Anyone know a more precise way to say that $z^2$ ($z$ is Complex) maps each point on the unit circle to a point "twice as far around the circle"?

@David it's a nice day out by me - so I guess nobody is by their 'puter to help us. :D

I am trying to decide if I should take calc 2 in the summer, its a 6 week class 10 hours a week

is that too fast for calc 2?

@Jordan how did you do at calc 1?

I failed it once and I might fail it again :P

why are you failing it? do you understand the concepts?

well I might get a C or a B but getting a C is like failing it

Not sure why I fail it, I understand the math better than most people in my class but I have trouble remember some things I guess but really I am a horrible tester

depending on your curriculum, calc 1 is often about derivatives, and calc 2 is then usually about antiderivatives. if you don't get the concepts of calc1, then calc2 in 6 weeks will be difficult.

can you do the homework problems assigned?

yes mostly

we are already into antiderivatives and such in calc1

well, even if so, calc 2 will build on what calc 1 taught you (sorta like the way you need addition to learn multiplication).

it just feels like I am wasting so much time and money in college

I have 6 semesters of school done and it feels like I am just starting, because I pretty much am

school is always such a big gamble

maybe if you take the calc course with no other courses you won't have any distractions and can do better in it

yeah, 5 hours of sleep... getting old

I mean if I take just calc 1 in the summer than I won't be able to transfer until 2014

@Ilya it sounds like you're getting young (unless you used to function on 3 or 4 hours of sleep) :D

@Jordan not sure what to say. i'd make sure to get past calc1 this semester first.

@Jeff nope, I've gone to bed at 23-00 and waken up at 4-00 without any chance to sleep again

never happened to me before

I mean I have a B right now in calc

but I have a test and a final left and if I get a C on either one of those I will get a C for the class and my college run will be over

a c in calc 1 is basically impossible to get rid of

@ilya bummer. sounds as bad as my nights (i have sleep apnea).

@Jordan then you'd better start studying! :D :D

doesnt seem to matter how much I study lol

I do try though

@Jeff: I hope for me it was just an occasion

Doesn't matter. Have some Feuerwasser.

@Ilya When you're back shall we visit a bar again? 8-).

@Jonas: I don't have it here (Feuerwasser) and I think I won't find it. Let me just smoke

Okay, and when you had some rest of course.

@Jonas: sure! a cool idea

how are you doing?

it's so hot and dry here, man

I'm okay.

Hot and dry.

Hah, so how are you except the hotness?

hot and wet - like on the beach

that was too much :)

:-).

@Jonas: ChatJax stopped working for me

@Ilya Hmm!

i'm guessing feuerwasser is some kind of distilled spirit?

@DavidWheeler any kind of it :)

so....approximately "rot-gut" as we say here in the states....

@Jeff that is exactly true....squaring "doubles the argument" in the complex numbers

of course, if you're more than halfway 'round, you could wind up less than halfway 'round after you double...

i suppose a better way of saying this, is that multiplying by a complex number represents a rotation-dilation

@DavidWheeler "rotation" sounds good. not sure what you mean by 'dilation'. thx

Stretch or shrink.

If someone has a moment, I could also use an answer to my last comment at math.stackexchange.com/q/128311/22544

you don't have to read the whole question, just the theorem at the top and then my comment.

@tb: hi - seems that I've done with that problem, thanks for the help

@Jeff it's just the length of the curve $C$, so $2\pi R$ in the specific example you ask about.

@tb but $C$ is stretched when it's put through $z^2$. In other words, there's $C$ in the domain - which has radius $R$ - and there's $C$ in the range, which has radius $R^n$

@Ilya no problem, glad to hear that! :)

@tb: I never worked with signed $\sigma$-finite measures - should I expect any difficulties with extension to that case?

Ilya has borrowed my precious Rudin and Folland.

@tb OK. I see now. I read the rest of your message in here (sorry, I replied to you after reading only up to the comma). :D.

So it's the curve in the domain. :D

@DavidWheeler Have you solved that roots of unity problem yet?

Say 2y = y. Obviously y = 0 is the solution but if instead I divide both sides by y I get 2 = 1....so why is dividing both sides by y not valid in this case?

