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12:00 AM
If I have a vector space $X$ with a translation invariant topology, does it follow that showing continuity of a linear operator $T : X \times X \to X$ reduces to showing continuity at $(0,0)$?
 
12:54 AM
@user193319: Yes.
 
Hey Ted!
 
Heya Demonark. How you doing?
How's geometry?
 
Decided not to continue auditing it, I didn't have much time to learn certain background things he thought we knew like tensors and Lie derivatives, and the first couple lectures weren't that good
 
aw, well, maybe when you get to grad school you'll do it.
heya Captain
 
1:01 AM
What kind of geometry was it Daminark?
 
Then again some friends of mine who kept going to the lectures said that they've been getting better. At the moment he seems to be doing curvature based on the pset for this week: math.uchicago.edu/~dannyc/courses/riem_geo_2019/homework3.pdf
It's grad differential geometry
 
oh cool
 
The colloquium he gave some years ago at UGA was very good.
 
danny is gud
 
It's tricky navigating a course when you don't know what people actually know. I always preferred to teach year-long sequences and be to blame for what they didn't know :P
heya ERic
 
1:03 AM
did u get a lot of complaints
when u taught?
 
Yeah the difftop course had 0 coordination with this one
 
CaptainAmerica: Only that I was too hard and made people work.
 
People who do that are lame.
 
I remember a bunch of my friends were struggling on the first pset because there were a bunch of words on there they just didn't know
 
Who do what?
 
1:04 AM
@Daminark that always seems to be what happens
 
If a course is too hard, the first person I blame is myself.
 
Well, Demonark, they should have gone to talk to Danny. No big deal.
 
Complain about the work in a course to the teacher
without a valid reason for doing so
 
Well, when a lot of teachers require nothing from the students to get an A (in advanced grad courses) and I require stuff, it's not unexpected to get some feedback :P
 
Thankfully there was someone there who knew a good bit of GR and independently saw stuff like tensors and vector bundles (the main stuff people struggled with) and they pulled it off
 
1:05 AM
Of course, students who actually want to learn appreciate the thought that goes into lectures and problem sets and grading ...
 
Did that ever get discouraging?
 
@TedShifrin tbf grad students at uchi dont even get letter grades
 
And I think people said that reading first chapter of Do Carmo mostly fixed the problems in that regard. The only person I asked about the second pset said that his main difficulty was in solving the ODEs
 
You get no sympathy from me re concrete computations.
 
Lol yeah here it's just pass-fail for the grad students
 
1:09 AM
Teaching two years for AoPS has made me appreciate the carrot/stick that is grades. Without it, the teacher is screwed.
 
the d in do Carmo is lower case btw
 
Just like the d in de Rham.
 
Ah I see
 
ya same exact reason
 
what do u mean
i guess without grades, students wouldn't work
 
1:11 AM
Yup.
In the ideal world, they would. But when there are all sorts of competing pressures, blah.
 
Yeah here there's the double whammy in grad school that every grad student has to take the full year of algebra/analysis/topology, while a number of them already don't care much for some subset, and then they only have to pass rather the class
 
I told my class yesterday that I was not going to teach any more for AoPS because I was too frustrated about the lack of responsiveness/work outside of class.
 
i guess the logic is that for grad students they should be able to find motivation within
 
The year I took analysis, only one grad student in the class was interested in analysis. So aside from that guy it was the undergrads who cared and the grads who didn't
 
You'd think so, @Érico, but nah.
 
1:12 AM
wow
 
Generally, grad students have a short-sighted view and don't care about well-roundedness/breadth, in general.
 
i wonder if my teachers feel bad when i don't pay attention in class
 
Unless their adviser or the department makes them.
I would, @CaptainAmerica. And I'd snap.
 
I would have suspected that grad students are in it for more serious research.
 
1:14 AM
Of course, if you were getting everything and working on hard extra credit problems, I'd cut you slack.
 
