# Strangely but closely related parametrized curves

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

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:

## 1 Answer

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 $$r = 1 / 3, k = 6$$).

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}