Mathematics - 2013-02-20 (page 2 of 6)

Orange Harvester

8:04 AM

@Ethereal You're the head of the tautology club.

user19161

Bleh and meh are very hard words. Often, I have to guess the intention of the speaker who utters them.

Orange Harvester

That is precisely the intention.

user19161

@ora I have now seen the second edition of Artin, seems he removed some exercises.

user19161

@GustavoBandeira Maybe recognition won't be important anymore after you get it.

user19161

It's weird but true, many things only seem important when you don't have them yet.

user19161

8:19 AM

@FrankScience You mean vacation, vocation is something else.

Orange Harvester

@JacobBlack Yes. He also reorganized some exercises.

user19161

@OrangeHarvester A nutcase just downvoted three answers here math.stackexchange.com/questions/307438/…

Orange Harvester

@JacobBlack seems harsh.

user19161

@OrangeHarvester Not harsh, simply ridiculous.

Orange Harvester

@JacobBlack the guy is basically asking you to prove that closed balls are complete and wants to use the convergence of cauchy sequence inside a ball to do so. A good proof would be by contradiction. Assume that the ball is not complete, then there is a limit point $x_l$of the ball outside the ball, and hence $||x_l-x_0|| > R$ and then show that there exists an open ball around $x_l$ which does not have a single point of $U$ and hence the proof.

user19161

8:31 AM

@OrangeHarvester Why are you telling me?

Orange Harvester

@JacobBlack I am trying to tell why the voter must have downvoted. He expects the answer I gave and feels other answers are just roundabouts. Its too harsh to downvote though.

user19161

@OrangeHarvester Well, then he is clearly weak in general topology.

Orange Harvester

@JacobBlack Yes.

user19161

@OrangeHarvester Therefore, he downvotes totally correct answers that do answer the question.

Orange Harvester

@JacobBlack Yes.

user19161

8:34 AM

@OrangeHarvester I think I should delete my account soon, there are too many such nutcases here...

Orange Harvester

@JacobBlack You might as well want to delete your account on earth then. (Too many nutcases there. What's worse is they are IRL, you can ignore the internet nutcases.)

user19161

@OrangeHarvester He didn't have the balls to comment when he downvoted, because he probably doesn't know what is wrong (since there is nothing wrong).

Orange Harvester

I don't understand this. If you do not like someone on internet, you can just turn it off. Its not like in real life where they can shout into your ear and make you listen to them. The most they can do is hack your accounts, but then even that is something which you can prevent easily.

@JacobBlack I wrote my answer, lets see if he downvotes that.

user19161

@OrangeHarvester I can tell if someone competent wants to downvote for a small reason, but this case is probably someone very incompetent.

Frank Science

@JacobBlack Thanks. My English is bad. vocation = profession (if this time I were not that stupid)?

user19161

8:41 AM

@FrankScience Yes, roughly speaking.

user19161

@OrangeHarvester In fact, the answer which I think he upvoted actually is the more roundabout way of doing things, hahahaha.

user19161

@OrangeHarvester Once again, not harsh. Simply ridiculous. =)

Orange Harvester

@JacobBlack I sometimes wish you would look at it from point of someone who does not know topology yet. It is possible to define cauchy sequences and completeness completely independent of closed/open/compact. May be he is learning on that route and will soon get to closed/open sets later. May be he will get to closed sets later.

Not everyone (very few) has a bird's eye view of mathematics from the very starting.

user19161

@OrangeHarvester I have not used any weird definitions in my thinking there, all very standard, so this is not the issue here.

user19161

@OrangeHarvester Do you know that it is so much harder to prove Heine Borel? =)

Orange Harvester

8:50 AM

@JacobBlack you are not using any weird definitions here. But it is possible that he has not encountered them. Instead of asking for clarification, he has downvoted the three answers. I am not condoning his behaviour btw.

user19161

@OrangeHarvester You mean 3 answers, not the question.

Orange Harvester

@JacobBlack Who said anything about Heine Borel? Look at my proof. It is not the most correct proof, but it is what would be intuitive at that level.

@JacobBlack Yes.

user19161

@OrangeHarvester I am referring to the answer the downvoter upvoted.

Frank Science

Heine-Borel?

A complete metric space is compact if and only if it's totally bounded?

Orange Harvester

@JacobBlack There is only one answer with two upvotes and it is not heine borel. How do you know he has upvoted the heine borel answer?

