7

It's known that there is no algorithm for deciding for any elementary function is it identically zero or not (http://en.wikipedia.org/wiki/Richardson%27s_theorem ).

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?

Mountain
  • 542
ptashek
  • 569
  • 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
  • 3
    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 '15 at 09:31
  • @KarolisJuodelė: Just to make clear, decidability of equality between real-valued expressions has nothing to do with the magnitude of their difference. You may want to look up Turing machines to learn about decidable sets. – user21820 Jul 28 '15 at 09:33
  • @user21820. The magnitude of their difference is relevant if you just hope to evaluate the expression numerically. – Karolis Juodelė Jul 28 '15 at 09:43
  • @user21820. It was 2 years ago, but I'm quite sure that my second comment was a continuation of my reply to Newb. – Karolis Juodelė Jul 28 '15 at 10:02
  • @KarolisJuodelė: Ahhh that makes complete sense. Then sorry my comments are irrelevant. – user21820 Jul 28 '15 at 11:55
  • I you think that this question will find more response on MO, I may try to post it there. – ptashek Jul 29 '15 at 11:14
  • Yes I think you should post it there and put a link in your question here. I'm interested to know the answer! – user21820 Jul 30 '15 at 09:19

2 Answers2

4

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.

Eric Towers
  • 70,953
  • 1
    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 '15 at 10:33
  • @user21820 : I did. The zeroes of $\sin \pi k$ are the integers. – Eric Towers Jul 31 '15 at 19:23
  • 1
    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 01 '15 at 03:20
  • @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 01 '15 at 23:05
  • @EricTowers : $:$ I asked that as "a follow-on question." $;;;;$ –  Nov 12 '15 at 21:57
  • @RickyDemer : Why did you not link to it in your comment here? For anyone who should land here in the future: http://math.stackexchange.com/questions/1382945/ . – Eric Towers Nov 13 '15 at 08:11
  • Uh, I completely forgot about that part. $:$ The more recent question is here. $;;;;$ –  Nov 13 '15 at 15:17
0

Having $\sin$ and $π$ does not make the problem any harder: $\sin(θ) = \frac{\exp(iθ) - \exp(-iθ)}{2i}$, and $i$ is algebraic.

In any case, there are existing literature on the topic e.g. https://link.springer.com/article/10.1007/s11786-007-0002-x giving an algorithm which:

  • assume Schanuel Conjecture is true, always terminates with the correct answer.
  • if it doesn't terminate, then Schanuel Conjecture is false and the input contains a counterexample.

Also, it is clear that a constant is nonzero is provably so: just compute it with sufficient precision.

Which means:

  1. If the statement "$C=0$" is undecidable for some constant expression $C$, then $C=0$.
  2. In this case, the aforementioned algorithm will not terminate, and $C$ contains a counterexample to Schanuel Conjecture, so Schanuel Conjecture is false.
user202729
  • 1,244