Talk:Omega constant

letter The fact that it is called the omega function makes me think someone mistook the letter &omega;, a vowel, for the letter w, because of visual similarity. Is that what happened? Michael Hardy 20:42, 12 Feb 2004 (UTC)

Comparison to the golden ratio
"It has properties that are akin to those of the golden ratio, in that


 * $$ e^{-\Omega}=\Omega,\,$$

or equivalently,


 * $$ \ln (1/\Omega) = \Omega.$$"

I'm not sure I can see any similarity to the golden ration here. Should this be obvious? :/ Woscafrench (talk) 01:00, 26 November 2007 (UTC)
 * One way to express the similarity in exact terms is as follows: if we replace f(x)=e^x with its linear approximation f(x)=x-1 then the solution to the equation f(x)=1/x would be golden ratio.  Grue   15:36, 26 November 2007 (UTC)

I also doubt that sentence. No one can explain this better; it is just a point of view. I already removed it. --Octra Bond (talk) 02:29, 3 November 2009 (UTC)

The Omega constant to larger approximation?
I used Windows calculator to evaluate the Omega constant. The value of Omega is:

0.56714329040978387299

99686622103555497485

40291138701110223886

36460700559427176005

70163043896301071636

09769866542711127501

81294450722431215290

864337423...

Garygoh884 (talk) 06:27, 15 July 2011 (UTC)

"due to Victor Adamchik"
Reference for part "due to Victor Adamchik"? — Preceding unsigned comment added by Reddwarf2956 (talk • contribs) 22:52, 13 December 2012 (UTC)

Interesting property of the Omega constant
Let k = 1/Ω, that is, the reciprocal of the Omega constant. Then it holds that:
 * $$\frac{d}{dx} k^x = k^{x-1}$$

That is, it imitates the power rule of derivatives (even though it's not a polynomial, but an exponential function with base k) in reducing the exponent by 1. The reciprocal Omega constant is the unique exponential base for which this is true.

Proof:
 * $$\frac{d}{dx} k^x = \frac{d}{dx} e^{x \ln k} = (\ln k)(e^{x \ln k}) = (\ln k)k^x$$

But since $$e^{-\Omega} = \Omega$$, it follows that $$e^{\Omega} = 1/\Omega = k$$, and therefore $$\ln k = \ln e^\Omega = \Omega = 1/k$$. Substituting into the above equation, we have:
 * $$\frac{d}{dx} k^x = (\ln k)k^x = (1/k)k^x = k^{x-1}$$

as claimed.&mdash;69.172.154.75 (talk) 04:53, 27 September 2015 (UTC)

Question
I was asking myself if Ω is the the absolute value of the solution to          $x+e^{x} = 0$ I've found it with the approximative form thanks to the comparison between y=x and y=-e^x. If I'm not wrong, we can add it to the principal page. — Preceding unsigned comment added by 87.0.114.210 (talk) 14:35, 15 January 2017 (UTC)

Other property
(1/omega)^(1/omega)=e — Preceding unsigned comment added by 191.111.6.32 (talk) 22:07, 31 May 2022 (UTC)


 * Yes, but not surprising (ie pretty simple to deduce) and nor more snappy. Unless this is especially useful for something (best shown with a souce), I don't think it needs to be mentioned in the article.
 * I do think the continued fraction of omega would be more interesting. Xario (talk) 12:34, 2 December 2023 (UTC)

OEIS Sequences A370490 and A370491
OEIS Sequences A370490 and A370491 are the denominators and the numerators of an infinite series that converges to the Omega constant. Dacicus Geometricus (talk) 17:10, 6 April 2024 (UTC)