@TedShifrin this is kind of a fear for mine for grad school
 
Still, @Rithaniel, you never know when you will need stuff "out of area" for your research or to teach.
 
i guess i get most of it
i kind of just ignore the stuff i dont
 
Not in my class, @CaptainAMerica :D
 
1:15 AM
since as u well know where im goin they seem to have a prejudice against intro classes lol
 
Yup.
Not that you need too many of those.
But note that I was prejudiced against your choice :P
 
Ah, that's fair enough. In my abstract algebra course this semester there are about 10 education majors, and you can tell they're not feeling the subject matter.
 
in the end i didnt feel like it was much of a choice anyway :/
 
Generally, sadly, math education majors/grad students don't have much love for math. I think that explains a lot of the problems with education in this country.
 
AA seems like one of the coolest subjects
 
1:17 AM
It can be, CaptainAmerica, or it can be ruined, like everything else.
 
Lol I dunno how grad students were feeling about algebra but last quarter was a bit of a mess
 
I think so, Cap. Like, we're getting into quotient fields and factorization now that we're at the end of the semester and I'm enjoying myself.
 
Basically the professor was trying to balance having a general interest that was more geometric vs commutative algebra for the people who were going that direction and it didn't work too well
 
Doing commutative algebra without giving insight into the algebro-geometric applications or number-theoretic applications is horrid.
 
that's weird cuz id imagine all the algebra grads wouldnt be taking the seequence anyway
 
1:19 AM
@Rithaniel That's awesome. I rly excited for it.
 
cuz that's what often happens for analysis and geo/top
 
I'm not sure how many people place out really. But yeah the class kinda flip flops when different people teach
 
u can test out of topology?
i thought that only applied for intro courses
 
They're talking PhD level, Captain.
 
at chicago grads can test out of the intro sequences if they know the material well enough
 
1:21 AM
oh ok
 
I didn't test out of anything at Berkeley. I just took analysis quals and took no basic analysis courses at all. Same with algebra.
 
I know 2 years ago apparently it mostly avoided commutative algebra, half because the professor himself doesn't seem to like it that much and half because he was like yeah the algebraists all place out so I'm assuming everyone here is an analyst and doesn't care about commutative algebra
 
I guess passing quals counts as "testing out."
 
if i go into math, i can't see myself ever specializing in analysis type stuff. I like to visualize things.
 
Then the year after another guy taught and made it mostly commutative algebra + a bit of varieties + Cech cohomology at the end from nowhere and everyone was like uhhh. Then apparently this year was more of an experiment, in part from requests to make things more geometric
But it ended up being a mess
 
1:24 AM
CaptainAmerica: You'd be surprised how few professors teach anything with visualization ...
 
@Ted did berkeley have like intro sequences and stuff when u went
 
Sure.
They still do.
 
lol i prospied there and dont even know how it works
 
I think the numbers are still basically like when I was there. 202, 215, 240, etc.
 
1:26 AM
Yeah I remember I was looking at them for a while and they actually had a nice class selection
 
Lots of classes ...
 
the math building is low key horrendous tho
it has nice views of the bay tho
 
Oh the math building at Madison is actually huge
It's got 3 "underground" floors (quotation marks because the place is on a very tall hill so the first 3 floors are a good bit above the the street), and then 9 floors above ground. The grad lounge is in the top floor and overlooks the city and lake, it's real nice
 
lol the view from the top of princeton's math dept makes it very clear how desolate central jersey is
 
Eric: The math building at Berkeley was basically new when I got there in 1974.
 
1:31 AM
The basement floors have the library and all the classrooms (each of them has a lot more area than the higher ones), floor 1 is basically just the entrance, I'm not sure what's on the second floor, 3-8 is all offices, and 9 has the ground lounge mainly
 
@TedShifrin oh wild
the campus was so lively generally, there were like a billion people out and about doing random stuff
 
A billion? That's like our supreme president's tweet about the number of people killed in Sri Lanka.
 
