Mathematics - 2016-05-15 (page 1 of 2)

Akiva Weinberger

01:05

@GridleyQuayle @TedShifrin Perhaps it's d/dy !

@Dr. Shifrin, sorry to hear about your friend :(

Akiva Weinberger

01:19

@BalarkaSen I only realized that your name worked for it halfway through writing the "algebra" post

I was watching something about Hopf algebras (taking a break from algebraic topology for a bit), and I was thinking that I'd call them Hopf algebrae

Kenny Lau

01:56

1. 1 ∈ X
2. (n ∈ X) ⇒ (4n+1 ∈ X)
3. (6n+1 ∈ X) ⇒ (4n-1 ∈ X)
4. (6n-1 ∈ X) ⇒ (8n+1 ∈ X)

Prove that X contains every odd number

:)

Balarka Sen

@AkivaWeinberger Hopf algebras are probably not so distant from algebraic topology. I think homology of an H-space has the structure of a Hopf algebra.

Sorry, I think I mean cohomology.

Qiaochu Yuan

02:43

both of those statements are true, although for homology you need to work over a field

Balarka Sen

02:53

@QiaochuYuan Ah, yes, true.

All we need is to look at the map $X \times X \to X$ and the diagonal inclusion $X \to X \times X$ I suppose. In homology I get a map $H_*(X) \times H_*(X) \to H_*(X \times X) \to H_*(X)$ where the first map comes from Kunneth's theorem (this is where I need field coefficients, I believe).

And the coalgebra map $H_*(X) \to H_*(X) \times H_*(X)$ goes the other way around, induced from the diagonal inclusion.

I haven't studied a lot about this structure; I vaguely remember it mentioned in Hatcher somewhere. Thanks.

Balarka Sen

03:29

hi @ForeverMozart

Forever Mozart

helloo

Balarka Sen

What's up?

Jim Wilson

05:01

6

Q: Prove that between any two roots of $f$ there exists at least one root of $g$

$$f(x)= 1 - e^x\sin(x)$$ $$g(x)= 1 + e^x\cos(x)$$ Prove that between any two roots of $f$ there exists at least one root of $g$. I know Rolle's Theorem and the Intermediate Value Theorem (I think) need to be used. Can someone show a step by step proof for this?

Can anyone explain this to me a little further? The accepted answer is good. But I still don't know how to proceed.

Mats Granvik

@BalarkaSen I have been listening to that Tom Lehrer song Lobachevsky over and over again since you posted it here.

Balarka Sen

05:43

@MatsGranvik It's catchy isn't it?

Mats Granvik

06:26

@BalarkaSen Yes it is. I later read about Tom Lehrer. He does not have russian accent naturally, instead he practiced russian accent with the help of a real russian just for this song.

idonutunderstand

06:40

"Harvard University: Scientists Reportedly Meet at School Privately to Discuss Human Genome Synthesis" anyone following this topic?

Brody

@MatsGranvik native Russian accents can sound very ... coarse. Lehrer's is relatively pleasant, if artificial

Balarka Sen

Hi @Brody.

@MatsGranvik Yeah.

Brody

elo thar, balarka

Balarka Sen

What's up?

Brody

my serotonin levels

i think that's the stuff that makes you sleepy??

wbu?

Balarka Sen

06:52

tries to recall what that is/does from past life

Brody

lol

Balarka Sen

@Brody "nothing" is probably the appropriate answer to that.

Brody

@BalarkaSen often the only appropriate answer

likewise... it's a dull hour

hmm speaking of Russian, declension is god awful. makes learning the thing an endeavor

Balarka Sen

I don't know Russian, so admittedly have no opinions on that.

I don't know many languages either.

Brody

any language that requires you to inflect every single component of a sentence

should die

Balarka Sen

07:03

I take your word for it.

Is there any math we could discuss about? Mostly looking for conversations in which I can actively participate in :)

Language is not a fair topic of discussion for me.

