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21:01
@Pedro: You summoned me ... A while ago?
Yes @Twink: That's actually the definition.
Discovering a definition is like re-inventing the wheel :(
that would be the definition if the sum converges to zero
and taking it from 1 to N
hi pal
hi pal
No: Write down limit minus partial sum $s_{N-1}$.
Hi @cyber and @Kevin.
@TedShifrin hi
21:08
@Ted Good evening
Just back from getting a broken filling replaced ... So extra martini for me ;)
$|\sum_{k=1}^\infty a_k - \sum_{k=N}^\infty a_k|<\epsilon$
$|\sum_{k=N+1}^\infty a_k |<\epsilon$
Right. Definition :)
but how do we know $|\sum_{k=1}^\infty a_k - \sum_{k=N}^\infty a_k|<\epsilon$?
@Twink actually, that is $\left|\sum\limits_{k=1}^{N-1}a_k\right|\lt\epsilon$
21:11
That's what convergence mean (given $\epsilon$, there exists $N$ ...)
if we take $a_1=1$ and $a_k=0$ for all $k \geq 2$
and $N=2$
then $|\sum_{k=1}^\infty a_k - \sum_{k=N}^\infty a_k|=|1-0|=1$
not $< \epsilon$
@Twink as I said above, that is not what you want
You didn't write down the definition if $s_N$, did you?
21:13
The Cauchy criterion in the post is ok. The statement in the title is not. — Did 5 mins ago
I haven't looked at the post. I responded to your original question.
it's exactly the same
as my original question
20 mins ago, by Twink
Is it true that if $\sum_{n=1}^\infty$ converges then for every $\epsilon>0$ there exists $N$ such that $|\sum_{n=N}^\infty|<\epsilon$?
I just put iff
in the post
So, what did I tell you to write down?
Yes, iff is right.
@Twink what you write in the linked comment is true, but its converse is not.
now we have a debate
21:15
@TedShifrin Didn't know you were a martini man! I've never had one myself. Keep meaning to.
@Ted Gin or vodka?
You don't even know that $\sum_{k=N}^\infty$ exists if the series does not converge
The Cauchy condition is that beyond a certain point, the partial sums do not differ by more than $\epsilon$
We'll have to meet for martinis, @Kevin. I prefer gin but I can mix either :)
shaken or stirred?
@Ted I generally prefer gin to vodka, so I think I'd be the same
I do both, @cyber :)
OK, have to give the cat antibiotics. Back in 5-10.
21:20
. o O ( cat = dude = Ted )
. o O ( antibiotics = medicine = drugs )
Hmm
Nah
Sorry, just being punchy
Been sleepless of late, @robjohn? :)
@TedShifrin Not really. The afternoon does funny things...
Well, at least you're not at my alma mater, Berkeley, where things exploded!
Your "fostering mother" exploded? :-)
21:33
You're too literal, @cyber.
@TedShifrin math does that to me sometimes...
...sorry
If we have an integral, say $\int_{-\infty}^{\infty} f(k,k') dk'$, is it possible to interpret this integral as an integral along a straight line in a $k$, $k'$ plane. We might then make a change of variables $k = e^{\rho} \cos{\theta}$, $k' = e^{\rho}\sin{\theta}$ where we are now integrating a function $g(\rho, \theta)$ along some other contour in the $\rho$, $\theta$ plane that can be parameterized in terms of whichever variable we chose. Does this all sound reasonable?
I hope you do it well on homework/tests.
thanks :-)
Not at all, @Kevin.
Then $k$ is in fact not an (independent) parameter.
21:36
Good. (I was going to say good either way, because part of me wants it to be reasonable and another part not)
You're de facto constraining analog and digital to a linear relation. Schizophrenic, anyone?
So interpreting the integral as being along a straight line contour in the $k$, $k'$ plane isn't problematic, is it? Its just the change of variables which is unreasonable
Yes; no.
Sorry, not sure how to parse that
Everything is problematic.
21:41
What part of "yes; no" don't you understand?
See my de facto remark.
@cyber: Do your homework!
Yes sir.
10 mins ago, by cyberskull
...sorry
again.
@TedShifrin I don't understand your analog/digital analogy. It seems like the problematic idea though is that the original integral is defined for independent parameters $k$ and $k'$. The idea of a plane with contours, though, suggests that along most paths they are in fact not independent.
when I said a straight line, I actually meant a horizontal/vertical line such that $k$ is a constant and $k'$ varies, if that was unclear
but it seems to not matter
Kawabonga.
Oh, the @Pedro who summoned me hath returned.
21:50
@TedShifrin Helloes.
Ah, sorry, @Kevin. What's the point of all that if you're just doing the original problem with no change?
i guess I'm confuzled.
@TedShifrin Well, if the interpretation as a horizontal/vertical line in a plane makes sense, it just confirms that my weird idea had at least solid foundation. If the change of variables is unreasonable, and your point has convinced me that it is (as I sort of always expected), then there is no point to it
So, @Pedro, I had a notification you'd summoned me, but perhaps it was 6 hours ago.
@TedShifrin Yeah. I cannot remember what it was for.
Probably to tell you that I might devote more time to number theory. Dunno.
Ah, well, then you shan't be bothering me. I know nothing.
21:54
I found myself with no math to study, or at least found myself with no interest.
Regarding Analisis II, it is all computation, and since all the theoretical cool stuff is missing I find it really dry and boring.
@Kevin, you're basically reparametrizing $\Bbb R$ by $e^r$, which, of course, won't work.
And Linear Algebra is well, it is interesting, but I am finding it very routinary.
I hope you're not depressed, @Pedro.
@TedShifrin Just on the verge.
Routine? :)
21:56
@TedShifrin Ah?
Yikes, that's not good. You need to get into hard classes.
Are you addressing the typo?
@TedShifrin Oh yes, of course. I should have written $\int_{0}^{\infty}$. But if I understand your point correctly, the specific transformation isn't the problem. Its the idea of doing a transformation
My students typically get depressed because I'm too hard on 'em. You need that.
Why are you so hard on them?
21:58
No, I was addressing your state of mind.
@TedShifrin How so?
Because education is going down the tubes and I believe in making people think and learn, not just throwing high grades at them the way most want (along with their parents).
@TedShifrin Yiss.
@Pedro: My not good comment was addressing your state of mind.
Students feeling like they deserve a high grade, or even a passing grade, just for showing up has been a consistent problem for us as a department
22:00
The most talented students here who take my class 1st year or 2nd year often end up in grad classes their 3rd year.
There have been a handful of really misguided students who have yelled and screamed at us for getting low grades
@Pedro: I have been trying to make sure you can actually do concrete, interesting computational stuff. I think you should take my advice and take much more advanced classes.
@TedShifrin But, in all fairness you can't rush the learning process too much, right?
@KevinDriscoll Did you try to bitchslap them?
@TedShifrin I don't feel like rushing things like that, though.
@Pedro The only one who tried to yell at me got thrown out of my office under threat of being escorted out by campus police
22:03
@KevinDriscoll "Power is power."?
Oh, we get phone calls from damn parents, @Kevin. Not to mention the parents who yank kids out of hard classes for feae they won't get As.
What does that mean @Pedro
@TedShifrin Why do people worry so much about grades?
@Pedro: you have to be aggressive, or you'll be bored, unhappy, and talking to me.
We have state lottery which gives people tuition with a good enough average. i voted against this 20 years ago for two reasons.
That's terrible @Ted. Luckily I'm not far enough up the chain to get calls from parents. I feel like graduate programs in medicine and law are at least partly to blame. For physics, your GPA matters but it isn't a deciding factor. And we put a reasonable weight on where you went as an undergrad. From what I've seen though for med and law school it doesn't matter if you went to Yale or Southwestern Kentucky State, you GPA is treated the same
22:06
@TedShifrin Oh?
@TedShifrin Well, I don't know what to do.
First, I knew we'd have grade inflation and all the things I'm complaining about. Second. It's mostly poor people who play the lottery, and they're subsidizing mostly rich families.
Good points^
@TedShifrin I don't think lottery is fair at all.
True, @Kevin, so when Harvard and Stanford give everyone As and we don't ...
@Pedro: As long as you make sure you learn basics, go for challenging stuff, because you're unhappy in standard courses. At MIT and Berkeley and here, the exceptional students don't take the same classes that the weaker students take.
Thanks, @cyber
@Ted Indeed. Grade inflation is certainly a big problem. And what's worse is that its a Nash equilibrium. If you try and unilaterally deviate from inflation, you potentially hurt you students or your career
22:11
hi @anon
@Ted I'm inclined to agree. It's a hard position to take because it just sounds wrong, but I've made quite a career of taking unintuitive positions
@cyberskull hello
Well, I am still tough and give real grades, but lots of students shun me. My career has never been in jeopardy for being a great but demanding teacher. :) If I had worries, it was about the quantity of my research, not the difficulty of my classes. But most people sell out.
@Ted I can attest that the same thing happens at Duke. There is a whole Math-1??X series in additon to the usual Math-1?? series for 1st and 2nd year students who are exceptional
@TedShifrin What do you call "exceptional" and "weak"?
22:13
Yes, @Kevin. I even taught that class for a week back in 88 :)
I work hard. @anon is witness of that, I guess.
@Ted you were teaching calculus before I was even born! :-P
@Pedro: i can give a detailed description, but I'd rather not do that now and here.
Sometimes there are no secrets, just curiosity and dedication.
Um. @Kevin, I first taught calculus in 1970, when I was still in high school :(
22:15
:O
seriously?
@Pedro: You know I have a lot of respect for your knowledge and talent. Go for it!
@TedShifrin "I was born with a chalk in my hand. For realz."
@Pedro I don't think @Ted is saying you're a prodigy. Its just that your level of knowledge in many areas seems far beyond the equivalent coursework.
Yes @cyber ... Seriously. And I taught an MIT vector calc class when I was a senior.
@KevinDriscoll I know. I wouldn't call myself that either.
22:19
No, not a prodigy at all, but you don't need to be a prodigy to be on the fast track. Talent + hard work! If it turns out overwhelming, then drop back a step.
