Mathematics - 2016-02-15 (page 1 of 4)

Ted Shifrin

12:00 AM

Enigma

Akiva Weinberger

They made a nice movie about that.

L33ter

@BalarkaSen if I work through the problems in hatcher and munkres would you like to check my proofs and providing feedbacks ?

Ted Shifrin

You can only brag if they're done well, Balarka.

skull petrol

The "right" of bragging?

Balarka Sen

@L33ter I can help depending on what the problem is.

Akiva Weinberger

12:03 AM

My problem happens to be related to plumbing, at the moment :(

Balarka Sen

@TedShifrin I think I have done them well :) I will write the proofs up after I finish up revision to chapter 6, which should get done by tomorrow.

Ted Shifrin

Don't forget the extra chapter 6 problems ...

Balarka Sen

Yep, I haven't.

Oh, and remember I asked you whether if a multivariable function $f$ admits a 2nd order Taylor polynomial then $f$ is $C^2$? I think the answer is yes.

L33ter

which book is that ?

Balarka Sen

It seems like it's really a corollary of the fact that Hessian is the same as derivative of $Df : \Bbb R^n \to L(\Bbb R^n, \Bbb R) \cong \Bbb R^n$.

@L33ter "Multivariable Mathematics".

Ted Shifrin

12:07 AM

I vote no, Balarka ... In any event, you only mean C^2 at the point ...

L33ter

oh yeah that reminds me I will buy that book btw @TedShifrin

Ted Shifrin

I'm flattered, Karim, but I'm not trying to make you spend money.

Balarka Sen

Yes, sorry, $C^2$ at the point. Well, note that by assumption $f(a + h) = f(a) + Df(a)h + 1/2 h^T \cdot Hf \cdot h + o(h^2)$. Now differentiate both sides w.r.t $h$ to get $Df(a + h) = Df(a) + Hf \cdot h + o(h)$. That says $Hf$ is the derivative of $Df$, doesn't it? So it's the Hessian matrix.

Ted Shifrin

Still doesn't prove the second derivative is continuous?

Balarka Sen

Oops, I didn't mean $C^2$. I meant the second partials exist. My bad.

L33ter

12:10 AM

I want to learn multi-variable analysis and this semester I am marking for vector calculus, so it makes sense to use the money on your book :D .

Balarka Sen

Of course, $C^2$ should be dead wrong even in the single variable setting.

Ted Shifrin

It is ...

Karim, there's lots for you to learn ...

Balarka Sen

Many continuations today, @Ted.

Ted Shifrin

Analytic continuations?

Balarka Sen

I bet some sin(1/x) type thing gives me a counterexample.

Ted Shifrin

12:14 AM

I bet that's right.

Balarka Sen

What's the point of wagering if there are no opponents? :)

Brennan.Tobias

can someone help me with analytic number theory please?

Balarka Sen

I mean, we're both betting for the same thing ...

Ted Shifrin

@Brennan: No number theory experts here ...

Brennan.Tobias

ok thanks anyway

Balarka Sen

12:16 AM

@Brennan.Tobias Ping mixedmath, although he's rather irregular in here.

@TedShifrin Well, we're essentially just looking for double differentiable functions such that 2nd derivative is not continuous, really.

L33ter

wait @BalarkaSen

your argument is wrong

Balarka Sen

So $x^3 \sin(1/x)$ for $x \neq 0$ and $f(0) = 0$ works fine.

L33ter

I just thought about it

you constructed the U and V to be disjoint

it can be modified to be right though

Balarka Sen

@L33ter Where is it wrong?

L33ter

You constructed them to be disjoint, so $X - (U \cup V) = (X - U) \cap (X - V) = \emptyset$

which is open

If you modify it however, so you will get the same argument that I have.

Balarka Sen

12:22 AM

@L33ter Why is it true that $U^c \cap V^c$ is empty??

Ted Shifrin

A better quesuon, Balarka: If $f(x)=o(x^2)$ does that mean $f''(0)=0$?

Balarka Sen

$[0, 1]$ and $[2, 3]$ are disjoint in $\Bbb R$. But their complements intersect.

Indeed, $U^c \cap V^c = (U \cup V)^c$ which is $A$, not empty.

This is set theory.

L33ter

yeah I am sorry my mistake my brain got confused with something else.

Somehow I recalled that you constructed V and U to be disjoint infinite sets of X.

Balarka Sen

Indeed. Draw pictures, you think too formally and end up being confuzzled.

@L33ter U and V are disjoint. But that doesn't mean their complements are too.

