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12:27 AM
Hello @ted, have you had dinner?
Hello @mike, how was your dinner?
 
12:50 AM
Hi @datalava, I am bored, lol.
 
@JasperLoy Hey. Me too. I've been working all day & getting sick of it, lol
@JasperLoy What are you up to
 
@datalava I have shortlisted my choice of language books to Teach Yourself Complete X, Living Language X Complete Edition, BBC Talk X Complete, Berlitz X in 30 days and Berlitz Basic X.
 
@JasperLoy I'm not familiar with any of those, I'm afraid. I learned all my foreign languages in school
 
@datalava It's hard to decide because some of them don't allow you to look inside on amazon. Going by the reviews alone is not good. I speak English and Chinese, lol.
 
@JasperLoy Isn't Chinese a foreign language option?
 
12:55 AM
@datalava You mean in US math grad school? Well, it is not the case in most schools. It is an option for example in MIT and Cornell currently though.
 
@JasperLoy Yeah that's what I meant. I only looked at the Cornell one because you mentioned it.
 
@datalava 99 per cent of schools require French, German or Russian. Some require only one, but very few require two.
 
@JasperLoy I see
@JasperLoy You should just go live in France, Germany, or Russia, lol That's the best way to learn it
 
@datalava I actually considered applying for grad school in Europe instead of America. But it seems not easy to get funding there. Maybe I am wrong. Maybe I should look again and think again.
 
@JasperLoy Hmm, I don't know
 
1:02 AM
I am better from the flu now. I think I will go out for a walk soon.
 
@JasperLoy That's good. I feel like crap, lol.
 
@datalava Why do you feel like crap?
 
@JasperLoy I'm tiiired. I'm running on caffeine and it's only 8 pm and I have so much work to do T-T
 
@datalava When I was working as a teacher for a short period, I never enjoyed it. They gave me too many classes. I had to prepare 5 different sets of material, which I feel is absurd.
 
@JasperLoy That is a bit crazy. My cooperating teacher teaches 5 classes, but only has to prepare two different lessons
 
1:07 AM
@datalava I might want to go into teaching in the US after I finish grad school. But do they require me to do a teacher course first, or can I do it concurrently or maybe not at all?
 
@JasperLoy If you want to teach at a public school you need to at least complete a certification program.
@JasperLoy Some of the programs can be completed concurrently, while teaching
 
@datalava OK. And I take it that I have to pay for the course fees all by myself?
 
@JasperLoy Yes, but some programs will pay you a stipend for working for them. This often means committing to teach in urban/high needs schools for a period
 
@datalava OK. I think I will try to find such a program then, when the time comes.
 
@anon
HERRO. Maybe this interests you.
 
2:01 AM
Hello friends.
 
3:00 AM
anybody know any combinatorial game theory
 
not a single person
 
Wouldn't that be exciting? A totally unexplored frontier..
 
i'm trying to figure out if there's an optimal strategy for ultimate tic tac toe
 
what's ultimate tic tac toe
if it's what I think it is it's solved
 
3:17 AM
@MikeMiller it's a 3x3 grid of 3x3 grids
 
I know
 
so it's solved?
 
is this the one where you're forced to one of the grids based on the previous move?
 
in your variant, can you play in grids that have already been taken?
i.e. if X takes the center but there are untouched squares, can X or O still play in there
 
3:20 AM
@MikeMiller i'd say no, you can't still play there
once it's won it's won
 
then the strategy I know doesn't work
 
what's your strategy?
 
(presumably instead, if O goes in the center box in another grid, X gets to choose his grid but cannot choose the center
X goes first; X picks the center
in every move until he's not allowed to anymore X picks the center
once he's no longer able to he's stuck in some other square based on O's move; henceforth he forces O back down to that square
i.e. if O's last move in the center square is bottom left, then he plays bottom left til there's none left
continue in this manner until X wins
 
ok im a little confused
So X plays 2,2 in 2,2
 
let me just find an app online we can play on
try this
is it letting you play in the center square, or anywhere?
ok, my strategy doesn't work for this variant
the variant my strategy works for is the one in which you would have been forced to play in the center square
 
3:29 AM
right, yeah, this is not what i was thinking of
this is just double layer tic tac toe
there is probably a tying strategy like in regular tic tac toe in this one
 
try this
the ultimate tic tc toe you're thinking of is "if X plays in the top square in his grid, then O has to play in the top grid"
right?
 
