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Let be $$\forall n > 0, S_n = \dfrac{1}{n} \sum\limits_{k=0}^{n - 1} \exp(2i\pi u_k),\quad \forall k \geq 0, u_k = \left(\dfrac{3 + \sqrt{5}}{2}\right)^k$$

I would like to prove or disprove the convergence of $S_n$ as $n \to +\infty$.

What I have tried:

  • First, I tried to express $u_k$ as $(\phi^{2k})_k$ with $\phi$ the golden ratio and use $\phi^2 = 1 + \phi$ in the exponential, but with no success.
  • Second, I tried to establish lower / upper bounds of $S_n$ or study $S_{2n}, S_{2n + 1}$ with no success.
  • I think I could make use of the irrationality of $\phi$ but would prefer to avoid a proof based on equipartition (as this is what I'm proving in the end).
  • Also, this problem is whether $(\exp(2i\pi u_k))_k$ is Cesaro-summable.
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  • $\begingroup$ Please make the title take up less vertical space -- this is a policy to ensure that the scarce space on the main page is distributed evenly over the questions. See here for more information. Please take this into consideration for future questions. Thanks in advance. $\endgroup$ – GNUSupporter 8964民主女神 地下教會 May 14 '18 at 20:01
  • $\begingroup$ Not an answer, but intuitively, I would expect $u_k$ to behave ergodically mod $1$, in which case the sum would be close to the average of the unit circle, namely $0$. I don't expect there to be anything special about the golden ratio here. $\endgroup$ – Jair Taylor May 14 '18 at 20:10
  • $\begingroup$ Did you do some numerical experiments? I expect that the sequence $S_n$ converges to $1$. $\endgroup$ – i707107 May 14 '18 at 20:33
  • $\begingroup$ @i707107 Yes, I expect that $S_n$ converges to $0$ as Jair Taylor expect it. An efficient way to compute $S_n$ is to use $\phi^k = F_{k - 1} + F_k \phi$ and it becomes a matter of computing Fibonacci sequences. $\endgroup$ – Raito May 14 '18 at 20:39
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    $\begingroup$ For further reading, see this: en.wikipedia.org/wiki/Pisot%E2%80%93Vijayaraghavan_number $\endgroup$ – i707107 May 14 '18 at 20:55
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Let $v_k = \left(\frac{3-\sqrt{5}}{2}\right)^k$ and $w_k = u_k + v_k$. The sequence $w_k$ satisfy recurrence relation: $$w_{k+2} = 3w_{k+1} - w_k$$ Together with $w_0 = 2, w_1 = 3$, one can conclude $w_k$ is an integer sequence. This leads to

$$\exp(2\pi u_k i) = \exp(2\pi( w_k - v_k )i ) = \exp(-2\pi v_k i)$$ Since $\left|\frac{3-\sqrt{5}}{2}\right| < 1$, we have $$\lim_{k\to\infty} v_k = 0 \quad\implies\quad \lim_{k\to\infty} \exp(-2\pi v_k i) = 1\quad\implies\quad \lim_{n\to\infty} S_n = 1$$

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  • $\begingroup$ What led you to introduce $w_k$? Intuition? $\endgroup$ – Raito May 14 '18 at 20:46
  • $\begingroup$ So, I guess there is something special about $\phi$ in this case. Am I right in thinking if you replace $(3 + \sqrt{5})/2$ with $(3 + \sqrt{5})/2 + \epsilon$ the sum will converge to $0$ (assuming $\epsilon$ is a small random number)? $\endgroup$ – Jair Taylor May 14 '18 at 20:51
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    $\begingroup$ Since $w_k$ are integers, then $e^{i2\pi w_k}=1$. So, $e^{i2\pi u_k}=e^{-i2\pi v_k}\ne (-1)^{\epsilon_k}e^{-i2\pi i v_k}$. $\endgroup$ – Mark Viola May 14 '18 at 20:51
  • $\begingroup$ @MarkViola ahh, you are right. I'll fix. $\endgroup$ – achille hui May 14 '18 at 20:52
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    $\begingroup$ @JairTaylor It is known $\{\alpha^n\}$ is equidistributed mod $1$ for almost all $\alpha > 1$. One known family of exceptions are something called PV number. $\frac{3+\sqrt{5}}{2}$ is one of them. $\endgroup$ – achille hui May 14 '18 at 21:04

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