TITLE: Why do we use the natural base for the logarithm in the Kullback–Leibler divergence?
QUESTION [2 upvotes]: Well known formula of KL divergence when we have a discrete probability distributions.
$$D_{KL}(P \parallel Q)=\sum\limits_i \ln \left(\frac{P(i)}{Q(i)}\right) P(i)$$
Can someone explain why the natural base of the logarithm? That will probably not yield the information in bits as a result?
Do I need to change the base of the logarithm to 2 in order to get the relative entropy in bits?
Or there is another way?
Thank you.
M.

REPLY [1 votes]: $$
\log_2 x = \frac{\log_e x}{\log_e 2} = \frac{\ln x}{\ln 2}
$$
Expressing entropy in bits means using base-$2$ logarithms.  Just divide the base-$e$ logarithms by $\ln 2$ and you've got it.


TITLE: Do equal angles necessarily mean a polygon is regular?
QUESTION [20 upvotes]: In a polygon, if all the sides are equal, it doesn’t necessarily mean that the polygon is regular (eg. a rhombus). Is this also true for angles? Meaning can you draw a polygon whose interior angles are equal, but the shape is still not regular? I couldn’t think of any examples, but I’m sure there is one.

REPLY [0 votes]: The side lengths need not be equal. It is amenable to a vector representation.
If  segments of length of an isotropic force set $F_i$  acts on a particle ( components ${F_{xi},F_{yi}}$ of consecutive positive integers $i$ )  then the vectors sum up to zero:
$$ \Sigma_i^n F_{x}=0 $$
$$ \Sigma_i^n F_{y}=0 $$
Each vector joins tail to head of vector and the force polygon comes  back to point of start after all $n$ rotations of $\dfrac{2 \pi}{n}$.
So.. irregular polygons of arbitrary parallel translation of each vector displaced from regular polygons because of force equilibrium.


TITLE: What weathering mechanism leads to reduction of Carbon Dioxide in the atmosphere?
QUESTION [4 upvotes]: Wikipedia says:

The global temperature of the Earth will climb because of the rising luminosity of the Sun, the rate of weathering of silicate minerals will increase. This in turn will decrease the level of carbon dioxide in the atmosphere. Within the next 600 million years from the present, the concentration of $CO_2$ will fall below the critical threshold needed to sustain C3 photosynthesis: about 50 parts per million. At this point, trees and forests in their current forms will no longer be able to survive.

What is mechanism of this removing of $CO_2$ from the atmosphere?

REPLY [4 votes]: Carbon Dioxide dissolves in water to form Carbonic acid.
This combines with silicate minerals to form insoluble clays and limestones which are buried by tectonic plates, volcanoes etc


TITLE: Proof By Induction for function
QUESTION [0 upvotes]: I am an undergrad Computer Engineering Student that is struggling through a class in discrete mathematics. One question in particular from a recent assignment has me stumped. Assuming that $T$ is a total non decreasing function that maps from Z+ to R+, also that $k \in Z^{+}$, and $T(2^{k}) = T(2^{k-1})+5$, and that $T(1)=7$ prove by induction that
$T(2^k) = 5k + 7$
I am confused about where to begin this. I would take my base case as k=0 from the expression im asked to prove, but then what is the inductive hypothesis and where would i go from there for the inductive step?
I appreciate any kind of help. Thanks

REPLY [1 votes]: Step one:
Find all elements up to k-1
By assuming T(2^k-1) = 5(k - 1) + 7
Step two:
Then add the kth term (2^k) to a closed form of the above equation, 5(k - 1) + 7
If the equations are the same you have proved by induction.
Base case as described in question for completeness of answer:
T(1) = 7
T(2^(0)) = 5(0) + 7
T(1) = 7


TITLE: Finding a chord length from other given chord lengths.
QUESTION [0 upvotes]: Let $E,$ $F,$ $G,$ and $H$ be points on a circle such that $EF = 22$ and $GH = 81.$ Point $P$ is on segment $\overline{EF}$ with $EP = 12,$ and $Q$ is on segment $\overline{GH}$ with $GQ = 6.$ Also, $PQ = 15.$ Line segment $\overline{PQ}$ is extended to the circle at points $X$ and $Y.$ Find $XY.$

At first I thought that getting the arc length of the different sectioned arcs would help me to find the chord length by working from the arcs, but now I am unsure if that is the correct way to approach this problem with the information given.

REPLY [2 votes]: HINT.
Intersecting chords theorem:
$$
\cases{
x(15+y)=10\cdot12\\
y(15+x)=75\cdot6\\
}
$$


TITLE: Example of Schauder basis to $\ell^2$
QUESTION [2 upvotes]: Let $\alpha, \beta \in \mathbb C - \{0\}$ such that $0 < |\frac{\beta}{\alpha}| < 1$ and
$$
x_1 = (\alpha, \beta, 0,...), x_2 = (0, \alpha, \beta, 0,...), x_3 = (0, 0, \alpha, \beta, 0,...),...
$$
Show that $(x_n)_{n \ge 1}$ is a Schauder basis of $\ell^2$.
I know that each $x_i$ could be written as follows
$$
x_i = \sum^{\infty}_{k=1} \alpha e_i + \beta e_{i+1}
$$
where $e_i$ represents the canonical vector with input 1 in the $i$ th coordinate and 0 in the rest, but I don't know how to use the condition of $0 < |\frac{\beta}{\alpha}| < 1$.

REPLY [2 votes]: Hint:
Let $\mathbf{e}_j=\delta_j$ ($1$ in the $j$-th component, zero otherwise).
Since for any  $\mathbf{x}\in\ell_2$, $\mathbf{x}=\sum^\infty_{j=1}\langle \mathbf{x},\mathbf{e}_j\rangle \mathbf{e}_i$,
it is enough to show that each $\mathbf{e}_j$ is of the form $\sum_k a_{jk}\mathbf{x}_k$.
\begin{aligned}
\mathbf{e}_1&=\frac{1}{\alpha}\mathbf{x}_1-\frac{\beta}{\alpha}\mathbf{e}_2\\
\mathbf{e}_2&=\frac{1}{\alpha}\mathbf{x}_2-\frac{\beta}{\alpha}\mathbf{e}_3\\
&\ldots\\
\mathbf{e}_j&=\frac{1}{\alpha}\mathbf{x}_j-\frac{\beta}
{\alpha}\mathbf{e}_{j+1}\\
\ldots
\end{aligned}
By induction
$$ \mathbf{e}_1 = \frac{1}{\alpha}\sum^N_{j=1}\big(-\frac{\beta}{\alpha}\big)^{j-1}\mathbf{x}_j +\big(-\frac{\beta}{\alpha}\big)^N\mathbf{e}_{N+1}
$$
From this and the assumption $0<|\beta/\alpha|<1$, one can see that $\mathbf{e}_1$ has the desired expression. Similar procedure for other $\mathbf{e}_j$'s.


TITLE: Let $\mathcal{F} = \left\{\bigcup_{i\in J}A_{i}:J\subset\mathbb{N}\right\}$. Show that $\mathcal{F}$ is a $\sigma$-algebra.
QUESTION [1 upvotes]: Let $\Omega$ be a nonempty set and let $\mathcal{A} := \{A_{i}:i\in\mathbb{N}\}$ be a partition of $\Omega$. Let $\mathcal{F} = \left\{\bigcup_{i\in J}A_{i}:J\subset\mathbb{N}\right\}$ where, for $J = \varnothing$, $\bigcup_{i\in J}A_{i} = \varnothing$. Show that $\mathcal{F}$ is a $\sigma$-algebra.
MY ATTEMPT
To begin with, let us notice that $\Omega\in\mathcal{F}$: it suffices to take $J = \mathbb{N}$.
Moreover, we do also have that
\begin{align*}
A\in\mathcal{F} \Rightarrow A = \bigcup_{i\in J}A_{i} \Rightarrow A^{c} = \bigcup_{i\in\mathbb{N}\backslash J}S_{i}\in\mathcal{F},\,\, S_{i}\in\mathcal{A}.
\end{align*}
Now, if $A\in\mathcal{F}$and $B\in\mathcal{F}$, we conclude that
\begin{align*}
A = \bigcup_{i\in J}A_{i}\,\,\text{and}\,\,B = \bigcup_{i\in J'}B_{i} \Rightarrow A\cup B = \bigcup_{i\in J\cup J'}S_{i},\,\,S_{i}\in\mathcal{A}.\
\end{align*}
Then $\mathcal{F}$ is an algebra. To prove the countable union property, we shall proceed as follows
\begin{align*}
A_{n}\in\mathcal{F} \Rightarrow A_{n} = \bigcup_{i\in J_{n}}A_{i} \Rightarrow \bigcup_{i\in\mathbb{N}}A_{n} = \bigcup_{i\in J}A_{i}\in\mathcal{F},\,\,\text{where}\,\,J = \bigcup_{j\in\mathbb{N}}J_{n}
\end{align*}
Is the wording of my proof correct? Can it be improved? Any contribution is appreciated.

REPLY [1 votes]: It’s fine apart from a fairly serious notational error in the last line. When you write $A_{ni}$, you’re talking about the member of $\mathcal{A}$ whose index is the product $ni$. The problem began because at the beginning of the line when you used $A_n$, which is already the name of a particular member of $\mathcal{A}$, as the name of an arbitrary member of $\mathcal{F}$. What you really meant is that if $F_n\in\mathcal{F}$ for each $n\in\Bbb N$, then for each $n\in\Bbb N$ there is a $J_n\subseteq\Bbb N$ such that $F_n=\bigcup_{i\in J_n}A_i$, and therefore
$$\bigcup_{n\in\Bbb N}F_n=\bigcup_{n\in\Bbb N}\bigcup_{i\in J_n}A_i=\bigcup_{i\in\bigcup_{n\in\Bbb N}}A_i\in\mathcal{F}\;.$$


TITLE: How to find lim sup and lim inf
QUESTION [1 upvotes]: let $\Omega = \mathbb{R}$ and define $A_n$ as
$$ \begin{cases} (-1/2,1] &\text{if } n  \text{ is odd} \\
(-1,1/n] & \text{if } n \text{ is even} \end{cases} $$
Find the $\limsup A_n$ and $\liminf A_n.$
If $n$ is odd then I'll have $(-1,1]$ $(-1/2,1]$ etc and the way that I see it is that the term on the right will be getting closer and closer to one and my set will be getting smaller and smaller and therefore ill have an infinite number of sets.
How does this translate into finding the lim sup or lim inf?
Similarly for n is even the set $(-1,1]$ $(-1,1/2]$ is getting smaller and smaller. How can I then choose find ONE lim inf and ONE lim sup in an infinite number of sets to choose from. Unless I misread the question and i'm looking at this all wrong.

