/sci/ - Science & Math

>>5964375
And that's a fail... Damn'it. 0/10 Peter.
Re try :
<span class="math">E \subset R^m, \alpha \geq 0[/spoiler]
is called a "meeting-point" if
<span class="math">
\forall n \in N^*, \forall (p_1,...,p_n) \in E^n, \exists (p,q) \in E^2
[/spoiler]
such as
<span class="math">
\frac{1}{n} \sum_{i=1}^n ||p-p_i|| \geq \alpha and \frac{1}{n} \sum_{i=1}^n ||q-p_i|| \leq \alpha
[/spoiler]

------
Question : Show that if E is finite, then there exists a meeting point. Good luck. (extract from a real oral contest).

First to solve can repost, if proof is not discussed.

we're not gonna do your homework

>>5964435
lol, just say you're too bad to do it...

It's called Science & Math, but except from baby-HS calculus, I don't see a lot of math on his board (no, "who's your favorite between Euler and Gauss" is not math, it's pop/sci/).

>>5965144
*this

Quelle école, pédé ?

>>5965165
I'm not french, "grenouille de caca".
But yes, I expected than a french guy from "prepa" could answer, because murikans are the best at pretending they are the smartest guys, but sage when there is serious shit coming.

Meanwhile in th catalog,
...
baby HS questions
pop/sci/
I love Carl Sagan
I am a master of quantum mechanics and group category and I have 3 PhDs. I'm a self-learner you know.
MMUhhh my IQ is bigger than yours

I don't even want to post the links

I'm struggling.
Is it a constructive proof?

OP is very aggressive to mask his insecurities about not being able to solve his homework which this actually is, as evidenced by his response to the claim that it was.

I'm not really sure what you are asking.
Do you mean "If E is finite, show that it IS a meeting point"?

>>5965308
His wording is ambiguous, but it means "alpha is a meeting-point if etc."

>>5965308
i think it is " if E is finite show that there exists a real alpha so that (E,alpha) is a meeting point "

>>5964375
what norm are you using ? and don't give me that "muh they're all equivalent in finite dimension duh" bullshit. If it's meant to be valid for every norm of R^m you should state it clearly in the problem. ( plus <span class="math">\frac{1}{n} \sum_{i=1}^n ||p-p_i|| \geq \alpha and \frac{1}{n} \sum_{i=1}^n ||q-p_i|| \leq \alpha[/spoiler] varie with the norm you're using )

>>5965187
what was the level/country of the contest ?

>>5965366
I'm assuming he means the euclidian norm.

All I have so far is the question is equivalent to:
"When E is finite, for any two subsets P and Q of E, there are p and q in E such that the average distance from P to p is larger than the average distance from Q to q."
I can't prove it, and it's making me very disappointed.

>>5965374
well yeah but the norm 1 and \infty are used so often ...

also i don't think that's equivalent.

>>5964384
take m=2 (and therefor i'm going to use C instead of R^2), for E we take the three roots of X^3-1 and n=3, take p1=1 p2=j=exp(2iPi/3) and p3=j^2=exp(4iPi/3) (the three elements of E)

each point is at the same distance of 0 and each point is at the same distances of the others. so for every p,q in E^2 you have <span class="math">\frac{1}{n} \sum_{i=1}^n ||p-p_i|| = \frac{1}{n} \sum_{i=1}^n ||q-p_i||[/spoiler] so alpha must be equal to |1-j|.

now take E={0}. <span class="math">\frac{1}{n} \sum_{i=1}^n ||p-p_i|| = \frac{1}{n} \sum_{i=1}^n ||q-p_i||=0[/spoiler] so alpha must be equal to 0...

Seriously OP, you're not gonna throw us a bone?

Not OP, but french :
sauce :
http://www.normalesup.org/~rose/maths/oraux/or-ens.pdf
see 1.2)
I didn't find any other similar exercise on the web.

A) >>5965366
>what was the level/country of the contest ?
France. 2 years after High-School. You do the "prepa", and you can try the ENS contest.
This is ENS Ulm. God-Tier school for math. (field medals there)

>what norm are you using ?
That's Ulm. When it is not stated clearly in the problem, you deal with it, it's on purpose.

B)>>5965278
Why do you reply, then ? And saging on /sci/...
Plus, it's from 2004, see (http://www.normalesup.org/~rose/maths/oraux/)), it is an extract from oral contest, and in France it's summer holidays.
So your "homework"-thing...
You perfectly fit his answer here >>5965187

C) I tried, I'm stucked, but thanks too autistics anons like >>5964435
or >>5965278 (maybe the same), I guess OP is just on mathoverflow or some website were you can do math without being judged.

