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I'm told that I can express the following sum of complex exponentials:

$$\sum_{n=0}^6 e^{-j\Omega n}$$

As 1 plus 3 cosine terms. I'm having a really hard time arriving at this, I see where the 1 comes from, when n = 0 and I consider:

$$e^{-j\theta}=cos(\theta) - jsin(\theta)$$

But I feel as though I'm missing something, and I'm failing to arrive at just 3 other cosine terms. Is it possible somebody can point me in the right direction? This problem originates from a class I am taking in digital signal processing.

Any help would be humbly and greatly appreciated.

AlwaysLearning
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1 Answers1

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This is only true if you factor out $e^{-j\Omega 3}$. Then you obtain $$\sum_{n=0}^6 e^{-j\Omega n} = e^{-j\Omega 3} \left(e^{j\Omega 3} + e^{j\Omega 2} + e^{j\Omega} + 1 + e^{-j\Omega} + e^{-j\Omega 2}+e^{-j\Omega 3} \right).$$ To evaluate this, you can use that $$2 \cos(x) = e^{x} + e^{-x}.$$

Raclette
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