Mathematics - 2012-09-20 (page 3 of 3)

user19161

16:00

@MeAndMath Bad, thanks for asking.

Parth Kohli

Hmm

user19161

@ParthKohli I see you changed your pic again.

MeAndMath

@WillHunting Why?Put a smile in that face!You will see that life still worthwhile...if you just smile!

Parth Kohli

Yeah, I am trying something out.

MeAndMath

@GustavoBandeira Você tem outra graduação?

user19161

16:01

@MeAndMath That seems to be a song! Who is the singer?

Gustavo Bandeira

@MeAndMath Não. Eu fazia economia e fiz eletrônica.

MeAndMath

@WillHunting this is a song for "modern time" ,by Chaplin.Many people sang it.Chaplin only made the arrangement.someone else wrote it.

►

@GustavoBandeira Interessante.Vai engajar na mat?

Gustavo Bandeira

@MeAndMath Sim. Eu num te falei?

MeAndMath

@GustavoBandeira o que ,exatamente?

Gustavo Bandeira

Eu te falei que vou fazer bacharelado aqui em Recife.

MeAndMath

16:07

@GustavoBandeira sim,isso falou.

quando é o vestibular?

Gustavo Bandeira

Próximo ano. Porque eu perdi nesse. =/

MeAndMath

por quê?

Gustavo Bandeira

Burrice.

MeAndMath

?

perdeu a data?

Gustavo Bandeira

Sim.

MeAndMath

16:09

ai ai...

Gustavo Bandeira

Então.

Eu tenho impressão de que o diretor do DMAT me curtiu.

MeAndMath

que bom1

Gustavo Bandeira

Eu fui lá e nem tinha entrado no curso, ele disse: "Seja bem vindo"

Ele me emprestou dois livros - só me viu duas vezes na vida.

Tô pensando em pedir pra ficar assistindo aula.

MeAndMath

gente fina

ora,vai lá

user19161

Most questions here are answered so quickly.

Gustavo Bandeira

16:11

Vou pegar as aulas do próximo semestre - que é quando começa o novo ano letivo.

@WillHunting Yep.

MeAndMath

@WillHunting :D

@GustavoBandeira aproveita!

Gustavo Bandeira

@MeAndMath Curso de matemática aqui é uma delícia.

Tem três pessoas na sala. xD

MeAndMath

vish

dependendo da matéria,tem umas 60...

Gustavo Bandeira

Yep.

MeAndMath

estat chega a ter 180

Gustavo Bandeira

16:15

Sim. Estatística é mina de ouro, né?

MeAndMath

mas a a aplicada costuma ter 20,15

@GustavoBandeira vários cursos precisam..

Gustavo Bandeira

Eu sei, mas o curso de estatística mesmo é mina de ouro, né?

MeAndMath

depende,o que você quer dizer?

Gustavo Bandeira

De ser um curso que dá muita grana.

Parth Kohli

Hmm... now it's good.

MeAndMath

16:18

@GustavoBandeira é o que dizem

@ParthKohli what is good?

Parth Kohli

My Gravatar.

Not changed yet, seems so.

Jonas Teuwen

Damn Gravatar.

MeAndMath

not yet

user19161

Is the site slow for you guys now?

user19161

This seems to happen once or twice a month.

MeAndMath

16:22

just gravatar

@Gustavo?

Parth Kohli

Hartnn?

Jonas Teuwen

@WillHunting A shocking treatment.

Gustavo Bandeira

@MeAndMath Oi. Tava no banho.

Parth Kohli

@hartnn How are you here?

MeAndMath

@GustavoBandeira ah,ok

Gustavo Bandeira

16:25

O povo tem uma preocupação absurda com dinheiro, já percebeu que não é tão importante?

Uma das poucas coisas que eu acho que o dinheiro traz de muito bom é plano de saúde decente.

Mas de resto.

MeAndMath

@GustavoBandeira não.

Gustavo Bandeira

O povo fica rico pra trocar os móveis de casa de 6 em 6 meses...

MeAndMath

fazer o que se gosta...daí o resto vem

Gustavo Bandeira

@Lembrei de um professor da faculdade.

