Mathematics - 2015-02-03 (page 3 of 3)

Axoren

20:00

If A is your starting rule, then it allows the following strings: $\varepsilon, 0, 00$

Consider using a second rule.

evinda

A->000S
S>0S| empty

Like that?

Co Koder

I have a very simple question. Please excuse me if it is so silly:

Axoren

Yeah, that works, I think.

Co Koder

User-1: 8/10 = 80%
User-2: 3/5 = 60%
User-3: 4/4= 100%

All: 15 /19= 78%

All User average = (80% + 60% + 100%)/3 = 80%

Axoren

Did we trigger some sort of chat script?

What is that @CoKoder?

Co Koder

20:03

why these two means are different? 80% vs 78%

evinda

@Axoren Thanks a lot!!! :-)

rschwieb

@MikeMiller Thanks for the help :)

Co Koder

@Axoren I am just trying to calculate the mean for the data that I have, but I realized different mean values different meaning

Axoren

Well, hold on.

I see what the problem is.

@CoKoder how many times does 80% occur? It occurs 8/19 times.

Co Koder

@Axoren you can think of these are test results of different users.

I just want to get overall accuracy

Axoren

20:07

I hate LaTeX sometimes

Co Koder

but if i get the overall accuracy out of per user accuracy, it gives me different results.

Chris's sis

@Axoren However, the fact that $$\lim_{\epsilon \to0} \int_{\epsilon}^{\infty} \int_{\epsilon}^{\infty}\neq\lim_{\epsilon \to0} \int_{\epsilon}^{\infty} \left(\lim_{\varepsilon \to0} \int_{\varepsilon}^{\infty}\right)$$ doesn't mean the values of the integrals will be different.

Axoren

@Chris'ssis You're right, it CAN be equal.

@Chris'ssis However, I'm skeptical that the values will ever be the same for integrals that are discontinuous at 0.

Chris's sis

@Axoren that's why I addressed that question to @TedShifrin. He's pretty (I mean very) experienced in multivariable calculus.

Axoren

@Chris'ssis I know, lol. I'm thinking of reading his book.

infinitesimal

20:13

Check out his YouTube videos too :-)

Chris's sis

@Axoren This is my particular case $\displaystyle f(x,y)=\frac{\sin(x) \sin(y)}{xy (x+y)}$.

Mike Miller

@rschwieb You've corrected my commutative algebra enough times, I'm glad to give back. :)

Exterior

Can anyone help me out with the connection between chain homotopies and homotopies in the sense of cylinders? math.stackexchange.com/questions/1132051/…

Axoren

@Chris'ssis There are too many $\sin \text c$s in your kitchen. That function is horrifying.

Chris's sis

@Axoren hehe, kind of. :-)

Axoren

20:17

LaTeX ruins the timing of another joke.

And the spacing of it.

Co Koder

@Axoren, i am still confused.

Axoren

@CoKoder It's an issue of which probability space you're taking the average in.

Co Koder

@Axoren so, it is not a silly question?

Axoren

It's not silly, it's actually complicated.

Lots of people make this mistake:

Co Koder

I am just trying to take very simple average

so, which one is more accurate? 80% vs 78%

Axoren

20:22

Remember the definition of the discrete expected value: $\frac{\sum_{i=0}^N n_i x_i}{n}$ where $N$ is the number of types of events, $n$ is the total number of measurements of events, $n_i$ is the number of times the event $i$ occurred, and $x_i$ is a numeric event.

In your first case: $x_i = 1$ (they got 100% of a question right) or $x_i = 0$ (they got the question 0% right), $n = 19$ (there were 19 questions asked total)

So, your first average is $\frac{8 + 3 + 4}{19}$.

Your second average is them taken separately.

Mary Star

Is there someone familiar with the german language??