REPLY [1 votes]: By the concrete expression for $\limsup$ and $\liminf$, we should find $\bigcap_{k\geqslant n} A_k, \bigcup_{k\geqslant n} A_k$. Clearly,
$$
\bigcap_{k \geqslant 2n-1} A_k= \bigcap_{k \geqslant 2n} A_k = (-1/2,1] \cap \bigcap_{k\geqslant n}(-1, 1/2k] = (-1/2, 1] \cap (-1, 0] = (-1/2, 0],
$$
then
$$
\liminf_n A_n = \lim_n \bigcap_{k \geqslant n} A_k= (-1/2, 0];
$$
and
$$
\bigcup_{k \geqslant 2n-1} A_k= \bigcup_{k \geqslant 2n} A_k = (-1/2,1] \cup \bigcup_{k\geqslant n}(-1, 1/2k] = (-1/2, 1] \cup (-1, 1/2n] = (-1, 1],
$$
thus
$$
\limsup_n A_n = \lim_n \bigcup_{k \geqslant n} A_k = (-1, 1].
$$


TITLE: Why are potential differences equal across two capacitors in parallel, but charge on each capacitor is not?
QUESTION [1 upvotes]: Two or more capacitors are said to be connected in parallel if each one of them is connected across the same two points. In a parallel combination of capacitors potential difference across each capacitor is same but each capacitor will store different charge. Why is this true? Why is the potential difference across each capacitor in parallel the same and why will each capacitor store different charge?
Also, in series, why are Potential difference across each capacitor different, while charge is the same?

REPLY [1 votes]: Why are potential differences equal across two capacitors in series, but charge on each capacitor is not?

This is based on a false premise. There is no rule that says that "potential differences are equal across two capacitors in series".

In a parallel combination of capacitors potential difference across each capacitor is same but each capacitor will store different charge. Why is this true?

In a lumped circuit model, any two devices in parallel must have the same potential across them. This is because of Kirchoff's voltage law (KVL) which says that the net potential drop around any loop in a planar circuit is zero. To put it in more basic terms, if the potential at point 'A' is $V_a$ and the potential at point 'B' is $V_b$, then the potential difference between points A and B is $V_a - V_b$ no matter what path you take through the circuit between those points.
As arvindpujari's answer points out, since the potential differences are equal across the two parallel capacitors, this means that if the capacitance values aren't equal then they must have different charges since a linear capacitor is defined by the equation $V=Q/C$.

Also, in series, why are Potential difference across each capacitor different, while charge is the same?

Imagine you start with two capacitors in series with no charge ($V=0$). Then start driving a current through them. The same current will flow through both capacitors (that's what it means for two elements to be in series), so the same charge will accumulate on each capacitor's plates.
Since they have equal charge, if the capacitance values are not the same, then the potentials must also not be the same.


TITLE: Number of monotonically increasing functions from the set $\{1,2,3,4,5,6\}$ to itself with the property that $f(x)\ge x$
QUESTION [5 upvotes]: While working on P&C problems, I stumbled upon this problem.
The number of monotonically increasing functions from the set $\{1,2,3,4,5,6\}$ to itself with the property that $f(x)\ge x$  for all x, is equal to $\frac{2}{k}.C^{11}_5$, where k equals...?
Using dummy variables I could figure out number of non-decreasing functions as $C^{11}_6$, but it doesn't take into consideration the condition $f(x) \ge x.$ How to proceed further?

REPLY [2 votes]: Imagine a "tree" of possibilities. At the top, there is all the possibilities for $f(6)$ (which is just $6$). There are branches coming from that one option for $f(6)$, going down to all the options for $f(5)$, etc.
Below is the tree if we replaced $3$ with $6$.

Now, define $g(n,k)$ to be the number of possibilities when $f(k+1)=n$. For example, in the above tree, $g(2,1)=2$, since the possibilities from $f(2)=2$ are $1,2$. Also, $g(3,2)=5$.
Now, we want to find $g(6,5)$. But:
$$g(6,5)=g(5,4)+g(6,4)$$
$$=2g(4,3)+2g(5,3)+g(6,3)$$
$$=5g(3,2)+5g(4,2)+3g(5,2)+g(6,2)$$
$$=14g(2,1)+14g(3,1)+9g(4,1)+4g(5,1)+g(6,1)$$
$$=42g(1,0)+42g(2,0)+28g(3,0)+14g(4,0)+5g(5,0)+g(6,0)$$
By definition, $g(n,0)=1\forall n$, hence $g(6,5)=132$. So $k=7$ in your question

Now, that was interesting and all, but it required a lot of casework. Also, doing smaller cases, you get the Catalan numbers every time! What's going on?  I'm leaving this as an exercise for you to figure out, I'll just leave you with two hints:

Look at the "bash" that I had above. Note that $1,2,5,14,42,132$ are all Catalan numbers

Examine the recursive definition(s) of Catalan numbers.


TITLE: 2D data fitting
QUESTION [1 upvotes]: I have some numbers as a function of 2 variables: $(x, y) \mapsto z$.
I would like to know which function $z=z(x,y)$ best fits my data.
Unfortunately, I don't have any hint, I mean, there's no theoretical background on these numbers. They're the result ($z$) of some FEM simulations of a system, being the simulation a parametric sweep over two parameters ($x$ and $y$) of the system.
Here's my data:
x = [1 2 4 6 8 10 13 17 21 25];
y = [0.2 0.5 1 2 4 7 10 14 18 22];
z = [1 0.6844 0.3048 0.2124 0.1689 0.1432 0.1192 0.1015 0.0908 0.0841;...
1.000 0.7096 0.3595 0.2731 0.2322 0.2081 0.1857 0.1690 0.1590 0.1529;...
1.000 0.7451 0.4362 0.3585 0.3217 0.2999 0.2797 0.2648 0.2561 0.2504;...
1.000 0.7979 0.5519 0.4877 0.4574 0.4394 0.4228 0.4107 0.4037 0.3994;...
1.000 0.8628 0.6945 0.6490 0.6271 0.6145 0.6027 0.5945 0.5896 0.5870;...
1.000 0.9131 0.8057 0.7758 0.7614 0.7531 0.7457 0.7410 0.7383 0.7368;...
1.000 0.9397 0.8647 0.8436 0.8333 0.8278 0.8228 0.8195 0.8181 0.8171;...
1.000 0.9594 0.9087 0.8942 0.8877 0.8839 0.8808 0.8791 0.8783 0.8777;...
1.000 0.9705 0.9342 0.9238 0.9190 0.9165 0.9145 0.9133 0.9131 0.9127;...
1.000 0.9776 0.9502 0.9425 0.9390 0.9372 0.9358 0.9352 0.9349 0.9348];

I tried with MATLAB with the Curve Fitting app, but I didn't succeed. The 'polynomial' fitting doesn't work well. I would like to use the 'custom equation' fitting, but I don't know what equation to start. I don't have much practice in data analysis.

REPLY [0 votes]: What about an equation like:
$\frac{1.05055 \text{x} \text{y}-0.0645537 \text{x}+3.30721}{\text{x} \text{y}+3.13029 \text{x}}$
It has a RMSE of 0.0102. If you overlay the original data and the model you get the below figure where the orange is the original data and the red is the model.


TITLE: Find alternating series that converges to $ \int_0^{1/2}x\log(1+x^3)dx $
QUESTION [2 upvotes]: I need to find the alternating series that converges to $ \int_0^{1/2}x\log(1+x^3)\,dx $
Here's what I did:
$$
\frac{d}{dx}[\log(1+x^3)]=\frac{1}{1+x^3}=\frac{1}{1-(-x)^2}=\sum_{n=1}^\infty(-x^3)^{n-1}=1-x^3+x^6-x^9+-...
$$
$$\begin{align}
f(x)&=x\log(1+x^3)=x\int(1-x^3+x^6-x^9+-...)\\\\
&=x\left[x-\frac{x^4}{4}+\frac{x^7}{7}-\frac{x^{10}}{10}+-...+ C\right]
\end{align}$$
$$
f(0)=0; C=0
$$
$$
f(x)=x^2-\frac{x^5}{4}+\frac{x^8}{7}-\frac{x^{11}}{10}+-...
$$
$$\begin{align}
\int_0^{1/2}x\log(1+x^3)dx&=\int_0^{1/2}(x^2-\frac{x^5}{4}+\frac{x^8}{7}-\frac{x^{11}}{10}+-...)\,dx\\\\
&=\left.\left[\frac{x^3}{3}-\frac{x^6}{6*4}+\frac{x^9}{7*9}-\frac{x^{12}}{10*12}+-...\right]\right|_0^{1/2}\\\\
&=\frac{1}{2^3(3)}-\frac{1}{2^6(6)(4)}+\frac{1}{2^9(7)(9)}-\frac{1}{2^{12}(10)(12)}+-...
\end{align}$$
Is my method correct?

REPLY [0 votes]: I would do it this way:
Using
$\ln(1+z)
=\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}x^k}{k}
$,
which converges for your range,
$\begin{array}\\
\int_0^{1/2}x\ln(1+x^3)dx
&=\int_0^{1/2}x\left(\sum_{k=1}^{\infty} (-1)^{k-1}\dfrac{x^{3k}}{k}\right)dx\\
&=\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k}\int_0^{1/2}x^{3k+1}dx\\
&=\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k}\dfrac{x^{3k+2}}{3k+2}\big|_0^{1/2}\\
&=\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k(3k+2)2^{3k+2}}\\
&=\dfrac1{5\cdot 2^5}-\dfrac1{2\cdot8\cdot 2^8}+...\\
\end{array}
$
More generally,
$\begin{array}\\
\int x^a\ln(1+x^b)dx
&=\int x^a\left(\sum_{k=1}^{\infty} (-1)^{k-1}\dfrac{x^{bk}}{k}\right)dx\\
&=\sum_{k=1}^{\infty} \dfrac{(-1)^{k-1}}{k}\int x^{a+bk}dx\\
&=\sum_{k=1}^{\infty} \dfrac{-1)^{k-1}x^{a+1+bk}}{k(a+1+bk)}\\
\end{array}
$


TITLE: Given two touching circles, find position of a third circle of known radius so that it touches them
QUESTION [3 upvotes]: To give an idea of what my end goal looks like visually:

You start with one circle. You add a second circle, making it touch the first circle at some point. For each successive circle (for which we know the radius) to be added, there should be exactly two ways for it to touch the previous two (for which we know the coordinates of their centers and their radii). Is there a formula that expresses these solutions?

