Mathematics - 2015-01-25 (page 1 of 4)

Ted Shifrin

12:00 AM

Hmm ... I'll think about that over dinner.

Pedro Tamaroff

@TedShifrin Why do you keep changing my statement!?!?!? =D

Clarinetist

FIS, I have. Let me rephrase what I'm looking for...

Pedro Tamaroff

I mean what I say, and I say what I mean, @Ted.

Ted Shifrin

Because I don't like you, @Pedro? :)

Mike Miller

It's okay, @Pedro, he's been doing the same to me.

Ted Shifrin

12:01 AM

<--- resigns and goes off in a huff, never to return

Mike Miller

Morning, @Ted. Enjoy dinner.

Pedro Tamaroff

@TedShifrin Wonders if the kids will follow Ted to his exile.

Ted Shifrin

I don't need kids, thanks, @Pedro.

Mike Miller

ahaha

Ted Shifrin

besides, you and Mike are kids.

Mike Miller

12:03 AM

Ah... I'm getting too old for this stuff.

Ted Shifrin

which stuff?

Mike Miller

Nothing. It's a cliched line that 50-60 year old action movie heros say.

Clarinetist

Here's the book I have (don't worry, it's on the official publisher's site): download.springer.com/static/pdf/560/… . TOC starts in p. 13 of the PDF... I am looking for a text which rigorously covers parts I and II.

Ted Shifrin

you love Clint Deadwood too much.

Mike Miller

Huh?

Ted Shifrin

12:04 AM

error @Clarinetist

You aren't acquainted with old action movie hero Clint Deadwood, @Mike?

Pedro Tamaroff

@MikeMiller The proper word is "shit."

Clarinetist

Try link.springer.com/book/10.1007%2F978-0-387-70873-7 @TedShifrin . Under Table of Contents (12 chapters), click the "Download PDF" under the very first Front Matter.

Ted Shifrin

LOL ... censors @Pedro and bans him

Mike Miller

I guess that's an Eastwood joke. Then yes, I'm acquainted with. I prefer his earlier stuff.

Ted Shifrin

So what's wrong with FIS, @Clarinetist?

Clarinetist

12:07 AM

It covers a good amount of it, but not everything. Like for example, generalized inverses (which I don't think are covered in here).

or what "full rank" means (at least I don't think).

Ted Shifrin

oh, you meant parts I and II. I misunderstood

Clarinetist

Yep, I and II.

I have yet to find a theoretical treatment of these topics.

Ted Shifrin

full rank? that means rank = $\max(m,n)$.

Clarinetist

Oh really? Haha. I didn't realize it was that simple.

Ted Shifrin

No, you're not going to find a theoretical treatment of all the statistical stuff. You might look at some older applied linear algebra texts. Look at Wilkinson. Also look at a cool modern book by Lorenzo Sadun (a former student of mine), called Applied Linear Algebra. It's more math-physics oriented than math-stat.

Noble is another old classic

Clarinetist

12:10 AM

Thank you!

Ted Shifrin

OK, bubye, @Pedro, @Mike, @clarinetist, et al

Mike Miller

Bye.

Clarinetist

Adiós @TedShifrin

user105491

12:37 AM

@Julian hi

Julian Rachman

Hello

user105491

how's life?

user105491

i saw you in the htpy theory chat room

Julian Rachman

Good. Just got my lecture notes and homework from CSUDH

user105491

classes started?

user105491

12:38 AM

already?

Julian Rachman

You? And I was just checking it out.

No. The "everything" was online.

Class starts MONDAY!

user105491

ah, ok

Pedro Tamaroff

No one solved my problem.

user105491

what problem @Pedro?

Pedro Tamaroff

42 mins ago, by Pedro Tamaroff

Prove there is a nonzero group morphism $\prod_p C_p\to \Bbb Q$.

user105491

12:40 AM

C_p is the cyclic group?

Julian Rachman

@Sanath I decided to stick with topology 1. I got my inspiration earlier.

Pedro Tamaroff

@SanathDevalapurkar Yes, it is the cyclic group of order $p$.

