Has anyone interest to establish a room for prime numbers and integer factorizations of several kinds of numbers ? I would coin in the expression (9n)!+n!+1. Is there a positive integer n, such that this expression is prime ? Upto n = 800 , this is not the case.
ok well if they delete the priv room I need to put here so I look discuss why its wrong or disgusting For any $\mathcal H={\{h_k}\}_{k=1..N} \subset \mathbb N \land N \gt 1$
I know it has something to do with residue classes modulo $n$ and Lagrange's Theorem for finite groups and that eventually explaining why the division by two holds I think well because the totient does anyway im sleep
@MadSpaceMemer In your first example, you're never going to get there with that substitution!! As Thorgott suggested, restrict to an interval on which $\cos$ is one-to-one (say $[0,\pi]$) so that you have an inverse function. Let $t=g(x)=\arccos x$. Then $\int (1-x^2)^{1/2}\,dx = \int (1-\cos^2 t)^{1/2} \frac{dx}{dt}\, dt = \int \sin^2t\,dt$.
What is the easiest way to show that
$$ n > 2 $$
$$ \pi(n) > \frac{n}{2 \ln(n)} $$
Where $\pi(n)$ is the prime counting function.
I read a proof of the PNT with the zeta function but this statement is much weaker !!
What is the shortest proof ?
The simplest ?
The most elementary ?
Do we u...
@Ted well the claim is equivalent to $n(n+1)/2 = a^2$ for some $a \in \Bbb N$, which is equivalent to $n^2 - 2a^2 = -n$, i.e. the elements $n + a\sqrt{2} \in K:=\Bbb Q(\sqrt{2})$ such that $N_{K/\Bbb Q}(n+a\sqrt{2}) = -n$
Can I ask a question? Given a Bi-Cubic interpolation. If all the values of it are in a range from 0 to 1. Is possible to get a result higher than 1 or lower than 0?
For an abelian topological groups $G$ and $H$, any continuous homomorphism $G\to H$ induces a continuous homomorphism of their completions $\widehat{G}\to \widehat{H}$.
What kind of topology is defined on the completion? I think there is a natural one from the subspace of the countable product topology of $G^\mathbb Z$.
@TedShifrin It's not my answer, it's just an answer to my question. But it used to display properly, without any changes being made to it.
I thought maybe answers that receive a bounty get special permissions to have extra large pictures or something, and that I may have caused it by awarding that bounty, but no idea haha.
@TedShifrin When you have a fundamental solution to a Pell equation you need to take powers of it to get the others iirc, so exponential growth of the solutions seems to be expected
Did you see my message in chat regarding the second game? Your position wasn't as bad as it looked, you could have avoided both mate and losing the bishop
Assuming this lemma here's a simple proof of the theorem, followed by a neat half page long argument. We now prove the technical lemma, followed by three ugly pages of messy computations
Well in our modular forms course we showed that some series $P_n$ satisfies the property that for any $g(z)$ (satisfying certain relevant properties), we have $a_n(g) = \langle g, P_n \rangle$ where $a_n(g)$ are the Fourier coefficients of $g$ and $\langle \cdot, \cdot\rangle$ is an inner product
I don't know if that's right though, maybe you've heard of something similar and could state it better
I think what's going on here is an application of the Riesz representation theorem. It says that if $H$ is Hilbert and $f\in H^\ast$, then there is a $y\in H$ such that for every $x\in H$, $f(x)=\langle x,y\rangle$
I suppose that $g$ lives in some kind of Hilbert space
I am sorry to raise a naive question: is it true that for the splitting fields $K$ of a polynomial $f\in k[X]$ of degree $n$, it is generic that $K/k$ is a Galois extension and the Galois group is $S_n$?
Let $\mathbb{Z}[x]_n$ be the set of polynomials in $\mathbb{Z}[x]$ of degree at most $n$.
Then, consider some sensible increasing filtration $$A_0 \subset A_1 \subset A_2 \subset \cdots$$
of $\mathbb{Z}[x]_n$ by finite sets, i.e., a sequence of nested, finite sets $A_k$ such that $\bigcup_k A_...
@WilliamSun It is not a subspace of the countable product: you need to mod out the convergence relation. The correct one should be the Hausdorff completion of a uniform space
What is the easiest way to show that
$$ n > 2 $$
$$ \pi(n) > \frac{n}{2 \ln(n)} $$
Where $\pi(n)$ is the prime counting function.
I read a proof of the PNT with the zeta function but this statement is much weaker !!
What is the shortest proof ?
The simplest ?
The most elementary ?
Do we u...
Let $\mathbb{Z}[x]_n$ be the set of polynomials in $\mathbb{Z}[x]$ of degree at most $n$.
Then, consider some sensible increasing filtration $$A_0 \subset A_1 \subset A_2 \subset \cdots$$
of $\mathbb{Z}[x]_n$ by finite sets, i.e., a sequence of nested, finite sets $A_k$ such that $\bigcup_k A_...
