Mathematics - 2014-12-11 (page 3 of 5)

Hello!!! Can someone help at the following exercise??

Let $p$ an odd prime and $F=F_{p^n}$ the finite field with $p^n$ elements.

1. Show that the set $F^2=\{a^2, a \in F\}$ has $\frac{p^n+1}{2}$ elements. Conclude that , if $t \in F$ the set $t-F^2=\{t-a^2, a \in F\}$ has $\frac{p^n+1}{2}$ elements.
2. For $t \in F$ show taht the set $F^2 \cap (t-F^2)$ is non-empty and conclude that each element $c$ of $F$ can be written in the form $c=a^2+b^2, a, b \in F$.
3. Show that the equation $x^2+y^2+z^2=0$ has a non-trivial solution in $F$.

Balarka Sen

I've been trying to find a way to compute Cech fundamental group of the solenoid I mentioned. I think for reasonably nice open covers, the 1-skeleton of the nerve of the covering should look like the inverse limit of circles with nodes being p^n-th roots of unities and edges being the arcs joining the consecutive nodes in the category of graphs.

@anon

Khallil

3:28 PM

Can somebody explain the difference between letting $\epsilon >0$ and choosing an $\epsilon$ to be equal to some number in proofs?

anon

@Khallil can you give an example of the latter?

Balarka Sen

@anon Have you studied Dessins De'nfants?

I've been planning to learn some of that for a while.

anon

nope

Huy

@Khallil: If you let $\varepsilon > 0$, you choose $\varepsilon$ to be a fixed number $>0$.

Khallil

When proving that every convergent sequence is bounded, I used the definition of the convergence of $(a_n)$ to state that for each $\epsilon>0$, there's a natural $N$ such that $|a_{n} - a|<\epsilon$ for all $n>N$. This is equivalent to saying that $a_n \in \left( a - \epsilon, a+\epsilon \right)$ for all $n>N$ and I concluded that if I define $\ell = \min \{ a_1, a_2, a_3, ..., a_N, a-\epsilon \}$ and $N = \max\{ a_1, a_2, a_3, ..., a_N, a +\epsilon \}$, then $a_n \in (\ell, N)$ hence bounded.

@anon

N3buchadnezzar

3:34 PM

@TheArtist I answered the thing math.stackexchange.com/questions/765198/…

Mike Miller

hi anon

Huy

@Khallil: And where is the problem?

Khallil

I thought that the whole point was to let $\epsilon >0$ so that it works for all $\epsilon >0$, not just one of our choosing, @Huy.

I was advised to choose $\epsilon = 1$ and docked marks for not doing so which is why I was confused, @Huy.

Huy

@Khallil: The point is that you choose $\varepsilon > 0$ but know nothing about the number, except, well, that it is greater than 0, and your proof works no matter which value you chose, hence for all $\varepsilon > 0$.

anon

you let $\epsilon>0$ when you're going to prove continuity. you can choose a value for $\epsilon$ when you wish to use continuity proving something else.

hi Mike

Mike Miller

3:37 PM

how's things

anon

oh, you know, procrastinating

Mike Miller

good stuff

killing time before I teach at 8 right now

ready to grade finals?

Khallil

I haven't learnt anything about continuity yet, @anon. ^_^"
That's in the second half of Analysis I/II I'll be doing next term.

Mike Miller

what the hell do you do in analysis I

Khallil

We look at sequences, completeness and infinite series, @Mike.

Continuity is the first thing we'll do next term (/semester).

Mike Miller

3:39 PM

ah, makes sense

fair enough

Huy

@Khallil: Do you already know what you will cover in Analysis 2?

anon

@Khallil sorry, I mean existence/value of a limit not continuity

I am not in analysis mode

Sawarnik

@BalarkaSen Really, it becomes annoying sometimes :D

Balarka Sen

Hello admiral Yu.

Khallil

Is it a fundamental mistake to let $\epsilon$ be any number $>0$ as opposed to choosing a single value of $\epsilon$ in my argument, @anon?

Alexander Gruber

3:41 PM

Ahoy.

Khallil

Yep, @Huy. It's all here.

skullpatrol

Ahoy.

Sawarnik

@BalarkaSen yeah, i know.

Ahoy.

