Mathematics - 2017-03-26 (page 1 of 5)

Daminark

00:00

Mostly just procrastination

Akiva Weinberger

00:11

Hi?

@MeowMix What was your last problem? Show that $V^\top$ is always closed?

Meow Mix

I'm back

Akiva Weinberger

Ohayo

Meow Mix

@AkivaWeinberger Yeah, and that $(V^\perp)^\perp$ is the closure of $V$

By the way, does the location for mathcamp change?

Akiva Weinberger

Yeah, every year

Meow Mix

I hope it's somewhere near me next year....

Akiva Weinberger

00:16

They're approximately on a three-year cycle

mr eyeglasses

hi @columbusatemyhw

Meow Mix

So, in that cycle, what would next year be?

Balarka Sen

Hi @Akiva

Daminark

Hey @Akiva!

Akiva Weinberger

@MeowMix Well, so far it's been ME, OR, WA, ME (last year), WA again (this year), IIRC.

skipping Oregon

so I don't know what's happening next year

Meow Mix

00:18

I really hope I can attend.

Akiva Weinberger

("IIRC" is not a state)

(to be clear :P )

Meow Mix

But with the way I've been fucked as of late, it seems unlikely

The International Independent Rebublic of Cauchy

Balarka Sen

language man

Daminark

(Also known as complex analysis)

Akiva Weinberger

Con el modo que había estado jodido recientemente… @BalarkaSen

(I'm sorry, horrible pun, I should leave)

Balarka Sen

00:20

can't read spanish

Meow Mix

You can't?

Akiva Weinberger

@BalarkaSen Just a pun on "language"

Meow Mix

I was going to make a joke about how he can't "read it" meaning he can't understand the letters and what not

Akiva Weinberger

@MeowMix Did you solve these?

Meow Mix

I did.

Akiva Weinberger

00:22

I'm not actually sure how to show that $V^{\bot\bot}$ is the closure…

Daminark

Oh did you also figure out officially why $V$ is dense in $V^{\perp\perp}$?

@Akiva What is this heinous notation?

Oh nevermind

Meow Mix

LOL he used the same notation as you did.

Akiva Weinberger

I mixed up my $\top$s and $\bot$s for a second there

Meow Mix

Oh, nevermind.

Daminark

No he started with a T and I'm like wyd???

Meow Mix

00:23

@Daminark All the other points in our vector space are limit points??

Akiva Weinberger

We were talking about $V^{\vdash\vdash}$, right? :P

Daminark

Eh... $V^{\mathbb{T}}$

Meow Mix

The orthogonal complement of the orthogonal complement

Akiva Weinberger

$V^{\vdash\top\dashv\bot}$

Daminark

@Meow I dunno if you can just assert that, you ought prove this fact

Meow Mix

00:25

@Daminark I was asking if that is what you meant

Balarka Sen

@AkivaWeinberger Well, for one, it's a closed subspace which contains $V$. All you need to show is it's the smallest one, da?

Akiva Weinberger

Yerp

Daminark

Oh yeah, that

Meow Mix

Oh, wait, I realize

Akiva Weinberger

Hm, if $a\in V^{\bot\bot}$, then $\lambda a\in V^{\bot\bot}$…

Meow Mix

00:25

Isn't it obvious because $V^{\perp\perp}$ is just $V$ with its limit points?

Daminark

There is a slightly overpowered argument that I may eventually spill, but for now you should figure this out.

Akiva Weinberger

…which might not actually help me

Balarka Sen

@MeowMix How do you know more stuff than the limit points aren't there?

Daminark

@Meow Why is there nothing which is orthogonal to everything orthogonal to $V$ which isn't a limit point?

Like we know that $\overline{V} \subset V^{\perp\perp}$

We need to find the reverse direction, and this is not a trivial task

Akiva Weinberger

He said he already proved that it's the closure, right?

