TITLE: Question on maximization over functions
QUESTION [0 upvotes]: Suppose I want to solve the following problem
$$\max_{f(\cdot)} \int f(x) g(x) d\mu(x)$$
where the maximization is over measurable functions $f:X\to [0,1]$ ,  $\mu$ is a finite measure and $g$ is measurable.
Can someone come up with an example such that the problem does NOT have a solution? I imagine one can try to maximize point by point and obtain something that is not measurable and then the problem would not have a solution?

REPLY [1 votes]: If $g$ is not integrable the problem can be unbounded: Let $\mu$ be the Cauchy distribution on the real line. Let $g(x) = |x|$. Then, we can take $f(x) = 1$ for all $x$ and we have $\int f g d\mu = \int |x| d\mu(x) = \infty$.


TITLE: Decoupling coupled differential equations in dynamically coupled two state system
QUESTION [2 upvotes]: Consider the following dynamically coupled two state hamiltonian, $$H=-B\sigma_z-V(t)\sigma_x.$$Taking the eigenfunctions of $\sigma_z$ ($|+>$ and $|- >$) as basis vectors, we have the wave function to be $$\Phi=c_
1|+>+ c_2|->$$ and we get coupled differential equations for the time evolution of these two coefficients.
$$\left[ \begin{array}{c} \frac{dc_1}{dt} \\ \frac{dc_2}{dt} \end{array} \right] = \begin{bmatrix} -B & -V(t) \\ -V(t) & B \end{bmatrix} \times \left[ \begin{array}{c} c_1 \\ c_2 \end{array} \right]$$
To decouple the equations I tried diagonalyzing the Hamiltonian involved. But, then the eigenvectors themselves involve time dependence due to $V(t)$ and thus, i'm not able to decouple the differential equations. So, is there any other method do it? Any hints are welcome.

REPLY [1 votes]: The system can be separated, but not necessarily in nice form. For instance, the time derivative of the first eq. reads
$$
i\hbar {\ddot c}_1 = - B {\dot c}_1 - {\dot V}c_2 - V {\dot c}_2
$$
Now remove $c_2$ using again the first eq.,
$$
c_2 = -\frac{i\hbar}{V} {\dot c}_1 - \frac{B}{V} c_1
$$
and ${\dot c}_2$ using the second eq., ${\dot c_2} = \frac{i}{\hbar}Vc_1 - \frac{i}{\hbar}B c_2$:
$$
i\hbar {\ddot c}_1 = - B {\dot c_1} + i\hbar \frac{d\ln V}{dt} {\dot c}_1 + B \frac{d\ln V}{dt} c_1 - \frac{i}{\hbar}V^2c_1 + \frac{i}{\hbar} BV\left(-\frac{i\hbar}{V} {\dot c}_1 - \frac{B}{V} c_1\right) = 0
$$
Simplify, rearrange, and obtain
$$
{\ddot c}_1 - \frac{d\ln V}{dt} {\dot c}_1 + \left[\frac{i}{\hbar}B\frac{d\ln V}{dt} + \frac{B^2 + V^2}{\hbar^2}  \right]c_1 = 0
$$
Similarly for $c_2$.
Better way:
Change from $c_1$, $c_2$ to
$$
c_+ = c_2 + c_1\\
c_- = c_2 - c_1
$$
such that the system becomes
$$
i\hbar {\dot c}_+ = -V(t) c_+ + B c_-\\
i\hbar {\dot c}_- = B c_+ + V(t) c_-\\
$$
Applying the same elimination procedure for $c_-$, this time using
$$
c_- = \frac{i\hbar}{B}{\dot c}_+ + \frac{V}{B}c_+\\
{\dot c_-} = - \frac{i}{\hbar}Bc_+ - \frac{i}{\hbar}Vc_- = - \frac{i}{\hbar}Bc_+ - \frac{i}{\hbar}V\left[\frac{i\hbar}{B}{\dot c}_+ + \frac{V}{B}c_+\right] = \frac{V}{B}{\dot c}_+ - \frac{i}{\hbar}\frac{B^2 + V^2}{B}c_+
$$
yields a much simpler looking eq. for $c_+$:
$$
{\ddot c}_+ - \frac{i}{\hbar}{\dot V}c_+ - \frac{i}{\hbar} V {\dot c}_+ + \frac{i}{\hbar} B {\dot c}_- = 0 \\
{\ddot c}_+ - \frac{i}{\hbar}{\dot V}c_+ - \frac{i}{\hbar} V {\dot c}_+ + \frac{i}{\hbar} V {\dot c}_+ +\frac{B^2 + V^2}{\hbar^2}c_+ = 0\\
{\ddot c}_+ + \left[\frac{B^2 + V^2}{\hbar^2} - \frac{i}{\hbar}{\dot V} \right]c_+ = 0
$$


TITLE: The infimum over all partitions is the same as the infimum over all partitions including a fixed partition $Q$
QUESTION [0 upvotes]: Take an interval $[a,b]$ on the real line and a bounded function $f:[a,b] \to \mathbb R$. For a partition $P =\{ a = t_0, t_1, \dots, t_n = b\}$ of said interval we define the Upper Darboux Sum of $f$ on the partition $P$ as
$$U(f,P) := \sum_{i=1}^n \sup_{t \in [t_{i-1}, t_i]} f(t) \cdot (t_i - t_{i-1})$$
We know that Upper Darboux Sums are decreasing with respect to refinements, so $P \subseteq P' \Rightarrow U(f,P) \ge U(f,P')$
I want to prove that for a fixed partition $Q$ of $[a,b]$, it's true that
$$\inf_P \;U(f,P) = \inf_P \;U(f, P \cup Q) = \inf_{P \supseteq Q} U(f,P)$$
The result is pretty "intuitive" but I don't know how to prove it formally
From the inequality $U(f,P) \ge U(f, P \cup Q)$ we immediately get the inequality
$$\inf_P \;U(f,P) \ge \inf_P \;U(f, P \cup Q) = \inf_{P \supseteq Q} U(f,P)$$
I'm struggling to either prove the other inequality or prove they must be equal

REPLY [1 votes]: The set of all partitions finer than $Q$ is a subset of all partitions. Taking the infimum over only some partitions is $\ge$ than taking the infimum over possibly more partitions


TITLE: Fluid dynamics equations, number of variables and number of equations
QUESTION [4 upvotes]: Continuity and Navier-Stokes equation for fluid are,
\begin{eqnarray}
\frac{\partial \rho}{\partial t} + \nabla\cdot (\rho \mathbf{u}) &=& 0 \\
\rho\left(\frac{\partial \mathbf{u}}{\partial t} + \mathbf{u}\cdot \nabla u \right)  &=& -\nabla p + \nabla \cdot\sigma + \mathbf{{F}_{ext}},
\end{eqnarray}
where letters have the usual meaning. In total we have 4 equations (one continuity equation and 3 momentum balance components of Navier-Stokes) available and number of unknowns are 5 (pressure+ density, 3 velocity components).
How can we then determine all the 5 variables in general?

