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Sam
6:02 PM
@TedShifrin Ah i think I understand
The 2 is treated as a constant
 
Yes, it is a constant ... and so is any power of $2$. :)
 
Sam
So, so far I've got to $2^{\frac{1}{3}}(\frac{4}{3}\frac{(x^{\frac{1}{3})}}{1})$
 
Your original question had $1/\root3\of{2x^4}$. Have you changed the question? Your answer is correct if the question is $\root3\of{2x^4}$ ... not in the denominator.
 
Sam
Ah right yeah I had the 1 / which for some reason I've emitted in my calculation here
Bare with me
I'm on my 4th page of scribbles for this 1 question XD
Wouldn't I just account by adding negatives to my fractions: $2^{-\frac{1}{3}}(\frac{4}{3}\frac{(x^{-\frac{1}{3})}}{1})$
 
Do it carefully.
 
6:11 PM
hullo
I'm stuck on a seemingly obvious integral here and feel retarded
 
Hi @Perturb and @Andrii.
 
I need to calculate a closed curve integral of 1/R dR over a circle R^2 = x^2 + y^2
so I do the usual substitution, x = Rsin(t), y = Rcos(t)
 
Whoa.
Stop.
 
if I get this correctly I should treat it like a regular u-subst and calculate dR/dt
 
Hey everyone!
 
6:12 PM
Do not use $R$ in both places.
 
Hey @TedShifrin :)
 
@TedShifrin yeah, formality. I get that it's a constant radius, that's not my question
 
@Andrii: When you write $dR$, you really mean $d\vec R$. And using $R$ for the radius of the circle is totally confusing. Now what does $1/\vec R$ mean?
 
@TedShifrin what?
 
Precisely. Your question makes no sense because you're using $R$ in at least two or three different meanings.
A curve integral should be of the form $\int \vec F\cdot d\vec R$, if that's your notation.
What is $\vec F$ here?
Better yet, please give us the question exactly as it was given to you.
 
6:15 PM
@TedShifrin am I missing some addon or something because that looks like raw mathjax to me
 
Well, we type in mathjax in here. See the link at LaTeX in chat above and to the right.
 
Sam
How would I remove the 1/ from $\frac{1}{2^{\frac{1}{3}}x^{\frac{4}{3}}}$
Do both exponents negate?
 
ahhhh now it works, sorry
 
Sam
Or, negate the whole expression in the denominator
 
So my question is
 
6:18 PM
Yes, @Sam, both exponents negate.
 
Sam
Great, thanks
 
Hi @Ted, @Perturbative and everyone
 
Heya demonic @Alessandro.
 
Hey @AlessandroCodenotti
 
Enjoying running over people this summer, @Alessandro? :D
 
6:19 PM
I can't drive the car we go on holiday with
 
So Uber it is? @Alessandro
 
Oh, what a relief for the bystanders, Alessandro :) It has a manual shift or they don't have it insured for you? :)
 
We have a limit for the first year after getting the licence forbidding to drive cars that have a too high engine power/weight ratio
 
Oh wow. How interesting.
 
6:21 PM
This still makes no sense, @Andriii.
 
We learn to drive with the manual shift at the driving school in Italy! (And most of Europe I believe?) Automatic shift is a rarity here
 
Sam
OK I've got $2^{-\frac{1}{3}}(-\frac{4}{3}\frac{x^{-\frac{7}{3}}}{1})$
 
Let's remove the radius from the equation and have a different letter. $x^2+y^2=a^2$. I still don't understand the question.
Correct, @Sam! :)
 
Sam
Ted is running on multiple cores right now
 
@TedShifrin hmm. That is literally how the question is presented
 
6:22 PM
Is one of the $R$'s boldface?
Does your book use boldface for $\vec R$ and then write $R = \|\vec R\|$?
 
@AndriiKozytskyi take a picture
 
A line integral (or you're calling it a curve integral) is ordinarily for the work done by a vector field along a path, so one writes $\int_C \vec F\cdot d\vec R$.
(I've only taught this stuff for more than 45 years and written a textbook on it.)
Good idea @Rudi.
 
not possible 45 y ago ;-)
 
Sam
Now I just need to simplify :|
Should I apply the power rule on $x^{-\frac{7}{3}}$ also
 
@Sam: It ain't going to simplify much.
What do you mean? You're not differentiating again.
 
6:26 PM
@TedShifrin well I honestly think this is sloppy notation, you're probably right, it should have an arrow
 
@Rudi: Nor was internet chat :P
 
@Ted so true!
 
But then $1/R$ makes no sense. You need a vector field there, unless they're using the notation that $R = \|\vec R\|$.
 
or emailing the professor ...
 
We need to know the textbook's conventions .. or the professor's.
As it stands, everything is very sloppy.
 
6:27 PM
@AndriiKozytskyi Bold face R like R?
now we guess the photo ...
 
