Mathematics - 2013-02-20 (page 4 of 6)

Chris's sister and pals

19:00

@PeterTamaroff: this is what I was thinking about. This way seems just nice. Thanks! :-)

Peter Tamaroff

@robjohn can probably help.

@Chris'ssisterandpals Welcome. Have a nice one.

Chris's sister and pals

It's OK. Your help was a really help.

I'm gone.

see ya later

Jenna Magic

Hey guys

Peter are you Peter Tamaroff?

Ben W.

Hi @Jenna. You can see someone's full screen name by hovering over their icon.

Charlie

Is Halmos the only good book for set theory?

Jenna Magic

19:10

OH ok

Hey Peter!

Tobias Kildetoft

@Charlie set theory at what level?

NBP

introduction to set theory

Ben W.

@Charlie Azriel Levy's and Thomas Jech's books are good. Jech's is more advanced I think.

NBP

What would you recommend to an undergrad taking introduction to set theory?

Tobias Kildetoft

I like the book Chapter 0 by Schumacher (I think)

Charlie

19:13

@TobiasKildetoft ZF axioms

Arkamis

gr

Ben W.

Levy's book is more introductory, so I'd go with that.

Arkamis

MATLAB is so fucked up sometimes.

Tobias Kildetoft

@Charlie ahh, no idea about good books for that

Jenna Magic

Can someone help me out with a discrete problem?

NBP

19:14

thanks benjamin

Charlie

@BenW. thanks, I will take a look at both :)

Ben W.

No problem!

Jenna Magic

Guys what do they mean when they ask are these two functions of the same order?

Ben W.

Not sure, sorry.

Peter Tamaroff

@JennaMagic I answered.

Jenna Magic

19:21

Peter

so for the first instance

amWhy

@robjohn You there?

Jenna Magic

f/g would mean f(x) / g(x) ?

Peter Tamaroff

@JennaMagic Yes.

Dominic Michaelis

hi

Jenna Magic

so in the first instance..3x + 7 / x?

do we take derivative of both?

or is it rather just infinity/infinity = infinity

Peter Tamaroff

19:23

Well, you can. But note that $$\frac{3x+7}{x}=3+\frac 7 x$$

What does $7/x$ approach to?

Jenna Magic

0

Peter Tamaroff

So your limit is $3$.

Jenna Magic

but how is that showing that they are the same 'order' ?

user58512

what is order

Jenna Magic

Peter, what about floords?

floors?

user58512

you there?

Peter Tamaroff

19:34

@JennaMagic I don't understand what you mean by floors.

Jenna Magic

you know, floor and ceiling

where 1.2 = 1

1.7 = 1

1.9 = 1

that's a floor

Peter Tamaroff

But why would you want that?

Jenna Magic

that's a problem

as in the problem should have been Floor of that function..not sure how to solve

Peter Tamaroff

What problem?

Jenna Magic

0

Q: How to show that pairs of functions are of the same order?

If we have these pairs of functions, how can we show that they are of the same order? a) $3x + 7,\quad x$ b) $2x^2 + x − 7,\quad x^2$ c) $x + 1/2,\quad x$ d) $\log(x^2 + 1),\quad \log_2 x$ e) $\log_{10} x,\quad \log_2 x$ Thanks guys!

Peter Tamaroff

19:36

You mean you want $$\lfloor 3x+7\rfloor /\lfloor x \rfloor$$?

Jenna Magic

C

no

c) x+1/2,x

floor of x+ 1/2

?

Peter Tamaroff

Could you be more specific?

Jenna Magic

How ?

Floor of (x+(1/2)), x

that's what "C" should be

in the question

Peter Tamaroff

No, $C$ is a constant.

Not a function.

Jenna Magic

ok

I listed it as A,B,C,D,E

Peter Tamaroff

19:46

Oh.

Jenna Magic

in the original question

Peter Tamaroff

You want to edit that?

Jenna Magic

Ok....

Or I can just ask here?

Peter Tamaroff

Look now. Is it OK?

Jenna Magic

Yeah!

Just wondering though how I owuld solve that one

Peter Tamaroff

19:47

Just use the squeeze theorem

Squeeze the floor between two common linear polynomials.

Jenna Magic

dont really remember tha tfrom calc 1

Peter Tamaroff

Then read it again =)

Jenna Magic

...

great.

Peter Tamaroff

I really mean it.

And I'm not being mean.

Really.

Jenna Magic

Yes you are.

Peter Tamaroff

19:52

You don't remember what the squeeze theorem is?

Jenna Magic

I do, I just don't know what to show it with here.

Nimza

Hi again!

Charlie

@Nimza Hi again!

