The Riemann zeta function and tuning/Vector's derivation

We start with the mu function:

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

Now, this is nowhere differentiable, 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:

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

Let's clean up the function by removing the scale factors on x:

$$ \mu_{c} \left( x \right) = \sum_{k=1}^{\infty}\frac{\cos (\ln\left(k\right)x)}{k^{2}} $$

By the complex exponential theorem, we know that

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

so that cos(x) can be rewritten as Re(e^ix).

$$ \mu_{c}\left(x\right)=\sum_{k=1}^{\infty}\frac{\operatorname{Re}\left(e^{i\left(\ln\left(k\right)x\right)}\right)}{k^{2}} $$

For now, we will ignore the Re() function as a sum of real parts is the same as the real part of the sum (by the rules of complex addition), and the denominator is just a real number.

$$ \mu_{d}\left(x\right)=\sum_{k=1}^{\infty}\frac{e^{i\left(\ln\left(k\right)x\right)}}{k^{2}} $$

e^ln(n)x = n^x, so:

$$ \mu_{d}\left(x\right)=\sum_{k=1}^{\infty}\frac{k^{ix}}{k^{2}} $$

Thus: