`Bernardo Meurer \hfill Maria Adelaide Ambrósio \hfill Inês Coelho\\`

@BernardMeurer There are probably several different ways to do this, but there's one. Use the package tabularx and do \begin{tabularx}{\textwidth}{ccc} Name1 & Name2 & Name3 \\ Number1 & Number2 & Number3 \end{tabularx}

and so on.

And damn, but chat dos wierd things with sequences of \. 1: 1 2: \\ 3: \\\ 4: \\\\

but only in backticks?

`a: \ 2: \\ 3: \\\ 4: \\\`

\centering
\begin{tabular}{ccc}
    Bernardo Meurer & Maria Adelaide Ambrósio & Inês Coelho\\
    86242 & 87064 & 87022
\end{tabular}

@dmckee I think it's the ChatJax script running amok ;)

Got it :)

Argh.

Thanks folks :3

\begin{table}[H]
\centering
\begin{tabular}{@{}ll@{}}
\toprule
Tarefa      & Tempo Gasto \\ \midrule
Planeamento & 20min       \\
Compra      & 120min      \\
Construção  & 65min       \\
Testes      & 15min       \\ \midrule
Total       & 220min
\end{tabular}
\hfill
\centering
\begin{tabular}{@{}ll@{}}
\toprule
Peça      & Custo \\ \midrule
Íman      & 1,60€ \\
Pilha     & 4,00€ \\
Terminais & 1,50€ \\
Fio       & 0,50€ \\ \midrule
Total     & 7,60€
\end{tabular}
\end{table}

Also, if I have two tables like this, how can I get the closer together?

How it looks

Don't do hfill? :P

I's like them closer to each other in the middle

@ACuriousMind tehn they're too close >:(

(yes those sexy arms are mine in the pic)

@BernardMeurer So manually adjust the gap by...\hskip{xpt}, I guess, where x is an empirically determined number

it says pt is illegal

You could also center each of those tables within their own minipage spanning half of the full page, but that's probably overkill and an abuse of minipage :P

@BernardMeurer Ugh, can never remember that, try em

and it also complained that I shouldn't be using primitive TeX

@ACuriousMind em is also kaput

Then it's not the em or the pt, it's the command itself

Fixed, removed {}

@BernardMeurer Hm, for the elegant or "correct" solution, you should probably ask the people at TeX - LaTeX

Were just hacks here ;)

@ACuriousMind I just need it to work, and that did the trick :)

Thanks ACM!

@BernardMeurer My question is why are your arms in that picture? :P

@ACuriousMind Sloppy photographer

Can I embed video in LaTeX? :P

I fear you can

Jesus

Some madman wanted to it here for his thesis, IIRC

Had to be a mathematician

Heh, top answer is by our very own @DavidZ, as I see now

@DavidZ Why are you financing the dreams of a madman?

@BernardMeurer wat

@ACuriousMind $V\cdot A$

@BernardMeurer wat

@ACuriousMind Look deep and you will see the joke there

What's $V\cdot A$?

Watt.

:')

Still doesn't tell me what you meant by "financing the dreams of a madman"

@BernardMeurer You wouldn't believe how many variants of this joke I've endured. My physics school teacher was very fond of it

He also was kind of a madman :P

My new favourite meme, @0celo7

@ACuriousMind Well, David is endorsing embedding videos on LaTeX, that was my point

@BernardMeurer "financing" means "giving money"

@ACuriousMind But he's financing with brain money

...brain money. Okay.

@acuriousmind do you know if this is the right approach for a problem like this: find points on $xy^2z^3 = 2$ that are closest to origin. I tried $xy^2z^3 = 2$ constrain by $g(x,y,z) = \mid \langle x,y,z \rangle \mid = \sqrt{x^2 + y^2 + z^2}$ so that it's a minimum/max when they share a common orthogonal vector i.e when $\nabla f = \lambda \nabla g $?

I tried solving it but it got so bad I just gave up

I don't know what "constrain by $g(x,y,z) = \sqrt{x^2+y^2+z^2}$" is supposed to mean.

Don't use the square root.

@ACuriousMind He meant $xy^2z^3 = 2$ is the constraint; $g$ is what he's trying to minimize.

like the closest to origin is when the vector from the origin to the point has minimum length which is given by the square root

@BalarkaSen Yes, that would be the correct approach.

However, my experience with @Obliv tells me to be careful in assuming he meant something other than what he wrote ;P

yeah I meant what balarka said lol

@Obliv A minimum of $g$ is the same as a minimum of $g^2$.

@Obliv Then that's the correct approach, and Balarka also already told you how to make it easier.

