On "compensation functions"? The circle $\alpha (t)=r(\cos t,\sin t)$ has the following property:

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*If we have a halfline from $(0,0)$ pointing to $(\cos t_0,\sin t_0)$, then it crosses the circle at $\alpha (t_0)$.

Which is obvious but this doesn't happen to other curves, for example if we take $\beta(t)=(2 \cos t, \sin t)$, a halfline from $(0,0)$ pointing to $(\cos t_0,\sin t_0)$ doesn't (in general) crosses $\beta (t_0)$ which can be seen in the following figure:

For $t_0=1,2,4,6$. But then I thought: Perhaps we could write $\beta(c t)$ in a way that it matches, ie we want
$$\beta(c t)=\lambda \alpha (t)\tag{$\star$}$$
for some $\lambda>0$ and all $t$. I tried to experimentally find such $c$ when I used $c=1.26$, I got:

Ie: The halflines from $(0,0)$ to $\alpha (1)$ and from $(0,0)$ to $\beta (c t)$ almost match! But the other one is missing. And when I tried $c=0.9$ I got:

Ie: The other pair of halflines almost match but the previous one doesn't. So $c$ must be a function. With this in mind, I tried to find solutions for $(\star)$: That is, fixing $t$ and finding solutions for $\lambda >0$ and $c$. I found some values and interpolated them. The plot of the interpolation is:

And I have no idea what function is that but If I take the halflines from $(0,0)$ to $\alpha(t)$ and from $(0,0)$ to $\beta(c(t)\cdot t)$, I get:

They all match now! So I have the following

 Questions:


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*What function is that? I tried to ask Mathematica to show me the values for $t=1,...,20$ and I got this:


Which seems really crazy for me, I know the sequence of coefficients of $\pi$ is $s_n =0,1,1,-1,0,0,0,1,1,-1,0,0,0,0,1,-1,-1,0,0,0,...$
But I don't know if that will hold when we pick other $t$ different of the $t$'s I picked and also, that would complicate a lot the construction of that function.

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*Is there a well-known process for finding $c(t)$? The one I'm using relies on numerical approximations, I know that by having:

$$\beta(c t)=\lambda \alpha (t)\implies (\beta_1(c(t)t),\beta_2(c(t)t))=(\lambda \alpha_1(t),\lambda \alpha_2(t))$$
And then we obtain, assuming that $\beta_i$ have inverse functions in a certain interval, we get:
$$c(t) =\frac{\beta_1^{-1}(\lambda(t) \alpha_1(t))}{t} \qquad c(t) =\frac{\beta_2^{-1}(\lambda(t) \alpha_2(t))}{t}$$
But this doesn't looks easy to solve. I am confused to even think what would be a solution of it.
 A: You have a function $\vec{\beta}(s)=(f(s),g(s))$, describing a curve in the plane.  For any $s$, the point $\vec{\beta}(s)$ is also vector from the origin to $\vec{\beta}(s)$; the ray from the origin through $\vec{\beta}(s)$ is the set of points $$\ell(s)=\{a\vec{\beta}(s):a\in[0,\infty)\}$$
A point point $(x,y)\in S^1$ has angle $\theta$ such that $\tan{\theta}=\frac{y}{x}$.  In particular, the right-hand side is invariant under a common scaling, so that the angles associated with $(x,y)$ and (say) $2(x,y)$ are the same.  Thus every point on $\ell(s)$ has angle $\theta_\beta(s)$ such that $$\tan{\theta_\beta(s)}=\frac{g(s)}{f(s)}\tag{1}$$
In particular, if $\vec{\alpha}(t)=(\cos{t},\sin{t})$, then $$\tan{\theta_\alpha(t)}=\frac{\sin{t}}{\cos{t}}$$ so that $\theta_\alpha(t)=t$.
You want to choose $s(t)$ such that $$\theta_\beta(s(t))=\theta_\alpha(t)=t$$  For some choices of $f$ and $g$, this can be done by inverting (1), although there may not be a nice formula for that inverse (because inverses in general do not exist, and when they do, rarely have nice formulas).
In the specific case of $\beta(s)=(2\cos{s},\sin{s})$, (1) is $$\tan{t}=\frac{\sin{s(t)}}{2\cos{s(t)}}=\frac{1}{2}\tan{s(t)}$$  Thus $$s(t)=\tan^{-1}\left(\frac{1}{2}\tan{t}\right)$$  AFAIK, that expression cannot be simplified further.
