Mathematics - 2012-01-28 (page 1 of 3)

Fortuon Paendrag

00:01

@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.

anon

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).

Srivatsan

01:12

@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.

Kannappan Sampath

@Srivatsan Hello!

Srivatsan

hi Kannappan. How do you do?

Kannappan Sampath

@Srivatsan Just answered one question the whole day man.

Srivatsan

@KannappanSampath Tell me about it.

:)

Kannappan Sampath

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?

Srivatsan

01:21

I am trying to hunt which post you are talking about.

Kannappan Sampath

Let you know about in a while. Editing to add! @Srivatsan

@Srivatsan Is there word limit for answers?

Srivatsan

Not afaik

Kannappan Sampath

@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!

robjohn

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

Srivatsan

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

Kannappan Sampath

01:26

Oh, I see!

Srivatsan

@robjohn How do we do this?

Kannappan Sampath

His tool on his home page helps @Sri

robjohn

@Srivatsan meta.math.stackexchange.com/a/3485

Srivatsan

I see, thanks, Rob.

Kannappan Sampath

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

Srivatsan

01:34

what is Poincare's theorem?

robjohn

@Srivatsan which one? there are several

Kannappan Sampath

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

In a fashionable way!

Srivatsan

Ah

robjohn

ah in context of Kannappan's post

Srivatsan

What is the fashionable way mean?

robjohn

01:35

@Srivatsan Not wearing a polyester leisure suit.

Srivatsan

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

Kannappan Sampath

It says, Let $H$ be a proper subgroup of finite index.

Then $|G| \mid n!$

Srivatsan

n?

robjohn

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

Kannappan Sampath

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

Srivatsan

01:38

@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

Kannappan Sampath

@Srivatsan LOL!

Srivatsan

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

robjohn

@Srivatsan I could write a proof that starts that way...

Then write the whole thing in ascii-art

Srivatsan

@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.

Kannappan Sampath

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

Srivatsan

01:41

@KannappanSampath No, I am not. Do you think I have written some?

robjohn

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

Kannappan Sampath

@Srivatsan Yes you've! Quite a lot!

Srivatsan

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

Kannappan Sampath

My random clicks lead me to this,this.

@Srivatsan

Srivatsan

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

robjohn

01:47

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

Srivatsan

@robjohn Ha, that one.

Kannappan Sampath

@robjohn How so sure?

robjohn

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

Kannappan Sampath

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

Srivatsan

I don't know if it was long, but it was impassioned.

@KannappanSampath Well, I tend to keep mumbling things. :=)

robjohn

01:50

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

Kannappan Sampath

Do you type with userid: or simply number!

robjohn

    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.

Kannappan Sampath

Why Am I called jon.doe.7297

Crazy database!

robjohn

Just a minute

@KannappanSampath I don't get anything for you...

Kannappan Sampath

I want to know how unsung I am?

So, I clicked that quer

I put in 21436

And, db says, division by zero error!

robjohn

02:00

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

Kannappan Sampath

Is that my longest post? Strange!

That's my first post!

robjohn

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

Kannappan Sampath

Atleast, it knows I exist :)

Thank You Rob!

robjohn

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

Brian M. Scott

@robjohn Page not found.

ymar

02:09

@BrianMScott When you add "queries" at the end of the http address, it works.

Hello everybody.

Brian M. Scott

@ymar Ah, okay; thanks.

Kannappan Sampath

Hello! @ymar

List of Stingy People here

robjohn

@BrianMScott which page?

Kannappan Sampath

parameters 30000; 300!

robjohn

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

Kannappan Sampath

02:16

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!

robjohn

@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]()

Kannappan Sampath

I see!

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

Brian M. Scott

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

Srivatsan

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.

Brian M. Scott

02:39

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

Srivatsan

03:14

Ah, I see.

Rajesh D

06:40

hi @all

anon

slow night it seems

Rajesh D

its 12 pm in here

anon

oh. 1am ish for me.

Rajesh D

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

anon

Yes mother.

Daniil

07:06

Hi guys.

anon

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

Daniil

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

anon

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.

Daniil

Yeah, sorry.

anon

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

Daniil

07:18

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?

anon

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.

