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12:00 AM
our analysis sequence was split into different parts and in that part we didn't even define Riemann integrals I think
 
But it's assumed that you know 'em, duh.
How else do you compute any integrals, like Fourier transforms?
 
yeah true
you use the definition of the Lebesgue integral and take the sup over all simple functions ... nevermind
in our first analysis course we did integration before differentation for some reason, so didn't have FTC, thus we basically couldn't compute any integrals for quite some time
 
The limit notions for integrals are more basic than those for continuity and differentiability :P
But you do need the FTC eventually :)
But, nevertheless, @Mathein, I think that you've established you're pretty knowledgeable and clever.
 
thanks lol
 
12:22 AM
@ParthKohli You should use this as your personal website link in the MSE profile instead of what you already have.
 
@BalarkaSen this sounds better than it should: youtube.com/watch?v=YkzevB2vYrs
 
Shrek me up inside
There was this video which demonstrated that song works with any tune of every possible pop hit ever
And the Smash Mouth themselves tweeted it on their twitter account
 
amazing
 
Retires from the chat.
 
12:50 AM
hmm. what's the simplest way to describe a set of r.v.'s $\{X_j\}$ such that $\langle X_j\rangle =0$ and $\langle X_j X_k\rangle =\delta_{jk}$?
independent r.v.s with zero mean and unit variance i guess?
i guess one could gloss 'zero mean' as 'centered r.v.'
 
@TedShifrin your already retired AFTER THAT COMES THE WHITE LIGHT... DON'T GO
caps was a happy accident
 
The beauty of music theory is that many things can be forced to work with each other.
Any two notes can sound good together as long as you play the right third note.
 
intresting
 
thanks, @Fausty
 
fausty is feisty
 
1:01 AM
Hi
 
blah, I was mixing up 'independent' and 'uncorrelated'
 
Balarka, do you play an nstrument?
 
Unfortunately nope
I wish I could
 
Yeah, @Semiclassic, that ain't independence.
 
mumble
 
1:02 AM
Product of random variables isn't their intersection :P
 
@BalarkaSen you should just pick one up! Better late than never!
 
and anyhow what I said is slightly off .. . $P(X\text{ and } Y) = P(X)P(Y)$.
@anakhronizein: He's likely to drop it.
 
Independent random variables are uncorrelated though.
 
But not vice versa.
 
yeah, but I meant uncorrelated
I may want independence for something later, but not to start with
 
1:04 AM
I have a counterexample in my notes.
 
Wikipedia has one as well on the 'uncorrelated random variables' page
 
@anakhronizein Maybe, yeah
What do you play?
 
I used to play cello, but I don't have time anymore
 
uncorrelated r.v.'s with unit variance and zero mean
i sorta would like a nicer name than that
 
I think you need all the conditions, Semi.
 
1:07 AM
(doesn't mean I'll get one tho)
@TedShifrin I don't follow.
 
I play bass guitar, @BalarkaSen
 
You have three conditions there. They're all independent.
 
Oh, sure.
 
Well, actually, 5 conditions.
 
@MatheinBoulomenos @anakhronizein Very cool
 
1:08 AM
I'm just looking for a name which summarizes all of that at once.
 
Anyhow, time for me to go cook dinner.
 
@TedShifrin Five?
 
Enjoy dinner, @TedShifrin!
 
2 for variance, 2 for mean, 1 for correlation
Thanks, @anakhronizein.
 
oh. I had in mind n such variables
so it's really a lot more than 5 :P
 
1:09 AM
Oh, hell.
 
but pairwise uncorrelated
 
$2n + \binom n2$.
 
right.
 
On that note ... bye.
 
later
 
1:09 AM
@anakhronizein What is the shortest acceptable abbreviation of your name?
 
the empty string
Bye @Ted
 
Call me whatever you like. ana is fine, anak is fine, an is fine, anakhr is fine, asndghny is fine.
 
