Mathematics - 2014-04-16 (page 2 of 5)

685252

12:00

@MattN. may I ask you why you hate skullpatrol so much?

Will Hunting

@685252 You are skullpatrol

Matt N.

@685252 I don't hate him at all. I just consider him not intelligent enough to be worthy of my time.

2

Khallil Benyattou

@MattN. No arrogance detected whatsoever.

Will Hunting

@MattN. 685252 is skullpatrol

Matt N.

Anyway. Who cares whom I hate?

@WillHunting I don't care.

I have so many people on ignore that it doesn't matter if I add one more.

@DanielFischer Does it make the previous argument invalid?

The one with padding the $2$ by $2$ matrix.

685252

12:03

Thank you for your honest answer :-)

Matt N.

@KhallilBenyattou : )

Daniel Fischer

@MattN. Makes it a different question. Given a row-finite (I suppose that's "finite in each row" indeed) invertible matrix $M$, does there exist an invertible $P$ such that $PMP^{-1}$ is normal?

Khallil Benyattou

@MattN. Beautiful rhyme :)
I have so many people on *ignore* that it doesn't matter if I add one *more*

Will Hunting

t.b. has vanished from SE

Matt N.

@WillHunting But now just in case I added them to my ignore list : )

@KhallilBenyattou Thank you! Btw, your equally beautiful pick up line has been removed by a mod.

Khallil Benyattou

12:05

@MattN. Mods ... shudders :(

Daniel Fischer

@WillHunting A long long time ago, as far as I gathered. Why do you mention it now?

Matt N.

@WillHunting I know. What can I say.

Will Hunting

@DanielFischer Well, I just like to mention random things now and then

Daniel Fischer

@WillHunting $4$

(chosen by fair die roll)

685252

@WillHunting but why now?

Will Hunting

12:06

@meer2kat I thought you were gonna email me, but nvm

Matt N.

@DanielFischer You know (of) t.b.? My heart just jumped. He was my favourite user. Doing maths with him on SE was exhilarating. He was my speed and since he's left I reverted back to normal maths mode.

@DanielFischer Wait isn't that what the original is asking?

Daniel Fischer

@MattN. Well, I have coma across a couple of his answers and comments, and he was mentioned here and there on main and mets. I don't know him aside from that. He was a capable guy, that's for sure.

Will Hunting

@685252 No particular reason

Matt N.

He was.

Daniel Fischer

Well, I guess he still is, if nothing bad happened to him. Just not here anymore.

Will Hunting

12:09

Jonas has also vanished. I emailed him a few days ago

Khallil Benyattou

Are there any SE chats that can help me out with my Android problem?

Will Hunting

Isn't there an Android site?

Matt N.

@DanielFischer Probably. Btw, you sort of remind of him sometimes. Your answers are of equal clarity.

Khallil Benyattou

Downgrading my phone is becoming very cumbersome.

Alyosha

@GabrielR. What is the definition of regularised here?

Khallil Benyattou

12:10

@WillHunting Which one?

Matt N.

So, to get back to what I was trying to figure out: A matrix is similar to a normal matrix if and only if it is row finite and invertible.

Will Hunting

@KhallilBenyattou I don't know, I just thought I saw it before

Daniel Fischer

@MattN. No, only the one direction.

Row finite and invertible implies similar to normal.

Matt N.

Ah! I knew you were the right person to ask : )

Gabriel R.

@Alyosha if $x$ is a point of discontinuity of $f$ then the arithmetic mean between right and left-hand limits of $f$ at $x$ is exactly $f(x)$

Alyosha

12:13

Ah, that. Thanks, @GabrielR.

Will Hunting

@GabrielR. Wow, you used arithmetic mean instead of simply average, hehe

I often wonder who is starring and flagging things in this chat while not in this chat

Matt N.

@DanielFischer Of course: if it's row finite and invertible it is basically an $n$ by $n$ matrix and that is similar to the identity.

Daniel Fischer

@MattN. No, on the one hand, infinitely many columns can be nonzero, on the other, $$\begin{pmatrix} 1&1\\0&1\end{pmatrix}$$ is invertible but not similar to the identity.

685252

@WillHunting life is full of "wonders" :-)

Gabriel R.

btw @DanielFischer what's your favorite inequality ? Mine would be $(x_1+\ldots+x_n)(1/x_1+\ldots+1/x_n) \geq n $

Will Hunting

12:18

@GabrielR. I hate inequalities. They are usually very hard to prove

Matt N.

