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Explain yourselves, comp sci. (mander.xyz)

submitted 6 months ago by [email protected] to c/[email protected]

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[–] [email protected] 0 points 6 months ago* (last edited 6 months ago) (1 children)

It still can be, just not on infinite precision as nothing can with fp.

[–] [email protected] 3 points 6 months ago* (last edited 6 months ago)

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.