Mathematics

12:01 PM

I wanted to use that bookmarklet on Google Groups some time ago. Didn't work either :-( (Which is a pity - Google Groups is, in my opinion, the way to discuss things with selected groups of people which is easiest to create. Although it is less organized than some kind of bulletin board/forum.)

Algebra

12:13 PM

@robjohn Thank you for that comment, that makes sense. Facebook comments don't work I am sure, I haven't tested other aspects(just saying so you know).

Daniel Fischer

@IceBoy I told him he should do that, but it's more likely that the upvote went to Noam Elkies's (really good, as usual) answer to Thursday's own question.

2

Algebra

This is actually the first time in the existence of Thursday that I have seen him drop off the first page of monthly rep.

Ice Boy

@DanielFischer interesting

Martin Sleziak

There should be a badge for this :-)

Ice Boy

what do we call it?

Martin Sleziak

12:25 PM

No idea.

Ice Boy

super negativity

Algebra

I am working with generating functions and am required to prove that the generating function for the sequence $\{a_k\}$ where $a_k = (-8)^k$ for all integers $k\geq0$ is $\frac{1}{1+8x}$ and I have gotten it to:

$g(x)[1+8x] = \lim \limits_{n\to\infty} 1 - (-8)^{n+1}x^{n+1}$ I can see that I want to take both of those ${n+1}$ terms to be $0$, but I don't know how I can say this is alright.

Is it appropriate for me to even ask that on chat here?

It works for $0\lt x\lt\lt\lt1$ I suppose

Martin Sleziak

Isn't it sufficient to show that $(1+8x)\sum_{k=1}^\infty (-8)^k x^k = 1$, when viewed as formal power series?

Algebra

I suppose so, would this be easier?

Than using the geometric trick($g(x) - xg(x)$)? Or is it that I can't move forward from what I have above

Martin Sleziak

To be honest, I haven't used generating functions for quite a long time. So maybe I should let someone else answer this. (I don't want to say you something incorrect.)

Algebra

12:39 PM

That's okay xD. I shall ask on the actual stack exchange

Martin Sleziak

This post seems related: geometric series and generating functions

Algebra

@MartinSleziak Thank you for that, it does seem to be what I am looking for. I just posted my question unfortunately.

Algebra

12:57 PM

http://math.stackexchange.com/questions/927548/showing-that-a-generating-function-is-equivalent-to-some-fraction?noredirect=1#comment1914690_927548

"but it has no meaning mathematically", surely this isn't true?

Or he is still talking about $x$ being indeterminate, I guess I will clarify

Chris's sis

1:18 PM

@robjohn It's OK. Indeed, it's pretty hard.

robjohn

@Chris'ssis I would be interested to see an answer if it exists

Chris's sis

@robjohn If there exists one, that will appear in my book.

My philosophy is this (it may sound pretty crazy): I'll publish my book only if I manage to create the nicest book on this topic from all times. Each page is meant to be absolutely amazing.

Algebra

Can someone read the comments under the answer here http://math.stackexchange.com/questions/927548/showing-that-a-generating-function-is-equivalent-to-some-fraction/927553?noredirect=1#comment1914748_927553

And tell me how we are miscommunicating?

Chris's sis

1:39 PM

@robjohn For this one I can tell you almost I'm almost 100% sure that there exists a nice form, but if we choose a certain type of $n$.

@robjohn math.stackexchange.com/questions/925092/…

@robjohn perhaps it's about the form $n=2k-1$

Algebra

@Chris'ssis I have $g(x)[1+8x] = \lim \limits_{n\to\infty} 1 - (-8)^{n+1}x^{n+1}$ for question of showing $a_k = (-8)^k$ for all integers $k\geq0$ is $\cfrac{1}{1+8x}$

You seem to know a thing or two about series. Am I allowed to touch the indeterminate $x$ at all? E.g. I want to say it is confined within $\frac{-1}{8}\lt x\lt\frac18$

Chris's sis

@Algebra I don't know what you mean by "Am I allowed to touch the indeterminate x at all? "

Algebra

Am I allowed so state it has to be inbetween those two bounds?