L33ter

Yeah I think to analytically I need to stop that

Balarka Sen

12:25 AM

@TedShifrin Hum. Let's see.

L33ter

No, I mean I thought how you constructed U and V is that you started with infinite disjoint subsets of X. I forgot about A.

Ted Shifrin

Not analytically, Karim. More formally/symbolically.

L33ter

yeah

Mike Miller

@TedShifrin What is $f$?

L33ter

One thing my topology professor told me to do is that whenever I have argument I started doing something wrong, then I should see where it leads me to get better understanding of what I did wrong.

and also understand more

Ted Shifrin

12:28 AM

$f\colon \Bbb R\to\Bbb R$, say, @MikeM. We're discussion whether having a Taylor polynomial at $0$ implies the appropriate number of derivatives exist at $0$.

Mike Miller

It's just any function of sets?

Ted Shifrin

Huh?

Mike Miller

You've told me nothing about what you're assuming about $f$.

Balarka Sen

It's twice diffable. Otherwise $f''(0)$ doesn't make sense.

Ted Shifrin

NO, @Balarka.

Balarka Sen

12:30 AM

Edited.

Ted Shifrin

I'm assuming $f(x)= o(x^2)$ at $0$. That's all.

I'm asking whether that even implies $f$ is twice differentiable at $0$.

Mike Miller

Which implies continuous at 0, but a priori, no other assumptions. OK.

That's what I was looking for.

Balarka Sen

@TedShifrin Ar, alright.

Ted Shifrin

I had that in the question you linked, @MikeM, but ok.

Mike Miller

@Ted: It was not clear that you weren't assuming anything further.

Balarka Sen

12:31 AM

@TedShifrin I doubt this.

Ted Shifrin

Are we betting? :)

Balarka Sen

You bet.

[disclaimer] that was an intentional pun [/disclaimer]

Just to make sure this time I am recognized.

Ted Shifrin

That's not even a pun.

Using the same word in the same context does not qualify as a pun.

Balarka Sen

Well, there was a missing "ing".

Ted Shifrin

glares

Balarka Sen

12:53 AM

@TedShifrin Hm. I can't seem to find a counterexample. If you give me till tomorrow, I can come up with a counterexample (or a proof, though I highly doubt if it's really true).

It's about 6 AM here and I am pretty sleepy :)

Ted Shifrin

You were supposed to be in bed hours ago.

Balarka Sen

I am naturally nocturnal.

Ted Shifrin

You should change your name to The Owl.

Balarka Sen

Wait a minute, I am confusing a few notations. $f = o(g)$ means $f(x)/g(x) \to 0$ as $x \to 0$ or $f(x)/g(x) \to 0$ as $x \to \infty$?

Ted Shifrin

We're talking about at $0$, of course.

Balarka Sen

1:00 AM

Right, phew.

Ted Shifrin

It can mean either, depending on context.

Balarka Sen

Of course, the latter doesn't make much sense in the given context. This means I should head to bed.

Uh huh.

G'night.

G'night.

2:13 AM

I finally got my Cayley table right, I think wow! It's actually quite simple once you get it.

Sanic Hodgeheg

4:04 AM

Hello!

skull petrol

4:25 AM

hi

Sanic Hodgeheg

4:46 AM

When proving the limit of some sequence in the form, $\frac{p(n)}{q(n)}$ I can just prove the limit of a fraction with the upper bound of $p(n)$ as the numerator, and the lower bound of $q(n)$ as the denominator. Are there any constraints as to which upper/lower bounds I can choose?

Mambo

hello

skull petrol

hi

Mambo

5:02 AM

do you have knowledge of topological groups

Albas

What's your question @Mambo

Mambo

Consider a square integrable function on circle

haar measure is considered

Say f

Let $A = \{f_g : f_g (x) = f(g^{-1}x , g \in \mathbb{S}^1)\}$

Albas

Sorry , but I have no knowledge of haar measures. You can ask Daniel Fischer I guess.

Mambo

no problem. Do you know any measure on circle

Albas

5:18 AM

Actually I just did intro of topological groups from munkres. So I don't know much. I just thought of looking at the question @Mambo

Mambo

What do yo do

@Albas

Akiva Weinberger

So, I know that, in $\Bbb R^4$, one can link two spheres (just like one can link two loops in $\Bbb R^3$). What would the "3D shadow" (i.e. the projection onto three coordinates) of that look like?

