Craig
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 Oct 7 comment Given the area of $1000$ sided polygon ,how could we find the maximum length between any two of its vertices? @FoolForMath, I'm pretty sure you haven't. I would guess the problem specifies a regular 1000-gon. If it can be any 1000-gon, Mark is right. Oct 7 answered Confusing notation in “Introduction to the Construction of Class Fields” Oct 7 revised How close to zero can a Dirichlet series get? added 4 characters in body Oct 7 comment Confusing notation in “Introduction to the Construction of Class Fields” I think they mean 4 * 4, so that 4.$4^t$ actually means $4^{t+1}$. Oct 7 revised How close to zero can a Dirichlet series get? added 15 characters in body Oct 6 comment How close to zero can a Dirichlet series get? I do not believe it is possible to keep it small for all Im($s$). I am looking for statements of the form, "for $|Im(s) - \sigma | \approx O(g(N))$ the following construction gives $|f(s)| \approx O(N^{-(Re(s) + h(N))})$". As for the construction I alluded to, consider that $N^{-s} - (N+1)^{-s} = s N^{-(s+1)} + O(s^2 N^{-(s+2)})$. Furthermore if we pick any two $k,k'$, we can find a linear combination of $N^{-s}$, $(N+k)^{-s}$ and $(N+k')^{-s}$ that will be of order $s^2 N^{-(s+2)}$. We can keep doing this approximately $ln N$ times at least before the error terms start blowing up. Oct 6 asked How close to zero can a Dirichlet series get? Oct 6 comment Confusing notation in “Introduction to the Construction of Class Fields” I think they are assuming t >= 0, as opposed to t > 0. Either way, they should make the assumption explicit in the statement (1.8). Oct 6 comment lower bounds for maximum computing times for integer factorisation Why is that? I'm not seeing that as being immediately obvious. Oct 5 comment With what probability is this polynomial equal to zero (mod a prime $p$)? N.B. This matrix is just $1/2 I + 1/2 S^{x^N}$, with $S$ being the cyclic-shift matrix. Oct 5 answered With what probability is this polynomial equal to zero (mod a prime $p$)? Oct 5 comment lower bounds for maximum computing times for integer factorisation Um, how exactly is it reducing the number of possible factors by 1/2 each step? No algorithm I know of does this. Oct 3 comment Primes sum ratio You will likely also be interested in the following paper: arxiv.org/abs/math/0408319v1 Sep 21 comment Solving a two-dimensional system of conservation laws If you know that $u(x,t) = f(x-t)$ and $u(x,0) = 0$, what does that tell you about $f$? Sep 21 revised Could someone prove they had a halting oracle? added 12 characters in body Sep 20 comment How can I get this tricky sum? So specifically you have the following expression: $c_{0,0}*c_{1,0}*...*c_{n,0} \space + \space c_{0,1}*...*c_{n,1} \space + ... + \space c_{0,m} * ... * c_{n,m}$ and you would like to evaluate it with fewer than $(m+1)n$ multiplies and $m$ adds? Sep 19 comment If the product of $x$ positive integers is $n!$ What is the smallest possible value their sum can have? To be specific, I am suggesting a greedy algorithm where you run thru the primes in order of decreasing size, each time multiplying the smallest of your $x$ numbers by said prime. Sep 19 comment If the product of $x$ positive integers is $n!$ What is the smallest possible value their sum can have? This is a knapsack-ish problem -- the values you are trying to fit into the $x$ boxes are the logs of the primes. In regards to this, the large primes are not as much of a problem as you might think; greedy algorithms tend to work well. And since you have large numbers of small primes (2 shows up ~n times in n!), you have a fair amount of "filler" to even things out. Sep 19 comment Lie algebra of $GL_n(\mathbb{C})$ Well, for one thing, a Lie algebra is an algebra -- which means that it is also a vector space over the underlying field $\mathbb{C}$. So you might first show that there is an isomorphism respecting the vector space strucure. (Very easy.) Being a Lie algebra, it also has a binary operation, the Lie bracket. In this case you want to argue that the commutator on $M_n(\mathbb{C})$, under the natural vector space isomorphism, actually obeys the properties you want the Lie bracket to obey. Sep 19 comment Order of growth proofs? In general, this is a difficult problem. However, for your specific case, it is relatively tractable. "$f(n) << g(n)$" is equivalent to "$lg f(n) << lg g(n)$" for your examples. And a constant grows slower that $lg n$, which grows slower than $n^k$, which grows slower than $2^n$.