0

Q: Prepare for the german examination

Ich will mich für die Prüfung von das Große Deutsche Sprachdiplom vorbereiten. Wisst ihr wo ich online Prüfungsmodulare finden kann? Kennst du das Buch: "Fromme/Guess, Fit fürs Goethe-Zertifikat C2. Lehrbuch mit integrierter Audio-CD: Großes Deutsches Sprachdiplom" ? Wisst ihr ob es auch ein...

standard-german

Axoren

$\frac{\frac 8{10} + \frac 3 5 + \frac 4 4}{3}$

Since everyone was given a different sized test, this average isn't the same average.

evinda

@Axoren I want to calculate the limit $\lim_{n \to +\infty} \frac{1}{(4n)^{\frac{1}{2n}}}$ ? Can we just say that it is equal to 1 because $\frac{1}{4^{\frac{1}{2n}}} \to 1$ and $\frac{1}{n^{2n}} \to 1$?

Co Koder

@Axoren so, the first one seems to make more sense (8 + 3 + 4 /19=78%)

Axoren

@evinda I believe so, but you should double check with L-Hospital's rule.

@CoKoder The first one "The average number of questions correctly answered." While the second one is "The average score across all tests."

Chris's sis

20:33

If $\varepsilon$ is defined as $\varepsilon=f(\epsilon)$, $f(0)=0$, under the assumption of the absolute convergence we should have that $$\lim_{\epsilon \to0} \int_{\epsilon}^{\infty} \int_{\epsilon}^{\infty}=\lim_{\epsilon \to0} \int_{\epsilon}^{\infty} \left(\lim_{\varepsilon \to0} \int_{\varepsilon}^{\infty}\right)$$. Say that the last integral in the right side can be rewritten to get the integral in the left side plus something that vanishes.

Axoren

Normally, those two are the same, but only because the tests have the same number of questions in them when we do this.

Mathy Person

[text=link]math.stackexchange.com/questions/1132334/… . Can anyone clarify what the answer meant?

Co Koder

@Axoren Thanks a lot. You made it very clear to me.

Axoren

@MathyPerson $^T$ means transpose.

Mathy Person

@Axoren: What is the transpose, then?

Axoren

20:35

$[x_1, x_2]^T = \left[ \begin{array} xx_1 \\ x_2 \end{array}\right]$

iwriteonbananas

Will the real @TedShifrin please stand up ???

Axoren

That's really odd...

Apparently, the array format ignores the first token...

evinda

@Axoren Nice... Thanks :D

Axoren

@evinda Did using L-Hospital's rule show the same result?

evinda

I haven't applied it yet... @Axoren

Mary Star

20:38

@DanielFischer are you familiar with the german language??

Mathy Person

@Axoren: What are the || || on each side of the matrices? I haven't used matrices in a very long time.

Axoren

@MathyPerson The norm. Usually means Euclidean distance unless some other norm is specified.

Chris's sis

@Axoren absolute convergence assures us that there is no slice of volume that blows up to infinity. The $|\epsilon-\varepsilon|$ will make the undesired integral disappear, and the both sides of the initial equality will hold.

This is said in short. (I mean we won't have to deal with the case $0\cdot \infty$)

Mathy Person

@Axoren: That means if there is $(x,y)^T = \sqrt{x^2+y^2}$?

Axoren

Nope, @MathyPerson transpose just means "turn the vector sideways", essentially.

$||[x, y]|| = \sqrt{x^2 + y^2}$

Norm (mathematics)

In linear algebra, functional analysis and related areas of mathematics, a norm is a function that assigns a strictly positive length or size to each vector in a vector space, other than the zero vector (which has zero length assigned to it). A seminorm, on the other hand, is allowed to assign zero length to some non-zero vectors (in addition to the zero vector). A norm must also satisfy certain properties pertaining to scalability and additivity which are given in the formal definition below. A simple example is the 2-dimensional Euclidean space R2 equipped with the Euclidean norm. Elements in...

Mathy Person

20:42

@Axoren: What if there are two vectors, then? ||[2, k] - [1, -3]|| ? Would that be simplified to: ||[2,k]|| - ||[1, -3]||?

Axoren

@MathyPerson There aren't two vectors.

There is a difference of two vectors which is a vector.

Chris's sis

However things might hold under weaker assumptions ...

Mathy Person

@Axoren Alright, then

Axoren

@Chris'ssis I can see absolute convergence can be applied to each integral one at a time.

But we still run the risk of getting a $1 * 2 \not = 1*1$ scenario.