Brody

i'm thinking i was too pressed on the russian thing, so moving to math would be ideal

user 1618033

@BalarkaSen Sure. Calculate in closed form $$\sum_{n=1}^{\infty} \frac{H_n^{(2)}H_n^{(3)}H_n^{(4)}}{n^2}$$

@BalarkaSen do you know even to start such a series?

Balarka Sen

now that we both agree on moving to math: what kind of math have you been learning lately?

Brody

@BalarkaSen the kind listed on my course syllabi, and not much else sadly

Balarka Sen

e.g.?

user 1618033

07:11

Anyway. Just be careful about what you wish. :-)

Brody

rote cookie cutter problems you'd find in rote cookie-cutter texts on basic ODEs and multivariate calc

though i peruse math.se to get some food for thought (not that it suffices)

Balarka Sen

:( multivariable calc can be fun; sad to hear your course does it in a dry way

maybe i can tell you something interesting if you tell me what you have learnt so far/find interesting.

if you want, that is

Brody

it was my uni's Calc IV class, which in reality covered only 2/3 of the projected course description

no mention of differential forms, lagrange multipliers, (generalized) stokes theorem, etc. etc.

@BalarkaSen I'm inclined to that

Balarka Sen

alright, let's see.

you know what a manifold is?

Brody

lol

I do not

Balarka Sen

07:18

great, then let's try to understand what they are.

Brody

hmm i've actually been anticipating this

Balarka Sen

yeah, I mean manifold are one of the things that originate from multivariable calculus and go a long way. it's worth knowing them.

Brody

it's a peculiar feeling having seen the word a billion times in many contexts, and having no clue what one really is

Balarka Sen

so, what is a manifold? It's a subset $M \subset \Bbb R^n$ such that at each point $p \in M$ there is a neighborhood $U$ of $p$ in $\Bbb R^n$ such that $U \cap M$ is graph of a function. That is, there is a function $\phi : A \subset \Bbb R^k \to \Bbb R^{n-k}$ from an open subset $A$ of $\Bbb R^k$ such that graph of $\phi$ is precisely $U \cap M$.

Mike Miller

There's a fairly simple general description.

Balarka Sen

07:22

The $k$ here has to be fixed. For all point $p$. (we say $M$ is a "$k$-dimensional manifold").

The intuition is that a manifold is "locally graph of some function". (whenever I say function here, I mean smooth function)

Mike Miller

It's a wide pipe (think "output") for which there are many pipes leading into it. You've seen one if you've ever opened up a car.

idonutunderstand

What is the meaning of the word graph in this context?

Balarka Sen

Exactly what it means. Subset of $A \times \phi(A)$ consisting of points $(x, \phi(x))$.

Brody

@BalarkaSen i've yet encountered "neighborhood" but I have the impression it's intuitively the "locality" around something?

Balarka Sen

A neighborhood around a point is just an open set around a point.

Open ball, if you prefer.

Brody

07:27

and the point need be center in $U$ huh

Balarka Sen

if $U$ is a ball, sure. for arbitrary open set $U$, "center" does not make sense: we just require $U$ to contain $p$.

Brody

ah that clarifies it

musing over the definition, will post thoughts...

Balarka Sen

You should try to think of some examples.

I claim that the unit circle (level set of $x^2 + y^2 - 1$) around $(0, 0)$ in $\Bbb R^2$ is a 1 dimensional manifold. Can you tell me why?

Mike Miller

@Balarka I find it extremely unlikely that anyone would look at what you said and walk away inspired with what a manifold is. Nobody needs a formal definition right now; I doubt you would have benefited from that before you had any idea what a manjfold was.

I certainly would not have gotten anything from what you said.

idonutunderstand

because a circle looks like a line in a small enough ball around a point?

Balarka Sen

07:32

I wondered whether telling the definition of being locally diffeomorphic to $\Bbb R^k$ would be more intuitive - I believe it indeed is - but I don't know of @Brody knows and understands what a diffeomorphism is.

