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

From Xenharmonic Wiki
Jump to navigation Jump to search
No edit summary
No edit summary
Line 5: Line 5:
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>$$ \mu_{b} \left( x \right) = \sum_{k=1}^{\infty}\frac{\cos\left(\log_{2}\left(k\right)\tau x\right)}{k^{2}} $$</nowiki>
 
 
Let's clean up the function by removing the scale factors on x:
 
<nowiki>$$ \mu_{c} \left( x \right) = \sum_{k=1}^{\infty}\frac{\cos x}{k^{2}} $$</nowiki>

Revision as of 01:46, 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:

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


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

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