaystyle \exp(F_s(x)) = \abs{\zeta(s + 2 \pi i x/\ln 2)}</math>
 so that we see that the absolute value of the zeta function serves to measure the relative error of an equal division.
@@ Line 118: / Line 118: @@
 <math> \displaystyle
-\left| \zeta(s) \right|^2 = \left[\sum_n n^{-(\sigma+it)}\right] \cdot \left[\sum_d d^{-(\sigma-it)}\right]</math>
+\abs{ \zeta(s) }^2 = \left[\sum_n n^{-(\sigma+it)}\right] \cdot \left[\sum_d d^{-(\sigma-it)}\right]</math>
 where d is a new variable used internally in the second summation.
@@ Line 125: / Line 125: @@
 <math> \displaystyle
-\left| \zeta(s) \right|^2 = \sum_{n,d} \left[n^{-(\sigma+it)} \cdot d^{-(\sigma-it)}\right] = \sum_{n,d} \frac{\left({\tfrac{n}{d}}\right)^{-it}}{(nd)^{\sigma}}</math>
+\abs{ \zeta(s) }^2 = \sum_{n,d} \left[n^{-(\sigma+it)} \cdot d^{-(\sigma-it)}\right] = \sum_{n,d} \frac{\left({\tfrac{n}{d}}\right)^{-it}}{(nd)^{\sigma}}</math>
 <span style="line-height: 1.5;">Now let's do a bit of algebra with the exponential function, and use Euler's identity:</span>
 <math> \displaystyle
-\left| \zeta(s) \right|^2 = \sum_{n,d} \frac{e^{-it \ln\left({\tfrac{n}{d}}\right)}}{(nd)^{\sigma}}
+\abs{ \zeta(s) }^2 = \sum_{n,d} \frac{e^{-it \ln\left({\tfrac{n}{d}}\right)}}{(nd)^{\sigma}}
 = \sum_{n,d} \frac{\cos\left(-t \ln\left({\tfrac{n}{d}}\right)\right) + i\sin\left(-t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}}
 = \sum_{n,d} \frac{\cos\left(t \ln\left({\tfrac{n}{d}}\right)\right) - i\sin\left(t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}}</math>
@@ Line 139: / Line 139: @@
 <math> \displaystyle
-\left| \zeta(s) \right|^2 = \sum_{n=d} \left[ \frac{\cos\left(t \ln\left({\tfrac{n}{d}}\right)\right) - i\sin\left(t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}} \right] +
+\abs{ \zeta(s) }^2 = \sum_{n=d} \left[ \frac{\cos\left(t \ln\left({\tfrac{n}{d}}\right)\right) - i\sin\left(t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}} \right] +
 \sum_{n>d} \left[ \frac{\cos\left(t \ln\left({\tfrac{n}{d}}\right)\right) - i\sin\left(t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}} \right] +
 \sum_{n< d} \left[ \frac{\cos\left(t \ln\left({\tfrac{n}{d}}\right)\right) - i\sin\left(t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}} \right]</math>
@@ Line 148: / Line 148: @@
 <math> \displaystyle
-\left| \zeta(s) \right|^2 = \sum_{n=d} \left[ \frac{\cos\left( t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}} \right] +
+\abs{ \zeta(s) }^2 = \sum_{n=d} \left[ \frac{\cos\left( t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}} \right] +
 \sum_{n>d} \left[ \frac{\cos\left(t \ln\left({\tfrac{n}{d}}\right)\right) - i\sin\left(t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}} \right] +
 \sum_{n< d} \left[ \frac{\cos\left(t \ln\left({\tfrac{n}{d}}\right)\right) - i\sin\left(t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}} \right]</math>
@@ Line 169: / Line 169: @@
 <math> \displaystyle
-\left| \zeta(s) \right|^2 = \sum_{n=d} \left[ \frac{\cos\left(t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}} \right] +
+\abs{ \zeta(s) }^2 = \sum_{n=d} \left[ \frac{\cos\left(t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}} \right] +
 \sum_{n>d} \left[ \frac{\cos\left(t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}} \right] +
 \sum_{n< d} \left[ \frac{\cos\left(t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}} \right]</math>
@@ Line 176: / Line 176: @@
 <math> \displaystyle
-\left| \zeta(s) \right|^2 = \sum_{n,d} \frac{\cos\left(t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}}</math>
+\abs{ \zeta(s) }^2 = \sum_{n,d} \frac{\cos\left(t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}}</math>
 Finally, by making the mysterious substitution {{nowrap|''t'' {{=}} {{frac|2&pi;|ln(2)}} ''x''}}, the musical implications of the above will start to reveal themselves:
 <math> \displaystyle
-\left| \zeta(s) \right|^2 = \sum_{n,d} \frac{\cos\left(2\pi x \log_2\left(\tfrac{n}{d}\right)\right)}{(nd)^{\sigma}}</math>
+\abs{ \zeta(s) }^2 = \sum_{n,d} \frac{\cos\left(2\pi x \log_2\left(\tfrac{n}{d}\right)\right)}{(nd)^{\sigma}}</math>
 Let's take a breather and see what we've got.
