yeah, while ionic radius has some periodic trend of getting smaller as you move across the period, exceptions are too many for this heuristic to be useful
and anions tend to get worse in predicting via heuristics compared to cations due to electron electron repulsion, which explains things like H- having a very large ionic radius to the point it behaves like a very reactive halide ion
heuristics are useful, however for most organic reactions, cause unless you deal with highly electronically complicated systems, most organic chemical reactions can be estimated quite well via heuristics like nucleophiles and electrophiles and the princple that electron densities tend to flow towards mroe electronegative elements, while slightly more advanced onces can be done well via frontier MO theory
But once you introduce just one metal in the system (or your organic molecule has a lot of delocalisation and other nontrivial things like radicals and excited states), then pretty much he…
or at least, says things like "and we're going to assume switching the order of integration works, and we'll worry about it later if we run into a problem"
@Secret on the metallic side it's usually the other way around: everything is delocalized due to the periodicity of the lattice, but impurities/disorder introduce localized states
if it's finite for one order of integration, it's finite for both (and finite for integration over the product measure etc, but nobody really uses that anyway)
I am not sure how familar I am with metallic periodic systems (such as bulk alloys or metal organic frameworks), the systems I work with are coordination or organometallic complexes, where it is molecular with 1-3 metal ions coordinated or covalently bonded to some organic molecules (known as ligands). But even these have highly delocalised electronic states in general, and for lathanides it is even worse because electronic levels are so closely spaced together that its ground state wavefunction has significant excited state character
I would like to recommend to all users the following talk: Construyendo tu sueño: Mario Alonso Puig at TEDxGranVia, by an spanish surgeon from the official channel TEDx Talks, in YouTube. He tell us higher thoughts. Thus if you've a spanish friend you could see this nice video.
@aminliverpool The easiest thing to think about is probably $\Bbb R\mathrm P^1$. There you can actually draw the picture that justifies the name "affine charts" for the standard charts on projective spaces.
@aminliverpool Then the standard charts are given by $[a:b:c]\mapsto (b/a,c/a)$ on the open subset where $a\neq 0$, and similarly for $b,c$. Now clearly you can also view the copy of $\Bbb R^2$ you obtain by this map as the subset of points of the form $(1,b/a,c/a)$ in $\Bbb R^3$. This is an affine copy of the plane.
Now use this same description and draw a picture for the case of $\Bbb R\mathrm P^1$.
@MikeMiller :( Hope you're OK
By the way, Mike, do you have any idea about existence results for Leray coverings?
Can I typically find a Leray covering for a sheaf (over a Riemann surface, if that makes things easier?)?
I am having trouble understanding quantifiers at the moment., could anyone here give an explanation as to why $\forall x (Ax)$ is not the same as its counterpart in a finite domain of discourse $Ax_1 \land Ax_2 \land ... Ax_n$?
See this question for a better idea of what I am talking about: https://math.stackexchange.com/questions/423656/were-logical-quantifiers-not-primarily-motivated-by-infinite-domains-of-discours
user84215
1:49 PM
@Danu That choice of charts comes from the fact that the points in each class are scalar multiples of each other ?
Consider the cross cap representation of $\mathbb{RP}^2$ where the line AB is the double line corresponding to the Whiteney's umbrella
Consider the followin injective mapping which embed $\mathbb{RP}^2$ into $\mathbb{R}^4$ (source of $f$ found here) hence remove all self intersections
$$f(x,y...
@parvin Closed under countable union means that if you take countable many things in your $\sigma$-algebra and union them together (unite them together?), the result is also in the $\sigma$-algebra.
I am having trouble understanding quantifiers at the moment., could anyone here give an explanation as to why $\forall x (Ax)$ is not the same as its counterpart in a finite domain of discourse $Ax_1 \land Ax_2 \land ... Ax_n$? See this question for a better idea of what I am talking about: https://math.stackexchange.com/questions/423656/were-logical-quantifiers-not-primarily-motivated-by-infinite-domains-of-discours
@user400188 I still don't see what you mean after seeing that question. Yes, any forall statement with a finite domain can be translated to such an "and" statement
What I mean, is what changes in our definition of "forall" when we move to an infinite domain of discourse? And how do we represent it? @TobiasKildetoft
@user400188 Ahh, I see. Well, what changes is that since we are only allowed to have finite strings for our expressions, we can no longer write down the forall as an "and"
I've seen the sentence: "for all x, or exists not x" taken to be a tautology, however I can;t for the life of me see how it could be verified. Could you elude to how it might be done, (or not verified)?
@Semiclassical I was about to write that myself. (the true or false bit) A or not A (where A is for all x (Bx) and exists x is defined as the negation of forall x not (Bx))
On to the empty set now....
I have a trick of remembering the empty set property of for all x
I define for all x Ax as this. If x has a domain of discourse, then it has property Ax.
The way I think of truth for implication is one is true when both sides of the arrow have the same term (or if one side can be simplified to the other with boolean algebra)
I mean you could contradict that its false. But by contradiction, we normally mean to logical and it with another statement (which happens to be its negation). But doing this operation will always result in another false.
I would love to use induction; but I am asking whether a thing is true in both the finite and infinite domain of discourse. Unfortunately induction does not work in a finite domain of discourse becuase the last element (the one at the end) pops off so to speak.
I wouldnt consider myself knowageable on quantifiers if I could not answer questions about them in infinite domains. I want to be knowlagable.... $(k[(\lnot I)\implies (\lnot K)])\implies I$ This tautologically implies that I need to learn about quantifiers in infinite domains.
sorry, couldn't help myself
Anyway: Thanks for the help @Semiclassical @TobiasKildetoft @SteamyRoot > I'm going to bed now.
Meanwhile in chemistry: That computing cluster I am doing my calculations on have a lot of path refering errors and inconsistency and I have been spending the last 6 hours working alongside the help centre to get them all ironed out so that my orbital calculations can finally work
Still working on it...
O btw speaking about clusters, metal clusters have very crazy electronics compared to both periodic structures, coordination complexes and organometallics, which makes DFT calculations on them very difficult
(I am working with organometallic molecules, thus I am alright except for those programming issues)
Well, it is not a periodic structure, but an isolated molecule, thus all you need is to compute its electron density and orbitals, which gaussian 16 can do it no problem
The gas phase calculations agrees quite well with the x ray structures and energetics that my supervisor have obtained, thus benchmarking suggests the modelling will be fine
Metal clusters (something that other groups done and they came to present about in a seminar some weeks ago) on the other hand, don't really have a notion of a geometry as there are many configurations of the metal atoms with very closely spaced energy levels, making the calculation quite nontrivial
Consider the cross cap representation of $\mathbb{RP}^2$ where the line AB is the double line corresponding to the Whiteney's umbrella
Consider the followin injective mapping which embed $\mathbb{RP}^2$ into $\mathbb{R}^4$ (source of $f$ found here) hence remove all self intersections
$$f(x,y...
