god help us my daughter wants to drum on her drum.

drum

drumm

drrumm

drumm

drum

get her an electronic drum set with a headset.

Nope

A cheeki breeki hardbass drum

With a daily supply of chebureki

that would be a crimea

Also cold kompot

You're an idiot for giving her a drum.

ted is right, as always

it turns out she didn't want to drum on her drum, as much as she wanted to store the drum and the drumstick in containers. but we did get a pretty good beat for about 60 seconds.

seems you got off lucky this time then :P

hide the drum

then she was upset that the cat was sitting on one of her dresses, and pulled the dress out from under the cat, and got swatted in the head.

the fun never stops

it was pretty close to a hardbass beat

THINK......right Ted? Instead of just doing things like a robot.....

never truly "thought" about the losing of information

I have no idea what robot you’re talking about, @dc3rd …

I’ve made that comment numerous times … to robots other than you.

@copper.hat Almost as bad as a Bob Johnson

. o O ( had to break the silence )

i used to live in johnson county, iowa

must have been a fine place ;-)

boris has not vaccinated my daughter yet :-(

Hi ... Is it possible to get different results when we use inequalities of Arithmetic and geometric mean ? If yes than which should be used ....

@ronakjain There is not enough information in your question to even think about answering.

Good night!

Good night copper

ddmmyy

@MadSpaces No, it's not cgt., it doesn't satisfy the necessary condition for convergence of a series.

@Koro How come?

Also unrelated question but if we have two sequences $ a_n , b_n $ i know that if $b_n > a_n $ and $b_n $ converges then must $a_n $ as well converge. Can we say if $ a_n$ does not converge then must $b_n $ not converge as well?

Now that i have wrote it, i believe its stupid, obviously thats true. otherwise we would have a contradiction.

@MadSpaces calculate the limit of nth term. What do you get? Surely you don't get 0

For n goes to infinity we get zero

@MadSpaces if $a_n, b_n$ are sequences of positive real nos.

@MadSpaces how?

Sorry it should be n squared as exponent

$\frac{|x+3|+x}{x+2} >1$

How can we solve for x ?

Got it

$|x+3|+x>x+2$?

$|x+3|>2$?

$x>-1$?

Is that the process?

Also @leslietownes when you wake up, I need you to know I made a terrible financial decision. I took out a loan at 13% interest. Worse there is a clause that early repayment of the loan is met with a fee equal to 7% of what would have been the interest earned multiplied by the remaining number of months on the contract.

13% ??

13%

I feel like Ive been screwed

@AndrewMicallef O_o

But I got a "chin up, have some chocolate" from the lender....so tgat is something

Early repayment clause is valid for entire loan duration? It should be valid if early repayment is done within say 1 year . No?

@Wolgwang cant i multiply both sides by $x+2$?

@Koro nah, looks like entire duration

In 72/13 =6 years (approx), you''ll be paying almost double of the loan amount. That sounds very bad.

@Thorgott: I managed to solve the group of order 15 problem

without using Cauchy theorem, class equation or Sylow's theorems

@Koro yeah, and the contract has already been signed...though I feel I didnt get a chance to read it. The lender showed up at my work,just after my break, was like "thisll just be a minute, just what we discussed, sign sign sign" and I was busy trying to get back to work...

@Xnero like fewer steps?

@Koro 72/13. ??

@AndrewMicallef What if $x+2<0$?

@AndrewMicallef But there should be some document submission from your end to the lender. If you delay it, the loan sanctioning may get delayed or may not happen at all! But if all documents have been submitted and it's not too late already you can request them to stop sanctioning the loan by withdrawing your application.

@AndrewMicallef calculating approximate years taken to double money: if rate of interest is $r$ % annually (compounded) then it takes approx. 72/r years to double the money. $r$ should be small though.

@Xnero huh, will have to grab some paper and go over that

@Wolgwang dunno, I guess you get what Xnero just got ;)

@AndrewMicallef this simplifies to: $x+3\gt 2 $ or $x+3\lt -2$.

@Koro Well I took the loan to buy a car, I havent actually signed for the car. So the loan hasnt come into effect. But if I reneg on the loan now I am still up for admin fees...so I should try to accept it, or think about the whole thing as a sunk kost

Ahhh, got it!