Nasser

6:46 AM

I am doing a HW: is polytope the same as convex hull? i.e. given set of generators. The problem says P is polytope of these generators. Is this the same thing as saying the convex hull of the generators?

Riggs

7:42 AM

Does a "approximately equal" to b mean a is close to b but not strictly not equal to it? Or does it mean a is close to b and may be equal to it?

Nasser

8:01 AM

@Riggs I would think it means close to b but not strictly equal to it.

Riggs

thanks

skull petrol

I agree

Approximation

An approximation is anything that is similar but not exactly equal to something else. The term can be applied to various properties (e.g., value, quantity, image, description) that are nearly, but not exactly correct; similar, but not exactly the same (e.g., the approximate time was 10 o'clock). Although approximation is most often applied to numbers, it is also frequently applied to such things as mathematical functions, shapes, and physical laws. In science, approximation can refer to using a simpler process or model when the correct model is difficult to use. An approximate model is used to...

Balarka Sen

9:20 AM

@AkivaWeinberger I am no knot theorist, but I'd think in general the projection need not even be an embedding of disjoint union of $S^1$'s in $\Bbb R^3$, let alone a link. There are things called spun knots, though, for which I believe the slices are all knots.

Danu

9:33 AM

Hi @BalarkaSen

Funny question from intro to analysis: Do you know any sequences in the extended reals such that $\lim (a_n b_n)\neq \lim a_n \lim b_n$?

There has to be some canonical example

Balarka Sen

10:18 AM

@Danu I haven't really worked with extended reals, do you have $+\infty \cdot 0 = 0$ there? In that case, $a_n = 1/n$ and $b_n = n$ works of course.

Actually, $a_n = 1/n$ and $b_n = n$ works anyway, because $\lim a_nb_n = 1$ and $\lim a_n \lim b_n$ wouldn't be defined otherwise.

However, I am certain it is true that if $\lim a_n, \lim b_n$ are both finite, or one is finite and the other is $\pm \infty$ or both are $\pm \infty$, then $\lim a_nb_n = \lim a_n \lim b_n$.

Danu

11:02 AM

Yes, yes

Albas

@Mambo I am a high school student

Danu

@BalarkaSen I of course already thought about the $n^{\pm 1}$ example, but I wasn't sure how to really prove that this cannot be given any meaning---though there may be nothing deep to it.

Even simpler is then $0$ and $n$

Balarka Sen

@Danu There is nothing to prove. It depends on how arithmetic works on the extended reals, which I am unfamiliar with. Either way, if $\pm \infty \cdot 0 = 0$, or $\pm \infty \cdot \infty$ is undefined, in both cases the example works.

(Typo, I meant $\pm \infty \cdot 0$).

Tobias Kildetoft

@BalarkaSen I don't think arithmetic works at all in the extended real actually (or rather, we lose too many of the basic rules to call it arithmetic)

So however you define the multiplication of stuff involving infinity, you can find sequences where the product fails to converge to the product

Silent

12:38 PM

Please give an example of partial ordered set with unique maximal element which has no maximum element

Balarka Sen

Uh oh I have forgotten how to prove the inverse function theorem.

Tobias Kildetoft

@Silent consider adding an extra element to the integers, and let it be not related to any integers.

Silent

@TobiasKildetoft, oh !!! Thank you so much.

@TobiasKildetoft, but isn't that extra element greater than every integer and hence maximum?

Balarka Sen

That extra element isn't related to any integer.

Tobias Kildetoft

@Silent No, as it is not related to any integers, it is not greater than any of them

Silent

12:47 PM

Oh sorru, I thought of $\omega_+=\omega\cup\{\omega\}$

Balarka Sen

No need to apulugize.

Silent

:)

iwriteonbananas

1:42 PM

hey @BalarkaSen

Balarka Sen

@iwriteonbananas Hi.

iwriteonbananas

@BalarkaSen Have you discovered anything new regarding the Hawaiian problem?

Balarka Sen

Nope, I haven't thought about it.

iwriteonbananas

Ok, I have but with no success. I'm gonna post a question

Balarka Sen

Sure, as you wish.

iwriteonbananas

1:47 PM

Btw. I randomly remembered the problem that Mike gave me a while ago: determine the smallest number which is a degree of a map $\Bbb RP^3\to L(3,1)$

Managed to figure it out now

answer is 3.

Balarka Sen

We had solved it, I think.

iwriteonbananas

No, we couldn't figure out whether there exists a degree 2 map

Balarka Sen

Wasn't that done by just looking at $\pi_1$?

Poincare duality plus that.

iwriteonbananas

No, that was for degree 1.

