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

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

7:36 PM

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

7:46 PM

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

7:47 PM

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

7:52 PM

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

7:56 PM

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

8:01 PM

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

8:05 PM

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

8:07 PM

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

8:08 PM

@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

8:10 PM

@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

8:12 PM

@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

8:14 PM

@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

8:16 PM

@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

8:18 PM

@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

8:21 PM

@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

8:21 PM

@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

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

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

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

8:24 PM

@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

8:26 PM

@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

8:28 PM

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

8:29 PM

@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

8:35 PM

@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

8:39 PM

@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

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.

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

8:48 PM

@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

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,

Matt N.

@PeterTamaroff Yeah -- got bored : )

@PeterTamaroff Never heard of that.

Peter Tamaroff

8:56 PM

@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.

8:57 PM

@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

8:58 PM

@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.

Transcript for

$Mathematics$ Mathematics