@FrankScience that and subset of $\mathbb{R}^n$ is compact if it is closed and bounded.

user19161

8:52 AM

@OrangeHarvester I look at the rep history of everyone there.

Ilya

@Victor depends on the question

user19161

@FrankScience A subset of Euclidean space is compact if and only if it is closed and bounded.

Orange Harvester

@JacobBlack okay.

user19161

@ora I think I know why he chose that answer to upvote: because of the magic word "Heine Borel" LOL.

Orange Harvester

@JacobBlack I will stop wondering about that and go back to my studies. :P

Read this. Its cool.

Frank Science

8:58 AM

It's not hard to show that a subset of $\mathbb R^n$ is totally bounded if and only if it's bounded.

Garbage Collector

@JacobBlack: Hi Jasper Loy!

user19161

@GarbageCollector Hi! I am not ignoring you. Sometimes, there is no need to reply so I don't say anything.

user19161

I am very upset with people who keep saying I am ignoring others.

2

user19161

@FrankScience That is correct.

Frank Science

Now I want to skip chapter 8~10 and go through chapter 11 of Baby Rudin about Lebesgue's theory.

That's attractive.

user19161

9:11 AM

@OrangeHarvester Sorry, but your proof there seems incorrect to me.

Orange Harvester

@JacobBlack How?

skullpatrol

@GarbageCollector Hi

user19161

@OrangeHarvester If the ball is not complete, why should there be a limit point of the ball outside it?

Garbage Collector

@skullpatrol Hi @skullpatrol, how are you?

user19161

@ora If one negates the definition of completeness, one just has the fact that some Cauchy sequence in the ball does not converge in the ball, that is all.

Orange Harvester

9:16 AM

@JacobBlack A metric space M is called complete (or a Cauchy space) if every Cauchy sequence of points in M has a limit that is also in M. So, if a ball is not complete, there exists a cauchy sequence with its limit outside the ball. We can show that the limit of a cauchy sequence has to be a limit point of the ball (because the limit of the cauchy sequence has atleast one point of the ball in every open sphere around it.)

peoplepower

E.g. it may not converge at all...

Frank Science

But it's Cauchy in a complete metric space so it does converge.

Orange Harvester

@JacobBlack Also,

user19161

@OrangeHarvester First, you have implicitly used the completeness of Euclidean space there.

Orange Harvester

27 mins ago, by Orange Harvester

@JacobBlack Who said anything about Heine Borel? Look at my proof. It is not the most correct proof, but it is what would be intuitive at that level.

user58512

9:18 AM

hello

mathematician valentines day: You cauchy complete me

user19161

@OrangeHarvester Second, if it converges to a point outside the ball, there may not be a limit point.

Orange Harvester

@user58512 :D

Frank Science

@JacobBlack No worry about them. 1000 persons have 1000 different opinions.

Orange Harvester

@JacobBlack If every other point of cauchy is inside the ball, then the limit has to be a limit point, no?

user19161

A limit point is different from a limit.

user19161

9:20 AM

A constant sequence converges but has no limit point.

Orange Harvester

@JacobBlack Yeah, but that cauchy sequence is not the sequence I would consider for my completeness then.

(I do assume that a cauchy sequence converges in a euclidean metric space.)

Frank Science

@JacobBlack How do you do who downvoted a post?

Orange Harvester

@FrankScience he looked at the reputations and changes in them,

user19161

@OrangeHarvester That is not the issue, the issue is my two points above which shows the gap and flaw respectively.

Frank Science

@OrangeHarvester Ugh?

Orange Harvester

9:23 AM

@JacobBlack For the first point, I agree I am using the completeness of the euclidean metric space. But that is not an issue. Since, it does not trivialize the proof in any way I feel.

user19161

@OrangeHarvester That is just a thing you used without mentioning, not an error like I said.

Orange Harvester

@JacobBlack Yes.

user19161

Of course, I am guessing your thought processes here...

Orange Harvester

For the second point.
If all points of cauchy sequence are inside the ball, then the limit of the cauchy sequence has to be a limit point of the ball right? Am I wrong here?

user19161

If you are supposing that some Cauchy sequence in the ball converges outside the ball to get a contradiction, limit point does not come in at all.

user19161

9:25 AM

Like I said, a limit point is not the same as limit.

awllower

Hello there.
There is an interesting question. Are you interested?
Thanks for any attention at all. :)
http://math.stackexchange.com/questions/308939/commutative-ring-and-its-group-algebra-and-abelian-group-algebra-as-a-commutati

Orange Harvester

@JacobBlack How can we do it without getting limit point in the picture?

user19161

@OrangeHarvester If you want to use limit point, then there are many gaps there to fill, which I am not trying to guess here.

user19161

@OrangeHarvester Simply use limit instead of limit point.

user19161

(0,0,0,...) converges to 0 in R but has no limit point.

