I was more thinking about trying to factor $x^4+1$ explicitly, Demonark. Certainly what you're doing doesn't work in characteristic $2$.

when an exercise looks dull, I try to put it in context, give background and explain why it is interesting. I guess that's over their head sometimes.

Ah right. Well, char 2 can be done as a separate case since $(x+1)^4 = x^4 + 1$

Well, if they're struggling to do basic stuff, a sophisticated context will not engage them, @Mathein. But if you throw it out for the best students and tell them to come talk to you, that's a good idea.

OK, Demonark. I haven't really examined your argument carefully to see if I buy it.

if you char 2 separately it's fine

@TedShifrin I only do that after I have persented the whole elementary solution of course

@TedShifrin yeah, it's more about throwing it out

like for example, they had to prove that $2,3,1\pm \sqrt{-5}$ are irreducible in $\Bbb Z[\sqrt{-5}]$

I told them that this is interesting, because this gives two distinct factorizations of $6$ into irreducibles, so we see that unique factorization fails in some rings

Actually, I think that's a good general comment for them all. Not over the top.

I used this to talk about Kummer's work on FLT (technically Dedekind's work to understand Kummer), since this failure of unique factorization is what started ring theory

Talking about ideal class groups ... over the top :)

But I was very vague and didn't go into details. I just told them that some mathetician before Kummer just assumed unique factorization which would give an easy proof of FLT, but that somehow the reason why FLT is so hard is because unique factorization fails sometimes. and that to investigate this further, Kummer and Dedekind basically introduced ring theory

I didn't mention ideal class group

Math history and perspective are woefully lacking in most courses.

yeah, I think historical perspectives can be great to put things into context

so do you think that was over the top or not?

I think a minute digression on this history is great.

okay good

oh and since we were working with norms all the time, at the end of the session I said something like "you can define norms in a general way through Galois theory, if you don't know this, don't worry, you don't need it for this course! And then I gave the general definition of a norm and showed how it agreed with that we had done. (that took maybe 2 minutes) I did this because I know that there are some students who already know Galois theory

I'm fond of the occasional (optional) exercise for such students ... But I know you're not writing those.

I did that partially because it wasn't really clear how one would come up with the norm function which somehow magically has all the properties we need

Well, it does come naturally from $|z|^2$ in the case of $z\in\Bbb C$. :)

we also had real quadratic fields

Well, sure.

Meow !!

hello

how has everyone been

Actually my algebra TA does have some supplementary problems from problem session. They're often not totally doable, he tells you this up front and gives them just as an optional "Here's something to play with if you're bored" type of thing

Haven't seen you in months!

I never object to things like that, Demonark.

oh theres one question i had about stat but the stat teacher wasnt here today

in my diff geo class, some regular problems accidentally ended up not totally doable ...

My graduate real analysis prof when I was in college was a very famous mathematician and used his own book. His take-home final was half unsolvable problems. I was so pissed.

It's probably too hard for us, Meow.

"the prof didn't really realize that you need some nontrivial facts about transversality that you don't know unless you've taken diff top (where the last course was offered 4 years prior or something like that)"

so we were doing sample distributions and part of the proof that $\sigma_{\bar{x}} = \sigma/\sqrt{n}$ involves showing that $\mathrm{Var}[\Sigma x] = n\sigma^2$

i have no idea how to show it

Well, this is my complaint about not being "aware" of what your students have as background, Mathein.

What happens if you just use the definition, @Meow?

it wasn't about the background really, if the prof were aware that you needed this stuff, he wouldn't have done it

definition of what

variance?

But wait a minute. We want variance of a random variable. What does $\Sigma x$ represent?

the sum of the sample

so you pick a sample of size $n$

you look at the variance of the sample sum across all samples of size $n$

and prove that it equals $n$ times the population variance

i was thinking like

you choose a sample of size $n$, so there's $n$ of the variance but i dont know if variance works like that (it probably doesnt)

So what's $\text{Var(X+Y)}$? It works linearly for independent things, not for non-independent things.

I put a question on that on my final.

thats what i was thinking

You need to assume your different samples are independent, and then it's easy.