@@ Line 225: / Line 225: @@
 <math> \displaystyle
-\left| \zeta(s) \right|^2 = \sum_{n,d} \frac{\cos\left(t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}}</math>
+\abs{ \zeta(s) }^2 = \sum_{n,d} \frac{\cos\left(t \ln\left({\tfrac{n}{d}}\right)\right)}{(nd)^{\sigma}}</math>
 Note that since there's no restriction that n and d be coprime, the "rationals" we're using here don't have to be reduced. So this shows that zeta yields an error metric over all unreduced rationals, but leaves open the question of how reduced rationals are handled. It turns out that the same function also measures the error of reduced rationals, scaled only by a rolloff-dependent constant factor across all EDOs.
@@ Line 234: / Line 234: @@
 <math> \displaystyle
-\left| \zeta(s) \right|^2 = \sum_{n',d',c} \frac{\cos\left(t \ln\left({\tfrac{cn'}{cd'}}\right)\right)}{(cn' \cdot cd')^{\sigma}}</math>
+\abs{ \zeta(s) }^2 = \sum_{n',d',c} \frac{\cos\left(t \ln\left({\tfrac{cn'}{cd'}}\right)\right)}{(cn' \cdot cd')^{\sigma}}</math>
 Now, the common factor ''c''/''c'' cancels out inside the log in the numerator. However, in the denominator, we get an extra factor of ''c''<sup>2</sup> to contend with. This yields
 <math> \displaystyle
-\left| \zeta(s) \right|^2 = \sum_{n',d',c} \frac{\cos\left(t \ln\left({\tfrac{n'}{d'}}\right)\right)}{(c^2 \cdot n'd')^{\sigma}}
+\abs{ \zeta(s) }^2 = \sum_{n',d',c} \frac{\cos\left(t \ln\left({\tfrac{n'}{d'}}\right)\right)}{(c^2 \cdot n'd')^{\sigma}}
 = \sum_{n',d',c} \left[ \frac{1}{c^{2\sigma}} \cdot \frac{\cos\left(t \ln\left({\tfrac{n'}{d'}}\right)\right)}{(n'd')^{\sigma}} \right]</math>
@@ Line 245: / Line 245: @@
 <math> \displaystyle
-\left| \zeta(s) \right|^2 = \left[ \sum_c \frac{1}{c^{2\sigma}} \right] \cdot \left[ \sum_{n',d'} \frac{\cos\left(t \ln\left({\tfrac{n'}{d'}}\right)\right)}{(n'd')^{\sigma}} \right]</math>
+\abs{ \zeta(s) }^2 = \left[ \sum_c \frac{1}{c^{2\sigma}} \right] \cdot \left[ \sum_{n',d'} \frac{\cos\left(t \ln\left({\tfrac{n'}{d'}}\right)\right)}{(n'd')^{\sigma}} \right]</math>
 Finally, we note that on the left summation we simply have another zeta series, yielding
 <math> \displaystyle
-\left| \zeta(s) \right|^2 = \zeta(2\sigma) \cdot \left[ \sum_{n',d'} \frac{\cos\left(t \ln\left({\tfrac{n'}{d'}}\right)\right)}{(n'd')^{\sigma}} \right]</math>
+\abs{ \zeta(s) }^2 = \zeta(2\sigma) \cdot \left[ \sum_{n',d'} \frac{\cos\left(t \ln\left({\tfrac{n'}{d'}}\right)\right)}{(n'd')^{\sigma}} \right]</math>
 <math> \displaystyle
-\frac{\left| \zeta(s) \right|^2}{\zeta(2\sigma)} = \sum_{n',d'} \frac{\cos\left(t \ln\left({\tfrac{n'}{d'}}\right)\right)}{(n'd')^{\sigma}}</math>
+\frac{\abs{ \zeta(s) }^2}{\zeta(2\sigma)} = \sum_{n',d'} \frac{\cos\left(t \ln\left({\tfrac{n'}{d'}}\right)\right)}{(n'd')^{\sigma}}</math>
 Now, since we're fixing &sigma; and letting ''t'' vary, the left zeta term is constant for all EDOs. This demonstrates that the zeta function also measures cosine error over all the reduced rationals, up to a constant factor. QED.