@Jeff But you're just doing the following: you parameterize it $\gamma: [0,1] \to C$ and write $$\left| \int_{C} f(z)\,dz \right| = \left| \int_{0}^1 f(\gamma(t))\,\dot{\gamma}(t)\,dt\right| \leq M \int_{0}^1 |\dot{\gamma}(t)|\,dt = ML$$

(using that $|f(z)|\leq M$.)

@Ilya extension of what?

@tb of the equality with integration of slices we have discussed, i.e. when for each $x$ the kernel $Q(x,\cdot)$ yields a $\sigma$-finite signed measure

@tb Oh, I see now. $\dot{\gamma(t)}$ is the derivative, and so the derivative of the integral is original curve. Sometimes I don't understand the what I'm reading until the 2nd (or more-th) time I see it or someone highlights it like (like here). Thx @tb.

Any know why I cannot divide both sides of "2y = y" by y?

@Ilya You would probably want to make an assumption ensuring that you can still integrate fibrewise (and essentially reduce to the case where you have probability measures). That is, you want a measurable function on the product space such that $\int f(x,\cdot) \,dQ(x,\cdot) = 1$ for (almost) all $x$.

(at least that's the assumption I've often see people use)

@tb got it, so it just restricts the class of kernels, right?

@ymar...it's an easy problem, i just want to know if an induction proof is feasible...i don't think so....

@DavidWheeler I don't see how it could be either, but I don't know. But I think more can be proven. I think this product without the absolute values is equal +/- n. But I don't know how to solve even your problem. Is there a geometric proof?

@Ilya well, basically it expresses that the the measures on the fibres over $x$ vary measurably with $x$. Else you'll probably run very quickly into integrability issues

@tb thank you - maybe I also will find a way not to deal with this case, but just wanted to now which problems to expect and should I try to avoid it

@Jeff I'm not sure if I understand what you mean by saying "so the derivative of the integral is original curve." Do you mean the last integral I wrote is the arc-length integral? If so, then yes.

@Ilya well, usually passing from finite measures to $\sigma$-finite measures is just a matter of an exhaustion argument (of which there are a few). If you find yourself needing it, it shouldn't be too hard. However, if you don't, stick to probability spaces :)

@ ymar...there is an EASY solution: factor $1-x^n$ over the complex numbers

I love probability spaces like monkey's.

@tb: I can't: the work I'm doing now is showing that small perturbations of Markov kernels don't lead to large deviations for measures on finite trajectories. There probability measures are enough. But I've also mentioned that the same approach can be applied to finding bounds on the error while using numerical methods - and there kernels are not necessarily stochastic, so measures are not necessary probabilities

@JonasTeuwen what's this obsession with monkeys you have there, lately?

@tb wait. do you mean that $\gamma$ is the arc length? i seem to be misunderstanding some notation

Not sure, I also like sheep and cows.

Sunday I saw a sheep and I did "Bêêêh", the sheep raised its head and said "Bêêêh" too. I felt this deep connection...

:))

Like we understood eachother.

@Jeff $\int_{0}^1 |\dot{\gamma}(t)|\,dt$ is the arc length of the parameterized curve $\gamma: [0,1] \to C \subset \mathbb{C}$

Try it for the circle of radius $R$!

@Ilya and small perturbations lead you to $\sigma$-finite measures?

@DavidWheeler Right, thanks! I would still like to see if this can be seen geometrically though. :-)

@tb nope, but numerical methods can in general. If they are not good :D

Fourier analysis is beautifully connected to complex analysis. I'm exploring this as we speak!

Fixee seems quite fixated on this question... Nomen est omen?

I don't have arms, but I do have a plant.

@ymar, well i think if you try to see it geometrically, it gets kind of ugly, since subtracting 1 from an n-th root of unity shifts it 1 to the left, and expressing what this does to the modulus...it doesn't look pretty....

@DavidWheeler Exactly, that's how I tried to see it, but I think someone smarter than I am might come up with a neat visualization. I don't know though. For induction, perhaps something like this could be an idea. (It's not strictly an induction, but something similar.) You might try to show it for prime $n$ and then show that muliplying numbers you that are already shown to be alright gives numbers that are alright too.