I got it to $g(x)[1+8x] = \lim \limits_{n\to\infty} 1 - (-8)^{n+1}x^{n+1}$ using that geometric $g(x)-xg(x)$ but now I have it only equalling $\frac{1}{1+8x}$ if that limit takes $-8^{n+1}x^{n+1}\to 0$

Which is only true for that interval of $x$

robjohn

1:56 PM

@Chris'ssis I will look at that, but I have to give an exam at UCLA today. I may not get much time until this later this afternoon.

Chris's sis

@Algebra You asked in a comment how is this step achieved $\displaystyle \sum \limits_{n=0}^\infty (-8x)^n$ to $\displaystyle \frac{1}{1+8 x}$. This is nothing but a a geometric series with some restrictions.

@Algebra do you know the geometric series?

Algebra

Yeah, but isn't that achieved by taking $g(x)=1+x+x^2+\dots+x^n$
$g(x)-xg(x)=1-x^n$, and then take $g(x)(1-x)=1-x^n$ goes to $\frac{1-x^n}{1-x}$

I tried that for this case, where I have $(-8)x$, and I get that limit problem. Maybe I will go back and try it all again

Chris's sis

@Algebra Choosing the proper bounds, that limit goes to $0$ as some already suggested in your answer, or do you refer at something else?

@robjohn could you take a look at @Algebra's question, I might misunderstand his point?

@robjohn OK

Algebra

I suggested that, but Snufsan says that " but it has no meaning mathematically" "The fact that we write it is just writing it, we dont care about it's mathematic meaning"

I will leave it at that, thank you for your efforts I will just keep at it

Will Hunting

Hi @Anastasiya-Romanova

Anastasiya-Romanova

2:08 PM

Hi @WillHunting! You're always kind to me

Will Hunting

@Anastasiya-Romanova Why did you change your username?

Chris's sis

@Algebra by the way, to be more exact, what is $$1+x+x^2+\dots+x^n$$?

anon

@Algebra the variable is an indeterminate, you don't bound it at all

one speaks of formal power series

Will Hunting

It is only in form.

Anastasiya-Romanova

@WillHunting Today I studied history about her

Algebra

2:09 PM

But then how does $g(x)[1+8x] = \lim \limits_{n\to\infty} 1 - (-8)^{n+1}x^{n+1}$ lead me to proving $g(x) = \frac{1}{1+8x}$

Will Hunting

@Anastasiya-Romanova Did you watch Good Will Hunting?

anon

@Algebra since (1+8x)(1-8x+64x^2-...) expanded equals 1, we know the latter is the inverse of 1+8x in the formal power series ring

by definition $x^n\to0$ as $n\to\infty$ in the formal power series ring (technically, using what's called the $(x)$-adic topology it has)

Anastasiya-Romanova

@WillHunting I become her fan now!

Algebra

Wait so $x^n \to 0$ as $n\to\infty$, doesn't that imply $0\lt x\lt\lt 1$? That was my confusion in the first place

Anastasiya-Romanova

@WillHunting Yes!

anon

2:12 PM

@Algebra $x$ is not a number

Anastasiya-Romanova

But parent didn't know. lol

Algebra

@Anon But is treated like a number being multiplied by itself $n$ times, and somehow equalling $0$, can you see why that is confusing to me?

anon

the product of $\sum a_nx^n$ and $\sum b_mx^m$ is defined to be $\sum(\sum_{n+m=k}a_nb_m)x^k$ in the formal power series ring

@Algebra everything in a ring is treated "like a number": you can add and multiply them

Will Hunting

I think formal power series ring does not make sense to him at all.

anon

that doesn't mean they are numbers

do you know what a ring is?

Will Hunting

2:14 PM

A ring is what you put on a finger at a wedding.

4

Algebra

Abelian group with two operations, commut etc

anon

have you heard of a polynomial ring?