Quaxton Hale

So I'm new to Arxiv.org. Just created an account. Do you think a not-so-serious paper on applying financial engineering to Team Fortress 2 would be welcome? It

Mathy Person

20:44

@Axoren: Would it be: $||[1, k+3]|| = \sqrt{1^2+(k+3)^2}$?

Axoren

@MathyPerson Yes, exactly.

@Chris'ssis Actually, no we don't. I'm stupid right now.

Mathy Person

@Axoren: Would "k" be $-3+2\sqrt{6}, -3-2\sqrt{6}$, then?

Chris's sis

@Axoren :D

Axoren

@Chris'ssis Both limits converge to the same function if taken in either order. Could it be that due to this symmetry we can just take the diagonal path?

@Chris'ssis Does this work for other brands of functions?

Chris's sis

@Axoren Well, in this case it works, but I need to think some more of other cases.

evinda

20:51

@Axoren I verified it... And something else..
The growth rate of a tree in cm per year can be described with the function f(x) = 90*0,87x , where x is the time die Zeit in years after the planting. The tree is at the moment of the planting 90 cm.
How can we find the average growth rate over the first 20 years?

Studentmath

I'm Strugling with some question. Assume for every $1\le i \le k$, $A_i$ is ideal in $R$, commutative Ring with identity. Also, for every $i\neq j$, $A_i+A_j=R$. I want to show that for every $j$, $A_1\cap \cdots \cap A_{j-1} \cap A_{j+1} \cdots \cap A_k + A_j =R$

Axoren

@Chris'ssis Try using a function that has no symmetry in any dimension.

Hippalectryon

I just put ferrofluid on my window, gg -__-

Studentmath

I've already shown that $A_m\cap A_n= A_mA_n$, when $m\neq n$. I have no idea how to proceed - I tried induction on $k$, but that leads me no-where

Axoren

@Hippalectryon you were performing this experiment at home?

Hippalectryon

20:54

@Axoren Yep

@Axoren In my room

Axoren

@Hippalectryon We have a proverb in English: "Don't try this at home."

Hippalectryon

@Axoren I put some ferrofluid in an old pen (without ink in it), but it turns out that it leaked through the bottom :c

And that things stains quite a lot

nullgeppetto

Hi people! I have a problem in evaluating the indefinite integral $\int \cos^n{t}dt$. Could you help me a bit? Actually, it concerns this question! If you like, take a look! thanks!

Axoren

@evinda The keywords "average" and "growth rate" should be enough to answer that question, :P

evinda

I am confused now... Could you give me a hint? @Axoren

Axoren

20:56

@evinda the average of the first derivative along the interval [0, 20].

@Hippalectryon How did it even end up on the window?

evinda

So you mean that it is $$\int_0^{20} f'(x)=f(20)-f(0)$$ ?

Axoren

Magnetic repulsion propulsion?

Hippalectryon

@Axoren I used tape to fix the pen to the window

Axoren

@evinda That would be the sum, to perform the average, you need something else, too.

Chris's sis

@Axoren Well, when having symmetry all seems to be just fine (as expected).

Axoren

20:59

@Chris'ssis Well of course. Integrating one face of a cube and then cubing it will be the same as integrating a cube over every dimension.

Well, in your case, it was more integrating the area along the diagonal of a cube.

Or more like expanding the volume of the cube along the diagonal until you reached it.

All of which are fancy things that are possible with symmetry.

Hippalectryon

@Axoren And now I have powdered silicon (SiO2) on my hands e_e

Chris's sis

@Axoren Yeap. At any rate, I'm interested to see a theorem in multivariable calculus that contains things I said above related to the absolute convergence., and then being enough to use the same speed for both limits. @TedShifrin can help here ...

evinda

@Axoren So is it equal to (f(20)-f(0))/20 ?

Axoren

@evinda That sounds right. It's the average of all growth rate values within 20 years.

Wait.

The function you were given was already the first derivative of a population function, @evinda

evinda

@Axoren So is it equal to $\frac{(\sum_{x=1}^{20} f(x))}{20}$ ?

Axoren

21:09

@evinda You would still use an integral, just not an integral of $f'(x)$.

It would be of $f(x)$

evinda

So is it equal to $\frac{\int_0^{20} f(x)}{20}$ ? @Axoren

Axoren

It should be, yes.

evinda

@Axoren Could you explain me why? I am a little confused now...

