&c

Then $\ker \eta=m_f$.

yes

@MarianoSuárez-Alvarez Right, that.

Now I need some kind of isomorphism theorem, yes?

who knows

I never know those theorems

$K[X]/(m_f)$ has dimension $n$.

you got that if C(f) is generated by f, then C(f) is iso to k[x]/(m_f)

and you can compute the dimension of that quotient

@MarianoSuárez-Alvarez That follows from some kind of $C(f)=K[f]\simeq K[X]/(m_f)$ thingy.

yes

So I am sure that is true, but I am just mimicking the theorem for rings that $K[X]/(g(X))\simeq K[u]$ where $g=m_u$.

well, it is that theorem

Right.

:-)

I'm just saying it is a little bit different.

Or not...

it is the same thing

if R is any k-algebra and u an element of u, you have the minimal polynomial m_u

and the subalgebra of R generated by u is isomorphic to k[X]/(m_u(X))

"subalgebra generated" = "polynomials in u"

in our case, we can take R = endomorphisms of V

k-algebras are rings with additional vector space structure yes?

yup

OK.

@MarianoSuárez-Alvarez And the proof of this is just another "isomorphism theorem"?

i don tknow how you call it

you haave a map k[X] --> R

its image is iso to the quotient of its domain by its kernel

by definition, the image is k[u], the set of polynomails in u

so you get what you want

@MarianoSuárez-Alvarez Right, ${\rm im}\; \eta\simeq R/\ker \eta$ I call the first iso theorem.

naming that is like naming breathing :-)

Well, OK =P

So, everything went better than expected.

@MarianoSuárez-Alvarez , maybe this is kind of a cliche question. Do you feel like there are some questions interesting in themselves, and worth pursuing, and others that are not so attractive?. Do you have any criteria to tell them apart?

heh

sure

there are some things that interest me personally

and there are others which I know are interesting or important but which do not interest me

there are things which are not very important nor interesting :-)

the first group I know because after some tie you get to know what interests you

the second group you recognize because of "general culture" you acquire also with time

the third group, well, it depends

@AlexanderGruber

Thanks. I still find it hard to find meaning and deep motivation (it seems i have a long way to go), and some things just look like "this is what we do to justify our grants" (but most probably, I'm speaking from ignorance)

haha

Darn, I forgot how to physics.

Ask Germán

@Bill Doesn't seem to be awake.

He sleeps early.

I should sleep now, though.

Two tennis classes today, kinda tiring.

The first guy played pretty well, which is good. I get to have fun in the class.

Was he older than 5?

@Bill 50.

Hay encuestas de profesor de tenis?

@Bill Jeje, no creo.

@Bill, D'Alembert used to tell his students «Allez en avant, et la foi vous viendra», which is something like «keep going on and faith will catch up with you»

Pero hay una pagina que se llama "Tenis Registrado" y es algo asi como una base de datos e intercambio para profes registrados.

Yo no renove el carnet.

No le vi mucho uso.

@MarianoSuárez-Alvarez IIRC Landau's PhD thesis revolved aroud proving $$\sum\mu(n)/n=1$$

I am a big fan of Landau.

=D

is that true?

the series $\sum\mu(n)/n^s$ is $\zeta(s)^{-1}$ for $s>1$

and $\zeta(s)$ is like $1/s$ near $1$.

@MarianoSuárez-Alvarez I guess I'll have to trust D'Alambert. Cynicism is strong lately.

@Bill :-)

pick something nice to read, and read it

@MarianoSuárez-Alvarez By bad. I just meant he proved it converges.

It is relatively easy to show it is bounded above in absolute value by $1$.

@Bill, «Winning Ways for Your Mathematical Plays» for example, by Conway, et al.

you can find it online or on the library

Landau says "The infinite series $\sum \mu(n)/n$ therefore either converges or oscillates between finite limits. The questoin as to which f these two alternatives holds does not interest use at the moment; (...) Gordan used to say something to the effect that "Number Theory is useful since one can, after all, use it to get a doctorate with."

he says that in his little book?

In 1899 I received by doctorate by answering this question.

really?

@MarianoSuárez-Alvarez Yes, "Elementare Zahlentheory."

that $\sum \mu(n)/n^s$ is $\zeta^{-1}(s)$ was known to the greeks or so

Which is the first volume of a big set of books.

