Mathematics - 2013-05-18 (page 1 of 8)

DanZimm

00:04

@Faust7 once it 'clicks' itll be delicious :D

Faust7

I really really hope so!

Peter Tamaroff

@DanZimm Arzela-Ascoli, say?

DanZimm

hrm?

oh right yea

@PeterTamaroff I'm just getting started with functional analysis/measure theory so I don't exactly know the analysis to solve difficult DEs but I do know it requires soime pretty difficult analysis

Peter Tamaroff

@DanZimm I basically says when a sequence of continuous functions has a (uniformly) convergent subsequence, i.e. when a set in $C[a,b]$ is compact.

Jonas Teuwen

Damn.

I am so. Wasted.

I need a drink.

DanZimm

00:10

@PeterTamaroff ah ok cool!

@JonasTeuwen makes sense

Peter Tamaroff

@DanZimm The proof is quite awesome, also.

Faust7

@JonasTeuwen Time for Rum?

DanZimm

:D

my favorite proof thus far is proving $\lim_{n \to \infty} \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} = \ln 2$

Jonas Teuwen

Yes.

Peter Tamaroff

@DanZimm You have to get to know more proofs! =)

DanZimm

00:17

@PeterTamaroff my first analysis course ever was last year, so I'm a bit of a newb in all this

Peter Tamaroff

What about Cantor's $|P(A)|>|A|$, say?

DanZimm

I haven't proved it yet, but I think I know how

but for some reason I liked the ln2 proof

dropbox.com/s/9p0upaj4wfyk8do/a.pdf

Charlie

@peter how are you tonight?

DanZimm

@Charlie I thought you moved up to beta whores?

Charlie

@DanZimm noooo,

@anon I solved my problem

Faust7

00:33

Alright i have been ispired to have a rum and coke

sadly i havent any coke

Charlie

Hi @Skullpatrol how are you?:?

DanZimm

@skullpatrol no way

Charlie

@Faust7 why rum?

Jonas Teuwen

Rum was a good idea.

Faust7

i wonder how run and DR pepper cherry would taste?

Jonas Teuwen

00:34

You must be gay if you do that.

2

Faust7

uhh it doesnt have to rum...

Tobias Kildetoft

@Faust7 or you could drink the rum neat (assuming it is good rum)

Charlie

@JonasTeuwen hi mr. Teuwen

Jonas Teuwen

Hi.

Even bad rum gets worse when you adulterate it with shit like coke.

Faust7

I have mor ethen one bottle of good rum, nice bottle of mount gay's 30yr thats is a neat or on the rocks only bottle =)

Jonas Teuwen

00:37

Son of a cow...

Gustavo Bandeira

@JonasTeuwen XD

Faust7

good stuff

hmm think its broken

Gustavo Bandeira

Faust7

wth

looks like the link works if u click on it

just won't link in the chat?

anon

@Charlie good

your avatar looks like a certain yu-gi-oh monster

Gustavo Bandeira

00:49

@Faust7 Odd.

@anon Did you watch it?

I had a collection of cards, ~700 cards.

anon

I played the card game. show was ultracheesy even though I was 11.

Gustavo Bandeira

@anon Yeah. Did you play magic?

anon

nope

Gustavo Bandeira

It seems to be also a nice cardgame.

Faust7

Any idea what TFAE might mean

itswritten beside

Suppose f is defined on a nhood of a.

Gustavo Bandeira

00:54

@Faust7 The Fate of Armies of E...

Faust7

rofl

Peter Tamaroff

@Charlie What did you remove there?

Peter Tamaroff

01:08

@Faust7 Ah?

Faust7

The norm $||P|| =\text{ max} \{ x_{i} - x_{i-1} | i=1,2,3,...,n \} $ a refinement of a partition P is and partition $Q= P \cup {more points}$

it then states $||Q|| \leq ||P||$

why?

Peter Tamaroff

@Faust7 Think about it. You're adding more points to $P$.