(which I have not yet done but I'm thinking about it)

Meow Mix

00:27

Let me think

Daminark

Well, that's what he's trying to prove, and he's gotten one direction of the subset argument

Hint: See what you can prove if $V$ is already closed

Eric

@Daminark do you know how this orthogonality relation works in Banach spaces?

Daminark

I mean, what does $\perp$ even mean in Banach spaces?

Balarka Sen

@Daminark Ah good hint.

Eric

$V^{\perp} = \{f \in E^{*}: \langle f, x \rangle = 0, \forall x \in V\}$ if $E$ is Banach

Meow Mix

00:30

Oops

Daminark

... Yeah I should've seen that coming

I'll try it

Akiva Weinberger

$+\!\!\!\!\!\times$ $~~\gets$Close enough

Eric

You can also define $F^{\perp} = \{x \in E: \langle f, x \rangle = 0, \forall f \in F\}$ where $F \subset E^{*}$. But you'll notice this isn't the same definition as the previous one I gave.

Meow Mix

Nice star

Balarka Sen

Actually it's not crazy hard to prove that by hand. I figured it out.

Daminark

00:35

One direction is clear, like if $f_n \to f$ and $f_n(x) = 0$, then $f(x) = 0$

So anything$^{\perp}$ is closed, and we can embed $V$ into its double dual by evaluation so that's fine

Eric

one direction is Hahn-Banach

if $F \subset E^{*}$ the double dual is not just the closure of $N$! that's the interesting case

Akiva Weinberger

So I want to show that if something is far away from $V$ (i.e. in the interior of its complement?) then it's not in $V^{\bot\bot}$, I guess

Balarka Sen

Right, @Akiva.

Daminark

@Eric That's neat

Balarka Sen

On that track of thinking, you could just try to pick something from $V^{\perp \perp}$ which is far away from $V$ and try to contradict.

Eric

00:41

Give a little thought as to what the double perp of $F \subset E^{*}$ is. It's like quasi-interesting @Daminark

Daminark

Aight, I shall do so

Balarka Sen

I guess the point is if it's far away from $V$, then dotting with stuff from $V$ should never give $0$.

Daminark

As for the Hilbert one, this way is not what you should be using @Meow but basically, you can prove that given some closed subspace of a Hilbert space direct sums with its double dual

So $\mathcal{H} = V \oplus V^{\perp}$

But since $V^{\perp}$ is closed always, you have $\mathcal{H} = V^{\perp\perp} \oplus V^{\perp}$

Thus giving that $V = V^{\perp\perp}$ in this case

Then you use continuity of the inner product in the case where $V$ isn't closed to show that $V^{\perp} = \overline{V}^{\perp}$

So then you're good

That's the 3 liner that you use if you have this direct sum decomposition at your disposal, but now continue your way

Balarka Sen

@Daminark Yeah, but you can also do it by hand. If there was something in $V^{\perp \perp}$ but not in $\text{cl}(V)$ then you could subtract off it's component along $\text{cl}(V)$ to get something perp to $\text{cl}(V)$... but that can't happen as then it'd be in $V^{\perp}$ but it's also in $V^{\perp \perp}$ etc etc

Wait. I think in that component logic I am implicitly using that decomposition.

Daminark

Womp womp

Thomas Rasberry

00:50

Hello everyone!

Daminark

I mean is this true in a pre-Hilbert space?

Balarka Sen

Yeah. Whoops-a-daisy. I need that projection of $V^{\perp\perp}$ to $\text{cl}(V)$.

Daminark

Hey @ThomasRasberry

Balarka Sen

@Daminark I don't think so.

You frequently use completeness of the inner product to prove the decomposition I think

Daminark

I'm suspicious because the fact that it works in a Banach space and the proof uses completeness suggests the generalization is in that direction

Meow Mix

00:54

I think I have it.

