It is currently Wed Dec 19, 2018 7:28 am


All times are UTC [ DST ]




Post new topic Reply to topic  [ 5 posts ] 
Author Message
 Post subject: A limit
PostPosted: Fri Apr 08, 2016 1:25 pm 
Administrator
Administrator
User avatar

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 837
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 $\mathscr{L}=\lim \limits_{n \rightarrow +\infty} \cos \left ( \frac{\sqrt{1-\xi^2}}{\prod \limits_{k=1}^{n} x_k} \right )$.

Hidden Message- Answer
As strangely as it sounds the limit evaluates to $\xi$.

_________________
Imagination is much more important than knowledge.
Image


Top
Offline Profile  
Reply with quote  

 Post subject: Re: A limit
PostPosted: 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.


Top
Offline Profile  
Reply with quote  

 Post subject: Re: A limit
PostPosted: Sun Jul 30, 2017 7:22 pm 
Team Member
User avatar

Joined: Mon Nov 09, 2015 1:36 am
Posts: 454
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.

_________________
Grigorios Kostakos


Top
Offline Profile  
Reply with quote  

 Post subject: Re: A limit
PostPosted: Wed Aug 09, 2017 2:48 pm 
Administrator
Administrator
User avatar

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 837
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!

_________________
Imagination is much more important than knowledge.
Image


Top
Offline Profile  
Reply with quote  

 Post subject: Re: A limit
PostPosted: Wed Aug 09, 2017 3:16 pm 
Team Member
User avatar

Joined: Mon Nov 09, 2015 1:36 am
Posts: 454
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)\\
&=\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


Top
Offline Profile  
Reply with quote  

Display posts from previous:  Sort by  
Post new topic Reply to topic  [ 5 posts ] 

All times are UTC [ DST ]


Mathimatikoi Online

Users browsing this forum: SemrushBot and 1 guest


You cannot post new topics in this forum
You cannot reply to topics in this forum
You cannot edit your posts in this forum
You cannot delete your posts in this forum
You cannot post attachments in this forum

Search for:
Jump to:  
cron
Powered by phpBB® Forum Software © phpBB Group Color scheme created with Colorize It.
Theme created StylerBB.net