Mathematics - 2012-07-16 (page 3 of 4)

anon

5:04 PM

call for votes

Peter Tamaroff

@anon BELEETED

anon

baleeted means deleted

Peter Tamaroff

@anon BTW, I'm trying to prove Bolzano Weierstrass in $\Bbb R$.

@anon Yeah, I know. But it is funny.

J. M.

Man, we sure have an efficient hit squad here...

anon

I just thought of a fun proof of BW I think.

Peter Tamaroff

5:10 PM

@anon Fun? Shoot.

t.b.

Just a quick hi there... I'll have to run again shortly

anon

Hey tb.

t.b.

Hey, anon, how's life?

J. M.

@t.b. hi and hello!

anon

pretty good

t.b.

5:13 PM

Great to hear!

@J.M. hello, nice gravatar you have there (as usual). :)

user19161

@PeterTamaroff What result are you using?

anon

eh, nevermind

Gigili

Hello, Tee-Bee.

J. M.

@t.b. Yes, I'm back to constant curvature surfaces...

Peter Tamaroff

@JasperLoy Monotone Convergence Theorem, which I already proved.

robjohn

5:15 PM

Wow, I step away for a bit and everybody shows up :-)

Peter Tamaroff

@Gigili Your gravatar is rather gloomy.

t.b.

@MattN. Pontryagin duality tells you that the dual of a compact group is discrete and vice versa. So, thinking of the compact group $S^1$ as $(0,1)$ is a pretty bad idea :) You wouldn't think of $S^1$ as Hilbert space, now would you? They have the same cardinality, after all.

Old John

Hi folks

t.b.

... and hi

user19161

@PeterTamaroff Oh you came up with the proof yourself?

t.b.

5:16 PM

@Gigili 'ello Gigili!

Gigili

@PeterTamaroff Umm, is it?

Peter Tamaroff

@JasperLoy I "just" used the Completeness of $\Bbb R$.

t.b.

@J.M. I noticed :) Very nice.

@J.M. Btw. does that ping you now?

Peter Tamaroff

@JasperLoy Proved that the limit of the sequence is the supremum of the set ${a_1,\dots,a_n,\dots}$ by a $\epsilon$-nbhd argument.

robjohn

@J.M. Is there a family of constant curvature surfaces that would make a nice movie by varying a parameter?

J. M.

5:19 PM

@t.b. It does now. So it seems balpha has done it...

robjohn

@JM I assume you need the periods?

t.b.

@J.M. interesting. He could have spared himself this ~~ugly~~ funky orange, though :)

J. M.

@robjohn I seem to recall there being a modification of Kuen's surface that fits your bill. I don't have my refs now, but I'll get back to you on that if I come across them.

Mark Dominus

Oh, hey! Now I see who anon is.

J. M.

@robjohn Yes, that didn't ping me

t.b.

5:21 PM

$\mathbb{Z}_2$?

anon

@MarkDominus can you tell me?

J. M.

@t.b. There certainly are better highlighting colors than orange... :)

Mark Dominus

Are you sure you want me to spoil the surprise for everyone else?

robjohn

@J.M. Thanks. Too bad gravatars can't animate :-)

J. M.

@robjohn That's be too distracting... :D

anon

5:22 PM

@MarkDominus Wait, you seriously know (of) me?

t.b.

@anon I was assuming Mark was referring to this quite the multi-persona-anon

robjohn

@anon hopefully, not in the biblical sense...

anon

oh.....

Mark Dominus

@t.b. That is exactly what I meant.

J. M.

@MarkDominus His family is the colorful sort, no? :)

user19161

5:23 PM

@robjohn Geezis.

t.b.

@anon pheew..., right? :)

anon

I live to see another day.

J. M.

@Mark, you sure raised anon's blood pressure right there... ;)

robjohn

@JasperLoy I see that anon inserted the (of) into his question :-)

Mark Dominus

You are Lev Pontryagin and I claim my five pounds!

robjohn

5:26 PM

@J.M. Yeah, he does try so hard to stay, um, hidden, obscure, unknown... what is the word I am looking for...