That's why I thought "locally graph of some smooth function" would have been more appropriate and accessible.

Brody

@MikeMiller might be a fair criticism but I personally don't mind it. formality is a refresh for me having come out of a computational factory of a semester

and it's a brain exercise nonetheless. it's constructive for me in some way

Balarka Sen

@idonutunderstand has a good intuition! listen to him.

Brody

i get his gist, it's just a matter of reconciling the intuition with the provided definition

this is honestly my first real encounter with anything formal-ish

Balarka Sen

Yes, one should definitely try to do that at some point.

Generalizing what @idonutunderstand said: a k dimensional manifold is a thing which locally looks like $\Bbb R^k$.

It's probably not immediately clear why this agrees with being locally graph of some function over $\Bbb R^k$ but it will clear up eventually, once you know what "locally looking like" means.

Brody

yeah, i don't see how the two correspond exactly

i.e. locally graph... and locally looks...

Balarka Sen

07:42

Think of graph of a function $\phi$ over $A \subset \Bbb R^k$ a the result of "distorting, without tearing or gluing" $A$ by $\phi$.

Mike Miller

@BalarkaSen Do you need fancy words? "Locally looks like Euclidean space." When (and if) there is confusion about what this means, one then starts to provide further intuition for the meaning of 'looks like'; and if it seems like it's the best approach for providing understanding, then throw out some definitions.

But it's past my bedtime.

Balarka Sen

@MikeMiller Hmm, I suppose that is a good approach. Intuitions-first is probably more useful than a definitions-first learning. I was trying to simultaneously provide the intuition along with the definition; but I suppose that's not a good idea.

Mike Miller

I got none of the first and only the latter.

Brody

ohhh my brain kept seeing cups instead of hats. hold up, reworking some thoughts lol

Mike Miller

I googled a bit to find something helpful. Try reddit.com/r/explainlikeimfive/comments/mfswi/….

But now I'm going to sleep. The local expert is here if you have further questions.

Brody

07:47

@BalarkaSen i claim the entire sine curve $y=\sin x$ is a manifold in $\mathbb{R}^2$

because why not the sine curve? >.>

Balarka Sen

I agree with you.

Can you give an example of something which is not a manifold?

Brody

if i'm not mistaken, it seems like many familiar objects are indeed manifolds

Balarka Sen

G'night, Mike. Also, lol at "local expert".

idonutunderstand

the integers in the real line are not a manifold i am guessing

Balarka Sen

Well, a point is a 0-dimensional manifold.

And $\Bbb Z$ is just a bunch of points :)

Brody

07:51

suppose n-M denotes a non-manifold...

idonutunderstand

oops :)

Brody

n-M in $\mathbb{R}^2$ is some subset of $\mathbb{R}^2$ such that for some point in n-M, there is no function mapping some subset of the real line to the real line, whose image is $U$ intersect n-M

mmm?

Balarka Sen

Yeah.

idonutunderstand

ok the unit interval is not a manifold?

Balarka Sen

But it's also worth keeping "locally looks like $\Bbb R$" intuition in mind, @Brody, if you want to find an example of a non-manifold.

@idonutunderstand Which unit interval? Open or closed?

idonutunderstand

07:54

closed

Balarka Sen

OK, you're correct (but not quite the example I wanted - there are much more obvious ones). Why not?

Brody

the set $\{-1,1\}$ perhaps

in the real line

Balarka Sen

It's still a bunch of points. Points are 0-dimensional manifolds.

I mean, tautologically. A point is $\Bbb R^0$.

OK, I'm going to give you an obvious example. Consider the figure eight curve. "8".

Why is it not a manifold?

Brody

i want to say the crossroads of the lemniscate is problematic

Balarka Sen

Yep. Around that self intersection point, it looks like an "X". That's very not like a line.

Intersection of any neighborhood around the self intersection point is like an "X".

idonutunderstand

08:01

why is that more obvious than the closed unit interval?

