Riemann zeta function/Vector's derivation - Revision history

Fredg999: Fredg999 moved page The Riemann zeta function and tuning/Vector's derivation to Riemann zeta function/Vector's derivation: (Fake) parent page was renamed

2026-07-16T19:25:26Z

Fredg999 moved page The Riemann zeta function and tuning/Vector's derivation to Riemann zeta function/Vector's derivation: (Fake) parent page was renamed

← Older revision	Revision as of 19:25, 16 July 2026
(No difference)

ArrowHead294 at 13:39, 11 April 2025

2025-04-11T13:39:26Z

← Older revision		Revision as of 13:39, 11 April 2025
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	{{breadcrumb}}		{{breadcrumb}}{{texops}}
	We start with the generalized mu function, which sums up the relative error on all integer harmonics weighted by an inverse power of the harmonic:		We start with the generalized mu function, which sums up the relative error on all integer harmonics weighted by an inverse power of the harmonic:

	[https://www.desmos.com/calculator/4zcynoue8s <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>]		[https://www.desmos.com/calculator/4zcynoue8s <nowiki>$$ \mu \left(\sigma, x \right) = \sum_{k=1}^{\infty} \frac{\abs{ \operatorname{mod} \left( 2\log_{2} \left( k \right) x, 2 \right) - 1 }}{k^{\sigma}} $$</nowiki>]

	Now, this is a rather annoying function to work with for math reasons, 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 a rather annoying function to work with for math reasons, 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:

FloraC: External links should be made known!

2025-04-11T13:23:40Z

External links should be made known!

ArrowHead294 at 12:23, 11 April 2025

2025-04-11T12:23:52Z

← Older revision		Revision as of 12:23, 11 April 2025
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	{{ext\|https://www.desmos.com/calculator/deafikrhvg\|<nowiki>$$ \sum_{k=1}^{\infty}\frac{\cos\left(\log_{2}\left(k\right)\tau x\right)}{k^{\sigma}} $$</nowiki>}}		{{ext\|https://www.desmos.com/calculator/deafikrhvg\|<nowiki>$$ \sum_{k=1}^{\infty}\frac{\cos\left(\log_{2}\left(k\right)\tau x\right)}{k^{\sigma}} $$</nowiki>}}

	Let's clean up the function by removing the scale factors on x. This just scales the function's inputs from EDO to [[Zetave\|EDZ]], and these can be added back later to go back to EDO.		Let's clean up the function by removing the scale factors on ''x''. This just scales the function's inputs from EDO to [[Zetave\|EDZ]], and these can be added back later to go back to EDO.

	{{ext\|https://www.desmos.com/calculator/26ypbwbglg\|<nowiki>$$ \sum_{k=1}^{\infty}\frac{\cos (\ln\left(k\right)x)}{k^{\sigma}} $$</nowiki>}}		{{ext\|https://www.desmos.com/calculator/26ypbwbglg\|<nowiki>$$ \sum_{k=1}^{\infty}\frac{\cos (\ln\left(k\right)x)}{k^{\sigma}} $$</nowiki>}}

ArrowHead294 at 12:23, 11 April 2025

2025-04-11T12:23:33Z

← Older revision		Revision as of 12:23, 11 April 2025
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	We start with the generalized mu function, which sums up the relative error on all integer harmonics weighted by an inverse power of the harmonic:		We start with the generalized mu function, which sums up the relative error on all integer harmonics weighted by an inverse power of the harmonic:

	[https://www.desmos.com/calculator/4zcynoue8s <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>]		{{ext\|https://www.desmos.com/calculator/4zcynoue8s\|<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 a rather annoying function to work with for math reasons, 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 a rather annoying function to work with for math reasons, 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:

	[https://www.desmos.com/calculator/deafikrhvg <nowiki>$$ \sum_{k=1}^{\infty}\frac{\cos\left(\log_{2}\left(k\right)\tau x\right)}{k^{\sigma}} $$</nowiki>]		{{ext\|https://www.desmos.com/calculator/deafikrhvg\|<nowiki>$$ \sum_{k=1}^{\infty}\frac{\cos\left(\log_{2}\left(k\right)\tau x\right)}{k^{\sigma}} $$</nowiki>}}





	Let's clean up the function by removing the scale factors on x. This just scales the function's inputs from EDO to [[Zetave\|EDZ]], and these can be added back later to go back to EDO.		Let's clean up the function by removing the scale factors on x. This just scales the function's inputs from EDO to [[Zetave\|EDZ]], and these can be added back later to go back to EDO.