D)>>5965165
Ca sonne moins bien en français, "pédé".

>>5965511
>1 hour
Oh hey and I thought it would be some simple 15 minutes question.
I'm less ashamed to struggle.

>>5965535
>1 hour
Actually, it depends on your efficiency.
The guy just had this exercise during is exam.

But some crazy guys, during the same-time will have a sheet with 2/3/4... exercises.

Every student can get a different problem, it's not normalized.

The teacher can also discuss with you about extra-question, quite "out of the subject". Trust me, Ulm can be really odd.

My oral (long time ago...) was miore like a "discussion" with the teacher. It was like we were both trying to solve the thing, and he was excited like a kid when we found something... (I guess it was the first time he gave that problem)

>>5965535
yeah it's a 1 hour problem of math but it's for ulm, the ammount of fields medal that came from ulm is astonishing, the contest is really difficult. Plus you got a teacher to discuss with you/give you hints.

like for example you can ask him what norm is used in the problem... but i doubt op can answer that question.

>>5965374
Bump for interest.
It seems so "simple", but I get nothing interesting...

I just studied m=1 and m=2 with some success, but it doesn't help me for the general case.

>>5965924
Bump.
Only valuable thread on the catalog.
No answer, no discussion

Face it, /sci/ has nothing to do with Math.

You all suck

FYI, I had help in minutes on usenet.

So I'm gonna begin another thread about Carl Sagan, you deserve it.
And then, I'm gonna claim that my IQ is so high that I'm depressed.
And in the end, a little homework-thread about how to solve a quadratic equation with no real roots, OMG

Oh, and math are flawed, also.

>>5968299
>I had help

What a surprise, it turned out be a homework thread.

Enjoy your 3 day cool-off, surrender frog.

>>5968337
read this, dumbass :
>>5965511

>>5968299
>Face it, /sci/ has nothing to do with Math.

If that's what you come for then you are either new or a little slow. /sci/ is high school kids making stupid threads, a few trolls asking what happens when a sun made of ice collides with a sun made of lava, a bunch of undergraduates thinking they are the shit and who will do anything to boost their ego and to put others down, and the occasional curious person who genuinely wants to discuss a scientific topic or theory. If he is lucky he can find one or two guys willing to share and debate ideas in a civilized way, but most of the time he won't get any remotely interesting reply and then will leave never to come back, hence why there are so few of them.

If you come here to discuss and know more about science you are coming for the wrong reasons.

>>5968368
good sum up, thanks.
9/10

because 10 is for GOD, who EXISTS MOTHER FUCKER, IT IS PROVEN !

What does <span class="math"> E [/spoiler] being finite mean here? Does it mean that it's a collection of finitely many points?

>>5968405
And if so does what does <span class="math"> E^2 [/spoiler] mean?

>>5968430
E^2 means ExE (cartesian product)

for instance (1,2) is in N^2
(-3/4,5) is in Q^2

(p,q) in E^2 just means that p is in E and q is in E, that's all.

>>5968405
that means card(E)=n, for some n in N.

>>5968455
So finitely many points.
>>5968448
Ohh i see, i thought they meant some kind of 2D projection. Now i at least understand the problem.

>>5968299
>FYI, I had help in minutes on usenet.
Which group?

>>5965187
>>5968299

lol this fag was looking for homework help the whole time.

I glanced at OP's question earlier but I've used several meanings for some of that notation and didn't want to waste time figuring out what OP meant given that it was probably math homework. It's satisfying to know I was right to ignore.

>>5965366
>>5965374
>>5965402
>>5965402
>>5965511
>>5965794
maybe we don't care because in <span class="math">\mathbb{R}^m[/spoiler], all norms are equivalent.

>>5968626
Well it's not enough, but yeah in that case I think it works for all norms.
I think I'm not far from the solution, and all I used is the triangle inequality.

Maybe I'm missing something. If E is finite, just choose arbitrary p and q, maximize the first some over all k tuples and minimize the second sum over all k tuples. Since E is finite there are a finite number of k tuples so we can perform max and min, if the max of the first sum is larger than the min of the second choose arbitrary alpha between the max and min. Otherwise swap p and q and do the same.

>>5968749
>first sum

>>5968749
Actually I got those reversed but the idea is right, minimize the first sum and maximize the second. Unless I am misinterpreting the question which I feel I am.

