Mathematics - 2013-02-09 (page 3 of 4)

CBenni

6:02 PM

what does $\amalg$ stand for? I was just wondering :D

anon

disjoint union usually, or coproduct more generally when dealing with category theory flavored math

CBenni

Huh... disjoint union was $\stackrel{.}{\cup}$

Khromonkey

what happened to math .se?

CBenni

but there may be different notations...

user58512

that's just normal union with a marker saying you know a priori the sets are all disjoint

Khromonkey

6:08 PM

why is it in read only mode?

CBenni

@user58512 I know what a disjoint union is ;) The symbol is called \amalg and therefore I was wondering.

we have the amagamized (?) product in topology, but I found no notion that $\amalg$ is used for it

@Khromonkey maybe you read what it says on the page, it says the page is in maintainence mode

Khromonkey

What kind of mantainance?

Orange Harvester

@Khromonkey sometimes when database needs to be backed up and verified, it is important that there are no changes to prevent loss of integrity of data. At such times, we have read only sites.(it is one of the many possible scenarios)

Khromonkey

cool

user58512

I want to go through this whole book but I don't h ave time

CBenni

6:15 PM

Btw someone know a nice elementary proof of $\begin{pmatrix}a\\b\end{pmatrix}\in\mathbb{Z}$ ? I have tried but I failed miserably...

user58512

@CBenni, which definition do you want to prove it for? it's immediate from the definition I use

CBenni

Oh right, the $\frac{n!}{k!(n-k)!} =: \binom{n}{k}$ one

The factors seem to cancel out always, but not in any manner that I could have described systematically

Probably possible to show that $n!\mid k!(n-k)!$, but I dont know how ;)

Sanchez

@CBenni

user58512

for example in this one [1*2*3*4*5]*[6*7*8*9] / [1*2*3*4*5]*[1*2*3*4]

1|7, 2|6, 3|9, 4|8

Novice

Nice, now we have a read-only mode for all the impatient mathematicians out there. Well done!

Sanchez

6:20 PM

do you know that the prime power for $p$ dividing $n!$ is exactly $[n/p] + [n/p^2] + \cdots $?

CBenni

no I have yet to hear a number theory reading

Sanchez

Well, it's something you may be able to figure out

CBenni

There definitely is a elementary proof, but is there a obvious/nice/short one?

user58512

see the idea from what I wrote?

it doesn't work nevermind

CBenni

I thought so

I tried it for rather large fractions

Novice

6:23 PM

Guys, my discovery is right.

CBenni

but I never found any kind of system

@Novice what discovery?

Novice

@CBenni For every base $b$, there is only 1 three-digit number $N$ such that the sum of digits of $N$ is $\dfrac{N}{b + 1}$

Nothing too big, I suppose.

Ah, and I have the proof.

user58512

@CBenni, well it's immediate that the coefficients of (1+x)^n are integers, and they satisfy the same recurrence relation as n!/(k!(n-k)!)

CBenni

@user58512 well yeah, 'course. When doing it via the recurrence relation, the proof is trivial

user58512

?

CBenni

6:30 PM

I meant if you can do it via divisibility properties

user58512

Sanchez showed how

just visualize the sum [n/p] + [n/p^2] + [n/p^3] + ..., [x] meaning integer part

CBenni

mmh... Then I will first have to look into that more specifically

but thanks ;)

mcb

hello, i’m not sure with this: is the n vs np problem basically the problem of proving the lower bound complexity of an np-complete problem?

Novice

And the number is always in the form $11(b + 8)_b$.

Orange Harvester

@Novice Cool result.

user58512

6:33 PM

@mcb, if you showed any NP algorithm cannot be computed in polynomial time you would be done yeah

CBenni

@Novice that is actually really cool :) How does such a proof work? I am currently trying to show it myself ^.^

user58512

@CBenni, it looks like this for p=2 and n=12: 1,[2],3,[[4]],5,[6],7,[[[8]]],9,[10],11,[[12]]

Novice

@CBenni I could not prove it, but my partner who was working on it with me did.

mcb

user58512: and the SAT problem is the most reduced known np problem?

