so im assuming you're defining $\mathbb{Z}_p$ as an inverse limit, and then take its fraction field to define $\mathbb{Q}_p$ and you want to find a way to get back $\mathbb{Z}_p$
apparently the multiplicative group $\mathbb{Q}_p$ is isomorphic to $\mathbb{Z} \times \mathbb{Z}_p \times \mathbb{Z}/(p-1)$. So I would imagine you can read off $\mathbb{Z}_p$ from here?
Prove that if $a,b\in \mathbb Z_+ \implies a+b \in \mathbb Z_+$
I can prove that $X=\{x: x\in \mathbb R \text{ and } a+x \in \mathbb Z_+\}$ is Inductive. We can easily prove that $1\in X$. Let $x\in X \implies a+x\in \mathbb Z_+.$ Then, $(a+x)+1=a+(x+1) \in \mathbb Z_+ \implies x+1\in X.$ Hence, $X$ is inductive. How do I prove that $\mathbb Z_+ \subset X$? Can you please help me?
@vidyarthi a reference can be found in theorem 3.2.2 of Serre's a course in arithmetic - the idea is that any $x\in \mathbb{Q}_p$ can be uniquely written in the form $p^a z$, where the p-adic valuation of $z$ is $0$ - then you have to figure out how to describe this guy
@N.Maneesh im not very sure how things are defined in your text and so I'm not too- but anyway if you proved that $1\in X$ and that if $x\in X$ then $x+1\in X$ then that already implies that $\mathbb{Z}_+ \subset X$
i guess you can look at $X\subset \mathbb{Z}_+$ (I'm assuming your $X$ is defined as a subset of $\mathbb{R}$, from your description), then argue by induction / if it's not $\mathbb{Z}_+$ there's going to be a least element $a$ in $\mathbb{Z}_+$ not contained in $X$ but then $a-1\in X$ and conclude
@Iza_lazet: Apparently Mochizuki and Fensenko have seen it, but I would not be aware of anyone else. I guess Mochizuki withholds it until S&S come out with their critics publically. Which would make sense. But since Mochizuki (concluding from what Fesenko writes) does not agree with the critics, it might well be, that its wrong or contains some larger problem in itself. Which all of course will further delay any progress ...
@TobiasKildetoft "PRIMS" can't tell it straight away, I read all kinds of rumors, would have to check it again. "invested" yes that's mentioned in the discussion of "not even wrong" by some very strong-opinioned person.
@TobiasKildetoft: "from my heart" was refering to his general implications on basic research.
Its in a way an odd choice. In a way I can understand him, though. He has his priorities and principles, he felt obliged to RIMS and so he submitted it to there despite the problem he is editor in chief --- not that I necessarily would have done the same. On the other side all depends on the refereeing process no matter what the "organ of publication" is. Perelman I think only "posted" it on ArXive, didn't he?
So maybe the "optics" is something which should be of least concern.
Also wait has that flared up again? I feel like every now and then there's some ridiculous new development with IUT and then at the end of the day the conclusion is still something to the effect of: shrug
@Mathein holy snap this is ridiculous. Are modular forms that formula-heavy?
@MatheinBoulomenos: looks mean, but don't underestimate your ability to "learn by mind". Just go and practice ... Its incredible how much stuff one can memorize.
@LeakyNun the p-adic valuation is the unique non-trivial valuation (up to equivalence) on $\Bbb Q_p$ that makes it complete (due to a theorem of F.K. Schmidt)
Ah I see. Yeah in some talks that my professor gave in the REU he seemed to emphasize the analysis a bit more but I've heard from at least one grad student that it doesn't really need to involve that much if the icky side you play your cards right
@Daminark it's a beautiful subject, no matter how you approach it (the analysis is just complex analysis, so that's alright), but our prof wants us to know every formula and every proof for the exam
I mean, I think most of my professors had the policy of, anything that happened in class or the psets is fair game for the final, so if you want to ensure you do well you need to know everything
Though the problem is that when the formulas are this bad... Yeah
(Also kinda random aside, someone I know brought this up once and it had me thinking: is math in the foreseeable future gonna reach a stage where it's "too hard"? Like, if the background required before you start doing research just gets too high?)