@DanielFischer Oops. You're right.

Daniel Fischer

@GabrielR. I don't think I have a favourite inequality. Cauchy-Schwarz-Bun'akovskij is terribly useful, so that might be a candidate.

Gabriel R.

@WillHunting In the scope of olympiads, yes they are sometimes very nettlesome. But anywhere else, they're of great use

685252

how about a favorite equation?

Khallil Benyattou

I've heard of Cauchy-Schwarz before. Could you explain it to me?
@DanielFischer

Will Hunting

12:20

@KhallilBenyattou It's best to read wikipedia

Gabriel R.

@685252 $div(E)=\frac{\pho}{\epsilon_0}$

685252

nice^

Khallil Benyattou

@WillHunting I'll give it a shot.

Will Hunting

@KhallilBenyattou My favourite proof of it is a one line completing the square proof

Gabriel R.

@DanielFischer do you mean $|\langle u,v| \rangle \leq ||u|| ||v||$ ? Because with the extra Bun'akovskij I feel you refer to something more general :)

Daniel Fischer

12:21

@685252 $$\frac{1}{2\pi i}\int_\gamma f(z)\,dz = \sum_{\zeta} n(\gamma,\zeta)\cdot \operatorname{Res}(f,\zeta)$$

685252

nicer^

Daniel Fischer

@GabrielR. Oui. For a general positive semidefinite hermitian form.

Gabriel R.

@DanielFischer hehe but maybe triangular inequality is even more "terribly useful"!

Khallil Benyattou

$$ \mathcal{L} \left[ f(t) \right] (s) = \displaystyle \int_{0}^{+\infty} f(t) e^{-st} \text{ d}t $$ Unilateral Laplace transform looks cool.

685252

$E=mc^2$
nicest^ imo

Daniel Fischer

12:23

@GabrielR. C'est vrai.

Will Hunting

@GabrielR. It is usually called triangle inequality

meer2kat

@WillHunting always :)

4 days until i can drink coffee again

i'm going nuts

Daniel Fischer

@meer2kat Why's that?

Will Hunting

@meer2kat So are you gonna email me or not?

Matt N.

Why can't you drink coffee?

Gabriel R.

12:24

@meer2kat I sent you a message on Facebook. No reply :(

Sawarnik

@WillHunting Forget it.

Will Hunting

@GabrielR. Haha, I don't have FB account

meer2kat

@DanielFischer lent will be over. i don't know why i gave something up. i'm not catholic.

685252

cold turkey?

meer2kat

@WillHunting dont have your email

Sawarnik

12:25

@meer2kat I sent you a message on Facebook. No reply :(

meer2kat

@GabrielR. havent been on yet

@Sawarnik havent been on yet

Will Hunting

@meer2kat Haha, it's [email protected], lol

meer2kat

@WillHunting send it to me at 3:45 PM today :)

Will Hunting

@meer2kat Send it to you? What do you mean?

meer2kat

need caffeine. need caffeine. need caffeine

Matt N.

12:26

I don't get it at all. Just make yourself a cup of coffee.

Gabriel R.

@DanielFischer If I were a member at the IMO committee, I would think of a very convoluted inner product, and ask contestants to prove the subsequent Cauchy Schwarz inequality

meer2kat

@WillHunting actually, just email me. [email protected] I'll answer you when i log on next

@MattN. Nah, it's been 36 days. i can go 4 more

Will Hunting

@meer2kat OK

Matt N.

Wow. Yeah, that's true, I guess.

Sawarnik

@WillHunting Why did you choose this email id?

Will Hunting

12:27

@Sawarnik What name?

meer2kat

@Sawarnik its his alter ego

685252

alter id

Will Hunting

@meer2kat Oh, you seem to know me very well lol

meer2kat

@WillHunting lol!

Will Hunting

I chose Will Hunting because I have similar problems as he does

685252

12:29

but you're not at an ivy league school

Khallil Benyattou

What problems does @WillHunting have. The character I mean.

Will Hunting

Nope, I am not. I don't think much of Harvard lol

Matt N.

@WillHunting What about Leonardo di Caprio?

Sawarnik

@KhallilBenyattou How's your phone now?

Will Hunting

@KhallilBenyattou You have to watch it yourself

685252

12:30

that's not the only school in the ivy league

Matt N.

No wait. Or was it Matt Damon?

Gabriel R.

