Maybe it is imposter syndrome, but I just feel like the people I know at a similar stage of their career are being productive and doing so well. I feel like I am constantly falling behind.
I did the best I could, @Michael. And I was very fortunate to find people to collaborate with pretty early on.
I was never that prolific, and I knew I was in academia more for the teaching than for the volume of papers. I don't know what I'd do today.
DogAteMy: I was never trying to be mean when I wrote my books, but I was trying to write a good book, and that means students would have to work to read it and do the exercises. So I suppose today's students think that's patently mean.
Yeah, I actually don't remember how I ended up with one when we were both grad students. I think I pointed out that I could do a bunch of stuff he'd done over $\Bbb R$ in the complex and projective case. ... Then at UGA collaborations grew naturally out of working together in seminar situations and talking.
I need some help with a paper I'm writing, what is some pretentious way to describe clamping very boring numbers? Is there an operator or something? For example, b=max(min(a,42),0) if I want to sandwich a number between 0 and 42
I'm writing some mathematical equations that describe some computations in my program and it's pretty important that it's written correctly. At one point, it clamps or truncates a value, $x$, into the range $[0, 1]$.
How would i best denote this in official mathematical notation? My best attempt...
Hello all. I've just finished Moise and Downs' Elementary Geometry. Which Analytic Geometry book should I read next before I read a book for a first course in linear algebra?
"Log-scale informs on relative changes (multiplicative), while linear-scale informs on absolute changes (additive). When do you use each? When you care about relative changes, use the log-scale; when you care about absolute changes, use linear-scale. This is true for distributions, but also for any quantity or changes in quantities." What does an exponential scale inform?
Define the expected value of a a divisor of any $1 \leq k \leq n$ to be $\sum\limits_{x=2}^{n-1} x\text{Pr}(x \mid k; 1 \leq k \leq n) =: E(n)$.
$\text{Pr}(x\mid k; 1 \leq k \leq \sqrt{n})$ is the probability that a randomly selected integer $1 \leq k \leq \sqrt{n}$ is divisible by $x$. This is...
In reading about the existence of an algebraic closed extension for any field, by S. Lang (credited to Artin)
He denotes the set of all letters $X_f$ for $f \in K[x]$ as $S$
(idk how to define this formally but it’s acceptable)
He considers $K[S]$ says that the ideal generated by $f(X_f)$ is not the unit ideal
If it were, of course there would be a finite combination of terms such that $\sum_i g_i(X_{f_1}, \ldots, X_{f_N})f_i(X_{f_i}) = 1$
But then he constructs an extension such that every polynomial $f_i$ has a root (it’s just the composition of the splitting fields) and he claims that this implies $0 = 1$. Why?
Take any field K. Let S = { pair (f,i) : f is a polynomial over K and i∈N }. Let T = { field (M,+,×) : M⊆S and +,×∈(M^2→M) and ∀(f,i)∈M ( f applied to (f,i) in (M,+,×) equals 0 }. Define binary relation ≤ on T such that M ≤ N iff M is a subfield of N. Then (T,≤) is a partial order that we can apply Zorn's lemma to. It suffices to prove that every non-empty chain has an upper bound. Indeed, for any chain C in (T,≤), the union U of fields in C is also a field in T.
@LucasHenrique No you cannot!! Are you working in a naive (inconsistent) set theory?
@LucasHenrique At least you are aware of that. Any rigorous treatment of field theory must show existence of algebraic closures to be able to deal with compositums in a general fashion.
For example, the theorem that any set of polynomials over a field K has a splitting field relies on either an ambient algebraic closure or transfinite recursion in the same straightforward manner as proving alg. cl. existence.
@LucasHenrique I know that, but you seem to want more than just as many symbols as polynomials, because you talk about some "algebra". It just doesn't work.
You cannot just say we want this and that symbol to have this and that algebraic relations. That was how ancient mathematicians extended from N to Q to R and then to C, and only in recent times do we have rigorous constructions that demonstrate that extensions with those desired properties actually exist. It's exactly why imaginary numbers were not accepted at first, because "Why should we believe that R can be extended just like that?".
You of course now know how to do a finite field extension, but that was not known back then.
Let me just say that there is no shortcut to the theorem (existence of alg. cl.). People who say or think there is are not going to get an actual proof of the theorem. Any claimed proof of the theorem will have the same ingredients as the one I gave you. If you're unwilling to learn, I cannot force you to.
You’re unwilling to teach as you assume (or treat me) like an idiot. And if you don’t like this proof, complain with Artin. I’m literally just trying to learn, and his proof is easier to understand, and uses Zorn’s lemma implicitly when talking about a maximal ideal.
I don't assume you're an idiot. You on the other hand assume that you understand a proof when you don't.
When I say you don't, it's based on my experience. It's not because I think you cannot understand.
If you want, I can give you an explicit concrete example to show why the naive idea of K[S] won't work. But other than that, you're going to have to work through the actual proof, because there isn't any way to show that an argument is wrong unless you write it formally in a system like ZFC.
