The n-leaf Rose
by Emily Kennedy

If a polar graph is cyclic, then its cycle c must be a multiple of π.

Why is this?

In the drawings below, the blue ray represents θ = θ₀, and the purple ray represents θ = θ₀ + c for some c. Any point with θ = θ₀ + c must lie either on the purple ray (if r > 0) or on the opposite ray (if r < 0). Note that when c is not a multiple of π, the purple ray (and its opposite ray) intersect the blue ray (and its opposite ray) only at the origin. Thus, if the cycle c is not a multiple of π, then the point (θ₀, r(θ₀)) and (θ₀ + c, r(θ₀ + c)) cannot coincide unless the point lies at the origin.

θ and θ + (2m+1)π

θ and θ + (2m)π

θ and θ + c (c not a multiple of π)

And the point r(θ) cannot always be at the origin since sin(nθ) is not identically zero (we are assuming n ≠ 0).

So c must be a multiple of π.

Furthermore, note that if c is an even multiple of π
(i.e., c = (2m)π for some integer k), then we have
r(θ + c) = r(θ)  θ

And if c is an odd multiple of π
(i.e., c = (2m + 1)π for some integer k), then we have
r(θ + c) = -r(θ)  θ

(The point must lie on the ray opposite the purple ray.)

If c = (2m)π, then

And if c = (2m + 1)π, then

So if the polar function r(θ) = sin(nθ) is cyclic,
then n must be a rational number.

That is, if n is irrational, the graph will never trace over itself,
no matter how high we let θ go.

Let , where gcd(α, β) = 1.

What is the cycle of r = sin(nθ)?

Let's find out!

We know c is either an even or odd multiple of π,
so let's try each and see what happens.

Try: c = (2m + 1)π for some integer m

Then we need
r(θ + c) = -r(θ)  θ

We want c to be as the minimum positive number such that
r(θ + c) = -r(θ)  θ

And by the above, the smallest positive value of j we can choose is α, so

So c = βπ.

Note that, by (1), we must have that α and β are both odd.

Otherwise, (1) gives a contradiction.

Try: c = (2m)π for some integer m

Then we need
r(θ + c) = r(θ)  θ

We want c to be as the minimum positive number such that
r(θ + c) = r(θ)  θ

And by the above, the smallest positive value of m we can choose is β, so

c = 2mπ = 2βπ.

So if either α or β is even, we have
c = 2βπ.

So, to summarize, we have found the following:

	c
α and β both odd	βπ
either α or β even	2βπ

Now, using this information, we can quickly come to some conclusions about the number of "petals" on the "rose" formed by the equation
r = sin(nθ) (n ≠ 0)

Click here to see that analysis.

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