REPLY [1 votes]: Assuming we want to calculate "standard" fluid dynamics - compressible single phase flow in 3D space - the variables to solve are:

Pressure
Density
Temperature
Velocity (3 components in 3 dimensions)

a total of 6 components.
The equations for conservative states are:

continuity  / mass conservation - one component)
momentum conservaiton (3 components - in 3D, more general 1 component per dimension)
energy conservation

a total of 5 equations, and (as pointed out before) the 6th equation will be the equation of state (could be ideal gas law..)
Depending on the task at hand, the total number of equation and states will change:

For lower dimensions (1D or 2D), the velocity vector reduces components by the same amount as the momentum equation does.
Incompressible flow removes one state and one equation
For more sophisticated calculations, often extra equations and states are added, e.g. for turbulence modelling (see k-epsilon), multiphase flow, or when adding chemistry


TITLE: Inability to divide because of initial value
QUESTION [2 upvotes]: I apologize in advanced for my confusion, however, in my ODE textbook, it gives the following example:
$$\frac {dy}{dt} = (1+y)t$$ with initial value of $y(0) = -1$.
I am confused with this question because although from my understanding, there is definitely a solution to this equation, $$y=ke^\frac{t^2}{2}-1,$$ taken by dividing both sides by $(1+y)$, multiplying both sides by $dt$, and finally integrating both sides. With $k=0$, $$y=0e^\frac{t^2}{2}-1.$$ Then, utilizing $y(0)=-1$, solving the equation, $$y=0e^\frac{0^2}{2}-1 = 0(1)-1 = -1.$$ Given this, ultimately the solution to the initial-value is $$y = -1.$$
However, my textbook states that even dividing both sides of the differential equation by $(1+y)$ is impossible because then the differential equation would be undefined at $y(0) = -1$, however, even looking the solution up online, it seems that the solution does indeed exist? I would appreciate any and all help on this. Thanks so much.

REPLY [3 votes]: It sounds like you're trying to use separation of variables based on your comment (also, that's consistent with separation of variables being taught much earlier than, for example, integrating factors).
Notice that If you plug in $y(0) = -1$ you get that $y'(0) = 0$ and thus $y$ is not changing at this instant. This is an equilibrium solution. That is $y(t) = 0$.
I think the thing you're supposed to take away from this problem is that for any separable equation, if the initial condition precludes division, then you have an equilibrium solution.

REPLY [2 votes]: Yes, because $y(0)= -1, y+ 1$ is $0$ AT $x= 0$ so we cannot divide by $y+ 1$ AT $x= 0$.  But we can do that to solve for $y(x)$ for all other $x$ and then extend the solution to $x= 0$ "by continuity".


TITLE: Is there a construction of the Wiener measure by discretization and limits which parallels the Physics ideas?
QUESTION [2 upvotes]: In Physics one constructs the path integral by a limiting process together with a discretization procedure.
Now, in order to better paralell with the Wiener measure, consider this in Euclidean signature. What Physicists find is that the Euclidean time evolution operator $K_E(q,t;q_0,t_0)$ can be written as
$$K_E(q,t;q_0,t_0)=\int e^{-S_E[x(t)]} \mathfrak{D}x(t)\tag{1}$$
where Physicists define the right hand side of (1) by a discretization procedure together with a limit. More precisely, it is something of the form
$$K_E(q,t;q_0,t_0)=\lim_{N\to \infty}C_N\int \prod_{k=1}^N dq_k e^{-S_{E,N}(q_0,\dots, q_N,\dot{q}_0,\dots, \dot{q}_N)} $$
Here $C_N$ is a normalization factor, $S_{E,N}$ is the discretized euclidean action.
At least from one heuristic point of view, the idea of Physicists seems to be something along the lines:

A path is comprised of infinitely many points. Now, pick $N$ of these, if $N$ is large this collection of points well approximate the path. If thus we pick $t_0,\dots, t_N$ values of the parameter and vary the associated $q_0,\dots, q_N$ along all possible value, we should obtain all paths, hence we should be integrating with respect to the paths there.

Although this seems to be the idea conveyed in all Physics texts I find it remarkable that someone is able to produce any mathematics whatsoever with this.
My question here is: is there some construction of the rigorous Wiener measure based on some limiting process together with discretizations, that in the end parallels the Physicist approach?

REPLY [1 votes]: There is a not only rigorous but also physically intuitive construction of this 1d Euclidean path integral in section X.11 "The Feynman-Kac formula" of Reed and Simon Vol. II. The key property, in the OP's notation is
$$
\int K_E(q_3,t_3;q_2,t_2)K_E(q_2,t_2;q_1,t_1)\ dq_2
\ = K_E(q_3,t_3;q_1,t_1)
$$
which is the semigroup property for the heat equation, or a Markovian case of the Chapman-Kolmogorov Equation. With that it is easy to construct the measure on $\mathbb{R}^{[0,\infty)}$ using the Daniell-Kolmogorov Extension Theorem.
The difficulty though is to restrict this measure to a space of continuous functions. This last step typically requires the dyadic grid mentioned by user10354138.


TITLE: Alternative Proof of Vandermode Convolution
QUESTION [2 upvotes]: I'm working through the book Discrete Calculus: Methods for Counting by Carlo Mariconda and Alberto Tonolo, and I had a question about one of the problems in the book.
The problem is 8.3 and says: Use generating formal series to give an alternative proof of the formula (2.57.a): $\sum_{l\in\mathbb{Z}} {r\choose m+l}{s\choose n+l} = {r+s\choose r-m+n}, \forall m,n,r,s\in\mathbb{N}$
Formula 2.57.a is Vandermone's Convolution, given as $\sum_{j=0}^k {m\choose j}{n\choose k-j} = {m+n\choose k}$.
I've convinced myself these two are equal but setting $j=r-(m+l)$ and then $k=r-m+n$ via substitution.
What I can't figure out, however, is the proof. I've looked at the solutions offered by Mariconda and Tonolo (which I am grateful for), which I've attached here.
Solution
I'm confused about two points with this solution. First, I don't see how it's equivalent to finding the $(m+n)$th coefficient of the convolution product. For instance, if I take r=s=m=n=1, and do the sum of above, I get 2, not 1 which is the $X^2$ coefficient of the given convolution product.
Secondly, I'm confused about the last step at the end. How do they go from ${r+s\choose m+n}$ to ${r+s\choose r-m+n}$?
I feel there's several steps missing and I can't figure it out and it's really causing me to go crazy. Are there perhaps typos? There's an errata page on their website for the book (https://discretecalculus.wordpress.com/), but there's no file there with errata. Or am I just missing something?
I'd appreciate any help, thanks!

REPLY [1 votes]: It seems the authors don't claim that $[X^{n+m}](1+X)^{r+s}$ is the solution of the formula (2.57a) but they rather want us to recall the Vandermonde identity and that the formula (2.57a) follows as a certain variation of it.
On the one hand, we can show formula (2.57a) is a variation of Vandermonde's identity. Since formula (2.57a) is symmetric in $m$ and $n$ we assume in the following wlog $0\leq m\leq n$. We obtain
\begin{align*}
\color{blue}{\sum_{l\in\mathbb{Z}}\binom{r}{m+l}\binom{s}{n+l}}
&=\sum_{l=-m}^{r-m}\binom{r}{m+l}\binom{s}{n+l}\tag{1.1}\\
&=\sum_{l=0}^r\binom{r}{l}\binom{s}{n+l-m}\tag{1.2}\\
&=\sum_{l=0}^{r+n-m}\binom{r}{l}\binom{s}{r+n-m-l}\tag{1.3}\\
&\,\,\color{blue}{=\binom{r+s}{r+n-m}}\tag{1.4}
\end{align*}
where the last step follows thanks to Vandermonde's identity.
Comment:

In (1.1) we set lower and upper limit of the sum by recalling the binomial coefficient $\binom{p}{q}$ is zero if the lower index $q<0$ and also if $q>p$.

In (1.2) we shift the index to start with $l=0$.

In (1.3) we change the order of summation $l\to r-l$ and set the upper limit of the sum according to the binomial coefficient with upper index $s$.

Generating functions:
Starting with (1.2) we derive formula (2.57a) using generating functions. We use the coefficient of operator $[X^n]$ to denote the coefficient of a series $A(X)$. We obtain
\begin{align*}
\color{blue}{\sum_{l=0}^r}&\color{blue}{\binom{r}{l}\binom{s}{l+n-m}}\\
&=\sum_{l=0}^r\binom{r}{l}[X^{l+n-m}](1+X)^s\tag{2.1}\\
&=[X^{n-m}](1+X)^s\sum_{l=0}^r\binom{r}{l}X^{-l}\tag{2.2}\\
&=[X^{n-m}](1+X)^s\left(1+X^{-1}\right)^r\tag{2.3}\\
&=[X^{r-m+n}](1+X)^{r+s}\tag{2.4}\\
&\,\,\color{blue}{=\binom{r+s}{r-m+n}}\tag{2.5}
\end{align*}
and the claim follows.