So you've plenty of options. (1) Write a proof in ZFC (or equivalent) and I will check and point out explicitly where you make a mistake (if any). (2) If you ask, I'll provide the concrete example. (3) Just work through the proof I gave you until you understand it. (4) Learn transfinite recursion and the theorem becomes obvious.
I agree it's hard to understand in the form I gave you, but that's because you wanted a proof using Zorn's lemma, so the easiest way to actually get a rigorous proof is by that method. I think it's easier if I gave you the proof sketch for the proof using plain transfinite recursion.
It’s utterly incomprehensible because of the formalism (it’s not that I dislike it, otherwise I wouldn’t study math, but I’d like to “really” understand what happening instead of making valid manipulations of logical symbols)
Let me begin by stating the basic stuff we need for transfinite recursion. A well-ordering is a linear ordering with no infinite strictly decreasing sequence, namely (W,<) such that < is a linear ordering on W and there is no infinite strictly <-decreasing sequence. Transfinite recursion is the notion that we want to construct a sequence that is 'as long as' a given well-ordering, namely a function on W.
Actually I read about the transfinite recursion, but I don’t know why it’s valid (I suppose it’s a consequence of well ordering) and I know nothing about ordinals
It goes as follows. Take any well-ordering (W,<) and set S. Let A[k] be the set of functions from W[<k] to S, for each k∈W, where W[<k] is defined as { i : i∈W ∧ i<k }. Basically, A[k] is like partial sequences. Take any recursive relation R, meaning R : { A[k] : k∈W } → S. Then there is a function F : W→S such that F(k) = R( F ↾ W[<k] ) for every k∈W.
It means restriction. F ↾ W[<k] is the restriction of F to the domain W[<k].
This is transfinite recursion. It says that if you have a way to define each point in an W-length sequence based on the previous terms, then you can define the entire sequence.
"previous terms" is captured by "F ↾ W[<k]".
Do you understand the statement of transfinite recursion? For ordinary mathematics, you do not need to know how to prove it, but it is very useful in conjunction with the well-ordering theorem: For every set W there is a well-ordering (W,<).
@user21820 oh, so that’s why there’s no infinite descending sequence? Like, you’ll always have a “new rank of infinity with $\omega$ much numbers until the next”?
The definition of well ordered set I know is that every non empty subset has a least element
Obviously, w.r.t the order that makes it well ordered
@LucasHenrique That's an equivalent definition under mild assumptions. It doesn't matter which one you use.
@LucasHenrique I'm not sure what exactly your asking about "rank of infinity", but I can at least tell you that not every element in a well-ordering is the 'next one after ω many previous elements'.
That's why I'm just giving you the theorems you need. You can if interested look at their proofs at another separate time.
So. These two theorems are all you need. Transfinite recursion plus well-ordering theorem.
Now here is the proof sketch of existence of alg. cl. of a field F. Construct a set S that is big enough to have enough elements to assign to each 'root' of a polynomial over F. Use transfinite recursion to go through each polynomial g over F one by one. At each step we have constructed a chain of fields, one for each previous polynomial, and their union U is also a field. If that union does not have a root of g, then we extend U to a new field that does, using unused elements of S.
Either way, we can assign a field to g that is either U or extends U, and the step is complete.
At the end of transfinite recursion we have a chain of fields, and their union can then be shown to have the root of every polynomial over F, which implies that it is an algebraic closure of F.
The "go through each polynomial g over F one by one" relies on the well-ordering theorem to provide a well-ordering for the transfinite recursion to use. That's it. No actual knowledge about well-orderings or ordinals is needed!
What other good ways could I show that the product topology on R^2 is the same as the standard euclidean metric topology? Other than showing that the identity is a homeomorphism?
@LucasHenrique You're not wrong. Some of the complications are unnecessary and can be reduced by using a more friendly system, but some are in some sense necessary. The reason is that we cannot use a foundational system that is known to be inconsistent, so the current standard one we use has some technical restrictions that we cannot evade.
@LucasHenrique The idea of 'partial' things is prevalent in rigorous constructions. Often, we cannot directly define each term of a (long) sequence, but we can define how to extend 'partial' sequences by one term.
So I defined A[k] so that I don't have to write out its definition everywhere I used it.
Sorry I made a typo... a recursive relation R is a function R : Union { A[k] : k∈W } → S, or equivalently R : { g : g∈A[k] and k∈W }.
Missed the "Union" there.
If you don't want to care about the set-theoretic issues, just remember that as long as you can extend any sequence (indexed along a well-ordering) by one term, you can extend all the way.
@topologicalmagician You are given two different bases for a topology: the product topology in terms of rectangles, the euclidean topology in terms of discs. It suffices to show that the rectangles are open (in terms of discs) and also that the discs are open (in terms of rectangles).
Yet another spherical question: Suppose I have two spherical triangles on the unit sphere, with sides $a,b,c$ and $a’,b’,c’$ respectively. (By sides I mean the geodesic distance between vertices.) Must there exist a spherical triangle with side lengths $(a+a’)/2,(b+b’)/2,(c+c’)/2$?
I have reason to think the answer is yes, but I really don’t see a geometric answer yet...
