Mathematics - 2013-05-18 (page 2 of 8)

DanZimm

02:00

do note that $10^k > k \; \forall k \in \mathbb{N}$

Peter Tamaroff

@DanZimm Yeah, you'll see that it can get cumbersome =)

And we don't want cumbersome.

DanZimm

I still don't see how to do the second part of Andrés answer directly

and yes as my answer demonstrates xD

@PeterTamaroff and wouldnt cumbersome be in the perspective of whoeevr is reading it?

you can call it cumbersome, I call it rigorous

Peter Tamaroff

@DanZimm Of course.

@DanZimm My point is that you won't be able to do that always, and sometimes we will need to be more practical.

DanZimm

I was just defending my answer :P

Faust7

If f is unbounded on [a,b], then f is not integrable on [a,b]

my book makes a statment

DanZimm

02:04

yea

Peter Tamaroff

@Faust7 I saw it, yes.

Faust7

we will show that if f is unbounded on [a,b], P is any partition of [a,b] and M>0 then there are reimann sums \sigma and \sigma^{'} of f over P such that |\sigma - \sigma^{'}| \geq M

DanZimm

@PeterTamaroff so would you consider my proof bad ?

Faust7

why is this true?

DanZimm

@Faust7 what do you mean why is it true?

isn't teh proof right there?

Faust7

02:06

why does |\sigma - \sigma^{'}| have to be bigger or equal to M?

Peter Tamaroff

@DanZimm Not at all.

It is good.

Faust7

no it tells you to prove it

DanZimm

ok just making sure :D

Peter Tamaroff

@DanZimm My point is, why not take $r$ of the form $10^m$ such that $r(x-y)>1$ in the first place?

DanZimm

@Faust7 well what does it mean for $\lvert \sigma - \sigma^{'} $\rvert \ge M$ ?

@PeterTamaroff because I find the exact value of what $m$ is

Gustavo Bandeira

02:08

DanZimm

thats the only difference

Faust7

thats what im trying to ask =)

Peter Tamaroff

@DanZimm There is no need, really.

Gustavo Bandeira

Dat's me. Yay!

Peter Tamaroff

@GustavoBandeira Who's that?

@GustavoBandeira I know who you are, yes.

DanZimm

02:08

@GustavoBandeira the guy in the blue shirt you mean? ;D

Gustavo Bandeira

@PeterTamaroff Me and my GF.

DanZimm

@GustavoBandeira ver nice

you look like a cute couple!

Faust7

ok how about this can you tell me what M is?

Peter Tamaroff

@DanZimm There is no "exact" $m$. My point is, anything good enough does the job.

Gustavo Bandeira

@DanZimm =D

DanZimm

02:10

@PeterTamaroff right but I give an $m$ that does do the job, it comes down to I'm just being more precise

amWhy

@Gustavo Nice pictures: your avatar, and you photo above!

Peter Tamaroff

@DanZimm Aha.

DanZimm

@PeterTamaroff do note that I am often called pedantic xD

so in most peoples POV I'm being unnecessarily precise there

Peter Tamaroff

@DanZimm Well, there is no pedantry here. I find it unnecessary to dwell into such "exactness".

amWhy

Okay, folks...let's be nice ;-)

Gustavo Bandeira

02:11

@amWhy =)

Peter Tamaroff

@amWhy We're being nice.

DanZimm

@PeterTamaroff isnt that what pedantry means? lol

@amWhy yes indeed its a general conversation

amWhy

@Gustavo $\langle \,:\circ )$

Peter Tamaroff

@DanZimm I take it as "excessively concerned with formalism".

amWhy

I know...I didn't mean you're being mean, just nitpicking one another...

