Mathematics - 2014-11-20 (page 1 of 4)

good night

@DanielFischer Yet another exercise asks me to show that for $a>1$, $z^n e^{a-z}=1$ has $n$ roots in $|z|<1$.

later

I have not yet realized how to apply Rouché.

12:21 AM

@RandomVariable: i like to think of those kinds of things in terms of a riemann sphere, mostly because then the keyhole contour looks more sensible

I feel the need for speed

:D

Hello Professor @TedShifrin

Hi @skull

@Semiclassical In the case of the dogbone contour, I like to attach it to the circle $|z|=R$ by two lines that are arbitrarily close together. Then one realizes that it's really nothing that special.

1:00 AM

@Semiclassical Have you been surprised by how early winter arrived this year?

@TedShifrin

yes, though not as much as i might've been. (i'm from minnesota, we're pretty fatalistic about winter)

@Semiclassical We only had 4 months this year without snow.

ugh, yep

fall was nice, mercifully

1:15 AM

hi semi-c

I can tolerate -20F in January. But it's hard to tolerate below zero in November.

I do bad in my exams :( I think I don't study correctly. How should I study correctly?

how do you study now?

I don't study

I'm very lazy :(

But I wanna start studying

but I don't know how to do it correctly

there is no real "correct" way that works for everybody

1:21 AM

u.u

do you rewrite your class notes?

no

what for?

I think I'm just stupid

to better organize them

that's the real reason I get bad notes :(

@TedShifrin Are you there?

1:35 AM

@RandomVariable: i can stand thsi right now, more or less. it's me looking forward to how long this winter is going to be that makes me gnash my teeth a bit. by the time march comes around, i'm not sure how much psychological stamina for winter i'll have left

No, @Pedro, not I.

Mary Star

Hello!!

I got stuck with the following exercise:

There are three armies A,B,C. Between the armies A and B there is a mountain. Only A and B together can beat the army C. A and B can only comminicate with a pigeon. But it is not sure that the pigeon gets to the other side of the mountain.
Which is the minimal number of successful communication??

Could you give me some hints what I am supposed t do??

@TedShifrin YAY.

I had to show that $\Gamma(z)$ is holomorphic in $\Re z>0$, so I took $\Gamma_n(z)=\int_0^n e^{-t}t^{z-1}dt$ and showed that $\Gamma_n\to\Gamma$ compactly in $\Re z>0$.

Oh, ok ... Or you could use dominated convergence to differentiate under the improper integral?

@TedShifrin We don't have Lebesgue. We integrate Riemann Stieltjes, I think.

Actually only Riemann, with smooth paths.

Real Analysis is next semester.

1:46 AM

Well, you can still prove it for Riemann. :) I assigned that as a challenge problem to my multivariable math class :)

But, ok, no prob.

I would use Morera's theorem.

@TedShifrin What? The dominated convergence theorem holds for Riemann integrals? I don't believe that.

Next semester Pedro can stop pretending he doesn't know measure theory.

@RandomVariable How?

Well, with hypotheses it does. But I meant you can prove differentiation under the integral sign for Riemann, even for improper if you have the appropriate uniform convergence hypothesis on partials.

Right @Mike. He'll know more than me in no time.

But @RandomVariable, then you have to justify Fubini with improper integrals.

But, yes, I love Morera. :)

1:49 AM

@TedShifrin Yeah, you would need to justify switching the order of integration.

@TedShifrin Yes, that I agree.

@TedShifrin Right, I truncated because I used Fubini and Goursat.

@Pedro: Here's the problem I give my students. You can even look up the relevant problem to which I refer. Suppose $f: [a,b] \times [c,\infty)\to\Bbb R$ is continuous, $f_x$ is continuous, $\int_c^\infty f(x,y)dy$ exists for all $x\in [a,b]$ and, moreover, $\int_N^\infty f_x(x,y)dy$ converges to $0$ uniformly as $N\to\infty$ for $x\in [a,b]$.

If Chris's sis was here, she would say this was "elementary" :D

Extend the result of Problem 7.2.20 to prove that, with these hypotheses, if we set $F(x)=\int_c^\infty f(x,y)dy$, then $F'(x) = \int_c^\infty f_x(x,y)dy$.

