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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?
 
NBP
introduction to set theory
 
@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?
 
I like the book Chapter 0 by Schumacher (I think)
 
7:13 PM
@TobiasKildetoft ZF axioms
 
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?
 
NBP
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.
 
@JennaMagic I answered.
 
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
 
Peter, what about floords?
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
Q: How to show that pairs of functions are of the same order?

Jenna MagicIf 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
 
Then read it again =)
 
...
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!
 
@TobiasKildetoft : i felt bad . i don't know what i should do about it .
isn't it bad ?
 
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 =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.
 
@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.
 

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