@DavidWheeler well, you add up the lengths of all diagonals from one corner of a regular $n$-gon

(I'm not saying this gives the result immediately, but there's a nice geometric interpretation)

Hm, a friend asked for a direct proof that tensoring is right exact and I'm not seeing anything that doesn't invoke category theory at some point...

Any know why I cannot divide both sides of "2y = y" by y? Seriously, why is it not valid to do this?

That must suck if you're category theoryly.

@ZhenLin There is a proof

@robjohn Wow, Stein's last book is really adorable.

I wish he told me all he knew about Hardy spaces 8-).

@tb I don't understand... Why add up and not multiply?

@ymar because I goofed :/

@tb Oh. :) But it still works as a visualisation?

My Fourier transform is more convergent than yours.

@ymar well, the lengths of the diagonals are certainly involved, yes.

hey guys

Hiya.

Hi Ben.

@tb the calculation seems...ugly

whoa! an unexpected guest!

Actually I think what I said before is a legitimate induction in the divisibility relation...

@tb Whoa!

@AsafKaragila Would you like some Feuerwasser?

@AsafKaragila hi there!

@tb Hey, after our discussion yesterday on direct limits and stuff

I tried to do more direct limits stuff

Hi Asaf!

and failed terribly, got so confused

No problem! :-).

@DavidWheeler I was more expecting a geometric argument, actually. Still thinking about it.

i was reading on direct limits, and apparently they are actually co-limits, which is confusing

@DavidWheeler that's why (almost) everybody calls them colimits nowadays.

@tb I had a phone call. So the curve is $z=R \cos 2 \pi t + Ri \sin 2 \pi t$, 0 \le t \le 1$?

@DavidWheeler That problem on stuff being "eventually zero"

I tried to set up something isomorphic to the direct limit

and then flopped

Turns out either you bash the algebra

or you construct those usual equivalence classes on $\bigsqcup M_i$

@Jeff I'd write it as $R e^{2\pi i t}$, but yes, (more or less)

the second cosine should be a sine and an $i$ is missing

@tb No wonder in Rotman (as well as in Osbourne) I can only find "colimit"

@JimCS I'm not sure exactly where you're coming from. Dividing by $y$ is perfectly fine in all situations where we can exclude the possibility that $y$ is equal to zero (because division by zero is commonly undefined).

Why is there cork in my wine which has a screw?

the original problem was: prove $\prod_{k=1}^{n-1} |\cos(2k\pi/n) + i \sin(2k\pi/n) - 1| = n$

my approach was to replace the LHS by $|\prod_{k=1}^{n-1} \cos(2k\pi/n) + i\sin(2k\pi/n) - x|$ and note that this is $|(1 - x^n)/(1 - x)|$

Hey did I see Asaf's avatar float away?

@KannappanSampath yes, you didn't dream.

which becomes $|1+x+\dots+x^{n-1}|$, and then i let x = 1.

@DavidWheeler And I would insist on proving $\prod_{k=1}^{n-1} (\cos(2k\pi/n) + i \sin(2k\pi/n) - 1) = (-1)^{n-1}n$

@JimCS But assuming we do not know anything about the value of $y$, if we are going to divide both sides by $y$, we are implicitly assuming that $y \ne 0$ (otherwise, we can't be sure the resulting equation makes sense).

@tb Thank you for confirming--it's a possibility that I wanted to rule out. (Its 4.30 am here. 8-))

@KannappanSampath How did jordan forms go?

@KannappanSampath I had a look at rotman's homological algebra I will look at direct limits again from there

@ymar, but, for example, look how complicated the arithmetic is just for n = 3....

@BenjaminLim Finally, I managed to settle them in my head. I am yet to bug teddy about this. I know he is angry with me but still....

@DavidWheeler I'm not sure I understand... Your proof solves this version as easily as the original one.

@KannappanSampath pick up man

@tb checks out. if $z=Re^{2\pi i t}$, then $\int \left| \dot{z} \right| = 2 \pi R$ (as expected). Confirms that $L$ refers to the arc length of $C$ (in the domain). TY.

@JimCS Is this helpful at all?