>>5968609
but you're dumb ?
>>5965511
>>5965535
>>5965577
>>5965794
-->obviously not homework...oral contest, 2004.
"I've used several meanings for some of that notation" oO : you don't know anything at math, obviously.

Stop projecting please. you are perfectly described here : >>5968368

>>5968552
I guess somewhere on sci.math.* , but there are a lot of posts every day, so good luck...

>>5968337
modos really exist on /sci/ ? I'm having a hard time believing that. Half of the catalog doesn't belong to /sci/... I thought the 3-days ban was only for serious things like porn on /lit/, CP on /b/, gore on /fit/,...

>>5968716
>>5968749
Could you please try to tex your solution ?
Because for a school like ULM, it seems to easy...there must be a trap somewhere

>>5968749
>Since E is finite there are a finite number of k tuples so we can perform max and min, if the max of the first sum is larger than the min of the second choose arbitrary alpha between the max and min.
It's not enough, could be that there are two different subsets (p_1,...,p_n) and (q_1,...,q_m) such that (1/n)sum(||p-p_i||,i=1..n) < (1/m)sum(||q-q_i||,i=1..m) for any p and q.

>>5968792
>Could you please try to tex your solution ?
I don't have a solution, but I'll tell you where I'm at, TeX it is then.

First, let's restate a bit, what we have to show is that for any two <span class="math">(p_{1},...,p_{n}) \in E^{n}[/spoiler]

and <span class="math">(q_{1},...,q_{k}) \in E^{k}[/spoiler], we can find <span class="math">(p,q) \in E^2[/spoiler] such that
<span class="math">\frac{1}{n}\sum_{i=0}^n\|p-p_i\| \geq \frac{1}{k}\sum_{i=0}^k\|q-q_i\|[/spoiler]

So we consider <span class="math">x[/spoiler] and <span class="math">y[/spoiler] in <span class="math">\mathds{R}^m[/spoiler] such that
<span class="math">\frac{1}{n}\sum_{i=0}^n\|x-p_i\|=min_{z\in \mathds{R}^m}\frac{1}{n}\sum_{i=0}^n\|z-p_i\|[/spoiler] and
<span class="math">\frac{1}{k}\sum_{i=0}^k\|y-q_i\|=min_{z\in \mathds{R}^m}\frac{1}{k}\sum_{i=0}^k\|z-q_i\|[/spoiler]

Then we pick <span class="math">(p,q)[/spoiler] so that
<span class="math">\|p-x\|=min_{r\in E^m}\|r-x\|[/spoiler]
<span class="math">\|q-y\|=min_{r\in E^m}\|r-y\|[/spoiler]

tbc

First, let's restate a bit, what we have to show is that for any two <span class="math">(p_{1},...,p_{n}) \in E^{n}[/spoiler]

and <span class="math">(q_{1},...,q_{k}) \in E^{k}[/spoiler], we can find <span class="math">(p,q) \in E^2[/spoiler] such that
<span class="math">\frac{1}{n}\sum_{i=0}^n\|p-p_i\| \geq \frac{1}{k}\sum_{i=0}^k\|q-q_i\|[/spoiler]

So we consider <span class="math">x[/spoiler] and <span class="math">y[/spoiler] in <span class="math">R^m[/spoiler] such that
<span class="math">\frac{1}{n}\sum_{i=0}^n\|x-p_i\|=min_{z\in R^m}\frac{1}{n}\sum_{i=0}^n\|z-p_i\|[/spoiler] and
<span class="math">\frac{1}{k}\sum_{i=0}^k\|y-q_i\|=min_{z\in R^m}\frac{1}{k}\sum_{i=0}^k\|z-q_i\|[/spoiler]

Then we pick <span class="math">(p,q)[/spoiler] so that
<span class="math">\|p-x\|=\underset{r\in E}{min}\|r-x\|[/spoiler]
<span class="math">\|q-y\|=\underset{r\in E}{min}\|r-y\|[/spoiler]

tbc

First, let's restate a bit, what we have to show is that for any two <span class="math">(p_{1},...,p_{n}) \in E^{n}[/spoiler]

and <span class="math">(q_{1},...,q_{k}) \in E^{k}[/spoiler], we can find <span class="math">(p,q) \in E^2[/spoiler] such that
<span class="math">\frac{1}{n}\sum_{i=0}^n\|p-p_i\| \geq \frac{1}{k}\sum_{i=0}^k\|q-q_i\|[/spoiler]

