Mathematics - 2014-10-07 (page 2 of 2)

sdf

18:04

If $f\colon X\rightarrow Y$ is a closed map of top. spaces and $U\subset X$ then is $\overline{f(U)}=f(\overline{U})$? I guess this is probably obvious?

Jasper Loy

@JayeshBadwaik Cinlar's book seems good, but I really like Oliver Knill's book Probability and Stochastic Processes which is also on his website.

sdf

Okay I figured it out... (it's false)

Tharindu

Can I ask a qucik question? :)

Jasper Loy

Just ask. Never ask to ask.

You may not be answered though.

Tharindu

No I asked if anyone is going to answer :p

Atul Gangwar

18:19

Somebody help ! math.stackexchange.com/questions/962497/…

Studentmath

18:30

@Tharindu we can't know before we know the question

Pedro Tamaroff

@Tharindu You already did.

Jasper Loy

@PedroTamaroff I capped today, lol

Studentmath

@Pedro how goes it?

Pedro Tamaroff

@Studentmath Pretty darn good.

Jasper Loy

@Studentmath is ignoring me lol

Pedro Tamaroff

18:32

*looks at questions
*judges Jasper

Studentmath

@Jasper I never ignore you!

I hate trying to guide someone to an answer via comments, and than someone pops in with the full answer

Jayesh Badwaik

@PedroTamaroff Wassup?

@JasperLoy I see. I am sticking to Cinlar for now.

Pedro Tamaroff

@JayeshBadwaik Not much. Going to tenis in a while, hoping to math at night.

Jayesh Badwaik

@PedroTamaroff Good, good. What are you doing in math now?

Pedro Tamaroff

@JayeshBadwaik Complex analysis.

Jayesh Badwaik

18:37

@PedroTamaroff I'm doing that too.

Nice, what are you doing in that?

Pedro Tamaroff

NO WAI

Jayesh Badwaik

Yes way!

Pedro Tamaroff

WAI WAI

What book are you using?

Jayesh Badwaik

Stein Shakarchi, Gamelin and Rudin

mostly Gamelin and Rudin

You?

Pedro Tamaroff

Oh. I'm using one book only. Remmert's.

Hopefully I'll get to read the second volume too.

I'll probably check out Lang rand Rudin.

Jayesh Badwaik

18:39

Well, its like our prof likes the order of topics in Stein, but the exposition in Gamelin, and I think Rudin is swell, so kinda three.

Where are you now?

Pedro Tamaroff

@JayeshBadwaik I'm jumping from chapters to chapters. Around chapter 7 I think.

Looking at the "fundamental theorems" like the identity principle, maximum modulus, minimum modulus, local minimum, &c.

Jayesh Badwaik

Ahh, okay I see. I am into Weierstrass Products now. I don't see a second volume for Remmert.

Pedro Tamaroff

Well, it is called "Classical Topics in Function Theory."

It is the continuation of the second one.

Jasper Loy

@PedroTamaroff Is it the analytic continuation?

Pedro Tamaroff

@JasperLoy No, they don't intersect.

Jasper Loy

18:43

@PedroTamaroff You mean the first one.

Pedro Tamaroff

YAS

Jayesh Badwaik

Ahh, okay, my remaining month of semester will now contain some of the second part now.

Jasper Loy

The best books on complex analysis I think are Freitag's 2 volumes

Jayesh Badwaik

some of those we've already done, Riemann Mapping and Montel and so.

Jasper Loy

@JayeshBadwaik You should go to the US and do your doctorate after this.

Jayesh Badwaik

18:46

@JasperLoy Okay.

Jasper Loy

@JayeshBadwaik Maybe I will see you there lol

Jayesh Badwaik

@JasperLoy May be.

Tharindu

@StudentMath I know I know ;)

@Pedro Yep I finished posting it

Chris's sis

19:01

@TheGame It's $\zeta(-1)=-\frac1{12}$. Note that series representation of Riemann zeta function only works for $s>1$.

The Game

@Chris'ssis Well, $1+2+3+...=1/12$ isn't that false at all

You, who likes to think out of the box, should understand well :D

Jasper Loy

I think I should put on a new avatar...

robjohn

19:29

@TedShifrin convolution and the Fourier transform work wonders

@TheGame you keep getting the sign wrong.

2 hours ago, by robjohn

@TheGame it's $\zeta(-1)=-\frac1{12}$

The Game

@robjohn Yeah i noticed that but it was too late to edit

Jasper Loy

@robjohn I should try using them on my mental problems.

robjohn

@JasperLoy There are many problems that it helps... I guess it couldn't hurt.

@TheGame it is false unless you say that you are using some regularization.