@DavidWheeler What arithmetic you have in mind?

for n = 3, we have $(-3/2 + i\sqrt{3}/2)(-3/2 - i\sqrt{3}/2) = 9/4 + 3/4 = 3$, it only gets worse from there

the trouble is, we have n in the denominator of what we're taking the trig functions of, so it's not clear how to do the "inductive step"

and even if we did have some trig identities to apply, they'd be horrendous

now the conjugate pairs will produce real numbers, so we're actually talking about absolute values, and not complex moduli

Haha, a browser that has a feature that it has no tabs. Ingenious!

I agree. This is why I think doing the "additive" induction (increasing $n$ by one) cannot work. There doesn't seem to be anything we could use there. But a "multiplicative" induction seems to give a slight chance of success.

perhaps if we reduce to prime cases

@DavidWheeler That's what I've been thinking, yes.

Not that I know how to do it.

i think if you go that route, you'd still involve invoking cyclotomic polynomials in some form, which amounts to what i did

That would be a failed attempt, yes.

@ymar Did you see the answer I posted to your question on a quick way to show that some element is in some field?

well, the first prime that might be illuminating is 5....

No need to invoke any $6 \times 6$ matrix

@BenjaminLim I only gave it a short look when you posted it, and I forgot about it later. Sorry. I'll take a look tomorrow, I'm too tired to understand field theory now.

ok

no worries @ymar

@DavidWheeler I think I see a nice geometric way to do it for the even case.

(n = 2k)

@tb, well, by using strong induction, we could use that if we had an odd case to go with it

What is the way? @tb

and. please, i'm all ears. yes. i am typing this with waxy lobes. go figure.

You remember how to multiply complex numbers?

in rectangular, or polar form?

But please, let me check again, maybe I've goofed again.

@BenjaminLim which question ( i want to read it, just because..ok, i'm nosy, alright?)

@DavidWheeler link

Bye all of you. A very productive day. Thanks @robjohn @tb @Ben @David . Thanks again to all of you.

hmmmm...

ohmigawd....i've seen that soooo many times. the length of the limit is NOT the limit of the lengths, 'kay?

@DavidWheeler my announced way didn't work as easily as I though, as you probably imagined by now. Sorry...

@DavidWheeler but, see, it's not about the length but about the area...

sometimes area behaves nicer than length...like with the koch snowflake, for example

@tb Could you elaborate? The area of what?

@ymar oh, it was about this question

Oh! sorry.

One of the best browsers! 800 lines of code!

as anyone can see, the area we will get at the end is approx. 0.78539816339744830961566084581988

anyone know anything about fafsa?

fafsa?

(never mind)

Jordan's gravatar reminds me of somebody...

@tb I have discussed that with people many times (the perimeter "paradox")

@robjohn it can be found in one of the top-votes questions/top-votes answers, if I remember correctly

@Ilya Who would that be ;-)

@Ilya I wouldn't be surprised.

@robjohn that one

@DavidWheeler math.stackexchange.com/q/132802/5783

@Ilya take some of the filling from the interior crosses and put it in the exterior circles... reverse the points along the border

looks like the two Gravatars kiss each other...

@tb Can I ask you for some advice?

Tests seem so pointless in math, they dont test math ability or knowledge they test test taking ability

of course.

(but don't take what I say too seriously)

@tb hey! don't even imagine it

@tb I feel that I am in quite deep water now

@tb that is the perimeter "paradox"; the question about the area is just ignorance.

@tb Some people reckon I'm biting off more than I can chew...

@Jordan For me exams are purely for administrative purposes

wow, I have gotten nothing done here today. :-(

so why not make them exceedingly easy?

@robjohn I found a very low hanging fruit :)

@Jordan then they don't even serve an administrative purpose.

@tb Yeah

@tb Maybe a bit of fear inside of me too

administrative?

@tb Matrix probability question?

@robjohn yep.

@tb Do you think I should continue slogging like this?

@tb some key information is in the comments; not that the answer couldn't have been gotten without that information, but...

@BenjaminLim stop slogging and be brilliant :-)

Wow! A Bill for an act introducing a new mathematical truth

@robjohn What do you mean?

@BenjaminLim just kidding, sort of.

@robjohn Well atm I am feeling a bit wobbly

@ymar new?