Mathematics - 2012-06-15 (page 1 of 4)

Zhen Lin

00:00

That's actually a deep question. For certain nice classes of ODEs, there is a unique solution for any given set of initial conditions.

Jordan

ODE?

Zhen Lin

ordinary differential equation.

Jordan

I am guessing that is what I am working with

skullpatrol

@Jordan Good job, you asked a "deep" question :D

Jordan

Completely accidental :P

skullpatrol

00:04

It is a start...

Jordan

I am a little confused here if I want to find $\int \frac{1}{tanx}$ it is just $log(sin(x))$ but how do I get there?

I know that I can't make a u substitution becausethe derivative of tanx is sec^2 x

user19161

00:37

@Jordan You mean how you get the integral of 1/tan?

user19161

Just write 1/tan as cos/sin and know that sin'=cos.

Jordan

oh of course

user19161

So you need to remember your trigonometric identities as well.

skullpatrol

00:55

@Jordan Yes, not only do you need to remember them; but as Jasper has shown, you need to be able to manipulate them.

Jordan

This is a weird question, Find an equation of the curve that passes through the point 0,1 and whose slope at (x,y) is xy

isn't that just 0 ,1?

I get it now, it is just like the old problems y(0)= 1

and y' = xy

skullpatrol

By George, I think he's got it :D

Jordan

01:11

so $\frac{1}{2} e^{x}$ is the same as $e^{\frac{x}{2}}$ right? Is there a preferred way to write that or is anything okay?

Also I do not quite understand the procedure to do this, and it is not explicitly stated in the book Solve the differential equation $y' = x + y$ by making the substitution $u = x + y$

or change or variable is the word they use

Joe L.

01:38

@Jordan $\frac{1}{2} e^x \ne e^{x/2}$

Try plugging in a value for $x$ and you'll get two different answers.

Peter Tamaroff

02:16

@MarianoSuárezAlvarez Estás? Si estas ocupado, no hay drama.

@Jordan Let $e^x=y$. You're asserting $\frac 1 2 y = \sqrt{y}$, do you realize that?

leo

02:48

je je

@PeterTamaroff como va todo?

Peter Tamaroff

@leo Todo tranquilo. Hoy me dieron la nota de mi primer examen parcial de epistemología y tuve un $8^{50}$, asi que puedo promocionar los tres cursos!

leo

@PeterTamaroff bien por ti. Me alegro

Peter Tamaroff

@leo Y a vos como te esta yendo?

leo

por ahora bien

Jordan

I have to learn spanish instead of math next semester, spanish is really hard

I just dislike spending a lot of time learning something I will never use

Peter Tamaroff

02:53

@leo Que estás cursando?

@Jordan What do you mean?

Why do you want to learn Spanish?

Jordan

Everyone here has to take a year of language in college pretty much

leo

@PeterTamaroff Análisi Real I, Grupos y anillos, ecuaciones diferenciales ordinarias

Peter Tamaroff

@leo Cooooooooooooool.

Jordan

I am taking spanish because I know a little bit already

Peter Tamaroff

Como van las ODEs? Que libro usan?

@Jordan Ah! OK.

leo

02:56

@PeterTamaroff este

Peter Tamaroff

@leo =D

leo

dice el profe que es el que usan en el Courant Institute. El estudió ahí

y es legal!

Jonas Teuwen

@BenjaminLim No, he is in Madrid and so am I!

Peter Tamaroff

@leo Hahahah Courant? Es un lugar importante no? Legal en que sentido?

leo

@PeterTamaroff hay permiso para que circule por ahí

Jonas Teuwen

02:58

It is bloody 5:02 AM.

Peter Tamaroff

@JonasTeuwen Stop complaining about the time, dude! =)

Jonas Teuwen

Have to get up in 3 hours man,.

Peter Tamaroff

@leo Ah! Muy bueno como esta formateado.

@JonasTeuwen Bah. I slept 3/4 hs today. Get some good coffee. You have that awesome machine. ;)

leo

@PeterTamaroff Courant Institute

Jonas Teuwen

I am in Spain!

I have been sleeping 3 hours for like two weeks!

Peter Tamaroff

03:00

@JonasTeuwen Olè!

@JonasTeuwen What are you doing there?

leo

@JonasTeuwen cool!

