umm

i don't think that limit is correct

there's no x

t is a dummy variable

i know

I mean that there's no x. Everything would be in terms of c

It's correct.

oh ok

so my question: we can have a limit exist but the point not, so how does that mean the function must be continuos at c

the derivative is lim h -> 0 (f(x+h)-f(x))/h

if f(x) does not exist, then how can you garner a value?

oh i see

sure, you could try to approximate a derivative if it's a fixably discontinuous point.

but that would be a garbage value

you know what fixable discontinuity is, right?

yes it means like y = (x-1)/(x-1)*x

exactly

those you might be able to get away with reducing somehow

but technically they are undefined

@JulianRachman Have you learnt say calculus and linear algebra?

@user19405892 I would say that it's a bit of a gray area and it all depends on whether or not you want that fixable discontinuity to be removed in your original graph. If the answer is "yes" then you can say the value is defined. Otherwise, you're forced to ignore it.

@DanielFischer With regard to my answer here, when I say that $\left|e^{- i \pi z} \csc(\pi z) e^{iz \alpha} \right| \sim 2 e^{-\alpha(N+1/2) \sin \theta} $ on the upper half of the circle with $N$ is large, is that technically correct? What if $\theta$ is very close to $0$ or $\pi$? I think I showed you this answer once before for some reason.

If I remember correctly, other than asymptotes and fractional discontinuity, there aren't any other times you get discontinuity with the standard functions, so unless you are doing something highly theoretical then you should be able to see when you have those issues. y-asymptotes carry over in differentiation anyway so you won't have to worry about juggling them.

@user19405892 Also, if there were one-sided derivatives then you could keep doing them repeatedly and get derivatives for the square root of negative numbers, which would be very undesirable.

yes i get this now

the derivative depends on the value of f(x)

i mean f(c)

yes

but as a further argument

if you kept defining values by left handed limits

if the two one-sided limits are equal we have proved differentiability at a point

right?

you could chain them over and over and build derivatives for square roots of negative values

No

The function has to be continuous at that point

it has to exist as well

(or you make it exist pre-differentiation but that's beside the point)

The point also cannot have sharp corners

meaning the derivatives limits have to be equal

obviously the function has to be continuous at the point or else the two one sided limits wouldn't be equal

no...

right?

oh

continuity has two parts

1. the point exists

2. the two limits equal the value within the point

is it possible the two one sided limits could be equal and the function not be continuous at the point?

yes

the point doesnt have a value

it's a hole

f(2) doesn't exist

but the graph approaches the same value on both sides

Hence, it's not continuous

That's not even a function $f : \Bbb R \to \Bbb R$.

It's a random image showing discontinuity. -_-

lol

are you guys still going on about this

In a more abstract sense, continuous means that that region of the graph can be drawn without lifting the pencil

@TheGreatDuck You said $f(2)$ does not exist. That means $f$ is not a function from $\Bbb R$.

@BalarkaSen That's not even relevant.

For functions $\Bbb R \to \Bbb R$, if the two sided limits of $(f(a+h) - f(a))/h$ as $h \to 0$ exists, then $f$ is differentiable hence is continuous.

Yes it is.

He's asking what continuity means

continuity is two-fold

@Balarka: Unnecessary pedantry. Define f(2), then. A starting calculus student does not need a careful discussion of domains and codomains.

1. the point must exist

2. the point must be equal to the two limits

He's essentially asking whether differentiability implies continuity, no? Which is true.

On $\Bbb R$.

no

he's saying continuity doesn;t need to be present to differentiate

he's saying the two limits imply differentiability

and that continuity isn't neccessary.

Isn't that the contrapositive of what Balarka just said

no

He's saying that if the two limits exist, then the function is differentiable

but if the point doesn't exist it's not continuos

o

so there has to be a contradiction somewhere

a few minutes ago he was trying to do one sided differentials.

Many of my students define continuous functions using the "pencil" definition. Often they think the definition of differentiable is "no sharp turns".

Maybe you can help @PVAL since you are a teacher. He's basically said (if I understand right) that the derivative is defined as "lim h -> x f'(h)" even for places the derivative doesn't exist.

He's implied that the existence of f(x) is irrelevant and I'm not sure how to get through to the kid. Granted, he's gone now so it's probably irrelevant.

This sounds like an argument about semantics no thanks.

:p

(Basically the kid is bootstrapping derivatives which while might lead to the derivative for negative square roots.... not good)

I think my example of an incorrect definition of derivative is much more damaging.

well, that may be true

but a few minutes ago, the kid was doing one sided limits for derivatives.

which is /very/ wrong

I actually disagree on that point. If a student wishes to use an informal definition that works in most cases that will pop up on their exam, so be it. However, if they attempt some sort of formality its damaging if they get it wrong.

Yeah, I suppose you're right.

They /were/ asking though.

the worst that could come up is fractions in polynomials

which I believe reduce anyway

Yesh, just glad I'm not TAing calc this semester.

hi

@TheGreatDuck: if you want to italicize, *use asterisks*.

The definition I gave isn't bad because it's informal. It's bad because it's plain wrong. There is a fixed informal definition that works. BUT THE ONE I SAID IS VERY WRONG.

@MikeMiller I see that Balmer is giving a talk at a university I'm applying to for gradschool. Small world.

the derivative is just defined using the two sides of the limits

He's been in Europe all year.

i don't see how that implies continuity

Oh. Not so small world.

Sabbatical?

Some grant.

Grant is a really nice guy to some people.

@PVAL Indeed.