Comment:

In (2.1) we use the coefficient of operator as in $\binom{p}{q}=[X^q](1+X)^p$.

In (2.2) we use $[X^{p-q}]A(X)=[X^p]X^qA(X)$.

In (2.3) we use the binomial theorem.

In (2.4) we apply the rule as in (2.2) and do some simplifications.

In (2.5) we select the coefficient of $X^{r-m+n}$.


TITLE: Conditional probability that a random spanning tree contains the edge e
QUESTION [4 upvotes]: Let $G$ be a connected simple graph with two distinct edges $e,f \in E(G)$. Choose a random spanning tree $T\subseteq G$, my question is whether there are any known upper bound for the following
\begin{equation} \frac{\mathbb{P}(e\in E(T))\mathbb{P}(f\in E(T))}{\mathbb{P}(e,f\in E(T)) } \end{equation}
for all possible choices of $e$ and $f$?

REPLY [3 votes]: I claim an upper bound on this ratio is  $O(D)$ where $D$ is the maximum degree of the graph. In a simple graph, this also implies the bounds $O(n)$ and $O(m)$ where the graph has $n$ vertices and $m$ edges.
Recall that the probability $\mathbb P(e\in T)$ is equal to the effective resistance between the ends of $e$, where we interpret the graph $G$ as a resistor network with unit resistances.
Given an edge $f \in G$, let $G/f$ denote the graph with $f$ contracted.
The conditional probability $\mathbb P(e\in T \,|\, f\in T)$ is equal to the probability that a random spanning tree in $G/f$ contains $e$, which is the effective resistance between the ends of $e$ in the resistor network $G/f$.

Thus, we are asked to find an upper bound on
$$ \frac{\mathbb P(e\in T)}{\mathbb P(e\in T \,|\, f\in T)} = \frac{Res_G(e)}{Res_{G/f}(e)} \leq \frac1{Res_{G/f}(e)},$$
since the ends of $e$ are connected by a unit resistor by assumption.
Thus it suffices to show that $Res_{G/f}(e) \geq \Omega(1/D)$.
We know the resistance is positive since we are not allowing multiple edges in $G$, and the question comes down to finding a lower bound on the effective resistance between any distinct nodes in a resistor network with maximum degree $O(D)$. (Note that the maximum degree of $G/f$ is bounded by $2D-2$.)
We claim that $\Omega(1/D)$ is such a lower bound.
The effective resistance between nodes $s$ and $t$ is equal to the voltage drop when 1 unit of current is sent from $s$ to $t$. Since $s$ has at most $D$ adjacent unit resistors, some resistor must receive at least $1/D$ units of current, so the voltage drop across this resistor will be at least $1/D$.
This gives a lower bound for the total voltage drop from $s$ to $t$, as claimed.

Note: I'm pretty sure this argument works for any graph (allowing loops and multiple edges) as long as the edges $e$ and $f$ are contained in some spanning tree.
I'm also pretty sure $2D-2$ works as an upper bound, but I wrote $O(D)$ to be safe. Maybe the $2D$ can be lowered further.


TITLE: A sequence $(t_{k,n})$ st $\forall n$, $\sum_{k=1}^n t_{k,n}=1$ but $\lim_n t_{k,n} \neq 0$.
QUESTION [2 upvotes]: I'm looking for an example of a sequence $(t_{k,n})$ such that for all $n$, $$\sum_{k=1}^n t_{k,n}=1$$ but $$\exists k: \ \lim_{n \to \infty}t_{k,n} \neq 0.$$ I've been looking at Toeplitz' lemma for sequences and I can't seem to think up an example that shows that all the assumptions are necessary. There's also another assumption, namely that  $$ \sum_{k=1}^n |t_{k,n}| \mbox{ is bounded.}$$
My guesses are that these assumptions are there for the case of negative $t_{k,n}$'s - is it true that if the weights were assumed to be positive, these assumptions would follow from the first one (that they sum up to 1)?
The Toeplitz lemma says that under these assumptions on $(t_{k,n})$,
$$\lim_{n \to \infty} a_n =a \implies \lim_{n \to \infty} \sum_{k=1}^n t_{k,n} a_n =a.$$

REPLY [2 votes]: $$\matrix{1&0&0&0&0&\dots\cr0&1&0&0&0&\dots\cr1&0&0&0&0&\dots\cr0&1&0&0&0&\dots\cr0&0&1&0&0&\dots\cr1&0&0&0&0&\dots\cr\vdots&\vdots&\vdots&\vdots&\vdots&\vdots\cr}$$


TITLE: Zeta-function of $x_0^3 + x_1^3 + x_2^3 = 0$ over $\mathbb{F}_4$.
QUESTION [6 upvotes]: Consider the projective curve over the finite field $\mathbb{F}_4$
$$C/\mathbb{F}_4\colon x_0^3 + x_1^3 + x_2^3 = 0.$$
Let's calculate the zeta function
$$Z (C/\mathbb{F}_q) = \exp \left(\sum_{n \ge 1} \frac{\# C (\mathbb{F}_{q^n})\,t^n}{n}\right)$$
(where $q = 4$ in this case).
I see two ways of doing that, but neither is elementary.

The curve is smooth of degree $3$, so its genus is $1$, and in this case the zeta function depends only on $\# C (\mathbb{F}_q)$: it is given by
$$Z (C/\mathbb{F}_q) = \frac{1 - at + qt^2}{(1-t)\,(1-qt)}, \quad a = 1 + q - \# C (\mathbb{F}_q)$$
(see e.g. [Silverman, GTM 106, Section V.2]). It is easy to count that $\# C (\mathbb{F}_4) = 9$, and therefore
$$Z (C/\mathbb{F}_4) = \frac{(1 + 2t)^2}{(1-t)\,(1-4t)}.$$
Alternatively, for any hypersurface of the form $x_0^m + x_1^m + \cdots + x_n^m = 0$ there is a formula in terms of Gauss sums (see e.g. [Ireland-Rosen, GTM 84, Section 11.3]), which in this case boils down to
$$Z (C/\mathbb{F}_4) = \frac{P (t)}{(1-t)\,(1-4t)},$$
where
$$P (t) = \left(1 + \frac{1}{4}\,g (\chi_1)^3\,t\right)\,\left(1 + \frac{1}{4}\,g (\chi_2)^3\,t\right),$$
and $\chi_1,\chi_2$ are the nontrivial characters $\mathbb{F}_4^\times \to \mathbb{C}$. A little calculation gives $P (t) = (1+2t)^2$, as expected.

However, once we have the zeta function, we see that the number of points is given by a rather easy formula:
\begin{align*}
\# C (\mathbb{F}_4) & = 9 = (2+1)^2,\\
\# C (\mathbb{F}_{4^2}) & = 9 = (2^2-1)^2,\\
\# C (\mathbb{F}_{4^3}) & = 81 = (2^3+1)^2,\\
\# C (\mathbb{F}_{4^4}) & = 225 = (2^4-1)^2,\\
\# C (\mathbb{F}_{4^5}) & = 1089 = (2^5+1)^2,\\
\# C (\mathbb{F}_{4^6}) & = 3969 = (2^6-1)^2,\\
 & \cdots
\end{align*}
So my question is:

could anybody give an elementary counting argument explaining this?

