you'd have $\sqrt{2}\leq |1/G_c|\leq |1/G_o|+1$

and so $|1/G_o|\geq \sqrt{2}-1$

so $|G_o|\leq \frac{1}{\sqrt{2}-1}=\sqrt{2}+1$

which is weaker than what you want, hrm

Yeah I don't get it either

I think a geometric pov is maybe helpful. Suppose you've got $w=z/(z+1)$. That's a linear fractional transformation, so it sends circles to circles

it's probably not true if the triangle inequality didn't work

Oh no, not a geometric approach!

because triangle inequality should be able to prove when a circle around 0 after translation is inside some other circle around 0

in particular, it'll send $|z|=1/\sqrt{2}$ to some circle (or possibly some line? I'm blanking on this)

@TedShifrin what did you think of the proof today?

banned

Did someone say geometric approach?

Hi Demonark ... Apparently Semiclassic did.

oh, I've got it backwards a little

@Semiclassic: You have to think about what maps to $\infty$ to decide if it's circle or line.

I guess here's a good challenge: Is there a $z$ with modulus $\sqrt{2}$ such that $w=z/(z+1)$ has modulus $\sqrt{2}+1$?

@TedShifrin right

@Faust: What proof? There isn't any proof.

@TedShifrin of the RH did you read it?

@mercio It should be true :P

There's no proof, @Faust. Just move on.

thats too bad

There's nothing there there

I just emailed you about this, perfect timing, but I'm working through chapter 1 of Hatcher right now, have about a week before class starts (minus time for packing and all), so I may not be able to do all the problems just yet.

Would you be willing to assign a pset of sorts, some problems to focus on at least for now?

Demonark: Whom are you talking to?

You

Oh, me. I just got an email.

Are you planning on taking a class for which this is a prereq?

Oh, I read the email.

OK, let me see what I can do.

Thank you very much! :)

@TedShifrin we havent deifned what a manifold or submanifold is in class yet so was alittle lost on your hw assinments

or an immersion

Demonark: So you've done some exercises in Chapter 0 to get ready?

Oh, wow, @Faust, so what have you done so far? Maybe I can find something.

@Lozansky Take $G_o(i\omega)=\sqrt{2}+1$. Then $G_c(i\omega) = \frac{1+\sqrt{2}}{2+\sqrt{2}}=\frac{1}{\sqrt{2}}$

ill email you the book and what we have covered

You already sent me the book, @Faust.

I had been using Rotman for the more preliminary stuff

Just tell me what.

Demonark: There's important stuff in Chapter 0, though.

in that case $|G_c(i\omega)|=1/\sqrt{2}$ but $|G_o(i\omega)|>\sqrt{2}$

@Semiclassical Hmm

So yeah, that bound is definitely too strong

Chapters 0-2, which went through some preliminary topological stuff (mapping cylinder, homotopy, some stuff on simplices and affine maps). I did every problem there

But yeah I'd be down for doing some of chapter 0 problems as well

@TedShifrin we have made it to section 4.4

I don't know Rotman's book, Demonark.

If you've got some other conditions on $G_c,G_o$ then you might be able to strengthen it

OK, @Faust. I'll get back to you after I deal with Demonark's Algebraic Topology.

but in general it's definitely not true

starting at the beggining

@TedShifrin also algebraic topology is kool!

watched some lectures on the weekend

Ah.

learned some kool things about the free group on 2 letters

everything is isomophic to a quotient group of it

@Semiclassical Closed/open transfer functions so they are in the laplace domain

got nothing, sorry

Np, I'll ask my prof - I don't buy it either

Changing my gravitar

@Lozansky yes you helped me a lot with expressing the solution to the wave equation

@Faust: Oh, 4.4 in chapter 1. Well, you could find my exercises on differential forms and vector fields. There are plenty of exercises in the book. Your prof can't be bothered to give you a list of ones to do each week? Have you tried picking some?

@Ultradark Np, hope it worked out

Got it, thank you very much!

After you spend 50 hours on 'em, you won't thank me, Demonark.

