I usually mess up tests, so I am in no position to judge them.

Nah, that's exaggrating. But I hate tests..

They hate you .l.

The test-graders probably hate me more than the tests

Do you grade yourself or do you have people that do it for you?

all me ...

That's a real pain..

It's my job.

I think most in the math department here grade their tests too, but I know it's never so in economy/computer science/ etc.

@TedShifrin You want me to solve it all in detail?

But I'm almost on vacations!

You made me do math on vacation :)

@PedroTamaroff Have you passed Real Analysis?

@TedShifrin Hehe, I know.

$\sqrt{1/n^2+x^2}$ is differentiable on $x=0$ but its limit isn't, right?

OK, I'll write down solutions.

@Studentmath Aha.

@FreeMind No, I'll take that next semester. But I know a little bit of it.

That came rather quickly, I am proud of myself.

Its limit doesn't exist at $0$

@TedShifrin He means $(x^2+n^{-2})^{1/2}$.

Oh, I misread anyhow

As a sequence

@PedroTamaroff Then how is it that your responds to some questions seem cool and rigorous :O

I kind of wrote it badly. It seems like the $n^2+x^2$ are both in the denominator

@FreeMind Well, I do hope they are rigorous. I study a lot by myself.

No, I just misread, @Studentmath. Mea culpa.

@PedroTamaroff When I see such responds I get motivation, keep on good job.

@DanielFischer But we found $count[3]=4$ and there is no position $4$..... :/ I am confused now....

Eitherway, I am off to bed. Would love to see the test after they all hand it back, @Ted! But I probably said it a hundred times by now. g'night!

@evinda No, but there are four elements. With $0$-based arrays, the cumulative count points one past the position.

@DanielFischer I haven't really understood it... Could you maybe explain it to me at this example?

Wow @BalarkaSen a "large Galois representation", that's not even on google haha

@evinda Write the array elements in base $4$, that gives $33,\, 13,\, 32,\, 01$. Count the numbers with last digit $k$, that gives 0->0, 1->1, 2->1, 3->2. Now take the cumulative sum of the count array: 0->0, 1->1, 2->2, 3->4. Now we fill array B, walking backwards from the end of array A. Last element is $01$ with last digit $1$.

We look up $1$ in the count array and find there is one number whose last digit is $\leqslant 1$, so that is placed at position $1-1 = 0$ in array B. And we decrease count[1] so after that, count looks 0->0, 1->0, 2->2, 3->4, and B looks 01,-,-,-. We move forward to the penultimate element of A, which is 32 (in base $4$).

Last digit is $2$, so we look up $2$ in count, the value is $2$, hence the element is placed at position $2-1 = 1$ in array B. We decrease count[2], and get 0->0, 1->0, 2->1, 3->4 in count, and 01,32,-,- in B. We move to the next element of A, which is $13$ ($= 7$), with last digit $3$.

We look up $3$ in count, that gives us $4$, and hence the element is placed at index $4-1=3$ in B. Also, count[3] is decremented. Now, count is 0->0, 1->0, 2->1, 3->3, and B is 01,32,-,13. We move the the first element of A, $33_4$. We look up the last digit: count[3] = 3. So that element is placed in B[3-1], and decrement count[3]. Final looks: count is 0->0, 1->0, 2->1, 3->2 and B is 01,32,33,13.

Can someone tell me if we have to prove this in that way?? math.stackexchange.com/questions/1047420/…

@DanielFischer I will think about it and will be on again tomorrow... See you... Good night!!! :)

@MaryStar there is a 2-line proof on the wiki, and the wiki does not construct a subsequence of g_n's (the wiki calls your g_n's as f_n's)

@bolbteppa So, is what I have written wrong??

@MaryStar it looks to me like you are using subsequences and the wiki is using closed sets, those methods of proof are usually equivalent e.g. in proving Heine-Borel,

I do not know if it's wrong, a test would be to read the wiki proof and see the logic of the proof using closed sets, then match your subsequences proof to that logic, I need to re-do all my measure theory :(

@robjohn should this be migrated?

@skullpatrol I wouldn't migrate it.

The answers seem to be pretty good there.

It has been nicely answered where it is @PedroTamaroff

@skullpatrol Yes, I think that too.

But it could also be thought of as a math education question to, right?

I am about to change my name

To?

Not really, @skull. Anyway, a question shouldn't be migrated if it would also fit on another site: it should be migrated if it would only fit on the other site, eg it's not a fit on the current site.

