12:01 AM
@anon So given an input of $n$, it needs to output product of all primitive roots modulo n?
never mind. I just saw your comment. My short attention span will be the end of me.

Lambda is the product of certain representatives of the primitive roots mod n - each representative between 1 and n - done in Z. There could be clever ways to compute it (computational algebraic number theory is intriguing but certainly not covered by my abilities and knowledgebase).

1 hour later…
1:12 AM
@anon I can't quite help you with the programming (and I am not really sure how programs can help with this). But I think it would make a good question.

@Srivatsan Hello!

hi Kannappan. How do you do?

@Srivatsan Just answered one question the whole day man.

:)

A group theory question. Given that I knew the solution, the OP exactly refused to accept it!
He said in his question, you CANNOT use that trick!
@Srivatsan Are you around?

1:21 AM
I am trying to hunt which post you are talking about.

@Srivatsan Is there word limit for answers?

Not afaik

@Srivatsan But, why is it that there is some error whenever I come upto a point. The page becomes non-responsive!
The third time I am writing the same blah!

@KannappanSampath You need to disable the immediate update of the MathJax

Not supposed to happen. Tip: Write it locally (save it), and copy-paste it here. That way you won't lose any data/effort

1:26 AM
Oh, I see!

@robjohn How do we do this?

I see, thanks, Rob.

@Srivatsan This is the post I've been talking about!

1:34 AM
what is Poincare's theorem?

@Srivatsan which one? there are several

It says, $G$ cannot be simple if there is a non-trivial normal subgroup :)
In a fashionable way!

ah in context of Kannappan's post

What is the fashionable way mean?

1:35 AM
@Srivatsan Not wearing a polyester leisure suit.

@robjohn Unless there is one that helps turn off MathJax. :-)

It says, Let $H$ be a proper subgroup of finite index.
Then $|G| \mid n!$

@Srivatsan I find those quite useful when writing very long answers.

Where $n$ is the index of the subgroup in question!

1:38 AM
@robjohn Does your proof start with the sentence "By Poincare's theorem, we can assume MathJax is turned off without loss of generality..."?
@KannappanSampath The index? Oh wow

@Srivatsan LOL!

If G is finite, then why bother saying finite index?

@Srivatsan I could write a proof that starts that way...
Then write the whole thing in ascii-art

@robjohn I haven't written one that needed this yet. Not sure if I should be happy, sad, or oblivious.
I want an ascii art proof here. I am telling you: it's a surefire way to glory here.

@Srivatsan You've not written long posts, are you kind of joking?

1:41 AM
@KannappanSampath No, I am not. Do you think I have written some?

@Srivatsan Some of the equation, integral, multi-fractioned, binomial filled answers bog down really fast.

@Srivatsan Yes you've! Quite a lot!

@KannappanSampath Then perhaps our definitions of long differ. Do you want to point me to something?

My random clicks lead me to this,this.
@Srivatsan

Hm, the second one was -well- interesting to type. Sorry: I meant the first.

1:47 AM
@Srivatsan: according to the perpetually out of date database, this is your longest post.

@robjohn Ha, that one.

@robjohn How so sure?

@Srivatsan You may have written one that is longer recently, but according to the DB, that was it a while ago.

I randomly clicked a few and got of hold of long one!

I don't know if it was long, but it was impassioned.
@KannappanSampath Well, I tend to keep mumbling things. :=)

1:50 AM
@KannappanSampath Go to data.stackexchange.com/mathematics and enter the right query :-)

Do you type with userid: or simply number!

    DECLARE @displayname varchar(100) = '##displayname##'
select top 10 a.id as ID,q.title as Title,len(a.body) as Length from posts a,posts q where a.parentid = q.id and a.owneruserid in (select id from users where displayname=@displayname) order by len(a.body) desc
That is the query I use to get the 10 longest posts.

Why Am I called jon.doe.7297
Crazy database!

Just a minute
@KannappanSampath I don't get anything for you...

I want to know how unsung I am?
So, I clicked that quer
I put in 21436
And, db says, division by zero error!