(@anakhronizein Challenge: Play the baseline here)
 
1:11 AM
I like anak, since then I would be able to describe you ranting as 'running anak'
(not that I've seen you rant, but hey. gotta be prepared)
 
this is the most difficult thing I played: youtube.com/watch?v=PCicM6i59_I
 
Heh Captain Beefheart
 
hmmm. now to figure out what a condition like $\langle Y_1 Y_2\rangle +\langle Y_2 Y_3\rangle +\langle Y_1 Y_3\rangle \geq -1$ could mean
 
lmao
So you already know about him
 
where these r.v.'s have unit variance and zero norm, but can be correlated however they want
 
1:14 AM
Yes.
The wikipedia page for this album has some quite notable quotes.
 
Trout Mask Replica is pure madness. I love it
A squid eating dough in a polyethylene bag is fast and bulbous. Pure poetry
 
It didn't live up to the rave reviews I found.
 
I wouldn't say it's the greatest experimental rock record ever, I guess, yeah.
It's pretty weird though
 
oh, standardized r.v's
so a set of pairwise uncorrelated standardized r.v.s
hmm. $\langle (Y_1+Y_2+Y_3)^2\rangle = 3+2\left[\langle Y_1 Y_2\rangle+\langle Y_2 Y_3\rangle +\langle Y_1 Y_2\rangle\right]$
So the above bound would amount to $\langle (Y_1+Y_2+Y_3)^2\rangle \geq 1$...okay
 
I didn't even find it all that weird, Balarka
 
1:25 AM
Really? Hah
 
Anyways, for my own reference: Let $X_1,X_2$ be uncorrelated standardized rvs. Then the r.v.s $Y_1=X_1,\;Y_2=-\frac12 X_1+\frac{\sqrt{3}}{2} X_2,\;Y_3=-\frac12 X_1-\frac{\sqrt{3}}{2} X_2$ form a set of standardized rvs with $\langle Y_1 Y_2\rangle=\langle Y_2 Y_3\rangle=\langle Y_1 Y_3\rangle=-\frac12$
 
The use of irrational non-periodic harmonics, oddball percussion, and stream of magic realism as lyrics seem to be the major three sticking points of the weirdness of the album to me
But hey, weirdness is relative!
 
(This doesn't satisfy the inequality I indicated before, so that bound is more restrictive.)
 
I think Merzbow is weirder.
 
Also unlistenable to me. But Merzbow is very recent.
So the comparison doesn't entirely make sense
 
1:30 AM
1979 is not that recent.
 
Which album do you have in mind? I'm thinking of the iconic Pulse Demon
That's like from the 90s
 
He's been doing noise music since 1979
 
Mm I see. I haven't listened to anything else other than Pulse Demon, which I somehow ended up listening in full while doing other stuff
It broke my ears completely
 
Like it's not experimental rock or anything. So not really on the same field. But I was expecting my mind to be blown by the weirdness, and I got something I considered even less weird than the Mars Volta.
 
Have a go at Scott Walker, "Bish Bosch" if you're up for weirdness. I think you'll enjoy that one.
 
1:36 AM
the closest thing I have to a weird album I like (and it's not that weird) is this: youtube.com/…
 
'>video blocked in your country. :(((((((((
 
same
 
huh
Robert Pollard's "Not in My Airforce"
 
What sort of stuff do you normally listen to, @Semiclassical?
 
i mostly stick to stuff I know at the moment tbh
I'm a fan of, let's see...
the mountain goats, explosions in the sky, and wolf parade come to mind?
 
1:40 AM
I somehow don't like Explosions in the Sky
 
Wolf Parade is a fave of mine. :)
A good canadian band
Though I didn't really get into their new album
 
tbh I haven't followed their most recent stuff
 
I was listening to "Heretic Pride" by Mountain Goats today
 
Heretic Pride?
 