Jonas Teuwen

Conference.

leo

@JonasTeuwen coool!

Peter Tamaroff

@JonasTeuwen Are you talking in Spanish?

Jonas Teuwen

No habla espagnol!

leo

03:01

ja ja

ha ha

Peter Tamaroff

@JonasTeuwen Hahahaha! I guess then you're horrified by the spaniard's english pronounciation right?

Jonas Teuwen

It is quite good.

The ones at the conference at least... Many others don't speak any English.

Peter Tamaroff

@JonasTeuwen Oh, sure. They must be good. Either, they'll put off the audience.

Jonas Teuwen

Usually the people giving talks are not Spanish!

Peter Tamaroff

@JonasTeuwen Oh! What is the conference about?

leo

03:09

the angles are dimensionless in the SI, right?

Peter Tamaroff

@leo Dimensionless.

Yes, since they are the quotient of two lengths.

leo

@PeterTamaroff Thanks

Jonas Teuwen

@PeterTamaroff Harmonic analysis.

leo

@PeterTamaroff and thanks!

Peter Tamaroff

@leo Have you discussed any operational approach to ODEs in class?

leo

03:14

@PeterTamaroff What do you mean by operational?

Peter Tamaroff

@leo Solving ODEs in terms of operators.

leo

@PeterTamaroff Differential operators, yes

Peter Tamaroff

@leo Right!

For example, the Cauchy-Euler equation.

It's solution is pretty operational.

leo

@PeterTamaroff Yes indeed. By the way, I did not know that they are the Cauchy-Euler equation

no sabía que se llamaban así =)

@JonasTeuwen why the torus is $\Bbb R/\Bbb Z$?

Jonas Teuwen

Why do you ask me this questions now? I'm off to bed! Good night. See you in ... a couple of hours.

Peter Tamaroff

03:22

@leo La ecuacion diferencial de Cauchy Euler es de la forma $$a_n x^n y^{(n)}+a_2 \cdots+x^2 y^{(2)}+a_1 x y' +a_0=0$$ si no recuerdo mal.

leo

@JonasTeuwen sleep wel!l then

@PeterTamaroff si correcto. Es un caso particular de ED de orden $n$ con coeficientes continuos

Peter Tamaroff

@leo Yeah. Pero me parece elegante como la tranformacion $x \mapsto z$ con $$x^n D^n = \mathcal D (\mathcal D-1)\cdots (\mathcal D-n+1) $$ la resuelve.

$$\frac{x^n}{n!} D^n = {\mathcal D \choose n}$$ :P

leo

@PeterTamaroff sí. Esta es linda por eso

Peter Tamaroff

Asi esta mejor!

@leo Y vieron algo en general, por ejemplo, la soluction de $$(D+a)y=f(x)$$?

Me imagino que si.

leo

@PeterTamaroff si

Peter Tamaroff

03:29

@leo Vieron algo de Heaviside?

leo

@PeterTamaroff Si digamos, $Ly=f(x)$ con $L=a_n(x)D^n+\ldots a_0$

Peter Tamaroff

@leo Ah, los coeficients son funciones de $x$?

leo

@PeterTamaroff si. Continuas

Peter Tamaroff

@leo Ah, espera.

Lo que dices es

escribir

$$a_0(x)+D^n y+a_1(x) D^{n-1}+\cdots a_{n-1}(x) D+a_n(x) y = F$$

como

$$F(x)=\sum_{k \leq n} a_k(x) D^{n-k} y$$

Luego poner $$ y \phi(D) = y \sum_{k\leq n} a_k D^{n-k} F$$

@leo Yo estudie de Spiegel.

leo

si

@PeterTamaroff Cómo se llama el libro?

Peter Tamaroff

03:35

@leo Applied Differential Equations.

Pero la verdad la teoría es excelente, aunque parezca que es todo aplicado, no es asi.

leo

No es cierto que ese libro lo usan para cursos de ingeniería?

Peter Tamaroff

@leo Si, puede ser. Pero me encató. Igualmente, no tiene las consideraciones teoricas mas ligadas al algebra como me imagino se analiza en matematica pura.

