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21:00
Now write it as (1+1/n)/ (1/n) and use lh
hey, also try expanding it to a convergent series
i think what i did works as an analysis type aproach
im just not sure i can ln the limit
Why do you introduce some random $c$?
it is random i dont know what it is just that it exists
$$\lim_{h \to 0} \frac{\ln(1+h) - \ln(1)}{h} = \left.\frac{d}{dx}\ln(x)\right|_{x = 1}$$
21:02
@SteamyRoot because we felt like it
i then prove it has to be 0
lol yes im using that fact but how do u know it has to be x=1
Because that's the definition of the derivative
The derivative at $x = a$ of some function $f(x)$ is $\lim_{h \to 0} \frac{f(a+h)-f(a)}{h}$
i do know that
Then you also should clearly see that in our case $a = 1$ ?
@BalarkaSen which uni?
21:06
Technically I'm not in one yet. Hopefully next year.
How to solve this question?
For $x \to 0$, do left and right limit separately.
For $x \to 1$, there should be no problem.
@fpqc Looks like you study algebraic geometry. That's one of the things I know nothing about (except the ez things I learnt from a month of reading Shafarevich part I). :P
@SteamyRoot I got 3 for both left and right
@SteamyRoot That's the problem
If left and right limit exist and equal eachother, that's the limit.
21:09
@SteamyRoot I got 3 as the limit for x tends to zero
@Daminark you awake bruh
If $x \to 1$, then you only need to consider what happens when $x$ is close to $1$. In that case clearly $x > 0$.
@Abcd it must be continuous function otherwise they wouldn't ask you to find the limit. Therefore, they must be f(0) and f(1).
Oh!
I understood my mistake.
I was foolishly calculating 0 (then 1) for both sub functions of the function; to yield 4 (3) values
Anyway, thanks for correcting me.
@SteamyRoot i finally understand i think i was trying to use the mean value theorem or something stupid. thanks
21:14
@abcd the function is continuous. It must be.
But here: there's no sub function value matching the x tends to
otherwise, they wouldn't give it to you.
@Typhon Which question are you talking about?
both
21:15
@Typhon That kind of reasoning will get you nowhere.
Could anyone help me with the 2nd one?
if the function is not continuous the limit doesnt exist
therefore, it must be continuous?
How to solve the 2nd one?
@Abcd 0?
f(0) = 0?
These are the kind of exercises you get before seeing a rigorous definition of continuity.
21:16
@Typhon No
No subfunction value matches the x tends to
huh?
If you don't have any proper help to offer, then please don't.
limit only exists if the function is continuous?
Therefore limit doesn't exist
@SteamyRoot i am helping. Why are you being rude.
21:17
well yes f(0)=0
do you know the concept of a right and left hand limit
@Abcd Again, try to work with left and right limits. It can be helpful to try draw the function.
Once you have left and right limit, compare to eachother and to the value of $f$ at $0$.
@SteamyRoot But my answer matches with the one in the answer key
And your answer is?
Limit doesn't exist
Indeed.
21:18
Actually @typhon was right.
But you still need to prove taht.
Function is discontinuous
@Faust7 Is that when you write the arrow backwards?
So limit doesn't exist
why is it discontinous?
21:19
Sigh...
thats important to get the question right
You can't just say "it's discontinuous".
@Faust7 There's no value assigned for less thanzero
You're calculating this limit exactly to prove that it's discontinuous.
not correct
there is
21:20
too many people at once.
@Semiclassical let us jump in too
what is $ \frac {x}{x} $
sure, let's go with that.
@Faust7 1 for all real x; x is not equal to infinity or 0
the function is 1 everywhere but 0
at 0 it is 0
21:21
then what is $\frac {-x}{x}$
therefore, it is discontinuous at 0 and the limit does not exist
duck please
x/x is not defined at 0. that's the point.
Guys something wrong happened with my chat
stop man
21:22
It extended to full screen'
@Typhon it is not 1 eveyrwhere
How do I reset it?
It doesn't feel normal this way
@Abcd Which browser are you using?
Chrome
F11 is often the fullscreen button, I think?
21:23
@Abcd if you want an inconvenient answer: press and hold down the power button on your device
Something different has happened steamy
what?
I can't press the reply button
ive actually had to close stuff down that way before
21:23
yeah, but that's not where you start.
I mean it's not exactly full screen but extended screen
press control alt delete?
@abcd you might be zoomed in. if so, there'll be a magnifying glass at the right end of your address bar.