MeAndMath

fala

fale,Gustavo

Gustavo Bandeira

16:28

De metodologia científica. Ele entregou um texto a importância do método científico e o texto fala que o método científico é bom - minimamente - e depois, falava sobre as dificuldades que o aluno da noite enfrenta... Aí ele veio pedir opinião. Eu: "Achei uma merda". xD

MeAndMath

noossa

Gustavo Bandeira

"O título do texto é esse - o conteúdo tem nada a ver com o título, é até deprimente a classe não ter notado isso."

MeAndMath

e ele/

Gustavo Bandeira

Ele perguntou se eu já tinha feito outro curso superior.

Porque pra notar esse tipo de coisa, preciso disso.

MeAndMath

e tu?

ce ve a mentalidade

Gustavo Bandeira

16:30

Eu: "Ué, mas isso num é interpretação de texto que todo mundo aprende na primeira série não?!"

MeAndMath

tome

Gustavo Bandeira

Eu fiquei indignado.

Mesmo.

MeAndMath

caracas...

já pensou em ir pra sp?

Gustavo Bandeira

Ensino superior serve pra notar besteira em texto. xD

Yep.

MeAndMath

e?

Gustavo Bandeira

16:32

Eu pensava a um tempo, mas por enquanto acho que num rola.

MeAndMath

por quê?

Gustavo Bandeira

Depois de um tempo eu percebi também que num é lá tanto problema morar aqui. Porque tem internet.

O stack e o MO tem gente do mundo todo respodendo coisa.

Pronto. Me arrumei.

@MeAndMath Quando ele falou isso, eu té lembro que eu fiquei triste. Eu pensando que ensino superior era pra formar gênios...

@ParthKohli If you change your pic again, I'm gonna ban you!

@MeAndMath Vô indo pra aula de piano. Até mais.

Parth Kohli

Heh, this was the last, I promise.

Nick Kidman

17:01

hey

math101

Hey

Nick Kidman

are there practical applications of algebraic geometry results past...1930

Mariano Suárez-Alvarez

of course

a minimal amount of googling will provide lots of examples

math101

lol

check this out and go from here

Mariano Suárez-Alvarez

from things in biology like the problem of phylotactic classification, to the study of nonlinear models in statistics, to the implicitation/parametrization of surfaces in computer graphics and industrial design to cryptography

people who work on controlling robots do algebraic geometry, people who do non-linear operations research do algebraic geometry, and so on

Zhen Lin

17:11

Algebraic geometry over $\mathbb{R}$ has a rather different flavour, though...

Peter Tamaroff

@MarianoSuárez-Alvarez Hola =)

Mariano Suárez-Alvarez

hola!

Peter Tamaroff

@MarianoSuárez-Alvarez Strange that you're around at this hours!

Mariano Suárez-Alvarez

heh

Peter Tamaroff

@MarianoSuárez-Alvarez I finally got my copy of Apostol back.

Mariano Suárez-Alvarez

17:12

never leave home without it

Peter Tamaroff

@MarianoSuárez-Alvarez =) How's it going?

Jonas Teuwen

@MarianoSuárez-Alvarez I always leave home without my copy of Apostol... what now?

Shall I resign from my position?

Mariano Suárez-Alvarez

nerdy YOLO

Peter Tamaroff

@JonasTeuwen "To the bonfire"

J. M.

I'm getting more out of touch with new stuff everyday... where is this "YOLO" coming from?

Peter Tamaroff

17:18

@MarianoSuárez-Alvarez I did a little thing about linear algebra yesterday.

@MarianoSuárez-Alvarez I also corrected the proof of the sine and cosine series, that we talked about.

Nick Kidman

okay, so regarding algebraic geometry the answer is "yes", but I still don't see how

the robot thing might be accessible to me, but why do you need more than usual geometry, coordinate stuff etc.

differential geometry

another question: Is Turing completeness a property of a programming language, the computer, or both?

