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Math Help - Flawed reasoning... but where?

  1. #1
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    Flawed reasoning... but where?

    I've written the question out on MathB.in for simplicity: Flawed Reasoning - MathB.in

    My gut feeling is that the answer is to do with the open interval in the hypothesis of the mean value theorem, but I'm not sure.
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    Re: Flawed reasoning... but where?

    Quote Originally Posted by Archie View Post
    I've written the question out on MathB.in for simplicity: Flawed Reasoning - MathB.in

    My gut feeling is that the answer is to do with the open interval in the hypothesis of the mean value theorem, but I'm not sure.
    f' does not exist at 0 so any interval containing 0 fails to meet the criteria of the mean value theorem.
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    Re: Flawed reasoning... but where?

    All derivatives, while possibly incredibly discontinuous, are Darboux continuous (which means they possess the Intermediate Value Property). Hence, the two results are not contradictory. What you are missing is that \theta is dependent upon h. So, the correct formula would be \lim_{h \to 0}f'(a+\theta(h)h) = f'(a). That does not imply that f'(x) is continuous at x=a. On the LHS, you have a composition of functions. On the RHS, you do not.
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    Re: Flawed reasoning... but where?

    Quote Originally Posted by SlipEternal View Post
    So, the correct formula would be \lim_{h \to 0}f'(a+\theta(h)h) = f'(a). That does not imply that f'(x) is continuous at x=a. On the LHS, you have a composition of functions. On the RHS, you do not.
    I agree that \theta is dependant on h, but 0 < \theta(h) < 1 so \lim_{h \to 0}{\theta(h)h} = 0. The only reason I can see that we would not have f'(x) being continuous is that \theta(h) is not necessarily continuous.
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  5. #5
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    Re: Flawed reasoning... but where?

    That is obvious that \theta(h) does not need to be continuous. The MVT guarantees the EXISTENCE of such a \theta, it does not give a limit for the NUMBER of such values. For a chosen h, there are, in fact, an infinite number of such values for \theta(h) that will satisfy the MVT. It is almost guaranteed that function will be discontinuous.

    Consider the topology of the space of all points \{h\}\times \psi(h) where \psi(h) are all possible values of \theta that satisfy the MVT for the specific h. If 0<h<1, this is a subset of (0,1)\times (0,1). That might give you an idea of the nature of a chosen function \theta(h).
    Last edited by SlipEternal; January 13th 2014 at 08:35 AM.
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