Snap, yeah
That
 
1:41 AM
yeah, it's a good one
both the song and the album as a whole
 
Agree
 
It's got some tracks I love, and some I'd love if they didn't make me feel sorta terrible
Autoclave, In the Craters on the Moon, Michael Myers Resplendent being examples of the latter
 
Darnielle can really write well
 
yeah
I feel like his storytelling is a bit sharper in his older stuff, but it's still there
"Beat the Champ" being a really really good example of his new stuff
 
I need to go through MG's discography
 
1:44 AM
("Goths", by contrast, i couldn't really get into past the first track...but the first track is incredible AF so it balances out)
@BalarkaSen lol, that'll take a bit of time
(the first track I mean is this one: youtube.com/watch?v=anS6bcPpvoQ)
 
@Semiclassical do you like Modest Mouse?
 
eh. i like some of what i've heard but i've never gotten deep into them
missed that phase somehow
 
They are probably my favourite band of all time.
They are how I found Wolf Parade.
 
I found Wolf Parade through the yearly album review that Questionable Content used to do
I'll Believe In Anything is a good starting point, lol
 
Heh. That was my first Wolf Parade song too.
 
1:58 AM
yeah
 
second album was sort of a miss for me, but the third one was awesome
Spencer Krug's own work has some good stuff, though occasionally it's been a bit too odd for me
@BalarkaSen triple-u tf
 
"Picture of Marilyn Munroe, flutters across the room, shaped like my ass"
10/10
Turns out this guy was a lover of Burroughs. Not surprised. Not surprised at all.
 
2:15 AM
Have a goodnight, guys!
 
2:28 AM
Hello, people. Can anyone tell me the definition of a subset of a topological space?
 
2:44 AM
@ParthaSarker its a subset of a space with a topology that inherits the superset's topology
in other words, it's a subset that's closed under the axioms for open sets
union and finite intersection
 
@GFauxPas your first sentence is correct, your second is not. If $X$ is a topological space, then if $Y \subset X$ is a subset, we can make $Y$ into a topological space by declaring the opens of $Y$ to be $Y \cap U$ where $U$ is open in X.
 
oh oops
agreed
 
3:03 AM
So just trying a sanity check
 
Dam you will always be sane in our hearts
 
Let's say $\varphi$ is a character $(\mathbb{Z}/p\mathbb{Z})^{\times} \to \mu_{p-1}$, then consider $\sum_{k\in \mathbb{F}_p^{\times}} \sum_{j\in\mathbb{F}_p^{\times}} \varphi(k)\overline{\varphi(j)}\zeta_p^{ak - aj}$
Should this equal $p$?
For reference, I'm trying to show that for any non-trivial character, $|G(\varphi)|^2 = p$
And the current method of attack is basically considering $\sum_{a=1}^{p-1} |\sigma_a(G(\varphi))|^2$ where $\sigma_a(\zeta_p) = \zeta_p^a$
On the one hand that's $(p-1)|G(\varphi)|^2$, and so I wanna show that it's $p(p-1)$
 
3:34 AM
Have you tried checking this for small $p$?
 
Have you tried turning it off, then turning it back on again?
 
3:45 AM
@Daminark this looks a lot like a Gauss sum
 
Yeah the $G(\varphi)$ is the Gauss sum
$G(\varphi) = \sum_{k=1}^{p-1} \varphi(k)\zeta_p^{k-1}$
@loch not quite
Oh no this approach is kinda bad
 
4:07 AM
So is $\varphi$ any nontrivial character or a irreducible character? you can scale a character by a positive integer and you still have a character
 
4:21 AM
@Mathein sorry I was out, and actually I realized that a different approach worked
Here it was any nontrivial character
 
what definition of character are you using?
I'm probably confusing myself by thinking about represention theory
 