	[https://www.desmos.com/calculator/26ypbwbglg <nowiki>$$ \sum_{k=1}^{\infty}\frac{\cos (\ln\left(k\right)x)}{k^{\sigma}} $$</nowiki>]		{{ext\|https://www.desmos.com/calculator/26ypbwbglg\|<nowiki>$$ \sum_{k=1}^{\infty}\frac{\cos (\ln\left(k\right)x)}{k^{\sigma}} $$</nowiki>}}





	By the complex exponential theorem, which relates trigonometric functions and the exponential function, we know that		By the complex exponential theorem, which relates trigonometric functions and the exponential function, we know that
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	$$ e^{ix}=\cos\left(x\right)+i\sin\left(x\right) $$		$$ e^{ix}=\cos\left(x\right)+i\sin\left(x\right) $$

	where i is the square root of -1, and e is the natural exponential constant.		where ''i'' is the square root of −1, and ''e'' is the natural exponential constant.

	~~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>) - but since we're only doing multiplication and addition and 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>$$ \mu_{c}\left(\sigma, x\right)=\sum_{k=1}^{\infty}\frac{e^{i\left(\ln\left(k\right)x\right)}}{k^{\sigma}} $$</nowiki>]~~





	~~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>$$ \sum_~~{~~k=1}^~~{~~\infty~~}~~\frac{k^{ix~~}~~}{k^{\sigma}} $$</nowiki>]~~		''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 {{nowrap\|''a'' + ''bi''}}.

	~~[https:~~/~~/www.desmos~~.~~com/calculator/6388kalfmq $$ \sum_{k=1}^{\infty}k^{ix}k^{-\sigma} $$]~~		With this knowledge, cos(''x'') can be rewritten as Re(''e''<sup>''ix''</sup>)—but since we're only doing multiplication and addition and this is the only place complex numbers appear, we can just ignore the Re() and add it back later.

	[https://www.desmos.com/calculator/~~l3q2dtd6xn~~ $$ \sum_{k=1}^{\infty}k^{-\sigma~~+ix~~} $$]		{{ext\|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>}}

	~~-σ+ix is just a complex number~~, ~~which we may write as -s~~:		Since {{nowrap\|''e''<sup>ln(''n'')''x''</sup> {{=}} ''n''<sup>''x''</sup>}}, the exponentials and logarithms cancel each other out (i.e. {{nowrap\|''e''<sup>ln(''n'')</sup> {{=}} ''n''}}), so:

	[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.~~		{{ext\|https://www.desmos.com/calculator/f4ojwn0an4\|<nowiki>$$ \sum_{k=1}^{\infty}\frac{k^{ix}}{k^{\sigma}} $$</nowiki>}}

	~~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.~~		{{ext\|https://www.desmos.com/calculator/6388kalfmq\|$$ \sum_{k=1}^{\infty}k^{ix}k^{-\sigma} $$}}

			{{ext\|https://www.desmos.com/calculator/l3q2dtd6xn\|$$ \sum_{k=1}^{\infty}k^{-\sigma+ix} $$}}

			{{nowrap\|−σ + ''ix''}} is just a complex number, which we may write as −''s'':

			{{ext\|https://www.desmos.com/calculator/esbdlxdoui\|$$ \mu_{d}\left(s\right)=\sum_{k=1}^{\infty}k^{-s} $$}}

			where, for {{nowrap\|''s'' {{=}} σ − ''ix''}}, {{nowrap\|Re(μ<sub>''d''</sub>(''s'')) {{=}} μ<sub>''c''</sub>(σ, ''x'')}}, our badness function.

			μ<sub>''d''</sub> is a mathematical function called the Riemann zeta function, so {{nowrap\|μ<sub>''d''(s) {{=}} ζ(''s'')}}, and re-adding the Re() function gives Re(ζ(''s'')) with {{nowrap\|''s'' {{=}} σ − ''ix''}}; ''x'' is the equal division and σ is the weight.