I dont know

But if somebody comes to this network looking for an answer to how to read math expressions, do you really think they're going to look for the answer on English Language Learners @MikeMiller?

Maybe, @skull. I think what's more likely is that someone would google "how to read math in english" and maybe stumble on that question, no matter which site it's on.

hi

@TomCruise raider nation has plenty of cool names ;-)

I want it to be sort of math related

Hey, can someone quickly explain to me why $x_{n+1} = \lambda x_n \forall n \in \mathbb{N}$ has the solution $x_n = \lambda^{n - 1}$.

plug it in

you get $x_{n+1}=\lambda \lambda^{n-1}=\lambda^n$

to get $x_{n+1}$ you know you muliplied $x_n$ by $\lambda$

Is it possible to solve for it a priori, ie not knowing what the solution is in the first place?

working with a few couple of terms, I do get that $x_3 = \lambda^2 x_1$, $x_4 = \lambda^3 x_1$

you can prove it by induction I suppose

but what is $x_1$?

you need to define the first term before the recursion can take place

dont know.. $x = {x_n}$ is a random sequence in $\ell^2$

well, see thats exactly where my confusion is :P

all I know is that $x_2 = \lambda x_1$.

I suppose you want to start at $x_1=1=\lambda^0$

That would be easy, but $x_1$ really is arbitrary

OH I READ IT WRONG lol.

We are LOOKING for a non-zero ${x_n} \in \ell^2$ such that $$(x_2, x_3, x_4) = (\lambda x_1, \lambda x_2, \lambda x_3)$$

so doing what I did, and setting $x_1 = 1$ a possible solution to $x_{n+1} = \lambda x_{n}$ is $x_n = \lambda^{n -1}$.

yes

So apparently this is in $\ell^2$, but I wonder what garuntees that?

you have to check

OHHH, lol Another mistake.

$|\lambda | < 1 $ so $x_n = \lambda^{n -1}$ is a convergent series.

the series of squares has to converge

yes!

You'd best have $|\lambda|<1$.

The $|\lambda| < 1$ is given to us :)

I have to read more carefully !

@masfenix: I tell my students frequently that reading is a prereq for math.

Just a FYI! I am trying to show that $\lambda \in C, |\lambda| < 1$ is an eigenvalue for the left shift operator $S_l$ defined by $$S_l(x_1, x_2, x_3, \ldots) = (x_2, x_3, \ldots)$$

By the way, @Ted, I asked you a long time ago whether Whitney's embedding was tight; it's not. All compact 3-manifolds embed into $\Bbb R^5$.

It's that time of the semester again: youtube.com/watch?v=mC8Vh76vy0w

@TedShifrin Teeeeeeeeeeeed.

Your final is too long.

How much time does a student have to finish it?

I did 1-5,7. I'm too lazy to do 6 now.

But the answer is $\frac{\pi}{2e}$.

What's his final?

I want to look at it and call you a huge weenie.

@MikeMiller He shared this.

8150? I thought the big numbers were for grad courses.

For example, writing the details of 9 is doable but tiresome. "Yadda yadda schwarz lemma yadda fixed points yadda $\eta\dfrac{z-w}{\bar wz-1}$."

I guess this is the graduate complex analysis class.

that is a pretty long exam

@MikeMiller I'd doubt it.

I figured they'd get through more than normal families, though.

I checked, it is.

although it should be doable in 2-3 hours for most students

@MikeMiller Oh.

@TomCruise No, that's not that much time.

It took me one hour and a half to do four exercises in my last midterm.

It is a 1-semester grad course. The final doesn't cover everything. We did Riemann mapping, order of growth of entire functions, Hadamard's Thm (see the last question on tne exam). @Pedro @Mike

@TedShifrin But it is looooooooooong.

I suppose that by $|dz|$ you mean $|f'(\zeta)|d\zeta$.

Awesome, @Ted

All my exams are long. My courses aren't for wimps :)

I mean $ds$, @Pedro

@TedShifrin AGH.

@TedShifrin I'm just saying I could tell you how to solve everything in there in less than an hour. But writing it down...

@TedShifrin What course is that though? Is it like Mike mentioned and it the high course number corresponds to a graduate course?

If by $f$ you mean a parametrization of the curve, yes, what you said is the pullback.

Yes, it's our first-year grad qualifying exam course. It's about comparable to the grad course I took at MIT as an undergrad.

@TedShifrin So you do mean to take that, I mean, if I take $\gamma(t)=2 e^{it}$, $|dz|=2$.

$2 dt$

hmmm is it easy to see that $\bar{\lambda} = \lambda$ for $|\lambda| < 1$?

@TedShifrin Yeah, sorry

@Pedro: The qualifying exam in algebra we just wrote is 8 questions, several with several parts. Three hours.

@TedShifrin Booooo!

=D

Huh? @masfenix

They're supposed to be hard...

That'

otherwise what would they qualify you for?

That's scary.. applying for PHd next year.

Many questions are very familiar from working past exams, @Pedro

Not that hard, @Mike, if one knows what one's doing.

Sure.

@MikeMiller My point is it is not hard.

It just takes time.

I claim to know something about the difficulty of qualifying exams, I guess.

@Pedro: For the line integral, you want to write $|dz|$ in terms of $dz$ and use Cauchy/residues.

@TedShifrin I did that, of course.

Ok ....

@TedShifrin The Laurent series part is a bit tiresome. How would you suggest doing it? I expanded them but then I have to consider some quite a lot of products.

Specially when negative terms appear.

You like to say tiresome!

Exhausting!

i don't remember the question ...

5.

I don't have it here. I'm on my iPad. Going to sleep :)

@Ted here's the exam

You should not yet know how to do the last question ....

@TedShifrin Me?

Perhaps. Hehehe.

I will try.

Yup.you.

@TedShifrin What about truncating the Taylor series and using Rouche?

Cool :)

Rouché to show infinitely many roots? What do you compare with?

Good night!

@TedShifrin I don't know!

Night ;)

I guess he wants to compare to an infinite series of $f_n$s, each of which have more zeroes.

Good luck :)

There is more depth to complex analysis ;)

Ok, tennis in the early AM. Night!

hi

:(

nobody likes me

I do.

I changed my name ;)

and my picture

burps

I am formerly Tom Cruise

cool

I will have to stick with this name now for a couple months

I like it.

you like Mozart?

his music is very simple yet elegant

easy on the ears

Rock me ammadaous

hah

Just kidding.

did you know Steven Hawking used to blast Wagner while he was studying at Cambridge?

that would disturb me

that is a grumpy cat

I did not know that about hawking.

did you just take that picture?

o________O

my latest attempt to become famous: math.stackexchange.com/questions/1067342/…

someone upvoted me, that makes me happy

Maybe in normal spaces zero dimensional is equivalent to totally disconnected... though I doubt it.

hi Hakim

@Hakim

Hi @ForeverMozart

let me reheat my coffee, and we'll chat

TomCruise = @ForeverMozart?

LEL

LEL =?= LOL

Laughed Enormously Loud.

icic

:)

@BalarkaSen

@ForeverMozart "Note that "every nondegenerate closed subset of X is disconnected" actually implies that every nondegenerate closed subset of X is totally disconnected." <-- this is false why should it be true?

but if you have a nondegenerate connected subset of a subset of $X$, it is also connected and nondegenerate in $X$

hmm oh wait. it is true

in fact more should be true

zero dimensional would be nice

that is stronger than totally disconnected of course

if $X$ is a normal connected space, then for every nondegenerate closed subset S of X which is disconnected it holds that S is also totally disconnected

yes

^ that is also true

but that is the same thing since we assume all such $S$'s are disconnected

yeah, more or less

I think my proof works if we say all such $S$'s are zero dimensional

but I'm not sure if there exists a space $X$ with that property...

thinking

if every proper subspace of a space is zero dimensional, then must the entire space be zero dimensional, contradicting connectedness of $X$?

Then every point of the space would be an explosion point

yes, but probably we can prove that if X is a normal connected space and S is a nondegenerate disconnected closed subset of X then S is zero-dimensional

I'm trying to think if every point in the space $W$ we talked about is an explosion point...

my hand hurts

damn long proofs

lol, what you proving?

uniqueness of the heat equation and for extra credit...why is it not unique

I'm typing all the proofs in Latex... f rewriting that shiz

it's not unique because we can't have a contained heat = insulated heat

can't have flux = temperature...doesn't make sense

@BalarkaSen we really just need some zero dimensional set $A$ whose closure contains $C$...

differential geometry? @usukidoll

thermodynamics?

PDES

partial differential equations

oh right

more like possessed demonistic bullshit

pee DEs, @usukidoll

ha! XD

that's how I feel about your PDEs

differential geometry requires lots of PDE's for existence and uniqueness of flows, geodesics, etc...

surprisingly I'm passing that shit with an above average grade, but dem calc iv students want me to tutor them next year because they are taking pdes omfg no

And then there's Ricci flow

I ain't tutoring that shit that's for damn shure

differential geometry is a beautiful area of math, but I quit once it became so linked with PDEs

@ForeverMozart have you studied algebraic topology?

Not really, I know a little about homotopy, fundamental group...

my homework for pissdes were 20+ pages wtf

There is a grad student in the physics chat room The h Bar, right now if you need help @usukidoll

link me @skullpartrol

ok I figured out why it's not unique... I mean suppose we have $u(0,t) = u_x(0,t)$ That means that at one end of the rod, our heat is contained and at the other end of the rod, our heat is leaving. How the hell can our heat being contained be equal to the heat leaving.. physically it doesn't make sense. It's either contained heat = contained heat or leaving heat = leaving heat. Can't have one and the other equal to each other

zero-dimensional looks hard @ForeverMozart

yes

Sorry pal, I'm on mobile @usukidoll just go to all rooms at the top right corner, the most active room show up first.

"The h Bar"

Do you understand what I'm doing in my "proof" @BalarkaSen ? I used the assumption that $X$ is connected in there somewhere, and also some properties of $\beta X$ for $X$ normal.

($X$ is connected means that an open subset of $X$ cannot be a singleton... that is how we know $A$ is nondegenerate)

and we use the standard open and closed bases for $\beta X$, disjoint closed sets in $X$ have disjoint closures in $\beta X$, etc...

More or less, @ForeverMozart, I haven't gone through the whole details.

yeah thats ok, I just sort of outlined my argument

hold onto your hat...

if I can find this paper I'm looking for...

I need to find assumptions under which zero dimensional is equivalent to totally disconnected

maybe that would be a good question to post

@ForeverMozart I think Hausdorff works

hmm

oh yeah they are equivalent in compact spaces, but that does me no good

@AlexanderGruber!

yo

How are you @AlexanderGruber?

finally done teaching. exhausting semester

Hmm...how's the back feeling?

Exhausting yourself can't help...

@AlexanderGruber Do you know of a good way to define \delta-hyperbolicity in uncountable groups?

@skullpatrol it's holding up.

i'm back to weightlifting, though I still can't deadlift

Get some well deserved rest pal :-)

@AlexanderGruber deadlifting -- lifting deads?

@BalarkaSen it just means lifting a barbell off the floor

it's a back exercise

and thighs

the best exercise is to sleep.

:P

That's the only exercise you're guaranteed to be well rested after :P

@BalarkaSen after a good deadlift routine, you sleep like a brick.

Do you use a belt?

Can someone quickly check my work on a limit? Someone already answered my question with Taylor series but I want to know if simply plugging in a value is acceptable ... math.stackexchange.com/a/1067427/111946

hi

anyone have any ideas about math.stackexchange.com/questions/1067443/estimating-the-mode ?

@AlexanderGruber Heya alex, you know python right? =)

Hey @DanielFischer, one question: Suppose $\{A_n\}_{n=1}^{\infty}$ is monotone decreasing sequence of positive bounded operators. For $f,g \in H$ (some Hilbert space) and $m \le n$ we find that $|\langle (A_m-A_n) f,g \rangle|^2 \le \|A_1 \| \|g\|^2 \langle (A_m-A_n)f,f \rangle$. And that is okay. But then, for $g=(A_m-A_n)f$ they say that it becomes $\|(A_m-A_n)f\|^2 \le \|A_1\|(\langle A_mf,f\rangle - \langle A_n f,f \rangle)$, and I don't understand how? I mean, why $\|g\|^2=1$?? Or..

@skullpatrol I think it's fine there

@robjohn thanks for having a look see :)

@Cortizol It's not $\lVert g\rVert = 1$. Note that in the first inequality you have $\lvert \langle (A_m - A_n)f,g\rangle\rvert^2$ on the left, and if you insert $g = (A_m-A_n)f$, you get $\lVert (A_m-A_n)f\rVert^4$. Then you can cancel $\lVert g\rVert^2$ (here, you can "cancel" that even if it's $0$, since the right hand side of the inequality is non-negative).

yay @robjohn!

@DanielFischer Oh...Silly me, I thought (in my head) that on left is $\| \|^2$ and not $\| \|^4$ when we plug $g$. Thank you!

@usukidoll cheering in general, or did I do something good?