# Thomae's function square

• Nov 12th 2010, 12:16 AM
aid
Thomae's function square
Let's take function given by a condition:

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

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

So, I am aware that if there was $\displaystyle q$ standing in the formula instead of $\displaystyle q^2$, 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?
• Nov 12th 2010, 01:28 AM
FernandoRevilla
If $\displaystyle q\in \mathbb{N}^*$ and $\displaystyle a\in \mathbb{R}-\mathbb{Q}$ consider the open interval:

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

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

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

and the endpoints of $\displaystyle I_q$ are irrationals.

Now analize:

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

considering the cases $\displaystyle q$ prirme and $\displaystyle q$ composite.

Regards.

Fernando Revilla
• Nov 12th 2010, 01:30 AM
emakarov
Hmm, this article thinks otherwise. Apparently, it was published in American Mathematical Monthly in 2009.
• Nov 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
• Nov 12th 2010, 02:22 AM
HallsofIvy
No, they are specifically talking about $\displaystyle T_{\frac{1}{n^2}}$ 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 $\displaystyle T_{\frac{1}{n^2}}$ is not continuous, and 0, of course, is rational, that would not affect the proof.
• Nov 12th 2010, 02:49 AM
FernandoRevilla
Thanks. I'm afraid I need a new computer. :-)

Regards.

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

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?
• Nov 12th 2010, 04:27 AM
aid
Quote:

Originally Posted by FernandoRevilla

Now analize:

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

considering the cases $\displaystyle q$ prirme and $\displaystyle q$ composite.

Regards.

Fernando Revilla

Dear Fernando Revilla,

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

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

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

Sorry if my questions sound rather silly.
• Nov 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
• Dec 17th 2010, 11:58 AM
lagrange
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