Balarka Sen

You still haven't told me why the closed unit interval is not a manifold :)

If you can tell me that, you'll see for yourself.

Brody

the endpoints are not very $\mathbb{R}$-like, whereas the open interval precludes this issue

Balarka Sen

@Brody Rigorously, why can't the shape "X" be graph of a function? More precisely, suppose I have $[0, 1]^2$, the unit square. Why can the union of the two diagonals in $[0, 1]^2$ be graph of a smooth function $[0, 1] \to [0, 1]$?

@Brody The neighborhood of the endpoints look like $\Bbb R^+$, yep, the positive half of the real line.

But it still looks like a "line". Whereas "X" is obviously not a line. That's why I said the latter is more obvious than the former.

Brody

do you mean "why can't the union..."?

Balarka Sen

Yeah, typo. Sorry about that.

Brody

08:07

i'm trying to go off your "distortion" advice, bear with me

Balarka Sen

Sure, no problem. Manifolds are not conceptually easy objects; but only as long as you don't understand them.

Brody

i mean it's trivial that the diagonals are non-functional. also, the interval $[0,1]$ cannot be "morphed" into the X-shape without injuring it quite a bit

fourier1234

Hello all

Balarka Sen

Yes, that is the intuition.

Brody

if that's along the lines of what you wanted, but i'm not sure

oh ok

Julian Rachman

08:13

It has been a while!

0

Q: Categorical Representation of the Set of All Strings

Let $\mathcal{A}$ be a preordered category for set $A$ such that the objects are the elements of $A$ and the morphisms are given $x,y\in A$, then $x\to y$ iff $x\leqslant y$ where $\leqslant$ is the subword order. How would we define a preordered category for all strings over $A$ (e.g., Kleene Cl...

abstract-algebra category-theory

Balarka Sen

@Brody Well, it's still worth understanding a bit more rigorously why "X" cannot be graph of a function [0, 1] --> [0, 1]. If it was, then consider two points lying over the same point in the x-axis in the figure "X" (away from the self intersection point).

Those two gets mapped to different points. So your function eats something and spits out... two different things?

Doesn't sound like a function to me. Contradiction.

Brody

@BalarkaSen yes, that was the "trivial ... non-functional" part I mentioned. sorry for not being more explicit

Balarka Sen

Ah.

Brody

that basically amounts to the "vertical line test" and I thought you were expecting more haha

Balarka Sen

No, I wasn't actually :) Glad that you have already figured it out.

OK, here's a bit less obvious example. Consider the shape "V". Is that a manifold?

Brody

08:16

mmm, tough one. the non-differentiability at the vertex is pausing me...

Balarka Sen

You noticed the right thing.

Brody

but i don't know the implication yet!

Balarka Sen

Think about the locally graph of a function definition. We required that in a manifold every point has a neighborhood which is graph of a [....] function. What was [...]?

user 1618033

It is a manifold, of course (intuitively speaking).

Brody

well it's continuous at least, so there's no issue. i agree with @user1618033

Balarka Sen

08:19

We required that it was graph of a smooth function...

"V" cannot be graph of a smooth function (why?).

Brody

we did?

Balarka Sen

Mhm.

Brody

oh, you did in the definition

it was a trailing remark, forgot about it ^^

Balarka Sen

No worries.

Brody

had to step away for a sec, i'm back

Balarka Sen

08:23

People in topology would call "V" a manifold, because they would drop the "smoothness" criteria. Continuity is enough. But people in differential topology/geometry would not call "V" a manifold, because they want smoothness. In fact you get different classes of manifolds by saying "locally graph of a C^k function" for different k's.

So "V" is an example of a C^0 manifold, but not a C^1 manifold (hence not a C^\infty manifold). Don't worry about this though. Just think smooth.

