Sum Series using De Moivres Theorem

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Toast210
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Struggling on a question at the moment. Appreciate any help.
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I'm on b at the moment. So far I've got
C = [1/2 cos theta - 1/4 - 1/4 cos theta(n+1) + 1/8 cos ntheta ] / (5/4 - cos theta)
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ghostwalker
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(Original post by Toast210)
Struggling on a question at the moment. Appreciate any help.
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I'm on b at the moment. So far I've got
C = [1/2 cos theta - 1/4 - 1/4 cos theta(n+1) + 1/8 cos ntheta ] / (5/4 - cos theta)
Really need to see the full working, but I suspect that when you summed the GP, you used a ratio of e^{i\theta} instead of \frac{1}{2}e^{i\theta}, and the 1/2 needs to be raised to the power of n as well in the summation formula - at a guess.
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Toast210
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(Original post by ghostwalker)
Really need to see the full working, but I suspect that when you summed the GP, you used a ratio of e^{i\theta} instead of \frac{1}{2}e^{i\theta}, and the 1/2 needs to be raised to the power of n as well in the summation formula - at a guess.
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ghostwalker
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(Original post by Toast210)
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As suspected, you've got r correct, but haven't raised it to the power of n corrrectly; should be \frac{1}{2^n}e^{in\theta}
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Toast210
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(Original post by ghostwalker)
As suspected, you've got r correct, but haven't raised it to the power of n corrrectly; should be \frac{1}{2^n}e^{in\theta}
Could you explain why?
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RDKGames
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(Original post by Toast210)
Could you explain why?
Because the sum of geo sequence result contains an r^n in there ... so what is (1/2 e^itheta)^n ?
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ghostwalker
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(Original post by Toast210)
Could you explain why?
As part of the formula for a finite sum of the geometric series the common ratio is raised to the power of n. In this case the common ratio is \frac{1}{2}e^{i\theta}, so raising that to the power of n gives us \displaystyle\left(\frac{1}{2}e^{i\theta}\right)^n=\frac{1}{2^n}e^{in\theta}
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