The Game

@robjohn It just depends of how one defines infinite summation, doesn't it ?

Jasper Loy

@TheGame Well, there is only one way to define it, lol

The Game

19:33

@JasperLoy No ?

@JasperLoy Usually we define it as the limit of the partial sums

@JasperLoy But that is just one chosen definition

@JasperLoy Other exists

Jasper Loy

@TheGame Thanks for the 4 pings. My name now looks impressive

The Game

@JasperLoy ? @JasperLoy Ping what @JasperLoy ping ? @JasperLoy What are you talking about ? @JasperLoy

Jasper Loy

Oh @thegame I forgot, how old are you?

The Game

16 (@Chris'ssis ^^)

Jasper Loy

Ah, I am about twice your age.

The Game

19:37

:-)

Jasper Loy

@TheGame Don't forget me when you win the Fields medal OK?

The Game

@JasperLoy How would that happen :/

Jasper Loy

@TheGame I will be the janitor at the ceremony, lol

The Game

@JasperLoy I'm in the PC (physics chem) section of prep schools, not in MP (math physics)

Jasper Loy

@TheGame Ah, then Nobel Prize then!

The Game

19:38

Uuuh

@JasperLoy Go for the Janitor prize ? :P

Jasper Loy

LOL

OK I am going to think of a new pic to use.

The Game

Another blue ?

Jasper Loy

No, I will use a photo of a superstar.

The Game

@JasperLoy You can join @Chris'ssis 's fan club and use a modified superman pic :D

Jasper Loy

@TheGame You seem in love with her, lol.

The Game

19:47

@JasperLoy $\huge>:c$

Jasper Loy

@TheGame I am in love with ... Well, that is my secret, lol

The Game

@JasperLoy That sentence was useless -__-

robjohn

@TheGame Unless you state it otherwise, it is the standard definition.

Jasper Loy

@TheGame By the way I don't understand all your smilies, lol

The Game

@JasperLoy I don't always either :P

@JasperLoy '-____-' can mean many things

Jasper Loy

19:53

@TheGame I think I should start using some too ---

The Game

@JasperLoy There's also the smiley 6IR64U B

It means 'oops my hand slipped on the keyboard'

Jasper Loy

I still can't believe I got 200 points today. That has not happened for a long time

The Game

0

Q: Suppose that A and B are nxn matrices with A row equivalent to B. Prove that if A is nonsingular, then B is row equivalent to In.

Suppose that A and B are nxn matrices with A row equivalent to B. Prove that if A is nonsingular, then B is row equivalent to In. I answered this on a test and it seemed right to me, but got zero credit. What did I do wrong? Was my logic incorrect? my answer: By definition, row equivalence me...

Isn't he right ?

Chris's sis

20:29

I just evaluated a cute integral.

What is the most marvellous way to compute this one?

$$\int_0^1\log\left(\frac{x}{2}+\sqrt{\frac{x^2}{4}+1}\right)\frac{\ dx}{x}$$

Jasper Loy

I think I will go to bed in 1 hour from now.

Not many people talking in chat today.

@Chris'ssis Do you work on your book every day?

Chris's sis

20:53

@JasperLoy Not really. For instance, today I was pretty busy, I didn't have much time to work on it.

Samuel Handwich

Can somebody help me prove an inequality?

Jasper Loy

Just ask. Never ask to ask.

Committing to a name

Noone can,

Samuel Handwich

I'm trying to prove that $P(|X|\geq t z)\geq (1-t)z^2$, given that $E(X^2)=1$, $0<z<E(|X|)$, and $0<t<1$.

($X$ is a random variable here, $P$ a probability, $E$ expected value)

I've been given the hint to consider Cauchy-Schwarz but I can't figure out how to use it.

Chris's sis

@JasperLoy How do you feel today?

Jasper Loy

21:03

@Chris'ssis I feel pretty good today. I walked for 4 hours, lol. Maybe I can go running next week.

@Chris'ssis I am trying to guess your age. I guess 25, lol.

nabla blah

How do I know if I should perform a proof by induction? The problem is to prove $j < k$ if $nj < nk$ in $\mathbb{N}$ (using only the definitions of multipication, addition, ordering, etc.) but I'm told induction doesn't help even though induction was used to prove $k = j$ if $nk = nj$

Chris's sis

@JasperLoy That's good.

@JasperLoy 25? Do you like the perfect squares? :D

Jasper Loy

@Chris'ssis Am I right? I think I am very close.

Samuel Handwich

No, she's 32.

Jasper Loy

@SamuelHandwich How do you know?

Samuel Handwich

21:11

@JasperLoy Psychic powers

Jasper Loy

@SamuelHandwich You like the Ham Sandwich theorem?