@Sawarnik The remove button should have a limited use

Will Hunting

@MattN. I think he is rather cute, but not as cute as Justin Bieber

Khallil Benyattou

@WillHunting I've seen the film, but a long time ago.

@Sawarnik Haven't done anything to it yet :(

@Sawarnik I'm going to give the new firmware a chance today. If I don't like it, I'll restore it.

Matt N.

@WillHunting What about this one? Hot or not?

Will Hunting

12:32

@meer2kat Actually, today is the third time I told you my email, lol, but it's OK!

685252

who's counting lol

Sawarnik

@WillHunting Means she isn't interested :P

Khallil Benyattou

@685252 @WillHunting certainly is.

Matt N.

Not the greatest actor though. But a pretty boy.

Khallil Benyattou

@Sawarnik You read my mind :P

Will Hunting

12:32

@MattN. Nope. Now my favourite is Steven Strait and Laura Ramsey in The Covenant

Sawarnik

@KhallilBenyattou :D

Matt N.

@WillHunting Ok, I get your type : )

Will Hunting

@Sawarnik I think I will ignore you, you are very irritating and mean

Matt N.

Doesn't match mine : )

Sawarnik

@WillHunting I know I am irritating but I am not a mean square!

Matt N.

12:34

I should go. See you all later!

Will Hunting

@Sawarnik I won't talk to you again, you have said mean things time after time

Sawarnik

Besides, you must have said that 3 times already.

Will Hunting

@MattN. Bye!

Matt N.

@DanielFischer Thanks for discussing the Scottish book problem with me!

Sawarnik

@WillHunting You lack the quality to recognize humor.

685252

12:35

he's not mean...

Daniel Fischer

You're welcome, @MattN.

685252

his humor is dry

Will Hunting

@sawarnik You have made many assumptions about me that are not true

@Sawarnik You think you know what I am thinking but you don't

685252

@WillHunting have you never heard of sarcastic humor?

Daniel Fischer

@685252 Contradictio in adiecto. But whose humour is dry?

Will Hunting

12:37

@685252 To be honest, maybe I should start ignoring you too

685252

@DanielFischer Sawarni

Daniel Fischer

k

685252

:D

Sawarnik

Evil.

Will Hunting

OK guys, I will now ignore several people in this chat, and I mean it

Sawarnik

12:38

@WillHunting You mean.

Karl Kronenfeld

:)

685252

@WillHunting the important thing is to be honest with yourself, pal

Sawarnik

@WillHunting Ok. I never think wat you are thinking LOL :P

685252

@WillHunting Don't go out of your way for me.

Khallil Benyattou

Break it up guys.

2

We're just here to chat and maybe look at some interesting maths, maybe.

I'm going to be off soon.

Peace out.

Stay sexy, people.

685252

12:40

later

3 mins ago, by 685252

@WillHunting the important thing is to be honest with yourself, pal

Will Hunting

@user127001 Hey Bart

685252

3 mins ago, by 685252

@WillHunting Don't go out of your way for me.

Chris's sis

Prove that

$$\lim_{n\to\infty}\left(\frac{\displaystyle \int_0^{\large \sqrt{2\cdot 1 /\pi}}\cos\left(\frac{\pi x^2}{2}\right) \ dx}{1^{5/2}}+\frac{\displaystyle \int_0^{\large \sqrt{2\cdot 2 /\pi}}\cos\left(\frac{\pi x^2}{2}\right) \ dx}{2^{5/2}}+\cdots+\frac{\displaystyle \int_0^{\large \sqrt{2\cdot n /\pi}}\cos\left(\frac{\pi x^2}{2}\right) \ dx}{n^{5/2}}\right)=$$

$$\sqrt{\frac{2}{\pi}}\left(\frac{\pi^2}{6}+\frac{1}{20} - \frac{\pi}{6} \right)$$

meer2kat

@WillHunting sorry

Gabriel R.

@Chris'ssis easy once dis-obfuscated (substitute $u=\sqrt{\pi/(2n)}x$)

Chris's sis

12:52

@GabrielR. I got your point.

robjohn

@GabrielR. I adjusted the latex :-)

@WillHunting someone is already ignoring me. I don't know who, but my chat profile says so :-)

Karl Kronenfeld

@robjohn Where do you see this?

Sawarnik

@robjohn How do you know this?

meer2kat

that looks much less confusing with mathjax hahaha

robjohn

@KarlKronenfeld maybe it's only in the room owner or moderator view

Karl Kronenfeld

12:57

Ah, perhaps moderator.

