Talk:Golden ratio/Archive 9

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Le Corbusier Modulor quotation

I can't find anything very closely matching the quotation in this article in Le Corbusier's The Modulor, a translation of which can be found at the internet archive (along with Modulor II) here: https://archive.org/details/moduloriii00leco/ even if I try pretty loose keyword searches under the assumption that the French original was being translated differently. Can anyone find the relevant passage (in English or French) or make the citation more precise? –jacobolus (t) 21:56, 29 November 2022 (UTC)

Octagrammum Mysticum and the Golden Cross-Ratio

(Split to other section.) —Nils von Barth (nbarth) (talk) 03:41, 27 January 2023 (UTC)

@Nbarth: While we are here, you may be interested in (and we may want to mention in this article): Evans, L. S., & Rigby, J. F. (2002). Octagrammum Mysticum and the Golden Cross-Ratio. The Mathematical Gazette, 86(505), 35. doi:10.2307/3621571, jstor:3621571. –jacobolus (t) 05:05, 26 January 2023 (UTC)

That's a pretty result, but a bit "tangential" to the content of golden ratio or cross-ratio. Maybe worth adding to Pascal's theorem, as a higher-order analog? —Nils von Barth (nbarth) (talk) 03:47, 27 January 2023 (UTC)

@Jacobolus —Nils von Barth (nbarth) (talk) 04:06, 29 January 2023 (UTC)

modular group

I think the section about the modular group has the kernel of something interesting/meaningful to it, but seems like it was previously incorrect, had no sources, and I'm not quite sure what it should say so I am temporarily removing it.

Here was the previous text:

The golden ratio and its conjugate $\varphi _{\pm }={\tfrac {1}{2}}{\bigl (}1\pm {\sqrt {5}}{\bigr )}$ have a set of symmetries that preserve and interrelate them. They are both preserved by the fractional linear transformations $x,1/(1-x),(x-1)/x$ – this fact corresponds to the identity and the definition quadratic equation. Further, they are interchanged by the three maps $1/x,1-x,x/(x-1)$ – they are reciprocals, symmetric about ${\tfrac {1}{2}},$ and (projectively) symmetric about $2.$ More deeply, these maps form a subgroup of the modular group $\operatorname {PSL} (2,\mathbb {Z} )$ isomorphic to the symmetric group on $3$ letters, $S_{3},$ corresponding to the stabilizer of the set $\{0,1,\infty \}$ of $3$ standard points on the projective line, and the symmetries correspond to the quotient map $S_{3}\to S_{2}$ – the subgroup $C_{3}<S_{3}$ consisting of the identity and the $3$-cycles, in cycle notation $\{(1),(0\,1\,\infty ),(0\,\infty \,1)\},$ fixes the two numbers, while the $2$-cycles $\{(0\,1),(0\,\infty ),(1\,\infty )\}$ interchange these, thus realizing the map.

This seems incorrect to me. Specifically, while $\varphi$ and $-\varphi ^{-1}$ are fixed by the identity and exchanged by the map $x\mapsto 1-x,$ they don’t seem to be fixed or exchanged by the other maps listed there. If we call these maps

$a:x\mapsto {\frac {1}{x}},\quad b:x\mapsto 1-x,\quad c:x\mapsto {\frac {x}{x-1}}$

then we have $a(\varphi )=\varphi ^{2},b(\varphi )=-\varphi ^{-1},c(\varphi )=\varphi ^{-1}.$

I’d like to get someone who is an expert here to explain what part of this is meaningful, interesting, relevant, etc., and ideally link some sources. –jacobolus (t) 02:48, 14 January 2023 (UTC)

This was added in February 2010 by Nbarth, and then has persisted since then without anyone ever really modifying it. Nbarth, maybe you can explain where you got this, what you were getting at, etc.? –jacobolus (t) 03:05, 14 January 2023 (UTC)

Overall if we apply these three maps to $\varphi$ we get 6 elements in total (as expected for something isomorphic to the dihedral group $D_{3}$). Arranging these elements in order alternating with the (projectively) equally spaced elements $0,{\tfrac {1}{2}},1,2,\infty ,-1$ for context, we have:

$0,\,\varphi ^{-2},\,{\tfrac {1}{2}},\,\varphi ^{-1},\,1,\,\varphi ,\,2,\,\varphi ^{2},\,\infty ,\,-\varphi ,\,-1,\,-\varphi ^{-1},\,0,\,\ldots$

This doesn’t seem especially noteworthy unless it can be related to other subjects, ideas, or theorems. If anyone cares about this we can try to make a figure showing the relation of these values projected onto a circle with $0,1,\infty$ equally spaced. I'm sure we could make it pretty to look at anyway.

