Number of Petals in Rose Curves

When you draw a sinusoid in polar coordinates, the results can be beautiful. Depending on the parameters of the sinusoid, we can obtain many different curves. The ones I want to focus on in this post are the rose curves which, as you could’ve guessed it, resemble a flower.

What we’ll consider a rose curve will be an equation of the form

r=a+bcos(mnθ)r = a + b \cos (\frac{m}{n}\theta)

For simplicity, we’ll assume that a=0a = 0 and b=1b = 1. Also, mn\frac{m}{n} is a positive rational number, where gcd(m,n)=1\gcd(m,n)=1. Irrational numbers are not considered because the equation would result in something resembling a disc instead of a curve. (it doesn’t have period, and it’s not even a full disc as it doesn’t go through all points of the disc). The cos\cos function can be replaced with sin\sin function, the only difference is that the curve will be rotated by πn2m\frac{\pi n}{2m}.

Feel free to play around with the below interactive GeoGebra applet to get the feel how the curves look like.

Click here to open the applet.

Simple Case

Consider first the simpler case when n=1n = 1.

By experimentation we can see that the number of petals is

This may be against intuition at first but starts to make sense once you recall what happens when a point in polar coordinates (r,θ)(r,\theta) has negative distance rr. We’re still drawing 2m2m petals over the range [0,2π)[0,2\pi), however, exactly half of them overlaps when mm is odd. Why is that so?

Before we start talking about the number of petals we should define what a petal is. Working with ranges will make things unnecessarily harder so we’ll define petal as a point (r,θ)(r,\theta) where r=1r=1 or r=1r=-1. In other words, a petal is a point the furthest from the origin. This is a nice definition because there’s only one value of 11 and one value of 1-1 in one cycle of the cos\cos (or sin\sin) function. What is more, these values are the maximum and minimum of the function so it’s easy to see that they are the vertices of the petals. The task now changes to finding the number of maximums and minimums of function cos(mθ)\cos(m\theta) over the range θ[0,2π)\theta \in [0,2\pi).

The period of cos(mθ)\cos(m \theta) is 2πm\frac{2 \pi}{m}. The graph starts redrawing at θ=2π\theta = 2 \pi, hence there are mm cycles before the graphs starts redrawing.

Note that it is not exactly the period of the graph as the redrawing can happen earlier. For example, the graph of equation r=cosθr=\cos \theta is the same for θ[0,π)\theta \in [0,\pi) and for θ[0,2π)\theta \in [0,2\pi). However, in this post, I’m going to simplify the situation and call 2π2\pi the period regardless if it starts redrawing at 2π2\pi or π\pi.

We established the number of cycles is mm. In each cycle there’s one minimum and one maximum, so the total number of petals drawn is 2m2m.

However, some of them overlap. To see that, consider a maximum point (1,θ)(1,\theta). On the graph, this point is equivalent to minimum point (1,θ+π)(-1,\theta+\pi). Therefore, the petal drawn around θ\theta will be redrawn at θ+π\theta+\pi if point (1,θ+π)(-1,\theta+\pi) belongs to the graph. This only happens when mm is odd as can be seen on the rectangular graph. For more formal proof, see the next section where we’ll show how it works for the more general case.

Note that when a0a \neq 0 is an integer, we no longer reach any negative values, hence the number of petals will be mm – the number of maximums.

More General Case

We’ll now take a close look at function

r=cos(mnθ)r = \cos (\frac{m}{n}\theta)

In rectangular graph, the function has period 2πnm\frac{2 \pi n}{m}. However, the range of θ\theta is no longer neccesarily [0,2π)[0,2\pi). We need to find the period of the function in polar coordinates. Let’s define the period as number of full rotations before the graph starts redrawing itself: the least integer k>0k > 0 such that for all θ\theta r(θ)=r(θ+2kπ)r(\theta)=r(\theta+2k\pi). Note that kk does not neccesarily has to exist, as would be the case if we were considering irrational numbers. The logic for the definition is that since we start from some point (r(0),0)(r(0),0), we want to end up at the same point (r(0),2kπ)(r(0),2k\pi). Anyway, in our case:


Since cos\cos is periodic and value for 11 is appearing once per cycle, we can substitue θ=0\theta = 0 to simplify calculations.


This equation is equivalent to the following for any Z\ell \in \mathbb{Z}:

mn(2kπ)=2π\frac{m}{n}(2k\pi)=2\ell \pi

After some algebraic manipulations we obtain

k=nmk = \frac{n}{m}\ell

Which will be the smallest positive integer for =m\ell=m, hence k=nk=n and the period of our polar function is 2πn2\pi n. Since each cycle of the rectangular function has length 2πnm\frac{2\pi n}{m}, there will be mm cycles. We can draw it on paper by repeating the cosine graph mm times and marking the last point as 2πn2\pi n.

With the interval established, we can count the number of petals as the number of solutions to two equations

cos(mnθ)=±1\cos(\frac{m}{n}\theta)=\pm 1

on interval θ[0,2πn)\theta \in [0,2\pi n).