Huy

@Khallil: Saying "Let $\varepsilon > 0$" is the same as "Choose $\varepsilon > 0$".

anon

@Khallil convergence means something has to hold for all $\epsilon$, so proving it for a single epsilon fails to prove convergence. but if you already have convergence of a sequence, you can set $\epsilon$ to be whatever you want and derive consequences from the inequality.

Sawarnik

3:42 PM

@Khallil I studied sequences one or two month ago but totally forgot it now. Do you have any idea how do I retain stuff in my mind?

anon

@Huy that's not what kha is talking about. see above where he mentioned setting $\epsilon=1$.

Khallil

Ahhhh, gotcha! Thanks, @anon.
(I know what you mean @Huy, but that's not what I was referring to!)

Honestly, I have no idea. I guess you just gotta look over stuff enough times so that you can command it with confidence.
I can hardly remember the stuff I ate for breakfast yesterday morning, @Sawarnik! I don't even think I had breakfast.

Mike Miller

I have no idea what i ate for breakfast yesterday morning

Sawarnik

Nor do I.

anon

that's called cognitive efficiency

Balarka Sen

3:46 PM

I didn't eat anything.

anon

prioritize what requires mental energy and what doesn't

Mike Miller

gotta save all that mental energy for smash

Sawarnik

@anon sequences took me time and energy... but had no real use so i forgot it all i think when my class exams came in between.

Mike Miller

does anyone recall how to define angles in euclidean geometry

Khallil

I think it's got to do with distances or something. Like an arc of the circumference of a circle with unit length maybe.

Balarka Sen

3:48 PM

LOL @MikeMiller. They're not well defined.

skullpatrol

strange... efficiency has two f's while deficiency has one?

Balarka Sen

Define $\text{acrtan}$s first, I guess.

Sawarnik

coordinates?

Mike Miller

that sounds believable, @Khallil.

Balarka Sen

yeah well @Khallil is right.

compare with the arclength of the unit sphere

but that is essentially equivalent to define arctan. to compute the arclength, you need to compute the corresponding integral, which in turn evaluates to arctan or whatnot. :P

Khallil

3:53 PM

Woah. Integrals, @Balarka?

Balarka Sen

well you need integrals to define arclength :P

Huy

@BalarkaSen: No, just a rope and a ruler.

Khallil

Oh, yea of course. By extending Pythagoras' theorem, right @Balarka?

Balarka Sen

Sort of @Khallil.

Mike Miller

not clear to me how one defines this (classically) in non-Euclidean geometry; it must be possible in order to talk about the defect of a triangle

Alexander Gruber

3:55 PM

Do not be deceived by the evils of the analysis

Balarka Sen

~~analysis~~

waits for smacks

Alexander Gruber

Algebra is the true path to salvation my friends

Balarka Sen

Exactly @Alexander.

Mike Miller

one could do this in some infinitesimal manner but I have my doubts this is the original way of speaking

Balarka Sen

All hail for Admiral Yu!

2

Mike Miller

3:56 PM

after all, I'm pretty sure it was (nearly) classical knowledge that the parallel postulate is equivalent to triangles having angle sum 180

So there should be some reasonable axiomatic interpretation of that sentence

Daniel Fischer

@MikeMiller Typically, your non-Euclidean model is embedded in something Euclidean, then you define the angles as Euclidean angles.

evinda

Hey!!!! :)

Let the algebraic curve $f(x_0, x_1, x_2) \in K[x_0, x_1, x_2]$. The inflection points are the non-singular points of the curve that are the intersection points with the hessian.

If we have the curve $x^3+y^3+z^3=0$ the hessian is equal to $216 \cdot x \cdot y \cdot z$. How can I find the non-singular points of the curve that are the intersection points with the hessian?

Balarka Sen

@DanielFischer It's not clear how you're embedding.

Daniel Fischer

Otherwise, @MikeMiller, you have a Riemannian manifold, and you know how to define angles there, don't you?

@BalarkaSen Conformally, of course ;)

Balarka Sen

Ah!

Mike Miller

4:00 PM

how do you propose we do this for hyperbolic geometry, @Daniel? And yes, I do. That's why I said classically.