So we consider <span class="math">x[/spoiler] and <span class="math">y[/spoiler] in <span class="math">R^m[/spoiler] such that
<span class="math">\frac{1}{n}\sum_{i=0}^n\|x-p_i\|=\underset{z\in R^m}{min}\frac{1}{n}\sum_{i=0}^n\|z-p_i\|[/spoiler] and
<span class="math">\frac{1}{k}\sum_{i=0}^k\|y-q_i\|=\underset{z\in R^m}{min}\frac{1}{k}\sum_{i=0}^k\|z-q_i\|[/spoiler]

Then we pick <span class="math">(p,q)[/spoiler] so that
<span class="math">\|p-x\|=\underset{r\in E}{min}\|r-x\|[/spoiler]
<span class="math">\|q-y\|=\underset{r\in E}{min}\|r-y\|[/spoiler]

tbc

First, let's restate a bit, what we have to show is that for any two <span class="math">(p_{1},...,p_{n}) \in E^{n}[/spoiler]

and <span class="math">(q_{1},...,q_{k}) \in E^{k}[/spoiler], we can find <span class="math">(p,q) \in E^2[/spoiler] such that
<span class="math">\frac{1}{n}\sum_{i=0}^n\|p-p_i\| \geq \frac{1}{k}\sum_{i=0}^k\|q-q_i\|[/spoiler]

So we consider <span class="math">x[/spoiler] and <span class="math">y[/spoiler] in <span class="math">R^m[/spoiler] such that
<span class="math">\frac{1}{n}\sum_{i=0}^n\|x-p_i\|=min_{z\in R^m}\frac{1}{n}\sum_{i=0}^n\|z-p_i\|[/spoiler] and
<span class="math">\frac{1}{k}\sum_{i=0}^k\|y-q_i\|=min_{z\in R^m}\frac{1}{k}\sum_{i=0}^k\|z-q_i\|[/spoiler]

Then we pick <span class="math">(p,q)[/spoiler] so that
<span class="math">\|p-x\|=\underset{r\in E}{min}\|r-x\|[/spoiler]
<span class="math">\|q-y\|=\underset{r\in E}{min}\|r-y\|[/spoiler]

tbc

First, let's restate a bit, what we have to show is that for any two <span class="math">(p_{1},...,p_{n}) \in E^{n}[/spoiler] and <span class="math">(q_{1},...,q_{k}) \in E^{k}[/spoiler], we can find <span class="math">(p,q) \in E^2[/spoiler] such that
<span class="math">\frac{1}{n}\sum_{i=0}^n\|p-p_i\| \geq \frac{1}{k}\sum_{i=0}^k\|q-q_i\|[/spoiler]

So we consider <span class="math">x[/spoiler] and <span class="math">y[/spoiler] in <span class="math">R^m[/spoiler] such that
1. <span class="math">\frac{1}{n}\sum_{i=0}^n\|x-p_i\|=\underset{z\in R^m}{min}\frac{1}{n}\sum_{i=0}^n\|z-p_i\|[/spoiler] and
2. <span class="math">\frac{1}{k}\sum_{i=0}^k\|y-q_i\|=\underset{z\in R^m}{min}\frac{1}{k}\sum_{i=0}^k\|z-q_i\|[/spoiler]

Then we pick <span class="math">(p,q)[/spoiler] so that
3. <span class="math">\|p-x\|=\underset{r\in E}{max}\|r-x\|[/spoiler]
4. <span class="math">\|q-y\|=\underset{r\in E}{min}\|r-y\|[/spoiler]

We have
<span class="math">\frac{1}{n}\sum_{i=0}^n\|p-p_i\| \geq \frac{1}{n}\sum_{i=0}^n | \|p-x\| - \|p_i-x\| |[/spoiler]
Because of 3., we can drop the absolute value, so
5. <span class="math">\frac{1}{n}\sum_{i=0}^n\|p-p_i\| \geq \|p-x\| - \frac{1}{n}\sum_{i=0}^n \|x-p_i\| [/spoiler]

We also have
6. <span class="math">\frac{1}{k}\sum_{i=0}^k\|q-q_i\| \leq \|q-y\| + \frac{1}{k}\sum_{i=0}^k\|y-q_i\|<span class="math">

So now we have to show that
\|p-x\| - \frac{1}{n}\sum_{i=0}^n \|x-p_i\| \geq \|q-y\| + \frac{1}{k}\sum_{i=0}^k\|y-q_i\|
i.e. it's about finding how much \|p-x\| is greater than \|q-y\

That's where I am.[/spoiler][/spoiler]