Of course the indexation is over primes.

user105491

p a prime, right?

user105491

and julian, what do you mean "stick with"? weren't you learning topology already?

Mike Miller

I did, @Pedro. I could email you the solution.

I'm going to take a nap.

Pedro Tamaroff

12:42 AM

@MikeMiller Sleep well.

Julian Rachman

@Sanath I know "stick with" in terms of staying with it.

user105491

that means you were considering an alternative

user105491

what was the alternative?

user105491

@Pedro i'll try it soon. i'm not in the mood for doing math now :-P

Pedro Tamaroff

@SanathDevalapurkar Oh, what got you?

user105491

12:43 AM

no clue

user105491

i woke up at 5:00 today and started doing some things with ring spectra

user105491

so now i simply can't think

user105491

:-P

user105491

@Julian can i ask you a question?

user105491

how much topology have you learned until now?

Pedro Tamaroff

12:44 AM

@JulianRachman Just study Apostol's Mathematical Analysis. That has enough topology to get you started.

Julian Rachman

Not now. Got to finish homework. Wait that is your question?

@Sanath

Pedro Tamaroff

Else you'll know the definition of ultrafilter and not know how to prove Heine-Borel.

user105491

that's one question

user105491

@Pedro Haha, i don't even properly know what filters are

Julian Rachman

@Pedro I am taking an Analysis course at CSUDH so it covers it. :)

user105491

12:45 AM

P.s. @Pedro does it have anything to do with the fundamental theorem of abelian groups?

user105491

@Julian do you know differentiation?

Pedro Tamaroff

@SanathDevalapurkar What does?

Julian Rachman

Yes

user105491

the question you asked

user105491

alright, cool @Julian.

Pedro Tamaroff

12:46 AM

Not really, no.

Julian Rachman

Why? You kinda scared me there

user105491

ah, ok

Julian Rachman

...

@Sanath

Pedro Tamaroff

@SanathDevalapurkar Do you know what an injective module is? That should help.

user105491

@Julian No, I was wondering whether you knew analysis because not knowing that would be bad when studying analysis

user105491

12:47 AM

what kind of analysis is it anyway?

Julian Rachman

Real

user105491

@Pedro i know what it is

user105491

@Julian proof based?

user105491

or no?

Julian Rachman

And we start with the axioms of $\mathbb{R}$

user105491

12:48 AM

Ok, i swear this thing doesn't allow me to post two things continuously

user105491

No, is it proof based or not?

Julian Rachman

and yes. I have to prove the propositions that satisfy the axioms

user105491

wait what?

user105491

you're building up a theory based on the axioms

user105491

so proving propositions based on the axioms seems a bit weirdly phrased to me

Julian Rachman

12:50 AM

basically the properties of $\mathbb{R}$

F M

@NachoDarago

user105491

@Pedro Q is an injective Z-module, right? i don't know what to do right now after that

user105491

(but don't tell me now, i want to work it out for myself)

user105491

@Julian like what?

Mike Miller

Morning, @FM

user105491

12:51 AM

i know you've studied some analysis by yourself

user105491

do you know compact spaces?

F M

@PedroTamaroff I know the definition of ultrafilter and I'm pretty sure I don't know the proof for Heine-Borel's

user105491

@Julian

Julian Rachman

I will show you on Monday. I dont want to writh the thing here

and some

user105491

@FM Ok, you need to study topology again! :-)

Pedro Tamaroff

12:51 AM

@FM Silly as usual.

user105491

@Julian tell me an outline of the definition

user105491

at least that

F M

I'm not being silly

I'm pretty sure I can prove it

Pedro Tamaroff

@FM Today Luis and I met.

Julian Rachman

definition? We are proving Propositions.

F M

12:52 AM

but I have no clue how the proof goes

Pedro Tamaroff

@FM That's what I mean.

Mike Miller

@FM I suggest delearning the definition of ultrafilter

user105491

@Julian how the heck can you prove something without knowing the definition of that thing?

F M

@Pedro ufa

user105491

come on, think straight

Julian Rachman

12:53 AM

It starts with a theorem on the field $\mathbb{R}$ !

not starting defn!

Listen to me before.