Faust7

why isn't $||Q|| \geq ||P||$

Peter Tamaroff

Won't the spaces get narrower? Make a drawing, for crying out loud! =D

Faust7

but your putting more stuff in it how does the max get smaller?

Peter Tamaroff

01:10

@Faust7 Make a drawing, I insist.

Faust7

Can you tell what to draw?

Peter Tamaroff

@Faust7 Say the initial partition is $$\{1,2,3\}$$

Then $||P||=1$

Say now I add $1/2$.

Then, what is $||Q||$?

Faust7

wait how come its 1?

why isnt it 3?

Q is 1/2 then but thats the min

not the max

Peter Tamaroff

@Faust7 Read the definition again.

It is the maximum of the differences of consecutive elements.

Faust7

how'd you figure that out? was that written above?

Peter Tamaroff

01:13

In this case $x_0=1,x_1=2,x_2=3$, $\Delta x_1=x_1-x_0=1,\Delta x_2=x_2-x_1=1$ so $\max{\Delta x_1,\Delta x_2}=1$.

Faust7

hmm ok

Peter Tamaroff

It says $\max (x_i-x_{i-1})=\max \Delta x_i$

You notes are assuming that $x_1<x_2<x_3<\cdots<x_n$

Faust7

well that makes a hell of alot more sense

DanZimm

@PeterTamaroff is there context as to why you can't just add in the point $10$ ?

Peter Tamaroff

You seem not to be paying good attention to the definitions.

@DanZimm Well, I am tacitly assuming the end points are $1,3$.

Faust7

01:15

were not really defineing them =(

Peter Tamaroff

In fact, in a partition, you always include endpoints.

Charlie

@PeterTamaroff it had a bizarre typo

Faust7

i think the class assumes too heavily that you have taken an analysis course before

Peter Tamaroff

That is $P=\{a=x_0<x_1<\cdots<x_n=b\}$

@Faust7 Just read the definitions carefully, it is nothing out of this world.

DanZimm

@PeterTamaroff but if you add in $10$ doesn't that just change the endpoints?

Charlie

01:16

@anon Ah yu-gi-oh.good times....

Faust7

why do you need to deifne P in that way?

Peter Tamaroff

@DanZimm You cannot add $10$. It is no longer a partition of $[1,3]$ then.

DanZimm

ah ok

Faust7

( sorry if imdriving you nutz with stupid questions)

Peter Tamaroff

@Faust7 In what way?

@Faust7 No, no problem.

DanZimm

01:17

@PeterTamaroff sorry, didn't realize that partitions aren't just normal sets

has never learned about partitions so he goes to learn

Faust7

Im assuming your ording the partition but why? ($P=\{a=x_0<x_1<\cdots<x_n=b\}$)

my notes i have the same statement

Peter Tamaroff

@Faust7 you're

@Faust7 Because you want $x_2-x_1$, $x_3-x_2$ etc to be positive.

(And that is not just a whim, it is important =) )

Faust7

and you can label them this way cause P is finite?

Peter Tamaroff

@Faust7 Yes.

Charlie

@anon Exodia, the forbidden one

Faust7

01:19

I understand why they use the word "refinement" now =)

DanZimm

@Faust7 what class is this for? I'm interested :D

Peter Tamaroff

@DanZimm Probably some sort of Analysis.

@Faust7 Well, don't they give you some pictures?

Read Spivak. He is also the master of us all.

Faust7

no pictures

Its for a 3rd year analysis class the textbook is here ramanujan.math.trinity.edu/wtrench/texts/TRENCH_REAL_ANALYSIS.PDF

Peter Tamaroff

@Faust7 Read Spivak. Read Spivak.

DanZimm

Spivak thats a familiar name!

Faust7

01:21

Spivak?

Peter Tamaroff

@Faust7 Yep. Spivak's Calculus.

DanZimm

@Faust7 what section?

Faust7

3.1

pg 114

Any chanew u got a link?

Peter Tamaroff

@Faust7 You do have pictures!!!

