Mathematics - 2016-12-02 (page 11 of 11)

Astyx

21:00

Bon appétit @Ted

Danu

@TedShifrin That's right.

Faraad Armwood

@Danu: I'm very naive. I really do think the best of people.

Danu

Don't be too disappointed :P

Hey-men-whatsup

see you Ted, thx for the advise

take care..

Astyx

Imma go too, bye everyone

Faraad Armwood

21:03

I'm also leaving. Cheers everyone. I guess I've found a place to talk with some people, that's really cool. Later :)

Aksel'sRose

how would you explain that the following is always positive: $s^2+st+t^2$

Hey-men-whatsup

See ya @FaraadArmwood

Aksel'sRose

its part of a proof so I dont want to do another proof, just need a way to describe it

Hey-men-whatsup

take care..

@Astyx, see ya take care..

Krijn

@Aksel'sRose $s^2 + t^2$ is larger then $st$ I think, in absolute value

Aksel'sRose

21:07

makes sense

Semiclassical

hi chat

Krijn

Hi @semi

Hey-men-whatsup

hi @Semiclassical

Semiclassical

@Null I feel conflicted about that.

Null

@Semiclassical i am allowed to prove stuff for myself. But i am not allowed to use, for example Pascal's rule out of thin air. That's how he meant it.

Semiclassical

21:10

On the one hand, it's hardly unreasonable to ask that a given problem be approached a certain; the point isn't just to offer a proof, but to develop your understanding of the method.

But if they don't say "do it without using X" I feel a bit awkward saying "you did it wrong."

Krijn

I finally fixed my computer :D

Null

@Semiclassical it involved a limit approximation of 1/n (which was 0). But we didn't even have limits yet in the seminar. Altho i seriously wonder how you determine the infimum without limits. Well..

Semiclassical

What was the question, more precisely?

Null

i look it up@Semiclassical

Semiclassical

mmkay

Null

21:14

@Semiclassical Determine the infimum of
$\{\frac{1}{2^k}+\frac{1}{n}|k,n\in\mathbb{Z}^{+}\}$, which has no minimum btw (easy to check)

and the set itself is obv a subset of R

Krijn

Why is the infimum not obviously 0?

Oh, without limits, right

For any other positive number you can easily give one that's beneath it, technically without limits

Semiclassical

Eh. You've got $1/2^k>0$ for all integer $k$, and similarly $1/n>0$. So the set has zero as a lower bound.

Null

@Semiclassical so it is left to show that for any positive number c, i can find one between c and 0?

Semiclassical

Yeah, that works.

That's not hard, either. There's some smallest integer $n$ such that $1/c>n$ i.e. $1/n <c$, so use that $n$

Null

that is Krijin statement lol^^

(but i came with it too)

Semiclassical

21:19

And then pick $k$ so that $2^{-k}<c-1/n$.

That gives you a value which is smaller than any given $c>0$, so no positive number can serve as minimum of this set.

Hence infimum is zero. (Not clearly stated here, but i'm too lazy to be precise.)

Null

@Semiclassical but how do we know then that 0 is the highest lower bound?

Fargle

@Null Because as has just been shown, any number $c > 0$ cannot be a lower bound, because there are $\frac{1}{2^k} + \frac{1}{n}$ lower than it.

TheGreatDuck

hi

Null

ah so it's actually a two in one argument. no minimum and a infimum at once :)

Fargle

Quite.

Null

21:24

@TheGreatDuck to your question about mathjax in MSE: you can use \newcommand.

Fargle

Hello @TheGreatDuck.

TheGreatDuck

huh?

meow-mix

yay! i proved something on my own :)

Null

@TheGreatDuck you asked wether it's possible to use packages in MSE. it's not possible. but with \newcommand you can mimic this

Semiclassical

Because $1/2^k+1/n$ definitely can't be negative.

TheGreatDuck

21:26

...when did I ask this?

Semiclassical

eh, but that just means it can't be higher. nm on that

Null

@TheGreatDuck some days ago. i made a thread about it even. but i deleted it because -rep haha

Semiclassical

better point is just that, for any possible lower bound $c>0$, the argument just sketched shows that you can find a value in the set which is smaller than this.

back later

TheGreatDuck

ok

Null

@Semiclassical see you later

TheGreatDuck

21:28

how can I use newcommand to create custom symbols through mini-images?

Null

@TheGreatDuck ah, I thought you only wanted to use normal symbols just with another syntax. i really don't know then.

TheGreatDuck

I was just curious about whether full latex could be used

:p

Null

@TheGreatDuck latex and mathjax are different things ;)

you can search the differences if interested

easiest example would be \\ to get a wordwrap

TheGreatDuck

I thought the chat posts were in latex

always?