REPLY [2 votes]: The idea in my latter comment (under main) leads to a "pure" counting argument. Unfortunately only in the case of an odd value of $m$. I abbreviate $q=2^m$.
Because we think projectively we can normalize the first non-zero homogeneous
coordinate of $[x_0:x_1:x_2]$ to be equal to one.
Let us first look at the points with one of the coordinates equal to zero.
The field $E=\Bbb{F}_{q^2}$ contains a primitive third root of unity $\omega$. If we for example assume that $x_0=0, x_1=1$, we see that $x_2$ must be a power of $\omega$. This shows that the projective curve has exactly nine points with the property that one coordinate vanishes, namely the cyclic shifts of $[0:1:1]$, $[0:1:\omega]$ and
$[0:1:\omega^2]$. For the rest of this post we consider the points with all the
homogeneous coordinates $\neq0$. In other words, we want to count the number of solutions
of
$$
x_1^3+x_2^3=1\tag{1}
$$
with $x_1,x_2\in E^*$.
The multiplicative group $E^*$ is cyclic of order $q^2-1=(q-1)(q+1)$. As $\gcd(q-1,q+1)=1$ we can split it into a direct product
$$
E^*=F^*\times S,
$$
where $F=\Bbb{F}_{q}$ is the unique subfield such that $[E:F]=2$, and $S$ is the
kernel of the relative norm map $N_{E/F}:x\mapsto x^{q+1}$.
If we assume that $2\nmid m$, then $3\nmid 2^m-1=q-1$ implying that cubing permutes the elements of $F^*$. Another key observation is that if $\eta_1$ and $\eta_2$ are two distinct elements of $S$, then their ratio is not in $F$, and hence $\{\eta_1,\eta_2\}$
is an $F$-basis of $E$. Observe that $3\mid 2^m+1$ implying that $\omega\in S$.
Anyway, we can uniquely write $x_i=y_i\eta_i$ with $y_i\in F^*$, $\eta_i\in S$. Plugging these in $(1)$ gives the equation $y_1^3\eta_1^3+y_2^3\eta_2^3=1$. Because cubing permutes the set of available $y$-components, and we are only in the interested in the number of solutions, we can equally well replace $y_i^3$ with $y_i$. So we want the number of solutions of
$$
y_1\eta_1^3+y_2\eta_2^3=1.\tag{2}
$$
On the other hand, cubing is a 3-to-1 mapping on $S$. This allows us to count the number of solutions of $(2)$:

Assume first that $\eta_1^3\neq1$. This holds for all choices of $\eta_1$ except powers of $\omega$, so there are $q+1-3=q-2$ choices for $\eta_1$. Assume further that $\eta_2^3\notin\{1,\eta_1^3\}$. To a fixed $\eta_1$ this leaves $q+1-6=q-5$ choices for $\eta_2$. Anyway, in this main case $\eta_1^3$ and $\eta_2^3$ for a basis of $E$ over $F$. Therefore $(2)$ has a unique solution $(y_1,y_2)\in F^2$. Here both $y_1,y_2$ are necessarily non-zero. For if, say $y_1=0$, then $(2)$ implies that $y_2\eta_2^3=1$ implying that $\eta_2^3\in F^*$ and thus forcing $\eta_2^3=1$ contradicting the choices we made.
If $\eta_1^3\neq\eta_2^3$ but exactly one of these cubes happens to equal to $1$, then the argument from the last line of the preceding bullet shows that one of $y_1,y_2$ vanishes, and we get no new solutions to $(2)$.
On the other hand in the case $\eta_1^3=1=\eta_2^3$ (three choices for both $\eta_1$ and $\eta_2$, so a total of nine combinations), the equation $(2)$ becomes
$$y_1+y_2=1.$$ As we prescribed $y_1$ and $y_2$ to both be non-zero, this equation holds for $q-2$ pairs $(y_1,y_2)$ in the field $F$.

The conclusion is thus that $(2)$ has
$$
(q-2)(q-5)+9(q-2)=q^2+2q-8
$$
solutions. Adding the nine points with a single vanishing homogeneous coordinate we see that
$$
\#C(E)=q^2+2q-8+9=(q+1)^2
$$
verifying the answer gotten by more high-powered methods.

Unfortunately I couldn't make this argument work in the case $2\mid m$ on either of first two attempts. This time cubing is no longer bijective in $F^*$. It is bijective in $S$, but the analogue of $(2)$:
$$
y_1^3\eta_1+y_2^3\eta_2=1\tag{2'}
$$
is not as easy to analyze:

Using linear dependence of $\mathcal{B}=\{\eta_1,\eta_2\}$ when applicable is not straight forward for it is difficult to analyze the occasions where the coordinates of $1$ with respect to the basis $\mathcal{B}$ should be cubes in $F$.
There is the further observation that in the special case $\eta_1=1=\eta_2$ the equation $(2')$ becomes $(1)$, but with the variables ranging over the smaller field $F$. This is not a problem per se, as we can use induction on $m$ to handle those points. However, it also implies that the number of solutions of the type $\eta_1\neq1, \eta_2\neq1$ will depend on the residue class of $m$ modulo four. I didn't spot a way forward reflecting this.
An alternative (initially my main line of attack) would be to use $(2')$ to solve one of the $\eta_i$s (the equation is linear!). All the elements of $S$ satisfy the equation $\eta^q=\eta^{-1}$. Together with $y_i^q=y_i$ an application of the Frobenius shows that the solution $\eta_2$ of $(2')$ is an element of $S$ if and only if the solutions (=the choices of $\eta_1$) of the quadratic
$$
\eta^2+\frac{1+y_1^6+y_2^6}{y_1^3}+1=0
$$
are not elements of $F$ (when they will necessarily be in $S$). This leads to a trace condition, and it looks like the number of solutions can only be countes with the aid of a character sum, which is something I wanted to avoid.

Anyway, if I come up with another way of handling $(2')$, I will edit this. Apologies for failing to deliver.


TITLE: find width of path
QUESTION [0 upvotes]: suppose we have following question
A square garden is surrounded by a path of uniform width. If the path and the garden both have an area of $x$, then what is the width of the path in terms of $x$?
so we have following picture right

because we dont know if   path of uniform width is square or not how can i find  width?if area of square is $x$,then  length is $\sqrt{x}$,but what about second figure?suppose it's length are $a$ and $b$,then $a*b=x$,then how can i continue?

REPLY [1 votes]: Define
$$\begin{align}
g &:= \text{width of garden (inner square)} \\
w &:= \text{width of path} \\
s &:=  \text{width of outer square} = g + 2 w
\end{align}$$
Then
$$\begin{align}
\text{area of garden (inner square)} &= g^2 \\
\text{area of path (outer sq., minus inner sq.)} &= s^2 - g^2 \\
&= ( g + 2 w )^2 - g^2 \\
&= g^2 + 4 g w + 4 w^2 - g^2 \\
&= 4 w^2 + 4 g w
\end{align}$$
We know that each of the two areas equals $x$. For notational simplicity, temporarily write "$y$" for "$\sqrt{x}$", so that $x = y^2$.
$$\begin{align}
g^2 &= x = y^2 &(1) \\
4 w^2 + 4 g w &= x = y^2 &(2)
\end{align}$$
From equation $(1)$, we get that $g = y$. Substituting that into $(2)$ gives
$$4 w^2 + 4 w y = y^2 \qquad \to \qquad 4 w^2 + 4 w y - y^2 = 0$$
Now, simply solve the quadratic equation for $w$:
$$w = \frac{- 4 y \pm \sqrt{(4y)^2-4\cdot 4\cdot(-y^2)}}{2\cdot 4} = \frac{-4y \pm \sqrt{16y^2+16y^2}}{8} = \frac{-4y\pm 4y\sqrt{2}}{8} = \frac{y}{2}\left(-1\pm\sqrt{2}\right)$$
We (presumably) want the positive root, so (since $\sqrt{2} > 1$) take "$\pm$" to be "$+$"; also, replace "$y$" with its defined value, "$\sqrt{x}$":
$$w = \frac{\sqrt{x}}{2}\left(\sqrt{2} - 1\right)$$


TITLE: How do you prove that the derivative $\tan^{-1}(x)$ is equal to $\frac{1}{1+x^2}$ geometrically
QUESTION [6 upvotes]: How do you prove that the derivative of $\tan^{-1}(x)$ is equal to $\frac{1}{1+x^2}$ geometrically?
I figured it out by working it out using implicit differentiation.
I also found how to plot a semi-circle using $\cos^2(x)+\sin^2(x)=1$ and found that $((x^2-1))^{0.5}$ plots a semi-circle because if you wanted to find $\sin(x)$ with $\cos(x)$ you would do $(\cos(x)^2-1)^{0.5}$. The reason only a semi circle is in order for it to work it must be both the positive and negative solutions.
I saw that $1$ and the $x^2$ and thought you could visually see that the derivative of $\tan^{-1}(x)$ is $\frac{1}{1+x^2}$ but I couldn't find any way so far.