I think you should be able to use the law of cosines to produce it geometrically. I guess you need to know that for some ordering of those lengths, you have $\cos a'' > \cos b''\cos c''$.
Hi professor @TedShifrin I read the preface of Spivak's book and I saw you mentioned there for having made a couple of exercises there, can I ask how was it back in the time with respect to the exercise business compared to now? sorry if this sounds like a very convoluted question
LOL ... I wrote probably several hundred, actually. This was back in the late 70s, although I did contribute some changes to editions after that. I don't understand your question. "Exercise business"?
I don't have an answer. A few questions were suggested by some other people. I honestly don't remember. In the meantime, I've written lots of books and thousands of exercises.
@TedShifrin the weird thing is that, if I replace those side lengths (aka central angles) with their cosines, then I’m entirely confident that convexity holds
But I want it on the angles, which seems more obscure
That said, I remembered a book which was useful for this kind of problem before, so while I was away I tracked it down. And, indeed, I think I’ve found a theorem of theirs which addresses this (and a lot more)
So that’s neat
Amusingly, it turns this into a graph theory problem :P
suppose you take your points on the sphere and you’re interested in some subset of the side lengths. I can use that to make a graph, with vertices corresponding to points on the sphere and edges corresponding to relevant side lengths
Their theorem applies so long as that graph doesn’t contain a $K_4$ minor
My cases of immediate interest correspond to the 3-cycle and 4-cycle graphs (ie some number of vertices connected along a circle) and those definitely don’t contain K4 minors
Suppose $q$ is a quotient map and $U\subseteq X$ is a saturated open set. Then $q|_U:U\rightarrow q(U)$ is continuous, and q(U) is open in $Y$ and further, $q$ is surjective.
I must show that $V$ is open in $q(U)$ $\iff$ $q|_U^{-1}(V)$ is open in $U$. If $V$ is open in $q(U)$ then because $q(U)$ is open in $Y$, it follows that $V$ is open in $Y$ and by continuity of $q$,
@Ultradark Suppose you get lost on your boat at sea. You have a device that indicates your distance from coast is 1km, but you have no idea the direction you should head to get to the shore. You know also that the coast is linear, being perpendicular to the line between your location and the coast itself. Q: what is the minimal length of a path which guarantees you meet the coast?
This reminds me of the police boat trying to chase the escaping prisoners in the fog, @anakhro. I gave that to my differential equations students when I was teaching in grad school.
The prisoners escape from an island at some point during the night and steal a boat that goes $5$ mph. Some time later, the staff notice and sound out a boat that goes $10$ mph (or whatever, but faster) to capture the prisoners. There is a dense fog and so they can't see the escapees. Assume only that the island is far enough from land that throughout this process everyone is still in the water. What path should the staff follow to be guaranteed of intercepting the escapees?
small series question: assuming the series of $a_n$ and $b_n$ are convergent, does it follow that $c_n=\sum_{k=1}^n a_kb_n+\sum_{k=1}^{n-1} a_nb_k$ is too?
@TedShifrin looks like $b_n\to 0$ and $a_n\to 0$, and $c_n=\sum_{k=1}^n a_kb_n+\sum_{k=1}^{n-1} a_nb_k=b_n\sum_{k=1}^n a_k+a_n\sum_{k=1}^{n-1} b_k$ so can I immediately say that it's convergent?
oops actually the questions asks whether $\sum_{n=1}^\infty c_n$ converges
my gut feeling is that $\sum_{n=1}^\infty c_n$ doesn't converge, but I wonder how I can put a convenient inequality with $\sum_{n=1}^m c_n= \sum_{n=1}^m b_n\sum_{k=1}^n a_k+\sum_{n=1}^m a_n\sum_{k=1}^{n-1}b_k$
I'm really sorry but I couldn't find any way because the double sum uses different indices, I know that if we had $\sum_{k=1}^n b_k\sum_{k=1}^n a_k$ then we could perhaps use that $$\sum_{k=1}^n b_k\sum_{k=1}^n a_k=n\sum_{k=1}^n a_kb_k - \sum_{1\leq j<k\leq n}(a_k-a_j)(b_k-b_j)$$
so this is as far as I could extrapolate $\sum_{n=1}|b_nA_n|>\sum_{n=1}|b_na_n|$ and so it diverges, but I couldn't reach anything concerning conditional convergence
@user21820 Pardon if I'm being naive, but I don't understand your claims that the construction of K[S] is the bulk of the proof and needs some kind of transfinite recursion. K[S] exists as a set, this has been agreed upon. But the additional ring structure required on K[S] can be defined and checked explicitly. Alternatively, constructing it as direct limit of the polynomial rings in finitely many of the variables (and these can be constructed using plain induction) should work.
Define the expected value of a a divisor of any $1 \leq k \leq n$ to be $\sum\limits_{x=2}^{n-1} x\text{Pr}(x \mid k; 1 \leq k \leq n) =: E(n)$.
$\text{Pr}(x\mid k; 1 \leq k \leq \sqrt{n})$ is the probability that a randomly selected integer $1 \leq k \leq \sqrt{n}$ is divisible by $x$. This is...