user58512

@mcb, I don't think "most reduced" means anything

Hagen von Eitzen

6:35 PM

@Novize Sure. $b^2x+by+z=(b+1)(x+y+z)$, so $(b^2-b-1)x=y+bz$. For $b\ge 3$, $b^2-b-1 > \frac12 b^2$, so there can be at most one solution. (and for $b=2$ we have $N=6=110_2$)

CBenni

@user58512 Well, I will have to learn more about it someday. Not now however. The fact there is no really easy proof suffices for now ;)

user58512

@mcb, all the NP complete problems are equivalent (you can solve one using a solver for an y other)

@CBenni, it is easy

CBenni

for someone who is into the subject, yes

user58512

@CBenni, count the brackets to see why 2^10 | 12!

CBenni

I guess it has to do with 10 brackets then? I dont see how that correlates however

user58512

6:37 PM

@CBenni, it is also immediate considering all permutations of n objects, and then quotient out by not caring about the order of the first k and order of the last n-k

@CBenni, you count all even numbers once [n/p], then you count all doubly-even numbers once more [n/p^2], etc..

CBenni

@user58512 mmh yeah that is obvious

Math Gems

@anorton MSE is on read only mode because SE is moving the data center from Oregon to New York, see here. Was there no notice about this?

2

user58512

I thought at first any sequence of k numbers would be divisible by k! "numberwise" like 1|7, 2|6, 3|9, 4|8 but then I remember about primes...

So I am not sure how easy to prove n!/(k! (n-k)!) is an integer directly.

mcb

user58512: why is it so difficult to prove P-NP?

user58512

@mcb, I don't know why but I think there are some metamathematical theorems that give serious obstructions to proving this - not something I have studied.. but even basic complexity class separation results are extremely difficult so it's not a big suprise

Hagen von Eitzen

6:41 PM

@MathGems Hm, but "soon" after 17.30 UTC is quite a while ago ...

Math Gems

@HagenvonEitzen Indeed, but that's usually par for the course for software.

Hagen von Eitzen

@MathGems nod Hofstadter's Law: It always takes longer than you expect, even when you take into account Hofstadter's Law.

mcb

user58512: my intuition says that in all of these problems you have to go through all possible permutation of the input, and that’s why the complexity is beyond polynomial. does my intuition go into the right direction?

Hagen von Eitzen

@mcb Yes, the intuition is fine - but how do you know that it can't be done faster? For example sortinmg could be described as testing all $n!$ orderings for correctness, and yet sorting can be done in $n\log n$.

user58512

@mcb, pretty much. The idea is that it's quick to check if a solution is correct - but, like you said, to create such a solution seems to require going through exponentially many possible cases.. but maybe there is a magic algorithm that solves them all very quickly! No one has been able to find one or disprove the existence

Orange Harvester

6:45 PM

@MathGems I thought they already moved it a while back (when they talked about keeping generators alive during the Huricane that was going on, I forget the name)

Math Gems

@HagenvonEitzen Hah. Also northeast USA (including New York) is in the middle of one of the biggest blizzards in years, so that probably has some impact.

Orange Harvester

@HagenvonEitzen I thought that was Murphy's second law.

user58512

I'm really bothered about my classes going faster than they should, I dont have to time to think about things properly :/

Hagen von Eitzen

@MathGems "There's no bad wheather, there's just inadequate clothing" ;) But in principle they seem to have been aware of the blizzard in their planning (with go/nogo checks and all), and who would have thought they carry the data by car? ;) - Enough of ranting.

Math Gems

@HagenvonEitzen Probably all the data is on a usb stick in Jeff Atwood's pocket. Let's hope the TSA doesn't confiscate it...

Hagen von Eitzen

6:50 PM

@OrangeHarvester It's probably the recursive nature that makes it Hofstadter's. For security I looked it up under Murphy's law at Wikipedia, though :)

@MathGems On one single USB stick? They probably needed to punch that stick (Anybody here still remembers that floppy disk capacity once could be doubled with a hole punch?)

Math Gems

@robjohn How rare is snow in whim?

user58512

7:08 PM

do you have any advice about things going too fast?

