Mathematics - 2014-04-08 (page 2 of 3)

Pedro Tamaroff

06:05

@AlexYoucis

Alex Youcis

Hey.

Pedro Tamaroff

would you be willing to join a Google hangouts with Mike?

@AlexYoucis i can give you the link on facebook

Alex Youcis

@PedroTamaroff i'mnotevenentirelysurewhatagooglehangoutis

Pedro Tamaroff

@AlexYoucis like a videocall.you can choose not to use video, just mic

=)

you just need a gmail account or sth of the kinds

Alex Youcis

@PedroTamaroff Why is this happening? haha

Pedro Tamaroff

06:08

@AlexYoucis Just because, dunno.

I was chatting with some friends here in Argentina

and we invited mike

because were rad cool guys

one just left

so it is only mike and me now

Alex Youcis

doesthismeanI'maradcoolguy

Pedro Tamaroff

YAS

is your spaecebarbroken

Alex Youcis

yaaaaaywhatI'vealwayswanted

Pedro Tamaroff

see

Alex Youcis

No, It's just something I do sometimes to emphasize things.

Pedro Tamaroff

06:09

now we're all happy ppl

@AlexYoucis it hurts my brains

Alex Youcis

@PedroTamaroff Part of the reason. I like my conversations to be taxing.

Pedro Tamaroff

i just left the link on your chat

Alex Youcis

I don't think I'm down tonight. Maybe in the future though.

Mike

Good choice, @AlexYoucis

Pedro Tamaroff

@Mike boo mike

boo

Mike

06:12

@AlexYoucis What's a good commutative diagram package?

Alex Youcis

@Mike tikz

Mike

@AlexYoucis I need to prove that two categories are equivalent. I wonder which definition/equivalent form I should use.

Both are equally easy and tedious.

Alex Youcis

@Mike Usually it's easiest to prove fully faithful and essentially surjective.

Unless the homotopy inverse is obvious.

Mike

I haven't heard the term homotopy inverse, but it is.

It's just tedium.

@AlexYoucis Do you like discrete math?

Alex Youcis

@Mike Sorry, I meant equivalence inverse. I sometimes think about equivalences as homotopy equivalences.

I do.

Why?

Mike

06:15

You should come to BAD Math day this Saturday.

Alex Youcis

what'sthat

Mike

bay area discrete math

Pedro Tamaroff

Hilbert's Hotel

@Mike i am hoping for the combinatorics course not to get too tough =D

it's an optional course

but we'll have two midterms

usually doesn't happen

Mike

it's combinatorics bro.

Alex Youcis

@Mike That sounds kind of terrible haha.

Mike

06:17

I was saying that to Pedro.

Pedro Tamaroff

@Mike combinatorics can get tough.

Mike

@AlexYoucis The abstracts are pretty neat.

Alex Youcis

@Mike Gunnar Carlson gave that talk like two weeks ago at Berkeley haha

@Mike I know Satyan, and have heard him give that talk before.

Mike

I'm not particularly interested in his tbh.

Ah, how was it?

Alex Youcis

@Mike Good, good. Satyan is a good presenter. Really nice dude.

Mike

06:19

His was one of the three I was particularly interested in.

Alex Youcis

@Mike Isn't this at your university?

Mike

Yeah.

Which is why I'm not too lazy to go.

Alex Youcis

Ah, I See your angle.

Mike

I'm not getting paid to advertise.

Alex Youcis

Yeah, I bet. Rube.

Mike

06:21

Man I'm too tired to do this shit.

It's not interesting.

The equivalence is kind of obvious in one direction.

Alex Youcis

@Mike Your life seems full of tedium involving homological algebra.

Mike

I'm gonna shower and sleep.

@AlexYoucis it is.

Alex Youcis

Good. I'm glad.

It's good for the soul.

Mike

That explains why yours is black as night

Alex Youcis

@Mike You're just a snollygoster.

Mike

06:24

And I think you're a cockalorum.

Alex Youcis

@Mike I'm one of those two things.

Mike

I know very well which one.

eXtremiity

06:40

Can someone give me the definition of what it means for a sequence to converge in the $L_1$ and $L_2$ sense?

Given that these sequences are functions.

Alex Youcis

@eXtremiity It just means that there is a function for which your sequence minus that function converges to zero in either the L^1 or L^2 norm.

eXtremiity

Ahh the norm.

And $L^{p} = [\int_{X} \vert f(x) \vert ^{p} dx]^\frac{1}{p} $?