@MikeMiller right
 
that's this one
 
Oh i think I see
 
in this one, you'll still be forced to use the center square even though it's been taken
 
3:30 AM
so you're just going to make me keep playing in the center square even after I win
 
yeah
it's quick to run through it
 
yeah let's do it
 
i think i messed something up
ohhh
ok i did
when you went into the right square, i was supposed to force you to keep going there i think
 
that may have been a blunder, i should have picked 2,3 after i won
right right
 
ok
let's restart, I got the strategy slightly wrong but I've figured out how to fix it
 
3:34 AM
ok, i think you have to send me the link over again
 
ok now here's where I got it wrong last time
I should never, ever, let you have choice
 
Dafaq is that @MikeMiller
 
hmm
 
Hi @AlexanderGruber
 
what did I do wrong
 
3:35 AM
@PedroTamaroff hi
@MikeMiller Maybe it's not solved!
 
let me go take a shower and think on this
 
@AlexanderGruber it definitely is, I played this a couple years ago at an REU and we all were heartbroken when one of us announced a solution
 
@MikeMiller hahahaha
We played Tichu at my REU
which is an amazing game.
 
@AlexanderGruber Is this tic tac toe on a torus?
Wait, on three tori?
WAT
 
3:37 AM
@PedroTamaroff it's a tic tac toe grid of tic tac toe
 
OH. OK.
 
the rule is, the next player has to move in the grid corresponding to the move of the previous player
so if i go 1,2 on any board, you have to make your move in board number 1,2
and then i have to make my move on whatever board corresponds to your move, etc
@PedroTamaroff wanna try?
 
im in
i didn't understand the follow up rule
now I do
 
it's weird right?
in Mike's variant, you can keep moving on a board that's been won, until the thing is full. in my variant, once it's been won, it can't be moved on anymore.
 
3:44 AM
brb
shoot i gotta go, we should play next time, if Mike doesn't sovle it.
 
I think I've got it @AlexanderGruber
yup
 
 
1 hour later…
5:30 AM
First exam tomorrow
 
Good evening everyne
**everyone
 
Good evening danlyBG
**daOnlyBG
 
what's tomorrow's exam on for you?
 
5:46 AM
Discrete Mathematics
 
oh nice. introductory discrete math?
 
Including Cryptography, Enumeration, Coding theory, Graph theory and design theory
It is Discrete Math II
 
I see. Good luck!
 
Thank you, I should go and study more on that note :)
 
see ya
 
 
2 hours later…
7:29 AM
wow! the map has 52 people now...
 
@AlexanderGruber I need to sleep but do you want to see the strategy first
 
@MikeMiller sure
 
ok, so here's how it goes from here
now I pick either the left box at all times, or if not available, the right box
 
ok
wait
so did i just win, or not?
 
i had already owned the far right square
 
7:37 AM
I see, so you claim it
it's not just who has 3 boxes in 3 boxes
 
yeah
 
interesting
 
I don't have a proof written down that this strategy works every time but I think it's clear that it does
 
Yeah it definitely does
Because starting first you can decide never to let me choose where to go
 
yeah
 
7:39 AM
so, if you can win at all, you can make yourself win
 
you get to force me into a second box; and then if I can't go into that, I go into the one opposite it
this strategy actually does require the center box, interestingly(?)
 
because it's the only box s.t. no matter what second box you put me in, I can still make a winning row out of it
your "claimed boxes are claimed, period; no using them again" variant is interesting and shouldn't be susceptible to similar strategies
 
@MikeMiller it would also be interesting to see the no claims version
three boxes in three boxes
 
yeah, i'd never thought of that possibility before
 
8:37 AM
@AlexanderGruber!
Whew... been some time since I have last talked in here.
 