Too bad Balarka isn't around to help when you get stuck :P

What does it mean for a wave to intersect another wave

@Ultradark The waves are present in the same place at the same time.

from a mathematical perspective I hope you can just treat them as passing through eachother and not interacting

thanks

can you write a differential equation to capture the intersecting waves

But they do interact (by interference), the interaction just doesn't change the original propagation of the wave.

I guess that would just be the wave equation equal to another wave equation, and solving this differential equation would give the locations of the waves intersecting each other

heya @Fargle

@Faust: You have mail.

Howdy @Ted

so there's some localized interference pattern at the intersection point...?

I'll find someone to bug, what I lack in math I make up for in precisely this :P

Demonark: We miss Balarka. :) ... I've talked about a lot of those (but not all) with grad students over the past decade ... The one about the fundamental group of the identification space built out of the cube I did with one of my advisees, and it was a blast. (I told you to get colored pencils/pens.)

I guess @Fargle was working through part of Hatcher. He might be fun to corner on some.

That was chapter 2 stuff.

And I didn't do my due chapter 0 diligence.

ohhh ... Demonark is concentrating on chapter 1.

But I suggested a few in chapter 0.

I'll still gladly be cornered about it.

Anything to escape diff. top. :P

Or that algebra question ... Demonark, did you solve my algebra question?

Remind me what the polynomial is again. I want to make sure I haven't made a sign error.

$x^4-10x^2+1$

Which one again?

which, btw, is the irreducible polynomial for $\sqrt2+\sqrt3$.

To show that polynomial is irreducible over $\Bbb Q$ but reducible over every $\Bbb F_p$.

It follows from some algebraic number theory stuff that this is, in fact, the generic situation :P

@Semiclassical Actually, the Mobius transformation may work. So say we got $w(z) = \dfrac{z}{z+1}$. So we have $|w(z)| = |\dfrac{z}{z+1}| \leq \dfrac{1}{\sqrt{2}}$ or, with $z=x+iy$, $x^2+y^2 \leq 0.5((x+1)^2+y^2) \Leftrightarrow x^2-2x+1+y^2 \leq 2 \Leftrightarrow (x-1)^2+y^2 \leq 2$, i.e. $|z-1| \leq \sqrt{2}$

Oh I guess if we know that $\sqrt{2} + \sqrt{3}$ is a root, then we can use that $\mathbb{Q}(\sqrt{2} + \sqrt{3}) = \mathbb{Q}(\sqrt{2}, \sqrt{3})$ is degree 4

Yeah, I had made a sign error. I finally found a factorization mod 7

so you're changing from being interested in a bound on |z| to a bound on |z-1|

OK, cool, @Fargle.

yeah, that could make a difference

Only a few more $p$ to go! :P

This gives irreducibility over $\mathbb{Q}$

I'll buy that, Demonark.

@TedShifrin howdy

Oh wait hmm I might have an idea how to do mod p using Galois theory

It's a trick I've seen in an algebra pset

Well, you know fancier stuff, of course. I meant it as an elementary problem. Takes only a tiny bit of group theory.

@TedShifrin I did all the ones in the textbook that we covered

Oh, did you, @Faust? You should have told me that. Did you do them right? :)

God only knows. I did however say originally that I did them all

Oh, I missed that. Sorry.

It was before u sent the assinments

Well, if you're not sure about the ones I listed, you can type some of them up and send them to me if you want.

They weren't very hard

Let me see what I can find in mine on these things.

The Galois idea is that if $F$ is a finite field, and $m,n\in \mathbb{Z}$, then the degree of $F(\sqrt{m}, \sqrt{n})$ is either 1 or 2, because the Galois group is cyclic and if it were degree 4 it'd be $(\mathbb{Z}/2\mathbb{Z})^2$. But then you can say that it's a polynomial of degree 4 with a root that's an element of degree at most 2, so it's a multiple of the linear/quadratic minimal polynomial

I'll ty0e up solutions if your willing to read them?

@TedShifrin

Not zillions, but a few interesting ones, sure, @Faust. I'm sending you a few from my problem sets to look at. Hang on.

OK, @Faust, I sent another email.

@Fargle what factorization did you get?