DanZimm

02:12

tis true

@PeterTamaroff ah well I suppose we have two different views then :P

Peter Tamaroff

@amWhy Meh.

amWhy

@PeterTamaroff being pedantic about being pedantic

DanZimm

@amWhy I noticed that then realized I was being silly xD

Faust7

rofl

DanZimm

metapedacism?

amWhy

02:14

@DanZimm Now you've got it ;-)

DanZimm

unfortunately pedacism isnt a word hehe

Faust7

well you're both pretty good at translating mathimagic into english =)

DanZimm

@PeterTamaroff by far knows more math than me so I probably should just shush :P

Peter Tamaroff

@DanZimm Wrong.

amWhy

@DanZimm It sounds like "bias against children" as in racism and sexism...pedacism

Peter Tamaroff

02:15

=D

@DanZimm You are in your right to defend your proof.

DanZimm

@amWhy thats exactly what I thought once I said it, and then I realized I added meta hahahahhaa

@PeterTamaroff well thank you very much sir :D

@PeterTamaroff and whatever you say mate, I'm sure you have more than 4 months of analysis experience (thats where I'm at currently)

Faust7

speaking of analysis wth does an unbounded integral have to have the existence of 2 unique sums over it?

Peter Tamaroff

@Faust7 Come again?

DanZimm

@Faust7 well

how do you understand an unbounded function?

Faust7

@PeterTamaroff thats what i said to my textbook lol.

DanZimm

02:19

@Faust7 what page?

Faust7

i relised i was using the wrong idea of bounded

pg 119

but even so im not sure why 2 unbounded integrals need to have a unique sums over the same interval

Peter Tamaroff

@Faust7 You can say $f$ is unbounded in $[a,b]$ if for each $M$ there exists $x\in [a,b]$ such that $f(x)>M$.

Faust7

hmm thats following form the defintion of mean value theorem or something?

DanZimm

@Faust7 what do you mean? are you trying to show that theorem 3.1.2 shows that $f$ is then not integrable?

Faust7

no im trying to understand why | \sigma - \sigma^{'}| \geq M

DanZimm

02:22

thats what they prove there!

Faust7

cause as we take the limit we get sigma = L and sigma' = l

L-L=0

so M=0

but its says M>0

Peter Tamaroff

@Faust7 Well, first note the following.

Faust7

@DanZimm the argument that is listed below proves that the second case is just the case stated in 3.1.7

DanZimm

the second case?

Faust7

+( =S

Peter Tamaroff

02:25

Oh, sorry I have to go.

Faust7

Hav fun m8

Peter Tamaroff

But, never mind about unbounded functions.

We don't like them!

=)

Faust7

rofl

DanZimm

@PeterTamaroff take it easy!

@Faust7 so which step are you in teh proof?

Faust7

honestly i just realised im a moron

the statement |\simg - \sigma^{'}|

they show that thats true

DanZimm

02:27

yea exactly

Faust7

if f is unbounded

DanZimm

yep

Faust7

which is completly ubsurd

i was just confused i thought they said

take this as true

and then we will prove that its not inegrateable

DanZimm

oh yea no

Faust7

and i just didnt belive them

DanZimm

02:29

no thats what youre supposed to do

haha yea

Faust7

=)

DanZimm

the exercise is to show that since unbounded functions follow this theorem, show that f is now non riemann integrable

Faust7

if im understanding the rest of it correctly

it is showing the the Limit is not unique

DanZimm

essentially yea it looks it

Faust7

ie | \sigma -L| < \epsilon$

doesnt hold

DanZimm

02:30

I'm a bit confused on the proof myself

yea

well it shows its not cauchy so its not convergent

you could say that

Faust7

yeah the middle bit of that page is pretty confusing

once they start talk about two diffrent tags in that interval

and then i think that they show each tag goes to a diffrent value

or they show that it does so it must be bounded? not really sure

im sure there talking about the left most tag and the right most tag

in that interval

DanZimm

ah i get it

one sec

Faust7

glad someone does =p

DanZimm

sec

sorry

Faust7

np man i think actualyl follow some of it

DanZimm

02:40

:D

the thing im stuck on is the last algebra step actually haha

bad at math

Faust7

which step?

DanZimm

ah ok i see

the last one

Faust7

|f(x)-f(c_{j})| step?

