Thomae's function square

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November 12th 2010, 12:16 AM
aid

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?
November 12th 2010, 01:28 AM
FernandoRevilla

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
November 12th 2010, 01:30 AM
emakarov

Hmm, this article thinks otherwise. Apparently, it was published in American Mathematical Monthly in 2009.
November 12th 2010, 01:49 AM
FernandoRevilla

Quote:

Originally Posted by emakarov

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
November 12th 2010, 02:22 AM
HallsofIvy

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.
November 12th 2010, 02:49 AM
FernandoRevilla

Thanks. I'm afraid I need a new computer. :-)

Regards.

---
Fernando Revilla
November 12th 2010, 03:04 AM
aid

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?
November 12th 2010, 04:27 AM
aid

Quote:

Originally Posted by FernandoRevilla

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.
November 12th 2010, 07:08 AM
FernandoRevilla

Quote:

Originally Posted by aid

What does Fernando's reasoning prove then?

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

Regards
December 17th 2010, 11:58 AM
lagrange

a