3

user19161

@robjohn I was also wondering about the meaning of "went into her" in the bible. It seems that it does not refer to "knowing" but to entering the tent of.

J. M.

@robjohn Reminds me of this German friend I had: "I wish there was a word to describe the pleasure I feel at your misfortune."

t.b.

@robjohn good ol' robjohn humor :) hey there! Unfortunately I have to go now. So, hi and bye.

Bye all

J. M.

See you, @t.b.

robjohn

@t.b. Nice to see you :-) see you when you get back.

user19161

5:31 PM

@J.M. Is that word schadenfreude?

robjohn

:5385246 as $n\to\infty$

Peter Tamaroff

@robjohn OK.

I leave that implicit for sequences.

Matt N.

NO!!

NOOOOO. :,(

Peter Tamaroff

@MattN. ?

Matt N.

I missed the teddy :,(

robjohn

5:33 PM

@PeterTamaroff or $\inf\limits_n\sup\limits_{k>n}a_k$

J. M.

@JasperLoy You tell me, Jas. You tell me. :)

Gigili

@t.b. Have fun.

Matt N.

That's more troll than I can take.

Peter Tamaroff

@robjohn Why is that equivalent to the other?

J. M.

@MattN. Not your day... :( maybe next time.

robjohn

5:33 PM

@PeterTamaroff think about it

Peter Tamaroff

@robjohn Hehehe, OK. =D

Matt N.

@t.b. I was afk. I missed you so much and then you appear in chat and I miss you. That's almost like trolling. Oh well. I would not have had anything interesting to say anyway.

robjohn

@PeterTamaroff It's always nice to define the $\limsup$ in terms of an $\inf$

Peter Tamaroff

@robjohn Its mildly confusing.

Matt N.

@J.M. </3

robjohn

5:35 PM

@PeterTamaroff but it can be useful.

user19161

@PeterTamaroff Rudin also has a very weird definition of the lim sup and lim inf.

robjohn

@JasperLoy I thought that was the one I was using there.

Matt N.

@t.b. Bye troll bear. x

@J.M. I noticed that chat messages don't ping me anymore unless I'm in the chat room. Is that just me or is that a recent change?

Peter Tamaroff

@JasperLoy Well, now I want to prove that $v_n=\inf \{a_n,\dots\}$ is monotone increasing and that $u_n=\sup \{a_n,\dots\}$ is monotone decreasing.

J. M.

@MattN. I'm not sure. It seems balpha has just made a number of changes to the chat software today, so I don't know if that was part of what was changed.

Jonas Teuwen

5:39 PM

Son of a cow! I missed tb.

2

user19161

@PeterTamaroff Quite trivial...

J. M.

@JonasTeuwen That gives you and Matt something in common... oh well.

Peter Tamaroff

@JasperLoy I know. I mean, I know it is true. I short "word argument" explains why.

Jonas Teuwen

8-).

user19161

@JonasTeuwen Holy monkey!

Peter Tamaroff

5:40 PM

But I want to make a good proof.

robjohn

@PeterTamaroff what happens to an inf when taken over a smaller set? or a sup when taken over a smaller set?

user19161

@MattN. It might be the internet connection or your browser too. I have problems with chat in Chrome but not in FF.

Peter Tamaroff

@robjohn It is either the same, or the element taken out was the $\inf$.

Mark Dominus

@J.M. My co-worker was once asked my his German grandmother "Rik, how do you say in English, when you feel happy, because someone else is sad?"

unNaturhal

Excuse me if I'm annoing again... If a function has as domain $\forall x \in \mathbb{R}\backslash\{0\}$, and in $x=0$ I restored the continuity putting
$$y = f(x) \text{ } \forall x \in \mathbb{R}\backslash\{0\}$$and
$$y = 0 \text{ for } x = 0$$
Could we say that the functions is now differentiable in $x=0$?

user19161

5:42 PM

@MarkDominus Sadism.