First, let's restate a bit, what we have to show is that for any two <span class="math">(p_{1},...,p_{n}) \in E^{n}[/spoiler] and <span class="math">(q_

{1},...,q_{k}) \in E^{k}[/spoiler], we can find <span class="math">(p,q) \in E^2[/spoiler] such that
<span class="math">\frac{1}{n}\sum_{i=0}^n\|p-p_i\| \geq \frac{1}{k}\sum_{i=0}^k\|q-q_i\|[/spoiler]

So we consider <span class="math">x[/spoiler] and <span class="math">y[/spoiler] in <span class="math">R^m[/spoiler] such that
1. <span class="math">\frac{1}{n}\sum_{i=0}^n\|x-p_i\|=min_{z\in R^m}\frac{1}{n}\sum_{i=0}^n\|z-p_i\|[/spoiler] and
2. <span class="math">\frac{1}{k}\sum_{i=0}^k\|y-q_i\|=min_{z\in R^m}\frac{1}{k}\sum_{i=0}^k\|z-q_i\|[/spoiler]

Then we pick <span class="math">(p,q)[/spoiler] so that
3. <span class="math">\|p-x\|=\underset{r\in E}{max}\|r-x\|[/spoiler]
4. <span class="math">\|q-y\|=\underset{r\in E}{min}\|r-y\|[/spoiler]

We have
<span class="math">\frac{1}{n}\sum_{i=0}^n\|p-p_i\| \geq \frac{1}{n}\sum_{i=0}^n | \|p-x\| - \|p_i-x\| |[/spoiler]
Because of 3., we can drop the absolute value, so
5. <span class="math">\frac{1}{n}\sum_{i=0}^n\|p-p_i\| \geq \|p-x\| - \frac{1}{n}\sum_{i=0}^n \|x-p_i\| [/spoiler]

We also have
6. <span class="math">\frac{1}{k}\sum_{i=0}^k\|q-q_i\| \leq \|q-y\| + \frac{1}{k}\sum_{i=0}^k\|y-q_i\|<span class="math">

So now we have to show that
\|p-x\| - \frac{1}{n}\sum_{i=0}^n \|x-p_i\| \geq \|q-y\| + \frac{1}{k}\sum_{i=0}^k\|y-q_i\|
i.e. it's about finding how much \|p-x\| is greater than \|q-y\

That's where I am.[/spoiler][/spoiler]

First, let's restate a bit, what we have to show is that for any two <span class="math">(p_{1},...,p_{n}) \in E^{n}[/spoiler] and <span class="math">(q_

{1},...,q_{k}) \in E^{k}[/spoiler], we can find <span class="math">(p,q) \in E^2[/spoiler] such that
<span class="math">\frac{1}{n}\sum_{i=0}^n\|p-p_i\| \geq \frac{1}{k}\sum_{i=0}^k\|q-q_i\|[/spoiler]

So we consider <span class="math">x[/spoiler] and <span class="math">y[/spoiler] in <span class="math">R^m[/spoiler] such that
1. <span class="math">\frac{1}{n}\sum_{i=0}^n\|x-p_i\|=min_{z\in R^m}\frac{1}{n}\sum_{i=0}^n\|z-p_i\|[/spoiler] and
2. <span class="math">\frac{1}{k}\sum_{i=0}^k\|y-q_i\|=min_{z\in R^m}\frac{1}{k}\sum_{i=0}^k\|z-q_i\|[/spoiler]

Then we pick <span class="math">(p,q)[/spoiler] so that
3. <span class="math">\|p-x\|=\underset{r\in E}{max}\|r-x\|[/spoiler]
4. <span class="math">\|q-y\|=\underset{r\in E}{min}\|r-y\|[/spoiler]

We have
<span class="math">\frac{1}{n}\sum_{i=0}^n\|p-p_i\| \geq \frac{1}{n}\sum_{i=0}^n | \|p-x\| - \|p_i-x\| |[/spoiler]
Because of 3., we can drop the absolute value, so
5. <span class="math">\frac{1}{n}\sum_{i=0}^n\|p-p_i\| \geq \|p-x\| - \frac{1}{n}\sum_{i=0}^n \|x-p_i\| [/spoiler]

We also have
6. <span class="math">\frac{1}{k}\sum_{i=0}^k\|q-q_i\| \leq \|q-y\| + \frac{1}{k}\sum_{i=0}^k\|y-q_i\|[/spoiler]

So now we have to show that
<span class="math">\|p-x\| - \frac{1}{n}\sum_{i=0}^n \|x-p_i\| \geq \|q-y\| + \frac{1}{k}\sum_{i=0}^k\|y-q_i\|[/spoiler]
i.e. it's about finding how much <span class="math">\|p-x\|[/spoiler] is greater than <span class="math">\|q-y\[/spoiler]

That's where I am.

