Mathematics - 2013-05-18 (page 3 of 8)

Faust7

04:22

seems the whole world is afk

DanZimm

@Faust7 whats up

Faust7

Just having difficulty with integration

nothing new lol

just having trouble wrapping my head around the fact that u can integrate a function that is not continous

DanZimm

@Faust7 well, whats confusing about it?

Faust7

never been able to integrate a jump discontinuos funtion before

i read the proof but i don't why its true still

pg 133 thrm 3.2.9

DanZimm

like intuitively?

Faust7

04:37

well intuitively and i don't understand the proof

basically there saying that the upper integral and the lower integral converge to the same value

so it must be integrable

( i think)

but ay time u start talking about the sup on the inf im going to get confused...

* sup of the inf*

DanZimm

well idk the proof but intuitively, its just area under a curve

Faust7

i think there saying the sup of the inf is greater then or equal to the inf of the sup of any 2 diffrent partitions

but if i define 2 disjiont partitions on the interval

then you can't refine one partition to get the other

DanZimm

i dont understand the sup of the inf

Faust7

well there taking the Sup of the partition

and the inf of the partition

the inf is the smallest largest diffrence

and the sup is the smallest of the largest diffrence O.o

i think and it makes my head hurt

perhas i should stop thinking =)

DanZimm

:P

i would help but im busy studying other things

hey

04:46

please help me, i am going insane

i want to understand, but i just cant

0

Q: How to prove that a set of connectives aren't adequate

I guess we have to prove it somehow by an induction as I saw a few examples online. But it just makes absolutely no sense to me... Can somebody explain it in human language? Thank you very much.

DanZimm

@hey understand what

Gustavo Bandeira

►

DanZimm

I would help but I dont understand what that is saying lol

and @GustavoBandeira ?

hey

@GustavoBandeira That song and video really doesn't make me more sane...

Frank Science

Hi, mathematicians!

Has anybody read Terence Tao's article?

Gustavo Bandeira

04:55

@hey What have you tried?

Gustavo Bandeira

05:09

@hey I'm not being annoying. I'm just telling you that users are more willing to cooperate if you show what you have done.

DanZimm

05:23

very much so

otherwise we have no idea where to come from

Gustavo Bandeira

Yes.

@hey And users will reply with their random guesses of what's the meaning of human languange.

DanZimm

yep

@Ethan hey!

Ethan

lol

you finish that thing?

DanZimm

hows it going?

yep!

@Ethan hows it going?

Ethan

good i guess lol

DanZimm

05:38

go to the other room

Ethan

i can't talk lol

DanZimm

dafuq

Ethan

lol

DanZimm

OH you meant like cant talk right meow

xD

Ethan

I can't talk in the chat

DanZimm

05:47

rly?

it doesnt even say youre in there hrm

@Ethan chat.stackexchange.com/rooms/info/8806/…

Gustavo Bandeira

@DanZimm unlock for him.

DanZimm

I thought I did...

lol

robjohn

@Ethan some other chatroom, I assume?

DanZimm

is there a way to delete chatrooms?

Gustavo Bandeira

@DanZimm Or to get the chat rooms (removed) - I believe Ethan has the proeficiency for it!

DanZimm

05:53

is this possible?

@Ethan what does it say when you try to talk?

Gustavo Bandeira

@DanZimm The mods have the power even to make us knell before them. Deleting a chat room is easy.

DanZimm

?

Gustavo Bandeira

Presuming I know the meaning of that word - of course.

Yes. I discovered I don't know! =D

robjohn

@Ethan: I like $\displaystyle\frac1{\log(q)}=\sum_{n=-\infty}^\infty\frac{2^n}{q^{2^n}+1}$. It uses a limit I like.

DanZimm

@robjohn thats pretty sweet!

robjohn

06:07

$\displaystyle\lim_{n\to-\infty}\frac{2^n}{q^{2^n}-1}=\frac1{\log(q)}$

DanZimm

lol thats awesome

robjohn

The usual statement is $\displaystyle\lim_{n\to\infty}n(x^{1/n}-1)=\log(x)$

Ethan

@robjohn thanks I have a bunch of other identities like that

It works for all $q>1$

DanZimm

@Ethan thats awesome, I'm jealous!

robjohn

@Ethan Yeah, I noticed it failed for $q\le1$

Ethan

06:10

$$q=\sum_{n=0}^\infty\frac{(-1)^nq^{2^n}}{(1-q)(1-q^2)(1-q^4)(1-q^8)...(1-q^{2^n})}$$

DanZimm

do these things take you a while to prove?

