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Thread: Pathological example

  1. #1
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    Pathological example

    Will someone help me on this problem?
    Let $\displaystyle f(x):R \rightarrow R$ be a continuous but differentiable nowhere ( for example the Weierstrass's function). Define $\displaystyle F(x)=\int_0^x f(r)dr$. Why does F(x) has the first derivative everywhere?
    Does F(x) have a second derivative?

    I think F(x) has the first derivative since that's why the FTOC part 2 says, but I'm stuck on part 2.
    Last edited by jackie; Sep 14th 2009 at 10:50 PM.
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  2. #2
    MHF Contributor chisigma's Avatar
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    By definition of 'integral function' if...

    $\displaystyle F(x)= \int_{0}^{x} f(\tau)\cdot d\tau$ (1)

    ... then $\displaystyle F(*)$ has prime derivative everywhere $\displaystyle f(*)$ is continous and is...

    $\displaystyle F^{'} (x)= f(x)$ (2)

    But $\displaystyle f(*)$ in continous $\displaystyle \forall x \in \mathbb{R}$ so that $\displaystyle F^{'}(*)$ exists $\displaystyle \forall x \in \mathbb{R}$...

    Now from (2) we derive that...

    $\displaystyle F^{''} (x) = f^{'} (x)$ (3)

    ... so that, because $\displaystyle f^{'} (*)$ nowhere exists, the same is for $\displaystyle F^{''} (*)$...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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  3. #3
    MHF Contributor chisigma's Avatar
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    According to...

    Weierstrass Function -- from Wolfram MathWorld

    ... the 'Weierstrass function of degree a' is defined as...

    $\displaystyle f_{a} (x) = \sum_{k=1}^{\infty} \frac {\sin \pi\cdot k^{a}\cdot x}{\pi\cdot k^{a}} $ (1)

    ... and it is everywhere continous but it's derivative exists only in a set of measure zero. More precisely in recent years it has been demostrated that the derivative exists and is $\displaystyle f^{'}_{a} (x)= \frac{1}{2}$ only for $\displaystyle x= \frac{2A+1}{2B+1}$, $\displaystyle A$ & $\displaystyle B$ integers. The Weierstrass function written in the form (1) is an example of Fractal Fourier Series...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
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