The Riemann zeta function and tuning/Vector's derivation: Difference between revisions

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 We start with the generalized mu function, which sums up the relative error on all integer harmonics weighted by an inverse power of the harmonic:
-{{ext|https://www.desmos.com/calculator/4zcynoue8s|<nowiki>$$ \mu \left(\sigma, x \right) = \sum_{k=1}^{\infty} \frac{\operatorname{abs} \left( \operatorname{mod} \left( 2\log_{2} \left( k \right) x, 2 \right) - 1 \right)}{k^{\sigma}} $$</nowiki>}}
+[https://www.desmos.com/calculator/4zcynoue8s <nowiki>$$ \mu \left(\sigma, x \right) = \sum_{k=1}^{\infty} \frac{\abs{ \operatorname{mod} \left( 2\log_{2} \left( k \right) x, 2 \right) - 1 }}{k^{\sigma}} $$</nowiki>]
 Now, this is a rather annoying function to work with for math reasons, so it might be useful to replace the "zigzag" that we use as our error function with a "smoother" alternative. The most obvious answer is cosine:
-{{ext|https://www.desmos.com/calculator/deafikrhvg|<nowiki>$$ \sum_{k=1}^{\infty}\frac{\cos\left(\log_{2}\left(k\right)\tau x\right)}{k^{\sigma}} $$</nowiki>}}
+[https://www.desmos.com/calculator/deafikrhvg <nowiki>$$ \sum_{k=1}^{\infty}\frac{\cos\left(\log_{2}\left(k\right)\tau x\right)}{k^{\sigma}} $$</nowiki>]
 Let's clean up the function by removing the scale factors on ''x''. This just scales the function's inputs from EDO to [[Zetave|EDZ]], and these can be added back later to go back to EDO.
-{{ext|https://www.desmos.com/calculator/26ypbwbglg|<nowiki>$$  \sum_{k=1}^{\infty}\frac{\cos (\ln\left(k\right)x)}{k^{\sigma}} $$</nowiki>}}
+[https://www.desmos.com/calculator/26ypbwbglg <nowiki>$$ \sum_{k=1}^{\infty}\frac{\cos (\ln\left(k\right)x)}{k^{\sigma}} $$</nowiki>]
 By the complex exponential theorem, which relates trigonometric functions and the exponential function, we know that
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 With this knowledge, cos(''x'') can be rewritten as Re(''e''<sup>''ix''</sup>)—but since we're only doing multiplication and addition and this is the only place complex numbers appear, we can just ignore the Re() and add it back later.
-{{ext|https://www.desmos.com/calculator/e7wn17tzjf|<nowiki>$$ \mu_{c}\left(\sigma, x\right)=\sum_{k=1}^{\infty}\frac{e^{i\left(\ln\left(k\right)x\right)}}{k^{\sigma}} $$</nowiki>}}
+[https://www.desmos.com/calculator/e7wn17tzjf <nowiki>$$ \mu_{c}\left(\sigma, x\right)=\sum_{k=1}^{\infty}\frac{e^{i\left(\ln\left(k\right)x\right)}}{k^{\sigma}} $$</nowiki>]
 Since {{nowrap|''e''<sup>ln(''n'')''x''</sup> {{=}} ''n''<sup>''x''</sup>}}, the exponentials and logarithms cancel each other out (i.e. {{nowrap|''e''<sup>ln(''n'')</sup> {{=}} ''n''}}), so:
-{{ext|https://www.desmos.com/calculator/f4ojwn0an4|<nowiki>$$ \sum_{k=1}^{\infty}\frac{k^{ix}}{k^{\sigma}} $$</nowiki>}}
+[https://www.desmos.com/calculator/f4ojwn0an4 <nowiki>$$ \sum_{k=1}^{\infty}\frac{k^{ix}}{k^{\sigma}} $$</nowiki>]
-{{ext|https://www.desmos.com/calculator/6388kalfmq|$$ \sum_{k=1}^{\infty}k^{ix}k^{-\sigma} $$}}
+[https://www.desmos.com/calculator/6388kalfmq $$ \sum_{k=1}^{\infty}k^{ix}k^{-\sigma} $$]
-{{ext|https://www.desmos.com/calculator/l3q2dtd6xn|$$ \sum_{k=1}^{\infty}k^{-\sigma+ix} $$}}
+[https://www.desmos.com/calculator/l3q2dtd6xn $$ \sum_{k=1}^{\infty}k^{-\sigma+ix} $$]
 {{nowrap|−σ + ''ix''}} is just a complex number, which we may write as −''s'':
-{{ext|https://www.desmos.com/calculator/esbdlxdoui|$$ \mu_{d}\left(s\right)=\sum_{k=1}^{\infty}k^{-s} $$}}
+[https://www.desmos.com/calculator/esbdlxdoui $$ \mu_{d}\left(s\right)=\sum_{k=1}^{\infty}k^{-s} $$]
 where, for {{nowrap|''s'' {{=}} σ − ''ix''}}, {{nowrap|Re(μ<sub>''d''</sub>(''s'')) {{=}} μ<sub>''c''</sub>(σ, ''x'')}}, our badness function.