Ethan

No I found the first one during history class lol

DanZimm

xD

Ethan

$$q=\frac{q(1-q)}{(1-q)}=\frac{q}{1-q}-\frac{q^2}{1-q}=\frac{q}{1-q}-\frac{1}{1-q}(\frac{q^2}{1-q^2}-\frac{q^4}{1-q^2})...$$

Iterate it enough times you will see the pattern lol

DanZimm

heh gotcha

Ethan

06:12

@DanZimm have you seen ramanujans iterated root

DanZimm

i remember reading about it but forgot what it actually is

Ethan

@DanZimm $$(x+1)=\sqrt{1+x(x+2)}=\sqrt{1+x\sqrt{1+(x+1)(x+2)}}...$$

$$3=\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+...}}}}}$$

DanZimm

the first makes sense the second im not seeing at first glance

Ethan

It should be (x+3)

my bad

Just resubstitute the expression for (x+1) with the argument shifted over 1

For the (x+2)

DanZimm

im still confused there no $x$ in the second equation

Ethan

06:19

?

$$(x+1)^2=x^2+2x+1=1+x(x+2)$$

$$(x+1)=\sqrt{1+x(x+2)}$$

shift $x$ over 1 we see that

$$(x+2)=\sqrt{1+(x+1)(x+3)}$$

Substitute this in to the $(x+2) appearing in the first expression

$$(x+1)=\sqrt{1+x\sqrt{1+(x+1)(x+3)}}$$

shift $x$ over 2 in the first expression so that

$$(x+3)=\sqrt{1+(x+2)(x+4)}$$

DanZimm

yea no I understand that

Ethan

substitute this into the (x+3) appearing

DanZimm

OH substitute $x=2$ into the first equation to get the $3= \sqrt{1 ...}$

Ethan

Yes

$$(x+1)=\sqrt{1+x\sqrt{1+(x+1)\sqrt{1+(x+2)\sqrt{1+...}}}}$$

DanZimm

I was trying to see that like by some sort of recursive thing without the first formula

Ethan

06:23

There are a remarkable number of very nice algebraic identities that can be obtained by nothing but repetitive substitution

DanZimm

huh cool

Dota time

Ethan

?

what?

@DanZimm know any good movies

DanZimm

mnatrix

Ethan

seen it lol

DanZimm

inception

prestige

Ethan

06:31

seen it, movie was some what silly

music is great though

hans zimmer

@danzimm prestige?

DanZimm

yea

the prestige

great movie

Ethan

whats it about?

DanZimm

magic hehe

but not really

its a good movie IMO

Ethan

lol you know of anything else

DanZimm

intellectually stimulating

Parth

06:38

Guys

What is the power series of ${\sqrt{1+2x+2x^2 + 2x^3\cdots}}$?

Ethan

write the term under the radical in terms of a geometric series

I think its the generating function for the central binomial coeiffients

let me check

Frank Science

$(1+(2x+2x^2+2x^3+\dotsb))^{1/2}$

Ethan

@Parth $$\sum_{n=0}^\infty \binom{2n}{n}\frac{x^n}{2^n}$$

Parth

@Ethan How did you get that?

@Ethan And is that the series for the stuff under the radical only?

Ethan

No its the series for the whole thing

Parth

06:41

Oh.

Ethan

@Parth the series under the radical is $$\frac{1}{1-2x}$$

Parth

I was writing an answer. I'll mark it as community wiki and give the credit for the last step to ya

But can you tell me how you did it?

Or you could just type it down with your account there.

Ethan

@Parth write out the maclaurin series for $(1+x)^a$

twist this around until you get what your looking for

Parth

@Ethan I may sound stupid, but isn't that just the binomial theorem?

BenjaLim

@robjohn Hey

are you around?

I have a quick question

If I have a function $f$ that is Lebesgue integrable on $[0,1]$ say then is it finite valued?