Anything she knows how to do is "elementary," @skull.

true dat

2:27 AM

@MikeMiller I can measure using a ruler.

The Substitute

hi, off topic. I am in the process of showing that $Aut(D_8)= D_8$, where $D_8$ is the group of symmetries of a square. I have 8 automorphisms, and I have two automorphisms $f,g$ which satisfy $f^4=g^2=1$ and $gfg^{-1}=f^{-1}$. Is this enough to claim that I have an isomorphism or do I have to show that the multiplication table of my automorphisms is the same as the multiplication table of $D_8$?

2:41 AM

@TheSubstitute That's actually pretty on-topic.

@TheSubstitute youtube.com/watch?v=c-VlZK2GLJg Check this out.

The Substitute

@PedroTamaroff will do, thanks

@TheSubstitute It is a very nice video. It should answer all your questions.

3:29 AM

@robjohn Are you there?

robjohn

3:41 AM

@PedroTamaroff I am now

@robjohn I am trying to find the integral $$\int_{|z|=2} \frac{e^{z+z^{-1}}}{1-z^2}dz$$

Note that by $z=1/w$ one gets $$\int_{|z|=1/2} \frac{e^{z+z^{-1}}}{1-z^2}dz$$

So I avoid two singularities out of three.

Now I know that $$e^{\lambda\frac{z+z^{-1}}2}=a_0+\sum_{\nu\geqslant 1}a_{\nu}(z^\nu+z^{-\nu})$$

Where $a_\nu=\frac 1 \pi\int_0^\pi e^{\lambda\cos t}\cos(\nu t)dt$

So I am trying to collect $z^{-1}$ terms.

Since $\frac{1}{1-z^2}=\sum_{\nu\geqslant 0}z^{2\nu}$, the terms picked should be $a_1,a_3,a_5,\ldots$.

But I think that $a_1+a_3+a_5+\cdots$ diverges.

@robjohn

@PedroTamaroff: probably isn't helpful for your purposes, but it's interesting to note that $$\exp(\frac{t}{2}(z+z^{-1}))=\sum_{k=-\infty}^\infty I_k(t)z^k$$

@Semiclassical That's what I said above.

doh

ah, except without the explicit linkage to modified bessel functions (the I_k's)

robjohn

@PedroTamaroff I would just try to find the residue of $e^{1/z}$ at $z=0$ first...

3:50 AM

@robjohn Oh. But where have I gone wrong above?

robjohn

@PedroTamaroff I'm not sure. I am busy doing several things at once here. Sorry.

I nominate Pedro for co-owner of this room @robjohn, since he is so concerned about people formatting their messages in here :-)

@robjohn No problem.

@skullpatrol (I'm the user with most messages here too, almost 90k)

Not that that's a good thing.

All the more reason pal.

The most active user should have some privileges.

Integrator

4:13 AM

@skullpatrol Then how'd you define most active?

most messages

@Integrator how would you define it?

hi

Welcome back

i must be crazy, it is 5:26 in the morning

yawn

Have you learned about residues at infinity, @Pedro?

4:28 AM

@TedShifrin I am not sure what you mean by that.

I know that ${\rm Res}(f,\infty)$ is the residue of $-f(1/z)1/z^2$ at $0$.

Oh, that 's effectively what you did.

I am getting a "value" for the integral, but I don't see a closed form.

Yes, because it's really 1-forms for which we compute residues, not functions.

I mean, if we let $g(z)= e^z/(1-z^2)$, then I am computing the integral $$\int_{|w|=1/2} g(w) e^{1/w}dw$$

And this basically sums the derivatives of $g$ at $0$.

By Cauchy, of course.

Since $e^{1/w}=\sum_{\nu \geqslant 0}w^{-\nu}/\nu!$.

i have not solved the problem with the covering map

4:33 AM

I don't have pencil here. What if you substitute $w=z+1/z$?

@TedShifrin Yes, $dw=(1-1/z^2)dz$ is nice.

Let's see.

@TedShifrin I know that that sends $\Bbb C$ minus the ray $|x|\geqslant 1$ to the upper half plane.

Bijectively, of course.

Well, it's a degree 2 rational map ... But does the integral simplify if we pull back?

I don't know.

I am trying to sort this out.

What's our goal?

Compute $$\int_{|z|=2}\frac{e^{z+1/z}}{1-z^2}dz$$

4:38 AM

Doesn't seem great :(

Oh...