2:00 AM
@KannappanSampath If I push in your ID, I get math.stackexchange.com/questions/92516

Is that my longest post? Strange!
That's my first post!

@KannappanSampath As I said, the database is perpetually out of date

Atleast, it knows I exist :)
Thank You Rob!

@KannappanSampath Ah, I was using Kannappan, not your full display name

2:09 AM
@BrianMScott When you add "queries" at the end of the http address, it works.
Hello everybody.

@ymar Ah, okay; thanks.

Hello! @ymar
List of Stingy People here

@BrianMScott which page?

parameters 30000; 300!

DECLARE @displayname varchar(100) = '##displayname##'
select top 10 a.id as ID,q.title as Title,len(a.body) as Length,u.displayname as Author from posts a,posts q,users u where a.parentid = q.id and u.id=a.owneruserid and a.owneruserid in (select id from users where displayname like @displayname + '%') order by len(a.body) desc
That will match the beginning of the displayname and since several may be matched, it also gives the author

2:16 AM
This is the question that occupied the whole night!
And, the OP said, "Thanks, Kannappan :)" (shouts aloud, never write answers here)
@robjohn Why was that removed!

@KannappanSampath I was saying wow to the empty link that you spent most of the night on. It was a snide "wow" so I removed it when you filled in the link
when you wrote [This]()

I see!
I thought you liked the question that kept me on for the night ? =)

@robjohn The data.MSE page, but ymar solved the problem for me.

Apparently, we need to update our SEModifications every now and then.. Who would know? :)
(@Ilya: This is regarding the "history" link that did not work for us. It seems that the SEModifications tool we both are using is out-of-date. Update it, and everything works.)
@Brian: I am currently watching some comedy series that is set in Cleveland. =)
Apparently it was originally called Cle-A-veland.

2:39 AM
@Srivatsan That’s correct: it was named after Moses Cleaveland.

3:14 AM
Ah, I see.

3 hours later…
6:40 AM
hi @all

slow night it seems

its 12 pm in here

oh. 1am ish for me.

@anon : Go to sleep, a fresh morning with fresh ideas awaits you!

Yes mother.

7:06 AM
Hi guys.

Hi. (Don't tell Rajesh I'm breaking fresh-ideas curfew.)

Why is $F_p/(x^n)$ not a model for $F_{p^n}$?

What is $x\in F_p$ specifically? Do you mean $F_p[x]$? If so, then the ring has zero divisors (e.g. $x\cdot x^{n-1}=0$) so it isn't a field.

Yeah, sorry.

That's why you need the $f(X)$ in $\mathbb{F}_p[X]/(f(X))$ to be irreducible.

7:18 AM
Ok, so let's say that $f(X)$ is irreducible. Then for any $g(x), h(x) \in \mathbb{F}_p[X]/(f(X))$ if $g(x)h(x) = 0$ then $g(0)h(0) = 0$ which is true only when $g(0) = h(0) = 0$. Now, how do I use irreducibility to prove that $g(x)$ or $h(x)$ is zero?

No, only true when $g(0)=0$ and/or $h(0)=0$.
Can't think of a proof of the top of my head, bah.

Is $g(x) h(x) \equiv 0 \mod{f(x)}$?

@Daniil, what does $gh=0$ in the quotient ring mean?

I guess that my guess was right then.

Note that $g(X)h(X)=0$ in $\mathbb{F}_p[X]/(f(X))$ implies that $g(X)h(X)=f(X)k(X)$ for some $k\in\mathbb{F}_p[X]$. Now use irreducibility...

7:26 AM
there you go
stylistic observation: it rarely serves any purpose to refer to polynomials as $f(X)$ instead of $f$
(sometimes it does, of course...)
it always looks funny when someone says «the function $f(x)$»

next step might be to use Euclid's lemma.

and the same funniness applies to polynomials :)

at the moment I like looking at big X's as formal variables. I'm feeling whimsical.

but the polynomial is just $f$

Hm, I somehow thought it was more common for polynomials to be denoted f(X) than simply f.