Frank Science

9:31 AM

Suppose $\{p_n\}\subset E$ and $E$ is a closed set.

If $p\not\in E$, there's a neighborhood of $p$, say $V_p$, such that $V_p\cap E=\emptyset$.

user19161

@ora I saw your edit. Again, limit point is not the same as limit.

Orange Harvester

@JacobBlack I know and I disagree how using only limits can do the proof.

The concept of limit point is important for proving the exclusion thing.

Frank Science

@OrangeHarvester Prove what?

user19161

@OrangeHarvester First, if you use limit there, the proof can work. Second, it is incorrect to use limit point there because there may be no such thing.

Orange Harvester

@FrankScience Here

Frank Science

9:36 AM

@OrangeHarvester It's not necessary to introduce the concept of limit points, yeah.

user19161

@ora In fact, if you replace limit point by limit there the proof becomes correct (if you understand it), but if you use limit point there the proof is wrong.

Orange Harvester

Hmm.

user19161

First, limit point may not exist.

user19161

Second, if a point is the limit, then eventually all terms of the sequence will lie in an open ball around it.

user19161

End of story, QED.

Frank Science

9:38 AM

$p_n\to p\implies\forall\epsilon>0,\exists N,d(p_N,p)<\epsilon$.

user19161

@FrankScience That is correct.

Frank Science

That's all.

user58512

yo

awllower

@OrangeHarvester Might I ask you a question about topological groups and projective limits?

user19161

Yo.

user58512

9:40 AM

if fermat didn't mention it, would other mathematicians have discovered fermats last theorem?

Orange Harvester

@user58512 wiles.

@awllower I know nothing about that.

awllower

Oh! My fault, due to bad memory...

Maybe Jacob knows something about the topic?

user19161

@awllower I only know 1+1=2, sorry.

Orange Harvester

@JacobBlack Yup. You are correct. But now do you see what kind of proof the OP might be looking for?

Frank Science

@JacobBlack What do you mean by general topology.

awllower

9:43 AM

Hm, Jacob is very humble and humerous!

user19161

@awllower I am only a banana.

user19161

@FrankScience It means point-set topology. I asked a question about what "general" means.

awllower

Haha.
I thought that bananas are good at such topics---banana limit...

user19161

I asked the question because I have seen 9000 general topology books, but none mention what "general" means.

user58512

general topology

Frank Science

9:45 AM

@JacobBlack Well, but the properties of metric spaces are discussed more in analysis, not topology.

user58512

Frank Science

For example, completeness is not a topological property.

user19161

@FrankScience Well, all branches of math are related, not too important what we call it.

awllower

Yes

user19161

Classifications are just for convenience to study a particular topic.

awllower

9:47 AM

I am wondering : if the completeness follows from compactness?

user19161

Completeness of R follows from the completeness axiom of R.

user58512

lol

user19161

A nonempty subset bounded above has a least upper bound.

awllower

So there is no relations between the two topics?

user58512

a compact space is complete

awllower

9:48 AM

the two notions.

Oh

user19161

Well, a metric space is compact if and only if it is complete and totally bounded.

user58512

but $\mathbb R$ is not compact

awllower

Oh!
Thanks very much!

Frank Science

It's locally compact.

user58512

snow dog youtube.com/watch?v=7muB4tVp8d0

awllower

9:51 AM

I am told that, if a series of homomorphisms has compact kernels, then the limit of isomorphisms is also an isomorphism, and so does the induced homomorphism on the quotient group. Can completeness along serve this?

user58512

oh is that an inverse limit like from category theory?

awllower

I referred to projective limits, but they are similar I suppose.

user58512

ive' got no idea how to even prove that

awllower

Oh
I see it in a book about CFT
There it is referred to Bourbaki, general topology...

I am trying to avoid reading that book though...

Orange Harvester

@JacobBlack I deleted the answer.

Mariano Suárez-Alvarez

10:04 AM

@awllower completeness of what?

awllower

of the kernel?

Mariano Suárez-Alvarez

I don;'t know how to prove that but I would be surprised if it worked with non-compact kernels

awllower

Ok
Thanks for the opinion!