@Daminark

Daminark

Do tell

MATH ASKER

if the problem says that an airplane is bearing at an angle of 340 at 325 mph would I draw the line going 340 degrees at the first quadrant

Thomas Rasberry

I have been wondering - due to some very unfortunate events outside of my control, my advisor asked me to get a "third, independent" way to have the last proofs on my dissertation checked for vaidity and for organization; it seems I only have three weeks left to do this important task.

MATH ASKER

because bearing problems go clockwise?

Meow Mix

@Daminark So, suppose we have some vector $\mathbf{v}$ outside the closure of $V$

Let us have some $\mathbf{w} \in \text{cl}(V)$

And, a $\mathbf{x} \in V^\perp$

It is known that $\langle \mathbf{x},\mathbf{w} \rangle = 0$

Now, for $\mathbf{v}$ to be in $V^{\perp\perp}$, we require that $\langle\mathbf{v}, \mathbf{w}\rangle = 0$

Akiva Weinberger

01:00

@MATHASKER Is bearing clockwise from north?

MATH ASKER

I think so

Meow Mix

Now, that's the same as $\langle\mathbf{v}-\mathbf{x},\mathbf{w}\rangle = 0$

MATH ASKER

would it be in the 2nd quad @AkivaWeinberger?

Akiva Weinberger

@MATHASKER I think so

$\begin{array}{c|c}II&I\\\hline III&IV\end{array}$

I think

Meow Mix

Yeah @Akiva

OH

I see

MATH ASKER

01:05

why thought because if you go clockwise wouldn't 90 be on the bottom line

Meow Mix

Wait, no I don't

MATH ASKER

Akiva Weinberger

That would be clockwise from east, not north

(remember that 360 is the same as 0)

MATH ASKER

ohhhhhh because 90 degrees is north

nvm thanks

Akiva Weinberger

For bearing, north would be 0

For like any other math thing, north would be 90, though, yeah

I just learned about the bearing thing just now

MATH ASKER

01:07

so in my case I would take <325cos20, 325sin20>?

Meow Mix

Ugh

Oh hey @Ted

I swear I wasn't ugh-ing at you :P

MATH ASKER

oh ok that was messing me up cause I didn't know why it went to the 2nd quad

Balarka Sen

@MeowMix No, why?

Ted Shifrin

Oh hey Zach ... DogAteMy, I think from helping @MATHASKER the last time that to measure bearings we go clockwise.

You still here, @Balarka? cluck

Meow Mix

Oops

Ted Shifrin

01:08

I'm only here for a few minutes. Guests arriving soon.

Balarka Sen

@TedShifrin Yeah...

MATH ASKER

@TedShifrin yes I remember you helping me I just forgot that we went clockwise from north

Ted Shifrin

@MATHASKER: I asked you if that was right. I'm not 100% sure, but I'm 90% sure. :)

Meow Mix

I meant $\langle \mathbf{x}, \mathbf{v} \rangle $

Daminark

@Meow How does this go on? I'm slightly concerned but continue

MATH ASKER

01:09

@AkivaWeinberger so If I have to find my component form I would do <325cos20, 325 sin20>

Akiva Weinberger

@TedShifrin Google says clockwise from north

Daminark

(Re)$^2$hi @Ted!

Ted Shifrin

OK, DogAteMy, so I am now 100% sure.

Meow Mix

@Daminark I gave up. I'm going to think about it a little more, then get back to you.

Daminark

Aight

Ted Shifrin

01:10

rolls 6 eyes at Demonark

Akiva Weinberger

(Or clockwise from wherever you're facing, if you want "relative bearing" rather than "absolute bearing")

Ted Shifrin

See, Zach, now you have the whole room picking on you. It's good that the onus isn't all on me! :P

MATH ASKER

absolute bearing is always from north?

Akiva Weinberger

Yeah

Daminark

Wait first you're a vampire and now you have 6i's?

Ted Shifrin

01:10

I never said I was a vampire. If you've paid attention in here, sometimes I have up to 9 eyes.