@AlexYoucis ?

Alex Youcis

@eXtremiity sure.

eXtremiity

I am very unfamiliar with these $L^{p}$ senses.

sigh.

So much work.

Anthony

Hallo

For delta epsilon proofs, our delta can be a function of.... x and epsilon?

eXtremiity

06:56

$\delta(x,\varepsilon)$?

I don't think.......so.

Anthony

Then what can it vary with respect to?

Mike

one fixes $\varepsilon$ and a point $x_0$ and finds an appropriate $\delta$

Anthony

Hmph. Can it be a functions of the second value we are evaluating at, say $t$?

so $|x-t|<\delta(\epsilon,t)$?

Hawk

07:44

@robjohn sorry for disturbing so late night...are you willing to help on a problem now? if not, I will ask you later...

r9m

@Hawk hi .. I hear 11th may is an interesting day :)

Hawk

@r9m you hear right!!

r9m

@Hawk are you taking correspondence or mock exams ?

Hawk

@r9m nothing of the sorts...but I am attending classes from the people of that interesting place...and now brushing things up...will start previous papers soon...

r9m

@Hawk cool .. who's taking class nowadays ?

Hawk

07:50

@r9m Samples were just over...was feeling them to be quite simple...

@r9m who have already underwent that interesting day...and is in the place of interest...do you want to know their names???

r9m

@Hawk sure ..

Hawk

@r9m do you know anyone from there??

r9m

@Hawk yea .. 4 schoolmates

Hawk

@r9m okay, don't know you know them or not... "Indranil Bhattacharya" and "Anirban De"

r9m

are they 1st year ?

in that case I wouldn't know ..

Hawk

07:55

@r9m masters first year...they have already done their bachelors there...

r9m

ooh .. okay ..

Hawk

@r9m did not get the admit card yet...continuously postponing the dates...

r9m

ya .. saw the official page ..

@Hawk are you on FB ?

Hawk

@r9m was...deleted the account...

r9m

baap re ... serious chele !!

eXtremiity

08:10

So I am investigating the domain of $f(x)=\sum_{k=1}^{\infty}\frac{x^{k}}{k^{2}}$. By the ratio test I see that this function converges and has a domain for $|x|<1$. This implies that $Dom(f) \in [-1,1]$. However, using a computer program called Maple, I have graphed this function and it shows values for $x<-1$. My question is why?

Hawk

@r9m no..nothing as such...will be back...need to go... :)

Mike

08:23

@eXtremiity the keyword is 'analytic continuation'

the function you get will no longer be defined by that power series, which is convergent only in $[-1,1]$

eXtremiity

T_T.

So what do I conclude?

Mike

what do you want to conclude?

is the question about the convergence of that power series?

eXtremiity

The domain.

Mike

what is $f$?

eXtremiity

That below.

It is back to front.

Just pretend $f$ is on top of part (i).

Mike

08:27

well, $f$ is not "the analytic continuation of the function given by this power series"

which are words you shouldn't know yet, but at the same time what maple was showing you

that power series converges for specific values of $x$. you've already found exactly what those values are.

eXtremiity

Ok, I won't lie - have not learnt this analytic continuation stuff.

Does this only apply when $f:\mathbb{C} \to \mathbb{C}$?

Mike

you shouldn't have, dude

eXtremiity

OH. You were 1 step ahead of me haha^^

Mike

that's my point, this question isn't asking about it

eXtremiity

Yes yes, you wrote it before I could read it :p.

Mike

08:29

ah, gotcha

eXtremiity

Ok, I see Mike.

Makes sense.

sigh. I was stressing over this :(.

Waste of energy.

Mike

anyway, i only brought that up because you asked why maple was giving you that picture. but the picture it gave you wasn't quite the same as $f$ - it was the "natural continuation" of $f$ into a smooth function that has a bigger domain

but $f$'s domain is exactly where that power series converges

which you've already found :)

eXtremiity

Fantastic.

Now, Mike. I also have shown that this function converges. Does this allude to a specific type of convergence? I.e. Uniform, Pointwise, L1, L2?

Mike

yikes

eXtremiity

Or am I required to go to their definitions find out.

Mike

08:31

that's above my paygrade

eXtremiity

Hahahah xD.

Mike

you know that it converges pointwise, certainly

that's what the ratio test business does

actually no. i rephrase.

you know that that series converges for every value of $f$

eXtremiity

Yes.