Sanchez

@HagenvonEitzen, just want to say that the wikipedia page you quoted on Riemann's Xi function is probably wrong. There must be an exponential factor to ensure convergence of product.

user58512

@Sanchez, I think the point is you can cancel it out with the functional equation and the assumption of finitely many zeros

Sanchez

no

For example, $e^z$ is nonvanishing.

and is of finite order.

How do you know that there's no nonvanishing entire function in front?

user58512

hmm I don't

Hagen von Eitzen

@Sanchez and it didn't even have a "citation needed"

anon

7:12 PM

@Sanchez In general, the exponential factor can simply be 1 identically for a Weierstrass factorization, which is the case I believe for $\xi$. (Mathworld also has this factorization.) Note that $\sum\frac{1}{\rho}$ actually converges when you do it in a P.V. sense (which is also the sense in which the product is taken I think). See mathoverflow.net/questions/91280/… for instance.

Theorem

what's happening with mathstack

Antonio Vargas

@Theorem stackstatus.net

Andreas Caranti

And... it's back!

Hagen von Eitzen

Ah! Then it's good-bye for me here ;)

skullpatrol

@HagenvonEitzen Later, thanks for dropping by :)

Theorem

7:22 PM

is it true that the element of the group $G$ can be either in the center or in the commutator group . ie can we have something like $|G|= |Z(G)| + |G'|$ ?

user58512

I think not

anon

It is unclear to me how those two statements are equivalent (and if your "either" is exclusive or non-exclusive or).

skullpatrol

Hi @JacobBlack

Theorem

@user58512 : are u saying no to my equation

user58512

yes

Theorem

7:28 PM

i guess it should be -1

anon

It is certainly possible for $Z(G)+G'=G$ and $Z(G)\cap G'=1$ to hold simultaneously; any abelian group will do.

Theorem

@anon : are u sure ? how ?

anon

If $G$ is abelian, then $Z(G)=G$ and $G'=1$, so...

user58512

oh G' = 1

Theorem

@anon : ah , abelian , thats ok but does it hold in general

user19161

7:31 PM

@skullpatrol I won't be talking in the OS room, no need to add me but thanks.

skullpatrol

@JacobBlack Why not?

user19161

@skullpatrol Well, just wanna keep the conversation in one room.

anon

@Theorem If $G$ is a nonabelian simple group then $Z(G)=1$ and $G'=G$ (since both $Z(G)$ and $G'$ are normal, and $G'$ cannot be trivial as $G$ is nonabelian). This also fulfills the conditions $Z(G)G'=G$ and $Z(G)\cap G'=1$.

skullpatrol

@JacobBlack You are welcome anytime to drop by :-)

anon

I would be interested in seeing something in between abelian and nonabelian simple, though.

Theorem

7:38 PM

@anon : Same here .

i have been wondering about it since yesterday .

user58512

I found this

gap> SmallGeneratingSet(SmallGroup( 28, 1 ));
[ f1, f3 ]
gap> Center(SmallGroup( 28, 1 ));
Group([ f2 ])
gap> DerivedSubgroup(SmallGroup( 28, 1 ));
Group([ f3 ])

anon

I am going to lunch and then my office, I'll ponder it. If it is false that there are extraneous examples, then perhaps we may suppose $G$ is nonabelian with a normal subgroup $N$ and derive a contradiction.

Theorem

@user58512 : whats that , i don't know gap and don't know how to read

@anon : Bon appettite :) come with some insight and inform me :)

user19161

Wow, now the whole world knows I said "I am only a banana" because of the post on meta.

user19161

0

Q: Why aren't flags in chat anonymous?

How is this user finding out information about flagging? http://chat.stackexchange.com/transcript/message/8039304#8039304 what is breaking the anonymity of the flag system?

user19161

7:41 PM

Thanks for the publicity.

user19161

Also, thanks for calling Hagen rude as well for helping you with his comment.

user19161

Also, thanks for calling me rude.

skullpatrol

Thank you for being a friend?

user19161

To keep calling people rude when they are not is an insult, and that is abuse.

user19161

Thanks for stabbing me in the back when I did nothing wrong.

skullpatrol

7:44 PM

Don't let it bother you pal ;-)

Life goes on...

...roll with the punches.