And then there's one weird area called the math bunker that's trickier to access, you have to leave the building from the first floor, head outside (still walking on the roof of the basement floors), go to this other structure, and then get in. Some number of grad student cubicles are there (other grad students get offices in the main building)
 
Sorta like the secret underground passages at MIT :P
 
lol what
 
1:36 AM
@TedShifrin ya but more benign and less weird and insane
 
LOL
 
why are there underground passages at MIT
 
I can't answer "why."
 
figured that would be the answer
i was hoping there would be a cool story tho
 
There may well be.
 
1:37 AM
I'd appreciate some underground passages so I could walk to class without being in the heat/cold
 
Along with the cars reassembled on top of the great dome.
 
I mean it won't be that much better
But at least you won't walk in the snow/sun
 
The passages at MIT are on the main campus ... there aren't any from where the dorms are.
 
Actually underground might be worse for the heat so I might have to eat my words
 
And on the main campus, you can do most within buildings, anyhow.
 
1:39 AM
the way buildings are numbered there threw me off when i was there last year
v weird
 
Yeah, weird, indeed.
 
there = MIT or Berkeley?
 
MIT
 
Underground stuff is usually pretty static in it's temperature.
 
MIT numbers majors and buildings.
 
1:40 AM
Oh yeah I remember courses actually, when I was looking over MIT their classes were all 18.xyz
So presumably math is major 18
 
weirdos
 
yup
 
MIT sounds like a fun place
 
the neighborhood it's in is pretty cool and they have cool ppl in math around there
 
I dunno that there's a neighborhood ... I suppose several ... Kendall Square, Central Square, etc.
 
1:44 AM
Yeah their department seems to be on the larger end which was a plus for me
 
i meant neighborhood as in "the vicinity" but ok lol
 
i'm mostly interested in the math department, although i highly doubt i'd ever make it in
 
It's hard to get a feel for which places are good at undergrad math. Highly ranked places are known for having good researchers but there's no "How well does this place teach?" ranking which is kinda more relevant if you're an undergrad
 
These days undergrads in science/engineering are obsessed with undergraduate research opportunities.
I think the obsession is too much.
 
i doubt it's due to pure interest
 
1:48 AM
Nope, it's pressure, just like the ridiculous extracurricular pressures now in high school.
 
I think interest might have started the trend, though it is true that grad admissions now is starting to make it closer to an expectation (friends of mine say that for experimental physics, classes and all definitely don't cut it anymore)
 
I wonder how my current university stacks up in regard to "teaching quality."
 
the expectations never end
 
things have gotten much more competitive and university funding is becoming more weird and disastrous and making it in the sciences doing research has become insanely hard and opaque
so there's a lot of resume padding that happens bc if everyone is doing it ull be uncompetitive if you dont do it too
even if it doesnt reproduce the internal benefits it should, it all becomes a game
 
I really hate all this.
 
1:50 AM
In math I don't have a clear picture. It seems there are a lot of Mickey Mouse projects that people seem to not help people much, but more and more people seem to do more serious things and that seems to become a bonus
 
@TedShifrin unfortunately the ppl most hurt by this have the least power to do anything about it (as is usually the case)
 
Sounds right.
 
"Mickey Mouse project" is a new term to me.
 
I had truly super-talented students who didn't get into any REUs. Does that make them worthless? Of course.
 
I spend so much time thinking about something obvious ... I feel so dumb
 
1:53 AM
Mathein: It's the middle of the night.
 
mayb ur sleepy
 
I've done that before.
 
We all have.
 
i do that almost everyday
 
One of my professors said it to describe a bunch of REUs, basically boils down to problems that some of these give their students which nobody really cares about but which undergrads could work on and get a paper out of
 
1:54 AM
@TedShifrin i think universities have been ostensibly a game of credentialism for a long time, they just used to be gated off to a lot more people than they are now (see: ppl from backgrounds like mine) and now that budgets shrink to nothing (while administrative costs balloon) the problem gets harder and harder for students
 
Ted, i made a big life decision recently and i'm starting to regret it
 
Yeah, Eric. I am glad I retired, but I am not optimistic about the future.
Not again, @CaptainAMerica.
 
I got my ears pierced and it freakin hurts
 
@Daminark im kind of glad that my letter writers were very anti dumb projects and very pro learning experiences
 
0
Q: Proving Asymptotic stability in Lyapunov's theorem

johnny09 In order to show that $x=0$ is asymptotically stable, one needs to show that $$\forall \varepsilon > 0, \; \exists\, T > 0 \; \mathrm{s.t.} \; t > T \implies || x ( t ) - 0 || < \varepsilon.$$ The intuitive sketch of the proof is that one has to fit a sublevel set of continuous functions $...

@TedShifrin can i get your input about this question of mine?
 
1:57 AM
Yeah same, even my REU papers were just, learn and write about something that's already known
 
LOL, vanity, @CaptainAMerica.
 
i can't stop touching it and they said i cant take them out for 2-3 months
I thought i would look cool ;-;
 
That's not what REUs are supposed to be, Demonark.
Sorry, @johnny09, I'm about to leave for dinner. Perhaps another time.
 
that's how the chicago reu works tho, it's about writing an expository thing on something u learned and it seems pretty cool and good i think
 
of course!
 
1:58 AM
Yeah in general no, I think it's just that Peter (who organizes the one here) very much doesn't believe in the usual REU format, he just wants it to be 8 weeks of learning math casually
 
That's not what the model of an REU is. It's good to get practice writing, of course.
OK, bye, all.
 
tchau tchau
 
Bye @Ted
 
bubye @Mathein
 
Cya Ted, enjoy dinner
 
1:59 AM
See you around!
Side note: woot tomorrow it's back down to 55
 
"If $U$ is a domain in $\Bbb C$ and $K$ is a compact subset of $U$, then for all holomorphic functions on $U$, we have $\sup_{z \in K}|f(z)| \leq C_K \|f\|_{L^2(U)}$ with $C_K$ depending only on $K$ and $U$" this took me way longer than it should have
 
Bye @Ted
 
@Daminark thank god
 
@ÉricoMeloSilva if a sequence of harmonic functions on an open set in $\Bbb R^n$ converges uniformly on compact subsets, is the limit harmonic?
 
mean value property my man
 
2:17 AM
In this proof, we do not use $f(1)=0$ anywhere, right?
Mathein! you are here.
@MatheinBoulomenos Can we say $\frac23$ minimum and $\frac95$ maximum there?
That was areal nice argument. I was trying to relate to ellipse, maybe.
 
2:33 AM
@Silent I'm not sure. I think the maximum and minimum of $||Av||/||v||$ will depend on the inner product between the eigenvectors for the two eigenvalues
 
Ok.
 
it is true if the eigenvectors are orthogonal (i.e. if $A$ is normal)
 
How to argue for or against of (a) and (b)?
 
3:42 AM
Well, $A$ has these two dictinct eigenvalues meaning that $A$ can be diagonalised to a diagonal matrix with these two values as its diagonal. What will that mean when multiplied to a given vector (x,y) and how will the magnitude of that vector changed?
Alternately, compute the operator norm of $A$ and see if it is larger or smaller than 2, 1/2
 
Question: Is there an easy way to show that multiplication in $\mathbb{Q}[\sqrt{\delta}]$ is associative (where $\delta$ is a square free integer)?
 
Aren't multiplication in any field always associative?
 
Currently writing it out and I have a really cumbersome line of variables. I'm hoping there is a more tidy way to show it.
Yeah, it is, but I'm being tasked with showing that $\mathbb{Q}[\sqrt{\delta}]$ is a field.
 
ah
 
So I can't refer to the fact that it's a field, clearly.
"It's a field cause it's a field. QED."
 
3:50 AM
Are elements in $\Bbb{Q}[\sqrt{\delta}]$ have the general form: $a p + b \sqrt{\delta}$?
o wait nvm, it has to
hmm...
 
$a+b\sqrt{\delta}$ where $a,b\in\mathbb{Q}$
 
ok
 
Generally, speaking, given. $\alpha=a+b\sqrt{\delta}$, $\beta=c+d\sqrt{\delta}$ we have that multiplication (which I am writing as $\otimes$) is $\alpha\otimes\beta=(a\cdot c+b\cdot d\cdot\delta)+(b\cdot c+a\cdot d)\sqrt{\delta}$
 
yup
so you basically want to check whether $(\alpha \otimes \beta) \otimes \gamma = \alpha \otimes (\beta \otimes \gamma)$. Bruteforcing it works but is super messy looking
 
Yep, the reason I am exploring alternative routes of showing associativity is because writing out three elements worth of variables is taking up more than a single line in Latex, and that is really bugging my desire to keep things straight.
 
3:57 AM
hmm... I wonder if you can argue about the rationals forming a ring (hence using commutativity, associativity and distributivitity). You cannot do that for the field you are calculating, but you might be able to take shortcuts by using the multiplication rule and then properties of the ring $\Bbb{Q}$
for example writing $x = ac+bd\delta$ and $y = bc+ad$ we then have $(\alpha \otimes \beta) \otimes \gamma = (xe +yf\delta) + (ye + xf)\sqrt{\delta}$ and then you can argue with the ring property of $\Bbb{Q}$ thus allowing you to deduce $\alpha \otimes (\beta \otimes \gamma)$
$(ace+bd\delta e + bcf\delta + adf\delta) + (bce+ade + acf + bd\delta f) \sqrt{\delta}$
 
Yeah, alphabet soup.
 
another thing you can try is to show each component in the multiplication has a unique identity and inverse, as a group must be associative
the rule for $\otimes$ all shows how the $\sqrt{\delta}$ itself never appeared in the expression, meaning that each component is a rational
 
So there's a question I've been thinking about a lot for a long time. Maybe about a year. The question is: out of all arithmetic statements, which ones are provable?
Of course, the answer is determined once you choose a working theory.
 
and $\Bbb{Q}\setminus \{0\}$ do form a group
 
4:13 AM
I feel like there's a vague consensus that an arithmetic statement is "provable" if and only if ZFC proves it. But I wonder what makes ZFC so great, that it's the standard working theory by which we judge everything.
I'm not sure if I'm making any sense. Let me know if I should either clarify what I mean or shut up. :D
 

 Logic

This room is meant for discussion about logic, including found...
well the experts are in there. While I love set theory, I have not dig into arithmetic statement proof theory yet
 
I've thought about alternate sets of axioms. But establishing anything worthwhile from a given set of fundamental axioms is a lot of work (at least, I get that impression)
 
Associativity proofs in general have no shortcuts for arbitrary algebraic systems, that is why non associative algebras are more complicated and need things like Lie algebra machineries and morphisms to make sense of
 
@Secret I'll poke my head in over there, thanks!
 
(So, in a sense, the reason we use ZFC is because mathematicians are lazy)
 
4:21 AM
Well, assuming the existence of an actual infinity simplifies a lot of proofs because of the nice properties these infinite sets has
assuming the existence of at least one uncountable infinity simplifies things even further because some notion of continuity is implicit in it
 
Also, axiom of choice makes things easier as well.
 
Axiom of choice make things like SUPER EASY, as otherwise you will have to demonstrate a procedure to get you to the end of a proof, this often means to check every single case
whereas axiom of choice allows you to just find some well ordered set of any cardinality, without need to worry about the details within, and it will does all the counting for you
 
Axiom of choice might make things "SUPER EASY", but the axiom of triviality makes it stupidly easy.
 
I don't know the axiom of triviality.
 
> Everything can be made isomorphic to a trivial proof
lol
 
4:25 AM
Basically, if you want to prove a property about a set you can reduce the number of cases to just one by saying this proof is obvious and left as an exercise to the reader.
 
XD
I like this axiom.
 
that is lazy
 
Fermat used this extensively in his last theorem.
 
Now I am starting to explosively wonder:
What do a theorem that is not isomorphic to a trivial proof look like...
(i.e. Let T be a theorem such that axiom of triviality is false, is it even definable under the language L?)
 
11
A: What are the strengths and weaknesses of the Isabelle proof assistant compared to Coq?

user48801One aspect, which I will illustrate, of the "push-button" efficacy of Isabelle/HOL is its automation of the classic "diagonalization" argument by Cantor (recall that this states that there is no surjection from the naturals to its power set, or more generally any set to its power set). theorem ...

The axiom of triviality is also used extensively in computer verification languages... take Cantor's Diagnolization theorem. It is obvious.
(but seriously, the best tactic is over powered...)
 
4:30 AM
Extensions is such a powerful idea. I wonder if there exists algebraic structure such that any extensions of it will produce a contradiction. O wait, there a maximal algebraic structures such that given some ordering, it is the largest possible, e.g. surreals are the largest field possible
4
Q: Proof that all ordered fields are in the Surreals

PyRulezIt says on Wikipedia that any ordered field can be embedded in the Surreal number system. Is this true? How is it done, or if it is unknown (or unknowable) what is the proof that an embedding exists for any ordered field?

 
Here's a question for you: We know that no set of axioms will ever decide all statements, from Gödel's Incompleteness Theorems. However, do there exist statements that cannot be decided by any set of axioms except ones which contain one or more axioms dealing directly with that particular statement?
"Infinity exists" comes to mind as a potential candidate statement.
 
Well, take ZFC as an example, CH is independent of ZFC, meaning you cannot prove nor disprove CH using anything from ZFC. However, there are many equivalent axioms to CH or derives CH, thus if your set of axioms contain those, then you can decide the truth value of CH in that system
@Rithaniel That is really the crux on those rambles about infinity I made in this chat some weeks ago. I wonder to show that is false by finding a finite sentence and procedure that can produce infinity
but so far failed
Put it in another way, an equivalent formulation of that (possibly open) problem is:
> Does there exists a computable proof verifier P such that the axiom of infinity becomes a theorem without assuming the existence of any infinite object?
Or also equivalently:
> Is there exists a predicative actual infinity?
 
If you were to show that you can attain infinity from finite things, you'd have a bombshell on your hands. It's widely accepted that you can't. If fact, I believe there are some proofs floating around that you can't attain infinity from the finite.
 
(potential infinity is often assumed to exist in the predicative framework by using the collection of natural numbers as some primitive object)
I don't really knew, but I do know some subsets of that problem do have a positive solution, such as this:
and Terry Tao in one of his blog mentioned something about some infinitistic proposition has a one to one correspondence with a finitistic proposition (let me dig that up...)
This one
Though moments after that, there are also papers in the literature that criticise on it
My philosophy of infinity however is not good enough as implicitly pointed out when many users who engaged with my rambles always managed to find counterexamples that escape every definition of an infinite object I proposed, which is why you don't see my rambles about infinity in recent days, until I finish reading that philosophy of infinity book
 
4:50 AM
Yeah, the foundations of math and logic are murky areas.
It's difficult to really get a good grasp of anything, even when you know entirely what you're doing.
 
I like the weirdness in those places though
I mean, e.g. infinite dedekind finite sets are onions you can never finish skinning and yet it will keep on shrinking forever, that's pretty cool
 
Indeed. I enjoy talking about weird spaces, too.
 
and amorphous sets are like a loaf of alien bread which you can always get finite amount of bread indefinitely no matter how you slice it
Having said that, there are also cool things in ZFC universes:
Nonmeasurable sets change size and shape depending on your orientation
and they are important in some idealised models of stochastic dynamics
 
