1. Differentiable Prob

Dirichlet defined a func. g into a piecewise function depending on whether the input $x$ is rational or not. That is:

$g(x) = 1\ \text{if}\ x \in \mathbb{Q} ... 0\ \text{if}\ x \notin \mathbb{Q}$

Using the same idea that Dirchlet did with his constructions, make a function on $\mathbb{R}$ such that it's diferentiable at a single point.

2. Not sure what this prob is really looking for..would this maybe work:

$f(x) = x^2\ \text{if}\ x \in \mathbb{Q} ... -x^2\ \text{if}\ x \notin \mathbb{Q}$

3. Can you give a proof that the derivative at zero is zero?

4. Hmm well $f$ is continuous everywhere except at $x = 0$. So, $f$ can't be diff for any $x \neq 0$... I guess we can look at:

$\dfrac{f(h)-f(0)}{h} = h\ \text{if}\ h \in \mathbb{Q} ... -h\ \text{if}\ h \notin \mathbb{Q}$

Limit is 0.. not quite sure how to show. So we know $f'(0)$ exists and is $0$.

Looks like it needs a lot of work.

5. Can anyone help me w/ this proof? Or is my function wrong in the first place?

6. Define a function to be $x^2$ at irrationals and $0$ at rationals.

This is Mine 76th Post!!!

7. In either case don't you have $\left| h \right| \to 0$?

8. Originally Posted by Plato
In either case don't you have $\left| h \right| \to 0$?
Hmm, I think TPH's is easier. That is,

$f(x) = \begin{cases} x^2 &\text{if\ }\ x \notin \mathbb{Q}\\
0 &\text{if\ }\ x \in \mathbb{Q} \end{cases}$

How do I prove it's diff at a single point, namely 0?

Question says nothing abt a proof, but I thought I should be thorough or justify why this works.