Jonas Teuwen

@MarkDominus Schadenfreude? 8-).

robjohn

@PeterTamaroff so the inf increases and the sup decreases as the set decreases.

J. M.

@MarkDominus :D Beautiful.

robjohn

@unNaturhal that depends on how $f$ behaves near $0$

Jonas Teuwen

@JasperLoy Schadenfreude is more subtile. It is "epicaricacy".

user19161

5:44 PM

@JonasTeuwen I learnt schadenfreude recently on ELU!

unNaturhal

@robjohn It's sx and dx limits give 0, and $f(0) = 0$

Jonas Teuwen

Holy cow, someone made his exam using a purple marker 8-).

robjohn

@unNaturhal that means it is continuous, but how about $f(x)/x$?

Peter Tamaroff

@robjohn Yeah, I wrote that. It seems just an "observational" proof, let's say.

user19161

@JonasTeuwen Made? You mean wrote his answers in?

Jonas Teuwen

5:45 PM

Yeah.

Peter Tamaroff

@JonasTeuwen He should be crucified.

user19161

@JonasTeuwen Holy cow! Clearly a nutcase.

J. M.

@JonasTeuwen I've once had to check a paper where the student used a glittery pink ink for her answers...

Jonas Teuwen

He got a 2.5 is that also okay?

unNaturhal

@robjohn $\displaystyle\lim_{x\to\infty}{\frac{f(x)}{x}} = 1$

robjohn

5:46 PM

@unNaturhal near $0$?

Matt N.

@JonasTeuwen Same here.

unNaturhal

@robjohn What do you mean?

Jonas Teuwen

@MattN 8-)).

Peter Tamaroff

@unNaturhal You're saying $y=x$ is an oblique asymptote?

@JonasTeuwen Yes.

robjohn

@PeterTamaroff if $A\subset B$ then $\sup\limits_{A}f\le\sup\limits_{B}f$ and $\inf\limits_{A}f\ge\inf\limits_{B}f$

unNaturhal

5:48 PM

@PeterTamaroff Nope, sorry. $y = x - 1$ is the oblique asymptote

Matt N.

@JonasTeuwen Hey Jonas. Do you know how to obtain beta blockers?

user19161

@J.M. I wrote Happy New Year at the end of all my papers one year when I took the exams in December!

Matt N.

I think I need. These exams are really important to me.

J. M.

@MattN. You have a heart condition now?

Jonas Teuwen

@MattN GP.

@MattN. But they mess with your memory...

Matt N.

5:49 PM

@J.M. Actually I don't know it it counts. But I wake up early and have a heart race because I'm stressed out.

Jonas Teuwen

Or at least propranolol does.

robjohn

@unNaturhal you gave the limit near $\infty$

J. M.

@JonasTeuwen That's the cheapest out of the lot of them, though.

Peter Tamaroff

@MattN. Then avoid coffee for one thing...

Matt N.

@JonasTeuwen In what way? If it won't let me remember theorems then that's bad during the exam.

J. M.

5:50 PM

@MattN. No, not that drastic.

user19161

@MattN. Try to regulate your breathing too.

Jonas Teuwen

Hmm, usually I have a near perfect recall for many things, and then my memory seemed so... normal.

J. M.

(In any event, I wouldn't recommend taking β-blockers willy-nilly.)

unNaturhal

@robjohn Wait, do you want to know $\displaystyle\lim_{x\to0}{\frac{f(x)}{x}}$? And what it represents?

Matt N.

@J.M. Meaning: if I take one before the exam the beneficial effects will probably outweigh? The answer must be yes since there are students taking them for exams.

J. M.

5:52 PM

@MattN. Just don't take it for every exam. One pill per day ought to do...

