Results 1 to 5 of 5

- Apr 3rd 2012, 09:20 AM #1

- Joined
- Aug 2010
- Posts
- 130

## Why does r=cos((2n+1)A) have 2n+1 petals, while cos((2n)(A)) has 4n petals?

In graphing the polar equation, (letting A be the angle, being too lazy to put theta every time) r=r(A)=cos(nA), for n a positive integer, one has n petals if n is odd, and 2n petals if n is even. (Letting the domain of A be sufficiently large.) It appears that this is because the petals are repeated more often in the odd cases, but I cannot figure out why. I would be grateful for any clarification. Thanks.

- Apr 3rd 2012, 01:07 PM #2
## Re: Why does r=cos((2n+1)A) have 2n+1 petals, while cos((2n)(A)) has 4n petals?

- Apr 3rd 2012, 08:19 PM #3

- Joined
- Aug 2010
- Posts
- 130

## Re: Why does r=cos((2n+1)A) have 2n+1 petals, while cos((2n)(A)) has 4n petals?

Thanks, skeeter, but apparently I did not express myself clearly enough, and you merely repeated my question in other terms: what you said about superimposition is what I mean by the petals being repeated more often (that is, letting the domain of theta be all reals, also the polar roses of cos(2A) have superimposed petals, merely with less frequency than the polar roses of cos(2A+1).) So my question is, why exactly is this difference between odd and even: why are the odd ones superimposed more frequently than the even ones?

- Apr 4th 2012, 03:21 PM #4
## Re: Why does r=cos((2n+1)A) have 2n+1 petals, while cos((2n)(A)) has 4n petals?

Rose (mathematics) - Wikipedia, the free encyclopedia

scroll down to the topic**How the parameter k affects shapes**

- Apr 4th 2012, 07:42 PM #5

- Joined
- Aug 2010
- Posts
- 130

## Re: Why does r=cos((2n+1)A) have 2n+1 petals, while cos((2n)(A)) has 4n petals?

Again thanks, skeeter, but that was one of the first sources I checked (good ol' Wiki...). But it basically tells me that there is the overlap, not why. Here is an example of what I meant: (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.