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8:00 PM
Oh, wait, maybe I misinterpreted and you wanted $2p(x)$. Hrmf.
 
@SteamyRoot what are you trying to tell me? A user sent me a random post and I replied to it...
i dont want anything
i was told to answer it
 
I know. And I'm saying your solution looks wrong.
 
@SteamyRoot what I'm saying is that I really don't care. I'm honestly confused as to why I was even sent the problem to begin with. It seems trivial.
 
Fair enough.
 
@TedShifrin hello
 
8:07 PM
howdy
 
@SteamyRoot would you say this is a duplicate post? Seems a bit of a stretch to me
5
Q: Are all solutions to an ordinary differential equation continuous solutions to the corresponding implied differential equation and vice versa?

TyphonNow I have to heavily emphasize the fact that I have never studied differential algebra or the concept of other types of differentiation (which is what I believe is the concept behind a differential algebra). So, if I am abusing the terminology a little bit, please forgive me. Let us define a di...

 
hi chat
 
heya ERic
 
Hi Ted
 
i must be missing something here?
they're two different conjectures...?
 
8:09 PM
Hi @Astyx ... done with more exams?
 
4 days to go
(and a half on the 18th)
 
how long do your exams last
 
Too long
I've had 2 weeks already
 
oh my god
 
And I'm supposed to take two more weeks
 
8:10 PM
so a marathon as opposed to a sprint like it is here
 
But I think I'll skip the last
Rather a series of 4 sprints over 3 miles
 
oh god sounds awful
 
It is
Luckily I have my morning tomorrow
I'll be able to catch up all the sleep I'm missing
 
im historically a good tester but honestly i hate exams
 
I used to like 'em
Grading exams and listening to panic attacks for the last 40 years was way worse than my taking them.
 
8:12 PM
i reeeaaally like take home exams though
 
Any take-home exams I wrote were way harder than any in-class exam.
 
yeah they gotta be
like i would like it if exams were like really difficult psets
Neves made his final a take home and put really hard problems on it
but i guess the drawback is that it's probably way harder to grade my 15 page take home final than it was to grade a bunch of 4 page exams
 
Exams seemed alright until now
 
indeed
 
I'm very proud of how well i did on that final
i got a 98 and the average was in the 60s
 
8:16 PM
I made my decision about X/ENS btw.
 
plus, except for a few advanced courses, I wanted a final exam to check that students knew the definitions and the basics, not super-tricky stuff like on problem sets.
that's pretty good, Eric ...
 
that's a super fair point @Ted
ive never done as well in any class as i did in Neves Riemannian geo
 
despite my reputation, I tried to be demanding but fair
 
@Astyx what did you decide?
 
you were more into his class than probably most others
 
8:17 PM
that's def true, I was suuuuper into it
i spent lots of my free time thinking about the course material
 
To pursue my maths interest and not care about other people/money
Thus the ENS
 
(that's if I get the opportunity obviously)
 
lol
 
well, ok, at least you done did made a decision
 
8:18 PM
cool
 
i just learned why you don't reuse a paragraph at the intro of two questions
because then lazy people will think they are duplicates
 
what's the curriculum like at ENS
 
Well, frankly, if you have nearly identical question names and they start in a nearly identical way, it's kind of to be expected?
 
i imagine it's really intense
 
no
heck no
the paragraph that is the same is the definition of an operator
everything else is totally different
 
8:20 PM
Make sure that's apparent in the title and you should be fine @Typhon
@EricSilva You have a choice between multiple courses in a broad spectrum of subjects
 
Or write at the top of your question what the difference is between your previous question and this one
And why this merits to be its own question and not an edit of the former question.
 
dude
the question that was closed
is the older one
 
(maths, physics, litterature, economics, philosophy etc)
 
@SteamyRoot that is always a horrible thing to do
you shouldn't have to justify a question
or prove it is different
 
Well, the moderation backlog is also quite the horrible thing.
 
8:22 PM
@SteamyRoot it's not moderation
it's user's closing posts
 
@Astyx i meant more along the lines of: What's the standard stuff all the math students have to do
 
and the user that initiated it hates my guts anyways
chances are they already voted to delete like normal
eye roll
 
@EricSilva That's the beauty of it : you don't have to do anything :p
(unless I'm mistaken)
 
really?
 
