Mathematics - 2016-12-05 (page 4 of 6)

9:00 PM

But you can choose a sequence of different $f$'s, each of which are analytic and equal on some portion of the complex plane

Ted Shifrin

<--- totally lost ... what does this have to do with the original question?

GFauxPas

Sorry I didn't specify

Ted Shifrin

Ah, there's DogAteMy

Akiva Weinberger

I actually need to leave in a moment sorry

Ted Shifrin

I only need two words, DogAteMy. Taylor's Theorem.

GFauxPas

9:01 PM

If I wanted to see if $(x+iy)^{2+3i}$ is analytic, I'd check whether $\left({re^{i\theta}}\right)^{2+3i}$ satisfies the Cauchy Riemann equations

Null

"read every book under the following aspects:
the text is wrong, don't believe anything, until checked.

maybe the text, if fully right, can still be stripped from unneccessary parts" our advice in my university

Ted Shifrin

But checking Cauchy-Riemann is a local question. Whether the function is well-defined is a totally different issue.

hi @ted

Ah, I see

Hi @Alessandro

9:03 PM

@Astyx The place where that reasoning does make sense is if, say, $\vec{F}=f(r)e_\phi$

Mike Miller

@Null Sounds like a good way to never get through anything.

Astyx

@Semiclassical Could you define these notations please ? Not sure I follow

GFauxPas

so in order to make it well defined, I'd have to choose a branch cut, and then I'd have no choice but to make it discontinuous. But CR equations would allow me do define some branch of the function if I need it to be analytic in any simply connected set not containing the origin?

Ted Shifrin

@GFauxPas: The Cauchy-Riemann equations assume that you have a well-defined $C^1$ function to begin with!

Semiclassical

Sure. $e_\phi$ is the unit vector $e_\phi=(-\sin \phi)e_x+(\cos \phi)e_y$

GFauxPas

9:04 PM

:o

I apologize for my naivete here, I'm teaching myself

they dont offer this course in my university

Semiclassical

so $\vec{F}$ here is some vector field that simply swirls around the origin, and which depends only on the distance $r$ from the origin.

GFauxPas

im undergrad

Ted Shifrin

You needn't apologize. I wasn't yelling at you :)

GFauxPas

:)

Semiclassical

@Astyx That clearer?

GFauxPas

9:05 PM

So here's why I'm fussing over it

Astyx

Yes, a lot :)

Semiclassical

mmkay

GFauxPas

I was using R to graph contours to try to visualize some integrals

Null

@MikeMiller it forces to actually understand something, and not simply adept it. At least for me as a beginner I try to follow this. Also I won't use stuff I don't understand no more.

Ted Shifrin

I thought R was only good for statistics stuff.

GFauxPas

9:06 PM

And I was plotting $t^q e^{-st}$ for various constant $q \in \mathbb{C}$, $t > 0$; this integrand comes up during Laplace transforms

nah man it's great, it originally was for stats but since it's open source and free people use it for so many things

Semiclassical

in that case we get $\nabla\times \vec{F}=\frac{1}{r}\frac{d}{dr}(f(r)r)e_z$, I think

GFauxPas

check it out: stackoverflow.com/questions/40938416/…

Ted Shifrin

Cool, @GFauxPas.

Astyx

@Semi I'm going to trust you on this one
Anyway I don't really have that kind of information in my homework. And I'm sure the reasonning I stated earlier is the one I'm expected to make. It just bugs me since it really seems very shallow to me

Semiclassical

which, setting aside the difference between $f'(r)$ and $f(r)/r+f'(r)$, is something like a rate of change in $f(r)$.

Ted Shifrin

9:08 PM

I'll stick to Mathematica, since I know what I'm doing with it (and have free access, fortunately).

GFauxPas

anyway, $t^q e^{-st}$ for fixed $q$ and $s$ were making curves that wrapped around the origin before settling into a very tightly wound spiral that shrinks at $t$ increases without bound

which is really cool

Ted Shifrin

$q$ with negative real part, I imagine [pun intended], based on what you just said

Semiclassical

What's the context of the HW?

