@AsafKaragila I keep trying, but I just do not understand enough to upvote your answers :-( If you go after some LHF, I might be able to understand your answer :-)
@robjohn I wonder what it is happens for the first half to whole hour of each day such that fresh votes are not shown. It doesn't seem to make software-engineering sense, given my mental model of how the site works.
@AsafKaragila I actually came up with something close to your answer to the first one while I was reading it, so that is definitely LHF (but a good answer). I will have to look up equipotent to vote on the second.
Yikes. So far I have like... six pages of material for tomorrow. I would roughly have time to talk about four of those. This is good news for me, in general, as it means that there will not be much time for many proofs.
The point is this: the listeners don't know anything about set theory.
So I am not going to even begin presenting the proof by Shelah.
Instead I am going to explain the problem itself.
So I have to explain a bit about Ext of abelian groups, which would take about ten minutes; and then give some definitions which would take another ten.
Then some claims without proof in between, and a few claims with proofs which would get me another 15 minutes into the lecture.
So far we're at roughly 35 minutes.
This gives me 10 more minutes to talk about two big theorems, and thus leaves me with very little time to discuss technical lemmas.
@AsafKaragila I said spruced up, that means it is not the same, but adjusted to work in the new case. If I am misunderstanding, then I will have to take my upvote back :-)
@robjohn My understanding was that it doesn't really prove omega+1=omega, but just shows how to lift X+1=X to P(X)+1=P(X) if you already know it for X (which may be omega or something wilder).
@AsafKaragila as I was looking through some old questions, for an unspecified reason, I came across this. Isn't it true that $A$ is the union of all the elements of $P(A)$?
There are many other problems: branch cuts, zero-recognition, keeping track of domains of validity, etc. A good place to find such information is to browse the web pages of leading researchers, and conference proceedings (ISSAC,SYMSAC,Sigsam,Eurosam, etc). For example, see Richard Fateman's paper...
Does someone have access to a computing environment like Mathematica, Maple or whatever handy?
I need to get an approximation of $$\int_0^\infty (e^x-\lfloor e^x\rfloor) e^{-x/2}\cos(14.134725141\cdot x)dx$$ if possible. (The 14... constant is the imag part of the first nontrivial zero of R. zeta.)
brb picking someone up
btw my hypothesis is approx $(2\pi)^{-1/2}(1/4+14.134725141^2)^{-1/2}$
another route would be to compare the graph of $$\int_1^\infty (u-\lfloor u\rfloor) u^{-3/2} \cos(\omega \log u)du$$ with that of $$\sqrt{2/\pi} \mathrm{Re}\left(\frac{1+\zeta(1/2+i\omega)}{1/2+i\omega}\right) $$
The answer is almost, but not quite. Why they are so close will be made clear momentarily.
We begin by partially evaluating the usual Fourier transform:
$$F_N(\omega)=\int_0^{\log (N+1)} \big(e^x-\lfloor e^x\rfloor\big) e^{-x/2} e^{ix \omega}dx $$
$$=\int_1^{N+1}\big(u-\lfloor u\rfloor\big)u^{...
The graph of my answer appears to have all of the zeros as the first graph in the OP, though his appears to be scaled, but I still wanted a bit more empirical confirmation. That's why I was asking.
I hit at this problem: If $E$ is a non-compact space, then prove that there is a unbounded continuous map from $E$ to $\mathbb R$. No spoilers please. I am just telling you. No Not even a hint right now. :-)
Never mind! But are you sure the exercise is true as it stands? Because, there may be a misprint and you may be trying to prove it and thus disprove Godel's inconsistency theorem....
Hold on. In a metric space you have that the space is compact if and only if it is sequentially compact i.e. if every sequence has a convergent subsequence.
@KannappanSampath Let every continuous $f$ be bounded. And let $e_n$ be a sequence in $E$. We want to show that we can find a convergent subsequence. We have that $f(E)$ is bounded hence $f(e_n)$ is bounded and by Bolzano-Weierstrass we have that $f(e_n)$ therefore has a convergent subsequence.
Let's denote that convergent subsequence $e_{n_k}$.
No, we started with an arbitrary sequence in $E$ looked at its image under $f$. Boundedness of $f$ gave us a convergent subsequence in $\Bbb R$ but why should it converge in $f(E)$. Am I missing something obvious?
We started with a sequence $e_n$ in $E$. Then we noted that $f(E)$ is bounded and hence $f(e_n)$ which is a sequence in $f(E)$ contains a convergent subsequence $f(e_{n_k}) \to f(e)$.
So we have a convergent (sub-)sequence in $f(E)$.
Now we want to use this to get a convergent subsequence $e_{n_k}$ in $E$.
A map $f: X \to Y$ is said to be proper if the inverse image of any compact set in $Y$ is compact in $X$. Prove that $f$ is a closed map if it is continuous and proper. What about the other implication?
@MattN Shall I go have dinner and think about this problem? I'll be back in less than 20 minutes. Sorry! : (
The answer is almost, but not quite. Why they are so close will be made clear momentarily.
We begin by partially evaluating the usual Fourier transform:
$$F_N(\omega)=\int_0^{\log (N+1)} \big(e^x-\lfloor e^x\rfloor\big) e^{-x/2} e^{ix \omega}dx $$
$$=\int_1^{N+1}\big(u-\lfloor u\rfloor\big)u^{...