Balarka Sen

If you say so. I am not going to think about it right now because I am doing calculus :)

iwriteonbananas

1:48 PM

sure, don't wanna drag you into it then

Balarka Sen

2:00 PM

@iwriteonbananas damn you got me curious. what's your proof?

iwriteonbananas

:D

$\pi:S^3\to L(3,1)$ be the covering map

Assume there's a degree 2 map $f:\Bbb RP^3\to L(3,1)$

the composition $S^3\to \Bbb RP^3\stackrel{f}{\to}L(3,1)$ can be lifted

Balarka Sen

Oh, that's a cute idea.

iwriteonbananas

call the lift $g$

then use multiplicativiy of degree and we get

$4=3\cdot \deg(g)$

Balarka Sen

Right, which is crap.

iwriteonbananas

yep.

Balarka Sen

2:03 PM

Very nice.

I did wondered if lifting would help, but couldn't find the correct map to lift. Excellent observation.

@iwriteonbananas I have been thinking a bit about the subtle difference between $C^k$ and $k$ times differentiability.

iwriteonbananas

@BalarkaSen It took me forever to figure out that we could do that

@BalarkaSen What've you been thinking?

Balarka Sen

@iwriteonbananas :) Well, you figured it out eventually, that's what matters. I am keeping this trick in my head.

@iwriteonbananas It seems lots of classical results can be generalized by weakening $C^k$ by some easier differentiability conditions.

E.g.

It's a standard result that if $f : \Bbb R^n \to \Bbb R$ is $C^2$ then mixed partials of $f$ of order 2 are the same, right?

iwriteonbananas

Indeed

Balarka Sen

Turns out a weaker result is true: assume $f$ is twice differentiable, that is, $Df : \Bbb R^n \to L(\Bbb R^n, \Bbb R) \cong \Bbb R^n$ is differentiable as a matrix-valued function. This is equivalent to assuming $\partial f/\partial x$ and $\partial f/\partial y$ are differentiable. Then mixed partials of $f$ of order 2 are the same.

iwriteonbananas

That's a stronger result, isn't it?

:D

Balarka Sen

2:16 PM

I meant a result with weaker condition :)

iwriteonbananas

right

Balarka Sen

Here's something even more interesting.

The inverse function theorem says if $f : \Bbb R^n \to \Bbb R^n$ is a $C^1$ function such that $Df(a)$ is invertible for some $a \in \Bbb R^n$, then $f$ is a local homeomorphism around $a$.

You can replace $C^1$ by everywhere differentiability and the result still remains true.

Semiclassical

morning

Balarka Sen

Morning, @Semiclassical.

Semiclassical

i forget, did i ever explain what i meant re: that one optimization problem? the one i suggested doing a $x=t^2$ substitution on, if memory serves

Balarka Sen

2:19 PM

nope, I think you didn't.

iwriteonbananas

@BalarkaSen Ok, I didn't know that

Semiclassical

mmkay

iwriteonbananas

Hey Semi

Semiclassical

hiya

if i remember right, after the change of variables, you had something like $(y-2t)(y-t)$ for $t\geq 0$ and real $y$

Balarka Sen

@iwriteonbananas I find it pretty surprising.

iwriteonbananas

2:21 PM

@BalarkaSen Haven't thought about that kind of stuff since my real analysis course over a year ago

Semiclassical

which has a saddle point at the origin, and vanishes identically along the lines $y=t,y=2t$

Balarka Sen

@Semiclassical Uh, the function was $(y - 2z)(y - z^2)$, iirc.

Semiclassical

$(y-2z^2)$, i think?

Balarka Sen

Oh, sorry, you're right.

Yes.

Semiclassical

right. hence $t=z^2$ is the relevant substitution

Balarka Sen

2:23 PM

Right. I am not sure what you want to do with that sub though.

Semiclassical

well, that's a strictly increasing function for $z\geq 0$, so it preserves extrema in the right half-plane

so if i know how things work in $(t,y)$ space i know about $(z,y)$ as well

Balarka Sen

well, the problem was to prove that the origin is a critical point restricted to any line through origin, but I already told you a proof of that.

does switching to $(t, y)$ tells you something more insightful?

Semiclassical

depends on what you mean by more insightful, perhaps?

Balarka Sen

I mean, I am not sure what your goal is.

What do you want to prove?

Semiclassical

well, here's where i was going. you've got that the gradient in $(t,y)$ space is $(4t-3y)e_t+(2y-3t)e_y$

(where those are the unit basis vectors)

Balarka Sen

2:28 PM

Yes.

