Talk:Golden ratio

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mvar template displaying variable name with odd line breaks in infobox on mobile

On my android mobile device, the title of the initial infobox takes up three lines because of what seems to be some odd behavior of the mvar template, adding line breaks within the parentheses for the variable name. Can anyone replicate this and figure out how to fix it? It displays perfectly on a computer browser... Willmskinner (talk) 03:11, 27 August 2024 (UTC)[reply]

Semi-protected edit request on 7 September 2024

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In the "Music" section of the article it talks about Debussy's "Reflets dans l'eau" which actually translates to "reflections in the water" instead of just "reflections in water" This is a very small change so I apologize for the inconvenience. 2A02:A58:8291:BC00:C82B:18DC:E9C4:DB34 (talk) 10:41, 7 September 2024 (UTC)[reply]

"reflections in water" is the correct translation: English and French grammars are not the same, and it is without "the" that the meaning is kept. However, I fixed the capitalization. D.Lazard (talk) 16:46, 7 September 2024 (UTC)[reply]

By the way, would reflets en eau also be correct? —Tamfang (talk) 02:44, 17 December 2024 (UTC)[reply]

Semi-protected edit request on 16 December 2024

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Above the definition of Rogers-Ramanujan function R(q) change "For , let" to "For example, let". 2A00:1028:8388:6446:AD6C:8E02:2C5A:FBD4 (talk) 16:48, 16 December 2024 (UTC)[reply]

 Fixed: This was a format error resulting that a fomula was not displayed before the comma. D.Lazard (talk) 18:21, 16 December 2024 (UTC)[reply]

Semi-protected edit request on 15 March 2025

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Please change:

The decimal expansion of the golden ratio ⁠${\displaystyle \varphi }$⁠^[1] has been calculated to an accuracy of ten trillion (⁠${\displaystyle \textstyle 1\times 10^{13}=10{,}000{,}000{,}000{,}000}$⁠) digits.^[2]

to:

The decimal expansion of the golden ratio ⁠${\displaystyle \varphi }$⁠^[1] has been calculated to an accuracy of twenty trillion (⁠${\displaystyle \textstyle 2\times 10^{13}=20{,}000{,}000{,}000{,}000}$⁠) digits.^[2]

as the referenced y-cruncher link indicates Jordan Ranous computed 20,000,000,000,000 digits of the golden ratio on November 27, 2023. Qqid (talk) 10:00, 15 March 2025 (UTC)[reply]

 Done Warriorglance(talk to me) 06:16, 16 March 2025 (UTC)[reply]

References

1. Cite error: The named reference a001622 was invoked but never defined (see the help page).
2. Cite error: The named reference ycruncher was invoked but never defined (see the help page).

Unreasoned reversion

The imaginary and negative values of the Golden Ratio is an important part in terms of number theory.

Imaginary Golden Ratio

Discussions on the negative value and the imaginary value of the Golden Ratio emerged in modern and contemporary mathematics. The negative value of the function ${\displaystyle {\frac {1-{\sqrt {5}}}{2}}}$ and the imaginary solution ${\displaystyle {\frac {1\pm i{\sqrt {3}}}{2}}}$ are both considered. With reltion to Euler's identity, it is proposed ${\displaystyle \Phi _{i}=e^{\pm {\frac {i\pi }{3}}}}$. ^[1][2] Quinhonk (talk) 13:09, 19 June 2025 (UTC)[reply]

I don't think there is anything in the 2014 reference that is not already in our article, Math Forums is not a reliable source by Wikipedia standards, and your "imaginary value" is not the golden ratio (it is a different complex number) and does not satisfy the usual definitions or equations for the golden ratio. —David Eppstein (talk) 18:23, 19 June 2025 (UTC)[reply]

By definition of the Golden Ratio, it never implicated its numerical value, but only the satisfaction of the functional predicate. The Golden Ratio is defined by the proportions, not the numerical solutions. Quinhonk (talk) 08:01, 20 June 2025 (UTC)[reply]

