TITLE: How to show that a one-dimensional subspace of $(\mathbb{Z}/2\mathbb{Z})^3$ is contained in exactly three two-dimensional subspaces?
QUESTION [0 upvotes]: Let $\mathbb{Z}/2\mathbb{Z}=\{0, 1\}$ be the two element field. We know that there are 8 vectors in $(\mathbb{Z}/2\mathbb{Z})^3$ given by $(a, b, c)$, $a, b, c \in \{0, 1\}$. There are $(8-1)/(2-1)=7$ one dimensional subspaces of $(\mathbb{Z}/2\mathbb{Z})^3$. The number of the two dimensional subspaces of $(\mathbb{Z}/2\mathbb{Z})^3$ is also 7 by duality of one dimensional subspaces and two dimensional subspaces.
How to show that a one-dimensional subspace  of $(\mathbb{Z}/2\mathbb{Z})^3$ is contained in exactly three two-dimensional subspaces of $(\mathbb{Z}/2\mathbb{Z})^3$? Thank you very much.

REPLY [1 votes]: A hint: It is enough to consider the one-dimensional subspace $\langle(1,0,0)\rangle$.


TITLE: Maximization of a certain ratio for a concave positive function
QUESTION [1 upvotes]: Assume $g : (0, 1) \to (0, +\infty)$ is a concave twice continuously differentiable function.
We want to make the ratio
$$
f(x) := -\frac{g''(x)}{g(x)}
$$
grow as $x \to 0$ as fast as possible. Some examples:
$$
g(x) = x^p,\ p \in (0, 1),\ \mbox{and}\ f(x) = \frac{p(1 - p)}{x^2}.
$$
This is maximal when $p = 1/2$, then $f(x) = 1/(4x^2)$. Also,
$$
g(x) = -x\log x\ \mbox{and}\ f(x) = -\frac1{x^2\log x},
$$
which does not grow as fast as $1/(4x^2)$ as $x \to 0$.

REPLY [1 votes]: Sure. Just do the most natural things.
Assume that you can have $g''<-(\frac 14+\delta)x^{-2}g(x)$ on a short interval starting at $0$ with some $\delta>0$. Note that the inequality is invariant under scalings $g(x)\mapsto ag(bx)$, so you can stretch the interval as much as you want and normalize to $g(1)=1$. Now use the compactness of concave functions normalized in such way (harnessing them with a fixed multiplicative convolution to avoid any issues with the differentiablity of the limit if you do not feel like working with generalized derivatives at the moment) and get a solution of the same differential inequality on the whole positive semi-axis.
Since you already suspect that $\sqrt x$ is the worst you can have, write $g(x)=u(x)\sqrt x$, differentiate honestly, and arrive at the inequality
$$
u''+x^{-1}u'+\delta x^{-2}u\le 0\\,.
$$
Now rewrite it as
$$
(xu')'+\delta x^{-1}u\le 0
$$
Since $u>0$, this means at the very least that $xu'$ is decreasing. If $u'$ gets negative anywhere, then you get $u'(x)\le -\frac cx$ at infinity. But then the integral of $u'$ diverges to $-\infty$, so $u$ gets negative itself somewhere, which is impossible. Thus $u'\ge 0$ all the way. But then $u\ge c$, so $(xu')'\le -\frac {\delta c}x$, whence (using the divergence of the harmonic integral again) $xu'$ tends to $-\infty$ making $u'$ eventually negative. Thus, "Kuda ne kin', vezde klin" (I surmise you understand Russian).
Of course, you can achieve $\frac{g''}g<-\frac{1}{4x^2}$ by considering $g(x)=\sqrt x-x^2$ on $(0,1)$ or something like that. So, strictly speaking, $\frac{1}{4x^2}$ is unbeatable only asymptotically, but I assume that this is what you meant from the beginning when asking the question.
I should confess that you arose my curiosity by claiming that you need a solution urgently and reviving an old post of mine to attract my attention. Normally you should understand that people visit MO in their free time and have no obligations whatsoever as to how fast to respond or whether to respond at all. So, I naturally wonder why it couldn't wait for a few hours or days :).


TITLE: There is no conformal mapping from $\mathbb C\setminus\{0\}$ to $\mathbb C\setminus \overline D$
QUESTION [1 upvotes]: I struggle solving the following exercise:

Are the sets $U := \{z\in \mathbb C: |z| > 1\}$ and $V := \mathbb C\setminus \{0\}$ conformally equivalent?