GFauxPas

yes

but those spirals wrap all the way around the origin

meaning that you can't use one definition of $t^q$ for the entire contour

that's what I was thinking about this weekend

Ted Shifrin

Yeah, that's clear when you write down $t^q = e^{q\log t}$ and realize ...

Astyx

9:10 PM

@Semi the magnetic field of celestial bodies and dynamo theory

Semiclassical

Ah, nice.

Ted Shifrin

Hi @PVAL

GFauxPas

Complex Analysis is amazing, but there is something lost by all these functions becoming multifunctions and losing the orderings

PVAL-inactive

@Ted Hi.

Semiclassical

So something like estimating the width of a current source by knowing the strength of the magnetic field it produces.

Ted Shifrin

9:11 PM

Multivalued functions is the nature of the beast, @GFauxPas.

Astyx

@Semi yes, it's quite interresting, but I find it very badly put

Semiclassical

It becomes simpler once you assume some sort of model.

e.g. a uniform volume current density

But if you're going to do that, it's more proper to go ahead and use Ampere's law.

Astyx

I have the Maxwell-Ampere equation giving me $\vec{\text{rot}}\vec B_1 = \mu_0 \sigma {\partial\over\partial t}\vec v \land \vec B_0$

Semiclassical

Yeah.

Astyx

And I must conclude that $B_1$ and $B_0$ are proportionnal

Semiclassical

9:14 PM

I think $B_0$ here would be the electric field, actually

GFauxPas

can you have an integral well defined on a contour that requires a countably infinite number of changing branches?

Semiclassical

...maybe.

Astyx

Not here

Semiclassical

Hmm.

should have $\nabla\times \vec{B}=\mu \vec{J}$.

GFauxPas

like $t^q$ and the spiral gets tighter as it goes around the origin until it picks somewhere outside the origin to settle down to

Ted Shifrin

9:15 PM

@GFauxPas: I'm not sure you even have a well-defined integral if you have to change branches at all.

GFauxPas

My book has one example of it

let me find it

Brown and Churchill

Astyx

Yes, but $\vec E$ is created by $\vec B_0$ and equal to $\vec v \land \vec B_0$ here

GFauxPas

found it

Semiclassical

Ah. So the changing magnetic field creates an electric field which drives a current which generates a new magnetic field.

Astyx

(and I shouldn't have edited and added partial derivative on my previous message)

Exactly

And so on

Semiclassical

9:17 PM

Yeah.

GFauxPas

$\displaystyle \int_{\mathcal C} \frac{z^{-a}}{z+1} \, \mathrm dz, 0 < a < 1$

Semiclassical

$\vec{J}=\sigma\vec{E}$ and all that.

Astyx

Yes

GFauxPas

$\mathcal C$ starts on the positive real axis, goes in towards the origin and makes a clockwise circle, then goes back to its original point and goes counterclockwise around a circle

Ted Shifrin

What's $C$? But they have to specify a branch of the function or I'll get countably many different answers.

Semiclassical

9:19 PM

Good old keyhole contour

Ted Shifrin

So they have actually made a branch cut here.

But you do need to specify which branch of the function you're using, or we'll multiply our answer by different numbers.

GFauxPas

And he says that if you choose the branch $0 < \theta < 2\pi$ in some places, and the branch $-\dfrac \pi 2 < \theta < \dfrac {3\pi}{2}$ in other places

you can sum the integral over two halves of the contour and get $2\pi i \operatorname{Res}(f,-1)$

Ted Shifrin

I don't follow. You cut along the positive real axis, and the contour avoids that (slightly above and slightly below).

Semiclassical

Hmm. Anyone know how to convert a list in Mathematica into an array of specified dimensions?

e.g. start with a list of 30 elements and get an array that's 5 by 6

Astyx

I knew that once

Ted Shifrin

9:22 PM

I've never tried to do that, @Semiclassic, but I'm confident you can do it.

Semiclassical

Yeah.

Ted Shifrin

Maybe something like MatrixForm ... but not quite.

Semiclassical

Sort've like the opposite of a Flatten

Astyx

reference.wolfram.com/language/ref/ArrayReshape.html

@Semi

Semiclassical

Ahah, ArrayReshape

Ted Shifrin

9:22 PM

I'm very suspicious about this @GFauxPas. I no longer have that book to look at, however.