Axoren

It goes to the definition of the average.

Studentmath

21:28

I've a nice one for you all:
5+3+2=151012
9+2+4=183662
8+6+3=482466
5+4+5=202504
7+2+5=1435??

Chris's sis

@Axoren consider the function $$f(x,y)=\frac{x y}{x^2+y^2}$$ and see that symmetry doesn't assure the existence of the limit at $(0,0)$.

@Axoren For my problem above, my bet is to make use of the fact that the integral converges absolutely, and then I have that $\displaystyle \lim_{\epsilon \to0} \int_{\epsilon}^{\infty} \int_{\epsilon}^{\infty}=\lim_{\epsilon \to0} \int_{\epsilon}^{\infty} \left(\lim_{\varepsilon \to0} \int_{\varepsilon}^{\infty}\right)$.

evinda

@Axoren Ok, thank you :)

Chris's sis

@Axoren My problem is pretty complex, and this discussion we had today is like a bit of sand in an ocean to the whole thing I have to do.

My aim is to turn my proof into a brilliant one.

brb

Pedro Tamaroff

21:49

@Studentmath Potato.

Studentmath

@Pedro!!

How's the semester going?

Pedro Tamaroff

I'm on summer vacations.

robjohn

@Chris'ssis it can take any value in $\left[-\frac12,\frac12\right]$ in any neighborhood of $(0,0)$

Studentmath

It's summer now?

Pedro Tamaroff

Yes, I live in South America, remember?

Studentmath

21:51

Oh, different pole

That's so odd

Chris's sis

@robjohn Did you use Mathematica?When considering $y=x$, we immediately see the issue there.

@robjohn btw, how do I compute double limits with Mathematica? Maybe it's not that easy.

robjohn

@Chris'ssis did I use Mathematica for what?

@Chris'ssis I'd have to look up the documentation. I have never done double limits in Mma

Chris's sis

@robjohn For visualizing the graph that is very suggestive here.

robjohn

@Chris'ssis No, I just know that $|2xy|\le x^2+y^2$

Chris's sis

@robjohn Right.

Studentmath

21:55

@Pedro Did you go for the combinatorics test after all?

Pedro Tamaroff

For the final? No.

Chris's sis

@robjohn I'm pretty depressed I lost so much time with a single question ... it's unbelievable. To make that proof nice, clear and short is not that easy. At least now I know how to do it!

Studentmath

You still have another shot at it, no?

@Chris how should I feel, being stuck for a couple of hours on a regular-book question?

Chris's sis

@Studentmath I worked for days on a single question. (not exclusively, but I assigned some time every day)

robjohn

@Chris'ssis I''ve done that many times

Studentmath

21:59

@Chris I did it too on some research question. Amounted to month, actually.

Chris's sis

@Studentmath Was about an integral? :D

Studentmath

@Chris I'm afraid not - I am sure it could be somehow converted to a question about an integral, but that wasn't how I approached it :P

Chris's sis

:D

Studentmath

I wonder if there's another aproach to this question rather than induction. I can't figure out how to do it with induction. It seems wrong.

It seems false, generally.

Hippalectryon

22:42

@Emrakul Oh, a foreign mod :P o/

evinda

What is a foreign mod? :/ @Hippalectryon

Hippalectryon

@evinda Wrong ping

user61230

wh-h-- Who hath summoned me from my lair!?

evinda

@Hippalectryon A ok :D

Pedro Tamaroff

@Emrakul Are the boys misbehaving?

user61230

22:46

@Pedro All of them. Every last one. Suspensions for everyone!

user61230

[Nah, everything's fine. Was just pinged s'all.]

Pedro Tamaroff

Ah, OK.

infinitesimal

Nooo not a mod attack

You'll never find me

user61230

@infinitesimal Ping! Found you!

infinitesimal

-_-

Hippalectryon

22:48

@infinitesimal run !!

Kevin Driscoll

@Emrakul Please, sir, allow me to keep my permanents in play

infinitesimal

runs

user61230

@Kevin Fear the wrath of the Strionic Resonator!

Kevin Driscoll

@Emrakul Actually, nevermind. It's not a 'may' trigger. I have to sac my permanents whether you want me to or not.

