7:00 PM
@PeterTamaroff: this is what I was thinking about. This way seems just nice. Thanks! :-)

@robjohn can probably help.
@Chris'ssisterandpals Welcome. Have a nice one.

It's OK. Your help was a really help.
I'm gone.
see ya later

Hey guys
Peter are you Peter Tamaroff?

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

Is Halmos the only good book for set theory?

7:10 PM
OH ok
Hey Peter!

@Charlie set theory at what level?

introduction to set theory

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

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

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

7:13 PM

gr

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

MATLAB is so fucked up sometimes.

@Charlie ahh, no idea about good books for that

Can someone help me out with a discrete problem?

7:14 PM
thanks benjamin

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

No problem!

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

Not sure, sorry.

7:21 PM
Peter
so for the first instance

@robjohn You there?

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

@JennaMagic Yes.

so in the first instance..3x + 7 / x?
do we take derivative of both?
or is it rather just infinity/infinity = infinity

7:23 PM
Well, you can. But note that $$\frac{3x+7}{x}=3+\frac 7 x$$
What does $7/x$ approach to?

So your limit is $3$.

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

what is order

floors?
user58512
you there?

7:34 PM
@JennaMagic I don't understand what you mean by floors.

you know, floor and ceiling
where 1.2 = 1
1.7 = 1
1.9 = 1
that's a floor

But why would you want that?

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

What problem?

0

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!

7:36 PM
You mean you want $$\lfloor 3x+7\rfloor /\lfloor x \rfloor$$?

C
no
c) x+1/2,x
floor of x+ 1/2
?

Could you be more specific?

How ?
Floor of (x+(1/2)), x
that's what "C" should be
in the question

No, $C$ is a constant.
Not a function.

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

7:46 PM
Oh.

in the original question

You want to edit that?

Ok....
Or I can just ask here?

Look now. Is it OK?

Yeah!
Just wondering though how I owuld solve that one

7:47 PM
Just use the squeeze theorem
Squeeze the floor between two common linear polynomials.

dont really remember tha tfrom calc 1

...
great.

I really mean it.
And I'm not being mean.
Really.

Yes you are.

7:52 PM
You don't remember what the squeeze theorem is?

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

Hi again!

@Nimza Hi again!

@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 $f(n)\to 0$.

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

7:56 PM
Peter, you never answered me in math.stackexchange.com/questions/309361/…
It goes to 0..now what?

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

Makes sense
ok last question
for the same order question
how can i do lhopitals
for the LOGS
and do i need to?

hi

how are you sweetie ?

8:01 PM
good, and you?

anyone here know discrete?
i have 1 question

my parents are so annyoing

@DominicMichaelis oh! why you say?

i am running mad because of them ...

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

8:05 PM
i am trying to think but they call me every 5 minutes

@DominicMichaelis they must be worried

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

@DominicMichaelis really?

So you are at the same place?

8:07 PM
fgfd

Does ANYONE know DISCRETE
OMG

@DominicMichaelis Oh!

@JennaMagic, what is "DISCRETE"

8:08 PM
@Arkamis lkçl?

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

Not that
I need to show that Floor(x + 1/2) is the same order as "x"

@JennaMagic, what is order

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

ok so do it

8:10 PM
@jenna the limit where ?

@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)

@Theo :D

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

@Theorem I'm good, and you?

just after coming out of my exam , i solved it :( :P
that was really straight forward

8:12 PM
@Theorem that tends to happen to a lot of people

i am good @Charlie

@Theorem I feel awful after exams
@Theorem good!

not much to do about it now

and also time factor .
@TobiasKildetoft : did u ever come across such situations :P

8:14 PM
@Theorem who never...

why x-1 instead of just x?
curious choice

@anon Hi.

yo yo yo

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

8:16 PM
@anon take x=1/10

@Theorem noo....

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

@DominicMichaelis ah, duh

@Theorem it happens to me all the time

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

8:18 PM
@anon lets see if i get any upvotes for this one ...

