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12:02 AM
@robjohn Thanks for the update!
12:58 AM
I'm thinking about offering a math prize like the able prize but significantly less money
1 hour later…
2:01 AM

Is this a good question? 🤔
2:30 AM
Huh, pretty neat. $\left(\ln w+iw\right)e^{\ln w+iw}=z\ln z$ shows why $W_n(z \ln z) = \ln z$ over the reals yet again.
I will never get bored of the complexes
Quite satisfyingly, because of getting into complex arithmetic, I also solved $r=\theta$ as $-iW_n(ix)$.
There. The complexes are finally well-ordered.
is this intersection an open interval? I feel like it should be $(-1, 0]$, but visualizing the intersection makes me think it should be $(-1, 0)$
but this set is supposed to be closed, but maybe there was a typo in the notes or something
Sorry I misspelled "canonical ordering" as "well-ordered".
Also clearly all $r=b\theta$ for $b\neq 1$ are fanfics. True fans recognize the canon lore.
2:47 AM
@SillyGoose Ask yourself, "is $0$ in the intersection?" Is there any of the sets being intersected which do NOT contain $0$?
3:11 AM
hm i see so i guess they all should contain $0$
3:22 AM
So the intersection contains $0$.
why is the euler mascheroni constant so important?
it is the limiting difference between the harmonic series and the natural log
@JohnZimmerman what do you mean by "so important"?
It naturally pops up in a few places---particularly in analytic number theory---but I am not sure that makes it important.
@XanderHenderson fair enough
3:39 AM
Hoping for someone to give me a hint as to how to proceed with the following:
The latter is what I have as yet. I've not been able to figure out how to use the hypotheses in order to show that the $k_i$ must be the same, which would be enough to prove my claim
But maybe my whole strategy is ugly and there's a much quicker and more elegant way to do this
Does the Paley-Weiner theorem apply to closed surfaces?
paley wiener applies to gaussian (rapidly decreasing)
gaussian is not closed.
closed=compact w/o boundary
sphere can be ruled out for obvious reasons
2d gaussian btw**
3 hours later…
7:06 AM
@EE18 why k_i
@EE18 lots of ways to organize it. maybe prove by induction on N that if i and j are naturals with min(i,j) = N and h(i) = h(j), then i = j. for the induction step, if min(i,j) = N+1 and h(i) = h(j), then writing i' = i-1 and j' = j-1 we have min(i',j') = N and from h(i) = h(j) one can deduce h(i') = h(j') [why?] so that i' = j' by an inductive hypothesis and hence i = j.
By definition $h(n) = f^n(c)$. There is nothing to prove
Nothing to prove that such $k_i$ exist
2 hours later…
8:56 AM
Quora, when seeing MSE: internal screaming
9:12 AM
let $M$ be a smooth manifold. let $x \in M$. i want to define the tangent space $T_xM$ as the space of all derivations over smooth functions at $x$ (with appropriate addition and scalar multiplication defined on the derivations).
I am a bit confused because wikipedia states that a derivation as $D: C^\infty(M) \to \mathbb{R}$, so it sends a smooth function $f: M \to \mathbb{R}$ to a real number.
Is that right? I would have thought that it should send the smooth function evaluated at the point $x$ to a real number, so $f(x = x') \mapsto Df(x) \lvert_{x = x'}$
Expressing the zeta labyrinth in terms of its automorphic representations and canonical geometric evolution dynamics
I think i dropped the ball with that title
oh well - at the end of the day it's just a question
@YourLordJoyBoy That was the idea behind the name. I had just finished that book when I created this account and that book was my whole personality for almost a month.
9:38 AM
@SillyGoose what does what you just said mean?