>>5968948
And by <span class="math">\|q-y\[/spoiler] I mean of course <span class="math">\|q-y\|[/spoiler], sorry for the LaTeX sperging.

>>5968948
uhm...

Your first line (you restatement) is different from the http://www.normalesup.org/~rose/maths/oraux/or-ens.pdf
there are no q_i ...

>>5968962
Yes, I skipped a few lines, here, do you want me to fill the holes or did you figure it out?

>>5968965
yes, how do you get
<span class="math">\frac{1}{n}\sum_{i=0}^n\|p-p_i\| \geq \frac{1}{k}\sum_{i=0}^k\|q-q_i\|[/spoiler]
from
<span class="math">
\frac{1}{n} \sum_{i=1}^n ||p-p_i|| \geq \alpha[/spoiler]and
<span class="math">\frac{1}{n} \sum_{i=1}^n ||q-p_i|| \leq \alpha[/spoiler]
please.
(It's obviously too hard for me, but I try)

>>5968980
I mean, where do the q_i come from ?
they are not in the subject.

>>5968980
>>5968983
Sure. The q's are just another array of elements of E.

Direct (original problem to my formulation)
If <span class="math">\alpha[/spoiler] is a "meeting-point" then we take <span class="math">p[/spoiler] such that
<span class="math">\frac{1}{n}\sum_{i=0}^n\|p-p_i\| \geq \alpha[/spoiler]
and <span class="math">q[/spoiler] such that
<span class="math">\frac{1}{k}\sum_{i=0}^k\|q-q_i\| \leq \alpha[/spoiler]
(these are two different arrays, I'm applying the property of the meeting-point twice. In the first case I just take p, in the second I just take q.)
From which follows
<span class="math">\frac{1}{n}\sum_{i=0}^n\|p-p_i\| \geq \frac{1}{k}\sum_{i=0}^k\|q-q_i\|[/spoiler]

Reverse (my formulation to the original problem)
From my formulation follows
<span class="math">\underset{(n,(p_1,...,p_n)}{inf}\underset{p}{max}\frac{1}{n}\sum_{i=0}^n\|p-p_i\| \geq \underset{(k,p_1,...,p_k)}{sup}\underset{p}{min}\frac{1}{k}\sum_{i=0}^k\|p-p_i\|[/spoiler]
If we take any <span class="math">alpha[/spoiler] in that range, we see easily that it's a meeting-point.

Hence the problems are equivalent.

>>5969011
Holy fuck why does 4chan lateX refuses \underset sometimes and not other times.
Let's try again: from my formulation follows:
<span class="math">\underset{(n,(p_1,...,p_n)}{inf}\underset{p}{max}\frac{1}{n}\sum_{i=0}^n\|p-p_i\|[/spoiler] <span class="math">\geq \underset{(k,p_1,...,p_k)}{sup}\underset{p}{min}\frac{1}{k}\sum_{i=0}^k\|p-p_i\|[/spoiler]

>>5969011
Holy fuck why does 4chan lateX refuses \underset sometimes and not other times.
Let's try again: from my formulation follows:
<span class="math">\underset{(n,(p_1,...,p_n)}{inf}[/spoiler]<span class="math">\underset{p}{max}\frac{1}{n}\sum_{i=0}^n\|p-p_i\|[/spoiler] <span class="math">\geq \underset{(k,p_1,...,p_k)}{sup}[/spoiler]<span class="math">\underset{p}{min}\frac{1}{k}\sum_{i=0}^k\|p-p_i\|[/spoiler]

>>5969018
If I may suggest something.
Do your tex somewhere else, and put it in the picture.
Tex on 4chan sucks.
(I'm not >>5968980
>>5969011
just lurking)

>>5969029
Thanks for the advice, I probably should.

>>5968948
>>5969011
>>5969027
Ok I got it. My mind wasn't clear.
Existence of max and min comes from E finite right ?

>>5969040
They do, it would work without that though but would require more writing.

>>5969048
ok. Feel free to post again if you find stgh new, I'll let this tab open.

I'm interested because this problem is very "easy" to understand, but seems hard to solve, even if you use "basic tools" (you didn't use "big theorem" until now).
Thx
(I'm>>5965924)