I think I could only take litterature courses even though I never got good grades in French in my life and that I come from the scientific door into the ENS
I could also do Design
 
8:24 PM
It doesn't have to be a proof or anything. A short justification "this question differs from question blah blah in that here we limit ourselves blah blah" whatever.
 
at my institution all the math students have to take a year of algebra + a year of analysis + 2 upper division math electives (topology, algebraic topology, PDE, functional analysis, complex analysis) and so on
 
@SteamyRoot exactly
those edits only clog up question
in fact, whenever I see those things I remove them
 
Sigh. Nevermind.
 
also
the newer question does have such a statement
 
@EricSilva That's probably what I'm going to do, but that's only because I want to
 
8:25 PM
oh they have to take a linear algebra class too now i guess, I skipped that
 
and yet the older one was closed
oh and here's something interesting for ya
when it was closed
that statement was edited out by the user that closed the post
 
@Astyx cool
 
in other words, they removed it so it would get closed
 
If you think that's unfair, contact a moderator.
 
i did
 
8:26 PM
idt I have a good sense of what all math students who want to try to pursue academic things should know
 
mods dont do shit
 
My sister is having an existential crisis about math and won't talk to me
Someone please answer her question, I'm letting her talk
 
@Eric "idt" ?
 
i don't think
 
Oh right, that completely changes the meaning :p
 
8:27 PM
@SteamyRoot they were asked a month apart.
 
Anyway, high hopes
 
Hi, quick question. I'm in calculus and when dealing with limits, we often have to simplify or rewrite functions to solve them
 
I probably wont be accepted in the school anyway :p
 
lol @Astyx im only a rising third year undergrad, im still a neophyte who has a lot to learn
i wish you luck on your admissions though
 
I think I'll always be a neophyte who has a lot to learn
Thanks !
 
8:29 PM
@Astyx Oi, don't be so negative! :P
 
But how can a function give me indeterminate form when written one way, but give me a proper limit when written another way?
 
@Typhon Which question is this? I could not find which one you are referring to on your list of questions
 
Shouldn't both forms be identical?
 
Truth is, I have no idea wether I'm among the best that are taking the oral exams.
 
5
Q: Are all solutions to an ordinary differential equation continuous solutions to the corresponding implied differential equation and vice versa?

TyphonRegarding the duplicate. Yes, I know the other one has a lot of shared text, but those were just definitions/setup and I was being lazy. The core questions are still different unless you believe derivatives are weak derivatives in which case you might need to read up on them. I don't know the exa...

3
Q: Are all weak solutions to an ordinary differential equation continuous solutions to the corresponding implied differential equation and vice versa?

TyphonYes,this is very similar to a previous question I asked. That was about normal solutions and not weak solutions. We define the operator known as the implied derivative denoted as $I(f)(x)(g)$ to be: $$I(f)(x)(g) := g(x) \left(\lim_{h\to 0^+} \frac{f(x+h)-f(x)}{h} \right) + (1-g(x)) \left(\lim_{...

 
8:30 PM
For polytechnique I have some idea, for the ENS I have absolutely no clue
 
how many people are vying for admission?
 
@sir what is a "proper limit"
 
They only take 40
 
@SirCumference I'm guessing your problem is about, like, why $\lim_{x \to 0} \frac{x}{x} = 1$ by cancelling out $x$, even though pluggin in $x = 0$ gives $0/0$ ?
 
8:32 PM
Sorry, I mean to actually get the limit and avoid indeterminate form
 
@TobiasKildetoft i posted them. Definitely in the same field, but not the same conjectures.
one is false
 
Yeah, steamy
 
one is questionably true
 
I wrote a page on this, one sec
 
Aren't the two mathematically identical?
 
8:32 PM
However since some chose polytechnique over the ENS, the last called is among the 90 first
 
Technically, they aren't. $x/x$ is undefined in $x = 0$
 
proofwiki.org/wiki/Limit_of_Functions_that_Agree Sir this is the formal reasoning why, but i dont think it gives you intuition
 
@Astyx only 40? how big are these schools
 
87 to be precise
 
But on every other $x \in \mathbb{R}$, they are identical.
 
