TITLE: $p$ and $q$ not having $2$ as a common factor, thus $q$ must be an odd natural number?
QUESTION [2 upvotes]: So, I was looking at this proof of the irrationality of square root $2$ and came across this alternate proof. I am familiar with the common proof of the irrationality of square root 2 presented in most books.  $(p/q)^2 = 2$
Since $p^2 = 2q^2$, we see that $p^2$ is even. This implies that $p$ is also even.
Which is fine but then I got stumped by this - since $p$ and $q$ do not have 2 as a common factor, then $q$ must be an odd natural number. Can someone please explain this specific line? I thought we were looking at the square of $p$ equalling $2$ into the square of $q$. How does he/she find out q is an odd natural number.

REPLY [1 votes]: Since you have proved that p is even, p will have 2 as a factor. But since p and q don't share 2 as a common factor, q wont have 2 as a factor. If a number isn't divisible by 2, it is odd. Hence q is odd


TITLE: A basic question on the definition of order
QUESTION [2 upvotes]: In the first chapter of Rudin's analysis book "order" on a set is defined as follows :
Let $S$ be a set. An order on $S$ is a relation, denoted by $<$, with the following two properties :
(i) If $x \in S$ and $y \in S$ then one and only one of the statements
          $$ x < y, x=y, y<x $$ is true.
(ii) If $x,y,z \in S$, then $x < y$ and $y < z$ implies $x<z$.
How is this different from the usual partial/total order notation. This looks like total order. Why is defining "order" like this ? Moreover, he has not defined $=$ here.

REPLY [3 votes]: The root of your difficulty seems to be that people use different conventions for defining "order". The first issue arises in the defintion of "partial order", by which some people mean a reflexive, transitive, and antisymmetric relation $\leq$, while other people mean an irreflexive, transitive relation $<$.  Given a partial order on a set $A$ in either sense, one can easily define the corresponding order in the other sense, by adjoining or removing the pairs $(a,a)$ for all $a\in A$.  So people don't usually worry too much about the distinction, but technically they are two different notions of order.  I'm accustomed to calling the reflexive version a "partial order" and the irreflexive version a "strict partial order", but there are people who prefer to use the shorter name "partial order" for the irreflexive version.
Total (or linear) orders are then defined in the reflexive version by requiring $a\leq b$ or $b\leq a$ for all $a,b\in A$, and in the irreflexive version by instead requiring $a<b$ or $a=b$ or $b<a$.
Again, one can convert either sort of total order to the other sort, just as with partial orders.
To further increase the confusion, some people use "order" (without any adjective) to mean a partial order, while others use it to mean a total order.  So the final result is that "order" can have any of four meanings.  You just have to get used to that; you'll get no suport for denouncing Rudin just because he chose a convention different from the one you learned first.


TITLE: A uniformly continuous semigroup has a unique infinitesimal generator
QUESTION [2 upvotes]: I'm studying the theory of semigroups from Pazy's book. I'm struggling to understand the proof of a theorem stating that

Theorem 1.2. Let $X$ be a banach space. A linear operator $A:X\rightarrow X$ is the infinitesimal generator of a uniformly continuous semigroup iff $A$ is a bounded linear operator.

The proof of $\Leftarrow$ is pretty straight forward by setting
\begin{equation}
T(t)=e^{tA}:=\sum_{n=0}^{\infty}\frac{(tA)^n}{n!}.
\end{equation}
But I don't understand a part in $\Rightarrow$ proof, where we let $T(t)$ be a uniformly continuous semigroup of bounded linear operators on $X$ and some $\rho>0$ be such that $\|I-\rho^{-1}\int_0^\rho T(s)ds\|<1$ so that $\int_0^\rho T(s)ds$ is invertible. We can show that the infinitesimal generator $A:=\lim_{h\searrow 0}h^{-1}(T(h)-I)$ converges in $\mathcal{L}(X)$:
\begin{align}
h^{-1}(T(h)-I)\int_0^\rho T(s)ds = h^{-1}\bigg(\int_\rho^{\rho+h}T(s)ds-\int_0^h T(s)ds\bigg)
\end{align}
and therefore
\begin{equation}
h^{-1}(T(h)-I) = h^{-1}\bigg(\int_\rho^{\rho+h}T(s)ds-\int_0^h T(s)ds\bigg)\bigg(\int_0^\rho T(s)ds\bigg)^{-1}.
\end{equation}
We have $h^{-1}(T(h)-I)\rightarrow (T(\rho)-I)(\int_0^\rho T(s)ds)^{-1}$ in $\mathcal{L}(X)$ as $h\searrow 0$.
By definition of a semigroup, the infinitesimal generator of $T(t)$ is uniquely defined. But it seems to me that there are several ways to construct a different $\int_0^\rho T(s)ds$ by fixing $\rho>0$ at another value.
So my question: how can we show that $(T(\rho)-I)(\int_0^\rho T(s)ds)^{-1}$ is independent of $\rho$ described above?