@Rudi_Birnbaum here's an image
for context
 
Sam
$2^{-\frac{1}{3}}(-\frac{4}{3\sqrt[3](x^7)})$ and its derivative is just $(-\frac{4}{3\sqrt[3](2x^7)})$
Because the derivative of a constant is 0
 
Sam, why are you differentiating again?
@Andrii: This has NOTHING to do with what you wrote originally.
 
Sam
@TedShifrin I'm learning how to differentiate :), for absolutely no reason other than personal gain I guess
 
Then write everything with exponents again and apply the power rule.
So, first of all, @Andrii, the notation in the problem is sloppy, but they're integrating with respect to arclength around the circle. But the function is very complicated.
 
Sam
6:30 PM
Is my derivation not correct? The answers provided seem to match
 
@TedShifrin and its horribly sloppy notation, even wrong I guess ...
 
@Sam, when you said "and its derivative is just ..." you're just rewriting the same thing.
 
@TedShifrin I got to the point where I was nearly finished, I had $(r^2 -1) \int \frac{1}{R} dR$
 
@Andrii: The function in question involves distance from a point on the curve to a fixed point $(x,y)$ and they want you to compute the thing as a function of $(x,y)$.
 
Sam
The derivative of the function $y = \frac{1}{\sqrt[3](2x^4)}$ is ${-\frac{4}{3\sqrt[3](x^7)}$
 
6:32 PM
Correct, @Sam, except for the missing $2$ in the root.
 
Sam
Yup I just tried changing it and broke the formula XD. Thanks for your help, much appreciated
 
@Andrii: Your notation is awful. What has the course covered? Do you know things about harmonic functions?
Why are they writing it as $\ln(1/r)$? There's a good reason for that.
There's no way you can compute this integral explicitly by parametrizing. It will be a horrendous mess.
 
@TedShifrin it's a voluntary course with not many people and out prof is an old russian with weird notations all over the place
 
isn't there $dr$ missing?
 
It is a $ds$ (arclength integral of a scalar function).
He's taught you about harmonic functions, @Andrii, I bet.
They have an important property called the mean value property.
 
6:37 PM
@TedShifrin nah, the name doesn't ring a bell
 
Have you studied Green's identities?
 
Yes, that we did
 
And from the Green's identities you deduce some things when the Laplacian of your function is $0$. That's a harmonic function.
You literally cannot compute this integral by hand explicitly for arbitrary $(x,y)$.
 
How do I determine if a fixed point $(x_0, y_0)$ is repelling or attracting for a 2D-map $F : R^2 \to R^2$ with $F(x, y) = (f(x, y), g(x, y))$?
 
Linearize at $(x_0,y_0)$ @G.Ünther.
 
6:40 PM
@TedShifrin Ted :D
It has been like a century since we last spoke :D
How are you sir ?
 
@Kasmir!! You exist!!
 
:D :D
Yes I have been working over summer time
 
I was worried about you!
Good, you doing OK?
 
$\exists$
 
God dag @KasmirKhaan
 
6:40 PM
not too much study :(
@LeakyNun Leaky :D
 
du lever
 
Long time no see guys :D
Så klart jag lever :D
Kasmir took break from math on summer
 
Mathein told me he'd emailed you and that you were alive, @Kasmir.
 
and worked instead :/
 
That's a good thing.
 
6:42 PM
Yes yes Indeed! I told him ill be back very soon :D
really Ted?
That is good?
 
Yes. A break for you was a good idea.
Now you miss math and want to do it ... even though you've forgotten some things.
 
haha okay good then :D
I sart with new energy , yes it will be slow at start
am taking linear analysis
on september and commutative algebra
 
Whoa ... It's gonna be challenging.
Not quite sure what the linear analysis course is ...
 
I guess Ill just take the class of algebra to get an idea not to do exam
 
@TedShifrin you might be able to do it by hand by appealing to the identity $$\sum_{n=0}^\infty T_n(x)\frac{x^n}{n!} = \ln \frac{1}{\sqrt{1-2tx+t^2}}$$ where $T_n(x)$ is the $n$th Chebyshev polynomial of the second kind
but...
 
6:43 PM
it should be easy because its second / third year class
 
@Semiclassical: Not I.
 
@TedShifrin So I was doing this entire thing wrong the whole time, huh. I get that I need to do a lot of reading on the mean value property then, thanks, for now at least
 
@Semiclassical semi :D
 
What are the topics, @Kasmir?
 
6:44 PM
@KasmirKhaan hiya
 
@TedShifrin let me check very fast! =p
 
@Andrii: Sometimes the problems have a development as you go from one to the next. I'd be curious to see what problems were assigned before this one.
But, yes, you need to use some theory that I suspect the professor showed you.
 
@TedShifrin I can link you the whole list, 35 of 'em
 
You're welcome :) Sorry I was stubborn about the meaningless notation. But, as you see, I was right to be stubborn.
I'm curious, @Andrii. Sure.
 
@TedShifrin what is written on the page its about fourir for the most part
 
6:45 PM
I think it's just as likely one would prove the above Chebyshev polynomial identity using the mean value theorem
 
Ah, interesting. Assuming you know Lebesgue integration or not, @Kasmir?
 