8:33 PM
nvmd
 
@EricSilva They are very selective (note that I'm only talking about the ENS Ulm, there are 3 others) which partially explains their internationnal success
 
@SteamyRoot are you real good with math and stuff?
i need help writing out better logic
 
The intuitive idea is pretty much that a limit of a function in a point isn't (just) determined by that point itself, but largely by the surrounding points.
 
@SteamyRoot but for example, you sometimes need to rewrite a function into something equivalent, before you can solve it
 
So not very big
 
8:36 PM
@Typhon Dunno. Depends on what area of math.
 
my school has like 5000 people in the undergrad @Astyx, but we accept something like 7% of people
 
I think the proof for either the product or quotient derivative rules requires you to add a term to the numerator, which evaluates to zero
 
so 30000 people usually apply in recent years but like 2000 something get in
 
@SteamyRoot sets/integrals
 
But adding it allows you to cancel things out and find the limit
 
8:37 PM
0
Q: For all ordinary differential equations, does there exist a corresponding integral equation?

TyphonMy second conjecture here does have some logic issues. The problem is that I am not sure how to express the idea of looking at integral equations under equivalency classes based on the solution sets being equal. Please see the comments of the current answer. Such logic is beyond my set theory abi...

 
and then there's attrition
so not everybody comes
 
im trying to use set theory to convey the idea that out of all sets corresponding to integral equation solution sets, there is only one that is a superset of the solution to a differential equation while having the smallest set magnitude.
and I know that has to have at least one loophole. XD
but if someone could please help rewrite the logic so that works correctly and makes sense, please edit my post
 
@SirCumference That might be the "$\lim f(x) + \lim g(x) = \lim f(x)+g(x)$ in disguise, then?
 
There were 1600 persons taking the written exams last year, they only kept 10% for the oral exams, and only 25% of those actually got in
Which explains my stress :p
 
wow
it's very different
 
8:40 PM
Wait, it was the product rule
 
we have no exams but we submit college applications and they get reviewed by like teams of people holistically and honestly it seems like it's mostly just luck if you've got the stats to get in
 
At some point, you add f(x+h)g(x) -f(x+h)g(x), if you are taking the limit as h goes to 0
 
my school is one of the most selective ones in the states but honestly it seems like admissions is like a black box that you dont really undesrtand unless youve seen directly how it works
 
Any admission strategy relies on luck at some point
 
sure
 
8:46 PM
hi chat
 
hi Semi
 
Any good math today?
 
not really
just weird close votes
 
If $f : \Bbb R^2\to \Bbb R^2$ is $1$-lipschitz for the eculidean norm, prove ${1\over n}f^{\circ n}(x)$ converges to a limit that does not depend on $x$
 
@Semiclassical you could probably help me out
0
Q: For all ordinary differential equations, does there exist a corresponding integral equation?

TyphonMy second conjecture here does have some logic issues. The problem is that I am not sure how to express the idea of looking at integral equations under equivalency classes based on the solution sets being equal. Please see the comments of the current answer. Such logic is beyond my set theory abi...

not sure how to fix the troublesome logic
 
8:53 PM
Bye chat
 
hi all
Hi @TedShifrin
I spoke with my supervisor he is awesome.
my new supervisor, it is easy to talk to him and he is very knowledgeable as well.
 
9:09 PM
@Typhon …No
$q(x^2)$ would equal $x^2p(x^2)\ne0$
@Typhon It was only "way too easy" because you solved it wrong
 
Hmmmm. Now I find myself wondering: Given a matrix $M$ with determinant 1, how would I construct matrices $A,B$ such that $M=A^{-1}B^{-1}AB$?
I think we agreed earlier that such matrices generically have to exist, but I'm not sure how one actually obtains them.
 
@Semiclassical I am not certain $M$ has to be a commutator, just an element in the commutator subgroup
 
It does happen to be the case that all elements in the derived subalgebra of a semisimple Lie algebra are commutators though, so it might follow from that
 
A product of commutators is also a commutator, isn't it?
 
9:17 PM
No
 
Not in general, no
 
(I think)
 
Yeah, thinking about a product of commutators there's not a great reason to expect it to be a commutator as well.
 
@AkivaWeinberger honestly, I didn't care much.
why did you even give it to me?
 
