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a teacher at university in china who is not a mathematician.:)
2d

comment 
Decidable theorem or result that is not weaker than Tarski's theorem
@EmilJeřábek, you are so severe on any questions, like an excellent judge in supremcourt. But let us list the decidable result near or above any theorems relating Tarski. 
2d

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Decidable theorem or result that is not weaker than Tarski's theorem
Yes, excellent, David, although I had read their paper several weeks ago. I hope we can have a complete or almost complete list of such answers so as to have an overview of the decidable problem 
Nov 25 
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Decidable theorem or result that is not weaker than Tarski's theorem
@Wojowu, thank you for your comments, which one of his theorems is about decidability? 
Nov 25 
asked  Decidable theorem or result that is not weaker than Tarski's theorem 
Nov 22 
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What is the computational complexity to compute the integral numerically?
Actually, I have been suspecting that there exists a polynomial time complexity algorithm, but one paper Steve give in the comment has claims that the computational complexity is $NP$hard. I have to carefully read the article or think about the question again. 
Nov 22 
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What is the computational complexity to compute the integral numerically?
@BrunoLeFloch, thank you for your comments. I am wondering whether the antiderivative of the integrand is relevant to the complexity. In fact, your suspicion may be relating the kind of antiderivative. 
Nov 22 
comment 
What is the computational complexity to compute the integral numerically?
@SteveHuntsman, thank you for the reference. 
Nov 22 
asked  What is the computational complexity to compute the integral numerically? 
Nov 19 
comment 
Given a formal power series ,decide whether there exists a polynomial the series satisfies and if it exists,how to write it down?
So, by a compactnesslike theorem, there is a semialgorithm that decide whether a power series is algebraic. 
Nov 19 
comment 
Given a formal power series ,decide whether there exists a polynomial the series satisfies and if it exists,how to write it down?
So, there is a semialgorithm that decide whether a formal power series is algebraic. 
Nov 19 
accepted  Can any bounded area defined by polynomial inequality in $\mathbb{R}^n$ be partitioned into simply connected finite area such that 
Nov 17 
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Why is the Hodge Conjecture so important?
Is it possible, that Hodged conjecture is undecidable? 
Nov 17 
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The definition of computational complexity or complexity measure of computing reals
There are so many interesting voters having downvoted and closed the question. But I have thought about it for a long time and still think it is not a simple question. What makes me puzzled is that so many persons have misunderstood it. 
Nov 15 
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Can any bounded area defined by polynomial inequality in $\mathbb{R}^n$ be partitioned into simply connected finite area such that
Yes, I have just found a lot of material online. Thank you again. 
Nov 15 
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Can any bounded area defined by polynomial inequality in $\mathbb{R}^n$ be partitioned into simply connected finite area such that
Thank you, I will read the article, and search with the keywords. 
Nov 15 
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Can any bounded area defined by polynomial inequality in $\mathbb{R}^n$ be partitioned into simply connected finite area such that
@DimaPasechnik, thank you very much. 
Nov 15 
revised 
Can any bounded area defined by polynomial inequality in $\mathbb{R}^n$ be partitioned into simply connected finite area such that
added 1 character in body 
Nov 15 
revised 
Can any bounded area defined by polynomial inequality in $\mathbb{R}^n$ be partitioned into simply connected finite area such that
edited body 
Nov 15 
comment 
Can any bounded area defined by polynomial inequality in $\mathbb{R}^n$ be partitioned into simply connected finite area such that
@QiaochuYuan, Sorry, a mistake, thank you again, let me edit again. 
Nov 15 
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Can any bounded area defined by polynomial inequality in $\mathbb{R}^n$ be partitioned into simply connected finite area such that
@QiaochuYuan, thank you for your patience. Let me give a example in 1dimension, $\int_a^b f(x)dx=\int_a^c f(x)dx+\int_c^b f(x)dx $, [a,b] is partitioned into [a,c],[c,b]. 