The Riemann zeta function and tuning/Vector's derivation

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:

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

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:

$$ \mu_{b} \left(\sigma, x \right) = \sum_{k=1}^{\infty}\frac{\cos\left(\log_{2}\left(k\right)\tau x\right)}{k^{\sigma}} $$

Let's clean up the function by removing the scale factors on x. This just scales the function's inputs from EDO to EDZ, and these can be added back later to go back to EDO.

$$ \mu_{c} \left(\sigma, x \right) = \sum_{k=1}^{\infty}\frac{\cos (\ln\left(k\right)x)}{k^{\sigma}} $$

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 a+bi.

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

$$ \mu_{c}\left(\sigma, x\right)=\sum_{k=1}^{\infty}\frac{\operatorname{Re}\left(e^{i\left(\ln\left(k\right)x\right)}\right)}{k^{\sigma}} $$

e^ln(n)x = n^x, because exponentials and logarithms cancel each other out (i.e. e^ln(n) = n), so:

$$ \mu_{c}\left(\sigma, x\right)=\sum_{k=1}^{\infty}\frac{\operatorname{Re}\left(k^{ix}\right)}{k^{\sigma}} $$

Thus:

$$ \mu_{c}\left(\sigma, x\right)=\sum_{k=1}^{\infty}\operatorname{Re}\left(k^{ix}\right)k^{-\sigma} $$

k^-σ is a real number, so its real part is equal to itself. Thus we can simplify this as follows:

$$ \mu_{c}\left(\sigma, x\right)=\sum_{k=1}^{\infty}\operatorname{Re}\left(k^{-\sigma+ix}\right) $$

-σ+ix is just a complex number, which we may write as -s:

$$ \mu_{d}\left(s\right)=\sum_{k=1}^{\infty}\operatorname{Re}\left(k^{-s}\right) $$

By the rules of complex addition, the real part of a sum is the same as a sum of real parts, so we can make the following change:

$$\mu_{e}\left(s\right)=\sum_{k=1}^{\infty}k^{-s} $$where Re(μ_e(s)) = μ_d(s), our badness function.

μ_e is a mathematical function called the Riemann zeta function, so μ_e(s) = ζ(s), and re-adding the Re() function gives Re(ζ(s)) with s = σ-ix; x is the equal division and σ is the weight.

Summary of the derivation: