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17:00
What I said above is false in a stupid way. I meant to say compact.
Of course $\Bbb R$ is a topological ring...
Oh
Yeah you have a whole bunch of topological vector space out there.
None of them are compact.
Banach algebras are, of course,
important examples of topological rings. Anyway.
Good morning to you, fine gentlemen!
None of us are fine, much less gentlemen.
I was actually meaning to say that, but thought you'd be offended :P
17:11
Most things you do offend me; that one would only make me smile.
Algebraic topology sucks.
Was that offending?
Every branch of mathematics is equally beautiful and fascinating imo
Number theory > Geometry. Boo geometers. Algebra > Analysis. Boo analysts.
Of course, @Balarka; but no more offensive than usual.
I am totally offending everyone now.
17:14
Please see the smacks on the side of chat. They should still sting.
OK I gotta go. I'm sure there'd be more smacks awaiting for me when I come back.
At least you have something to look forward to when you return :D
17:33
0
Q: Dimension of $C(M)$

pourjourSuppose that $$M=\begin{pmatrix} 1 & 0 & 0 & . & . & 0 \\ 0 & 2 & 0 & 0 & . & . \\ 0 & 0 & . & 0 & . & . \\ . & 0 & . & . & . & . \\ . & . & . & . & . & 0 \\ 0 & . & . & . & 0 & n \end{pmatrix}$$ We note that $C(M)$ the subspace formed by the matrices which commute with M. I proved that $C(M)$ ...

Any help please
Sorry, I haven't taught myself any linear algebra yet :(
I wish I could help
@teadawg1337 pure math is algebra what you didn't learn it
@pourjour I'm currently teaching myself real and complex analysis, I just meant that I haven't read any books purely about linear algebra
I've taught myself vector calculus, so I know some of its concepts
17:48
@teadawg1337 ahh ok
I'm particularly weak with matrices, my apologies
Anyone familiar with Algebraic Topology: link
4
Q: Understanding Hatcher's proof of Borsuk-Ulam theorem for $n=2$

mathreaderI am trying to understand the proof of the Borsuk-Ulam theorem for $S^2$ given in Hatcher's "Algebraic Topology" (Th. 1.10), as another person does here, but we are stuck at different places, so I hope this question is not considered a duplicate. Borsuk-Ulam: If $f:S^2\rightarrow\mathbb R^2$ ...

18:09
hehehe, I'm an explosion of creativity these days :-)))))))))))
(I just put down another titan)
It's 01 Dec here. 1 more month to the new year.
It's a quarter past noon on 30 Nov here
@WillHunting yeah, and then it's time to work hard on math. :D
@Chris'ssis Yes. I will try to finish my 12 holy math books next year and also solve all my mental problems completely next year, so that I wake up completely well on the first day of 2016.
Mental problems? Sorry, nvm. None of my business to ask
18:18
@teadawg1337 Yes, it is well known in this chat that I have OCD, and many other problems.
@WillHunting I'm a newbie to this chat, I wasn't aware
@WillHunting I'd love if I found more people like you in this world, that sick as you say you are. I'm honest with you now.
@WillHunting I prefer the disease to body than the one to the spirit. The health of this world lies on the shoulders on the former ones I think (if there were a way to choose between the 2).
I am very troubled today because I keep thinking of how one problem is affecting some people in my country.
I will have to let go and accept that I cannot change everything I want to.
18:51
I have to say this piece looks like a masterpiece.
$$\int_0^1 \left(\frac{\log(1+x)}{1-x}+\frac{\log(1-x)}{1+x}\right)\operatorname{Li}_2(x) \ dx$$ Well, it is a masterpiece.
Without some research is hard to touch it.
@Chris'ssis Looks similar to one of the questions I've posted
@teadawg1337 Which one? What I post here are my creations (most of the time).
@Chris'ssis I encountered something similar to that while trying to solve this
@teadawg1337 I see.
@Chris'ssis Sorry, didn't mean to rain on your parade...