DanZimm

so first the show, hey, look here, at some partition $i$ there is some number $c$ so that $f(c) - f(c_i)$ is unbounded ($> M$)

Faust7

they assume that

yes then prove it must be so

DanZimm

02:43

no they show that

(proof by contradiction, because if this isnt true ....)

Faust7

i follow the rest of that until |f(x)| \leq max |f(c_{j)| + M

why does that imply that there assumption was true?

DanZimm

Well do you see how that's showing that $\exists \; i , c : \lvert f(c) - f(c_i) \rvert > \frac{M}{x_i - x_{i-1}}$ ?

its like a subclaim in there

oh yea you do understand it

sorry heh

Faust7

oh x is in that interval

i get it =)

that proves that its not less than it

so it must be equal or bigger than it

DanZimm

yep!

Faust7

and c is n the interval we assumed x to be in

in*

DanZimm

02:47

the confusing thing is they're doing a sort of implicit iteration over $j$ and then they say lets take the max

Faust7

but why does the final statement contradict f being unbounded?

DanZimm

well actually no, so $x$ is arbitrary

Faust7

because the right would be a bound?

DanZimm

yea

exactly

Faust7

tricky bastards.

DanZimm

02:48

hehe

thats how analysis is, my friend described it as "witty math"

Faust7

the implication over the interval

DanZimm

hrm?

Faust7

there taking the biggest one on the partition

DanZimm

yea

technically they could take any of them

@Faust7 the only weird thing, is $M$ is supposed to be arbitrary

Faust7

but it has to be >0

DanZimm

02:54

but the definition of bounded says its a particular value

yea

but still

sec lemme think

I'm used to a different style of proof, so sorry its taking me so long heh

Faust7

i have actually managed to make ti though the first section without looking at the dam thing with 2 heads, its agood day

DanZimm

they dont use quantifiers

:D

Faust7

time for more rum =)

almost out of the 12 yr stuff gf went to get me more

DanZimm

heh

good gf xD

so yea otherwise itd be bounded

@Faust7 does the whole book lack quanitifers?

Faust7

define quantifiers

DanZimm

03:01

$\forall , \exists$

Faust7

they usually use words even in proofs

DanZimm

well I mean for example they do $1 \le j \le n$

Faust7

yes

DanZimm

rather than $\forall \; j : 1 \le j \le n$

I'm just not used to that

so like I wasn't sure if it's $\exists \; x : x_{i-1} \le x \le x_i$ or $\forall \; x : x_{i-1} \le x \le x_i$

when they say $x_{i-1} \le x \le x_i$

Faust7

ohh

each partition is finite

and order in such a way that

x){i-1} is <x < x_{i}

DanZimm

03:05

but they define $x$

Faust7

basically its like x_{i-1} = inf

x_{i} = sup

DanZimm

and they say $x_{i-1} \le x \le $x_i$

Faust7

and x is in the interval

DanZimm

right, but is it there exists or is it for every x in the interval?

Faust7

for every x in that interval

DanZimm

03:06

right, thats what I didnt realize for a while hehe

Faust7

=)

DanZimm

cuz im used to having a $\forall$ in there like $\forall x \in [x_{i-1} x_i]$

anyways, yea makes sense

Faust7

now that i look at that its a bit wierd

DanZimm

what is the book or what i say?

Faust7

in our class notes he deifnes it more intuitivly

the book

DanZimm

03:07

ya

oh well!