Mike

ohh. are you asking about the convergence $f_n(x) = \sum_{k=0}^n x^k/k^2$ to $f(x)$? (if there's a sine in there I don't care, pretend its there)

eXtremiity

Yes I am.

Mike

08:33

then yes, what you've proven is that it converges pointwise

it may or may not converge uniformly, or with respect to some nice norm, but you haven't proven that yet

eXtremiity

I see. Ok - so I got to do some more work.

Thanks. I'm just so relieved about the analytic continuation stuff.

I was wondering and wondering and wondering.

Mike

heck no you're not supposed to know about that

that's serious stuff that requires some serious complex analysis to talk about

eXtremiity

Hahahah, you almost seem afraid of it :p .

r9m

08:51

@Sawarnik sounded like poetry .. that I definitely wouldn't want to be there on exams !

@eXtremiity can't we say $f$ is Uni cgt from the fact that the interval of convergence is a closed interval ?

eXtremiity

Sorry, @r9m. What is your definition of a closed interval?

$[a,b]$. The interval of convergence is $(a,b)$.

And that interval is open.

Sawarnik

@r9m poetry? its a dilemma that i faced just now.

eXtremiity

@Sawarnik. Wrestle with them without hurting them :) .

r9m

@eXtremiity $f$ converges on $[-1,1]$ ?

eXtremiity

That is the ratio test I used to find the interval of convergence. It is convergent if $|x|<1$.

That excludes -1 and 1, does it not?

r9m

08:56

@eXtremiity ya .. ratio test is inconclusive in that case .. but dosent mean it cant converge at 1 or -1

eXtremiity

How did you find that it converges in [-1,1]?

Mike

by checking the endpoints

r9m

$f(1) = \frac{\pi^2}{6}$ and $f(-1)=-\frac{\pi^2}{12}$ :)

Mike

@r9m convergences is uniform provided a few conditions. one is that the limit is continuous

one first needs to prove that the limit is continuous to use that theorem

r9m

ya

eXtremiity

08:59

@r9m. How did you evaluate $\sum \frac{sin(k)}{k^{2}}$.

Mike

@eXtremiity he didn't, he ignored that part

eXtremiity

Oh I see.

r9m

oops .. the second function .. sorry .. I was thinking about the dilog part only

eXtremiity

Ahh I see - But it doesn't matter because that clearly converges

No no, that is fine. It converges anywhere so it's like adding a constant $L$ to it.

I see. So the ratio test got me $(-1,1)$, but I had to check the end points.

Mike

yep

eXtremiity

09:00

That was close. Thanks @r9m :p.

Mike

and the ratio test also gave you that it diverges for $|x|>1$

eXtremiity

AHHHH

Roland

hey everyone

I'm looking for a question which was posted on math.SE a while ago, and I can't find it with the search function

Mike

i can link you plenty

Roland

:D

it was on the definition of continuity

Mike

09:04

what about it

Roland

there were about 9 different versions of the permutations of the for-each and for-all quantors

and the task was to describe the sets of functions which fulfill these (partly non-sensical) definitions

acutally, I just realized that I liked the question then

and I went to check if I liked it that much that I made it a favorite

and indeed, I did.

28

Q: A game with $\delta$, $\epsilon$ and uniform continuity.

UPDATE: Bounty awarded, but it is still shady about what f) is. In Makarov's Selected Problems in Real Analysis there's this challenging problem: Describe the set of functions $f: \mathbb R \rightarrow \mathbb R$ having the following properties ($\epsilon, \delta,x_1,x_2 \in \mathbb R$) : ...

sometimes, it really helps just to have someone to talk to!

Mike

dear god you liked that?

Roland

this sounds like you didnt

Mike

you are a smart man

Roland

why is that so?

(the not-liking)

r9m

09:21

I have a problem .. that asks to find a function $f:[0,1] \to \mathbb{R}$ such that $\displaystyle \int_0^1 x^nf(x)\,dx = 0$ for $n=0,2,3,\cdots$ but $\displaystyle \int_0^1 xf(x)\,dx \neq 0$ ?

Karl Kronenfeld

@Mike dat imperative. Btw, I dislike the answer but I am also too lazy to find something better.

kahen

@r9m you might find this useful: en.wikipedia.org/wiki/Hausdorff_moment_problem

r9m

@kahen thanks :) .. but I have zero knowledge of measure ..

eXtremiity

09:39

By converge can we imply equality?

For e.g, $\sum^{\infty} X $ converges to a, does this mean $\sum^{\infty} X = a$?

