@Jason I think I've seen this proof, something to the effect of how you can always unitarily triangularize a matrix, but then if you start with a normal one you get a normal one
I envisioned an exchange like, me: "uhhhh the Congo?" someone else: "WHICH ONE? AND BESIDES THE DRK WAS BELGIAN AND THE REPUBLIC OF CONGO WAS PARTLY GERMAN AND PARTLY FRENCH"
Maybe I'm just too used to the Internet being as it is.
@Fargle it is astounding how few people in south Florida know that Brazilians don't speak spanish given that there's a huge Brazilian population in south Florida
Fun fact: though Spanish and Portuguese are closely related enough to be considered distant dialects of one another, Spanish and Italian are even closer
It's pretty hard to live in south florida without hearing spanish though. And if you know what spanish sounds like you know that portuguese sounds nothing like spanish
I've been stopped by a policeman for speeding and he spoke to me in broken spanish for two minutes before I just looked at him blankly and said "I am not spanish dude"
I was on a train ride where people near me were talking in what sounded like Spanish, but I couldn't understand any of it, so I was unsure if it was just an accent I'm unfamiliar with (aka anything other than my Spanish teacher) or actually Portuguese
I understand Spanish quite well and I speak Italian and Portuguese, so I could be a possible guy who could have deciphered the mysterious language they, the unknowns, were speaking
So a Paraguayan tells a Chilean, "Vení aquí desde Paraguay para matarte" and the Chilean, who wasn't really listening, says "¿Para que?" to which the Paraguayan replies "Paraguay"
Anyway what I said is that spanish is closer to portuguese than portuguese is to italian. I'm pretty sure in terms lexical similarity spanish is still closer to portuguese than it is in italian.
Hi. Simulations show that matrices of this type mathb.in/143097 always seem to have a negative eigenvalue. While I might see how to prove it for small $K$, the proof is difficult to generalize. Any ideas?
There are still Confederate flags flying out here. Just today I passed a truck that was flying the American flag and the Confederate flag at the same height and angle out of the truck bed.
First of all, stop flying the Confederate flag, you just look dumb. But second of all, if you really do care about the "heritage of this great nation and all its traditions", you'd know that the American flag should always be flying above any other flags in the vicinity.
By flying a flag that represents an open, armed rebellion against your country at the same height as the flag of your country, all you do is demonstrate hypocrisy.
Well, the first point is something that can't be used in an argument because if you are in the right location, many people around you will affirm it. But yeah I don't see the logic in literally flying a flag of treason
Though I've become habituated after seeing it enough times
I mean, this was from when I was 9 to 14, it kind of already happened a bit too deeply, so at some point those flags just became a regular part of the highway experience
They kind of become like general bad drivers that you see
And sometimes more like crosslights, they're just always there, as if their presence was synonymous with that of the road itself
My point is that when you're a kid it just about can't be helped
It happened with so many things, like the prospect of Muslims not having to deal with the whole terrorism stuff is just alien at this point, etc
I don't like it at all but at some point it takes the place of everything else which seems like a necessary element of existing, that isn't easily shaken
@TedShifrin what's your opinion on this For doing proofs should one automate any tedious steps such as algebraic manuiplations via a computer algebra system
Think it depends on what "doing proofs" means. Well written exercises shouldn't have enough tedious steps to justify using a CAS (unless using a CAS is one of the skills being exercised). For REAL problems in mathematics (e.g. research problems) using a computer to compute a giant number of examples (that can't possibly be done by hand) can be a very useful tool.
Most analysis proofs that undergraduates do involve very little algebra, but — rather — require learning how to think about estimates in a productive and intelligent way.
@PVAL: I am reminded that Hoffman-Meeks many years ago discovered a deep theorem re minimal surfaces because of sophisticated computer graphics the drew them the surfaces ... and then they could figure out a proof. So I'm certainly not dismissing intelligent use of technology!
I wanted to see what $\mathbf{Q}$-automorphisms do to the square root of that thing (if we assume its a square) and get a contradiction but for some reason I wasn't getting anywhere
Namely: Suppose $f(x), g(x)$ are polynomials of degree $n<m$. Then every derivative of $f(x)/g(x)$ is a rational function. What's the minimal degree of the numerator of the $k$th derivative?
