Good night guys

Never seen that before... This is just...

nite @JonasTeuwen

Good night.

@BenjaminLim I think that's putting it a bit harshly, but I do think you're closer to the upper limit of what you can grasp at the moment. I don't mean that in any negative way, but maybe some less "highfalutin" stuff would be more worth your time and energy. You do seem to have trouble with examples. That's always a sign that the concepts aren't properly digested. Maybe slow down a bit. That doesn't mean you should do less, but maybe something slightly less advanced?

@Benjamin sorry. I am not even sure what you're talking about. It was simply a glib remark.

@tb You're probably right.

@tb more foundation? it makes the structure less wobbly

@robjohn yes :)

@tb Let me look at Rotman's text today. There were a few computational things in there and perhaps I can try it out.

@tb For example with tensor products I thought I was in deep water, and then I read KCd's notes and a lot of stuff clicked

@BenjaminLim what topic are you working on?

@robjohn Atm direct limits

don't know what they are myself.

reading

Ack! they are too algebraic...

diagrams all over the place. :-)

@tb Well I kinda felt like a little wobbly after yesteday's flop attempt to construct something isomorphic to the direct limit

I learned a lot from Arturo though

@robjohn Think of the space of test functions: it is the direct limit of the $C_{c}^\infty (B_{R}(0))$ as $R \to \infty$.

@tb You told me yourself to be fearless and not be afraid simply because people toss terms around.

@tb depending on the topology of the convergence, yes

@BenjaminLim knowledge of those terms will overcome the fear. At least the fear that comes from the unknown.

That is the biggest fear I have when starting a new topic.

@robjohn I know. Thanks for the support. I guess part of the fear is the doubt of my capabilities to handle such things.

@robjohn don't be scared by people using those terms and don't be impressed. The point of mathematical words is that they should make things simple, not more complicated!

Once I start reading a bit, the anxiety lessens.

That's why I am not going to just give up like that

tensor product took some time to understand

so will this one

@BenjaminLim Hey, if you can hang on to the world hanging from your feet... :-D

@robjohn what do you mean?

@BenjaminLim It's a down-under joke?

hahahaha

I'm sorry if those are offensive at all.

nah

@tb I will certainly take in what you told me.

thanks

no problem. It's hard to give good advice. Don't be discouraged! You've got plenty of time and you don't have to learn everything right now.

but "I" don't have plenty of time...if it takes me 20 years to learn something...i'm screwed

@DavidWheeler Sounds faster than I learn :D

@robjohn I can't decide whether I find this tongue-in-cheek comment hilarious or mean, probably both.

I like things like that they are fake proofs that people in basic math classes can solve

@tb I don't see it as mean. I see it as someone perhaps showing off a little.

I wish instructors would let us work out proofs like that on our own more, instead of just doing math on a chalkboard for 2 hours

@tb At least that would be the intent I would have had if I'd made that comment.

@Jordan there is nothing barring you from proving things on your own

it is difficult on my own and somewhat pointless when I see the instructor do it, most of the calculus proofs are incredibly difficult

@Jordan Have you tried doing the proofs they do in the book, trying one step at at time and looking back to the book when you're stuck?

Yes they are way too hard for me, I can follow the authors work though

practice this until you at least understand the proofs

@Jordan once you've followed it, try doing it yourself to see if you understand it

@Jordan some things are like that. it's hard to explain (in this limited venue here) why calculus can be hard. the ideas are genuinely sophisticated, and it's rather a shock to the system: ok, i hope you know your algebra because now we're taking it for granted, and heading into OUTER SPACE!!!!!

I mean I can do the more basic ones but the ideas behind them seem a little far out from what we are learning

@Jordan it is a steep set of stairs. You need to get fully on one step before climbing to the next.

continuity, for example...it seems like such a simple concept...but the thing is: intervals like (a,b) on the real line are SPECIAL

Understanding the previous material is quite important in calculus

@tb Thanks so much, I really appreciate your words.