@Daminark I don't think it will. Because at the same time that the background expands, we also get a much more conceptual understanding of many of the underlying things, which allows us to better understand them without necessarily first going through all of the original way of doing things
@Daminark I think it's unlikely that this will happen without parts of the background being blackboxable. I talked to a grad student in number theory and he told me basicallly that it would take forever to get a proper foundation in all the stuff that is used in modern number theory like abelian varieties, harmonic analysis, étale cohomology, crystalline cohomology, motivic cohomology, K-theory etc. but he said that you can just take some results as a black box and work with that. Like for étale cohomology you just accept that there's a cohomology theory with some properties and you can wor…
@MatheinBoulomenos Yeah, I read up on $l$-adic cohomology sufficiently to understand the idea behind Deligne-Lusztig theory in a few hours. Actually learning that would take years
@LeakyNun year 3 is kinda tight, you need an intro course on ANT before that and for that you need Galois theory and a bit of commutative algebra, so I think year 3 is the earliest you can reasonably do it without skipping stuff
Oh Neukirch has a proof of that. Basically take an integral element, the characteristic polynomial of the map x to ax is actually gonna be a power of the guy that annihilates it
and the fact that $\lim_{n \to \infty } \varphi(n) = \infty$
if $L/K$ is a finite Galois extension you can show that the characteristic polynomial of $\alpha$ is the minimal polynomial to the power $n/d$ where $n=[L:K]$ and $d=[K(\alpha):K]$ you can get the equivalent definitions of norms from that
I am working through the proof that, given any vector $x$, closed convex subset $C$ of a Hilbert space has a unique vector closest to $x$. Here is the part confusing me: "Uniqueness is clear--if two vectors $y$ and $z$ in $C$ minimize the distance to $x$, then $x,y,$ and $z$ lie in a (real) plane so any vector on the line segment between $y$ and $z$ would be strictly closer to $x$."
The claim seems to be that $||x-ty-(1-t)z||$ is strictly smaller than $||x-y||$ and $||x-z||$ for any $t \in [0,1]$, but I don't see why this is true. I tried the triangle inequality...but that didn't seem to go anywhere.
@user193319 since the theorem doesn't hold in general Banach spaces, just using the triangle equality can't be enough
@user193319 let's just work with $t=\frac{1}{2}$ for simplicity. Clearly we must have $\|x-y\|=\|x-z\|$ since both $y$ and $z$ minimize the distance to $x$. Thus we get by the parallelogram identity: $4\|x-y\|^2=2\|x-y\|^2+2\|x-z\|^2=\|2x-y-z\|^2+\|y-z\|^2$, so $\|2x-y-z\|^2=4\|x-y\|^2-\|y-z\|^2$ which implies that $\|x-\frac{y+z}{2}\|^2=\|x-y\|^2-\frac{1}{4} \|y-z\|^2$, so if $\|y-z\| \neq 0$, then $\|x-\frac{y+z}{2}\| < \|x-y\|$ which contadicts the fact that $y$ minimizes the distance to $x$
in general it's often easier to work with the square of the norm than the norm itself in a Hilbert space
Hello all. I'm trying to implement an algorithm from a book about recommendation systems, and the math is a little over my head in some parts. I'm trying to parse this formula. It says that phi represents the density and distribution functions of the standard normal, but it seems they only take one parameter? The algorithm models click rates for two items over two intervals (current and next).
The parameters: x: fraction of user visits to item, theta: state vector with clicks and views for item, q0: known click rate of other item, q1: known click rate of other item in next interval, N0: nu…
forgive me not using latex
the p-hat values are for estimated click rates of the item
I've just watched a promotional video for the undergrad maths course at nottingham and there's a guy talking about how much he loves the applied side of mathematics
and then the video cuts to him flicking through a copy of dummit and foote
Do you mean $L_{\omega_1,\omega}$? Or what's that $\infty$? Anyway I'm somewhat familiar with model theory and know next to nothing regarding infinitary languages
ok, i want to show that given the constant symbols $d,c_0,c_1,\dots$ then the following set is finitely satisfiable but not satisfaible : $$\{d \ne c_i : i\in \omega\} \cup \{ \forall v \lor_{i \in \omega} v = c_i\}$$
how does a finite subset looks like ? $ \{d\ne c_0 , \dots, d\ne c_n\} \cup \{ \forall v \lor_{i\in \omega} v = c_0 , \dots ,\forall v\lor_{i\in \omega} v = c_m\}$ ?
so to show any finite set is satisfiable, then if $d\ne c_0 , \dots, d \ne c_n$ are the formulas in it( with the one from the other set) then we can take any set of $n+1$ elements and this set will satisfy this finite set? @AlessandroCodenotti
Intuitively it is finitely satisfiable because you can have every element be a $c_i$ while some element $d$ is different from finitely many of the $c_i$
Can every element be a $c_i$ while an element is different from all the $c_i$?
Find the number of ways in which a person can buy 6 chocolates 🍫, if there are 3 types of chocolates 🍫 available. (Chocolates of same type are identical)
do I have to solve for x+y+z=6 for positive integer solution? But then how to remove identical cases?
@MatheinBoulomenos Could you help me understand how almost-dedekind domains fit in with everything else? It looks like Prüfer domains are almost-Dedekind, and I suppose almost-dedekind implies some other conditions I already have?
