@user112495 Just to give a second approach here. If $f(z) = \sum a_n z^n$ is your function, Daniel was suggesting you look at $g(z) = f(z) - a_0 - a_1 z$.
Then $g(z)/z^2$ is again a holomorphic function. You should verify that, because of your hypothesis on $f$, $g(z)/z^2$ is a bounded entire function that goes to $0$ as $z \to \infty$...
Right, @user112495 - you know it's a constant, and you also know that it goes to 0 at infinity... and there's only one constant that goes to 0 as we go off to infinity.
@DanielFischer So since $x_k \in (\mathbb{Z}/p^{k+1}\mathbb{Z})^{\star}$ it implies that there is a $y_k \in \mathbb{Z}/p^{k+1}\mathbb{Z}$ such that $xy=1$, right?
@evinda At the moment, we're only at $x_k y_k \equiv 1 \pmod{p^{k+1}}$, or $\overline{x_k}\cdot\overline{y_k} = \overline{1}\in \mathbb{Z}/p^{k+1}\mathbb{Z}$. But we have that for all $k$, so it remains to see that $y = (\overline{y_0},\overline{y_1},\overline{y_3},\dotsc) \in \mathbb{Z}_p$.
@DanielFischer So, is it better to write it like that: $x=(\overline{x_0}, \overline{x_1}, \overline{x_2}, \dots)$ or like that $x=(x_0, x_1,x_2, \dots)$ ? Alao, does it hold $y_k \in \mathbb{Z}/p^{k+1}\mathbb{Z}$ or $y_k \in (\mathbb{Z}/p^{k+1}\mathbb{Z})^{\star}$?
Also, why does it remain to see that $y = (\overline{y_0},\overline{y_1},\overline{y_3},\dotsc) \in \mathbb{Z}_p$ ?
@evinda How you write them is unimportant. What matters is that one somehow distinguishes between elements of $\mathbb{Z}$ and elements of $\mathbb{Z}/n\mathbb{Z}$ when using both. And I made an unfortunate choice above, since you used $x_k,y_k$ to denote elements of $\mathbb{Z}/p^{k+1}\mathbb{Z}$ just before. So let's keep the simpler notation without overlines for the components of $x\in \mathbb{Z}_p$.
Since $y_k$ was chosen so that $x_k\cdot y_k = 1$ in $\mathbb{Z}/p^{k+1}\mathbb{Z}$, it follows that $y_k \in (\mathbb{Z}/p^{k+1}\mathbb{Z})^\ast$. But $(\mathbb{Z}/p^{k+1}\mathbb{Z})^\ast\subset \mathbb{Z}/p^{k+1}\mathbb{Z}$. And we need to check that the $y_k$ satisfy the compatibility relations to conclude that $y\in\mathbb{Z}_p$.
@MikeMiller Don't, just don't ;) No, sorry, I have fortunately forgotten what I once knew about nonlinear PDEs.
@DanielFischer We know that $x_{k+1}y_{k+1} \equiv x_ky_k \mod{p^{k+1}}$ and $x_ky_k \equiv 1 \mod{p^{k+1}}$. So, we conclude that $x_{k+1}y_{k+1} \equiv 1 \mod{p^{k+1}}$. Which other relations could we use?
@evinda We can. For we also have $x_k y_k \equiv 1 \pmod{p^{k+1}}$, so $x_k y_{k+1} \equiv x_k y_k \pmod{p^{k+1}}$, and hence $x_k(y_{k+1} - y_k) \equiv 0 \pmod{p^{k+1}}$.
@DanielFischer I see :) But why have shown in that way that $y \in \mathbb{Z}_p$? We have shown that $y_{k+1} \equiv y_k \mod{p^{k+1}}, \forall k \in \mathbb{N}$, right? But doesn't it need to hold also that $y_k \in \mathbb{Z}/p^{k+1}\mathbb{Z}$ ?
@evinda Have we already shown that $y_{k+1} \equiv y_k \pmod{p^{k+1}}$? Or is there still a little to do? Concerning the latter, we have $y_k \in \mathbb{Z}/p^{k+1}\mathbb{Z}$ by construction.
@DanielFischer I thought so, because $x_k(y_{k+1} - y_k) \equiv 0\pmod{p^{k+1}}$ and since $x_k \in (\mathbb{Z}/p^{k+1}\mathbb{Z})^{\star}$, it has to hold that $y_{k+1} - y_k \equiv 0 \mod{p^{k+1}}$... :/
@DanielFischer From which relation do we have that $y_k \in \mathbb{Z}/p^{k+1}\mathbb{Z}$?
@evinda No relation, we chose $y_k$ to be that element of $\mathbb{Z}/p^{k+1}\mathbb{Z}$ such that $x_k\cdot y_k = 1$ in $\mathbb{Z}/p^{k+1}\mathbb{Z}$.
@evinda That follows directly from $y_{k+1}\equiv y_k \pmod{p^{k+1}}$.
@DanielFischer So since it holds that $y_k \in \mathbb{Z}/p^{k+1}\mathbb{Z}$ and $y_{k+1}\equiv y_k \pmod{p^{k+1}}$ $\forall k \in \mathbb{N}$ we conclude that $y \in \mathbb{Z}_p$, right? :)
@DanielFischer To show that the units of $\mathbb{Z}_p$ are $\mathbb{Z}_p^{\star}=\mathbb{Z}_p \setminus p \mathbb{Z}_p$, i.e. the powerseries with constant term different from 0.
@DanielFischer We took a $x \in p\mathbb{Z}_p$ and showed that it cannot be a unit... Then, we took a $x \in \mathbb{Z}_p \setminus p\mathbb{Z}_p$ and we showed that we can find a $y$ such that $xy=1$, so each element in $ \mathbb{Z}_p \setminus p\mathbb{Z}_p$ is a unit, right?
