Compare the following two parametrized curves for $k \in \mathbb{N}^+$:

$$x_r(t) = \cos(t)(1 + r\sin(kt))$$ $$y_r(t) = \sin(t)(1 + r\sin(kt))$$

with $0 \leq t < 2\pi$ and $0 \leq r \leq 1$ (being the plot of the sine function with amplitude $r$ over the circle instead of the real axis) and

$$X_R(t) = R\cos(t/2)\sin(kt/2)$$ $$Y_R(t) = R\sin(t/2)\sin(kt/2)$$

(which give Grandi roses) with $R = 2\sqrt{r}$.

Find depicted the outer ("sine") curve $(x_r,y_r)$ thinner, the inner ("rose") curve $(X_R,Y_R)$ thicker, and the argument circle (for $t$) only looming.

The curves are coloured according to the argument $t$ that gives rise to them, the color ranging from black (for $t=0$) over red (for $t=\pi/2$), white (for $t=\pi$) and blue (for $t=3\pi/2$) back to black (for $t = 2\pi$).

Here for $k=1,3,5,7$:

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My questions are:

  1. Why doesn't this work for even $k$?

  2. What happens when $r = 1$, that is when also $(x_r,y_r)$ exhibits an $k$-fold intersection point – like $(X_R,Y_R)$ always does?

  3. Especially: How are the curves $(x_1,y_1)$ and $(X_2,Y_2)$ related (topologically)?

  4. Which other pairs of curves $(x,y)$, $(X,Y)$ behave in a similar way?

For $r < 1$ the curves$(x_r,y_r)$ and $(X_R,Y_R)$ are obviously not homeomorphic. On the other hand $(x_1,y_1)$ and $(X_2,Y_2)$ are homeomorphic as point sets – but not as parametrized curves, because there is no continuous bijection $f: [0,2\pi] \rightarrow [0,2\pi]$ such that $x_1(t) = g(X_2(f(t)))$, $y_1(t) = g(Y_2(f(t)))$ with $g$ the homeomorphism that maps the two curves as points sets.

Is this the right way to say it – "homeomorphic as point sets, but not as parametrized curves" – and is that all there is to say?

To see what goes wrong for even $k$, find here the cases $k=2,4,6$. I didn't try to "fix" them:

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First, notice that by reparameterizing the Grandi roses by replacing $t$ with $2 t$ (so, doubling the speed), we can write the curves respectively as graphs of polar functions $\rho, \textrm{P}$ in the angular variable (which I'll still call $t$):

\begin{align} \rho &:= 1 + r \sin k t \\ \textrm{P} &:= R \sin k t . \end{align}

(1) This viewpoint quickly explains both of the differences we notice in the plots with $k$ even: First, we see that if we plot both $\rho, \textrm{P}$ over a full period (an angular interval of $2 \pi$), the asymmetry of the plots with $k$ even disappears: With the original parameterization, we only trace for half a period of $\textrm{P}$, which is anyway less natural. Plotting both curves over a full period for even $k$ gives this more symmetric graph (for $r = 1 / 3, k = 6$).

enter image description here

We can also see immediately which this issue didn't occur for odd $k$, even with the slower parameterization. Expanding using the usual sum formula gives $$\textrm{P}(t + \pi) = R \sin k(t + \pi) = (-1)^k R \sin k t = (-1)^k \textrm{P}(t).$$ But the polar coordinates $(t + \pi, \alpha)$ and $(t, -\alpha)$ represent the same point, which (for $r > 0$) tells us exactly that the parameterization $\textrm{P}(t)$ traces the graph of $\textrm{P}$ with period $\pi$ iff $k$ is odd. So, for $k$ odd we still get the complete graph by plotting it over an angular interval of length $\pi$, or equivalently over an angular interval of length $2 \pi$ (rather than $4 \pi$) in the original parameterization of the Grandi rose. The same identity also immediately more-or-less explains the more essential difference between the situations for $k$ even and odd, namely that for $k$ even (but not odd) the lobes of the curve $\textrm{P}$ extend outside the graph of $\rho$.

(2) We can see that for $r = 1$, $\rho$ has a minimum of zero---and so its graph intersects the origin---at integer multiples of $\frac{\pi}{k}$. When $r < 1$, $\rho(t) \geq 1 - r > 0$, in which case this behavior doesn't occur.

(3) It's really not clear to me what is meant here. But NB for $r < 1, k > 1$, the polar graphs of $\rho, \textrm{P}$ (as subsets of $\Bbb R^2$) are topologically inequivalent: The polar graph of $\rho$ is topologically equivalent to a circle, whereas the polar graph of $\textrm{P}$ is topologically equivalent to a bouquet of $k$ circles (for $k$ odd) and of $2 k$ circles for $k$ even. For $r = 1$, the polar graphs are topologically equivalent iff $k$ odd.

(4) Notice that we can preserve some of the qualitative behavior of the examples if we replace $\sin kt$ with any other odd function with period $2 \pi$. In particular, taking the first two interesting terms of the Fourier series of any such function gives pairs

\begin{align} \tilde \rho &:= 1 + r (\sin k t + a \sin 3 k t ) \\ \tilde{\textrm{P}} &:= R (\sin k t + a \sin 3 k t) . \end{align}


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