@UserX Is it really called a differential? Some people call it the variable of integration.

@Nick It reminded me of an Iron Maiden song. Be quick or be dead, nothing more nothing less.

@Nick well it makes sense to call it a differential if you take into consideration what happens in multivariable calculus

@N3buchadnezzar And I thought you were pissed off at me for that Sympathies for Slender Man song I linked the last time we met.... yeah, not a good song :(

Then use $x \mapsto 1/u$ on the last integral =)

@Nick Dont remember

@N3buchadnezzar Yay :D Well, I've never listened to Iron Maiden either. Great to get that off my chest :D

@N3buchadnezzar Then what?

@UserX thanks for clarification!

@UserX I might regret asking this but out of curiosity... what do you mean by what happens in multivar?

@Integrator $x \mapsto 1/u$ on the last integral.

@Nick well, multivariable calculus clears things up a lot

@Integrator I know 3-4 ways to evaluate that integral.

Best two are feynman's trick and laplace transform recognition

@UserX that's great!

Or maybe using dirchlet function properties but that's close to cheating

@UserX What was method to cheat on integral using complex numbers called?

@Nick complex analysis? Using the residue theorem isn't cheating...

@N3buchadnezzar $$ \int_1^0 \frac{\log \left(\frac{1}{u^2}\right)}{\frac{1}{u^2} + \frac bu + 1}\frac{1}{u^2}\,\mathrm{d}u$$

$$\int_1^0 \frac{\log \left(\frac{1}{u^2}\right)}{1+bu+u^2}\,\mathrm{d}u$$

I think that the complex analysis method uses the Sokhotski Plemelj theorem which isn't that basic compared to the other methods

@UserX The residue theorem. Interesting, I'll look into it. Also, I'm the guy who considers L'Hospital's Rule cheating in limits :D

@Integrator 100% correct. Awesome now $\log 1/a = - \log a$ and $-\int_b^a = \int_a^b$. Can you use these two ?

@Nick depends on the context. L'hospital's rule can't be considered cheating if you're dealing with limits of unknown functions

$$\int_0^1\frac{\log \left({u^2}\right)}{1+bu+u^2}\,\mathrm{d}u$$

@Integrator the derivative is $\mathrm{d}u = - x^2 \mathrm{d}x$. You forgot a minus sign =)

So, $0$ ?

@Integrator Y!

fun problem ?

@UserX Ah, well, I was referring to the context of known functions. It's nice to think about all the applications mathematical tools have.

@N3buchadnezzar Thanks! You gave me a strong reason to learn these things!

@Integrator Being clever saves you quite a bit of work

@N3buchadnezzar a lot of!

@Nick God knows!

@Nick it's undefined unless you define it as a limiting behavior. Then it's minus infinity

@Integrator Actually it's not defined and when it is it's not finite. I'm sorry, I forgot you didn't integrate that yet. lol

@UserX yes limiting behavior. good word.

@N3buchadnezzar Same here $\int_0^{\infty}\frac{ln(x)}{1+x^2}dx$?

@Integrator Well thats just $b=0$ is it not? =)

@N3buchadnezzar oops I didn't noticed!!

=)

@N3buchadnezzar and why $b>-1$?

@Integrator Well that is wrong. We need $b>-2$. Why do you think it has to be bigger than $-2$?

@Integrator two simple poles at $\pm i$

Have you tried plotting the function (or just the denominator)?

π/4 as residues on each

I guess you know what that implies

@UserX in simple baby words, define a pole.

After i started hanging in chat I got rep'ed. Should advertice that you can get rep'ed on math.exchange much faster than in the gym.

Got it!

@Nick hmm. When you have a point that's excluded by the complex domain but is among other points in the domain I guess

$\pm i$ make the denominator 0 in our case.

@Integrator =)

@UserX Well, atleast I won't imagine polar bears the next time someone says it. thnks =)

@N3buchadnezzar Yes

@Nick :P

@UserX Please see meta.stackexchange.com/questions/9902/…

@N3buchadnezzar (y)

oops... No smiley?

@N3buchadnezzar It's very awkward how an expression like that has to be evaluated inorder to make sense.

