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Is there a way to perform the finite sum $\sum_{m = 1}^n \exp(2 \pi i k (\sqrt5) ^m)$?, m even.

I am trying to show a specific sequence is not equidistributed, and so I'd like to show that Weyl's criterion fails, but I am not sure how to perform this sum, since it is not a geometric series.

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Currently your index does not appear in your summand. Did you mean to write $$\sum_{k=1}^{n}\exp\left(2\pi i k(\sqrt{5})^n\right)$$ rather than $$\sum_{m=1}^{n}\exp\left(2\pi i k(\sqrt{5})^n\right)$$? –  000 Mar 26 '12 at 21:31
    
Yes, that is what I meant. –  Mary Mar 26 '12 at 21:34
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I am confused. Are you trying to evaluate $$\sum_{m=1}^{n}\exp(2\pi i k(\sqrt{5})^m)$$ or $$\sum_{k=1}^{n}\exp(2\pi i k(\sqrt{5})^n)$$. You just stated it was the latter that you meant, yet your edit provides the former as the series you intend. Also, if $m$ must be even, your series cannot begin at $1$. –  000 Mar 26 '12 at 21:45
    
Right, ack. I've edited the post. It's the first thing you typed,and m should start at 2. –  Mary Mar 26 '12 at 21:59
    
Maybe you could just tell us what sequence is the one whose distribution you are investigating? –  Gerry Myerson Mar 26 '12 at 22:24

2 Answers 2

As written, the sum is $\left\lfloor\frac{n}{2}\right\rfloor$ since $\exp(2\pi ik5^{m/2})=1$ for $m$ even.

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Oh. Duh. ${}{}$ –  anon Mar 26 '12 at 22:25

An elaboration on what has been stated: $$ \begin{align} \sum_{m=2, m \text{ even}}^{n}\exp\left(2\pi i k \left(\sqrt{5}\right)^m\right)&=\sum_{m=2, m \text{ even}}^{n}e^{2\pi ik \cdot 5^{\frac{m}{2}}}\\ &=\sum_{m=2,m \text{ even}}^{n}\left(e^{2\pi i}\right)^{k\cdot 5^{\frac{m}{2}}}\\ e^{2\pi i}&=1\\ \therefore \sum_{m=2,m \text{ even}}^{n}\left(e^{2\pi i}\right)^{k\cdot 5^{\frac{m}{2}}}&=\sum_{m=2, m \text{ even}}^{n}1^{k\cdot 5^{\frac{m}{2}}}\\ &=\underbrace{1+1+1+\dots}_{\frac{n}{2} \text{ ones}}\\ &=\left \{ \begin{array}{cl} \frac{n}{2}&n \text{ even }\\ \left\lfloor\frac{n}{2}\right\rfloor &n \text{ odd} \end{array} \right. \end{align} $$

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