I was typing.......... :(

user105491

that's because the definitions are trivial and therefore you don't need to restate them

Pedro Tamaroff

@SanathDevalapurkar How are definitions trivial?

Julian Rachman

@Pedro that is what I am saying!

user105491

but when you don't know what the definiiton of compactness is, how do you show that a particular topological space, or in this case, the real line, is compact?

user105491

@Pedro Ah, what i meant was for example the definition of an element of the set of real numbers

Julian Rachman

12:55 AM

Wait. are you talking about analysis or topology?

Pedro Tamaroff

@SanathDevalapurkar I don't know what you mean by that.

user105491

@Julian analysis is topology

Pedro Tamaroff

Dedekind cuts are not trivial.

bolbteppa

Anybody know integral equations (e.g. Volterra, Fredholm?)

Julian Rachman

I am talking about specifically analysis right now.

user105491

12:55 AM

@Pedro what i'm talking about is a basic definition that the reader is assumed to know

user105491

@Julian i told you, analysis is the topology of the real numbers

Pedro Tamaroff

@SanathDevalapurkar That depends on the reader.

Don't smother Julian over a definition.

F M

@SanathDevalapurkar [citation needed]

2

user105491

@Pedro that's why courses have prereqs! :-)

Julian Rachman

@Pedro Again, thank you

Pedro Tamaroff

12:56 AM

@SanathDevalapurkar And no, that's far from true.

F M

analysis has much more to do with the differential structure of $\mathbb{R}$ than with its topology

Julian Rachman

Thnak you!

Here is an example:

Pedro Tamaroff

@FM I'm writing up notes on the structure theorem over PIDs.

F M

for the final?

Pedro Tamaroff

Mainly because I have too much free time now.

And for the final, yes.

user105491

12:57 AM

@FM and @Pedro I was talking about the definition of the usual topology on the real line used in analysis

user105491

not what analysis is studying

user105491

though i should have made that explicit

user105491

:-P

F M

yep whatev

user105491

@Julian what's the example?

F M

12:58 AM

how far into it are you @PedroTamaroff?

Pedro Tamaroff

@FM Just started. Also because Luis convinced me G.C. might not be a kind feller.

Julian Rachman

I am typing it. hang on

F M

haha seriously?

Pedro Tamaroff

@FM Yes.

F M

I don't think he's harsh, but I don't know him

Pedro Tamaroff

12:59 AM

I'm just kidding though. I'm doing it for the Lulz.

user105491

@Pedro I was wondering if an exposition of simplicial sets would be ok for the mse blog

F M

I know that tochi got a 9 with him in algebra III after messing up the galois correspondence

user105491

perhaps grad level?

F M

which is pretty much the most important theorem

Pedro Tamaroff

@FM Luis told me he technically didn't mess up. Just stuttered.

Julian Rachman

12:59 AM

$(\exists!a\in\mathbb{R})(\forall b\in\mathbb{R}) \ a+b=b+a=b.$

F M

he may be right, I'm not 100% sure

user105491

@Pedro like the exposition I wrote up here: docs.google.com/…

user105491

@Julian in words please

user105491

i don't want to read math symbols now

user105491

ok, let me attempt it:

Julian Rachman

1:01 AM

a+b=b+a=b for there exists exactly one a in R and for all b in R.

user105491

there exists a unique real a for all real b such that a+b=b+a=0

user105491

what's this a?

user105491

do you know?

user105491

no thinking necessary

Julian Rachman

a is an additive neutral element

i would respect it if you would wait

user105491

1:02 AM

it's trivial

user105491

but that's fine

Pedro Tamaroff

@FM Today Luis and I called you.

You didn't respond.

Julian Rachman

you know i have to typr right?

user105491

it's a single number

user105491

there's honestly not that much to type

F M

1:03 AM

@PedroTamaroff I was literally taking a shit

Julian Rachman

well. i gave it formally

F M

I called you back later and you didn't respond

user105491

formally?

F M

o(

user105491

the number a is ...

Pedro Tamaroff

1:03 AM

Dies of laughter.

