There are 7 portable mini suites (a.k.a. cages) in a row at the Paws and Claws Holiday Pet Resort. They are neatly labeled with their guests' names. There are 3 poodles and 4 tabbies. How many ways can the "suites" be arranged if: a) there are no restrictions.
b) cats and dogs must alternate.
c) dogs must be next to each other.
d) dogs must be next to each other and cats must be next to each other.
Ah, right, the row space is a subspace of $\Bbb R^m$ and the column space is a subspace of $\Bbb R^m$, @Astyx. [We think of the row vectors as vectors, not transposed vectors.]
(Recall that I said you could think of it as the image of the transpose map.)
@Kasmir: Think of the three dogs as one clump and the four cats as another clump. Where can you put the clumps?
@TedShifrin Haven't progressed a lot; got a fever. But I started reading, so that counts for something. I can define $\omega_i$ and $\omega_{ij}$, currently reading ahead.
Sounds like you are packaging the Ist and the IInd fundamental form.
Well, if I say, ask Kasmir why this works, they'd say, "Kasmir, why does this work?" So instead they say "self." :P It's a silly joke, but they like it.
1111000 and 0001111 1 is cat and 0 is dog cat must be in the middle since there are 4 of them, so 4 ways to chose witch cat in the middle and 3! ways to arrange to rest of cats and 3 ! ways to arrange the dogs
Plus, one thing I know for sure about the general 4-by-4 matrix considered in the paper is that it should only be 'solvable' when certain conditions are fulfilled. So the hope is to find something interesting about the characteristic polynomial, considered as a Riemann surface in an appropriate sense, in these special cases.
Sure, @Balarka. All the $\omega_{ij}$ are $1$-forms; they're just components of $de_i$ with respect to the orthonormal basis. (There's officially a tensor product going on here, of course.)
If [A,B]=0, then $i\dfrac{d}{dt}\Psi=(A+Bt)\Psi$ could be done just from the matrix exponential since the matrix on the RHS would commute with itself at all times.
The idea, roughly speaking, is that one can compute things about the full time-dependent equation by knowing the characteristic polynomial for the adiabatic problem (the above replacement).
No, no, @Balarka. But you would get the second fundamental form by taking $\sum \omega_{i3}\otimes\omega_i\otimes e_3$. (Similarly for higher codimension.)
Hey everyone! I have a question that didn't get attention (probably because it links to another question) It's really stupid and it's driving me nuts, I'm desperate can anyone help? It's: math.stackexchange.com/questions/2100482/…
I'm pretty we sure we always assumed that principal minors are those you get starting from the upper left corner, for example we'd say that a matrix is positive definite iff it's principal minors are positive
@Alessandro: But most people say that the coefficients of the characteristic polynomial are the sums of the principal minors, and this includes all the ones I described.
For that to be true, you'd need the number of different outcomes with at most 4 heads to be the same as the number of different outcomes with at least 4 heads.
The number of different outcomes with at most 4 heads is the same as the number of different outcomes with at least 4 tails. (If you've got 8 coins and you flip 3 heads, then the rest are tails.)
regarding my question from yesterday I noticed that actually the trivial topology is enough to make addition and multiplication continuous functions so probably the correct question to ask is whether the euclidean is the coarsest topology with which $\Bbb R$ is a topological field @Ted
But the probability shouldn't depend on heads versus tails. The two outcomes are equally likely. So therefore switching tails to heads in that what I just said shouldn't change a thing.
yep, I'm confused now, I'll have to check the terminology in my linear algebra book because I'm pretty sure we call principal the leading principal and I have no idea how we call the principal
A boy has 3 red , 3 yellow and 2 green marbles. In how many ways can the boy arrange the marbles in a line if: a) Marbles of the same color are indistinguishable?
11! / ( 3!*3!*2!)
b) All marbles have different sizes?
where they have different sizes they are all different now right ?
@TedShifrin I've been trying to find him, as he doesn't frequent the Mod's chat. I need to ask him if he can attach a "wedge of cheese" to the top of my gravatar... He's very skilled at doing such things. (It's a Green Bay Packers thing).
This quesiton came into my mind : What is the goal of MSE ? Is it only to make it easy for people to discuss maths, or is it to build some kind of huge pile of questions/answers to make it easy for people to answer questions they might have ?
Because a lot of questions have a lot (too much) in common. For instance I've seen dozens of diophantine equations that are not all the same, but the method for solving them is the same. It seems a bit silly to keep all those questions since all it does imho is drown more relevant questions/answers in a flood of questions that are fundamentaly the same
@Astyx Look for meta math posts about "abstract duplicates". Indeed, you have good company with respect to your question, and so "a duplicate" has come to mean "not necessarily identical to"....
@barto Let's say it is defined as a finite union of convex polytopes. The problem is I don't know how to formalize the geometric intuition about vertices.
@Astyx Glad to help. And yes, it does get tiring to read the same-old same old as answers to the same-old same hold, especially when there are six duplicate questions with answers, listed to the right of the question/answers.
