# If $V_n(a)$ counts sign changes in the sequence $\cos a, \cos2a,\cos3a,\ldots,\cos na,$ show that $\lim_{n\to\infty}\frac{V_n(a)}n=\frac{a}\pi$

Let $$0\leq\alpha\leq \pi$$. $$V_n (\alpha)$$ denote the number of sign changes in the sequence $$\cos\alpha,\cos2\alpha,\cos3\alpha,\ldots,\cos n\alpha$$. Then prove that $$\lim\limits_{n\to\infty}\dfrac{V_n (\alpha)}{n}=\dfrac{\alpha}{\pi}.$$

I saw a hint where $$\dfrac{V_n (\alpha)}{n}$$ is considered as the probability. I mean how this expression is a probability of something. If it is, how can I progress further in this way?

Update: I have a solution to this problem

In $$n\alpha$$ rotation the number of times full circle rotation occures $$=\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor$$

In one full circle rotation sign change occures 2 times. Hence in $$\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor$$ full rotation sign change occures $$=2\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor$$

Now the rest angle is $$n\alpha-\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor\times2\pi$$

If we consider 0 as a change of sign in case of $$\cos\left( \dfrac{\pi}{2}\right)$$ and $$\cos\left(\dfrac{3\pi}{2}\right)$$ then:-

(1) If $$0\leq n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi<\dfrac{\pi}{2 }$$ sign changes 0 times

(2) If $$\dfrac{\pi}{2 }\leq n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi<\dfrac{3\pi}{2 }$$ sign changes 1 times

(3) If $$\dfrac{3\pi}{2 }\leq n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi<2\pi$$ sign changes 2 times

Let $$f$$ be a function such that $$f\left(\left\lfloor \dfrac{n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi}{\dfrac{\pi}{2}}\right\rfloor\right)=\begin{cases}0,\text{ when }\left\lfloor \dfrac{n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi}{\dfrac{\pi}{2}}\right\rfloor=0\\ 1,\text{ when }\left\lfloor \dfrac{n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi}{\dfrac{\pi}{2}}\right\rfloor=1\\ 1,\text{ when }\left\lfloor \dfrac{n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi}{\dfrac{\pi}{2}}\right\rfloor=2\\ 2,\text{ when } \left\lfloor \dfrac{n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi}{\dfrac{\pi}{2}}\right\rfloor=3\end{cases}$$

Therefore $$\dfrac{V_n(\alpha)}{n}=\dfrac{2\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor+ f\left(\left\lfloor \dfrac{n\alpha-\bigg\lfloor\dfrac{n\alpha }{2\pi }\bigg\rfloor\times 2\pi}{\dfrac{\pi}{2}}\right\rfloor\right)}{n}$$

Hence $$\dfrac{V_n(\alpha)}{n}\geq \dfrac{2\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor}{n}$$ and $$\dfrac{2\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor+ 2}{n}\leq \dfrac{V_n(\alpha)}{n}$$

$$\lim\limits_{n\to \infty}\dfrac{2\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor}{n}=\dfrac{\alpha}{\pi}$$ and $$\lim\limits_{n\to\infty} \dfrac{2\bigg\lfloor\dfrac{n\alpha}{2\pi}\bigg\rfloor+ 2}{n}=\dfrac{\alpha}{\pi}$$

Hence by Sandwich Theorem We get $$\lim\limits_{n\to \infty}\dfrac{V_n(\alpha)}{n}=\dfrac{\alpha}{\pi}$$ [Proved]

Is this correct?

• Hint: Think of hopping around the unit circle. The probability is an "area" (really an angle or arclength of some sort) Commented Aug 18, 2020 at 5:52
• I thoughy of that to...unit radius circle..then $\alpha$ becomes the arc length...but no progress..if you have something please tell me Commented Aug 18, 2020 at 5:53
• @Ninad Munshi area of what Commented Aug 18, 2020 at 6:07
• Area in this context means length. So the "area" (arclength) of the circle Commented Aug 18, 2020 at 6:23
• Nope cant do it...please upload the solution Commented Aug 18, 2020 at 7:31

• I will assume that the case where $$\alpha\in \pi \mathbb{Q}$$ is easy, because the sequence $$\big(\cos(k\alpha)\big)_{k\ge1}$$ is periodic in this case, and if we consider $$0$$ as positive number then $$V_{2q}(p\pi/q)=2p\pm 1$$ and the result holds in this case.
• Now we assume that $$\alpha\notin \pi\mathbb{Q}$$. This implies that the sequence $$\big(k\alpha \mod(2\pi)\big)_{k\geq 1}$$ is equidistributed in $$[0,2\pi]$$. See Equidistributed sequences.
Now, let $$f$$ be the $$2\pi$$ periodic function defined by $$f(\theta)=\cases{0, & if \cos\theta \cos(\theta+\alpha)\geq0,\\ 1,& if \cos\theta \cos(\theta+\alpha)<0.}$$ With this definition, $$V_n(\alpha)=\text{card}\left\{k\in\{1,\ldots,n\}:f(k\alpha)=1\right\}$$ But if we define $$\mathcal{I}=\cases{\left(\frac{\pi}{2}-\alpha,\frac{\pi}2\right)\cup \left(\frac{3\pi}{2}-\alpha,\frac{3\pi}2\right) ,&if 0<\alpha<\pi/2,\cr \left[0,\frac{\pi}{2}\right)\cup \left(\frac{3\pi}{2}-\alpha,\frac{3\pi}2\right)\cup\left(\frac{5\pi}{2}-\alpha,2\pi\right] ,&if \pi/2<\alpha<\pi.}$$ Then for $$\theta\in[0,2\pi]$$ we have $$f(\theta)=1\iff \theta\in\mathcal{I}$$ So, equidistribution of the sequence implies that $$\lim_{n\to\infty}\frac{V_n(\alpha)}{n}=\frac{\text{length}(\mathcal{I})}{2\pi}=\frac{\alpha}{\pi}$$ Done.$$\qquad\square$$
HINT: let $$b_n\equiv n a \pmod {2\pi}$$ indicate the angle formed with the $$x$$- axis in the $$n^{th}$$ term of the sequence. Assume that $$b$$ is uniformly distributed in the range between $$0$$ and $$2\pi$$.
Now firstly consider the case in which $$0 or $$3\pi/2. In the next step, a change of sign will occur only if $$b_{n+1}>\pi/2$$. What is the probability that this occurs, given that $$b_{n+1}=b_n+a$$?
Then repeat the same considerations for the case in which $$\pi/2. A change of sign will occur only if $$b_{n+1}>3\pi/2$$. What is the probability that this occurs?