You do not need two branches. You do need to specify which branch we're taking to compute $(-1)^a$.

GFauxPas

hmm you're right, he doens't have to define the function twice, he can just make the inner and outer circle not connect all the way around

Ted Shifrin

Think about a simpler case. How do you define $\int_C \sqrt z\,dz$ for $C$ the unit circle?

GFauxPas

Um

$e^{1/2 \operatorname{Log} z}$?

Ted Shifrin

First you need to decide what I even mean. The two answers will be $\pm$.

GFauxPas

Ugh MATH

Ted Shifrin

9:25 PM

OK, so basically, pull out the point $z=1$ from $C$ and do the standard branch, with $\sqrt{-1} = +i$. Agreed.

The fact that $\log z$ is multivalued is essential for a ton of mathematics (topology, geometry, analysis, ...) and physics ...

GFauxPas

What's the nature of the pole of $\operatorname{Log}$ at the origin, I try to avoid it

Ted Shifrin

It's not a pole.

Semiclassical

hmm, this isn't perfect buuut

Ted Shifrin

Branch points are different from poles.

@GFauxPas: If you had a pole, you'd have a convergent Laurent series on $0<|z|<1$, for example. You ain't got none.

Semiclassical

@akiva i.stack.imgur.com/9AU2o.png

GFauxPas

9:27 PM

Ah yes, I've felt the pain of not having a Laurent series of $\log$ at the origin before...

my old nemesis

Ted Shifrin

Multivalued functions can't have Taylor/Laurent series :P

A branch can ... :)

Semiclassical

Would prefer for those to join up better, and for the lines to be darker, but it's that kind of thing? @Akiva

GFauxPas

So how do I integrant $\sqrt{z}$ on a closed contour around the origin?

I only know how to integrate things like rational functions

Ted Shifrin

You do what you said, @GFauxPas, once we agree on the sign. $\int_0^{2\pi} e^{i\theta/2} (ie^{i\theta})\,d\theta$.

Or you use FTC. Once you've chosen a branch on the slit plane, you have a well-defined antiderivative.

Astyx

How complicated is Fourier theory ?

GFauxPas

9:31 PM

Well usually for $\sqrt{z}$ we want to pick the $\theta$ smallest in magnitude

Astyx I'm going through "Fourier Analysis and Its Applications" by Vretblad and it's very accessible

I'm undergrad and I understand what he says

Lozansky

GFauxPas, what country are you from?

GFauxPas

US

Lozansky

Oh

I thought only us swedes use that book

Astyx

@GFauxPas Right thanks

GFauxPas

I found it while googling for pdfs and I liked it and bought it online, I didn't know it was well known

Lozansky

9:34 PM

Since Vretblad is swedish

Ted Shifrin

hi @Lozansky

Lozansky

Heya @TedShifrin

Ted Shifrin

I always love to recommend Körner's book on Fourier Analysis. Beautiful, beautiful book. And also an accompanying problem book.

Lozansky

The important thing is if it's available online (for free)

Semiclassical

@akiva and an improvement:

Ted Shifrin

9:36 PM

Dunno. I'm a dying breed who actually bought/read books.

Semiclassical

Ted Shifrin

Hmm, when DogAteMy comes back, somebody mumble "Taylor's Theorem" at him :P

GFauxPas

so its $\displaystyle \frac{i}{\frac i 2 + i} \exp\left({(\frac i 2 + i)\theta}\right) |^{2\pi}_0$?

Semiclassical

Not sure why they don't join up perfectly

Kind've annouyed by it.

Ted Shifrin

What are you drawing, @Semiclassic? [annoyed]

GFauxPas

9:38 PM

Right Ted?

Ted Shifrin

I think you can simplify $\frac i2+i$ and $i$ divided by it :P

Semiclassical

lol. I know why I slip into doing british english, though it's not worth explaining

Ted Shifrin

Do the British really spell annouy?

GFauxPas

yeah I know I can I'm feeling lazy

Ted Shifrin

I've never seen any such thing, @Semiclassic.

