The Riemann zeta function and tuning/Vector's derivation: Difference between revisions
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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>$$ \mu_{\cos} \left( x \right) \sum_{k=1}^{\infty}\frac{\cos\left(\log_{2}\left(k\right)\tau x\right)}{k^{2}} $$</nowiki> | <nowiki>$$ \mu_{\cos} \left( x \right) = \sum_{k=1}^{\infty}\frac{\cos\left(\log_{2}\left(k\right)\tau x\right)}{k^{2}} $$</nowiki> | ||
Revision as of 01:45, 10 April 2025
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:
$$ \mu_{\cos} \left( x \right) = \sum_{k=1}^{\infty}\frac{\cos\left(\log_{2}\left(k\right)\tau x\right)}{k^{2}} $$