REPLY [2 votes]: Realized I was basically writing an answer in comments, so here's the comments as an answer.
Note that $\rho$ appears in two places in your formula (in $T(\rho)$ and in the integral you constructed, to the $−1$ power); so it is not formally inconceivable that the $\rho$- dependence might "cancel out." Running the construction with $T(t)=e^{At}$ for bounded $A$ will show this more explicitly.
For example in the one dimensional case if $T(t) = e^{at}$ is the semigroup, and writing $r$ in place of $\rho$, then $\int_0^r T(s) \, ds = (e^{ar} - 1)/a$, so $(T(r)-1) * (\int_0^r T(s) ds)^{-1} = (e^{ar} - 1) * a/(e^{ar} - 1) = a$.
And of course in general, the very presence of $\rho$ in as a limit of something an operator limit involving only $T(h)$ and $h$ (which do not depend on $\rho$) also shows that the choice of $\rho$ won't matter, subject to the hypotheses. It's just uniqueness of the generator.


TITLE: How to test convergence for a tetration series slightly below the harmonic series?
QUESTION [1 upvotes]: I have the following series to test convergence
\begin{align}
S_{\infty}=
\sum_{n=1}^{\infty}
\dfrac{1}{n}
\left(
\dfrac{1}{n}
\right)^{
\left(
\dfrac{1}{n}
\right)
}
<
\sum_{n=1}^{\infty}
\dfrac{1}{n}
\ ,
\end{align}
however it has several annoying properties that make it difficult to test convergence such as its derivatives create $\ln(n)$ terms, its terms are non-integrable in terms of a known function, and its finite sum is slightly below the harmonic series.
How does one test convergence for a tetration series slightly below the harmonic series?

REPLY [1 votes]: Hint: $\displaystyle\lim_{n\to\infty}\sqrt[^n]n=1$.


TITLE: Is the canonical height of a totally p-adic point on an abelian variety bounded away from zero?
QUESTION [2 upvotes]: Inspired by the result of Schinzel and Smyth that a totally real number other than $0$ and $\pm 1$ has height at least $\frac{1}{2}\log \Big( \frac{1+\sqrt{5}}{2} \Big) = 0.240659\ldots$, Bombieri and Zannier discovered that totally $p$-adic algebraic numbers which are not roots of $1$ likewise have height bounded away from $0$. More precisely, the proved that the lim inf of the height over the totally p-adic algebraic numbers is between $\frac{1}{2}\frac{\log{p}}{p+1}$ and $\frac{\log{p}}{p-1}$; cf. Ch. 4.6 of the Bombieri-Gubler book (Heights in diophantine geometry). Both results (real and $p$-adic) can be explained as consequences of a Galois equidistribution property for algebraic numbers of small height.
Similarly, for the totally real points of an abelian variety, the Archimedean Galois equidistribution of Szpiro-Ullmo-Zhang implies (S. W. Zhang, Equidistribution of small points on abelian varieties, Corollary 2) that the height of a totally real (non-torsion) point on an abelian variety is bounded away from $0$.
Now, there are various non-Archimedean equidistribution results in the literature in the spirit of the Szpiro-Ullmo-Zhang theorem. Does any of them imply a lower bound on the height of a totally $p$-adic non-torsion point on an abelian variety? Is there anything otherwise published on this problem?
Added a little later. By W. Gubler's "tropical equidistribution theorem" (which is really a statement about the equidistribution of $p$-adic valuations mod 1), we do know that the answer is positive for abelian varieties which do not have potentially good reduction at some place above $p$ (that is to say, they acquire some $\mathbb{G}_m$-part at some place above $p$).
But what about the proper case?
An example. The level $N$ division field $K_N := \mathbb{Q}(A[N])$ of the abelian variety contains the level $N$ cyclotomic field $C_N = \mathbb{Q}(\mu_N)$, so a prime which splits completely in $K_N$ is congruent to $1 \mod{N}$, hence $> N$. Hence, for a given $p$, the abelian variety contains only finitely many totally p-adic torsion points.

REPLY [3 votes]: In my paper Mesures et équidistribution sur les espaces de Berkovich, Crelle, 2006,
I had proved an equidistribution theorem in the good reduction case. The limit measure is a Dirac mass at a single point of the Berkovich space (the one whose reduction is the generic point of the special fiber). Since the Galois orbit of a totally $p$-adic point is contained in a compact subset disjoint from that point, this implies that there exist a dense Zariski open subset $U$ and a strictly positive real number $c$ such that the Néron-Tate height of any totally $p$-adic point of $U$ is at least $c$.
For related results, see also the paper of Baker-Hsia (Crelle, 2005).