@Jordan Try tackling direct limits!!

or don't :-) not while trying to understand calculus.

@robjohn hahahahahahahaha

what is a direct limit?

hey! .-)

@tb you still in here? what was the name you gave that thm? $\int_0^1 \left| \dot{\gamma(t)} \right|dt=$ length of the curve?

@Jeff the Length of a Curve Theorem :-)

@b is there a name for it?

@Jeff it is pretty much the definition.

i was curious what he called it. i found it in my book and wanted to write that name.

it's a theorem in my friend's complex book

@Jeff I called the left hand side the arc length integral.

(I think)

@Jeff try $\dot{\gamma}(t)$

@tb :-)

@Jordan you may enjoy reading this: jedidiah.stuff.gen.nz/narrow_road.pdf

arc length integral!

it won't let me edit it :(

@Jeff you only have 2 minutes to edit a comment :-)

sorry :-)

@robjohn nice editing job :D

@Jeff If you click on the down pointing triangle to the left of the comment, you can see the long history.

okay, I should really go now. Good night, all!

@tb good night!

@tb goodnight (if i didnt' miss you)

gate, gate, paragate, parasamgate, bodhi svaha!

"Why is math mode delimited by dollar signs? My contemporaries who chose not to stay in math mode made more dollars." - I friggin' love André...

loose translation: gone, gone, gone beyond, utterly gone beyond, hail the enlightened traveller

that is: elvis has left the building

This is going to be quite interesting...

Arc length theorem again: First, the book says $\Lambda(C)=\int_a^b ||r\prime(t)||dt$. Then it says this becomes $\Lambda(\gamma^*)=\int_a^b |\gamma\prime(t)| dt$, where $\gamma\prime$ is the image of curve $\gamma$. So first time it seems to say it evals to the length of the curve, the second time seems to say it's the length of the image of the curve. What's wrong?

@Jeff you're familiar with the usual arclength integral for a parametrically defined curve?

well, that's what i'm working on.

well... if $\mathbf r(t)$ is a vector, then you have to take the norm of the derivative, and integrating that gives the arclength integral.

@JM ok

if you have the complex representation $\gamma=f(t)+i\,g(t)$, you take the absolute value of the derivative. That yields the arclength integral again.

Different formulae for different representations, and all that.

OK. i get that. i'm really asking about the left side. if $||r(t)||=|\gamma(t)|$, and $||r'(t)||=|\gamma'(t)|$. Then how come the first integral seems to be saying it calculates the integral of the curve and the second integral is calculating the length of the image?

@Jeff what is a curve?

@DavidWheeler it's a parametric curve $\gamma(t)=f(t)+i g(t)$.

I don't know why you're considering it as an "image", whatever that means...

no...i mean, what IS a curve...i'm not looking for you to give me a definition, i'm talking about what is your conception of a curve?

@JM that is the way the book is defining it

@DavidWheeler a line (possibly closed) in a plane

ok, you have a book. that's fine. but what is YOUR idea of a curve?

is a circle a curve? a parabola? a helix?

@david yes, yes, and yes

...you aren't ever integrating the curve. Note that what's within the integral sign is nothing more than an arclength element, suitably adapted to the representation you're interested in...

how many dimensions does a curve have (this is sort of a trick question)?

If you make the proper translations, you see the correspondence.

@DavidWheeler well, the curve itself is 1 dimension (no width or height). but it lies in a plane

What can I do to prevent the constant and unintentional pings that I receive that are meant for other users? Or do all the Davids get them?

normally, you deal with what are called "parametrized curves" something like: $r(t) = (x(t),y(t),z(t))$

@DavidK sorry, (i guess that was me). one think you can do is ask us to stop :D

@DavidK we all get them, lol

Everyone does it. I always log in and have 2 or 3 notifications in my MSE mailbox.

@DavidWheeler yes. working in 2 dimensions atm

I see.

Ping ALL THE DAVIDS!