@rschwieb the other way around: almost-Dedekind domains are Prüfer domains
by theorem 9.4 in Larsen-McCarthy, if $R$ is an integral domain that is not a field, $R$ is almost-Dedekind iff it is a Prüfer domain of Krull dimnesion 1 that has no idempotent maximal ideals
@Fawad the required number is the coefficient of $x^6$ in $(1+x+x^2+x^3)^3$, which is equal to $\binom{8}{6}$, as you could choose 0,1,2 or 3 chocolates of each kind,nwhich correspond to $1,x,x^2,x^3$ in the formula respectively.
@MatheinBoulomenos Ah, i overlooked the "discrete" in "discrete valuation domain" in your definition. I knew that Prufer domains have localizations that are valuation domains. I was searching for domains characterized by that, but I couldn't find anything clear
@MatheinBoulomenos So Dedekind --> AD--> Prüfer and normal . I did a little bit of searching but I couldn't find a non-normal Prüfer domain. Are you aware if that's possible?
so you want to draw something that is a prototype of such, in the sense that we can follow the pattern set by it in order to obtain the true limiting curve
conceivably, you could take your curve to be: a horizontal line to the right, a short vertical line up, then a horizontal line to the left, and another short vertical line up
and then repeat
the shorter the vertical segment, the more times you'll repeat that portion over a given vertical range
Actually I would love some kind of sketch proof that can show how to prove what I just said. Sure it intuitively makes sense, but continuum cardinals are well known to play havoc to our intuition
It also does not help I cannot start with rationals and then complete the process, because horizontal lines at every rational definitely does not form one connected component
though clearly the limiting case is simply-connected
that's missing a little context: the full quote from Wikipedia is "Wiener pointed out in The Fourier Integral and Certain of its Applications that space filling curves could be used to reduce Lebesgue integration in higher dimensions to Lebesgue integration in one dimension."
@rschwieb another thing is that AD implies regular (since DVRs are regular local rings), this doesn't hold for Prüfer domains since regular local rings are by definition Noetherian
Hi, I have trouble proving that the language $L_1=\{a^ny, y \in \{a,b\}^*, |y|= n, n\in \mathbb{N} \}$ is not regular using the pumping lemma. My issue is, when I try to pump a $x \in L_1, x =uvw$ I can always construct $uv$ such that it contains only $a's$ and after pumping $x'$ is even, then $x' \in L_1$
@Abcd Let $H(s)$ be the antiderivative of $h$, i.e. $\frac{\mathrm d}{\mathrm ds} H(s) = h(s)$. Then, left hand side = $\frac{\mathrm d}{\mathrm dx} [H(g(x)) - H(f(x))] = H'(g(x)) g'(x) - H'(f(x)) f'(x) = h(g(x)) g'(x) - h(f(x)) f'(x)$
@rschwieb forget what I said about AD => regular. The definition of regular ring I had in mind was "every localization at a prime ideal is a regular local ring", but it seems common at least to require regular rings to be Noetherian
Looking at rigid body transforms it looks like inverting the matrix gives you -Rtrans*t for the translation component, I guess I was expecting -t and (Euler) (-Rx, -Ry, -Rz); is there any mathematical term for just "negating" a rigid body transform like this? I think I must have gotten confused somewhere. I can just multiply t by -1 but doing the same thing with R doesn't work, I'm not sure what to search for on that.
I guess a more rigorous definition of what I'm thinking of is, a reflection over a plane drawn at the origin with the orientation of the rotation matrix? Does that sound right? I have a hard time googling this because a reflection is itself a rigid transform
You shouldn't be able to have reflections according to the construction they provide
"Supposedly first you align the axes by rotating the frame O around x, then translate the frame O to O′ - and finally align the two coordinate frames."
Prove that for any sequence $(a_n)$, we have $\liminf_{n\to\infty} a_n \le \limsup_{n\to\infty} a_n$. Can someone help me with this? Just to clarify, my definitions for limsup and liminf are in this question: math.stackexchange.com/questions/2861356/…
@Semiclassical did it on paper a little while ago, I solved it by making three matrices (one for Rx, one for Ry, one for Rz) then doing R = Rx * Ry * Rz
I think part of the confusion here is, I might have been expecting a rigid body transform to be a translation, followed by a rotation (where the two operations are totally independent), doesn't seem like that's the case because R affects t when doing the inverse
If you suppose that the inverse T' of a rigid body transformation is another rigid body transformation, then one should have T'(v)=R'v+t' such that T'(T(v)) = v
so that imposes R'(Rv+t)+t' = R'Rv +R't+t' = v
in which case it looks like one wants R'R = 1 (so R' is the inverse of R) and t'=-R't
