Mathematics - 2015-10-28

Paradox 101

00:53

can anyone explain this?

How do we get f(x)?

Antonio Vargas

01:08

@Paradox101 For any $x \neq 0$ do you believe that $|x| > 1/n$ for $n$ sufficiently large?

Clarinetist

Can someone tell me what's wrong with my answer?

0

A: Prove that if $f:[a,b]\to[a,b]$ is continuous then there is a $c\in[a,b]$ such that $f(c)=c$

If $f:[a,b] \to [a,b]$ is continuous, by the Extreme Value Theorem, it attains a minimum and maximum in $[a, b]$, say at $x$-values $x_m$ and $x_M$ respectively. So for all $x \in [a, b]$, $$f(x_m) \leq f(x) \leq f(x_M)\text{.}$$ Let $c \in [f(x_m), f(x_M)]$. Then $f(x_m) \leq c \leq f(x_M)$. How...

Antonio Vargas

@Clarinetist, the second 'c' you have found is not the same as the first.

not necessarily, at least

Paradox 101

@AntonioVargas yes

Clarinetist

@AntonioVargas Sorry, I'm not understanding your explanation here

Antonio Vargas

@Paradox101 Then do you believe that $f_n(x) = |x|$ for $n$ sufficiently large?

@Clarinetist You use the name $c$ to refer to two distinct quantities, one in the range of $f$ and one in the domain, but these two quantities may not be equal as numbers. Maybe if I write out your argument slightly differently...

Clarinetist

01:14

@AntonioVargas AH, I see what you're saying

Antonio Vargas

cool :)

Clarinetist

@AntonioVargas Is there any way that I can fix my argument?

or would I have to essentially do what the other answer did

?

@AntonioVargas I think what you're getting at is that I should use a different symbol upon applying the IVT

Antonio Vargas

The other answer is the argument I'm familiar with. I'm not sure whether yours could be easily modified into a proof. It's definitely worth trying though.

Clarinetist

so something like, there is a $c^{\prime} \in [a, b]$ such that $f(c^{\prime}) = c$

Antonio Vargas

Yeah, exactly

Ted Shifrin

01:17

@Clarinet: Your answer is flawed. You're basically just saying it's true because it's true. You need to apply the intermediate value theorem to $f(x)-x$.

Clarinetist

So now it suffices to somehow show that $c^{\prime}$ must be equal to $c$

Paradox 101

@AntonioVargas i suppose so but given only $f_n(x)$ how do we get to $f(x)$ as n approaches infinity?

Clarinetist

Meh, I will just delete it for now

Antonio Vargas

@Clarinetist, but, in general, it won't be.

Clarinetist

True

Ted Shifrin

01:18

You have to pick the right $c$ in the range in the first place, or it won't work.

Clarinetist

@TedShifrin I shall remember that

Ted Shifrin

What? :D

Antonio Vargas

@Paradox101 For any given $x \neq 0$ you agree that $f_n(x) = |x|$ for all $n$ sufficiently large. Therefore, for all $x \neq 0$ you have $$\lim_{n \to \infty} f_n(x) = |x|.$$ There's your $f(x)$.

Ted Shifrin

@Paradox. You take $\lim\limits_{n\to\infty} f_n(x)$.

Never mind, I shouldn't butt into it when @Antonio always does a superb job.

Antonio Vargas

you need to check $x=0$ separately, and show that $f_n(0) \to 0 = |0| = f(0).$

@TedShifrin Always appreciate the help :)

Paradox 101

01:29

@AntonioVargas but what about when x is less than 1/n? We take the limit of n goes to infinity for it and the answer comes out to be infinty

Antonio Vargas

@Paradox101 For a fixed $x$, its absolute value $|x|$ will only be less than $1/n$ for a few $n$. Once $n$ gets large enough we'll have $|x| > 1/n$, and at that point $f_n(x) = |x|$.

(again, assuming $x \neq 0$)

Paradox 101

@AntonioVargas so since we have a large n we ignore the less than 1/n part?

Antonio Vargas

@Paradox101, Pick a number, as small as you'd like. Maybe $x = 1/100$. Then it's true that $|x| < 1/n$ for all $n=1,2,\ldots,99$, but as soon as $n = 101$ we have $|x| > 1/n$, and so $f_n(1/100) = |1/100|$ for all $n > 101$. Then we use the definition of a limit.