We do have a surfeit of Davids on the entire SE network...

Also Matts and Jonases.

hopefully not too many @Jeff's :D

so there's 2 "competing" ways you can view a curve, as a function (which in your case would be 3-dimensional, one for t, one for y(t), and one for z(t)), or as the image of some part of the real line in two-dimensional space

Suppose I have a subset $B\subset\mathbb{R}^{n}$ that is bounded in the Euclidean metric topology. What can I conclude about the boundedness or unboundedness of the complement of $B$ ?

@DavidWheeler ok (i think i see where you're going).

@DavidWheeler (...but continue)

there is an inherent ambiguity in mathematics between a function and its image

@DavidWheeler OK.

for example, people will talk about "the function $x^2$" when that is actually the image of the function $f = [\ ]^2$

@DavidWheeler Ah, I see where Jeff's having trouble with now...

now, you're integrating over i presume curves in the complex plane

@DavidWheeler yes

we want to "turn these into regular (one-variable) integrals"

ok

that's where "t" comes in

it's our "one-variable thingy"

Anyone feel like helping me with some vector calculus? :)

@DavidWheeler right

Now we have a lot of blue dudes...

to be more precise, in this case the "curve $\gamma$" is actually the image $\gamma(t)$ provided we have a suitable way to parametrize it

@DavidWheeler i think you're heading towards that $C$ is a curve and $\gamma*$ is the image of some function which creates the curve $C$ (but I still want to hear the end of your explanation).

@DavidWheeler ok

almost $\gamma$ is the function, and $\gamma(t)$ is the curve C

@DavidWheeler @DavidWheeler your saying that the "curve $\gamma$" is just some line in the plane. whereas $\gamma(t)$ is the math equation of describing the curve?

line is a bad word...the path you're taking the integral along usually isn't "straight"

@DavidWheeler Right, that's why one usually uses the word contour.

ok, better is that $\gamma$ is just some function (any old math function) and $\gamma(t)$ is the curve traced out by $\gamma$ over the interval $t=[a,b]$?

but...we are going to "adjust it" so that we in fact wind up integrating over a line (a real interval)

@Jeff Yes, like that.

@Jeff...yes, that's a good way to put it

@jm @davidw OK. so next is the "adjustment"

well we're integrating over "little bits of curve"

so we need to know how the "curviness" of the curve affects its length

ok

basically, what we're doing is applying the pythagorean theorem on a really, really small scale

btw, you have my full right now (just so i don't have to keep typing "ok" the way i would not my head if we were talking in person).

@DavidWheeler right

On that note: Jeff, have you seen the usual proof for the arclength integral expression in your calculus studies?

@JM i have, don't recall all the details. but i know it was basically the pythagorean theorem (aka distance equation) summed over small intervals as the limit of the those intervals approaches zero.

"applying the Pythagorean theorem on a really, really small scale" - in parametric form in two dimensions, you'll see something like $ds^2=dx^2+dy^2$.

@JM right

@Jeff So yes, something like that. In the complex plane, you have the absolute value function playing the role of distance.

now, for a "polygonal" approximation, we would have $S = \sqrt{(\Delta x)^2 + (\Delta y)^2}$

@DavidWheeler right

or, in terms of t, $ds = \sqrt{(x'(t))^2 + (y'(t))^2}\ dt$

wait. how did $\Delta x$ become the derivative $x'$?

if $\gamma(t) = x(t) + i y(t)$ this is just $||\gamma'(t)|| dt$

@Jeff You took limits. Differences become derivatives.

@JM ok

Geometrically, you started with a piecewise linear approximation, and consider what happens as the segments become tinier and tinier.

@DavidWheeler you mean that the rhs of $ds=$ is just $||\gamma'(t)|| dt$?