19:00
@teadawg1337 I don't know what your expression means. There is no parade. No worry, no matter the meaning. :-)
@teadawg1337 hehe, thanks! :-) Good to know it! No problem, you're welcome anytime.
Integration involving polylogarithms is surprisingly difficult imo
@teadawg1337 Especially when you wanna finish them in the spirit of the art. I'm not only interested in solving things since everybody does it, the math sites are full of solutions.
I'm interested in the math art.
I guess math is similar to art in the sense of how all the numbers come together and form different answers
It's better than art, since there's infinitely many ways to combine numbers to form different solutions
@Chris'ssis Back to the integral you posted, does $\log(x)$ represent the natural logarithm?
19:12
@teadawg1337 Yeah.
Hmm.... Looking back in my notebook, I found that $$\int_1^0\frac{\ln(1+t)}{1-t}\mathbb{d}t=\frac{\pi^2}{12}+\frac12\ln^2(2)-\ln2\‌​text{Li}_1(1)$$
And Li$_1(1)$ is $\infty$
@teadawg1337 Maybe you wanted to write $$\int_{-1}^0\frac{\ln(1+t)}{1-t}\mathbb{d}t$$?
No, the original integral was $$-\int_0^1\frac{\ln(1+t)}{1-t}\mathbb{d}t$$
So I reversed the interval of integration to remove the negative sign
@Chris'ssis I don't see any errors in my work...
19:33
@teadawg1337 $$-\int_0^1 \underbrace{\ln(1+t)}_{\large \text{near} \space 1 \space \text{is} \approx \log(2)}\cdot \underbrace{\frac{\log(t)}{1-t}}_{\large \text{near} \space 1 \space \text{is} \approx-1} \cdot \overbrace{\frac{t}{t\log(t)}}^{\large \text{numerator is}\approx 1 \space \text{near} \space 1}\mathbb{d}t$$
Q.E.D.
I did it too complicated though!
Again, in a different style
$$-\int_0^1\overbrace{\frac{\ln(1+t)}{1-t}}^{\large \text{numerator is}\approx \log(2) \space \text{near} \space 1}\mathbb{d}t$$ and hence the integral diverges.
Initially I had in mind the problem I attended here that is really cute!
3
A: How Prove this integral is diverge $\int_{0}^{1}\dfrac{dx}{\ln{x}\ln{(1-x)}}$

Chris's sisNote the simple fact the integrand is positive and $$\int_{0}^{1}\dfrac{1}{x\ln{x}}\cdot \underbrace{\frac{x}{\log(1-x)}}_{\large\text{near 0 it behaves like $-1$}}dx\longrightarrow \infty$$ Q.E.D.

It's amazing to see the difference between proofs as length. I cannot imagine a better proof than this one. Well, yeah, it's mine, but still, it's definitely the best way in my opinion (however, not upvoted enough).
And then you get problems like the one I linked where there's no way to get around writing a long proof xD
@teadawg1337 My philosophy is that there are always shortcuts, ways to improve things a lot, but one needs to do research for that. I don't see a better way than doing research.
19:53
@Chris'ssis My particular question is limited by the currently known properties and identities of dilogarithms/polylogarithms
eventually, there may be a way to simplify it
puddi puddi sugoku dekai gigapuddi
English, por favor
I'm aware of the meme, I was just making a joke :P
No meat, no sugoku dekai puddi
20:06
@Jasper: Damnation.
I sense an imminent slap coming...
^Comment brought to you by the Department of Redundancy Dept.
@TedShifrin!
hi @Anthony
who are you, teadawg?
I'm a newbie here
So why are you anticipating a slap?
20:14
@BalarkaSen Not for me, but for Balarka
@TedShifrin Are you terribly busy, or could I ask you a measure theory question?
I'm not terribly busy (although I have 3 more recommendations to write for people applying to grad school), but I am not a measure theory expert.
I will just ignore Balarka, @teadawg. He ignores my smacks.
For all intents and purposes, has Balarka on ignore
heya @Karl
20:16
@TedShifrin I think the problem isn't very hard, I'm just terribly behind...
Goofed off too much for Thanksgiving break, @Anthony? What's the question?
My family made me!
Let α be the non-decreasing function on R defined by α(t) = 0 if t ≤ 0
and α(t) = 1 if t > 0. Let $µ_α$ be the corresponding premeasure as discussed
in lecture, with $µ^∗α$ the corresponding outer measure. Determine which sets
are in $M(µ^∗α)$, that is, measurable for this outer measure.
Apparently so did a number of my students, @Anthony. They have web-based homework due tonight at 11 PM and an exam on Tuesday.
Ooof, laziness is not a good trait to have when taking advanced mathematics
20:18
I mean, I haven't seen my family in forever, I felt bad saying "I can only do homework".
0
Q: Disk method and integrals with absolute value

MathHArdAbsolute values in calc 1 kill me, I don't even know how I'd draw this: Find the volume of the solids you get by rotating the following graph around the x-axis $$f(x)=(1+x^2)^{-1/2}, |x| \le 1$$

Unfortunately, I don't know what corresponding premeasure was discussed in your lecture.
what would be different in the integral if he rotated around x axis?
You speak from years of experience, @teadawg?
Oh, sorry. I don't think I do either. Let me think for a moment.
There are so many words.
20:19
@UserX What you said makes no sense
@TedShifrin Nah, just years of facepalming at fellow students in math classes
You said "advanced mathematics," @teadawg.
Oh I think it's just the premeasure generated from evaluating the function on the end points, and I think our semi-ring is the half open half closed intervals.
Would that make sense?
I haven't thought about this sort of stuff since grad school, @Anthony. Sorry. Just not my cup of tea.
Yeah, like advanced undergrad courses/grad courses
I just recently graduated from high school, experience is something I lack
20:22
LOL ... This room is often full of people speaking of things of which they know not.
@TedShifrin Mine neither. Thanks for considering.
DanielFischer can certainly help you, @Anthony, if he shows up.
I will eagerly await him. I feel like this isn't suppose to be hard, I just got so lost... rings, semi-rings, measures, pre-measures, outer measures....
I'm an expert in measure theory, @Anthony. I think about it as I go to sleep every night.
@MikeMiller I'm dubious.
20:26
I'm currently studying real analysis on my own, I dunno if that counts as advanced mathematics... I'm young, I've got my whole life ahead of me
I'm on the International Committee for Measure Theory.
Woah!
I don't know what to say.
He is quite prolific. Just look at his Vixra profile.
Stay dubious, @Anthony.
Haha.
20:30
Is there some good reference for solving differential equations with Green function?
@Ted @Mike \o
@Chris'ssis: whee! ;-)
@robjohn ;)
Speaking about things you don't know is what Humans do
That statement included?
20:42
@Chris'ssis It's been a busy weekend here. Thanksgiving and guests and another party last night. I've had less time than I wanted for math this extended weekend.
It's been a busy weekend for everyone, I'd imagine
How are people able to format math equations nicely on this site?
@robjohn I noticed that. I posted above a very nice question.
@teadawg1337 Depending... it is not Thanksgiving everywhere, though
I have the following equation: (81)^3/4. But I do not know how to format it properly on this site.
20:43
@Karl of course, considering I am (hopefully) a Human
@Chris'ssis I shall have a look
@Studentmath You don't see the paradox?
@robjohn OK
@robjohn Oh, that is true...
@KarlKronenfeld Ahhh, I get it. That was clever
@JohnMerlino It could be 1/4 * (81)^3, which would be formatted as $(81)^3/4$ or 81 to the 3/4 power: $(81)^{\frac{3}{4}}$, or if you prefer $(81)^{3/4}$
20:45
Hi @Studentmath ... You saw the solution of the prob question?
@Karl Making paradoxes is Human as well
@Ted Yes! I think I commented on it yesterday already
It wasn't trivial for me, the technique
I left immediately after.
I wonder where that question is from, I would like that set of exercises
It was the way to use the inequalities on the given region ....
I wouldn't think about integrating it that way
20:48
Professors do make up their own exercises. That might come from a more theoretical course ...
@Chris'ssis Does this converge near $x=1$? $\mathrm{Li}_2(1)=\zeta(2)$, and $\frac{\log(1-x)}2$ is nowhere near cancelling $\frac{\log(2)}{1-x}$.
Probably, seeing his other questions it seems so indeed
Interesting... there were 3362 messages since I left, and I got pinged 116 times
You are too popular, @Mike, and too bored if you counted to 116 and 3362.
20:52
@robjohn hmmm, I might have written it wrongly from my papers ...
@Karl when I thanksgiving vacationed I left my computer behind. 3362 was the number of messages the tab showed, and 116 was the number of pings it showed I had.
I didn't count... I promise
@MikeMiller that'll teach you for leaving MSE behind ;-)
I don't ordinarily get counts ...
@TedShifrin I thought I read around the line y=x
Also, seems like my request will be approved and I will be able to teach my seminar to high-schoolers from bilanguagual school - after I present the work to the university
20:53
and the integral was for rotation around the x axis
ah, you've been chatting over the phone all thanksgiving break? @MikeMiller
but yea just misread
Instead of presenting it to a class in the university..
Aye, @Karl, which is why my LaTeX was frequently bad (and why I never answered questions)
@TedShifrin if you switch tabs when people are chatting, this tab will show a number in the title
20:54
@robjohn ahhhh, it's $1/2$ on top
@MikeMiller I don't have chatjax on half the time, so I didn't notice that.
I thought I read x-axis everywhere, @UserX
@robjohn That is $$\int_0^{1/2} \left(\frac{\log(1+x)}{1-x}+\frac{\log(1-x)}{1+x}\right)\operatorname{Li}_2(x) \ dx$$
@Venus the correct version ^^^ (initially I considered the wrong upper limit)
Yeah, I don't have it on my phone, so I don't know it's bad unless someone tells me. I'm also less willing to read TeXnically dense stuff.
So what was the meatball verdict, @Mike?
20:56
@Chris'ssis That does indeed avoid the problem near $x=1$
@Chris'ssis That looks like it should converge at first glance
We forgot to use breadcrumbs, so they were very quick to fall apart. But they were still tasty. The sauce was great.
I brought some home with me, so I'll probably have more pasta tonight.
Breadcrumbs, but egg more important for binder.
We used enough egg :)
Question: $inf\{d(x,a)|a\in A\}$; it could be equal to $d(x,y)$ for some specific $y\in A$, but it could also 'not-be-achieved'
Is there any theorem from calculus that tells me that in that case there is some series $<a_n>$ so that $lim_{n\to \infty} d(x, a_n)=inf$?
21:07
Sure, let $A' \subset \Bbb R$ be the set $\{d(x,a) : a \in A\}$. Then there's a sequence of points in $A'$ converging to the inf, by definition of the inf. Now pick corresponding $a_n$s.
Endorsed by the ICMT
@Mike right, thanks!
@KarlKronenfeld There are 3 of us, and I'm the only one who sees it, so only a plurality of us endorse it, not a majority.
I can't help but wonder who the other members are.
A pox on you, @Karl.
21:14
I don't intend to tell you, @KarlKronenfeld. Also, sure, poxes, whatever Ted said
I feel strange for knowing what that phrase means...
Well pox is a disease, one can infer
I just feel like a walking encyclopedia of idioms today, lol
Actually, a sitting encyclopedia of idioms
Hello, please if $g(x)=x^2$ then $g(]-1,1[)=[0,1[$ or $g(]-1,1[)=]0,1[$ ? please
21:30
Are you asking for a clarification on its range using interval notation, or am I completely misinterpreting the question?
if $-1<x<1$ then $0\leq x^2<1$ or $0<x^2<1$
Well, @Vrou, g(0)=0 and $0\in ]-1, 1[$
So which is true?
@robjohn this one is amazing $$\int_0^1 \frac{\log(1-x)}{1+x} \operatorname{Li}_2(x) \ dx$$
21:34
[0,1[
Unless I am mistaken, you are right :)
@Studentmath thank you,
@Chris'ssis $$\int_0^1\frac{\log(1-x)}{1+x} \operatorname{Li}_2(x)\mathbb{d}x=\int_1^0\frac{ \operatorname{Li}_0(x) \operatorname{Li}_1(x) \operatorname{Li}_2(x)}x\mathbb{d}x$$
please for a function $\frac{1}{1+x^2}$, i want to proove that this function is continuous @Studentmath
@teadawg1337 are you sure?
21:43
@Vrou what do you know about continuous functions? And about discontinious ones?
@Chris'ssis $$\log(1-x)=- \operatorname{Li}_1(x)$$$$\frac1{1+x}=\frac{ \operatorname{Li}_0(x)}x$$
so i must find $\forall \varepsilon >0,\exists\delta>0$ such that $\forall x\in\mathbb{R} |x-x_0|<\delta\Rightarrow |\frac{1}{1+x^2}-\frac{1}{1+x_0^2}|\leq \varepsilon$
@Studentmath
You forgot an $ there
Right, epsilon-delta proof.. Let's see if I am of use here. What have you tried thus far?
@teadawg1337 maybe you have a wrong sign there.
But that negative can be moved in front of the integral, and thus the interval can be reversed(?)
21:48
@teadawg1337 I was referring to $\operatorname{Li}_0(x)$
@Studentmath i calulated $|\frac{1}{1+x^2}-\frac{1}{1+x_0^2}|=|\frac{x_0^2-x^2}{(1+x^2)(1+x_0^2)}|=|\frac‌​{(x_0-x)(x+x_0)}{(1+x^2)(1+x_0^2)}|$
The missing } is in the middle there, before the second =
Oh, I see my mistake... $$\frac{x}{1-x}= \operatorname{Li}_0(x)$$
@teadawg1337 Yeap.
@Studentmath I dont know how to do after
21:54
Obviously in the last one it's a fraction
Hrm, to be honest I only did a couple of such proofs, but let me see if I can be of use..
@Chris'ssis I do know that $$\int_0^1\frac{\log(1-x)}{1+x}\mathbb{d}x=\frac12\log^2(2)-\frac{\pi^2}{12}\;,$‌​$ but the added $ \operatorname{Li}_2(x)$ complicates things a bit...
Prove something more general: if $f, g$ are continuous with $g(x) \neq 0$ for all $x$, then $f/g$ is continuous.
@teadawg1337 that integral is trivial
I've only learned of polylogarithms within the past week, I haven't quite gotten the hang of them yet :/
@teadawg1337 It's $$\sum_{n=1}^{\infty} (-1)^n \frac{H_n}{n}$$
22:00
trivial is a fighting word
@MikeMiller you speak with me ?
To both of us
@Anthony In what sense it is a fighting word?
@MikeMiller why g\neq 0 ?
@r9m note the connection of $$\int_0^1 \frac{\log(1-x)}{1+x} \operatorname{Li}_2(x) \ dx$$ with your question here math.stackexchange.com/questions/1023022/…
22:04
@Vrou because if $g(x)=0$ for some $x$, then $f/g$ won't be continious at that $x$!
ohhh yes i reade it \leq
but how to prove this ?
@Chris'ssis The word "trivial" seems to have a negative connotation to me, I dunno why... I wasn't offended, just slightly taken aback
@Vrou okay, think I have it. How would you try to prove that if $f,g$ are continious (and $g(x)\neq0$ for all $x$), that $f/g$ is too?
@teadawg1337 I didn't know that.