So, we define a mod p character as a homomorphism $(\mathbb{Z}/p\mathbb{Z})^{\times} \to \mu_{p-1}$
 
ohh
okay
nevermind
 
But yeah turns out you just kinda do it directly and recall that if $\chi:G\to S^1$ is a homomorphism where $G$ is finite, then $\sum_{g\in G} \chi(g) = 0$
 
So how did you do it? I think I did this computation once for $\varphi$ the Legendre symbol
 
4:40 AM
If I have a category C and a subcategory C’ and I define a functor F: C —> C’ and I show that F composed with the inclusion functor i: C’ —> C is naturally isomorphic to the identity functor on C, is that enough to conclude and equivalence of categories between C and C’
It seems to make sense because the thing remaining to check would be that the inclusion composed with F is naturally isomorphic to the identity functor on C’, but that would just be the restriction of the identity functor on C? So essentially I would be checking that F restricted to C’ and identity restricted to C’ are naturally isomorphic but doesn’t that follow from the fact that they were naturally isomorphic over C
 
Let $f:\Bbb R\to\Bbb R$ be a polynomial such that $f(0)>0$ and $f(f(x))=4x+1$ for all $x\in \Bbb R$. Then $f(0)$ is .......
I found by trial and error that $f(x)=2x+\frac 13$ is such a polynomial. But is there another way for this exercise?
 
@PrinceM I think the statement is true. i is faithful because it's the inclusion of a subcategory. The fact that the composition of F with the inclusion functor is naturally the identity functor gives you that the inclusion functor is essentially surjective and full
 
4:55 AM
@silent you get immediately that degree of f is at most 1 so f can be written in the form f = a_o + a_1x
 
ok
 
Composing f with itself yields f(f(x)) = a_0(1+a_1)+a_1^2x
Immediately we see a_1 must be +/- 2, but the choice of -2 contradicts the fact that f(0) > 0, so the only solution is the one you found
 
thank you!
 
Thanks @MatheinBoulomenos
 
5:18 AM
@MatheinBoulomenos I'm about to get on my computer and I'll post a screenshot
 
5:35 AM
@Daminark ah yeah makes sense
 
Hmm, why is it true that if $q$, $r$, and $p$ are prime, then $q = r^ic^p$ and $r = q^j d^p$ don't have solutions for $c,d\in \mathbb{Q}(\zeta_p)$? (where $i$ and $j$ are coprime to $p$)
(Of course $r\ne q$)
(/is it true? Should be)
Actually nvm lmao it's ez
Or maybe not
:(
 
5:56 AM
$N(\frac{q}{r^i})=\frac{q^{p-1}}{r^{(p-1)i}}$, that's clearly not the $p$-th power of any rational number
Are you familiar with the norm map?
 
Not quite
But I now think I have something, maybe you can confirm it?
 
Suppose $L/K$ is finite Galois, then you can define a map $N_{L/K}:L \to K$ given by $N_{L/K}(x)=\prod_{\sigma \in \operatorname{Gal}(L/K)} \sigma(x)$
Note this satisfies $N_{L/K}(xy)=N_{L/K}(x)N_{L/K}(y)$ and if $x \in K$, then $N_{L/K}(x)=x^{[L:K]}$
you need Galois theory to show that this actually maps to $K$
 
So I said aight, assume $x^p - \frac{q}{r^i}$ has a solution in $\mathbb{Q}(\zeta_p)$. Well, that's irreducible over $\mathbb{Q}$ since $r^ix^p - q$ is Eisensteinable (right?)
 
But then $\mathbb{Q}(\zeta_p)$ is the splitting field of that guy since if you have one root, you can get all the other roots by multiplying by $\zeta_p$
 
6:08 AM
sure
you could also say that this is impossible, because $\Bbb Q(\zeta_p)$ has degree $p-1$
 
But that's shit since the degree of $\mathbb{Q}(\zeta_p)$ is $p-1$ while it's supposed to contain the splitting field of a degree p irreducible polynomial
Yeah exactly
 
you don't need the splitting field part. A field of degree $p-1$ can't contain a single root of an irreducible degree $p$ polynomial
 
True
 
the part that $i$ is coprime to $p$ is not necessary as your proof shows (it also follows from my proof)
 
Not related to math: I can't understand which word he uses here before Julia Sweeney, it sounds like 'sererera ohivelum'
Does anyone know?
 
