TITLE: How to prove or disprove that $\mathbb{Z}_p[x]$ have unique factorization theorem?
QUESTION [0 upvotes]: How to prove or disprove that $\mathbb{Z}_p[x]$ have unique factorization theorem, where $p$ is a prime number.
My idea is to first show that every polynomial in $\mathbb{Z}_p[x]$ can be factored as $a(x-x_1)(x-x_2)\dots(x-x_n)$. Showing it is true in degree $0$ (basically UFT in integers), and then prove using induction for higher degree.
However, I seemed failed to show that $(x-x_1)$ is irreducible.
For example $x-4 \equiv 2(3x-2)$ in $\mathbb{Z}_5[x]$.

REPLY [0 votes]: Hint: Given any field $k,$ $k[x]$ has UFT. Here is an outline of a proof.

Define the gcd of two polynomials.
Define division of polynomials.
Prove Bezout's lemma (if $\gcd(f, g) = 1,$ then there are polynomials $a, b$ so that $af + bg = 1$).
Using Bezout, prove unique factorization into primes.

This is analogous to integer unique factorization.


TITLE: Reflexivity: How can something be related to itself?
QUESTION [1 upvotes]: Background: I'm a philosophy student. I'm comfortable with math, but don't have much of a background in it. One of the topics I'm writing about (I-relation in theories of identity) closely mirrors concepts in math. One of those is reflexivity.
My attempt to answer the question:
I read the Reflexivity article on Wikipedia, but I'm still foggy on the idea. I get that 1=1 is reflexive, and 1<2 is not. I understand that 1=1 is relating one to one, but it seems so redundant that I can't imagine it being often used in math - but I believe that it is - so I must be missing something. Also, I read that the 'divides' relationship (2 divides 4) is reflexive. I don't see how that is reflexive.
Question

How can something be related to itself?
Why is reflexivity a useful concept?
How is the divides relationship reflexive?

An abstract explanation and concrete example would be helpful.
Thank you.
-Hal

REPLY [3 votes]: A name is needed for this property whether it is intuitive or not, since many relations happen to satisfy it, the need to prove or use the property often comes up, and there is a similarity between these use cases that makes it worth abstracting as a concept of its own.
It helps to have a uniform treatment of different cases ("consider any line parallel to $L$"), rather than constantly making annoying exceptions ("let $L'$ be any line parallel to $L$, or equal to $L$").   Or "two lines are parallel if and only if they have the same slope" versus "...same slope and different height". If you had defined the relation of parallelism to exclude the case of self-parallel lines, you would find that all these sentences that explicity include it as an additional case are really talking about a different equivalence relation, parallel-or-equal, and introducing a name or notation for the new relation simplifies so many statements that it is easier to use the more uniform thing as the definition of parallel. [This is similar to allowing an 'empty set' and applying definitions like 'number of functions from set $A$ to set $B$' to that case whenever they make sense.]
You might be proving that $x \sim y$ for a whole set of $y$ at once, without knowing whether the set contains $x$ or not.  It would not make sense to introduce artificial distinctions about hypothetical possibilities just to match a language preference that relations be between different objects.
A more basic difficulty, that actually happens in mathematics, is that sometimes you can prove all kinds of relations between a specific pair of things $a$ and $b$, without knowing that they are equal.  Maybe you will eventually identify enough similarities to prove $a = b$, but that should not invalidate the earlier proofs that $aRb$ .
For some relations it is paradoxical to exclude self-relatedness. "Near", "as tall as", "as fat as", etc.
In many situations you want to be able to substitute $b$ for $a$ in a formula or statement (preserving its validity) when they satisfy a particular relation.  It disrupts this to not be able to use $a$ more than once.
Generally the tendency to be inclusive rather than exclusive about trivial cases (such as "motions" of a space including the one thats keep every point in place, empty sets, digraphs without edges) is forced by the need to have the special cases handled automatically and uniformly by the language when dealing with more and more complicated situations.