Srivatsan

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

Mariano Suárez-Alvarez

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

Srivatsan

I guess that my guess was right then.

anon

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...

Mariano Suárez-Alvarez

07:26

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 $f(x)$»

Srivatsan

next step might be to use Euclid's lemma.

Mariano Suárez-Alvarez

and the same funniness applies to polynomials :)

anon

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

Mariano Suárez-Alvarez

but the polynomial is just $f$

Srivatsan

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

Mariano Suárez-Alvarez

07:30

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 $X$

Srivatsan

Ah, I see. Point taken.

Mariano Suárez-Alvarez

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 $\sin x$»... :)

anon

trifles

Mariano Suárez-Alvarez

greatness is in the details

anon

details like letters and parentheses ;-)

Gigili

07:36

Hello , I've a question again =\

anon

Don't we all.

Srivatsan

@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".

Mariano Suárez-Alvarez

yeah

Gigili

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

anon

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?

Gigili

07:38

Like $x^2y^6$ in the expansion of $(x+y)^n$

I can determine these kind of questions

Mariano Suárez-Alvarez

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

Srivatsan

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

Mariano Suárez-Alvarez

Man, I hate this chat thing :/

3

Gigili

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

Srivatsan

@Gigili Well, the question is not so interesting: the answer is zero.

There are at least two ways of doing it:

anon

07:44

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

Srivatsan

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

anon

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

Srivatsan

Nice. exactly what I had in mind.

anon

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

Gigili

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

Mariano Suárez-Alvarez

07:51

what papers?

Gigili

Papers I wrote my questions on them

anon

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 $0$.

Gigili

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

anon

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

Gigili

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

anon

07:56

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

Gigili

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

Srivatsan

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 $a-b=3$?

Gigili

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

Srivatsan

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.

Gigili

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

Srivatsan

08:02

What is $y$?

Gigili

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

anon

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.

Gigili

Got it, except $[x^3]$

anon

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.

Srivatsan

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

Gigili

08:12

Got it, thanks a lot @anon, @Srivatsan.

anon

LOL

Gigili

Wow! I lost my mind learning maths.

Srivatsan

@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.

anon

Ah, thanks.

Srivatsan

You're welcome.

Srivatsan

08:27

Heh.. "Dude, these are questions off my problem sets, so you're a student in my class. If you use answers you get off math.SE in the problem set you submit tomorrow, please be sure to cite them as such, OK? I won't deduct marks..."

He seems like a cool prof.

anon

At least the OP said what they tried.

Srivatsan

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

Martin Sleziak

Just a quick tagging question.

Do you think introducing symmetric-polynomials tag would be a good idea?

Or would it be better to use tags that already exist and tag them by both symmetric-functions and polynomials?

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)

anon

That's a good question.

Put http:// in front of the urls and see if it works

Martin Sleziak

[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.

Test: Number of terms in a monomial symmetric polynomial

It's probably just again the problems with multiline posts.

Daniil

08:41

@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)

anon

I don't follow you at all mate.

Are you using $a,b,k$ from my comment above?

Mariano Suárez-Alvarez

@Srivatsan, Daniel M is cool :)

anon

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

Martin Sleziak

@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.

Daniil

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

Srivatsan

08:47

@MartinSleziak In fact, post it in Jury Duty anyway.

Martin Sleziak

ok

Srivatsan

But what does the [symmetric-functions] tag contain (other than the polynomial questions)?

I don't like that symmetric-functions tag. :=)

Martin Sleziak

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

anon

@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.

Srivatsan

@MarianoSuárezAlvarez Noted. :)

Daniil

08:49

@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!

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.

Asaf Karagila

Morning.

Martin Sleziak

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

Hi Asaf!

Daniil

Hello, @AsafKaragila

Matt N.

Morning everyone.

It's snowing again.

@MartinSleziak Done. : )

Going to bakery for breakfast, bbl.

Asaf Karagila

08:52

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

anon

I predict fuss over the mathematical status of this question.

Mariano Suárez-Alvarez

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

Srivatsan

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

Martin Sleziak

@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.)

Srivatsan

@MartinSleziak Ah, ok. :=)

Martin Sleziak

09:00

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

Srivatsan

@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.)

Martin Sleziak

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

Srivatsan

Yes, I agree with the creation of the new tag.