Mariano Suárez-Alvarez

compactness is a finitness assumption

completeness... isn't

awllower

Oh
This explains the reason of this compactness assumption!!

Mariano Suárez-Alvarez

10:07 AM

in any case, go look at the bourbaki book

which is the best of the bourbaki books!

awllower

Really!
I have never heard of that!
Thanks for the suggestion!

I will grab the book and digest I think, haha

Frank Science

Bourbaki?!

awllower

Ja

Frank Science

Recalled me of something called New Math in U.S.A.

awllower

Oh

user58512

10:13 AM

see you later

awllower

m8er

L8er

user58512

:)

skullpatrol

Sorry I was called afk @GarbageCollector Fine thanks, how are you?

@MarianoSuárez-Alvarez Have you seen this link on the starred comments?

3 @Khromonkey Possibly interesting: ams.org/notices/200410/fea-grothendieck-part2.pdf - 6h ago by MJD ▼

awllower

10:29 AM

@skullpatrol very interesting indeed!

To have dinner now.
Later thus!

Orange Harvester

10:45 AM

spikedmath.com/544.html

peoplepower

@OrangeHarvester Darn, they crossed out my suggested price.

Orange Harvester

@peoplepower :-)

Dominic Michaelis

11:08 AM

huhu

mh is my answer so stupid that it is worth to downvote is withouth a comment [combinatorics]math.stackexchange.com/questions/308986/…

Gam Erix

hey guys I have to derrive f(x) = 43.3/x to f'(x)= , Iforgot how to do that, anyone knows? ; o

Dominic Michaelis

whats the deriviative of 1/x ?

Gam Erix

1/(2(x)^0.5) ?

oh okay thanks xd

Dominic Michaelis

its -1/x^2

Gam Erix

okay thanks :$

Orange Harvester

11:23 AM

@DominicMichaelis many people downvote without explanation. but I see that your problem has already been solved.

Dominic Michaelis

well i still think my solution is correct ...

Alexander Gruber

anybody here know how to use cauchy's integral formula?

Dominic Michaelis

for what ?

Alexander Gruber

to evaluate integrals. i can't figure out how i am supposed to use it.

first problem is $\oint z^{-1}\operatorname{Log}(z+e)dz$

where the integral goes counterclockwise around the unit circle

Frank Science

Why is it called $\sigma$-ring?

Alexander Gruber

11:32 AM

@FrankScience probably comes from $\sigma$-algebra? I don't know why that has that name, though.

awllower

@AlexanderGruber I think the formula intends to make use of the residues, so you can calculate the residues , thus obtaining the integral.

David Robert Jones

hello

awllower

@DominicMichaelis The question asks for the largest value, while your answer gives a lower bound. I guess this is why someone downvoted.

@DavidRobertJones Hello :)

David Robert Jones

today i was looking the questions and i found this: Show this equation has at least one root in (0, 1)

i'm very impressed with Babak solution

but i'm not sure if i understand why he suggested that function

that function allows you to evaluate f(0) = 0 and f(1) = (1/3)*a + (1/2)*b + c, which is equals to 0 since 2a + 3b + 6c = 0, and so you can apply the Lagrange's Mean Value Theorem and conclude it has to have a root in (0, 1) right?

Frank Science

I think it's only a trick.

David Robert Jones

11:40 AM

yes, but for what?

awllower

Yes your arguments are correct.

David Robert Jones

@awllower: that function will have one more root (x = 0) than the original function + in the (0, 1) interval the "original roots" will be kept, right? it's just a trick that allows to apply the theorem. am i correct?

awllower

What function will have one more root?

David Robert Jones

the function Babak suggested

the question started with ax^2 + b^x + c and he suggested a third grade polynomial

awllower

I think the trick is that the derivative of that function will be the original function.

David Robert Jones

11:48 AM

how is that related to the question? what is the relationship between the question and a function whose derivate is the same as the original function?

awllower

When the derivative is 0, you get a solution to the original one.

Since f(0)=f(1), there is a point in the interval (0,1), with derivative 0; this is what we want exactly!

user19161

@DominicMichaelis There are many downvotes on correct answers and upvotes on wrong answers here...

Dominic Michaelis

oh ok

user19161

I am not talking about careless mistakes we all make.

user19161

I am talking about voters who have poor judgment because they have little mathematical ability.

Frank Science

11:57 AM

@JacobBlack Yeah. In fact, some thoroughly wrong answer, which is wrong ON PURPOSE, gets some upvotes.

user19161

In fact earlier today, three correct answers on the same question each got a downvote for no reason. No comments at all.

user19161

Going to eat...