I have a question where I implementing a math formula in excel which is not producing the expected result, is such a question best asked on Mathematics forum or Stack Overflow?
@DanielFischer Isn't the other definition that if $x \in \mathbb{Z}_p$ then it can be written as a sum of powers of $p$: $\sum_{n=0}^{\infty} a_np^n, a_n \in \{0,1,2, \dots, p-1\}$ ?
Hi. If f is continuous on the complex plane, and for each z, we have have f(z+2)=f(z) and |f(z)| <= exp(pi*|z|), then I want to show |f(z)| <= c*exp(pi*|Im z|) for some constant. Any suggestions on where to begin?
@Behaviour: If you click on the unanswered tab and then select 'my tags', it sorts them by date. Do you know if there is a way to get them sorted by votes? You seem to know these sort of things.
@MichaelAlbanese There is no way to sort them on that tab. In case of questions with 0 answers, you can search for answers:0 intags:mine to get a list that can be sorted in any way. For questions with un-upvoted answers one needs SEDE.
@MikeMiller I saw it. Did not vote on it. I guess it's in scope. I thought of referring the user to Computer Science, but they would likely close it as an abstract duplicate of this
So I picked up a reprint of the first edition of Coolidge's Treatise on Algebraic Plane Curves.
But there's some weird terminology in the very beginning, that I can't find used anywhere else, at least on the Internet.
"If the polynomial have an infinite root, that is to say, if the function $x^n f \left(\frac{1}{x}\right)$ have a root 0, then $a_0 = 0$ and conversely."
That's actually a good starting exercise @beginner. Assuming you've encountered the terminology, can you show that all finite integral domains are fields?
@beginner Well, the author defined what a ring was.
Then he says "let's call rings fields."
That's suspicious.
An altogether different matter is something like "Suppose A is a ring, and moreover suppose it has property X. Then it is a field." For example, Artinian commutative domains are fields.
So a field with n elements has characteristic n, since 1*n=0? Else it is characteristic 0 if this isn't possible? Why would this ever not be possible? If you don't have the element 1?
It's all relative. I look at some of the algebra quals out there and I'm definitely humbled. Ultimately I guess what matters is that we are constantly improving.
I'm actually curious why they say "characteristic 0" and not "characteristic $\infty$"
@MichaelAlbanese For a couple of those questions, a compilation of comments made a decent CW answer. Consider this as one of the options, when comments are of value. A closed question with no answers may well be deleted -- either by user votes, or through an expanded auto-deletion process that SE might implement in the future.
I suppose for the first one, my motivation to close was the question itself which I think is fairly broad; do you disagree? However, this does not take into account the useful contributions made in the comments.
On one hand yes, it's too broad. But the threat of "too many possible answers" has not materialized. And while it'd be good to narrow the question down, the OP is long gone, so we are left with what's there.
@MichaelAlbanese There isn't an argument against closing, except the extra time spent reviewing and voting; compare to posting a CW answer (5 user actions vs 1). A matter of convenience mostly.
Hi. I want to install MathJax on my website. I have read that there are two possible ways to do this. The first uses the CDN and the second is manually installed the files onto my server. Which method is best and why?
Let $f(x)$ be an irreducible integer polynomial of degree $k$.
Let $x_1,x_2,...,x_j$ be some zero's of $f(x)=0$ where $j<k$.
How do I find identities of type $P(x_1,x_2,...,x_j) = 0$ where $P$ is an integer polynomial that has at most degree $m<k$ in every variable $x_1,x_2,...x_j$ ?
Im not an...
a major problem students seem to have is that if they don't know what to do they simply don't do anything at all, as if you don't have permission to do anything unless you know what needs to be done first. this is the wrong mindset, students aren't just workers, they can be explorers, albeit with guidance.
Quantum mechanics (QM; also known as quantum physics, or quantum theory) is a fundamental branch of physics which deals with physical phenomena at nanoscopic scales, where the action is on the order of the Planck constant.
$\left( \left| a_n \right| \right) \to a$ if for each $\epsilon > 0, \ \exists N \in \mathbb{N}$ s.t. $\left| |a_n| - a \right| < \epsilon$ for all $n > N$.
@Mike so what we'd have to prove is that given CW complexes $X = \{D_i, f_i\}$ and $Y = \{D_i, g_i\}$, the CW complex obtained from $\{D_i \times D_j, f_i \times f_j\}$ is precisely $X \times Y$, with the topology and all.
looks scary.
@Alizter Mike says the nonsense stuff you did on $S^3$ is equivalent to the Hopf fibration.
now idea how that's true.
:P
can you recite me what you did? I'll try to figure out.
My definition of a Hopf fibration is a fiber bundle $S^3 \to S^2$ with fibers $S^1$.
yeah i can't see how these two are equivalent
but let's think.
@Alizter the actual problem (roughly) is to produce a map from $S^3 \stackrel{p}{\to} S^2$ such that for any point in $S^2$, there is a neighborhood $U$ around it such that $p^{-1}(U)$ (the fibers) look like $S^1 \times S^2$.
in less rigorous terms, you have to find a way to stick a circle at each point of a sphere such that locally it looks like S^1 cross S^2 but globally S^3.
Let $f(x)$ be an irreducible integer polynomial of degree $k$.
Let $x_1,x_2,...,x_j$ be some zero's of $f(x)=0$ where $j<k$.
How do I find identities of type $P(x_1,x_2,...,x_j) = 0$ where $P$ is an integer polynomial that has at most degree $m<k$ in every variable $x_1,x_2,...x_j$ ?
Im not an...