Balarka Sen

@Daminark I think his maximum was 11 or something

Ted Shifrin

Did it get to 11?

Akiva Weinberger

And absolute solfege is always from C, as opposed to relative solfege which is always from the tonic

Ted Shifrin

tonic.

Nothing to do with gin & ...

Balarka Sen

Apparently 14 : chat.stackexchange.com/transcript/message/17307602#17307602

Ted Shifrin

01:11

LOL.

I don't want to look.

Well, at least I keep you entertained.

Or maybe eye-tertained.

Akiva Weinberger

@TedShifrin And yet you shall!

Aug 23 '14 at 15:53, by Ted Shifrin

rolls all fourteen eyes

Meow Mix

How can you tell when a @Ted is cold?

When he's SHIFRIN (shivering)

Ted Shifrin

LOL, pretty good, Zach :)

MickLH

@TedShifrin Sorry I'm late, Please let me into the gin & tonic exam, I have studied so hard for this.

Balarka Sen

@TedShifrin Nah, I have done the statistics before.

Akiva Weinberger

01:12

J**** ******* ******, Zach

Ted Shifrin

Behave, DogAteMy.

Meow Mix

JFC is short and concise @Akiva

Except you might confuse it with ZFC

Akiva Weinberger

Or KFC

Ted Shifrin

Anytime you show up, sure, @Mick.

MickLH

<3

Akiva Weinberger

01:13

(Or JFK? Which is kinda the same thing)

Ted Shifrin

Well, I'm glad I came for such momentous mathematics. Back to the kitchen with me.

Bye again, all.

MickLH

peace

Daminark

"Kentucky-Fraenkel Choice"

MickLH

lol

Meow Mix

"Jesus-Fraenkel Kennedy"?

Thomas Rasberry

01:14

I asked SO Meta if doing so was in the guidelines of the Exchange community; I since deleted it after realizing that post contained personal data on a forum, but the main spirit of the answers seemed to be "no," and I think it was because the commenters misunderstood that I did not mean checking my whole thesis- just a few proofs I am unsure of is all I'm checking for.

However, my central proof itself is quite long, and posting it may be a big no-no here due to its potential length. I am a bit stuck at the point of what the heck I am panning to do. I am kind of "in between" Mathematics SE …

Akiva Weinberger

♫ Re(2) Do(1/2)Fa(1)Sol(1/2) Re(4) ♫

Balarka Sen

What's that

Meow Mix

What do you call a commutative scam? An Abelian Dupe.

Akiva Weinberger

@BalarkaSen Music?

Balarka Sen

huh

MickLH

01:17

@ThomasRasberry I think you'll have a smoother time fitting the Stack Exchange paradigm if you instead of asking for direct review, ask for assistance with rigorously supporting the claims you may be unsure of.

Balarka Sen

never seen those notations

Akiva Weinberger

It's solfege

Thomas Rasberry

I guess I'm just afraid of anything I post either getting ignored or else downvoted out of exisence. Is there a website, person, etc. who could provide helpful emphasis on the organization, validity, and thorough-ness of a handful of proofs I've made but feel uneasty about

Akiva Weinberger

I dunno how to say how long each note should be so I just wrote it in beats in parentheses

MickLH

The only SE I've seen do well with outright review requests is Code Review.SE

Balarka Sen

01:18

Ah ok

You should try to write down the solfege for "Culture Shock"

Meow Mix

Here's one. ♫ Re($\zeta(x)$) (1/2) ♫

Daminark

Also @Meow cheers at the abelian dupe

I'm not abel to come up with puns of that caliber

Lol but really it sounds closer to "group" than the classic "Abelian grape"

Balarka Sen

The best I heard along those lines is "What do you get when you cross a citrus fruit with a bull?"

"A trivial lime bundle on a taurus"

MickLH

lol

Akiva Weinberger

O_o

MickLH

01:27

Look I'm just a really good person. Ok?

