Wikipedia:Reference desk/Archives/Mathematics/2018 April 7

= April 7 =

Name for ellipse-like shape that resembles a dumb-bell
Hi, I'm looking for the name of a curve (and its mathematical parameterization) that amounts to a generalization of the ellipse -- one that is pinched a bit along the minor axis so that it can resemble a dumb-bell. Thanks. Attic Salt (talk) 17:40, 7 April 2018 (UTC)


 * Do you possibly mean the Cassini oval...? --CiaPan (talk) 18:55, 7 April 2018 (UTC)
 * That will do very nicely. Thank you. Attic Salt (talk) 18:56, 7 April 2018 (UTC)


 * The Cassini ovals are not "a generalization of the ellipse", though: that is, no Cassini oval is an ellipse. —Tamfang (talk) 01:43, 8 April 2018 (UTC)
 * Okay, thank you. Attic Salt (talk) 19:26, 8 April 2018 (UTC)

Is there a constant?
Is there a constant that can make this relation $$ln(x)/ln(c)=x$$? — Preceding unsigned comment added by 37.98.231.36 (talk) 19:17, 7 April 2018 (UTC)


 * I'm not sure what you're asking exactly, but depending on the value of c, your equation may have 0, 1, or 2 (real) solutions. In general, I think you can solve for x in terms of the Lambert W function.  –Deacon Vorbis (carbon &bull; videos) 19:44, 7 April 2018 (UTC)
 * He's asking for a constant c such that $$c^x = x$$, i.e. a fixed point of the exponential function with base c. First off, the only c that satisfies this for all x is 1. Secondly, the exponential function for real x is always positive, so only positive x can be fixed points. There is only one value of c such that there is only one fixed point and it satisfies both $$c^x = x$$ and $$\ln(c) c^x = 1$$ (as the graph of the exponential function there is tangent to the graph of the identity function), from which we conclude that $$x = \frac{1}{\ln(c)}$$, $$c^\frac{1}{\ln(c)} = \frac{1}{\ln(c)}$$, $$c = \ln(c)^{-\ln(c)}$$. Fixed point iteration of the latter equation yields $$c \approx 1.4446678610098$$ from which we recover $$x = \frac{1}{\ln(c)} \approx 2.718281828459$$. This looks uncanningly close to the value of e so I conjecture that $$x = e$$ and $$ c = e^\frac{1}{e}$$. For any c below this we have two solutions.--Jasper Deng (talk) 20:10, 7 April 2018 (UTC)


 * By inspection of the curves, we can see that $$c^x = x$$ has either 0, 1 or 2 solutions, depending on the value of $$c$$. As Jasper notes, the case where there is exactly 1 solution corresponds to the case where $$c^x$$ is tangent to $$x$$. That is $$\frac{d(c^x)}{dx}=1$$. So $$c^x{\ln(c)}=1$$. It's easy to see that Jasper's conjectured values for $$c$$ and $$x$$ satisfy this, proving Jasper's conjecture. RomanSpa (talk) 00:57, 8 April 2018 (UTC)
 * Jasper: I think your sentence starting "First off..." is false and should be deleted: in general $$1^x$$ is not equal to $$x$$. RomanSpa (talk) 00:57, 8 April 2018 (UTC)
 * Brain fart. --Jasper Deng (talk) 07:04, 8 April 2018 (UTC)