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Joined 1 year ago
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Cake day: July 2nd, 2023

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  • But the vector space of (all) real functions is a completely different beast from the space of computable functions on finite-precision numbers. If you restrict the equality of these functions to their extension,

    defined as f = g iff forall x\in R: f(x)=g(x),

    then that vector space appears to be not only finite dimensional, but in fact finite. Otherwise you probably get a countably infinite dimensional vector space indexed by lambda terms (or whatever formalism you prefer.) But nothing like the space which contains vectors like

    F_{x_0}(x) := (1 if x = x_0; 0 otherwise)

    where x_0 is uncomputable.



  • Depends on the kind of blur. Some kinds can indeed be almost perfectly removed if you know the used blurring function, others are destructive. But, yes, don’t take that chance. Always delete/paint over sensitive information.

    Source: we had to do just that in a course I took a long time ago.