9:20 PM
'Cause it was kinda related to stuff we were discussing with the rational powers puzzle
 
fair enough
 
Eh, whatever, doesn't matter
@Semiclassical $abca^{-1}b^{-1}c^{-1}$ seems to be a commutator, though
because it's $(ac^{-1})(cb)(ac^{-1})^{-1}(cb)^{-1}$
(it expands to $(ac^{-1})(cb)(ca^{-1})(b^{-1}c^{-1})$)
 
Hmm, yeah.
But this is a bit afield of what I was after.
Namely: If $M$ has determinant 1, does $M=ABA^{-1}B^{-1}$ for some $A,B$, and how are these matrices found if so?
 
9:45 PM
is there a name or notation for the group of $n \times n$ matrixes with determinant $\pm 1$?
 
$U(n)$, I think, but there's a decent chance I'm wrong
 
$\operatorname{SL}(n,V)$ is $\det = 1$?
where V is a vector space
entries of the matrix
 
(I'm wrong)
 
happens to the best of us
 
10:02 PM
can anyone check my argument to this limit?
seems like it lacks some rigorosity
$\lim_{n \to \infty} \prod_{i=1}^n (1 - \frac{i}{n^2-1}) = 1$
Argument: $1 - \frac{1}{n^2 - 1} = 1$ and $1 - \frac{n}{n^2 - 1} = 1$. But the sequence of the factors is strictly decreasing; therefore the limits of the factors must all be equal. Thus the limit equals to $1^n = 1$
(Note: $i^2 \neq -1$, it's just an index)
tbh it's pretty much mean-value theorem
 
That doesn't work
By the same argument, $\lim_{n\to\infty}\prod_{i=1}^n(1+\frac1n)$ equals $1$, but it really equals $\lim_{n\to\infty}(1+\frac1n)^n=e$.
@LucasHenrique
 
It's not the same argument
Note that $a_1 = 2$ and $\lim_{n \to \infty} a_n = 1$
 
@Semiclassical Some Googling shows that every determinant 1 matrix is in the commutator subgroup (unless your field is $\Bbb Z_2$, in which case it fails for two dimensions)
 
Now I see that my argument is actually correct
 
Wait, why's that different?
It it because the stuff approaches $1$ from below rather than above in yours?
Just take the reciprocal, then
 
10:15 PM
Basically I'm using sandwich theorem but with a sequence
 
$\lim_{n\to\infty}\prod_{i=1}^n(\frac n{n+1})=e^{-1}$
 
It's easy to see that it's a decreasing sequence
 
$1-\frac1{n^2-1}\ne1$
It just approaches $1$ in the limit. But the same is true for my examples
 
Therefore $a_n \leq a_i \leq a_1$ if $ 1 \leq i \leq n$
but as $n \to \infty$, $a_n \to a_1 \implies \lim_{n \to \infty} a_1 = a_2 = \dots = a_n = 1$
(ugh, the notation, jeez)
 
So? The same is true for my thing!
$\lim_{n\to\infty}a_1=1$, etc
Note that the expression for each factor does not contain $i$
 
10:19 PM
Well, isn't this just a direct application of mean value theorem?
 
@LucasHenrique this
 
You have two competing forces. One is that each term goes to $1$. The other is that the amount of terms grows.
You don't know which one wins.
 
So my argument is wrong. Meeh.
 
(I don't doubt that $\lim_{n\to\infty}a_1=\lim_{n\to\infty}a_2=\dotsb=1$. I'm just saying that doesn't prove it.)
 
10:21 PM
Either way, the product is, in fact, 1. How do I prove this?
@AkivaWeinberger Isn't the limit of products the products of the limits?
If the limits converge, obviously
 
@LucasHenrique That's only if the number of factors doesn't change
Like, you have $\lim AB=\lim A\lim B$
 
what's the product-sequence we're analyzing? I'm late to the game
 
22 mins ago, by Lucas Henrique
$\lim_{n \to \infty} \prod_{i=1}^n (1 - \frac{i}{n^2-1}) = 1$
He was arguing that since each individual term goes to $1$, so must the product
I countered that $(1+\frac1n)^n\to e$ provides a counterexample to that principle
 
both limits of multiplication in the multiplicand? that's weird
 
Also there's a thing here (I think it won't change the result): it is $i-1$, not $i$
 
10:27 PM
What do you mean?
 
what about fixing $n = N$, setting $y = \prod(1-\text{stuff})$, and then taking $\ln$ of both sides?
then you have $\sum_i^N \ln\left({1-\frac{i}{n^2-1}}\right)$
 
Since we're all into this, the original problem was "prove that $u_n = (1 + \frac{1}{n^2})(1+\frac{2}{n^2})\dots(1+\frac{n}{n^2}) \sim \sqrt{e}$"
wait, where is the tilde
 
\sim
hm, that doesnt help very much, does it?
 