Daniel Fischer

@MikeMiller You take the disk-model, or half-plane-model [or the higher-dimensional analogues], and you have your hyperbolic plane [space] as an open subset of the Euclidean plane [space]. Or take the hyperboloid model and you have a hypersurface in a Euclidean space.

skullpatrol

how did euclid define an angle

So, usually, elementary geometry texts count on students to have an intuitive idea of angle, introduce angle through the common definition and then often stealthily expand the notion of angle to angles that the definition does not cover. Doing otherwise requires much more serious mathematics than could be safely handled by both students and teachers.

N3buchadnezzar

Uses two hours to write an answer Recieves one upvote. Sigh

@BalarkaSen It's a trap!

Khallil

Is that in the "Some users are great at integrating things" thread, @N3?

Daniel Fischer

4:15 PM

@N3buchadnezzar Give it time. If it took long to write, it also takes long to gather upvotes.

N3buchadnezzar

@DanielFischer Yeah that seems to be the trend here. Quick answers to simple answers are usually awarded more. ^^

teadawg1337

The most upvotes I've gotten on answers were answers to super popular questions.

Sawarnik

@DanielFischer You look quite different from your previous avatar.

teadawg1337

The vast majority of people don't seem to appreciate the more complex questions that require an in-depth answer, which is saddening to me

Daniel Fischer

@Sawarnik A couple of years older.

The Artist

4:22 PM

@N3buchadnezzar I read it :D cool :D (Upvoted!)

Sawarnik

Hmm.

Khallil

I mean, you can't just make a sweeping statement like that, @teadawg1337.

skullpatrol

@DanielFischer have you ever watched any of Professor N J Wildberger's videos on youtube?

N3buchadnezzar

@TheArtist I can easilly double that list, but no time for now.

Balarka Sen

I have seen one @skullpatrol.

He speaks nonsense most of the time.

Daniel Fischer

4:27 PM

@skullpatrol No. I have watched some of Kate Bush's videos on youtube, though. I think that's a much more sensible use of my time.

teadawg1337

@Khallil I was referring to the MSE community as a whole, not the people in chat here. The simpler questions get more attention than the complicated ones

The Artist

@N3buchadnezzar It's okay :) you wrote a lot :) hmmm y don't u love "interesting integral" , I love it :)

Khallil

Ah, sorry! I misread that one, @teadawg1337! ^_^

Daniel Fischer

A deep question requires many answers

7

Balarka Sen

LOL

Another answer!

teadawg1337

4:31 PM

Does anyone know if there's any other pairs of factorials which, when added together, produce a perfect square? The one I found is $4!+5!=24+120=144=12^2$

Daniel Fischer

I can't believe that 100k+ member's first reaction is not to look for a duplicate :-( — Jyrki Lahtonen 5 mins ago

Balarka Sen

2!+2! = 2^2

teadawg1337

Well, and $4!+1!$, but that one is trivial...

Balarka Sen

Not at all trivial.

Daniel Fischer

@teadawg1337 $7! + 0!$

Balarka Sen

4:33 PM

I think there's a conjecture on the finiteness of the solutions of $n! + 1 = m^2$

It's still open, the last time I saw.

teadawg1337

Challenge accepted

Are there any other pairs that don't involve adding 1, though? Besides $2!+2!$?

Balarka Sen

Ah it's called Brocard's problem

Sawarnik

@DanielFischer Hmm.. the normal long divison answer is missing, shall I add one? :D

Balarka Sen

@teadawg1337 PARI/GP says none upto $(m, n) \leq (10^3, 10^3)$ where the equation is $m! + n! = p^2$

Daniel Fischer

@Sawarnik You don't see the two deleted answers yet.

Sawarnik

4:39 PM

Aww.

skullpatrol

What's wrong with good old fashion long division?

Sawarnik

@skullpatrol Yup, exactly. There is need of more answers.

Tom Cruise

@DanielFischer are you topologist?

Balarka Sen

not again @TomCruise :P

teadawg1337

Long division with polynomials can produce alternating power series, which is pretty neat

Tom Cruise

4:42 PM

maybe someone else to share the problem with :)

it is open problem in topology number 521

skullpatrol

He is the top chat ologist.

Hahahaha!

Hi there

Hey, @Kelly.

I see his posts a lot but not here

Balarka Sen

4:43 PM

I topologize for the inconvenience.