REPLY [1 votes]: Let $C_1$, $r_1$ and $C_2$, $r_2$ be the centers and radii of the two existing circles, and $r$ the radius of the tangent circles. We want to find the center $C$ of the tangent circle that’s counterclockwise from $\overrightarrow{C_1C_2}$.
Consider the triangle $\triangle{C_1C_2C}$. Let $B$ be the foot of the altitude from $C$.

By the Pythagoran theorem, $CC_1^2-BC_1^2=CC_2^2-BC_2^2$. Setting $d=BC_1$ and substituting the known side lengths, this equation becomes $$(r+r_1)^2-d^2=(r+r_2)^2-(r_1+r_2-d)^2.$$ Solving this for $d$ gives $$d={r_1(r+r_1)-r_2(r-r_1)\over r_1+r_2}\tag1$$ and the altitude $BC=h=\sqrt{(r+r_1)^2-d^2}=\sqrt{(r+r_2)^2-d^2}$. Set $\vec u={C_2-C_1\over r_1+r_2}$ (the unit vector that points from $C_1$ to $C_2$) and $\vec v=\operatorname{rot}(\vec u)=\langle -u_y,u_x\rangle$ ($\vec u$ rotated 90° counterclockwise). Then $$C=C_1+d\vec u+h\vec v.\tag2$$ If you’re coding this up, it might be more convenient to absorb the normalization factor $r_1+r_2$ into the triangle side lengths, i.e., $C=C_1+{d\over r_1+r_2}(C_2-C_1)+{h\over r_1+r_2}\operatorname{rot}(C_2-C_1)$.


TITLE: Definitory question about tensor product
QUESTION [0 upvotes]: Let $M,N$ be modules over a a ring $R$, and consider the tensor product $M\otimes_RN$. From what I understood $M\otimes_RN$ is the quotient of the free $R$-module over $M\times N$ by the relation $\sim$ defined as follows: $$(m+m',n)\sim (m,n)+(m',n);$$ $$(m,n+n')\sim (m,n)+(m,n');$$ $$(rm,n)\sim (m,rn).$$ In particular $(m,n)$ is in relation with another pair $(m',n')$ if and only if one these two situations is verified: $$\exists r\in R:m=rm', n'=rn;$$ $$\exists r\in R:n=rn', m'=rm.$$ Now, supposing that the  first situation occurs, it can happen that $m,n\neq 0$ and $rn=0$, so that $(m,n)\sim (m',0)$; it  is also easy to see that $(m',0)\sim(0,0)$. However I'm having troubles in understanding how come $(m,n)\sim (0,0)$, that as $\sim$ is transitive, must be true. In fact  since $m\neq 0$ also $m'\neq 0$, meaning that  $(m',n')$ cannot be $(0,0)$. I hope is just a misunderstanding that can be solved easily, but personally I've spent more than an hour on this problem, and I still don't understand.

REPLY [4 votes]: It is not true that $M\otimes_R N$ is the quotient of the free module by the equivalence relation $\sim$ you describe.  Indeed, $\sim$ is not even an equivalence relation--it is not typically transitive or symmetric.  Instead, $M\otimes_R N$ is defined as the quotient by the submodule of the free module generated by all elements of the form $$(m+m',n)-(m,n)-(m',n),$$ $$(m,n+n')-(m,n)-(m,n'),$$ $$(rm,n)-r(m,n),$$ or $$(m,rn)-r(m,n).$$  Or intuitively, you take a quotient by the relation $\sim$, but you make all further identifications that are forced on you in order for the quotient to still have a natural $R$-module structure (including both the identifications needed to get an actual equivalence relation, but also more since that equivalence relation additionally must respect the $R$-module operations).
(Also, the third case in your definition of $\sim$ is wrong: you need to not just identify $(rm,n)$ with $(m,rn)$, but also identify these with $r(m,n)$.  Alternatively, your third case would be correct if you were starting with the free abelian group on $M\times N$ instead of the free $R$-module.)


TITLE: Calculate this sum $\sum \frac{1}{(4n+1)(4n+3)}$
QUESTION [5 upvotes]: I'm having troubles to calculate this sum: $\sum \frac{1}{(4n+1)(4n+3)}$. I'm trying to use telescopic series, without success:
$\sum \frac{1}{(4n+1)(4n+3)}=1/2\sum \frac{1}{(4n+1)}-\frac{1}{(4n+3)}$
I need help here
Thanks a lot

REPLY [1 votes]: Just consider the series
$$\sum_{n\geq1}\dfrac{\mathbb{i}^n}{n}=-\ln(1-\mathbb{i})=-\dfrac{\ln 2}{2}+\dfrac{\pi}{4}\mathbb{i}$$
hence,
$$\sum_{n\ge 0}\left(\frac1{4n+1}-\frac1{4n+3}\right)=\sum_{n\geq0}\dfrac{(-1)^n}{2n+1}=\Im\left(\sum_{n\geq1}\dfrac{\mathbb{i}^n}{n}\right)=\dfrac{\pi}{4}$$


TITLE: Do randomness and indeterminacy in Quantum Physics mean the same?
QUESTION [1 upvotes]: I have been trying to learn about the randomness in Quantum Physics. But of the many sources I referred to, some say about "Randomness in Quantum physics" and some others say about "Quantum indeterminacy". So my question is do these two mean the same or are they completely different? Is randomness because of that indeterminacy or that randomness is actually "there" and that causes the indeterminacy in Quantum Physics? Or are they just two ways of referring to the same thing and if yes then what are they actually referring to?

REPLY [1 votes]: You might be interested in this question or this question (and some others I cannot track down now).
The basic problem is this: It is not clear what we exactly mean by "deterministic". If you mean that we can in principle determine the future state of a system solely from initial conditions, then the time evolution given by the Schrödinger equation is indeed deterministic.1 However, if you prescribe that "deterministic" means that we can in principle determine the result of any measurement ahead of time, then you run into Bell's theorem: Either your theory is non-deterministic in this sense or it has non-local dependencies.
The or is the vexing thing here, since it means that saying "QM is non-deterministic" could always be countered by "Or it is deterministic and non-local!" (the latter theory constructions are called non-local hidden variable theories).
So, the answer to your question is: These terms are not used perfectly consistently. Determinacy of time evolution is something different from determinacy of measurement results, and whether or not you think QM is deterministic/non-deterministic depends on your interpretation of it.
Randomness is encapsulated by non-determinacy of measurement results, since the Born rule of the Copenhagen interpretation gives you probabilities for each possible result. Non-determinacy of time evolution would not be random, since collapses of the wave function occur during sharp events, i.e. interactions/measurement processes, at not according to a probability distribution in collapsy interpretations.
1There's another pitfall here, namely if you subscribe to a collapse or a non-collapse interpretation of QM. Some collaspy types may say that QM is not deterministic even in the time evolution sense.


TITLE: $\mathscr{C}^{k-1}$-manifold structure on the cotangent bundle
QUESTION [0 upvotes]: Let $X$ be a $\mathscr{C}^k$-manifold of dimension $n$ and atlas $(U_\alpha,\varphi_\alpha)$ (i.e. its atlas is such that the transitions maps $\varphi_\beta\circ\varphi_\alpha^{-1}$ are $\mathscr{C}^k$).
In this thread, we endowed the tangent bundle $TX$ with a $\mathscr{C}^{k-1}$-manifold structure of dimension $2n$. I re-did the proof entirely to make sure I had everything correct in my mind. Now, I'm onto the cotangent bundle.
I looked in the book the answer adviced (W. D. Curtis & F. R. Miller, Differential manifolds), and I noticed a few things :

This time, they work with smooth manifolds, and therefore end up proving that $T^\ast X$ is a smooth manifold of dimension $2n$.
The notations used seem a little strange. They mix up subscript and superscript asterisks, which confuses me quite a lot. I kept writing the pushforward with a subscript asterisk, whereas they sometimes seem to use it for linear maps between dual spaces (that they denote using superscript asterisks...).

I'm unsure where to continue my research, I cannot find lots of data about this, since everyone either 1) works with smooth manifolds, or 2) don't even bother and assume that $T^\ast X$ is a manifold.
I was lucky enough to have some previous lecture notes about the $\mathscr{C}^{k-1}$-manifold structure on the tangent bundle, but in these lectures, the cotangent bundle was completely avoided and differential forms were defined as formal objects, meaning I have no support about this.
I am in hope that a similar result holds, mainly that $T^\ast X$ has $\mathscr{C}^{k-1}$-regularity.
Can anyone give me hints about what to do ?
Are there references dealing with this subject ?
(One last note perhaps : I defined the tangent space algebraically, i.e. with operators acting on locally-$\mathscr{C}^1$ real-valued functions.)

REPLY [1 votes]: If $x \mapsto \varphi_i \varphi_j^{-1}x$ are the transition functions of your atlas of $M$, then the transition functions of $TM$ are $$(x,y) \mapsto (\varphi_i \varphi_j^{-1}x, \text{Jac}(\varphi_i \varphi_j^{-1})y).$$
The Jacobian is a matrix of first partials, so because $\varphi_i \varphi_j^{-1}$ is a $C^k$ function, the function above is a $C^{k-1}$ function. It seems you are already comfortable with this point.
Now the atlas on the cotangent bundle has transition functions given by $$(x,y) \mapsto (\varphi_i \varphi_j^{-1} x, \left(\text{Jac}\left(\varphi_i \varphi_j)^{-1}\right)^T\right)y).$$ The inversion and transposition operations are smooth, so again these transition functions are $C^{k-1}$.


TITLE: Deriving the Lorentz factor $\gamma$
QUESTION [3 upvotes]: Recently, I learned how to derive $E_0 = m_0c^2$. However, to do so, one must accept the relativistic mass equation. How does one derive the Lorentz factor $\gamma$?