Oh.. duh. thanks

@Obliv The Hessian gives the best quadratic approximation to the function: $f(a+h) = f(a) + Df(a)h + 1/2h^T Hf(a) h+ \epsilon(h)$ where the error satisfies $\epsilon(h)/\|h\|^2 \to 0$, so upto second order it doesn't matter (this is exactly the multivariable Taylor's theorem). If $Df(a) = 0$, you can say $f(a + h) - f(a)$ looks like $h^T Hf(a) h$. WLOG assume the critical point is at $a = 0$ and $f(0) = 0$.

Then that's saying $f(x)$ looks like $x^T Hf(0) x$, which is really a quadratic form in $x_1, x_2$ - the components of $x$. So write $f(x)$ as $ax_1^2 + bx_1 x_2 + c x_2^2$. This is the canonical local model for a function near a critical point. If $b^2 - 4ac > 0$, then that's a normal parabola or an upside down parabola depending on $a > 0$ or $a < 0$. If $b^2 - 4ac < 0$ that's a saddle surface. So you get information about the local nature of the critical points of $f$ likewise.

I am handwaving the error off here but I think you physics people won't sew me for that. It can be handled appropriately, but the analysis is unimportant.

what is ^T?

Transpose.

Since the Hessian Hf(a) at a is a symmetric matrix, I'm using it as a quadratic form, is all.

is that in general for f(a) where a \in r^n?

What're you referring to? The Taylor's theorem I wrote down?

i mean, is the approximation you set up only for functions of 2 variables?

oh, no, it holds in all dimensions, as long as your function is C^2. I just used n = 2 for simplicity.

I mean, it's exactly the second order approximation to the Taylor series for n = 1.

hmm wait then how is $h^T$ read? isn't $h$ just some increment $(\Delta x,\Delta x_2 ,...)$?

$h$ is just an arbitrary vector in R^2. Note that I added an appropriate error term, which becomes small if h is small.

why do you transpose h?

wait nvm

so that Hf(a) * h^T returns a 1xn matrix?

for whatever amount of variables f has

It's h^T * Hf(a) * h I am looking at. By doing the transpose all of that makes sense, yes.

And it's an actual real number.

@Obliv Where do you see $Hf(a)\dot h^T$ in there? You may think of $h^T Hf(a) h$ simply as the dot product of the vector $Hf(a) h$ with the vector $h$, if you're that bothered by the transpose.

yeah okay makes sense. then you're looking at this approximation as a quadratic form?

@acuriousmind no i don't mind it I just didn't think of it that way before.

yes, because that's what it is. a symmetric matrix is the same thing as a quadratic form.

is that the same thing as a bilinear form?

i heard that term mentioned before and i didn't know what it meant

The point is this gives a 2nd order approximation, or better rephrased, a conic which is the best approximation of the graph of your function at that point, like the tangent space is the best linear approximation.

and you observe the properties of that approximation to understand whether it's a saddle point, max, min, etc.

So when you have a critical point, you can just read off whether it's a max/min or a saddle just by looking at what that conic is: an upside down paraboloid/ a normal paraboloid or a saddle.

yep

@Obliv It's a generalization. Don't worry about that.

so for larger n, though, you'd have a polynomial form as the approximation, along with the quadratic form? @balarka

You can always cut out a 2nd degree polynomial, for any dimension whatsoever. But yes, in general you get a power series.

like 3 vars? you'd have a 2nd non-zero term in the approx

@Obliv No. For any dimension and any matrix $A$, $x^T A x$ is always just quadratic in the variables.

(Multiply it out in components if you don't believe it)

Oh, he's talking about that.

i like how you guys have to try and interpret me. lol

@acuriousmind currently doing that

We don't, though. Work on your mathematical communication abilities.

you were right @acuriousmind

and sorry @balarka that's a long-standing issue of mine.

I really like this point of view on the second derivative test. It's also the gateway to Morse theory; the proof of the Morse lemma is just a bit more careful analysis with the error term I shrugged off.

Anyway, I said what I had to say, and I'm going to bed. Feel free to badger @ACM with further questions :P

thanks again @balarka cya around.

@Obliv I know. :P

good evening @obe

@obe do you still remember the inverse trig integrals & all the other random trig integral techniques

my next exam will have at least double integrals in it and im wondering if i should memorize them again

its ok @obe i'm sure you can derive them/figure it out

all you really need is a set of numbers to work with and a finite amount of time to be able to reconstruct modern math and to be able to solve a problem

yeah i remember the integral u linked in chat was integration by parts

yeah lol the 12 dimensional one

yea nvm that was w.r.t. y first not x.

and even if I did have to integrate w.r.t y first to solve it, i'd just switch the integrals because of fubini's theorem.

yea ik i'm dumb.

@obe i have a curse where i have to write down something and post it before i recognize how dumb it sounds/the point i'm missing.

what's bad about it being 1. ?

reading is different from memorizing all the problems lol

no, i'm stressing over procrastinating studying for 3-4 exams next week , though. but this time, i'm studying over the weekend. still probably not enough time lol

yeah but I have to take care of some business first..and i can apply in the spring probably.

yea

thanks