Somewhere in the world, there exist those who switch the order of summation, limits, differentials, etc. without verifying if doing so is allowed. We call these vile creatures engineers.

well assuming the $a_i$ are positive, any of those expressions diverges iff any other diverges too

These manipulations wouldn't be justified, for instance, if $\sum_i a_i$ or $\sum_i ia_i$ aren't convergent.

Everything is zero or positive so no problems with reordering yeah.

@SteamyRoot i thought some of them are also called physicists

I call them lower species

The distinction there is probably how much you get paid to ignore said limits :P

yes, probably

sub-humans

Sub-humans with good job satisfaction and retirement plans :>

That is why they are sub-humans

Suffering is exsistence :p

exsistence is a good word, i like it

i have always been curious about how some people can write things like 2^{sqrt(2)} and when you ask them "what does that mean?", they throw their hands up and say "idk"

I was like that when i was very young o.o

If, given enough time, I can compute approximations to that, then I'm fine with just saying "it is what it is."

playing with volumes of spheres and thinking "hey wouldn't it be really cool if $(-1/2)! = \sqrt \pi$ it would make the formula very simple"

@mercio do you mean sqrt(pi)/2

I'm not actually sure

I forget which one has the factor of 1/2.

I think it's (1/2)! that's sqrt(\pi)/2.

yeah

and then i asked my calculator about $(-1/2)!$ and it gave the result I wanted

my calculator was pretty good it seems

ya, okay, i just can't do u-substitution

yes, very good calculator

i never had a calculator like that, but i have wolfram alpha on my phone

so i suppose that's fine

well well well what have we here... Oh roasting physicists?

@Secret perhaps you should phrase this as a coherent question, and post it somewhere on the website

then link back here

that one is not really a question. more like progress so far

Guys i have a problem

what's the problem?

the derivative of this function looks like the gamma function but kind of shifted in both axis

@SteamyRoot relevant youtube.com/watch?v=rp8hvyjZWHs

excuse me

Hi.

hmm

@PearlSek the term in the parantheses looks very much like a form for the sum of a geometric series

@Starfall yes i know

I guess it can be extended to a function on $\Bbb R$

with the integral formula

oh, yes, sure

well what would happen if you could differentiate under the integral

@Sophie Lol

@Starfall that's precisely the reason it's the sum of $k!$'s isn't it

it comes from there :

@BalarkaSen indeed

write down the geometric series, integrate termwise

well, it's more likely it was obtained the other way around, as seen in this picture

oh wait a sec i'll write something

well, if you can differentiate under the integral, you pick up a factor of 1/(1-t) in addition to the usual derivative of the gamma function

inside the integral

You can also differentiate it when it's still in geometric sum form, i.e. $\sum_{k=1}^n t^k\mapsto \sum_{k=1}^n k t^{k-1}$.

But then you've gotta resum it, etc.

Though if you return to the gamma function at that point, you have $\sum_{k=1}^n k \Gamma(k) = \sum_{k=1}^n k(k-1)!=\sum_{k=1}^n k!$...somehow, I feel like I've made an algebra error.

but you're not differenting with respect to $t$

Oh, derp. I did $t^k\mapsto kt^{k-1}$, which is differentiating w/r/t $t$ not $k$.

Yeah.

Should be $t^k\mapsto (\ln t)t^k$.

ln(k) still sounds wrong

isn't it ln(t)

yeah

siiigh.

You kids and your silly insistence that I say things that aren't utter nonsense :P

In which case the integrals are $\int_0^\infty (\ln t)t^k e^{-t}\,dt$...hrm.

well you're a physicist aren't you

ducks

mutter

Happy Boxing Day, chat!

i found this

is not British, Canadian, or Australian

no

Hi @Fargle @Semiclassic @Balarka @mercio @Pearl ... et al.

@PearlSek Yeah, that's basically what I just did.

semiclassical you forgot the denominator

Hi @Ted

hi @Ted

I think the $1-t$ should still be there

Hi @TedShifrin

Oh, and hi, Alessandro

hi ted

I'm just talking about the individual terms, not the resummed version.

ah

(wait what)

i.e. what the individual integrals in eq (2) become upon $k$-differentiation...wait.

Yeah, that don't make sense. Need to differentiate the final sum for that to make any sense.

but equation 2 is only defined for $n \in \Bbb N$

yes, that's rather troublesome

Yeah.

You can't differentiate that with respect to $n$

Yeah. Have to wait until it's summed.