Khallil

^_^

Daniel Fischer

@TomCruise I can distinguish a tea cup from a doughnut, so not really.

Tom Cruise

oh well I'm only concerned with totally impractical spaces

skullpatrol

But do you dip your doughnut in your tea :P

Balarka Sen

@DanielFischer Is every nondegenerate closed subset of the solenoid disconnected? I want to verify my suspicion :P

Tom Cruise

4:46 PM

I thought we decided there were little intervals?

Balarka Sen

@TomCruise There are?

where?

Tom Cruise

cut a little cylinder from each $T_i$ so that they are nested, then take the intersection

Daniel Fischer

@BalarkaSen What was "the solenoid" again?

Balarka Sen

If you cut $T_1$, the slice of $T_2$ embedded inside is disconnected.

You don't really get an interval, but the cantor set cross [0, 1]

Tom Cruise

right but then you choose one piece

Balarka Sen

4:49 PM

@TomCruise how is that a subset?

@DanielFischer Take a bunch of tori, one wrapped n times in another. Take the intersection.

That's the n-adic solenoid.

Tom Cruise

you slice a cylinder out of $T_1$. Then you have cut $T_2$ into two cylinders. Choose one of these. Then there are two disjoint cylinders inside of this one. Choose one of these. Continue and take the intersection

Balarka Sen

@TomCruise hmm. the cantor set cross [0,1] has connected subsets. hmm.

right. darn it.

Tom Cruise

yes it is sort of like cantor cross [0,1

Balarka Sen

however there is some hope

Tom Cruise

yes you could possibly construct one of these spaces from the solenoid like we did with the Knaster

Balarka Sen

4:53 PM

what if you wrap $T_2$ $p$ times around $T_1$ but instead of just having an unknotted braid structure, construct an uncountably knotted braid.

eugh.

Kelly Blunie

If there is set of vectors S and we want to prove span(S)=V , but the determinat of the coeficient matrix of the system formed by c_1s_1+..+c_ns_n =u where s_i are the vectors in set S and u is any vector in V , is zero, then there is at least one choice of u for which this system will not have a solution and hence can not be written as a linear combination of these vectors?

The determinant is zero what i meant

Balarka Sen

like consider the dyadic solenoid and wrap T_2 twice inside T_1 by first making a loop, pulling up the endpoint a little above to not to make a loop and then wrap around the first loop like crazy and then meet with the starting point.

that is, some kind of uncountable knot structure.

not even sure if there is something like that :P

Tom Cruise

maybe you could weave the tube like we weaved the line in the Knaster

but I'm not sure how this would look

Balarka Sen

right we are essentially closing in at your construction using the Knaster continuum.

Tom Cruise

oh I think I see

once the tube has completed one loop you let it enter inside the beginning (you must shrink the tube to do this of course)

picture a rotated cantor set inside the tube

Balarka Sen

5:02 PM

ah!

Tom Cruise

you weave the tube back and forth through this rotated cantor set

like with the Knaster

Balarka Sen

that is a cool idea @TomCruise

Tom Cruise

this is not exactly like a solenoid though... more like a 3D Knaster

Balarka Sen

right, indeed.

Tom Cruise

I'd have to work out the details to make sure this is possible

Balarka Sen

5:08 PM

I think you can do it in another way @TomCruise. Call $X_1$ the torus. $X_2$ be the tube which after completing the loop enters inside the tube and then loops again and the end pastes with the beginning. Of course, you can't do it without intersections in $\Bbb R^3$. Now $X_n$ be the same thing done after $n$ loops inside. Consider $\bigcap X_n$

Tom Cruise

yeah the intersections are possibly problems

Balarka Sen

But $\bigcup X_n$ doesn't really have intersections.

It's the tube which goes inside the tube and goes inside that one and this continued ad infinitum.

interesting : if you cut it, each cylindrical piece is cantor set cross $S^1$.

Tom Cruise

yeah

Balarka Sen

sheesh. we are so fiddling with pathological spaces.

Tom Cruise

yes pathological spaces are fun though :)

Balarka Sen

5:16 PM

fair enough

Kelly Blunie

Is there anyone who can help me with one question algebra?

