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It's known that there is no algorithm for deciding for any elementary function is it identically zero or not ( ).

But if I consider only constants - is there some algorithm for deciding for any constant expression composed from elementary functions (e. g. $\ln (\sin 1 - \tan (\pi^2))$), is it equal to zero or not?

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For constants? Just evaluate them. You can use some heuristics, like $e^x \neq 0 \forall x$. – Newb Dec 24 '13 at 19:39
@Newb, computers use only finitely much memory. Real numbers don't. – Karolis Juodelė Dec 24 '13 at 19:47
Try deciding whether $\tan p - q = 0, p, q \in \mathbb{Z}$. I'm sure that this number can be made arbitrarily small. How will a computer decide if you can't? – Karolis Juodelė Dec 24 '13 at 19:50
Also there are expressions like $\sqrt[3]{2 + \sqrt{5}} + \sqrt[3]{2 - \sqrt{5}} - 1$ (it's actually zero). – ptashek Dec 24 '13 at 19:55
If no one can answer it here, should it be posted to Math Overflow? – user21820 Jul 28 at 9:31

1 Answer 1

The page you cite eventually (at least as of 30 July 2015) links to an answer to your question about decidability.

[Edit:] Following thinking about comments: In fact, any function with a (real-)periodic level set (with positive minimal period) would do. Every non-empty level set of all six trig functions qualify. The easy case, for a warm-up, is to consider a function with a periodic discrete level set.

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Your answer is too vague on the third point. Can you at least give an outline of why the undecidability has to do with the periodicity? – user21820 Jul 31 at 10:33
@user21820 : I did. The zeroes of $\sin \pi k$ are the integers. – Eric Towers Jul 31 at 19:23
That's obvious. What is vague is why undecidability of statements over PA implies undecidability of constant expressions involving $\sin$ and $\pi$. Constant expressions do not have quantifiers. – user21820 Aug 1 at 3:20
@user21820 : The OP is answered. Sounds like you have a follow-on question. That's what the "Ask Question" link is for. Alternatively that's what all the embedded references are for. – Eric Towers Aug 1 at 23:05
@EricTowers : $\:$ I asked that as "a follow-on question." $\;\;\;\;$ – Ricky Demer Nov 12 at 21:57

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