Craig
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 Oct7 comment Confusing notation in “Introduction to the Construction of Class Fields” I think they mean 4 * 4, so that 4.$4^t$ actually means $4^{t+1}$. Oct7 revised How close to zero can a Dirichlet series get? added 15 characters in body Oct6 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. Oct6 asked How close to zero can a Dirichlet series get? Oct6 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). Oct6 comment lower bounds for maximum computing times for integer factorisation Why is that? I'm not seeing that as being immediately obvious. Oct5 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. Oct5 answered With what probability is this polynomial equal to zero (mod a prime $p$)? Oct5 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. Oct3 comment Primes sum ratio You will likely also be interested in the following paper: arxiv.org/abs/math/0408319v1 Sep21 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 $f$? Sep21 revised Could someone prove they had a halting oracle? added 12 characters in body Sep20 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? Sep19 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. Sep19 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. Sep19 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. Sep19 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 $2^n$. Sep19 comment Is there an analytical solution to this nonlinear ODE? Well, under the substitution $z = (1-y^2)$, one gets $4z(1-z) dx/dz = \sqrt{ x^2 - 4z^2 }$. One can then substitute $x = 2z \sec \phi$ and get $4z(1-z) * (2\sec \phi + 2z \sec \phi \tan \phi d\phi/dz) = 2z \tan \phi$. This simplifies to $(\csc \phi + z \sec \phi d\phi /dz) = 1/4(1-z)$. Not sure how much good that does you, but it gets rid of the square root. Sep16 comment Help needed solving for bounded random walk expectation; problem involves strange (to me) Factorial/Gamma-function summation You're basically saying that each step is an independent random normal variable drawn from $N(0,\sigma)$. So the probability density for $x_1 + x_2 = X$ is $\int_{-\infty}^{\infty} dx_1 \frac{1}{\sqrt{2\pi \sigma^2}} e^{-x_1^2/2\sigma^2} * \frac{1}{\sqrt{2\pi \sigma^2}} e^{-(X - x_1)^2/2\sigma^2}$ = $\frac{1}{\sqrt{2\pi (2\sigma^2)} e^{-X^2/2(2\sigma^2}$. And the probability density for the position after 3 steps is the same with $(2\sigma^2)$ replaced by $(3\sigma^2)$. I'm not sure where you are getting a "folded Pascal triangle" from. Sep15 comment Unbiased (random?) selection algorithm I think the problem is still underspecified. How do you know that there is any subset of $S$ that sums to $B$? How do you know that there is more than one? Do you really require that the sum is no more than $B$?