Timeline for Finding ellipse-ellipse intersections in $\mathbb R^2$

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Dec 2, 2019 at 23:29	comment	added	Alexandre Eremenko		@Zach Teiler: it is an algebraic function of the angle, but not an algebraic function of the points on the circle ("corners" of the sector). Newton's argument is of course applicable to a disk as well.
Dec 2, 2019 at 21:20	comment	added	Zach Teitler		Ah, suppose the ellipse has area $A$, fix a ray from the center, and let $f(\theta)$ be the area of the sector swept by angle $\theta$ from the initial ray (counting area multiply, or negatively if $\theta<0$). Let $g(\theta) = f(\theta)-A\theta/(2\pi)$. Then $g$ is periodic, hence not algebraic, so $f$ is also not algebraic; except if $g=0$, which is the case for a circle. - Just a minor change from what you wrote: the area function is not infinitely-valued, it’s the difference $g$ that is.
Dec 2, 2019 at 19:26	comment	added	Zach Teitler		Please forgive a stupid question, but doesn't that proof idea apply to the area of a circular sector? But the area of a circular sector is an algebraic function... of the central angle of the sector. Unless perhaps you mean to consider it as a function of something else? Or perhaps, to ignore multiple counting of areas, e.g., the sector from $0$ to $5\pi/2$ is the same as the sector from $0$ to $\pi/2$?
Dec 2, 2019 at 19:09	history	edited	Alexandre Eremenko	CC BY-SA 4.0	added 3 characters in body
Dec 2, 2019 at 19:03	history	answered	Alexandre Eremenko	CC BY-SA 4.0