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I got the following function:

[itex]\int_{0}^{x} \left(-1 \right)^{\lfloor 2^{i} \cdot t \rfloor} \ dt, \quad x \in \left[0,1 \right][/itex]

I want to show it is not differentiable at [itex] x= 2^{-i} k [/itex] where k is a natural number greater equal 0.

I already calculated the right derivate by considering those x with [itex]2^{-i}k \leq x < 2^{-i}k + 2^{-i}[/itex]. What I got was [itex](-1)^{k} [/itex].

Now when it comes to the left hand derivate I have the following:

[itex]\lim\limits_{x \rightarrow 2^{-i}k} \frac{\int_{0}^{x} \left(-1 \right)^{\lfloor 2^{i} \cdot t \rfloor} \ dt - \int_{0}^{2^{-i}k} \left(-1 \right)^{\lfloor 2^{i} \cdot t \rfloor} \ dt}{x - 2^{-i}k} [/itex]

But since I consider those x approaching from the left side [itex]x \leq 2^{-i} k[/itex] and [itex]x \geq 0[/itex].

That is why the above integral will be

[itex]\lim\limits_{x \rightarrow 2^{-i}k} \frac{\int_{0}^{x} \left(-1 \right)^{\lfloor 2^{i} \cdot t \rfloor} \ dt - \int_{0}^{2^{-i}k} \left(-1 \right)^{\lfloor 2^{i} \cdot t \rfloor} \ dt}{x - 2^{-i}k} = \lim\limits_{x \rightarrow 2^{-i}k} \frac{\int_{x}^{2^{-i}k} \left(-1 \right)^{\lfloor 2^{i} \cdot t \rfloor} \ dt }{x - 2^{-i}k} [/itex]

I know that for the last integral [itex]x \leq t \leq 2^{-i}k[/itex] which is equivalent to [itex]2^{i}x \leq 2^{i}t \leq k [/itex].

But that doesn´t help me when it comes to calculating [itex]\lfloor 2^{i} \cdot t \rfloor [/itex]

How can I proceed from there or is there any mistake in my way of thinking I don´t see at this moment?

Thanks.

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# Not differentiable function

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