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I have the equation

$\sum_{i=0}^{\infty} 2^{i}[0 \leq x - 2^{i}][x - 2^{i + 1} < 0]$

and I would like to convert the Iverson brackets to the Heaviside function.

I've read this post but I'm still a bit confused as to how this would work inside the summation.

EDIT:

If it helps, I believe I've reduced it to

$u_1(x) + \sum_{i=1}^\infty 2^{i-1}u_{2^i}(x)$

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sdasdadas
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    Iverson brackets definition for the Heaviside function $H(x)$ is $H(x) = [0\leq x]$ (except for $x=0$ where the Heaviside is usually defined to be $1/2$). Ignoring the $x=0$ issue then $H(x-2^i)H(2^{i+1}-x)$ should do the trick. – Winther Oct 05 '14 at 23:42
  • @Winther I understand - it seems very simple when looking at your response. Thanks very much! – sdasdadas Oct 05 '14 at 23:45
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    btw you can remove the summation all togeather. Given $x$ there is only one integer $n$ s.t. $2^n \leq x < 2^{n+1}$ and the only non-zero term in the series is the $n$'th term so: $\sum_{i=0}^\infty 2^i [0\leq x - 2^i][x-2^{i+1} < 0] = 2^{n} = 2^{\lfloor\log_2 x\rfloor}$ where $\lfloor \cdot\rfloor$ is the floor-function. – Winther Oct 05 '14 at 23:50
  • @Winther That's what I just started! Thanks again, this has been very helpful! – sdasdadas Oct 05 '14 at 23:55

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