On linear algebra

r9m

@Khallil gawd ,, the fillers bored me to hell :(

Khallil

Really? I liked how the animators developed the relationship between Hinata and Hanabi, @r9m.

Also, Hanabi was such a qt$\pi$. ^_^

r9m

@Khallil that's really not interesting to me (I'm more attracted to the stronger characters ;) .. )

Balarka Sen

i working with solenoids from a different point of view at the moment @TomCruise. a solenoid can be thought of algebraically as inverse limit of a bunch of circles, which sits inside the infinite-torus. i am trying to find a way to define the right fundemental group of that space, and hoping to use this to understand the absolute galois group over Q geometrically, a problem originated in Grothendieck's Equisse d'un Programme.

just pure algebraic interest though.

Khallil

5:20 PM

It's all about the Senpou: Bijuu Rasenshurikens, @r9m!

Balarka Sen

rasenshurikens are practically nothing compared to spirit bombs, apparently @Khallil

ducks

Khallil

The time taken to charge a spirit bomb is extremely long compared to that of a Rasenshuriken.

Balarka Sen

that's untrue. think of the spirit bomb made by cell.

Khallil

He never made it in the anime, nor the manga I don't think.

That's just like all of those insane moves people can do in the Naruto games.

Balarka Sen

or am i confusing with cooler's giant bomb?

forget about what i said :P

ok, i gotta go

Khallil

5:27 PM

See ya later @BalarkaSen!

Mike Miller

@DanielFischer I dunno, I'd call the hyperboloid model in Lorentzian space...

Behaviour

5:48 PM

Winterbash Countdown

3

Hippalectryon

@Behaviour What's that ?

Behaviour

@Hippalectryon Countdown to HAAAAAATS

Hippalectryon

Oh I see :DD starred

@Behaviour It's a bit buggy though

Behaviour

Blame @balpha, but he's not in this room.

Hippalectryon

Actually, it has an awful flaw

Behaviour

5:52 PM

If we delete a question every second, the site will still have some left by the time Winterbash begins.

Hippalectryon

Enough to make my computer lag for some seconds

Behaviour

It works fine as long as it's in the active tab. But don't leave it running in a background tab.

Hippalectryon

Exactly

I made that error

And also I just had a 9 on a 0

The 0 didn't go away

Now a 7 on a 6 :/

The old number doesn't go away if it changes in a background tab

Anyway I gtg

I'll be back

Mike Miller

@Behaviour I hope you'll at least put on a hat before hiding the rest from your eyes.

@Hippalectryon I hope my answer helped (and I also hope you filled in the details :) )

Behaviour

@Hippalectryon Already reported: Please don't snow when I'm not there

Kelly Blunie

6:06 PM

Hi there, is there anyone who can help me with a linar algebra question?

MarcGato

@Behaviour I am trying to prove that if $A=f^{-1}(B)$ is open for any open set $B$ of $(F,\gamma)$ (where $f:E\rightarrow F$) then $f$ is continuous

Just writing I have for all $\epsilon >0$ $B(f(a),\varepsilon)$ is an open set

robjohn

@TheArtist you don't nee to say who a bounty is for. If there is a good answer, there is a reason already "Reward existing answer: One or more of the answers is exemplary and worthy of an additional bounty." Just use that and give it to whomever you want.

MarcGato

Does $a$ necessarily in the open ball?

we don't necessarily have $d(a,f(a))\le \varepsilon$

Chris's sis

Greetings

Anastasiya-Romanova 秀

6:21 PM

Mr. @robjohn I wanna ask about what kind of stuffs which I can put/ write in my profile page?

Hippalectryon

@MarcGato Tu fais de la topo ?

@Chris'ssis Hello

Anastasiya-Romanova 秀

Is there a meta reference(s) which discuss this?

MarcGato

@Hippalectryon oui ^^

Chris's sis

@Hippalectryon Hi

robjohn

@Anastasiya-Romanova秀 Unless it is offensive, I don't think there are many restrictions.

Chris's sis

6:23 PM

@robjohn Is that offensive "A Romanian that one day is going to be like Ramanujan or far beyond."? :-)

I think Ramanujan would be proud of all people wanting to become like him or far better.