Semiclassical

...now i'm forgetting where I was going with this.

Balarka Sen

@iwriteonbananas I think what I said above could be misleading. I meant if $f$ is everywhere differentiable and the derivative is everywhere invertible, then $f$ is a local homeomorphism.

@Semiclassical Were you trying to prove that gradient descent around the origin leads me down the origin, or something of that sort?

Which would be nice, because the origin isn't really a minimum, even though it is restricted to each line.

Semiclassical

well, it's a saddle point

Balarka Sen

No.

Semiclassical

in $(t,y)$ space it is

Balarka Sen

2:31 PM

In the $(t, y)$ plane, yes.

But not in the $(z, y)$ plane. Indeed, the Hessian is degenerate at $(0, 0)$.

Semiclassical

and given that it's just a quadratic in $(t,y)$ there should be lines of steepest ascent and descent

which, returning to $(z,y)$, imply parabolas of steepest ascent and descent

Balarka Sen

Right.

Semiclassical

unfortunately, my brain is being silly about something

there's also the whole $t\geq 0$ thing with that mapping, but that's just symmetry

Balarka Sen

I am not sure what you are referring to.

Kaustabha Ray

A function f : N
+ → N
+, defined on the set of positive integers N
+, satisfies the following
properties:
f(n) = f(n/2) if nis even
f(n) = f(n+5) if nis odd
Let R = {i|∃ j : f(j) = i} be the set of distinct values that f takes. What is the maximum possible size of R?

Semiclassical

2:36 PM

oh, just the fact that the mapping $t=z^2$ is two-to-one

Balarka Sen

ah, ok.

Semiclassical

but that's easy in this case since knowing the behavior for $z\geq 0$ gives the same for $z\leq 0$

Balarka Sen

@Semiclassical Speaking of, did you see the question Ted posed yesterday?

Semiclassical

the o(x^2) one?

Balarka Sen

Yep. If $f$ is a C^0 single variable function with $f(x) = o(x^2)$ near $0$, then it's twice differentiable at $0$.

I think I proved that it is true.

Semiclassical

2:38 PM

i think he posed it as "is it true"?

Balarka Sen

Yeah.

Semiclassical

i was wondering about that myself

i couldn't see a counterexample, but being able to imagine a counterexample by myself is a sufficient but not necessary condition for there to be one :)

Balarka Sen

I searched a bit for a counterexample this morning, but when I didn't find one I tried to prove it and it turned out it's true :P

Semiclassical

neat

i gotta go for now, though

Balarka Sen

bubye! have a nice day.

L33ter

3:40 PM

hi @BalarkaSen

Balarka Sen

4:01 PM

Hi @L33ter.

Akiva Weinberger

Hi all

GPhys

I have an exam on metric spaces and stuff in 7 hours QQ

A lot of it isn't really fundamentally hard; it's just a lot of new precise terminology at the same time

Balarka Sen

Hi @AkivaWeinberger.

Mike Miller

4:22 PM

@GPhys: You'll do fine. Or maybe you won't. In either case, it's much too late to worry.

4

Albas

4:32 PM

Hello@Balarka

Semiclassical

4:47 PM

hi chat

Balarka Sen

Hi @Albas.

Akiva Weinberger

So I'm trying to wrap my head around homology

Balarka Sen

Homology is good stuff.

Akiva Weinberger

Confusing, though.

Balarka Sen

Which bit, @Akiva?

Akiva Weinberger

4:54 PM

The beginning

I've only just started

TheCoolDrop

Hello people,a major question from me to you all.Is anybody willing to partake in a project to create open mathematical books which are on the par with quality and rigor to the "standard" aka Spivaks Calculus,Jech Set theory and so on ?

Akiva Weinberger

There's this big theorem about long exact sequences and homologies

Balarka Sen

You need to be specific than that if you want help, @Akiva.

Akiva Weinberger

@TheCoolDrop Wikibooks exists

TheCoolDrop

Yes but their quality is questionable and not on the par with "standard" textbooks

Akiva Weinberger

4:56 PM

So how would your project be any different?

Theoretically, you could just ask us to help improve WikiBooks pages.

Balarka Sen

@AkivaWeinberger Are you referring to the long exact sequence of homologies of short exact sequence of chain complexes? Where do you feel confused about it?