Como en el ensayo de esta pregunta mia

Bill Dubuque explica que hay conexiones con el Algebra de Lie. Me interesa bastante todo eso!

leo

Es bien interesante

por cierto Argentina tiene buenos especialistas en EDs

=)

Peter Tamaroff

@leo Si!!! Jonas estaba contandome eso. Caffarelli es un maestro a nivel mundial.

El nombro dos mas.

Pero no me acuerdo ahora como se llaman.

leo

bueno también está un Rodrigues, de la fórmula de Rodrigues para los polinomios de Legendre

y el gran Calderón en análisis, que era de Mendoza

Peter Tamaroff

03:44

@leo En serio!?!?!?!? La fromula de Rodrigues es de un Argentino!?! WOW!

Me parece tan elegante esa formula. Es increiblemente bella.

leo

Creo que sí

voy a revisar

Peter Tamaroff

Alberto Pedro Calderón

Alberto Pedro Calderón (1920-1998), fue un relevante ingeniero y matemático argentino. Calderón se destacó como investigador y docente en el campo de la matemática pura. Es conocido por sus trabajos sobre la teoría de las ecuaciones diferenciales en derivadas parciales y sobre los operadores definidos por integrales singulares. Este concepto su vez ha dado origen a la actual teoría de operadores pseudo diferenciales. También son importantes sus trabajos sobre la interpolación de operadores y sobre los problemas inversos. Las técnicas desarrolladas por Calderón son de importancia fundame...

@leo Rodrigues era Francés

leo

@PeterTamaroff Jaja la de Rodrigues no. Es decir, ese Rodriguez era francés: en.wikipedia.org/wiki/Olinde_Rodrigues

Peter Tamaroff

@leo Jjajaj si.

Igual creo que lo de Calderon es mucho mas relevante.

Jonas debe de conocer su trabajo bastante bien.

leo

@PeterTamaroff En algún lado escuché que era argentino

@PeterTamaroff si, seguramente

Peter Tamaroff

04:03

@DylanMoreland Hey.

@Leo Maybe this is what you were asking Jonas? here

Dylan Moreland

Hi Peter.

Peter Tamaroff

@DylanMoreland What are your graduate studies?

@DylanMoreland The properties Bill and you exploit here seem to resemble those of the $\gcd$. Is that a generalization?

leo

So the torus is $\Bbb R/\Bbb Z$ as groups

robjohn

@leo That is the unit circle, too :-)

Dylan Moreland

@PeterTamaroff Well, I'm not sure. My answer just uses what prime ideals are, really.

leo

04:14

are they isomorphic groups?

Dylan Moreland

@leo Yes. Use the exponential map!

leo

ok I see.

Peter Tamaroff

@DylanMoreland Well, primes are a special case of prime ideals right? Or something of the sort?

Dylan Moreland

@PeterTamaroff Like it's true that the ideal generated by $a, b \in \mathbf Z$ will be $\gcd(a, b)\mathbf Z$.

leo

Is that the way as the torus is defined?

Peter Tamaroff

04:16

What does $$\gcd(a, b)\mathbf Z$$ mean?

leo

@PeterTamaroff Thanks!

btw

Peter Tamaroff

@DylanMoreland Is it something like $\mathbf Z / n \mathbf Z$?

Dylan Moreland

The ideal generated by that number.

It's just Bézout's identity.

Is there an equality or identity on that page that screams out to you?

Peter Tamaroff

@DylanMoreland I just know basically nothing about abstract algebra. It just rang a bell, that's all.

I'm just a freshman =)

robjohn

@tb Even though you are on break, I thought you might want to look at this question. You are after all, the highest ranking answerer for inversive-geometry :-)

Dylan Moreland

04:22

Well you're learning some number theory now. I think that's a good avenue for understanding this sort of thing. Kummer and Dedekind invented these concepts to try to deal with rings of algebraic integers.

And in Dedekind domains, which is what those rings are, the analogy is really tight. Every ideal factors into products of prime ideals. And you have g.c.ds of ideals and all this stuff.

robjohn

@DylanMoreland Ooh! Bezout! :-)

@DylanMoreland Bill says I overuse Bezout.

Peter Tamaroff

@DylanMoreland Nice!

I'll keep rowing.

Dylan Moreland

But in general... I mean as a dumb example you have in $\mathbf C[x, y, z]$ ideals $(xy)$ and $(xz)$. So the g.c.d. of those elements is $x$, but I don't think that's even in the ideal that those two elements generate.