@MarianoSuárez-Alvarez XD

and that $\zeta$ has a pole at 1 was proved by Riemann in his one paper on $\zeta$

I am missing something

What year was Riemann's paper published on?

Landau was born on 1877

no idea

So he was 22 on 1899.

You're smart Pedro

How did you do that?

@Bill ¬¬

=)

I am just thinking maybe it was a "smaller" title.

@PedroTamaroff the first line of Doudy's thesis is: «The purpose of this thesis is to provide its author with a Ph. D. title.» (in French, though)

@MarianoSuárez-Alvarez Heh. I had heard about that one, but I don't know who Doudy is.

Yay!

$\uparrow$ Proof.

Douady

So it was a "new" proof.

cool

Darn, people got PhD at such young ages. =D

that shows how awsome Riemann was :-)

The old times had a faster pace.

his proof is mostly trivial :-)

You are young Pedro, don't cry

@MarianoSuárez-Alvarez What are the "Opponents" in Landau's paper?

hm?

@MarianoSuárez-Alvarez "This statement was first expressed by Euler,"

And Euler does it again.

it was probably never proved by Euler, though

@MarianoSuárez-Alvarez In the paper it says "Opponents"

@MarianoSuárez-Alvarez Right.

They are the members of the jury

who were they?

Zeigel and Hartoge

"Students of Mathematics"

Steinitz, a PhD.

steinitz?

the chessmaster?

Ernst?

The chess player wasn't a mathematician. Sorry

ah, Steinitz's very cool :-)

he is one of the inventors of field theory as we know it

Reading he was the one who constructed $\Bbb Q$ as a quotient of $\Bbb Z\times \Bbb Z^{*}$.

"En 1910 Steinitz publicó su influyente trabajo Algebraische Theorie der Körper (en alemán: Teoría algebraica de cuerpos), donde estudió axiomáticamente las propiedades de los cuerpos y definió conceptos importantes como cuerpo primo, cuerpo perfecto y grado de la extensión de un cuerpo."

Wow.

=)

@MarianoSuárez-Alvarez Landau is a trickmaster.

His proofs are sometimes jaw dropping, really.

For example, I already told @Fernando his proof of $L(1,s)\neq 0$ (for real $L$) is amazing.

A direct proof.

Pedro, can you count cards?

@Bill As in Blackjack? I don't know how.

"Landau wrote over 250 papers on number theory which had a major influence on the development of the subject."

"According to our current on-line database, Edmund Landau has 31 students and 4553 descendants."

I should really sleep now,.

@Bill @MarianoSuárez-Alvarez Byes,.

@AlexanderGruber Someone needs to create software to make interactive, animated, colored, three-dimensional commutative diagram. The arrows and objects can be colored differently to highlight which are organized together. The arrows could be shown in order of their creation by universal properties or factorings or whatnot, and various arrows could be highlighted right before a new arrow comes about to emphasize which morphisms are being UP'd over.

Three dimensions just has more room for more stuff than two dimensions, and the "interactive" part means that users could click and drag to rotate the figure around and see it from all angles, thereby making the three-dimensionality viable.

Hey guys, I have an algebra test tomorrow...if prof assigns me to solve a quartic polynomial, is that just evil? Test will probably be 5 questions in 75 mins

I understand cubic

but quartic is f*ing unfair don't you guys think?

modern algebra or college algebra?

wow if they assign these problems to college algebra kids then i feel sorry for them lol

we had to solve quartics in my abstract algebra class, but we covered the techniques and it wasn't really that hard

for quartics, you have to know how to solve a cubic AND something else

but i dont know what the something else is

@DonLarynx there are many "nice" quartics that you are expected to factor in say a college algebra class, but which may scare students

Let me clarify: Will it be sadistic if one of the problems is: Use Cardano's method to solve this quartic.

On an in-class exam, perhaps. On a take-home, no. If you want to know how to do it, see here (this is the method my class used).

If your class didn't cover solving quartics in class though and say it would be on the test that does weigh in on the sadistic side if if were to then appear on it.

I'm doing cubics right now and it's really tolerable

like i could do it in 10 min

maybe less

I have to use @anon Ferrari's equation right above your method

you have to? blegh

now you know my feels

who makes students solve cubics?!