If a smooth function is evaluated, then this is just a composition with a function $\mathbb{R}\to\mathbb{R}$?
hm i see i guess that would not be what we want
is there a definition of pullbacks similar to the shown definition of a pushforward? The text I am using writes a definition like $(3)$, but it utilizes an inner product and that seems to be extra data than just the smooth manifold and its natural constructions
@SillyGoose what is the star supposed to mean
dual space?
the star $T^*_xM$ means cotangent space at $x \in M$.
dual to $T_xM$
then $\alpha\in T_xM$
and $\beta\in T_{f(x)}^*M$
there's typos in the text
this is not an inner product but a dual product
$\langle T, x\rangle = T(x)$
its not called dual product, a second
oh oopsies that is my bad i made the typo
9:48 AM
In mathematics, a dual system, dual pair, or a duality over a field K{\displaystyle \mathbb {K} } is a triple (X,Y,b){\displaystyle (X,Y,b)} consisting of two vector spaces X{\displaystyle X} and Y{\displaystyle Y} over K{\displaystyle \mathbb {K} } and a non-degenerate bilinear map b:X×Y→K{\displaystyle b:X\times Y\to \mathbb {K} }. Mathematical duality theory, the study of dual systems, has an important place in functional analysis and has extensive applications to quantum mechanics via the theory of Hilbert spaces. == Definition, notation, and conventions == === Pairings === A pairing or pair...
found it
anyway its a very canonical pairing between a vector space and its dual and it has nothing to do with Hilbert spaces or inner product spaces
you can consider it for Banach spaces no problem (for example)
is this literally just like if i start with a vector space $V$ then construct $V^*$ as the space of linear functionals over $V$, then by construction there is a product defined $v^*v$ where $v^* \in V^*$ and $v \in V$?
other than, indeed an inner product on a inner product space is a dual pairing in some sense, where the dual is isomorphic to the vector space
so its just a more specific context of this more general thing
@SillyGoose well, its literally evaluation of that functional at a vector
but the notation $\langle x, y\rangle$ is sometimes helpful because it does behave a lot like an inner product
good god a smooth manifold comes with so many natural constructions :P
9:52 AM
I don't know if it has anything to do with smooth manifolds, its just vector spaces really
oh i am just trying to learn a bit about smooth manifolds right now for a physics thing. and i am trying to construct all the necessary structures and see if each is a natural construction or if they require more data than just the manifold to specify
so far all are natural :P
well the list is not so long: (co)tangent bundle, pushforward and pullbacks of smooth maps, and exterior algebras of differential $k$-forms
@SillyGoose so here the mapping $f^*$ would be the transpose of the map $f_*$
there's a natural way to define, given $T:V\to W$, a map $T^*:W^*\to V^*$ given by $$\langle T^* \alpha, \beta\rangle = \langle \alpha, T\beta\rangle$$
this is known as a transpose in linear algebra
ah :D
or dual map maybe
i am trying to build up to learning about the holonomy of a connection...I am just about at starting to learn what connections are
9:57 AM
Then you can say, do it again and obtain $T^{**}$ and so on
and if $V, W$ are finite dimensional then $V\cong V^{**}$ and $W\cong W^{**}$ in a natural way, and we have $T^{**} = T$ for an appropriate interpretation of that equality
@SillyGoose I don't know any of that. I never had a course from differential.. well, anything
10:39 AM
@Jakobian It's not a typo, writing $\langle-,-\rangle$ for a pairing of vector spaces is something we do on occasion in topology or geometry.
Though agree it can be potentially confusing if it's not explained
11:29 AM
@Thorgott uhhh. I never said it was?
and pairing of vector spaces is literally something I've been explaining
ok, I didn't get what you mean then
two messages above me saying "there's typos in the text" is what I was referring to
11:54 AM
Q: Understanding where my approach is incorrect

John ZimmermanThe gist of this approach is to encode the zeta function into a space, similar to how number theorists like to encode sequences into generating functions to leverage analytic properties. I will refer to this construction in general as the $\zeta$-labyrinth, and the main idea here is to link the a...

I will remember that downvote
weird how the previous question got +15 lol
And I will take no prisoners.
I think it's just so foreign looking that people don't know how to react
12:20 PM
@JohnZimmerman You wrote "There must be flaws in my reasoning perhaps fatal ones. Which part of this seems the least rigorous? I want to identify where to put my energy first."
The purpose of this site is not to peer review your proposed solutions
yeah i think it's how i phrased the ending there
and it's not a solution
by the way
1:15 PM
You'll get at least one downvote immediately if your question post includes the word "Zeta function"
1:58 PM
And this is why I'm sceptical about posting questions now. I'd rather ask my questions in here.