Semiclassical

terribad connection + MSE insisting on a CAPCHTA verification = 5 minutes of laptop rage

Samuel Handwich

@JasperLoy I'm told it's very nice.

Jasper Loy

@SamuelHandwich Yes, it uses top grade ham.

Semiclassical

but! @TheGame: hope you like my answer for your determinant question

The Game

21:12

@Semiclassical I upvoted it :) Now i'll have to find a good way to generalize it

Semiclassical

woohoo

i suspect the key is to find a way to write the generic case compactly

Samuel Handwich

$X=X1_{(-\infty,-tz)}+X1_{(tz,\infty)}+X1_{(-tz,tz)}$

ech that doesn't work

Jasper Loy

No emails in my inbox, I am bored, lol.

Samuel Handwich

how the hell do i relate these conditions on expectations to $P(|X|\geq tz)$

the only think I can think of that involves that are Markov's and Chebyshev's inequalities but they point the wrong way

Cauchy Scharz says $E(XY)^2\leq E(X^2)E(Y^2)$.

Khallil

Hey, all.

Samuel Handwich

21:19

so, $E(XY) \leq E(Y^2)$

But that gets me nowhere

Studentmath

What do you mean point the wrong way, Samual?

Khallil

Can Partial Differential Equations have an order like Ordinary Differential Equations do?

Studentmath

Ah I get it.

What's the exact question @Samuel?

Samuel Handwich

@Studentmath $P(|X|\geq t z) \leq \frac{E(X^2)}{t^2z^2}$ but that doesn't get me any closer to $P(|X|\geq t z) \geq (1-t)z^2$ at least in any way i can see

18 mins ago, by Samuel Handwich

I'm trying to prove that $P(|X|\geq t z)\geq (1-t)z^2$, given that $E(X^2)=1$, $0<z<E(|X|)$, and $0<t<1$.

that is it exactly

Studentmath

@Samuel will give it a look, not sure I can help but will try.

Samuel Handwich

21:22

I've also got Markov's inequality, $P(|X|\geq tz)\leq \frac{E(|X|)}{tz}$ but again doesn't go the right way

hint is to use Cauchy Schwarz but I have no idea how

$0<t<1$ so $0<(1-t)z^2<z^2$. If we could make $z^2\leq P(|X|> zt)$ we'd be in business. We know that $P(|X|\geq zt)\leq \frac{E(X^2)}{z^2t^2}=\frac{1}{z^2t^2}$.

Jasper Loy

Just post the question on the site, it will get more responses

Samuel Handwich

@JasperLoy my friend did, it didn't get any.

Studentmath

@Samuel right.

But again, all these things are in the wrong direction. It doesn't help us.

Samuel Handwich

@Studentmath they're the only inequalities I know relating $P$ with $E$

i'd just try to carry it out from the definition of $P$ but that can be really hard

Studentmath

What do you mean from the definition of P?

Huh, even the fact that $z<E(|X|)$ helps in the 'wrong' way.

Samuel Handwich

21:33

@Studentmath i mean making a sequence $X_n\rightarrow X$ of simple random variables and using $E(X_n)\rightarrow E(X)$. since each $X_n$ is simple it's $X_n=\sum_{m=1}^Mc_m1_{A_m}$ for $A_m$ in the $\sigma$-algebra, so $$E(X_n)=E(\sum_{m=1}^Mc_m1_{A_m})=\sum_{m=1}^ME(c_m1_{A_m})=\sum_{m=1}^Mc_mP(A_‌m)$$

Studentmath

Woha

Samuel Handwich

and... idk, trying to pull some kind of inequality out of the convergence of that to $E(X^2)=1$?

i have no idea it's the only other thing i can think of

Ted Shifrin

heya @Studentmath

Studentmath

@Ted!

Ted Shifrin

Weird how much probability shows up in here now that I'm supposed to be learning it :P

@khallil: Of course

Studentmath

21:36

It always works that way :P

Samuel Handwich

@TedShifrin have any idea how to prove this?

Studentmath

@Samuel as a mattar of fact, all the inequalities I know on P work the 'wrong' way. I'm afraid it just might be the way you suggested.

Samuel Handwich

@TedShifrin It's that Baader-Meinhof phenomenon.

Studentmath

Second moment method

In mathematics, the second moment method is a technique used in probability theory and analysis to show that a random variable has positive probability of being positive. More generally, the "moment method" consists of bounding the probability that a random variable fluctuates far from its mean, by using its moments. The method is often quantitative, in that one can often deduce a lower bound on the probability that the random variable is larger than some constant times its expectation. The method involves comparing the second moment of random variables to the square of the first moment. ��2�...