I cannot think of many things more useless than that

a number theorist dressed as my algebra professor

haha, my algebra teacher too was a number theorist

in fact the whole class has been about numbers

we have not discussed groups yet

:o

Hi guys

@PaulRS I don't quite understand the example you gave

Greetings the great ones!

I just created another beautiful series, it's so divine.

$$\sum_{n=-\infty}^{\infty}\frac{(-1)^n \cos(n)}{1+n^2}$$

Um, could someone please explain to me why this happens in the complex plane

This is the graph of $\lim_{x \to 0} x^x$

What is happening there?

I get that the imaginary part vanishes but why does this guy say that there's more than one way to aproach x^x in the complex plane.

I don't understand that.

Hey

@Nick Branches.

Probably.

Much like $\log z$ is multivalued, or $z^{1/n}$, and so on.

@Pedro: I have no idea why they're multi valued. Pweez tell me more about them.

@Nick Well, there are $n$ values we can assign to $z^{1/n}$.

Much like if $x=y^2>0$ we can assign the value $y$ or $-y$ to $\sqrt x$.

XD

Ok, I get that for that

but how is it multivalued in x^x

@Nick Well, it is not multivalued for real numbers.

But if you look at complex numbers you need to start making choices.

One would define $z^z=e^{z\log z}$

(I think?)

And then $\log z$ is MV in the complexes.

So you need to chose.

Oh, you mean the natural log when you say $log (x)$ !

@Nick Yes, usually that is what it means for math people =)

really?

I thought it was ln()

You can use $\ln z$. Some people prefer $\log z$.

I do.

In fact for complexes one writes ${\rm Log}\; z$ to denote the "principal" logarithm.

oh... interesting

@robjohn @DanielFischer ?

Hola, @PedroTamaroff, 'sup?

@DanielFischer Just waking up. Sleepy, though =P

Didn't sleep that much.

Yers?

@PedroTamaroff Cold.

@DanielFischer Ah, summer is coming here.

What temperature?

Summer, that was a Tuesday this year. Temperature about 5 degrees, here, outside, but inside is only about 16, which is why I feel cold.

The radiator isn't too efficient :(

@DanielFischer Ah, drats.

Well, I was re-reading some notes on Lebesgue's integral last night. And I realized I never looked at something with care.

@Daniel: Is that in celsius or fahrenheit?

@Nick Kelvin? (Nah, Celsius)

@Nick Daniel is smart enough to be using Celsius!

XD

@PedroTamaroff Although sometimes I use Reaumur, just for the heck of it.

@DanielFischer It is a theorem of Beppo Levi that if $s_n$ is an increasing sequence of step functions defined over an interval $I$, such that $$\int_I s_n$$ converges, then $s_n$ converges almost everywhere to an upper function $f$ and $\int_I f=\lim\int_I s_n$

Now, the proof is not too long.

@Daniel: But I think it'd be best for the universe if you were cryogenically preserved for future use.

@DanielFischer But there is a detail I am missing.

@Nick No, I'm best preserved when kept warm and dry.

@PedroTamaroff What are you missing?

@Daniel: That's what einstein said before they froze his brain.

@PedroTamaroff And by the way, what is an "upper function"?

A lower semicontinuous one?

@DanielFischer Heh, Apostol defined the Lebesgue integral in a different way. We define the usual step functions and their integrals. Then we say $f$ is an upper (superior?) function if it is the increasing limit of a sequence of step functions with finite limit of integrals.

hey guys

Then we define $\int_I f=\lim\limits_{n\to\infty}\int_I s_n$

Okay, unusual terminology, but fine. So what detail are you missing?

And that's the class of upper functions, ${\mathscr U}(I)$. The class of Lebesgue integrable functions is ${\mathscr L}(I)=\{u-v:u,v\in{\mathscr U}(I)\}$.

@DanielFischer Well, let $\varepsilon>0$ be given.

Since the sequence of integrals converges, it is bounded above by a positive constant, call it $M$.

We define $$t_n=\left\lfloor \varepsilon\frac{s_n}{2M}\right\rfloor$$

Then $t_n$ increases too.

Moreover, $t_n(x)$ is eventually constant if $s_n(x)$ converges, and $t_n(x)$ is such that for infinitely many $n$, $t_{n+1}(x)-t_n(x)\geqslant 1$ if $s_n(x)$ diverges.