@JacobBlack

MSEoris

hey can an element be its own additive inverse in a vector space?

user58512

yes 1 + 1 = 0 in F_2

which is a field so you can make vector spaces out of it

MSEoris

k

wasnt sure if the elements invierse need be distinct fmo the element

skullpatrol

How about 0?

Theorem

@MSEoris take some matrix such that $A^2 =I $

Alexander Gruber

7:57 PM

@Theorem so your question is whether it is true that $x\in G\Rightarrow x\in Z(G) \text{ or } x \in G'$?

MSEoris

i see

Theorem

@AlexanderGruber : yes exactly .

MSEoris

i was just cehcking vector space proerties when we have an 'addition' operation where we really just pairwise ssubtract elements

in that case an element is its own inverse

user19161

@skullpatrol Yeah, I will stay cool. By the way, there were four more flags just now.

skullpatrol

@JacobBlack Great...let the games begin.

user19161

8:00 PM

@skullpatrol No, I am not playing this game.

Theorem

@AlexanderGruber : does that mean if the $Z(G) =1$ then the $G' =G$

Alexander Gruber

@Theorem it isn't true. You can have centerless groups (i.e. $Z(G)=1$) that are not perfect ($G\not= G'$).

user19161

I don't want to make a commotion, but sometimes I need to express myself a little @skull

skullpatrol

@JacobBlack That is why you are my pal ;-)

Theorem

@AlexanderGruber : for example $S_3$

Alexander Gruber

8:01 PM

@Theorem exactly

Theorem

@AlexanderGruber No $S_3$ is not centerless

Alexander Gruber

yeah it is.

Theorem

@AlexanderGruber : yup , right . got confused

@AlexanderGruber : then what kind of elements can be not neither in the center nor in the commutator . Any insights .

skullpatrol

@JacobBlack Here.

Alexander Gruber

@Theorem a partial result is a theorem of thompson that says that $G=Z(G)G'$ under some conditions, but even this is not always true (e.g. $Z(G)=1$ and $G'$ proper in $G$). i'm trying to remember the conditions but it's not coming to me.

@Theorem well you could multiply elements of the center and commutator which which end up being in neither. In particular, if $x\notin Z(G)\cap G'$, then $xG'$ is a whole class of examples.

Theorem

8:07 PM

@AlexanderGruber : its true when $G$ is simple .

Alexander Gruber

yes - nonabelian simple, anyway.

Theorem

@AlexanderGruber : I suppose u meant $\cup$ instead of intersection .

user19161

@skullpatrol Yes, you are right. We should choose happiness because we deserve it. QED.

Alexander Gruber

well, in general, we could also just look at $\langle Z(G), G' \rangle$ and if that is not equal to $G$ we can take anything outside of that.

skullpatrol

@JacobBlack True that.

Let the haters hate.

Theorem

8:14 PM

@AlexanderGruber : thats quite interesting , that non commuting doesn't mean that it will have some form $x^{-1}y^{-1}x y$ . that means if $x not in G'$ and $y\in G$ , then $xy not in G'$

MSEoris

is Jax supposed to work in chat?

CBenni

no

MSEoris

kk

CBenni

@MSEoris see the most upper highlighted chat post

Etiquette Guidelines | LATEX support for chat <--

MSEoris

sorry i think im slow, where is this :D

Alexander Gruber

8:19 PM

@Theorem $G'$ is a dual notion to $Z(G)$ in the way that $Z(G/G')=G/G'$, but it is not the set of noncommuting elements. in a sense we want to "take out" the noncommuting elements by modding by $G'$, but it is certainly not literally a set-minus. :)

MSEoris

aha found it thanks @CBenni

CBenni

YAAY Math.SE is out of read-only mode!

party time \o/

user58512

@CBenni, math.stackexchange.com/questions/2158/division-of-factorials/…

Theorem

@AlexanderGruber : yup , although i cannot see things clearly whats happening when u take mod that makes the whole thing commuting .

CBenni

@user58512 why thank you! ;)

Alexander Gruber

8:25 PM

@Theorem for $x,y\in G$ $$(xy)N=(yx)N\Leftrightarrow (x^{-1}y^{-1}xy)N = N \Leftrightarrow x^{-1}y^{-1}xy\in N$$ so $G/N$ is abelian if and only if $G'\leqslant N$.

skullpatrol

@CBenni I came to PARTY!!!