Ethan

06:45

@Parth The maclaurin series for $(x+1)^a$ can be expanded using the binomial theorem only when $a$ is an integer

BenjaLim

for all $x \in [0,1]$?

Parth

@Ethan Ah, that clarifies it.

robjohn

@BenjaLim yes

Gustavo Bandeira

@robjohn What do you think of my proposal?

BenjaLim

ok then my lecturer is wrong

Parth

06:46

let me bring in my paper.

robjohn

@BenjaLim no, it is not. The yes pointed to your question about me being around.

BenjaLim

@robjohn finite valued for almost every $x$?

Frank Science

@BenjaLim For $-1<x<1$

Parth

1 + ax

+ a(a-1)/2! x^2

robjohn

@BenjaLim a function must be finite almost everywhere to be integrable, but there is no bound on the finiteness...

anon

06:48

@Ethan http://en.wikipedia.org/wiki/Binomial_series

BenjaLim

@robjohn My lecturer claims that if $f(t)$ is integrable on $[0,1]$ then $g(x) = \int_{[0,x^2]} \frac{f(t)}{1-t} dt$ is integrable on $[0,1]$

robjohn

For example $\frac1{\sqrt{x}}$ is integrable on $[0,1]$, but it is not bounded. It is finite everywhere except at $0$

Parth

Phew, that was right.

BenjaLim

But if I put $f(t) = 1$ then at $x = 1$ $\int_0^1 \frac{1}{1-t} dt$ is not finite valued

@robjohn Ok so if you have a function which is not finite valued at just a point then it is integrable?

Parth

OH! I got it.

Learned something new today, thanks @Ethan

robjohn

06:50

@BenjaLim no.. $\frac1{x^2}$ is not integrable on $[0,1]$

BenjaLim

@robjohn hmmmm

ok

Because my argument to show that $g(x)$ is integrable rests on $g(x) \leq C\cdot \int_{[0,x^2]} \frac{1}{t-1} dt$

where $C$ is a constant

robjohn

@BenjaLim what is $C$ (and don't say "a constant")

BenjaLim

basically $C=\int_{0}^1 f(t) dt$

$f$ is integrable by assumption. WLOG we may assume it is positive

Parth

@Ethan Eh, wait a second.

How did you say it was $\frac{1}{1 - 2x}$ under the radical? The series under the radical is $1 + 2x + 2x^2 + 2x^3\cdots$ and not $1 + (2x) + (2x)^2 + (2x)^3\cdots$

BenjaLim

@robjohn

Ethan

06:55

@Parth oh damn I made a mistake lol

@Parth it should be $$\sqrt{\frac{1+x}{1-x}}$$

robjohn

@BenjaLim That all depends on what $x$ is.

Parth

@Ethan lol, that is returning me back to what the question was.

@Ethan That is the question; I have to find the power series of that.

BenjaLim

@robjohn But because wlog $f(x)$ is positive, $\int_{0}^{x^2} f(t) dt \leq \int_0^1 f(t)dt$

robjohn

@Ethan nvm

Parth

math.stackexchange.com/questions/395219/…

I got till $\sqrt{1 + 2x + 2x^2 + 2x^3\cdots}$

But that is no power series.

robjohn

06:58

@BenjaLim The limit will depend on how close to $1$ $x$ is. $x$ must be less than $1$

BenjaLim

@robjohn That's why I think my lecturer is wrong, and that it is integrable only on $[0,1)$

Ethan

@Parth The answer provides a good way to go, use the maclaurin series we just used for $$\sqrt{1-x^2}$$

robjohn

@Parth That is the same thing

Ethan

And multiply this by the power series $$\sum_{n=0}^\infty x^n=\frac{1}{1-x}$$

Parth

Arhmagerd.

robjohn

06:59

@BenjaLim If a function is integrable on $[0,1)$ it is integrable on $[0,1]$

Parth

So I shouldn't even be typing an answer right now.

Ethan

@Parth $$\sqrt{1-x^2}=\sum_{n=0}^\infty a_nx^n$$ $$\sqrt{\frac{1+x}{1-x}}=\sum_{n=0}^\infty (\sum_{k=0}^na_k)x^n$$

BenjaLim

@robjohn Right. So in your example above why was $1/x^2$ not integrable on $(0,1]$?