And then you can pretend like $n$ isn't an integer and differentiate formally.

indeed

Heya @Ted. How goes it?

Swimmingly, and you?

Getting ready for my trek across the country. Looking forward to experiencing Spirit Airlines for one time. (Maybe not so forward.)

Alright. Among many other things, I got pens and pencils for Christmas.

...it's like my family knows me.

All the better to math with?

I got a beautiful mechanical pencil as a thank you gift from the high school kid I taught the Spivak calculus course to individually ... last I heard, he's now doing well as a math major at Yale.

Indeed. Pilot G2 pens are my favorite easily-found writing utensil with which to do math.

I feel so darn tired today

I got a fancy mechanical pencil that rotates the lead while you write

@BalarkaSen you should sleep more

I liked Pilot Precise V7 for grading, @Fargle, but otherwise I use fountain pen (which I've had for 50+ years) or pencils.

@Sophie: I thought @Ted would be the first.

Sometimes Ted leaves it to other people.

@Sophie I did

I had the use of pens drilled into me by my pre-cal/cal teacher.

Whereas I still have an electric pencil sharpener on my desk :)

In general, pencils are better for math. I don't like getting papers with messy cross-outs, which is what pen leads to.

I like gel pens more than fountain pens, even though it messed up my handwriting

hi @TedShifrin

hi Zach

i got a raspberry pi, which included mathematica, so i've been drawing some conics :P

I kind of agree, but I tend to do my thoughts for HW separately on scratch paper, and then write up a solution page.

what have you been doing for fun, Zach?

playing a new game i got

and programming my raspberry pi

currently waiting for the resistors and LEDs i bought to come

is the "t" in Lambert silent

@GFauxPas oui

merci byukopf

taking suggestions as to how to mark the branch point in an aesthetically pleasing manner:

the chat doesnt want to upload my image :/

there we go

i saw it

the point $\left({-e^{-1},-1}\right)$ is a branch point

how should I mark it

with a point maybe

how about what I did

I don't like it

or a horizontal line

what kind of point. solid color? two colors?

oh and highlight the axises

I'm not sure if its helpful to draw the axes if the axis ticks are on the border of the graph

i dont know

Hey everyone, I'm having some trouble with this problem : "Show that if $\mathcal{A}$ is a basis for a topology on $X$, then the topology generated by $\mathcal{A}$ equals the intersection of all topologies on $X$ that contain $\mathcal{A}$"

"Now" is in the present. "Just now" is in the past. I hate English.

The thing is every basis generates a unique topology $\mathcal{T}$

and that would imply that the basis $\mathcal{A}$ is an element of $\mathcal{T}$

What you just said makes no sense, @Perturbative.

$\mathcal A$ is a collection of sets. So it can't be an element of a collection of sets.

What does it mean for $\mathscr T$ to be a topology that contains $\mathcal A$?

I realize it makes no sense, which is why I'm confused....

@TedShifrin $\mathcal{A} \in \mathcal{T}$

No, you're saying the same wrong thing again now. You were right the first time, but it means more than that.

You need to unscramble all the definitions of everything. What does it mean for the topology $\mathcal T$ generated by $\mathcal A$ to be the intersection of all topologies containing $\mathcal A$? You need to translate this into "English."

So "contains" here is supposed to mean, subset, as opposed to element? If so then that removes any confusion I should have had. I've only ever seen contains used to refer to an element of a set (e.g $A$ contains $b \implies b \in A$)

yes, the collection of sets $\mathcal T$ contains the collection of sets $\mathcal A$.

Every $A\in\mathcal A$ is also an element of $\mathcal T$.

I've seen "contains" to mean "is a superset of" in topology rather often, but I don't think it comes up too much outside of it. Just a weird problem with imprecise language.

Hi.

Well, we're working with elements of a set $X$, open sets of $X$, collections of sets $\mathcal T$, etc.

Hello @Mahmoud.

@TedShifrin, Thanks for clarifying that, you're always a great help!

Hi @Mahmoud

@Perturbative: Except when I'm not :D

Hello, @Fargle @TedShifrin

Thank you for the help yesterday.

@Mahmoud: As an application of your result on even and odd functions yesterday, if I hand you a polynomial $f$, how do you get the $E$ and the $O$? For example, if $f(x)=2x^3-x^2+7x+5$, what are $E$ and $O$?

@Balarka: Cool question

@TedShifrin Arrange all the $x^{2n}|n\in \Bbb Z, $they compose the even $E$, then all the $x^{2n+1}|n\in \Bbb Z,$ their sum compose the odd $O$ ?