J. M.

14

Q: "Immediate" Applications of Differential Geometry

My professor asked us to find and make a list of things/facts from real life which have a differential geometry interpretation or justification. One example is this older question of mine. Another example my teacher presented is proving that on a soccer ball which is made of regular pentagons and...

Jayesh Badwaik

@NickKidman I do not know much about other fields, and I do not know much about what I am going to say either, but I have read that Algebraic Geometry is terribly useful in describing solitons.

MeAndMath

@JayeshBadwaik Hello ,Jayesh!

Jayesh Badwaik

@MeAndMath Hello!

How are you?

MeAndMath

17:26

good,good

you?

Parth Kohli

@MeAndMath Y U SO CHAPLIN?

MeAndMath

:DDD

Love him

Parth Kohli

Oh. :)

Jayesh Badwaik

I am good. Trying to finish some notes.

MeAndMath

At last:someone is good around here!

Lews Therin

17:39

Hello

John Junior

hi

Lews Therin

Is this chatroom as stackoverflow? :) I may be asking some noobish questions in the future :O

Jayesh Badwaik

@MeAndMath :-)

@LewsTherin Ask away.

Lews Therin

I have to go out now, but when I get back get ready to be pissed off by my ignorance

Are you guys programmer/mathematician or just purely mathematicians?

J. M.

@LewsTherin Me, neither.

@LewsTherin If it's pretty short, you can ask here. If you think an answer would involve a long explanation, ask on main.

Lews Therin

17:45

Well, the question I am asking shouldn't be too complex

Like getting the sense of vectors, rotation and stuff

J. M.

@LewsTherin I've a feeling that has certainly been asked before in main...

Lews Therin

I will check thanks

J. M.

Try searching around and report back.

hhh

$\dot y=x+y^2$, how to check linear conditions i.e. $f(a*x)=a*f(x)$ and $f(a+b)=f(a)+f(b)$?

It is non-linear ODE, according to my material, but I want to verify it.

Let $f(x)=y$. So

$f(a*x)^2=\dot y(a*x)-x$

Now $f(x)^2=\dot y(x)-x$ so $a f(a)=a\left( \dot y(x) -x\right)$.

Matt

J.M.! I lost my teddy!

How are you, btw?

hhh

17:56

Suppose $f(a*x)^2=af(x)$ then

J. M.

@Matt Oi, really?

@Matt Somewhat stressed, but I'm holding up.

Matt

@J.M. Yes :,( Look at the profile.

Lews Therin

Why is there so much formulas used? If I ask a question will I get an answer without the formulas?

Matt

@J.M. Why? Or secret stuff? (my situation has been resolved ftw)

hhh

$a\left( \dot y(x) -x\right) = \dot y(a*x)-x$

J. M.

17:57

@Matt Checked just now. Somewhat like Arturo, it seems.

hhh

So $x(1-a)=\dot y(a*x)-a \dot y(x)$ where $\dot y=x+y^2$ by the definition.

Matt

@J.M. Yes, but this one I miss! : )

J. M.

@Matt Glad to hear you're now okay. Well, part of the stress is due to the elections in the site I'm modding. The other... well, can't say here, except it's work-related.

@LewsTherin Sometimes, the formulas say things that words have a hard time saying.

Matt

@J.M. Eew, sorry to hear. Will it be resolvable?

Lews Therin

@J.M. It won't help me. I'm no mathematician :(

J. M.

18:00

@Matt Fingers crossed.

@LewsTherin Well, then you can ask about the formulas, sure, but you should try to understand them also.

arete

I'm trying to determine the probability of rolling at least a pair of sixes in 24 independent dice rolls with a pair of die.

From what I can glean, it will most likely be easier to compute the complement. That is the probability that one six will be thrown in 24 rolls.

Peter Tamaroff

@arete You mean only one, right?

If more than one is thrown, you're done. So you want to find the prob that only one is thrown.

arete

@peter using two die, I'm looking to find the probability of rolling a pair of sixes within 24 rolls.

Yes.

So, I want to find the complement of the above situation, which is the probability that only one six is thrown when rolling a pair of dice 24 times.

Peter Tamaroff

@arete Oh, two die.

