TITLE: Question about inverse with respect to convolution product.
QUESTION [1 upvotes]: Let $\mathcal{I}(X)$ be the collection of real valued functions  $f:X\times X\to \mathbb{R}$ with the property that $f(x,y)=0$ when $x>y$. The convolution product $f*g$ for $f,g\in \mathcal{I}(X)$ is defined by $$ f*g(x,y)=\begin{cases}
      \sum_{\lbrace z:x\le y\le z\rbrace}f(x,z)g(z,y) & x\le y \\
      0 & \text{otherwise} \end{cases}$$
From page 14 of http://www.math.ust.hk/~mabfchen/Math232/Inclusion-Exclusion.pdf, I don't follow the last step.

If the latest summ where $\delta$ is the Kronecker Delta, is equal to $\delta(x,y)$, wouldn't that imply the sum representing $g(x,y)$ is 0?

REPLY [1 votes]: No. It means, if $x<y$ then
$$\sum_{x\le z\le y} g(x,z)\,f(z,y)=0\,.$$
And, also $g(x,x)\,f(x,x)=1$.
For example, if $y$ covers $x\ $ ($y>x$ and there is no element in between), we have
$$g(x,y)=-\frac1{f(y,y)}\big(g(x,x)\,f(x,y)\big)=-\frac{f(x,y)}{f(x,x)\,f(y,y)}\,.$$
Then
$$g(x,x)\,f(x,y)\ +\ g(x,y)\,f(y,y)\ =\ \frac{f(x,y)}{f(x,x)}\ +\ \frac{-f(x,y)}{f(x,x)}\ =\ 0$$
indeed.


TITLE: Geodesics on $S^2$ with specific Riemannian metric
QUESTION [2 upvotes]: This is a revision of the statement in the question Geodesics on $S^2$

Suppose a Riemannian metric on $S^2$ satisfies the condition that there exist points $N,S \in S^2$ such that all geodesics originating from $N$ pass through $S$. Then the arc-lengths of all these geodesics are the same, assuming each geodesic connects $N$ to $S$ and does not pass through $S$ in the middle.

I started with picking a $v_0 \in T_NS^2$ such that $\exp_N(v_0) = S$, and let $\gamma_0 \colon [0,1] \to S^2$ be defined by $\gamma_0(t) = \exp_N(tv_0)$.
Then let $v \colon (-\epsilon,\epsilon) \to T_nS^2$ be a smooth map such that $v(0) = v(0)$, and $\exp_N(v(u)) = S$.
Now we have a one-parameter family of geodesics $\gamma_u(t) = \exp_N(tv(u))$.
If we define $\alpha \colon (-\epsilon,\epsilon) \times [0,1] \to S^2$ by $\alpha(u,t) = \gamma_u(t) = \exp_N(tv(u))$, we get a variation of $\gamma_0$ through geodesics, and this variation fixes endpoints $N$ and $S$.
So we have the following facts:
(1) $\displaystyle v(u) = \frac{d}{dt} \gamma_u(0) = \frac{\partial}{\partial t} \alpha(u,t) \Big|_{t=0}$
(2) $\displaystyle \frac{D}{\partial t} \frac{\partial}{\partial t} \alpha(u,t) = 0$
(3) $\displaystyle \frac{\partial}{\partial u} \alpha(u,0) = 0$, and $\displaystyle \frac{\partial}{\partial u} \alpha(u,1) = 0$.
If we can show that $\displaystyle \frac{d}{du} \|v(u)\| = 0$, we can use a compactness argument on $S^1$ to say that the arc-length of $\gamma_u$ does not depend on $u$.
My trouble is in the following argument:

\begin{align*}
\frac{d}{du} \langle v(u),v(u) \rangle
&= \frac{\partial}{\partial u} \langle \frac{\partial}{\partial t} \alpha(u,t), \frac{\partial}{\partial t} \alpha(u,t) \rangle \Big|_{t=0}
= 2 \langle \frac{D}{\partial u} \frac{\partial}{\partial t} \alpha(u,t), \frac{\partial}{\partial t} \alpha(u,t)\rangle\Big|_{t=0} \\
&= 2 \langle \frac{D}{\partial t} \frac{\partial}{\partial u} \alpha(u,t), \frac{\partial}{\partial t} \alpha(u,t)\rangle\Big|_{t=0} \\
&= 2 \frac{\partial}{\partial t} \langle \frac{\partial}{\partial u} \alpha(u,t), \frac{\partial}{\partial t} \alpha(u,t) \rangle \Big|_{t=0} - 2 \langle \frac{\partial}{\partial u} \alpha(u,t), \frac{D}{\partial t} \frac{\partial}{\partial t} \alpha(u,t) \rangle \Big|_{t=0} \\
&= 2 \frac{\partial}{\partial t} \langle \frac{\partial}{\partial u} \alpha(u,t), \frac{\partial}{\partial t} \alpha(u,t) \rangle \Big|_{t=0}
\end{align*}