Peter Tamaroff

@DylanMoreland Are you studying for a Master's?

Dylan Moreland

I'll pick up a masters at some point this year. But that's some formality on the way to a PhD. I hope.

@robjohn I think I might too. It's the only trick I ever remember. I'm sort of bad at elementary number theory.

robjohn

04:30

@DylanMoreland Usually I invoke Bezout when there is a threat that big guns are being aimed at elementary results. I worry about circular arguments.

Often Bezout can be used as a replacement for using the FTA (a big gun).

Peter Tamaroff

@robjohn Right. I us FTA a lot. I think the proof is so simple, the theorem can be used with no worries. I don't know if it is a big gun actually.

robjohn

@PeterTamaroff If proven correctly (which uses Bezout) then worries are minimal. However, I never know how FTA was proven to people.

Peter Tamaroff

@robjohn First, the prime decomposition theorem is proven.

Then, by induction, the FTA is proven.

FT of Arithmetic, that is.

Eugene

04:56

huh. interesting that all this conversation on number theory so happened on my day off.

@robjohn is your mother in law alright?

Peter Tamaroff

@Eugene 'sup dawg.

Eugene

nothing much.

Peter Tamaroff

@Eugene Is Wilson's theorem important? And Wolstenhome's?

Eugene

i haven't heard of the second one.

Peter Tamaroff

$$\sum_{k=1}^{p-1}\frac{(p-1)!}{k}\equiv 0 \mod p^2$$

Eugene

05:02

nope never heard of it.

Peter Tamaroff

@Eugene BTW I've been meditating about this

I think I can try and enunciate one sufficient condition.

Which might then apply to the Harmonic series.

However, I can't find the relevance of the result.

Eugene

i don't see why this is interesting even.

Peter Tamaroff

@Eugene That's what I'm saying! =D

Eugene

well if you enjoy it that's all that matters i guess.

Peter Tamaroff

@Eugene That's a bloody good reason, dude!

How are you doing?

Eugene

05:08

ok. i've been doing mostly my own stuff today. also i'm heading off to bed now cause i'm quite tired. bye!

leo

I'll go to sleep as well

@PeterTamaroff buenas noches

good night all

robjohn

@Eugene It was not a stroke, but she does need to have a pacemaker. I am not sure how serious a surgery that is.

anon

05:47

The profinite completion of the integers can be realized as a direct product of the p-adic integers. I wonder if this can be generalized. Is any inverse limit (indexed by a countable, locally finite poset I) decomposable as a product of inverse limits (over linearly ordered subposets of I)?

robjohn

@anon I see the words on the screen, but they are bouncing off my brain.

anon

turn your reflectors off

oh, I thought Dylan was in here too

Dylan Moreland

@anon Interesting question.

What's a locally finite poset?

anon

meaning for any $x,y$ with $x\le y$ the set of elements $u$ with $x\le u\le y$ is finite

Dylan Moreland

Aha. Thanks.

anon

06:01

then again that might be superfluous

Dylan Moreland

That isomorphism is more or less the Chinese remainder theorem, so I guess you'd have to figure out when that will get you something in other cases.

anon

I was thinking linearly ordered meant locally finite for some reason when I asked the question

@DylanMoreland For that approach I'm worried might be too inevitably tied to the objects' structure in the system, which we don't have access to in general.

What I was thinking is that we can "cut out" linearly ordered posets from the index set, throw away scraps without changing the limit, and then use some kind of inductive reasoning on this process.

After reading some of keith conrad's group theory notes, whenever I see the letters "xkcd" I'm going to think it's some kind of edgy alter ego of KCd.

J. M.

Hey guys.

anon

yo JM

J. M.

Just came from the post office. The swag's neat.

anon

06:06

indeed, I wore it for like a week straight

J. M.

The mug's huge; I'd definitely be buzzed for the whole day with that much coffee...

Matt N.

Morning : )

J. M.

Hi Matt.

Matt N.

I see, the doughnut is gone.

J. M.

@MattN I'm doing CMC surfaces for the time being. :)

Matt N.

06:18

: )

anon

Actually locally finite would be necessary, because otherwise we can have empty inverse limits.