So is this system analytically solvable? $y = A\sin(kx)$ and $x = B\sin(ky)$ for some $A$ and $B$. The intersections are given by: $x = B\sin(kA\sin(kx))$ and $y = A\sin(kB\sin(ky)) $

@Ultradark: I would bet money the answer is no. Of course, you can do it numerically with Newton's method.

@Semiclassical $$x^4 - 10x^2 + 1 \equiv x^4 - 3x^2 + 1 \equiv (x^2 - 1)^2 - x^2 \equiv (x^2 + x - 1)(x^2 - x - 1).$$

Ah, good, @Fargle.

Nice

@TedShifrin how can you just tell on first sight?

Because most transcendental equations can't be solved in closed form. You can't solve $\sin x = x/10$ other than numerically. I suppose I don't know a proof, though.

@Fargle: So how many $p$ are left to do? :D

Either infinitely many or none.

LOL

Demonark: There's also the result that (unless there's wild ramification or something like that), the Galois Group mod $p$ injects into the Galois Group over $\Bbb Q$.

@Fargle: How 'bout $p=3$ and $p=5$? Same trick or different trick?

@TedShifrin 3: $(x^2 + 1)^2$, 5: $(x^2 + 2)(x^2 - 2)$

Oh hmm, actually now I see what you're trying to do

The idea I’m playing with on paper is to start with a generic factorization into quadratics

Interesting: mod $3$, we're really doing $(x^2+1)^2 - 0^2$ :P

Or wait no the idea I had in mind was dumb

And huh, that's pretty nifty actually

Semiclassic, Demonark: The hint I gave @Fargle was to try to write it as a difference of squares three different ways.

@TedShifrin Ahhhh. For $7$ I could have also done $(x^2 + 1)^2 - 2x^2$, giving $(x^2 + 3x + 1)(x^2 - 3x + 1)$.

Right, cuz over $\Bbb Z_7$ we have $2$ is a perfect square.

Dang it, this is gonna require quadratic reciprocity.

whistles in the dark

But really it's just a very simple group theory statement.

I don't know nothing fancy in number theory.

I solved a transcendental equation

it was fun

Me either, @Ted.

But I think this problem is sooo cooool.

Ted do you mean closed form as in what

expressible in terms of elementary functions, roots, exponentials, logs, etc.

I solved one in closed form and got 5 upvotes

I'm happy about that

must have been a spectacularly special equation

$e^{1/ln(x)}=x$

Oh, I remember when you were playing with that thing.

But that's easy, right? $1/\ln(x) = \ln(x)$, so $\ln(x) = \pm 1$.

That is indeed spectacularly special.

well I added a t

in the exponent

in the actual question

so the solution is

$x=e^{\sqrt{t}}$

Still spectacularly special.

or $x=e^{-{\sqrt{t}}}$

Try it with $e^{\sin x}= x$ !! :P

Yuck

I can't solve that one exactly

I would bet that no one can.

I posted a question about the Black Scholes model here: math.stackexchange.com/questions/2929478/…

I am concerned that it is not on topic.

Is it?

How do I adjust the font size for the next X amount of symbols of a total of T symbols required to write some algebraic expression out, and don't worry about the sum bound thing I just retyped it

@Bob What do you mean by "first four given last two unknown"? only volatility and r are unknown?

If so why does it matter that both T's are the same

anyway, @Ultradark 100 points, participated in a "capitula" that appeared for no apparent reason when I attempted to log on with a tor browser RH solved QED

@user525966 What I want to be able to do is write a program that will take two option quotes with the same expiration and solve for both the volatility and r. I was doing this in SciLab and I got at least one crazy answer.

I suppose it does not mater if you have one T or two T s.

if you have two equations and two unknowns you can solve for a unique solution, assuming these variables only make sense to compare against each other when their T's are equal

@user525966 Are you sure about that?

i suppose if all the variables were equal you'd be unable to

@user525966 T represents the time until the option expires

@MikeM: Didn't see you sneak in. I'm busy chatting on FB about one of my best friends in GA who just had a stroke, trying to inform people. ... Did your lecture go well?

So Riemann hypothesis still looms?

Yup. Indubitably.

Oh well

Better than Regensburg. But someone told me I need to learn how to cut a little bit for the less expert audiences.

That's probably not bad advice. Not quite like colloquium, but still metered technicality.