The domain of A for r=cos(kA) before the petals start to overlap

[nomadreid]

I know that, for k an integer, the rose polar curve expressed by r=r(θ)= cos(kθ) has a period of π if k is odd and 2π if k is even (usually expressed as saying that there are k distinct petals if k is odd, and 2k petals if k is even). However, I have yet to pin down the reason for this difference. Can anyone point out the reason (or calculations) for this conclusion? Thanks.

 

[chiro]

I know that, for k an integer, the rose polar curve expressed by r=r(θ)= cos(kθ) has a period of π if k is odd and 2π if k is even (usually expressed as saying that there are k distinct petals if k is odd, and 2k petals if k is even). However, I have yet to pin down the reason for this difference. Can anyone point out the reason (or calculations) for this conclusion? Thanks.

Hey nomadreid.

I'm not going to give an algebraic explanation, but one suggestion I have is that you get some kind of animated plot over time for the actual object itself if possible.

In other words you have a movie where you plot the function where 1 second corresponds to so many radians and then look at how the function evolves after seeing the animation.

My guess algebraically though is that the one with lesser petals creates an overlap of some kind whereas the other one does not create an overlap. In other words the one with less petals has a situation where the function repeats itself on the petals that it has already drawn later on instead of drawing a new petal that is offset to existing petals.

 

[nomadreid]

Thanks, chiro, I had indeed looked at some of the videos doing precisely that (there are several on the Internet), but I think I figured out an algebraic explanation, at least partly, that goes like this (Being a bit sloppy, and too lazy to put the Greek symbols in):
It is obvious that {(A, cos(kA)):0<A<2(pi)} and {(A, cos(k(A)):2(pi)<A<4(pi)} will be the same. In fact, point by point, (A, cos(kA))= (A+2pi, cos(k(A+2pi)). Now, however, to check (pi) in the same role:
(A, cos(kA))
with
(A+(pi), cos(k(A+(pi))))

If k = 2n, this becomes
(A+(pi), cos(2n(A+(pi)))) = (A+(pi), cos(2nA+2n(pi)))=(A+(pi), cos(2n(pi))) = (A+(pi), cos(kA))
which is definitely different to ((A, cos(kA)), so no overlap.

If k = 2n+1, this becomes
(A+(pi), cos((2n+1)(A+(pi)))) = (A+(pi), cos(2nA + 2n(pi)+A+(pi))) = (A+(pi), cos(2nA +A+(pi)))= (A+(pi), cos((2n+1)A +(pi)))=
(A+(pi), -cos((2n+1)A)) = (A+(pi),-cos(kA))=(A,cos(kA))

That could be made more rigorous (and/or more elegant), but this is the main idea, I think. Any opinions?

 

[chiro]

Thanks, chiro, I had indeed looked at some of the videos doing precisely that (there are several on the Internet), but I think I figured out an algebraic explanation, at least partly, that goes like this (Being a bit sloppy, and too lazy to put the Greek symbols in):
It is obvious that {(A, cos(kA)):0<A<2(pi)} and {(A, cos(k(A)):2(pi)<A<4(pi)} will be the same. In fact, point by point, (A, cos(kA))= (A+2pi, cos(k(A+2pi)). Now, however, to check (pi) in the same role:
(A, cos(kA))
with
(A+(pi), cos(k(A+(pi))))

If k = 2n, this becomes
(A+(pi), cos(2n(A+(pi)))) = (A+(pi), cos(2nA+2n(pi)))=(A+(pi), cos(2n(pi))) = (A+(pi), cos(kA))
which is definitely different to ((A, cos(kA)), so no overlap.

If k = 2n+1, this becomes
(A+(pi), cos((2n+1)(A+(pi)))) = (A+(pi), cos(2nA + 2n(pi)+A+(pi))) = (A+(pi), cos(2nA +A+(pi)))= (A+(pi), cos((2n+1)A +(pi)))=
(A+(pi), -cos((2n+1)A)) = (A+(pi),-cos(kA))=(A,cos(kA))

That could be made more rigorous (and/or more elegant), but this is the main idea, I think. Any opinions?

That's pretty much what I was thinking would be shown graphically and you've given enough to illustrate your point that you were trying to make. Your question did not require anything too specific in terms of a proof so for this purpose I think it is ok.

It sounds like this is more just a question out of curiosity to answer something less formal rather than something informal and it's a good reminder for you to realize that many things in mathematics, even the ones that end up formal end up this way (an informal curiosity will lead to something more rigorous), so don't think that informal things in mathematics don't have their place because they do in the most subtle and important ways.

 

[nomadreid]

Thanks again, chiro. Yes, you are right, the matter was out of my own curiosity. Although it was clear that the petals would rotate as their numbers changed, that they would fit into place so neatly according to even/odd was not intuitively clear to me; in these cases, the formal proof helps my informal intuition, rather than the other way around.