The golden ratio is the positive solution of the equation ${\displaystyle x^{2}-x-1=0}$. Both solutions of that equation are real. There is no imaginary solution and no solution with an imaginary component. If you want a root that is not real, you need to start with a different equation. There is no "imaginary golden ratio". Stepwise Continuous Dysfunction (talk) 15:39, 20 June 2025 (UTC)[reply]

There's no such thing as an "imaginary golden ratio", or an "imaginary value of the golden ratio". The golden ratio is a real number by definition. Stepwise Continuous Dysfunction (talk) 20:10, 19 June 2025 (UTC)[reply]

It is only expressed as a number, and the normatively used one is only one of the derivatives. Quinhonk (talk) 08:02, 20 June 2025 (UTC)[reply]

The complex number has an interesting analogy to the golden ratio (some identities correspond with a changed sign), but is not itself golden. With language suited to this caveat it might be worth mention. —Tamfang (talk) 02:06, 20 June 2025 (UTC)[reply]

I think you're on something. Indeed, from the definition of the Golden Ratio, many identities can be derived and look quite fun. But the positive solution is only one of the identities that was accepted, depending on the number line. Quinhonk (talk) 08:04, 20 June 2025 (UTC)[reply]

From the Chinese history of mathematical philosophy, Taoism never conceived the notion of negative numbers, and zero is literally reserved in the wording of Tao. Quinhonk (talk) 08:06, 20 June 2025 (UTC)[reply]

Your number is a sixth root of unity and a unit among the Eisenstein integers. There are many analogies between Eisenstein integers and the "golden integers" (and various other quadratic integers), between the numbers in the corresponding quadratic fields, etc. For example the norm of a number ⁠${\displaystyle a+b\omega }$⁠ where ⁠${\displaystyle \omega =\exp {\bigl (}{\tfrac {2}{3}}\pi i{\bigr )}}$⁠ is ⁠${\displaystyle a^{2}-ab+b^{2}}$⁠ and the norm of a number ⁠${\displaystyle a+b{\hat {\omega }}}$⁠ where ⁠${\displaystyle {\hat {\omega }}=-\omega ^{-1}=\exp {\bigl (}{\tfrac {1}{3}}\pi i{\bigr )}}$⁠ is ⁠${\displaystyle a^{2}+ab+b^{2}}$⁠ in the field ⁠${\displaystyle \mathbb {Q} (\omega )}$⁠. Analogously, the norm of a number ⁠${\displaystyle a+b\varphi }$⁠ is ⁠${\displaystyle a^{2}+ab-b^{2}}$⁠ and the norm of a number ⁠${\displaystyle a-b\varphi }$⁠ is ⁠${\displaystyle a^{2}-ab-b^{2}}$⁠ in the field ⁠${\displaystyle \mathbb {Q} {\bigl (}{\sqrt {5}}~\!{\bigr )}}$⁠. But these are distinct topics which should be largely discussed separately in their own articles. Conceivably a whole new article could be written about the field ⁠${\displaystyle \mathbb {Q} {\bigl (}{\sqrt {5}}~\!{\bigr )}}$⁠, and there might be places on Wikipedia where various direct comparisons between the behavior of different quadratic fields, their units, etc. is appropriate, but I don't think this article can grow much more about this topic without getting a bit out of scope. –jacobolus (t) 21:00, 20 June 2025 (UTC)[reply]

Update: Stepwise Continuous Dysfunction created an article about ⁠${\displaystyle \mathbb {Q} {\bigl (}{\sqrt {5}}~\!{\bigr )}}$⁠ at the title Golden field. –jacobolus (t) 23:38, 21 June 2025 (UTC)[reply]

Thanks for the explanations. It looks like it can be furthered to the Möbius Band again... I'll leave the Golden Ratio consensus as is then... Quinhonk (talk) 14:18, 22 June 2025 (UTC)[reply]

References

1. García-Caballero, Esther M.; Moreno, Samuel G.; Prophet, Michael P. (2014). "The golden ratio and Viéte's formula". Teaching Mathematics and Computer Science. 12 (1): 43–54. doi:10.5485/TMCS.2014.0351.
2. Agno. "Imaginary Golden Ratio". Math Forums.

