public-notes/Petals in a Rose Curve.md

#Math #Trig #Algebra

Question

Show there are n petals in a rose curve with odd n and 2n petals in a rose curve with even n.

Solution

WLOG the rose curve has the form r = \cos n\theta. Then there are 2 cases to get a maximum point on a petal: either \cos n\theta = 1 or \cos( n(\theta + \pi)) = -1

Case 1

\cos n\theta = 1

For all integers i:

n\theta = 2i\pi

\theta = \frac {2i\pi} n

Case 2

\cos( n(\theta + \pi)) = -1

For any arbitrary integer i:

n(\theta + \pi) = 2i\pi + 1 \theta = \frac {(2i + 1)\pi - n\pi} n

For \theta \in [0, 2\pi) both sequences create

Even Case

Factor out \frac \pi n from both sides. Case 1 has \frac \pi n multiplied by an even factor, while case 2 has \frac \pi n multiplied by an odd factor. Therefore the sequences are unique, creating 2n unique petals.

Odd Case

Take case 2:

\theta = \frac {(2i + 1)\pi - n\pi} n \theta = \frac {2i\pi + (1 - n)\pi} n

Since n is odd, 1 - n is even, and therefore the numerator is an even factor of \pi.

\theta = \frac {2i\pi} n

Therefore case 2 produces the same \theta as case 1, resulting in n petals.