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

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


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:
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>$$ \mu_{b} \left(\sigma, x \right) = \sum_{k=1}^{\infty}\frac{\cos\left(\log_{2}\left(k\right)\tau x\right)}{k^{\sigma}} $$</nowiki>]
[https://www.desmos.com/calculator/deafikrhvg <nowiki>$$ \mu_{b} \left(\sigma, x \right) = \sum_{k=1}^{\infty}\frac{\cos\left(\log_{2}\left(k\right)\tau x\right)}{k^{\sigma}} $$</nowiki>]
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Let's clean up the function by removing the scale factors on x:
 
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>$$ \mu_{c} \left(\sigma, x \right) = \sum_{k=1}^{\infty}\frac{\cos (\ln\left(k\right)x)}{k^{\sigma}} $$</nowiki>]
[https://www.desmos.com/calculator/26ypbwbglg <nowiki>$$ \mu_{c} \left(\sigma, x \right) = \sum_{k=1}^{\infty}\frac{\cos (\ln\left(k\right)x)}{k^{\sigma}} $$</nowiki>]
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By the complex exponential theorem, we know that
 
By the complex exponential theorem, which relates trigonometric functions and the exponential function, we know that


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


so that cos(x) can be rewritten as Re(e<sup>ix</sup>).
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<sup>ix</sup>).


[https://www.desmos.com/calculator/e7wn17tzjf <nowiki>$$ \mu_{c}\left(\sigma, x\right)=\sum_{k=1}^{\infty}\frac{\operatorname{Re}\left(e^{i\left(\ln\left(k\right)x\right)}\right)}{k^{\sigma}} $$</nowiki>]
[https://www.desmos.com/calculator/e7wn17tzjf <nowiki>$$ \mu_{c}\left(\sigma, x\right)=\sum_{k=1}^{\infty}\frac{\operatorname{Re}\left(e^{i\left(\ln\left(k\right)x\right)}\right)}{k^{\sigma}} $$</nowiki>]


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.


[https://www.desmos.com/calculator/92wnbnh9x7 <nowiki>$$ \mu_{d}\left(\sigma, x\right)=\sum_{k=1}^{\infty}\frac{e^{i\left(\ln\left(k\right)x\right)}}{k^{\sigma}} $$</nowiki>]






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


e<sup>ln(n)x</sup> = n<sup>x</sup>, so:
[https://www.desmos.com/calculator/f4ojwn0an4 <nowiki>$$ \mu_{d}\left(\sigma, x\right)=\sum_{k=1}^{\infty}\frac{\operatorname{Re}\left(k^{ix}\right)}{k^{\sigma}} $$</nowiki>]
 
[https://www.desmos.com/calculator/f4ojwn0an4 <nowiki>$$ \mu_{d}\left(\sigma, x\right)=\sum_{k=1}^{\infty}\frac{k^{ix}}{k^{\sigma}} $$</nowiki>]


Thus:
Thus:
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But this is the definition of the Riemann zeta function, so μ<sub>e</sub>(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.
This is the definition of a mathematical function called the Riemann zeta function, so μ<sub>e</sub>(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:

Revision as of 08:32, 11 April 2025

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(eix).

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



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

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

Thus:

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

so

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

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

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


This is the definition of a mathematical function called the Riemann zeta function, so μe(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:

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