SQB

I don't know if this is the right place to ask, but a question I answered has been deleted. Is there any way for me to see what my answer was?

math.stackexchange.com/questions/684726/surface-area-and-volume/…

r9m

@eXtremiity does this answer your question math.stackexchange.com/questions/345945/… ?

robjohn

@Hawk sorry for taking so long to respond. Do you still want help?

eXtremiity

Hmmmmmmmmmmmm

So the answer is yes :D ? @r9m. Would you agree?

And thanks for sharing that. It was beautiful !

r9m

I guess so :)

Hawk

09:53

@robjohn Yes, I do...I am having difficulty in proving this...

SQB

Or is there a different room where all the mods hang out?

Hawk

@robjohn $x_1=2,x_2=2^{x_1},x_3=2^{x_2},\ldots$. How do I prove that, if $x_is$ are divided by any integer, after a certain point, all the remainders of division will be equal.

eXtremiity

Hmmmmmmmm

robjohn

@Hawk It seems that all the $x_i$ are increasing powers of $2$...

Oh, I see what you are asking...

Hawk

@robjohn yes...that is true...

Karl Kronenfeld

10:00

You just need two successive $x_i$s to be equal.

robjohn

@Hawk So you are saying that $\mathrm{mod}\ m$, the sequence becomes constant...

Karl Kronenfeld

So, we're solving $x=2^x\pmod n$, meh not really eh.

Hawk

@robjohn yes, $\mod m$,the sequence becomes constant after some time...

r9m

@Hawk one of my juniors asked the same question a few days ago :D .. what a coincidence ! math.northwestern.edu/~mlerma/problem_solving/solutions/…

Hawk

@r9m really? what a coincidence...thanks for the reference and solution...

@robjohn I am still open to your solution...

eXtremiity

10:17

@r9m. how'd you evaluate f(-1) ?

I know f(1) = \pi/6. Not sure about f(-1) though.

Oh no. I found it :).

r9m

hehe :D

eXtremiity

The proofs are not very short, are they.

See, I have to present to my teacher a proof of it's interval of convergence.

Do you think I'd have to show him why f(1) = blah and f(-1) = blah.

Or just jump into the conclusion that they equal those numbers we discussed before.

r9m

I dont think you have to write f(1) or f(-1) .. just show that the series converges at the two end points should suffice

eXtremiity

How would you show that it converges at the endpoints?

"show"

Ratio test shows for (-1,1), and the only way I know to show that -1 and 1 exist is by plugging them into f and spitting those pi results.

r9m

the f(1) case is easy .. just show $0 <f(1) < 2$

the f(-1) case also shouldn't be a problem once you figure out how to show the first one .. :D

eXtremiity

10:26

Hmmm, ok.

robjohn

@Hawk The only way I have seen this proven is in the pdf cited by r9m.

Hawk

@robjohn are there no other better way? the proof provided by the pdf by r9m is a bit extensive...

robjohn

@Hawk There are many things that are simple to state, yet have no simple solution.

@Hawk That one is pretty simple, actually.

Hawk

@robjohn and this is one of them...okay...thank you for the effort and time spent by you upon my question...

@robjohn yes, the solution is simple but extensive...

robjohn

@Hawk and seeing as the only way I can think of to show that $2^x\equiv2^{2^x}\pmod{m}$ is that $x\equiv2^x\pmod{\phi(m)}$, I can't think of how any other proof would go (this doesn't mean there is no other proof).

Hawk

10:35

@robjohn okay...please let me know if you come up with any other proof...the problem seemed quite interesting to me...

eXtremiity

10:45

Yay, I think I proved uniform convergence! I can do this using the Weierstrass M-test
correct?

Ramana Venkata

I am reading Topology From differentiable view point by Milnor. He uses strange symbol looks like capital A while defining the when a manifold is "smooth" in page 1. Can somebody tell me what it means

Karl Kronenfeld

$\forall$?

@RamanaVenkata I looked it up, still not finding what you're referring to.

Ramana Venkata

See the third paragraph

He writes U A X

at the end

Karl Kronenfeld

I see this:$U\cap X$.

That means intersection.

Ramana Venkata

Sorry I am using PDF version

the djvu version had intersection

Karl Kronenfeld

11:01

ah, yeah I see what you mean

that's an ugly symbol

Ramana Venkata

but pdf version somehow managed to change a few of intersections to A

eXtremiity

I have Introductory Real Analysis by Kolmogorov and the Analysis book by Rudin.