@arete Well, the first is $1/6 \times 5/6$, then $(5/6)^2$ 23 times.

arete

Ok, so you're taking the probability of rolling a single six on one die and the probability of rolling all but a six on the other

Then you square it, because you have to take into account that there are two die.

Leaving you with $\frac{25}{36}$.

Problem is, I know the final answer and that complement doesn't get me there.

Matt

18:16

@J.M. crosses fingers

arete

I'm not taking something into account.

Peter Tamaroff

@arete Yes, but you need

$$\frac{1}{6}\frac{5}{6}{\left[ {{{\left( {\frac{5}{6}} \right)}^2}} \right]^{23}} = \frac{1}{6}{\left( {\frac{5}{6}} \right)^{47}}$$

arete

@peter

Where did that come from?

Peter Tamaroff

18:30

@arete Didn't we say we need $1/6\times 5/6$ and then $5/6\times 5/6$ 23 times?

Gustavo Bandeira

19:45

Where's John Senior?

Last time seen: Sep 15 at 14:44

John Junior

19:58

$\Huge\text{Vacation}$

Ed Gorcenski

20:27

Good morning.

Jayesh Badwaik

Good morning. Good morning. (Beatles)
Its 2 am here, and I have had a lot of beer.
Good morning. Good morning. (Beatles)

:P

Ed Gorcenski

It is 4:31 PM here but my brain feels like it is morning.

Link

hi

John Junior

hi

Ed Gorcenski

hi

John Junior

20:43

lo

lo

lo

Link

I was doing some physics HW, and came across a problem

As you left your home one morning, your bus was 20.0m away and has just strating to ppull away from the bus stop at 4.00m/s^s. You start running at a constant speed towards the bus, catching up with it 5.00 seconds later. How fast did you run?

I just wanted to know how I would start doing this problem?

I have no idea..

Ed Gorcenski

So you have two things that are equal

The first, is the position of the bus as a function of time

Let's denote that as B(t), let's also assume you're standing at the origin.

Then, B(0) = 20. (The bus is 20 meters away when you start running.)

Link

okay

Jayesh Badwaik

@JonasTeuwen Do you use the auto-complete package of emacs?

Jonas Teuwen

@JayeshBadwaik No.

Ed Gorcenski

20:46

Then, the formula for the position of the bus, B(t), given constant acceleration is B(t) = 1/2at^2+vt+B(0)

Jayesh Badwaik

@JonasTeuwen Okay. What about brackets and stuff? You type out the closing brackets yourself? Or is there something for that?

Ed Gorcenski

v is the initial velocity of the bus, but that's zero (the bus is leaving it's stop)

Jonas Teuwen

@JayeshBadwaik No, use AucTeX.

C-c C-e for environments.

Ed Gorcenski

And a = 4 m/s^2, as given in the problem statement.

Jayesh Badwaik

@JonasTeuwen Okay.

Ed Gorcenski

20:47

Thus, the distance the bus travels in t seconds is B(t) = 1/2*4*t^2+20 meters

Jonas Teuwen

@JayeshBadwaik I don't like brackets when not needed.

Ed Gorcenski

Plug in 5 seconds, and you get B(t) = 1/2*4*5^2+20 = 2*25+20 = 70 meters.

So that's how far the bus traveled in 5 seconds.

Now, in those same 5 seconds, you, running at constant velocity, catch up to it.

Link

okay

Ed Gorcenski

Consider your position as a function of time: It's the same formula, let's call it P(t)

Link

ah

what is the formula called?

Ed Gorcenski

20:49

P(t) = 1/2*a*t^2 + v*t+P(0)

Link

okay

thanks

Ed Gorcenski

But, you're running at a constant velocity, which means your acceleration is zero: a=0. Also, you're starting at the origin, so P(0) = 0

Jayesh Badwaik

@JonasTeuwen Yup, same problem with me when I used Kile. But for some, they are needed all the time, so it was a decent compromise. Anyway, let me try C-c C-e.

Link

i got it now

Ed Gorcenski

And, your position is the same as the bus's. SO 70 = P(5) = v*5. Solve for v.

