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Fourier expansion of boolean functions

by Dragonfall
Tags: boolean, expansion, fourier, functions
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Dragonfall
#1
Mar4-14, 09:05 PM
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P: 993
Any boolean function on n variables can be thought of as a function

[tex]f : \mathbb{Z}_2^n \rightarrow \mathbb{Z}_2[/tex]

which can be written as

[tex]f(x) = \sum_{s \in \mathbb{Z}_2^n} \hat{f}(s) \prod_{i : x_i = 1} (-1)^{x_i}[/tex]

where

[tex]\hat{f}(s) = \mathbb{E}_t \left[ f(t) \prod_{i : s_i = 1} (-1)^{t_i} \right][/tex]

This is the Fourier expansion of a boolean function. But this uses the group [itex]\mathbb{Z}_2^n[/itex]. Why not use [itex]\mathbb{Z}_{2^n}[/itex]? Characters of the latter would be nice looking roots of unity on the complex circle, instead of points on an [itex]n[/itex]-cube.

EDIT: You know what, nevermind. I don't even understand my own question anymore.
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