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Can a non trivial computable function have an uncomputable root, and how would you show this.

Formally,

Given $f$ a computable function and $r$ a real uncomputable number.

Then $f(r) \neq 0 \vee ( \exists a,b. a \neq b \wedge \forall x \in ]a,b[. f(x) = 0 )$

For differentiable function, the existence of root finding algorithms such as Newton-Raphson method seem to proof the theorem. However for the broader class of computable functions, I cannot think of a proper proof. Even though I still expect to theorem to hold.

Sam Coutteau
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  • Consider "distance to the Cantor set." – Noah Schweber May 15 '24 at 17:54
  • Basically, the issue is that all that Newton-Raphson gives (even for differentiable or better functions) is that there are no isolated non-computable zeroes. But this doesn't prevent having a nontrivial, analytically "nice" function with a non-computable zero which is the limit of a sequence of computable zeroes (and so "hidden" by them, in a sense). – Noah Schweber May 15 '24 at 18:07
  • I assume "Distance to cantor set" can be simplified to "Element of Cantor set" or to be exact: 0 if an element otherwise 1. However it is not immediately clear that, that function is computable, as this involves checking infinite digits in the base 3 expansion of a number if done naively. – Sam Coutteau May 15 '24 at 18:17
  • No, that's not what it means. See e.g. https://math.stackexchange.com/questions/83505/distance-to-a-closed-set. – Noah Schweber May 15 '24 at 18:47

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Let $f(x)=\min\{\vert x-c\vert: c\in C\}$, where $C$ is the Cantor set, be the "distance to the Cantor set function" (note that this is not just the characteristic function of the Cantor set!). This $f$ is computable in the sense of computable analysis and nontrivial, but has uncountably many zeroes. With a bit more care we can "make $f$ smooth," that is, there is a smooth computable $g$ with $g^{-1}(0)=f^{-1}(0)$; it's a fun exercise to do this (see here).

Newton-Raphson-style arguments do show something in this setting, but not what you hope for; they show that there can be no isolated noncomputable zero. But every noncomputable element of $C$ is a limit of computable elements, so there's no obstacle.

Noah Schweber
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