TITLE: Is the map $\pi$ continuous?
QUESTION [1 upvotes]: Consider $\mathbb R$ endowed with its usual topology $\mathcal{T}_{\mathbb R}$ and let $\mathbb R\cup \{\infty\}$ be endowed with the following topology:
$$\mathcal{T}=\{\emptyset\}\cup \{\mathcal{U}\cup\{\infty\}: \mathcal{U}\in\mathcal{T}_{\mathbb R}\}.$$
Consider the subespace $X=\mathbb R\times \{0\}\cup \mathbb R\times (0, +\infty)$ of $\mathbb R^2$ with the induced topology. Define $\pi: X\rightarrow \mathbb R\cup\{\infty\}$ setting:
$$\pi(x, y)=\left\{\begin{array}{clc}
\infty & \textrm{se}& y\in (0, +\infty)\\
x & \textrm{se}& y=0
\end{array}\right..$$
Is it true that $\pi$ is continuous?
Notice that if $\mathcal{U}$ is open in $\mathcal{T}$ then $\mathcal{U}=\mathcal{V}\cup\{\infty\}$ where $\mathcal{V}\in\mathcal{T}_{\mathbb R}$. Then:
$$\pi^{-1}(\mathcal{U})=\pi^{-1}(\mathcal{V})\cup \pi^{-1}(\{\infty\})=\pi^{-1}(\mathcal{V})\cup \mathbb R\times (0, +\infty),$$ hence it suffices to show $\pi^{-1}(\mathcal{V})$ is open in $X$. Here lies the problem, because:
$$\pi^{-1}(\mathcal{V})=\mathcal{V}\times \{0\}$$ and this does not seem open in $X$, right?
Did I make any mistake or $\pi$ is not continuous?
Thanks.

REPLY [1 votes]: Suppose that $\varnothing\ne U\in\mathcal{T}$. As you say, $U=\{\infty\}\cup V$ for some $V\in\mathcal{T}_{\Bbb R}$, so
$$\pi^{-1}[U]=\big(V\times\{0\}\big)\cup\big(\Bbb R\times(0,\to)\big)\,,$$
but this actually is open in $X$: $X=\Bbb R\times[0,\to)$, and
$$\pi^{-1}[U]=\big(V\times[0,\to)\big)\cup\big(\Bbb R\times(0,\to)\big)\,,$$
which is an open subset of $\Bbb R\times[0,\to)$, because it’s the intersection with $X$ of $(V\times\Bbb R)\cup\big(\Bbb R\times(0,\to)\big)$, an open set in $\Bbb R^2$.


TITLE: Direction of displacement
QUESTION [1 upvotes]: Displacement is defined as the vector obtained by joining the final position to the initial position (head towards the final position). Well,i know this is silly but what are these final and initial positions?Common sense says that initial position must be the one from which we started observing the motion and the final position should be the one where we are asked to end our observation as per the problem and it works almost everywhere except in SHM. Consider a body performing SHM about it's mean position O. Say, one of it's extreme positions is E (somewhere on positive x axis) and M be a point somewhere in between O and E. Now what is the direction of displacement as the body moves from E to M? Using my 'common sense defintions of final and initial position' displacement must be in negative x direction but i was told that this is not the case (it is towards positive x) because the mean position is taken to be the initial position no matter where the body starts from. So my question is how do we know what are the initial and final positions? Are there any conventions?

REPLY [0 votes]: Instead of initial and final position, I think it's better to think of reference position and current position. You're free to choose any reference position that suits your taste, and then you measure the current position relative to that. Essentially, you're choosing the origin of your coordinate system.
For a harmonic oscillator, choosing its resting or mean position is sensible, because that point is meaningful to the problem at hand. But we could have some fun and decide to use a different reference point, say, one that is $1$ unit to the left and $1$ unit above the resting position. Then the displacement of an oscillator oscillating vertically would be
$$\vec r=\begin{pmatrix}1\\0\\-1+A\cos(\omega t+\varphi_0).\end{pmatrix}$$
But it's a lot easier on the eyes to choose the resting point as a reference, because then it's just $(0,0,A\cos(\omega t+\varphi_0))$. But displacement doesn't have to be relative to some physically meaningful point. They're just more convenient.