The real definition went like this(I think); Let U be an open subset of C, and take an element "p" in U, and a function f:U\{a}->C

Yeah... about the links in chat. I do not care, anarchy woo...

@UserX If that's a pole, then I'm lost in the arctic.

Then define a function g:U->C and a positive integer n. If f is holomorphic on its domain and $f(z)=\frac{g(z)}{(z-p)^n}$ then p is a pole of order n

Order 1 is a simple pole

@UserX ... when is something holomorphic?

But my above explanation might be more intuitive

Yes! Much more!

@N3buchadnezzar Anarchy; that reminds me of someone.

A function is holomorphic in an open set U if it's complex differentiable everywhere in $U\subset D_f$

"Voilà! In view, a humble vaudevillian veteran cast vicariously as both victim and villain by the vicissitudes of Fate. This visage, no mere veneer of vanity, is a vestige of the vox populi, now vacant, vanished. However, this valorous visitation of a bygone vexation stands vivified and has vowed to vanquish these venal and virulent vermin vanguarding vice and vouchsafing the violently vicious and voracious violation of volition!"

@Nick I have this weird feeling that my bounty will go wasted

@UserX did the link explain nothing. Ofcourse it will be wasted. I can't say for sure, but most likely.

The only verdict is vengeance; a vendetta held as a votive, not in vain, for the value and veracity of such shall one day vindicate the vigilant and the virtuous.

It seems a hard problem to tackle so it's ok.

@Nick `?

@Nick Who is Community

It has two active bounties!

@Nick you know what I like

@Integrator Community is a bot.

@Nick So, Bounties?

@Integrator a user opened 1750 rep worth of boundies and deleted his/her acc

$$ \int_0^{\infty} \frac{\mathrm{d}x}{1+x^3}=\int_0^{\infty} \frac{x}{1+x^3}\,\mathrm{d}x = \frac{1}{2}\int_0^{\infty} \frac{1+x^{\phantom{3}}}{1+x^3}\,\mathrm{d}x = \frac12\int_0^{\infty} \frac{\mathrm{d}x}{1-x+x^2}$$

That's why the community bot has bounties

@Integrator Community "Owns bounties from deleted users"

@Nick oops, didn't noticed

@Integrator Please refer to meta.stackexchange.com/questions/19738/…

@UserX Oh, I see

@N3buchadnezzar you can solve that by immediately using partial fractions and reducing it to easily evaluated integrals

@UserX I hate partial fractions

@N3buchadnezzar I remember doing an integral where it was the only way. Do you know of such integrals?

Hi @Nick

@N3buchadnezzar Or define an analytic function under the integral. It then has 3 simple poles. The rest are trivial(I think)

@Nick is there an integral that can only be done by partial fractions?

@Sawarnik Like you, I appreciate the comforts of everyday routine—the security of the familiar, the tranquility of repetition. I enjoy them as much as any bloke.

:D

Sawarnik! Long time no see

Yup, I had gone to Bombay :D

@UserX Indeed

Great place :)

I also met a team from Kochi @Nick

Anyway I'm out to watch a movie

Byes :)

I m happy :)

I hate how sundays always turn out. Headache till 8 pm, fever till monday afternoon

@Sawarnik Aw, a football team?

@Nick No, a school quiz team :P

@UserX Is fever a codename for something? mhh?

@Sawarnik Ofcourse, and did you crush them brutally?

@UserX Well there are different levels to attack problems. You can generalize it further, and use deeper theory, using a rocket launcher to kill a mockingbird. I like that as well. But sometimes the beauty lies in the simplicity. Using the basics to handle seemingly difficult problems.

For that one above one way is contur integration (as you mentioned) or simply rewriting it in terms of the beta function. But where is the fun in that?

@N3buchadnezzar It is illegal in many US states to kill a mockingbird.

@N3buchadnezzar saving time...

@Nick fever is a codename for getting sick from hangovers

@UserX Then use wolfram ffs ;)

@Nick land of the free

@N3buchadnezzar saving time WHILE doing it myself would be a better explanation

@UserX I knew it. Stop doing those. i know you're young but that's no excuse. One day, you're going to wake up old and regret it ruining your health.