7:30 AM
usually the variable is fixed as soon as you write the ring itself; if you are working in $R=k[X]$, that fixes the name of the variable used to write out all elements of $R$.
because $k[X]$ is the set of polynomials in the variabe $X$

Ah, I see. Point taken.

When dealing with functions, this is different, because functions are not formal expressions, but simply rules
Of course, this is a matter of taste, and I've had a Fields medalist say to my face «the function $\sin x$»... :)

trifles

greatness is in the details

details like letters and parentheses ;-)

7:36 AM
Hello , I've a question again =\

Don't we all.

@MarianoSuárezAlvarez Was that F.m. an American? I remember Pete Clark mentioning somewhere that people trained in the Am. system are more prone to say something like "f(x) is an even function".

I'm a bit confused about questions asking the coefficient of a statement.

of a statement? surely the questions mean the coefficient of one side of the statement, and they only put the statement there to condense the material?

7:38 AM
Like $x^2y^6$ in the expansion of $(x+y)^n$
I can determine these kind of questions

good math writing is like good cooking: when you are reading it, it is very very difficult to pin down exactly where does perfection come into it, but you surely know when it is there, and the tiniest pinch of a spice can do wonders

@Gigili You mean the coefficient of a term in a polynomial?

Man, I hate this chat thing :/
3

But there's a question asking about the coefficient of $x^4$ in the expansion of $(x+1/x)^7$
@Srivatsan I think so, yes .. Sorry I got "time out" errors and couldn't answer

@Gigili Well, the question is not so interesting: the answer is zero.
There are at least two ways of doing it:

7:44 AM
The cheapest way is to note that $a+b=7\implies a-b\not\equiv0\mod2$ hence the coefficient is $0$.

Method 1. Hypothetically expand $(x+\frac{1}{x})^7$ using the binomial theorem. What does a general term look like?

Method 2 I guess is then looking at the $x^{4+7}$ term of $(x^2+1)^7$; also easy to see this is $0$.

Nice. exactly what I had in mind.

Now perhaps we should get Gigili up to speed here...

I'm so sorry, I was looking for the question among the papers.
It shouldn't be like this.
I know the main idea is to use binomial theorem, but this one was a bit different

7:51 AM
what papers?

Papers I wrote my questions on them

The first approach I summarised is this: the nomials popping out of the expansion of $(x+x^{-1})^7$ will be of the form $\text{something}\cdot x^a x^{-b}$ for $a+b=7$. The issue is that $a-b\equiv a+b=7=1\mod2$ so therefore can't equal $4$, ergo the $x^4$ term of this expansion does not exist i.e. has coefficient $0$.

I don't understand this part : $a-b\equiv a+b=7=1\mod2$

In mod 2 we have -1 = +1 so I just switched the sign.

$a-b \equiv a+b mod 2$

7:56 AM
$a+b=7$ because the count of $x^1$ and $x^{-1}$ involved equals $7$, and of course 7 is 1 mod 2.

Right, could you determine a coefficient which is not zero?

In other words: $a+b$ is $7$, and so $a+b$ is odd. Then $a-b$ is also odd, since $(a+b)+(a-b) = 2a$, which is even. Therefore $a-b$ can never be $4$. Thus the coefficient of $x^4$ is zero.
Ok, let us find the coefficient of $x^3$ instead; that will not be zero.
As before, the general term looks like $x^a \cdot \frac{1}{x^b} = x^{a-b}$, with $a+b=7$.
Now to find the coefficient of $x^3$, we need $a-b=3$. Can you now solve the pair of equations
$a+b = 7$ and $a-b=3$?

$x^5$ \cdot $1/x^2$?

Yeah, a=5 and b=2.
Ah, I made a mistake: the general term looks like $\binom{7}{a} x^a \cdot x^{-b}$. I forgot the binomial coefficient.

Got it, thank you. So whatever $x$ and $y$ are, we can use binomial theorem, right?

8:02 AM
What is $y$?