Respawns.

user105491

0

Julian Rachman

ok. then that is hwo you approached it.

user105491

@Julian a=0

Pedro Tamaroff

@FM We talked about some type theory with Luis.

Actually he explained some to me.

user105491

@Julian there's honestly no approaching

Julian Rachman

1:04 AM

I approached it in a different way

user105491

it's trivial as i said

Julian Rachman

the proof!

user105491

how can you approach it differently?

Julian Rachman

I am talking about the proof!

user105491

it's not a proof

bolbteppa

1:04 AM

So nobody's studied integral equations?

Pedro Tamaroff

@bolbteppa Not me, sorry.

user105491

it's a freaking computation

user105491

@bolbteppa I don't know enough integral eqns

user105491

sorry

Julian Rachman

We ahve to prove it without it!

the next defnition shows that it is denoted by 0!

user105491

1:05 AM

ok

Julian Rachman

see. you have not seen the lecture notes yet!

bolbteppa

Okay cool, it looks to me like you can view Sturm-Liouville & Green function solutions more naturally as part of this theory rather than an explicit ODE topic

user105491

that's really weird

Julian Rachman

That is what i have been meaning this whole time!

user105491

@Julian

Julian Rachman

1:06 AM

Please understand me...

user105491

I don't care about the lecture notes

user105491

it's stupid if you don't understand that a=0 without having to read the next definition or whatever

user105491

@julian ok, one more question

Julian Rachman

Well. I knew it was 0 ok. It makes me look stupid when I am jsut following by how they teach it

user105491

you know what limit points are right?

Julian Rachman

1:08 AM

user105491

@Julian don't read blindly from the textbook

user105491

develop your own thoughts

Julian Rachman

look. here it is

user105491

but do you know what limit points are?

Julian Rachman

I did! I knew it was 0. I wrote it in my own notes!

and yes

user105491

1:08 AM

ok

user105491

can you explain to me how limit points relate to convergence (you should know about convergence too then i suppose)

bolbteppa

@SanathDevalapurkar all these a + b = b + a = 0 or a + b = b theorems require heavy justification to be quite honest, if you haven't gone back to the depths of axiomatic set theory one can never be sure one's not missing some logic in order to appreciate such seemingly simple questions, it's not fair to give out to someone for needing a textbook

user105491

@bolbteppa I know what you mean

user105491

i know julian personally, and i know up to what depth of set theory he has studied

bolbteppa

ok

Julian Rachman

1:10 AM

I understand what I am capable of.

user105491

but given that you're working in the familiar setting of real numbers you should be able to develop your own intuition on what such eqns mean

Julian Rachman

@bolbteppa thank you

user105491

@Julian I was not saying you're not capable of anything; i was saying that you should develop your own intution without having to read futher

user105491

@Julian thanking him makes it seem like we're in a war

bolbteppa

The only book I've found worth studying on all this stuff is Inder K Rana's From Numbers to Analysis, then you have no excuse haha

Julian Rachman

1:11 AM

we are not.

bolbteppa

Oh and Enderton's set theory book

user105491

stop and think about what you're reading before saying you know what it means

Julian Rachman

because i was jsut fustrated that I keep telling you about the notes and you didnt lsten and then i look stupid in front of many higher-level mathematicans

user105491

@Julian can you answer my question on how limit points relate to convergence

Julian Rachman

Yes. i will. I actually want to go back and compare my notes to the new ones

hang on.

user105491

1:13 AM

@Julian you didn't tell me about the notes

Julian Rachman

let me type

I even ust sent you a photo

and yes i did.

user105491

you told about it to me once

user105491

but that's fine

Mike Miller

@PedroTamaroff OK, now give me an example without using the axiom of choice.

Pedro Tamaroff

@MikeMiller An example of what?

Mike Miller

1:17 AM

A nonzero map from $\prod_p \Bbb Z/p\Bbb Z$ to $\Bbb Q$.

user105491

@Julian found the answer yet?

user105491

if not you can think about it for some time

user105491

but do you know what boundary points are?

user105491

Hey @Mike can i confirm whether you've recieved my email?

