bio | website | |
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location | China | |
age | ||
visits | member for | 4 years, 1 month |
seen | May 1 at 14:03 | |
stats | profile views | 2,188 |
a teacher at university in china who is not a mathematician.:)
Mar 30 |
awarded | Yearling |
Dec 8 |
comment |
Counting path generating sentences in a specific formal language
@BenjaminSteinberg, thank you for comment, I want the generating function for derivations of words using the grammar, for language there is a theorem that is Schutzenberger-Chomsky theorem hard to be extended |
Dec 8 |
asked | Counting path generating sentences in a specific formal language |
Dec 1 |
comment |
Solution to system of polynomial equations
@JasonStarr, thank you very much for your comment.But I think your comment can be extended to an answer. By the way, I cannot get your comment completely due to my background of knowledge. Would you give an expository answer in a more concrete way? |
Nov 30 |
revised |
Solution to system of polynomial equations
added 52 characters in body |
Nov 30 |
asked | Solution to system of polynomial equations |
Nov 26 |
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. |
Nov 26 |
comment |
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 |
comment |
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 |
comment |
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 |
comment |
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 anti-derivative. |
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 compactness-like theorem, there is a semi-algorithm 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 semi-algorithm 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 |
comment |
Why is the Hodge Conjecture so important?
Is it possible, that Hodged conjecture is undecidable? |
Nov 17 |
comment |
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 |
comment |
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. |