Frank Science

I should admit that I have accepted a wrong answer.

Alexander Gruber

ok so i got the first one.

Frank Science

But I did not accept it deliberately.

Alexander Gruber

11:59 AM

now i am working on $\oint_{|z-2|=2}\frac{\operatorname{Arctan}(z)}{z^2-1}dz$

is $\operatorname{Arctan}$ analytic on $\mathbb{C}^*$?

David Robert Jones

@awllower: thanks

awllower

NP

Orange Harvester

@JacobBlack I am sorry for that limit point I stuck to. It was careless I guess.

user58512

12:34 PM

hi

user19161

1:08 PM

@user58512 Are you better at algebra or analysis?

user58512

i dont know

awllower

Then how about your interests?

user58512

I like them both

vkeles

Do you have an idea about $\sum_{n=1}^{\infty } \frac{n! x^{n}}{3^{n}}$ ?

Dominic Michaelis

it konverges

iff x \leg 3/e

sry striclty lower 3/e

is my first idea

let me check that

vkeles

1:14 PM

Answer : for x [-2,0) diverges

user58512

the n! dominates 3^n, so x must be negative so you have an alternating sum.

Dominic Michaelis

oh i read n^n

i need glassed :D

glasses :D

user58512

@JacobBlack, how do you choose an area?

vkeles

I've tried the root test but i couldnt solve

user19161

@user58512 Meaning an area of research for yourself?

user58512

1:16 PM

yes

Dominic Michaelis

@vkeles as n! is about n^n /e^n

you find the minorant n^n/9^n x^n

vkeles

n^n =? n! @Dominic

user58512

stirlings approximation

Dominic Michaelis

its called stirling formel

user19161

@user58512 Well, I am not qualified to speak on this, because I am not even a graduate student. But I would say it depends on a few factors: interest, ability, current research topics, supervisor.

Dominic Michaelis

1:17 PM

n! is about \sqrt{2\pi n } \cdot \frac{n^n}{e^n}

vkeles

After replacing it with n^n what should i do?

Dominic Michaelis

you have about n^n /(e^n \cdot 3^n ) * x^n

as e is about 3 g you say this one is greater than n^n/9^n *x^n

so you get (nx/9)^n

where e comes from?

stirling

I didnt write e

1:20 PM

i now

i know

wait let me search something

mathbin is so slow nowadays

vkeles

Can you ask this question? I reached the limit :)

Dominic Michaelis

i know that it only converges if x=0

because if x isn't 0 the sequence isn't a 0 sequence

user58512

@vkeles, I doubt the sum converges for x < -1 though.

vkeles

:/

@u

@user58512 can you ask this?

user58512

oh good point. n! (x/3)^n does not tend to zero because n! dominates (x/3)^n

Dominic Michaelis

1:24 PM

or lets take the good old ratio test

not ratio

user58512

@vkeles, @DominicMichaelis just solved it

Dominic Michaelis

root test

but as i said, its not a sequence with limit 0

you want a rigorous proof ?

user19161

I'll give you proof! punches

Dominic Michaelis

it's funny we did have the stirling approximation in ana 2 but no other student still remebers at :D

(from my course)

awllower

Interesting article!!

jstor.org/stable/2690261

vkeles

1:29 PM

@DominicMichaelis I need a rigorous proof :/

Orange Harvester

@JacobBlack hi

Dominic Michaelis

oh just say that :)

user19161

@OrangeHarvester Wow, first time you said hi to me...

Orange Harvester

@JacobBlack :P

skullpatrol

Yay and you have never said "hi" to me...

Frank Science

1:31 PM

@vkeles See this

user19161

I don't like to say hi in a chat room.

skullpatrol

why?

user19161

Why? That's just me.

skullpatrol

but you say bye???

Orange Harvester

And huzzah.

user19161

1:32 PM

You must understand that people can be very different from one another in small things like that.

user19161

We don't need to read too much into these things.

skullpatrol

Oh but we do need to be consistent

Orange Harvester

My computer tells me "your battery is very low, either plug it up or shut it down" :P

Frank Science

I usually act just as a phantom without saying hi or bye.

user19161

Well, we can be inconsistent with what we do too, no problem.

user19161

1:34 PM

Today I feel like drinking tea, tomorrow coffee.

Orange Harvester

@FrankScience That's frank.