Brody

there's no smooth mapping from the open interval around the abscissa of the vertex to the V-graph around the vertex

following the distortion approach

Balarka Sen

I agree. More specifically: if there was a smooth function whose graph is like "V", then the derivative has to be discontinuous around "V" because the tangents approach different slopes approached from different directions towards the vertex.

Contradiction. It cannot even be differentiable.

Brody

i like the differential topologist's version. it's actually less trouble due to the stronger constraint

Balarka Sen

Yes. Smoothness is a useful constraint which actually makes life simpler in many ways.

So, seems like you now know and understand what a manifold is :)

Brody

I can so far determine if something is or isn't one

hesitant to say I already understand manifolds

but that was very neat and enlightening. thanks for that

idonutunderstand

08:31

yes indeed, thanks

Balarka Sen

No problem. Let me know when you learn implicit and inverse function theorem (or, if you already do, when you want to talk about them). Those are the key tools to work with manifolds.

user 1618033

If the integrals and series I daily do were so simple, it would be cool. I learned a long time ago the story with the manifolds, it was a matter of minutes, but I didn't care about it later.

Back to some serious stuff.

Brody

Balarka, haven't seen those yet but I will look into them

user 1618033

(I lied, it would not be cool - I actually like the very hard stuff)

Balarka Sen

Sure thing.

user 1618033

08:36

BBL

Balarka Sen

Slightly surprised that your course hasn't gotten to those yet. Those are probably the most important theorems in whole of multivariable calculus.

Brody

like I said, it was rote cookie-cutter computational stuff. no proofs or theory, just the application of theorems

and even the use of the few theorems was poor. never cared if the hypotheses were satisfied or what the theorem itself meant. they were basically reduced to crank formulas

Balarka Sen

Sorry to hear that: seems like a horrible course to me.

user 1618033

@BalarkaSen I almost forgot to ask. I see you talking about all kind of area in math, but I have a simple question to you: have you ever brought your contribution (say, a significant one) in any of these areas by publishing some stuff?

Brody

the course was decent for the engineer, mediocre for the (lazy) physicist, and underwhelming for the prospective mathematician

Eric Stucky

08:46

sounds like a multivariable course, yeah.

Balarka Sen

Understandably so.

Brody

scratch that. we barely covered real-world applications. just computing vanilla integrals and such, so even the engineers were scheisted in some sense

user 1618033

@BalarkaSen We know you don't ignore me, but it seems you don't like what I ask you ... (I see)

Balarka Sen

@EricStucky I know of multicalc course which I personally found are great. Is it generally the case that they are horrible?

Eric Stucky

I mean, I only speak from experience

But yeah I've seen four at this point and they were all bad.

user 1618033

08:47

@BalarkaSen Hope one day you might like to learn to do some elementary calculations of integrals and series.

Balarka Sen

Yikes.

user 1618033

:-)

Brody

if i were at UGA in some past year, I'd be taking Ted's class, which looks golden

Balarka Sen

yeah, Ted's course is awesome cool.

user 1618033

@BalarkaSen Or even some struff in middle school I'm sure you have no idea how to do it ...

(before talking about manifolds)

Brody

08:49

i recall his course description trying to solicit prospective lawyers and pharmacists to take it

which is fine I suppose. just sounded kinda funny

Balarka Sen

oh well

you should definitely have a look at inverse function theorem at some point. I find it quite fascinating and powerful tool, myself.

Brody

does the single variable case suffice? or should I look at higher dimensions as well?

Balarka Sen

Ideally the higher dimensional case. The single variable inverse function theorem is not very hard to prove.

Evinda

Hello!!!

I have applied a lot of times the euclidean division of $x^6-1$ with $x^2- \alpha^{a+1}x+ \alpha^{2a+3}, a \geq 0$, $\alpha$ a primitive $6$-th root of unity.

We are over $\mathbb{F}_7$.