@Mahmoud: OK ... but an answer in English or French would be fine (rather than math symbols) :)

But good.

@Fargle, The problem with imprecise language pops up far too often from some authors .....*sigh*

@TedShifrin Sounds like this relates to the problem Mike gave me earlier

I don't see the relation with the result we got yesterday..

Work it out for an example, @Mahmoud.

Which problem was that, @Balarka?

What book are you working out of, @Perturbative?

@Fargle Munkres' book. For the large part its a joy to read, but every now and then things like this happen

No, Munkres is extremely careful with his language.

He didn't do anything wrong.

You just need to be careful to pay attention to who is what.

@Perturbative As I recall he tries to be as pedantic as possible about the different levels of sets.

Note who used the word "pedantic," @Fargle :D

@TedShifrin, Understood, I should have realized that "subset" was implicitly implied in this context

@TedShifrin I'm a fan of the word myself, I just like to give you crap >_>

You'll get yours, @Fargle.

I don't remember when he sets it up--might be in the prelim chapter--but whenever he talks about elements, they're lowercase. Sets are uppercase, collections of sets are fancy uppercase, and collections of collections of sets are yet fancier.

@TedShifrin hi

hi Karim

@TedShifrin Which functions on a compact surface appear as Gaussian curvature wrt some metric? (I think any function which doesn't have everywhere wrong sign relative to the sign of the Euler char does)

But actually this asks about surfaces in R^3 so nah

Ohhh, yeah, his favorite surface. Right, his question was for abstract Riemannian metrics.

It is really cool that in arbitrarily category if something is both monic and epic then it doesn't have to be an isomorphism

little bit counter intuitive.

meh, @Adeek

(He loves the theorem of Kazdan-Warner, yes, @Balarka.)

That's what you get when you're not working with sets and functions, Karim.

Ah, so that's what it was

@Adeek Yeah, that's the price of generalizing functions between sets. The true generalization of one-to-one + onto = bijective is monic + retraction = epic + section = iso.

that sounds like a poem

haha @Fargle

Ah, you're looking at Ali's question? I'm sure it's false.

I've actually never seen anything resembling this. The curve theorem (which is what the OP is generalizing) you read in my notes ... or you didn't. It's just basic single-variable calculus. I have no idea what's true for Gaussian curvature. All the examples I know have way too much symmetry.

@Fargle that is pretty cool stuff

Do you know him, @MikeM?

I'be seen him on SE.

@Fargle did you see the my blog btw ? I will write a post today on some stuff on those things.

I didn't, @Adeek.

I think it's a great question, but I have no idea of how I'd find a surface whose $K$ has only a max and a min. I keep thinking we should try to do the Morse theory for $K$, but it gets me nowhere.

@Ted: My picture is of a top-heavy sphere-like object.

Radially symmetric.

We know we should get this at the end : $f(x)=(5-x^{2})+(7x+2x^{3})|E(x)=(5-x^{2}) \land O(x)=(7x+2x^{3})$ $\\ E(x)=\frac 12(f(x)+f(-x))=\frac 12(10-2x^{2})=5-x^{2}$ $\\ O(x)=\frac 12(14x+4x^{3})=7x+2x^{3}$, $\\ f(x)=E(x)+O(x)$, @TedShifrin

With no symmetry?

@Fargle hey

Oh, I see.

Yes?

So work it out, @Mahmoud.

Didn't I ?

Oh yeah, @MikeM, in fact a shape like that is what you get when you try to maximize the gravitational attraction of a body (uniform density) on a point on its boundary.

You said "we know we should get this at the end," @Mahmoud.

@Fargle I just saw your post about geometry of transitivity. I think I have a picture way of explaining it. I was looking at some problem similiar to what you are looking at. The reason we need reflexivity is that to produce sets that are semi disjoint.

But I guess you worked it out for yourself. OK, done. Good.

@Ted You can probably write down a formula...

Aren't there multiple maxima?

YaY, many problems to go ..

I have one for that surface, in fact, @MikeM. But, yeah, I was being dopey. Hats off to you. You should write an answer.

@Fargle I guess it is old post but interesting though

I like it.

No, @Balarka, if you shape the curve right (figure this out), you get a max at the bottom and a min at the top.

yeah

I'm out today. Since you have the formula, you should, @Ted. You can thank me if you feel strongly about it.

Oh, wait, that's not clear.

I'm gonna have to look more carefully at the formula for $K$ on a surface of revolution.