Faust7

what page? =)

Charlie

01:24

Good night, all.

@anon good night.

anon

night

Charlie

:)

Peter Tamaroff

@Faust7 After 114, for sure. 116, 117.

DanZimm

@Faust7 so it can make $a = x_0 < x_1 < \cdots < x_n = b$ since it's saying "we can create a partition of the set $[a,b]$ in the following ways

Peter Tamaroff

@Charlie Nighty night.

Faust7

01:25

Is that why the rectangles in fig 3.1.2 are all messed up because the distance between 2 points doenst have to be uniform?

Peter Tamaroff

@Faust7 Right, we do not require that distances be equal.

Faust7

ohh i thought the picture was just messed up =)

DanZimm

@PeterTamaroff so I didnt realize $\{1,2,3\}$ was the partition of $[1,3]$ - i apologize

Peter Tamaroff

@DanZimm There is nothing to apologize about.

DanZimm

@PeterTamaroff I was wondering, now that I've taken analysis I can pretty quickly pick up any newer low level math concepts, is this normal?

Peter Tamaroff

01:28

@DanZimm Well, experience helps, yes.

DanZimm

for instance I had never knew what a partition was before tonight, but now I understand how it's applied and is used to define the riemann integral

(before that I just called them subintervals xD)

Peter Tamaroff

You are more used to "speaking in mathematics".

DanZimm

well it feels awesome!

Peter Tamaroff

@DanZimm Good.

Faust7

it says$ f(c_{j})(x_{j}-x{j-1})$ where $ x_{j-1} \leq c_{j} \leq x_{j}$ f(c_{j}) is the height of those so called rectangles but why can we take any hieght in the interval?

does the height become unique i guess is what i am asking?

DanZimm

01:31

hrm?

what don't you understand, meaning from the base, what are you trying to understand?

Peter Tamaroff

@Faust7 Yes, intuitively, when $\Delta x\to 0$, the different $c_j$ get nearer and nearer.

@Faust7 $c_j$ is called a "tag" in $[x_{j-1},x_j]$.

We choose one point inside and take $f(c_j)$ as the height.

DanZimm

^ this discussion is why this chat room might be my favorite place on the internet

Faust7

ok that makes some sense

DanZimm

what are you confused on?

Faust7

not really sure i can explain what i was asking better DanZimm but he answered my question =)

DanZimm

01:34

ah ok

well thats good :D

Faust7

the f(c_{j}) is kind of floating in the interval so i was wondering how we had the correct one

Peter Tamaroff

@Faust7 There is no "correct" one. We just choose one.

Faust7

but as the refinement becomes better evenuallt f(c_{j}) become a point not an interval

if i understood what he said anyway

Peter Tamaroff

@Faust7 $f(c_i)$ is a value, what you means is that $c_k,c_j$ get really close for any two tags in $[x_{j-1},x_j]$

DanZimm

@PeterTamaroff I'm curious, why are they called tags?

Peter Tamaroff

01:36

@DanZimm Well, you "tag" a point inside each interval, you "mark" it. In Spanish it is called "marcas", which is "marks".

DanZimm

ah ok

Faust7

Can i interpret this as, if we are in the interval if the interval shrinks to $\delta$ thickness then all of the choice in that interval become a point?

Peter Tamaroff

@Faust7 Well, in the limit, "yes", but if $\delta >0$ then there are infinitely many tags to choose from =)

Faust7

intresting

and surprisingly makes sense

DanZimm

got an upvote on my density answer :D

Faust7

01:40

ok im still abit lost on the def of ||p||

and ||Q||

if we have p= {1,2,3}

and Q = {1,2,2.5,4}

why isnt ||Q|| =1 still?

cause it says max

wouldny you ahve to add points until

Peter Tamaroff

@Faust7 You have $2.5-2=0.5$

Faust7

Q={1,1.5,2,2.5,3} now ||Q||=0.5

Peter Tamaroff

@Faust7 No, do not add $4$.