(Talking to myself.)

actually that breaks countableness too, nevermind

Is \frac12 really an "improvement" over \frac{1}{2}? I think the latter is more readable myself.

Matt N.

I agree. But when I want to type a question and I'm keen to get an answer then I'm in a rush and I use the former.

anon

some user named Frank Science made a couple of edit suggestions (still pending) with these improvements. the only other thing touched was ln-->\ln (which is an improvement, but I'm not sure if I should decide it's "substantial" enough to approve..)

Matt N.

I'm not sure either. If it changes nothing else though I think I'd approve it.

Ok, see you all later!

anon

06:36

nah

that just means they are pushovers that will agree to do things asked of them

rather than vain and appealing to each other's egos or whatever

J. M.

@anon Not worth the extra rep, methinks.

anon

I upvoted the two top answers, I think I'm going to accept mutual admiration society.

what is with skull and removing messages

In other news I was recently suspended. That was intriguing.

3

J. M.

I remember being there when that extended rant happened; the only mods at the time were Mariano and Bill...

But I don't think they're that touchy.

anon

Jasper Loy swooped in and told Jordan off after the ban, so I assume the flag hit people from other chatrooms and they're the ones who made the decision.

I always get flags from the Gaming chatroom for example.

And it's always the same user that gets flagged ...

J. M.

@anon That place is about as noisy as the bar at Happy Hour...

anon

06:46

whoah, when did gaming.se get renamed?

J. M.

Speaking of Mister "waah I can't do math!!!1!", I finally found the psychological term of art: "learned helplessness". Yeesh.

skullpatrol

When did $E = (mc)^2$ become a true statement?

anon

It's not a true statement. The OP of that question is simply wondering how one would say $(ab)^2$, as opposed to $ab^2$, in oral English.

skullpatrol

The product of $a$ and the square of $b$.

anon

Or you could say "$a$ $b$ squared" for the latter and "the square of $a$ $b$" for the former.

skullpatrol

06:53

You may want to put a comma in $a$, $b$ squared :D

anon

but I don't pronounce commas so...

J. M.

I don't remember commas ever being pronounced in mathematical stuff...

skullpatrol

I have heard people pronounce "periods" as in "End of argument period "

I just noticed I didn't put a "." at the end of that^ sentence.
;-)

Martin Sleziak

Wikipedia: Learned helplessness can also be a motivational problem. Individuals who have failed at tasks in the past conclude erroneously that they are incapable of improving their performance. This might set children behind in academic subjects and dampen their social skills.

@JM Where did you learn about that term?

anon

@J.M. I have a question I'm thinking about asking on main. What do you think? That is, given a map $f:M_n(k)\to k$ (with $k$ some field) such that $f(AB)=f(A)f(B)$ for all matrices $A$ and $B$, is it necessarily the case that $f$ factors through the determinant, i.e. does there exist a multiplicative map $g:k\to k$ such that $f=g\circ\det\,$?

Martin Sleziak

07:10

If you know the value of $f(A)$ for matrices of row operations and for upper triangular matrices, the map $f$ id determined uniquely by the choice of these values, right?

@anon IMO it seems interesting. I would say it is worth posting a question on main.

anon

By Guassian elimination that sounds right.

Can we decompose upper triangular matrices as a multiplicative monoid in some straightforward fashion?

Martin Sleziak

IIRC Strang defines determinant by listing a few properties and than saying that these properties define determinant uniquely. So I was thinking perhaps something along the lines could be useful, if we are trying to prove this.

anon

Alternating multilinear sending the identity matrix to unity - this determines $\det$ uniquely I believe.

Martin Sleziak

Right.

anon

Also you could define it coordinate-freely as the trace of the $n$th exterior power.

(Of a linear map.)

J. M.

07:20

@MartinSleziak I'd been reading The Psychology of Everyday Things again, and those words just stuck out from the section I was reading on why people feel helpless about using complicated devices. Which is what math is to a number of people, I guess.

@anon Yes, sounds like a good one. I don't have an answer offhand.

anon

I posted this in IRC a couple years ago, and I just remembered it.

Someone uploaded a pdf with a (supposed) answer but I don't remember where to find it, and I never read it.

