I can do this, but there's an additional constraint.
My wall is not a normal wall, it admits a foliation. The constraint is throughout this transfering, the height of the terrain doesn't change along this foliation as it moves through the wall
That is to say at every instance of time $t$, the height of the terrain (which is a function) is constant along each leaf of the foliation that foliates the wall
Idea: Consider the initial height of the terrain function, look at it's level sets. Draw a vector field along these level sets flowing along which will move the wave out of the wall
Regarding this question: math.stackexchange.com/questions/519161/… can't one just take the long exact sequence (this should be reduced but I'm on mobile) $$...\rightarrow H_n(\sqcup_\alpha x_\alpha) \rightarrow H_n(\sqcup_\alpha X_\alpha) \rightarrow H_n(\sqcup_\alpha X_\alpha, \sqcup_\alpha x_\alpha) \simeq H_n(\vee_\alpha X_\alpha) \rightarrow ...$$
Since that disjoint sum of points is a deformation retract of the disjoint sum of the spaces if I'm not mistaken
Ah lol thats exactly what the person asking the question does
Should've read that instead of the answer
Agh no it's not what they did but I get now why the answer is the quickest fix to the asker's attempt. The notation $H_n\left(\bigsqcup_\alpha (A_\alpha,B_\alpha)\right)$ is something to get used to
Finally, a radical Trump’s foreign policy would destabilise America’s alliances and escalate tensions with rivals. His protectionist stance could incite a global trade war, and his insistence that allies pay for their own defence could lead to dangerous nuclear proliferation, while diminishing American leadership on the world stage.
But it is actually more likely that Trump will pursue pragmatic, centrist policies.
Betraying the kurdish fighters that were allies in the fight against ISIS. To please a dictator buddy. As with Putin. Estranging european powers, waging trade war against the EU, pulling troops from germany and trying to separate from NATO
Climate deals
Oh, and he reincited the west sahara conflict recently? To make morocco open up to israel
@BalarkaSen but the policy in syria wouldn't have been a hillary thing I suppose. Couldn't imagine her being fond of erdogan. She would've stayed more involved in libya as well, since the current state there is directly connected to her actions
(basically on anything related to EU foreign policy, trump made our lives harder, so I've been looking forward to a more sensible US policy for a while)
i s'pose you could make a case that his republican ghouls modified the courts forever. climate crisis is a real thing but it seems to be transcending political affairs extremely fast; soon any political action will either hasten it or do nothing.
but i think the most significant thing that happened is crackpots are coming out of the woodworks to babble about their opinions. the pro-lifers, the climate-deniers, ...
@copper.hat I often think about garbage separation, but not from the end. It's crazy how many different types of plastic are used in some supermarket products
@copper.hat estimated 150 million plastic bottles raining down in microplastics onto yellowstone each year. There's also microplastic in babies placentas now
$f:D^n\to D^n$ has no fixed point, then $f$ can be thought of as a map from the northern hemisphere to itself. Here $D^n=\{x\in R^n:||x||\leq 1\}$ why this is true?
What does "can be thought of" mean here? $S^n$ is homeomorphic to the union of two $D^n$ with common intersection $S^{n-1}$, so if you want this is fulfilled for every map from a disk to a disk
(where each $D^n$ stands for one hemisphere)
@Thorgott grr
That actually elucidates it though. And makes sense with my earlier question about reduced relative homology
not really tbh the only reason I'm taking this course is that if one day I will use algebra in my prospective research
it'll probably be slightly categorical and I can't think of anything more probable than (co)homology
Okay, there's some hopf-algebras stuff in SPDE. But I don't intend on doing anything explicitly geometrical so I don't expect Lie groups or other related stuff to show up
The worst thing when I look at it now, is that all the diagrams that are actually really bad to justify look totally innocent, while the diagrams I understood pretty well look terrifying
A woman is walking home with distance $d$ and speed $v$.
The dog is happy and runs at speed $\frac{3v }{ 2}$ always between woman and House back and forth.