Pedro Tamaroff

@MikeMiller We have thought about it with a friend. Note that since $\Bbb Q$ is torsionfree, the map is zero on the subgroup $\sum_p C_p$, which is exactly the torsion of that.

The kernel is in fact uncountable.

Such morphism is pretty pathological. I doubt you can construct one explicitly.

Mike Miller

1:20 AM

@SanathDevalapurkar I don't check my email on weekends, usually. If you sent it to the email in my profile I received it.

user105491

@mike ah thanks

user105491

just confirming that i didn't send it to the wrong email

Mike Miller

@PedroTamaroff One can construct things implicitly without using choice. I'm wondering whether "$\text{Hom}(\prod \Bbb Z_p, \Bbb Q) = 0$" is possible in models of ZF.

Pedro Tamaroff

@MikeMiller Well, you would need a model where $\Bbb Q$ is a noninjective $\Bbb Z$-module.

One proves this using Baer's criterion which uses Zorn.

Mike Miller

Yes, I know.

Hence why I brought up choice in the first place.

Pedro Tamaroff

1:22 AM

I know you know, I just like to lay things clearly.

@MikeMiller Sure.

Eric Gregor

1:40 AM

Does anyone understand this terse answer: math.stackexchange.com/a/1118201/28480

Fallen Apart

If you are working with universal property, do you have construction in mind or mayby, as time go on you forget it and work only with arrows?

bolbteppa

@EricGregor I think it means ||x||^2 = <x,x> = x^t A x with A diagonal, i.e. you are re-expressing a norm in terms of an inner product and then re-expressing an inner product, which is a symmetric bilinear form, in terms of matrices, since every bilinear form has an associated matrix

Eric Gregor

interesting, @bolbteppa. Do you think he is right that this will help me solve the integral?

even if I get it in that form, I'm not sure how it will work

bolbteppa

1:58 AM

Just quickly looking at it I think the problem is the x^t∑x matrix, you want to diagonalize this so it's a sum of squares too, then it becomes a product of e^-x^2 Gaussian integrals, it's like the standard method of integrating "multivariable Gaussian integrals" www-biba.inrialpes.fr/Jaynes/cappe1.pdf I didn't look too close but you probably just need this method

Eric Gregor

@bolbteppa, When you say diagonalize do you mean take a change of variables? Or do you mean just re-express this exponential? I'm afraid my calculus background is a bit weak, not very familiar with integrating forms

bolbteppa

I mean x^t∑x is not a diagonal matrix but it can be converted into one by a change of variables on x, there is some theory on this, check my link and google multivariable gaussian integrals or even the Gaussian integrals videos here perimeterinstitute.ca/training/perimeter-scholars-international/…

pourjour

what is the equivalence of $\exp(-nx)$ as $x\in [0,1[$?

Eric Gregor

2:28 AM

@bolbteppa thanks. i will check it out

Eric Gregor

2:41 AM

@bolbteppa, i am about to go and i will look at your videos. but in your opinion this integral will be straightforward?

bolbteppa

2:59 AM

The guy's answer was saying that once you diagonalize everything and get it into a sum of squares it's just a product of Gaussian integrals, assuming that is actually what the problem required of you then once you understand the diagonalization part of this problem it'll be straightforward, but it's a method worth spending time on regardless of this problem tbh

user105491

@Pedro @Mike Hey guys, I know I made quite a few mistakes earlier (you may remember commenting about me here on chat), but I would like to tell you that I have learned from my mistakes. Please do not hold that as a grudge against me. I was much younger then, and I have learned. Apologies if it such a post is inappropriate here.

user105491

@Julian I have sent you an email.

Philip Hoskins

3:41 AM

Is there an English translation of Alice Roth's paper "Approximationseigenschaften und Strahlengrenzwerte meromorpher und ganzer
Funktionen"?

Caddyshack

Could anyone please suggest an introductory combinatorics book with lots of problems?

@PhilipHoskins I know German is renowned for its long words, but wow! O_o

Philip Hoskins

@Caddyshack Haha, yeah. I think it translates to something like "Approximation characteristics and radial limits of meromorphic and entire functions"

Kaj Hansen

4:57 AM

Hey guys

Pedro Tamaroff

5:18 AM

Hello Kaj.