Constructible Number

It looks like no one wants to change the status quo here with the revision (mainly I see that this page is entirely written in Python style but computing is not maths):

The infinite power root of ${\displaystyle (\phi ^{2}-\phi )^{\frac {1}{n}},n\in N}$ is bounded by ${\displaystyle i^{-{\frac {1}{2}}}}$.^[1]

References

1. Cao, Yang I. "Defining Zero According to the Definition of the Golden Ratio". ResearchGate.

Inconsistent expressions for constants involving the golden ratio

I wonder if there's a way to clarify that some of the different-looking constants which appear in § Geometry and § Other properties are equal (or related), or perhaps try to pick some more consistent conventions. Letting ⁠${\displaystyle {\overline {\varphi }}=-\varphi ^{-1}=1-\varphi }$⁠: $${\displaystyle {\begin{aligned}{\sqrt {5}}&=-1+2\varphi \\[5mu]\varphi {\sqrt {5}}&=\varphi (-1+2\varphi )=2+\varphi ,\\[8mu]\varphi ^{-1}{\sqrt {5}}&=(-1+\varphi )(-1+2\varphi )=3-\varphi =2+{\overline {\varphi }},\\[6mu]{\sqrt {\frac {2+\varphi }{5}}}&={\sqrt {\frac {\varphi }{\sqrt {5}}}}={\frac {1}{\sqrt {\varphi ^{-1}{\sqrt {5}}}}}={\frac {1}{\sqrt {3-\varphi }}}={\frac {1}{\sqrt {2+{\overline {\varphi }}}}},\\{\sqrt {\frac {3-\varphi }{5}}}&={\sqrt {\frac {\varphi ^{-1}}{\sqrt {5}}}}={\frac {1}{\sqrt {\varphi {\sqrt {5}}}}}={\frac {1}{\sqrt {2+\varphi }}},\\{\frac {15\varphi }{\sqrt {3-\varphi }}}&=15\varphi {\sqrt {\frac {\varphi }{\sqrt {5}}}}=3{\bigl (}\varphi {\sqrt {5}}~\!{\bigr )}^{3/2}=3(2+\varphi )^{3/2},\\{\frac {5\varphi ^{3}}{6-2\varphi }}&={\frac {1}{2}}{\frac {5\varphi ^{3}}{\varphi ^{-1}{\sqrt {5}}}}={\frac {1}{2}}\varphi ^{4}{\sqrt {5}}={\frac {1}{2}}(4+7\varphi ),\\{\frac {5}{6}}(1+\varphi )&={\frac {5}{6}}\varphi ^{2}={\frac {1}{6}}(\varphi {\sqrt {5}})^{2}.\end{aligned}}}$$

It's weird to have inconsistent examples such as:

$${\displaystyle R(e^{-2\pi })={\sqrt {\varphi {\sqrt {5}}}}-\varphi ,\quad R(-e^{-\pi })=\varphi ^{-1}-{\sqrt {2-\varphi ^{-1}}}}$$

When we could instead write these more consistently as one of: $${\displaystyle {\begin{aligned}&R(e^{-2\pi })=-\varphi +{\sqrt {\varphi {\sqrt {5}}}},\quad R(-e^{-\pi })=\varphi ^{-1}-{\sqrt {\varphi ^{-1}{\sqrt {5}}}}\\&R(e^{-2\pi })=-\varphi +{\sqrt {2+\varphi }},\quad R(-e^{-\pi })=-1+\varphi -{\sqrt {3-\varphi }}\end{aligned}}}$$

Cheers, –jacobolus (t) 12:57, 11 July 2025 (UTC)[reply]

IMO, every golden number must be written in a canonical form, either as ⁠${\displaystyle a+b\varphi }$⁠ or ⁠${\displaystyle a+b{\sqrt {5}}}$⁠, with ⁠${\displaystyle a}$⁠ and ⁠${\displaystyle b}$⁠ rational numbers. So, expressions such as ⁠${\displaystyle \varphi {\sqrt {5}}}$⁠ and ⁠${\displaystyle \varphi ^{-1}{\sqrt {5}}}$⁠ must definitively be avoided. For elements of a quadratic extension of the golden numbers, I would prefer ⁠${\displaystyle A+B{\sqrt {C}}}$⁠ where ⁠${\displaystyle A,B,C}$⁠ are golden numbers in canonical form and ⁠${\displaystyle C}$⁠ is square free for the prime factorizaation of golden numbers. So, for the last formulas, I would prefer $${\displaystyle R(e^{-2\pi })={\sqrt {2+\varphi }}-\varphi ,\quad R(-e^{-\pi })=\varphi -1-{\sqrt {3-\varphi }}}$$

(the colons around the formula, now fixed, seem a bug of Phabricator which, in automatic replies, inserts colons at the beginning of every line inside latex formulas).