Anastasiya-Romanova 秀

@robjohn Could you be more specific about the word offensive since every culture/ country can be subjective?

robjohn

@Chris'ssis Has someone claimed that that is offensive? It seems a bit presumptuous, but offensive, I don't think so

@Anastasiya-Romanova秀 For those very reasons, I cannot give a definitive rule

Hippalectryon

@MikeMiller Hum I was wondering if that way could work : $f^{-1}(U^c)$ is closed hence $a\in f^{-1}(U^c)$ but $U^c$ is open so there exists a ball around $a$ in $U^c$ which contradicts the continuity of $f$.

@MikeMiller $a=\inf f^{-1}(U^c)$

Kelly Blunie

Hi there, is there anyone who can help me with a linar algebra question?

skullpatrol

Good question @Anastasiya-Romanova秀

robjohn

6:26 PM

@KellyBlunie just ask the question. If there is someone who can answer, they will

Chris's sis

@robjohn I see. Well, it's not presumptous, it's just the truth told in advance.

Kelly Blunie

Ok

evinda

If we have the curve $x^3+y^3+z^3=0$ the hessian is equal to $216 \cdot x \cdot y \cdot z$. How can we find the non-singular points of the curve that are the intersection points with the hessian?

Hippalectryon

@MarcGato what is $\gamma$ ?

Kelly Blunie

0

Q: If the det of a set of vectors is zero, why does not span a vector space?

If we want to see if a set of vectors spans a vector space $V$, then lets say the set $A$ spans a vector space $V$. If every linear combination of $A$ produces $V$, then $\text{Span}(A) = V$ Edit: if we forme the coeficient matrix of the system formed by $c_1s_1+..+c_ns_n =u$ where $s_i$ are the...

Chris's sis

6:27 PM

Do we have a solution in one line for $$\int_0^1 \space _3F_3(1,1,1;2,2,2;\log(x)) \ dx=\frac{3}{4}\zeta(3)$$?

Hippalectryon

A solution in one line

Anastasiya-Romanova 秀

@robjohn For example: "I hate user xxx and I will ignore him/ her in every posts in Math SE", is that offensive?

Hippalectryon

Here you go ^ @Chris'ssis :D

Kelly Blunie

Adding to that my question is if we have det A non-zero then every vector u has exactly one solution what happens when determaniant is zero and why?

Chris's sis

@Hippalectryon :-)

MarcGato

6:28 PM

@Hippalectryon I made a mistake I juste need to take $f^{-1}(B(f(a),\varepsilon)$. $\gamma$ is the distance for $F$. I am working with metric space.

robjohn

@Anastasiya-Romanova秀 I would stay away from stating hatred. You can say currently ignoring xxx.

Hippalectryon

@MarcGato Distance=norm right ?

evinda

Hi @robjohn Do you have maybe an idea how can we find the non-singular points of the curve $x^3+y^3+z^3=0$ that are the intersection points with the hessian?I found that the hessian is equal to $216 \cdot x \cdot y \cdot z$.

MarcGato

@Hippalectryon no. If you have a norm then you can define a distance but the converse is not true.

Hippalectryon

@MarcGato Hm so $\gamma(a,b)=N(b-a)$ ?

MarcGato

6:31 PM

@Hippalectryon yes

robjohn

@evinda Your terminology is quite confusing to me. You talk about a curve whose equation look like that of a surface. The question you were asking me about that was on main did similar things.

MarcGato

@Hippalectryon but not if you are working on metric space. math.stackexchange.com/questions/21792/…

Chris's sis

@Venus I have a nice question for you. Compute without pen and paper, using only high school knowledge $$\lim_{n\to\infty} n\left(H_n-\log(n)-\gamma\right)$$

Anastasiya-Romanova 秀

@skullpatrol Hi. Long time no see :)

robjohn

@evinda What is the hessian of a curve? I have heard of the hessian of a coordinate transform, but not a curve.

evinda

6:33 PM

@robjohn I edited my post.. I added a theorem: math.stackexchange.com/questions/1059418/flexes-of-cubic-curve and I wantes to know how I could apply it.. Do you have an idea?

skullpatrol

Hi, how are you @Anastasiya-Romanova秀?

Anastasiya-Romanova 秀

@robjohn Suppose that I say "I am currently ignoring A" and then user A raise an objection to that statement, will I get suspended because of that?