Matt N.

@J.M. My plan is to take 2 in total: one before the CA exam and another one before the FA exam.

Now I need a GP.

J. M.

@MattN. Those are on separate days, I hope? Are you taking anything else at the moment?

Matt N.

I'm never sick and the old lad that I go to for vaccinations might not give me beta blockers...

robjohn

@unNaturhal well, it represents the derivative of $f$ at $0$ (since $f(0)=0$)

user19161

@MattN. You may see a different doc for a different opinion too.

Matt N.

5:54 PM

@J.M. No except alcohol. : ) Well and sometimes I have a headache then I take some paracetamol. But that's like twice a month or so.

J. M.

@MattN. Yeah... don't drink on those days you're on propranolol.

Matt N.

@J.M. One is on the 9th of August and the deadly one is on the 20th of August. So: yes : )

@J.M. Obviously not. Although if I cannot get pills then I might try alcohol.

unNaturhal

@robjohn LOL This is the limit of the incremental.. :p

Matt N.

I'm quite sure that the negative impact is more than cancelled by its beneficial effects.

J. M.

@MattN. I know. I was just saying those two interact not too well in the liver, so don't have them on the same day.

Matt N.

5:56 PM

Then chewing gum so the lecturer won't notice...

@J.M. Oh, thanks for pointing that out. I'm quite likely to have a drink after the exam. But if I can get those pills then I won't.

unNaturhal

@robjohn So it give 0 :P

Matt N.

@J.M. I think I need a beta blocker to get over the fact that I missed troll bear.

robjohn

@unNaturhal what is the function?

unNaturhal

@robjohn $$y = e^{-\frac{1}{|x|}}$$

robjohn

@unNaturhal Ah, yes, that is $C^\infty$

Gustavo Bandeira

6:01 PM

A zero of a polynomial $p(t)$ is any number $r$ for which $p(r)$ takes the value $0$. What does $r$ stands for? The set of the real numbers?

J. M.

@GustavoBandeira $r$ is any complex number, if the coefficients of your polynomials are complex.

user19161

@GustavoBandeira In this case r is just the number you substitute into the expression.

robjohn

@MattN. Do you mean Trolli Bear?

unNaturhal

@robjohn o.o

Matt N.

@robjohn : )

robjohn

6:04 PM

@MattN. For $8 US, you can get 5 pounds :-)

Gustavo Bandeira

@J.M. Wouldn't it be better to use $C$ for complex numbers?

Jonas Teuwen

Done. Grading. Yey.

J. M.

@GustavoBandeira Well, I suppose, but it's the generic variable you were using...

Peter Tamaroff

@JasperLoy DO you think I can use the theorem that given a subset $A$ of a metric space $X$ and a point $x\in X$, then there is a sequence of points in $A$ that converges to $d(x,A)$?

Matt N.

@robjohn Did you mean gain 5 pounds for 8$? : )

Gustavo Bandeira

6:06 PM

@JasperLoy Yep. I'm just not very sure why the $p(t)$ became $p(r)$

J. M.

@MattN. That too.

user19161

@GustavoBandeira r is just a dummy letter. One can use almost any letter there really.

J. M.

@GustavoBandeira The $r$ there is just any of the $n$ complex numbers that make your polynomial of degree $n$ zero.

user19161

@JonasTeuwen Congrats. Have a bottle of whisky!

Peter Tamaroff

@robjohn To prove that there is a subsequence of $a_n$ that converges to $\lim \sup\limits_{k \geq n} {a_k}$

Jonas Teuwen

6:07 PM

@JasperLoy Yea...

Gustavo Bandeira

@JasperLoy Ok, thanks.

@J.M. Thanks

Matt N.

@J.M. I'm not worried, I was just joking. : )

user19161

@GustavoBandeira The author in that context wanted to distinguish the polynomial say $1+t+t^2$ from the actual concrete value you substitute for $t$. That's all.