E.g., $f(x) = x^{1/3}$. No sharp turns, yet not differentable at $0$.

In fact if the inverse of any differentiable map is differentiable (as that definition would imply), there would be no exotic spheres of higher dimension.

Agree.

That shouldn't be obvious.

This sharp-turn thing is a good point and should be raised in calculus I classes. I like it.

@RandomVariable Technically, it depends on what precisely you mean with $\sim$ there. For $N \to\infty$, you have asymptotic equality for every $\theta\in (0,\pi)$, and it's uniform for $\theta\in [\delta,\pi - \delta]$ for every $\delta \in (0,\pi/2]$. If you intended it to mean that you have a uniform asymptotic equality for $\theta\in (0,\pi)$, that would be incorrect, since the relevant bit is $\operatorname{Im} z$. But the local uniformity suffices, since everything is bounded on the small arcs.

@PVAL I haven't dealt with diff structures a lot, but if $S^n$ has two smooth structures, then call the resulting diff manifolds $X$ and $Y$. The identity map gives me a bijection $X \to Y$. This is differentiable, not? Then the inverse would have been differentiable, giving me a diffeomorphism.

Why should that map be differentiable?

I mean, if that was differentiable, its inverse is also the identity map, and is differentiable.

oops, you are right. I have forgotten what a diff structure means.

It's past your bedtime.

There's like a sketch of a proof on MO that uses some limiting thing with PL structure.

So for any two exotic spheres I want a smooth bijection between them.

I actually proved it a different way using a C^1 version of Bing's shrinking lemma.

This would be a good thing for a blog..

@PVAL: I don't know how to do it via your second proof. I would read it.

We should cowrite a blog so that it gets updated more than once a year.

I am going to read it, but I don't remember anything about charts.

The best example of a smooth manifold I know right now is $\Bbb R^n$.

I'm convinced the first one would be really annoying to write up, and the proof I did isnt.

I was actually trying to prove that Schoenflies was true for smooth homeos for a few weeks but I gave up.

I don't know what that means.

For any smoothly embedded $S^3$ in $S^4$, do there exists smooth homeomorphisms from the complementary regions to $D^4$ (and maybe vice versa)?

Oh I would probably give up too.

I gave up not necessairly due to lack of progress, but realizing that literally no one was thinking about similar questions.

Is the answer to this not known?

Full Schoenflies is going to be false so I would probably be pretty apathetic about such a result.

LOL

What makes you so sure of that?

It's obvious, of course.

@BalarkaSen I don't think anyone really has seriously thought about this.

Ah, ok. Interesting problem.

@PVAL If 4-dimensional topology is any fun there should be a zoo of smooth structures on $S^4$ and on subthings of $S^4$ etc.

Do we know this in higher dimensions?

I think h-cobordism can answer that.

Go to bed.

Not sure, haven't thought about it carefully.

@MikeMiller :(

In higher dimensions the diffeo. version is known (its in Milnor's notes for instance).

hi everybody

im in season 4 of six feet under now

@PVAL

he just moved back into the house

this show is insane man

There's 5 seasons or 6?

or 7?

6

oh wait

no only 5

Ah

I think I'm going to drop out of grad school and just watch IAS videos all day.

the new house of cards season is out today also

IAS?

Ya nothing at all happens in season 4 (something actually does happen).

Princeton?

@MikeMiller Nice idea.

@ForeverMozart Yes\

OK, I am really going to bed now. Can't stay awake. Night.

lord they must record every speech there

I think I've quit grad school and now I just try to prove theorems.

@PVAL: I'm not sure how that's different than grad school.

but most are so specialized I would not understand them

well you have to take some courses, pass some qualifying exams too

so I guess you could quit that part

I'm taking 3 courses. None of which I have went to this week.

I thought graduate students always go to class

I didn't take anything this quarter. Maybe you just have better classes than we do.

I'm going to something next quarter but mostly because I want an excuse to come in every week.

I will watch this one: video.ias.edu/mini-symposium-topology-2015

@DanielFischer I meant the latter, which I realized made no sense. What if I simply state that $\left|e^{- i \pi z} \csc(\pi z) e^{iz \alpha} \right| $ decays like $2 e^{-\alpha \text{Im}(z)}$ as $\text{Im}(z) \to +\infty$ and is bounded on the rest of the contour?

@RandomVariable That would work fine.

@DanielFischer It looks like I'm going to need to edit 4 answers.

@RandomVariable No problem with that. Four is not a large number.

@MikeMiller I think I have proven the construction of something implies Poincare is false.

Something that is potentially easy to check whether such a construction is valid or not.

Such a construction may not exist, and if it does it might be very very hard to find.

You should disprove the Poincare conjecture.

I'd give you $20 if you dod.

I'm beyond 100% sure I can't construct any of these things.

There's someone here who I'm only about 100% sure cant do it.

Is it Bob

Idk he might think what I've proven is really interesting, but he might still not consider it a valid attack on Poincare.

Idk if hed try.

I might be able to convince another grad student (who has spent a lot of time trying to construct things which these are a proper subset of).

I wonder if a valid attack on Poincare appeared, if he would drop everything to work on it.

Are your things certain kinds of Stein manifolds?

ya

wew

how does this question have so much attention

It might have popped up on the sidebar.

It's an impossibly beautiful question.

Look, the guy even knew the answer before it was posted. He comments to one of the answerers: "Your answer is not useful. I think checking the eigenvalue of the Hessian matrix maybe a good approach"

Gotta get dem internet points