Is there any effective argument that says complex numbers does not include real numbers
 
@Integrator So you essentially want to prove that $\sqrt{-1}$ is not real, is that right?
wait a sec
 
i misread
@Integrator complex numbers do include real numbers.
 
@BalarkaSen Have you seen that answer with one down-vote and @Henry 's comment on it?
@BalarkaSen There?
 
8:45 AM
sorry the internet's going off and on in here.
i got to go
 
Fine!
 
hi @integrator @balarka
@BalarkaSen I stayed in the Salt Lake area there :D
It was nice.
I even went to the stadium :D
@BalarkaSen Oh bye :(
 
9:03 AM
@Sawarnik Hi
 
@Integrator too late.
 
@Sawarnik That's probably bug in notification system
 
Ok.
@Integrator our avatars look similar.
ok i have gtg too byes :)
 
@Sawarnik I'm behind proxy now Can't see
 
9:28 AM
What is an orthogonal latin square?
 
10:10 AM
@Committingtoachallenge I got a small question from U :) shall I ask? It's nothing got to do with math btw :p
 
10:27 AM
@TheArtist Perhaps I will answer, I am certainly curious
 
@Committingtoachallenge sure :p
@Committingtoachallenge so I have two offers, one from UOW and other from univ of adelaide. I have a scholarship that reduces my tuition fee from 25% in UOW. No scholarship from univ of adelaide. I want to go with univ of Adelaide Becoz of its ranking, is this worth it ? I mean does goibg to a G8 univ have benefits over univs like Uow ?
@Committingtoachallenge since UR in australia and going to a G8 univ. I think U know better than me, so Im asking what are your thoguhts :)
@Committingtoachallenge tuition fee in adelaide : 31,900 , in UOW i have to only pay 22,000.....is paying the extra 9900 dollars for adelaide worth it ?
 
11:04 AM
@TheArtist I can't give you a full answer because I am really busy. But take Adelaide in my opinion
Adelaide and HECS-HELP
I can give you a better answer if it is still relevant on the 14th I suppose
I have went to three universities and only the G8 one was good if that matters
 
11:31 AM
@TheArtist Do not choose a university solely based on its ranking. Many rankings don't mean much.
@TheArtist I think an important factor in choosing a uni for undergrad is its curriculum. Look at the course descriptions to decide. Does it have the breadth and depth of the topics you want to learn?
 
12:02 PM
Greetings
0
Q: 2 logarithmic integrals twins

Chris's sisThis question (I'd never close or put on hold) also has the value of an answer to the first integral here How to evaluate $\int_0^1\frac{\ln(x)\ln(1+x^2)}{1-x}dx$ and $\int_0^1\frac{\ln(x)\ln(1+x^2)}{1+x}dx$ $$\int_0^1\frac{\ln(x)\ln(1+x^2)}{1-x}dx$$ $$=\int_0^1\frac{(1+x)(1+x^2) \log(x)\log(1-x...

People that post nice questions shouldn't be discouraged. I'm not here for judging people, I'm only interested in questions.
 
Can someone help me (one who neither remembers inverse trigonometry nor dealing with series) how to evaluate the following: $$\sum_{k = 0}^{\infty}\tan^{-1}\left(\frac{1}{k}\right)$$
or in simpler words, $$\lim_{N \to \infty} \sum_{k=0}^{N} \cot^{-1}\left(k\right) $$
 
I only hope no one will think of the fact that @user178256 is me since I answered the question in less than 1 min from the moment I saw it.
Sometimes I'm incredibly fast ... (never at typing, my fingers do not listen to me)
 
@Chris'ssis No, no one will think that. Your secret is safe :D
 
@Nick Well, I saw that user has some questions witn integrals only, and you never know what people think of.
 
@Chris'ssis I have a typing issue as well, it's not anything wrong with my fingers but my luck in getting functioning keyboards
@Chris'ssis Oh, people are judging and scrutinizing everything about you, my dear. From the way you walk to how you breath, every morsel of your being is a potential target of someone and everyone's mindless criticism. I'm quite surprised that you care what people think. You must be very strong.
 
