The Riemann zeta function and tuning/Vector's derivation: Difference between revisions
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We start with the mu function: | We start with the generalized mu function: | ||
<nowiki>$$ \mu \left( x \right) = \sum_{k=1}^{\infty} \frac{\operatorname{abs} \left( \operatorname{mod} \left( 2\log_{2} \left( k \right) x, 2 \right) - 1 \right)}{k^{ | <nowiki>$$ \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}} $$</nowiki> | ||
Now, this is nowhere differentiable, so it might be useful to replace the "zigzag" that we use as our error function with a "smoother" alternative. The most obvious answer is cosine: | Now, this is nowhere differentiable, so it might be useful to replace the "zigzag" that we use as our error function with a "smoother" alternative. The most obvious answer is cosine: | ||
<nowiki>$$ \mu_{b} \left( x \right) = \sum_{k=1}^{\infty}\frac{\cos\left(\log_{2}\left(k\right)\tau x\right)}{k^{ | <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> | ||
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: | ||
<nowiki>$$ \mu_{c} \left( x \right) = \sum_{k=1}^{\infty}\frac{\cos (\ln\left(k\right)x)}{k^{ | <nowiki>$$ \mu_{c} \left(\sigma, x \right) = \sum_{k=1}^{\infty}\frac{\cos (\ln\left(k\right)x)}{k^{\sigma}} $$</nowiki> | ||
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so that cos(x) can be rewritten as Re(e<sup>ix</sup>). | so that cos(x) can be rewritten as Re(e<sup>ix</sup>). | ||
<nowiki>$$ \mu_{c}\left(x\right)=\sum_{k=1}^{\infty}\frac{\operatorname{Re}\left(e^{i\left(\ln\left(k\right)x\right)}\right)}{k^{ | <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. | 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. | ||
<nowiki>$$ \mu_{d}\left(x\right)=\sum_{k=1}^{\infty}\frac{e^{i\left(\ln\left(k\right)x\right)}}{k^{ | <nowiki>$$ \mu_{d}\left(\sigma, x\right)=\sum_{k=1}^{\infty}\frac{e^{i\left(\ln\left(k\right)x\right)}}{k^{\sigma}} $$</nowiki> | ||
e<sup>ln(n)x</sup> = n<sup>x</sup>, so: | e<sup>ln(n)x</sup> = n<sup>x</sup>, so: | ||
<nowiki>$$ \mu_{d}\left(x\right)=\sum_{k=1}^{\infty}\frac{k^{ix}}{k^{ | <nowiki>$$ \mu_{d}\left(\sigma, x\right)=\sum_{k=1}^{\infty}\frac{k^{ix}}{k^{\sigma}} $$</nowiki> | ||
Thus: | Thus: | ||
$$ \mu_{d}\left(\sigma, x\right)=\sum_{k=1}^{\infty}k^{ix}{k^{-\sigma} $$ | |||
$$ \mu_{d}\left(\sigma, x\right)=\sum_{k=1}^{\infty}k^{-\sigma+ix} $$ | |||
-σ+ix is just a complex number, which we may write as -s: | |||
$$ \mu_{e}\left(s\right)=\sum_{k=1}^{\infty}k^{-s} $$ | |||
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. | |||