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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:
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