	Summary of the derivation:		Summary of the derivation:

ArrowHead294 at 12:13, 11 April 2025

2025-04-11T12:13:20Z

← Older revision		Revision as of 12:13, 11 April 2025
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			{{breadcrumb}}
	We start with the generalized mu function, which sums up the relative error on all integer harmonics weighted by an inverse power of the harmonic:		We start with the generalized mu function, which sums up the relative error on all integer harmonics weighted by an inverse power of the harmonic:

VectorGraphics at 08:57, 11 April 2025

2025-04-11T08:57:01Z

← Older revision		Revision as of 08:57, 11 April 2025
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	Now, this is a rather annoying function to work with for math reasons, 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 a rather annoying function to work with for math reasons, 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:

	[https://www.desmos.com/calculator/deafikrhvg <nowiki>$$ ~~\mu_{a} \left(\sigma, x \right) =~~ \sum_{k=1}^{\infty}\frac{\cos\left(\log_{2}\left(k\right)\tau x\right)}{k^{\sigma}} $$</nowiki>]		[https://www.desmos.com/calculator/deafikrhvg <nowiki>$$ \sum_{k=1}^{\infty}\frac{\cos\left(\log_{2}\left(k\right)\tau x\right)}{k^{\sigma}} $$</nowiki>]


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	Let's clean up the function by removing the scale factors on x. This just scales the function's inputs from EDO to [[Zetave\|EDZ]], and these can be added back later to go back to EDO.		Let's clean up the function by removing the scale factors on x. This just scales the function's inputs from EDO to [[Zetave\|EDZ]], and these can be added back later to go back to EDO.

	[https://www.desmos.com/calculator/26ypbwbglg <nowiki>$$ ~~\mu_{b} \left(\sigma, x \right) =~~ \sum_{k=1}^{\infty}\frac{\cos (\ln\left(k\right)x)}{k^{\sigma}} $$</nowiki>]		[https://www.desmos.com/calculator/26ypbwbglg <nowiki>$$ \sum_{k=1}^{\infty}\frac{\cos (\ln\left(k\right)x)}{k^{\sigma}} $$</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:		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>$$ ~~\mu_{c}\left(\sigma, x\right)=~~\sum_{k=1}^{\infty}\frac{k^{ix}}{k^{\sigma}} $$</nowiki>]		[https://www.desmos.com/calculator/f4ojwn0an4 <nowiki>$$ \sum_{k=1}^{\infty}\frac{k^{ix}}{k^{\sigma}} $$</nowiki>]

	~~Thus~~:		[https://www.desmos.com/calculator/6388kalfmq $$ \sum_{k=1}^{\infty}k^{ix}k^{-\sigma} $$]

	~~[https://www.desmos.com/calculator/6388kalfmq $$ \mu_{c}\left(\sigma, x\right)=\sum_{k=1}^{\infty}k^{ix}k^{-\sigma} $$]~~		[https://www.desmos.com/calculator/l3q2dtd6xn $$ \sum_{k=1}^{\infty}k^{-\sigma+ix} $$]

	[https://www.desmos.com/calculator/l3q2dtd6xn $$ ~~\mu_{c}\left(\sigma, x\right)=~~\sum_{k=1}^{\infty}k^{-\sigma+ix} $$]

	-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:

VectorGraphics at 08:56, 11 April 2025

2025-04-11T08:56:05Z

← Older revision		Revision as of 08:56, 11 April 2025
Line 5:		Line 5:
	Now, this is a rather annoying function to work with for math reasons, 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 a rather annoying function to work with for math reasons, 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:

	[https://www.desmos.com/calculator/deafikrhvg <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>]		[https://www.desmos.com/calculator/deafikrhvg <nowiki>$$ \mu_{a} \left(\sigma, x \right) = \sum_{k=1}^{\infty}\frac{\cos\left(\log_{2}\left(k\right)\tau x\right)}{k^{\sigma}} $$</nowiki>]


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	Let's clean up the function by removing the scale factors on x. This just scales the function's inputs from EDO to [[Zetave\|EDZ]], and these can be added back later to go back to EDO.		Let's clean up the function by removing the scale factors on x. This just scales the function's inputs from EDO to [[Zetave\|EDZ]], and these can be added back later to go back to EDO.