12:15 PM
@JasperLoy Thank you very much :) but the problem is they just tell the topics, not the depth of it , and plus most topics I don't know about
@Committingtoachallenge ok :) thank you a lot too :)
 
@Nick Well, it's not about care, but I'd like people know that it is possible to answer such questions in less than 1 min without cheating. You only need to be natural, you see the problem and (almost) instantaneously know the answer.
 
@Chris'ssis Well, do you know the answer to my series instantly ?
 
@Nick Yeah ... what is the behaviour of $\arctan(1/k)$ when $k$ is large?
 
@Chris'ssis My only point was not to let what anyone says distinguish who you are. You're awesome .. not because I say so... but because you are:D
 
@Nick :D
 
12:21 PM
@Chris'ssis $\arctan(1/k) \to 0$ as $k \to \infty$
 
@Nick When $k\to\infty$ you have $1/k \to 0$, but $\lim_{x\to0} \frac{\arctan(x)}{x}=1$. Thus the large terms behave like $1/k$
 
Right :D I'm with you.
The entire series diminishes. I understand that it's convergent but Idk how to evaluate it
But cwap, wolfram says it diverges
 
@Nick Your series behaves like harmonic series that we know it diverges.
 
@Chris'ssis Ah....
Well, that was educational :D
@Chris'ssis I tried to extend the 3 square problem into an $\infty$ square problem. That's how I got the series. It would have been nice if it did converge.
 
@Nick Cute problem.
@Nick I have a mind blowing proof for this problem (one without words). I think I posted it somewhere in one of my answers here.
 
12:37 PM
@Chris'ssis I actually got the above problem (3 square) in 8th grade during a junior math competition. I found the sum of angles using a protractor because my geometry skills were and still are limited. It was nice to find a video after such a long time.... and the problem is introduced by a person who knew of the problem since her childhood. I thought that was a very serendipitous coincidence.
 
@Nick I don't like the solution in the video. Too much writing, too much talk.
 
@Chris'ssis Well, it was meant for fifth graders. It's not the best nor easiest solution. It's the one meant for kids.
 
@Nick See here, I'll undelete the answer and then delete it again. No one upvoted my absolutely brilliant answer.
 
@Chris'ssis I'd love to see the answer :D
 
45
Q: Why does $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi$?

Clayton KershawPlaying around on wolframalpha shows $\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3)=\pi$. I know $\tan^{-1}(1)=\pi/4$, but how could you compute that $\tan^{-1}(2)+\tan^{-1}(3)=\frac{3}{4}\pi$ to get this result?

 
12:42 PM
@Chris'ssis Wait, aren't you in risk of losing points if you do that?
eh no, I don't get it. It seems too abstract to me.
 
@Nick Yeah, that's all. :-)
@Nick Have in mind the line at the bottom, all things will happen there.
 
I understand the following version of it.
 
@Nick lol, in my picture you find the same thing.
 
@Chris'ssis Sometimes little things matter...
 
@Nick :D
 
12:51 PM
Well, this was fun. I've got to go build a light dependant resistor and a light bulb now. See ya later :D
 
@Nick See ya :-)
 
 
I'm gonna fucking bomb my room. It's filled with flies.
 
@UserX Good luck.
 
@JasperLoy man it's november, I left the window open this morning and all hell broke loose
 
1:03 PM
@Nick build a light dependent resistor? I didn't know those were in the realm of home build.
 
hehe, finally someone gave me an alternative approach to my questions
2
A: An integral by O. Furdui $\int_0^1 \log^2(\sqrt{1+x}-\sqrt{1-x}) \ dx$

M.N.C.E.Let $\sqrt{1+x}-\sqrt{1-x}\mapsto2\sin{x}$. We get, for the first integral, \begin{align} \color{#6F00FF}{\int^1_0\ln^2\left(\sqrt{1+x}-\sqrt{1-x}\right)\ {\rm d}x} =&\int^\frac{\pi}{4}_02\cos{2x}\ln^2(2\sin{x})\ {\rm d}x\\ =&\sin{2x}\ln^2(2\sin{x})\Bigg{|}^\frac{\pi}{4}_0-\color{#FF4F00}{\int^\f...