Jonas Teuwen

20:50

@JayeshBadwaik There is keybinding to make {} and put the cursor between.

Link

14!

Jonas Teuwen

@JayeshBadwaik For some when they are needed, adjust that for them...

Or only use autocomplete for those.

Jayesh Badwaik

@JonasTeuwen Yup, already using that keybinding here for those things as of now. Damn, this auctex thing is really good. It partially auto-labels stuff, nice!

Ed Gorcenski

What's auctex?

Jayesh Badwaik

Emacs extension to $\TeX$.

Ed Gorcenski

20:53

Ah

when you say auto-labels, do you mean it auto inserts \label{} commands for you?

Jayesh Badwaik

No, I was not clear. Whenever you say \label{}, it will determine the environment you are in and assign a keyword for it. So for section, it automatically adds section: to the label, so now your label is \label{section: <something you type>}

Ed Gorcenski

Ah nice

Jayesh Badwaik

@EdGorcenski a major reason to use emacs for LaTeX is this atleast for me.

Ed Gorcenski

That is a nice feature.

Jonas Teuwen

21:09

@JayeshBadwaik I don't use preview-latex.

@JayeshBadwaik Would we able to have it use MathJax instead?

Jayesh Badwaik

@JonasTeuwen I guess so. I am not sure. Probably the svg renderer of mathjax.

Jonas Teuwen

@JayeshBadwaik Can also use it as a browser so I guess should work.

Would work much better.

Then I might consider using it.

N3buchadnezzar

Hey

Jonas Teuwen

Sup ho.

N3buchadnezzar

Problems with super easy problem

Jonas Teuwen

21:11

@N3buchadnezzar That's the time you should go to the fridge bro.

If you start having trouble with easy problems.

It means you ran out of fuel.

N3buchadnezzar

Rather I do not understand the problem?

No, I am sick. I just ate, and need to get this done =/

Ed Gorcenski

Here's an extremely frustrating "simple" problem that I'd love to figure out.

Let $x_1 > \sqrt{\alpha}$ and define $x_{n+1} = \frac{1}{2}\left(x_n + \frac{\alpha}{x_n}\right)$. Show that $x_n > \sqrt{\alpha}$ for all $n > 1$.

N3buchadnezzar

Consider the set $\mathbb{R}^2$ and define addition and scalar multiplication
on it as follows.
$$(x_1, x_2) + (y_1, y_2) = (x_1 + 2y_1; x_2 + 3y_2) \qquad (x_1, x_2) = (x_1, x_2)$$
Prove or disprove that $\mathbb{R}^2$ under these operations is a vector space.

@EdGorcenski Its just newton bro, draw a figure and do some asymptotic juggling.

Jayesh Badwaik

@EdGorcenski By definition $\sqrt{\alpha}$ is a positive root of alpha. After that it is easy to see that $x_n + \frac{\alpha}{x_n} > 2 \alpha$.

Ed Gorcenski

I know it's Newton but I need a rigorous proof.

N3buchadnezzar

21:16

I was thinking that it was a Vspace if it was closed under multiplication, and addition. But the problem is states that it is, so I do not know what is left to prove?

Jayesh Badwaik

@EdGorcenski $\frac{(x_{n} - \alpha)^{2}}{x_{n}} >0$

Ed Gorcenski

@JayeshBadwaik I don't follow the latter. I know it should be "easy to see" but somewhere I get crossed up

Jayesh Badwaik

52 secs ago, by Jayesh Badwaik

@EdGorcenski $\frac{(x_{n} - \alpha)^{2}}{x_{n}} >0$

Are you referring to this?

Ed Gorcenski

No, I see that now.

That makes sense

Jayesh Badwaik

Are you doing those bifurcation maps?

Ed Gorcenski

21:18

But, I don't have that $x_n$ > 0 necessarily.

Jayesh Badwaik

@EdGorcenski You have for the first term, and then induction.

Ed Gorcenski

No, this is just a tiny part of a problem I had to solve, and I know i've done it before.

Jayesh Badwaik

@EdGorcenski Okay.