REPLY [6 votes]: We can use the fact that the speed of light is the same in every inertial reference frame:
$$c=\frac{dx}{dt}=\frac{dx'}{dt'}$$
From that we can construct the space-time interval:
$$(ds)^2=c^2dt^2-dx^2$$
The space-time interval is invariant in every inertial reference frame.
$$(ds)^2=(ds')^{2}$$
In proper frame, the space-time interval has the following form:
$$(ds)^2=c^2dt^2$$
where $dt$ is usually written as $d\tau$ and it is called the proper time.
We know that the space-time interval is the same in every inertial reference frame, so we can write that the space-time interval in our proper frame is equivalent to any other inertial reference frame:
$$c^2d\tau^2=c^2dt^2-dx^2$$
Divide both sides by $c^2dt^2$:
$$\frac{d\tau}{dt}=\sqrt{1-\frac{v^2}{c^2}}$$
so:
$$d\tau=dt\sqrt{1-\frac{v^2}{c^2}}$$
or:
$$dt=\frac{d\tau}{\sqrt{1-\frac{v^2}{c^2}}}=\gamma\cdot d\tau$$


TITLE: How to apply Gronwall's lemma on a variational system
QUESTION [2 upvotes]: I am trying to figure out how to get a bound using Gronwall's lemma while reading a paper. Assume that $u(\cdot)$ and $v(\cdot)$ are the solutions, with initial condition $u(0)=v(0)=0,$ of the variational system
$$\begin{pmatrix}\dot u(t)\\\dot v(t)\end{pmatrix}=\begin{pmatrix}0&f'(q(t))\\1&0\end{pmatrix} \begin{pmatrix} u(t)\\v(t)\end{pmatrix} + \begin{pmatrix}\frac{1}{12} \dddot p(t)\\ -\frac{1}{6} \dddot q(t)\end{pmatrix}.$$
Here, $p(\cdot)$ and $q(\cdot)$ are real valued functions with the bounds
$|\dddot p(t)|\le K(|p(t)|^2 + |f(q(t))|$), and $|\dddot q(t)|\le K|p(t)|$, where $K$ is some constant and $f$ is a $C^3$ function for which, $f', f^{(2)}, f^{(3)}$ are uniformly bounded by a constant, say $B$ in particular and $K$ depends on $B$.
For a fixed $T>0$, and an arbitrary function $g$: we define $$\Vert g(\cdot)\Vert_\infty := \sup_{0\le t \le T} |g(t)|.$$
Under the above settings, how can we apply Gronwall's lemma and the bounds for $\dddot p$ and $\dddot q$ to obtain the bound (where $K$ is some constant that may not be equal as the one above):
$$\Vert u(\cdot)\Vert_\infty + \Vert v(\cdot)\Vert_\infty \le K(\Vert p(\cdot)\Vert_\infty ^2 + \Vert f(q(\cdot ))\Vert_\infty )? $$
I am only aware of the Gronwall's lemma for a single function, and not on a variational system.
Gronwall lemma for system of linear differential inequalities this post is the closest kind of result I found but here the additional terms are constants and not functions.
I would greatly appreciate if anyone could show me how to apply Gronwall's lemma in this case and also what form the constant $K$ would be (does it depend on the time $T$ or is it just a constant independent of the rest and just some constant multiplication of the bound $K$ for $\dddot p$ and $\dddot q$?)

REPLY [1 votes]: You have
\begin{align*}
\dot{u}(t)  & =f^{\prime}(q(t))v(t)+\frac{1}{12}\dddot{p}(t),\\
\dot{v}(t)  & =u(t)-\frac{1}{16}\dddot{q}(t).
\end{align*}
Integrating
\begin{align*}
u(t)  & =u(0)+\int_{0}^{t}\left[  f^{\prime}(q(r))v(r)+\frac{1}{12}\dddot
{p}(r)\right]  dr,\\
v(t)  & =v(0)+\int_{0}^{t}\left[  u(r)-\frac{1}{16}\dddot{q}(r)\right]  dr.
\end{align*}
Define $g(t)=|u(t)|+|v(t)|$. Then,
\begin{align*}
|u(t)|  & \leq\int_{0}^{t}\left[  |f^{\prime}(q(r))||v(r)|+\frac{1}{12}
|\dddot{p}(r)|\right]  dr\leq\int_{0}^{t}\left[  |f^{\prime}(q(r))|g(r)+\frac
{1}{12}|\dddot{p}(r)|\right]  dr,\\
|v(t)|  & \leq\int_{0}^{t}\left[  |u(r)|+\frac{1}{16}|\dddot{q}(r)|\right]
dr\leq\int_{0}^{t}\left[  |g(r)|+\frac{1}{16}|\dddot{q}(r)|\right]  dr.
\end{align*}
Summing the two inequalities you get
$$
g(t)\leq\int_{0}^{t}\left[  |f^{\prime}(q(r))|+1\right]  g(r)\,dr+\int_{0}
^{t}\left[  \frac{1}{12}|\dddot{p}(r)|+\frac{1}{16}|\dddot{q}(r)|\right]  dr.
$$
You can now apply Gronwall's inequality to get
$$
g(t)\leq\int_{0}^{t}\left[  \frac{1}{12}|\dddot{p}(r)|+\frac{1}{16}|\dddot
{q}(r)|\right]  dr\exp\left(  \int_{0}^{t}\left[  |f^{\prime}(q(r))|+1\right]
\,dr\right)  .
$$
Now you can apply the bounds on $\dddot{p}$ and $\dddot{q}$ to obtain
\begin{align*}
|u(t)|+|v(t)|  & \leq\int_{0}^{t}\left[  \frac{K}{12}|p(r)|^{2}
+K|f(q(r)|+\frac{K}{16}|p(r)|\right]  dr\exp\left(  \int_{0}^{t}\left[
|f^{\prime}(q(r))|+1\right]  \,dr\right)  \\
& \leq\int_{0}^{t}\left[  \frac{K}{12}\Vert p\Vert^{2}+K\Vert f(q)\Vert
_{\infty}+\frac{K}{16}\Vert p\Vert_{\infty}\right]  dr\exp\left(  \int_{0}
^{t}\left[  \Vert f^{\prime}(q)\Vert_{\infty}+1\right]  \,dr\right)  \\
& \leq T\left[  \frac{K}{12}\Vert p\Vert^{2}+K\Vert f(q)\Vert_{\infty}
+\frac{K}{16}\Vert p\Vert_{\infty}\right]  dr\exp\left(  \left[  \Vert
f^{\prime}(q)\Vert_{\infty}+1\right]  T\right)  .
\end{align*}
You could have used also $h(t)=\sqrt{|u(t)|^{2}+|v(t)|^{2}}$.


TITLE: Linear programming problem
QUESTION [1 upvotes]: Some additional information:
In the next season the harvesting amount is estimated at 900 for farm A, 1200, 1500, 1800 for farm B,C and D respectively.
In this scenario I'm asked to minimize the maintenance costs.
Unfortunately I can't solve it.
Denoting $s_i$ as the harvesting machine; senior 1,2,3

I've made the obvious constraint:
$$s_1,s_2,s_3 \geq 0$$
But I'm not sure how to handle the fact there are 6 machines of each type and how to involve the estimated amount....
I want to minimize these costs:
$$16s_1+13s_2+15s_3$$
$$20s_1+29s_2+25s_3$$
$$40s_1+38s_2+42s_3$$
$$37s_1+49s_2+45s_3$$
Can someone help me out?
Thanks in advance!
Edit: and what is in this case parameter/variable? I seem to mix them up quite often...

REPLY [4 votes]: I can help you with the constraints.
You can't have more than 6 of any machine:
$$0 \leq s_1,s_2,s_3 \leq 6$$
The machines must be able to harvest at least the estimated amount:
$$250s_1+200s_2+230s_3 \geq 900$$
$$270s_1+350s_2+300s_3 \geq 1200$$
$$490s_1+470s_2+520s_3 \geq 1500$$
$$460s_1+630s_2+550s_3 \geq 1800$$

It seems that a better way to approach this problem is to use a different variable for each machine on each farm. I am going to use the simplex algorithm. So, if $w$ is our total cost, we will be trying to minimize the equation:
$$w = 16x_1+13x_2+15s_3+20x_4+29x_5+25x_6+40x_7+38x_8+42x_9+37x_{10}+49x_{11}+45x_{12}$$
Where $x_1$ is machine type 1 on farm A, $x_2$ is machine type 2 on farm A, etc. By adding "surplus" variables, we can get rid of the inequalities we had above, so:
$$250x_1+200x_2+230x_3-s_1 = 900$$
$$270x_4+350x_5+300x_6-s_2 = 1200$$
$$490x_7+470x_8+520x_9-s_3 = 1500$$
$$460x_{10}+630x_{11}+550x_{12}-s_4 = 1800$$
And since we still can't use the same machine on more than one farm at a time, we have to restate our other constraint, this time with "slack" variables:
$$x_1+x_4+x_7+x_{10}+s_4 = 6$$
$$x_2+x_5+x_8+x_{11}+s_5 = 6$$
$$x_3+x_6+x_9+x_{12}+s_6 = 6$$
We use our original equation with all of the constraints to produce our tableau. It's a big matrix; I'm sorry if it doesn't render correctly.
1   0   0   1   0   0   1   0   0   1   0   0   1  0  0  0  0  0  0  0  6
0   1   0   0   1   0   0   1   0   0   1   0   0  1  0  0  0  0  0  0  6
0   0   1   0   0   1   0   0   1   0   0   1   0  0  1  0  0  0  0  0  6
250 200 230 0   0   0   0   0   0   0   0   0   0  0  0  -1 0  0  0  0  900
0   0   0   270 350 300 0   0   0   0   0   0   0  0  0  0  -1 0  0  0  1200
0   0   0   0   0   0   490 470 520 0   0   0   0  0  0  0  0  -1 0  0  1500
0   0   0   0   0   0   0   0   0   460 630 550 0  0  0  0  0  0  -1 0  1800
16  13  15  20  29  25  40  38  42  37  49  45  0  0  0  0  0  0  0  1  0

By using one of several online calculators, we get the optimal solution $w = \frac{5306}{13}$
when: $x_1 = \frac{14}{9}, x_2 = {23}{9}, x_3 = 0, x_4 = \frac{40}{9}, x_5 = 0, x_6 = 0, x_7= 0, x_8 = 0, x_9 = \frac{75}{26}, x_{10} = 0, x_{11} = \frac{20}{7}, x_{12} = 0$
However, in reality we can't have fractions of a machine, so this answer will have to be further refined by using a cutting plane technique (you can't necessarily just round up). Unfortunately, I did not find a calculator that does this.


TITLE: Criteria for representability of a functor from schemes to sets
QUESTION [3 upvotes]: I have read in several texts that in order to show that a given functor from the category of schemes (over, say an algebraically closed field $k$) to $\mathsf{Set}$ is representable, it suffices to check that it is representable in the subcategory of affine schemes. How is this the case? I have not been able to find a proof of this fact, and it may be rather trivial, but I do not see the connection.