If they're not, you get $n^2$ instead of $n$.

independent

so wait

before you say it can i try and show its linear

I'm not saying nothing.

double negative

I know you've missed me, Meow.

wait

$X$ and $Y$ are both from the same population right

Yeah, but I think you assume they're independently chosen.

These are these subtleties with $n$'s and $(n-1)$'s that show up in stat.

Hello all :)

i hate stat

Don't hate. I wish I had started teaching probability decades ago. It's neat and important.

i really liked analysis

i want to allocate time to it over the summer

You mean Spivak stuff?

god knows i have no fucking time right now with finals and stuff

yes

Well, stop stressing.

I'm suffering from the Imposter Syndrome at the moment, I swear. I was awarded a scholarship for my PhD but I don't feel good enough :/

Shaun, I've told you this before. Stop beating yourself up ... we all go (went) through this. If you really don't want to do math, then quit and go get a real job that you'd like. But otherwise, ride the waves.

I don't think imposter syndrome is terribly specific to math, regardless of the context one can always feel like their accomplishments are not out of any real competence so much as luck or riding on others or something. But the message still seems holds that these are feelings everyone has to contend with at some point or another, and to power through and not let your confidence get destroyed too much

I'm going to sleep. Bye everyone!

When my book discusses double integrals, it seems that whenever they change order of integration they take the inverse of the function that moves to the inner integral. But, they never come out and say it. Is it sometimes the case that they don't do that?

Night, Mathein.

It's trickier than that, @JoeStavitsky. You have to draw pictures of the region and think.

Yeah, I know; I'm sorry, @TedShifrin. I feel like I should be able to answer more questions on MSE than I do currently though. It's not that I don't want to study Maths. I mean: "A man must love something very much to practice it not only without hope of fame or fortune, but without hope of doing it well," Chesterton.

LOL ... I didn't mean for you to apologize. You do need a thick skin to make it.

@TedShifrin, it's just that I can't come up with an example that I would need to do anything else.

Hold on, @JoeStavitsky. I'll give you one from one of my exams.

Damn, how did that happen?

i saw that one

LOL

it appeared, and disappeared.

I thought I had a premonition of you, Ted

It's called a nightmare.

can you come to me in my final

there you are again

This is getting ridiculous.

and there you go

this is getting fun

No, it's not.

what are you trying to do

is your computer plugged in, sir

This could have been very embarrassing ...

oh boy

"easily, and as efficiently as possible"

This Feynman quote is soothing: youtu.be/ABx55cEop-o

read: I reserve the right to arbitrarily take off points

No, it's a warning to think and not be a dope.

@JoeShmo, here's a final exam from the last time I wrote one for the manifolds course.

me^

Anyway, goodnight :)

I hope you're happy now.

thanks

im terrified

11 questions?

You asked.

I don't grade on a typical 90-80-70 scale. I'm sure students got A's if they scored 65 or better.

I don't berember.

ok im pretty sure if this is our final im getting a round 0

and so is the rest of the class, if there's any justice in the world

Now you know why I told you to stop pestering me.

and now i know why you told me to stop pestering you

Now, in fairness, I did a few weeks of curves/surfaces differential geometry in that course.

Ok, I'm pretty sure I could do number (3)

Lie groups, no

What range of birthyears is the procedural generation

That wasn't comprehensive, but you can see my courses are more down-to-earth than most.

@TedShifrin, 36 years, my goodness. Have you heard the one about the calculus professor at the pearly gates?

LOL, I don't want to, @JoeStavitsky. And I'm still ruining lives.

Ted, that's the final for the undergrad class? 3510?

No, no, grad diff geo.

First term.

oh oh oh

Manifolds, etc. Second term was more Riemannian and other stuff.

@TedShifrin it's not offensive (at least I don't think so).

LOL

@TedShifrin but I'm not dickish enough to force my "humor" on anyone".

That's encouraging.

Did you solve my double integral?

our "graduate" class is closer to 3510

without any of the rigor ;)

Oh good grief.

Don't get me upset.

@TedShifrin I'll need to get back to you because I'm prepping for an exam

oh ive got so much to say

I think I'm tougher than most, but I make it possible for people to learn a ton if they work.

agh ... put that all in past tense.

wanna come to guest lecture at my school? :D

are you from around by chance

you made a reference to the LIR in one of your lectures

not me

I've driven on the LIE

ah

I have no recollection of mentioning the LIR, but I defer to you.

maybe the LIE

i defer back

I have no idea why I would have mentioned it, except for California-like traffic jam.

i vaguely recall the LIE (/LIR) being mentioned, and then a pun followed soon after

so perhaps you were talking about Lie groups in passing

I doubt that. But i'm not going to watch 112 videos to find out.

Brouwer's proof is neat

very neat

Just cohomology :)

by the way, for the unit normal of the sphere im getting n(x) = 2 (x1, ..., xn)^T, what did i miss?

not unit

in which case the proof still holds, 2n*vol(D_n) > 0 still

right ok, not unit

well, the volume form on the disk is mu_(D_n) = n . *dx

No, $\star 1$.

thats the volume form on R^n

mu_(R^n)

Same thing.

The volume form on any Riemannian manifold is $\star 1$.

imgur.com/a/Hb40Fmb

It makes no sense to have $n$ as unit normal when you're talking about the unit disk.

I have to interpret that as $n$, not the normal. And it makes no sense.

You meant sphere, not disk.

yes

You still lose 50%.

Precisely.

I didn't solve the double integral. watching lectures and doing work instead. also i dont know how to integrate fractions. im not wolfram alpha

arctan something

Does anyone know how to show that any graph with 5 vertices and 7 edges has diameter 2?

I know how to show $\text{diam}(G) \geq 2$ but I'm having trouble proving the inequality in the other direction.

there are two possibilities: you win the lottery or you dont. therefore there is a 50% chance of winning

@user76284 You may find it useful to determine possible complements of a given graph with those properties.

I was thinking about taking the complement as well (and comparing its number of edges), but I'm not sure how to translate the diameter of $G$ into a property of $\overline{G}$.

I know $\overline{G}$ has only 3 edges.

The problem, then, is to find a 2-path for each edge of the complement.

Sorry, what do you mean by a 2-path?

Sequence of vertices $v_1v_2v_3$ such that $v_1v_2$ and $v_2v_3$ are edges.

There are only two complements up to isomorphism, so there should be only two graphs to check up to isomorphism as well.

I found three distinct complements

Oh, yeah, you're right. Still, only 3 graphs to check.

(I forgot about $K_3$ somehow)

There might be a slicker proof, though.

I wouldn't doubt it.

My brain has stopped working after 4 final exams in the last 4 days :(

I know how to proceed by cases given the complements, but I assume you had something else in mind? What do the vertex sequences you're referring to imply?

You mean find a 2-path in the original graph $G$ for each edge in $\overline{G}$?

Yeah, it should not differ at all from your approach.

Oh yeah, that makes sense.

Also, I think there are actually 4 complements. Whoops.

Ah, yes.

What's the fourth? I have $P_4$, $K_3$, and $P_3 + P_2$.

A star.

Ah of course.

That's really it, for real this time.

>_>

I suspect there's a proof that doesn't proceed by cases then. Something like the Moore bound.

I still had to use a case construction in a proof I just cooked up, but the cases weren't the possible complements.

I don't want to spoil it for you though. :P

Ok, let's see. Suppose $\text{diam}(G) \geq 3$. Then $\exists a,b \in V(G)$ such that $d(a,b) \geq 3$.

Let the other vertices be $c,d,e \in V(G)$.

I think you're on to a better proof than I had.

$\{a,b\} \not\in E(G)$ for this would mean $a$ and $b$ are distance 1 from each other.

{a,c} and {c,b} cannot both in the graph (this would give a path length of 2). Same for {a,d} and {d,b}, or {a,e} and {e,b}.

Nice

Thus there must be at least 4 edges not in $G$. Thus $G$ has at most $\binom{5}{2} - 4 = 10 - 4 = 6$ edges. But $G$ is supposed to have 7 edges. Contradiction.

Nicely done.

anyone?

@Semiclassical, if $A$ is $m\times n$ matrix of rational numbers and $b$ is $m\times1$ vector of rational numbers, then $Ax=b$ has a rational solution. But it may have non-rational solution, too right? Also, if $A$ is square, then do we get rtional solutions only?

@Silent: If $m>n$ you may not have a solution at all. If $A$ is square and nonsingular, then the unique solution is rational, yes. But if it's singular, you'll have either zero or infinitely many solutions (most of which aren't rational).

Thank you very much!

Standard Deviation formula predicts that the variance is inversely proportional the number of observations. Which means that the larger number of data points we have, the smaller is the variance and lesser is the uncertainty in the data. But, are there any conditions there this may fail? I mean for eg, let we have highly duplicated data points. One physical example may be the case of an optics experiment where one may exchange u and v (obj and img dist). Let we have one observation u=5,v=10.

We can readily get a second data with u=10,v=5 (by principle of reversibility of light). So now we have created two data points from a single one but its basically duplicated logically. Is this two data going to reduce by uncertainty or variation? I doubt because both are same and ideally it shouldn't. So what will happen?

Can I arbitrarily duplicate n number of data into 2n no. of data without any conditions and make the variance arbitrarily small?

the only such standard deviation formula i know is based on the central limit theorem, and that's definitely not always valid

an obvious issue is that not all pdf's have well-defined variance

so if that's what you're sampling from, you shouldn't expect to get a sample variance which converges to 0 as the number of samples goes to infinity

What is pdf?

probability distribution function

I didn't understand the last part-"so if that's what you're sampling from, you shouldn't expect to get a sample variance which converges to 0 as the number of samples goes to infinity"

Actually I don't know much stats

well, note that the central limit theorem has a few assumptions

[Random] Pr(skull patrol active) propto extent of how dead a chat is. Cause: No idea

one of which is that the variance of the underlying random variable is finite

Is the example I provided violating that?

dunno.

i don't know so much about optics

Do you know anyone round here who might help?

but it seems plausible that you could arrange for physical examples for which the CLT won't hold

Thanks... But actually I am not being able to correlate the maths and physics here

then i really have no idea what you're hoping to get out of this

can you come up with examples where the sample variance doesn't go to zero as the number of observations goes to infinity? definitely

do i know any firm examples off the top of my head? not really.

@Semiclassical, if a real matrix has real eigenvalue, how do we show that it contains real eigenvector?

ok, got it.

:S

@Daminark, is it true that there is a field of order $p^n$ for every $n\in \Bbb N$ and prime $p?$

Yeah

@Daminark can you please provide some link which contains proof? I googled some keywords but could not find a thing

I'd expect it to be on proofwiki, but it's not actually that hard to prove. Do you know about splitting fields (in particular, their existence)?

ok, i will learn about that today.

That's how you would go about proving it. Basically, you show that every polynomial has a splitting field, and the splitting field of $x^{p^n} - x$ must have $p^n$ elements

ok. thanks

@Silent because the theory works in any field, and Q is a field

@LeakyNun Thank you! Will you please check this: an odd order group with order $n$ is isomorphic to a subgroup of $A_n$, and in general, any group of order $m$ is isomorphic to a subgroup of group $A_{n+2}$

I meant $A_m+2$ in last line, too late to edit

eh, are you asking me how to prove it?

@LeakyNun no, no! just check it! if i have understood them correctly.

maybe

I think so

it feels right to me

ok

you see, Cayley gives a map $G \to Sym(|G|)$

$G \to S_{|G|}$

and then we have a map $S_{|G|} \to C_2$ whose kernel is $A_{|G|}$

but for any $g \in G$, its image in $C_2$ being $h$, satisfies $h^{|G|}=e$, i.e. $h=e$

so the image in $S_{|G|}$ lies inside $A_{|G|}$, proving the claim

as for your second claim, Cayley gives a map $G \to S_{|G|}$, but $S_{|G|}$ is a subgroup of $A_{|G|+2}$

by transposing the last two elements depending on whether the permutation is odd or even

thank you so much

hello

Anyone here has an idea about JEE?

what do you need to know?

How to prepare for JEE?

@Silent You know JEE exams? They are one of the hardest exams on earth

oh, there are other online forums, which can guide you better,

Or try asking someone here

Phd exams are quite easier than JEE, because, the chances of selection is 18%, but in JEE it's 0.0023%, in MIT entrance it's 5.3% and in Oxford it's 3.4%

I have ML Khanna's IIT mathematics, it looks pretty good, and flawless. But i would say the TMH book on math is not so good.

TMH?

what's it?

@AbhasKumarSinha tata mac graw hill

But, TMH is one of the reputed book out there.....

You've done IIT?

@Silent mean, have you cracked IIT?

@AbhasKumarSinha Well, i think these numbers do not mean much, you see, almost anyone can apply for jee since it is very easy to fill a form and sit for an exam

so, selection rate has to be low

@Silent really? Most of them are really bad students?

same goes with ias aspirants, only 10 percent applicants are real competitors, others are just to create hype

@AbhasKumarSinha not very good

@AbhasKumarSinha no

@Silent Are you a Phd?

i have lived and am living with some of IITians, though, and i must say that if you get good guidance, then this is an ordinary exam

@AbhasKumarSinha no! i am preparing for msc math

entrance

@Silent I've been state toppers here multiple times in exams like Olympiads and stuff and I'm reading in class - 11th now (session not started, but coaching already)

@Silent You want to become a teacher?

no plans :)

@Silent why?

@AbhasKumarSinha best of luck

@Silent Thanks :)

@Silent How's your mathematics side?

I am learning analysis and algebra.

Do you feel mathematics sometimes harder?

Yes! it is hard, but amazing. No other subject, except for logic and philosophy has attracted me this much

Once, you'll start understanding Logic, then mathematical world becomes more beautiful than real ones

You know we have 2 different mathematics teachers in our school, one of them says that we should practice about 200+ questions a day, and other says, just practice NCERT, and understand the concepts, later everything will become easy

But, I only do NCERT, it's enough for me with Coaching Materials. NCERT is really an interesting Textbook out there

hello can some one tell me why this question is closed ?

@Abra001 you've to add details and also, you've to add some details on how you tried and tell how much you've got it proved

i would, if i were the author.

@Abra001 You've got an answer

Maybe this person has no idea where to start, genrally the question bears an interesting add-up for the topic of modular arithmetics. I wonder why such a question gets closed.

scontent.fbbi1-1.fna.fbcdn.net/v/t1.0-9/fr/cp0/e15/q65/…

Can anyone mathematically prove this?

@AbhasKumarSinha we have no information about rotation degree.

@Abra001 this can be proved without that hint: area is constant

Area in blue and red ones are constant, now go ahead ;)

@Abra001 you figured out the way, i figured out 2 ways till now and working on 3rd

triangle ratios ?

@Abra001 I don't think that'd work, but you try

that would always work, but not elegantly.

try try try, it's not that hard,

or should I tell you my way

oh dude

don't tell me they are .... equal!

@AbhasKumarSinha right, if the blue rectangle is a square, areas are equal right?

@Abra001 yes

you want to hear how I proved it?

there is two equal right triangles of area A

minus A plus A is always same value

@Abra001 That's also another way. Good :)

But i wonder if the square steps over the right angle.

an easy way is to draw 2 perpendiculars in red shape to the side of the square

which means $\theta > pi/4$

@Abra001 No need of using angles

just overlap blue over red shape, you'll see how?

use your imagination ;)

the triangles have added and subtracted the same area on the shape

got it?

or shall I draw an diagram to make you understand?

dude i know what are you talking about

just in mathematics, everything needs a proof

yes

you can't tell this area is equal to this one because it's obvious, you need palpable proofs.

@Abra001 hold an second

that's interesting for a problem that exists in FB.

Overlap red shape over the blue and you'll see a similar figure

XD

@Abra001 Sorry, it's just an rough figure, you understood what I say?

i did from the beginning.

overlap red shape formed between the squares over the blue shape formed by the same squares, then you'll see it

@Abra001 specifically $x^2$

another way of doing this is by taking the area of two squares constant then using intersection formula as similar in sets theory

I just saw another good question about collatz, put on hold to be sold.

hmm

@Balarka are you here?

@MatheinBoulomenos math.stackexchange.com/questions/2775215/… this was the question i was talking about yesterday, if you want to help me and make some rep ahahahahaha

What is the boundary asymptotic of a surface?