@Emrakul LOL. I never played M14, so I was not aware such a card existed.

Its like twincast for triggers

user61230

@Kevin The wording on that would be very odd if it were optional. And yeah, it's horrifying.

Kevin Driscoll

22:51

@Emrakul You're right because you don't get to pick which ones are sacrificed, I do. So the 'may' wording there would be quite unusual.

user61230

It'd have to be something weird like "Whenever this creature attacks, you may have target player sacrifice up to [X] permanents." Which is... really strange.

Kevin Driscoll

@Emrakul Indeed! I play Runed Halo, naming Emrakul, the Aeons Torn

user61230

But yeah, in the right deck Strionic Resonator is up there with Isochron Scepter w.r.t. capacity to break decks.

Kevin Driscoll

:-P

Ah wait shit, the way you worded it, you are targetting me not Emrakul...... BLAST

user61230

Oh, yeah, annihilator isn't a targeted ability. Muahaha! Technical gotchas win again!

Alessandro

22:55

@KevinDriscoll the runed halo prevents the damage, but you still have to sacrifice 6 permanents. An oblivion ring gets rid of him

user61230

@Alessandro I play blue. ;)

Kevin Driscoll

@Alessandro Emrakul has protection from colored spells

Oblivion ring can't target him, can it?

Axoren

gatherer.wizards.com/Pages/Card/…

Kevin Driscoll

Wait, no it does

Alessandro

@KevinDriscoll it does target, but it's not on the stack anymore, so it isn't a spell but a permanent and emrakul (thank god) doesn't have protection from permanents

Axoren

22:57

@KevinDriscoll gatherer.wizards.com/Pages/Card/…

Kevin Driscoll

@Alessandro Pretty sneaky, sis.

Axoren

@Alessandro It has protection from White, though.

mtg.wikia.com/wiki/Protection

Kevin Driscoll

@Axoren only white spells, I think

Axoren

You need things that force your opponent to Sacrifice a creature of their choosing. (Doesn't target.)

Alessandro

@Axoren no, it has protection from white spells

Axoren

22:59

Oh, I see.

user61230

Colorless destroy target permanent works, too.

Axoren

The wording on Oblivion Ring is weird.

The spell is to put it into play.

Kevin Driscoll

@Axoren That's why they changed it

Axoren

What is it now?

Ted Shifrin

@Studentmath: If you were my student, you'd be used to being stuck on a problem for hours :P

Kevin Driscoll

23:01

@Axoren Banishing Light

Still works on Emrakul though

Ted Shifrin

@Chris'ssis: Other than needing Fubini's Theorem, improper integrals in more variables are just like in one variable. If you have existence of the integral, then you can do limits any way you want. If you have absolute convergence (or existence of the Lebesgue integral), then you're in good shape. But I haven't looked at your particular question.

hi @Kevin @Axoren

Anyone want to help me grade exams tonight?

Axoren

@TedShifrin Yo.

@TedShifrin No.

user61230

Another one that works is Ensnaring Bridge. Doesn't get rid of Emrakul, but makes me relatively useless until it's removed.

Axoren

lol

Ted Shifrin

ignores @Axoren

Axoren

23:03

javelin through the heart

Kevin Driscoll

@Ted Howdy

@TedShifrin Got a question for ya! If $f$ and $g$ are complex analytic in $U$ and their values agree everywhere on a line which runs through $U$, are $f$ and $g$ the same function?

Chris's sis

@TedShifrin If we consider the integral with different speeds at limits $\displaystyle \lim_{\epsilon \to0} \int_{\epsilon}^{\infty} \left(\lim_{\varepsilon \to0} \int_{\varepsilon}^{\infty}\right)$, then I suppose that under the assumption of absolute convergence, we have that the previous variant equals the variant $\displaystyle \displaystyle \lim_{\epsilon \to0} \int_{\epsilon}^{\infty} \int_{\epsilon}^{\infty}$ where the speed of the limits is the same.

Alessandro

@TedShifrin !from xkcd.com that's about everything I know about grading papers, also I don't know how to put images in the chat.

Ted Shifrin

@Kevin: If they agree on a set with a limit point, they agree period.

@Chris'ssis: For the improper integral to exist, you need the limit to exist independently. If it does, then of course you can specify $\varepsilon = \epsilon$.