Can anyone recommend me others that I can get in pdf?

r9m

@eXtremiity Apostol's Real Analyisis ?

eXtremiity

Awesome, Ill look ijnto it

skullpatrol

11:50

Hi @somaye it is nice to hear from you again after such a long time :-)

Chris's sis

Greetings

@robjohn I just created another marvellous question.

skullpatrol

Greetings greatest one :-)

3

Sawarnik

@Chris'ssis You should consider writing a book on your marvellous integrals and series and their techniques.

Chris's sis

@Sawarnik I really want to do that one day...

@robjohn $$\sum_{n=1}^{\infty}\frac{1-2^{2^{1-n}}}{2^{n+1}(1+2^{2^{1-n}})}=\frac{5\log(2)‌-3}{6\log(2)} $$

eXtremiity

Is the L1 Norm for functions defined to be $\int |f(t)-g(t)|dt$?

I can't find ANY information on it :|.

robjohn

11:55

@Chris'ssis The $2^{2^{1-n}}$ should make things difficult...

skullpatrol

that is what she does best :-)

Chris's sis

@robjohn yeah

robjohn

@eXtremiity Try this page

eXtremiity

The word norm can be interchanged with what in this sense?

space?

Wait.....a norm is just a type of metric space with additional properties.

But that's for a vector space.

I'm in a "measure space"

Chris's sis

@robjohn I think the terms in numerator must be interchanged. Could you do it pls? (because it gives a result with minus)

robjohn

12:01

@eXtremiity That isn't written too well, $$\|f\|_p=\left(\int_S|f|^p\,\mathrm{d}\mu\right)^{\frac1p}$$ should be noted as the norm.

eXtremiity

Is $S$ the set in where $f$ is defined?

robjohn

yes

eXtremiity

What on earth is $\mu$?

robjohn

the measure you are integrating against

it can be $\mathrm{d}x$ if you are using Lebesgue measure

eXtremiity

See, I am required to show convergence in the L1, L2, senses.

And I am trying to grasp exactly what it is, and how it differs from uniform and pointwise.

And now there are Lp of them...

robjohn

12:04

So you'd use that norm with $p=1$ and $p=2$

eXtremiity

I see. Can I use 3 and 4 and etc ?

robjohn

That is, the distance between two functions $f$ and $g$ would be $\|f-g\|_p$

eXtremiity

These values of p....how do they change meaning to these measurements.

robjohn

@eXtremiity not if you are looking for $L^1$ and $L^2$ convergence

eXtremiity

Well the question says "In what sense does the series converge".

So it has not said L1 and L2. But that's all we've been..skimming through in class.

robjohn

12:06

@eXtremiity functions can be in $L^p$ for one $p$ and not another.

eXtremiity

But now that I know there are a whole range of them, Lp...

REALLY

I would really like to know more about these.

robjohn

@eXtremiity we'd have to see the sequence to know how to answer that. There are many kinds of convergence.

eXtremiity

There are.

This is my function.

I have found its domain to be [-1,1], and I believe I have shown that it converges uniformly via the W-M test.

robjohn

@eXtremiity I assume the domain is $\mathbb{R}$

eXtremiity

No, it is [-1,1]

skullpatrol

12:11

what is the W-M test?

robjohn

Okay, I was just about to say that it only converges for $|x|\le1$

eXtremiity

Ignore the part less than -1. Maple has used analytic continuation to get that.

Weierstrass M-test

In mathematics, the Weierstrass M-test is a test for showing that an infinite series of functions converges uniformly. It applies to series whose terms are functions with real or complex values, and is analogous to the comparison test for determining the convergence of series of real or complex numbers. The Weierstrass M-test is a special case of Lebesgue's dominated convergence theorem, where the measure is taken to be the counting measure over an atomic measure space. Statement Weierstrass M-test. Suppose that {fn} is a sequence of real- or complex-valued functions defined on a set A,...

skullpatrol

thanks :-)

robjohn

@eXtremiity It is using the dilogarithm function to extend this

eXtremiity

Yes, I believe so. I was so frustrated a few hours ago. I could not understand the extension. Then Mike sorted it out for me.

So I am relieved. But now I have all this Lp stuff that is driving me crazy.

robjohn

12:17

@eXtremiity I get a similar thing with Mathematica

eXtremiity

Interesting, Wolfram on the other hand.

God, it gave something so ugly.

That hook irritates me.

robjohn

@eXtremiity Ah, it plotted the part where it is no longer real

eXtremiity

Oh hold on....no wolfram is ok.