6:20 AM
Okay this is convenient, things are fitting together
Also I met with my algebra prof today and he showed me how to prove quadratic reciprocity with cyclotomic business
 
you mean with Gauss sums?
or with splitting of primes?
 
Gauss sums
 
that proof is just magic
 
He presented two slightly different proofs
So with our notation $p^*$ was $\pm p$ in order to be 1 mod 4
So that the statement was $(\frac{p^*}{q}) = (\frac{q}{p})$
The first just asks whether $\sqrt{p^*} \in \mathbb{F}_q$
The second asks whether $\sigma_q$ is trivial or not on $\mathbb{Q}(\zeta_p)$
 
What's $\sigma_q$?
 
6:26 AM
(Here $\sigma_q:\zeta_p\mapsto \zeta_p^q$)
 
that's never trivial
 
Sorry there was a typo earlier
 
I assume you mean trivial on the quadratic subfield $\Bbb Q(\sqrt{p^*})$?
 
On $\mathbb{Q}(\sqrt{p^*})\subset \mathbb{Q}(\zeta_p)$
Yeah
 
that's basically also the algebraic number theory proof, but you can skip the stuff with the Gauss sums
 
6:31 AM
So we get the action of $(\mathbb{Z}/p\mathbb{Z})^{\times}$ on $\mathbb{Z}[\zeta_p]$, and pass through to $\mathbb{F}_q[\zeta_p]$
 
(or one possible algebraic number theory proof)
 
So in this case you're looking at $\mathbb{F}_q \subset \mathbb{F}_q[\sqrt{p^*}] \subset \mathbb{F}_q[\zeta_p]$. The middle guy is $\mathbb{F}_q[x]/(x^2 - p^*)$
Hmm, okay we're reached the part where I'm a bit less sure of what he said
(Also his handwriting is tricky)
So the middle and top guys are products of fields, choose them compatibly
So you'll get fields $\mathbb{F}_q \subset \mathbb{F}' \subset \mathbb{F}''$
The action descends and $\mathbb{F}'$ is the fixed field of the squares
Err, of $\langle q\rangle \cap \text{squares}$
So now $\langle q \rangle = \langle q \rangle \cap \text{squares}$ iff $q$ is a square mod $p$
Which is true iff $(\frac{q}{p}) = 1$
 
yeah, that's pretty cool
I thought you meant the proof were you just do a couple random calculations with Gauss sums and the result magically falls out
 
He gave another Gauss sum proof
Write $\sqrt{p^*} = \sum (\frac{a}{p})\zeta_p^a$, so you raise each side to the $q$
The left hand side obviously scales by $(\frac{p^*}{q})$
Right hand side scales by $(\frac{q}{p})$
Since that's $\sum (\frac{a}{p}) \zeta_p^{aq} = (\frac{q^{-1}}{p})\sum (\frac{aq}{p}) \zeta_p^{aq}$
So yeah it's pretty nice
I'll need to look over the first proof more carefully and be sure I understand it but it's quite nice
 
7:35 AM
Anyone?
1 hour ago, by Silent
Not related to math: I can't understand which word he uses here before Julia Sweeney, it sounds like 'sererera ohivelum'
 