The advantage of these canonical forms is to make immediately apparent which numbers are algebraic integers. D.Lazard (talk) 15:10, 11 July 2025 (UTC)[reply]

One alternative would be to give specific symbolic names to a few of these numbers someplace, e.g. ⁠${\displaystyle \sigma ={\sqrt {\varphi {\sqrt {5}}}}={\sqrt {2+\varphi }}}$⁠ and ⁠${\displaystyle {\overline {\sigma }}={\sqrt {\varphi ^{-1}{\sqrt {5}}}}={\sqrt {3-\varphi }}}$⁠, and then write some of the other constants in those terms, like: ⁠${\displaystyle {\tfrac {1}{6}}\sigma ^{4}}$⁠ instead of ⁠${\displaystyle {\tfrac {5}{6}}(1+\varphi )}$⁠; ⁠${\displaystyle 3\sigma ^{3}}$⁠ instead of ⁠${\displaystyle {\tfrac {15\varphi }{\sqrt {3-\varphi }}}}$⁠; and ⁠${\displaystyle \sigma ^{-1}}$⁠ instead of ⁠${\displaystyle {\sqrt {\tfrac {3-\varphi }{5}}}}$⁠ or ⁠${\displaystyle {\tfrac {1}{\sqrt {2+\varphi }}}}$⁠. These particular constants show up a lot because ⁠${\displaystyle {\overline {\sigma }}}$⁠ is the ratio between the side length and circumradius of a regular pentagon and ⁠${\displaystyle \sigma }$⁠ is the ratio between the diagonal length and circumradius. –jacobolus (t) 15:46, 11 July 2025 (UTC)[reply]

I do not like introducing too many notations.

However, I should moderate my above comment: in some cases, especially for radicands, the prime factorization is a canonical form that is more informative. This leads writing ⁠${\displaystyle {\sqrt {2+\varphi }}={\sqrt {5\varphi }}}$⁠ and ⁠${\displaystyle {\sqrt {3-\varphi }}={\sqrt {-5{\overline {\varphi }}}}}$⁠. This gives a much more symmetric formula for the above example:

${\displaystyle R(e^{-2\pi })={\sqrt {5\varphi }}-\varphi ,\quad R(-e^{-\pi })={\overline {\varphi }}-{\sqrt {-5{\overline {\varphi }}}}}$

(except for possible sign errors by myself). The advantage is that the natural symmetry is not hidden. D.Lazard (talk) 17:53, 11 July 2025 (UTC)[reply]

(Your 5s are missing sqrts, so it should be something like: $${\displaystyle {\begin{aligned}&R(e^{-2\pi })=-\varphi +{\sqrt {\varphi {\sqrt {5}}}},\quad R(-e^{-\pi })=-{\overline {\varphi }}-{\sqrt {-{\overline {\varphi }}{\sqrt {5}}}}\quad {\text{or}}\\&R(e^{-2\pi })=-\varphi +{\sqrt {2+\varphi }},\quad R(-e^{-\pi })=-{\overline {\varphi }}-{\sqrt {2+{\overline {\varphi }}}}\end{aligned}}}$$ ... but more or less; and ⁠${\displaystyle {\overline {\varphi {\sqrt {5}}}}=-{\overline {\varphi }}{\sqrt {5}}}$⁠. Also I realize my ⁠${\displaystyle {\overline {\sigma }}}$⁠ symbol is confusing, as it isn’t conjugation in ⁠${\displaystyle \mathbb {Q} {\bigl (}{\sqrt {5}}~\!{\bigr )}}$⁠; but you get the idea.) –jacobolus (t) 21:32, 11 July 2025 (UTC)[reply]