I am not sure how to conclude that the expression evaluates to be zero. But I assume I need to make use of fact (3) afterwards, since I have not used it anywhere in my argument. Any help is appreciated! Thanks!

REPLY [1 votes]: I think I have figured it out.
To show that $\displaystyle \frac{\partial}{\partial t} \langle \frac{\partial}{\partial u} \alpha(u,t), \frac{\partial}{\partial t} \alpha(u,t) \rangle \Big|_{t=0} = 0$, I only need to show that $\displaystyle \frac{\partial}{\partial t} \langle \frac{\partial}{\partial u} \alpha(u,t), \frac{\partial}{\partial t} \alpha(u,t) \rangle$ is constant in $t$, i.e., $\displaystyle \langle \frac{\partial}{\partial u} \alpha(u,t), \frac{\partial}{\partial t} \alpha(u,t) \rangle$ is linear in $t$, and conclude that it must be $0$ by fact (3).
Now,
\begin{align*}
\frac{\partial}{\partial t}\frac{\partial}{\partial t} \langle \frac{\partial}{\partial u} \alpha(u,t), \frac{\partial}{\partial t} \alpha(u,t) \rangle
=& \frac{\partial}{\partial t} \langle \frac{D}{\partial t}\frac{\partial}{\partial u} \alpha(u,t), \frac{\partial}{\partial t} \alpha(u,t) \rangle \\
=& \langle \frac{D}{\partial t}\frac{D}{\partial t}\frac{\partial}{\partial u} \alpha(u,t), \frac{\partial}{\partial t} \alpha(u,t) \rangle \\
=& \langle \frac{D}{\partial t}\frac{D}{\partial u}\frac{\partial}{\partial t} \alpha(u,t), \frac{\partial}{\partial t} \alpha(u,t) \rangle \\
=& \langle \frac{D}{\partial u}\frac{D}{\partial t}\frac{\partial}{\partial t} \alpha(u,t) + R(\frac{\partial\alpha}{\partial u},\frac{\partial\alpha}{\partial t})\frac{\partial \alpha}{\partial t}, \frac{\partial}{\partial t} \alpha(u,t) \rangle \\
=& \langle R(\frac{\partial\alpha}{\partial u},\frac{\partial\alpha}{\partial t})\frac{\partial \alpha}{\partial t}, \frac{\partial \alpha}{\partial t} \rangle = 0
\end{align*}


TITLE: Calculating elastic energy constant
QUESTION [2 upvotes]: I ran into a kinetic physics problem:
"A spring is hanging on the ceiling. Let's place an object 'M' at the end of the spring. Yet hold 'M' so the spring doesn't stretch. The distance between the floor and 'M' is 'h'. Now let 'M' be free in the air. The spring stretches as much as 'x'. What is the value of the elastic potential energy constant 'k'? ( k from $F = kx$ and $E = \frac{1}{2}kx^2$ )"
Here are constants for the problem.

$m$ : the mass of 'M'
$g$ : gravity acceleration
$h$ : The initial distance between the floor and 'M'
$x$ : Stretched length of the spring after 'M' is attached.

My approach was:
$$m g h = m g (h - x) + 1/2  kx^2 $$
subtract $mgh$
$$-mgx + 1/2  k x^2 = 0$$
add $mgx$
$$ mgx = 1/2 kx^2$$
factor by $(1/2  x^2)$
$$k = (2mg)/x$$
However the official answer says I'm wrong.
The official answer says I should get $k$ in the relation of Force.
$$mg = kx$$
factor by $x$
$$k = (mg)/x$$
Why do I get two different values of k? I asked my school assistant and he doesn't know why but you can use two different methods separately depending on the situation. He was also puzzled because,
$$d/dx (mgx - 1/2 kx^2) = mg - kx$$
So the my logic seems to make sense.
Could someone kindly explain why please?
Thank you!