I withdraw my convincedness.

@TedShifrin I guess the curve is slightly skew to the top?

I still believe it.

The curve you rotate looks like an ear.

Yeah, got it

Right, but monotone $\kappa$ for the curve doesn't necessarily mean monotone $K$ for the surface.

This is like stuff Balarka and Fargle and I were just discussing a few days ago.

Yep, that's what I was thinking

The picture I see appears to have the right curvature. But I'll leave it to you.

So, Balarka, assuming arclength parametrization of the profile curve, $K=-f''/f$, where $f$ is the distance from the axis.

But that has singularities when $f=0$ (which is what Mike was envisioning).

I don't see this geometrically. what does $K$ being singular mean?

curvature blows up?

So we need to think carefully about what will make a smooth surface when $f=0$ and ....

but what if you make the top really flat?

yeah, I mean, it's not actually surface of revolution

It's not?

Actually, if you think about a flat top, this corresponds to a point where $f$ isn't differentiable — think $f(t) = t^{1/n}$, $n\ge 2$. But we need to think about the chain rule correction for arclength parametrization here.

Now I want to know the answer.

Can't you do it upside-down? So that the profile curve is flat ($\alpha' = 0$) at the top?

I don't know what $\alpha'=0$ means. Sure, I don't care about that. I'm working on the arclength param correction.

Or we could do it with principal curvatures more geometrically, I suppose.

You need your curve to be orthogonal to the y-axis at the ends.

Yeah, that's what I meant

Yes, for a C^1 surface you need that.

But, even away from the poles, critical points of $K$ are nonobvious from my formula. You can't just get $K$ from $\kappa$, because the distance from the axis comes into the picture for $K$.

The top has to be really flat, the bottom really curved. But I am not sure if I can make the curvature always go in between them in the middle; although I more or less want to believe a slightly skew curve should do it

Whereas $\kappa$ is very easily seen, $K$ is more subtle.

@TedShifrin Why don't we try to look at the product of principal curvatures? If the profile curve curves less than at one end (where the surface is supposed to have a maximum), why should that not mean the product is less than it's purported max?

principAl, dammit

fixed

sorry

So remember that we need the product of $\kappa$ with $(\cos\phi)/x$.

Agreed.

So whether you have a critical point or not isn't easy to see.

In particular, at the "fat" latitude (max $x$), $\cos\phi = 1$, so you have $\kappa/x_0$. $1/x$ has its minimum there, but with the $\cos\phi$ coming back into play $(\cos\phi)/x$ may not. And then there's the growth of $\kappa$ to consider. Agh.

Hm

I'm going back to the chain rule computation for the moment.

Given some monotonically growing funcion $f:[0,\infty) \rightarrow [0,\infty)$, is there any way to show that $\Theta(\sum_{i = 1}^{n}f(i)) = \Theta(nf(n))$?. I know for a fact that this holds for the identity function $f(n) = n$. Proof: $\sum_{i = 1}^n i = n(n + 1)/2$. What about other monotonically growing functions of $f$?

cool question Salehan

blegh. I have no energy/brain today.

certainly the LHS is big O of the RHS

the question is can we show the RHS is big O of the LHS

All I find myself wanting to do is take a nap.

thinking

I tried to fiddle with the principal curvature approach for a while, @Ted, and I get the seriousness of your objection. We got to handle both $\phi$ and $x$ to be smaller than $K_{max}/\kappa$, aka, handle (1) the spacing between the circles and (2) the size of the circles and how quickly to decrease/increase it

That gets complicated too soon

Yup ...

at the moment I'm kicking around the following AMM question

You solve it?

$$\sum_{k=1}^\infty\left(1+\frac{1}{2}+\cdots+\frac{1}{k}-\log k-\gamma-\frac{1}{2k}+\frac{1}{12k^2}\right)=?$$

Nah. There's a geometry one in there which I have an analytic geometry solution for, but that seems inelegant.

In other words, take the harmonic numbers and subtract off the terms occurring in the asymptotic expansion to order 1/k^2, and sum that up.

Hell no, @MikeM. I'm trying to compute $f''/f$ for the $x=y^{1/n}$ case.

Since it's an asymptotic series, one can't just write down the rest of the terms and resum them term-by-term.

@SalehenRahman I can only think of one direction. @ me if you think of the other direction.

Nevertheless it should certainly converge.

Maybe you need to keep the surface very very very flat until it gets to the curved end.

You can't.

Let me know when you do. I don't really have time to dig into this calculation when I have my own calculations to do.