REPLY [10 votes]: The statement

to show that a given functor [$F\colon \mathsf{Sch}^{\mathit{op}}\to\mathsf{Set}$] is representable, it suffices
  to check that it is representable in the subcategory of affine schemes

isn't exactly true: for example, if $X$ is a scheme that's not affine, $\mathrm{Hom}(-, X)$ is representable in
$\mathsf{Sch}$, but not in the subcategory $\mathsf{AffSch}$ of affine schemes.
What is true, though, is that any representable functor is determined (up to natural isomorphism) by its
restriction to $\mathsf{AffSch}$. This is because, for any schemes $X$ and $Y$, $U\mapsto \mathrm{Hom}(U,X)$ (for
$U\subset Y$ open) is a sheaf on $Y$ in the Zariski topology, and every scheme $Y$ admits an affine open cover.
Hence we can determine $\mathrm{Hom}(Y, X)$ from the collection of $\mathrm{Hom}(U, X)$ as $U$ ranges over the
affine open subschemes of $Y$.
This leads to the statement that you might have seen: if you're defining a scheme by first defining a functor
$F\colon\mathsf{Sch}^{\mathit{op}}\to\mathsf{Set}$, then checking that $F$ is representable, you actually only need
to define $F$ on affine schemes. This is particularly useful for defining group schemes, e.g. defining
$\mathrm{GL}_n$ by checking that the functor $\mathrm{Spec} R\mapsto \mathrm{GL}_n(R)$ is representable.
(Note: I might have misinterpreted you when you wrote “it suffices to check that it is representable in the
subcategory of affine schemes” – if you didn't mean the incorrect statement, then sorry about that!)


TITLE: global complete intersection and independence of $l$
QUESTION [1 upvotes]: Hello,
I remember reading that if $X/\mathbf F_q$ is a projective smooth global complete intersection, then the characteristic polynomial of the $\mathbf F_q$-linear Frobenius of $X$
on $H^i_{et}(X\otimes\overline{\mathbf F_q},\mathbf Q_l)$, $l\nmid q$, has integer coefficients and is independent of $l$.
How does one prove this fact?
Thanks

REPLY [1 votes]: As was pointed above, this statement was proven by Deligne for general smooth projective varieties. Yet in your case you don't require the theory of Deligne's weights, since you have only one cohomology group (the middle one) that could vary. Using the latter statement, you can easily deduce your fact from the properties of etale cohomology proved by Grothendieck and Artin.
You can probably find more detail here:
http://www.jmilne.org/math/CourseNotes/LEC.pdf


TITLE: Infinite Sequences on Arbitrary Field - Countable Basis?
QUESTION [5 upvotes]: Please note that this is a homework question, I've been thinking about this for a couple days and I haven't had much luck! Thought I'd get some other input.
Let $V$ be the vector space of all infinite sequences with entries in an arbitrary field $\mathbb{F}$.  (The goal is to show it doesn't have a countable basis)
First show that if $V$ admits a countable basis then there is a nested sequence of subspaces $W_{1} \subset W_{2} \subset ...$ such that $dim(W_{j}) = j$ and $V = \cup W_{j}$.
This proof was fairly obvious because you are given the fact that there is a countable basis so you can just take spans.  However, then ext part says

We can represent an infinite sequence as a concatenation of sequences $(s_{2}, ... )$ where $s_{j}$ is a sequence of length $j$.  For each $j \geq 2$ show that there exists a sequence $S_{j}$ with length $j$ and the property that
$s_{j} = S_{j} \rightarrow (s_{2}, ... S_{j}, .. ) \not\in W_{j-1}$

For me what this is trying to get at that given any element in your vector space you can find a sequence of length $j$ such that if you shove it in the $j$th spot of the sequence representation of the element in the vector space than it isn't in $W_{j-1}$.
My first idea was thinking about picking our "countable basis"(which we know doesn't exist) to be the element whose ith entry is 1 and the rest is 0.  Then we could just pick up a diagonal argument to ensure that it wouldn't be in $W_{j-1}$ but I can't quite wrap my head about how to apply the same idea to an arbitrary countable basis where we don't know what the elements look like!
Lastly it is to show that there's an element which doesn't belong to any of the $W_{j}$ which follows from the above just by picking all those finite length sequences and putting them together.
So what gives, I feel as though I'm getting too focused around the idea that I know this not to be the case that I haven't been able to quite get it.
Thanks!

REPLY [2 votes]: Given a field $\mathbb K$, we can show that any basis of $\mathbb K^\mathbb N$ has size at least $\mathfrak c=|\mathbb R|$. I'll prove this as a consequence of the following combinatorial statement:

Lemma. There is a collection of infinite subsets of $\mathbb Q$ of size $\mathfrak c$ with the property that any two of them have finite intersection.

Such sets are called almost disjoint. To see the lemma, note that to each real we can associate (the range of) a strictly increasing sequence of rationals converging at the real. Any two such sequences have only finitely many terms in common.
Given an almost disjoint family of subsets of $\mathbb Q$, we obtain an almost disjoint family $\mathcal F$ of subsets of $\mathbb N$ of the same size simply because there is a bijection between $\mathbb Q$ and $\mathbb N$. Now, given such a family $\mathcal F$, consider its characteristic functions as vectors in $\mathbb K^\mathbb N$. More precisely, if $A\subseteq \mathbb N$, we associate to $A$ the vector $v_A=(a_1,a_2,\dots)$ where each $a_i$ is $0$ or $1$, and it is $1$ iff $i\in A$.
We are done if we can prove that these vectors $v_A$, $A\in\mathcal F$, are linearly independent. To see this, consider any finitely many of them, say $v_{A_1},\dots,v_{A_n}$. Note that (since $A_1,\dots,A_n$ are almost disjoint) for each $j$ with $1\le j\le n$ there is an $\displaystyle i\in A_j\setminus\bigcup_{\substack{k=1\\ k\ne j}}^n A_k$ since $A_j$ is infinite, but each intersection $A_j\cap A_k$, $k\ne j$, is finite. This means that $v_{A_j}$ has a $1$ at the $i$th position, while the other $v_{A_k}$ have a $0$ there. But then, any combination of the $v_{A_k}$, $k\ne j$, also has a $0$ at the $i$th position, and cannot equal $v_{A_j}$.

By the way, note that the same argument shows that the $v_A$, $A\in\mathcal F$, in spite of how many there are, do not form a basis. For instance, none of the standard vectors $e_i$ is a combination of them.


TITLE: Strategy to beat the casino with unlimited amount of money (Martingales)
QUESTION [1 upvotes]: Brzezniak and Zastawniak's book on stochastic processes shows that that there is no way to beat the casino by having a finite amount of money available:
Let $(X_1,X_2,\cdots)$ be independent random variables and let $(\alpha_1,\alpha_2, \cdots)$ be a non-negative bounded sequence of previsible random variables.
$$E(\sum_{i=1}^{n+1}\alpha_iX_i|X_1,\cdots,X_n)=\sum_{i=1}^n\alpha_iX_i+\alpha_{n+1}E(X_{n+1}|X_1,\cdots,X_n)=\sum_{i=1}^n\alpha_iX_i+\alpha_{n+1}E(X_{n+1})\leq\sum_{i=1}^n\alpha_iX_i$$
(Assuming $E(X_{n+1})\leq0$)
Proving that $\{\sum_{i=1}^n\alpha_iX_i\}_{n\in\mathbb{N}}$ is a supermartingale. Therefore there is no possible strategy that can make the game favorable to a gambler.
I also know that if $(\alpha_1,\alpha_2, \cdots)$ was not bounded then the theorem is no longer true. For example, the martingale system where we double the bet each time we lose until we win once has positive mean.
My question is, what is wrong with the previous proof if $(\alpha_1,\alpha_2, \cdots)$ is not bounded?

REPLY [1 votes]: The process $\{ \sum_{i=1}^n \alpha_i X_i\}_{n \in \mathbb{N}}$ can still be a super-martingale if the $(\alpha_i)$ are unbounded, and will be under the doubling strategy you mentioned, but it may no longer be uniformly integrable.  This means that if you only play for a finite time (such as expecting to not live for more than $200$ years) the strategy will be expected to lose money over that time frame, but if you could continue the strategy forever you will make a profit.
The proof can also break down if $\mathbb{E}[|\alpha_i X_i| ] = \infty$ for some $i \in \mathbb{N}$ (say if your strategy involves placing an infinitely large bet at some time).  Then $\mathbb{E}[\alpha_i X_i | \mathcal F_{i-1}]$ may not be well-defined, so the proof won't work.


TITLE: Map that sends one map to another
QUESTION [1 upvotes]: Question asks to prove that $(g  \circ h)_* = g_* \circ h_*$ and $(g \circ h)* = h^* \circ  g^*$, where $f_*=f \circ$ and $f^* = \circ f$. So it is basically some operation that sends one map to other. I did not understand the meaning $(g_* \circ h_*)$. I can see that $(g \circ h)_*(X \rightarrow Y))$ Just takes a $g(h(X \rightarrow Y))$. But how do we define $g_* \circ h_*$ for $X \rightarrow Y$ as an element?

REPLY [0 votes]: I think you need to reconsider your notation!
You have given $2$ maps:
$$X \xrightarrow{h} Y  \xrightarrow{g} Z$$
Now lets consider $Set(\_ , M)$, thwe set of maps from some set to a given set $M$ (I call it Set(/_ , /_) because it is the morphism set in the category of sets).
Now every map $f: Z \to M$ induces a map $f\circ g : Y \to M$. So we get a map $$h^*:Set(Z,M) \to Set(Y,M)\\ f \mapsto f \circ h$$ And analogously for $Y,Z$ and $h$ (observe that here the arrows "flip"). This map is referred to as $g^*$ respectivel $h^*$ so you need to prove, that $$g^* \circ h^* (f:Z \to M )=g^*(h^*(f: Z \to Y))\stackrel{!}{=}(g\circ h) ^*$$ which just boils down to unfolding the definition.
Now gfor the $g^*$ we just do the same thing dually:
I.e. we now consider $Set(M,\_)$ here we now get the maps $h^*$ $g^*$ in exactly the same way (only that now the arrows don't get "reversed") since we postcompose. I.e.
$$h_*:Set(M,Y) \to Set(M,Z)\\ f \mapsto h \circ f$$
I hope that helps your understanding, writing down more details might be even more confusing for you, and with these thingys it really helps to once wrap your head around yourself!


TITLE: solving this laplacian
QUESTION [0 upvotes]: given the laplacian $ -y^{2}( \partial _{x}^{2} + \partial _{y}^{2} )f(x,y)=Ef(x,y) $
can we find a solution in the form
$ f(x,y)= \sum_{n,m}C_{n,m} \phi (x,E) g(y,E)$
if we impose the extra boundary conditions $ f( \frac{ax+b}{cx+d})=f(x)$ for the modular group
how can we calculate the energies and the eigenfunctions ?? , thanks.