5:45 AM
The knapsack problem or rucksack problem is a problem in combinatorial optimization: Given a set of items, each with a weight and a value, determine the number of each item to include in a collection so that the total weight is less than or equal to a given limit and the total value is as large as possible. It derives its name from the problem faced by someone who is constrained by a fixed-size knapsack and must fill it with the most valuable items. The problem often arises in resource allocation where there are financial constraints and is studied in fields such as combinatorics, computer science...
O great, given a transcendental $s$, computing $\min_P(|P(s)|)$ is a knapsack problem
hmm...
By the fundamental theorem of algebra, every complex polynomial $P$ can be expressed as:
$$P(x) = \prod_{k=0}^n (x - \lambda_k)$$
If the coefficients of $P$ are natural numbers , then all $\lambda_k$ are algebraic
Thus given $s$ transcendental, to minimise $|P(s)|$ will be given as follows:
$$|P(s)| = \left|\prod_{k=0}^n (s - \lambda_k)\right| = \prod_{k=0}^n |s - \lambda_k|$$
Thus minimisation of $|P(s)|$ means every factor $|s - \lambda_k| < 1$
However, there are countably many such $\lambda_k$ to bring $|s - \lambda_k| < \epsilon$ for some $\epsilon > 0$, thus $|P(s)|$ can be as small as we want
In other words:
Apr 18 at 11:50, by Secret
likewise if $r$ is algebraic, there exists a unique polynomial $P$ such that $\lim_{m\to \infty }P([r]_m) = 0$
is not useful unless we demand some kind of bound on $|s-\lambda_k|$ which if it fall outside this bound, then it is transcendental
 
6:10 AM
Hi.
 
hi
 
Any hint on how to show the uniform convergence of $\dfrac{-z^2}{(1+z^2)^{n}}$?
 
not really except working through the usual definitions?
 
Tried to find an upper bound but...meh.
 
Hmm...
$\lim_{n \to \infty} \frac{-z^2}{(1+z^2)^n} = 1$
So we want $\min_n |\frac{-z^2}{(1+z^2)^n} - 1| < \epsilon$
$|\frac{-z^2 - (1+z^2)^n}{(1+z^2)^n}| < \epsilon$
are we working in the reals or complex numbers?
 
6:20 AM
Complex numbers
Uhm
 
$|-(z^2 + (1+z^2)^n)| < \epsilon |1+z^2|^n$
$|z^2 + (1+z^2)^n| < \epsilon |1+z^2|^n$
 
I don't see what I can get with that. Uhm...
 
I am trying...
Take $z$ in the unit circle, the largest of the LHS is $|1+2^n|$ and the smallest is any number in the neighbourhood of $|-1|$
 
6:35 AM
Whenever I see a function like that, my first instinct is to try and identify things I can replace such that the value of the function increases for all $z$.
 
$|1+z^2|^n = k$ implies $n = \dfrac{\log(k)}{\log(1+z^2)}$ so if $|z|<1|$... Uhm
 
The first thing I think of with that particular one is to replace the $(1+z^2)$ with $z^2$. Though, this is just at a cursory glance, so it would be worth checking to make sure that such a replacement doesn't have any ugly corner cases.
 
$z$ is complex makes it more complicated, if $z$ is real, then it is easy to find the required inequality and hence bound n in terms of epsilon
 
If we take $n = \dfrac{\log(\dfrac{1}{\epsilon})}{\log(1+z^2)}$ then $k>n$ implies $|f(z)-1 | < \epsilon$ I think
 
yeah that will work
 
7:00 AM
actually you need $\log (\frac{2}{\epsilon})$ otherwise $|f(z)-1| < \epsilon$ will fail if $z^2 < 0$
 
Hmm...
In number theory, a Liouville number is a real number x with the property that, for every positive integer n, there exist integers p and q with q > 1 and such that 0 < | x − p q | < 1 q n . {\displaystyle 0<\left|x-{\frac {p}...
Do these still exist if the axiom of infinity is blown up?
Hmmm...
Under a finitist framework where only potential infinity in the form of natural induction exists, define the partial sum:
$$\sum_{k=1}^M \frac{1}{b^{k!}}$$
The resulting partial sums for each M form a monotonically increasing sequence, which converges by ratio test
therefore by induction, there exists some number $L$ that is the limit of the above partial sums. The proof of transcendentally can then be proceeded as usual, thus transcendental numbers can be constructed in a finitist framework
7
Q: Mean value theorem and the axiom of choice

