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A limit

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Tolaso J Kos
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A limit

#1

Post by Tolaso J Kos » Fri Apr 08, 2016 1:25 pm

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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As strangely as it sounds the limit evaluates to $\xi$.
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Re: A limit

#2

Post by dr.tasos » Wed Jul 26, 2017 12:33 pm

I think you need to give an initial value though.
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Grigorios Kostakos
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Re: A limit

#3

Post by Grigorios Kostakos » Sun Jul 30, 2017 7:22 pm

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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Re: A limit

#4

Post by Tolaso J Kos » Wed Aug 09, 2017 2:48 pm

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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Re: A limit

#5

Post by Grigorios Kostakos » Wed Aug 09, 2017 3:16 pm

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)\\
&=\cos \Big(\tfrac{\sqrt{1-\xi^2}}{α} \Big)=\xi\quad\Rightarrow\\
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.
Grigorios Kostakos
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