Faust7

i just didnt realize cause my prof did a good job at playing to the books weakness

DanZimm

yea heh, sounds like you have a prof who knows what up!

believe it or not I've never learned this stuff heh

so I give you this as a warning: I have the possibility of being horribly wrong

Faust7

yeah this isnt really suppose to be a 1st analysis class

one of the pre reqs for the class is a second year analysis course

DanZimm

I'm taking it in a year, gunna be my third one technically

ah ok cool

Faust7

or enough calculus and algerbra

and they will let you into it

im in the second boat claerly

and have nfi whats going on

DanZimm

03:09

heh

have you taken a proof based class before this ever

Faust7

not really

calculus at higher levels reqauires some proof ish stuff

DanZimm

ah well thats the issue

Faust7

thats all the experience i have

DanZimm

I did the same thing, but you'll get it eventually

Faust7

or rage quit

DanZimm

03:10

my school offered a "number theory" course which was supposed to introduce you to proofs

hahaha

Faust7

=)

DanZimm

but i skipped it (took it at same time as analysis)

Faust7

i know how to prove using analysis continity and bounded / compact etc

in a fairly redumentry way

DanZimm

in what space?

meaning, like in a general metric space, or in the reals?

Faust7

and i can get around alot of other proofs with limits and etc using tricks

euclidean and some stuff in C

DanZimm

03:13

so like $\mathbb{R}^n$?

Faust7

yeah r^n is euclidean space =)

DanZimm

just making sure im not making any assumptions and end up being an idiot heh

Faust7

isnt not really focused on

DanZimm

so whats the definition of compact?

Faust7

that there exists a finite open covering of S

Jordan

03:14

math is too hard, I took a 6 month break from math and I forgot everything. I need to go back to the 6th grade and story 10 years of math again but I only have a month to laern it all

it was an incredibly relaxing sixth months though

DanZimm

ok just making sure its not just "closed and bounded" heh

Faust7

well there isnt a diffrenc in r^{n}

DanZimm

@Jordan what kind of math do you need to know?

@Faust7 yea thats true, but thats a result, not the definition

Jordan

@danzimm Factoring, algebra, trig, calculus, geometry

DanZimm

@Jordan you'll be fine, I'm sure at some point stuff will click again

Faust7

03:15

=)

DanZimm

I have faith! xD

Jordan

@DanZimm I am not that smart

Faust7

im a moron and i figured it out after a 3 year break

DanZimm

if you think the then itll be true

hahaha

Faust7

and u would have to add de's and linear algerbra to that list

DanZimm

03:16

de's are soooo easy lol

Faust7

more rum brb

DanZimm

well

low level easy ones with closed solutions xD

Faust7

non-linear dynamics can be a pain

Jordan

factoring is so hard

so much memorization

DanZimm

yes but still usually has some pretty intuitive approaches

@Jordan tbh I don't know how to factor very well atm

Jordan

03:17

I dont even know how to google the answer, what is it called when I havea thing that has a power of 3?

DanZimm

and I'm apparently taking upper level math courses

exponent?

like $x^3 + 3$ ?

Jordan

@DanZimm I am too bad to get to higher level classes, taken calculus like 3 times

yeah

x^3+x^2+x+4

DanZimm

erm yea no idea lol

theres some sort of trick

Faust7

i dont think u can factor that

not over the reals

try a -4 instead

Jordan

I just meant something in that form

a third degree polynomial

Faust7

03:20

anything of full order liek that

i cant factor

it need to be missing a term

like the x^{2} or x or the constant

or its a pai for anyone to factor

Jordan

I am trying $x^3 - 3x^2 - 4x + 12$ but I am too stupid to get it on my own

Faust7

you remove all the tricks

ok not that one

x=2 is a solution so use syntetic division to facto out

DanZimm

well $x^3 - 3x^2 - 4x + 12 = x ( x^2 - 3x - 4 ) + 12 = x(x-4)(x+1) + 12$

Jordan

I am trying some weird stuff

Faust7

x-2 form the it

Jordan

03:21

well

I know I can pull an x out of the front

but what happens when I have two seperate terms?