J. M.

Well, I should be going. See y'all later.

Jonas Teuwen

@J.M. Babai :-)))).

Peter Tamaroff

6:13 PM

@robjohn Silly me, that follows from the MCT.

user19161

@PeterTamaroff What are you trying to prove now?

Peter Tamaroff

@JasperLoy Bolzano Weierstrass in $\Bbb R$.

user19161

@PeterTamaroff And you already proved that every sequence in R has a monotone subsequence?

Peter Tamaroff

@JasperLoy No.

user19161

@PeterTamaroff I thought you said you did just now?

user19161

6:15 PM

Myabe I misunderstood your monotone theorem thingie.

Peter Tamaroff

@JasperLoy MCT is Monotone convergence theorem. A bounded montone sequence in $\Bbb R$ is convergent.

Matt N.

@J.M. See you later!

Peter Tamaroff

Now with that and using

$\limsup$ and $\liminf$ I have to prove Bolzano Weierstrass

This is, prove that given a bounded sequence, then it contains a convergent subsequence.

See this

user19161

@PeterTamaroff Ah since you already have MCT, you just need to show that every sequence in R has a monotone subsequence and you are done.

Peter Tamaroff

It is EXERCISE 5

@JasperLoy Could that be proven by a simple "ordering" argument?

user19161

6:20 PM

@PeterTamaroff Oh I see, so that gives another proof of the BW.

user19161

@PeterTamaroff It happens that this approach is found here. en.wikipedia.org/wiki/Bolzano%E2%80%93Weierstrass_theorem

Peter Tamaroff

@JasperLoy The "peaks" approach?

user19161

@PeterTamaroff Yes, the link is one proof of BW. Your exercise is another proof of BW. Two proofs.

Peter Tamaroff

@JasperLoy Right. Well, I need to see how to wrok out my proof now... ?XD

I have to show there exist subsequences of $\{a_n\}$ that converge to $\lim \sup_{k \geq n}\{a_k\}$ and to $\lim \inf_{k \geq n}\{a_k\}$

I have that $u_n= \sup_{k \geq n}\{a_k\}$ and $v_n=\inf_{k \geq n}\{a_k\}$ are monotone decrasing and increasing respectively so by the Completeness of $\Bbb R$ I have that

$\lim u_n= \inf \sup_{k \geq n}\{a_k\}$ and that $\lim v_n= \sup \inf_{k \geq n}\{a_k\}$

Maybe I should ask on main....

user19161

@PeterTamaroff I think you should spend more time thinking about it first...

Peter Tamaroff

6:31 PM

@JasperLoy I think I understand what $v_n$ and $u_n$ do.

They "scoop" through the tail of the sequence and find the peaks and the mins

user19161

@PeterTamaroff Drawing pictures or visualizing things in your mind helps greatly in proving analysis stuff.

Peter Tamaroff

So if the $\inf$s and the $\sup$s at each $n$ were elements of $a_n$ then that would find the required subsequence.

But it might be the case $\inf$ or $\sup$ are not in the sequence.

Henry T. Horton

$\{u_n\}$ and $\{v_n\}$ might not be subsequences though

Peter Tamaroff

@HenryT.Horton That's what I'm saying

user19161

@PeterTamaroff In which case you have to use the definition of the inf and sup to pick a point sufficiently close to its value.

Peter Tamaroff

6:40 PM

@JasperLoy I'm at loss.

Henry T. Horton

►

Peter Tamaroff

@HenryT.Horton I have an idea though.

user19161

@PeterTamaroff Maybe think about it over a day first before posting on main... Just a suggestion.

Peter Tamaroff

@JasperLoy I'm thinking about it now

user19161

You won't improve unless you try to push yourself.

user19161

6:46 PM

Of course depending on the difficulty one thinks the problem has and the ability one has to solve it, the waiting time will vary.

user19161

@PeterTamaroff And also you don't have to get it immediately. You can do other things first and come back to it later too.