Chris's wise sister

Greetings people

robjohn

@BenjaLim because $\int_\epsilon^1\frac1{x^2}\,\mathrm{d}x=\frac1\epsilon-1$, which blows up as $\epsilon\to0$

Chris's wise sister

07:04

Today I think I'll continue the work on $$\int_0^{\infty} \frac{x^2 \sin x}{\cos x + \cosh x} \ dx=\sum_{n=1}^{\infty} \frac{H_{2n}-H_{n}}{n^2}$$

Parth

:9489506 Oh, you're fast at clicking.

BenjaLim

@robjohn nvm. So if we look at $h(x) = \int_0^{x^2} 1/(1-t)dt$ this is not Lebesgue integrable?

Ethan

@Chris'swisesister what is it your working on?

Chris's wise sister

@Ethan trying to prove the relation above.

robjohn

@BenjaLim That is $-\log(1-x^2)$ and that is integrable.

Ethan

07:05

@Chris'swisesister also sums of that form have a nice interpretation using veitas formulas, and the product formula for the sine function

BenjaLim

oh yes sorry I was getting mixed up

between $h(x)$ and its integral @robjohn :)

So $g(x) = \int_0^{x^2} f(t)/(1-t)$ is indeed integrable.

robjohn

@BenjaLim under what conditions on $f$?

BenjaLim

$f$ is integrable sorry :D @robjohn

@robjohn because $f$ is bounded by a constant times $h$

robjohn

@BenjaLim what constant?

Ethan

@Chris'swisesister $$\sum_{n=0}^\infty \frac{H_n}{n^m}=\frac{(m+2)}{2}\zeta(m+1)-\frac{1}{2}\sum_{n=1}^{m-2}\zeta(m-n)\zeta(n+1)$$

For integers $m\ge 2$

BenjaLim

07:09

@robjohn $|g(x)| \leq \int_0^{x^2} |f(t)| \int_0^{x^2} \frac{1}{1-t} \leq \int_0^1 |f(t)| \int_0^{x^2} \frac{1}{1-t} = C \cdot \int_0^{x^2} \frac{1}{1-t}$ where $C = \int_0^1 |f(t)|$

robjohn

@BenjaLim how do you justify the first inequality?

BenjaLim

@robjohn Couldn't I use $\int fg \leq \int f \int g$

Chris's wise sister

@Ethan I also think you wanna see this. $$\lim_{n\to\infty} \sum_{k=0}^{\infty}\frac{\cosh{(k \pi/n})}{(1+k)^2}$$

robjohn

@BenjaLim Ack! no!

BenjaLim

really? shit!

Ethan

07:11

@Chris'swisesister you showed me this before ;p

Chris's wise sister

@Ethan some ideas?

Ethan

@Chris'swisesister for what?

robjohn

@BenjaLim consider $f=g=\frac1{\sqrt{x}}$

BenjaLim

ok

Chris's wise sister

@Ethan for some possible ways to evaluate it. :D

BenjaLim

07:12

@robjohn Right I will have to think why $g$ is integrable then

Ethan

@Chris'swisesister the first integral or the second sum?

Chris's wise sister

@Ethan for the 2nd one

Ethan

The hyperbolic cosine sum?

Chris's wise sister

Yeap.

Ethan

Re-write the hyperbolic cosine in terms of exponentials, and use the geometric series formula

You should get the whole sum in closed form, then take the limit, and see what happens lol

BenjaLim

07:15

@robjohn hmmm

robjohn

@BenjaLim It is integrable, but the reason is more subtle

BenjaLim

ok

robjohn

@BenjaLim and the bound is $\int_0^1|f(t)|\,\mathrm{d}t$

BenjaLim

right I know I want to get a bound like that

Chris's wise sister

@Ethan I think I met it in the past.

robjohn

07:17

@BenjaLim try integrating the absolute value...

BenjaLim

ok

absolute value of $f(t)/(1-t)$ @robjohn

Chris's wise sister

brb

DanZimm

@Ethan still here?

robjohn

@BenjaLim yes

DanZimm

@Ethan that what you worked on tonight? heh thats awesome

Ethan

07:23

no thats not mine

I found it here mathworld.wolfram.com/Series.html.