GFauxPas

9:38 PM

but where's the sign ambiguity

Semiclassical

1 hour ago, by Akiva Weinberger

If you take a grid of squares, and replace each square by a pair of disjoint line segments each joining midpoints of adjacent sides of the square, and make the choice for which pairs of sides randomly, you get interesting patterns

Ted Shifrin

We chose the branch of $\sqrt z$ before we started, @GFauxPas. There was a sign ambiguity before we chose.

Oh @Semiclassic.

Semiclassical

It's not perfect, alas.

GFauxPas

ahh, okay, so when I decided I wanted $\theta$ smallest in magnitude I was choosing the branch $(-\pi..\pi]$?

Ted Shifrin

Maybe you need to specify a larger PlotPoints command, @Semiclassic.

Semiclassical

9:40 PM

hmm.

GFauxPas

Thanks a lot Ted

Mahmoud

@TedShifrin

Ted Shifrin

I chose a branch cut on the real axis, @GFauxPas, and I specified $\sqrt{-1}=+i$. Whatever.

Semiclassical

Ahah, fixed it

GFauxPas

So here's a question @TedShifrin

Sometimes when I graph parametric equations of two variables

I introduce a third variable arbitrarily and "stretch out" the curve along the $z$-axis

Can I, instead of a branch cut, introduce a third variable?

Ted Shifrin

9:42 PM

Oh, you mean parametric curves? I thought with two variables you meant parametric surfaces, @GFauxPas.

Maks

Hi ! Good ... afternoon ?

Semiclassical

i.stack.imgur.com/sGYSP.png @akiva

GFauxPas

well theyre similar but I meant curves

so can I do something like

Ted Shifrin

Basically you're going to draw the Riemann surface of the multivalued function, @GFauxPas.

It's like a multi-sheeted parking garage ramp.

Akiva Weinberger

@TedShifrin Yes?

Ted Shifrin

9:43 PM

DogAteMy!!!

GFauxPas

for $z<0$ I'll choose one branch cut, and $z > 0$ I'll choose another branch cut, and maybe there's something I can do at $z = 0$ to make the line continuous or even differentiably

Semiclassical

Figured out how to do it for squares, @akiva

GFauxPas

that's called a Riemann surface?

Ted Shifrin

So the mumble is: Taylor's Theorem, DogAteMy.

Vrouvrou

Can someone tel me how we get this two inequalities

Akiva Weinberger

9:43 PM

@Semiclassical Nice! Maybe the squares are a pixel too big or something

Vrouvrou

Ted Shifrin

yes, @GFauxPas. You want a more mathy book eventually than what you're reading :)

Semiclassical

Well, the latest version doesn't even have that issue

Ted Shifrin

How did you fix it, @Semiclassic?

GFauxPas

well after this semester i finished all my mandatory classes so I'll have free time :)

Semiclassical

9:44 PM

PlotRangePadding->0

And now that I look at it, it's not -perfect-

GFauxPas

@Vrouvrou what's the relationship between $\psi$ and $\Psi$, antiderivative?

Semiclassical

there's like a pixel missing, as @akiva just noted

Akiva Weinberger

Some sort of off-by-one pixel error

Semiclassical

Yeah.

Ted Shifrin

Take off your glasses :) @Akiva @Semiclassic

Akiva Weinberger

9:45 PM

I have no idea whether I have any idea what I'm talking about

Ted Shifrin

DogAteMy: So you finally heard my mumble? I've said it 4 times.

GFauxPas

@Vrouvrou I think the first one is a triangle inequality

Mahmoud

I want to express a function $y=f(x)$ with a vector, but I only know how to do so with polynomials, to be honest I know so little about how this works, can you tell me where can I learn this ?

Akiva Weinberger

@TedShifrin I was kind of hoping for details but OK

Astyx

Retina screens is more of an incovenience it seems

Semiclassical

9:46 PM

Maks

Hey, If I want to prove a series $ a_n $ converges, I have to find convergent $ b_n $ such as $ a_n > b_n $ right ?
And If I want to prove it diverges then $ a_n < b_n $, and $ b_n $ has to diverge

GFauxPas

@Mahmoud "Vector" is used in lots of different ways in different types of math, you'll have to be more explicit

Ted Shifrin

DogAteMy: Write down Taylor's Theorem with remainder. I'll let you decide which order to use.