Now I am pretty sure the answer is "no". My try:
Assume there existed a biholomorphism $g:V \to U$.
$U$ is conformally equivalent to $D' :=D\setminus \{0\}:= \{z\in \mathbb C : \lvert z|< 1, z \neq 0\}$ (for example via the map $z \mapsto 1/z$), so it suffices to show that $g: V \to D'$ can't be a conformal mapping.
To this end, distinguish between the following cases:
Case 1: $0$ is a removable singularity of $g$. It follows easily by Liouvilles theorem that then $g$ has to be constant, a contradiction.
Case 2: $0$ is an essential singularity of $g$. As an easy application of the Casorati-Weierstraß theorem one sees that $g$ can't be injective near $0$, a contradiction.
Case 3: Here's where I am stuck. I don't know how to get to a contradiction here and hope someone of you can give me a hint (or correct my proof to this point if it has flaws).

REPLY [2 votes]: Hint: $1/z$ is a bounded analytic function on $U$.  Is there a non-constant bounded analytic function on $\mathbb C \backslash \{0\}$?


TITLE: Limit of floor function and sine function
QUESTION [0 upvotes]: for $$\lim_{x\to k}\lfloor x\rfloor \sin\frac{π x}2$$
find the limit for $k=0,1,2,3$
i started with
$$x-1<\lfloor x\rfloor\le x$$
$$\Downarrow$$
$$x\sin\frac{π x}2-\sin\frac{π x}2<\lfloor x\rfloor \sin\frac{π x}2\le x\sin\frac{π x}2$$
for $k= 0$ and for $k=2$
$$\lim\limits_{x\to k}x\sin\frac{π x}2-\sin\frac{π x}2=\lim\limits_{x\to k} x\sin\frac{π x}2= 0$$
and because of squeeze theorem $$\lim\limits_{x\to 0}\lfloor x\rfloor \sin\frac{π x}2=0$$
for $k=1$ and $k=3$ $$\lim\limits_{x\to k}x\sin\frac{π x}2-\sin\frac{π x}2=0$$ and
$$\lim\limits_{x\to k} x\sin\frac{π x}2= 1 , -1$$
so it means there is no limit at all? can i use sqeeze therom on the opposit and say that it represent the limit on the right and the limit on the left?

REPLY [2 votes]: You can sometimes use the Squeeze Theorem to show that a limit does exist, but never to show that a limit does not exist.  So your solutions are fine for $k=0$ and $k=2$ but you need to find an alternative for $k=1$ and $k=3$.
Hint.  Try taking the limit from the right and left separately.  If $3<x<4$ then $\lfloor x\rfloor=3$, so
$$\lfloor x\rfloor\sin\frac{\pi x}{2}=3\sin\frac{\pi x}{2}\ .$$
If $2<x<3$ then $\lfloor x\rfloor=2$ and so
$$\lfloor x\rfloor\sin\frac{\pi x}{2}=2\sin\frac{\pi x}{2}\ .$$
You can now find the limit for both of these as $x\to3$.  If the two limits have the same value then it is the value of
$$\lim_{x\to3}\lfloor x\rfloor\sin\frac{\pi x}{2}\ ,$$
if they are different (or if either does not exist) then this "two-sided" limit does not exist.


TITLE: Jordan Canonical Form of linear operator on vector space of polynomials
QUESTION [0 upvotes]: I'm working on the following problem:
Let $V_{100}$ denote the $\mathbb{C}$-vector space consisting of all polynomials in $\mathbb{C}[x]$ of degree $\leq 100$. Let $T$ be the linear operator on $V_{100}$ defined by $T(f(x)) = 2f(x) - f'(x)$. Find the Jordan canonical form of $T$.
In order to find the Jordan canonical form of $T$, I know that I need the minimal polynomial of $T$ and the characteristic polynomial of $T$. To this point, I tried to find an annihilating polynomial of $T$ first. But, the annihilating polynomial for $T$ that I found is $f(x) = x^{101} - 2^{101}x$. This seems to be too complicated of an annihilating polynomial to deduce what the minimal and characteristic polynomials have to be. Certainly, it doesn't seem to be a nilpotent operator, which is an easier case.
Is there another approach that can tell me what the minimal and characteristic polynomials should be ? I feel as if my approach (sort of brute force) is giving me lots of trouble.
Thanks!

REPLY [2 votes]: Since $Af:=(T-2\operatorname{Id})f=-f'$, we have $A^{101}=0\ne A^{100}$. Therefore the minimal polynomial of $A$ is $x^{101}$, the minimal polynomial of $T$ is $(x-2)^{101}$ and the Jordan form of $T$ is the $101\times101$ Jordan block for the eigenvalue $2$.