Just be thankful you never had me for a class, @Alessandro :D

Chris's sis

@TedShifrin I considered the absolute convergence.

Kevin Driscoll

23:09

@TedShifrin A set with a limit point..... I am not sure a line qualifies

user61230

@Ted You recommended Spivak's Calculus to me a while back - does the version matter much?

Ted Shifrin

Um, every point of the line is a limit point, silly @Kevin

user61230

(Sorry to add a third voice to respond to >_>)

Ted Shifrin

@Emrakul: IMHO, the third and fourth editions are better, and the fourth has text contributed by me :P

@Chris'ssis: You can prove absolute convergence in the improper sense?

Chris's sis

@TedShifrin Well, in some cases the limit in $(0,0)$ might not exist, but still, the improper integral exists.

Exterior

23:11

@TedShifrin could you recommend a quick introduction to connections (diff.geo)?

user61230

Hah! Well, I'll have to see if I can find it. Library has Spivak v1, but nothing more recent locally.

Kevin Driscoll

@TedShifrin But neighborhoods of point on the line all contain many points which are not on the line. Does this not matter?

Ted Shifrin

When you take limits independently, @Chris'ssis?

Um, depends on your background, @Exterior.

user61230

@Ted Does this always happen to you? :P

Ted Shifrin

Relearn your definition, @Kevin.

Kevin Driscoll

23:12

@Ted Oh I see. This does not matter, my mistake inr eading the definition

Ted Shifrin

I dunno, @Emrakul, but I think grading tests would be less stressful.

Ah, 1st edition is still a great book, @Emrakul, but we improved it a lot :P

Chris's sis

@TedShifrin You may consider this example $$\lim_{(x,y) \to (0,0)} \frac{{x{y^2}}}{{{x^2} + {y^4}}}$$

Ted Shifrin

I don't need to consider such examples, @Chris'ssis :P We're talking about integrals.

Exterior

@TedShifrin I was hoping for something categorical in spirit, maybe like Kolar, Michor and Slovak's Natural Operations in Differential Geometry but not as terse. As far as background is concerned I haven't taken any formal courses in diff.geo, but I sneaked peaks in Lee and Boot & Tu

Ted Shifrin

I am totally anti-categorical, so I'm exactly the wrong person to ask, @Exterior. I prefer to do differential geometry with differential forms, but I guess the question is ... what is your ultimate goal?

Chris's sis

23:15

@TedShifrin My point was that even when the limit doesn't exist in $(0,0)$, the integral still exists. I thought you asked me about that.

Hippalectryon

@TedShifrin to conquer the universe

Ted Shifrin

No, @Chris'ssis, I asked if you'd proved that the improper double integral exists.

smacks @Hippa (encore une fois)

Hippalectryon

@TedShifrin calls 119

Ted Shifrin

911 aux États-Unis

user61230

@Ted Fair enough! I've really liked the first couple chapters. I'll see if I can grab v3 or v4 on inter-library loan.

Hippalectryon

23:16

@TedShifrin Nope, 119

@TedShifrin Police in France is 17

Exterior

@TedShifrin I don't have a grand goal. I do have a more appropriate question though. Here (mathoverflow.net/questions/193639/…) I asked about Palais's paper which characterizes the exterior derivative as 'the' linear map which commutes with pullbacks. Palais himself promised to answer but I suspect he might have forgotten. Could you shed some light on the paper?

Ted Shifrin

The exterior derivative seems removed from connections on bundles, though.

I've looked at Palais's paper many years ago. I won't have time to deal with this for about a week, if then.

Exterior

Is it alright to email you in a week?

Chris's sis

@TedShifrin Yes, by inequalities, I bounded the initial double integral with another one.

Ted Shifrin

Sure, @Exterior, but no promises. I'm buried with my classes and trying to get ready to retire and sell my house. So totally busy.

With absolute values, @Chris'ssis? Not doing iterated integrals?

Michael Albanese

23:19

Hi @TedShifrin. I have to run but I just wanted to let you know I said hi to Lawson for you and he said to say hi back.

Ted Shifrin

Oh, cool, @MichaelA ... he's one of my all-time favorite teachers, friends, and mathematicians.