GFauxPas

the other way around Maks

Ted Shifrin

You have it backwards, @Maks.

Semiclassical

9:46 PM

Weirdly, when I look at it in Mathematica it doesn't have those missing pixels

Astyx

@Maks very weirdly put

Mahmoud

@GFauxPas Linear Algebra.

Semiclassical

Must be a resolution issue

GFauxPas

give me a function as an example

Semiclassical

But I'm okay with that.

Maks

9:47 PM

For convergence I have to find a ... bigger one then ?

Akiva Weinberger

That would be nice if I could remember what the remainder term was… something like $f^{(n)}(c )/n!$? @TedShifrin

Semiclassical

For reference:

Vrouvrou

@GFauxPas $\Psi$ is primitive of $\phi$

Ted Shifrin

You don't even need that, actually, DogAteMy. You only need to know that the error of the $n$th degree polynomial is $o(x^n)$.

@Vrouvrou: Do you know that $\psi$ is $\ge 0$?

GFauxPas

@Vrouvrou then I think the first one is the triangle inequality

Semiclassical

9:48 PM

`diag1 = Graphics[{Line[{{1/2, 0}, {0, 1/2}}],
Line[{{1/2, 1}, {1, 1/2}}]}, ImagePadding -> None,
PlotRangePadding -> None];
diag2 = Graphics[{Line[{{1/2, 0}, {1, 1/2}}],
Line[{{1/2, 1}, {0, 1/2}}]}, ImagePadding -> None,
PlotRangePadding -> None];
With[{m = 13, n = 21},
GraphicsGrid[
ArrayReshape[
If[# == 1, diag1, diag2] & /@ RandomInteger[1, m n], {m, n}],
Spacings -> {0, 0}, Frame -> None, ImageSize -> 600]]`

Mahmoud

@GFauxPas We need to set the basis functions.

Ted Shifrin

@GFauxPas: I think you need $\psi\ge 0$.

Akiva Weinberger

@Semiclassical By the way, I think total lines minus loops is constant, since it's the amount of lines that hit the edge, aka the amount of points on the edge over two

Vrouvrou

@TedShifrin yes

Semiclassical

Sigh. Not sure why that's not outputting nicely, but there you are.

Akiva Weinberger

9:48 PM

Or you could also prove that with V+E-F=1 probably

Ted Shifrin

Then write down Fundamental Theorem of Calculus and use triangle inequality, as @GFauxPas suggests.

GFauxPas

@Mahmoud you need to find a basis, there are infinitely many bases, you want to choose one convenient to your problem

Astyx

@Semi I like it a lot

Null

line, plane, what comes next?

Semiclassical

Next would be to do the hexagon version

Akiva Weinberger

9:49 PM

Right

Astyx

hyperplane

GFauxPas

@Null "hyperplane"

and after that, it's still called a hyperplane

Astyx

@Semi Please keep me informed !

Ted Shifrin

No, @GFauxPas ... hyperplane has dimension $1$ less than the space you're in.

GFauxPas

interestingly enough, if you call a line a "hypoplane" people probably won't know what you're talking about

really?

Mahmoud

9:50 PM

@GFauxPas I only how to find a basis for polynomials.

Maks

How can I tell if $ \int_1^\infty \dfrac {cos(x^2)} {x} $ Converges or diverges ? What can I compare it to ??

Semiclassical

will do

Ted Shifrin

You actually have to call it a $k$-dimensional plane (or subspace) for appropriate $k$.

GFauxPas

@Mahmoud different functions have different bases

There are some that are more commonly used than other

Ted Shifrin

@Maks: That's a bit tricky. Try integrating by parts.

Vrouvrou

9:50 PM

@GFauxPas how we use trianglar inequality in integral ?

Mahmoud

@GFauxPas I'm interested in $f(x)=\frac{ax+b}{cx+d}$

Null

so to not confuse the reader, subspace is the way to go, and only refer to line and plane in, well spaces where it makes visual sense

Semiclassical

@AkivaWeinberger The trick I did for this won't work for the hexagons, alas.