I did not see "real part"

Ahhhhh.

robjohn

@eXtremiity In any case, that series converges absolutely in all $L^p$ spaces, and pointwise

eXtremiity

How about uniformly?

robjohn

12:22

@eXtremiity that, too

eXtremiity

I know that IF uniform, therefore pointwise. So all I need to understand is this $L^{p}$ business.

"functions can be in Lp for one p and not another."
"series converges absolutely in all Lp spaces"

Chandan

Hi, I have a problem in complex analysis, can anyone help me out pls?

eXtremiity

"Just ask; don't ask to ask."

Chandan

0

Q: One-one analytic functions on unit disc

Is the following statement true? Suppose, $ f:D\to \mathbb C $ is an analytic function where $ D $ is the unit disc of radius $ 1 $ around $0 $. Suppose, $ f $ is analytic on the boundary of $ D $ as well. Then prove that, if $ f $ is one-one on the boundary of $ D $, then $ f $ is one-one on $ ...

complex-analysis

eXtremiity

Unfortunately I am not sure about this.

Chandan

12:28

Can you give a counter-example?

eXtremiity

Well, if it says:
" Then prove that, if f is one-one on the boundary of D, then f is one-one on D."
Then a counter-example should not exist.

Oh wait. No it is asking whether it is true or false.

Missed that.

Chandan

Actually I made up the problem, I got a proof, but not sure.

eXtremiity

I see. Yeh not sure either sorry.

Chris's sis

12:49

This series is also very nice $$\sum_{n=0}^{\infty} \frac{1}{2^n}\tan\left( \frac{1}{2^n}\right)$$

@robjohn have you met this series so far?

I also wonder if $$\sum_{n=0}^{\infty} (-1)^{n+1} \frac{1}{2^n}\tan\left( \frac{1}{2^n}\right)$$ has a nice closed form.

N3buchadnezzar

Heya

skullpatrol

13:05

heya

N3buchadnezzar

Bah

Wikipedia uses complicated estimates for things that should be simple

eXtremiity

Agreed.

And I hate the typesetting.

Proofwiki and wolframathworld are nicer

MickLH

mathworld is too fine of print for me

I always have to zoom in and then the latin gets ugly

N3buchadnezzar

http://en.wikipedia.org/wiki/Jordan%27s_lemma
Why not bound $e^{- a r \sin \theta}$ by $1$ instead of $e^{- a r 2 \theta/\pi}$ ?

It is a sloppy estimate sure, but if $f \to 0$ as $R \to \infty$ who cares?

skullpatrol

not me

eXtremiity

13:27

Hey, the series converges uniformly for $x \in [-1,1]$

Since $f_{k}(x)$ is continuous for all x in its domain, can I conclude that f(x) is continuous ?

Daniel Fischer

@N3buchadnezzar Because you have a factor $RM_R$ before the integral. You need something to get rid of the $R$, so need to bound the integral by $\frac{C}{R}$. If you estimate $e^{-aR\sin\theta}$ by $1$, the bound for the integral you get is $\pi$, not depending on $R$, that's not sufficient. If $f$ decays fast enough, so that $RM_R\to 0$, then you don't need the better estimate indeed.

N3buchadnezzar

@DanielFischer I see =) I was experimenting with smaller conturs, for integrals like the fresnel. So in order to shrink the contour and still ensure that the line integral goes to zero then one need that $f$ decays at least as $1/R^p$, with $p>1$. Seems like one trades a smaller contour against the need for faster convergence.

Daniel Fischer

I don't understand what you mean with "smaller contour".

N3buchadnezzar

Something like

$$C_R=\{z : z=R e^{i \theta}, \theta\in [0,\pi/4]\}$$

Daniel Fischer

13:45

Ah, smaller angle. The integral over the circular arc still goes to zero if $f(z)\to 0$ for $\lvert z\rvert \to \infty$, if your integrand is $f(z)e^{iaz},\; a > 0$. Of course the integral over the ray $\arg z = \frac{\pi}{4}$ generally isn't $0$.

N3buchadnezzar

But you need a faster convergence right? for f

Daniel Fischer

No. You estimate the integral by taking absolute values. The resulting non-negative integrand is the same, and when you integrate a non-negative function over a smaller interval, the integral certainly isn't larger.

MWarsi

Hello, everybody. Sorry if this is a newbie question but I just wanted to make sure my answer was correct: what is the div of (b.x) a, where x is the radial position vector and a, b are constant vectors?

MickLH

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