6:02 PM

hi

@robjohn :-) Its a really nice problem.
@Charlie Hello!!

@Charlie I'll try Ms. Brown.

@JayeshBadwaik my answer generalizes $\det(I-AB)=\det(I-BA)$

6:06 PM
@robjohn Hmm, I am not sure what you mean. Tell me more. Do you use Sylvesters?

@JayeshBadwaik I show that $\lambda^m\det(\lambda I_n-AB)=\lambda^n\det(\lambda I_m-BA)$

@robjohn Ahh, you use eigenvalues. :-)

Let $a,b\geq 0$ and $p,q>0$ s.t :$\frac {1}{p} + \frac {1}{q}=1$, then

$$ab\leq \frac {a^p}{p} + \frac {b^q}{q}$$
under what conditions the equality holds?

@robjohn ohh. Hmm. Let me think over it.

6:09 PM
@JayeshBadwaik It is a very simple, but highly non-intuitive proof.

help!

@Charlie Proving Hölder?

@Anonymous I need somebody..not just anybody..heelp!
@robjohn no
My prof solved this one..but i didn't understand...

@Charlie Ah, that inequality is used in one of the simplest proofs of Hölder's inequality

@Charlie it is holder, or rather a lemma for holder.

6:12 PM

@Charlie solved what?

it's Young's inequality

@Charlie for convolutions?
@Charlie Oh, I see... that inequality is also called Young's inequality

looks like an interesting inequality
for p=q=2 we have 2ab <= a^2 + b^2

@Charlie Anyway, try doing something similar to AM GM inequality.

6:16 PM
@Charlie Look for the critical points: $b=a^{p-1}$ and $a=b^{q-1}$

my main trouble is finding numbers where 1/p + 1/q = 1

@robjohn All simple proofs are non-intuitive. :-|. I think I grasp the outline of the proof, but I don't have enough machinery to get to the bottom of it.

@JayeshBadwaik which proof? Did you find my post?

@robjohn No, not yet. (You have too many posts!)

I'm not good at inequalities

6:20 PM
That's was his idea:
If $a\neq 0 \neq b$, let $ab=\alpha >0$, then
$\frac {a}{\alpha ^{1/p}} \frac {b}{\alpha^{1/q}}$,
We have that:
$1\leq \frac {1}{p}\left (\frac {a}{\alpha ^{1/p}} \right ) ^p +\frac {1}{q}\left (\frac {b}{\alpha^{1/q}} \right ) ^q$

@cassandra0 Hmm. There was a time when I was bad at inequalities too. Then I got introduced to Mr. Holder and the grand daddy aand all was good in the world.
@robjohn Ahh found. I searched for "robjohn determinant \lambda" :P

@JayeshBadwaik simple but the motivation is hard to see

@robjohn yeah. Nice proof though. I should add it to my collection.

does anyone know van der wardens theorem?
I was trying to understand the proof with Hayes-Jewett
it's really hard to understand

6:27 PM
the idea is simple..

@cassandra0 Haven't gotten into Ramsey theory

it says they made a simpler proof

@robjohn :-) Get to 1000 fast!!

@JayeshBadwaik That is a milestone :-)

oh this is about a more difficult "density version" of the theorem
I just want a simple proof
of ackermann function bound
II just got interested from the question
0

In the following paper http://www.math.ucsd.edu/~ronspubs/74_01_van_der_waerden.pdf, just in the first paragraph the author defines what $l$-equivalence for two m-tuples $\in [0,l]^m$ means. Can somebody please give me a more precise definition of what he means? I am not even sure what $[0,l]^m$ ...

6:36 PM
in wolfram how do i designate variables are integers

@robjohn, it seems to be a generalized version of pigeonhole principle,

Is DHJ Polymath the pseudo-author for the polymath project?

Heeeeeeeeeey........

yes

@RajeshD Hello...