GFauxPas

@Vrouvrou use the FToC but in the other direction from what you usually do

Ted Shifrin

fun watching GFauxPas be on the battle field instead of me :)

GFauxPas

9:51 PM

:)

Akiva Weinberger

By the way, it occurred to me that if one could make all squares within a distance $r$ from the origin (for $r$ large) "colored" one way (//, say), and everything else "colored" the other way (\), then we would essentially have hidden a circle in a picture with only vertical and horizontal lines.

That's actually what started me on thinking about this in the first place. @Semiclassical

Ted Shifrin

DogAteMy: You saw the $o(x^n)$ remark?

GFauxPas

hmm that's a rational function, so you might want to use a power series

Akiva Weinberger

Yeah. I was thinking about slanty lines, sorry

GFauxPas

rational functions often have pretty good power series

Maks

9:52 PM

@TedShifrin Integrate $x$ and derivate $ cos(x^2) $ ??

Ted Shifrin

OK, just wanted to be sure you saw.

Maks

Because seems like integrating cos(x^2) is a bit tricky

Akiva Weinberger

So $f(x)=0+0x+o(x^2)$?

Mahmoud

@GFauxPas What could be the basis then ?

Ted Shifrin

Yes, DogAteMy.

Akiva Weinberger

9:52 PM

And I want $\sum f(1/n)$ to diverge?

Ted Shifrin

Oh wait.

GFauxPas

@Mahmoud $(1,x,x^2,x^3,\cdots)$

it would be an infinite basis

Null

is there some way to find easy questions on MSE, that are still viable questions?

Ted Shifrin

That's not right, DogAteMy.

You have only first derivative.

Mahmoud

@GFauxPas Isn't that for polynomials ?

Ted Shifrin

9:53 PM

@Maks: First of all, it's $x^{-1}$ and $\cos(x^2)$.

Semiclassical

There's not a closed form result for $\int_1^\infty \frac{\cos(x^2)}{x}\,dx$, for reference.

GFauxPas

@Mahmoud You can do a polynomial with a finite sequence of powers. The one I suggested goes on forever

Akiva Weinberger

Isn't the second $0$ the first derivative?

Vrouvrou

@GFauxPas i must apply it for the firt integrale ?

Maks

@TedShifrin Second of all ?

GFauxPas

9:54 PM

let me google if there are good bases for rational functions

Ted Shifrin

Right, DogAteMy, so the error in that case is $o(x)$.

Akiva Weinberger

Oh

Let me find the original problem by the way…

22 hours ago, by Ted Shifrin

Suppose $f$ is continuous near $0$ and let $a_n=f(1/n)$. Prove that (a) $\sum a_n$ converges $\implies f(0)=0$. (b) If $f'(0)$ exists and $\sum a_n$ converges, then $f'(0)=0$. (c) If $f''(0)$ exists and $f(0)=f'(0)=0$, then $\sum a_n$ converges. (d) Suppose $\sum a_n$ converges. Must $f'(0)$ exist? (e) Suppose $f(0)=f'(0)=0$; must $\sum a_n$ converge?$

There.

Ted Shifrin

@Maks: No, you want to differentiate $x^{-1}$ and integrate $\cos(x^2)$. Of course, you can't. So write this as $$\frac{x\cos(x^2)}{x^2}.$$ Now try integration by parts.

Which part are we talking about, DogAteMy? You were trying to use L'Hôpital.

GFauxPas

I honestly don't know of any other than infinite series, and then you're going to have to deal with radius of convergence, and that's Calculus not linear algebra, so maybe I'm giving you bad advice

Maks

@TedShifrin I integrate $ x cos(x^2) $ and differentiate $ x^{-2} $ ?

Ted Shifrin

9:56 PM

No, you differentiate $x^{-2}$.

Semiclassical

@TedShifrin That's slick.

Ted Shifrin

I don't see what else to do, @Semiclassic.

Yes, @Maks :P

Semiclassical

I don't either, but that just means I wouldn't have an answer :P