After some basic algebraic manipulations we have θ=2knmπ\theta = \frac{2kn}{m}\pi or θ=nm(2k+1)π\theta=\frac{n}{m}(2k+1)\pi. They both lie in the interval when k{0,1,,m1}k \in \{0,1,\ldots,m-1\}, so there are 2m2m solutions in total.

We’re not done yet, however, since we’re also counting the overlapping petals. There are two ways the two petals can overlap.

  1. The first case is when point (1,θ)(1,\theta) is equivalent to (1,θ+2kπ)(1,\theta+2k\pi) for some 2kπ<2nπ2k\pi<2n\pi. However, that’s impossible since we already determined the period to be 2nπ2n\pi.
  2. The other case is when point (1,θ)(1,\theta) is equivalent to (1,θ+(2k+1)π)(-1,\theta+(2k+1)\pi). Note that since the period is now greater than 2π2\pi we have to consider shifting by odd multiples of π\pi which we can interpret as 2k+12k+1 180180 degree rotations ending at the same point (remember what r-r means in polar coordinates!).

Notice that θ+(2k+1)π\theta+(2k+1)\pi is one of the solutions to the second equation (for minimum). Therefore, if the equivalent point to a maximum exists, it is a corresponding minimum, and vice versa. For this reason, we’ll only consider the points (1,θ)(1,\theta) when finding equivalence, as the steps are very similar for (1,θ)(-1,\theta).

Anyway, two petals are equivalent if the following condition is met:


In this case θ=2πnma\theta = \frac{2\pi n}{m}a for a{0,1,,m1}a \in \{0,1,\ldots,m-1\}, as we’re moving from point (1,θ)(1,\theta) and we already solved for θ\theta earlier. However, we can’t shift too much to not exceed the period 2πn2 \pi n, but we’ll worry about that later.

cos(mn(2πnma+(2k+1)π))=1\cos(\frac{m}{n}(\frac{2\pi n}{m}a+(2k+1)\pi))=-1

mn(2πnma+(2k+1)π)=π+2π\frac{m}{n}(\frac{2\pi n}{m}a+(2k+1)\pi)=\pi+2 \ell \pi

for some Z\ell \in \mathbb{Z}. After some simplification steps we end up with the following equation:

2a+(2k+1)mn=2+12a+(2k+1)\frac{m}{n}=2\ell +1

Since gcd(m,n)=1\gcd(m,n)=1, the term (2k+1)mn(2k+1)\frac{m}{n} is the only one that has risk of not being an integer. Therefore, 2k+12k+1 must be a multiple of nn. However, it can’t be any other multiple than 1n1\cdot n otherwise we would end up at point beyond 2πn2\pi n. So


It immediately follows that nn must be odd. Notice that this must mean there is only one candidate which can be our equivalent petal: θ+πn\theta+\pi n. I find it interesting that we’re shifting by exactly half the period. This means the equivalent points can happen only in the second half of the period, hence the graph can only start redrawing itself after either nπn\pi or 2πn2\pi n.

However, we’re not done yet. We still don’t know which of our mm maximums have equivalent petals. After substituting n=2k+1n=2k+1 we are left with

2a+m=2+12a+m=2\ell +1

So, the equivalent points can exist only if 2a+m2a+m is odd. If mm is even, the whole expression becomes even, so mm can’t be even. Therefore mm must be odd and indeed it is easy to check that 2a+m2a+m is odd for odd mm.

Last thing we need to do is to check when we exceed the period. a{0,1,,m1}a \in \{0,1,\ldots,m-1\} but we can’t shift the angle too much or we’ll exceed the period 2πn2\pi n and the equivalent petal won’t matter to us. So

2aπnm+nπ<2nπ\frac{2a\pi n}{m}+n\pi < 2n\pi

a<m2a < \frac{m}{2}

Remember that mm is odd, so am2a \le \lfloor \frac{m}{2} \rfloor.

Summing up, we have a bijection from m2+1\lfloor \frac{m}{2} \rfloor + 1 maximums in the first half to minimums in the second half of the period. By following similar steps, we also have a bijection from m12\lfloor \frac{m-1}{2} \rfloor minimums from the first half to the maximums in the second half. These describe two disjoint sets of equivalent petals which together sum up to mm. \blacksquare


Rose curve r=cos(mnθ)r = \cos(\frac{m}{n}\theta) where gcd(m,n)=1\gcd(m,n)=1, m,n>0m,n>0:

  1. Has period nπn\pi when both mm and nn are odd, and it has mm petals.
  2. Has period 2nπ2n\pi when either mm or nn is even, and it has 2m2m petals.

Think about what happens to the equation

r=a+bcos(mnθ)r = a + b \cos (\frac{m}{n}\theta)

when a0a \neq 0 or b1b \neq 1.

Now, don’t cite me on that but here’s my intuition: bb only stretches the values (including maximums and minimums), so the number of petals should remain the same, at least according to the definition of the furthest points from origin.

Note, however, there start to appear “smaller petals” as we play with values of aa and bb. When aa is large enough to remove xx-intercepts in the rectangular graph, the number of petals stops to double regardless of value of mm and nn. This might be worth further investigation in the future but for now that’s all.