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Math Help - Help with proving deffferentiable

  1. #1
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    Help with proving deffferentiable

    Define h : R - R by
    h(x) = {x^2, x rational
    h(x) = {0, x irrational


    1. Prove that h is di erentiable at 0.


    2. Prove that h is not continuous at c not = 0. You may use the fact that any
    interval in R contains both rational and irrational points.
    (Hint: Split the proof into two cases: one for c rational, the other for c irrational.)

    3. Prove that h' is a function whose domain is {0} and that h" does not
    exist.
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  2. #2
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    Quote Originally Posted by 450081592 View Post
    Define h : R - R by
    h(x) = {x^2, x rational
    h(x) = {0, x irrational


    1. Prove that h is di erentiable at 0.


    2. Prove that h is not continuous at c not = 0. You may use the fact that any
    interval in R contains both rational and irrational points.
    (Hint: Split the proof into two cases: one for c rational, the other for c irrational.)

    3. Prove that h' is a function whose domain is {0} and that h" does not
    exist.

    Ok, what've you done so far? Did you try the definition of h'(0), taking limits when x --> 0 and all x's are rational, or all are irrational? What happened? Where are you stuck in the rest?

    Tonio
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    ok, now I have proved part 1 then I know h is continuous and differentiable at 0, now hoe do I prove that h is not continuous at c not = 0, how do I define the interval in R? I know I need to prove that exist a |f(x) - L| > absola, how do I start this.
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    Quote Originally Posted by 450081592 View Post
    ok, now I have proved part 1 then I know h is continuous and differentiable at 0, now hoe do I prove that h is not continuous at c not = 0, how do I define the interval in R? I know I need to prove that exist a |f(x) - L| > absola, how do I start this.

    Do as they tell you: split in tow cases, rational and irrational: if c\neq 0 then there exists a sequence of rational points that --> c, but ALSO a sequence of irrational points that --> c . If f were continuous at c then f(x) --> f(c) no matter how we choose to make x --> c, but over rationals and over irrationals we get two different results.

    Tonio
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  5. #5
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    Quote Originally Posted by tonio View Post
    Do as they tell you: split in tow cases, rational and irrational: if c\neq 0 then there exists a sequence of rational points that --> c, but ALSO a sequence of irrational points that --> c . If f were continuous at c then f(x) --> f(c) no matter how we choose to make x --> c, but over rationals and over irrationals we get two different results.

    Tonio
    I still dont understand what that means, can you tell me how to do it please?
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  6. #6
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    Tonio was assuming that you knew that a function is differentiable at x= 0 if and only if \lim_{h\to 0}\frac{f(h)- f(0)}{h} exists. And that will be true if and only if \lim_{n\to \infty}\frac{f(h_n)- f(0)}{h_n} exists and is the same for all sequences \{h_n\} that converge to 0. Since a sequence will converge as long as subsequences converge, it is sufficient to consider sequences consistin only of rational numbers and only of irrational numbers.
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  7. #7
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    Quote Originally Posted by HallsofIvy View Post
    Tonio was assuming that you knew that a function is differentiable at x= 0 if and only if \lim_{h\to 0}\frac{f(h)- f(0)}{h} exists. And that will be true if and only if \lim_{n\to \infty}\frac{f(h_n)- f(0)}{h_n} exists and is the same for all sequences \{h_n\} that converge to 0. Since a sequence will converge as long as subsequences converge, it is sufficient to consider sequences consistin only of rational numbers and only of irrational numbers.
    so should I use the absola and delta argument to prove it, my friend hinted me to take absola = to c^2/2, will that work, how do I implement it?

    Can you show me the solution please, I am really confused now
    Last edited by 450081592; November 3rd 2009 at 11:07 AM.
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