@Koro grats

I guess Im just peeved by how this all transpired. I went to buy a car, asked for a loan, was shown a loan at 3%. Signed to buy the car. Was told I couldnt have that interest rate, got told itll be closer to 6%. Said: sure sounds fine. Got asked to pay for car, said: am waiting on loan. Loan guy shows up says sign here, oh wait it is now 13%

And I thought I coukd trust all the actors because, this is the car dealership I work at

/end rant

there is competition among financial institutions, you should have known which ones give the loan at reasonable interest rate. @Andrew.

13% seems very high to me...By reasonable, I mean not exceeding 8%

Yeah. Turns out I dont have a good credit rating huh

and the problem is: even stock markets aren't giving good returns these days (to me at least)... I doubt such returns can beat your 13%

13% is a lot to ask a random set of stocks

I'm in a bind, I cant afford a decent car outright; have shitty credit, and need a car to earn more money

So Im hoping I can do enough work travel for the tax payer to cut the check for the interest :/

Also the lender had the balls to tell me with a straight face "cars dont lose their value; this is an investment!"

@MadSpaces I am sorry if someone else has answered, but I did not see it. Note that $\left(\frac{n-1}{n+1}\right)^n=\left(\frac{1-\frac1n}{1+\frac1n}\right)^n=\frac{\left(1-\frac1n\right)^n}{\left(1+\frac1n\right)^n}$

hey chat

so my topology professor made a few proofs about nets and the neighborhood directed set and one of them is the "i've seen that in metric spaces!" one: $f:X \to Y$ is continuous iff $\forall \{x_\lambda\}_{\lambda \in \Lambda} \to x$ convergent net, ${f(x_\lambda)}_{\lambda \in \Lambda} \to f(x)$

i had a question he didn't know how to answer. is there any countability property in metric spaces such that this proof is exactly the proof for metric spaces? i mean: is the neighborhood directed set in metric spaces a countable one? or can you at least use something like a local basis or something

not sure if that answers your question, but in a metric space, you always have the countable neighborhood basis (which is naturally directed) by open balls of radius $1/n$ about the point

and, modulo phrasing, I think that's pretty much the point of the proof in metric spaces too

thanks @Thorgott, i think that does it

how are you doing?

I'm doing fine, yourself?

Yes first countability is enough to get the equivalence of continuity and sequential continuity

fine too, struggling a bit with basic representation theory because i missed a few classes

nothing too hard but there are like 40 pages of content i need to read before today's classes

Hm what's a space wich is not first countable but in which every sequentially continuous function is continuous? Maybe $\omega_1+1$ works

aren't you asking for a non-first countable Frechet-Urysohn space

Ah now that you mentioned it I think that the equivalence of sequential continuity and continuity holds for a space iff the space is sequential

(or maybe Frechet-Urysohn was the right notion, I always get super confused between those two)

Apparently the Fort space is an example of a space that is sequential but not first countable, while omega1+1 is not sequential

ah no, you're right, it holds precisely for sequential spaces

the difference is dumb and super confusing

I think we discussed this confusion before, the issue is that taking sequential closure is in general not an idempotent operation

since you can't necessarily diagonalize in a sequential space

looking at Wikipedia again, a Hausdorff sequential space is Frechet-Urysohn iff you can diagonalize, which makes sense and is a somewhat funny characterization

Yeah I think F-U means you take sequential closure and you're done

While sequential means you keep taking sequential closures and for some ordinal you'll be done

right

Polynomials $f,g\in k[x]$ have a common zero iff their resultant (being defined as the determinant of the Sylvester matrix of $f$ and $g$) is zero. The resultant of $f$ and $g$ being zeros implies (as far as I know) that there exist polynomials $s,t\in k[x]$ (or certain degrees) s.t. $sf+tg=0$.

I tried showing this holds without using the resultant, i.e., if $f$ and $g$ have a common zero $\alpha$ (where $\alpha$ comes from an algebraic closure $L$ of $k$), then we can write $fs+gt=0$ for polynomials $s,t\in k[x]$. Clearly we can write $f=(x-\alpha)\cdot f/(x-\alpha)$, but the problem is that $f/(x-\alpha)$ lies in $L[x]$. So it seems I would have to do something else. Anyone any idea? ;v

Never mind

found the solution on stack

donno how commendable that is, but whatevs

@robjohn sir can you please come to our room

how can we integrate:

$$\int_{0}^{t} sin(t-x)f(x)dx$$

where f(x) =0 in $[0,\pi) $ and sinx in $[\pi, 2\pi)$, and is periodic with period $2\pi$

What is difficult about doing the integration?

i mean the trouble is with the piecewise function

we dont know where t is

i think we can proceed by writing t= $2\pi k + b$, where k is an integer and b is in $[0,\pi)$

this will reduce the entire integral to

$$k*\int_{\pi}^{2\pi} sin(x)sin(t-x)dx$$. The integral is trivial, and we can write k as floor(t/2pi) to get everything as a function of t. But the answer doesnt match

i get the answer as $$[t/2\pi] *{-\pi cos(t)}$$

where have I gone wrong

?

@copper.hat do you have any idea?

@satan29 You need to compute the integral for each $[k\pi, (k+1) \pi)$ separately.

Once you have done it for $k=0,1$ you are basically finished.

I have the following expression $(Cx^m+o(x^m))^2$, where $C$ and $m$ are fixed. $o()$ denotes the little oh notation. I'd like to write this expression out, so $C^2x^{2m}+2Cx^mo(x^m)+o(x^{2m})=C^2x^{2m}+o(x^{2m})+o(x^{2m})$. Now, is it correct that $o(x^{2m})+o(x^{2m})=o(x^{2m})$?

@copper.hat i know thats what I did

@satan29 $\sin(t-x)\sin(x)=\frac12(\cos(t-2x)-\cos(t))$

Howdy @robjohn

Hey there

The A/C in my car just gave out. I took my car in in March because the A/C gave out. I guess whatever they did, did not last.

Well, and today is effectively the first day of summer.

What did they say they had done? New compressor?

New freon?

hey chat

Hi Lucas

how are you doing @Ted?

Still bumbling along, and you?

@TedShifrin new compressor, new receiver drier, A/C oil, R134

total over $1300

a bit tired, trying to make up the lost ground on representation theory

Wow ... They ought to stand behind that!!

I never studied representation theory, @Lucas, although I encountered a few smidgeons.

my teacher made a few unjustified claims in her proofs so i'm trying to fill in the gaps by my own

@TedShifrin oh, it's an obligatory discipline here

@LucasHenrique I took a quarter of representation theory from Chari while I was in graduate school. On the one hand, I was entertained that the author of the book we worked out of had my last name.

On the other hand, Chari scared the hell out of me.

I wonder how many of my former students have said that about me over the years :P

@TedShifrin To be clear, I have tremendous respect for Chari. She is an amazing researcher, and has done well by her advisees. But she still scares the hell out of me. She is blunt to the point of rudeness, values the solitude of her office, had an office directly across the hall from me, and I am a gregarious person with a voice that carries.

I am under the impression that rep theory is a subject that is very rarely taught in an appropriately comprehensible manner

On more than one occasion, she interrupted my office hours to demand that I shut the hell up. ;D

@Thorgott Her class was good. I feel like I learned something.

I have forgotten most of it, but I used to know what a Dynkin diagram was. :\

There is a limit to rudeness.

aha, so it was rep theory of lie groups or lie algebras in particular

judging from what little in the literature I've read, those somehow usually seem more well-exposited than purely group-theoretic rep theory

@Thorgott Lie algebras, yes.

I'll be honest---I have never heard of pure group theoretic representation theory.

@XanderHenderson When my office was up a few floors with the university research staff, they repeatedly complained that my students and my office hours kept them from doing their jobs. I had to remind them that without students (and faculty) they wouldn't have jobs. They celebrated when I moved to a different floor.

Ha!

I am curious to see how things shake out when we get to go back to campus. My office is next to a historian's office.

I think that, right now, we have one mathematician at each campus.

Wow.

it's just studying homorphisms $G\rightarrow\mathrm{GL}(V)$, $V$ a finite-dimensional vector space and (usually) $G$ a finite group

M'kay, I'm out. Gotta make dinner.

serre's linear presentations of finite groups is very good.

I staunchly disagree

not only does the book contain very little motivation, it also contains very little conceptual insight, even though the latter readily exists in the subject

he proves important results like Schur orthogonality by some dumb matrix calculations (and his indices are horrible) and proves the crucial fact that the irreducible characters span the space of class functions using a completely ad hoc trick

you're just going to have to fight me on this.

i did read it after reading other books which may have supplied the stuff you say is missing. i may not have noticed the ad hoc trick.

i just love serre.

have you ever looked at fell and doran? i don't own it but i remember liking it a lot. it was my first introduction to a lot of stuff.

mackey i think also had a good book? or set of notes.