I got that $x^6-1=(x^2- \alpha^{a+1} x+ \alpha^{2a+3}) (x^4+ \alpha^{a+1} (\alpha+1) x^3+ \alpha^{2a+2}[(\alpha+1)^2- \alpha] x^2+ \alpha^{3a+3}(\alpha+1)[\alpha^2+1]x+ \alpha^{4a}[\alpha^4(\alpha+1)^2(\alpha^2+1)-[(\alpha+1)^2- \alpha]])+ \alpha^{5a}(\alpha+1)[\alpha^2+ \alpha+1+\alpha^5]x-2 \alpha^4-2\alpha^3-2 \alpha^8-\alpha^7+\alpha^4+\alpha^3-1$.

Brody

ok, that will take some time then. until then i'll meditate on our late convo

Balarka Sen

08:54

Calculus/analysis on $\Bbb R^n$ is generally harder than $\Bbb R$ because $\Bbb R$ has a natural order on it, whereas $\Bbb R^n$ doesn't.

Indeed, to prove the inverse function theorem on $\Bbb R$, you'd use that property.

That doesn't generalize to higher dimensions.

Brody

the statement alone seems to have a lot of vocabulary I don't know

so it's definitely gonna take some time before I get there

hmm, i actually know all the vocabulary. just not super comfortable w/ them yet

Balarka Sen

Really? It just says if $f : \Bbb R^n \to \Bbb R^n$ is a $C^1$ function such that $Df(a)$ is invertible then there are nbhds $U$ of $a$ and $V$ of $f(a)$ such that $f : U \to V$ is a bijective $C^1$ function with $C^1$ inverse.

Brody

ah, wikipedia tripped me up mentioning Jacobians

Balarka Sen

Well $Df(a)$ is the Jacobian here.

Brody

oh no :(

Balarka Sen

09:02

Are you not familiar with derivative of multivariable functions?

That's what $Df(a)$ is. It's also known as the Jacobian.

Brody

was taught up to partial and directional derivatives

Balarka Sen

Ah, OK.

Brody

i hope you see more and more how incomplete my course was :P

Balarka Sen

:(

Brody

no worries, I'll catch up on the preliminaries when I have time

Balarka Sen

09:05

I'd recommend grabbing hold of Ted's multivariable calculus book at some point and devouring through it but people complain about it being too expensive, so...

Brody

i essentially bought Spivak's Calculus twice, with the answer book, so you'd be surprised

Balarka Sen

huh

Brody

09:19

@user1618033 i most easily find beauty in math when I associate its concepts with space/nature or by picturing some advanced space-faring civilization

@user1618033 but maybe this is more "feeling" beauty than perceiving the intrinsic beauty of math. it's a motivator. these textbook exercises I'm doing are driving spaceships across the galaxy

Eric Stucky

:)

I appreciate this picture, but I don't think she will XD

Brody

yes, i suspect the possibility that she deems it misguided lol

but I felt the need to bother them

Balarka Sen

there are more constructive ways to waste ones time...

Brody

maybe for you. when i waste time, i really waste time

behold, son of perdition. i am become death. i contain multitudes.

Balarka Sen

lol "i am become death"

Brody

09:28

indeed! to SLUMBER the weary traveler goes. silent and SINISTER, like a ghost. pHanTOm AWaAaaAayyyYyyyyyyyyyyyyyyyyYyy

Balarka Sen

G'night.

Eric Stucky

night :)

Balarka Sen

i should develop a habit of posting captain america memes whenever i understand references

user 1618033

@Brody First, there is no guarantee that the math you described above will match the math you wanna work on. Initially, yes, you may say to yourself this is the math you wanna do, but after working a lot on it you might not like it that much. Only working much on an area you will realize its beauty. So, it's hard to have a big picture of a certain area from the very beginning.

Secondly, hope you'll meet people that will make nicer introductions to the math concept than the one you received today from @BalarkaSen (which seemed to me not even MikeMiller agreed with that much). The figure-8 you saw is from a link posted by Mike (just to know that).