By inverting one gets the same, but this time around $|w|=1/2$.

Where's Daniel when one needs him?

Doesn't sound so fun, so I'll leave it to the experts.

Kaj Hansen

@TedShifrin, what is your opinion on this? redandblack.com/uganews/…

trust me I'm an expert

4:40 AM

Am I being stupid or can't you just do this by writing down the Laurent series?

I think Athena's crap is driving academic policy @Kaj

@MikeMiller Well, you get something, but I am not getting anything nice.

Ah, I see.

How can I find the circle or line in $\Im z>0$ that maps to $|z|=1/2$ by $\frac 1 2 (z+z^{-1})$?

It's completely determined by where 3 points go.

Find where 3 points go and find the only possible circle it could have gone to.

4:42 AM

@MikeMiller So I have to model an arbitrary line or circle?

$A(x^2+y^2)+Bx+Ay+C=0$.

Seems ugly.

bye bye

i have to go to the univerity now

later

I'm wrong, since your thing isn't a Mobius transform.

Why do you think the inverse image of $|z| = 1/2$ is a circle?

have a good night/day (depends on country)

Oh, sorry.

I have no idea what the thing is.

4:44 AM

The inverse image is going to be a potato or something.

@MikeMiller Cannot I find it with Mathematica or something?

note that if $|z|=1$ then $z+1/z$ is real and between -2 and 2

Probably @Pedro but that doesn't sound interesting.

@Semiclassical Yes, that thing sends $|z|=1$ to $[-1,1]$.

If you give me an example with an actual Mobius transform and circle, I can show how to find what the inverse image is.

4:46 AM

@MikeMiller OK. Consider the circle $|z-1|=1$ and the line that passes through $2i$ in the direction $1+i$.

I want to die.

I want to find a Möbius transformation that sends the line to the circle.

And that send $-2\to 1+i$ and $3i\to 1$.

Sure.

That was in an exam of mine.

I didn't do it.

There's an actual algorithm to do this but you can do this entirely algebraically.

Like the term is cross-ratio.

4:47 AM

Give me the algorithm, please. =D

here

if i'm remembering more generically, it maps the interior (and exterior) of the unit disk to the complex plane cut from -1 to 1

I've forgotten how to do this, but most books on complex analysis explain it. I know Conway does, and I'm pretty sure Ahflors does.

Kaj Hansen

Well it looks like your WF's from this semester are now just "W"s @TedShifrin

it's common in conformal mappings for reasons which i'm forgetting how to explain

4:49 AM

This article looks worthless for explaining how to use this to find Mobius transforms. Just look at Conway or Ahlfors or whoever.

forgot about this: en.wikipedia.org/wiki/Joukowsky_transform

see? circles to potatos

@PedroTamaroff But seriously, here, just solve for $a,b,c,d$ s.t. $f(z) = \frac{az+b}{cz+d}$ has $f(-2) = 1+i, f(3i) = 1$, and, say, $f(0) = \infty$.

The last one makes it super nice.

You can even assume WLOG that $c=1$.

Pedro pointed out to me that this is all garbage, since $3i$ isn't on the circle. :P

@Pedro: My hunch was right. My substitution wins ... Then we get a very simple integral and only need to figure out the contour.

@TedShifrin Did you work it out?

The 1-form, yes. Very nice.

5:02 AM

OK @Pedro the right way of doing this is almost certainly to think about cross-ratios. I don't want to.

@MikeMiller Yes, that's what's supposed to be done. I hate this...

I said this elsewhere but I've found it valuable in the past.

And I'm just a kid.

@TedShifrin I am missing something, since from $w+1/w=z$ I get $(1-1/w^2)dw=dz$, but then I get the $1-w^2$ in the numerator and not the denominator.

Keep working.

@TedShifrin I'm not seeing it.

5:11 AM

Heyo.

It turns into $e^w dw/(4-w^2)$. It's past my bedtime! Check with you tomorrow.

Night @TedShifrin.

Hi/bye @Anthony

Oh no, it's Anthony.

Awh.

5:12 AM

Mark Anthony?

@MikeMiller Can I ask you a really easy measure theory problem?

Mark Twain?

@TedShifrin What substitution are you making?