If it's easier, consider that $f,g$ are continuous at $a$ and $g(a)\neq0$
22:13
yes why not
Then prove continiouty of $f/g$ at $a$
@teadawg it reminds me my linear algebra 2 prof., stating about every question how trivial it is - where it wasn't at all to me, demotivated me..
@Studentmath but how please ?
that's my problem
@Studentmath I'm terrified of asking people for help, I fear of being tossed aside or ridiculed...
I know it's irrational to think that way, but it's difficult to fight the subconscious...
You shouldn't care - anyone that isn't a genious went through the same difficulties you do
@Vrou let me try to guide you -
@Vrou let's work first at your example
I just computed $$\int_0^1 \frac{\log(1+x)}{1-x} \operatorname{Li}_2(1-x) \ dx$$ too (in a brilliant way).
22:21
You were at $\frac{|x^2-y^2|}{|(1+x^2)(1+y^2)|}$, right? @vrou
I use $y$ instead of $x_0$ as I am lazy
r9m
r9m
@Chris'ssis okay !! :-) I was just reading this wonderful solution of Mirek OlÅ¡ák to my combinatorics problem here .. :D
I like this site very much !!!! :-) I wish I knew about this a few years ago too !!
@r9m I might have seen that question somewhere ...
r9m
r9m
@Chris'ssis I answered it yesterday (in another question) in a different way (non combinatorial proof though) :-) .. but I was not able to complete a combinatorial proof of the same ..
@r9m It's not a hard question though.
r9m
r9m
@Chris'ssis I'm a beginner ^_^ .. I am learning :)
can't say I'm kid anymore .. I have a moustache :P LOL
22:29
Which reminds me, I haven't shaved in a few days....
@r9m :-))))))))
r9m
r9m
@teadawg1337 ah .. its Dec 1st here .. so end of the no shave month ! ;)
@r9m I don't practice "No Shave November," I look terrible with facial hair xD
@r9m Did you know there is a very nice way to compute $$\int_0^1 \frac{\log(1-x)\log(x)\log(1+x)}{1-x} \ dx$$ ?
r9m
r9m
@teadawg1337 yea !! I shaved too ... but I look like a child when I shave :P ..
@Chris'ssis How ?!! :D
22:36
@r9m I'll show you when I write up my proof. By the way, have you ever seen it before?
r9m
r9m
@Chris'ssis :D Okay !!! :D .. In the mean time I try :)
@Chris'ssis never ! :O maybe I have .. but I haven't given particular attention to it before .. :)
@r9m Take your time (then). :-)
r9m
r9m
@Chris'ssis I have exams now .. I try later once my exams are over ! :-)
@r9m OK
@r9m Anyway, it's good to know that all flows naturally and at a certain point you also have to use the series you posted on main and I answered it.
22:58
Rather talkative today, hm? /sarcasm
The silence is deafening...
Consider $S_F$, the space of real sequences $a=(a_n)_{n=1}^{\infty}$ such that all but finitely many of the $a_ns$ are zero. Let the linear map $T:S_F\rightarrow \mathbb{R}$ be defined as $Ta = \sum_{j=1}^{\infty}a_j$. If we equip $\mathbb{R}$ with the usual Euclidean norm, and $S_F$ with the norm $||a||_w = max_{n\geq 1}|na_n|$. Determine whether or not T is continuous.
I think that T is not continuous, but i'm struggling to show this.
I'm trying to show it by showing that the set $\{\sum_{j=1}^{\infty}|y_j| : max_{n\geq 1} |ny_n| \leq 1\}$ is bounded.
@DanielFischer Have you got any hints?
23:18
@user112495 by definition, what does it mean that $T$ is continious?
I'm not hinting anywhere, just wanna know if I can help
@Studentmath We have a theorem that gives us three different conditions that are all equivalent. But the general definition is as follows:

@Studentmath If $(V, ||.||_V), (W, ||.||_W)$ are normed vector spaces, then a map $f:V \rightarrow W$ is said to be continuous at $x \in V$ if $\forall \epsilon > 0, \exists \delta > 0 : \forall y : ||x-y||_V < \delta \implies ||f(x)-f(y)||_W < \epsilon$.