I feel like eating cake. (I am eating cake)

user19161

This week pepsi, next week coca cola.

skullpatrol

Ok Ok bye :''(

consistently bipolar

user19161

@ora Where do you get your definition of limit point from?

Orange Harvester

@JacobBlack I was mistyping my limit point as limit when I meant the point which is a limit. Major brainfade.

I get my "correct" definition from Simmons by the way. A point x is a limit point of set A if every open sphere centered on the point x has atleast one point from A which is not x.

user19161

1:39 PM

@OrangeHarvester OK, same as Rudin then.

Orange Harvester

@JacobBlack Yes. Everyone has the same definition I suppose.

user19161

@OrangeHarvester More or less. In a metric space, this is equivalent to every open ball centred at x has infinitely many points of A.

Orange Harvester

@JacobBlack Yes.

user58512

I need to get to work soon

peoplepower

Work avoidance lemma: If you and some friends each avoided work on some day then the reason you still your jobs is that some guy quit.

Dominic Michaelis

1:54 PM

this question is a nice one math.stackexchange.com/questions/309073/…

Frank Science

@JacobBlack It's true according to a $T_1$ space.

vkeles

Can anyone ask this question for me? --> \sum_{n=1}^{\infty} \frac{(x-4)^{n^{2}}}{n!}

Orange Harvester

@peoplepower that or.... your manager also took an off day.

awllower

Why not ask yourself?

vkeles

@awllower I reached the limit

Dominic Michaelis

1:58 PM

@vkeles konvergent for all x

its the same as exp((x-4)^2

ah not the same

user58512

this is great math.stackexchange.com/questions/308683/…

complex analysis

vkeles

@DominicMichaelis I need to see the solution so can you ask this?

Frank Science

@DominicMichaelis WLOG, consider $f(x)=0$ for $x\neq c_k$, $k=1,2,\dotsc$

awllower

I did not know there was a limit for asking question.

Orange Harvester

@vkeles The limits are there for a reason. Unless it is very urgent, you might want to post it in around 10 hours?

vkeles

2:02 PM

@OrangeHarvester I have a midterm,its urgent for me :/

Orange Harvester

@vkeles Apply root test.

vkeles

@OrangeHarvester I've tried but I couldnt solve

Dominic Michaelis

@vkeles writ it on mathbin

and post it t ous

vkeles

@DominicMichaelis \sum_{n=1}^{\infty} \frac{(x-4)^{n^{2}}}{n!}

Dominic Michaelis

na waht you tried i mean

Frank Science

2:08 PM

It's unnecessary to post on mathbin.

We can use ChatTeX.

See the bulletin right.

$\LaTeX$ support for chat.

vkeles

Question : $\sum_{n=1}^{\infty} \frac{(x-4)^{n^{2}}}{n!}$

Tobias Kildetoft

@vkeles that is not a question

vkeles

@TobiasKildetoft I mean for what values of x it converges

Tobias Kildetoft

@vkeles well, what are your thoughts?

Frank Science

It's just a power series and apply Cauchy-Hadamard theorem.

user58512

2:10 PM

substitute v = x-4 first

$$\sum_{n=1}^{\infty} \frac{v^{n^{2}}}{n!}$$

vkeles

@FrankScience We didnt see this theorem

Frank Science

It's just a by-product of root test, which indicates that the convergence radius is certainly determined by root test.

Orange Harvester

@vkeles You know the root test right? Write down the expression that needs to be evaluated.

user58512

I finally managed to answer a group theory question

Orange Harvester

@user58512 coool.

off to dinner guys

bye

skullpatrol

2:19 PM

later

(bye)

;-)

vkeles

@OrangeHarvester \lim_{n\rightarrow \infty } \frac{(x-4^{n}))}{n!^{\frac{1}{n}}}

Why dont you ask this question :/ ?

amWhy

@user58512 Nice answer, too!

user58512

thanks

Tobias Kildetoft

@vkeles for future reference, you might want to take a bit longer in writing the question on the main site. Your current questions all lack some motivation and mention of what you have tried yourself. (I mention this because a couple of them have received a fair number of upvotes, which might indicate that there were no problems with them)

amWhy

@user58512 Are you going to accept my answer to your "conversation" question?

user58512

2:22 PM

yes

amWhy

@user58512 ;-)

Tobias Kildetoft

@user58512 you might want to add an "since" or similar after the "so" to make it clearer what is meant (just after the first displayed math)

Dominic Michaelis

is there a highlight for mathjax ? -.-

Tobias Kildetoft