	[https://www.desmos.com/calculator/26ypbwbglg <nowiki>$$ \mu_{c} \left(\sigma, x \right) = \sum_{k=1}^{\infty}\frac{\cos (\ln\left(k\right)x)}{k^{\sigma}} $$</nowiki>]		[https://www.desmos.com/calculator/26ypbwbglg <nowiki>$$ \mu_{b} \left(\sigma, x \right) = \sum_{k=1}^{\infty}\frac{\cos (\ln\left(k\right)x)}{k^{\sigma}} $$</nowiki>]

VectorGraphics at 08:55, 11 April 2025

2025-04-11T08:55:31Z

← Older revision		Revision as of 08:55, 11 April 2025
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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.		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>) - but since ~~(among other things)~~ this is the only place complex numbers appear, we can just ignore the Re() and add it back later.		With this knowledge, cos(x) can be rewritten as Re(e<sup>ix</sup>) - but since we're only doing multiplication and addition and 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>$$ \mu_{c}\left(\sigma, x\right)=\sum_{k=1}^{\infty}\frac{e^{i\left(\ln\left(k\right)x\right)}}{k^{\sigma}} $$</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>]

VectorGraphics at 08:54, 11 April 2025

2025-04-11T08:54:44Z

← Older revision		Revision as of 08:54, 11 April 2025
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	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.		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>$$ \mu_{c}\left(\sigma, x\right)=\sum_{k=1}^{\infty}\frac{e^{i\left(\ln\left(k\right)x\right)}{k^{\sigma}} $$</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>]

← Older revision		Revision as of 12:23, 11 April 2025
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	{{ext\|https://www.desmos.com/calculator/deafikrhvg\|<nowiki>$$ \sum_{k=1}^{\infty}\frac{\cos\left(\log_{2}\left(k\right)\tau x\right)}{k^{\sigma}} $$</nowiki>}}		{{ext\|https://www.desmos.com/calculator/deafikrhvg\|<nowiki>$$ \sum_{k=1}^{\infty}\frac{\cos\left(\log_{2}\left(k\right)\tau x\right)}{k^{\sigma}} $$</nowiki>}}

	Let's clean up the function by removing the scale factors on x. This just scales the function's inputs from EDO to [[Zetave\|EDZ]], and these can be added back later to go back to EDO.		Let's clean up the function by removing the scale factors on ''x''. This just scales the function's inputs from EDO to [[Zetave\|EDZ]], and these can be added back later to go back to EDO.

	{{ext\|https://www.desmos.com/calculator/26ypbwbglg\|<nowiki>$$ \sum_{k=1}^{\infty}\frac{\cos (\ln\left(k\right)x)}{k^{\sigma}} $$</nowiki>}}		{{ext\|https://www.desmos.com/calculator/26ypbwbglg\|<nowiki>$$ \sum_{k=1}^{\infty}\frac{\cos (\ln\left(k\right)x)}{k^{\sigma}} $$</nowiki>}}

← Older revision		Revision as of 13:23, 11 April 2025
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	We start with the generalized mu function, which sums up the relative error on all integer harmonics weighted by an inverse power of the harmonic:		We start with the generalized mu function, which sums up the relative error on all integer harmonics weighted by an inverse power of the harmonic:

	~~{{ext\|~~https://www.desmos.com/calculator/4zcynoue8s\|<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>}}		[https://www.desmos.com/calculator/4zcynoue8s <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 a rather annoying function to work with for math reasons, 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 a rather annoying function to work with for math reasons, 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:

	~~{{ext\|~~https://www.desmos.com/calculator/deafikrhvg\|<nowiki>$$ \sum_{k=1}^{\infty}\frac{\cos\left(\log_{2}\left(k\right)\tau x\right)}{k^{\sigma}} $$</nowiki>}}		[https://www.desmos.com/calculator/deafikrhvg <nowiki>$$ \sum_{k=1}^{\infty}\frac{\cos\left(\log_{2}\left(k\right)\tau x\right)}{k^{\sigma}} $$</nowiki>]

	Let's clean up the function by removing the scale factors on ''x''. This just scales the function's inputs from EDO to [[Zetave\|EDZ]], and these can be added back later to go back to EDO.		Let's clean up the function by removing the scale factors on ''x''. This just scales the function's inputs from EDO to [[Zetave\|EDZ]], and these can be added back later to go back to EDO.