6:40 PM
@Jayesh did you see LoP?

anyone

$$N (k, m) â‰¤ 2^{2^{m^{2^{2^{k + 9}}}}}$$

@RajeshD What is that?

@cassandra0 after the second 2 i can't read

Life of $\pi$ !

6:43 PM
@RajeshD that movie?

yepp

@RajeshD Ahh, there is already an acronym for that.. Nice. :-) Not yet. I am dying to see it though. Hopefully, when the rush in theatres reduces a little bit?

$$N (k, m) â‰¤ 2^{\Large 2^{\huge m^{\huge 2^{\huge 2^{\large k + 9}}}}}$$

haha

does anyone know wolfram any?

6:44 PM
@KaliMa You might be better off asking about it in the mathematica.SE?

I'm trying to read Shelah's proof which gives a primitive recursive bound
rather than ackermann

@KaliMa One of my attempts would be just writing $a,b \in \mathbb{Z}$ I have found that wolfram understands LaTeX somewhat well.

@anon much better!

hi all, anything special about a polynomial where for every complex root there is a conjugate other than that it factors into (x^2+ax+b)(x^2+cx+d)... for real a,b,c,d...?

Van der Waerden's theorem is a theorem in the branch of mathematics called Ramsey theory. Van der Waerden's theorem states that for any given positive integers r and k, there is some number N such that if the integers {1, 2, ..., N} are colored, each with one of r different colors, then there are at least k integers in arithmetic progression all of the same color. The least such N is the Van der Waerden number W(r, k). It is named after the Dutch mathematician B. L. van der Waerden. For example, when r = 2, you have two colors, say red and blue. W(2, 3) is bigger than 8, because you...

6:49 PM
@PeterSheldrick this is true of every real-coefficient polynomial. I'm unsure of what you're asking.

0

If $L$ is a finitely generated abelian group, then is $Hom(L,\mathbb{Z})$ a finitely generated abelian group? Thank you

If L is a finite group what are the homomorphisms into Z?

trivial if L is finite
finitely generated is another story of course

@anon, does every real-coefficient polynomial only have complex roots where both the root and the conjugate are included?

@PeterSheldrick yes: http://en.wikipedia.org/wiki/Complex_conjugate_root_theorem (a nonreal root and its conjugate have the same multiplicity)

@anon, so basically doesn't that imply only the freely generated part of L will give rise to nontrivial homomorphisms?

6:51 PM
alright thanks

@PeterSheldrick It is an easy exercise to show that the conjugate of a root of a real poly is also a root

@cassandra0 correct, by the fundamental theorem of finitely generated abelian groups, it suffices to compute $\hom(\Bbb Z^n,\Bbb Z)$, which is quite easy to do.

thanks
I think Hom(Z,Z) is isomorphic to Z isn't it? Because every homomorphism is a |-> n*a for some n?

right

wonder if Hom(Z^n,Z) is just Z^n then. Probably is but seems a bit harder

6:55 PM
not much harder. a hom is determined by its effect on the n generators.
so let e_n be the projection in the nth coordinate. an element of hom(Z^n,Z) is just a linear combination of the e_n's.

so it's actually Z^{n^2} ?
nxn matrices

no
There are only n different e_n's here, not nxn.
hom(Z^n,Z^n) would in fact be nxn integer matrices.

cool

note that for G an abelian group, hom(G,G)=End(G) is not just an abelian group under pointwise addition, it is also a ring under functional composition

I posted it here math.stackexchange.com/questions/247468/… and there's an advanced answer which doesn't make sense
quite interesting I guess I can see what it means by just hanging the proof we came up with over it

7:04 PM
the advanced answer seems to be a slick homological algebra version of the easy stuff. the OP tagged the question that way after all.