Foool

07:42

Guys, is it wrong to say $$ \sqrt{x + 4} \times \sqrt{x+4} =\sqrt{16+8 x+x^2}=\sqrt{(4+x)^2} = \pm(x+4)$$?

\sqrt{(4+x)^2} = \pm(x+4)

But, $\sqrt{x + 4} \times \sqrt{x+4} \neq (x+4)$$ right?

anon

the square root function on reals is always nonnegative, not multi-valued

Jaydon Zhao

hey, I have a question

anon

shoot

Jaydon Zhao

if $Q$ is a nilpotent matrix, i.e. $Q^2 = 0$, and $S$ is another matrix (but invertible), if $QS^{-k}Q = 0$, does it follow that $k$ can be any integer?

anon

07:58

you can edit your comments within a 5 min (ish) window

Jaydon Zhao

thanks

experimentX

Hi

how to evaluate this

anon

@JaydonZhao Do you mean (a) that if $QS^{-k}Q=0$ for some $k$, then $QS^nQ=0$ for all $n$, or (b) that just knowing that $QS^{-k}Q=0$, it's impossible to determine what $k$ can or can't be?

@experimentX Expand $u/(1+u)^2$ in a power series (with $u=e^x$), use linearity to make it a sum of integrals, scale each integral so that you get Riemann zeta times a gamma function.

Jaydon Zhao

@anon: I meant (a) i.e. it will apply for any $k$

experimentX

Oh ... Haven't worked with Riemann zeta function!!! Thanks ... i'll try and report

Jaydon Zhao

08:03

specifically I was wondering if it is valid to premultiply and postmultiply by $Q$

and then observe that the statement will hold for any $k$?

since the appearance of $Q^2$ will reduce it to $0$.

anon

multiply what by Q?

Jaydon Zhao

$QQS^(-k)QQ = Q0Q$

anon

That doesn't change the power of $S$ though.

Jaydon Zhao

sorry I mean

(edited previous equation)

ugh, let me repost

anon

someone's new to $\LaTeX$ :)

Jaydon Zhao

08:07

$QQS^{−k}QQ=Q0Q$

haha, yeah I sometimes forget about {} and ()

anon

Still, that doesn't change the power of $S$ and so proves nothing, right?

Jaydon Zhao

hmm...I guess so

so you would say that (b) is true?

just from that statement alone

anon

I would guess that - in a situation where neither $Q$ or $S$ are known - no alternative value for $k$ could be ruled out, but that's a gut feeling ...

@experimentX Erm, I forgot that won't work because of e^x being really big and all. :) You'll have to move it around so you get a convergent power series first.

experimentX

Hmm ...

Doesn't this work out

Jaydon Zhao

further, if we were also given that $S^{-1}Q = QS^{-1}$ through a process of reforming $QS^{-k}Q$to $S^{-1}QS^{-k+1}Q$ until $-k+1 = -1$, where we have $S^{-k + 1}QS^{-1}Q = S^{-k}Q^2 = 0$

anon

08:16

@experimentX nah, not when $|u|>1$, which is the case when $u=e^x$ and $x>0$.

experimentX

Oh ... so what do we do in this case??

Jaydon Zhao

would that prove it is true for any integer $k$?

anon

@experimentX $$\frac{e^x}{(1+e^x)^2}=\frac{e^{-x}}{(1+e^{-x})^2};$$ it turns out to be a symmetric function :)

experimentX

Hmm ... , can we expand it now??

anon

yes

experimentX

08:19

$ |e^-x| $ is < 1 for x>0

thank you ... i'll try and let you know where i get stuck

anon

@experimentX when you get stuck, refer to the line here with $\displaystyle \left(1-\frac{2}{2^s}\right)\zeta(s)$ in it. :)

experimentX

thank you!! :)

You are farsighted :D

anon

@JaydonZhao Yes that works. $S^{-1}Q=QS^{-1}$ says that $S^{-1}$ and $Q$ commute, hence powers of $S$ commute with $Q$, hence $QS^{-k}Q=Q^2S^{-k}=0$.