$(x+y)^n$ ... $\binom{n}{a} x^a \cdot y^{n-a}$

For a general method of determining nonzero coefficients, consider the following. Let $[x^m]\text{blah}$ denote the coefficient of $x^m$ in $\text{blah}$. Then $[x^m]x^k P(x)=[x^{m-k}]P(x)$ (exercise!). Whence $[x^3](x+x^{-1})^7=[x^3]x^{-7}(x^2+1)^7=[x^{10}](x^2+1)^7$. The $x^{10}$ term of course is $$\binom{7}{5} (x^2)^5 1^2$$ so our answer is 7 choose 5 or 21.
That's probably too high-level, actually. If you can't understand that just ignore it.

Got it, except $[x^3]$

It's notation from Enumerative Combinatorics. You can think of it as an operator that hones in on the coefficient of $x^3$ of a polynomial or other power expansion, e.g. $[x^3](x^{-1}+x+2x^3+x^4)=2$.
Actually the notation might be from Generatingfunctionology and I'm confusing it with EC, or it's in both. Can't remember.

Gigili: You can think of it as a shorthand for "Coefficient of $x^3$ in this polynomial".

8:12 AM
Got it, thanks a lot @anon, @Srivatsan.

LOL

Wow! I lost my mind learning maths.

@anon It's certainly in gfology. I just checked EC, and didn't find it anywhere. Not sure if I was looking at the wrong places, but it seems that Stanley does not use it in his book.

Ah, thanks.

You're welcome.

At least the OP said what they tried.

I don't have anything against the op. My statement was independent of this.

Just a quick tagging question.
Do you think introducing tag would be a good idea?
Or would it be better to use tags that already exist and tag them by both and ?
Some examples of questions at MSE related to symmetric polynomials:
[Number of terms in a monomial symmetric polynomial](http://math.stackexchange.com/questions/14461/number-of-terms-in-a-monomial-symmetric-polynomial)
[Linear Algebra of Symmetric Sums](http://math.stackexchange.com/questions/30491/sum-of-cubed-roots)
[Algorithm(s) for computing an elementary symmetric polynomial](http://math.stackexchange.com/questions/30807/algorithms-for-computing-an-elementary-symmetric-polynomial)

That's a good question.
Put http:// in front of the urls and see if it works

[Roots of elementary monomials](http://math.stackexchange.com/questions/81471/roots-of-elementary-monomials)
[Elementary symmetrical polynomial equations, whose solutions are known to be natural numbers](http://math.stackexchange.com/questions/90170/elementary-symmetrical-polynomial-equations-whose-solutions-are-known-to-be-nat)
[Symmetric polynomials have unique expressions as polynomials in symmetric elementary functions](http://math.stackexchange.com/questions/96278/symmetric-polynomials-have-unique-expressions-as-polynomials-in-symmetric-elemen )
@anon http is there, as you can check in transcript. But that's not the important thing here.
It's probably just again the problems with multiline posts.

8:41 AM
@anon so $f(X)k(X) = 0$ in $\mathbb{F}_p[X]/(f(X))$. If f(X) were reducible we could write it as $a(x)b(x)k(x) = 0$ where a, b and k could be non-zero (since a < f and b < f)

I don't follow you at all mate.
Are you using $a,b,k$ from my comment above?

@Srivatsan, Daniel M is cool :)

@Martin: I'd take the symmetric question up to meta. I don't really have feelings either way.

@anon I'm not sure that is important enough to raise it at meta. But in case I get no reaction here, I'll try to post it in Jury Duty, to get a little more exposure.

@anon no sorry, I was still talking about my stupid problem with irreducible polynomials

8:47 AM
@MartinSleziak In fact, post it in Jury Duty anyway.

But what does the [symmetric-functions] tag contain (other than the polynomial questions)?
I don't like that tag. :=)

There are 8 questions. I think one or two are not about polynomials.

@Daniil: Yes, but I'm not sure why you put them altogether for $abk=0$. If $f(X)$ is reducible then $f=u\cdot v$ with lesser degrees hence we have zero divisors and no field. If $f$ is irreducible then assume $ab=0$ whence $f|ab$ whence $f|a$ or $f|b$ by Euclid's lemma (as Sri noted) whence either $a$ or $b$ is $0$ in the ring, so we have no zero divisors.

