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Bounded sequence

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Papapetros Vaggelis
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Bounded sequence

#1

Post by Papapetros Vaggelis » Wed Oct 26, 2016 6:51 pm

Let \(\displaystyle{x\in\mathbb{R}}\). Prove that the sequence

\(\displaystyle{s_n=\sin\,x+...+\sin\,(n\,x)\,,n\in\mathbb{N}}\) is bounded.
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Grigorios Kostakos
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Re: Bounded sequence

#2

Post by Grigorios Kostakos » Thu Oct 27, 2016 8:17 am

For $x\neq2m\pi\,,\; m\in\mathbb{Z}$, maybe the formula
\[s_n=\mathop{\sum}\limits_{k=1}^n\sin(kx)=\frac{\sin\big(\frac{(n+1)\,x}{2}\big)\,\sin\big(\frac{nx}{2}\big)}{\sin\big(\frac{x}{2}\big)}\] and that \[|s_n|=\Bigg|\frac{\sin\big(\frac{(n+1)\,x}{2}\big)\,\sin\big(\frac{nx}{2}\big)}{\sin\big(\frac{x}{2}\big)}\Bigg|\leqslant\frac{1}{\big|\sin\big(\frac{x}{2}\big)\big|}\] could be useful.




edit: 28/10/2016, 07:40. The condition $x\neq0$ replased by $x\neq2m\pi\,,\; m\in\mathbb{Z}$.
Last edited by Grigorios Kostakos on Fri Oct 28, 2016 5:41 am, edited 1 time in total.
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Re: Bounded sequence

#3

Post by Riemann » Thu Oct 27, 2016 9:38 am

What Grigorios has written is the one way to go I think!

Let us see something interesting here. Let us see what happens to infinity. We have seen that $s_n$ is bounded. Note that there are some values of $x$ for which the series $\sum \limits_{n=1}^{\infty} \sin nx$ converges and others that does not.
  1. In a general sense , including Cesaro summation sense and Abel summation sense the following holds:

    $$\sum_{n=1}^{\infty} \sin nx = \Im \left(\frac{e^{ix}}{1-e^{ix}}\right) = \frac{1}{2}\cot\frac{x}{2}$$
  2. The series converges if we assume that we are in the upper half plane , that is if $\Im (x) >0$, diverges if $x$ is real unless $x \in \pi \mathbb{Z}$.
  3. In general

    $$\sum_{n = 1}^\infty \sin nx \neq \Im \bigg[\sum_{n = 1}^\infty e^{inx}\bigg]$$

    Why?
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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Re: Bounded sequence

#4

Post by S.F.Papadopoulos » Thu Oct 27, 2016 2:14 pm

It is not bounded near zero.
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Re: Bounded sequence

#5

Post by Riemann » Thu Oct 27, 2016 6:36 pm

S.F.Papadopoulos wrote:It is not bounded near zero.
Yes, you are right. I , myself, oversaw that. Thus we need to compe up with another stategy to attack this.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
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Re: Bounded sequence

#6

Post by Papapetros Vaggelis » Fri Oct 28, 2016 11:46 am

Here is a solution.

If \(\displaystyle{x=0}\) or \(\displaystyle{x=2\,\pi}\), then \(\displaystyle{s_n=0\,,n\in\mathbb{N}}\)

and we are done.

It is sufficient to prove the argument if \(\displaystyle{x\in\left(0,2\,\pi\right)}\) since the \(\displaystyle{\sin}\) - function

is a \(\displaystyle{2\,\pi}\) - periodical function.

So, for every \(\displaystyle{n\in\mathbb{N}}\), we have that

\(\displaystyle{\begin{aligned} s_n&=\sum_{k=1}^{n}\sin\,(k\,x)\\&=\dfrac{1}{2\,i}\,\sum_{k=1}^{n}\left(e^{i\,k\,x}-e^{-i\,k\,x}\right)\\&=\dfrac{1}{2\,i}\,\sum_{k=1}^{n}(e^{i\,x})^k-\dfrac{1}{2\,i}\,\sum_{k=1}^{n}(e^{-i\,x})^k\,\,(I)\end{aligned}}\).

Since \(\displaystyle{x\in\left(0,2\,\pi\right)}\), we get \(\displaystyle{e^{i\,x}\,,e^{-i\,x}\neq 1}\),

so the relation \(\displaystyle{(I)}\) gives us

\(\displaystyle{\begin{aligned} s_n&=\dfrac{e^{i\,x}}{2\,i}\,\dfrac{e^{i\,n\,x}-1}{e^{i\,x}-1}-\dfrac{e^{-i\,x}}{2\,i}\,\dfrac{e^{-i\,n\,x}-1}{e^{-i\,x}-1}\\&=\dfrac{e^{i\,x}}{2\,i}\,\dfrac{e^{i\,n\,x}-1}{e^{i\,x}-1}+\dfrac{1}{2\,i}\,\dfrac{e^{-i\,n\,x}-1}{e^{i\,x}-1}\end{aligned}}\)

Therefore,

\(\displaystyle{|s_n|\leq \dfrac{2}{2\,|e^{i\,x}-1|}+\dfrac{2}{2\,|e^{i\,x}-1|}=\dfrac{1}{\left|\sin\,\dfrac{x}{2}\right|}\,,\forall\,n\in\mathbb{N}}\).
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Re: Bounded sequence

#7

Post by Grigorios Kostakos » Fri Oct 28, 2016 1:44 pm

Papapetros Vaggelis wrote:Let \(\color{red}{x\in\mathbb{R}}\). Prove that the sequence

\(\displaystyle{s_n=\sin x+\dots+\sin(n\,x)\,,\; n\in\mathbb{N}}\) is bounded.
I think there is a misunderstanding here about boundedness. Let me try to elucidate. If we consider $x$ as a fixed real number, then the sequence $s_n(x)$ is a sequence of real numbers and a boundary for this sequence is $\frac{1}{\sin({x}/{2})}$.
But, if we think $s_n(x)$ as an sequence of functions, then boundedness, means uniform boundedness, that is there exists a real number $M$ such that \[(\forall\, x\in\mathbb{R})(\forall\, n\in\mathbb{N}) \quad |s_n(x)|\leqslant M.\]
So, $s_n(x)$ is bounded for every $x\in\mathbb{R}$ by $\frac{1}{\sin({x}/{2})}$, but it is not uniformly bounded(*), according to the definition above.

(*) Proof left as an exercise!

P.S. I don't blame anyone here! I think that Vaggelis was quite clear by putting $x\in\mathbb{R}$ in the first line.
Grigorios Kostakos
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