Chris's sis

@Hippalectryon One of the worst things that can happen to a human being is to face people that are trying to kill his/her dreams. Never discourage people in their dreams, but encourage them to achieve any dream. If one says he wants to build a rocket and land on the moon, that I'd accept. It's simply that person's dream.

Mike Miller

@Hippalectryon Sure, as long as you know that closed, bounded sets in $\Bbb R$ have a minimum, that works perfectly.

Chris's sis

@Hippalectryon Out there are so many people with limited dreams, they have never ever had the courage to achieve a great dream. I'm not one of them.

MarcGato

6:36 PM

@Hippalectryon Le lien n'a rien à faire là je me suis trompé de réponse.. Cependant une norme est définie sur un espace vectoriel c'est beaucoup plus "spécialisée". (I am sorry I am better to explain myself in French)

Hippalectryon

@MikeMiller Thanks

Anastasiya-Romanova 秀

@skullpatrol I'm good. You? Anyway, where's Jasper? I've not seen it since yesterday.

robjohn

@Anastasiya-Romanova秀 I don't see how that is offensive. The person may not like it, but you are just stating the fact that you are ignoring them; that is just information to let them know why you don't reply to them in chat.

Hippalectryon

@MarcGato Ah je n'ai pas vraiment étudié les normes en dehors des EV :) désolé

Mike Miller

@Anastasiya-Romanova秀 Jasper has left the site to focus on his studies.

Hippalectryon

6:37 PM

@Chris'ssis Landing on the moon though >.>

skullpatrol

Fine thanks @Anastasiya-Romanova秀

MarcGato

@Hippalectryon Justement une norme c'est sur un ev, mais pas une distance. De ce fait on pourrait dire que espace vectoriel normé est inclus dans les espaces métriques.

evinda

@robjohn The Hessian is:

$$H(x,y,z)=\det \begin{bmatrix}
\frac{\partial^2 f}{\partial x^2} & \frac{\partial^2 f}{\partial x \partial y} & \frac{\partial^2 f}{\partial x \partial z} \\
\frac{\partial^2 f}{\partial x \partial y} & \frac{\partial^2 f}{\partial y^2} & \frac{\partial^2 f}{\partial y \partial z} \\
\frac{\partial^2 f}{\partial x \partial z} & \frac{\partial^2 f}{\partial y \partial z} & \frac{\partial^2 f}{\partial z^2}
\end{bmatrix}$$

Hippalectryon

@MarcGato Oui j'avais compris :) je voulais juste dire que si dans ta question $E,F$ ne sont pas des EV alors je ne vais pas pouvoir aider

skullpatrol

Yes, he needs time to focus on his plan for the coming year @Anastasiya-Romanova秀

MarcGato

6:40 PM

@Hippalectryon Ah ok d'accord, au temps pour moi :-). Mais la démonstration est globalement la même pour les EVN.

robjohn

@evinda Oh, I was thinking Jacobian. Yes, that is the Hessian. Sorry, there are many things going on at once today.

Chris's sis

@Hippalectryon Yeah, I'd only say go for it, land on the moon. :-) The limits are there only to be broken, again and again and again ...

Hippalectryon

@MarcGato D'ailleurs, si ce n'est pas impoli, tu fais des maths dans quel contexte ? (études sup, .....)

Anastasiya-Romanova 秀

@MikeMiller @skullpatrol That's a bad news for me (╥︣﹏᷅╥)

Hippalectryon

@Chris'ssis I'd have expected you to say only the Moon ? That's too mainstream! Try Mars !

evinda

6:41 PM

@robjohn Do you maybe have an idea how I could apply the theorem?

MarcGato

@Hippalectryon J'ai fait un bac S puis L1 de physique chimie, et là L2 de maths je galère un peu donc ^^.

Chris's sis

@Hippalectryon Sure, why not? :-)

Hippalectryon

@MarcGato Bonne chance alors :)

MarcGato

@Hippalectryon Merci!

Anastasiya-Romanova 秀

@robjohn How if I put picture or use ava while sticking out my tongue or show my middle finger?

robjohn

6:42 PM

@evinda what theorem?