@MarianoSuárezAlvarez Noted. :)

8:49 AM
@anon yeah, that's what I was trying to say; I just used a and b instead of u and v
Thank you for help, @anon!

Some background information that my be useful: know why Euclid's lemma holds in these polynomial rings (the base ring has to itself be an integral domain), and know why finite integral domains must be fields.

Morning.

@Srivatsan This one How do I find out the symmetry of a function? is not about polynomials.
Hi Asaf!

Hello, @AsafKaragila

Morning everyone.
It's snowing again.
@MartinSleziak Done. : )
Going to bakery for breakfast, bbl.

8:52 AM
What a bunch of jerks. They all answered that AC question when I finally got to rest.

Is this chat thing deleting my «goodnight!» ?!?!

@MartinSleziak Wait: how is that a symmetric function? "Symmetries of a function" is not quite the same thing, is it?

@Srivatsan Well, I did not claim that the question is tagged correctly. (But it's not entirely clear what is the intended use of that tag.)

@MartinSleziak Ah, ok. :=)

9:00 AM
But I can imagine that some symmetric functions that are not polynomials might be useful too. So there might be questions about them.

@MartinSleziak Actually, it does sound useful. (For instance, I come across "boolean symmetric functions" a lot. Even though one could potentially write such a function as a polynomial, calling it a symmetric polynomial would be quite a stretch.)

@Srivatsan So we agree that and the proposed tag will be different, if the latter is added.

Yes, I agree with the creation of the new tag.
My prediction is that [symmetric-functions] will be underused.

ok, I'm not sure I will have time to create the tag right away; I wanted just to know the opinion of some other users on it. But it's not an urgent matter, I'll get to it - hopefully later today.

Sure, thanks for this.

9:29 AM
$65454\ldots5455 = 6 \cdot 5454\ldots5455 + 65454\ldots545 \cdot 5$

I contributed an answer anyway. I think that Ricky's answer is good, but does not cover any details.

9:40 AM
Hm.

10:26 AM
@robjohn I also don't know why. Maybe because my answer uses the most elementary methods? (Jensen is at least here not as known as CS).

2 hours later…
12:15 PM
...@ (<-- tumble weeds)

Hi hi

12:33 PM
In fact tumbleweeds are here

Hello. I've found two questions which basically ask the same thing. What do I do?

@Skullpatrol Obviously, yes.

@MartinSleziak Don't you think that is odd?

@Skullpatrol I don't think that. You don have to take all badges that seriously. And awarding a badge can help drawing attention to a question that went unnoticed; which is IMHO a good thing.

12:46 PM
@MartinSleziak It is sort of like that tree that falls in the forest and nobodies around to hear it ... If I ask a question and nobody votes, answers, comments, or even views it for a week ... I get to earn a badge for doing that?

@Skullpatrol That's the way I understood the description of that badge.

@MartinSleziak ...@ (<-- a tumble weed indeed) ;-)
"In fact"

@ymar You write a comment with a link to the duplicate and then vote to close or flag for moderator attention.

1:05 PM
@MattN Thanks. I can't vote to close with my reputation. I've flagged it.

@MattN Have you heard about the math war in education?

@Skullpatrol No.

@MattN Are you able to see the link Matt?

1:30 PM
@Skullpatrol Yes.

1 hour later…
2:55 PM
@ymar Are you around?

@KannappanSampath yep. Hi!

Thank You for the edit, @ymar
I had to just write an answer and go for my dinner!

@KannappanSampath You are welcome. At first I couldn
't understand what you meant, so I edited in case somebody else would have trouble too.

I wanted to add this divisible groups as well, but then, I .....
@ymar

@KannappanSampath Shouldn't I have added that?

3:02 PM
@ymar No, It's good you did! I was in a dilemma!

@KannappanSampath What was the dilemma?

@ymar If I added that, I need to write down the definition, or else it will be a stand alone link!
Or should I prove the general case for these groups and take $\mathbb R$ as a consequence!