Peter Tamaroff

@Nimza Man. Can you help me with something?

It's about finding how fast a sequence goes to zero.

That is, $O(f(n))$, for some $f(n)$.

And $f(n)\to 0$.

Nimza

@PeterTamaroff oh, sorry, not now, I'm prepairing for test now :(

Jenna Magic

19:56

Peter, you never answered me in math.stackexchange.com/questions/309361/…

It goes to 0..now what?

Peter Tamaroff

@JennaMagic Well, that means that $n!$ is much bigger than $2^n n^2$ for large $n$.

Jenna Magic

Makes sense

ok last question

for the same order question

how can i do lhopitals

for the LOGS

and do i need to?

Dominic Michaelis

hi

Charlie

hi

Dominic Michaelis

how are you sweetie ?

Charlie

20:01

good, and you?

Jenna Magic

anyone here know discrete?

i have 1 question

Dominic Michaelis

my parents are so annyoing

Charlie

@DominicMichaelis oh! why you say?

Dominic Michaelis

i am running mad because of them ...

Charlie

@DominicMichaelis what could they have possibly done for such thing?

Dominic Michaelis

20:05

i am trying to think but they call me every 5 minutes

Charlie

@DominicMichaelis they must be worried

Dominic Michaelis

they call me for stuff like, could you reach me that bottle over there ?

Charlie

@DominicMichaelis really?

Dominic Michaelis

yeah

Charlie

So you are at the same place?

Arkamis

20:07

fgfd

Jenna Magic

Does ANYONE know DISCRETE

OMG

Dominic Michaelis

yeah

Charlie

@DominicMichaelis Oh!

user58512

@JennaMagic, what is "DISCRETE"

anon

http://en.wikipedia.org/wiki/Discrete_mathematics

Charlie

20:08

@Arkamis lkçl?

Dominic Michaelis

@jennaMagic as n! is about n^n/e^n it grows much faster than 2^n n^2

Jenna Magic

Not that

I need to show that Floor(x + 1/2) is the same order as "x"

user58512

@JennaMagic, what is order

Jenna Magic

i think that i should divide the first function by the second

right?

i just have to prove that the limit of the ratio is a non zero constant

user58512

ok so do it

Dominic Michaelis

20:10

@jenna the limit where ?

anon

@JennaMagic observe x<=floor(x+1/2)<=x+1; divide it through by x.

note that floor(x+1/2)/x is different from floor((x+1/2)/x) (the latter is what the title of your question intimates)

Charlie

@Theo :D

Theorem

Today i had an exam , and i had to show that group of order 16 , whole nilpotent degree is 3 has center or order 2 .

@Charlie

wassup

:D

Charlie

@Theorem I'm good, and you?

Theorem

just after coming out of my exam , i solved it :( :P

that was really straight forward

Tobias Kildetoft

20:12

@Theorem that tends to happen to a lot of people

Theorem

i am good @Charlie

Charlie

@Theorem I feel awful after exams

@Theorem good!

Theorem

@TobiasKildetoft : i felt bad . i don't know what i should do about it .

isn't it bad ?

Tobias Kildetoft

not much to do about it now

Theorem

and also time factor .

@TobiasKildetoft : did u ever come across such situations :P

Charlie

20:14

@Theorem who never...

Dominic Michaelis

@jenna math.stackexchange.com/questions/309445/…

anon

why x-1 instead of just x?

curious choice

BenjaLim

@anon Hi.

anon

yo yo yo

Theorem

@Charlie : i think its just nature of bad student like me

Dominic Michaelis

20:16

@anon take x=1/10

Charlie

@Theorem noo....

BenjaLim

@Theorem Same thing happened to me after my algebraic topology exam last semester

anon

@DominicMichaelis ah, duh

Charlie

@Theorem it happens to me all the time

Theorem

@BenjaLim : its so easy to miss some small trivial stuff and lose so many points . i hate it x-(

Dominic Michaelis

20:18

@anon lets see if i get any upvotes for this one ...

Charlie

@Theorem Think like this: you solved!! Brilliant, you are capable! you are good

Theorem

but after exam @Charlie

Charlie

@Theorem you solved it anyway

@Theorem you should be proud of yourself

Peter Tamaroff

@anon Do you think you can help me with something calculish?

anon

maybe

Peter Tamaroff

20:21

@BenjaLim Or maybe you?

Theorem

i hate myself for being quite an idiot :D

@Charlie

Peter Tamaroff

@anon Well, it is the following.

Theorem

Quora is awesome

Charlie

@Theorem no, no, no

Theorem

anyone there in Quora ?