From looking at modular group a bit, it does seem like the Hecke group $H_{5}$ is based on $\varphi .$ Maybe someone who is an expert can write something about that here? –jacobolus (t) 03:55, 14 January 2023 (UTC)

Sorry, I made a sign error when writing this! The points that are preserved by the 3-cycles of the anharmonic group are $e^{\pm i\pi /3}=(1\pm {\sqrt {3}}i)/2$, the solutions to $x^{2}-x+1$ (the primitive sixth roots of unity), not the golden ratio $(1\pm {\sqrt {5}})/2$, which are the solutions to $x^{2}-x-1$. I made this mistake in Projective linear group [1], then copied it to Golden ratio in [2]. I have fixed the original error now too ([3]).

There's no relation to the golden ratio; this is just about the cross ratio and projective linear group; see those for interesting (and correct!) details.

Thanks for catching this and asking me!

—Nils von Barth (nbarth) (talk) 05:39, 14 January 2023 (UTC)

Thanks for clearing that up. I'm amazed nobody else ever looked carefully at this paragraph after almost 13 years, despite millions of page views.

There is something at least a little bit interesting about the 6 elements generated from $\varphi$ by the maps $x\mapsto 1-x,\,x\mapsto x^{-1}$ being all powers of $\varphi$ or their negatives,

$\varphi ^{-2},\,\varphi ^{-1},\,\varphi ,\,\varphi ^{2},\,-\varphi ,\,-\varphi ^{-1}.$

If instead we apply the maps generated by (square dihedron symmetry, corners at $0,1,\infty ,-1$) $x\mapsto -x$ and $x\mapsto (1-x)/(1+x)$ to starting element $\varphi ,$ we get the set:

$\varphi ^{-3},\,\varphi ^{-1},\,\varphi ,\,\varphi ^{3},\,-\varphi ^{3},\,-\varphi ,\,-\varphi ^{-1},\,-\varphi ^{-3}.$

I didn’t really try to figure out what Hecke groups are, but that probably is probably more relevant still, meriting some discussion on this page.

@Nbarth when looking at this I found that it was helpful to explicitly draw the projection of the projectively extended real line onto a circle (the inverse stereographic projection centered at ${\tfrac {1}{2}}$). I think that makes it easier to see what is going on than File:PGL2_stabilizer_of_3_points_on_line.svg which draws them in a straight line (P.S. you may want to edit that image description).

That other page may also benefit from a reference to a source or two if you can track them down. –jacobolus (t) 07:17, 14 January 2023 (UTC)

jacobolus Oops, good catch about the image description! Fixed in [4]; thanks!

The pattern you pointed out seems interesting; the numbers ⁠$\pm \varphi ^{k}$⁠ are the units in the ring of quadratic integers with the golden ratio, ⁠$\mathbf {Z} [\varphi ]$⁠ (see Quadratic integer § Examples of real quadratic integer rings, Golden ratio base, and Golden ratio § Other properties), so maybe there's something going on with these symmetries of them?

I tried seeing if there was anything obviously interesting for the Hecke group ⁠$H_{5}$⁠, or the corresponding triangle group (2, 5, ∞), but this isn't my expertise, and Google didn't return anything promising.

A better diagram would be welcome, but I don't have the time (or probably artistic skill ;) – I was just writing a quick schematic, which hopefully gives some geometric insight to a mostly algebraic point. I suspect that a circle might be bulky (due to needing to label the point in the middle, as well as several points on the circle), and take up a lot of space on the page. Good for a book on complex geometry, but a bit distracting for a small point in these articles. —Nils von Barth (nbarth) (talk) 20:37, 14 January 2023 (UTC)

If cross-ratios of the six permutations of four points are mapped by a Möbius transformation onto a circle, they have the geometric symmetries of a regular triangular dihedron.

@Nbarth: I mean something like the image shown here to the right. (You’ll have to figure out the clearest / most accurate way to write this caption though [e.g. pointing out that these reflections also swap the interior and exterior if the circle]. And feel free to edit the wiki commons description page if you want to use this image.) –jacobolus (t) 21:36, 14 January 2023 (UTC)

jacobolus Thanks, that's very helpful! I added your diagram (and my explanation) to cross-ratio in [5], and updated the file description in [6] (initially as triangular bipyramid, but you're right that trigonal dihedron is more correct; also added a note about the point at infinity).

It's a bit big, but appropriate to clarify the geometry, thanks!

—Nils von Barth (nbarth) (talk) 17:01, 16 January 2023 (UTC)

Looks good. Let me know if you want any changes to the diagram. –jacobolus (t) 22:51, 16 January 2023 (UTC)

Regarding the diagram, there's one serious problem: the involutions are not reflections - they are rotations! Rotations by 180 degrees, most obviously because they are complex maps, so orientation-preserving, but also because they don't fix $e^{i\pi /3}$ (as the reflections would, but instead switch it with $e^{-i\pi /3}$! Thus I think it's important to replace the double-headed arrow with a "rotation symbol" like ↺. (I only noticed this when looking at it closely.)

