Crazy cool. But what happens at x ≈ sqrt(2)? Does the discontinuity in sign(x^2 - 2) somehow automagically wash away?
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these bounds 𝑥 ∈ [0,2]
at square root of 2 you can see that all of the f's vanish leaving behind just the trivial root(2)=root(2)
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Oh duh, I see. For some reason I misread your previous tweet as sign(0):= 1 . This is quite mysterious (in a good way)
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So interesting aside.. if 𝑥 ∈ Q (instead of R) then here we don't need a definition for sign(0)
End of conversation
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