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Math Help - Differentiable Prob

  1. #1
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    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.
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  2. #2
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    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}
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    Can you give a proof that the derivative at zero is zero?
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    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.
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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 x^2 at irrationals and 0 at rationals.

    This is Mine 76th Post!!!
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    In either case don't you have \left| h \right| \to 0?
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    Quote Originally Posted by Plato View Post
    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}\\<br />
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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