Might be of use?

Ted Shifrin

No idea, offhand, @Samuel.

Samuel Handwich

21:40

hmmmm

Studentmath

I can get $(1-t)^2*z^2$. Close.

Ted Shifrin

Is that inequality even right for a normal random variable with the appropriate $\mu$ and $\sigma$?

Studentmath

But not there just yet.

Samuel Handwich

Hmm how did you get that part?

Studentmath

Using Paley-Zygmund inequality. Though it only works if the random variable has finite variance, which we don't know.

Ted Shifrin

21:42

realizes @Studentmath knows a hell of a lot more probability than he does

Studentmath

@Ted it's just because of this seminar paper :P I got familliar with lots of inequalities. But I have no clue when it comes to sigma-algebra and measure theory definitions.

Ted Shifrin

you can learn that in grad school, if you're so inclined (well, you'll definitely have to learn real analysis)

Studentmath

I hope to, I guess.

@Samuel are you bounded by the inequalities you may use, pardon the pun? Because I am sure I can find one fitting.

Samuel Handwich

@Studentmath i have never taken measure theory. The sigma algebra stuff makes sense to me when it works... just not the counterexamples when things don't work

@Studentmath Well he gave Cauchy-Schwarz as a hint so I assume that anything we find can possibly be reduced to that

anything would help really

Studentmath

Hm

I will give it a look

Ted Shifrin

21:45

LOL @Studentmath: kudos.

Samuel Handwich

$E(X1_{(-\infty,-tz]\cup [tz,\infty)})\leq \sqrt{E(X^2)P(|X|\geq tz)^2}$

Studentmath

@Ted about the pun? I was quite proud of it

Gahah @Samuel

Samuel Handwich

bleh

that does nothing

Studentmath

Ah. Actually it's nice, if you can figure out that $E$.

Samuel Handwich

$E(X1_{(−∞,−tz]\cup[tz,∞)}) \leq P(|X|\geq tz)$

by the condition on $E(X^2)$

Studentmath

21:50

Yep

What can you say about $E(X1_{[tz,\infty )})$

Don't really know if you can say anything, but could be useful.

Samuel Handwich

i still havent used that $0<a<E(|X|)$

Studentmath

I can actually bound it by $(1-t)^2E(|X|)^2$. The inequality used is proveable using cauchy-schwartz only

But it doesn't help

Can we bound $E(X1_{(-\infty ,-tz]})$ by $-t*E(|X|)^2$ somehow?

Samuel Handwich

i can't think of a way

gosh this is just a tough son of a bitch. i will be right back i need a cigarette.

Studentmath

I know the frustruation. Ask @Ted how I grumble at every exercise here.. Sure you will figure it out. Off with the dog, if by the time I get back you are still on it, will try to see how I can help

Alizter

22:06

@TedShifrin

not here?

oh well

Studentmath

@Samuel if it helps you, we can get by Markov that $P(X<tz)\ge (1-t)$

Not sure if it does.

Samuel Handwich

22:39

why is $E(Z1_{Z<\theta E(Z)})\leq \theta E(Z)$?

oh duh.

(wow.)

barznjy

Hi
how to ask a specific question to specific person that I know?
I mean I know his name here, but when I open his profile I can't see any contact detail, such as email, skype, ... etc
is there any way to contact a member here?

Samuel Handwich

@barznjy there isn't any way. You can ping him in chat if he is on chat.

barznjy

@SamuelHandwich Thanks. He is not ON now. and I don't know when he will be on.
Can I create a private chat room?

Samuel Handwich

@barznjy i think it's possible but it is a little difficult

barznjy

@SamuelHandwich if for example, on your profile I click on (start a new room with user) this will be a private room or public room ?

Samuel Handwich

22:53

@barznjy yes that's what you'd have to do with him

barznjy

@SamuelHandwich the problem is that he don't have this option in his profile ....

Samuel Handwich

@barznjy because he's not online

barznjy

@SamuelHandwich so when anyone is online this option will be active automatically?

Samuel Handwich

@barznjy i believe so

barznjy

@SamuelHandwich Thanks

Samuel Handwich

22:57

no sweat

Semiclassical

23:31

sometimes, the various historical conventions of math can't help but make me laugh

for example, the fact that the Binet-Cauchy identity in algebra is a special case of...the Cauchy-Binet formula for determinants in linear algebra

evidently the generality of a theorem is not invariant under reflection of its authors

Julius

Hi, where is the last inequality in oi60.tinypic.com/2uz8di1.jpg coming from?

Studentmath

23:59

@Samuel managed it?

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$Mathematics$ Mathematics