Oh I think I see how to do that possibly
Wait no never mind
Oh, uh, wait I might have it again
I have to think for a bit
But I'm thinking a sort of "Gauss trick" works here
 
my first approach was to write $(1+\frac{i}{n^2})$ as $(1+\frac{1}{n^2})\dot \frac{1 + \frac{i}{n^2}}{1 + \frac{1}{n^2}}$
jesus christ what happened here
Can you guys understand it ?
 
10:34 PM
like you know how, to sum $1+2+\dotsb+100$, you rearrange it into $(1+100)+(2+99)+\dotsb+(50+51)$
 
$\displaystyle \left({1+\frac 1 {n^2}}\right)\cdot \frac{1 + \frac i{n^2}}{1 + \frac 1 {n^2}}$
 
@AkivaWeinberger Yes, I know this trick
 
there we go
 
@GFauxPas this
 
it was the \dot
 
10:36 PM
I'm not a god with LaTeX
 
you wanted \cdot
 
Oh, so DogAteMy hasn't vanished!
\dot puts a dot over the letter
 
tbh I'm just a dumb high schooler that wants to be a mathematician
 
just in time to help us, Ted
 
Yeah I don't actually know if I have the necessary self-control to leave this place @TedShifrin
 
10:37 PM
It did occur to me to wonder, DogAteMy.
 
7 mins ago, by Lucas Henrique
Since we're all into this, the original problem was "prove that $u_n = (1 + \frac{1}{n^2})(1+\frac{2}{n^2})\dots(1+\frac{n}{n^2}) \sim \sqrt{e}$"
 
So you need to speak to us only in Spanish?
 
@AkivaWeinberger can't my argument be proven using induction?
 
Uh… tal vez…
 
that got us $\displaystyle \ln y = \sum_i^N \ln\left({1−\frac i {n^2-1}}\right)$
 
10:38 PM
@LucasHenrique How?
 
wait no
 
I would take logs immediately.
Then I should be able to recognize it as a Riemann sum for some integral ... quite likely.
 
Ooh, probably.
I should work through my Gauss trick idea, though, because if it works, I'll feel really clever
 
You're usually far cleverer than I, DogAteMy.
 
not a Riemann sum at first glance. could be I made a silly mistake though
locked out of editing before I couild finish cleaning it
 
10:41 PM
So it's $[(1+\frac1{n^2})(1+\frac n{n^2}][(1+\frac2{n^2})(1+\frac{n-1}{n^2})]\dotsb$
 
$\ln u_n = \displaystyle \sum_{i=1}^N \ln \left({1 + \frac i {N^2}}\right)$
 
Yeah, the $n$ versus $n^2$ is disquieting.
 
Hey everyone!
 
I guess I can assume $n$ is even and equals $2k$
 
Hi Demonark.
 
10:41 PM
@Daminark Hi!
 
How's everything going?
 
there we go, pretty sure that's right
 
Let $n$ be the number of functions in the product $\lim_{x \to a} f_1(x) \cdot f_2(x)\dotsb f_n(x)$. Suppose by induction that it works for $n$. Then for $n+1$, it follows from case $n=2$ that $\lim_{x \to a} f_1(x) \cdot f_2(x)\dotsb f_{n+1}(x) =(\lim_{x \to a} f_1(x) \cdot f_2(x)\dotsb f_n(x)) \cdot \lim_{x \to a} f_{n+1}(x)$. I think you can see my point
 
Or actually, better, $n$ is odd and equals $2k-1$
 
I don't think you can get a proof like that, @LucasH. For one thing, you'd need to know the limit exists.
 