Plus, like Chris'sis I love one-line solutions

@N3buchadnezzar free da birds man!

@UserX Yeah, the one above is a one liner

@Nick to be honest, I don't plan having hangover.

@N3buchadnezzar the real one line is probably the beta function(do we have to take limits too?)

@UserX Nah, you probably don't plan what causes the hangover either but man, you have a choice. And we can always choose to be the best that we can be.

@Sawarnik When is thing going to air on TV. Did I miss it? Did you win 1st place!?

$$ \int_0^\infty \frac{\mathrm{d}x}{1+x^3} = \frac{1}{3}\text{B}\left( \frac{1}{3} , 1 - \frac{1}{3}\right) = \frac{\pi/3}{\sin \pi/3} = \frac{2\pi}{3\sqrt{3}}$$

@Nick whatever. When I don't have the time to go out for 7 days straight and then I suddenly have it I just grab it

@UserX Is seven days a long time not to go out?

@N3buchadnezzar yes

@UserX Mmm. but yeah. Its not that hard constructing integrals complex integration will have problems with

I know

Just construct a function with an infinite amount of poles or use a totally discontinous function.

@N3buchadnezzar No, not if your indoor living conditions are good enough.

$$ f(x) = \left\{ \begin{array}{ccc} 1 & \text{if} & x \in \mathbb{Q} \\ 0 & \text{if} & x \not\in \mathbb{Q} \end{array} \right. $$

@N3buchadnezzar Once when my mental problems were very bad, I did not go out for a month. That happened last year in December.

@UserX Going out as in drinking. Not leaving the house, dorm, appartment etc. Geehsh

Bad internet :<

:<

@JasperLoy One week, I got depressed and didn't leave my bed... I didn't leave it for anything... (yes, even for what you're thinking)

@Nick No, obviously I didn't. Regarding the telecast date, I am not too sure too but

I can't remember the last day I didn't leave my house at all.

in on 10-11 am starting from 29 Nov.

@Nick I am sure you went to pee, even if you did not poo.

@N3buchadnezzar I suspect you can help me. Can you give me intuition on why $\displaystyle\int_a^b f(x)\mathrm{d}x=\displaystyle\int_a^b f(a+b-x)\mathrm{d}x$?

@Sawarnik Aw, it's a sunday. I'll be sure to catch it. 10 am on DD. kay. So, which position did you bag.

@UserX Yeah. I have it in my book

@JasperLoy Not to disgust you.... but, 3 empty bottles of Aquafina.

@Nick Well, 3rd. In the semifinals. Better you don't catch it!

@Nick What happened to you then?

@UserX I like to think of it as a reflection. We just rotate the function around $x = (a+b)/2$.

@N3buchadnezzar what book?

Rotations change integrals(right?)

So in some sense we are integrating backwards.. just draw the functions f(x) and $f(a+b-x)$ then the symmetry line $x = (a+b)/2$.

Hmm wait

Have you all seen this map :D

@UserX It is the same function, just reflected. So the area should stay the same right?

@TheArtist I definitely live north

@JasperLoy teenage hormones? I have no idea. I was depressed without a reason to be depressed. I just felt angry at myself because I was sad. I promised myself I wouldn't leave the bed until I was better again....

@Nick Better bed than dead :D

@TheArtist I think this was starred a while back.

@UserX did you understand it?

@N3buchadnezzar tried with f(x)=sin x, a=0, b=π/2. Didn't make much sense

@N3buchadnezzar not too sure about that... the human body does not function properly after such long periods of unhealthy inactivity. Yes but you're right :D

@UserX I can try to draw you an picture, here is the book thingy (folk.ntnu.no/oistes/Diverse/Integral%20Kokeboken.pdf). The part about symmetry and symmetric functions starts on page 22. But I see I forgot to add the images.

Would be glad

@UserX :D

@Nick it's so beautifully drawn, what a creative work

@TheArtist Lots of math guys dwell on art. Not suprising.

@TheArtist Not complex enough...

It's so sad that art guys seem to hate math though... @TheArtist Any thoughts?