REPLY [2 votes]: Another proof comes from
the atan addition/subtraction formula
$\arctan(a)\pm\arctan(b)
=\arctan(\frac{a\pm b}{1\mp ab}
$.
Then,
with a little hand-waving
as $h \to 0$
and assuming that
$\lim_{z \to 0} \dfrac{\arctan(z)}{z}
=1
$,
$\begin{array}\\
\arctan(x+h)-\arctan(x)
&=\arctan(\dfrac{x+h-x}{1+x(x+h)})\\
&=\arctan(\dfrac{h}{1+x(x+h)})\\
&\to\arctan(\dfrac{h}{1+x^2})\\
\text{so}\\
\dfrac{\arctan(x+h)-\arctan(x)}{h}
&\to\dfrac{\arctan(\dfrac{h}{1+x^2})}{h}\\
&=\dfrac1{1+x^2}\dfrac{\arctan(\dfrac{h}{1+x^2})}{\frac{h}{1+x^2}}\\
&\to\dfrac1{1+x^2}\\
\end{array}
$


TITLE: Why are there so few Euclidean geometry problems that remain unsolved?
QUESTION [17 upvotes]: Stillwell mentions in his book Mathematics and its History that:

Most of the really old unsolved problems in mathematics, in fact, are
  simple questions about the natural numbers...

What is it about Euclidean geometry that makes it have so few classical unsolved problems, relative to number theory?
*** Inspired by this question, Why are so many of the oldest unsolved problems in mathematics about number theory? and motivated by this comment.

REPLY [8 votes]: Because Euclidean geometry is currently not fashionable, most people do not study topics in it or discuss problems in it, and so you simply hear of fewer problems, solved or unsolved.
Any claims that all Euclidean geometry problems are decidable, as given in the comments to the question, will depend on some restricted definition regarding the form that a "problem" can take.
There are plenty of unsolved geometry problems.  I recommend the book Unsolved Problems in Geometry by Croft, Falconer, & Guy (1991).  In addition to hundreds of problems, the book even points to 17 other collections of unsolved geometry problems.
Many of these unsolved problems are fairly easy to state.  For example: Among all configurations of $n$ points not all on one line, what is the minimum number of lines they might determine?  Explainable to a small child, yet unsolved.
There are many, many beautiful unsolved geometry problems, including ones with Erdös rewards, and I cannot do them justice in this answer.


TITLE: Simplify Fraction with a Square Root?
QUESTION [0 upvotes]: I'm trying to simplify a problem that I have the answer to but when I check it with Wolfram it seems a bit more simplified and I can't think of what steps are taking place to get that result.
Somehow this: $\frac{1}{2\sqrt{1-\frac{x^2}{4}}}$
Becomes this: $\frac{1}{\sqrt{4-x^2}}$
And I'm not sure how. I'm guessing its some simple Algebra operation that I'm totally forgetting right now. Any help breaking it down is much appreciated.

REPLY [0 votes]: We have $$\frac{1}{2\sqrt{1-\frac{x^2}{4}}}=\frac{1}{2\sqrt{\frac{1}{4}(4-x^2)}}$$
$$=\frac{1}{2\sqrt{\frac{1}{4}}\sqrt{4-x^2}}=\frac{1}{\sqrt{4-x^2}}.$$


TITLE: CFG where the number of 0s is the same as the number of 1s and there is exactly one 2.
QUESTION [3 upvotes]: So to create a CFG where the number of 0s and 1s are the same, I have:
$$ S \rightarrow SS \ |  \ 0S1 \ |  \ 1S0 \ | \  \epsilon $$
However, I don't know how to inject one single 2 to create the CFG that the number of 0s is the same as the number of 1s and there is exactly one 2. Any help is appreciated! Thanks!

REPLY [1 votes]: Here's one example grammar:
$$\begin{align*}S &\mapsto \mathtt{2} | \mathtt{0}S\mathtt{1} | \mathtt{1}S\mathtt{0} | ST | TS \\
T &\mapsto TT | \mathtt{0}T\mathtt{1} | \mathtt{1}T\mathtt{0} | \epsilon\\
\end{align*}$$
It turns out that, just as there's a simple but counterintuitive grammar for "strings with an equal number of zeroes and ones", there is a simple but counterintuitive grammar for strings with an equal number of zeroes and ones, and exactly one 2.

Here's one approach: you might realize that you can modify the grammar for the language "strings with equal numbers of zeroes and ones" by allowing just one of those productions to put in a two somewhere.
Starting from this grammar for your original language:
$$\begin{align*}S &\mapsto SS | \mathtt{0}S\mathtt{1} | \mathtt{1}S\mathtt{0}|\epsilon\\\end{align*}$$
How about this grammar for your new language:
$$\begin{align*}S &\mapsto \mathtt{2} | \mathtt{0}S\mathtt{1} | \mathtt{1}S\mathtt{0} |  \mathtt{20}T\mathtt{1} | \mathtt{02}T\mathtt{1} | \mathtt{0}T\mathtt{21} | \mathtt{0}T\mathtt{12} | ST | TS\\
T &\mapsto TT | \mathtt{0}T\mathtt{1} | \mathtt{1}T\mathtt{0}|\epsilon\\
\end{align*}$$
Here, $T$ behaves exactly like our old start symbol used to. $S$ behaves similarly, except $S$ means "You must have exactly one 2 in here somewhere". So we had to modify the old production $S\mapsto SS$ to either $S\mapsto ST| TS$.
This grammar suffices, but you can simplify it if you introduce a production like $S \mapsto \mathtt{2}T | T\mathtt{2}$ which allows some of the other productions to simplify. Then you get the grammar showed above.

If you don't have a flash of intuition, you can also construct this grammar in a more mechanical way, too. It's useful to see what that process looks like:

You can confirm that your language $L$ is context-free because it's the intersection of a context-free language ($L_A$, the language of strings with equal numbers of 0 and 1, with any number of 2s interspersed) with a regular language ($L_B$, the language of strings with exactly one 2 in it). Therefore, you can be confident that there is a context-free grammar for it.

$$L = L_A \cap L_B$$

One proof that context-free languages are closed under intersection with regular languages involves building a pushdown automaton that keeps track of the states of both machines and accepts if both accept. So one approach to building a grammar for $L$ is to build machines for $L_A$ and $L_B$ and follow the recipe for forming the intersection machine.
This turns out to be tedious, but by following the recipe, you can suddenly see how to make a grammar for $L$.

First, you build a grammar for $L_A$ (strings with equal numbers of 0 and 1, with any number of 2s interspersed). Here's an example:
$$\begin{align*}S &\mapsto SS | TS | ST | T | \mathtt{0}S\mathtt{1} | \mathtt{1}S\mathtt{0}\\ T &\mapsto \mathtt{2}T | \epsilon\end{align*}$$
This is basically your grammar, with a new symbol $T$ that means "you can put any number of twos here".