Jaydon Zhao

@anon: okay, thanks! :)

experimentX

@anon what to do with these coefficients?? link

Series form of u(1+u)^-2 is $ \sum_{n=1}^{\infty} u^n(-1)^n n $

can i integrate this?? as $$ \int_{1}^{\infty} u^n (-1)^n n dn $$

besides what to do with that (-1)^n

anon

08:36

@experimentX $$\begin{array}{c l}\int_0^\infty x^2\frac{e^x}{(1+e^x)^2}dx & =\int_0^\infty x^2\sum_{n=1}^\infty n(-1)^{n-1}e^{-nx}dx \\ & = \sum_{n=1}^\infty n(-1)^{n-1}\int_0^\infty x^2e^{-nx}dx \\ & = \sum_{n=1}^\infty n(-1)^{n-1}\int_0^\infty \left(\frac{x}{n}\right)^2e^{-x}\frac{dx}{n} \\ & = \left(\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^2}\right)\int_0^\infty x^2e^{-x}dx \\ & = \left(1-\frac{2}{2^2}\right)\zeta(2)\Gamma(3) \\ & = \frac{\pi^2}{6}. \end{array}$$

experimentX

Oh ... amaizing!!

Thanks ... you so much ... though it wasn't my question ... I learned a new thing !!!

anon

no prob

user19161

@anon The way I say it is quite different, infinitely more refined than yours!

anon

Infinitely you say...

skullpatrol

... he said infinitely more...

refined.

anon

08:47

I guess the way Jasper says it must so refined as to be dense in the reals.

skullpatrol

@anon How about "intellectual co-masturbation"?

anon

comasturbation is not a real word, and if you asked a category theorist to guess what it should mean I'd be afraid to hear the answer.

skullpatrol

@anon Well, the more I try to refine it the further away I get from the intended meaning... until it becomes infinitely refined as to be dense in the reals ;-)

anon

09:02

this is hilarious

Jaydon Zhao

do you receive rep for edits? :S

anon

low-rep users receive +2 for suggested edits that are approved by higher-rep users. (this does not include Michael - we can make whatever edits we want freely with no incentive except badges and our names on others' posts) I don't remember what the specific threshold is.

skullpatrol

what is ∫dx/x, ∫du/u, ∫dz/z, ∫da/a, and then, as the clincher, I'd ask about ∫d(cabin)/cabin. Some of them would grin amiably and shout out "log cabin", and they were surprised when I told them that I didn't agree. The right answer (as I learned when I was learning calculus) is "house-boat", "log cabin plus sea".

anon

I remember that one.

skullpatrol

That's a good example of a teacher being one-step ahead of the students.

While teaching them how to walk ;-)

Martin Sleziak

09:19

@anon It seems to be 2k, see here.

I thought it was 4k; so I am not sure whether they changed it, or I remembered it wrong.

anon

thanks

Martin Sleziak

As I've checked now, I have earned 66 reputation from suggested edits.

anon

lol, I earned 2 I think

Frank Science

How to display math equations in chat?

Eh?

Martin Sleziak

@FrankScience See here.

anon

09:28

I normally just link straight to this, saves time.

Martin Sleziak

I guess you're right.

anon

Also it's starred on the right star panel (as well as stickied by robjohn, the room owner).

Martin Sleziak

I just wanted to write that the easiest way is to follow alternate installation link from robjohn's post.

Does the link from robjohn's page work for chorme too?

anon

@FrankScience I remarked on your suggested edit a little while ago.

@MartinSleziak Yes, it's what I use.

However if you've just installed chrome you need to check "Show bookmarks bar" in Wrench > Bookmarks.

Martin Sleziak

Since there used to be Matt N.'s post with a solution for Chrome. Now I see that it's deleted.

Frank Science

09:31

Sorry, I'm using Firefox.

anon

Not a problem.

Martin Sleziak

Even better.

Frank Science

and vimperator

anon

is that like a cross between an imp, a vampire and an operator?

ah, addon

Frank Science

It's vim-like

Let me try.

$\frac12$

$\TeX$

anon

09:37

you will most of the time need to click the bookmark manually to parse the latex (and have to repeat this ...)

Martin Sleziak

Well, unless you use Zhen Lin's version.

Frank Science

I'm looking for a keyboard macro in vimperator.

Can I use Geasemonkey?

Martin Sleziak

I don't use GM, so I can't help you there...

Frank Science

09:59

Let me test again.

$\frac13$

Transcript for

$Mathematics$ Mathematics