Perhaps useful too would be to putt a point for $e^{-i\pi /3}$ in the corner "at infinity", but I leave that to your discretion (it's a bit confusing because it should be on all the lines of symmetry). We could also mention it in the caption instead? —Nils von Barth (nbarth) (talk) 03:37, 27 January 2023 (UTC)

The reason I drew them as double-headed straight arrows instead of more obvious rotation-y looking symbols was that I made the picture with Desmos and that was a lot easier to draw. https://www.desmos.com/calculator/knthhtclga But I could probably open it up in Adobe Illustrator to make a symbol that looks more properly like rotation. I’ll see what I can do tomorrow. –jacobolus (t) 04:39, 27 January 2023 (UTC)

@Jacobolus Thanks for explaining! Understood that it's easier, but it's pretty misleading; hopefully easy to just add a semicircle arrow, which should convey it? —Nils von Barth (nbarth) (talk) 04:09, 29 January 2023 (UTC)

Blacksmith scroll use?

Was researching how to make a 'master' Scrolling jig for not only repetitive work such as a property fence with scrolls six inches apart, but also one that uses the ratio in sharp 90° bends, or for that matter, golden ratio bend angles. Just a curious person who respects Wikipedia's immense knowledge base. Don't really want my Gmail out there for all to gander. Will look here tomorrow. Thank you. 2600:8800:700F:C500:868:55A2:45CD:8E2C (talk) 11:12, 25 May 2023 (UTC)

This page is about improving the Golden Ratio article. You might want to ask your question there instead: WP:Reference_desk/Miscellaneous. Dhrm77 (talk) 21:00, 25 May 2023 (UTC)

thank you for the help. I am using a small android with onscreen keyboard, so do I just 'click' the light blue highlighted text? Thanks again for your kindhearted understanding and help in my quest. 🙏⚒️🔥⛹️ And a tad off topic, which really did come first, the 🐔 or the 🥚. 2600:8800:700F:C500:505A:8382:3D5B:80EC (talk) 12:38, 26 May 2023 (UTC)

If you want my answer to the chicken or egg question, either create a user-page for yourself by registering an account (and tell me about it), or come to my talk page. JRSpriggs (talk) 20:04, 26 May 2023 (UTC)

Semi-protected edit request on 19 June 2024

The Golden Ratio is wrong because -1=b^2+2a+a. meaning -1=(-1)^2+2(-2)+(-2) (-1)^2+2(-2)+(-2)=1-4-2=-5 — Preceding unsigned comment added by 108.35.38.31 (talk) 18:44, 19 June 2023 (UTC)

Where did you come up with the equation $-1=b^{2}+3a$ and why did you set $b=-1$ and $a=-2$? I don't understand what you are trying to say. –jacobolus (t) 19:26, 19 June 2023 (UTC)

Reference to "Zometoy" product

The reference to "Zometoy" is erroneous; the actual product is "Zometool". 2603:8080:4E05:7812:0:0:0:BA4 (talk) 12:44, 8 July 2023 (UTC)

"Zometoy" was Steve Baer's original (very low production) toy made of wooden dowels and drilled (?) nodes. Take a close look at the picture. Zometool was a later company by Marc Pelletier and Paul Hildebrandt which took the same concept and figured out how to make a better version of it out of injection molded plastic. For some about the history, see http://people.tamu.edu/~ergun/hyperseeing/2018/06/FASE/buehler2018.pdf –jacobolus (t) 15:38, 8 July 2023 (UTC)

Summing the Contained Fraction

It is probably quite, but not sufficiently, well known, that the continued fraction expression, Ø = 1+1/(1+ 1/(1+1/(1+1/(1+1/( ... infinitely, is easy to evaluate as a short algebraic expression. The part to the right of the gap has value Ø, so we can curtail the expression to Ø = 1+1/(1+Ø); multiply to a quadratic ... .

That may be worth mentioning. 94.30.84.71 (talk) 13:47, 7 November 2023 (UTC)

Or not worth mentioning. JRSpriggs (talk) 00:56, 8 November 2023 (UTC)

When the article says "The formula $\varphi =1+1/\varphi$ can be expanded recursively to obtain a continued fraction for the golden ratio" that is the converse of your statement. But notice that each step in going from one form to the other is reversible, so these are essentially the same idea. If you like you can write this explicitly as $$\varphi =1+{\cfrac {1}{1+{\cfrac {1}{1+\ddots }}}}\iff \varphi =1+{\frac {1}{\varphi }}.$$ (Under the assumption that $\varphi >0.$) I think the version in the article is fine though. –jacobolus (t) 01:01, 8 November 2023 (UTC)