TITLE: Is it possible to solve $k = \frac{x}{\ln(x)}$ for $x$?
QUESTION [4 upvotes]: Is it possible to solve $k = \frac{x}{\ln(x)}$ for $x$? My suspicion after a fruitless hour of manipulation is that it is not.

REPLY [1 votes]: Consider the function $f(x)=x-k\ln x$, defined for $x>0$. We have
$$
\lim_{x\to\infty}f(x)=\infty
$$
and
$$
\lim_{x\to0}f(x)=
\begin{cases}
\infty & \text{if $k>0$}\\[4px]
-\infty & \text{if $k<0$}
\end{cases}
$$
We already see that a solution for $f(x)=0$ (which is the same as your equation) exists when $k<0$, but this is not sufficient.
Consider the derivative
$$
f'(x)=1-\frac{k}{x}=\frac{x-k}{x}
$$
Case $k<0$
If $k<0$, the derivative is positive everywhere, so the function is increasing and we get a single solution.
Case $k=0$
There is no solution, because this would imply $x=0$, but the equation makes no sense for $x\le0$.
Case $k>0$
So let $k>0$. We see that the function has a minimum at $k$, with
$$
f(k)=k(1-\ln k)
$$
The number of solutions then depends on whether $1-\ln k$ is positive (no solution), zero (one solution) or negative (two solutions). Clearly, $1-\ln k>0$ if and only if $0<k<e$.
Summary

no solution if $0<k<e$;
one solution if $k=e$ or $k<0$
two solutions if $k>e$.

With numerical methods you can approximate the solutions to the required accuracy.


TITLE: Evaluate $\int \frac{xe^{2x}}{(2x+1)^2}$
QUESTION [0 upvotes]: Evaluate
$$\int \frac{xe^{2x}}{(2x+1)^2}$$

I tried
$$\int \frac{xe^{2x}}{(2x+1)^2}=-\int\frac{e^{2x}}{2}\frac{(2x+1)^2 - 4x(2x+1)}{(2x+1)^4} dx+\frac{xe^{2x}}{(2x+1)^2}$$
I also tried
$$\int \frac{xe^{2x}}{(2x+1)^2}=-\int\frac{-1}{4x+2}(2xe^x + e^{2x}) dx + \frac{2xe^x + e^{2x}}{(2x+1)^2}=\int\frac{xe^x}{2x+1}dx+\int\frac{e^{2x}}{4x+2}dx+\frac{2xe^x + e^{2x}}{(2x+1)^2}$$
They don't seem promising.

REPLY [0 votes]: Let $u=xe^{2x}, \;du=(2x+1)dx$ and $\displaystyle dv=\frac{1}{(2x+1)^2}dx,\; v=-\frac{1}{2(2x+1)}$ to get
$\displaystyle\int \frac{xe^{2x}}{(2x+1)^2}\,dx= -\frac{xe^{2x}}{2(2x+1)}-\int-\frac{1}{2}e^{2x}dx=-\frac{xe^{2x}}{2(2x+1)}+\frac{1}{4}e^{2x}+C=\color{blue}{\frac{e^{2x}}{4(2x+1)}+C}$


TITLE: How to determine a reaction force?
QUESTION [1 upvotes]: An object sits on an inclined plane. The weight of the object will have a normal and parallel component. I always thought that the reaction of the plane was simply the negative of the normal component of the weight.
Similarly, an object swings on a pendulum. The weight of the object can be decomposed into a radial and tangential component. I assumed that the reaction (tension) of the string must be the negative of the radial component.
But several examples have made me doubt my assumptions. One exercice in my textbook involves determining the reaction on a mass as it slides down a parabolic surface, as a function of $\theta$, the angle the vertical makes with the surface at any given point. Under my assumptions about reaction force, the answer is so trivial as to make the exerice pointless (since they give you $\theta$), but instead the exerice launches into a lot of complicated reasoning involving the Frenet-Serret base and comes out with a very different answer to just $-mg\ sin(\theta)$.
Also, consider the second example I gave. If the object is swinging in a circle (to make it even simpler, say it has uniform circular motion), then it must have radial acceleration (centripetal, specifically). But that's impossible if the reaction is exactly the negative of the radial weight.
But then... how can I determine a reaction force? In the absence of the simple rule I gave in the first paragraph, what else is there? Can it be done without making assumptions about the motion (eg. if you suppose the motion of an object is uniform circular, you have a formula all ready for the centripetal force and may thus be able to deduce the reaction force).