@Anthony I don't know any measure theory, ask Pedro.

@PedroTamaroff?

@Anthony Hello.

5:13 AM

Hola.

I'm just wondering why if we define a function from the left closed right open intervals on $\mathbb{Q}$ that is just $\mu([a,b)) = b - a$, it isn't countably additive?

@Anthony Countably additive where?

Woops. On the family of those intervals in $\mathbb{Q}$.

I need to make use of the fact that I'm in $\mathbb{Q}$, but I really don't know how that changes the left closed right open intervals.

@Anthony But the family of half open intervals isn't a sigma algebra.

Oh. True.

Is this a textbook question or something?

5:20 AM

Yeah, I think I just misphrased it.

Hold on.

You do that every other time you ask a question. ;P

@MikeMiller I know.

I don't know how I manage. Really, it astounds me.

I do that everytime too.

That's why I have no friends.

I don't. Know what my secret is?

Assume now that $\alpha(r) = r$, which is the function that leads to Lebesgue
measure. But now take P to be the family of intervals $[a, b)$ in $\mathbb{Q}$, the field of rational numbers (with a, b ∈ Q). I.e. pretend you have never heard of the
real numbers. Define $\mu_{\alpha}$ as above. Show by example that $\mu_{\alpha}$ is not countably additive on P.

OH

The semi ring $P$. In the reals.

hmph.

Also $\mu_{\alpha}$ what defined as evaluation at the end points.

You know what, just ignore me. I'm rambling and fudging this problem more and more.

Thanks for trying.

5:24 AM

Chill, bro.

That's a good idea.

5:40 AM

@TedShifrin @MikeMiller With the slightly better substitution $w=\frac 1 2(z+z^{-1})$ one gets the integral $$\frac 1 2\int_\gamma \frac{e^{2z}}{1-z^2}dz$$ where $\gamma$ is a countour inside the punctured unit disk. Since the integrand is holomorphic there, the integral should vanish.

:D

Here I am using $q(z)=\frac 1 2 (z+z^{-1})$ is a bihilomorphism from $\Bbb E^\times$ to $\Bbb C\smallsetminus \{x\in \Bbb R:|x|\leqslant 1\}$.

@MikeMiller AW YISS.

wtf is $\mathbb E^\times$

$B(0,1)-\{0\}$-.

Hi @ThomasAndrews

5:42 AM

E is for some German word Daniel knows.

Enkrhautstein.

Hehehe, no.

Ehrstrolkst.

Einheitskreis

What Mobius transformation maps the circles |z-1/4| = 1/4 and |z|=1 onto two concentric circles centered at w=0?

5:44 AM

@flapjackery Möbius transformations are taboo here.

aaaaaaaa

AAAAAAAAAA

flapjackery

how about "linear fractional transformation"? =)

@flapjackery HAIL MÖBIUS.

BURN THE NONBELIEVER.

lol

I will make you a believer

Burn the Non-Believers

hi pal @Twink

hi pal

I'm a believer

5:52 AM

good :-)

:D

:)

pal do you like lady gaga?

sure

what's not to like

:)

6:32 AM

@skullpatrol ~~Are~~ Were you IceBoy?

yes

I think my cognitive biases dislike the new name identicon combo

The identicon is in protest to my teams current girl like losing streak.

0-10

Once they win a game I'll return to the silver & black

pal

do you ever feel like a plastic bag?

in what way?

pal

6:43 AM

drifting through the wind, wanting to start again

I prefer the old name Iceboy :)

sometimes @Twink

So the chatroom is dying is it? We lost chris's sis, we will soon lose JasperLoy

@Committingtoachallenge why will we lose Jasper?

@Twink He said he is quitting at the end of the year :\

6:49 AM

People come and go in cycles, but yes losing chris's sis was sad :(

:(

why did we lose chris's sis?

is she in a better place now?

Pedro told her to format her message.

that's why she left forever?

apparently, check her profile

which message?

6:54 AM

Her profile message. She will probably be back soon, but she was getting serial downvoted apparently and apparently conflict with a few users.

I read once that there's a person that downvotes her answers

I just saw many -2 in her answers list

Did anyone want to review my draft challenge 2.0?

2

It may be a little excessive, 15 textbooks in two years

Any changes?

Do any of your courses use these books?