	~~{{ext\|~~https://www.desmos.com/calculator/26ypbwbglg\|<nowiki>$$ \sum_{k=1}^{\infty}\frac{\cos (\ln\left(k\right)x)}{k^{\sigma}} $$</nowiki>}}		[https://www.desmos.com/calculator/26ypbwbglg <nowiki>$$ \sum_{k=1}^{\infty}\frac{\cos (\ln\left(k\right)x)}{k^{\sigma}} $$</nowiki>]

	By the complex exponential theorem, which relates trigonometric functions and the exponential function, we know that		By the complex exponential theorem, which relates trigonometric functions and the exponential function, we know that
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	With this knowledge, cos(''x'') can be rewritten as Re(''e''<sup>''ix''</sup>)—but since we're only doing multiplication and addition and this is the only place complex numbers appear, we can just ignore the Re() and add it back later.		With this knowledge, cos(''x'') can be rewritten as Re(''e''<sup>''ix''</sup>)—but since we're only doing multiplication and addition and this is the only place complex numbers appear, we can just ignore the Re() and add it back later.

	~~{{ext\|~~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>}}		[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>]

	Since {{nowrap\|''e''<sup>ln(''n'')''x''</sup> {{=}} ''n''<sup>''x''</sup>}}, the exponentials and logarithms cancel each other out (i.e. {{nowrap\|''e''<sup>ln(''n'')</sup> {{=}} ''n''}}), so:		Since {{nowrap\|''e''<sup>ln(''n'')''x''</sup> {{=}} ''n''<sup>''x''</sup>}}, the exponentials and logarithms cancel each other out (i.e. {{nowrap\|''e''<sup>ln(''n'')</sup> {{=}} ''n''}}), so:

	~~{{ext\|~~https://www.desmos.com/calculator/f4ojwn0an4\|<nowiki>$$ \sum_{k=1}^{\infty}\frac{k^{ix}}{k^{\sigma}} $$</nowiki>}}		[https://www.desmos.com/calculator/f4ojwn0an4 <nowiki>$$ \sum_{k=1}^{\infty}\frac{k^{ix}}{k^{\sigma}} $$</nowiki>]

	~~{{ext\|~~https://www.desmos.com/calculator/6388kalfmq\|$$ \sum_{k=1}^{\infty}k^{ix}k^{-\sigma} $$}}		[https://www.desmos.com/calculator/6388kalfmq $$ \sum_{k=1}^{\infty}k^{ix}k^{-\sigma} $$]

	~~{{ext\|~~https://www.desmos.com/calculator/l3q2dtd6xn\|$$ \sum_{k=1}^{\infty}k^{-\sigma+ix} $$}}		[https://www.desmos.com/calculator/l3q2dtd6xn $$ \sum_{k=1}^{\infty}k^{-\sigma+ix} $$]

	{{nowrap\|−σ + ''ix''}} is just a complex number, which we may write as −''s'':		{{nowrap\|−σ + ''ix''}} is just a complex number, which we may write as −''s'':

	~~{{ext\|~~https://www.desmos.com/calculator/esbdlxdoui\|$$ \mu_{d}\left(s\right)=\sum_{k=1}^{\infty}k^{-s} $$}}		[https://www.desmos.com/calculator/esbdlxdoui $$ \mu_{d}\left(s\right)=\sum_{k=1}^{\infty}k^{-s} $$]

	where, for {{nowrap\|''s'' {{=}} σ − ''ix''}}, {{nowrap\|Re(μ<sub>''d''</sub>(''s'')) {{=}} μ<sub>''c''</sub>(σ, ''x'')}}, our badness function.		where, for {{nowrap\|''s'' {{=}} σ − ''ix''}}, {{nowrap\|Re(μ<sub>''d''</sub>(''s'')) {{=}} μ<sub>''c''</sub>(σ, ''x'')}}, our badness function.