Ed Gorcenski

Ah, right, fair enough

N3buchadnezzar

Now onwards to the next problem!

Ed Gorcenski

21:20

Jeez I should have seen that last night

Link

hey Ed, how do you get maximum speed from accerlation?

Ed Gorcenski

What do you mean?

The general formula for distance as a function of time under constant acceleration is: x(t) = 1\2*a*t^2+v_0 * t + x_0. The general formula for velocity as a function of time given constant acceleration is v(t) = a*t+v_0.

Here, v_0 is initial velocity (velocity at time t = 0) and x_0 is initial position (position at time t = 0.

Using these formula, you can solve for all problems with constant acceleration.

@N3buchadnezzar What axiom are you having trouble showing?

Link

okay

thanks

N3buchadnezzar

@EdGorcenski Which ones do I need to prove? I thought it was enough to show that it was closed under addition and multiplication?

Ed Gorcenski

Yeah, basically. But you also need to show the existence of a zero vector and an additive inverse for the vector elements

N3buchadnezzar

21:35

It is closed under addition right because it is stated in the problem that $\lambda(x_1,x_2)=(\lambda x_1 , \lambda x_2)$

It is closed under addition since $(x_1,x_2) + (y_1,y_2)$

Ed Gorcenski

That's not a complete statement.

What you need to show is that $\mathbf{x}+\mathbf{y} \in V$.

(for every x and y)

So $\mathbf{x} + \mathbf{y} = \left(x_1+2y_1, x_2+3y_2\right) \in \Bbb R^2$?

Well yes, because $\Bbb R$ itself is closed under multiplication and addition, so there is no combination of operations that can lead to an element being "outside" of $\Bbb R$.

And the same argument works for scalar multiplication.

Link

hmm, I have a car going at 2.00m/s^2, for 1.50s, then slows down at 1.00m/s^s, how long does it take to stop?

I tried to solve with two eqautions and intersection point

but that doesn't seem to work

Ed Gorcenski

Then you have to show that the zero element exists, and that the inverse element exists.

Zero element is easy. You can show inverse element by construction.

@Link you don't need two equations for this because you only have one thing in motion.

Well, you kind of do, but in a different way.

When do you start to hit the brakes? At t = 1.5

Link

yes

N3buchadnezzar

@EdGorcenski How does one show that the zero element is unique?

Link

21:46

i mean 1.00m/s^2

Ed Gorcenski

@N3buchadnezzar by contradiction.

@Link So your position for the first 1.5 seconds is modeled by the position equation with a = 0: x(t) = 1/2*a*t^2+v_0 * t +x_0

a = 0, and assume x_0 = 0

So for 0 <= t <= 1.5, x(t) = v_0 t = 2*t

So when you start to hit the brakes, then you have gone 3 meters.

Link

k, i got that far

Ed Gorcenski

So remember that for a second.

Now, let's look at just the phase where braking starts.

Link

okay

Ed Gorcenski

use the velocity equation: v(t) = a*t+v_0, with a = -1, and v_0 = 2

We want to figure out when the car stops? What is v(t) when the car is stopped?

Link

21:51

0?

Ed Gorcenski

@N3buchadnezzar Assume that there exist two zero elements, $z_1$ and $z_2$, $z_1 \neq z_2$. Then, $x+z_1 = x$, $x + z_2 = x$. But, $(x+z_1)+z_2 = x+z_1 = x$.

@Link Yes. So solve for 0 = -1t+2 for t.

Link

hmm okay

is it 2?

Ed Gorcenski

Yes

Link

but my book shows it as 3.00s to slow down and stop, not 2

Ed Gorcenski

@N3buchadnezzar But $x+(z_1+z_2) = x+z_1 = x$, so $z_1+z_2 = z_1$. A similar argument shows that $z_1+z_2 = z_2$. Thus, $z_2 = z_1+z_2 = z_1$, then transitivity proves that the element is unique.

@Link I'm sorry, I read your equation wrnog

I thought your initial velocity was constant.