Do you agree that, for any $\epsilon > 0$, $$\Big||1/100| - |1/100|\Big| < \epsilon$$ for all $n$? (Hint: yes, since there is no $n$.) Then, for any $\epsilon > 0$, do you agree that $$\Big|f_n(1/100) - |1/100|\Big| < \epsilon$$ for all $n > 101$?

(Hint: yes, since the left-hand side is $=0$ for all $n > 101$, and $0 < \epsilon$.)

Paradox 101

01:45

@AntonioVargas ok i think i get it. thanks :)

Remember me

07:43

@TobiasKildetoft Are you there?

Tobias Kildetoft

@Rememberme Yeah

Remember me

This is a question I have asked but I am asking this to you. Balarka told me that there is an algorithmic way to find the covering spaces of a given space $X$. Can you tell me in general what is the way to find all the covering spaces of a given space $X$. Does this algorithmic method also apply if we are finding universal cover of a space $X$@Tobias

Tobias Kildetoft

@Rememberme I don't really recall enough algebraic topology for that

Remember me

Oh,okay.

Balarka Sen

09:27

@Remember No, I didn't tell you that.

I told you there is an general method to find covering spaces of $X$, but that is not effective.

And yes, it works for the universal cover too.

user281591

11:25

@DanielFischer Can you help? Let $ u_0, u_1 \in \mathbb{R}$ and $u_1/u_0 \in \Bbb{Q}$. It is asking to study $u_{n + 2} = \vert u_{n + 1} - u_n\vert$

I was thinking about Euclid's lemma

Martin Sleziak

They grow up so fast these days! Did we always have so many users who are barely above 100 reputation points and come to meta to suggest changes in the software or blame more experienced users for their bad behavior?

@user281591 You only said something about $u_0$ and $u_1$. You should say what $u_n$ is if you expect some kind of reasonable answer.

But maybe it is a continuation of some previous discussion which I am unaware of.

user281591

@MartinSleziak nop, it''s a new 'discussion', but the exercise is as I wrote it.

I need to study $(u_n)$ defined by the recurrence relation

Tobias Kildetoft

@user260674 what does "study" mean here?

@MartinSleziak Do you have a specific meta post in mind?

Martin Sleziak

Oh, I see.

@TobiasKildetoft There were at least two I am aware of, which were posted recently by users between 100 and 200 rep points.

user281591

@TobiasKildetoft I think it's to say if u_n converges or not..

Tobias Kildetoft

11:38

@user260674 Well, only you can know

Martin Sleziak

@user281591 You could take $v_n=\frac {u_n}{u_0}$. Then you have the same recurrence $v_{n+2}=|v_{n+1}-v_n|$ with you have $v_0=1$, $v_1\in\mathbb Q$. In this way, you always stay in the rational numbers. (If it is of some advantage for you.)

Tobias Kildetoft

@MartinSleziak I have a feeling such posts come up fairly frequently, probably about as frequently as answers on site policy by people with experience on some other SE site and who feel that makes them well qualified to answer meta questions here

Martin Sleziak

If $v_1>1$, then the sequence will be periodic: $1, v_1, v_1-1, 1, v_1, v_1-1, \dots$

user281591

yes I was thinking about that :))

Martin Sleziak

In fact, maybe we get $1, v_1, |v_1-1|, 1, v_1, |v_1-1|, \dots$ for $v_1\ge 0$.

user281591

11:51

if it can be reduced to a sequence of integers, it makes me think about Euclid algorithm

crl

13:36

in JavaScript on Stack Overflow Chat, 10 mins ago, by Kendall Frey

There are two 5-digit numbers, A and B, which are the reverse of each other, e.g. 12345 & 54321. There is a constraint that A * 4 = B. There are two solutions to this. One is A = 00000. What is the other one?

Razieh Noori

14:17

May i ask a question?

Martin Sleziak

Feb 8 at 20:32, by anon

> just ask; don't ask to ask

Maybe nobody will respond. Maybe somebody will. It is best to ask the question. You will see whether somebody in the room has time and knowledge to answer.

