Limit NO. 1 : The limit points of the sin n sequence

14Nov08

The proof for the limit points of the sequence a_n = \sin{n} haven’t given me a calm dream until today.

Of course the limit points of that sequence (almost intuitively) \in [-1;1]

Lemma.

\mbox{There can be found a sequence } n_k \in \mathbb{N} \mbox{ such that:}

\lim_{k\to\infty} \lfloor n_k \rfloor_{2\pi} =\lim_{k\to\infty} n_k - \lfloor \frac{n_k}{2\pi} \rfloor 2\pi = 0

Lemma proof.

\lfloor n \rfloor_{2\pi} = n-\lfloor \frac{n}{2\pi} \rfloor 2\pi \quad n \in \mathbb{N}

from floor function definition web have the following inequality:

0 < \lfloor n \rfloor_{2\pi} < 2\pi

Now we will prove that

\forall_{n_0 \in \mathbb{N}} \Rightarrow \exists_{n_1 \in \mathbb{N}} \lfloor n_{1} \rfloor_{2\pi} < \lfloor n_{0} \rfloor_{2\pi}

Let \lfloor n_{0} \rfloor_{2\pi} = \alpha \quad \lfloor \frac{n_0}{2 \pi} \rfloor = k_0 \in \mathbb{N} where \alpha \in (0, 2 \pi)

So we have \alpha = n_0 - 2k_0 \pi

n_0=2k_0 \pi + \alpha

Let p = \lfloor \frac{2\pi}{\alpha} \rfloor \quad p > 1

so of course p\alpha < 2\pi < \alpha(p+1)

Lets consider n_{0}(p+1) = (2k_{0}\pi + \alpha )(p+1) = 2k_{0}(p+1)\pi + \alpha(p+1) =

= 2k_{0}(p+1)\pi + 2\pi + \alpha(p+1) - 2\pi = 2\pi(k_{0}(p+1) +1 ) +\beta

and look closer at \beta

The following is true 0 < \alpha (p+1) -2\pi = \beta < \alpha

It’s also simple to show (for example geometricly), that we can find such s \in \mathbb{N} that the following inequality is satisfied \frac{\alpha}{2} \leq s(\alpha-\beta) < \alpha

Now we will consider n_{1} = n_{0} + sn_{0}p \in \mathbb{N}

n_{1} = 2k_{0}\pi + \alpha + s(2\pi ( k_{0} (p+1) +1)+\beta - 2k_{0} \pi -\alpha)=

= 2\pi(k_{0}(sp+1)+s) + \alpha -(\alpha-\beta)s

From which \lfloor n_{1} \rfloor_{2\pi}= 2\pi(k_{0}(sp+1)+s)+\alpha - (\alpha-\beta)s - \lfloor (k_{0}(sp+1)+s)+ \frac{\alpha - (\alpha-\beta)s}{2\pi} \rfloor 2\pi =

2\pi(k_{0}(sp+1)+s)+\alpha-(\alpha-\beta)s-2\pi(k_{0}(sp+1)+s)

=\alpha-(\alpha-\beta)s \leq \alpha-\frac{\alpha}{2}=\frac{\alpha}{2}

And that means that for all k \in \mathbb{N} \lfloor n_{k} \rfloor_{2\pi} \leq \frac{1}{2^k} \lfloor n_{0} \rfloor_{2\pi} so \lim \lfloor n_{k} \rfloor_{2\pi} = 0  ■

Proof.

We’ll show that \forall_{s\in [-1,1]} \exists_{l_k\in\mathbb{N}} \lim_{k\in\infty}\sin{l_k} = s

Lets take \lfloor m \rfloor_{2\pi} < \beta - \alpha where m \in \lbrace n_k \rbrace and \alpha \in [0;2\pi), \beta \in [0;2\pi)

Then \exists_{k\in\mathbb{N}}{k}\lfloor{m}\rfloor_{2\pi}\in(\alpha,\beta) ( because the step is smaller then the range)

So when’ll take n=mk then

\sin{n}=\sin{mk}=\sin{\left(\lfloor m\rfloor_{2\pi} + \lfloor\frac{m}{2\pi}\rfloor 2\pi \right) \cdot k}=\sin{k\lfloor m \rfloor_{2\pi}}\in (s-\epsilon,s+\epsilon)

which satisfies the limit definition. ■



One Response to “Limit NO. 1 : The limit points of the sin n sequence”

  1. Thanks for your Solution and I appreciate you for having a serious fight with this problem as I have spent some days like you.
    Cheers


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