TITLE: Similar to Goldbach Conjecture
QUESTION [2 upvotes]: I have thought of a conjecture similar to Goldbach Conjecture. I have shown the result to be true with a program in C++ up until $n=30000$.

$\forall n>2$ with $n$ even,there exists two primes $p,q$ with $n<p,q<2n$ $\;$ and $\;$  $p+q=3n$

We know Betrand's postulate:

For any integer $n > 3$, there always exists at least one prime number $p$ with
  $n < p < 2n - 2$.

But I don't know if this postulate can help me.
I wanted to place it somewhere public so people can think about it.
Are there any results similar to this?

REPLY [1 votes]: Sure, of course this conjecture is similar to the Goldbach Conjecture.
GC: If $n>2$ is even then there are primes $p,q$ with $p+q=n$.
Yours: If $n>2$ is divisible by 6 then there are primes $p,q$with $p+q=n$ and $\frac{n}{3} < p,q < \frac{2n}{3}$.
I'm certain that the following strengthening of your conjecture is also true:
Yours++: if $n>2$ is even then there are primes $p,q$ with $p+q=n$ and $\frac{n}{3} < p,q < \frac{2n}{3}$.
The only difference is that I'm not requiring that $n$ is divisible by 6, only that it's even. Now, this stronger version is actually the same as GC except that it makes a further claim about the size of the primes.
We believe that not only GC is true, it is in fact true "in abundance": there are likely many, many pairs of primes that add up to any given even number. In particular, there's no reason to believe that such a pair cannot be found in the interval between one third and two thirds of n. Nevertheless, this is a stronger conjecture, and it is conceivable that we will prove GC without proving Yours++, and it's even possible (though extremely unlikely) that GC is true while Yours++ is false.
Now, your original conjecture (Yours) does not imply GC because it doesn't make any claim about even numbers that don't also happen to be divisible by 3. I find it very unlikely that this restriction makes any difference; in other words I would regard Yours and Yours++ as essentially equivalent.
The main question I would propose is - what do you expect would be the value of exploring this variation of GC? It's fine to suggest refinements or versions of open problems, but this is usually done with an eye towards making things clearer or more specific or easier to prove.
For example, if you had reason to believe that for large, even numbers, the only possible primes that add up to those numbers are in the middle-third interval you suggest, that would be quite sensational and an important thing to focus on. But I'm fairly certain this isn't true, so that for all even numbers there are sum-of-primes representations that lie all over the place. Therefore I'm afraid it seems unlikely that these variations would help in the quest of resolving GC itself.


TITLE: Conservation of momentum 1-D problem: track is not level
QUESTION [0 upvotes]: This problem is a standard 1-D problem. You have 2 equal carts approaching each other and collide on linear air track.
BUT the person setting up the track did NOT do a good job at making the track Level(or NOT perfectly horizontal).
THE QUESTION is : HOW "UNLEVEL" is the track?
The collision took t= 0.750 seconds.
And the data collected is below:
DATA

m1=m2= 1kg
  Initial velocities:
  v1i= +0.450m/s
  v2i= -0.505m/s
  Final velocities:
  v1f= -0.510m/s
  v2f= +0.420m/s

From the data we can easily calculate the momentum before and after the collision and
see that the momentum of this system is NOT Conserved.
To me this makes sense because the track is unlevel, we have made the track to be like
an inclined plane which means that gravity now has influence on this system!
SO we have now gone into 2-D space, and we have acceleration happening due to gravity.
I was able to figure out this much.
BUT how do I calculate, based on the given information, say the angle that the track makes with respect to the horizontal.(This is what I guess the question is asking).
BUT, I also don't understand how they can measure the velocities(in the data given) if
we have acceleration, which means that the velocity is changing(velocity is no longer constant).
SO I see some contradictions here in this question.
I found this question in an old book, and there are no answers to the questions.
BUT this one got me interested.
Hope somebody has seen this kind of question before.
Your help is greatly appreciated.
Palu

REPLY [1 votes]: Variation of momentum means force: $F = \dot p$. In a very non-rigorous way, we have:
$$
F = \frac{dp}{dt} \quad\Longrightarrow\quad
Fdt = dp \quad\Longrightarrow\quad
F = \frac{\Delta p}{\Delta t}
$$
The force in this case, it is likely assumed to be the gravitational force: $F = mg\sin\theta$, where $\theta$ is the angle of the ladder and the horizontal axis, and $\Delta t$ is the time taken by the collision. Hence:
$$
\sin\theta = \frac{\Delta p}{\Delta t}\frac{1}{mg}
$$