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6:02 PM
@JayeshBadwaik I found the answer, but I won't link to it here :-)
@robjohn :-) Its a really nice problem.
@Charlie Hello!!
@JayeshBadwaik :D
@JayeshBadwaik Mr. Badwaik could you give me a hand here?
@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?
@JayeshBadwaik not really
@robjohn ohh. Hmm. Let me think over it.
6:09 PM
@JayeshBadwaik It is a very simple, but highly non-intuitive proof.
@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
@JayeshBadwaik OKAY
@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
@JayeshBadwaik 977 undeleted answers :-)
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
Q: definition of $l$-equivalence

florekIn 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?
@RajeshD Hello...
6:40 PM
@Jayesh did you see LoP?
$$N (k, m) ≤ 2^{2^{m^{2^{2^{k + 9}}}}}$$
@RajeshD What is that?
@cassandra0 after the second 2 i can't read
Life of $\pi$ !
6:43 PM
@RajeshD that movie?
@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?
$$N (k, m) ≤ 2^{\Large 2^{\huge m^{\huge 2^{\huge 2^{\large k + 9}}}}}$$
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...?
@cassandra0 Have you looked at the Wikipedia page?
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.
Q: For a finitely generated abelian group are the homomorphisms into the integers finitely generated

user39947If $L$ 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.
I think Hom(Z,Z) is isomorphic to Z isn't it? Because every homomorphism is a |-> n*a for some n?
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
There are only n different e_n's here, not nxn.
hom(Z^n,Z^n) would in fact be nxn integer matrices.
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
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
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
@aDangerousIdea hiiiiiiiiii!!!!!!!!
4 hours ago, by Charlie
@aDangerousIdea :))
@aDangerousIdea you missed me!
How are you?
8:22 PM
@aDangerousIdea Fine
Fine thanks.
@aDangerousIdea why did you leave for so long?
I just needed to get away...
@aDangerousIdea why?
why not?
8:27 PM
@aDangerousIdea i thought i had done something to you
@Charlie No, you are very sweet to me :-D
You have been gone for awhile to?
@aDangerousIdea yep
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 bad
not that good hey
too mellow
8:41 PM
and repetitive
house music is usually pretty repetitive
Hi everybody!
@Argon Hi Aaron!
@Charlie Hi Marilia!
@aDangerousIdea Hello!
8:44 PM
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
@aDangerousIdea hehe
8:46 PM
who you gonna call?
@aDangerousIdea GHOSTBUSTERS!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
@Argon wouldn't that be Trollbusters?
Skull pa TROLL buster
8:48 PM
@Argon Aaron!
@aDangerousIdea i want Skullpatrol back
@aDangerousIdea Haahahaha!
@WillHunting Will!!!
@WillHunting Hi Will!
Hi @WillHunting
8:49 PM
Hello everyone!
@WillHunting hello!
@WillHunting Wait... Why JB again?
@Argon Well, old habits are hard to die.
@WillHunting So you do still like him?
@WillHunting Marilia is back!
@Argon Back to where?
8:50 PM
@WillHunting Back to here
back to the future
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...
@Charlie Who? Melvin?
8:53 PM
@WillHunting yes
@Charlie Oh I see. Well, just a casual remark.
The person who sends me positive energy is XXX. =)
Why are some chat rooms locked?
@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.
However there are no private rooms as everything is publicly visible except the moderators room.
@JayeshBadwaik Hahaaha Oh noes, Argon jokes :)

How correct was I?
@Argon :D
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 ...

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