2:10 PM
Speaking of, may I have some help with a problem?
$2t^2 + 17t + 35$
@JohnZimmerman Beware
@Swan Was unaware that Black Swan had a book, I recall the film which I watched for my psychology class several semesters back.
@Pizza returns
With his anchovies spinach and pineapple pizza
i have a question for this
calculate the definition set of the function:
$$f(x,y) = \sqrt{(\log(x))^2 - y^2}$$
@Pizza Immediately wondering if you know where that is from
And I'm asking out of amusement
do i have to solve
$(\log(x))^2 - y^2 >= 0$?
2:20 PM
@Pizza Hoping someone else can help you with this one since I'm unable to.
@YourLordJoyBoy it's just pineapple!
@Pizza You know I was kidding right? :P
And my joke has a fragment of history behind it.
@YourLordJoyBoy yess
@YourLordJoyBoy dont worry
Thought so mate :P
I hate needing help with these. But I need it. :( If I can just get what I'm doing.
2:44 PM
I'm confused about a remark in my notes following the dominated convergence theorem (DCT), which is proved using Fatou's lemma. The author states that an alternative proof may be obtained from Egoroff's theorem and the absolute continuity of the integral. However, Egoroff's theorem, as stated in my notes, works on finite measure spaces. Now suppose the sequence of functions in the DCT have domain $\mathbb R$ and $\mu$ is Lebesgue measure.
Then this is not a finite space, right? So can we apply Egoroff's theorem to prove the DCT nonetheless?
i need to solve $(\log(x))^2 - y^2 >= 0$ and $x > 0$
so $(\log(x))^2 >= y^2$ ?
And I need to know how to solve $2t^2 + 17t + 35$ correctly.
@JohnZimmerman The question seems to boil down to a lot of exposition, followed by a very vague kind of "what is wrong with this? or how can it be better?". Such a question isn't really a good fit for the SE format.
@psie It has been many years since I played with the nitty-gritty of these things, and my analysis books are all at home, so I don't have a good statement of Egoroff in front of me, but my recollection is that you can use Egoroff to get DCT on a sigma-finite space by looking at the finite bits and kind of gluing them together or taking a limit in a clever way.
E.g. use Egoroff to show that DCT "works" on $[-n,n]$ for any $n$, and think carefully about how you can take $n$ to infinity.
But, again, this is not something that I have at my fingertips right now.
I would bet dollars to donuts that a Google search like egoroff DCT site:math.stackexchange.com will come up with results.
ok, sounds sensible
I can't even do down votes yet
3:06 PM
Can someone help me
Q: How can I find the argument of this complex number?

user123234 I am working in the unit disk where the boundary is parametrized by $e^{it}$ for $0\leq t<2\pi$ and I define $h_0([e^{i\theta_1},e^{i\theta_2}]):=\frac{\theta_2-\theta_1}{2\pi}$ where the $0$ in the index set means that my referenz point is zero. Now I want to compute $h_\omega([0,t])$ where $z=...

Also seeking help with a problem of my own.
Let $(I,\leq)$ be a directed set, so for any pair of $a,b\in I$ there exists $k\in I$ such that $a\leq k$ and $b\leq k$. I was wondering what would be an example of a directed set if we add a stronger condition that for any countable family of elements $\{a_n\}\in I$ there exists $k\in I$ such that $a_n\leq k$ for all $n$.
We can create examples that includes $\omega_1$ but I am thinking if there is one without it.
Assuming that's for user?
They don't have a username while we do.
3:36 PM
@robjohn Hey Mean Box!
nvm, he's gone
4:23 PM
@YourLordJoyBoy no
Any set bounded from above
not really sure what kind of example you are expecting
Oh right that works
What if the set is not bounded above?
there is still massive amount of examples
again, what are you expecting
@Jakobian like?
like any limit ordinal $\alpha$ with uncountable cofinality
in some sense this should be the only totally ordered example since you should be able to find a copy of such ordinal in any unbounded from above totally ordered example you can think of
@Jakobian No it didn't have anything to do with that user?