@KannappanSampath Well, you did, actually. You only used the fact that $\mathbb R$ is divisible, so it proves the general case just as well. You only refrained from mentioning it. If the OP wants to the definition, they will follow the link I've provided and learn that. It's a simple notion and I think there's no need of additional explanation.

True @ymar!
As you said, the dilemma was in whether or not to mention!

@KannappanSampath Well, fair enough.
OK, I need to smoke. Where's my lighter?!

1 hour later…
4:09 PM
@KannappanSampath Good fix on that "160" answer.

hi Dylan
hi Clash & m.k.

Hey @Srivatsan.

hello there

@Martin: Your suggestion sounds fine. Just that it is unlikely that a beginner would know to search for that tag name.

The question is how to name a tag similar to , but for arbitrary base.

4:19 PM
Yes, I am favoring base-b expansion since it sounds similar. (Also this is the first time I am coming across "g" used for this purpose. :))

Also, the tag name should be chosen in such way, that it's not confuse with p-adic numbers.

Yes, that too.

I think base-b expansion sounds reasonable.
I tried to find Wikipedia article on this and steal the name from there, but the closest thing I was able to find was Positional notation

There should be a separate application for this chat. That's the only reasonable thing I can think of.

Done. I tagged it base-b-expansion temporarily. We can change it, of course.

4:26 PM
@AsafKaragila What do you mean by separate application?

Like there is a separate application for gTalk.
With its own window, and without all the browser stuff; but the links would open in the browser.

Like using chat.SE in pidgin, miranda or your favourite software...?

Probably something based on a browser window is better, to keep the ChatJax compatibility.

@Srivatsan Sound reasonable, too. BTW do you want them to be separate tags, or synonyms?

My thought was to make [decimal-expansion] a synonym, but I don't have too strong a feeling either way

4:33 PM
The question Uniqueness of the infinite expansion in base b is similar, it is tagged .

Hm, that fits too.

If you have a look at the tag wiki and the tagged questions, it seems precisely the same thing for which you wanted to create this base-b tag.

@Srivatsan: I amended my answer to reformat the ugly table and give a simple recurrence for every other $c_d$.
We have to go to a Bat-Mitzvah today, so I will be in and out of chat.

@robjohn Ah, wow, that's a simple recursion. (I wanted to suggest the table thing myself actually.)
@Kannappan: Are you here?
@robjohn What is Bat-Mitzvah? I have heard of Bar-Mitzvah? Is it for boys and girls or something like that?

Hello guys

4:47 PM
Quick sanity check: Americo Tavares' answer to this question uses an algebraic identity derived in another answer, but the identity looks wrong to me -- the values for the expressions differ at x=0 and x=1. Am I mistaken?

Hello people, what is the difference between numerical analysis and optiminaztion also the difference between the measure theory and functional analysis?

@Srivatsan I just came in!
@Dylan Thank You!

But I have a few stylistic comments: should I mention them?

Yeah Sure.
I am always willing to know any comment on my answers here!
Will @AlexYoucis get pinged by this?

(a) I got stuck a little bit at the very beginning since I wasn't sure what you were trying to prove. Maybe instead of saving it for the punchline, you could start off declaring it. "No, there is no proper subgroup of R with finite index."

4:53 PM
OK.

@Srivatsan - What is optiminaztion and do you have to learn it in computer science

@Victor I know what it is about, but I don't know it well enough for me to provide a brief executive summary. (Besides: I'm in between a conversation right now.)

Welcome Matt!

@KannappanSampath (b) This is minor: probably even I would have done the same thing if I were you. You are using both multiplicative and additive notation for groups, back to back. Since this is the abelian case, maybe you can do away with the multiplicative notation.

Hello there.

4:56 PM
@Srivatsan In this line: $(rH)^n = H$ and then $nr \in H$?

@KannappanSampath The very same.

Thank You, Srivatsan. I'll fix it!

But I know: the additive notation is not natural (for general groups) once we get used to the multiplicative one. I am not sure what to do in this case.

What is that (c)?
Why did it disappear? =)
Please tell me that as well!

Not sure if it is important, but I will say it now anyway.