Razieh Noori

14:59

0

Q: trace norm and tensor product

Let $(M_n (\mathbb{C}), n\|.\|)$ , $(M_n (\mathbb{C}), n\|.\|)$ and $(M_{nm} (\mathbb{C}), nm\|.\|)$ be three Banach algebras. where $$\|A\| = \mathrm{tr}\sqrt{(A^* A)}. $$ What is the norm of $\phi$ and $\phi^{-1}$ defineded $$\phi: M_n (\mathbb{C})\otimes M_m (\mathbb{C})\to M_{nm} (\mathbb{C...

any idea will be helpful, thanks in advanced

Tien-Cheng Huang

15:52

Could someone give me an idea that in what order to prove these 6 statements is the easiest way? math.stackexchange.com/q/1500345/275935

Probably I beter find by myself the easiest order to prove them...

Jessy Cat

16:16

0

Q: Showing that if $f=g$ a.e. on a general measurable set (for $f$, $g$ continuous), it is not necessarily the case that $f=g$.

This is related to a question I just asked, that I now think was based on wrong assumptions. It is true that if $f=a$ a.e. on the interval $[a,b]$, then $f = a$ on $[a,b]$. However, apparently it is not true for a general measurable set $E$ with $m(E) \neq 0$, which confuses me greatly. I jus...

real-analysis measure-theory lebesgue-measure

I hate smartasses.

Szabolcs

I asked this question in a Mathematica-specific way, but I would also like to know in general what is known about this problem (keywords, etc.):

2

Q: Smallest connected subgraph containing all given vertices

Given a connected graph $G$ and a subset of its vertices, $S$, I need to find the smallest connected subgraph of $G$ containing all of $S$. How would you approach this in Mathematica? I am asking this here on Mathematica.SE because I am looking for the most convenient way to use the functions...

graphs-and-networks algorithm

Any pointers will be appreciated.

Balarka Sen

16:43

@Szabolcs What do you mean by smallest? Relevant keyword could be "maximal tree".

Szabolcs

@BalarkaSen I only see one possible interpretation in this case, as least vertices and least edges is the same for trees.

@BalarkaSen Let us say: least possible edges. That implies that it has to be a tree, and the number of vertices will be just one more than the number of edges.

Antonio Vargas

@Szabolcs I think the term is minimal spanning tree.

Szabolcs

@AntonioVargas I think that's not going to be good. Here's why. Consider this graph:

Balarka Sen

Ok. I am no programmer, but I am sure such a thing exists. You have to choose a connected subgraph of $G$ with vertex set = $S$. And then you look at it's maximal tree.

Antonio Vargas

@Szabolcs Done.

Szabolcs

16:49

Let's say I want the smallest possible connected subgraph that contains {2,3}. That will be just 2, 3 and the edge connecting them.

If I start with a minimal spanning tree of the graph, and go from there, it might be this:

Antonio Vargas

no no no

you take the subgraph first

Szabolcs

Antonio Vargas

then calculate its minimal spanning tree

Szabolcs

@Antonio Take which subgraph exactly?

Balarka Sen

lots of subgraphs there with vertex set containing {2, 3}

Antonio Vargas

16:51

My mistake, I thought you were starting with a subgraph

Szabolcs

If my vertices are {2,4}, the subgraph containing only them is not going to be connected.

A good solution would be 2 - 1 - 4.

Balarka Sen

@Szabolcs I think you want geodesics in your graphs.

Szabolcs

@BalarkaSen That works if I have two nodes in my starting set. If I have more, what do I do?

Balarka Sen

Give your graph the standard geodesic metric space structure by letting each edge to be of length 1.

@Szabolcs Hm.

This is a cool question.

Antonio Vargas

Isn't this travelling salesman?

Szabolcs

16:52

Possibly related.

Also, I don't really need this. I thought I did but I realize I don't. However, it is a cool question so I am still interested! :)

Antonio Vargas

I guess that's about finding a path (or cycle), not a tree

Szabolcs

Yes.

evinda

Hello!!! Could you take a look at my question?

0

Q: Description of the Simplex algorithm

I am looking at the description of the Simplex algorithm. Let $\overline{x_0}$ be a non degenerate basic feasible solution. We suppose that $\overline{x_0}=(x_{10}, x_{20}, \dots, x_{m0},0, \dots,0)$ with $x_{10}, x_{20}, \dots, x_{m0}>0$. Thus the first $m$ columns of $A$, i.e. $P_1, P_2, \dots...

optimization simplex

Balarka Sen

ok, I resign. I don't know :)

Szabolcs

This is related, but it is not the same thing:

Steiner tree problem

The Steiner tree problem, motorway problem, or minimum Steiner tree problem, named after Jakob Steiner, is a problem in combinatorial optimization, which may be formulated in a number of settings, with the common part being that it is required to find the shortest interconnect for a given set of objects. The Steiner tree problem is superficially similar to the minimum spanning tree problem: given a set V of points (vertices), interconnect them by a network (graph) of shortest length, where the length is the sum of the lengths of all edges. The difference between the Steiner tree problem and the...