@@ Line 260: / Line 260: @@
 So far we have shown the following:
-* Error on prime powers: <math>\log \left|\zeta(\sigma + it)\right|</math>
+* Error on prime powers: <math>\log \abs{\zeta(\sigma + it)}</math>
-* Error on unreduced rationals: <math>\left|\zeta(\sigma+it)\right|^2</math>
+* Error on unreduced rationals: <math>\abs{\zeta(\sigma+it)}^2</math>
-* Error on reduced rationals: <math>\frac{\left|\zeta(\sigma+it)\right|^2}{\zeta(2\sigma)}</math>
+* Error on reduced rationals: <math>\frac{\abs{\zeta(\sigma+it)}^2}{\zeta(2\sigma)}</math>
 Since the second is a simple monotonic transformation of the first, we can see that the same function basically measures both the relative error on just the prime powers, and also on all unreduced rationals, at least in the sense that EDOs will be ranked identically by both measures. The third function is really just the second function divided by a constant, since we only really care about letting <math>t</math> vary&mdash;we instead typically set <math>\sigma</math> to some value which represents the weighting "rolloff" on rationals. So, all three of these functions will rank EDOs identically.
@@ Line 270: / Line 270: @@
 It turns out that using the same principles of derivation above, we can also derive another expression, this time for the relative error on only the harmonics&mdash;i.e. those intervals of the form <math>1/1, 2/1, 3/1, ... n/1, ...</math>. This was studied in a paper by Peter Buch called [[:File:Zetamusic5.pdf|"Favored cardinalities of scales"]]. The expression is:
-Error on harmonics only: <math>\left|\textbf{Re}\left[\zeta(\sigma + it)\right]\right|</math>
+Error on harmonics only: <math>\abs{\textbf{Re}\left[\zeta(\sigma + it)\right]}</math>
 Note that, although the last four expressions were all monotonic transformations of one another, this one is not&mdash;this is the 'real part' of the zeta function, whereas the others were all some simple monotonic function of the 'absolute value' of the zeta function. The results, however, are  very similar&mdash;in particular, the peaks are approximately to one another, shifted by only a small amount (at least for reasonably-sized EDOs up to a few hundred).
@@ Line 277: / Line 277: @@
 The expression
-<math>\displaystyle{\left|\zeta\left(\frac{1}{2} + it\right)\right|^2 \cdot \overline {\phi(t)}}</math>
+<math>\displaystyle{\abs{\zeta\left(\frac{1}{2} + it\right)}^2 \cdot \overline {\phi(t)}}</math>
 is, up to a flip in sign, the Fourier transform of the unnormalized Harmonic Shannon Entropy for {{nowrap|''N'' {{=}} &infin;}}, where &phi;(''t'') is the characteristic function (aka Fourier transform) of the spreading distribution and {{overline|&phi;(''t'')}} denotes complex conjugation.
@@ Line 630: / Line 630: @@
 <math>
-\displaystyle\left|1 - p^{\frac{1}{2} - it}\right| = \sqrt{1 + \frac{1}{p} - \frac{2 \cos(t \ln p)}{\sqrt{p}}}
+\displaystyle\abs{1 - p^{\frac{1}{2} - it}} = \sqrt{1 + \frac{1}{p} - \frac{2 \cos(t \ln p)}{\sqrt{p}}}
 </math>