$t_n$ is a sequence of integers.

Right.

So, we define for each $n$ the set $D_n=\{x\in I:t_{n+1}(x)-t_n(x)\geqslant 1\}$

By the above, if $D$ is the set where $s_n$ doesn't converge, $D\subseteq \bigcup\limits_{n\geqslant 1}D_n$.

So it suffices the union has measure zero.

The first claim is that each $D_n$ is a finite union of intervals.

And the rest follows pretty easily. I didn't prove that.

I guess that you want to show that $$\bigcap_{k=1}^\infty \bigcup_{n=k}^\infty D_n$$ has measure zero.

@DanielFischer Why?

Because there will be some $D_n$ that have positive measure in general.

@DanielFischer Apostol shows that $\sum\limits_{n\geqslant 1}|D_n|<\varepsilon$.

Right (well, looks right).

@DanielFischer So, what I need to show is that in fact each $D_n$ is a union of finite intervals.

$s_{n+1} - s_{n}$ is a non-negative step function.

Aha.

So we can find a partition of $I$ so that both $s_n$ and $s_{n+1}-s_n$ are constant on the (open) partition intervals.

$D_n$ is a union some of these partition intervals (plus some of the end points).

@DanielFischer Does this depend on $s_{n+1}-s_n$ being positive?

Not really, but if the $s_n$ weren't increasing, $t_n$ could drop sometimes.

@DanielFischer Yes, yes.

So the point is: give $t_n,t_{n+1}$ a common partition.

Then we're done.

Yes, that's most of the time the thing to do when dealing with several step functions / simple functions, find a common partition.

Heh, I had the picture in my head, I'm OK with this.

I don't care about a more detailed proof., though.

I mean, not much else can be said. =)

@DanielFischer Thanks.

De nada.

@DanielFischer Oh! The end of the proof:

c.q.f.d.

For the proof, Apostol writes $$\int_I{t_{n+1}-t_n}\geqslant \int_{D_n}(t_{n+1}-t_n)\geqslant \int_{D_n}1=|D_n|$$ Then $$\sum_{n=1}^m|D_n| \leqslant \int_I t_{m+1}-\int_I t_1\leqslant \int_I t_{m+1}\leqslant \frac{\varepsilon}{2M}\int_I s_{m+1}\leqslant \frac{\varepsilon}2<\varepsilon$$

=)

I don't believe what I'm seeing, but is Vieta's substitution legit?? Because Cardano's method is F*ING HUGE. en.wikipedia.org/wiki/Cubic_function#General_formula_for_roots

@DonLarynx Brazil

I have a test in 30 mins and i just found out about vieta's method, holy moly

@DonLarynx I wouldn't try to learn a new method in 30 minutes. =)

@DonLarynx @nick and I were playing with it days ago

It's a horrible method

I was doing this last night while you were asleep @PedroTamaroff

@Charlie, cubics and quartics are horrible in general.

@DonLarynx a pain in our asses

yes and a sadistic teacher

Yes

@TheNotMe I gave you the ideas needed to do it, now, if you just want a plain example take $L = \{a^i : i \text{ not a power of }2\}$ or $L = \{a^i : i \text{ not prime}\}$, and the alphabet $\Sigma = \{a\}$.

Computing $L\cdot \{\epsilon,a\}$ should be very simple, and to see that it is regular.

However, neither of those $L$ is regular.

Well, i'm off now. Cya on the other side @Charlie

To see those $L$ are not regular, use the same argument we always do (hidden under the name of "pumping lemma"), of using pidgeon-hole principle.

@Charlie: Hi charlie! :D

Bye @don

@Nick Hi nick!

@Charlie: Watcha upto?

@TheNotMe More generally, given an increasing sequence $b_k$ in ${\mathbb N}$, you can show that if the "gaps" satisfy $\limsup_k b_{k+1}-b_k = \infty$, then $L=\{a^i : i\text{ not a }b_k\}$ is not regular.

@Nick I don't know :) and you?

@Charlie: Trying to find the sum of a sequence

@Nick :-O

@Charlie Try to find the sum of a sum.

What's the best book for brushing up on aptitude questions?