@Faust7 Well, yes, but $\Vert\{1,2,2.5,3\}\Vert=0.5$ also.

Faust7

whats Vert?

DanZimm

\lVert = $\lVert$

\rVert = $\rVert$

@Faust7 not sure if this is what you're struggling on, but the partitions must still lie within some interval

for instance lets take some partitions of the interval $[1,3]$

Peter Tamaroff

01:44

@DanZimm I must confess your answer on density seems really farfetched.

DanZimm

Put $P = \{1,2,3\}$

Faust7

$||P||= \text {max} \{ x_{i}-x_{i-1}| i= 1,2,...,n \} $

DanZimm

@PeterTamaroff explain?

Peter Tamaroff

Unorthodox.

DanZimm

why do you say that? I don't understand

Peter Tamaroff

01:45

@DanZimm That your solution seems a little unorthodox to me.

Faust7

i want to take that as the maxium length between any 2 points of all the possible points next to each other

DanZimm

@Faust7 yes...

Peter Tamaroff

@DanZimm See André's answer for instance.

DanZimm

@PeterTamaroff I've never seen a direct proof done any other way

his answer is a contradiction proof

Faust7

well then then why isnt ||Q||= {1,2,2.5,3} = 1?

DanZimm

01:46

@Faust7 it is

Peter Tamaroff

@DanZimm Let me see what I can suggest.

Faust7

O.o

Time for a dirnk brb

Peter Tamaroff

@Faust7 Oh, sorry. Yes, it is.

DanZimm

$2-1=1$ i think :P

Faust7

ok why not?

DanZimm

01:47

@PeterTamaroff how is it not? the max of $\{2-1,2.5-2,3-2.5\}$

Peter Tamaroff

@DanZimm Yes, yes, I was looking at something else =)

DanZimm

or $\max\{1,.5,.5\}$

@Faust7 so it is :P

@PeterTamaroff although unorthodox, is it still valid?

Peter Tamaroff

@Faust7 It is, sorry about the confusion.

DanZimm

@Faust7 so if you keep adding points in that interval, the distance between the points can only go down

Faust7

I understand now =)

DanZimm

01:48

:D

Faust7

Ty both so much.

DanZimm

@PeterTamaroff has the real smarts here lol

Peter Tamaroff

@DanZimm Well, look at Andrés idea.

DanZimm

@PeterTamaroff yea just did a quick google search, they just use the archimedian property rather than use the floor function

Peter Tamaroff

First, you show you can take $n$ such that $\frac 1{10^n}<x-y$

DanZimm

01:52

@PeterTamaroff again that's the only direct proof I've ever seen

Peter Tamaroff

Then the idea is just to add that enough times so that it lays inside $(x,y)$

DanZimm

but how do you know how many times is enough

and how do you know you wont skip over the interval

Peter Tamaroff

@DanZimm Well, you know you won't skip precisely because $10^{-n}<x-y$

DanZimm

I think you mean $10^{-n} < y - x$

Peter Tamaroff

And you know there is an integer because we can take $k$ large enough so that $k/10^n >y$

@DanZimm I took $x>y$.

DanZimm

01:55

@PeterTamaroff chat.stackexchange.com/transcript/message/9486956#9486956

Peter Tamaroff

In both cases we use the Archimedean property.

@DanZimm Oh, well, I am making drawings pages and I have put $x>y$. Whatever.

DanZimm

eitehr way thats not the point xD

Peter Tamaroff

@DanZimm Yes, exactly.

DanZimm

so why can you find an $n$ that $10^{-n} < x - y$ ?

an integer n that is

Peter Tamaroff

@DanZimm Well, $10^n$ is a natural number.

Archimedes says for each $\epsilon >0$ there exists $n$ such that $\frac{1}{n}<\epsilon$

Now, take $k$ such that $10^k >n$.

DanZimm

01:59

I essentially do the same thing, but rather than rely on properties, I find the exact integer

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