CBenni

@skullpatrol mmh, mathematicians dont party, sorry :/

skullpatrol

6 mins ago, by CBenni

party time \o/

@CBenni excuse me?

Theorem

@AlexanderGruber , thanks

CBenni

@skullpatrol just trolling around, you know

skullpatrol

8:28 PM

@CBenni I know.... I caught you ;-)

skullpaTROL

:-D

CBenni

no pun intended

skullpatrol

Its all good pal.

CBenni

I dont enjoy learning theorems by heart

but I guess theres no other way to pass this exam >_<

skullpatrol

Do what you must do.

Sanchez

@anon (regarding the factorization of Xi function), fair enough, more believable now.

user58512

8:37 PM

I'm reading this proof ffor the product formula for Gamma

basically they guess a random function in product form with the same poles... divide them and out pops e^{-gamma s}. done!

this would never happen if i tried it...

"exp(x)/(1+x) = 1 + 1/2*x^2 - 1/3*x^3 + 3/8*x^4 - 11/30*x^5 + ..." can you tell me what small x means "exp(x)/(1+x) = 1 + O(x^2) for small x"?

is that just |x| < 1?

anon

@user58512 In general, for small x means for x in some neighborhood (equivalently, some interval) containing 0. Also see "Big O notation" en.wikipedia.org/wiki/Big_O_notation

user58512

ok thank you & why does Gamma(s) never equal 0?

$$\int_0^\infty t^{z-1} e^{-t}\,{\rm d}t $$

I can't read that off the integral

well that integral is only value for Re z > 0 I think

anon

you can use one of the reflection identities for $\Gamma$ if you wish

user58512

ty

CBenni

what does $\overline{1010...01}$ mean (it is supposed to be some kind of integer)?

user58512

8:52 PM

where did you see it?

CBenni

-1

Q: Primes $n=\overline{10101\cdots01}$ with $k$ ones.

Find all primes $n=\overline{10101\cdots01}$ with $k$ ones. The number is in standard base 10.

elementary-number-theory prime-numbers

anon

I think the OP means 1010...01 as an integer. There appears to be no purpose to the overline.

CBenni

I guess so

I thought there might be a notation I havent seen before

skullpatrol

Or no purpose to the "..."

anon

Note that when you presented your question without any context, you made it seem like the notation was established, when in fact I don't think it is. In general, when you ask questions without providing context you risk using up other people's effort and resources needlessly.

CBenni

8:56 PM

@anon Whom are you telling that ^.^

skullpatrol

Yay whom?

-_-

\o/

Party!

or drowning

all day, all night

drowning = ~\o/~

user58512

8:59 PM

The integrand is $e^{(z-1)log(t) - t}$ and we sum this for growing t

skullpatrol

;-)

anon

personally, I see it as the top of a margarita

CBenni

Why is there no ASCII-art.SE

anon

that would be troll central

CBenni

totally. How about Troll.SE?

2

anon

9:00 PM

that would be lame station

skullpatrol

It would attract "tolls", no?

CBenni

It wouldnt. Trolling isnt fun if allowed

user58512

how to prove an integral isn't zero? :(

integral e^{i b log t} t^{a-1} e^{-t} dt, the only trouble is the phase

CBenni

show that its absolute value is >0 :P

user58512

how?

|integral f(x) dx| <= integral |f(x)| dx

I need >

CBenni

9:03 PM

you mean $\int_a^be^{(z-1)log(t)-t}dt$?

anon

why write latex if you're not going to put dollar signs around it? we have chatjax functionality (see the link on the right)

user58512

@CBenni, integral from 0 to infinity

anon

It would probably be cooler and cover ground more efficiently to establish a reflection formula for gamma and then use that to prove it has no zeros (pretty sure the reflection formula is for arbitrary complex numbers).

user58512

@anon, oh I thought that I needed to show no zeros for Re z > 0 and then apply reflection to get no zeroes anywhere

anon

@user58512 nope; $1/\sin$ is never zero, so if it is equal to a product of two things then can either one of those things be zero?