Kaj Hansen

Hey there @PedroTamaroff. Just thinking about some complex & topology problems.

Here's Pete's problem set for this week: math.uga.edu/~pete/4200HW_two.pdf

Pedro Tamaroff

@KajHansen Mind sharing? I'm off to snooze in a few though.

@KajHansen How does Pete define a partition?

@KajHansen Did you do 2.6 c)? That's nice, though not too complicated.

Kaj Hansen

@PedroTamaroff, sorry I was afk

Pedro Tamaroff

@KajHansen No problem.

Kaj Hansen

A partition of a set $X$ is a collection of sets $A_i$ such that $A_i \cap A_j = \emptyset$ for all $i \neq j$, and further $\displaystyle \bigcup_i A_i = X$.

Pedro Tamaroff

5:27 AM

The exercises are all pretty doable, although I haven't thought about the Cantor Bendixon thingy. I have no idea how to do that one.

Kaj Hansen

Further, $A_i \neq \emptyset$ for all $i$.

Pedro Tamaroff

@KajHansen Oh. =P

Kaj Hansen

Is there a more sophisticated idea that you have in mind @PedroTamaroff ? Regarding "partition" that is?

Pedro Tamaroff

@KajHansen No, I was missing that the parts be nonempty.

Which he doesn't ask for in a "prepartition."

Kaj Hansen

Indeed. Prepartition nixes the empty condition

Pedro Tamaroff

5:29 AM

So... this is for topology right?

Kaj Hansen

I've not done any of these problems yet. This was just posted.

Pedro Tamaroff

Ah, OK. They shouldn't give you any trouble, I think. =)

Kaj Hansen

Actually, I think I've done some of them regarding the cantor-bendixson because that problem was inspired by a question I asked in lecture.

Yeah, this is topology. He provided some justification for him reviewing some real analysis stuff first @PedroTamaroff

Pedro Tamaroff

Well. I'm going now. Say hi to Pete for me!

Kaj Hansen

Sure thing! Sleep well man

Ilya_Gazman

6:21 AM

Is here any experts in Integer Factorization?

0

Q: Integer Factorization: Possible progress

I build an algorithm for solving Integer Factorization problem when the number to factor is selected by multiplication of two prime numbers with the same bit count. It can factor RSA1024 within $10^{73}$ seconds on my laptop. Please tell me if this considered to be a good time. Next I want to s...

factoring

Chris's sis

6:46 AM

Greetings

Kaj Hansen

Hey there

Chat's always dead whenever I get on :/

A Beautiful Mind

7:04 AM

Hi.

Kaj Hansen

Hey there @ABeautifulMind

A Beautiful Mind

@KajHansen I am very sad. I wonder when I will get well. I hope to get well and enter grad school by 40. That would be my final deadline, so I still have 6 more years.

Kaj Hansen

Are you studying math right now?

A Beautiful Mind

No, I will study after I get well, because I cannot focus right now. But I can do other things till then, like go work as a waiter or something.

I will try new ways to sort out my thoughts. That is my way of getting well.

It's painful for me to think of all these lost years. But I try to tell myself not to regret anything, because I think I have been doing the best I can to deal with my thoughts, even though it is imperfect.

@KajHansen Did you manage to get her number?

Kaj Hansen

I haven't seen her since I last talked to you. I'm not pessimistic though. Things are looking up I feel like.

Getting in shape has given me some more confidence.

A Beautiful Mind

7:16 AM

@kaj You know what upsets me the most now? I don't understand what I have been doing these few years. I don't understand why I am taking so long to get better. I am really confused.

Kaj Hansen

Sometimes I think I will always have depression. But I try to distract myself from it with my hobbies, and that helps.

A Beautiful Mind

I am now pretty convinced that nobody really understands mental illness, nobody.

N3buchadnezzar

7:35 AM

we know that the composition of an integrable function with a continuous function is integrabl

Kaj Hansen

Are you asking @N3buchadnezzar ?

N3buchadnezzar

@KajHansen I kmow it is true, I just have some difficulties proving it through the definition

Kaj Hansen

Hmmm. So we want that $g \circ f$ is integrable if $g$ is integrable and $f$ is continuous?