@robjohn see the answer above :-)
 
@Chris'ssis I added a second approach to your classroom problem. It only requires $\log(1+x)\le x$
 
@robjohn Really???
 
@Chris'ssis I was about to look
 
@robjohn You should have posted that separately such that I might be able to upvote again. Very creative!
@robjohn why do you think just you and china math answered my question? Is it a hard question?
 
1:09 PM
@Chris'ssis I think all your questions are hard, lol.
 
@JasperLoy lol, but look at the answers of @robjohn, he brought everything at the high school level.
@JasperLoy btw, how is your condition now?
 
@Chris'ssis well, when being asked for not only an answer, but also a particular focus on the answer, it might be more than people want to deal with.
 
@Chris'ssis Still having the flu. My mental problems, OK.
 
@nick
 
@JasperLoy Wait, hopefully you didn't catch it from me. The flu I had put me in a state of depressing hoola hoop of terror.
@Sawarnik hey buddy :D
 
1:15 PM
:D
I would be off the internet for a relatively long period of time from tomorrow :/
@Nick Wow, how much would you sell it for?
 
@robjohn No, apparently it is not. I'm struggling with the light bulb right now (because of a lack of good tools) and I hoped to find some instructions on how to make a giant ldr for my giant lightbub but alas, your words have shattered me.
 
heehaha..
 
@Nick I didn't say that you couldn't do it; I just didn't know that one could do it.
 
@Sawarnik You see when he asserts that those two big green triangles are congruent. That's where he's lied to you.
 
@Nick Why? :/
 
1:21 PM
@robjohn No, but what you said led me to discover that there is no documentation of building it by hand... that's what shattered me. Now, why I am I wasting my time with this bulb if the other thing can't be hand made as well. :(
 
evil mean square!
 
@JasperLoy @TedShifrin suggested I'm obsessed with the integrals, series and limits, maybe I also have OCD. :-)
 
@Sawarnik Because the sides are not equal. You're overlooking that point to say that they are.
 
@Nick Which sides?
B * X and C * X?
 
@Sawarnik ...ugh, I'm busy. You figure out what I meant.
 
1:24 PM
@Nick I cant, busy boy.
Wait.
 
See that $AB^* \neq AC^*$
 
@Nick :/ Isnt that proved by congruency?
 
You assume it is though.
 
Why did I remove it?
@Nick How really?
 
@Sawarnik Look i didn't pay attention to much of it but you neglect the fact that those sides (may or may not be) unequal, that's why you could use congruency
 
1:28 PM
:/
Ok lets try GeoGebra.
@Nick I never see it assumed. Can you clarify where?
Yes, I thought right.
 
Oy vey, I'm merely the village idiot here so I can't explain this to you.
 
The perpendicular point is not outside the side.
 
Anyhow, it's a very famous fallacy.
 
@Nick The famous one was when they assumed the point met inside the triangle.
This is new to me.
 
1:32 PM
Ok, the one I had seen must be a variation.
 
@Sawarnik Absolutely right :D
 
@Nick See, it proves my point lol.
 
Well, I guess I learned something else today :D
Thanks
 
@Nick :D
 
@Sawarnik I'm multitasking... I have no logic.
Dang, this bulb isn't working.
 
1:34 PM
@Nick Famous fallacies don't have such silly mistakes. The mistake is almost always in the location of points.
@Nick I knew.
 
@Sawarnik Famous fallacy
In mathematics, certain kinds of mistaken proof are often exhibited, and sometimes collected, as illustrations of a concept of mathematical fallacy. There is a distinction between a simple mistake and a mathematical fallacy in a proof: a mistake in a proof leads to an invalid proof just in the same way, but in the best-known examples of mathematical fallacies, there is some concealment in the presentation of the proof. For example, the reason validity fails may be a division by zero that is hidden by algebraic notation. There is a striking quality of the mathematical fallacy: as typically presented...
 
In mathematics, certain kinds of mistaken proof are often exhibited, and sometimes collected, as illustrations of a concept of mathematical fallacy. There is a distinction between a simple mistake and a mathematical fallacy in a proof: a mistake in a proof leads to an invalid proof just in the same way, but in the best-known examples of mathematical fallacies, there is some concealment in the presentation of the proof. For example, the reason validity fails may be a division by zero that is hidden by algebraic notation. There is a striking quality of the mathematical fallacy: as typically presented...
^see this
Oh you posted it!
 