REPLY [1 votes]: When using "your aproach" you are assuming that the mass is going to move exactly into the equilibrium position and stay still there. Instead, the mass will descend with some velocity till that point and overpass it, and then bounce back and forth like a harmonic oscilator. In real life the oscillation will vanish due to friction after some time, and the mass will stop at the equilibrium point x, but there will be a transient period in which this phenomenon must be accounted.
The expression $\ E = kx^2/2 $, the energy stored in the spring, comes from integrating the variable force F= kx along the downward path, but the work done into the system comes from the gravity acting on the mass. That (constant) force is bigger than kx for any x smaller than the equilibrium point. So there is an extra work done, not stored in the spring, but in the form of kinetic energy of the mass.
Ideally, assuming no friction or damping, your expression should be refrased like
$\ mgh = mg(h-x) + kx^2/2 + mv^2/2 $
The final situation when all the oscillation is damped, is determined only by the equilibrium of forces, so kx = mg. The energy is then less than in the previuos situation: energy is not conserved in the mass-spring system, part of its initial energy is lost into the surrounding environment.


TITLE: Chains of forking extension in stable theories
QUESTION [6 upvotes]: Let $T$ be an stable theory.
Further we work in the monster model of $T^{eq}$.
We say that a chain of types of the form
$$tp(a_1/A_1)\subset tp(a_2/A_2) ... \subset tp(a_n/A_n)$$
is a forking chain if for every $1< i \le n$
the type $tp(a_i/A_i)$ forks over $A_{i-1}$.
What can we say about the length of forking chains?
For example in strongly minimal theories such a chain (in the home sort) has an maximum length of 1. Moreover, there exists no chain of length $|T|^+$ if and only if $T$ is simple. This is since forking has local character (every type does not fork over a set of size $<|T|$) if and only if $T$ is simple.
For theory of Morley rank $N$ such a chain is bounded by $N$. But what about the lower bounds?
Can we find a chain of length $n$ in a theory with Morley rank $\ge n$?
Does any type $p$ of Morley rank $n$ start an forking chain of length $n$?
Do theories without Morley rank have chains of length $|T|$?

REPLY [3 votes]: The answer is no to all three questions.
First notice that if there exists a maximal finite forking chain,
 then its length is the SU-rank
 (or U-rank in the stable context) of $tp(a_1/A_1)$.
Since there are theories of Morley-rank $>1$ and U-rank $=1$,
 this gives negative answer to 1 and 2.
Then note that in a supersimple theory there is no infinite forking chain.
As $\bigcup_i tp(a_i/A_i)$ would not fork over some finite $B$
 and this would be contained in $A_N$ for $N$ sufficiently big.
Hence strictly superstable theories are a counter-example to question 3.


TITLE: Permutations with limited repetition and various constraints
QUESTION [1 upvotes]: you have a string of 360 letters: 180 x 'A' and 180 x 'B'. I (hope I) understand that the number of ways this string can be permuted is $$\frac{360!}{180!180!} = \binom{360}{180}$$
What I have absolutely no clue about is how to calculate how this number changes if the following restrictions are both put into place.

The numbers of 'switches' ('AB' or 'BA') should be as close to equal as possible to the number of repetitions ('AA' or 'BB'). That would mean either 180 switches and 179 repetitions or vice versa.
The maximum number consecutive repetitions is 3 (thus AAAB would be valid, AAAAB not).

Thanks for any insights!

REPLY [0 votes]: Let's consider groups of one to three consecutive identical letters as building blocks and denote the numbers of such groups by $a_1$, $a_2$, $a_3$ for the As and by $b_1$, $b_2$, $b_3$ for the Bs, and let's first consider strings beginning with A and ending in B (and thus $179$ switches and $180$ repetitions). Then we have the following constraints:
\begin{align}
a_1+2a_2+3a_3&=180\;,\\
b_1+2b_2+3b_3&=180\;,\\
a_1+a_2+a_3&=90\;,\\
b_1+b_2+b_3&=90\;,\\
\end{align}
with solution
\begin{align}
a_2&=90-2a_1\;,\\
b_2&=90-2b_1\;,\\
a_3&=a_1\;,\\
b_3&=b_1\;.
\end{align}
Thus the number of admissible arrangements of this form is
$$
\left(\sum_{k=0}^{45}\binom{90}{k,k,90-2k}\right)^2=448581910545737950775083699792561795907509^2\;.
$$
The same number of arrangements begins with B and ends in A. You can carry out a similar analysis to find the number of admissible arrangements that begin and end with the same letter (and thus have $180$ switches and $179$ repetitions).