@Jeff this is just a "heuristic" explanation...rigorously, one should prove that this is actually true, and to do that, one needs certain assumptions on $\gamma$

@Jeff yep. more or less.

ds is often called "an increment of arc-length"

@DavidWheeler right. got that. :D

@DavidWheeler k

or "arclength element"

a fairly common condition on $\gamma$ is that it be "piece-wise smooth", which works for a lot of cases

otherwise...let's just say some curves aren't very "measureable" when it comes to arc-length

(e.g. space-filling curves)

if every point is a "corner"...it's a problem

nods as if listening (which I am) :D

but you're going to be mostly concerned with circles and lines, which behave quite nicely

in the complex plane, there's usually just isolated points you want to either avoid, or go around, and circles and lines provide most of the ways to do that

ok

Problem 1 on the page 940 here. What does $$\int_A f_{xy} (x,y)dx dy$$ mean? I mean what does $f_{xy}(x,y)$ mean?

In that regard, you'll encounter this particular theorem of Cauchy...

It is not differentiation $f_{xy}$ but something else?

@hhh it probably means the partial derivative of $f$ with respect to $x$, then partial derive w.r.t $y$.

@hhh it's a mixed second partial derivative i believe

@JM well, next is Cauchy's Thm. But I'm still curious to understand the notation of the arc length thm

$\int_A f_{xy} (x,y)dx dy=\int_A \partial_x \partial_y f (x,y)dx dy$ ?

@Jeff...use it to calculate the circumference of a circle

@DavidWheeler use the arclength integral to calc the circumference? right. i did that.

and what answer did you get?

@Jeff for $\gamma=\exp(it)$?

$2 \pi R$, where $R$ is the radius

@JM for $\gamma=Rexp(it)$.

Okay, seems you've gone through that ceremony... :)

ok, now try an ellipse

@DavidWheeler That's mean... :)

OK. that will take me a bit

@Jeff I'll save you the trouble; it can't be done elementarily.

@DavidWheeler is the ellipse of the form $\gamma=a\cos(t)+bi \sin(t)$

is it? one should be aware of the pitfalls involved in an endeavor....

Probably the lesson David is hammering at is that most curves don't admit neat arclength expressions.

@davidw before i do that, though, i thought I knew where you were heading. but now am getting lost...

"some" arc-lengths come out nice...and others are just "nasty"...it's instructive to see that even a simple curve can pose a difficult challenge

(For the ellipse in particular, one requires a special function to present the result.)

... is the point that $\gamma^*$, the image of the curve $\gamma$ equals $C$?

@Jeff what JM is saying is true, the arc-length of an ellipse is one of those integrals that is "too hard to do" (using the normal techniques)

$$\int_A f_{xy} (x,y)dx dy=\int_A \partial_x \partial_y f (x,y) dy dx$$, by which theorem this was possible?

for the exercise, i'm going to let $\gamma=2\cos(t)+3\sin(t)$.

where $f(x,y)=e^ {x^4-2xy-y2}$.

that's fine, in fact, you could let the "a" be 1, and just have the "b" be something besides 1, it wouldn't make much difference

OK. i'll let $a=1, b=2$... then $\gamma'(t)=-\sin(t)+2\cos(t)$

$\partial_x f(x,y)$ is not possible to calculate directly or?

@hhh sure it is: use the chain rule

@JM oh, hey there! I'll cross my fingers till Friday (you had left just before I entered the room yesterday).

then $|\gamma'(t)|=\sqrt{\sin^2(t)+4\cos^2(t)}$

@tb Oh, good thing you caught me today. I'm actually about to step out in a few... :) (Also, hi!)

i need to integrate that w.r.t $t$.

@Jeff, you can get rid of the sin, but you're stuck with some cos terms

why can i get rid of the sin?

@JM lucky me :) hey, what do you think of this? It's not the first time I see that happen (same user).

$4cos^2(t) = cos^2(t) + 3cos^2(t)$

@Jeff Or you can use the Pythagoran identity.

ok. now i have $\sqrt{1+3\cos^2(t)}$

@tb "Do you have any original thought to put into the matter, or are you waiting for a large spoon?" - :D

(couldn't have said it better myself...)

