@MikeMiller That would be the translation of Bourbakiste. I think the "chique" ending is related to "chic" and it's a play on words. Can we have context?
@MikeMiller Seems it isn't a word play, and just a softening of the k to get the ch. Still, it's about the choices of Bourbaki (the group), not of Bourbakists, so translating it as Bourbakist wouldn't be right. The choice of Bourbaki, or The Bourbakian choice, I'd think.
@AlecTeal "$x$ and $y$ act as coordinates on" some surface $S$ means that the restriction of the projection $(x,y,z) \mapsto (x,y)$ to $S$ is an bijection from $S$ to an open subset of $\mathbb{R}^2$ with continuous inverse (which then has the form $\Phi\colon (x,y) \mapsto (x,y,\varphi(x,y))$). Then $\Phi$ is a (global) parametrisation of the surface, and $\Phi^{-1}$ a coordinate chart. What you have to do is compute the Jacobian of $\Phi_\pi^{-1} \circ F \circ \Phi_A$.
@DanielFischer I can't view LaTeX can you post that as a comment (my bookmarks toolbar doesn't work and I also can't move tabs or drag links - I would restart firefox but there's a lot open ATM)
@ZachSaucier you should see that you have a segment of a circle, you literally just have to integrate from pi/4 to 0 with phi, or pi/4 to pi/2 depending on where phi=0 is, and theta from 0 to 2pi and p=0 to 6
What you have is actually a part of a sphere, if you were clever you could use work out what fraction you have and use volume for a sphere!
Seeing Jar Jar not get run down will upset you, and now you understand the story the character will make you wish Qui Gon had let natural selection take its course.
Haha @user130018 glad we agree, but also the first 3 were not safe from Lucas
@AlecTeal Main is read-only atm, so I post it here: The Jacobian depends on the used charts, but once you've chosen charts, you have "the" Jacobian in those coordinates. What is chart-independent is whether the Jacobian is $0$ [at a point on the surface] or not.
@user130018 youtube.com/watch?v=PiDRgDmXGi4 I cringe now, I also hate the "look, the droids are slipping!" added, which also ruined Revenge of the Sith
@user130018 that vid is Jabba's palace, also weirdly before Lucas sold to Disney
@r9m seriously watch them again, until I did I couldn't see what was wrong with Jar Jar, now I have these weird fantasies involving Jar Jar, Gordan Freeman and the guy from "Will it blend" and Gordon brings a spare crowbar <3
@DanielFischer thanks, I thought this was the case, but I didn't have the confidence to be sure, thanks for your time, really!
Yeah...
@PedroTamaroff you were right, I am a very angry person
@ZachSaucier do you know what a sphere is? Draw an isoc. triangle and get a ruler and put 0 at the apex, then sweep across, you'll find a triangle doesn't have constant distance from the apex to the opposite edge.
Then get a compass and draw part of a circle, you'll find the curve are points of equal distance from the centre.
For finding the volume bounded above by $z = 8-8\!\left(x^{2}+y^{2}\right)$ and below by $z = 4\!\left(x^{2}+y^{2}\right)^{2}-4$, I can break it up into find the volume from 0 to the top and then find the volume from the bottom to 0 (adding the two), correct?
So it's going to be the integral from 0 to 2pi of the integral from 0 to (largest radius where they intersect) the integral from 4(1-r^2) to 8(r^2-1) of 1 r dz dr d\theta @ZachSaucier
@Alizter Not really. The world is too big to say that. Hey if you want the IB books tell me, I found the perfect set. Only one such set exists now that covers everything.
My best friend has returned from Oxford. I will meet him on Mon.
@TedShifrin my line of thinking is that $\mathbf{Z}_p$ might have some connection with the inverse limit of $S^1$s with the covering maps $x \mapsto x^n$
thus analogizing this to arbitrary profinite groups
Vieta&quadratic formula, vieta couldn't get past the first step, quadratic I reached that the discriminant had to be a perfect square, and the numerator of the solution has to equal $2m$ for some $m \in \Bbb Z$
@justabrickinthewall If you look at the profiles of the askers, there is the curious coincidence that both (members for today) voted seven times, four times on questions, and three times on answers. Hmmm.
math.stackexchange.com/questions/1001423/… mmm unanswered question with no votes or favs.... but honestly what does it matter? Preventing someone getting 6 accounts so they can delete stuff seems... a lot of work for a rare event
A voting ring is good for more than just mass-flagging. If left uncaught, they expand and become more difficult to clean out. More common on SO than here, so far.
@justabrickinthewall are you the author of mathindex.wordpress.com ? Its really useful ! Nice !!! .. I just found nice links to a bunch of series I was looking for !! :-)
@justabrickinthewall I hate it, someone puts a homework question and people always answer before me. I post a question and because it's not easy stuff any more - no one answers.
@AlecTeal Your question about covering map is difficult to approach, because it's not clear what you know (open mapping theorem in complex analysis), and what you need (exactly what is your definition of covering map)?
The idea of writing $\mathbb C$ as the product of half-line with circle is good.
But attaching special significance to $\theta=0$ works against you. Instead, you could observe that the angle-doubling map can be inverted on any proper subarc, etc.
@justabrickinthewall covering map is a topological thing, also I need to give $\theta=0$ attention (which is easy enough) because then I can have a neigh. around a complex point of argument 0.
@r9m You were talking once about candies, but I never gave you my opinion about sos's questions. I think they are all candies too. I and sos think the same about our questions. :-)
hmmm, I'm annoyed that no one has come up with another solution to this candy
The following integral was proposed in a paper by O. Furdui, namely
$$\int_0^1 \log^2(\sqrt{1+x}-\sqrt{1-x}) \ dx$$
and then the generalization
$$\int_0^1 \log^2(\sqrt[k]{1+x}-\sqrt[k]{1-x}) \ dx$$
As regards the first integral, my approach was to combine the integration by parts and the va...
@Chris'ssis oh btw I added your+winther's solutions in my blog :-) that makes an excellent proof of the limit #(53) in wolfram catalan constant page :)
@alizter Wow, I just found out all about IB math. It seems that taking IB Math HL and IB Further Math HL is about the same as taking A Level Math and A Level Further Math.
@TedShifrin Well I would sure consider it a great achievement to EVER become a mathematician like you. I do respect mathematical intellect, regardless of branches of mathematics on which it is used on.
@TedShifrin It's not obsession, it's simply too much stuff around, there are tons of integrals, series and limits, and it's more than that, one needs years to bring the knowledge to the art level and nicely compute them.
@TedShifrin I can agree with that, sure, it's a matter of taste too, but I'm trying to point out that it's not about an obsession, it's a huge amount of stuff around.
@TedShifrin attacks ?! that's not an acceptable behavior in a community :| (although I have seen lots of great (great answerers) users leave behind attacking comments)
The following integral was proposed in a paper by O. Furdui, namely
$$\int_0^1 \log^2(\sqrt{1+x}-\sqrt{1-x}) \ dx$$
and then the generalization
$$\int_0^1 \log^2(\sqrt[k]{1+x}-\sqrt[k]{1-x}) \ dx$$
As regards the first integral, my approach was to combine the integration by parts and the va...