You can convert this grammar into Greibach Normal Form. In this form, every production must consist of a terminal symbol followed by zero or more nonterminal symbols. This form will make it easier to see which rules must change in order to make the grammar for $L_A$ into a grammar for $L_A\cap L_B = L$.  There's a recipe to follow to convert a grammar into gnf— but here, I just reasoned about what the first nonterminal symbol might be in each case.
Here is a grammar for $L_A$ in Greibach normal form:
$$\begin{align*}S &\mapsto \mathtt{2} | \mathtt{2}S | \mathtt{1}SO | \mathtt{0}SI | \mathtt{2}SS | \mathtt{0}SIS | \mathtt{1}SOS\\
O &\mapsto \mathtt{0} | \mathtt{2}O | \mathtt{2}OX | \mathtt{0}X\\
I &\mapsto \mathtt{1} | \mathtt{2}I | \mathtt{2}IX | \mathtt{1}X\\
X &\mapsto \mathtt{2} | \mathtt{2}X
\end{align*}$$
You can interpret the symbols as:

$S$  = a string of equal numbers of zeroes and ones, interspersed with any number of 2s,
$I$  = a string consisting of a single 1 and any number of 2s,
$O$  = a string consisting of a single 0 and any number of 2s,
$X$ = a string consisting of one or more 2s.

Now comes the interesting part. Because every rule in Greibach normal form reads a character first, it becomes possible to see how to modify this grammar for $L_A$ into a grammar for $L_A\cap L_B = L$.
We'll just need two sets of symbols— one set of symbols will be allowed to have a single 2 inside them, and another set of symbols will not be allowed to have any 2s inside them.  So we'll have a grammar with a symbols $S$ and $S^\prime$, $I$ and $I^\prime$, etc.  The unprimed symbols will mean "...and the single 2 in this string occurs here". The primed symbols will mean "...and there are no 2s in this string".
We can modify the rules of our grammar— by adding appropriate prime symbols and deleting inappropriate productions — to finally make a grammar for $L = L_A \cap L_B$. You can figure out what the rules must be based on the interpretation of the symbols written above.
$$\begin{align*}S &\mapsto \mathtt{2} | \mathtt{2}S^\prime | \mathtt{1}SO^\prime | \mathtt{1}{S^\prime}O | \mathtt{0}SI^\prime | \mathtt{0}S^\prime I | \mathtt{2}{S^\prime}S^\prime | \mathtt{0}S{I^\prime}{S^\prime} | \mathtt{0}{S^\prime}I{S^\prime} | \mathtt{0}{S^\prime}{I^\prime}S | \mathtt{1}S{O^\prime}{S^\prime} | \mathtt{1}{S^\prime}O{S^\prime} | \mathtt{1}{S^\prime}{O^\prime}S \\
O &\mapsto \mathtt{02} | \mathtt{20}\\
I &\mapsto \mathtt{12} | \mathtt{21}\\
&\\
S^\prime &\mapsto {S^\prime}S^\prime | \mathtt{0}S^\prime\mathtt{1} | \mathtt{1}S^\prime\mathtt{0}\\
O^\prime &\mapsto \mathtt{0} \\
I^\prime &\mapsto \mathtt{1}
\end{align*}$$

Clean it up and simplify, and you get:
$$\begin{align*}S &\mapsto \mathtt{2} | \mathtt{0}S\mathtt{1} | \mathtt{1}S\mathtt{0} | ST | TS \\
T &\mapsto TT | \mathtt{0}T\mathtt{1} | \mathtt{1}T\mathtt{0} | \epsilon\\
\end{align*}$$


TITLE: For what values of $a>0$, $x(-\ln(x))^a \geq (1-x)(-\ln(1-x))^a$ if $0<x<0.5$?
QUESTION [0 upvotes]: Suppose   $$f(x)=x(-\ln(x))^a-(1-x)(-\ln(1-x))^a,$$
where $a>0$ and $x\in (0,0.5)$. It is clear that $\lim_{x \rightarrow 0^+}f(x)=0=\lim_{x \rightarrow 0.5^-}f(x)$. The question is:
under what conditions for $a$, $f(x)\geq 0$ for  $x\in (0,0.5)$. I guess $f(x)\geq 0$ for  $a> \ln(2)$.

REPLY [1 votes]: I think you are right about your claim that "$a>\ln2$". You can see the antisymmtery at $x=0.5$ if you consider $f(x)$ on $(0,1)$. So, if you want $f(x) > 0$ on $(0, 0.5)$ then you must have $f(x)<0$ on $(0.5, 1)$.
That means you need $f$ to be decreasing around $x=0.5$ or in other words $f$ has to have negative slope at $x=0.5$ so $f'(0.5)<0$ must be satisfied (and this is also sufficient). Here is the derivative of $f$ given by WolframAlpha:
$f'(x) = -a(-\ln(1-x))^{a-1} + (-\ln(1-x))^{a} + (-\ln(x))^{a} - a(-\ln(x))^{a-1}$
$f'(0.5) = -a(\ln 2)^{a-1} + (\ln 2)^{a} + (\ln 2)^{a} - a(\ln 2)^{a-1} = 2(\ln 2)^{a-1}(\ln 2 - a) < 0$
$ \implies \ln 2 < a$.
Another approach may be considering concavity of $f(x)$ on $(0, 0.5)$ and convexity of $f$ on $(0.5, 1)$ again by the antisymmtery. That is, making $f''(x)<0$ on $(0, 0.5)$ and $f''(x)>0$ on $(0.5, 1)$ which implies $f''(0.5)=0$. However, it can be checked that $f''(0.5)=0$ holds whatever $a$ is. Thus, this approach will become far more complicated as you need to determine the sign of the second derivative.


TITLE: Prove that $\lim_{x\to0, y\to0}\frac{x^3+y^3}{x^2+y^2}=0$
QUESTION [3 upvotes]: Prove that $\lim_{x\to0, y\to0}\frac{x^3+y^3}{x^2+y^2}=0$
My Try

REPLY [0 votes]: HINT
Notice that
\begin{align*}
|x|^{3} = |x|\times x^{2} \leq |x|(x^{2} + y^{2}) \Rightarrow \left|\frac{x^{3}}{x^{2} + y^{2}}\right| \leq |x|
\end{align*}
Based on the same procedure, you can prove that
\begin{align*}
|y|^{3} = |y|\times y^{2} \leq |y|(x^{2} + y^{2}) \Rightarrow \left|\frac{y^{3}}{x^{2} + y^{2}}\right| \leq |y|
\end{align*}
Then apply the squeeze theorem.
Can you take from here?


TITLE: What happens to the momentum when 2 black holes collide and no matter is present?
QUESTION [3 upvotes]: Imagine 2 non-spinning black holes with identical mass collides with each other at the same speed in the absence of other matter, I think the gravitational waves can only carry a little of the momentum but how about the rest which cannot be imparted to falling matter since it is initially vacuum? Normally when 2 objects collides, the momentum is distributed according to Newton 3rd law and heat as well but on the case of black hole no heat is produced and gravitational wave cannot carry the balance momentum also black hole don't knock back each other.

REPLY [3 votes]: If two black holes with the same mass collide head on at the same speed, the total momentum is zero. What you may be asking is not momentum, but the kinetic energy of the black holes moving toward each other. After the collision, the mass of the resulting larger black hole will consist of the masses of two original black holes plus their kinetic energy minus the energy emitted in gravitational waves ($c=1$).
In a simpler example, if a small relativistic particle with a relatively large kinetic energy hits a black hole, the mass of the black hole is increased by the total energy of the particle or  the sum of its mass and kinetic energy (neglecting gravitational waves).


TITLE: if matrix multiplication $B*A=C*A$, does it mean $B=C$?
QUESTION [2 upvotes]: If matrix multiplication $B*A=C*A$, does it mean $B=C$?
If A is invertible, then I guess this should work.
If not, then?