BAYMAX

Hi all,actually what does this mean "...."

sorry

...

like 1 + 2 + 3 + 4 +...

MickLH

it means "do the obvious thing"

BAYMAX

there is some special word for it

MickLH

The symbol is called an ellipsis

I don't know what the notion of using it like that is called

BAYMAX

ok

Daminark

01:29

"ellipsization"

Akiva Weinberger

I think people used to write "… etc." instead of just "…"

(or whatever the old-times equivalent of 'etc.' was)

(probably '&c.')

BAYMAX

$\frac{1}{1-x} = 1 + x + x^{2} +...$

Meow Mix

That's only for $x \in (-1,1)$

BAYMAX

yes

Akiva Weinberger

In LaTeX, you either want \dots or \dotsb (or any of a bunch of these idk)

(usually \dotsb)

Daminark

01:31

\ldots is what I go with

BAYMAX

but I was thinking what are those called,in grammar they are ellipsis but here ?

Meow Mix

they don't have any name

Akiva Weinberger

Still just ellipses

Meow Mix

You just say "and so on"

BAYMAX

oh

Akiva Weinberger

01:31

(Singular ellipsis, plural ellipses)

BAYMAX

ok

Meow Mix

Don't mix up with actual ellipses

BAYMAX

:)

yu

p

MickLH

I looked it up and I guess the ellipsis can refer to the actual inferred content, or the dots that represent it

BAYMAX

ok

Akiva Weinberger

01:33

$\displaystyle\sum_{n=1}^\infty\frac1{n^3}={\cdots}$

Meow Mix

@Daminark I really am not sure how to show it

Tim The Enchanter

There are some really terrible math puns (that I steal from prof. B Sury)
"The prime number theorem has an one-line proof : the Riemann zeta function does not vanish on the `one' line (Re(s)=1) "

Akiva Weinberger

Guys I solved it what do I win

Meow Mix

Can I have a hint?

MickLH

time

Mike Miller

01:34

@MeowMix What are you trying to do?

Akiva Weinberger

Show that $V^{\bot\bot}$ is equal to the closure of $V$

Meow Mix

I showed that $\overline{V} \subset V^{\perp\perp}$

Akiva Weinberger

He needs the reverse inclusion

Mike Miller

What have you thought about?

MickLH

@AkivaWeinberger Did you get something that approximately equals 1.2?

(I'm trying to not spoil it for anyone lol)

Akiva Weinberger

01:35

@MickLH $\cdots{}\approx1.2$

Yeah I think so

Tim The Enchanter

Seems legit

Meow Mix

I stained my math notebook with a blueberry

LOL @MickLH

Balarka Sen

i can't even read what you're writing man

Mike Miller

@MeowMix Talk to me about the problem. Help me help you!

Akiva Weinberger

Well, to start with, he had a blueberry

and the problem is his textbook got stained with it

Mike Miller

01:41

@AkivaWeinberger makes me think of -

MickLH

Is it too radical if I were to use half a page at the very beginning of my paper just to define all the consistently used symbols? (Nobody objected!)

Meow Mix

@Mike Well, I know that all the limit points and $V$ itself are in $V^{\perp\perp}$

Mike Miller

OK, great, so $\bar V \subset V^{\perp \perp}$.

Meow Mix

Ugh

Could I show that $\bar{V} \cup V^\perp$ is the whole space?

Akiva Weinberger

It's not

not even for finite dimensions

Meow Mix

01:51

No, wait.

Akiva Weinberger

Consider the $x$-axis in $\Bbb R^2$.

Meow Mix

I meant something else

Akiva Weinberger

You want $+$, not $\cup$

Meow Mix

Yeah

Akiva Weinberger

I'mma go now

Meow Mix

01:54

It's about showing how given a point outside $V^{\perp\perp}$, you can't get arbitrarily close to it with a point inside $V$

Mike Miller

@MeowMix Yeah. Here, let's phrase this differently. It should be pretty easy to prove that $V^{\perp \perp} = \overline V^{\perp \perp}$. So it suffices to show that for $V$ closed, $V = V^{\perp \perp}$. So you don't need to talk about being arbitrarily close: just plain equal.