Hippalectryon

@MarcGato Je suis presque sûr que @MikeMiller a la réponse. J'utilise le "sens direct" (l'image d'un ouvert par une fonction continue est un ouvert) dans ma dernière question

skullpatrol

You can always email him @Anastasiya-Romanova秀

robjohn

@Anastasiya-Romanova秀 ava? middle finger might be considered offensive.

evinda

@robjohn "Let the algebraic curve $f(x_0, x_1, x_2) \in K[x_0, x_1, x_2]$. The inflection points are the non-singular points of the curve that are the intersection points with the hessian."

How can we find the non-singular points of the curve that are the intersection points with the hessian?

Mike Miller

@Anastasiya-Romanova秀 Well, he seems to think it's best for him. I hope he does well in his studies and his health.

MarcGato

6:44 PM

@Hippalectryon Oui j'ai plus au moins lu ta phrase s'adressant à lui mais je ne comprenais pas le contexte. Cependant j'ai réussi, je me suis trompé entre f^-1 et f..

Anastasiya-Romanova 秀

@skullpatrol Ah, I forget that his email is in my contact

@robjohn Avatar, my profile picture

@MikeMiller I hope he is always in good healthy

Hippalectryon

@MarcGato Quelle est la ligne directrice de la preuve ?

Mike Miller

@Hippalectryon If my French is correct, that's not true. The image of an open set under a continuous map is rarely open. It's the preimage of an open set that's always open.

Hippalectryon

@MikeMiller Sorry I meant preimage indeed

Kelly Blunie

if we want to prove that either A span V or not and we forme the coeficient matrix of the system formed by c1s1+..+cnsn=u where si are the vectors in set A and u is any vector in V , if the determinant is zero, then there is at least one choice of u for which this system will not have a solution and hence can not be written as a linear combination of thexse vectors?why is this?

Mike Miller

6:48 PM

Okie doke. (A continuous function $f$ such that $f(U)$ is open for every open $U$ is known as an open map.)

skullpatrol

Showing your middle finger to everybody in your avatar is offending everybody @Anastasiya-Romanova秀 don't you think?

Anastasiya-Romanova 秀

@robjohn Anyway, since I'm running for a moderator, may I promote or advertise my name in other sites?

I mean my nomination?

Kelly Blunie

If the determinat was non-zero it has exactly one solution for each u if not what happens?

Anastasiya-Romanova 秀

@skullpatrol How about pose of pictures some rock stars?

MarcGato

@Hippalectryon En fait j'essaye de montrer que f continue sur E équivalent à pour tout ouvert de $(F,\gamma)$ l'image réciproque est une partie ouvert de (E,d) équivalent à la même chose pour fermé équivalent à f(adh(X))inclus dans adh(f(X)). Je le fais par implication successives et la dernière est donc l'image réciproque est une partie ouvert de (E,d) implique f continue

et j'utilise que $f^{-1}(B(f(a),\epsilon))$ contient $a$.

robjohn

6:53 PM

@Anastasiya-Romanova秀 I don't know how people will take that. This is not a political popularity contest. We are concerned about the smooth running of the site and someone doing the mostly thankless work that the moderators do. If a lot of people from other sites start getting accounts here simply to vote for candidates, the community managers may have to stage an intervention.

Anastasiya-Romanova 秀

@skullpatrol Please don't get me wrong, I have no intention to use that kind of ava. I just use it as an example

skullpatrol

Rock stars @Anastasiya-Romanova秀 get away with that kind of stuff because they like to shock people and get them talking about them. That's how they stay in the news.

Hippalectryon

@MarcGato Je pensais passer par l'absurde, supposer $f$ non continue en $x_0$ et trouver un $B$ ouvert tel que $f^{-1}(B)$ soit fermé, mais je n'y arrive pas comme ça :/

MarcGato

@Hippalectryon a priori ça semble difficile car tu n'as aucune expression de $f$ donc je vois mal comment on pourrait trouver $f^{-1}(B)$ fermé. Je peux me tromper cependant.

Anastasiya-Romanova 秀

@robjohn Oh thanks God I ask you first this question. I almost do a stupid thing like promoting my nomination. So your suggestion is I shouldn't do it, should I?

robjohn

6:58 PM

@Anastasiya-Romanova秀 I wouldn't. It is about the running of math.SE. Other sites should not really be involved.

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$Mathematics$ Mathematics