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.)
@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).
@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.
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?
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.)
@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.
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
@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 $\frac{-1}{8}\lt x\lt\frac18$
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 $-8^{n+1}x^{n+1}\to 0$
@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.
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
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
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
@BalarkaSen Ah. I wondered about the union ;) Yes, it's okay. You can always have branches on the complement. If you pick a sensible path, the complement is simply connected, which makes it easy to define a branch.
@MikeMiller Well, but in that case, it's also homeomorphic to $S^n\times S^n$, and to $P^n\times P^n$, so that base case isn't too safe to base assumptions on here ;)
Let $a,b\in\mathbb{R}$, with $a<b$. Assume $a>0$. By the archimedian principle there is an $n \in \mathbb{N}$ such that $n > (b-a)^{-1}$. Note that $nb>nb-a>1+na-a\geq 1$; so $nb>1$
@N3buchadnezzar If $a, b$ are integers such that $b > a$ then there exists a $k \in \Bbb N$ such that there is a power of $2$ sitting inside $(a^{2^k}, b^{2^k})$. yep, that's what I meant.
A piece of string walks into a bar and orders a drink The barman says, "Sorry, we don't serve string" The piece of string leaves, disappointed. Just outside the bar, the piece of string is mugged, leaving it disheveled and somewhat tied up. It decides to go back into the bar. The barman asks, "are you a piece of string?" It replies, "I'm a frayed knot".
@rehband Sorry, I can't now. This one won't be seen until I publish the book. I also communicate some problems to the others once in a while, but the special ones are meant for my book only. :-)
@N3buchadnezzar If $a, b$ are integers such that $b > a$ then there exists a $k \in \Bbb N$ such that there is a power of $2$ sitting inside $(a^{2^k}, b^{2^k})$. yep, that's what I meant.
Lucas pseudoprimes and Fibonacci pseudoprimes are composite integers that pass certain tests which all primes and very few composite numbers pass: in this case, criteria relative to some Lucas sequence.
== Basic properties ==
Given integers P and Q, where P > 0 and , let Uk(P, Q) and Vk(P, Q) be the corresponding Lucas sequences.
Let n be a positive integer and let be the Jacobi symbol. We define
If n is a prime such that the greatest common divisor of n and Q (that is, GCD(n, Q)) is 1, then the following congruence condition holds (see page 1391 of ):
If this equation does not hold, then n...
Riemann zeta waves are also random, I would like to apply this waveform to that hypothesis, fully understanding that if i started doing math now, i wouldn't fully understand 10% of the concepts after 10 year. That's why i'm here