But I see that it's 2m/s^2 initially

So, we need to re-compute what the speed of our car is when we hit the brakes

Link

21:55

it was 3.00m/s

Ed Gorcenski

That's easy: use the velocity equation. v(t) = 2*t+v_0, let v_0 = 0. then $v(1.5) = 2*1.5 = 3

Now, solve 0 = -1t+3 for t.

I misread and misunderstood your question at first.

Link

okay

I get it now

thanks, and no problem

Ed Gorcenski

@N3buchadnezzar Are you coming along with it?

N3buchadnezzar

@EdGorcenski YEah, I just had to have a break and eat something. You helped me a lot =)

Ed Gorcenski

No problem.

N3buchadnezzar

22:07

math.ntnu.no/emner/TMA4145/2012h/exercises/oving4.pdf

Any hints on 3b?

I understand it, but it is hard to prove.

Jonas Teuwen

@N3buchadnezzar Meh.

$x = \frac12 x + \frac12 x$.

Peter Tamaroff

@N3buchadnezzar Well, do you understand what it is you want to prove?

Ed Gorcenski

Yes

Hint: assume ||x|| < 0

Peter Tamaroff

@N3buchadnezzar The axiom of non negativity is $||x||\geq 0$ for any $x$.

Ed Gorcenski

Then, look at ||x+x|| and ||x - x|| and see what they have to be

Then note that ||x-x|| = || 0 || = ||0 x|| = |0| ||x|| and show that that is impossible for ||x|| < 0

N3buchadnezzar

22:11

Gimme a few mintues, I am eating =)

Ed Gorcenski

No problem. I, however, am out for the night.

later all

achacttn

Hello all

Link

hi

Jeff

22:35

hey all, if a matrix (with complex entries) equals it's complex transpose, then it is self-adjoint, right? If it is self-adjoint, then it's inverse equals it's regular transpose (the transpose without taking each elements complex conjugate), right?

Jayesh Badwaik

mathworld.wolfram.com/HermitianMatrix.html

Do you mean this? conjugate transpose is the correct term.

Also, the inverse might be equal to its regular transpose only if the determinant is equal to 1.

Jeff

yeah, that.

hhh

22:50

Suppose a ODE such as $\dot y = x$, is it hard to check whether this is linear or not? $f(ax)=a f(x)$ and $f(a+b)=f(a)+f(b)$.

Now $y(x)=x^2 /2+C$ where $C\in\mathbb R$.

$y(ax)=(a^2/2) x^2 + C$ and $a f(x) = a( x^2 /2 + C)$.

$y_2(ax)=(a^2/2) x^2 + aC$

$y_1(ax)=(a^2/2) x^2 + C$

Evan

Have any of you used MyMathLab?

hhh

But now $y_2=y_1$ for some arbitrary constants $C_1, C_2 \in \mathbb R$, right?

So this ODE $\dot y =x$ is linear.

ERR
$a y_2(x)=(a^2/2) x^2 + aC_2$

$y_1(ax)=(a^2/2) x^2 + C_1$

Now $a y_2(x) = y_1(a x)$ for $C_1, C_2 \in \mathbb R$.

where $y_1 = y_2$.

- so one of the linear property holds.

Jeff

23:06

@Evan i have

hhh

$f(a+b)=a^2/2+ab+b^2/2+C_3$

$f(a)+f(b)=a^2/2+b^2/2+C_4$

So $g(a,b) := f(a+b)-(f(a)+f(b))=ab+C_3-C_4$.

Because $g(a,b)\not =0$ for every $a,b\in\mathbb R$, $f$ is not linear, correct?

Moved it to a question here.

N3buchadnezzar

23:43

hi

Link

hi

anyone good with basic kinmatics here?

me 7:14 PM
i have a problem
and have no idea
what to do
basically, a ball is dropped from an unknown height, accelrating at 9.81m/s^2 downward until it hits the ground 2.00s later. After hitting the ground, the ball rebounds at half the speed it had just prior to impact.
a (When does the ball first come to rest after hitting the ground, where the drop time was t =0s
b, how high does the ball rebound?

i don't actually understand how to even start

can someone point me in the right direction?

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