Exterior

@TedShifrin oh, good luck! I have to ask though, are you "satisfied" with the normal definition of the exterior derivative via all those games with minus signs and indices?

Chris's sis

\begin{aligned}
\displaystyle \left|\int_0^{\epsilon}\int_0^{\epsilon} \frac{\sin(\alpha x)\sin( \beta y)}{x y (x+y)} \ dx \ dy\right|&\le \int_0^{\epsilon}\int_0^{\epsilon} \frac{|\sin(\alpha x)\sin( \beta y)|}{x y (x+y)} \ dx \ dy \\
& \le\alpha \beta\int_0^{\epsilon}\int_0^{\epsilon} \frac{1}{ x+y} \ dx \ dy \\
&=2\log(2)\alpha \beta \epsilon
\end{aligned}

Ted Shifrin

No, @Chris'ssis, that integral does not exist.

Chris's sis

@TedShifrin Sorry? What do you mean?

Ted Shifrin

23:20

I define exterior derivative in coordinates, not by vector fields, @Exterior. And uniqueness is easy to prove.

Do you have a typo, @Chris'ssis? But you have to be careful. You cannot work with iterated integrals unless you know Fubini's Theorem applies, so you need to know that the function has a double integral, not just iterated integrals. You have to be very careful here.

Exterior

@TedShifrin thanks for your time. Good luck with retirement business!

Ted Shifrin

Thanks, @Exterior.

Studentmath

@Ted I appreciate hard questions on full stomach, when I am hungry being stuck just makes me irritated

Ted Shifrin

well, @Studentmath, if you didn't spend so much time here, you wouldn't be so irritated.

Studentmath

@Ted might be true - though I don't spend much time here anymore

Axoren

23:24

Oh my god... The product rule applied to an infinite number of products becomes a sum of an equal number of terms in which each term is just as wide.

Studentmath

And won't at all in 4 weeks, or so it seems

Ted Shifrin

@Chris'ssis: To give yuou an idea that you have a problem, think about that integral in polar coordinates.

In all seriousness, @Studentmath, I shall miss you.

That makes no sense, @Axoren.

Chris's sis

@TedShifrin Do I have a typo? Where? Sure I know to apply Fubini's theorem, but I'm afraid I hardly can catch your point. Is there something wrong in what I did?

Studentmath

I will keep the e-mail touch @Ted! Unless you switch e-mails, then you can get rid of me (or update me with the new one, whatever you so desire)

Ted Shifrin

Integrability means a lot more than existence of iterated integrals, @Chris'ssis. I'm telling you, in particular, to look at that integral in polar coordinates, and it will not converge.

Axoren

23:26

@TedShifrin If you remember my function $f(x)$, calculating $f'(x)$ using the product rule for derivatives.

Ted Shifrin

That doesn't make sense, @Axoren. All sorts of convergence issues.

Axoren

@TedShifrin Like how?

Chris's sis

@TedShifrin You mean, say, $$\int_0^{1}\int_0^{1} \frac{\sin(\alpha x)\sin( \beta y)}{x y (x+y)} \ dx \ dy$$ doesn't converge? I tell you that even $$\int_0^{\infty}\int_0^{\infty} \frac{\sin(\alpha x)\sin( \beta y)}{x y (x+y)} \ dx \ dy$$ converges.

Ted Shifrin

Any time you interchange a limit process with an infinite sum (or, worse, product), you are likely to have issues.

No, @Chris'ssis, I don't believe so.

Axoren

@TedShifrin I want to use the fact that $f(x)$ is finitely differentiable, but it's looking far more dangerous.

Ted Shifrin

23:28

I'm not even dealing with infinity, @Chris'ssis. You have problems at $0$.

Chris's sis

@TedShifrin $$\int_0^{\infty} \int_0^{\infty} \frac{\sin(\alpha x)\sin( \beta y)}{x y (x+y)} \ dx \ dy=\frac{\pi}{2}\log\left(\frac{(\alpha+\beta)^{\alpha+\beta}}{{\alpha}^{\alpha} {\beta}^{\beta}}\right), \space \alpha, \beta>0$$

Axoren

@TedShifrin Actually, nevermind, it's not even finitely differentiable. It just gets worse and worse. I thought it might be finitely differentiable.