TITLE: If $G$ is an open set in $\Bbb{R}^n$ and $T$ is a regular $n\times n$ matrix, then is $\{Tx\mid x\in G\}$ open?
QUESTION [2 upvotes]: Let $G$ be a subset of $\mathbb{R^n}$ and $T$ : $n \times n$ a regular matrix. Then $TG=\{ Tx | x \in G \}.$
Does this proposition hold?
$$G \text{ is an open set} \;\Longrightarrow\; TG \text{ is an open set}$$

My attempt
Let $G$ is open.
I want to prove $TG \subset TG^i$. Let $y \in TG$.
Then, I can write $y=Tx(x \in G)$
and $x=T^{-1}y \in G$.
Because $G$ is open, there exists $r>0$ such that $B(x, r) \subset G$.
If I find $\delta>0$ such that $B(y, \delta) \subset TG$, I can prove $TG \subset TG^{i}$. But I couldn't find such $\delta$.
Does anyone know what $\delta$ is? or this proposition doesn't hold?

REPLY [1 votes]: $T$ is a linear map so continuous (as the dimensions are finite), and being "regular", presumably means $T^{-1}$ also exists and is continuous (same reasons). So $T$ is a homeomorphism and in particular an open map. So your proposition holds.


TITLE: What is the equation of the line tangent to two equally spaced circles on the y-axis
QUESTION [0 upvotes]: This graph shows two circles $h$ units vertically either side of the origin with common radius $r$. What is the equation of the line that passes through the origin, and is a tangent to both circles, in terms of $r$ and $h$. Approximate picture:

REPLY [0 votes]: Ok so with ajotatxe's advice I was able to see that if the slope of the line was $tan(\theta)$ then the height $h$ can be expressed as $tan(\theta)*rcos(\pi/2-\theta)+rsin(\pi/2-\theta)$ see here. Thank you everyone.


TITLE: Example of a function $f:\mathbb{R}^n\to\mathbb{R}$ that is not a subversion for all open $U$
QUESTION [2 upvotes]: The definition of submersion is:

Let $f:U\to\mathbb{R}^n$ a differentiable function defined in the open $U\subset\mathbb{R}^m$, if for all $x\in U$, $f'(x)$ is an surjective linear transformation then $f$ is a submersion.

I have to find a open function $f:\mathbb{R}^2\to\mathbb{R}$ of class $C^\infty$, so that $f$ is not a submersion.
I thought in the function, $f(x,y)=x^3$.
If the open $U$ contains any point of the form $(0,y)$, we will have that $f'(0,y)$ is not surjective. But if $U=\{(a,b) : a>1 \}$ we will have that $f$ will be a submersion.
I was wondering if it would be possible to construct a open function $f:\mathbb{R}^2\to\mathbb{R}$ of class $C^\infty$ so that for every open $U\subset\mathbb{R}^2$, $f$ it would not be a submersion.
For another example see this link.

REPLY [2 votes]: This is not possible, as a consequence of Sard's theorem.  In fact, given a smooth open map $f:\mathbb{R}^2\rightarrow \mathbb{R}$, and given any non-empty $U\subseteq \mathbb{R}^2$, there is an open subset $V\subseteq U$ for which $f|_V$ is a submersion.
To see this, suppose $f:\mathbb{R}^2\rightarrow \mathbb{R}$ is a smooth open map and let $U\subseteq \mathbb{R}^2$ be any open subset.
Then $f(U)$ is open, so by Sard's theorem, it contains a regular value $r\in f(U)$.  Pick $x\in U$ with $f(x) = r$.  Saying $r$ is a regular value means that $(\nabla f)(x)\neq \langle 0,0\rangle$.  Since the partial derivatives $f_x$ and $f_y$ are continuous, there is some neighborhood $V$ of $x$ (which we can take to be a subset of $U$ if we wish) for which $\|(\nabla f)(y)\| \geq \|(\nabla f)(x)\|/2  > 0$ for any $y\in V$.
Because $\nabla f$ is non-zero on $V$, $f|_V$ is a submersion.