Anastasiya-Romanova

@WillHunting Why my nickname is still V-Moy?

Algebra

Not that I recall

Will Hunting

@Anastasiya-Romanova What do you mean?

anon

2:15 PM

look it up

Anastasiya-Romanova

@Chris'ssis and @robjohn Please take a look my new problem: math.stackexchange.com/q/927427/133248

anon

F[x] is the ring of all F-coefficient polynomials in a formal variable x

F[[x]] is the ring of all formal power series in the formal variable x

Anastasiya-Romanova

@WillHunting In my PC my display name is V-Moy in the chat room

anon

by definition (for your purposes anyway) equipped with exactly the multiplication I defined above

Will Hunting

@Anastasiya-Romanova You may need to log out, log in and refresh your browser for changes to be effected.

anon

2:16 PM

you don't speak of limits in F[[x]] unless you know what the topology is, and you don't know what the topology is so I don't think you should even be talking about limits in it

read about it on wikipedia: http://en.wikipedia.org/wiki/Formal_power_series

Algebra

Alright, so at my level I can just write, x^n goes to zero here, and this guarentees $8^{n+1}$ goes to zero

anon

I would not even write that

Chris's sis

@robjohn I think the core of that problems gets reduced to computing

Anastasiya-Romanova

Okay, I see. Good @WillHunting

anon

8^(n+1) does not go to 0, it's just as close to 0 as 1 or 1/10000 is. it's the x^n that's going to 0.

Chris's sis

2:18 PM

@robjohn see above

Will Hunting

@Anastasiya-Romanova What is your favourite calculus book?

Chris's sis

@Anastasiya-Romanova OK

Anastasiya-Romanova

@WillHunting Calculus 9th Edition - Dale Varberg, Edwin Purcell, and Steve Rigdon

Will Hunting

@Anastasiya-Romanova Never heard of. I like Marsden and Weinstein's Calculus I, II and III.

Anastasiya-Romanova

@Chris'ssis Thank you. Don't forget to upvote. hohoho

Chris's sis

2:21 PM

Sometimes I write in English as if I'm drunk, although I almost never drink. Today is just one of those days.

Will Hunting

@Chris'ssis Alcohol tastes terrible.

Anastasiya-Romanova

@WillHunting Never heard of.

@WillHunting I also visit Paul's Online Math to learn calculus

Chris's sis

@WillHunting Is your plan to become a professor in the future? :-)

Will Hunting

@Chris'ssis Yes, I hope to find a job in some university in the US. I don't want to return to my country.

Chris's sis

@WillHunting Nice.

Hippalectryon

2:25 PM

Not nice :c

'I don't want to return to my country'

Will Hunting

@Chris'ssis You should be one too.

@Hippalectryon There is nothing here for me. I don't like the people here and they don't like me. I am too different.

Anastasiya-Romanova

Good night everyone! Zai jian @WillHunting!

Will Hunting

@Anastasiya-Romanova See you in your dreams!

Hippalectryon

@WillHunting Come to France :D

Chris's sis

@WillHunting I don't manage to get a job in the automotive industry, and then I can barely imagine such a career in some university. :D

(it's also true I put the correct price on myself)

Hippalectryon

2:27 PM

@Chris'ssis Uh ? a job in the automotive industry ?

Will Hunting

@Hippalectryon I hope to be born in Germany or France in my next life. I believe in rebirth.

Chris's sis

@Hippalectryon Yeah, that was my last try. :-)

Hippalectryon

@WillHunting Hence the constant accounts change :P

@Chris'ssis really ??

@Chris'ssis What part of the automotive industry exactly ?

Well, as long as you don't apply for a Comcast job, I guess it's okay :c

Chris's sis

@Hippalectryon steering wheels manufacturing

Algebra

Before I sleep I just wanted to say thank you to users, Anon, Chris's sis and others who helped, I will think more about it over the next few days and hopefully all will be well. Goodnight

Will Hunting

2:36 PM

@Algebra See you in your dreams!