4:31 PM
@YourLordJoyBoy yes
@Jakobian Who was being answered then?
@YourLordJoyBoy no one
@Jakobian I see, it was a problem from @SoumikMukherjee then.
@SoumikMukherjee an explicit example would be $\omega_2$
@robjohn Mean box returns?
Again, @Swan I sadly didn't know of the book. I did watch the movie for a psychology project though.
4:41 PM
Thanks Jakobian
5:17 PM
Can someone recommend me a reference for the analogue of Teichmüller theory in higher dimensions? I.e. the classification of spaces of complex structures (or Kähler or Calabi-Yau or whatever) of (complex) dimension $n>1$
Anyone mind giving me some help in learning to do this problem? $2t^2+17t+35$
writing down an expression is not a problem statement
@leslietownes Thanks for the comment Leslie :) I as yet have not seen the order on $\omega$ defined so I am hoping not to use min and such which rely on that. I think what I am majorly missing is why the $c \notin \ran F$ is so important
5:33 PM
consider $f=id$
Why would people being anti-gun as to hold your gun? STUPID
ask not as >:(
@Thorgott what do you mean
that this gives an example of why the condition is important
ah. True
5:56 PM
Out here doing weird geometry stuff
Q: What is the largest possible diameter of a shape made of centers of bounding boxes?

Akiva WeinbergerGiven a shape in the plane (fixed in space) and an angle $\theta$, the shape's bounding box with orientation $\theta$ is the smallest rectangle containing the shape such that one of the sides is an angle of $\theta$ counterclockwise from horizontal. We may call the center of the bounding box with...

6:24 PM
How to find the cardinality of the set of all k rank vector bundles over S^1?
k rank vector bundles upto isomorphim *
6:40 PM
you have to know some things about vector bundles to answer this question
I'm not sure how much you've already covered
And what of a problem like this? $A rocket is launched straight up into the air from the ground with initial speed of 64 ft/sec. The height
of the rocket (in feet) is given by the equation
h = -16t^2 + 64t where t is the time in seconds after launch (t􀀃􀂕 0).
Find the time(s) when the rocket is at ground level.
A) The rocket is at ground level at 0 sec and 4 sec
B) The rocket is at ground level at 4 sec
C) The rocket is at ground level at 16 sec
D) The rocket is at ground level at 0 sec and ±4 sec$
And this came out completely wrong
Lemme redo it
hmm, so I know that 1) If f,g: M-->N are homotopic. And E is a vector bundle over N, then pullbacks of f and g are isomorphic bundles over M. (M,N are manifolds).
2) vector bundle over a contractible manifold is trivial.
A rocket is launched straight up into the air from the ground with initial speed of 64 ft/sec. The height
of the rocket (in feet) is given by the equation
$h = -16t^2 + 64t$ where t is the time in seconds after launch (t􀀃􀂕 0).
Find the time(s) when the rocket is at ground level.
A) The rocket is at ground level at 0 sec and 4 sec
B) The rocket is at ground level at 4 sec
C) The rocket is at ground level at 16 sec
D) The rocket is at ground level at 0 sec and ±4 sec
BTW the bit about t two squares and 0 is t is greater than or equal to 0
you solve h=0 to get the answer.
6:57 PM
@Thorgott this was for me I take it?
7:08 PM
@EE18 it was
@Koro great, this is a good starting point. the circle S^1 is the union of two semi-circles, so try applying 2) with that in mind.
7:48 PM
Talking with a friend, he was trying to say that for $f:[a,b]\to\mathbb{R}$ such that $f(x)>0$ for each $x \in [a,b]$ we have $\int_a^b f(x)dx>0$ because, if $F$ is the integral function of $f$ with $a$ as lower bound of integration, it is $F'(x)=f(x)>0$ and so $F$ is increasing hence $\int_a^b f(x)dx=F(b)-F(a)>0$ because $b>a$ and $F$ is increasing.