There are lots of numbers that you have to hover over to determine what they mean. I could just not do that and pretend that's the number of users ignoring me.

2

Gabriel R.

@Chris'ssis @robjohn So I get $$\sqrt{\frac{2}{\pi}}\sum_{n=1}^{\infty}\frac{1}{n^2}\int_0^1cos(nu²)du$$ but that last integral is not computable, and I'm not strong when it comes to asymptotics

Pedro Tamaroff

@KarlKronenfeld Well, overnight the problem has been completely settled.

Thanks.

robjohn

@GabrielR. Try computing $$\sum_{n=1}^\infty\frac{\cos\left(nu^2\right)}{n^2}$$ I think it involves $\mathrm{Li}_2$ however

Gabriel R.

actually @robjohn, which function has Fourier expansion $$\sum_{n=1}^{\infty}\frac{1}{n^2}cos(nx)$$ ?

robjohn

@GabrielR. Ah, so in this case $\mathrm{Li}_2$ has a closed form :-)

agent154

13:12

Greetings. I come with another question: How can I define a recurrence relation for the following problem?

"Write a recurrence relation for the number of n-letter words in which no consecutive pair of 'e's appear."

My initial guess (though I feel shakey about it) is that, assuming that the first letter is **not** an e, then there would be $25a_{n-1}$ sub-sequences for the remaining $n-1$ letters in the sequence (that satisfy the requirement). However, if it **is** an e, then there would be $a_{n-1}$ valid sub-sequences for the remaining letters. This would give me the relation $a_n=25a_{n…

Gabriel R.

@robjohn I don't know Li functions, so I don't know what you mean

meer2kat

@GabrielR. uh, clearly he's talking about lithium... ;)

robjohn

@GabrielR. $\mathrm{Li}_2$ is described here

Sush

How to find $\lim_{x\rightarrow\infty} \sqrt x(\sqrt{x-4}-\sqrt x)?$

Gabriel R.

@agent154 "However, if it is an e, then there would be $a_{n-1}$ valid sub-sequences for the remaining letters." nope. the $a_{n-1}$ may start with an $e$

@robjohn I'll take a look later, thanks

meer2kat

13:23

@Sush i'd start by bringing that first sqrt(x) into the other parts of it

Sush

@meer2kat, i.e multiplying sqrt(x) inside?

meer2kat

@Sush mhmmm

agent154

@GabrielR. OK then, so just subtract the single case where $a_{n-1}$ starts with e? So it'll be $a_2=26a_{1}-1$..

where $a_1=26$

meer2kat

@Sush this gives you sqrt(x^2-4x) - x

MickLH

@Sush Well the easy way to "find" it is to plot it and notice it clearly goes to $-2$ :P

agent154

13:28

But that doesn't seem right again... $a_3=26a_2-1=26(26^2-1)=26^3-26$... but there should only be $26^3-2$ such sequences

meer2kat

@MickLH lol that's cheating. i did it too though xD

Sush

@MickLH, haha

MickLH

hehe he said find not prove

meer2kat

@MickLH touche

Sawarnik

@Sush $$\lim_{x\rightarrow\infty} \sqrt x(\sqrt{x-4}-\sqrt x)=-4\lim_{x\rightarrow\infty}\frac{ \sqrt x}{\sqrt{x-4}+\sqrt x}=-4\lim_{x\rightarrow\infty}\frac{1}{\sqrt{1-\frac4{x}}+1}=-2$$

meer2kat

13:31

@Sawarnik well fine then. just solve it for him why don't ya

Sush

@Sawarnik, thanks!

Gabriel R.

@agent154 i would say $a(n+1)=25a(n)+25a(n-1)

MickLH

@meer2kat Most people seem to ask "how to find" when they really just want the solution

meer2kat

@MickLH yeah i know. still. maybe it's just the future teacher in me

Sawarnik

@meer2kat :P I am not a teacher, just a student.

meer2kat

13:33

@Sawarnik fair enough

Sawarnik

@Sush :D

meer2kat

@Sawarnik i do prefer bringing it all in first still. looks icky with that extra sqrt XD

agent154

@GabrielR. What's the combinatorial reasoning for that though?

Gabriel R.

@agent154 to build an n+1 letter word without consecutive $e$, you can either add any letter except $e$ at the beginning of an $n$ letter convenient word (that's 25a(n)) or you can start with an $e$ but then the second letter can be anything but e and the n-1 remaining letters are those of a convenient word of length $n-1$ (that's 25a(n-1)). Total a(n+1)=25a(n)+25a(n-1)

Sush

@Sawarnik @meer2kat, my book says it is $2$

!