REPLY [4 votes]: This question is about automorphic/modular forms, of course, because that "Laplacian" is the $SL(2,\mathbb R)$-invariant one on the upper half-plane, and, implicitly, the coordinate is $z=x+iy$, and the group action is on $z$, not $x$. Much more can be said in this particular case than for generic, non-compact, finite-volume Riemann surfaces.
The desired form of solution presumably dropped some subscripts... But, yes, $L^2$ solutions, and moderate-growth solutions, have expansions of the form
$\sum_n c_n\cdot W_\lambda(|n|y)\cdot e^{2\pi i n x}$, where $W_\lambda$ is (up to normalization) a Bessel function depending on the eigenvalue $\lambda$. This can be seen by noting that $SL(2,\mathbb R)$ contains the subgroup $\pmatrix{1&m\cr 0&1}$, which translates the real part $x$, thereby giving a Fourier expansion. The differential operator allows separation of variables, giving a differential equation in the imaginary part, $y$, which is essentially Bessel's equation.
Iwaniec' book Spectral methods of automorphic forms (AMS) talks about such stuff in an accessible form. My on-line modular forms class notes discuss such things, also.
It is non-trivial to prove that there is infinite point spectrum! Selberg did this c. 1956, by inventing his "trace formula". Still, even though now we know that the discrete spectrum does satisfy "Weyl's Law", the specific numerical features of the discrete spectrum seem mysterious. (Work of Phillips-Sarnak, Wolpert, and others shows that perturbations of the modular surface tend to kill off discrete spectrum.)
There is also significant continuous spectrum, spanned by "wave packets" of (non-$L^2$) Eisenstein series. For this modular curve, this is well-established and understood in the literature.


text:
\section{Smallest Cunningham Chain of the First Kind of Length 12}
Tags: Cunningham Chains

\begin{theorem}
The smallest [[Definition:Cunningham Chain of the First Kind|Cunningham chain of the first kind]] of [[Definition:Length of Sequence|length]] $12$ is:
:$554 \, 688 \, 278 \, 429$, $1 \, 109 \, 376 \, 556 \, 859$, $2 \, 218 \, 753 \, 113 \, 719$, $4 \, 437 \, 506 \, 227 \, 439$,
::$8 \, 875 \, 012 \, 454 \, 879$, $17 \, 750 \, 024 \, 909 \, 759$, $35 \, 500 \, 049 \, 819 \, 519$, $71 \, 000 \, 099 \, 639 \, 039$,
::$142 \, 000 \, 199 \, 278 \, 079$, $284 \, 000 \, 398 \, 556 \, 159$, $568 \, 000 \, 797 \, 112 \, 319$, $1 \, 136 \, 001 \, 594 \, 224 \, 639$
\end{theorem}

\begin{proof}
Let $C$ denote the [[Definition:Sequence|sequence]] in question.
We have that $554 \, 688 \, 278 \, 429$ is [[Definition:Prime Number|prime]].
First note that:
:$\dfrac {554 \, 688 \, 278 \, 429 - 1} 2 = 277 \, 344 \, 139 \, 214 = 2 \times 138 \, 672 \, 069 \, 607$
and so is not [[Definition:Prime Number|prime]].
Thus $554 \, 688 \, 278 \, 429$ is not a [[Definition:Safe Prime|safe prime]], and thus fulfils the requirement for $C$ to be a [[Definition:Cunningham Chain of the First Kind|Cunningham chain of the first kind]].
Then:
{{begin-eqn}}
{{eqn | n = 1
      | l = 2 \times 554 \, 688 \, 278 \, 429 + 1
      | r = 1 \, 109 \, 376 \, 556 \, 859
      | c = which is [[Definition:Prime Number|prime]]
}}
{{eqn | n = 2
      | l = 2 \times 1 \, 109 \, 376 \, 556 \, 859 + 1
      | r = 2 \, 218 \, 753 \, 113 \, 719
      | c = which is [[Definition:Prime Number|prime]]
}}
{{eqn | n = 3
      | l = 2 \times 2 \, 218 \, 753 \, 113 \, 719 + 1
      | r = 4 \, 437 \, 506 \, 227 \, 439
      | c = which is [[Definition:Prime Number|prime]]
}}
{{eqn | n = 4
      | l = 2 \times 4 \, 437 \, 506 \, 227 \, 439 + 1
      | r = 8 \, 875 \, 012 \, 454 \, 879
      | c = which is [[Definition:Prime Number|prime]]
}}
{{eqn | n = 5
      | l = 2 \times 8 \, 875 \, 012 \, 454 \, 879 + 1
      | r = 17 \, 750 \, 024 \, 909 \, 759
      | c = which is [[Definition:Prime Number|prime]]
}}
{{eqn | n = 6
      | l = 2 \times 17 \, 750 \, 024 \, 909 \, 759 + 1
      | r = 35 \, 500 \, 049 \, 819 \, 519
      | c = which is [[Definition:Prime Number|prime]]
}}
{{eqn | n = 7
      | l = 2 \times 35 \, 500 \, 049 \, 819 \, 519 + 1
      | r = 71 \, 000 \, 099 \, 639 \, 039
      | c = which is [[Definition:Prime Number|prime]]
}}
{{eqn | n = 8
      | l = 2 \times 71 \, 000 \, 099 \, 639 \, 039 + 1
      | r = 142 \, 000 \, 199 \, 278 \, 079
      | c = which is [[Definition:Prime Number|prime]]
}}
{{eqn | n = 9
      | l = 2 \times 142 \, 000 \, 199 \, 278 \, 079 + 1
      | r = 284 \, 000 \, 398 \, 556 \, 159
      | c = which is [[Definition:Prime Number|prime]]
}}
{{eqn | n = 10
      | l = 2 \times 284 \, 000 \, 398 \, 556 \, 159 + 1
      | r = 568 \, 000 \, 797 \, 112 \, 319
      | c = which is [[Definition:Prime Number|prime]]
}}
{{eqn | n = 11
      | l = 2 \times 568 \, 000 \, 797 \, 112 \, 319 + 1
      | r = 1 \, 136 \, 001 \, 594 \, 224 \, 639
      | c = which is [[Definition:Prime Number|prime]]
}}
{{eqn | n = 12
      | l = 2 \times 1 \, 136 \, 001 \, 594 \, 224 \, 639 + 1
      | r = 2 \, 272 \, 003 \, 188 \, 449 \, 279
      | c = which is $19 \times 119 \, 579 \, 115 \, 181 \, 541$ [[Definition:Prime Number|prime]]
}}
{{end-eqn}}
Establishing that this is indeed the smallest such [[Definition:Cunningham Chain of the First Kind|Cunningham chain of the first kind]] of [[Definition:Length of Sequence|length]] $12$ can be done by a computer search.
{{qed}}
\end{proof}

config: wiki


TITLE: What are the current breakthroughs of Geometric Complexity Theory?
QUESTION [16 upvotes]: I've read from Wikipedia about Geometric Complexity Theory (GCT) which (if I understood correctly) is a program for coping with the $ P=NP $ problem using algebraic methods.
That program seems promising, but I wonder what the current results of it  are.
So what are the current breakthroughs of Geometric Complexity Theory?

REPLY [18 votes]: Let $X=(X_{ij})_{1\le i,j\le n}$ be a generic $n\times n$ matrix and $F_1(X)={\rm det}(X)$ the degree $n$ homogeneous polynomial given by the determinant. Let
$$
F_2(X)=(X_{nn})^{n-m}\times {\rm perm}\left[(X_{ij})_{1\le i,j\le m}\right]
$$
which takes the permanent of an $m\times m$ submatrix and multiplies by one's favorite linear form in order to make another homogeneous polynomial of degree $n$ (one could also use the entry $X_{11}$ instead of $X_{nn}$).
This modification is called padding.
Then define the number
$$
c(m)=\min\{\ n\ |\ n\ge m\ \ {\rm and}\ \ \overline{G\cdot F_2}\subset \overline{G\cdot F_1}\ \}
$$
where $G$ is $GL(n^2)$ acting on the space $V$ of dimension $n^2$ where $X$ lives and therefore also on the space of degree $n$ polynomial functions of $X$. The $\overline{G\cdot F_i}$ are Zariski closures of orbits. The big conjecture in the area or Valiant's Hypothesis (a complex version of ${\rm P}\neq{\rm NP}$)
is that $c(m)$ grows faster than any polynomial in $m$.
Now if $\overline{G\cdot F_2}\subset \overline{G\cdot F_1}$, then one has a surjective $G$-equivariant map
$$
\mathbb{C}[\overline{G\cdot F_1}]_d\longrightarrow \mathbb{C}[\overline{G\cdot F_2}]_d
$$
between degree $d$ parts of the coordinate rings of these orbit closures.
So the game is to try to show that this does not happen, for $n$ insufficiently large relative to $m$, by proving the existence
of a multiplicity obstruction, i.e., an irreducible representation $\lambda$ for which multiplicities satisfy
$$
{\rm mult}_{\lambda}(\mathbb{C}[\overline{G\cdot F_1}]_d)<{\rm mult}_{\lambda}(\mathbb{C}[\overline{G\cdot F_2}]_d)
$$
or at the level of ideals
$$
{\rm mult}_{\lambda}(I[\overline{G\cdot F_1}]_d)>{\rm mult}_{\lambda}(I[\overline{G\cdot F_2}]_d)\ .
$$
An optimistic approach is to try to show there exist occurrence obstructions, i.e., $\lambda$'s such that
${\rm mult}_{\lambda}(\mathbb{C}[\overline{G\cdot F_1}]_d)=0$ and ${\rm mult}_{\lambda}(\mathbb{C}[\overline{G\cdot F_2}]_d)>0$. This hope has been squashed in the work of Bürgisser, Ikenmeyer and Panova mentioned by Timothy.
However, the possibility of multiplicity obstructions is still open.
I think the approach by Mulmuley is to try prove the existence of such multiplicity obstructions by leveraging all the tools available from representation theory for the computation of these multiplicities. Personally, I have never been a fan of this approach.
Having studied 19th century invariant theory in some depth, it seems more natural to me to approach the orbit separation problem using the explicit tools from that era. This article by Grochow seems to also point in a similar direction (I suspect the third article mentioned by Joseph is in the same vein).
In classical language (see Turnbull or Littlewood), one has to explicitly construct
a mixed concomitant which vanishes on $F_1$ but not on $F_2$. One also has to do this infinitely often (in $m$) in order to establish the superpolynomial growth property. Such a (degree $d$) concomitant is the same as a specific $G$-equivariant
map from your favorite model for the irreducible representation $\lambda$ to ${\rm Sym}^d({\rm Sym}^n(V))$ (Grochow calls that a separating module). Invariant theorists from the 19th century had two methods for generating such objects: elimination theory and diagrammatic algebra.
A very baby example where $F_1$ and $F_2$ are binary quartic forms under the action of $G=SL(2)$ (see this MO question) is say
$$
F_1(x,y)=x^4+8x^3y+24x^2y^2+32xy^3+16y^4
$$
and
$$
F_2(x,y)=16x^4-24x^3y+12x^2y^2-2xy^3\ .
$$
A separating concomitant (here in fact a covariant) is the Hessian of a generic binary quartic $F$
$$
H(F)(x,y)=\frac{\partial^2 F}{\partial x^2}\frac{\partial^2 F}{\partial y^2}-\left(
\frac{\partial^2 F}{\partial x\partial y}
\right)^2\ .
$$
It vanishes (identically in $x,y$) for $F=F_1$ but not for $F=F_2$.
In this case, the Hessian can be seen as an equivariant map form the irreducible given by the second symmetric power (of the fundamental two-dimensional representation)
into the coordinate ring for the affine space of binary quartics.
So a possible superoptimistic "plan" for GCT involves the following sequence of steps.