7:53 AM
[Random]
This is actually not a 4D maze. This is actually a 11D maze
The reason is that for a given room with coordinates xyz, depending on which room you are in, switching between the so called 8 levels of the 4th dimension in the game actually will give you different connectivity to the neighbouring rooms, while if this maze is truly 4D, then you expect neightbouring levels should be connected (and you won't be able to move from level 0 to 2 without being in 1 first
I dare anyone who attempt to map the maze, they will find neigbouring cross sections don't match up
Unless...
It is a 4D maze, but the connection along the 4th dimension is arbitrary thus there is a tunnel that connects between level 1 to level 5 for example
Still, this showed how just a little bit change in the topology on a maze, can make the maze inherently more complex despite having the same number of spatial dimensions
With this, I am starting to suspect the following:
What is space:
Space is a set of objects called points, plus relations between them
The simplest example is a topological space, where the relation is a topology
Our euclidean space obeys a remarkable number of relations for a space which is why it is so nice:
1. It has a linear ordering
2. The topology is Hausdorff
3. It is path connected
4. The euclidean metric is symmetric
5. It is complete
So in general, for an abstract space $X$, navigating between points is in general nontrivial (and could be unidirectional e.g. $a \to b$ but not $b \to a$)
Therefore, if a set of objects is not related by any relations, it is spaceless
 
8:20 AM
The number of points at which the function $f(x)=\left|\left(\left|x\right|-3\right)\sin\ \left(\pi x\right)\right|+\left|\left(x^2-1\right)\left(x^3-27\right)\right|$ takes zero value is ..... @LeakyNun, please help!
I see that it has at least three roots: $-1,1,3$
 
In general topology, a pretopological space is a generalization of the concept of topological space. A pretopological space can be defined as in terms of either filters or a preclosure operator. The similar, but more abstract, notion of a Grothendieck pretopology is used to form a Grothendieck topology, and is covered in the article on that topic. Let X be a set. A neighborhood system for a pretopology on X is a collection of filters N(x), one for each element x of X such that every set in N(x) contains x as a member. Each element of N(x) is called a neighborhood of x. A pretopological space is...
where everything are filters
and thus spaces can be understood as a collection of filters equipped with a notion of preservation of unions
In category theory, a branch of mathematics, a Grothendieck topology is a structure on a category C which makes the objects of C act like the open sets of a topological space. A category together with a choice of Grothendieck topology is called a site. Grothendieck topologies axiomatize the notion of an open cover. Using the notion of covering provided by a Grothendieck topology, it becomes possible to define sheaves on a category and their cohomology. This was first done in algebraic geometry and algebraic number theory by Alexander Grothendieck to define the étale cohomology of a scheme. It has...
and thus, spacelessness is a collection of mathematical objects lacking a category
In mathematics, specifically order theory, a well-quasi-ordering or wqo is a quasi-ordering such that any infinite sequence of elements x 0 {\displaystyle x_{0}} , x 1 {\displaystyle x_{1}} , x 2 {\displaystyle x_{2}} , … from X {\displaystyle X} contains an increasing...
 
8:41 AM
@Silent Well you have already found all roots. For it to be zero, both terms must individually be zero, and the rightmost term is zero only for $-1,1,3$
 
 
2 hours later…
10:24 AM
[Random]
Topologically discombobulated corsin
 
10:35 AM
13
Q: If you can be "discombobulated", is it possible to be "combobulated"?

Grant PaulI've often heard the word "discombobulated" used. But I've never heard of something being "combobulated", and it's not in any dictionary I've looked at. If "combobulated" is not word, where did "discombobulated" come from?

There exists no continuous deformation that maps between a topologically discombobulated domain to a topologically combobulated one
 
 
2 hours later…
12:49 PM
@samjoe Thank you very much
 
 
2 hours later…
2:35 PM
Discombobulated decomposed topologically disinstituted
 
2:47 PM
What is the asymptotic boundary of a surface in hyperbolic space $H^3$?
 
Update: AKS not dies not exist
 
A standard pack of 52 playing cards contains 13 kinds of card in four different suits.
The pack is shuffled and five cards drawn. What is the probability of drawing ‘four
of a kind’, i.e. the same kind of card from each of the four suits, plus any extra card.
Leave your answer in fractional form N/(52C5)
Can anybody help?
 
