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

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 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>).
+With this knowledge, cos(x) can be rewritten as Re(e<sup>ix</sup>) - but since (among other things) this is the only place complex numbers appear, we can just ignore the Re() and add it back later.
-[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>$$ \mu_{c}\left(\sigma, x\right)=\sum_{k=1}^{\infty}\frac{e^{i\left(\ln\left(k\right)x\right)}{k^{\sigma}} $$</nowiki>]
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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:
-[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>]
+[https://www.desmos.com/calculator/f4ojwn0an4 <nowiki>$$ \mu_{c}\left(\sigma, x\right)=\sum_{k=1}^{\infty}\frac{k^{ix}}{k^{\sigma}} $$</nowiki>]
 Thus:
-[https://www.desmos.com/calculator/6388kalfmq $$ \mu_{c}\left(\sigma, x\right)=\sum_{k=1}^{\infty}\operatorname{Re}\left(k^{ix}\right)k^{-\sigma} $$]
+[https://www.desmos.com/calculator/6388kalfmq $$ \mu_{c}\left(\sigma, x\right)=\sum_{k=1}^{\infty}k^{ix}k^{-\sigma} $$]
-k<sup>-σ</sup> is a real number, so its real part is equal to itself. Thus we can simplify this as follows:
+[https://www.desmos.com/calculator/l3q2dtd6xn $$ \mu_{c}\left(\sigma, x\right)=\sum_{k=1}^{\infty}k^{-\sigma+ix} $$]
-[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:
-[https://www.desmos.com/calculator/esbdlxdoui $$ \mu_{d}\left(s\right)=\sum_{k=1}^{\infty}\operatorname{Re}\left(k^{-s}\right) $$]
+[https://www.desmos.com/calculator/esbdlxdoui $$ \mu_{d}\left(s\right)=\sum_{k=1}^{\infty}k^{-s} $$] where, for s = σ-ix, Re(μ<sub>d</sub>(s)) = μ<sub>c</sub>(σ,x), our badness function.
-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.
+μd is a mathematical function called the Riemann zeta function, so μd(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.
-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.