@robjohn do you have ideas how to expand the map? :)

@Nick I guess

@N3buchadnezzar any translation?seems an interesting book

@Nick you can then tell math guys hate art too :p

@TheArtist There is not a single person in this room who hates art [citation needed]

I hate art with some exceptions

@TheArtist The general population of artists do not like math. I say this on the basis of the community of deviantart.

@UserX Hence it is a fallacy to say you hate art :D

@Nick I hate Art :p

Two people already in the chat room hate Art

The tooth pick fractal

I made this yesterday

the scan is a bit blurry

@TheArtist But you clearly liked the map of mathland. This is an exception to your statement. Hence you do not hate all art. Also, @TheArtist who hates art. How much of a hypocrite is he.

@robjohn What changes will you make to the map :D Tell me your ideas :D

@Nick xD

Bbl

$$\text{Math is Art and Art is Math}$$ There are $10$ kinds of people in this world: Those who realize this and those who don't. Also, those who didn't know I was counting in ternary.

@Nick Ok , so only 1 person in the chat room who hates art

fail

@TheArtist 1 person is not the vox populi. But out of curiosity, who's that? We've already ruled you out.

@Nick it's @UserX

@MatsGranvik Beautiful, Mat. Did you paint that by hand? What kind of paint?

@TheArtist

Exceptions cause a rule to fail.

@Nick you win :p

@Nick painted by hand yes.

@MatsGranvik Wait, literally painted with hand?

@Nick I used a brush and water colours, painted on metal covered with arabicum, then print through a press.

oooohh, shiny.

but this one is fake

I used a plastic snow flake and put it through the press on water soaked paper.

@MatsGranvik It seems to be convoluted divisor recurrence similar to series reversion giving Catalan numbers. Although, you already knew that since it was your algorithm that made it :D

xD

@MatsGranvik Very resourceful. Bravo.

@Nick it is the convouluted divisor recurrence yes closely related to the mobius function. If I put in {1, -a, a, a^2, a^3, a^4, a^5, a^6, a^7, a^8} I get row sums of the binomial coefficients.

@TheArtist I was making a joke about the lack of a mention to Complex Analysis... Perhaps the North Complex Pole?

Off to the park... BBL

@robjohn Im wondering if I can draw this in a much more creative way....

Along the lines of art in math, I want to recommend a beautiful book by my old friend Frank Farris. It comes out next spring.

Part of his incredible skill in teaching comes from his mastery in providing visual intuition, so I've no doubt the book will be very good.

@TheArtist The artist, have you a larger type of this image?

@UserX Translation ?

Still no sign of Chris's sis?

@MatsGranvik Sure it is 178 and not 174?

@robjohn Well after all she promised me she would tell me when her books come out, so she'll come back at some point :D

@MatsGranvik Could not find anything in OESIS

@N3buchadnezzar what tables?

oeis.org/…

Clever, did not find that. Must have typed in an error.

@MathMan Sorry nope :/

@MatsGranvik Is it possible to search for sequences based on author?

@MikeMiller You can search for words, so in that sense you can search for authors, in the OEIS.

hi

Champion of the sun

ahhhhhh

I'd say the chorus is more of an "ooaaaaah"

@N3buchadnezzar translation on that book you sent me to see an example. It was in Dutch, is there an English translation?

@MikeMiller I bet you'll know this, I keep meeting the word "measure" a lot, however wikipedia is nonsensical(I can't understand anything). Where can I get a simple explanation of what measure is?

Given a set (say, $\Bbb R$), a measure on that set is a coherent way of assigning a volume to subsets of it.

The problem is that you can't do this for all subsets; so you need to restrict to some class of "measurable sets".

Volume in $\Bbb {R^1}$?

This ultimately gives you a way to integrate really messy-looking functions.

Volume, area, whatever you want to call it. I guess you'd want to call it length here.

For the standard measure on $\Bbb R$, we have $\mu([a,b]) = b-a$, like one might like.

That makes sense

Why not $|b-a|$ though?

Same thing, since the notation $[a,b]$ supposes $a \leq b$ :P

Oh... why does measure 0 come up so often then?

hi @Mike @robjohn @UserX

@TedShifrin Good morning...