Find a way to generate tons of concomitants.

Identify some explicit candidates for the vanishing on $F_1$ and prove that property.

Show they don't vanish on $F_2$.

Step 1) is in principle solved by the First Fundamental Theorem for $GL(n^2)$ but there is a mismatch: the determinant is a natural object
in the invariant theory for $GL(n)\times GL(n)$ (acting on rows and columns) rather than $GL(n^2)$. One could try to repair the mismatch by expressing the basic building block for the invariant theory of $GL(n^2)$ in terms of the one for $GL(n)\times GL(n)$ (see this MO question for a similar reduction problem from $SL(n(n+1)/2)$ to $SL(n)$).
Guessing the right candidates for Step 2) looks hard to me.
Knowing beforehand that some multiplicities ${\rm mult}_{\lambda}(I[\overline{G\cdot F_1}]_d)$ are nonzero would definitely help.
Although, one could procrastinate and defer the proof of nonidentical vanishing of the concomitant to Step 3) which should show more than that anyway.
If one has such right candidates, showing they vanish on $F_1$
may be easy by arguments one could call Pauli's exclusion principle (contracting symmetrizations with antisymmetrizations),
high chromatic number property, or simply "lack of space".
However, I think the most difficult part is Step 3). For example, in my paper
"16,051 formulas for Ottaviani's invariant of cubic threefolds" with Ikenmeyer and Royle,
the guessing was done by computer search,
but with the right candidate in hand, the vanishing on $F_1$ was relatively easy to explain (it's a rather pretty example of chromatic number due to the global properties of the graph rather than some big clique).
The analogue of Step 3) in our article was done by brute force computer calculation and we still don't have a clue for why it is true.
The paradigmatic problem related to Step 3) is the Alon-Tarsi conjecture (see this MO question and this one too). In my opinion, one needs to make progress on that kind of question (the Four Color Theorem is of this type too, via a reduction due to Kauffman and Bar-Natan) before Valiant's Conjecture.
Since the question is about breakthroughs in GCT. I think this article by Landsberg and Ressayre also deserves some attention since it suggests that a reasonable guess for the exact value of $c(m)$ is
$$
\left(\begin{array}{c}2m\\ m \end{array}\right)-1\ .
$$
Note that a proof of concept for the explicit "Step 1),2),3) approach", on a much simpler problem, was given by
Bürgisser and Ikenmeyer in this article.
Finally, for more information on GCT, I highly recommend the review
"Geometric complexity theory: an introduction for geometers" by Landsberg.
PS: I should add that my pessimism is specific to the Valiant Hypothesis which is the "Riemann Hypothesis" in the field. Of course, one should not throw the baby with the bath water and denigrate GCT because it so far failed to prove this conjecture. There are plenty of more approachable problems in this area where progress has been made and more progress is expected. See in particular the above-mentioned article by Grochow and review by Landsberg.


TITLE: Understanding quaternions & gradient descent in a paper on inertial / magnetic sensor arrays
QUESTION [1 upvotes]: I hope this question is appropriate here!
I and a friend at work are trying to understand Sebastian Madgwick's paper, "An efficient orientation for inertial and inertial/magnetic sensor arrays" (paper linked here.) Our goal as I understand it is to implement the algorithm for a hardware bracelet in order to tell its orientation in space with a clear reference to earth i.e. canonical x, y, z axes. (As far as I know --  I'm just trying to help out with the math :)
The article uses quaternions as a way to model rotations on R^3 as well as to embed vectors in R^3. Sedgwick gives an analysis whereby he appears to use gradient dissent to minimize error. between I beleve the computed and actual position of the sensors, thus yielding a high accuracy result. What has confused us is how he arrives at equation (39). His gradient dissent algorithm seems to no longer need to account for the prior step in its algorithm. It seems like he is making an argument that the RHS of (39) is so large that we no longer need to account for the prior step in the algorithm, but intuitively this seems strange (couldn't you overshoot the minimum you are looking for by not accounting for the last value you were at? Especially when the increment is considered to be very large?)
I'd also find any references that make thinking about this sort of thing -- quaternions as rotations, gradient dissent and Jacobians / differentiation in this context -- easier. I woke up this morning reminded of Schey's work, "div, grad, curl and all that", which I used when teaching multivariable calc once, but I think it doesn't quite get into this detail? That was 10 years ago so I may just misremember :).
Any ideas would be welcome, either about the paper specifically or just references to better understand this domain. Also if this question is too applied for MathOverflow, please let me know and I will happily move it. (Also any suggested tags would be appreciated.)
Thank you!!!
-Michael

REPLY [2 votes]: Maybe this book will be helpful: Visualizing Quaternions, by Andrew J. Hanson


TITLE: Why am I getting three answers here for the same problem here?
QUESTION [0 upvotes]: Its probably a silly question. But , why am i getting three different answers for the same question, just by differentiating by three different methods?
P.S: Please do forgive my handwriting, i don't know how to use latex yet.

REPLY [0 votes]: Since the implicit functions for this curve equation are simple enough to extract, you might also view the matter this way.  The curve $ \ x^2y^2 = 1 \ $ can be expressed as $ \ y^2 \ = \ \frac{1}{x^2} \ \Rightarrow \ y \ = \ \pm \frac{1}{x} \ \ , $ which is the union of two "rectangular hyperbolas".  The derivatives for these two explicit functions are then $ \ \frac{dy}{dx} \ = \ \mp \frac{1}{x^2} \ \ . $  This expression tells us that $ \ \frac{dy}{dx} < 0 \ $ over the entire domain of the hyperbola $ \ y = \frac{1}{x} \ $  and $ \ \frac{dy}{dx} > 0 \ $ over $ \ y = -\frac{1}{x} \ \ . $
With these derivatives in hand, and that $ \ xy = 1 \ $ or $ \ xy = -1 \ , $ respectively for the two hyperbolas, we can manipulate the derivative expressions in various ways.  On $ \ xy = 1 \ \ , $ we then have everywhere
$$   \ \frac{dy}{dx} \ = \ - \frac{1}{x^2 · 1}  \ = \ - \frac{1}{x^2 · xy} \ = \ - \frac{1}{yx^3} $$ $$ \text{or} \ \ \ \frac{dy}{dx} \ = \ - \frac{xy}{x^2 }  \ = \ - \frac{y}{x} \ = \ - \frac{y·1^2}{x} \ = \ - \frac{y·(x^2y^2)}{x} \ = \ -xy^3 \ \ . $$
With the other hyperbola, the sign of the derivative and the sign of the product $ \ xy \ $ are reversed, so the same set of derivative expressions is produced (for instance, $   \ \frac{dy}{dx} \ = \ + \frac{1}{x^2 · 1}  \ = \ + \frac{1}{x^2 · \ (-xy)} \ = \ - \frac{1}{yx^3} \ ) \ \ . $  We see that all of these versions are equivalent (and we could create any number of others as well) and that we may choose any of them to work with for the complete curve.


TITLE: Relativistic Hamiltonian Formulations
QUESTION [5 upvotes]: Possible Duplicate:
Hamiltonian mechanics and special relativity?

The Hamiltonian formulation is beautifully symmetric. It's a shame that the explicit time derivatives in Hamilton's equations mean that the Hamiltonian formulation is not manifestly Lorentz-covariant. Is there any variant of the Hamiltonian formulation that is manifestly relativistic?

REPLY [2 votes]: The SHP (Stuckelberg, Horwitz, Piron) Hamiltonian formulation is manifestly covariant. The equations of motion are
$$\frac{\mathrm{d}x^\mu}{\mathrm{d} \tau} = \frac{\partial K }{\partial p_\mu}$$
$$\frac{\mathrm{d}p^\mu}{\mathrm{d} \tau} = - \frac{\partial K }{\partial x_\mu}$$
$K=K(x^\mu,p^\mu)$ is the covariant Hamiltonian and $\tau$ the invariant evolution parameter (in general it differs from proper time $s$). The basic monograph is Classical Relativistic Many-Body Dynamics


TITLE: About Poincare section for the double pendulum
QUESTION [4 upvotes]: I am reading Prof. Louis N. Hand's Analytical Mechanics. In the chapter about chaos, it introduces the concepts of Poincare section based on the example of double pendulum. Also, it plot the section when the initial angles for the upper and lower pendulum is small (about 1 degree for each). So in this case, the section contains some closed curves. I have few questions about the section.

Firstly, is it possible for some curves shown in the section crossing? Why and why not?
Secondly, in the plot it shows in the book (fig. 11.4), there are two fix points but they are not symmetric about the generalized momentum. Why is that?