Meow Mix

Because if you could it would be a limit point, and therefore it would be inside $V^{\perp\perp}$

Mike Miller

Sure, but now you need to prove that :)

Meow Mix

Prove what?

Mike Miller

That when $V$ is closed, $V = V^{\perp \perp}$.

Meow Mix

02:01

Well $V^{\perp\perp}$ is just $V$ with all its limit points

Wait no, that's what I need to prove. Shit.

Mike Miller

You need to start somewhere. Lay out what you know on the table and look at that.

I've got an inner-product space $U$ that everything lives inside. $V$ is a subspace of $U$ - I'm going to assume it's closed. I have another subspace, $V^{\perp \perp}$, that I know contains $V$. I want to show it doesn't countain anything else.

I know we just picked on you for writing down a proof by contradiction, but let's do that here - let's suppose there is something else, and work from there. After we have a finished argument we can go back and rewrite this without the contradiction.

Meow Mix

If there is anything else, obviously the infimum of the distance between it and any point of $V$ is not zero

But I suppose that's not helpful either

Mike Miller

Why is that true, and why is it unhelpful?

Meow Mix

Because if it were it'd be a limit point of $V$

Oops I need to fix that

It's too late to edit. I meant $V^{\perp\perp}$

Actually, I guess that's fine without.

Mike Miller

You should try to be careful with vocabulary and being very clear. Try not to mold the way you talk about math like people do in here, so informally - at least not quite yet.

"If $v \in V^{\perp \perp}$, but $v \not\in V$, then I know $d(v, V) \neq 0$. For if there was a sequence of vectors $v_n \in V$ with $d(v, v_n) \to 0$, then because $V$ is closed, $v \in V$." I know I'm being really pedantic, but clearer writing makes it harder to lie to yourself about whether or not something's true.

Meow Mix

02:16

Ugh I'm feeling incompetent right now

Mike Miller

You can't get anywhere by not saying anything (nor can I help). Sound things out.

Meow Mix

Well, so we have that for $w \not\in V^{\perp\perp}$, $\inf \;\; \langle v-w, v-w \rangle \neq 0$ for $v \in V^{\perp\perp}$,

Because that's just the distance function squared

Is that correct?

Mike Miller

Sounds good so far.

Meow Mix

Now, I can make this $\inf \langle v, v-w\rangle - \langle w, v-w\rangle$

Mike Miller

You want to write that as "for $w \in V$, $\inf_{w} \langle v - w, v - w \rangle \neq 0$.

(Up above. You wrote $w \not \in V^{\perp \perp}$.)

Eric Stucky

02:24

Noah Snyder is a person I could know from MSE, right?

Mike Miller

Yeah

Eric Stucky

is he a chat person?

Mike Miller

No, I don't think so.

Eric Stucky

okie

tyty

Meow Mix

That's what I wanted

It wouldn't be true if it was $w \in V$... right?

Mike Miller

02:26

Huh? The definition of $d(v,V)$ is $\text{inf}_{w \in V} d(v,w) = \text{inf}_{w \in V} \langle v-w, v-w \rangle$.

Meow Mix

Sorry, let me clarify what I meant.

I was assuming $w$ is a point not in $V^{\perp\perp}$ while $v$ is, and said that the infimum distance between the two is non-zero

Mike Miller

I don't understand why you're working with anything outside of $V^{\perp \perp}$.

Meow Mix

I thought we were doing contradiction?

Mike Miller

Yes, but I'm not sure you know what you're trying to contradict.

Meow Mix

Wait, no

I meant $w \not\in V$ and $v \in V$, and I was showing that $v$ must have a different dot product than $w$

Mike Miller

02:30

I don't understand what the last sentence means.