Steve

Hi @Alessandro, I'm having some trouble with this problem:

We observe the subspace known as $U\subset \mathbb{R}^{3\times 3}$ that is spanned by the following vectors:
$
\begin{bmatrix}
1 & 0 & 0 \\
0 & -2 & 0 \\
0 & 0 & 3 \\
\end{bmatrix}
$, $
\begin{bmatrix}
0 & -3 & 0 \\
0 & 2 & 0 \\
0 & -1 & 0 \\
\end{bmatrix}
$, $
\begin{bmatrix}
0 & 0 & 1 \\
0 & -2 & 0 \\
3 & 0 & 0 \\
\end{bmatrix}
$, $
\begin{bmatrix}
0 & 0 & 0 \\
-1 & 2 & -3 \\
0 & 0 & 0 \\
\end{bmatrix}
$.
Choose a basis for $U$ and show that $

Null

21:32

it just happens to be that mathjax is quite latex compatible

TheGreatDuck

oh

Kasmir Khaan

21:59

hi chat

anyone kind enough to help me check my work on this exercice ?

1

Q: Surface integral

1) Find the area of the surface: $Z= x^{2}-2y^{2} ,x^{2}+4y^{2}\leq 1$ I need detailed solution because it is the first problem in my problem set,if I understand it well I can figure out the rest by myself. the more detailed answer the better it is. Thanks in advance

multivariable-calculus

Chris6657456456

22:24

if I have a quarter and want to know how many attempt I need before it lands on heads x times in a row, what math function am I supposed to use? I completely forgot what its called

nm

alan2here

22:40

I was not expecting W/A to be so stumped: m.wolframalpha.com/input/?i=find+k+in+2+choose+k+%3D+m

Alessandro

@steve are you still there?

alan2here

Hmm, pretty: m.wolframalpha.com/input/?i=find+n+in+n+choose+3+%3D+m&x=0&y=0

Physics Guy

Hey, what's up ?

Semiclassical

Hi chat

Physics Guy

Hi

Fargle

22:54

Yo @Semi, chat

alan2here

Has Sqrt(3), Root(3, 3), 3^(2/3), 3^3, 3^0 and 3^(2+3) several times in the formula.

Physics Guy

What ?

alan2here

m.wolframalpha.com/input/?i=find+n+in+n+choose+3+%3D+m

A real ode to (0, 1, 2, 3) powers and roots

Physics Guy

Please explain

Null

@Fargle hi

Fargle

22:59

How goes it?

Null

i want to prove exponents grow faster then polynomials. But that is quite much to understand.

so im frustrated haha

Physics Guy

Hmmm

meow-mix

@Null show that at some point a polynomials $n$-th derivative becomes constant

Null

@meow-mix that was my thought

meow-mix

or, alternatively, use induction

on the degree of the polynomial

Null

23:03

but that is basicly l'hopitals argument or not?

Ted Shifrin

You can do it without calculus.

Look at the sequence $a^n/n^k$, e.g., $2^n/n^5$.

Null

@TedShifrin then observe that they diverge. And therefore the reciprocal has to converge to 0?

Ted Shifrin

Yeah, better even to work with the reciprocal.

Well, maybe it's not so obvious. I was thinking of a different question.

Liad

math.stackexchange.com/questions/2041057/… - someone could help ?

Null

@TedShifrin mmh, is there a geometric approach to this?

Physics Guy

23:09

@Liad

Ted Shifrin

Well, one standard thing is to prove $(\log n)/n\to 0$ using the integral definition of $\log$. What exactly do you know at this point?

Liad

@PhysicsGuy ?

Physics Guy

@Liad

Something's wrong

Liad

what?

Ted Shifrin

@Null: You have the binomial theorem, right?

Null

23:10

@TedShifrin yep

Ted Shifrin

Then $2^n = (1+1)^n \ge 1+n$, for example.

Then, if you're tricky, using $2^n = 2^{n/2}\cdot 2^{n/2}$, you should be able to prove that $2^n/n \to\infty$. You can modify this for $2^n/n^k$.

BTW, hi @Fargle

Null

@TedShifrin well, the first terms of $(1+1)^n$ are 1+n, so the inequality follows from that

Ted Shifrin

Right. Now figure out how to do the rest of my hints.

Null

@TedShifrin why exactly $2^{n}$ btw and not some base >1?

Ted Shifrin

Yeah, I don't care. I was giving you something more concrete to start with. Then you can easily generalize.

Null

23:20

@TedShifrin ah ok ;)

Ted Shifrin

Sometimes that's a good approach to figuring out mathematics, by the way.

Null

@TedShifrin yup =)

Maks

23:45

Hey, is there a formula to know to what number does a non-geometric series converges ??

Null

@Maks do you have a specific example?

Maks

@Null I cant find it :(

It was kindda like $ (\dfrac {1} {5})^n + 5n + 3 $

meow-mix

hi @ted :)

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$Mathematics$ Mathematics