Peter Tamaroff

@JasperLoy I have an idea, but I still haven't materialized it enough even to explain it to you.

user19161

@PeterTamaroff I think you should try to explain it to yourself first. I have the impression you are very eager to write things down before the thought crystallizes, but you should distill the thoughts more before expressing yourself.

Peter Tamaroff

@JasperLoy I think the theorem can follow from this:

Henry T. Horton

I have to go to Analysisfest 2k12, LaTeRz

Peter Tamaroff

6:53 PM

@JasperLoy THEOREM 5.9 Given a metric space $(X,d)$, let $A\subset X$ with $A\neq \varnothing$. Let $a\in X$. Then there is a sequence of points in $A$, $\{a_n\}$, such that $\lim d(a,a_n)=d(a,A)$

Given $(\Bbb R, d)$ with $d(x,y)=|x-y|$, we have that the bounded sequence $\{a_n\}$ is non-empty and is a subset of $\Bbb R$.

user19161

@HenryT.Horton Sounds wonderful. Have some whisky there!

Matt N.

@JonasTeuwen It sucks monkeyballs. : )

Peter Tamaroff

Sorry, I was talking stupid.

user19161

@HenryT.Horton BTW, I really hate this typing sTyLe.

user19161

@PeterTamaroff Is this Mendelson?

Peter Tamaroff

7:09 PM

@JasperLoy Yes.

@JasperLoy Why do you ask?

user19161

@PeterTamaroff I find it weird that he includes this calculus exercise in a topology book, quite a nut case I must say.

Peter Tamaroff

@JasperLoy "nut case"?

But Bolzano Weierstrass is pretty important in topology, isn't it? It it about "sequential compactness" or something of the sort.

user19161

@PeterTamaroff Well, "nutcase" is one of my favorite words. I like to use it liberally!

Peter Tamaroff

@JasperLoy What does it mean?

XD

Old John

@PeterTamaroff I would agree that Bolzano Weierstass is pretty much topological

user19161

7:12 PM

@PeterTamaroff I mean that this kind of thing that you are trying to do now should already be done in another course. It is only slightly related to the issue at hand.

Old John

Surely compactness is one of the most important basic concepts in topology

Peter Tamaroff

@JasperLoy Steven Abott uses the Nested Interval Property to prove BOlzano Weierstrass.

user19161

@PeterTamaroff My favourite proof is the one I linked to in the Wiki article. It is nice and sweet.

robjohn

@PeterTamaroff I am confused about where this came from. The link refers to a comment about gummi bears.

Peter Tamaroff

@robjohn Oh darn. I misslinked again

user19161

7:26 PM

@PeterTamaroff I thought you just used the last message of X to ping X randomly!

Peter Tamaroff

I was saying that the fact that $\lim \sup_{k \geq n}\{v_k\}=\inf \sup_{k \geq n}\{v_k\}$ follows from MCT

Mark Dominus

What's good to read if I want to learn a little bit about Pontryagin duality, dual groups, and theory of Fourier transforms on arbitrary groups?

Wikipedia has a short bibliography, but it is not descriptive.

Old John

@MarkDominus I once tried reading Rudin (Fourier Analysis on Groups) - but didn't get past the first chapter :(

Mark Dominus

Wikipedia listed that, but I guessed it might not be a great place to start a self-study. :)

Old John

although it is supposed to be a bit of a classic

Peter Tamaroff

7:35 PM

@MarianoSuárez-Alvarez Insists self-study is utopical. =D

Mark Dominus

"Classic" means that it has been hated by generations of students!

Old John

@MarkDominus Yep! - good point

user19161

@MarkDominus I don't know anything about these matters, but Folland and Hewitt/Ross come to mind.

user19161

And I can say that Rudin, Folland, Hewitt are all great writers!

Mark Dominus

Bleh, my network connectivity is too poor for this.