REPLY [1 votes]: For a given matrix $A$, we can say that $BA=CA \Rightarrow B=C$ iff $DA={\bf{0}}\Rightarrow D=\bf{0}$.
If the rows of $A$ are linearly independent, then the second condition will hold, since for any (row) vector $\vec{v}$, $\vec{v}A$ gives a linear combination of the rows of $A$, and only the trivial linear combination gives the zero-vector.  In particular this will hold for each row of $D$.
Conversely if the rows of $A$ are linearly dependent, we can find a nonzero row vector $\vec{v}$ such that $\vec{v}A=\vec{0}$.


TITLE: Substituting total derivative d for partial derivative \partial
QUESTION [3 upvotes]: In economic models it seems to be commonplace to substitute a total derivative derived from one equation, say $\frac{d k}{d \tau}$, for the partial derivative derived from another equation, say $\frac{\partial k}{\partial \tau}$.
Why is it allowed to perform this substitution? After all, $d$ and $\partial$ are different concepts, are they not?

Here is a (simplified) example from a model on tax competition by Köthenbürger (2002):
A firm maximizes profits from the production and sale of a single good. One of the resulting first-order conditions is $f_k(k) = r(\tau) + \tau$, where $f_k$ is the derivative of the production function with respect to the input capital $k$, $r$ is the interest rate, and $\tau$ is the tax rate of a tax on capital. Differentiating with respect to $k$ and $\tau$ on both sides, we can rearrange to arrive at
$$
\frac{d k}{d \tau} = \frac{\frac{\partial r}{\partial \tau} + 1}{f_{kk}}
$$
The model also contains the maximization problem of a government which yields the first-order condition
$$
\frac{d u}{d \tau} = u_c \left( - f_{kk} \frac{\partial k}{\partial \tau} k + \dots \right) + \dots \overset{!}{=} 0
$$
The analysis of the model continues with a substitution of $\frac{d k}{d \tau}$ from the first equation for $\frac{\partial k}{\partial \tau}$ in the second equation.

REPLY [1 votes]: In this particular case, the substitution actually makes sense, if you put on your economist's hat.
The first expression $\frac{dk}{d \tau}$ comes from differentiating firm's optimal condition, marginal product = marginal return of capital.
In the government's problem, $\frac{\partial k}{\partial \tau}$ is the rate of change of capital conditioned on all other variables controlled by the government being held constant. But presumably this is a general equilibrium model. Also, those other variables are exogenous to the firm. So in equilibrium, this $\frac{\partial k}{\partial \tau}$ must be the same as the firm's  $\frac{dk}{d \tau}$.
For the same reason, one assumes government's capital is the same as firm's capital, as is always done in decentralized equilibrium models.


TITLE: Logarithm of complex matrices in holomorphic families
QUESTION [4 upvotes]: Let $n,k\geq 1$ be integers, let $U \subseteq \mathbb C^n$ be a contractible open subset, and let $f:U\to \mathrm{GL}_k(\mathbb C)$ be a holomorphic function. Does there exist a holomorhpic function $F:U\to \mathrm{M}_k(\mathbb C)$ such that $\exp(F(u))= f(u)$ holds for all $u\in U$?

Here, $\mathrm{M}_k(\mathbb C)$ means complex $k$ by $k$ matrices. The answer is of course "yes" if $k=1$.
As soon as $k\geq 2$, the problem is that for some invertible matrices $A \in \mathrm{GL}_k(\mathbb C)$ the set of matrices $B\in \mathrm{M}_k(\mathbb C)$ with $\exp(B)=A$ is not discrete. This happens for example if $A$ is diagonalisable and has a double eigenvalue. If in the question we require that for all $u\in U$ the eigenvalues of $f(u)$ are pairwise distinct, the answer would again be yes.

REPLY [3 votes]: It might be worth mentioning that there is an analogous problem with $C^{\infty}$ functions, even when all the matrices are diagonalizable.
Let
$$F(x,y) = \begin{pmatrix} e^{ix} & 0 \\ 0 & e^{-ix} \end{pmatrix} \begin{pmatrix} \cos y & \sin y \\ - \sin y & \cos y \end{pmatrix}$$
Since $F(x,y)$ is unitary, it is always diagonalizable. We claim that $F$ does not have a smooth logarithm on any open neighborhood of $[0,\pi] \times \{ 0 \}$.
Proof: Let $f(x,y)$ be a logarithm of $F$. First, look at the situation with $y=0$. The continuous $\log$'s of $F(x,0)$ are of the form $\left( \begin{smallmatrix} ix+2k\pi i & 0 \\ 0 & -ix+2 \ell \pi i \end{smallmatrix} \right)$ for integers $k$ and $\ell$. So it is impossible that $f(0,0)$ and $f(\pi,0)$ are both multiplies of the identity. Without loss of generality, assume that $f(\pi, 0)$ is of the form $\left( \begin{smallmatrix} (2a+1) \pi i & 0 \\ 0 & -(2b+1) \pi i \end{smallmatrix} \right)$ with $a \neq b$.
The Jacobian of the exponential map at $\left( \begin{smallmatrix} (2a+1) \pi i & 0 \\ 0 & -(2b+1) \pi i \end{smallmatrix} \right)$ has rank one, and $\partial F/\partial y (\pi, 0)$ is not in the image of that rank one map. So it is impossible for $F$ to equal $e^{f(x,y)}$ for any smooth function $f$ near $(\pi, 0)$. If $f(\pi, 0)$ was a multiple of the identity, then there would be no problem at $(\pi, 0)$, but there would be a problem at $(0,0)$ instead.


TITLE: Is there a compact formula for this cross product involving $\nabla$?
QUESTION [1 upvotes]: Let $\textbf{u}=(u_1(x,y,z),u_2(x,y,z),u_3(x,y,z))$ and $\textbf{v}=(v_1(x,y,z),v_2(x,y,z),v_3(x,y,z))$ be two smooth vector fields in $\mathbb{R}^3$, is there a simplified form of the following difference $$\nabla\times((\textbf{u}\cdot\nabla)\textbf{v})-\nabla\times((\textbf{v}\cdot\nabla)\textbf{u})~~?$$
where $\nabla = \left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z}\right)$ is the del operator and  $(\textbf{u}\cdot\nabla)\textbf{v}=\left(u_1\frac{\partial}{\partial x}+u_2\frac{\partial}{\partial y}+u_3\frac{\partial}{\partial z}\right)\textbf{v}$.
My goal is to simplify above difference in the special case where $\textbf{v} = \nabla\times\textbf{u}$ and $\nabla\cdot\textbf{u}=0$, but it will be nice if there is a universal simplification to the difference when $\textbf{u}$ and $\textbf{v}$ are unrelated, then one could apply that to the special case.

