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seen May 9 '12 at 2:58

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 $2^n$.
Sep
19
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.
Sep
16
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.
Sep
15
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 $B$?