View source for The Riemann zeta function and tuning/Vector's derivation

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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:

[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: