Facebook Page
Twitter
RSS
+ Reply to Thread
Results 1 to 5 of 5

Thread: tough limit

  1. MHB Apprentice

    Status
    Offline
    Join Date
    Jan 2013
    Posts
    29
    Thanks
    20 times
    Thanked
    1 time
    #1
    $ \displaystyle \lim_{n\to\infty} \left (\lim_{k\to\infty} \cos (\left| n! \pi x\right|) ^{2k} \right) = \begin{cases} 1&x \in \mathbb{Q} \\0& x \not\in \mathbb{Q}\end{cases} $. How to prove it?
    Last edited by Opalg; January 17th, 2013 at 04:53. Reason: further LaTeX fixing

  2. MHB Master
    MHB Global Moderator
    MHB Math Helper
    MHB Ambassador
    Sudharaka's Avatar
    Status
    Offline
    Join Date
    Feb 2012
    Location
    Canada
    Posts
    1,608
    Thanks
    7,490 times
    Thanked
    2,524 times
    Thank/Post
    1.570
    Awards
    University POTW Award (Jan-June 2013)
    #2
    Quote Originally Posted by Lisa91 View Post
    $ \displaystyle \lim_{n\to\infty} \left (\lim_{k\to\infty} \cos (\left| n! \pi x\right|) ^{2k} \right) = \left\{\begin{array}{l} 1 x \in \mathbb{Q}\\1 x \not\in \mathbb{Q}\end{array}\right. $. How to prove it?
    Hi Lisa91,

    I think there is a typo in the right hand side of the equation. What do you mean by \(1x\)? Is it,

    $ \displaystyle \lim_{n\to\infty} \left (\lim_{k\to\infty} \cos (\left| n! \pi x\right|) ^{2k} \right) = \left\{\begin{array}{l} x \in \mathbb{Q}\\1 \not\in \mathbb{Q}\end{array}\right. $

    Kind Regards,
    Sudharaka.

  3. MHB Master
    MHB Site Helper
    MHB Math Helper
    chisigma's Avatar
    Status
    Offline
    Join Date
    Feb 2012
    Location
    Beograd, Serbia
    Posts
    1,704
    Thanks
    393 times
    Thanked
    3,370 times
    Thank/Post
    1.978
    Awards
    MHB Statistics Award (2014)  

MHB Statistics Award (Jan-June 2013)
    #3
    Quote Originally Posted by Lisa91 View Post
    $ \displaystyle \lim_{n\to\infty} \left (\lim_{k\to\infty} \cos (\left| n! \pi x\right|) ^{2k} \right) = \left\{\begin{array}{l} 1 \in \mathbb{Q}\\ 0 \not\in \mathbb{Q}\end{array}\right. $. How to prove it?
    With great probability Lisa intends the Diriclet function defined as in...



    $\displaystyle D(x)= \lim_{m \rightarrow \infty} \lim_{n \rightarrow \infty} \cos^ {2 n} (m!\ \pi\ x) = \left\{\begin{array}{l} 1 \in \mathbb{Q}\\ 0 \not\in \mathbb{Q}\end{array}\right.$ (1)

    The 'proof' is relatively easy because if $x \in \mathbb{Q}$ then $x=\frac{p}{q}$ with p and q integer coprimes. Now if $m \rightarrow \infty$ for a certain $m>m_{0}$ q divides m! and it will be $\cos (m!\ \pi\ x) = \pm 1$. Anyway I personally have more than one doubt on the 'logical architecture' of the definition of the Diriclet Function...

    Kind regards

    $\chi$ $\sigma$

  4. MHB Apprentice

    Status
    Offline
    Join Date
    Jan 2013
    Posts
    29
    Thanks
    20 times
    Thanked
    1 time
    #4 Thread Author
    $ \displaystyle \lim_{n\to\infty} \left (\lim_{k\to\infty} \cos (\left| n! \pi x\right|) ^{2k} \right) = \begin{cases} 1&x \in \mathbb{Q} \\0& x \not\in \mathbb{Q}\end{cases} $

    No, this '2k' has to be in the place I wrote.

  5. MHB Master
    MHB Site Helper
    MHB Math Helper
    chisigma's Avatar
    Status
    Offline
    Join Date
    Feb 2012
    Location
    Beograd, Serbia
    Posts
    1,704
    Thanks
    393 times
    Thanked
    3,370 times
    Thank/Post
    1.978
    Awards
    MHB Statistics Award (2014)  

MHB Statistics Award (Jan-June 2013)
    #5
    Quote Originally Posted by Lisa91 View Post
    $ \displaystyle \lim_{n\to\infty} \left (\lim_{k\to\infty} \cos (\left| n! \pi x\right|) ^{2k} \right) = \begin{cases} 1&x \in \mathbb{Q} \\0& x \not\in \mathbb{Q}\end{cases} $

    No, this '2k' has to be in the place I wrote.
    Also in this case if x is rational for any $n>n_{0}$ the term $n!\ x$ is an even integer so that for any k is $\cos \{(n!\ \pi\ x)^{2k}\}=1$. The problem however is when x is irrational because in this case [probably...] $\cos \{(n!\ \pi\ x)^{2k}\}$ has no limits in n and k...


    Kind regards


    $\chi$ $\sigma$

Similar Threads

  1. A limit
    By DigitalComputer in forum Calculus
    Replies: 1
    Last Post: November 23rd, 2012, 06:00
  2. [SOLVED] Limit x and y
    By dwsmith in forum Calculus
    Replies: 1
    Last Post: November 12th, 2012, 21:52
  3. Limit
    By Amer in forum Calculus
    Replies: 3
    Last Post: April 5th, 2012, 15:45
  4. Limit
    By anemone in forum Pre-Calculus
    Replies: 8
    Last Post: February 25th, 2012, 00:04
  5. Limit
    By Alexmahone in forum Calculus
    Replies: 7
    Last Post: February 23rd, 2012, 09:43

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •  
Math Help Boards