@Moses $[0,1]$ is connected. Let's say $f(0) = k$. Then $f^{-1}(k)$ is closed, and $f^{-1}(k)$ is also open, hence by connectedness $f^{-1}(k) = [0,1]$. Before the language of connectedness is established, that is usually dressed up in the way of "Let $a = \sup \{ x\in [0,1] : f(y) = f(0) \text{ for all } 0 \leqslant y \leqslant x\}$. Then $f(a) = f(0)$ by continuity, and there is a $\delta > 0$ such that $\lvert b-a\rvert < \delta \implies \lvert f(b)-f(a)\rvert < \frac{1}{2}$, hence ...
... $f(b) = f(a)$ for $\lvert b - a\rvert < \delta$. But if we had $a < 1$, then we'd have a contradiction to the definition of $a$, hence $a = 1$ and $f$ is constant."
a) one can use TeX in more contexts than math b) even if they do use it in a math context, just because they like to help with TeX stuff doesn't mean they like to help with math stuff
@Integrator They are guidelines for being welcome in this room. They are reasonable guidelines, so I would say that not observing them would be willful annoyance, and people will probably complain and the room owner or mods may ban you from chat either temporarily or permanently depending on how bad the behavior is. Minor inadvertent infractions will usually be overlooked.
@Integrator when someone asks "what happens if I disobey the rules", it indicates that they are considering not following them. This tends to indicate problematic behavior.
@DanielFischer Out of interest do you maybe know of a function which satisfies what I described but is nowhere differentiable? I'm thinking that it obviously not feasible to consider another solution using the Mean Value Theorem?
@Moses Uniform continuity on a compact set is just continuity, so it just suffices to find a continuous nowhere-differentiable function. And Daniel just said the standard example.
I'm having trouble determining ds and the bounds for this problem
> Integrate $\int_C (x^{3}+y^{3}) dy$ where C is the curve consisting of two edges of the rectangle in the first quadrant having opposite vertices at the origin and the point (6, 5), where the first edge traversed starts at the origin and is horizontal.
@N3buchadnezzar: I've barely finished the first 20 pages but you've really made me take a dive into the uncomfortable waters of integrals. I can't wait till my bath is over and I'm out of the shower as shiny and squeeky clean like a rubber ducky integral.
@MikeMiller Three votes to delete by 20k+ users. As for their reasons to vote that way, I could only guess. Probably along the lines of " A nonsensical question. Should be closed. But the bounty prevents it being closed. A pretty sneaky strategy! – GEdgar Nov 25 '12 at 1:30" and " This question is pretty much a deepity. – Michael Greinecker♦ Nov 26 '12 at 16:30".
I saw a question yesterday which was interesting, and now I can't find it. As I recall, it went like this: let $H$ be a complex Hilbert space, let $T_n$ be a sequence of bounded self-adjoint operators with $T_{n+1} \ge T_n$ (in the sense that $T_{n+1}-T_n$ is positive semidefinite) and suppose $\sup_n \|T_n\| < \infty$. Show that $T_n$ converges strongly to a bounded self-adjoint operator $T$.
It is related to math.stackexchange.com/questions/1007208/… but not the same. I am sure it was posted or at least bumped on Nov 9 or Nov 8. Maybe it was deleted but I can't find it in deleted questions either.
@DanielFischer: SOT. It's definitely not true for norm convergence (consider $T_n$ being orthogonal projection onto the span of $\{e_1, \dots, e_n\}$).
@N3buchadnezzar I have dreadful amount of practice with it. And it seems I need to hit the books for a bit more exercise. I'm too sloppy. Did I tell you about my lightbulb?
@N3buchadnezzar What do you mean? What are you talking about? I have the book on my comp but since it was made in latex. Google translate can't crack the norwegian code :D
@Nick Attatch a $\mathcal{C}$ at the end. Or just write "because of space restraints the integration constant has ben omitted". There are cases where the different constans play a role, but usually it does not matter.
Most of the time the integration constant is the like the higgs boson, it does not matter. But some very strange scientists and mathematicans do think it matter.
Hello everybody. I need help in solving a system of linear equations in 5 variables (a..e) x 6 equations ((a + c = 17) (a + d = 16) (a + e = 28) (b + c = 4) (b + d = 29) (b + e = 15)). I decided to try the Gauss method, but it requires n x n matrix, so I combined the last two equations, but it gave a wrong result. What can I do? P.s. sorry for my English.
An infinite amount of mathematicians walks into a bar. The first orders a beer, the second orders two beers and so forth. After a while the bartenders shouts enough! and pours -1/12 beers.
@DanielFischer Hi Daniel Fischer could you please give me an idea how your proof using uniform continuity would look for the question I mentioned earlier? The proofs that I have is the one I proposed which you said was using 'connectedness' and then another one using the Intermediate Value Theorem. I'm interested to know how the one you said using uniform continuity would look?
@Moses By uniform convergence, there is a $\delta > 0$ such that whenever $\lvert x-y\rvert \leqslant \delta$, then $\lvert f(x) - f(y)\rvert < 1$. For integer-valued functions, that then means $f(x) = f(y)$. So you reach $1$ starting from $0$ in finitely many steps.
A catholic teacher was having an argument about gay marriage with a student today. So in the raising of voice I decided to shout "pray the gay away" however the teacher did not share my satirical comment.
@UserX If the function is continuous, it is either $0$ or some $a^x$. If you don't require continuity, the answer usually begins something like "Let $\mathcal{B}$ be a Hamel basis of $\mathbb{R}$ over $\mathbb{Q}$ ..."
@UserX Okay. You can see that either $f\equiv 0$ or $f(x) \neq 0$ for all $x$ easily. So suppose the latter. You get $f(x) > 0$ for all $x$ by $f(x) = f(x/2)^2$. Take the logarithm, $g(x) = \log f(x)$. Then $g(x+y) = g(x) + g(y)$. Now google "Cauchy functional equation".
"This condition was weakened in 1875 by Darboux who showed that it was only necessary for the function to be continuous at one point." What does this mean?
@UserX If a function satisfying $g(x+y) = g(x) + g(y)$ for all $x,y$ is continuous at one point - actually, if it is bounded in a neighbourhood of one point - then it is globally continuous. Follows from the functional equation.
@UserX If its domain is only one point, it is continuous. On its whole domain. Which is exactly one point. So it's continuous at exactly one point. But that's boring. The thing with Cauchy/Darboux is that you have a function defined on all of $\mathbb{R}$, and if it satisfies the functional equation, continuity at one point as an additional requirement forces it to be analytic.
oops..thought my class was at 4 for some reason but it's actually at 3..might be my subconscious telling me i already failed the class anyway so why bother going
@hb20007 "Very close" may be impossible, depends on how you define the distance. But once you've reached infinity, you've left the realm of integers (or real numbers, or complex numbers).
@Saw i can't because without the notes from lecture, i can't memorize them (prof wants us to memorize everything word for word or he'll mark it incorrect)
Okay, I read some papers, they all include the following two things; Hamel Basis, Indexing Set. I have no idea what these are so I'll leave the problem as is
@UserX No. The Weierstraß function is very very tame. A discontinuous solution of the Cauchy functional equation is ugly. It is not measurable. Its graph is dense in the plane.
@MikeMiller If the circle group $\Bbb T$ is isomorphic to $\Bbb R/\Bbb Z$ is there any other way to prove this other than isomorphism theorem because of kernel of exponential map?
