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mrenney · Dec 7, 2010, 03:50 PM

Why an even n gives you 2n petals and an odd m gives you m petals on polar grids?

galactus · Dec 7, 2010, 04:36 PM

Think about
$a\cdot sin(n({\theta}+\pi))=a\cdot sin(n{\theta}+n{\pi})=\left\{\begin{array}{rcl}a \cdot sin(n{\theta}), \;\ \text{n even}\\ -a\cdot sin(n{\theta}), \;\ \text{n odd}\end{array}$

If theta starts at 0, then theta would have to increase by some positive

integer multiple of Pi radians in order to reach the starting point and

begin to retrace the curve. Let $(r, \;\ \theta)$ be the

coordinates of a point P on the curve for $0\leq {\theta}\leq 2\pi$ .

Now, $a\cdot sin(n({\theta}+2\pi))=a\cdot sin(n{\theta}+2{\pi}n)=a\cdot sin(n{\theta})=r$

so P is reached again with coordinates
$(r, \;\ {\theta}+2\pi)$ thus the curve is traced out either

exactly once or exactly twice for $0\leq {\theta}\leq 2\pi$ .

If for $0\leq {\theta}<{\pi}$ , $P(r,{\theta})$ is

reached again with coordinates $(-r, {\theta}+\pi)$ then

the curve is traced out exactly once for $0\leq {\theta}<{\pi}<br />$ , otherwise exactly once for $0\leq {\theta}<2\pi$ .

So, referring to the top of the post, the curve is traced out exactly once

for $0\leq {\theta}<2\pi$ if n is even, and exactly once for

$0\leq {\theta}<\pi$ if n is odd.

The cos case can be proved similarly.