But that this does not check that it is surjective yet, it can trace half the circle and back.

I thought about ArcTan[x[t], y[t]] to be monotous but I am hesitating because it is a not a continuous function on arbitrary interval. MMA can deal with it and returns correct answer to D[..., x] >=0 //Simplify.

But, for other than circle, basic curves which don't have obvious natural coordinate systems it won't work.

I also have many ideas how to approach this in an old fashioned discrete way but I would prefer to utilize MMA symbolic powers first.

Btw, nice to see that SE chat stopped blocking my VPN

So I guess my question boils down to how to verify that two parametric equations+intervals trace the same curves. A nice to have is an orientation information. Whether given eq traces circle clockwise or counter clockwise.

@SPPearce I'm not sure if Classify/Predict etc have support for Around yet - I'm not even really sure what the behaviour should be. Around is new though, so support might come along (if it doesn't already exist)

@Kuba Not easy, I think, if the parametrization is arbitrary. The following gives the net number of times the path winds around the origin (aka "winding number"):

NIntegrate[Evaluate@D[ArcTan[x[t], y[t]], t], {t, t1, t2}]/(2 Pi)

A result >=1 or <= -1 indicates the path went around the circle

If the path goes around, reverses, and comes back to the starting point, the winding number will be zero. I don't know an easy way right now how to determine whether it went all the way around, unless you can determine the points at which the path reverses direction.

There's also WindingCount[] in V12 for polygonal paths.

Somehow the general problem of determining whether two parametrizations have the same image seems similar to determining global extrema. So many things can go wrong. Perhaps under suitable hypotheses.

They way I think about it is: The term with the highest power of x in the first bracket is x^(2n+2) and the one after is x^(2n) y^2 * constant. That constant is bigger or equal to 2 (since n is a positive integer - that is smallest value of n is 1). Since z is real and smaller than one, then the coefficient of the second term in the inequality is between 0 and -1. Hence whatever happens, the x^(2n) y^2 * constant from the first bracket will be larger than that term.

Am I missing something?

or am I not nice enough to Mathematica to see it