I don't agree with this solution: we have no hypothesis about the continuity of $$, hence the fundamental theorem of calculus does not hold and so we can't say that $F'(x)=f(x)$ and that $\int_a^b f(x)dx=F(b)-F(a)$. Am I correct or my friend's
In the latter paragraph I meant: "We have no hypothesis about the continuity of $f$"
if you don't assume $f$ continuous, you indeed cannot argue like that
the claim, however, is true
granted, you should assume at the very least that $f$ is measurable
8:18 PM
@Thorgott: Thanks for the answer!
8:32 PM
@ZaWarudo Let $A_n = \{x : f(x) > \frac{1}{n}\}$, then $\int_a^b f(x) \geq \frac{1}{n}\lambda(A_n)$ where $\lambda$ is the Lebesgue measure. Since $[a, b] = \bigcup_{n=1}^\infty A_n$ we must have $\lambda(A_n) > 0$ for some $n$
here $f$ is measurable as Thorgott mentioned above
@ZaWarudo for this argument to work we have $F'(x) = f(x) > 0$ a.e. and not for all $x\in [a, b]$ and then you need to argue why $F$ is increasing from this
I think its clear that $F'(x) > 0$ implies that $F(x+h)-F(x) > 0$ for say, $h$ positive and close enough to zero
and since $F(x)-F(y) = \int_y^x f(x)dx\geq 0$ for $x > y$, then $F$ is weakly increasing
so clearly $F(b)> F(a)$
@Thorgott so you actually can argue like that, with some more effort
@Jakobian: Thanks for this example. I have two questions: if $f$ is only Riemann integrable, is this reasoning of yours still valid or the solution of my friend is wrong? Second question: isn't $F(x)-F(y)=\int_y^x f(x)dx$ only if $f$ is continuous because of FTOC? I mean, the relationship between the integral function $F$ and one of the antiderivative of $f$ is valid only because of the FTOC (at least for Riemann-integrable functions), or am I missing something?
@ZaWarudo This is valid for all Lebesgue integrable functions, this includes Riemann integrable functions
@Jakobian yeah, if you know the improved ftc
@ZaWarudo as for second question, define $F(y) = \int_a^y f(x)dx$
then $F' = f$ a.e. and $F$ is continuous
As Thorgott mentions this doesn't follow from classical fundamental theorem of calculus
8:50 PM
Ok, so that proof works but it is necessary to consider Lebesgue integrable functions and use results from a more general measure theory than the one we see in the first analysis courses. However, without explicitly saying this and using brutally using the FTOC without using a.e. arguments and similar things (sorry for the imprecise words, I still haven't studied measure theory), the proof is flawed
Did I understand properly?
@ZaWarudo sure yes.
Ok, thanks to both for the clarifications
The proof that $F' = f$ a.e. relies on a result called Vitali's covering theorem
so its slightly advanced than what you do in an analysis class, sure
for what it's worth, you should be able to provide an elementary proof in the case that $f$ is Riemann-integrable
but it's a subtle issue, I think
@Thorgott in the same way, you mean? I don't think such function even has to have an anti-derivative
8:55 PM
not in any particular way
Just learn Kurzweil integration theory
the parts I enjoyed best about it is with functions like $F(y) = \int_a^y f(x)dx$
Can I ask some advices about studying during college courses? I like to solve problems, but sometimes I get stuck and I lose a lot of time (and I can't afford that much time loss because I have exams). However, each problems book I read says that one should find the solutions of the problems without reading the solutions; how much hurtful is this practice of not reading solutions? And, at the same time, how much hurtful it is to read solutions? To me, is hard to find an equilibrium...
Actually, "often I get stuck" is more precise than "sometimes" :D
@YourLordJoyBoy Okay, that is a problem. Have you thought about the answer?
@ZaWarudo Mathematics is not a spectator sport. You learn by doing, not by watching other people, or reading what other people have done. If you are serious about learning mathematics, you need to get comfortable with (a) not knowing how to attack a problem, (b) trying and failing many times to solve any particular problem, and (c) solving problems without looking to see what other people have done in the past.
I would suggest that you should not look for a solution to a problem until you have given it at least a few days, during which you try several things.