Yeah I wanted something nicer than just doing out the algebraic manipulations, but they're probably doable (although it should be like a system of four homogeneous quadratics in $4$ variables which is kinda annoying) @PVAL
Meanwhile, I in my Galois ignorance have to settle with writing out $(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})^2$, expanding and matching coefficients, then check that the system has no rational roots :/
Ok so for the first part, thinking about the non-trivial $\mathbf{Q}$-automorphisms of $\mathbf{Q}(\sqrt{2}, \sqrt{3})$ which fix $\mathbf{Q}(\beta)$.. that's nothing so it has to be $\mathbf{Q}(\sqrt{2}, \sqrt{3})$, right, so $[\mathbf{Q}(\beta): \mathbf{Q}] = 4$.
There's a certain small parameter in the problem that lets them reduce said integral to a saddle point computation. That gives a condition for the saddle point, from which they obtain the function they're really after as a Legendre transform, i.e. some optimization problem over $z$.
However, once they get to said optimization they rather blithely observe: Hey, this optimization problem can be smoothly continued a bit into z<0!
$\text{Define f(x) for all positive integers such that f(1) = 1 and f(2) = 0}\; \; f(k^2) = 1 \text{and for other n, f(n) = 0 iff for all} k^2 \text{such that} k^2 < n \; \; f(n-k^2)=1 \text{or 0 otherwise}$
$\text{I am a SUPEEER GENIUS, a perfect crackpot}$. $\text{I declare}$ $\lim_{n \rightarrow \text{very big number}} \sum_{i =1}^{n} \frac{1}{i^2} > \frac{\pi^2}{6}$.
Yeah that's right. Well, I want to show that a certain operator is self-adjoint in $H\oplus H$, and I have a $2x2$ matrix that defines it. I am unsure how to approach it.
So I'm pretty sure that if $B=C*$ and both $A,D$ are self-adjoint, then [[A,B],[C,D]] is self-adjoint; otherwise you might have to do some real work and check the adjoint :P but what''s the matrix?
Hmm yeah that does look self-adjoint; if you (mumble mumble orthogonal basis) then the adjoint is represented in the same basis by the conjugate transpose.
How can I see that the derivative of the map $\text{SL}(n,\mathbb{R}) \to \text{SL}(n,\mathbb{R}), A \mapsto A^{-T}$, is given by $X \mapsto -X^T$ via the canonical identifcation of the tangent space of $\text{SL}(n,\mathbb{R})$ at the identity matrix with the $n \times n$ matrices of trace zero?
@BAYMAX Intuitively, you could think about the probability of, given a fixed vector in $\Bbb R^n$, randomly picking a vector that's orthogonal to the fixed one.
The set of all vectors orthogonal to some vector form an $n-1$-dimensional subspace of $\Bbb R^n$, which has measure $0$.
Absolutely. Generally when you want orthogonality, you're demanding that the dot product be exactly zero, which will usually produce a measure $0$ subspace.
But if you're simply demanding that the dot product be bounded by some quantity, that opens it up a lot more.
PVAL: I'm not sure if there's anything you can do about this, but when you were talking with other-Eric I ended up getting a few of those notifications, which doesn't normally happen. Perhaps you accidentally pressed "reply to this message" for one of my posts instead and then I got looped into the thread? I just figured I'd let you know.
Is there an easy way to show (with algebra) how many integers there are in the geometric sequence 54*(-2/3)^n-1 ? By testing I get it to be four elements.
Calculate the flow of the vector field $$\mathbf{A} = \nabla \dfrac{\mathbf{a}\cdot\mathbf{r}}{r^3}$$ from a cube with side length $1$, centered at origin and with one space diagonal parallel with the constant vector $\mathbf{a}$.
Attempted solution
Let the $z$-axis run parallel to $\mathb...
@Lozansky The reason to avoid Gauss's law here is that it'd require you to interpret $(\mathbf{a}\cdot\mathbf{r})/r^3$ at $r=0$.
(Physically, the point is that $\mathbf{A}$ would be the field of an ideal dipole. This has zero net charge, but this isn't so obvious just looking at it.)
As for the second point: If two surfaces enclose the same charge, they've got the same flux. So they proceed by using Gauss's law to equate the flux over the cube to the appropriate volume integral, and then use it again to equate both to the flux over the sphere.
If I have a chance this morning I'll type that up as a solution. (If someone beats me to it, oh well.)
@liad That said, there's really no reason to wait for Ted in stating the question. If someone can help you now, great; if not, you can point Ted back to this conversation.
Calculate the flow of the vector field $$\mathbf{A} = \nabla \dfrac{\mathbf{a}\cdot\mathbf{r}}{r^3}$$ from a cube with side length $1$, centered at origin and with one space diagonal parallel with the constant vector $\mathbf{a}$.
Attempted solution
Let the $z$-axis run parallel to $\mathb...
