Have you tried seeing a physical therapist instead?
Im 6' 5'' and have had terrible posture my entire life and quite a lot of pain in my back and chest over the last year, but it has improved remarkably with exercise and massage.
There was a point when I was trying to get an appointment to see a doctor on my university's health website about my muscular chest pain, and couldn't do it because on the forms to make an appointment chest pain would immediately direct to a dead end page telling you to go the ER.
@TedShifrin Often my job is to tell people formulas they have to memorize without giving any good reason to other than the scholastic one(usually I don't have these formulas memorized). It's hard to be terribly optimistic about this part of it.
Yeah, @PVAL, this is one on which I'm slightly conflicted. But shouldn't a college math student know basics like area of a circle and volume of a box ... and have some idea about units?
@TedShifrin Perhaps, but memorizing something like the general solution to the Euler's equation when the characteristic equation has a repeated root seems kind of meaningless if you don't know how Euler derived it.
Yeah, actually, I think most DE courses don't give any motivation for what happens with repeated roots of the characteristic equation, in general. That bugs me.
I have a question about sets. I think it's pretty hard for highschoolers. I was given a physics exercise that states $5t^2 -20t + 2d > 0$ where $0 \le t \le 4$ and $d > 0$. I must find the minimum value for $d$ given the conditions above
Just to state: We got a car at 20m/s and another car at 10m/s with a distance of d. If to not crash, car 1 have an acceleration of -5m/s², what is the minimum value of d?
@RudytheReindeer Hi, I see you around the site a lot because for some reason whenever I search for questions, you have always asked the same ones I have now
I think the question was: If the limit exists, then show that you can get the same limit first letting $y \to b$ and then letting $x\to a$, and oppositely.
Then I think the OP wanted to ask that. If the joint limit exists and for all fixed x the y-limit exists and for all y the x-limit exists then so do the iterated limits and are equal.
If I have vector spaces $V_0\subset V_1\subset V_2\subset V_3\subset\cdots$ then can you see that $\bigcup V_i$ is a vector space? It has nothing to do with what types of things the vectors are (besides the fact they are elements of vector spaces that are in a chain like that)
Hey @Ramanewbie and @Huy. I was just wondering whether it's valid to substitute the set $A$ with the logic formula $x \in A$ and work with an expression of sets as a logic formula instead.
@Huy I actually work on a couple of different things. Continuing on from my dissertation, I am working on figuring out how certain tensor products of representations for algebraic groups decompose
I am also working on understanding the 2-category of Soergel bimodules in type $B$, hoping to understand if all simple transitive representations come from left cells
and parallel to that I am working on understanding how a new description of special representations fits into various settings
I remember having an algebra lecturer who insisted on using it even though it is going out of style, since he had been part of introducing the notation
I am asking this question because according to Balarka there is a easier proof that a topological group is abelian using that $\pi_{1}$ is a functor@Tobias
I think we only used that set (and not the notion of integers) to prove that for any p-adic number exists a unique sequence $(a_n) \subset \{0, \dots, p-1\}^{\mathbb{N}}$ with $$x = \lim_{n \to \infty} |x|_p^{-1} \sum_{k=0}^n a_k p^k$$
I meant that the fundamental group of a topological group is abelian@Tobias
I mean to say that if $G$ is a topological group with $x_0$ as its identity then $\pi_{1}(G,x_0)$ is abelian where loops have $x_0$ as its base point@Huy
I think you can also argue with the universal cover very quickly that the fundamental group must be abelian, but my knowledge of covers isn't very good so I'd have to write it down on a paper. :P
@TobiasKildetoft: do you by any chance know whether there's some similar way to easily figure out the fundamental group of a quotient, similar to the product?
@Huy MY approach has been to take two loops , use concatenation of loops and then prove that it is abelian. I would like to see the universal cover method
Hi guys. I need some help proving the existence of $\sqrt{2}$. I understand that I need to look at the set $A =\{x \mid x < 2\}$ and use the least-upper bound axiom, though I'm not sure where to go from there.
I'm trying to show that the supremum of A is not less than two, and not greater than two. In doing so, I let $(S+\epsilon)^2<2$ and try to reach a contradiction, but I'm not succeeding.
I've been struggling with this for far too long now and I'm pretty tired of the question. It seems that the least upper bound property is going to imply that $S^2<(S+\epsilon)^2=S^2+2S\epsilon+\epsilon^2<2$, but I'm not sure if I'm on the right track
@Huy @Tobias It's not really that complicated. $\pi_1$ is a functor from Top to Grp, which preserves products. Group objects in Grp are abelian groups. Done.
But I realized it's just a fancy way to cover up Eckmann-Hilton.
@Adolfo I'd like some of the exercises from your book (as it seems picture based) and stuff, but my languages are English, and Welsh (iffy) not Spanish. If you see any could you let me know?
I have Skype, and my email address is [first letter of first name] dot [second name] at warwick.ac.uk
@AlecTeal Sure, I can send you the exercises there a couple of nice exercises related to gluing together some moebius bands and I'm sure there are some more that I haven't looked at yet.
Hey all. Question - I was curious as to whether there was already a question posed on MSE regarding the sequence where $a_{n} = \sqrt{n}$, and whether it is contracting/convergent. I recently completed it for an assignment, and just wanted to see if any MSE answers could offer some extra insight I didn't think about. I feel like it is most definitely already asked somewhere, but my searching skills are really failing me :( I was wondering if anyone can recall such a question.
@TobiasKildetoft By contracting I mean that the difference between adjacent terms in the sequence is getting smaller as $n$ gets larger. I just showed that the terms weren't in fact contracting, then showed the sequence was not convergent. I think the purpose of the question was to assert that the sequence was not contracting, mostly.
I showed that (sorry I can't seem to get LaTex working in chat I hope you can see it) $\frac{|\sqrt{n+2} - \sqrt{n+1|}{|\sqrt{n+1} - \sqrt{n}|} < 1$ for all $n$.
Oh no, I made a horrible mistake. I see now. Oh dear. Thanks.
Apologies, I've now managed to confuse myself. It appears to me that my presented solution was wrong, and that $a_{x} = \sqrt{x}$ is indeed a contractive sequence, since $|a_{x+2} - a_{x+1}| < |a_{x+1} - a_{x}|$. But then I have a theorem detailing that all contractive sequences in $\mathbb{R}$ are Cauchy and thus must be convergent in $\mathbb{R]$.... which doesn't make sense for $\sqrt{x}$. What am I missing here?