Third, if you're interedted much in "math when I associate its concepts with space/nature or by picturing some advanced space-faring civilization" then you should try my math. :-)

Eric Stucky

:P

I'm surprised by you

user 1618033

09:43

Einstein had a nice word "If you can't explain it simply, you don't understand it well enough." You never ever make such introductions to the math concepts as I saw today (excepting the case you wanna be seen by others - that know much more math than the simple student that wanna learn the new concept).

Eric Stucky

Oh, lay off him, u16.

Balarka doesn't understand a typical student's path.

It's a fairly common affliction.

user 1618033

It's just a matter of being nice with new people in a certain math area.

Jim Wilson

Hey guys, can someone here explain to me how the result f'(x) = f(x)-g(x) will help me prove the question?

0

A: Prove that between any two roots of $f$ there exists at least one root of $g$

You can consider that $$ \eqalign{ & f(x) = 1 - e^{\,x} \sin (x) \cr & g(x) = 1 + e^{\,x} \cos (x) = 1 - e^{\,x} \sin (x - \pi /2) \cr} $$ so that$$ \eqalign{ & f(x) = 0\quad \to \quad \sin (x) = e^{\, - x} \cr & g(x) = 0\quad \to \quad \sin (x - \pi /2) = e^{\, - x} \cr} $$ and...

Eric Stucky

No, it's really much more than that, u.

user 1618033

Well, I don't describe my tutoring sessions, but here I find it enough to be nice.

Jester Tran

09:49

What does $\mathbb{C}[G]$ mean if $G$ is a group?

Eric Stucky

Jest: You can think of it in two ways

Either, as complex-valued functions on $G$, that is, maps $G\to\Bbb C$

Or, as formal $\Bbb C$-linear combinations of elements in $G$

(Usually called the "group algebra" or "group ring" if you want to Google for other sources.)

Jester Tran

Both ways of thinking are equivalent?

Eric Stucky

Yeah.

Jester Tran

Many thanks!

Eric Stucky

npnp

Tobias Kildetoft

09:52

@JesterTran just note that the first way might indicate the wrong multiplication, which in that case is convolution rather than pointwise

Eric Stucky

Ooh, ty tobias :)

Well, except that, you would have to be willing to extend the functions linearly right?

Tobias Kildetoft

probably the comultiplication is a bit more natural to define in the first case though (even if it is easy enough in either)

@EricStucky Not sure what you mean

Eric Stucky

Actually, I'm not sure what I mean either.

How do you compose these functions?

Erm

Convolution :/

ignore me.

Tobias Kildetoft

right, no composition of functions here

Jim Wilson

Lol I'm the only one that got ignored. But I don't blame anyone, analysis is ugly and I hate it

Eric Stucky

09:58

XD not the only one

Just the only one recently :P

Tobias Kildetoft

(as an aside, always remember to write the group multiplicatively when using the second formulation, or you will confuse yourself a lot with two different plusses, say when considering $\mathbb{Z}[\mathbb{Z}]$).

Eric Stucky

After reading, though, I don't know what your question is, Jim.

Are you not satisfied with Steamy's second paragraph?

@JimWilson

Jim Wilson

It is a good explanation but I'm still unsure how to prove it

Eric Stucky

Hmm

The first paragraph is a little sketchy, but the second one seems pretty copy+pastable.

I think my usual method here is, to write down the thing I want to write, and then

start asking myself where reasonable doubts might happen

If I can't find anything specific, but just have a vague uneasiness about it, then I submit it anyway with a comment to the grader.

But if I can find some specific concern, I'll try to write down counter-arguments; rinse repeat.

Jim Wilson

10:15

Our professor did mention we'd need to use Rolle's theorem for that question

Eric Stucky

Hmm.

I think your professor had a different proof in mind.

Jim Wilson

Yea he said "(hint: use Rolle's theorem)" in an email he sent us

Eric Stucky

One thing I notice, when I look at the graph:

It seems that for $x>0$, the extrema of the graphs occur at their intersection points.