Daniela DiazThere's this theorem in Spivak's book of Calculus: Theorem 7 Suppose that $f$ is continuous at $a$, and that $f'(x)$ exists for all $x$ in some interval containing $a$, except perhaps for $x=a$. Suppose, moreover, that $\lim_{x \to a} f'(x)$ exists. Then $f'(a)$ also exists, and $$f'...

and neither Rolle nor mean value theorem need the axiom of choice
Thus under finitism, we can construct at least one transcendental number. If we throw away all transcendental functions, it means we can construct a number that cannot be reached from any algebraic procedure
Therefore, the conjecture is that actual infinity has a close relationship to transcendental numbers. Anything else I need to finish that book to comment
typo: neither Rolle nor mean value theorem need the axiom of choice nor an infinite set
 
7:27 AM
Morning all
 
7:38 AM
Morning, ÍgjøgnumMeg
 
How's it going?
 
Working on the last abstract algebra homework of the semester.
Current task is to show that in the field $\mathbb{Q}[\sqrt{\delta}]$ the function that takes $a+b\sqrt{\delta}$ to $a-b\sqrt{\delta}$ is an automorphism.
How about you?
 
Nice :) $\delta$ being a square free integer? Usually people write $\Bbb Q(\sqrt{\delta})$ for that field (though the notations are equivalent in this case)
 
Yeah, that's the one. "Rationals adjoined the square root of a square-free integer."
 
Right
I'm alright, first day back at work after holiday so I'm sad
 
7:45 AM
Ultimately it's going to be fairly easily, but I want to get the bijective-ness out of the way by saying that $a-b$ is unique for every $a+b$. Doing that cleanly has be stopped for the moment.
What work are you doing?
 
are there a finite number of pairs of prime numbers $a, b$ such that $ab$ is a palindrome?
 
@Rithaniel do you know what possibilities there are for the kernel of a field homomorphism?
 
There are probably infinitely many such prime pairs. What base are you working in?
I imagine that it's the same as with homomorphisms of rings. The kernel is an ideal(?) of the field? (Not sure if fields even have ideals, though)
 
@Rithaniel that's an interesting question; how do you define a field?
 
An abelian group $(R,+)$ equipped with a second binary operation $(\cdot)$ such that $(R\setminus\{0\},\cdot)$ is an abelian group and $(a+b)\cdot c=a\cdot c+a\cdot b, \forall a,b,c\in R$
 
7:53 AM
Right, so every element is a unit
 
Aaaaaaaah, good point
 
The category of fields is a full subcategory of the category of rings.
 
what does that mean for the ideals of a field?
Right, so what possibilities are there for the kernel?
 
Just the trivial ideals.
 
So either the kernel is either the entire preimage or just the additive identity.
 
7:54 AM
Right.
 
Right, if the kernel is the whole field then you have the trivial map so if you want something interesting you need the kernel to be trivial, but what does it mean for a kernel to be trivial?
 
That it's one-to-one.
 
> are there palindromes such that the explosion of palindromes is a palindrome nonstop palindrome explosion palindrome prime square palindrome explosion palirome prime explosion explosion palindrome explosion cyclone cyclone cyclone hurricane palindrome explosion palindrome palindrome explosion explosion cyclone clyclonye clycone mathphile palirdlrome explosion rexplosion palirdrome expliarome explosion exploesion
/Endrant
 
@Rithaniel exactly :)
so show that the kernel of that map is trivial and you've got injectiveness
surj. is easy
 
Not every epimorphism is surjective, though.
 
7:57 AM
I'd actually say that injectiveness in this case is as self-evident as surjectiveness.
 
What do you mean?
 
The function in question is taking an element in an algebraic field to it's conjugate.
Specifically $\mathbb{Q}(\sqrt{\delta})$ where $\delta\neq 1$ is square-free.
 
Oh, you were talking about the problem above. Sorry, didn't catch that.
 
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