DanZimm

ah ok

Ethan

I have no idea what sourcerey was used there lool

DanZimm

everytime a message is edited when you're mentioned in it, you're rementioned lol

ya no idea

@Ethan did you watch the prestige?

Ethan

@DanZimm Is it actually good? I don't want to watch a magic of tricks and crap

robjohn

@Ethan why do you delete your comments in chat?

Ethan

07:25

@robjohn Habbit sorry

DanZimm

its about a murder in a magic act and a competition of magicians

and about "how did he do his trick"

Ethan

lol sounds silly

robjohn

@Ethan where did you develop this habit?

DanZimm

@Ethan what movies do you like? I might be able to recommend something based on that

@robjohn he has an OCD

Ethan

@DanZimm go into a separate chat for a second

robjohn

07:26

@BenjaLim don't spend too much time on it, I can post it here.

DanZimm

@Ethan ok one sec

@Ethan you arent there?

robjohn

@BenjaLim let me know if you want me to.

Chris's wise sister

@ethan the last series was given on a math contest for high school.

DanZimm

@Chris'swisesister to find a closed form of it?

simplified*

Chris's wise sister

@DanZimm to evaluate it.

DanZimm

07:35

is the sum always to infinity or is the sum supposed to be to n?

i remember you posting that but the sum was from 0 to n

Chris's wise sister

$$\lim_{n\to\infty} \sum_{k=0}^{n}\frac{\cosh{(k \pi/n})}{(1+k)^2}=\frac{\pi^2}{6}$$

DanZimm

ok

Chris's wise sister

There was a mistake. Sorry.

DanZimm

np

you can try a squueze thm on it

Chris's wise sister

@DanZimm yeah, that could be a good idea.

DanZimm

07:38

what domain does k have

Ethan

@DanZimm back lol, do you watch any tv shows?

DanZimm

yea

ive watched a few

Ethan

@DanZimm which ones

DanZimm

pther room

@Chris'swisesister whats the domain on $k$?

Ethan

@Chris'swisesister $$\lim_{n\to\infty} \sum_{k=0}^n \frac{e^{-\pi k/n}+e^{\pi k/n}}{2(k+1)^2}$$

@Chris'swisesister nvm I am being stupid use this, $$\cosh(k\pi/n)=1+O(k^2/n^2)$$

this should get you pi^2/6 farely fast

@danzimm any commedies lol, have you seen the colbert report?

DanZimm

07:44

yes I have, good show

Ethan

wrong chat my bad

DanZimm

np

Ethan

yes lol, the dailyshow is good too

american dad is funny sometimes, archer is also good imo lol though they don't show it very often

DanZimm

youre in the wrong chat room but sure heres fine xD

yea I think thats a good one too

Zolomon

good morning!

DanZimm

07:50

good evening :P

how goes it?

Zolomon

New question! Which mathematical operators in the subset of calculus does not preserve/conserve an equality or statement? I know that ^, sqrt changes the range of an .. equality?

Ethan

questions about notation are usually boreing

:9

Zolomon

@Ethan: You have to learn the boring stuff first before you can get to the fun stuff! :)

DanZimm

what do you mean by range of an equality?

Zolomon

I mean, x^2 = 4 => x=2, but x^2=4 <=> x=±2

Should probably swap places on the sides of the statements.

Ethan

07:53

@Zolomon Yes I think this is the way it is with most academic disciplines, though I know very little

DanZimm

thats not true, when you take the squareroot you're supposed to do $\pm$

meaning $x^2 = 4 \implies x = \pm \sqrt{4} = \pm 2$

Zolomon

Ah yeah, sorry!

I just woke up - haven't had enough coffee yet.

But wait

Oh, need to find a latex cheat sheet first

Lord_Farin

@Zolomon You can have it rendered; follow the 11-star link on the right.

...And good morning, everyone.

Faust7

go back to bed

Zolomon

@Lord_Farin: Yeah, I have activated the bookmarks :)

DanZimm

07:57

xD

@Zolomon I always just google introduction to latex and use the "not so short introduction to latex"

@Faust7 lolol

Zolomon

Found codecogs.com/latex/eqneditor.php - even better

Faust7

well fine then, but im going bed eithier way c ya guys tomorrow

Lord_Farin

@Faust7 Bye.

Transcript for

$Mathematics$ Mathematics