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

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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)

Template:Re 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 • 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)