The Riemann zeta function and tuning/Vector's derivation: Difference between revisions
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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>, because exponentials and logarithms cancel each other out (i.e. e<sup>ln(n)</sup> = n), so: | ||
[https://www.desmos.com/calculator/f4ojwn0an4 <nowiki>$$ \mu_{ | [https://www.desmos.com/calculator/f4ojwn0an4 <nowiki>$$ \mu_{c}\left(\sigma, x\right)=\sum_{k=1}^{\infty}\frac{\operatorname{Re}\left(k^{ix}\right)}{k^{\sigma}} $$</nowiki>] | ||
Thus: | Thus: | ||
[https://www.desmos.com/calculator/6388kalfmq $$ \mu_{ | [https://www.desmos.com/calculator/6388kalfmq $$ \mu_{c}\left(\sigma, x\right)=\sum_{k=1}^{\infty}\operatorname{Re}\left(k^{ix}\right)k^{-\sigma} $$] | ||
so | k<sup>-σ</sup> is a real number, so its real part is equal to itself. Thus, by the rules of complex addition, we can simplify this as follows: | ||
[https://www.desmos.com/calculator/l3q2dtd6xn $$ \mu_{ | [https://www.desmos.com/calculator/l3q2dtd6xn $$ \mu_{c}\left(\sigma, x\right)=\sum_{k=1}^{\infty}\operatorname{Re}\left(k^{-\sigma+ix}\right) $$] | ||
-σ+ix is just a complex number, which we may write as -s: | -σ+ix is just a complex number, which we may write as -s: | ||
[https://www.desmos.com/calculator/esbdlxdoui $$ \mu_{ | [https://www.desmos.com/calculator/esbdlxdoui $$ \mu_{d}\left(s\right)=\sum_{k=1}^{\infty}\operatorname{Re}\left(k^{-s}\right) $$] | ||
By the rules of complex addition, the real part of a sum is the same as a sum of real parts, so we can make the following change: | |||
[https://www.desmos.com/calculator/esbdlxdoui $$\mu_{e}\left(s\right)=\sum_{k=1}^{\infty}k^{-s} $$]where Re(μ<sub>e</sub>(s)) = μ<sub>d</sub>(s), our badness function. | |||
μ<sub>e</sub> is 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. | |||