DanZimm

idk if that helps xD

Jordan

It isn't helpful I dont think is it

Faust7

nah guess a root

Jordan

I dont know divison like that

Faust7

now x-2 divides | (x^3 - 3x^2 - 4x + 12)

Jordan

03:22

I have tried to learn division like that dozens of times, I forget it every time I learn it

Faust7

its jsut long division

the first is x^{2}

Jordan

I dont know long division

Faust7

so subtract x^{2} ( x-2) form the above statement

Jordan

been using a calculator for a while now

Faust7

i don't know how to use a calculator =)

its ok being able to do ariemitic is def not pre req for getting a phd in math

Jordan

03:25

I cant do long division

need to look it up

Faust7

i know sevral that can't add or subtract half the time correctly

Jordan

it is such an abstract and memorization thing

DanZimm

@Faust7 raises hand

i said $-(-1 - 1) = 0$

at one point on a test

Faust7

=)

lol some stared it being gay to add dr pepper cherry and rum together!

Dominic Michaelis

@DanZimm I once made the same :D

But calculating in $\mathbb{F}_2$ is neater anyway

DanZimm

03:28

its so cool i dont even know what that is

lol

Faust7

me eithier

some wierd space im assumeing

Dominic Michaelis

$\mathbb{F}_2$ is the finite field with 2 elements

DanZimm

I wonder what $\lim_{n \to \infty} \frac{1}{n} + \frac{1}{n^2} + \cdots + \frac{1}{n^n}$ is

Faust7

it doesnt exist

oh i have actually seen f before

its 0 and 1

DanZimm

derp?

Faust7

03:30

i wonder if o and -1 would work

DanZimm

f?

Faust7

F= \{ 0,1 \}

its a field

you taken algrebra before?

DanZimm

not like an abstract algebra class but I do know what a field is

Faust7

addition multiplaction all hold over the finite field F

anon

@DanZimm $$0\le \frac{1}{n}+\frac{1}{n^2}+\cdots+\frac{1}{n^n}\le \frac{1}{n}+\frac{n-1}{n^2}\to0$$

Faust7

03:32

theres a unit and an identity

DanZimm

@anon yea just saw the same thing essentially

was sad it was so simple

@Faust7 didnt realize you were talking about what @DominicMichaelis said not what I said heh

Dominic Michaelis

@Faust7 as 1=-1 this absolutly works even though every number will work even though the multiplication will look a bit strange

Faust7

ah i just realized it doesnt work

-1 * -1 = 1 and its not in F

so it won't =(

DanZimm

define -1*-1 = -1

2

Faust7

rofl

DanZimm

03:33

xD

Dominic Michaelis

yeah that absolutly works

DanZimm

since theres only 2 elements you can pretty easily just create new binary ops

Dominic Michaelis

$-1$ will be the neutral element of multiplication

DanZimm

as if theres a sort of hashtable for the results

yea

Faust7

im going to tell that to my proof next time i mess up the orientation of a surface integral or something

prof*

DanZimm

03:35

hahaha

Faust7

citeing it

DanZimm

@anon I didn't use that $\frac{1}{n^2} + \cdot + \frac{1}{n^n} \le \frac{n-1}{n^2}$ though, where does that come form?

I notcied that $\frac{1}{n} + \frac{1}{n^2} + \cdots + \frac{1}{n^n}\le \frac{1}{n} + \frac{1}{n}$

Dominic Michaelis

@DanZimm becase $\frac{1}{n^k} \geq \frac{1}{n^{k+1}}$

@DanZimm that is wrong just that the case where $n=1$

DanZimm

when $n=1$ we get $\frac{1}{1} \le \frac{1}{1} + \frac{1}{1}$

Jordan

god I am bad at math and lazy

DanZimm

03:39

the lazy part is where itll get you

anon

@DanZimm $$\frac{1}{n^2}+\frac{1}{n^3}+\cdots+\frac{1}{n^n}\le\frac{1}{n^2}+\frac{1}{n^2}‌+\cdots+\frac{1}{n^2}=\frac{n-1}{n^2}$$

DanZimm

@anon AH ok cool

yea I couldnt prove exactly what I came up with xD

thanks!

@amWhy how do you make things blue in your answers?

ah $\color{blue}{\mathbb{R}}$ cool!

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$Mathematics$ Mathematics