What if you drop back a step and you're still overwhelmed? Is it time to drop out?
No, @cyber ... Drop back another step.
@Pedro If you're anything like me then taking classes where you know most of the material already is a recipe for disaster. Like low grades. I was required to take multivariable calculus a second time by Duke, even though I had taken it in high school and gotten an A. I ended up with a C in that class because I never did the homework.
But, if you're smart and doing badly in classes you should be doing well in, go get a job and come back when you are ready to excel.
@KevinDriscoll Heh. Did they grade "homework making"?
22:24
@Kevin: You should have been in my class — the book I wanted @Oedro to peruse.
@Pedro Sadly, yes. It was actually a HUGE shock to me when I went to Duke that they wanted to grade my homework. I sent the majority of my last 2 years of high school at a technical college where the only grades were midterms/finals/papers/projects
@TedShifrin Anyways I play tennis today, so maybe I can take some heat off.
@Ted Probably. I also should've pushed to take Math 103X instead. I only asked once and got some kind of "Well...... most students aren't ready and we prefer them to take....." speech
@Pedro In that class there were 3 tests and a Final. Homework was worth 1.5 tests, I believe
Ok, @Pedro. Play well. hugs.
@KevinDriscoll Well, our university as a particular characteristic: one really has to look after oneself. They publish the problem sets, we have a theoretical class (2hs) and then a practice class (3hs, but never really that long) where we can ask questions and where problems are solved for everyone. But there is no such thing as homework, we have two midterms which we have to pass (not graded) and the only grade is the final.
If you don't pass a midterm you have a makeup exam which is usually tougher, and if you don't pass them, you have to take the course again.
22:29
@Pedro That would be a little harsh for me. I'd be lazy enough to just pass the midterms and then try and catch up for the final. At least that's what I would've done as a freshman. But I would've preferred it to having to hand in homework twice a week
@KevinDriscoll The freedom is nice.
@TedShifrin Look @KevinDriscoll
@pedro: So why didn't you put that as an answer?
@TedShifrin Because it is not what the OP is asking. "My question is whether my proof is correct, because I have a little bit doubt. If it's not correct, what is the best I can do? "
@Pedro: You have the European system, which I'm really not fond of.
I just think that proof is both slick and illuminating.
@TedShifrin Ah?
22:39
Ok, dinner's ready. Play well, and let me know if we need to talk.
@TedShifrin Byes. I am off too.
guys, what is non-linear algebra called?
@DonLarynx Here
23:00
@DonLarynx What do you mean by non-linear algebra?
So, wikipedia, and my book, both define functionals as elements of B(E,F) (spaces of bounded linear mappings from a normed space E to a scalar field F). Is this correct? I.e. it would not be right to call the elements of a family of bounded linear mappings from a Banach space X into a normed space Y for functionals"?
Hmm. Let me retract part of that. The book certainly doesn't say anything else can't be called functionals as well, though they do make a point of the scalar field, so I figured that was central.
@MikaelÖhman I can't answer your question formally, but my informal answer is that the output of a 'functional' is always a scalar
mapping of vectors to vectors is not called a functional
@MikaelÖhman Does that answer your question?
@KevinDriscoll Thank you. It more or less confirms the interpretation I have gotten as well. I was trying to make sense of what a classmate told me earlier, but I think he was just wrong to call them functionals in that case.
@MikaelÖhman Sounds like it.
although there could always be conflicting terminology with a specialized field
What book are you using?
23:12
Introduction to Hilbert spaces with applications
Looks good.
Hilbert spaces are weird to me. We learn in Quantum Mechanics that our state vectors are members of this Hilbert Space etc etc etc
and then we spend all this time doing examples where they ARENT
I'm working with FEM and continuum mechanics, and in those cases, we almost always end up with the simplest, finite spaces, so I'm not sure I'll ever get much pratical use out of all this fancy theory
hi
@MikaelÖhman do you know how can I fix this problem: I wrote a eqnarray in LaTeX but it's very long, almost one page
and it goes to a new page
it starts in a new page
and leaves a lot of blank space in the precedent page
@Mikael FEM is finite element method?
23:21
@KevinDriscoll Yes
@Mikael Ah yeah. I'm inclined to agree then. It may come in handy though to help you realize when a problem is ill-posed.
@Twink Don't use eqnarray, it is problematic. align environment from amsmath replaces it. ( Also, there is a tex.stackexchange you know )
and which command should I use?
instaed of eqnarray?
@Twink \ begin{align}
you can search for the syntax
23:25
you need to know it, but it snot complicated
and &=& ?
thanks :D
Just google for amsldoc and look at the examples
I'm gonna try
I have to import amsldoc?
23:26
No
amsmath
amsldoc = documentation. A pdf you'll find on the net. Download it and keep it open. Tons and tons of examples in it for everything you might want to do
but do I need to use $ for the math symbols?
I alread changed to align
but the equations still stay on the next page :S
they don't move
@Twink I bet the manual has something on this...
I posted a question :)
I already solved the problem :D
I just put some equations together
to reduce the number of rows

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