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Thread: Differentiable Prob

  1. #1
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    Differentiable Prob

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

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

    Using the same idea that Dirchlet did with his constructions, make a function on $\displaystyle \mathbb{R}$ such that it's diferentiable at a single point.
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  2. #2
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    Not sure what this prob is really looking for..would this maybe work:

    $\displaystyle f(x) = x^2\ \text{if}\ x \in \mathbb{Q} ... -x^2\ \text{if}\ x \notin \mathbb{Q}$
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  3. #3
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    Can you give a proof that the derivative at zero is zero?
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    Hmm well $\displaystyle f$ is continuous everywhere except at $\displaystyle x = 0$. So, $\displaystyle f$ can't be diff for any $\displaystyle x \neq 0$... I guess we can look at:

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

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

    Looks like it needs a lot of work.
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    Can anyone help me w/ this proof? Or is my function wrong in the first place?
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    Define a function to be $\displaystyle x^2$ at irrationals and $\displaystyle 0$ at rationals.

    This is Mine 76th Post!!!
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    In either case don't you have $\displaystyle \left| h \right| \to 0$?
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    Quote Originally Posted by Plato View Post
    In either case don't you have $\displaystyle \left| h \right| \to 0$?
    Hmm, I think TPH's is easier. That is,

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

    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.
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