No semi
if you click that, you'll be able to reset to the default zoom. (if there's no such icon, then that's presumably not the problem.)
nuts.
21:24
@Abcd control-0
@Abcd if x>0 the $\frac {x}{|x|}= \frac {x}{x}$ if x<0 then $ \frac {x}{|x|} = \frac {-x}{x}$ where x>0
Let me delete history then try
no!
that is irrelevant
i doubt that'll help, yeah. (regarding the chat screen, I mean)
Thats all the hint im giving >.>
21:26
@Faust7 so the limit doesnt exist due to discontinuity all around?
Done
it's fine again
ah, good.
what did you do?
@Faust7 ?
@Typhon Deleted history, re-logged in
21:27
well it is continuous on $(0,\infty) $ or $(-\infty,0) $
Now how to solve that question?
no it isnt
it isnt continuous at infinity
lol
You can debate about this later :(
Sigh...
21:28
^^
what?
@Abcd if you scroll up to my last post with your name highlighted it should help you understand why the limit duoes not exist
@SteamyRoot How do I write the solution to this problem? (I understood it but I mean how to write it on paper)
to clarify, are we talking about x/x or about x/|x| ?
I understood that it doesn't exist.
21:29
@Abcd the left hand limit is -1 and the right hand limit is 1. Therefore since they are not equal, the limit does not exist.
@Semiclassical latter
just evaluate the left and right hand limits
In order for a limit to exist, you need that the left and right limit exist, and that they equal eachother and that they equal the value of the function
@SteamyRoot NO
21:30
the last one is only a requirement of continuity
I am talking about this question
@Semiclassical $ \frac {x}{|x|} $
If you know what the function is for $x < 0$ and $x > 0$
the function need not be continuous at 0 to have a limit
Then you can also check the left and right limits and show they don't equal eachother.
21:31
@SteamyRoot What to write on paper? I understood the solution.
3
Q: The Alaoglu's Theorem

Answer LeeI am trying to prove the Alaoglu's Theorem. But my professor told me there is something wrong with my proof. Can anyone help me? Thank you! The Alaoglu's Theorem: Let $X$ be a normed space. Then ball $X^*$ is weak-star compact. Proof:Let $D_x=\{\alpha\in\mathbb{F}:|\alpha|\leqslant1\}$ for each...

I cannot find an error with this proof
Moreover, i can't substitute
and prove
in an ideal chat moderation system these comments would be considered disruptive to the conversation and you would be banned
I would write down left and right limits, and then write a line "they don't equal so limit doesn't exist".
because there's no value of 0
21:31
There is a value at $0$ ?
@Abcd "the left hand limit is ____. The right hand limit is ____. Since they are not equal, the limit does not exist."
@Balakra o.o my post?
No, of course not your post
Your question literally says "$0 \;\;\;\;\; x = 0$"
lol
21:32
How can you write LHL/RHL
when 0 doesnt equal either value in the function
@BalarkaSen me???
@SteamyRoot :p. Sorry.
The left hand limit only depends on the "$x < 0$" part of your function.
The value of the function at zero is irrelevant to whether the limit at 0 exists. (at least according to the definition of limit taught in the US. The french have a different definition, I think?)
@BalarkaSen Yeah I work in moduli theory.
21:34
@fpqc Ah, fun stuff
Yes.
It's very stacky
Depends on what you read though
The kind of moduli spaces I am most familiar with are topological in nature
@Semiclassical I think that the french (and hence belgians) sometimes define the limit as just "left and right limit exist and equal", without caring about the value of the function at that point. At least that's how I learned it in school.
I don't know the algebraic side of things
I mean surely you care about the moduli space of riemann surfaces of genus
21:35
alright gunna pack n switch computers bbl
I do, I just don't know a lot about it :)
@BalarkaSen why do you think i should be banned?
@SteamyRoot hmm, maybe it's the french definition of continuity.
I know there's some issue that comes up.
Well, once you see epsilon-delta those school-definitions are abolished, of course.
21:36
@SteamyRoot what is epsilon-delta?
@BalarkaSen I think it's on of the triumphs of 20th century math, along with say the proof of the weil conjectures and fermat's last theorem
huh I see
@fpqc then why haven't I heard of it?
@BalarkaSen Of course in other fields too there are many great achievements
@SteamyRoot if memory serves, the issue extends to the epsilon-delta definitions as well.
21:39
Oh I understood that you meant in algebraic geometry
Really?
@BalarkaSen Sure.
this chat is getting way too advanced for me
D:
@BalarkaSen Even in number theory too.
