Suppose I have a system of equations. It's big - say, 16 equations in 12 variables. All eqns and variables are real. Is there an algorithmic way to find if there's a solution?
I mean, I'm asking for an algorithm to calculate whether or not there's a solution to the system of equations. This means I have actual honest to god equations with actual honest to god coefficients to plug into this algorithm.
How could we find x such that (37 * 34 +72* 5)*x=84 *34+76 * 5 without using a calculator? Does the prime factorization help? Or is the only way to calculate the products and additions?
Could you give me a hint how the clever looks like? The prime factorisation doesn't really help, right? At the one side we would have the primes 17 and 37 and at the other side the prime 19. What else could we do?
First of all dividing by 2 we get the equation $(37 \cdot 17 +36 \cdot 5)x=84 \cdot 17 + 38 \cdot 5 $. To find the first digit of a number do we not find the remainder of the division of x and 10? @KarlKronenfeld
There is this kid who went to my middle/junior high school and when Messi came to the States, he saw this kid playing and got off of the bus to talk to him cuz he liked his skills
unless you're wondering why there are two things ($H$ and $\varphi$) that make up a representation - that means $H$ is the actual space and $\varphi$ the map out of $A$
often representation theorists will refer to a representation by a homomorphism $\varphi$ or a space $H$ interchangeably (this is pretty abusive for newcomers)
You take the direct sum of all those Hilbert spaces with the induced inner product. This will usually not be complete. Take the completion. That's the Hilbert space direct sum.
That's what $\oplus_\lambda H_\lambda$ refers to here.
@RandomVariable no. They said the defense was playing lights out and the offense was in a rut. They were dominant today but still played kinda crappy. Carli Lloyd just had a monster day and put the team on her back. That midfield shot was crazy.
But they could have lost at various points in the tournament. But with a team like the US, they always have a chance usually
that's also probably unfair to dump on you. the result you want there is that when $H$ is a closed subgroup of $G$, $G/H$ has a natural manifold structure such that $G \to G/H$ is a submersion
@anon: if you cared, it seems like there's just absolutely not a general theory (even in the weakest sense) of representations of finitely presented groups. when you get past, like, two generators and one small relator, it becomes impossible to check in the obvious way that there's a nontrivial representation
@RandomVariable the crazy part wasnt just the shot but also how that was (1) her 3rd goal and (2) 16 minutes into the game. I was having flashbacks to the Brazil vs Germany game m.youtube.com/watch?v=Oo2_yFv1kmQ
the specifics depend on the school. you could probably find out about wisconsin's qualifying exam system on the school's website
once you pass you're done
at UCLA advisors (often) won't work with you until you've passed your qualifying exams, since you need to be studying for those instead of doing research, and they usually test fundamentals any researcher in the field should know
I'm planning the last week of a 5 week (very basic; essentially remedial algebra) class, and I'm considering a bonus question (I've had random bonus questions each test).
I'll tell the students in advance that the bonus question will be deriving the quadratic formula (we will go over completing the square and the formula, separately, in class). The hard version (more points) will be for $ax^2 + bx + c = 0$, the easy version (less points) $x^2 + bx + c = 0$.
That's exactly it! I figure I'm covering my bases, if I have the two versions from which they choose (at most) 1. And they have time to work on it in advance, to gauge their own skill
Because really, I just want them to learn "completing the square", that's my ulterior motive - but they'll think of it purely in terms of bonus points.
But, who knows, I don't know why I was looking for confirmation. How's summer treating you, @MikeMiller?
it's alright. in vancouver for a conference this week; then I go back to LA to put together my new IKEA legos; then I need to prep for the probability class I'm TAing in august
My girlfriend and I. 'next move' is indeterminate, I'll definitely be in Akron for the next academic year, but ideally I'll have found a PhD program that accepts me, faults and all, for the following fall.
I have come up with a brief "book of analogies" for covering spaces vs. galois theory. wanna hear?
fix some galois ext $L/k$. "points" - inclusion $L \hookrightarrow \bar{k}$, "paths" - morphisms $\bar{k} \to \tilde{k}$ between alg. closures, "fiber" over the point $k \hookrightarrow \bar{k}$ - either $\text{Hom}_k(L, \bar{k})$ or $L \otimes_k \bar{k}$ (haven't figured out which). "fundamental groupoid" - category $\mathsf{Gal}$ with objects being alg. closures and morphisms being isomorphisms between them.
rep theory is another thing I'm curious about. it has a bunch of physics applications. (I've heard) I may be Mr. Category Theory (Paul's appellation) but I'm a sucker for cool applications. :P
@BalarkaSen yeah, pretty psyched about learning actions.
This winter I am planning on teaching a small seminar (20 lectures 45 minutes each) for high school students. I was was given the freedom to choose the topic of the seminar, but it is supposed to be about some "advanced" mathematics in an elementary exposition.
I was thinking about lecturing on ...
@r9m I have seen a beautiful result as an exercise : Let $f$ be analytic and bounded in the right half plane $\Re(z)>0$. The result is that the series $\sum_{n=1}^\infty \Re(\frac{1}{z_n})$ is convergent, where $z_1,z_2,z_3,\cdots$ are the zeros of $f$ is this half plane.
@SohamChowdhury I've heard bad things about the film.
LOTR is good.
I hope to read The Silmarillion someday.
It's about the first and second age of the middle earth, I think. LOTR doesn't say anything much about the first age except a bit about Morgoth and breaking of Thagorodrim.
yes, the point is that you have to pick a path, i.e., choose a continuous function $f : [0, 1] \to X$ with $f(0) = x_0$ and $f(1) = x_1$, given $x_0, x_1 \in X$.
@Rememberme well, yes, like I mentioned above. but I am not sure how else you can prove it.
this is the geometrical translation : for any pair $x_0, x_1$ in $\Bbb R^2$, you take the straight line $f : [0, 1] \to \Bbb R^2$ given by $t \mapsto tx_1 + (1 - t)x_0$ joining $x_0$ and $x_1$ as your path.
not so fast, get well-acquainted with path connectedness first. can you prove that $\Bbb R^2 - \{(0, 0)\}$ is path connected now? (note that your argument doesn't work anymore. why?)
Note that you define a path for a given pair of points $x_0, x_1$ in your space $X$. $X$ is path connected if a path can be defined for any pair of points. Just making a note, since you seem to always confuse quantifiers (without which much mathematics cannot be done)
Lets see if we dont have a point in $\Bbb{R^2}$ then we can still form a continuous path because since we are on the plane I could just draw a path around it which will still be continuous
@BalarkaSen The first film was good. Then the quality slowly drops off, I agree. But, considering the crap that goes for film adaptations (I'm looking at you, HP), LOTR ones are nice.