10:43 PM
what about making it a recursive sequence and resolving the recurrence
 
Then $(n)(1)=k^2-(k-1)^2$, $~(n-1)(2)=k^2-(k-2)^2$, etc
 
@GFauxPas I thought that but I couldn't build it
@TedShifrin assuming all the limits do exist
 
$u_{n+1} = \left({1+\frac{n+1}{n^2}}\right) u_n$?
 
and it ends at $(k-1)(k+1)=k^2-1$ and then just a single $k$
 
@GFauxPas no
 
10:44 PM
@Ted Is it possible that tomorrow at some point I could take you up on the whole, practice lecture over Skype thing?
 
This thing converges really slowly to 1/2 ...
oh, sure, Demonark ...
 
So it's $(1-\frac{k^2-(k-1)^2}{4k^2})\dotsb(1-\frac{k^2-1}{4k^2})\cdot(1-\frac k{2k})$
 
@GFauxPas the denominator $n^2$ becomes $(n+1)^2$ for each factor
 
Like what time, Demonark?
 
If not it's alright, but I have now signed up to lecture on Wednesday (right now I'm still trying to make sure I have internalized the stuff)
 
10:46 PM
oh
 
I've got no specific time in mind, preferably not 12-2 or 3-4:30
 
Well, I can't do anything past 7 PM your time.
 
Hmm, given that it's about a one hour lecture, maybe 5 is good?
 
OK, I'm cool with that.
I'll try not to interrupt too much.
 
Oh wait that was stupid let me try again
What's $(1+\frac1n)(1+\frac n{n^2})$?
 
10:49 PM
Given how the lectures have been so far, interrupt as much as you see fit
 
It's like $1+\frac{n+1}{n^2}+\frac n{n^3}$, yeah?
 
Something like $(n+1)^2/n^2$?
 
There are definitely a good number of them, and if nothing else it'll help with pacing perhaps (some people before did not account for a lot of questions/comments, so were in a bit of a rush or didn't get through everything)
 
Demonark, you always have to allow for interruptions and some confusions, yup.
So typically if you prepare a half-hour lecture it fills an hour :P
Neophytes think they can cover way more than they actually can.
 
@TedShifrin Oh my god, this is beautiful
 
10:51 PM
But it's good to have more prepared than you'll cover, just in case.
 
Ted, you just solved it
 
And then $(1+\frac2{n^2})(1+\frac{n-1}{n^2})$ is
 
No, DogAteMy solved it.
 
$1+\frac{n+1}{n^2}+\frac{2n-2}{n^4}$
 
But he and I are disagreeing. Which one of us is right?
 
10:51 PM
Who's DogAteMy? Hahahah
 
looks up I did what?
@LucasHenrique Me
 
Oh, sorry hahah
I'm new here
 
I got a different answer than you did for that product, DogAteMy. Which one of us is right?
 
Oh, I wrote it wrong
It should be $n^2$ on the denominator on the left
$(1+\frac1{n^2})(1+\frac n{n^2})$
 
I still am off by a 2 ...
Oh, crap.
Now let's see ...
Now it's a mess.
 
10:53 PM
What? Just FOIL, don't simplify fractions
 
Now I get $\dfrac{(n+1)(n^2+1)}{n^3}$.
 
Crap
$(1 + \frac{j+1}{n})(1 + \frac{n-j}{n})$ is a function of j
 
There are still $n^2$'s in there.
 
$1+\frac{n+1}{n^2}+\frac n{n^4}$
which could be equivalent, I dunno
 
Yes, now we're agreeing on that, DogAteMy.
Oh, we're being dopey, I think.
This is way easy.
Dopey, dopey, dopey.
 
10:58 PM
And then $(1+\frac k{n^2})(1+\frac{n-k+1}{n^2})$ equals $1+\frac{n+1}{n^2}+\frac{nk-k^2+k}{n^4}$
 
Use $\log(1+x)\approx x$, and it falls out immediately. Then we only need to justify ignoring higher order terms (which is easy).
 
So if we can ignore that last term, this is essentially $\approx(1+\frac 1n)^{n/2}\to\sqrt e$.
 
$(1 + \frac{j+1}{n})(1 + \frac{n-j}{n}) = \frac{2n^2 + 2n + nj - j - j^2}{n^2}$
 
$$\sum_{k=1}^n \frac k{n^2} = \frac{n(n+1)}{2n^2}.$$
 
But I'm thinking this is probably not the intended solution.
 

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