(i) At what distances $(d_n)_{n\geq 1}$ do women and dogs meet?
(ii) Determine the total path length of the dog with the help of $(d_n)_{n\ge...
As in the usual proof of 'every nonzero ring has a maximal ideal', can we always find a maximal ideal in a nonempty collection of ideals with certain property?
@LeakyNun Well I thought that if you apply all $k$-algebra operations to your set, you get your $k$-algebra, but I wasn't sure if taking the empty product was part of it.
@LeakyNun This is fair, let me check in my ex
I think I was underestimating that taking products in $k^n$ would yield more stuff
If a collection $A_i:i\in I$ of events indexed by a nonempty index set I is mutually independent, then the collection is also pairwise independent. How can I prove that?
Hello @robjohn !! Could you take a look at my question about analysis? I saw that in the past you have answered a similar question. Do you have an idea also about my question?
Definition 6.3 (Independence of events). Let $I \neq \emptyset$ and $\left\{A_{i}: i \in I\right\} \subset \mathcal{F}$ be a collection of measurable events, indexed by $I$. - The events in the collection $\left(A_{i}\right)_{i \in I}$ are pairwise independent if for any $i, j \in I,$ $$ \mathbb{P}\left(A_{i} \cap A_{j}\right)=\mathbb{P}\left(A_{i}\right) \mathbb{P}\left(A_{j}\right) $$ - The events in the collection $\left(A_{i}\right)_{i \in I}$ are mutually independent if for any (i.e. for every) finite, nonempty subset $J \subset I$,
Something slightly bothering me: for a field $k$, with closure $\overline{k}$, polynomial rings $k[x,y]$ and $\overline{k}[x,y]$, say we have $\overline{M}$ a maximal ideal of $\overline{k}[x,y]$ and $M := \overline{M} \cap k[x,y]$.
Why does $[\overline{M} \neq \overline{k}[x,y]] \Rightarrow [M \neq k[x,y]]$ hold?
I guess because it not being equal and being maximal means modding by it is $\overline{k}$, so it must not contain $1$ (or any constant really) so its intersection with $k[x,y]$ will not contain a constant and thus cannot equal $k[x,y]$ either?
@BigSocks yes, but that that field is $\overline{k}$ is not entirely trivial
Here's the general scenario: if $f\colon R\rightarrow S$ is a ring homomorphism and $I\subseteq S$ is a proper ideal, then $f^{-1}(R)$ is a proper ideal in $R$. Prove this.
oh yeah, ok I buy that. there is also a piece of info I am omitting, that $\overline{M} = (x-a, y-b)$ for $(a,b) \in \overline{k}^2)$, but I don't think it should matter for the question (and I think it makes it clearer that you get $\overline{k}$ after modding, for this instance
In reinforcement learning (RL), the reward function (RF), which can be denoted as $r(s)$, $r(s, a)$, $r(s, a, s')$, $r(s, s')$ depending on its specific definition, provides the learning signal, which guides the RL algorithm/agent towards the desirable behaviour. As opposed to a loss (or cost) fu...
if you have the eigendecomposition, then that quickly gives the matrix exponential. if you have the jordan normal form (or more generally a decomposition into diagonal and nilpotent matrices) then that also gives a speedy result
if you have an arbitrary matrix, though, then that's going to be tough
so uh, not sure if i could ask this on Math.SE but, does anyone know an algorithm or formula to reference the position of specific characters in a string in the smallest notation possible?
Feel free to tell me if there any such thing. Already know of hamming distance and also using BCD to know the position of specific character but wondering if there others method i could use.
So say you have a unique monic irreducible element $f(y) \in K[y]$ that generates $K[y]$. I guess that means you have a primitive polynomial $f_0(y) \in A[y]$, $A$ the ring from which $K$ is the field of fractions.
$1.$ $f$ is monic, but might it have some other coefficients that are not elements of $A$, say, fractions of elements of $A$ $2.$ if this is true, wouldn't you get $f_0$ not monic?