CBenni

9:07 PM

@user58512 and where is your problem? $\forall z\in\mathbb{C}\quad e^z\neq 0$

user58512

the only possibility is Gamma(1-n)Gamma(n) pole and we already know the poles aren't zero!

great,thatproves it

anon

@CBenni the question is not about the exponential function being nonzero, though

CBenni

well... If the function IN the integral is >0, how would it be zero?

anon

@CBenni what does >0 mean for complex numbers?

CBenni

$|e^z|>0$

user58512

9:09 PM

@CBenni, the trouble is the phase factor e^{i b log t} could potential cause cancelling

anon

so? $|z|>0$ for all $z$ on the unit circle $\gamma$, but $\oint_\gamma zdz=0$ by symmetry. Indeed, more elementary, we could say that $|1|,|-1|>0$ but $|1+(-1)|=0$.

user58512

because we're summing up lots of these

CBenni

umm, yeah... waitasec

well... sorry, I was being stupid >_<

Orange Harvester

@user58512 I suspect you are jdoe aka spermer's lemma. Am I correct?

user58512

@Ethan, why did you ignore my answer about zeta?

Orange Harvester

9:14 PM

@JonasTeuwen I am using python to control i3. Ultra-Cool I am. Yo.Yo. Dream Come True$^{TM}$

Jonas Teuwen

@OrangeHarvester What are you controlling?

I own masteroftrolling.com

Ethan

@user58512 I didn't

user58512

ok

Orange Harvester

@JonasTeuwen Right now not much. But I am using this. And also writing my own stuff based on this. I only hope as the thing evolves, I will have one liner scripts for my personal use.

Jonas Teuwen

What would be an interesting thing to do this way?

anon

9:20 PM

@CBenni: I think one can show that, for any fixed prime $p$, of all products of $k$ consecutive integers, $k!$ minimizes the highest power of $p$ dividing it, so that $k!$ divides any product of $k$ consecutive integers. And hence $$\frac{n(n-1)\cdots(n-(k+1))}{k(k-1)\cdots2\cdot1}\in\bf Z.$$

2

This would be a direct and elementary argument.

user58512

that's a really nice argument

CBenni

it is

eventhough I would have to look itno the details

at the same time, people are pretending that my solution to another problem is wrong, eventhough it isnt.

user58512

@CBenni, let me see?

CBenni

Here is another lower bound: 42 (valid for every large enough $n$). "Therefore, my argument is correct." — Did 4 mins ago

1

Q: BigO sorting complexity help

Given a bit sequence of length a, what is the minimum number of comparisons needed to determine if it contains a pair of consecutive 1's in BigO notation

algorithms asymptotics

anon

"people are pretending that my solution to another problem is wrong, eventhough it isnt" is a rather crank way to characterize criticism, be it legitimate or illegitimate

CBenni

9:24 PM

I used 1/3n as a lower bound instead of 1/2n and TWO people are saying it is "wrong!!!11!1!!1!!oneoneeleven"

user58512

"Good job on reading my posting" this is very rude, I don't like whne people talk like this to me

Orange Harvester

@JonasTeuwen I am not exactly sure, just exploring right now. But some cool features can be to tile all windows with a certain combination tag , to get an overview of the current stuff. (Now tags are not implemented in i3wm, but I guess we can implement them ourselves?) I am not sure about that. Current project is to try that.

user58512

anyway your argument is clearly right

Alexander Gruber

i need a word

i am blanking on my writing over here

Orange Harvester

ambrosia

CBenni

9:26 PM

well yes. They even agreed that the final solution is correct, but they have not understood what lower bounds are...

user58512

turn it into a rigorous proof if you the criticism is making you unsure about it

CBenni

And Did didnt read my post. Simple as that

user58512

@CBenni, well you shouldn't write O(n) for a lower bound

Theorem

@AlexanderGruber : like what kind of ? will hunky dory work ? :D

CBenni

I didnt, I wrote O(1/3n), however this is equal to O(n)

Thats just how bigO works

user58512

9:27 PM

I'm saying you shouldn't use big O for a lower bound

CBenni

oh

Alexander Gruber

"We then introduce a class of groups whose commuting graphs ____ a certain property"

CBenni

aah that might be

Alexander Gruber

obey a certain property doesn't sound right

CBenni

iirc it was $\Omega(n)$?