Perhaps if you try to expand this into a proof, it would use Rolle.

Jim Wilson

ohh

fourier1234

hello

Eric Stucky

10:26

hello

fourier1234

Hi Eric I was just reading your blog

Eric Stucky

:D

I'm looking forward to really getting back into it this summer

fourier1234

Are you a post grad?

Eric Stucky

Nope, just a grad student.

fourier1234

Oh thats the same I think

Eric Stucky

10:36

okay :)

fourier1234

sorry I'm using British terms for these things

Eric Stucky

Ah, fair.

fourier1234

Are you doing a PhD in Maths?

I was reading your blog and I find the idea of qualifying exams interesting and very different to what we do here

Eric Stucky

So, yes, but

In US, it is typical for a student to go straight into a PhD program after undergrad.

So really, what I am doing right now, if I was doing it in Britain, would be a masters program.

fourier1234

Ah I see

Do you know what area you want to do your PhD in?

Eric Stucky

10:39

Nah

I spend a lot of time with combinatorialists, so that seems likely

But I want to give a good solid swing at analysis

user147690

Hey @EricStucky How is life?

Eric Stucky

And topology gets better and better

Haha Alex :D

user147690

Some tropical geometry quickly came up in something I was working on haha

Eric Stucky

I keep seeing you around, missing each other for by an hour on chat

user147690

Yeah I saw that too haha

Eric Stucky

10:40

:P

But yeah, life's good.

Easy to say after the semester ends :P

user147690

My adviser tropicalised something and said it was a bit of magic :P, didn't explain why it was a motivated choice :P

user147690

@EricStucky Definitely haha

Balarka Sen

@EricStucky Why so?

Eric Stucky

Bala weren't you going to sleep?

user147690

If anyone knows some AG: In AG we define the tangent space at the point $a$ on a variety $X$ as follows:

$T_aX : = V(f_1 : f\in I(X))\subset \Bbb A^n$ where $f_1$ is the linear part of $f$

Balarka Sen

10:42

No, I have already had sleep today.

Eric Stucky

But yeah I don't know what you're asking about, Balarka.

user147690

How exactly does this rely on $a$. This seems wrong

user147690

Seems only legit when $a=0$

Eric Stucky

Alex, "linear" means "linear at $a$", in this case.

Balarka Sen

@AlexClark You look at the linear part of $f(x - a)$.

user147690

10:43

@BalarkaSen So a translation

Balarka Sen

You scale it to origin.

Eric Stucky

Think Taylor expansion, and take the linear part.

Balarka Sen

That's right, @Alex.

@EricStucky "I'm surprised by you". No idea what you were referring to.

Eric Stucky

oh, wow that was a while ago

I was surprised that u16 agreed with the description of her work as so...

what's the word

Brody's description sounded to me like a "gorgeous trapping"

Balarka Sen

Oh I have been ignoring him/her, I thought that was directed to me.

Eric Stucky

10:46

Oh, no.

Balarka Sen

@user1618033 Bro, the figure 8 is a standard example of a non-manifold.

Also I don't really care for criticisms about how I explain someone something from someone who pretends to know that something but doesn't actually know that something. However I do agree with Mike that I probably didn't do a good job for an introduction.

In any case, dude, it was a casual discussion - I wasn't teaching anyone anything.

Eric Stucky

You lay can off too, Bala :/

u16 may be opinionated but she's also competent

Balarka Sen

I wasn't really replying to the stuff he/she said, so not sure why you're telling me to lay off. I just said what I had to.

user147690

I also have them on ignore, so I was confused by Eric above too :P

Balarka Sen

@EricStucky I am not contradicting that?

user147690

10:55

I thought he was surprised by your captain america memes :P

Eric Stucky

XD

Balarka Sen

Yeah, me too.

I was thinking of a meme-reply to that, that's why I replied to it.

Transcript for

$Mathematics$ Mathematics