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 Post subject: A limitPosted: Fri Apr 08, 2016 1:25 pm

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 841
Location: Larisa
Let $\xi \in (-1, 1)$. Define a sequence $\{x_n\}_{n=1}^{\infty}$ as:

$$x_{n+1}=\sqrt {\frac{1}{2} ( 1+x_n)}$$

Evaluate the limit $\mathscr{L}=\lim \limits_{n \rightarrow +\infty} \cos \left ( \frac{\sqrt{1-\xi^2}}{\prod \limits_{k=1}^{n} x_k} \right )$.

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 Post subject: Re: A limitPosted: Wed Jul 26, 2017 12:33 pm

Joined: Tue Nov 24, 2015 7:47 pm
Posts: 13
I think you need to give an initial value though.

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 Post subject: Re: A limitPosted: Sun Jul 30, 2017 7:22 pm
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Joined: Mon Nov 09, 2015 1:36 am
Posts: 460
Location: Ioannina, Greece
dr.tasos wrote:
I think you need to give an initial value though.
It's not necessary! Obviously assuming that $x_1\stackrel{(*)}{>}-1$, in any case the sequence $\{x_n\}_{n=1}^{\infty}$ is monotonic and bounded.

$(*)$ If $x_1=-1$, then the sequence is the zero sequence and the fraction $\frac{\sqrt{1-\xi^2}}{\prod \limits_{k=1}^{n} x_k}$ has no meaning.

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 Post subject: Re: A limitPosted: Wed Aug 09, 2017 2:48 pm

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 841
Location: Larisa
Unfortunately,

I do not remember where I had found this particular exercise and since I cannot recover the link this means I am unable to check for any particular typos that may have occured during typesetting.

Whoops!! Mea Culpa!

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 Post subject: Re: A limitPosted: Wed Aug 09, 2017 3:16 pm
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Joined: Mon Nov 09, 2015 1:36 am
Posts: 460
Location: Ioannina, Greece
Tolaso J Kos wrote:
...I am unable to check for any particular typos that may have occured during typesetting...

The above note comes after an interchange of private messages. Let's make it more clear:

There is some problem with limit $L$ : To be equal to $\xi$ (as been given), $\mathop{\lim}\limits_{n\to+\infty} \prod_{k=1}^{n} x_k$ must exists in $\mathbb{R}$ (otherwise, if $\mathop{\lim}\limits_{n\to+\infty} \prod_{k=1}^{n} x_k$ does not exists, then the limit in question does not exists also. If $\mathop{\lim}\limits_{n\to+\infty} \prod_{k=1}^{n} x_k=\infty$ then the limit in question equals to $1$).
So let $\mathop{\lim}\limits_{n\to+\infty} \prod_{k=1}^{n} x_k=a$. Then
\begin{align*}
\mathop{\lim}\limits_{n\to+\infty} \cos \Big(\tfrac{\sqrt{1-\xi^2}}{\prod_{k=1}^{n} x_k} \Big)&= \cos \Big(\mathop{\lim}\limits_{n\to+\infty}\tfrac{\sqrt{1-\xi^2}}{\prod_{k=1}^{n} x_k} \Big)\\
&=\cos \Big(\tfrac{\sqrt{1-\xi^2}}{\mathop{\lim}\limits_{n\to+\infty}\prod_{k=1}^{n} x_k} \Big)\\
a&=\frac{\arccos\xi}{\sqrt{1-\xi^2}}\,.
\end{align*}
Because the sequence $\{x_n\}$ is not related to $\xi$, the same must hold for the $\mathop{\lim}\limits_{n\to+\infty} \prod_{k=1}^{n} x_k$. Contradiction.

So, must exists a typo somewhere in the exercise.

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