TITLE: Find solution of differential equation $y'(t)=-2y(t)+1$
QUESTION [1 upvotes]: Could you help me explain how to find the solution of the differential equation
$$
y'(t)=-2y(t)+1,
$$
with
$$y(0)=1.$$
I know that the solution is
$$y(t)=\frac12 (1+e^{-2t}).$$
How about the IVP
$$y'(t)=1,\quad y(0)=1?$$

REPLY [1 votes]: This initial value problem has exactly one solution: If $y=y(t)$ is the solution, then
$$
\mathrm{e}^{2t}y'(t)=-2\mathrm{e}^{2t}y(t)+\mathrm{e}^{2t},
$$
or
$$
\big(\mathrm{e}^{2t}y(t)\big)'=\mathrm{e}^{2t}y'(t)+2\mathrm{e}^{2t}y(t)=\mathrm{e}^{2t}
=\left(\frac{1}{2}\mathrm{e}^{2t}\right)',
$$
or equivalently
$$
\mathrm{e}^{2t}y(t)=\frac{1}{2}\mathrm{e}^{2t}+c,
$$
where $c$ is some constant, and due to the fact that $y(0)=1$, the value of $c$ has to be $c=1/2$. Thus
$$
y(t)=\frac{1}{2}+\frac{1}{2}\mathrm{e}^{-2t}.
$$
Note. The procedure followed above establishes uniqueness as well, for if $z=z(t)$ were another solution, then following for $z$ the same procedure we would reach to the same answer.
Edit. For the IVP
$$
y'=1,\quad y(0)=1,
$$
the unique solution is $y(t)=t+1$.


TITLE: uniform convergence of $\sum\limits_{n=1}^{\infty}3^n\sin\left(\frac 1 {4^nx}\right)$ in $[1,\infty)$
QUESTION [2 upvotes]: I want to check the uniform convergence of $\sum\limits_{n=1}^{\infty}3^n\sin\left(\frac 1 {4^nx}\right)$ in $[1,\infty)$. The hint in the book was using Cauchy condition for uniform convergence. For questions of proving a series doesn't converge uniformally, we can take x values s.t the sum is bigger than epsilon.$\forall\epsilon>0 $ it's clear I need to find $N\in\mathbb N$ s.t $\forall n>N,p\in\mathbb N,x\in[1,\infty)$ the sum $$\left|\sum_{k=n+1}^{n+p}3^k\sin\left(\frac 1 {4^kx}\right)\right|<\epsilon.$$ Which $N$ can I choose?

REPLY [5 votes]: This follows immediately from geometric series and from the fact that $|\sin y|\leq |y|$ for every $y\in\mathbb{R}$. Which can be shown by the MVT, for instance.
Added: so it follows from the Weierstrass M-test. Or directly, from where you stopped with the Cauchy criterion:
$$\Big|\sum_{k=n+1}^{n+p}3^k\sin \left(\frac{1}{4^kx}\right)\Big|\leq \sum_{k=n+1}^{n+p}\left(\frac{3}{4}\right)^k=\frac{\left(\frac{3}{4}\right)^{n+1}-\left(\frac{3}{4}\right)^{n+p+1}}{1-\frac{3}{4}}\leq 4\left(\frac{3}{4}\right)^{n+1}$$
for every $|x|\geq 1$. I think you can see how to choose $N$ now.


TITLE: Reference request for limit representation of Riemann zeta function
QUESTION [0 upvotes]: I'm looking for more references for the following limit representation of the Riemann zeta function which I found on Wolfram's web site. Thanks in advance.
$$\zeta (s)=\lim_{n\to \infty } \, \left(\sum _{k=1}^n k^{-s}-\frac{n^{1-s}-1}{1-s}\right)-\frac{1}{1-s},\quad\Re(s)>0.$$

REPLY [1 votes]: $$n^{-s} - \int_n^{n+1} x^{-s}dx = \int_n^{n+1} \int_n^x s t^{-s-1}dtdx = \mathcal{O}(n^{-s-1})$$
So that
$$\sum_{n=1}^\infty (n^{-s}-\int_n^{n+1} x^{-s}dx) = \lim_{N \to \infty} \sum_{n=1}^N n^{-s}-\int_1^{N+1} x^{-s}dx=\lim_{N \to \infty} \sum_{n=1}^N n^{-s}-\frac{1-(N+1)^{1-s}}{s-1}$$
converges and is analytic for $Re(s) > 0$.
Clearly for $Re(s) > 1$ : $\lim_{N \to \infty} \sum_{n=1}^N n^{-s}-\frac{1-(N+1)^{1-s}}{s-1} = \zeta(s)-\frac{1}{s-1}$ and  by analytic continuation this stays true for $Re(s) > 0$.