Meow Mix

Like, there exists some element of $V^\perp$ where the dot product of it with $v$ is different than with $w$

Nevermind. :P

You were saying that if we have one inside of $V^{\perp\perp}$ but outside of $V$, then what?

Like, how would you contradict with that?

Mike Miller

You were on your way to something. Maybe a proof, maybe not. Why not chase it?

Meow Mix

Because you're better at stuff.

lol

Mike Miller

huh?

Meow Mix

I mean

I guess

I'll try

Now, does symmetry hold for dot products?

Mike Miller

02:36

Yes

Daminark

(Real)

Mike Miller

Buzz off unless you want to take over, pal.

Meow Mix

Sweet. Now we have $\inf_{v\in V} \langle v,v \rangle - 2\langle v,w \rangle + \langle w,w \rangle$ right?

Mike Miller

Sure.

Daminark

Lolol, aight, I concede. Just wanted to roflcopter for a second

Meow Mix

02:39

@Daminark Microsoft Sam says soi soi soi soi

Umm, wait. This doesn't seem so helpful

Actually, this might look like Cauchy-Schwarz

Maybe not.

Mike Miller

Nah, not particularly. That just tells you the infimum is nonnegative.

Meow Mix

Maybe what I had earlier

I get that $\inf_{v\in V} \langle v, v-w \rangle- \langle w, v-w \rangle \neq 0$

Mike Miller

That form feels a little more promising to me, since the space $w$ lives in is defined by a particular property.

Meow Mix

I don't know if I'm allowed to "distribute" the infimum

Mike Miller

Yeah sure why not

Meow Mix

02:45

So obviously $v-w \not\in V$

Otherwise $w \in V$ which is plain wrong

Mike Miller

Sure, but I'm looking at this and thinking it would be convenient if $w$ had a particular form.

Meow Mix

Umm

Mike Miller

Hm, ignore what I said. Here's a hint and I'll be off: show that there's a point in $V$ closest to $w$ and tell me everything you can about it.

Meow Mix

The closest one... I think they're orthogonal

Mike Miller

Is there a closest one?

Meow Mix

02:53

If there wasn't it would be open

Mike Miller

Huh?

MickLH

@TedShifrin On the gin & tonic exam... I think the last ansere 42 fhewewww cant remamber

deostroll

10

A: IMO 1988, problem 6

This is a famous problem, here is one of the solutions that I like the most that I read it in a book previously, but later in a topic on here I realized the importance of the problem (The credit goes to T. Andreescu & R. Gelca If I remember it correctly, but I'm not sure, since 11 individuals in ...

How does the minimality of the sum of roots guarantee that k is a perfect square? ^ imo 1986, Q6

Meow Mix

@Mike Well first thing I notice is that I can kind of reduce this to a minimum problem

Daminark

@Meow Remember that not closed doesn't equal open

Also are we assuming that $V$ is closed?

Meow Mix

03:01

Well let me just consider the point $x \in V$ such that $\langle w, x \rangle = 0$

@Daminark Yes

Daminark

Wait sorry for the interruption again, continue

Mike Miller

@MeowMix You've lost me again.

All we know is that $w \in V^{\perp \perp}$ is not in $V$, and that $d(w, V) > 0$.

Yes, @Daminark.

Don't prove this for him, but don't let me stop you from doing it yourself

Daminark

I've done this previously, he's basically hashing out a detail. But I think he did in fact have a good insight

Just needs a slight revision

Mike Miller

@Daminark I know :)

Meow Mix

Sorry my phone died

I don't have

time.

Thanks a lot for helping and sorry I was a pain in the ass, I'm new to this.

Good night

Mike Miller

03:13

@MeowMix You're not a pain.

Send me an email sometime.

Meow Mix

Alrighty.

Bye

Transcript for

$Mathematics$ Mathematics