@user112495 Right, will ping you back if I find myself useful
Won't there be some sequences $a$ such that $Ta$ will diverge?
@Studentmath Will there? Because all the sequences in $S_F$ only have a finite number of non-zero terms.
@Studentmath So won't all the sums be finite?
Oh, I misread. I read it as if all will be non-zero except finitely many, sorry. @user112495
Yes, all sums will be finite
If you manage to show that set is bounded, can you complete the proof?
r9m
r9m
@Chris'ssis okay :)
23:32
@Studentmath Yes. If I can show that that set is bounded, then we have a theorem which essentially completes the proof for us. That set being bounded is equivalent to T being continuous.
And all the $y_j$'s there are from a sequence $y$?
@Studentmath Yeah. So we're essentially summing the absolute value of each term in the sequence.
@Studentmath I've actually got five different norms i've got to determine the continuity of T on, but I'm struggling on this one (plus possibly a couple of the later ones, but I haven't tried those yet).
Well, we know that every summation is finite (as all the sequences are zero except for finitely many, and thus the summation is on finite number of non-zeros), and therefore is bounded
Isn't it so?
@Studentmath Sorry, I mean i'm trying to prove that the set isn't bounded.
Well, then if it isn't bounded what I said is wrong - is it? I think it's true, since if it wasn't, for every $M$ we would pick there would be some sequence that is greater than it, and thus that sequence would diverge.
23:38
@Studentmath I need to prove that T is discontinuous under this norm. This is equivalent to showing that the set above is unbounded.
@Studentmath But if we fix an upper bound, say U, then we can find an N large enough (where N is the number of non-zero terms) such that the sum is greater than U. I'm just not sure is the right way round for doing this type of question.
there's a question at qr.ae/lgats about what portion of a diagonal will pass through the black squares of an MxN checkerboard, and Alon Amit's answer is understandable except for the case where M and N are both odd. What symmetry could he mean should be exploited?
Right, the sequence of 0 and 1s where there are 1s in $N$ places
Hm, I'm not sure. What are the other conditions?
(If I am being completely unhelpful drop me off)
@Studentmath What i'm confused about is whether we're allowed to fix the U beforehand? Because if so, then any N>U will give the desired result. But does it make sense to assume an upper bound for a set, and then choose elements of the set afterwards?
And I think it would have to be a sequence of 1/n's
due to the condition in the set
But as N tends to infinty, the sum of 1/n diverges and so we can still find an N large enough
@Studentmath Actually, although it will need to be quite a bit bigger than U. But does any of this make sense? Because I'm not sure if what I'm doing is completely nonsensical.
It makes sense to me, You already have the set - you assume for contradiction it has a bound, you mark it U, and you show that the set has a sum larger than the bound - reaching the desired contradiction.
Whether or not the steps between are true, I am unsure
Well actually they also make sense to me.
I have an unrelated question. Is there such a thing as posting too long of an answer? Should I have whittled this down a bit before posting it?
23:50
But then there is something wrong with the idea of if it is unbounded, then for every $U$ we would pick there has to be a sum greater than $U$ - doesn't that lead us to an unbounded sum in the set, which means a diverging sum? @User112495
Obviously one of these ideas is wrong - probably mine
@Studentmath What do you mean. If we suppose there are N non-zero values in the sequence. Then we know that as N tends to infinity, the sum diverges. Is there anything wrong with having a diverging sum? The set is still unbounded which is all I need?
Because every seqence has all zeroes except in finitely many places, the absolute sum of it's parts should be finite and thus converge
I am afraid I am just confusing you - you better wait for DanielF, Mike or anyone who has better idea what he talking about than me
I got a naive question
What are some applications for solid of revolutions?

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