Well, maybe they stick to that other definition in France then. I learned the french definition in school, but in university we saw the "international" definition.
21:40
@BalarkaSen If you care about topology, maybe you'll care about \'{e}tale cohomology.
Basically Andre Weil realized that if we could have a "Weil Cohomology Theory", i.e. something that behaves like singular cohomology for varieties over a finite field
english uses $\epsilon>0$, french uses $\epsilon\geq 0$?
@fpqc I have been to a small talk on \'etale cohomology and heard it thrown around at places. It definitely intrigued me.
@BalarkaSen We would be able to prove the functional equation and rationality of zeta
@BalarkaSen The problem with learning etale cohomolgoy is that you need to know algebraic geometry. The proofs are often quite complicated and involve a lot of devissage
@Semiclassical Oh, hmm, not sure. I've never seen the french definition with $\varepsilon$
The thing which attracts me the most about Weil cohomology theories is that it satisfies the Lefschetz hyperplane theorem
yeah I dunno how it's constructed
This was this year's seminar on etale cohomology
@Faust: It's just continuity of the exponential that justifies it.
woo cool stuff
The proof of Grothendieck-Lefschetz is a long devissage using poincare duality, etc
21:43
of course, with my connection being an utter arse at the moment I'm having a hard time finding anything useful
@TedShifrin Hello!
@BalarkaSen At some point there's various annoying commutative diagrams that one has to check are true. And no one knows except Brian, de Jong and Gabber
Random fact: Lefschetz had no hands.
@fpqc lol
I'm looking at the notes right now
My talk on was on april 12th
I have only been around here for 3-4 years, @fpqc. Did we meet at Stanford?
21:45
@TedShifrin No. We met here!
I work in algebraic geometry.
Oh, you gave a talk there too? Noice
OK. I visit Stanford sorta regularly. Rafe Mazzeo is a former student and longtime friend.
@TedShifrin Really? Rafe taught the second course in the grad analysis sequence
@TedShifrin Wait wasn't rafe a student of melrose?
Now that I retired and live in CA I try to drive up once a year.
@BalarkaSen Yes. I'm in AG but also in the arithmetic things too.
21:48
Yes. He was a undergrad student of mine and wrote an undergrad thesis with me.
ahhhh
damn man
you probably know andrasz vasy too
He took diff top from me and then learned diff geo and wrote a thesis w/ moving frames.
hrinking tubes and variants of the Gauss-Bonnet formula
Hmm, I don't recognize that name.
@TedShifrin it was his undergrad thesis
@TedShifrin I think the quality of math.se has really gone down honestly
Used to be a lot of high quality stuff
21:50
Rafe's? Yeah, volume of tubes.
There's still some good stuff, but not much.
it's infested with homework help seekers
yeah
It crowds out people like me
plus people who just give them help
we're forced to go to MO
Yeah they want the reputation
21:52
I've answered real questons here ...
MO is a really nice crowd; i wish i could communicate to anybody in there though
@SteamyRoot Found this remark of Astyx from a while back
Nov 9 '16 at 19:56, by Astyx
I've been taught that a function has a limit at $a$ iff it is continuous at $a$
which, speaking of, that homotopy theory chat tho
So it's that kind of thing.
I hope Astyx wasn't taught that! Damn French!
21:54
Whelp, that's terrible.
huh I was under the impression that French definition of continuity would be something like, preimage of open sets are open
Bourbaki 101
alright guys
I'm going to answer a math.se question on the blow up
+1
blowup is good
see ya
blowups are great conversational material when boarding a plane :^)
21:58
@BalarkaSen The only real blow up calculation I've done is the one in Deligne-Mumford :)
It's the only one I really care about
what's a blow up
I just know it's complex algebro-geometric relevance :P
Guys, this is a proof I’ve never been able to grasp properly. So I’m guessing they mean that
$$
\left\vert
\frac{(g\circ f)(x_n)-(g\circ f)(a)}{x_n-a}
\right\vert=
\left\vert
\frac{g(y)-g(f(a))}{y-f(a)}
\right\vert\cdot
\left\vert
\frac{f(x_n)-f(a)}{x_n-a}
\right\vert.
$$
I don't really understand why they take $y\in I\setminus\{a\}$, instead of $y_n$? I understand that they kind of want to avoid dividing by $y_n-f(a)=0$, but why are we allowed to write the above then?
21:59
@EricSilva Take a complex n-dimensional manifold. Connected sum with \bar{CP^n}
you have ze blowup
oh ok i see
good name

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