Charlie

20:21

@Theorem me

Peter Tamaroff

We want to find $$\lim {\left( {\int\limits_0^1 {{{dx} \over {1 + {x^n}}}} } \right)^n}$$ @anon

Theorem

what's ur name :P i would like to follow you

@Charlie

Peter Tamaroff

Now, with a change of variables $x\mapsto x^{-1}$, we get that

anon

hmm

Charlie

@Theorem hahaha I almost never use it, used

Peter Tamaroff

20:22

$$\int\limits_0^1 {{{dx} \over {1 + {x^n}}}} = \int\limits_{1}^\infty {{{n{x^{n - 1}}} \over {1 + {x^n}}}{{dx} \over {nx}}} $$

I added the $n$s.

Integration by parts yields

$$\int\limits_0^1 {{{dx} \over {1 + {x^n}}}} = - {{\log 2} \over 2} + \int\limits_1^\infty {{{\log \left( {1 + {x^n}} \right)} \over {n{x^2}}}dx} $$

robjohn

@PeterTamaroff: sorry, I have been very busy looking into things on meta. Do you still have a question?

Peter Tamaroff

Look at the above @robjohn

anon

you could probably use some sort of local expansion of the digamma function

robjohn

@amWhy: I am now. Sorry about the delay.

Peter Tamaroff

Now we aim to use $(1+1/n)^n\to e$

Summing and negating $1$, we get

$$\int\limits_0^1 {{{dx} \over {1 + {x^n}}}} = 1 - {{\log 2} \over n} + \int\limits_1^\infty {{{\log \left( {1 + {x^n}} \right)} \over {n{x^2}}}dx} - 1$$

$$\int\limits_0^1 {{{dx} \over {1 + {x^n}}}} = 1 - {{\log 2} \over n} + \int\limits_1^\infty {{{\log \left( {1 + {x^n}} \right)} \over {n{x^2}}}dx} - \int\limits_1^\infty {{n \over {n{x^2}}}dx} $$

amWhy

20:24

@robjohn You've obviously been busy, np.

Charlie

@Theorem you like it?

Peter Tamaroff

$$\int\limits_0^1 {{{dx} \over {1 + {x^n}}}} = 1 - {{\log 2} \over n} + \int\limits_1^\infty {{{\log \left( {1 + {x^n}} \right) - n} \over {n{x^2}}}dx} $$

Theorem

@PeterTamaroff : it seems the limit is $0$ ?

Peter Tamaroff

Now I aim to show that $$\int\limits_1^\infty {{{\log \left( {1 + {x^n}} \right) - n} \over {n{x^2}}}dx} =O(\text{something appropriate})$$

@Theorem No, the limit is $1/2$

skullpatrol

@Charlie have you seen skull's partyzone?

Theorem

20:26

@PeterTamaroff : oh ok , what is the question ? haven't looked in your actual question :P

robjohn

@amWhy I am not expecting any repercussions from the meta thread. I am simply performing due diligence.

Charlie

@skullpatrol No!where??

Peter Tamaroff

So that $$\lim \left(1-\frac{\log 2}{n}+O(f(n)\right)^n=e^{-\log 2}=1/2$$

amWhy

@robjohn I understand. It's just discouraging, and demeaning...but I understand the need for due diligence.

Peter Tamaroff

@robjohn Do you see what I want?

It is immediate that the "$\log(1+x^n)$..." integral goes to $0$ by comparison to $log(x^n)=n\log x$, but I need to know it goes really fast so as to somehow neglect it.

Theorem

20:28

what is $f(n)$ ? @PeterTamaroff

skullpatrol

@Charlie click on "site rooms" in the top right corner

Peter Tamaroff

@Theorem What I want to get.

@anon Do you see what I want?

Theorem

@PeterTamaroff : Oh ok , from the above relatiion u are supposed to get $f(n)$ ?

robjohn

@PeterTamaroff do you mean $x^2$ in the denominator?

Charlie

@skullpatrol I saw :D

Peter Tamaroff

20:29

@robjohn Nope.

@robjohn Where?

@robjohn No, just $x$.

robjohn

@PeterTamaroff never mind, I see

Peter Tamaroff

I am on a good path, yes?

robjohn

@PeterTamaroff However, I get $-\frac{\log(2)}{n}$ as the endpoint term for the integration by parts

@PeterTamaroff see that it is corrected later

Ilya

@rob: hey :)

Peter Tamaroff

@robjohn Yes, I get the same.

Ilya

20:35

@Pete: hey :)

Peter Tamaroff

@robjohn Sawry.

I have a nice idea.

