Results 1 to 10 of 10

Math Help - Thomae's function square

  1. #1
    aid
    aid is offline
    Newbie
    Joined
    Nov 2010
    Posts
    3

    Thomae's function square

    Let's take function given by a condition:

    f(x) =  \begin{cases} \frac{1}{q^2} \leftrightarrow \ x \in \mathbb{Q} \wedge x = \frac{p}{q},\\ 0 \leftrightarrow \ x \notin \mathbb{Q} \end{cases}

    Prove the existence of the derivative of f in all points x \notin \mathbb{Q}.

    So, I am aware that if there was q standing in the formula instead of q^2, the derivative wouldn't exist (that would simply be the Thomae's function). The thing I couldn't figure out is, why would the replacement change anything and where should I start the proof?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45
    If q\in \mathbb{N}^* and a\in \mathbb{R}-\mathbb{Q} consider the open interval:

    I_q=\left(a-\dfrac{1}{2q},a+\dfrac{1}{2q}\right)

    This interval contains a rational x=p/q because:

    \dfrac{p}{q}\in{I_q}\Leftrightarrow \ldots \Leftrightarrow aq-\dfrac{1}{2}<p<aq+\dfrac{1}{2}

    and the endpoints of I_q are irrationals.

    Now analize:

    \dfrac{f(x)-f(a)}{x-a}\quad (q\rightarrow{+\infty})

    considering the cases q prirme and q composite.

    Regards.

    Fernando Revilla
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,545
    Thanks
    780
    Hmm, this article thinks otherwise. Apparently, it was published in American Mathematical Monthly in 2009.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45
    Quote Originally Posted by emakarov View Post
    Hmm, this article thinks otherwise. Apparently, it was published in American Mathematical Monthly in 2009.

    I can't see most of characters in the article, but surely they are no talking exactly about the same function.

    Regards.

    Fernando Revilla
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,957
    Thanks
    1631
    No, they are specifically talking about T_{\frac{1}{n^2}} which is almost exactly the same function except that the function in the article is defined to be 1 at x= 0, while the function here is not defined at x= 0. Since the conclusion is "there exist am uncountable set or irrational numbers, dense in the real numbers, on which T_{\frac{1}{n^2}} is not continuous, and 0, of course, is rational, that would not affect the proof.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45
    Thanks. I'm afraid I need a new computer. :-)

    Regards.

    ---
    Fernando Revilla
    Follow Math Help Forum on Facebook and Google+

  7. #7
    aid
    aid is offline
    Newbie
    Joined
    Nov 2010
    Posts
    3
    Thanks for all the answers.

    So, basically speaking, the theorem I stated was entirely false.

    What does Fernando's reasoning prove then? (I am going to look into it later but was unable to decipher it quickly.)
    And where the heck is the square of Thomae's function differentiable?
    Follow Math Help Forum on Facebook and Google+

  8. #8
    aid
    aid is offline
    Newbie
    Joined
    Nov 2010
    Posts
    3
    Quote Originally Posted by FernandoRevilla View Post

    Now analize:

    \dfrac{f(x)-f(a)}{x-a}\quad (q\rightarrow{+\infty})

    considering the cases q prirme and q composite.

    Regards.

    Fernando Revilla
    Dear Fernando Revilla,

    your message is entirely clear to me until this part. Isn't it true that, either q is prime or not, the expression:

    \frac{f(\frac{p}{q} )}{\frac{p}{q} - a} = \frac{1}{q} \cdot \frac{1}{p - aq}

    converges to 0 whenever q \rightarrow +\infty? (Since \frac{1}{q} \rightarrow 0 and the second element of the product is limited?)

    Sorry if my questions sound rather silly.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45
    Quote Originally Posted by aid View Post
    What does Fernando's reasoning prove then?

    Forget it, Fernando had an unfortunate intuition. :-)

    Regards
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Newbie
    Joined
    Dec 2010
    Posts
    1
    a
    Last edited by lagrange; December 17th 2010 at 04:39 PM.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. square root function
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: September 16th 2011, 09:32 PM
  2. Example of function not integrable and the square is
    Posted in the Differential Geometry Forum
    Replies: 7
    Last Post: July 6th 2011, 02:45 AM
  3. Replies: 1
    Last Post: December 3rd 2009, 08:45 AM
  4. function and completing square
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: August 8th 2008, 12:27 AM
  5. Square Root Function
    Posted in the Pre-Calculus Forum
    Replies: 8
    Last Post: May 17th 2008, 10:50 AM

Search Tags


/mathhelpforum @mathhelpforum