TITLE: Evaluate the limit $\lim_{x\to 0} \frac{(\tan(x+\pi/4))^{1/3}-1}{\sin(2x)}$
QUESTION [3 upvotes]: Evaluate the limit
$$\lim_{x\to 0} \frac{(\tan(x+\pi/4))^{1/3}-1}{\sin(2x)}$$
I know the limit is $1\over3$ by looking at the graph of the function, but how can I algebraically show that that is the limit. using this limit: $$\lim_{x \rightarrow 0} \frac{(1+x)^c -1}{x} =c$$? (without L'Hopital Rule)

REPLY [1 votes]: $\newcommand{\+}{^{\dagger}}%
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\fermi}{\,{\rm f}}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\half}{{1 \over 2}}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\ol}[1]{\overline{#1}}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}
&\lim_{x \to 0}{\tan^{1/3}(x + \pi/4) - 1 \over \sin\pars{2x}}
=\half\lim_{x \to 0}\bracks{{2x \over \sin\pars{2x}}\,
{\tan^{1/3}(\pi/4 + x) - 1 \over x}}
\\[3mm]&=\half\lim_{x \to 0}\bracks{{\tan^{1/3}(\pi/4 + x) - 1 \over x}}
=
\left.\half\,\totald{\tan^{1/3}\pars{x}}{x}\right\vert_{x\ =\ \pi/4}
=
\half\,\bracks{{1 \over 3}\,\tan^{-2/3}\pars{x}\sec^{2}\pars{x}}_{x\ =\ \pi/4}
\\[3mm]&= {1 \over 3}
\end{align}
$$\color{#0000ff}{\large%
\lim_{x \to 0}{\tan^{1/3}(x + \pi/4) - 1 \over \sin\pars{2x}}
= {1 \over 3}}$$


TITLE: If $K$ is the midpoint of $AH$, $P ∈ AB$, $Q ∈ AC$ and $K ∈ PQ$ such that $OK \perp PQ$, then $OP = OQ$.
QUESTION [2 upvotes]: Proposition. Let $ABC$ be a triangle with $H$ and $O$ as it’s orthocenter and circumcenter, respectively. If $K$ is the midpoint of $AH$, $P ∈ AB$, $Q ∈ AC$ and $K ∈ PQ$ such that $OK \perp PQ$, then $OP = OQ$.

To prove this, I wish to show that $\triangle AKQ \cong \triangle HKP$, which would yield $KP = KQ$. I’ve also considered the approach of proving that $H ∈ (PKE)$, where $BH \cap AC = D$ and $OK \cap (DQK) = E$. In both cases it suffices to have $PH \parallel AC$, though I have yet to find the crux move in order to back this claim. I suspect similar triangles or the fact that $AH$ and $AO$ are isogonal to come into play, though not knowing exactly how.

REPLY [2 votes]: Let $D$ be the intersection of $AO$ with the circumcircle. So, we have $BHCD$ is a parallelogram, and thus $BC$ bisects $HD$. Let $M$ be the intersection of $BC$ and $HD$, so $M$ is the midpoint of $HD$. Let $EF$ be the perpendicular line of $HD$ pass $H$. Since $EF\perp HD$, $BH\perp AC$, so $\angle BHC=\angle AFE$. Also, since $BD\perp AB$, $EF\perp DH$, so $\angle BDH=\angle AEF$. Thus, $\triangle AEF\sim \triangle BDH$. Also, since $M$ on $HD$ and $\angle HBM=90^\circ-\angle ACB-\angle FAH$, we also have $\triangle BHM\sim \triangle AFH$. So $HF/FE=HF/AF\times AF/FE=MH/BH\times BH/HD=HM/HD=1/2$. Therefore, $EF=2FH$, so $EH=FH$. Since $K$ is the midpoint of $AH$, $O$ is the midpoint of $AD$, we have $HD\parallel OK$. Also $OK\perp PQ$ and $HD\perp EF$, we have $PQ\parallel EF$. So $PK/KQ=EH/HF=1$. Thus, $PK=KQ$. Also, since $OK\perp PQ$, we can arrive at $OP=PQ$.