My prediction is that [symmetric-functions] will be underused.

Martin Sleziak

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.

Srivatsan

Sure, thanks for this.

m. k.

09:29

$65454\ldots5455 = 6 \cdot 5454\ldots5455 + 65454\ldots545 \cdot 5$

Asaf Karagila

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

Asaf Karagila

09:40

Hm.

Jonas Teuwen

10:26

@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).

Matt N.

12:15

...@ (<-- tumble weeds)

Zhen Lin

Hi hi

Martin Sleziak

12:33

In fact tumbleweeds are here

ymar

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

Skullpatrol

@MartinSleziak You can earn a badge for asking a question with no votes, no answers, no comments, and low views for a week?

Martin Sleziak

@Skullpatrol Obviously, yes.

Skullpatrol

@MartinSleziak Don't you think that is odd?

Martin Sleziak

@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.

Skullpatrol

12:46

@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?

Martin Sleziak

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

Skullpatrol

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

"In fact"

Matt N.

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

ymar

13:05

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

Skullpatrol

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

Matt N.

@Skullpatrol No.

Skullpatrol

books.google.ca/…;‌+or+numeral%22&hl=en&sa=X&ei=Lg0jT8PIIueUiAKsycWHCA&ved=0CDYQ6AEwAA#v=onepage&q‌=%‌22numerical%20expression%2C%20or%20numeral%22&f=false

@MattN Are you able to see the link Matt?

Matt N.

13:30

@Skullpatrol Yes.

Kannappan Sampath

14:55

@ymar Are you around?

ymar

@KannappanSampath yep. Hi!

Kannappan Sampath

Thank You for the edit, @ymar

I had to just write an answer and go for my dinner!

ymar

@KannappanSampath You are welcome. At first I couldn

't understand what you meant, so I edited in case somebody else would have trouble too.

Kannappan Sampath

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

@ymar

ymar

@KannappanSampath Shouldn't I have added that?

Kannappan Sampath

15:02

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

ymar

@KannappanSampath What was the dilemma?

Kannappan Sampath

@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!

ymar

@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.

Kannappan Sampath

True @ymar!

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

ymar

@KannappanSampath Well, fair enough.

OK, I need to smoke. Where's my lighter?!

Dylan Moreland

16:09

@KannappanSampath Good fix on that "160" answer.

Srivatsan

hi Dylan

hi Clash & m.k.

Dylan Moreland

Hey @Srivatsan.

m. k.

hello there

Srivatsan

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

Martin Sleziak

@Srivatsan Or g-adic representation and b-adic representation.

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

Srivatsan

16:19

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. :))

Martin Sleziak

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

Srivatsan

Yes, that too.

Martin Sleziak

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

Asaf Karagila

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

Srivatsan

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

@MartinSleziak That reminds me: how about "radix notation"?

Martin Sleziak

16:26

@AsafKaragila What do you mean by separate application?

Asaf Karagila

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.

Martin Sleziak

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

Asaf Karagila

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

Martin Sleziak

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

Srivatsan

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

Martin Sleziak

16:33

The question Uniqueness of the infinite expansion in base b is similar, it is tagged number-systems.

Srivatsan

Hm, that fits too.

Martin Sleziak

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.

robjohn

@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.

Srivatsan

@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?

Daniil

Hello guys

lewist

16:47

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?

Victor

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

Kannappan Sampath

@Srivatsan I just came in!

@Dylan Thank You!

Srivatsan

Well, your answer for finite-index subgroup of R is quite nice.

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

Kannappan Sampath

Yeah Sure.

I am always willing to know any comment on my answers here!

Will @AlexYoucis get pinged by this?

Srivatsan

(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."

Kannappan Sampath

16:53

OK.

Victor

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

Srivatsan

@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.)

Kannappan Sampath

Welcome Matt!

Srivatsan

@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.

Matt N.

Hello there.

Kannappan Sampath

16:56

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

Srivatsan

@KannappanSampath The very same.

Kannappan Sampath

Thank You, Srivatsan. I'll fix it!

Srivatsan

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.

Kannappan Sampath

What is that (c)?

Why did it disappear? =)

Please tell me that as well!

Srivatsan

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

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$Mathematics$ Mathematics