@anakin feel in the force...you will get the answer

$\sum\limits_{n = 0}^{infinity} \frac{2}{n(n+1)}$

@GustavoBandeira summing sums

@Charlie What if I'm on the dark side of the force? :P

@anakin choke someone to get the answer

@charlie: can you help me O_o?

@Nick I don't think so :(

@Nick why it starts at 0 ?

@anakin: Um, I like you a lot. So, I feel it my obligation to tell you. You're going to be killed by a guy named Luke... he's also going to be your son

@charlie: XD sorry, 1.

@Nick :)

@anakin: Also, there's a vent in the deathstar that's going to be a huge security problem.

@Nick hahaha

@Charlie: eh, it's ok. I know the answer is going to be $2n/(n+1)$. Too bad my question is to prove that

@Nick $$\frac{1}{n(n+1)}=\frac{1}{n}-\frac 1{n+1}$$

Hahaha good, @pedro

@Charlie Didn't want to give it away =)

@PedroTamaroff ;)

@Pedro: I don't get the telescope.

en.m.wikipedia.org/wiki/Telescoping_series @nick

@nick ?

@DanielFischer

@PedroTamaroff

@charlie

@DanielFischer Consider $$\int\limits_C {\left( {y + \sin x} \right)dx + \left( {\frac{{3{z^2}}}{2} + \cos y} \right)dy + 2{x^3}dz} $$ where $C=(\cos t,\sin t,\sin 2t)$.

I am hinted that to calculate that integral, I should use $C$ sits in $z=2xy$.

I would think you should first separate out the $\sin x\,dx + \cos y\,dy$ part.

And for the remainder, what happens if you replace $z$ with $2xy$ everywhere? Any obvious simplifications?

@charlie: I just did it 10 times. I keep getting a wrong answer. This is more harder than I initially thought.

@DanielFischer Can I also replace $dz$ with $2(ydx+xdy)$?

@PedroTamaroff $dz = z'(t)\,dt$, so yes.

@DanielFischer Well, I am getting save the trigonometric part: $(4x^3+1)ydx+(4x^4+6x^2y^2)dy$

So.

I need @TedShifrin =D

What for?

How do I get the RHS from the LHS of this ? $ \sum_{i=0}^{k-1}x_{i}2^{i}\leq\sum_{i=0}^{k-1}2^{i}=2^{k}-1<2^{k} $

@anakin What are the $x_i$?

@DanielFischer Solve this.

@DanielFischer $x_i$ can only take values of 0 and 1

@PedroTamaroff If one doesn't see a slick way (and I must admit I don't, off-hand), there's always brute-force.

@DanielFischer I will think a little more.

Though it is American Horror Story time.

@anakin So then it is clear, isn't it? $x_i 2^i \leqslant 2^i$, sum it, get the RHS.

That should give us 2k−1 in the RHS of 2k−1<2k also right? Then how 2k only?

@DanielFischer

morning

Let $F\subseteq E$ be fields of char $p$. If $a\in E$ is transcendental over $F$, is it true that $F(a)\subseteq F(a^p)$?

maybe @anon

@anakin You get $LHS \leqslant 2^k - 1$. And $2^k-1 < 2^k$.

@PedroTamaroff hello!

@leo If $a$ is transcendental, then $F(a^p) \subsetneq F(a)$.

Got it @DanielFischer

@leo Howdy.

@DanielFischer The trig part is zero.

@DanielFischer that inclusion is always true

@leo Yes. Well, except I said they're not equal, $\subsetneq$.

@PedroTamaroff oh I see. I didn't activate the MathJaX thing

so it's no true

True. If $F(a) \subset F(a^k)$ for some $k > 1$, then $a$ is algebraic over $F$.

@PedroTamaroff have you hear about the perfect closure?

@leo Nope.

@PedroTamaroff what are you taking this semester? :)

@leo Currently Linear Algebra, Analisis II and a course on sequences and series.

@PedroTamaroff cool

@PedroTamaroff pedro, whats the deal with stonehenge ?

@leo what is it?

@Charlie the perfect closure?

@leo yes

@charlie: Did you happen to see morgan freeman sort of sing The Fox ?

@Nick no

@Charlie given field $F$, its perfect closure is the smallest perfect field which extends $F$. A field is perfect if all its polynomials factors as product of different linear factor

@leo ah :)

@DanielFischer indeed. Thanks. Text with mistakes. Or bad exposition mostly