I would also suggest that mathematics is a team sport (setting aside the small number of famous people who have been successful individually), and that you should try to find a study group to work with. As an undergrad, there were multiple mini-lounges in the building which housed the math department. Each of these lounges had a large whiteboard and five or six comfortable chairs for folk to sit in (as well as a low table, typically).
There were several groups who consistently made use of these spaces. These groups tended to be formed of the people who did best in their coursework.
9:17 PM
@XanderHenderson: Yes, that's what I do mostly because I agree that it is hurtful not struggle with problems; that's why, even if there are the solutions in the back of the book, I always try to answer myself and look at them basically when I already found a solution and I checked it multiple times. I am only afraid that sometimes I miss some particular technique or some hard idea that is too much to find by myself.

About the last thing you said: isn't working in team somehow in contradiction with not being a spectator sport? I mean, if I have to struggle with the problems but a friend of
And thank you for the help
@ZaWarudo No, there is no contradiction.
If you are working in a team, you are contributing.
I am not suggesting that you work with people who already know the answer. The point is to find people who are at about the same level as yourself, and work through problems together.
Ok, I misunderstood you. Thanks for clarifying!
You talk about not knowing some particular technique. Maybe someone else in your team will know that technique, but won't know how to finish some problem; on the other hand, you might see the result of the technique, and know what to do next. By working with them, you learn the technique and the other person sees whatever it is that you know to finish off the problem.
When you are a spectator, you are passive. When you are playing the game, you are active.
it would be hard to make a paper ship out of toilet paper, right?
@XanderHenderson I agree. And for self study? I usually proceed like this: I try a problem, if I get stuck I try another problem. And so on, until I have not too much problems "open" to think about. Maybe the next day I solve, let's say, 2 of the 5 problems I open. If in the next days I cannot solve, let's say, 1 of them all 5, I cover the solution except for the first line and come back to the problem to see if I can solve it with that "first line suggestion". Do you think this is a good practice?
9:26 PM
@ZaWarudo Self-study is very hard.
Most people don't really make a lot of progress trying to study on their own.
But, like I said, don't go looking for solutions until you have spent several days working on a problem.
Ideally, give it a week.
(Assuming that there is a deadline.)
@XanderHenderson everyone has to self study to some degree, don't they
@Jakobian To a degree. But people who are outside of support systems (other students, tutors, TAs, professors, etc) tend to make much less progress, and cap out much earlier, than people with support.
@XanderHenderson I'm always thinking about the answer whenever I see a problem.
1 hour later…
11:03 PM
I must master synthetic division!!!!!!!!
Everyone concerns about quadratic residues, but what about cubic, quartic, or quintic residues?
Guess it depends on the math one is focused on. I'm LITERALLY going to be doing my algebra 1 final next week.
@XanderHenderson I tend to bomb when I try to study alone. I am much more prepared when studying with others.
One curious observation is that, in $\mathbb{F}_5$, $2^{1/3} = 3$ and $3^{1/3} = 2$ hold.
Of course, there should be three cubic roots, which means there are two other $2^{1/3}$. Something analogue to complex numbers for finite fields...?
And in $\mathbb{F}_5$, the quartic roots of $1$ are $1$, $2$, $3$, and $4$. That is exhaustive.
11:28 PM
Does anyone know if this theorem also holds if we replace the codomain with $\mathbb R$?
Isn't that just trivial? If $f_n : X \to \mathbb{R}$ pointwise converges to $f : X \to \mathbb{R}$, we can plug these functions in the $\overline{\mathbb{R}}$ version.
So, affirmative.
Ok, thanks!
The reason I was asking is because in one version of the Lebesgue dominated convergence theorem (DCT) we simply have a sequence of functions with codomain the reals that converges to some function $f$, but it is not specified whether the function $f$ is measurable, which is a necessary assumption in proving the DCT. So this theorem is probably the reason why.
11:52 PM
It turns out the two "complex" cubic roots of $1 \in \mathbb{F}_5$ are $2 + 2 \sqrt{2}$ and $2 + 3 \sqrt{2}$. Interesting.
Consider a principal bundle $P$. Define a connection on $P$. How do I prove that the choice of horizontal subspaces induced by the connection is precisely the kernel of the corresponding connection $1$-form?

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