N3buchadnezzar

@KajHansen Yeah. I mean it should obviously hold since continuous functions are Rieman integrable

Kaj Hansen

Indeed.

I don't think it's obvious necessarily. But I definitely am strongly inclined to believe it.

N3buchadnezzar

7:46 AM

Yeah, since the composition of two Rieman integrable functions is not neccarily integrable

Kaj Hansen

I might have something @N3buchadnezzar

N3buchadnezzar

@KajHansen =)

Kaj Hansen

So it will suffice to show that we can find a partition $P$ such that $U(P) - L(P) < \varepsilon$ for any given $\varepsilon > 0$.

We know that we can do this for $g$ since it is integrable.

N3buchadnezzar

@KajHansen Yeah thats just the definition of being Rieman integrable (well really Darboux, but they are equivalent so)

Kaj Hansen

Sure

Let me think for a sec about the best phrasing of what I want to say.

Ugh, I'm actually running into some problems. Let me think a bit more.

Alex Wertheim

8:02 AM

Hey @Kaj.

Kaj Hansen

Yeah, I think I might be doing something wrong @N3buchadnezzar, but consider this:

Let's say we have $g \circ f : [a, b] \rightarrow \mathbb{R}$. Let $\varepsilon > 0$ be given.

Breaking this down, we have $f:[a, b] \rightarrow X$ and $g:X \rightarrow \mathbb{R}$, for some $X \subset \mathbb{R}$. Now $f$ is continuous, and we know the continuous image of a connected set is connected. Therefore, $X$ is an interval, call it $[c, d]$ for some $c, d \in \mathbb{R}$.

Since $g$ is integrable, we can find a partition of $[c, d]$ such that $U(g, P) - L(g, P) < \varepsilon$. This partition is a collection of points $\{c, x_1, x_2, \cdots x_n, d \}$.

Now let's look at the preimage $f^{-1}$ of each of these points.

Let's see, we have to be careful now since $f$ isn't necessarily injective.

We can get from this a partition for $[a, b]$ with something like this: for each $x_i$ in the original partition of $[c,d]$, choose $w_i = \inf \{ x \in [a, b] : f(x) = x_i \text{ and } x > w_{i-1}\}$.

And the set of $w_i$ should yield a good partition so that $U(P) - L(P) < \varepsilon$. Really check that last part though.

Huy

8:21 AM

@AlexWertheim: Nice picture.

Kaj Hansen

Hey there @AlexWertheim

A Beautiful Mind

8:32 AM

@KajHansen You should go to bed.

Kaj Hansen

I will in a bit @ABeautifulMind

@N3buchadnezzar, for safe measure you may need to refine the $w_i$ partition so that the maximum distance between each $w_i$ is less than the minimum distance between each $x_i$

Chris's sis

@robjohn I also try to create an excel file with each problem and add comment to each one. For instance, why is important for the reader that question, about the difficulty level, ideas of improvements, mistakes, remarks to be made to the reader, and last but not least, the "wow" impact.

The problems and solutions must be definitely attractive.

robjohn

9:43 AM

@Chris'ssis I need to learn how to properly use Excel.

Chris's sis

@robjohn lol, you never used it? :-) It's easy to learn, no worry.

usukidoll

-_- anyone know how to save matlab scripts into .txt files

robjohn

@Chris'ssis I've used it, but just never learned to use it right

Chris's sis

@robjohn I'm not an expert in Excel either.

VividD

9:58 AM

hello, may I just ask any of you to consider math.stackexchange.com/questions/1102872/… for reopening?

the nature if the question is that it should remain open, so that if anybody finds an interesting term, it can enter the appropriate answer

Kaj Hansen

10:23 AM

@VividD, I was disappointed to see that closed.

Technically, the reason for closing is correct, but I don't think it should've been nevertheless.

Janko Bracic

I would like to offer a bounty for a question which is not mine, however I have given a partial answer to it. I cannot see the "start a bounty" link on the question page. Can anyone help me, please?

user91500

@JankoBracic When that question were asked?

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