TheCoolDrop

True,but wikipages do not offer high quality,older projects are harder to refactor ,they do not have pdf versions avilable,they do not have style of proper latex written books and honestly I do not think they enjoy quite high credibility among readers

Akiva Weinberger

The whole proof of that theorem on how $\dotsb\to\tilde H_n(A)\to\tilde H_n(X)\to\tilde H_n(X/A)\to\tilde H_{n-1}(A)\to\dotsb$ is exact

TheCoolDrop

The intent here is that everything should be community reviewed first and structured according to predefined standars,with room for improvement.Idea came from creation of HOTT book,which started on github and evolved into something quite great

Akiva Weinberger

HOTT?

Balarka Sen

5:00 PM

@AkivaWeinberger Your sequence is wrong, by the way.

Akiva Weinberger

?

Balarka Sen

Sorry, I misread the indices.

TheCoolDrop

Homotopy Type Theory

Akiva Weinberger

Ah

Well, I'm not enough of an expert in anything to partake in a project like that, but it certainly sounds interesting and I wish you luck

TheCoolDrop

github.com/VanioBegic/Micromathics is the link to the project.I think you are more of an expert than me,idea relies on community and not solo knowledge.Take a look around if you have time.Any input will be appreciated

Mike Miller

5:04 PM

Traditionally, books are written by experts, because experts alone tend to have the appropriate knowledge of the field and its pedagogy to decide how it should best be written (and to do so in a competent way). I see this as the bigger challenge of a WikiBooks-type project than the number of people who are willing to participate.

HoTT worked out well because Voevodsky was working on it, to my understanding; but in any case it had a number of people who were at the IAS program doing so.

Semiclassical

books need editors, and editors need expertise

Mike Miller

Certainly massive book collaborations exist: see Bourbaki or Besse. But I tend not to have much faith in a book written by and for non-experts.

Akiva Weinberger

I just noticed that Hatcher writes "If there was" rather than "If there were" at one point.

Balarka Sen

@AkivaWeinberger Hatcher's proof is rather algebraic, but you can think about it at the chain level. It's actually quite geometric.

The real deal is the construction of the snake map, the rest is more or less straightforward.

You can define the snake map $H_n(X, A) \to H_{n-1}(A)$ as follows. Take a relative cycle $\alpha$, i.e., a chain in $X$ with boundary $\partial \alpha$ inside $A$. Then define $[\alpha] \mapsto [\partial \alpha]$. This is well-defined, and is precisely the snake map.

Akiva Weinberger

I feel like I need to think of a few examples for it to click in

To see why it's exact, I mean

Why is it called "exact," anyway?

Balarka Sen

5:20 PM

Geometrically, $H_n(A) \to H_n(X) \to H_n(X, A)$ is exact because the kernel of the 2nd map is precisely the collection of cycles in $A$, because those are the ones which are rel-ed off in $H_n(X, A)$. But that's the same as the image of the 1st.

The snake map bit is nontrivial, again, but can be seen geometrically. Try thinking about it.

If you want an example, $(D^2, \partial D^2)$ is ideal.

@AkivaWeinberger Maybe because they are a special kind of chain complexes, where $\text{im} \partial$ is exactly equal to $\ker{\partial}$, rather than being contained in it.

Akiva Weinberger

You can say that about anything that involves two sets being equal, though

That they're exactly equal

Balarka Sen

No, I mean, in chain complexes, image is contained in kernel. But in this case they are exactly equal.

Akiva Weinberger

I guess

Balarka Sen

I have to go study right now but feel free to ping if you have questions.

Akiva Weinberger

OK

Good luck with whatever you're studying for

Balarka Sen

5:26 PM

I am studying multivariable calculus :)

Akiva Weinberger

GLHF

Brody

Hello!

skull petrol

5:51 PM

hi

Brennan.Tobias

hi

Michael

6:11 PM

hi

mr eyeglasses

6:27 PM

hi

Akiva Weinberger

chai

Balarka Sen

@SemiC: My proof has a flaw. I think what I have proved is that if $f$ is continuously differentiable around $0$, $f(x) = o(x^2)$, then $f''(0)$ exists and is $0$.

Semiclassical

so $f$ being assumed as $C^1$ at zero rather than $C^0$?

Balarka Sen

Right.

Not at $0$, but at a small interval around $0$.

Semiclassical

right

hmm. supposing one wanted to generate a counterexample, what we'd need

defining $g(x)=f(x)/x^2$, we require $g(x)\in C^0$ and $g(x)\to 0$.

Balarka Sen

6:36 PM

(1) $f'$ needs to be discontinuous at $0$ (2) $f(x) = o(x^2)$