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

but after exam @Charlie

@Theorem you solved it anyway
@Theorem you should be proud of yourself

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

maybe

8:21 PM
@BenjaLim Or maybe you?

i hate myself for being quite an idiot :D
@Charlie

@anon Well, it is the following.

Quora is awesome

@Theorem no, no, no

anyone there in Quora ?

8:21 PM
@Theorem me

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

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

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

hmm

@Theorem hahaha I almost never use it, used

8:22 PM
$$\int\limits_0^1 {{{dx} \over {1 + {x^n}}}} = \int\limits_{1}^\infty {{{n{x^{n - 1}}} \over {1 + {x^n}}}{{dx} \over {nx}}}$$
I added the $n$s.
Integration by parts yields
$$\int\limits_0^1 {{{dx} \over {1 + {x^n}}}} = - {{\log 2} \over 2} + \int\limits_1^\infty {{{\log \left( {1 + {x^n}} \right)} \over {n{x^2}}}dx}$$

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

Look at the above @robjohn

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

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

Now we aim to use $(1+1/n)^n\to e$
Summing and negating $1$, we get
$$\int\limits_0^1 {{{dx} \over {1 + {x^n}}}} = 1 - {{\log 2} \over n} + \int\limits_1^\infty {{{\log \left( {1 + {x^n}} \right)} \over {n{x^2}}}dx} - 1$$
$$\int\limits_0^1 {{{dx} \over {1 + {x^n}}}} = 1 - {{\log 2} \over n} + \int\limits_1^\infty {{{\log \left( {1 + {x^n}} \right)} \over {n{x^2}}}dx} - \int\limits_1^\infty {{n \over {n{x^2}}}dx}$$

8:24 PM
@robjohn You've obviously been busy, np.

@Theorem you like it?

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

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

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 $1/2$

@Charlie have you seen skull's partyzone?

8:26 PM
@PeterTamaroff : oh ok , what is the question ? haven't looked in your actual question :P

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

@skullpatrol No!where??

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

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

@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.

8:28 PM
what is $f(n)$ ? @PeterTamaroff

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

@Theorem What I want to get.
@anon Do you see what I want?

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

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

@skullpatrol I saw :D

8:29 PM
@robjohn Nope.
@robjohn Where?
@robjohn No, just $x$.

@PeterTamaroff never mind, I see

I am on a good path, yes?

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

@rob: hey :)

@robjohn Yes, I get the same.

8:35 PM
@Pete: hey :)

@robjohn Sawry.
I have a nice idea.
@Ilya Sire.

@Ilya howdy :-)

Roll back $1/x \mapsto x$
We get the integra

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

8:39 PM
@Ilya I leave that to you :-)

@robjohn in fact I've already decided

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

there is no point in such discussions

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?

@PeterTamaroff how meaningful is the question of one guy

8:44 PM
@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.

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

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

@PeterTamaroff your words worth more than gold

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

8:48 PM
@DominicMichaelis Yello.

@PeterTamaroff worth trying :)
@DominicMichaelis hi

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

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

@MattN. : well here only the subgroup is abelian .
$G$ is not abelian

I see.

8:54 PM
@MattN.: hi

Hi!

@MattN. Yes.

: )

@MattN. Hey. Long time since you dropped by.
I happen to be studying algebra from Jacobson's BAI,

@PeterTamaroff Yeah -- got bored : )
@PeterTamaroff Never heard of that.

8:56 PM
@MattN. Oh, probably I wasn't around.

: ) Probably!

@MattN. What are you up to now?

@Theorem Does ablienity have any effect at all?

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

someone said "probably"?

8:57 PM
@PeterTamaroff Stuff : )
@Ilya whistles

@Ilya HAHAHA yes.

I'm hungover.
And I fancy a drink. Better not.

@MattN. I like coffe plus aspirin on that.

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

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

8:58 PM
@MattN. Oh, noes.

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

@MattN. only 1 paracetamol pill left? apply Banach-Tarsky
8

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.