@TedShifrin they are saying the course is good for modelling for eg physics
@TedShifrin nope!
 
Ohhh, so probably plug-and-chug solving partial differential equations by separating variables and using Fourier series and boundary conditions.
 
lebesque is advanced analysis here
 
It's more of an engineering-style course usually.
Very computational. Maybe a good idea to balance out the theoretical commutative algebra.
 
6:46 PM
yes it is something to learn =P
I dont plan to do anything with it just to gain more knowledge =p
 
You know I believe in computations :P
 
haha yes :D
 
Did you enjoy your summer?
 
You got me to get an A on that course
Yes i enjoyed work more than just sitting like old summers
 
Good.
 
6:47 PM
what about you sir? :D
 
@TedShifrin How was your summer :D
I hope you got to play bridge with friends and cook :D
 
In fact a question on the Green's theorem came up right before this one, damn
 
I know for a fact you like those things =P
the rest I dont know what else you like ._.
 
You all speak like summer is over!
 
6:50 PM
There's lots of Green's Theorem stuff. But I don't see Green's formulas for harmonic functions. The other questions are reasonably standard applications of Green's Theorem.
@Alessandro: In the US school has already started lots of places.
 
Oh, in Germany lectures will begin on the 8th of October so I also have September free!
 
i dont start school till october
how strange
 
Quarter system is not so many places any more, Eric, but UCSD and Stanford are still quarters.
 
here it always starts just after labor day
 
@TedShifrin Here we start August 27th. But the final oral exam for those upgrading their math from B to A level may be as late as the 29th (so it seems like some people forgot to actually check the dates when they arranged the exams)
 
6:53 PM
@Kasmir: I was supposed to go on a trip to Stanford, Berkeley right now to see friends ... but there are fires and bad smoke in northern CA so I canceled.
 
@TedShifrin yeah I believe he didn't ever mention harmonic functions. Actually, I've made quite a bit of progress brute forcing this particular problem and he didn't point me out that this is a wrong approach entirely, which is why I was so confused
 
sad that CA is on fire
 
I do not know how to do it by brute force except when $(x,y) = (0,0)$.
 
Harmonic functions are the best functions, they have a lot of magical properties
 
that's because they're the analogue of analytic functions on the complex numbers
 
6:55 PM
id say it's the other way around but ok
 
well, @Andrii, to apply Green's Theorem you first have to write the $u ds$ integral as an appropriate $P\,dx + Q\,dy$ integral and then you can use Green's Theorem. I honestly do not know what he's taught you. I know what I teach my students :P
 
@EricSilva that's because you're a PDE person :P
 
Say we are given two CW complexes with the same cell structure, can we conclude that they are homeomorphic?
 
i'm a complex person
 
@Perturbative: Do you mean same boundary maps?
 
6:55 PM
@TedShifrin oh that sad :/ i hope it will be fine soon so you can make that trip :D
 
@Perturbative I'd say yes
 
@AlessandroCodenotti we start very soon in september =p
 
What exactly does "same cell structure" mean?
 
@AlessandroCodenotti i think to say that a lot of the nice properties of complex stuff are really coming from ellipticity isnt just a matter of perspective but a substantive point
 
Note my question!!!
 
6:56 PM
ive said this here before
 
Totally agree, @EricSilva: $\bar\partial$ is elliptic :P
 
I think one really might be able to brute-force the integral using the identity I quoted, but knowing that identity is a big assumption
 
So have I :)
 
Ted !
 
@TedShifrin Without taking into account the attaching maps
 
6:57 PM
No, no, Semiclassic. That's absurd. The rest of the problems are standard Green's Theorem exercises.
 
Fair, I actually agree, harmonic functions are more general
 
You never asked me what I worked :D
during summer :D
 
@Kasmir: I didn't want to pry. What did you do?
 
haha
I worked at a construction site for tunnels
data prossesing
 
@Perturbative: Then it's totally hopeless. You can do some simple examples with different attaching maps.
 
6:58 PM
It's an absurd way to solve the problem, sure. But I don't think the approach itself is absurd.
 
Yeah I'm trying to come up with a few now
 
@Kasmir: Oh, interesting. That's almost like that question you were asking me "for a friend" before you disappeared.
 
@AlessandroCodenotti So like the have the same skeletons
 
You get counterexamples even with two 0-cells and one 1-cell
 
@TedShifrin haha it was for a friend but then they told me to come for work :D
 
6:58 PM
Oh, how cool!
 
@TedShifrin not very high payment ofc, just to help out with some data :D
 
So you actually didn't do hard physical labor. You had to use your brains.
That's great experience for you.
 
Yes was not as hard as doing math but it was something good =p
 
Is there no representative of the mad algebraists crew tonight?
 
Well, I hope my remarks on that question were helpful :P
LOL @Alessandro. They're missing.
 
6:59 PM
Yes ofc :D
Ted I gotta buy you some present :D
 
And I wonder if we'll ever see Balarka again.
 

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