"Army appears in full control amid reports hundreds of MPs are being confined in government housing" says the guardian so yeh I could call this a coup attempt but I'm too lazy
It turns out that $\left\lfloor {n\over 2}\right\rfloor \ge {n-1\over 2}$, but $\left\lfloor {n\over 3}\right\rfloor$ isn't (always) greater than $ {n-1\over 3}$
There's a similar bound for $\left\lfloor {n\over 3}\right\rfloor$ which I believe you can find, given you understand how the bound for $\left\lfloor {n\over 2}\right\rfloor$ was found
@TedShifrin Right. I was more wondering what methods were used to find roots for the equations. Espacially since it gives exact solutions (integers) and not approximations given by, say, the newton method
Suppose $\{a_n\}, \{b_n\}$ are infinite sequences such that the series $\sum_{i=1}^\infty a_i = A, \sum_{i=1}^\infty b_i = B.$
Let $\{d_n\}$ be the sequence such that $d_n = a_n + b_n$ for all $n$. Prove that $\sum_{i=1}^\infty d_i = A + B.$
We are given that the series of $a_n$ and $b_n$ converge, so $\forall \varepsilon > 0, \exists N_1 \in \mathbb{N}, \forall n \in \mathbb{N}, n > N_1 \Rightarrow \left| \sum_{i=1}^n a_i - A \right| < \varepsilon/2$ and $\forall \varepsilon > 0, \exists N_2 \in \mathbb{N}, \forall n \in \mathbb{N}, n > N_2 \Rightarrow \left| \sum_{i=1}^n b_i - B \right|…
Guys, does anyone know how they arrived at the underscored iso? I would think that we have $$ \left(\oplus_{k\neq i}A_i dx_{ik}\oplus A_i dx^{-1}_{ij}\right)/(x_{ij}dx^{-1}_{ij}+x_{ij}^{-1}dx_{ij}) $$ instead
The reason for this is that $\mathcal O_{\mathbb P^n}(U_i\cap U_j)\cong k[x_{i0},\dots,\hat{x_{ii}},\dots,x_{in},x_{ij}^{-1}]/(x_{ij}x_{ij}^{-1}-1)$
@TedShifrin I like my natural numbers to be a monoid.
Although, I'm sure there are compelling reasons not to want 0 as a natural number. I don't know any of them, and if I heard them I would reject them without even giving them due consideration, but hey, "intellectual honesty" is just a twelve-letter word with a seven-letter word.
underpaid PHDs to 200 clueless students: so uhh if p if false and q is true then p => q is true and uhh I guess thats the way it is, and you just act as if that's obvious when anybody asks
Hey quick diff geo question - is there a name for the class of manifolds such that you could define the surface in terms of a parametrization S where all the partial derivatives of S with respect to each parameter at any point produces a set of orthogonal vectors? Is this just all differentiable manifolds?
in dimensions <=2, these are all differentiable manifolds, in dimensions >=4 these are the conformally flat manifolds, forgot what happens in dimension 3
So say you got 2 local PIDs, $R$ and $S$, with $R \subset S$. Apparently $R = S$. On the way there it is apparently provable that $S = T^{-1} R$ where $T$ is a multiplicative subset of $R$. Any hints?
right bc any other ideal would be maximal bc you can't have one generated by more than one element. there would be two of 1 generator. they would both be maximal
The point is that there is a discrete valuation (ie a multiplicative function $K^*\to \Bbb Z$) such that R is exactly the elements with positive valuation
I mean, if $R$ is a domain, $K$ its field of fractions and $R\subset S\subset K$ a subring, then let $T$ be the elements in $R$ that appears as denominators in $S$. THis gives a multiplicative subset of $R$, localizing at which yields back $S$. So $S=T^{-1}R$. If $R$ is a local PID, there is only the zero ideal, in which case $S=K$, or the maximal ideal, in which case $R=S$. $S$ being a local PID seems unnecessary,
Suppose that $X$ is locally compact and $f$ is an homeomorphism $X\to X$. Is it obvious that the set of points fixed by $f$ must be locally compact in the subspace topology?
@NicoA @Thor Actually, I did not know the answer to this. I don't think Thor is quite right here. We're talking about diagonalizing the metric in charts, and that's not conformal flatness. (Of course, orthonormal parametrization results in flatness, but that's much more strict.) Anyhow, here is the answer.
@Astyx I'm here- how to see? you can factor out the generator of $\mathfrak{m}$ from any element (represented as a sum of products) and... something more...
