@Danu Thanks for your response. If we extend that reasoning to the uncertainty principle then how does that imply that for large spread of $\phi(k)$ you would get well defined position by observing a sharper spike in the wave $\Phi(x,0)$?

@Danu Apologies if this is trivial...I'm just starting to learn some QM.

@JohnDoe It means that you can get a more precisely-defined position. Note, indeed, that a delta-function in position space corresponds to a very special function in momentum space (I'll leave it to you to work out which one; but it should be clear from the uncertainty what kind of a function it should be)

@Danu Oooo, I know which one it is!

@skillpatrol In my defense those go back to 2012. I make it an average of about 3 times per year. And I only type that when I did laugh out loud. Often than means I was very tired. I get punchy and then lots of things strike me as funny.

This is pretty cool: en.wikipedia.org/wiki/Monsky's_theorem

I can kinda see that. But I have to wonder when or how the subject even came up.

Sounds like a mathematician bar bet or something.

I has editted this question physics.stackexchange.com/questions/235329/….

So board

This question should be accept.

I have some problems with Feynman exercise.

@FenderLesPaul come down to Knoxville and we can hang

If you like BBQ I'll be at some BBQ festival in Memphis in May

@Danu this is much cooler:

en.m.wikipedia.org/wiki/Banach-Tarski_paradox

@yuggib No it's not, because it's not physically relevant!

@yuggib Are you in an algebraic mood

@0celo7 never :-þ

@yuggib You remember the whole "quotient out the ideal from the tensor algebra" thing?

@0celo7 vaguely

@yuggib it reared it's ugly head again

but more sophisticated this time

in what sense?

Beats me why $j(v)^2$ is constructed so it equals $Q(v)\cdot 1$.

@yuggib any ideas lol

I think this book is a bit much for me at this point in time.

I have no idea what $Q$ is

@yuggib A quadratic form on $V$

@FenderLesPaul so what have you found out on emailing professors you want to work with after admission?

@GPhys I got lucky

the professors emailed me

@FenderLesPaul with me, I know

@0celo7 that too

@FenderLesPaul oh

@FenderLesPaul wow

@yuggib I have an analysis test tomorrow p.p

help

@0celo7 what are the topics

uhhhh

Sequences and series for the most part

I'm sure some sup/inf stuff will be on there, too

@yuggib I'm having a hard time with this series

I see

$$\sum \frac{\mathrm{e}^{(-1)^n\sqrt{n}}}{2^n}$$

One can see that it converges by the root test

yes

But when I apply the ratio test, I get $$\frac{a_{n+1}}{a_n}=\frac{\mathrm{e}^{(-1)^{n+1}(\sqrt{n+1}+\sqrt{n})}}{2}$$

I don't know what the limit of that is, but it seems like the numerator is unbounded...

the sequence at hand has no limit

so you can't use the ratio test

Ok, so the ratio test can't be used at all the limit does not exist?

@yuggib Ok, next question...about the nested interval property

@yuggib Let $I_n=[a_n,b_n]$ such that $I_1\subset I_2\subset\cdots$. Then $\bigcap_{n=1}^\infty I_n\ne\emptyset$.

@0celo7 it can be used only if the limit exists and is different from one

In the proof of this we showed that $x:=\sup\{a_n\}$ is in that intersection. But to do this, we needed $x\le b_n$$\forall n$.

Why is this the case?

@yuggib Uh, those $\subset$s should be $\supset$s

tedcruzforamerica.com

#rekt

why the sup? the inf should be in the intersection

@yuggib Inf of what

inf of $a_n$ wrt $n$

What? A trivial counterexample shows that's not true

ah yes, sup you're right

Is it just that $b_n$ is an upper bound for $\{a_n\}$?

That's obviously true, but how do I prove it?

Contradiction?

It should work yes

@yuggib Ok, I don't think I have any more questions

Give me a problem pls

"'[W]atching his campaign go up in flames finally explains Cruz’s logo,' Trump replied in a statement."

which problem do you want?

@yuggib something interesting but challenging

But a German middle schooler should be able to do it.

Prove that a bounded sequence such that every convergent subsequence has the same limit converges (to the aforementioned limit)

@yuggib Ok.

Let's call the sequence $(a_n)$ and the supposed limit $a$. We will prove this by contradiction.

The negation of $\lim a_n=a$ is there exists and $\epsilon>0$ s.t. for all $N\in\mathbb{N}$, there exists $n\ge N$ s.t. $|a_n-a|\ge\epsilon$.

Let $(a_{n_j})$ be the subsequence of $(a_n)$ consisting of those $a_n$s which satisfy the above inequality. It is clear that $(n_j)$ can be made strictly increasing.

Then $(a_{n_j})\not\to a$.

ok

but you're not finished

By Bolzano-Weierstrass, there is a convergent subsequence $(a_{n_{j_k}})$ of $(a_{n_j})$. But $|a_{n_{j_k}}-a|\ge\epsilon$ for all $k$, so it cannot converge to $a$.

This contradicts the hypothesis that every convergent subsequence converges to $a$.

Therefore, $a_n\to a$.

ok

Problem set 3, problem 7.

:)

:-þ

@yuggib Can you give me one I haven't done?

I don't know what you have done...

Just give me some interesting problems!

@FenderLesPaul basically I'm planning to visit a uni outside of the normal open house, and I just want to email the person I'm interested in to see if they're available then

By the way @Qmechanic I went through that second list (and did a bunch of tagging).

Shhhhh!!!

lol

@0celo7 It's too late to think to problems

@Danu : Thanks.

@yuggib Huh?

@0celo7 it's 23:30 over here

@FenderLesPaul this planning is super complicated...

@yuggib So?

@GPhys that sounds like a good plan

@FenderLesPaul I'm like, hyper-analyzing the wording of this email...

like the statement of purpose but not nearly as bad :P

@0celo7 I think you have more deleted hbar comments than anybody else.

@DanielSank I also have more hbar comments than anybody else.

perhaps

I'm fairly sure that's a fact.

95% sure.

@DanielSank Does Alphabet pay for any books you need?