Alexander Gruber

9:27 PM

"conform to" doesn't sound right

admit isn't really the right word

Theorem

$possess $

3

@AlexanderGruber : Possess

anon

cue dramatic trailer music

Alexander Gruber

that's pretty good.

thanks

Theorem

or @AlexanderGruber : or endow

endow is not that suiting

anon

"endowed with"

Orange Harvester

9:32 PM

endowment is generally used for material possessions me thinks. Like real estate, money.

Theorem

@OrangeHarvester : we do use in mathematics , like a function space is endowed with certain norm .

anon

well, we use endow and endowed and endowing, but I don't think I've ever seen endowment in mathematics. tricky variants.

Theorem

How do the right ideal in $R \times M \times S$ look like . where $R$ and $S$ are Rings and $M$ is R-S bimodule .

anon

@CBenni Or, consider this algorithm.

CBenni

@anon I have seen that, Ive been linked to it earlier

thank you however ;)

anon

9:36 PM

ah

user58512

I don't understand this step: $$\frac{s}{s+1}\left(\exp \sum_{k=1}^N \frac{1}{k}\right) \prod_{k=1}^N \frac{k+1}{k+s+1} = s \frac{N}{N + s + 1} \exp\left(- \log N + \sum_{k=1}^N \frac{1}{k} \right)$$

nevermind I saw it after i typed

anon

~~your \prod is indexed by n instead of k~~

user58512

shoudl be k+s in the numerator

Theorem

is someone thinking about my question

:D

anon

is that the group theory one?

Theorem

9:47 PM

@anon : no diffferent one

did u see ?

anon

What I want to see is a proof of $\displaystyle{a+bp\choose c+dp}\equiv{a\choose c}{b\choose d}\bmod p$ (for $0<a,c<p$) using group actions, orbits and stabilizers. The usual proof is binomial expansion on $(1+x)^{a+bp}\equiv(1+x)^a(1+x^p)^b\bmod p$ I believe.

Theorem

@anon : where did u take this problem from

anon

One can show ${p\choose k}\equiv0\bmod p$ (for $0<k<p$) like this; let $C_p$ act on itself via translation, and let $X$ be the collection of all $k$-subsets of $C_p$. Then $X$ inherits a $C_p$-action having no fixed points, so it must be a disjoint union of orbits of size $p$, i.e. $p\mid\#X$.

@Theorem Personal curiosity.

user19161

@user58512 Stop calling everyone rude. That is ridiculous.

anon

The user did not call everyone rude in the comment you are replying to.

user19161

9:52 PM

I think I need to flag you for calling me and Hagen rude when we are not.

Theorem

@anon : by the way , i want to settle something here , i often get confused with $\mathbb Z_n$ and $C_n$ notation . i really can't distinguish very often .

user19161

@anon Everyone is a hyperbole, and also in reference to past events.

anon

I figured.

user58512

@Theorem, they are isomorphic

user19161

If you think people are rude, don't talk to them. I think they don't like to talk to you either.

user58512

9:55 PM

considering Z_n as additive group at least

Theorem

@user58512 whats the difference

anon

@Theorem In pure group theory, and elementary number theory, ${\Bbb Z}_n$ denotes the integers modulo $n$. In group theory, $C_n$ denotes the cyclic group of order $n$. Of course these are the same thing. One reason I avoid using Z, though, is because ${\bf Z}_p$ is also the $p$-adic integers, which is very different. Also ${\Bbb Z}/n{\Bbb Z}$ or ${\Bbb Z}/(n)$ both represent quotient rings in abstract algebra (sometimes ${\Bbb Z}_n$, too), which are the ring of integers modulo $n$.

Theorem

what does $C$ stand for ?

anon

C stands for cyclic

user19161

Also yes @user I can see your flags.

user19161

9:58 PM

There is no need to ask on meta, I can tell you all the answers you need.

Theorem

@anon Thanks , that has been troubling me a lot .

user19161