Balarka Sen

16:57

yes, it's not the same thing

ultimately, what you want to do is to join pairs of vertices cyclically by geodesics on the graph. and then you want to see if you can minimize the # of edges by pushing the whole subgraph off the center. i don't know how to do this.

Antonio Vargas

From wiki: The problem of finding the Steiner tree of a subset of the vertices, that is, minimum tree that spans the given subset, is known to be NP-Complete.

Balarka Sen

maybe take the subgraph which is bounded by that geodesic polygon, and look at it's homology. if homology is Z, then you know it's a circle and there is no smaller subgraph.

that's one possible case.

but it might as well get very complicated.

Szabolcs

You are right Antonio, I missed that part.

Sorry guys, I need to go, my wife is calling ...

I'll come back to this later (and will read chatlog)

Selfstudier

17:13

Hi guys. Does anybody know of any online notes on Riemannian geometry that give a good exposition of Jacobi Fields and some of their uses in calculations and proofs of theorems?

Pedro Tamaroff

17:26

@anon Do you have any notes on cyclotomic extensions?

evinda

18:01

Hi @AntonioVargas Are you familiar with the simplex algorithm?

Antonio Vargas

Nope

evinda

A ok... @AntonioVargas

Is anyone else familiar with it?

Xingdong

18:13

A short question, is Arnold actually okay for a first course about ODE ? I've heard that the it focuses on theory perspective with geometric point of view which is very good, but the calculation part also cannot be missed for one who is learning ODE

Ted Shifrin

18:56

@Balarka: F you. That is a classic problem that has videos and articles written about it. And my students are fascinated by it. So f you.

2

Balarka Sen

:P

I was only joking.

@TedShifrin By the way, I think I have found a typo in your book.

(I wonder who did that)

r9m

20:38

@evinda if I have ever known anything about simplex algos .. I have deleted it from my hdd :P that stuff scares me :P

Karim Mansour

22:19

Hi

@BalarkaSen suppose we have $f : X \rightarrow Y$ map of sets suppose Y is a topological space what is the coarsest topology on X making f continuous ?

Daniel Fischer

@KarimMansour The initial topology induced by $f$. Which consists of the sets of the form $f^{-1}(U)$ where $U$ is open in $Y$.

Balarka Sen

@Karim Well, obviously you want the open sets to be preimage of open sets in $Y$.

Ah, well, Daniel beat me to it.

Karim Mansour

Well, I had the following but prof told me you can have simpler answer I considered the intersection of all topologies on X that make f continuous, but he told me that you can do simpler solution

which is what daniel described

Balarka Sen

yeah, what you said was not very explicit.

Karim Mansour

and the reason this intersection isn't empty is because we know P(X) on X make f continuous

Balarka Sen

22:24

right.

hlapointe

Hi everyone!

evinda

@r9m I find it interesting...

hlapointe

Can I ask any mathematic questions in this room?

Ok. I'll take that silence for yes! ;)

PVAL

22:58

Yes, but you are not allowed to ask if you can ask. This rule you have already broken.

Mary Star

Could you give me some hints what I am supposed to do?

The formula looks like the formula of the tangent of a curve on a point. Does this related to this?

Karim Mansour

just a sec making sure of something

Suppose $X = {a,b}$ and $\tau_X = \{\emptyset,X,\{a\}\}$ this will be $T_0$ as we can create a open set that contain b and doesn't contain a.

but it is not $T_1$

since the points are a,b of this set

Mary Star

@Huy @robjohn do you have an idea about my question above?

Karim Mansour

and we don't have nbhds $U_a,V_b$ that satisfy $T_1$ condition.

actually this topology is one of the minimal counter example for something that is $T_0$ but not $T_1$

anon

23:31

@PedroTamaroff I remember reading about them, but didn't write down anything special. (except you can compute class numbers using resultants and mobius inversion IIRC)

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