REPLY [1 votes]: Here are two simple rules about reaction forces. 1) they do no work and 2) their job is to enforce a motion constraint.
In the simplest term when an object of mass $m$ is sliding on a horizontal plane with velocity vector $\vec{v}$, there is a motion constraint of the form $y=0$. In Pfafian form this constraint is $\dot{y}=0$ and $\ddot{y}=0$.
The reaction force vector $\vec{N}$ has to do no work so $\vec{N}\cdot \vec{v}=0$ or $N_x\,\dot{x} + N_y\,\dot{y} = 0$ and since $\dot{y}=0$ and $\dot{x}\neq 0$ you must have $N_x=0$ and $N_y \neq 0$. That is how you find the direction of any reaction force. The direction is always orthogonal to the motion.
To find the value of the reaction magnitude you do the force balance perpendicular to the motion
$$ N_y - m\,g = m \ddot{y} = 0 $$
With solution $N_y=m \, g$. Now I know this is simplistic example, but it illustrates the process exactly on how to deal with reaction forces for every case.


TITLE: Covariant Derivative of Metric Tensor
QUESTION [0 upvotes]: I'm an amateur studying General Relativity.  I'm reading some notes of lectures by Susskind.  In them, it is written that

"we know that [the covariant derivative of the metric tensor] is zero.  Why?  Because the ordinary derivative of the metric tensor in Gaussian coordinates is zero.  So, in any coordinate system, we have [that the ordinary partial derivatives of the metric tensor in arbitrary coordinates minus the two Chrisoffel correction terms] = 0."

I don't see how this follows.  Could someone explain?

REPLY [0 votes]: This is simply by definition of connection. The metric behaves as a constant under covariant derivative along any vector. This condition is  part of the definition of the connection and is called compatibility with the metric or metricity.


TITLE: A set of prime numbers
QUESTION [12 upvotes]: Consider a non-empty set $S$ of primes, with the property that, for every finite subset $S'\subset S$, all the primes dividing $\left(\prod_{k\in S'}k\right)+1$ are in $S$.
For instance, it can easily be proven that $2\in S$ (if not, then the smallest member $q$ of $S$ is odd, hence, $q+1$ is even, and thus $2\in S$). In fact, with this way, $2+1=3\in S$, $2\cdot 3+1=7\in S$, $2\cdot 7+1=5\in S$, and so on.
It seems this set must contain all primes, but I could not prove it.
I tried to use that if $p_n$ is the smallest prime that is not contained in $S$, then $p_1,p_2,\dots,p_{n-1}\in S$. $p_n\sim n\log n$ for $n$ large, and if I could somehow show that there is enough residues that can be constructed using subset-products of $p_1,\dots,p_{n-1}$ this should have been handled, but how? Can anybody see a way to do it?

REPLY [8 votes]: If I'm not mistaken, it is true that $S$ must contain all primes.
First of all it is obvious that $S$ is infinite -- indeed, as Euclid teach us, if $S$ is finite then $\left(\prod\limits_{p\in S} p\right) + 1$ is coprime to all primes in $S$ and therefore has at least one prime factor not in $S$.
Now let $p$ be a prime number. We will show that $p\in S$. Assume by contradiction that $p\notin S$. Let $R = \{r_1, \ldots , r_k\}$ be residues modulo $p$ that appear infinitely often as residues of primes from $S$ modulo $p$. Since $S$ is infinite $k \ge 1$. Now let $A = \{a_1, \ldots, a_n\}$ be all elements of subgroup of $(\mathbb{Z}/p\mathbb{Z})^*$, spanned by all $r_i$. Also denote by $b$ product of all other primes from $S$ whose residues modulo $p$ are not from $R$.
Note that we can form each $a_i$ as product of different primes from $S$ (and moreover we can use only primes congruent to numbers from $R$ modulo $p$). Then we can also have $ba_i$.
If some of $ba_i + 1$ is congruent to $0$ mod $p$ then we found $p$ and we are done. Thus we can assume that $ba_i + 1$ is never congruent to $0$. Also note that $ba_i$ never congruent to $0$ (since $p\notin S$ by assumption) and all $ba_i$ are pairwise distinct(since $p$ is prime). Therefore $ba_i + 1$ is never congruent to $1$. Finaly note that $1\in A$. From all this we can deduce that there is some $i$ such that $ba_i + 1\notin A$. Let $C$ be corresponding number. Let $q$ be any of its prime divisors. Note that $q$ can't be from exceptional set since we added all of them through $b$. Thus $q$ is congruent to some $r_i$. But then $ba_i + 1\in A$, as product of primes from $R$ -- a contradiction.