32 lines
1.1 KiB
Markdown
32 lines
1.1 KiB
Markdown
#Math #Trig #Algebra
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# Question
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Show there are $n$ petals in a rose curve with odd $n$ and $2n$ petals in a rose curve with even $n$.
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# Solution
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WLOG the rose curve has the form $r = \cos n\theta$.
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Then there are 2 cases to get a maximum point on a petal: either $\cos n\theta = 1$ or $\cos( n(\theta + \pi)) = -1$
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## Case 1
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$$
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\cos n\theta = 1
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$$
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For all integers $i$:
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$$
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n\theta = 2i\pi
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$$
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$$
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\theta = \frac {2i\pi} n
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$$
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## Case 2
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$$\cos( n(\theta + \pi)) = -1$$
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For any arbitrary integer $i$:
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$$n(\theta + \pi) = 2i\pi + 1$$
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$$\theta = \frac {(2i + 1)\pi - n\pi} n$$
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For $\theta \in [0, 2\pi)$ both sequences create
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## Even Case
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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.
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## Odd Case
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Take case 2:
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$$\theta = \frac {(2i + 1)\pi - n\pi} n$$
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$$\theta = \frac {2i\pi + (1 - n)\pi} n$$
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Since $n$ is odd, $1 - n$ is even, and therefore the numerator is an even factor of $\pi$.
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$$\theta = \frac {2i\pi} n$$
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Therefore case 2 produces the same $\theta$ as case 1, resulting in $n$ petals. |