These are sets that have some sort of "negligible" volume. e.g., $\Bbb Q \subset \Bbb R$ has measure 0 with the standard measure on $\Bbb R$.

Hey @TedShifrin

@MikeMiller but $1,2\in\Bbb Q\subset\Bbb R$ and $\mu([1,2])=1$

@UserX: $\{1,2\}\ne [1,2]$ !!

$[1,2]$ isn't the set containing 1 and 2, it's the closed interval $[1,2]$.

Which is certainly not a subset of $\Bbb Q$!

Oh [a,b] denotes a closed set

That makes sense, I thought it was notation to denote two numbers in this case

Hmmm. A question. If the cardinality of a set is less than $\mathfrak{c}$ (like Q) then it's measure is 0?

hi @robjohn

A subset of $\Bbb R$ with the standard measure? If it's countable, yes. This follows from the axioms of a measure.

anyone good at prob? math.stackexchange.com/questions/1023351/…

@MikeMiller does the measure change if we remove $\{a,b\}$ from the interval?

No, nor if you removed a countable set from it.

So the measure of an interval $[a,b]\in\Bbb R \setminus \Bbb Q$ is still $\mu ([a,b])$?

Your notation is wrong (you should write $[a,b] \cap (\Bbb R \setminus \Bbb Q)$.) But yes, its measure is still $b-a$.

The empty set has measure 0 I guess. Also, measure is positive or 0 and you can find the measures of unions of sets by adding their measures separately. Can we form a group that has elements measures? What will be the operation to satisfy the group axioms?

I don't understand your last question. And you can find the measures of disjoint unions by adding their measures separately.

I am really weak with limits, I just realised again.. Hi @Ted @Mike @UserX

@UserX Not yet. Need to use google translate for it.

@MikeMiller nevermind the last question, it's nonsensical

heya @Studentmath ... I was just typing up a handout for my probability class, since I made a mess of this at the end of lecture on Friday and don't want to take 10 minutes to do it over.

What mess?

@MikeMiller thanks for the explanation, that cleared things up

np

@Ted also, I am making a mess here because of something, probably silly.. What's the handout too, btw?

Any (very) interesting questions there?

@UserX The blue line is $f(x) = \sin x$. If you take a point on the blue line and reflect it over the dotted line you get the red line. Or $f(a+b-x) = \sin(\pi/2-x) = \cos x$.

I was upset about stuff with the students and my brain literally wouldn't work. I was trying to derive the expected number of different coupons you have (with 6 possible coupons) after $n$ trips to the store. @Studentmath

@TedShifrin Better ted than dead =)

What are you being silly about, @Studentmath? BTW, did you see @robjohn's picture of the probability density for $3\cdot$Uniform?

What's the probability of each numbers of coupons, what's the probability of $n$?

No! when did he post it?

And some probability in graph theory, but I got where my mistake was now.

@robjohn seen 10 hours ago ! she is taking break maybe ?

No, @Studentmath, after $n$ trips, what's the expected number of different coupons. (With 6, you expect 14.7 trips to get them all.) Here's the picture of 4.Uniform.

Here is 3

@Studentmath: Since the $n$-sum of the uniform distribution is just the continuous analogue of the binomial distribution, it's no surprise that averages of the uniform are super-well approximated by the normal r.v.

BTW, @robjohn, you made the Piecewise command way too complicated. :)

That's very normal indeed

Wait, I don't get that question at all.

@TedShifrin how is that?

Yeah, the 3-convolution is a cubic spline, basically, and the $n$-convolution is an $n$th degree spline.

Each trip to the store you might get a coupon?

Or how does it work?

@TedShifrin so I put both cases...

Yes, @Studentmath: You get randomly one of the six coupons each trip to the store. Ross does these coupon problems in his book, too.

LOL, yes, @robjohn. I will try today to code the integral, too, out of stubbornness, after I prepare a few lectures.

@TedShifrin code what integral?

the convolution integral ...

@TedShifrin that's the product of each integral

It might be that I was using Convolve command in my code, and it never complained that it was an already-defined command. That might have been part of the problem.

Since it's the convolution of functions with unit integral...