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

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We start with the mu function:

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:

<nowiki>$$ \mu \left( x \right) = \sum_{k=1}^{\infty} \frac{\abs{\operatorname{mod} \left( 2\log_{2} \left( k \right) x, 2 \right) - 1}}{k^{2}} $$</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:

[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.

[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

$$ e^{ix}=\cos\left(x\right)+i\sin\left(x\right) $$

where ''i'' is the square root of −1, and ''e'' is the natural exponential constant.

''i'' is the imaginary unit, which is on a line perpendicular to the real number line. A complex (two-dimensional) number may be written as {{nowrap|''a'' + ''bi''}}.

With this knowledge, cos(''x'') can be rewritten as Re(''e''''ix'')—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.

[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''ln(''n'')''x'' {{=}} ''n''''x''}}, the exponentials and logarithms cancel each other out (i.e. {{nowrap|''e''ln(''n'') {{=}} ''n''}}), so:

[https://www.desmos.com/calculator/f4ojwn0an4 <nowiki>$$ \sum_{k=1}^{\infty}\frac{k^{ix}}{k^{\sigma}} $$</nowiki>]

[https://www.desmos.com/calculator/6388kalfmq $$ \sum_{k=1}^{\infty}k^{ix}k^{-\sigma} $$]

[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'':

[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''d''(''s'')) {{=}} μ''c''(σ, ''x'')}}, our badness function.

μ''d'' is a mathematical function called the Riemann zeta function, so {{nowrap|μ''d''(s) {{=}} ζ(''s'')}}, and re-adding the Re() function gives Re(ζ(''s'')) with {{nowrap|''s'' {{=}} σ − ''ix''}}; ''x'' is the equal division and σ is the weight.

Summary of the derivation:

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-We start with the mu function:
+{{breadcrumb}}{{texops}}
+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:
-<nowiki>$$ \mu \left( x \right) = \sum_{k=1}^{\infty} \frac{\abs{\operatorname{mod} \left( 2\log_{2} \left( k \right)  x, 2 \right) - 1}}{k^{2}} $$</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:
+[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.
+[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
+$$ e^{ix}=\cos\left(x\right)+i\sin\left(x\right) $$
+where ''i'' is the square root of −1, and ''e'' is the natural exponential constant.
+''i'' is the imaginary unit, which is on a line perpendicular to the real number line. A complex (two-dimensional) number may be written as {{nowrap|''a'' + ''bi''}}.
+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.
+[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:
+[https://www.desmos.com/calculator/f4ojwn0an4 <nowiki>$$ \sum_{k=1}^{\infty}\frac{k^{ix}}{k^{\sigma}} $$</nowiki>]
+[https://www.desmos.com/calculator/6388kalfmq $$ \sum_{k=1}^{\infty}k^{ix}k^{-\sigma} $$]
+[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'':
+[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.
+μ<sub>''d''</sub> is a mathematical function called the Riemann zeta function, so {{nowrap|μ<sub>''d''(s) {{=}} ζ(''s'')}}, and re-adding the Re() function gives Re(ζ(''s'')) with {{nowrap|''s'' {{=}} σ − ''ix''}}; ''x'' is the equal division and σ is the weight.
+Summary of the derivation: