@robjohn when you have some time pls let me know how it seems to you that question as difficulty level.

@robjohn I sent this version to some students

$$\lim_{n\to\infty} (1+\cos^2(\gamma)+\cos(\gamma)^2 \cos(2\gamma)+\cdots+\cos(\gamma)^n \cos(n\gamma))$$

@Mike I could cheat and look it up. What's your formulation of the spectral theorem for compact normal operators? That $$N = \sum_{\lambda\in \sigma(N)\setminus\{0\}} \lambda\cdot P_\lambda,$$ where $P_\lambda$ is the projection onto $E_\lambda$?

I should clarify: For once, I'm just making conversation. But one that's simpler than that: every element of the spectrum is an eigenvalue, and the spectrum accumulates at 0 and only at 0.

We'll probably be talking about some such decomposition tomorrow.

@DanielFischer I take it the projection is the one given by integrating around eigenvalues?

@Mike That's an immediate corollary of the spectral theorem for compact operators on a Banach space.

Ah, but I don't have that.

@Mike Booo Mike.

Booo.

So I suppose I lied about the statement of the problem.

@Sush could the problem be done?

@Mike Theorem 4.25 in Rudin.

:P

My only special tool is that $C^*(N) = C(\sigma(N))$, and its immediate corollaries.

@Hawk, I think you say that $(\frac{2-3x}{1-3x+2x^2})^n$

@Mike But proving it for compact normal operators on a Hilbert space is very probably more elementary.

How can we expand just $\frac{2-3x}{1-3x+2x^2}$?

@Sush No, I do not...try to think of how the sum of GP looks like...compare this with the above expression.

@DanielFischer This is one of those "use your new tool to overkill this theorem" exercises.

@Hawk, which gp should i try?

I can demonstrate an elementary proof without too much difficulty.

@Sush Okay, would you please factorise the denominator, I do not have something to do upon nearby at present.

These are all things I should have said before trying to make conversation :P

@Hawk, yes. $(2x-1)(x-1)$

Okay, take seperately these as $\dfrac{1}{2x-1}$ and $\dfrac{1}{x-1}$

Yes.

Now, can you compare this with sum of infinite GP formula $\dfrac{a}{1-r}$?

Here, $a=1$ which you can understand.

and you can find out $r$

$r=2-2x$

@Hawk

and $r=x-2$

@Sush Why? $(2x-1)=-(1-2x)$, so $r=2x$

I officially give up !

On this one little question.

I have been defeated.

"People look at me like I'm a little bit strange," says Martin. "It's certainly a showstopper, but living in England, no one likes to point out the hippo in the corner."

It is time to head back home and see the kids. I can not win this war.

How do you train for running the Virgin London Marathon dressed as a hippo?

@Sush Did I make myself clear now?

So, $a=-1$, right?

@Hawk

@Sush Exactly...you are right

@Sush Now...treat those fractions seperately and then multiply the sums of GP...and the other term in numerator...can you do it now?

So, $-1-2x-4x^2-8x^3-...$ is the series@Hawk

@Sush Yes, it is. Now, find out the other

Other is $-1-x-x^2-x^3-...$@Hawk

@Sush This one is correct too...

So, you are done I suppose? Now, you are left with plain calculation.

@Chris'ssis you would think so, but it is highly dependent on the triangle being equilateral, which shows why both answers rely heavily on being equilateral. I could not find anything simpler.

@Sush Let me know if you are done with the problem or any other complications crop up.

@Hawk, sorry, but i can't go further.

@Sush Okay, so the expression you provided can now be written as $(2-3x)(1+2x+4x^2+8x^3+\ldots)(1+x+x^2+x^3+\ldots)$. Yes?

Ok! Now i can do it! I was mistaking the last two fractions in denominator! @Hawk, Sorry. And, thank you so much!

@Sush you are most welcome. Sorry for what?

Sorry, i am very weak at math. You tried to teach me everything though i could not understand at once @Hawk

@robjohn I think you're right. I initially felt I saw that somewhere before (maybe in some proofs with vectors?).

@Hawk, bye. Going to take dinner. Good night.

Namaste.

@Sush I was weak once too...how does it matter? There are several situations where I cannot understand at once, what is there to feel sorry. Only thing that matters is the effort to learn...which you had...that's all matters.

@Sush Good bye...Good night. Namaste...

@Chris'ssis I can find a lot of references to Ceva's Theorem, and since this is so similar in description, I would think there would be something about it.

@robjohn Yeap, that should work...

(I didn't make tries there yet)

@Sush Not that I am strong now...but I am not so weak now... :D

@Chris'ssis the one I am looking for is the one about the sum of the squares of the parts of the sides. That is the one that is true in general. That is the one I think should have a name.

Morning!

I guess that depends on where you are.

@robjohn hey

hello anthony

@Mike HAI MAIK

@Anthony Smartboy.

When will it be my turn to be smart

@PedroTamaroff :D

@PedroTamaroff Try this without using complex numbers $$\lim_{n\to\infty} (1+\cos(\gamma)^2+\cos(\gamma)^2 \cos(2\gamma)+\cdots+\cos(\gamma)^n \cos(n\gamma ))$$

@robjohn I see.

@Mike Hey!

How can I master linear algebra? I am doing the problem sets (done about 110 problems out of 218) but I feel like I don't understand the concepts completely. What else should I do?

@robjohn Which questions are you both talking about?

@user2357 study the main points with the intent of learning how to explain them in your own words

@anon thank you for the advice, I will do that. Any resources you recommend?

just get a good linear algebra book

Ok, thank you very much. I will probably use hoffman and kunze

"oh, just get a good lin alg book"?

(:

youtube.com/watch?v=5fP4emqw7O4 gotta love charlie

@anon Could I check something with you?

sure

oh good

the bold stuff is gone. yay

@anon OK. So, first I proved that ${\rm Spec}(A)=X$ is always (quasi)compact. Used AM's hint and $V(\mathfrak a)=\varnothing\iff \mathfrak a=A$ to show that if $X=\bigcup X_{f_i}$ the ideal $\sum (f_i)$ is all, so we have a finite sum of those $=1$; and $X_1=X$ so we're done. Then I showed that each basic open set $X_f$ is (quasi)compact by something similar.

Namely, $V(\mathfrak a)=V(\mathfrak b)\iff \sqrt{\mathfrak a}=\sqrt{\mathfrak b}$.

yeah, I don't actually know algebraic geometry

just sort of go with the flow

@anon But this is comm. algebra.

I think...?

I will check what @FernandoMartin did then.

He doesn't seem to be around.

@meer2kat yay.

@Sawarnik it was annoying :P

@meer2kat i know! it was really annoying!

I am tired

@Mike Get some coffee

Sleep.

that's my first stop when I get off the bus

@Mike Where do you buy your coffee?

out here there's nowhere that sells good coffee.

"that's my first stop when I get off the bus"?

so...Starbucks

that's fine

it's TERRIBLE

acceptable

Acceptable but nonetheless terrible.

I already forgot my new Yahoo password

LEL

@Sawarnik this question.

I guess @meer2kat missed my ping

sup

@Pedro:

@FernandoMartin YAO

I solved the problem about $\text{Spec} (k[x])/(x^n)$

That was the one you had a cool solution for?

Yes, @Mike helped.

it's just $(x)/(x^n)$ right?

for n>1

The nilradical of $k[x]/(x^n)$ is $(x)/(x^n)$.

Aha.

And the quotient is a field.

Hahah, slick

That is, the quotient is just $k$.

So by exercise 10 of AM I think.

I just did it by hand, that argument is really nice

There is only one prime ideal.

Yes

@FernandoMartin What does "by hand" mean?

Well, I tried to find all prime ideals of $k[x]/(x^n)$

that's the same as finding all prime ideals of $k[x]$ containing $(x^n)$

such an ideal must contain $x$

aha

just a keyword for the interested

@anon Yes, that's what I noted above.

The unique maximal (prime) ideal is $(x)/(x^n)$.

and $(x)$ is maximal

so that's the only one

it's maximal because it's the complement of the units

also because $k[x]/(x)\simeq k$

A local ring is a ring bought at the local supermarket.

@Pedro I really do like that theorem.

Then you should read Cohn. It is the best algebra book in the world.

@WillHunting Do you know what a bad joke is? =D

@PedroTamaroff Yes. It is what I wrote, lol.

@Mike Did your downvotes got reversed...?

what ping @micklh

Doesn't look like it.

Yeah, I got 58 rep back.

@meer2kat I forgot now, it was something about VBA :P no worries

@Mike Oh, OK.

@PedroTamaroff the downvotes stay in the rep history even after reversal

What have you been studying lately?

I really want to know who did it. But that's confidential.

I wish I can go back to the age of anon, Mike and Pedro.

@WillHunting Jasper, the sum of their ages might exceed yours. But I agree that would be interesting.

cute question

@ccorn Nvm, I will look forward to my next life. Perhaps this life is finished for me.

hi

hello, moonstar2001

what happened to the other 2000 moon stars?

they all paid tribute to jesus already

haha

@MickLH oh joy

:D

@EnjoysMath THAT IS HILARIOUS.

need 10 more rep

almost there

wait... hey mods. if you have 100 rep, can you offer all 100 rep as bounty? @robjohn

Where's your question?

@meer2kat Who would tell.

Your pic isn't fake.

@meer2kat I am not sure. I don't think you can go below 1 rep, so perhaps not.

Here's a question for you all. If you can imagine an abstract mathematically defined universe then do the beings inside of it exist for realz??

@meer2kat I see users with negative reputation, so perhaps I am wrong.

They exist in your head for realz

@enjoys

It would fall into the realm of mathematical philosophy maybe?

I hope more people vote for my lhf answers, hard to earn a living on this site.

lhf?

Low hanging fruit

Easy questions

Oh. -_-

Sometimes I don't understand why the other answers get votes and not mine. I must be doing something wrong.

@WillHunting À vaincre sans péril, on triomphe sans gloire

Do you usually post yours first

I usually post second or third

@EnjoysMath DAT SONG TITLES.

DAFAQ.

@GabrielR. Hi, never seen you before

I'd rather win and triumph regardless of risk and glory

@WillHunting Greetings then. I've not been around here for some time...

@GabrielR. OK, I just changed me username so you may not recognise me, lol

@user127001 Hey Bart, recognise me?

Hi Will, blue square

@GabrielR. I see you are the black square, I am usually the blue square

Are we still on for dinner tonight @WillHunting?

@user127001 Haha, yes!

@pedro I see that userXXX is getting many downvotes these days, lol

@WillHunting ORLYT

Let's a see.

@Pedro: I got stuck proving $V(I) = V(J)$ iff $\sqrt I=\sqrt J$

obviously $\Leftarrow$ holds

@WillHunting I've heard the mean square is around...

@FernandoMartin Yes.

But $V(I)=V(\sqrt I)$.

Yup but I'm not sure how to prove $V(\sqrt I)=V(\sqrt J)\Rightarrow \sqrt I=\sqrt J$

am I being silly?

@FernandoMartin Well, $\sqrt I=\bigcap V(I)$!

amirite

I'm not getting what you mean

alright, then i guess i need 11 more XD

Where are you taking your intersection over?

If I have a function defined as the root of an arbitrary trigonometric expression, how do I integrate this function?

@FernandoMartin I mean the intersection of the collection $V(I)$.

@meer2kat I hope you didn't get me wrong over my comment. Most people lie about their identities on the interwebs.

Ah, gotcha

$\sqrt I=\bigcap \{\mathfrak p:\mathfrak p\supseteq I\}=\bigcap V(I)$.

yes

@MickLH you can substitute $u=tan(x/2)$ and see what happens

@FernandoMartin I will probably check with you on how you do E12 of Alicia's guide, i.e. 17 of AM.

I think my work is correct.

I'm doing that one

But nevertheless.

@GabrielR. I'm unaware of how to set up the integral at all, I've only know how to work with explicit expressions

@PedroTamaroff sorry, what comment?

Ohhh that one

@MickLH you mean you don't know substitutions in integrals ?

No, I mean I'm using a trapezoidal approximation, and evaluating the function with newtons method

@meer2kat "Who would tell. Your pic isn't fake."

@PedroTamaroff That is indeed me. Someone has proven it before when they creepily took one of my pics and traced it back to my sorority's website.

OUCH.

What's a sorority again...?

@MickLH ohhh ok... I don't know such methods, sorry

Boys form a fraternity and girls a sorority

It's like a private club in college

Look up the dictionary

@PedroTamaroff I am very honest about who I am online. A sorority is a woman's "Greek" organization. A group of girls who go through a ritual and educational period, study, learn, and live life together, and pay ridiculous amounts of fees :)

@GabrielR. I guess I'll write up a question on MSE

@meer2kat Heh, I see. So those $\Psi\Gamma \rm blah$ thingies.

@meer2kat is it really an "educational" period ? :)

@MickLH You mean a question

@GabrielR. Yes. We have to learn the Greek system, our histories, our values, and every single chapter and when it started, and the rules of the sorority.

@PedroTamaroff Yes! Mine is Phi Sigma Rho, the engineering only sorority

Want to check?

@meer2kat Do the letters stand for anything?

ok @Pedro, I'm done with 17

@FernandoMartin "I'll show you mine if you show me yours."

ok

@PedroTamaroff Are you showing your XXX?

@PedroTamaroff Greek symbols. And then those symbols represent a specific value that the sorority stands for. I can't say beyond that. More than that is secret, ritual information.

Is it the compact part you want to check?

@meer2kat Do you mean $\Phi\Sigma P$ ?

@MickLH Yes, exactly right :)

@FernandoMartin How did you show each basic open set is (quasi)compact?

@meer2kat I know the Greek alphabet! =P

@PedroTamaroff I forgot the Greek alphabet

?

@FernandoMartin Each $X_f$ is quasicompact.

@meer2kat You program?

shit, I missed vi and vii

give me 5 minutes

@meer2kat I watched too many American Pie movies, so I'm way too biased about this. How do you refer to people in sororities ? I know there are so called "frat boys". What about girls ?

@FernandoMartin LE LULZ.

@GabrielR. HAHAHAHA.

@GabrielR. Fat boys and sore girls, lol

@WillHunting well thought lol

@PedroTamaroff Yes

@GabrielR. sorority girls. several stereotypes are true

@GabrielR. or srat girls

@meer2kat Do you think Ubuntu is a cool OS?

I am getting different opinions from different computer people.

@PedroTamaroff I don't like the Unity interface. I use Ubuntu GNOME or Mint or Debian.

@PedroTamaroff I don't program THAT much

@meer2kat Ubuntu is not about programming, lol

@robjohn here is the solution to the previous series I created today. I hope you like my way!:D

@Chris'ssis Do you know about the user Cleo? Just look at her answers. She is Ramanujan.