Ted Shifrin

I don't care about your formula, @Chris'ssis. What you're writing doesn't make sense. Do you think $\dfrac1{xy}$ is integrable on $[0,1]\times [0,1]$?

Axoren

But the first derivative is just as horrid as it sounded.

And infinite sum of infinite products.

Ted Shifrin

You have to worry about all sorts of uniform convergence issues to even begin to make sense of things, @Axoren.

Chris's sis

23:30

@TedShifrin But did you consider the behaviour of $\sin(\alpha x)/x$ near $0$?

Ted Shifrin

Yes. I'm complaining about other things.

Axoren

@TedShifrin Yeah, right now the only things I have outside of the definition of the function is that I have the intervals on which it's negative, positive, and I have an exact value for exactly one point.

@TedShifrin Other than that, it's a really volatile thing.

Oh, and I am sure that it's differentiable on all of $\mathbb R$. Doesn't mean it's easy to do.

Eric Gregor

Is this problem non-trivial: math.stackexchange.com/questions/1132417/…

I read it in a context which implies that it is

Hippalectryon

@TedShifrin i'm gonna try to go to sleep with all the awful chemical smell in my room e_e

Good evening/Night/Whatever

Julian Rachman

@TedShifrin Hello!

Ted Shifrin

23:39

night, smelly @Hippa

hi @Julian

Hippalectryon

@TedShifrin Blame ferrofluids and powdered Silicon

Julian Rachman

@Ted How is everything going at Georgia

in*

Kevin Driscoll

@JulianRachman Probably going to hell, the same way all thigns at UGA are going

Ted Shifrin

bumbling along, @Julian, and CA ?

Julian Rachman

@Kevin Haha

@Ted It is good. We got measles going around at the local Disneyland.

Ted Shifrin

23:43

Yes, I know, @Julian. in fact, the wife of one of my old friends/students diagnosed the first case up in Oakland.

Kevin Driscoll

@TedShifrin When I moved universities I almost didn't have to switch insults. At Duke, we say "Go to hell Carolina" and at Tech they say "To hell with Georgia". Not very original.......

Chris's sis

@TedShifrin don't let me go to sleep more puzzled ... I thought my work is perfect.

Julian Rachman

@Ted wow. That is disappointing to hear

Kevin Driscoll

@JulianRachman @TedShifrin It's pretty shameful. I'm of the opinion that if the parents responsible are identified that they should be liable in civil court.

Ted Shifrin

The integral $\int_R \dfrac1{x+y}\,dA$ does not exist on $R=[0,1]\times [0,1]$.

Julian Rachman

23:45

@Kevin True.... true....

Chris's sis

@TedShifrin You mean $$I=\int_0^1 \int_0^1 \frac{1}{x+y} \ dx \ dy \neq 2\log(2)$$?

Ted Shifrin

No, the integral does not exist, @Chris'ssis.

Put it in polar coordinates and try.

Kevin Driscoll

Indeed not, that pesky corner.

Chris's sis

@TedShifrin It depends on how you look at the integration limits.

Ted Shifrin

Iterated integrals is not the same thing as integrable.

Chris's sis

23:57

@TedShifrin I know your point, but still, you can look at that integral in the sense that lower limits are strictly higher than $0$ ...

Kevin Driscoll

@Ted @Chris'ssis I learned this the hard way. One can only write the integral over a region as an iterated integral if the previosu itnegral exists

Chris's sis

@KevinDriscoll I know, but it depends on how you look at the integration limits.

Ted Shifrin

For example, @Chris'ssis, $$\int_1^{\infty}\int_1^\infty \frac{x^2-y^2}{(x^2+y^2)^2}\,dx\,dy = - \int_1^{\infty}\int_1^\infty \frac{x^2-y^2}{(x^2+y^2)^2}\,dy\,dx !!$$

Chris's sis

$$\lim_{a\to 0^{+}}_{b\to 0^{+}}\int_a^1 \int_b^1 \frac{1}{x+y} \ dx \ dy$$

Ted Shifrin

But iterated integrals are supposed to represent a double integral, and the double integral does NOT exist.

Chris's sis

23:59

@TedShifrin ^^

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