I will be doing Functional analysis, complex analysis next semester (in 4 months)

and Topology

So they will help with my course work, and in a year I will be doing more related courses, so they should help me heaps for classes

Many of them should complement eachother, which is the only thing that makes me think this is possible at all. The previous challenge I considered impossible

I would start with the books most closely related to your course work :-)

7:04 AM

My Algebra is really bad, and I am doing Abstract Algebra next semester aswell haha, so I am

you're the one

you're the policeman?

nah

I am a raider fan

and what is your job?

I bet Chris's will make a new account if she doesn't come back

7:16 AM

I'll take that bet.

Meaning you are betting with or against me ;P

against

don't be like that, be the nice guy

Yeah pal :)

I bet she comes back with the same account.

7:19 AM

I did say 'if'

She will either be back within three days, or she will make a new account IMO

please listen to this youtube.com/watch?v=XZ3ydHuk6-o

I'll be starting that challenge tomorrow and won't change it until the end of the period, so if anyone does think of anything, let me know please

Listening now Twink

no you are the one

kimbo, c-a-m-p-b(as in boy)-e-l-l

I didn't find it very funny sorry Twink

The police are often overwhelmed with calls, atleast in Australia, so it can be problematic for them to waste any time

sounds like a troll

7:29 AM

The caller?

yep

the nonpolice

:(

?

I love the bacation lady

there are lovable trolls

7:32 AM

and the internet telemarketer too youtube.com/watch?v=W8m-8hcEz1A

jessie :)

and judge judy too

same idea

nullgeppetto

Good morning!

@JasperLoy Is it true that you're leaving at the end of the year? :(

@Twink Yes, I will be busy studying, every day from morning to night. I should not log in.

please log in sometimes :(

we'll miss you

7:41 AM

Well, lots of people here are sick of what I say in this chat.

For example, I keep talking about my 12 holy books and my mental illness.

I'm not sick of that, I like you

me too pal

@JasperLoy You won't be in isolation will you Jasper? That would probably be worse

@Twink Thanks. I hope you find a nice boyfriend and get married soon, lol.

@JasperLoy don't be an isolated point

7:43 AM

I didn't realise Twink was female(or gay[nothing wrong with that])

@Twink By the way, since you are in Europe, have you ever heard of Assimil?

Jasper have you seen my challenge 2.0?

no

@Committingtoachallenge No, what is it about?

It is an extension to the textbook challenge

7:44 AM

@JasperLoy will you be in isolation?

You will probably give me a strong critique

48 mins ago, by Committing to a challenge

Did anyone want to review my draft challenge 2.0?

The first challenge was going to be impossible, so I adjusted it and added some texts that fit better with my class work I will be doing

@Twink Maybe.

@Twink some people learn best in isolation with their textbooks

7:46 AM

@JasperLoy Let us be your social interaction :)

@JasperLoy don't be :(

@Committingtoachallenge My 12 holy books will never change. You are free to read whatever books you want for yourself and your classes.

@JasperLoy But you are an expert of textbooks, so what do you think?

@JasperLoy we like you here, please don't leave us

Are any redundant?

7:47 AM

@Committingtoachallenge I will give you a quick comment on all the other books now...

Okay, thank you :)

@Committingtoachallenge I guess you don't know what a twink is

@Committingtoachallenge Of course, I know about all the books listed there, and I think they are all pretty good for learning from. However, I dislike one of them, which is May's book. It is a very exotic book that goes deep into some topics yet not have the basic theorems. It is used as a text in University of Chicago though.

I thought it was a low level buff character on a game called World of Warcraft

Actually May's book does look very intense

But it is 2 cohn & rudin in

It is only 240 pages long, so maybe I wil just have to push through it

@Committingtoachallenge it's not that

7:53 AM

Having said that, do not choose books on impulse. Look through and think through first before deciding to read the whole book from cover to cover or to purchase it.

Otherwise, you will end up with lots of useless books on your shelf.

no book is useless

@JasperLoy That is very true. I have looked through all of these at a few local libraries and they all seem pretty nice. But Mays and Ddo's texts both looked quite rough

May's was really really dense block text on every page

and Ddo Carmo's book just looked very difficult

But Zorich looked like it might be nicer Calculus-wise for me than Marsden

@Twink What is a twink?

Oh my second assumption was correct?