Henry T. Horton

7:40 PM

Hi JaSpEr

user19161

@HenryT.Horton Hi! I noticed your style again...

Henry T. Horton

JuSt 4 U

user19161

@HenryT.Horton I am actually beginning to like it now that you use it.

Henry T. Horton

Thanks sweetie

Peter Tamaroff

for any $a\in \Bbb R$

Henry T. Horton

7:44 PM

No, let $a = y$

Peter Tamaroff

:5387452

►

@HenryT.Horton I have to prove that given $A\subset X$ where $(X,d)$ is a metric space, the function $f:X\to \Bbb R$ defined by $f(x)=d(x,A)$ is continuous. I already proved that the triangle inequality holds for distance to subsets of metrics, that is $x,y\in X$ and $A\subset X$ then $d(x,A)\leq d(x,y)+d(y,A)$. So now I intend to use that to prove that $|d(x,A)-d(a,A)|<\epsilon$ whenever $d(x,a)<\delta$ for some $\delta$.

robjohn

@PeterTamaroff where were you attempting to link? Did you want a $\delta$-$\epsilon$ proof of that fact?

Peter Tamaroff

@robjohn I was just telling you I understood why the definitions of $\limsup$ we were discussing aer equivalent. =)

@HenryT.Horton So I argue $$\left| {f\left( x \right) - f\left( a \right)} \right| = \left| {d\left( {x,A} \right) - d\left( {a,A} \right)} \right| \leqslant \left| {d\left( {x,a} \right) + d\left( {a,A} \right) - d\left( {a,A} \right)} \right| = d\left( {x,a} \right)$$

But that is wrong by what you just noted.

robjohn

@PeterTamaroff Ah, and here I was trying to find a way to explain why such a sequence always exists. Whew

Peter Tamaroff

@robjohn What were you trying to explain?

Henry T. Horton

7:57 PM

Piotr

Peter Tamaroff

@HenryT.Horton Yes?

Henry T. Horton

@PeterTamaroff By your triangle inequality, you have $d(x,A) \leq d(x,y) + d(y,A)$ and $d(y,A) \leq d(x,y) + d(x,A)$

Which together imply that $|d(x,A) - d(y,A)| \leq d(x,y)$

Right?

Peter Tamaroff

@HenryT.Horton Right.

Henry T. Horton

So let $\delta = \varepsilon$ and you're good

Milosz Wielondek

Hey guys, what am I missing here -> how do you go from $x=\frac{\sqrt{6}-\sqrt{5}}{\sqrt{6}+\sqrt{5}}$ to $x=11-2\sqrt{30}$?

Peter Tamaroff

8:02 PM

@MiloszWielondek Do you know what the conguate of a number is?

Milosz Wielondek

@PeterTamaroff Yes.

oh, wait...

Peter Tamaroff

@MiloszWielondek Well, then multiply and divide what you have by the conjugate of the denominator.

Henry T. Horton

$d(x,y) < \varepsilon \Rightarrow |d(x,A) - d(y,A)| < \varepsilon$

Milosz Wielondek

@PeterTamaroff Of course.. thanks!

Peter Tamaroff

@MiloszWielondek ;)

@HenryT.Horton This one is easy, but enlightening: "Prove that $d(x,A)=0$ with $A\subset X$ and $x\in X$ if and only if every neighborhood of $x$ contains a point of $A$."

Chuck Fernández

8:29 PM

what is the value of $$\sum\limits_1^{99} \frac{1}{(n)\sqrt{n+1}+(n+1)\sqrt{n}}$$

?