$=\sqrt{1+\frac 32 (1-\cos(2t))}$

@JM I wonder if that warrants mod attention (repeat offense, as I said). The questions are always of a very poor quality (look at the revision history in the deleted thread).

@JM you mean you won't do people's homework for them? that's just so....perfect....

$=\frac{\sqrt{2}}{2}(5-3\cos(2t))^{\frac 12}$

(Segue: jurors and executioner needed.)

@tb I'd go with a three strikes rule myself... that's the third, I guess?

@DavidWheeler I doubt I can directly calculate it that way. I earlier messed up with indefinite integrals so needs to be careful here... my guess is thtat one cannot calculate directly the derivatives but one can use some rule to calculate the integral just by knowing integral/derivatives cancel there...

and the integral of that is ...

@DavidWheeler I'm here to have fun... one thing I don't miss about studying in an institution is homework. :)

i don't have any homework anymore, so anything i ask here is from pure inquisitiveness

@Jeff when you give up, look at this: wolframalpha.com/input/…

@DavidWheeler yes precisely, the integral cannot be calculated by using the chain rule -- and calculating the innner things there. It is indefinite, the parts explode inside otherwise...

i tried $u$-subst and integrating by parts - no luck (the only advanced integral techniques i can remember) and i tried Maple and got a result too long to type in here which has a term called "EllipticE(cos(t),sqrt(3))".

@DavidWheeler Or this or this...

@DavidWheeler woflram alpha is pretty damn cool. i have to learn to use it.

But there is some theorem which allows me to change the order of integration...Fubious theorem?!

@Jeff yes, that elliptic integral is the special function I was alluding to.

@Jeff just type in what you want to know...or look at some examples if you want some formatting guidelines

@hhh Fubini's theorem

@DavidWheeler Yes, lovely -- I think I am on the right track :D

how do I learn algebra?

@DavidWheeler so that was a good excercise. but either i misunderstood something, or i'm still waiting for the conclusion to why curve $C$ is the same thing as the image, $\gamma*$, of function $\gamma$.

@Jordan get a book and tutor

What kind of book? I cant really afford a tutor

@Jeff often, people identify a function as the "complete graph" of the function....like viewing the upper semi-circle as the set of points $(x,\sqrt{r^2 - x^2})$

and I dont really have an interest in getting a tutor, I haven't ever seen a good tutor

@DavidWheeler so the semicircle is the complete graph of the function $\sqrt{r^2-x^2}$

but you can also view this as $(\cos(t), \sin(t))$ for $t \in [0,\pi]$

Except in that second case you now have a notion of direction...

I cannot understand a word about the $\alpha$ -measurable and $\mu$ -measurable with Fubini's theorem -- well I will integrate then with intuition :P

yes, because we introduced "a third dimension" what $t$ lives in

ok

@hhh it is sufficient for f(x,y) to be continuous on a rectangle [a,b] x [c,d]

(on that note: that bit about direction will be important once you start considering contour integrals...)

well it's important in "beginner calculus", too, it's usually just glossed over

$\int_a^b f = -\int_b^a f$

@DavidWheeler Back when I was teaching, I was surprised at the number of kids who had trouble with that identity...

@JM I'm pretty sure it is... On a different note: have you noticed this bug on meta that you can't look for users that weren't active for some time, e.g. Lugo or Rajesh can't be found using the search form here?

i think because it's "normal" to parse left-to-right, so integrating "forwards" seems natural

@davidw I think you and me are using different understandings of C and \gamma. Let me start over...

@tb Not until you just pointed it out. Odd. So, something different in how main and meta are set up...

Yeah. Only minimally annoying, but still :)

$C=\{r(t):t \in [a,b]\}$ is a curve, where $r(t)=(x(t),y(t))$. and $\Lambda(C)=$the arc length integral in terms of $r$.

@DavidWheeler How can I prove the continuity there?