@Semiclassical I have been able to adapt that Taylor expansion for my purposes. Thanks again. I'd like to include you in the acknowledgments section of the paper I am preparing. If you agree, please drop me a line with your real name; here you will find my e-mail
 
I got N as 4x58
But thats wrong
Oh wait nevermind I dont understand how cards work
 
I'd say N is 13*(52-4), isn't it?
There are 13 ways of obtaining the four significant cards, and for each of them there are 52-4 possibilities for the fifth card
 
3:38 PM
@Mancala I would guess that it is the "parts" of the surface at infinity. Without context though I am not sure
 
Is someone of you familiar with PDEs? I have posted a question in the main: math.stackexchange.com/questions/2795720/…
 
4:03 PM
Hello all, apologies for such a basic question, but I am proving that a given space V is (or is not) a vector space. It is closed under addition, but multiplication by a scalar is defined to be (c * (x,y)) = (c*x, 0) and I can clearly see that it is not closed, but I am having a hard time identifying which property fails
Oh, no, my brain was malfunctioning, that's all. Apologies for the clutter!
 
o..o
 
My only defense is that I'm a neophyte student of linear algebra... ha.
 
Asking for feedback: Is anything wrong with this (four-year-old) answer? math.stackexchange.com/a/702705/205 I just got two downvotes on it (went from 0 to -2) so although I'm fairly sure it's right and does answer the question, I wonder if there's something I've missed.
 
it is wrong because you are dismissing what he is asking as obvious
he could be asking a legitimate question on how to prove it formally with the definition of limits
 
4:11 PM
@mercio Interesting feedback, what about the answer suggests that I'm dismissing it as a obvious?
 
he also specifically sys to use the definition of the limit
you don't justify the middle equality
 
It's justified on the last line
 
no because he is asking a formal proof using the definition of limits
 
The OP is asking why the last line is true - using the definition of limits (i.e. you want to use some $\epsilon$ s)
 
@mercio @loch Thanks for your time and for sharing your opinion. I'll wait to hear from more others before I consider editing or deleting the answer. Obviously, as I wrote it, it seems fine to me and seems a fully worthwhile answer (and in fact may be helpful to someone who gets mystified by the notation and doesn't realize the question is about the same sequence…), but if enough people think otherwise then the communication has failed and I need to re-word.
 
4:14 PM
OP's problem is that he is not fluent enough when using quantifiers and the definition of limits that he is unable to make very basic proofs like this. And your answer is certainly not helping him with that
 
I think the main thing I wanted to get across is that sometimes one has to forget about the notation and qualifiers and think about what one is trying to prove — the arguments become simpler
 
Of course the statement is obvious :) but if one knows the content one should be able to write it down in terms of $\varepsilon$'s etc. - and I guess the point is OP doesn't know how to
 
I think my point is that one should not prove this using epsilons and deltas, because it obscures the statement being made: what is being said is two notations about the limit of the same sequence. It's not like trying to prove that, say, limit of a_n and limit of (a_n + 1/n) are the same, where we're talking about limits of two different sequences
(of course, if the question was asked as a homework exercise or something it's a different matter; I didn't consider that possibility actually…)
 
Well the problem is what is $\lim a_n$ then? If you're not using the definition
 
and one day OP will think something is obvious and he will just dismiss it as obvious without doing a formal proof and seeing it is in fact not obvious and even wrong
(and they are arguably not the same sequence if youdefine a sequence as a map from N to R)
I don't think I have ever seen someone define a sequence as an equivalence class of sequences under translation and changing a finite number of terms
 
4:26 PM
Interesting (and good points), thanks. I'll think about how to reword this while still retaining the main insight in words, and avoiding notation as much as possible. I'll edit if I can think of something. :-) Until then I'm happy to leave it at -2 or -N
 
4:44 PM
Iwonder what's my lowest scored answer
wow I would have thought I would have more than 4 negative-scored answers
 

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