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Math Help - A question on infinite derivative.

  1. #1
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    A question on infinite derivative.

    Definition of infinite derivative: Let E=\text{dom }f, c\in E, c is a limit point of E, real-valued function f is said to have a infinite derivative at c if f is continuous at c and the limit \lim\limits_{x\to c}\frac{f(x)-f(c)}{x - c} is +\infty or -\infty. We write in this case f'(c)=+\infty or f'(c)=-\infty.
    For finitely differentiable functions it is well-known that (f+g)'=f'+g'. But if both f and g has infinite derivative at c, what about the differentiability of the sum f+g at c? It may be still differentiable, e.g. f=x^{1/3} and g=-x^{1/3}, f+g=0 so (f+g)'=0 at c=0 even if f'(c)+g'(c)=(+\infty)+(-\infty) which is undefinable. Now I want to find an example such that f'(c)=+\infty, g'(c)=-\infty, but f+g is not differentiable at c, where c is a finite real number. I hope to find a function F satisfying 1)continuous at c, 2)has bounded derivative about c but except c and 3) F' oscillates like \sin\frac{1}{x} so that F is not differentiable at c. If I can find such F, My example can be easily constructed. But I failed to find such F, Can you help me find such F, or construct the example in another way? Thanks!
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  2. #2
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    Try two functions that behave more or less the same way as the functions you gave, but don't combine. How about f(x)=\sqrt[3]{x}, and g(x)=-\sqrt[5]{x}? Here's an example where, I think, there's too much symmetry, and the solution involves destroying symmetry.
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  3. #3
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    \sqrt[3]{x}-\sqrt[5]{x} does not meet my requirement: it has derivative -\infty at 0. We should not only destroy the symmetry, but also the destroyer is not differentiable at c. Since derivatives do not have first-kind discontinuities, I think the destroyer should have an oscillating derivative, that is, second-kind discontinuity, near c, as condition 3) in my first post indicated.
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  4. #4
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    Wow. It is not intuitive at all that my candidate actually had a derivative at 0. I must confess I didn't actually check it. I just figured you'd have a cusp at the origin, which would not have a definable two-sided derivative. If you plot my function, it does appear to have a cusp at the origin. But it must hang to the left of the origin just enough to get that negative slope.

    Anyway. You've got a tough assignment there. I would look into the construction of the everywhere continuous, nowhere differentiable function. I realize that that function, as is, does not solve your problem. But its construction might give you some ideas. You might also try cycloids which have bona fide cusps.
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