Nope, it's in the Wiki page for fallacy.
well, this is awkward.
...
 
@Nick just look at the isosceles triangle fallacy there.
its a nice one.
 
If a problem states that a lim exists in $\Bbb R$ does it mean it takes a value $a\in\Bbb R$?
 
1:36 PM
@Sawarnik That's why I linked it. I wanted to show you that it was famous.
 
@Nick oh gawd, have you even looked at it?
 
@Sawarnik ... busy with a lightbulb here also (Maxwell 1959, Chapter II, § 1) seems like a pretty relevant source. (We may have different notions of the word "famous")
 
just take a hammer and give a blow to that irritating bulb.
 
I just spent 3 hours getting it the way it is.
 
@Nick take a look at the fallacy there -_-
 
1:40 PM
ohk... :(
 
And can anyone give a correct explanation here, this chain of comments is completely absurd and hilarious: youtube.com/watch?v=Yajonhixy4g
 
"This fallacy has been attributed to Lewis Carroll" ... seriously, this is the difference between a raven and a writing desk?
 
What do you mean eh?
 
@Sawarnik Ok, I get it. You were right. "The error in the proof is the assumption in the diagram that the point O is inside the triangle. In fact, O always lies at the circumcircle of the â–³ABC (except for isosceles and equilateral triangles where AO and OD coincides " .. It's about placement of that point.
@Sawarnik You've never heard the mad hatter say that?
 
@Nick No :/
 
1:45 PM
@Sawarnik so, you've never read the book, seen the animation, watched the movie starring johnny depp?
 
No :/
 
Ugh, why do I bother making pop culture references!?
 
@DanielFischer I can see your post here in starred section
 
@Integrator Yes everyone can see, so?
 
Sorry, I'm a bit frustrated about the bulb. The glass jar just burnt up a bit when I connected the electricity... This thing is dangerous.
 
1:48 PM
@Integrator Hi.
 
@Integrator So?
 
Dang this screwey thing
Nick throws his homemade bulb onto the ground... and it smashes into a kajillion pieces
I'm going to go buy an actual light bulb and LDR now. Making things by hand is stupid.. especially when it's impossible.
 
@Nick Good.
 
@Chris'ssis You have no idea how serious OCD is. Do you?
 
@DanielFischer That's just because people can see how OP is desperate to accept an answer
 
1:57 PM
Now, I'm going to take Michael J Fox's advice and go own my burdens.
 
Hi @DanielFischer Could ask your opinion about an idea of a proof.
 
@Integrator I think it's more that they hope for upvotes, regardless of whether they have read the question, or whether the question is in a shape that answering it is adequate at all.
@Moses You can. I don't guarantee that I have an opinion, or if I have, that it is correct, though.
 
1 hour ago, by Nick
user image
 
Okay thanks @DanielFischer The question is if $f: [0,1] \rightarrow \mathbb{R}$ is a function continuous function which takes only values which are integers then show that $f$ is a constant function on $[0,1]$. I know how to prove this but I am interested in one particular type of proof. I will give my idea as briefly as possible now.
 
If a function is Riemann integrable does that mean it is continuous?
 
2:09 PM
@UserX No. Step functions are Riemann integrable, but not necessarily continuous.
Riemann integrability implies continuity outside a set of Lebesgue measure $0$, however.
 
@DanielFischer can you explain me simply what's the difference between continuous almost everywhere vs continuous?
 
@robjohn Do we have to follow any kind of rules here?
 
@DanielFischer Let $a_{1} = 0$ then if we take $\epsilon < 1$ it follows that there exists a $\delta_{1} > 0$ such that for all $y \in [0,1]$ such that $|a_{1}-y| < \delta_{1}$ $\implies$ $|f(a_{1})-f(y)| < \epsilon < 1$ therefore $f(a_{1}) = f(y)$.
Then take $a_{2} := a_{1} + \delta_{1}$, it follows then that again for $\epsilon < 1$ there exists $\delta_{2}$ such that for all $y \in [0,1]$ we have that $|a_{2}-y| < \delta_{2}$ $\implies$ $|f(a_{2}) - f(y)| < \epsilon$ therefore $f(a_{1}) = f(a_{2}) = f(y)$. If we continue this process we get that $f$ is continuous on all of $[0,1]$. Is this a complete proof?
 