Henry T. Horton

8:48 PM

@ChuckFernández $9/10$

It's a telescoping sum

$$\frac{1}{n\sqrt{n+1} + (n+1)\sqrt{n}} = \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}}$$

robjohn

3 hours ago, by Peter Tamaroff

@robjohn To prove that there is a subsequence of $a_n$ that converges to $\lim \sup\limits_{k \geq n} {a_k}$

Chuck Fernández

How did you get that? @HenryT. Horton

@HenryT.Horton

Peter Tamaroff

@ChuckFernández Some conjugate trickery, he made. Mmhh

@robjohn I will post some lhf in main

Chuck Fernández

9:07 PM

I don`t see how you found out $$\frac{1}{n\sqrt{n+1} + (n+1)\sqrt{n}} = \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n+1}}$$

Peter Tamaroff

@ChuckFernández Let me help.

Chuck Fernández

ok

robjohn

$$
\begin{align}
\frac{1}{n\sqrt{n+1}+(n+1)\sqrt{n}}
&=\frac1{\sqrt{n(n+1)}}\frac{1}{\sqrt{n}+\sqrt{n+1}}\\
&=\frac1{\sqrt{n(n+1)}}\frac{\sqrt{n+1}-\sqrt{n}}{(n+1)-n}\\
&=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n(n+1)}}\\
&=\frac1{\sqrt{n}}-\frac1{\sqrt{n+1}}
\end{align}
$$

Peter Tamaroff

$$\frac{1}{{\left( {n + 1} \right)\sqrt n + n\sqrt {n + 1} }} = \frac{1}{{\sqrt n }}\frac{1}{{\sqrt {n + 1} }}\frac{1}{{\sqrt n + \sqrt {n + 1} }}$$

robjohn

@PeterTamaroff for the lim sup?

Peter Tamaroff

9:14 PM

@robjohn Nay, another excercise. I gave up in Bolzano Weierstrass for the moment.

Old John

@PeterTamaroff Hi Peter - still working through Mendelson?

Peter Tamaroff

@OldJohn Yes. Posted something just now.

$$\eqalign{
& \frac{1}{{\left( {n + 1} \right)\sqrt n + n\sqrt {n + 1} }} = \frac{1}{{\sqrt n }}\frac{1}{{\sqrt {n + 1} }} \cr
& \frac{1}{{\sqrt n }}\frac{1}{{\sqrt {n + 1} }}\frac{1}{{\sqrt n + \sqrt {n + 1} }}\frac{{\sqrt n - \sqrt {n + 1} }}{{\sqrt n - \sqrt {n + 1} }} = \frac{1}{{\sqrt n }}\frac{1}{{\sqrt {n + 1} }}\frac{{\sqrt n - \sqrt {n + 1} }}{{ - 1}} \cr
& \frac{1}{{\sqrt n }}\frac{1}{{\sqrt {n + 1} }}\frac{{\sqrt n - \sqrt {n + 1} }}{{ - 1}} = \frac{{\sqrt {n + 1} - \sqrt n }}{{\sqrt n \sqrt {n + 1} }} = \frac{{\sqrt {n + 1} }}{{\sqrt n \sqrt {n + 1} }} - \frac{{\sqrt n …

AAAAAAAAAAAAA somemany $n$s!

Old John

@PeterTamaroff Eeeek - might have to look at that in the morning - had a bit too much wine tonight to sort through all those n's!

Oh - you mean on the main site ...

Peter Tamaroff

@OldJohn Yes. That above is for Mr. Fernández.

Old John

@PeterTamaroff OK - just worked that out - I'm a bit slow tonight

Peter Tamaroff

9:21 PM

@robjohn Rob, what is Norbert trying to tell me here?

leo

hi

I have $Y$ a closed subspace of $X$ and $f$ a continuos function on $Y$ with compact support on $Y$. Is there a continuous extention $F$ of $f$ continuos on $X$ and with compact support on $X$?

@robjohn

Peter Tamaroff

@leo Can't help you there!

leo

@JonasTeuwen

robjohn

@leo can't you just extend it to $0$ on $X\setminus Y$?

Chuck Fernández

wow, thanks where did you learn to do that?

leo

9:32 PM

@robjohn but then the extention can be dscontinuous at the boundary of $F$