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Prime numbers, diophantine equations, diophantine approximations, analytic or algebraic number theory, arithmetic geometry, Galois theory, transcendental number theory, continued fractions
5
votes
0
answers
330
views
On Galois' criterion for resolvents
Let $K$ be an algebraic number field, $f(x) = 0$ an algebraic equation over $K$ of degree $n \ge 2$ with only simple roots $x_1, \dotsc, x_n$, and $L \mathrel{:=} K[x_1, \dots, x_n]$ the splitting fie …
2
votes
What are immediate applications of the classification of connected reductive groups?
Come that far, one can begin to exploit the subgroups, turn one's attention to symmetric spaces, of which there are the compact and noncompact ones, the riemannian and the hermitian ones. And one can …
2
votes
1
answer
431
views
Proof for an explicit formula for the even Euler numbers
The EULER numbers $E_n$, $n \in \mathbb{N}$, are defined via the TAYLOR
expansion of the hyperbolic secant:
\begin{equation}
\text{sech}(x) = \sum_{n=0}^{\infty} \frac{E_n}{n!} x^n =
\sum_{n=0 …
2
votes
1
answer
230
views
Proof of a binomial identity
Computations with Maple suggest the following binomial identity
\begin{equation*}
\forall{p,j}: \sum_{k=j+1}^{p+1} (-1)^j \dfrac{1}{k}\binom{k-1}{j} =
\sum_{k=j+1}^{p+1} (-1)^{k-1} \dfrac{1} …
1
vote
Proof for an explicit formula for the even Euler numbers
@Sam Hopkins: The formula there reads
\begin{equation*}
E_{2n}=\sum_{k=1}^{2n }(-1)^k\frac{1}{2^k}\sum_{\ell=0}^{2k}(-1)^{\ell}\binom{2k}{\ell}(k-\ell)^{2n}
\end{equation*}
which is slightly diffe …