@Integrator There are the chat guidelines pinned to the sidebar
 
@UserX Continuous means "continuous at every point". "Continuous almost everywhere" allows for some discontinuities, e.g. isolated discontinuities are allowed. But they need not be isolated, for example the characteristic function of a Cantor set of measure $0$ is discontinuous at every point of the Cantor set, but these aren't isolated.
 
2:15 PM
@robjohn can you send screenshot? Maybe my screen is small!
 
@Integrator Chat guidelines there is a copy of the link that is pinned to the sidebar.
 
@DanielFischer how is "some" discontinuities defined?
 
It's too bad when someone puts a bounty on a question and then doesn't award it before they are deleted.
 
@Moses Sorry, misread first. You haven't proved that $$\sum_{k=1}^{?}\delta_k \geqslant 1.$$
@UserX In this case, "some" means "the set of discontinuities has Lebesgue measure $0$".
 
@DanielFischer Why is that important? Is it not enough to note that each $\delta$ is positive and so eventually we will cover $[0,1]$.
 
2:21 PM
@Moses You could use either uniform continuity of $f$ or connectedness of $[0,1]$ to show that in the most easy ways. But I suspect you want to avoid both of those arguments, in which case it becomes cumbersome.
 
@Moses There can be an infinite number of positive values and it might still not cover $[0,1]$
 
@DanielFischer Yes I see.
 
@Moses If $\delta_k$ decreases fast enough, say $\delta_k = 4^{-k}$, then you never reach $1$.
 
@robjohn Yeah
@DanielFischer Is continuity not enough to prove this? Do you need uniform continuity?
 
@robjohn oops!
 
2:34 PM
@Moses Uniform continuity or connectedness. Other arguments may be possible, but just continuity doesn't get you there, since if you consider $[0,1]\cap \mathbb{Q}$ instead of $[0,1]$, the conclusion doesn't follow.
 
Writing Laplace transforms in LaTeX is a pain in the ass but the result sure is beautiful
 
@UserX Do you mean $\mathscr{L}$, $\mathcal{L}$, or what?
 
The scr seems nicer, didn't know about it
 
2:49 PM
@DanielFischer How about the following: Assume $f$ is not constant on $[0,1]$.
Let $f(0) = c$ for some $c \in \mathbb{Z}$. Then by assumption there exists $x \in [0,1]$ such that $|f(x)| = c + n$ with $n \in \mathbb{Z}^{+}$. Let $a := \inf\{x \in [0,1]: |f(x)| = c + n \}$ then for every $\alpha > 0$ we have $|f(a-\alpha)| = c$.

Take $\epsilon = \frac{1}{2}$ then for every $\delta > 0$ we have some $y \in [0,1]$ such that $|y-a| < \delta$ $\implies$ $|f(y)-f(a)| > 1 > \epsilon$. This contradicts the assumed continuity of $f$. Therefore $f$ is constant.
 
@robjohn And what could be the consequences of not following those guidelines?
 
@Moses You shouldn't use $\lvert f(x)\rvert$, what if $f$ took the values $1$ and $-1$? Apart from that, you seem to assume that $f$ takes at most two distinct values when you say that $\lvert f(a-\alpha)\rvert = c$ for all $\alpha > 0$. But basically, that is the connectedness argument.
 
Yes!! I reached $1k$
 
r9m
@Integrator congrats !! ;)
 
I don't know enough math to get more rep :(
 
2:56 PM
@DanielFischer I see, so would you say this is a complete proof for the most part?
 
@user130018 On the one hand, you learn more math every month, on the other, you can also get rep by asking decent questions without knowing more math.
@Moses For the most part.
 
@r9m Thanks!
 
@DanielFischer Could I ask what exact connectedness result it used?
 

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