REPLY [1 votes]: Using the sum convention $$\nabla \times \vec{a} = \epsilon_{ijk} \frac{\partial a_j}{\partial x_i} \vec{e_k} = \epsilon_{ijk} \partial_i a_j \vec{e_k};$$
$$\nabla \times ((\vec{u}\cdot \nabla)\vec{v})−\nabla \times ((\vec{v} \cdot \nabla)\vec{u})= \nabla \times [(\vec{u}\cdot \nabla)\vec{v} - (\vec{v}\cdot \nabla)\vec{u}]$$
$$ = \epsilon_{ijk} \partial_i [(\vec{u}\cdot \nabla)\vec{v} - (\vec{v}\cdot \nabla)\vec{u}]_j \vec{e_k} $$
$$= \epsilon_{ijk} \partial_i [u_m \partial_m v_j - v_m \partial_m u_j]\vec{e_k} $$
It doesn't look like there is much more you can do.
If we assume as well that $\vec{v} = \nabla \times \vec{u}$ then $v_j = \epsilon_{lnj} \partial_{l} u_{n}.$
$$\nabla \times ((\vec{u}\cdot \nabla)\vec{v})−\nabla \times ((\vec{v} \cdot \nabla)\vec{u})= \epsilon_{ijk} \partial_i [u_m \partial_m \epsilon_{lnj} \partial_{l} u_{n} - \epsilon_{lnm} \partial_{l} u_{n} \partial_m u_j]\vec{e_k} $$
$$=\left(\epsilon_{ijk} \epsilon_{lnj} \partial_i [u_m \partial_m \partial_l u_n] - \epsilon_{ijk} \epsilon_{lnm}\partial_i \partial_l[u_n \partial_m u_j]\right)\vec{e_k}$$
$$= \left[\epsilon_{ijk} \epsilon_{lnj}\left(\frac{\partial u_m}{\partial x_i}\frac{\partial^2 u_n}{\partial x_m \partial x_l} + u_m\frac{\partial^3 u_n}{\partial x_m \partial x_i \partial x_l} \right) - \epsilon_{ijk} \epsilon_{lnm}\partial_i \left(\frac{\partial u_n}{\partial x_l}\frac{\partial u_j}{\partial x_m} + u_n \frac{\partial^2 u_j}{\partial x_l \partial x_m}\right)\right]\vec{e_k} $$
$$= \left[\epsilon_{ijk} \epsilon_{lnj}\left(\frac{\partial u_m}{\partial x_i}\frac{\partial^2 u_n}{\partial x_m \partial x_l} + u_m\frac{\partial^3 u_n}{\partial x_m \partial x_i \partial x_l} \right)  - \epsilon_{ijk} \epsilon_{lnm}\left(\frac{\partial^2 u_n}{\partial x_i \partial x_l}\frac{\partial u_j}{\partial x_m} + \frac{\partial u_n}{\partial x_l}\frac{\partial^2 u_j}{\partial x_i \partial x_m} + u_n \frac{\partial^3 u_j}{\partial x_i \partial x_l \partial x_m} + \frac{\partial u_n}{\partial x_i}\frac{\partial^2 u_j}{\partial x_l \partial x_m}\right)\right]\vec{e_k} $$
Which can possibly be simplified.
Kronecker delta and Levi-Civita symbol


TITLE: Ways of selecting 6 random balls from 6 blue, 9 red and 8 white balls
QUESTION [0 upvotes]: I'm studying probability from probabilitycourse.com and was reading the "Unordered sampling without replacement" section for a motivation on combinations. It's very nicely explained as follows: suppose there's $n$ items from which $k$ must be selected and order doesn't matter. Then the number of ordered samples would be $n(n-1)\ldots(n-k+1)$ and each unordered sample would've been overcounted by a factor of $k!$, so we divide by that, getting the usual $\binom{n}{k}$.
Now we turn to the question in the title and for the life of me, I can't apply the same kind of logic in this scenario. The answer seems to be $\binom{23}{6}$ but I can't understand why.
Another question relevant to this post is here: You have 6 red balls, 6 blue, and 6 white. You randomly select a sample of 5 balls. The number of ways to select $5$ balls from $6$ red, $6$ blue and $6$ white seems to be $\binom{18}{5}$
To make life easier I considered a simpler case - if there are $3$ red balls and $3$ yellow balls, and we have to randomly select $3$ from them.
Let's label the balls $a,b,c,d,e$ and $f$, and suppose the first three are red and the last three are yellow. The number of ordered pairs would be $6\times 5\times 4$. Adjusting for overcounting for triplets like $(a,d,e),(a,e,d),(e,a,d),(e,d,a),(d,a,e),(d,e,a)$, etc., we divide the whole thing by $3!$.
But haven't we still not adjusted for another kind of overcounting? Considering all triplets with, e.g., one red and two yellow as equivalent, triplets such as $(a,d,e)$ and $(b,d,f)$ are still overcounts, because all of them have one red and two yellow. So wouldn't $\binom{6}{3}$ be an overestimation? And is there a way to arrive at the correct answer without considering counts for individual cases like "two red, one yellow", "one red, two yellow", "three red", etc. and adding them up?

REPLY [3 votes]: It all depends on what a "way" is.  When do we consider two selections the same?  Take the case of selecting $6$ balls from $23$ mentioned in the title.  We might consider that all the balls are distinguishable, for example, they might be number from $1$ to $23$.  Then the colors are red herrings.  All we care about is the numbers, and the answer is indeed $\binom{23}{6}$.  You can use your unordered sampling logic to arrive at this answer.
On the other hand, perhaps balls of the same color are indistinguishable.  Then we simply want to know how many balls of each color we chose.  So, as long as we have at least $6$ balls of each color, as we do in this case, the question is, how many solutions in nonnegative integers has the equation $$b+r+w=6$$  The answer is $$\binom 82$$ given by stars and bars.
Imagine the we put the select balls in a line.  First the red balls then a divider, then the blue balls, then a divider, then the white balls.  We have $8$ objects, and the answer is completely determined by which two of them are the dividers.  Hence the answer.
If there were fewer than $6$ balls of one of the colors, then we'd have to make some adjustments.  So long as the number is close to $6$ we can use stars and bars.  If the discrepancy is larger, say we wanted to count the ways to choose $40$ balls from $30$ indistinguishable balls of each of $3$ colors, then a different method is advisable.


TITLE: Finding function $f$ of gradient field $F = \nabla f$
QUESTION [1 upvotes]: We have a gradient field $F = \nabla f = <(x+1)e^{x+y},\space xe^{x+y} >$ .
It's easy to find that $f = xe^{x+y} + c \space$  only by guessing, but I try to solve it as follows:
By the Fundamental Theorem for Line Integrals:
$\int_{(x_0, y_0)}^{(x, y)} F.dr = f(x, y) - f(x_0, y_0)$
We know that $f(x_0, y_0)\space is\space constant$, so we can rewrite that as follow :
$f(x, y) = \int_{(x_0, y_0)}^{(x, y)} F.dr + c$
And because $F$  is a gradient field, we know that it is path-independent so we can choose
a path $C$ that consists of two paths $C = C_1 + C_2$ :

So $C_1$ goes from $(0,0)\space to \space (x_1, 0)$
, and $C_2$ goes from $(x_1,0)\space to \space (x_1, y_1)$
And we have :
$\int_{(x_0, y_0)}^{(x, y)} F.dr = \int_{C}F.dr$
$\int_{C}F.dr = \int_{C_1}F.dr + \int_{C_2}F.dr$
And also :
$F.dr = M.dx + N.dy$, where $M = f_x, N = f_y$
On $C_1$ we have that : $y = 0 \implies dy = 0 \space, x \space goes \space from \space 0 \space to \space x_1 $, so:
$\int_{C_1}F.dr = \int_{0}^{x_1}M.dx = \int_{0}^{x_1}(x+1)e^{x+0} dx = x_1e^{x_1}$
On $C_2$ we have that : $x = x_1 \implies dx = 0 \ \, y \space goes \space from \space 0 \space to \space y_1 $, so:
$\int_{C_2}F.dr = \int_{0}^{y_1}N.dy = \int_{0}^{y_1}x_1e^{x_1+y} dy = x_1e^{x_1+y_1}$
But :
$\int_CF.dr = x_1e^{x_1} + x_1e^{x_1+y_1} = x_1e^{x_1}(1+e^{y_1})$
And because $x_1$ and $y_1$ is just x and y (we put the subscript to avoid confusion):
$\int_CF.dr = xe^x(1+e^y)$
$\implies f(x, y) = \int_CF.dr + \space c = xe^x(1+e^y) + c \neq \space xe^{x+y} + c $
Can anyone tell me what the wrong in my solution?

REPLY [1 votes]: The only mistake is that $$\int_{C_2} F\,\mathrm d(\text{curve})=\int_0^y x e^{x+t}\,\mathrm dt=xe^x(e^y-1)\color{red}\neq xe^{x+y}.$$