@Ilya Sire.

robjohn

@Ilya howdy :-)

Peter Tamaroff

Roll back $1/x \mapsto x$

We get the integra

Ilya

@robjohn fine, trying to understand whether it's worth proceeding

Peter Tamaroff

$$\int\limits_1^\infty {{{\log \left( {1 + {x^n}} \right) - n} \over {n{x^2}}}dx} = \int\limits_0^1 {{{\log \left( {1 + {x^n}} \right) - n\log x - n} \over n}dx} $$

$$\int\limits_1^\infty {{{\log \left( {1 + {x^n}} \right) - n} \over {n{x^2}}}dx} = {1 \over n}\int\limits_0^1 {\log \left( {1 + {x^n}} \right)} dx$$

Much more workable now.

$$\int\limits_0^1 {{{dx} \over {1 + {x^n}}}} = 1 - {{\log 2} \over n} + \int\limits_0^1 {{{\log \left( {1 + {x^n}} \right)} \over n}dx} $$

robjohn

20:39

@Ilya I leave that to you :-)

Ilya

@robjohn in fact I've already decided

Peter Tamaroff

So that $${1 \over n}\int\limits_0^1 {\log \left( {1 + {x^n}} \right)dx} < {1 \over n}\int\limits_0^1 {{x^n}dx} = {1 \over {n\left( {n + 1} \right)}}$$

Ilya

there is no point in such discussions

Peter Tamaroff

Yes!!!

@robjohn That is it, right???

$$\int\limits_0^1 {{{dx} \over {1 + {x^n}}}} = 1 - {{\log 2} \over n} + {1 \over n}\int\limits_0^1 {\log \left( {1 + {x^n}} \right)dx} = 1 - {{\log 2} \over n} + O\left( {{1 \over {{n^2}}}} \right)$$

@Ilya What are you discussing?

Ilya

@PeterTamaroff how meaningful is the question of one guy

Peter Tamaroff

20:44

@Ilya I see. I have no clue what that is.

@Ilya Do you see what I have done above? I think suffices to show what I want.

Since we can neglect the term that goes to zero like $n^{-2}$ in the limit.

Ilya

@PeterTamaroff I see it, but I have no clue what that is :)

Peter Tamaroff

@Ilya =P This is bad. We should learn more about what the other does

@Ilya Though your "stochastic processes" seem out of my league for the moment.

Ilya

@PeterTamaroff your words worth more than gold

Peter Tamaroff

@Ilya Hehe, then I should start cashing them for future research!

Dominic Michaelis

hi

Peter Tamaroff

20:48

@DominicMichaelis Yello.

Ilya

@PeterTamaroff worth trying :)

@DominicMichaelis hi

Theorem

i want to find a property (minimal) such that , a abelian subgroup is normal in $G$ .

Hi @MattN.

Matt N.

@Theorem If $G$ is abelian all subgroups are normal, no?

Hi, btw.

Theorem

@MattN. : well here only the subgroup is abelian .

$G$ is not abelian

Matt N.

I see.

Ilya

20:54

@MattN.: hi

Matt N.

Hi!

Peter Tamaroff

@MattN. Yes.

Matt N.

: )

Peter Tamaroff

@MattN. Hey. Long time since you dropped by.

I happen to be studying algebra from Jacobson's BAI,

Matt N.

@PeterTamaroff Yeah -- got bored : )

@PeterTamaroff Never heard of that.

Peter Tamaroff

20:56

@MattN. Oh, probably I wasn't around.

Matt N.

: ) Probably!

Peter Tamaroff

@MattN. What are you up to now?

Matt N.

@Theorem Does ablienity have any effect at all?

Tobias Kildetoft

@Theorem I am not quite sure what you mean by a minimal property

Ilya

someone said "probably"?

Matt N.

20:57

@PeterTamaroff Stuff : )

@Ilya whistles

Peter Tamaroff

@Ilya HAHAHA yes.

Matt N.

I'm hungover.

And I fancy a drink. Better not.

Peter Tamaroff

@MattN. I like coffe plus aspirin on that.

Theorem

@MattN. : i feel that its weaker than taking a subgroup which is not abelian .

Matt N.

@PeterTamaroff I did coffee$^2$ + paracetamol. But there was only one paracetamol left...

Peter Tamaroff

20:58

@MattN. Oh, noes.

Theorem

@TobiasKildetoft : minimal condition on the subgroup, some kind of relation between the abelian property of a subgroup and the group .

Ilya

@MattN. only 1 paracetamol pill left? apply Banach-Tarsky

8

Matt N.

I found a picture. It depicts a child drawing something. The drawing looks exactly like an image of the music coming out of the flute when the child upstairs plays it.

It's like ear-rape.

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$Mathematics$ Mathematics