Of course. I'll be traveling leaving 5:30 AM Wednesday, so it may take a while :P

@TedShifrin What goes wrong if the profile curve looks like a very long nose (appropriately curved near the top like a circle instead of hitting the axis of revolution at a bad angle) instead of the ear?

$\phi$ keeps close to $0$ mostly, not?

LOL ... I can hardly wait for another anatomical examples ...

@Balarka: Seems to me you're basically doing $y=x^n$ examples for large $n$.

Does the ear not work?

@TedShifrin dear no

"Take off your pants. I need to try a different surface of revolution."

@MikeMiller it the normal-to-the surface and normal to a circle might make a bad angle, so the curvature might not go as expected

@Balarka: To see something as subtle as critical points of $K$, I truly think we need more analysis and less intuition.

@MikeMiller We will soon be banned.

Yeah, I just want to make a guess of what I want to look at.

What is the problem you're discussing?

@Alessandro: This. Time for you to learn differential geometry.

I'm working on it :P

(Wait, what's the difference between differential topology and geometry?)

Geometry has metrics in it.

Curvatures.

Connections (Mike's favorite word).

and topology is stretchy stuff. not much rigidity.

Ah, I see

We'll let you do differential topology first. Did I send you exercises? There's one involving tubes around curves (for Morse functions), and the Frenet formulas show up there.

Oh yeah. I hate G&P's definition of Morse. When you get to transversality, I'll give you a better one.

Transversality is the real deal

@Balarka: I define Morse by gradient transverse to zero section. Then all the transverality theorems give results, rather than all the ad hoc stuff G&P do in Chapter 1. Never understood why Guillemin did that.

Yeah, you told me that. I struggled with it for a while to understand that.

Ah, at least I spread my suffering widely.

Hah

Mathematics can't compute the $f''$ by a limit. Hmm.

Good evening, DogAteMy.

Good evening. What's going on?

Balarka and I are stuck on this.

See my diff geo notes for the four-vertex theorem for curves :)

There was a Garfield strip I read once that I could never find. In the first panel, he said - with his usual half-sad and half-grumbly face "Oh, how I suffer!". In the second panel, he says, with a slightly happier face, "Oh, how I make others suffer!".

Have you seen Garfield Minus Garfield, @Balarka?

That must be how I've run my life :)

Nope, @Fargle

Yep, you sent me the exercises @Ted. No transversality in sight though, I'm still reading about the inverse function theorem

When we have that F is a field, does it follow that F[x] is a principal ideal domain?

@Alessandro: Skip Morse functions when you get to them in a few sections, @Alessandro, and we'll discuss later.

@BalarkaSen It takes Garfield out of Garfield strips (as one might guess), so it just paints Jon as a lonely depressed man.

Ok, I'll remember that

@Fargle That's rather... dark.

At the moment, if I believe Mathematica, @Balarka, I'm getting infinite limit for $f''/f$ at the axis.

@Balarka: yeah, it is, but it's funny.

I may have to give in and do the easy calculus by hand.

Strange, @Ted

I am rusty on my Mathematica. :(

@Alessandro Inverse function theorem is great.

68.media.tumblr.com/fSymsOGXO5e1ey3rpHp8vsnT_500.jpg @Balarka

@Fargle er

Yep, I didn't quite get the point when we saw it in real analysis last year, but in the context of manifolds it's beautiful

It's not for everyone. I find it hilarious in an absurd kind of way.

I won't bore you with more exercises, @Alessandro (from my multivariable course).

@Balarka: It seems like I'm getting $K=4$ for the $y=x^2$ curve (at the origin) and $0$ for the $y=x^3$ curve, so the computation seems correct.

@MaryStar I believe so.

But now to do it for the ear or nose curves seems not fun.

Lunch break for Ted.

@Ted Wait, if connections are geometry, why am I not a geometed? :)

@MikeM: Moishe just nailed it. Kazdan-Warner + Pogorelov (which i'd forgotten).

I also forgot Pogorelov, but I'd like the explicit example.

Ohh we all have burning logs on our pictures. That's nice.

Wait I don't. That's not nice.

it's a hat

@TedShifrin Great. So we're all happy.

I didn't say that.

Pogorelov is neat; just saw Cohen's comment

Hey everybody!

Oh we don't have an example yet. Sorry, @MikeMiller.

What do totally disconnected noncompact perfect sets in $\mathbb R$ look like? Are they homeomorphic to countably many Cantor sets in a row or can they have a more variegated structure?