I'm guessing "a" is the torsion part, "b" the free part

it's $\ne$ its
@cassandra0 I don't think so
b is the free part though
I think Z^a is the free abelian group we quotient by to get L (hence the term "presentation" for L)

that makes sense!
so 0 --> A --> B --> C says that A is a subgroup of B, B a subgroup of C
because kernel = image in each --> bit
i.e. kernel A-->B = im 0-->A = 0 so it's properly included in there

something tells me it should be Z^a->L->Z^b though
nevermind, that makes sense
the image of Z^a in Z^b are the things being quotiented by to get L
derp

how do we know that $\text{Hom}(-,\mathbb Z)$ is an exact contrafunctor?

7:18 PM
Dunno. I'd probably consult a text since I don't feel like checking it.

I feel like proving it
is is the category Hom(-,Z) goes into abelian?
I guess it's AbGrp/Z
oh it's Hom(-,B) : C → Set
It's only a contravariant left-exact functor
if 0 --> A --f--> B --g--> C --> 0 is exact do we have gf = 0?

7:44 PM
that is like the definition of exact
innit?

I do't have that in my definition of exact
it's just f = ker g and g = coker f

you mean im(f)=ker(g) for the first, right? what does g=coker(f) mean when g is a map and coker is a quotient thingie of B?
at any rate, im(f)=ker(g) implies gf=0.

im confused then

oh, you want im(g) isomorphic to coker(f)

oh I think my definition is just for abelian categories

8:20 PM

4 hours ago, by Charlie
hi

:)

How are you?

8:22 PM
you?

Fine thanks.

:D
@aDangerousIdea why did you leave for so long?

I just needed to get away...

why not?

8:27 PM

@Charlie No, you are very sweet to me :-D

hmm

You have been gone for awhile to?

a few days

Hi @OldJohn how are you?

8:35 PM
@aDangerousIdea Hi - fine, thanks - and you?

@OldJohn Fine thanks.
@Charlie What do you think of this?
I think they call it old skool house music...

from the year i was born

not that good hey

yeah...

too mellow

8:41 PM
and repetitive

house music is usually pretty repetitive

Hi everybody!

@Argon Hi Aaron!

@Charlie Hi Marilia!

8:44 PM
:D

we thought a troll ate you

@aDangerousIdea I ain't afraid of trolls

I ain't afraid of no troll

I ain't scared of no ghost

8:46 PM
who you gonna call?

TROLLBUSTER

@Argon wouldn't that be Trollbusters?

TROLLTBUSTERS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Skull pa TROLL buster

user19161
8:48 PM
@Argon Aaron!

@WillHunting Will!!!

@WillHunting Hi Will!

Hi @WillHunting

user19161
8:49 PM
Hello everyone!

@WillHunting hello!

@WillHunting Wait... Why JB again?

user19161
@Argon Well, old habits are hard to die.

@WillHunting So you do still like him?
@WillHunting Marilia is back!

user19161
@Argon Back to where?

8:50 PM
@WillHunting Back to here

back to the future

user19161
Wow @amwhy I see your rep is soaring day by day! In other news, I just joined the 7k club. =)

@WillHunting i saw that someone sends you good energy waves...

user19161
@Charlie Who? Melvin?

HAHAHA!

8:53 PM
@WillHunting yes

user19161
@Charlie Oh I see. Well, just a casual remark.

hahahahahahahahah

user19161
The person who sends me positive energy is XXX. =)

Why are some chat rooms locked?

user19161
@Argon Because the room owner created it so that only a few people can talk.

8:56 PM
@WillHunting That's mean!

@Argon Because they can turn so volatile that even Argon has an explosive reaction.

user19161
However there are no private rooms as everything is publicly visible except the moderators room.

@JayeshBadwaik Hahaaha Oh noes, Argon jokes :)

http://chat.stackexchange.com/transcript/message/7065740#7065740

How correct was I?
@Argon :D

user19161
Wow @old I see that your rep is also soaring day by day! =)

8:59 PM
@WillHunting I wouldn't say it is "soaring" more like creeping up at a steady rate :)
but I am in no rush to reach the dizzy heights ...