Sawarnik

13:36

@Sush You must have made a typo then.

Sush

@Sawarnik, No!

agent154

@GabrielR. OK, I think I get it now...

MickLH

no pretty sure it's still -2

Sawarnik

@Sush Are you sure it isn't $\lim_{x\rightarrow\infty} \sqrt x(\sqrt{x+4}-\sqrt x)$?

Daniel Fischer

@Sush Since $\sqrt{x-4} < \sqrt{x}$ for $x \geqslant 4$, the limit cannot possibly be positive. Someone made a typo, you or your book. You say you didn't, so it must have been the book.

Sush

13:38

@DanielFischer, thanks!

Chris's sis

@robjohn how are you doing?

meer2kat

@Sush it is negative 2.

@Sush that is a fact. check it in a graphin calculator.

Sawarnik

@meer2kat Its resolved: typo somewhere.

Sush

yes.

Charlie

@Chris'ssis HI CHRIS

Sawarnik

13:40

@Charlie She is not Chris.

Chris's sis

@Charlie Great CAT!!! :-))))))))) HI!!!

Charlie

@Chris'ssis :DDDDDDDDDDDDDDDDDDDDDD

Gabriel R.

@DanielFischer can you please check my reasoning for the combinatorics question ? I'm afraid I'm wrong :P

Chris's sis

@Sawarnik now worry about the nickname they use to me. ;-)

agent154

Somebody's disgustingly happy here...

Charlie

13:41

@Sawarnik I'm not a cat

3

Chris's sis

@Charlie :D

Gabriel R.

@Charlie wassup dawg ?

Charlie

@Chris'ssis how are you doing?

@GabrielR. not much, the ususal, and you?

Gabriel R.

@Charlie I'm fine, nice weather has returned

Chris's sis

@Charlie Besides the fact that I only slept 4 hours last night and I'm a bit tired, I try to create some beautiful stuff (like series and integrals). And you? :-)

Sawarnik

13:43

@Charlie Why were you in a hibernation for a long time?

Charlie

@GabrielR. great! here is less hot, so all fine

@Chris'ssis studying complex analysis right now

Sawarnik

@Chris'ssis Are you a bachelor in maths?

Chris's sis

@GabrielR. I liked the way you tackled my problem. You're good.

Daniel Fischer

@GabrielR. This? Sound.

Charlie

@Sawarnik when you stay here for too long you start to get a little bit irritated about a few things...

@Chris'ssis I'm surrounded by integrals now

Sawarnik

13:45

@Charlie Yup! (like me isn't it?)

Chris's sis

@Sawarnik I have no background in mathematics (in general, I'm self-taught).

@Charlie Cool!

@Charlie Some nice multiple integrals?

Charlie

@Sawarnik nah, there are worse things, I overcame our differences

@Chris'ssis no, not like the ones you do

Sawarnik

Wow, seeing Chris'sis in a chatty mood, after a long time!

Charlie

lot simpler

Gabriel R.

@Chris'ssis but I couldn't finish it. The sum $$\sum_{n=1}^\infty\frac{\cos\left(nx\right)}{n^2}$$ really looks like the Fourier expansion of a nice function, but I can't figure out which :) What's your solution if you have one ?

Chris's sis

13:47

@GabrielR. Use Clausen functions - see here mathworld.wolfram.com/ClausenFunction.html

Charlie

@Sawarnik it's my presence :)

user127001

Hi @will

Chris's sis

@GabrielR. More precisely $(7)$.

brb

Sawarnik

@Charlie I still like to think you are a cat. Ok, bye.

Gabriel R.

@Chris'ssis ah ok, good to know thanks !

Charlie

13:51

@Sawarnik okay,dear

meer2kat

@Sawarnik excellent

@Sawarnik Ummm...Charlie is a Unicorn.

Chris's sis

Yesterday I posted this limit. I wanna say again it's one of the most beautiful limits I've ever attended. $$\lim_{n\to\infty} \frac{1}{n+1}\left(\sum_{1\le k_1\le n+1}\frac{1}{k_1}+\sum_{1\le k_1< k_2\le n+1}\frac{1}{k_1 k_2}+\cdots +\sum_{1\le k_1<...<k_n\le n+1}\frac{1}{k_1 k_2...k_n}\right)$$

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$Mathematics$ Mathematics