Just check the differentials?

then it says "we translate $(\alpha(t),\beta(t))$ into function $\gamma:[a,b]\rightarrow \mathbb{C}$. so $\gamma(t)=\alpha(t)+i\beta(t)$.

We know that $e^{t}$ is continuous.

then it says "the image of $\gamma$ is the curve $\gamma^*=\{\gamma(t):t \in [a,b]\}$

there are 3 ways to see a curve in the plane: as the graph of f, or points (x,f(x)), as the image of a parameter (x(t),y(t)), or implicitly as a level-set f(x,y) = k...there are pros and cons to each approach, and the number of pieces of info you need to specify vary with each one

and finally $\Lambda(\gamma^*)=$ the arclength integral (but in terms of gamma).

@Jeff...yes, in its full glory $\gamma$ is a 3-dimensional set

@DavidWheeler ok, so there are different ways to specify the curve. one of them is $C$ and one of them is $\gamma$. in my description, is $C$ equivalent to $\gamma(t)$?

@DavidWheeler $y=f(x)$?

@Jeff isn't this translation just a passage from a function $[a,b] \to \mathbb{R}^2$ to a function $[a,b] \to \mathbb{C}$?

one part of the set (its domain) is one-dimensional (which is why we have a curve, instead of say, a surface)

the other part (its co-domain) is two-dimensional, it's a subset of the plane

o

ok

@DavidWheeler I am lost in the abstraction/vizualization, may I move this into a q?

you can have curves in 3-space, too, like the helix $r(t) = (\cos(t),\sin(t),t)$

@hhh well exp is continuous everywhere, so you just have to check the continuity of the exponent

@tb yes

Hi people!

Hey, Kannappan!

@DavidWheeler ok

@Jeff yes! that is what "C" is....one caveat, one has to count how many times $\gamma$ traverses $C$

Had a good start into the today with a question from group theory.,

sup kanna

(My group theory answer in a long while now.)

$g(x,y)=x^4-2xy-y^2$, is this continuous? Just differentiating?

@anon Hmm, been thinking about Jordan Canon forms and stuff like that...

$[-1,1]\times[0,2]$

hhh: all polynomial functions are continuous in R^n

How are you @anon?

So for instance $\exp(it)$ for $0 \leq t \leq 2\pi$ is, well, different from $\exp(it)$ for $0 \leq t \leq 4\pi$...

pretty good, im with friends in a restaurant

@Jeff because the parametrization $\gamma(t) = e^{it}$ might go around the circle more than once, for example

@hhh that is even better than continuous, it's infinitely differentiable

@anon Hmmm, I would advise talking more to them and less in the room!?!

@DavidWheeler so $\gamma$ is equivalent to $C$ just a different representation (except that it might overlap, or go around more than once).

@Jeff it's better to say $\gamma$ traces $C$, because $t$ has a direction, but $C$ doesn't, it's just a set.

@anon BTW, enjoy your time out there, dude!

@DavidWheeler ahhhh. i see.

...and with that, I should be stepping out. Later (and hopefully not too late)!

later jm

See you, J.M. Have a good day!

See you soon JM

so sometimes, they draw integrals with this little circle thingy with an arrow, to emphasize that we are choosing a kind of "orientation" by the direction which we parametrize in

@MarianoSuárezAlvarez, are you there?

@DavidWheeler so then what is the image of $\gamma$, called $\gamma^*$? what's the diff between $\gamma$ and $\gamma^*$ if $\gamma(t)$ is already tracing a curve?

sort of, but busy right now :)

@MarianoSuárezAlvarez quick question: how do you translate coset to spanish?

coclase

an image is a set....but we're talking about a "directed image"...there's a slight difference

@MarianoSuárezAlvarez Thanks :-)

Can someone help me parse Qiaochu's comment:

I am not seeing what is that abelianization map doing here?

@DavidWheeler you're still talking about $\gamma$, right? $\gamma(t)$ is a directed set of the curve $C$?

First of all, how is parity connected with this homomorphism at all?