Following the post found in https://mathematica.stackexchange.com/q/40122/, I have the code modified such that

REPLY [2 votes]: First. Yes, it is possible for the curves in Poincare section to cross. (I am assuming you mean generally). Remember, that Poincare section is a 2D projection of a 3D section of a 4D phase space. Regular, non-chaotic dynamics correspond to the winding of a $T_2$ torus embedded in this 4D space. Sections of such torus could be the couple of closed curves in the 3D hyperplane. Projecting this hyperplane on the 2D space could produce intersections. Additionally, Poincare section may include only points with a given directions of a hypersurface crossing, then the number of resulting curves would be reduced. Probably even then there still is a possibility for the intersecting curves in the Poincare section, although those would require highly nonlinear regimes (possibly after at least one bifurcation).
Second. Why would you expect the symmetry with respect to sign of the momentum? The Hamiltonian for the double pendulum contains the term proportional to the product of both momenta $l_\alpha l_\beta$. Therefore Hamiltonian and the torus in the 4D phase space are invariant only with respect to change of signs of both momenta,
but not with respect to a change of sign of a single momentum.  After making cross-section and projecting it onto 2D plane, no mirror symmetry w.r.t. momentum sign is to be expected in the resulting Poincare section.
Addition Upon discussion in the comments, I understand that the one unresolved issue is the set of initial data which generates the figure from the book.
In the book the figure is accompanied by the comment:

To produce Figure 11.4 and all subsequent Poincare sections,
  the initial conditions are varied. The total energy has the value $E = H(\alpha(0), 0, -\alpha(0), 0)$.

So, all those curves correspond to different sections arising from varied ICs with the fixed energy $E_0$. Here is my attempt at reproducing the figure:

Various colors correspond to an individual Poincare section defined by its own initial point. The point are chosen so that:
$$
\alpha=0, \quad \beta=0, \quad \dot{\alpha}>0, \quad H(0,l_\alpha,0,l_\beta)=E_0.
$$
The value of $l_\beta$ is chosen so that curves are evenly spaced between two limiting points corresponding to normal modes of oscillation. The last two conditions and known $l_\beta$ allow us to unambiguously calculate $l_\alpha$.
Finally, the limiting curve is defined by the equations:
$$
\alpha=0, \quad \dot{\alpha}=0, \quad H(0,l_\alpha,0,l_\beta)=E_0.
$$
The code for the picture is based on the linked question from Mathematica SE and can be seen here.


TITLE: Quantum chromodynamics - why are there no $rrb$ or $ggr$ terms?
QUESTION [0 upvotes]: $$\Psi_{colour}^{qqq} = \frac{1}{\sqrt{6}}(rgb + gbr + brg -grb - rbg - bgr)$$
My lecturer stated that there cannot be any $rrb$ or $ggr$ terms in the expression above. I would like to understand what the reason for this is?

REPLY [1 votes]: A physical hadron has to be colourless, to escape the strong force. This is more than just having net colour zero: it has to be a colour singlet.  What that means is that under any rotation in colour space, it must transform into itself (analogous to the S=0 M=0 combination of two one-half spins, as opposed to the S=1 M=0 triplet member, in 3D space).
Consider a small rotation $\epsilon$ about the $r$ axis. That causes $g \to g + \epsilon b, b \to b - \epsilon g, r \to r$ .  Put those into the 6 terms given and you will find that all the $\epsilon$ terms cancel.  The same is true for rotations about the $g$ and $b$ axes. That only happens because these specific terms, with correct minus signs, are written in the expression: including others like $rrg$ etc. will mess it up.


TITLE: When do infinitesimal deformations lift to global deformations?
QUESTION [1 upvotes]: Hi,
   I understand the notion of Infinitesimal and global deformations and the fact that global deformations lead to certain infinitesimal deformations. But I could not find any criterion or idea to understand when an infinitesimal deformation cannot be lifted to a global deformation.

REPLY [0 votes]: @Emerton and Paul: Thanks a lot for the answers. Sorry for being so vague. However the answers include what I was looking for. More specifically, I was interested in the following:
As I understand, infinitesimal deformation of a variety $V$ in $k^n$, means considering an ideal $I'(V)$ in $B:=k[X_1,...,X_n,t]/(t^2)$ such that the image of $I'(V)$ in $B/(t)$ is $I(V)$. So I guess if take the image in $B/(t-a)$ I should get ideals isomorphic to $I(V)$, hence with the same Hilbert polynomial. This is what I meant by global deformations coming from local deformations. But the confusion is that if this is true, then what happens when the family containing $V$ is not smooth (eg. non-reduced)? Would I still get an ideal isomorphic to $I(V)$?


TITLE: A dense orbit of the logistic map
QUESTION [2 upvotes]: In this question, it was determined in the answers that the logistic map $x_n: [0,1] \mapsto [0,1]$ defined by $x_{n+1} = 4x_n(1-x_n)$ has a dense orbit if and only if every finite binary string appears in the binary expansion of the point $x_0$. My question is, what is the equivalent condition for the sequence to be dense around the point $\dfrac{1}{2}$? In other words, for any arbitrarily small positive $\epsilon$, there exists an $n$ such that $|x_n-1/2| < \epsilon$.
For example, if $x_0 = \dfrac{1}{26}$ would it satisfy this property?

REPLY [2 votes]: Logistic map is semi-conjugate to the shift. If $F(t)=-(1/2)\cos 2\pi t+1/2$
then $F(2t)=f\circ F(t)$ where $f$ is your logistic map. To the point $1/2$
correspond $t_1=1/4$ and $t_2=3/4$, and $t\mapsto 2t$ just cuts the first digit
of the dyadic expansion. Therefore, if you want your orbit of $T_0$ to visit every neighborhood of $t_1$ or $t_2$ infinitely many times, it is necessary and sufficient that the dyadic expansion of $t$ contains sequences $001000000...$
or $011000000...$ with arbitrary many zeros in the tail.


TITLE: Chess Board and Knights
QUESTION [3 upvotes]: I can't solve this question. Anyone knows how I must solve?
You have a 6x3 chess board. How many forms exist to put a Knight in a square and with valid moviments pass in all squares but only one time in each one?
And with in a LxC chess board?

REPLY [4 votes]: In the book "Across the Board" (Watkins, 2004) there's a nice explanation why no knight's tour (i.e. starting and ending at the same square) exists on the $6 \times 3$ chessboard:

Suppose there is a knight's tour. One can connect the squares formed by the tour to form a necklace. If you remove one piece of a necklace, the necklace breaks, but is still in one piece. Taking another piece from the necklace then breaks the necklace into at most two parts. But removing the two squares marked $\ast$ above disconnects the two square marked $\times$ from the rest of the board by knight moves. So we get three components, which is a contradiction. So no knight's tour exists on a $6 \times 3$ chessboard.

This could be extended to show that no knight's path exists. If we have a chain (i.e. an open necklace), and we remove $6$ pieces of the chain, we should be left with at most $7$ components. But if we remove the six squares marked below:
$$\begin{matrix}
\cdot & \cdot & \times & \times & \cdot & \cdot \\
\cdot & \cdot & \times & \times & \cdot & \cdot \\
\cdot & \cdot & \times & \times & \cdot & \cdot \\ \end{matrix}$$
then we are left with $8$ components:
$$\begin{matrix}
1 & 2 & \times & \times & 5 & 6 \\
3 & 4 & \times & \times & 7 & 8 \\
2 & 1 & \times & \times & 6 & 5 \\ \end{matrix}$$
So there is no knight's path on a $6 \times 3$ chessboard either.


TITLE: Extension of isomorphisms of subring
QUESTION [4 upvotes]: Let $S$ and $R'$ be disjoint rings with the property that $S$ contains a subring $S'$ such that there is an isomorphism $f'$ on $S'$ onto $R'$.
Prove that there is a ring $R$ containing $R'$ and an isomorphism $f$ of $S$ onto $R$ such that $f'=f|_{S'}$.
So there are two parts here, first we need to find a ring $R$ containing $R'$, and next we need to construct an isomorphism between $S$ and $R$.
I was first thinking about $S\cup R'$ but this may not be a ring since their operations may be different. Next I think about $S\times R'$. Although it is a ring but $S$ is not isomorphic to it.

REPLY [1 votes]: What you want is $(S \setminus S') \cup R'$. The extension of $f'$ is the identity on $S \setminus S'$.
More details available if you need them.


TITLE: Tachyon black hole, would it repel gravitationally?
QUESTION [0 upvotes]: This question is not about tachyons entering normal matter BHs, nor is it about what would happen to tachyons in BHs. It is about a BH made of tachyons.
Tachyons on this site are considered quasi mainstream, so I will ask what would happen if we made a BH out of tachyons.
A tachyon has a spacelike worldline.

Nope, in both Newtonian physics and in general relativity, negative mass would be attracted gravitationally to positive mass, although negative mass would exert a repulsive gravitational effect on positive mass (but if the negative mass is small compared to the mass of the black hole this latter effect is negligible). In Newtonian physics this is not too difficult to derive, the Newtonian gravitational force law indicates the gravitational force vectors between a positive and negative mass would point away from each other, so the positive mass is obviously repelled, but for the negative mass the acceleration is in the opposite direction of the force due to the negative mass in F=ma, so the negative mass is attracted. In general relativity the analysis is obviously more complicated, but Hermann Bondi showed negative mass would have the same basic properties in GR, see this article. Note that if negative mass didn't fall downwards in gravity just like positive mass this would be a violation of the equivalence principle, since being in a chamber at rest in a gravitational field is supposed to be equivalent to being in a chamber accelerating in deep space, and if you let go of both a positive and negative mass in such a chamber they should naturally just move inertially while the floor of the chamber accelerates up to meet them.

What would happen if a negative mass crossed the event horizon of a black hole?
So basically both Newtonian and GR would suggest that a BH made of tachyons (negative mass), should repel gravitationally.

Since in GTR there is no gravitational stress-energy tensor, the gravitational energy of a body is determined indirectly. For example, one method is when by the given mass density and body size the mass of the body is calculated, first in the absence of the metric’s influence, and then taking into account the metric’s influence and the corresponding change in the volume differential in the integral of the mass under influence of the gravitational field. The difference of the mentioned masses is equated to the mass-energy of gravitational field as to the manifestation of the metric field.

Basically, as per GR, the stress-energy of this BH would be determined indirectly, given mass density (of the tachyons) and body size the mass of the BH is calculated, first in absence of the metric's influence, and then considering the metric, with the corresponding volume differential.
https://en.wikiversity.org/wiki/Gravitational_stress-energy_tensor
Gravitational coupling of tachyons
Based on this, the stress-energy of this BH could be so that the BH would repel gravitationally.

In special relativity, a faster-than-light particle would have space-like four-momentum,[3] in contrast to ordinary particles that have time-like four-momentum. Although in some theories the mass of tachyons is regarded as imaginary, in some modern formulations the mass is considered real,[6][7][8] the formulas for the momentum and energy being redefined to this end.

https://en.wikipedia.org/wiki/Tachyon
The mass of the tachyon could be regarded as real, allowing compatibility with GR.

Now, the momentum 4-vector of a Tachyon is
  necessarily space-like and consequently, the sign of energy no longer Lorentz-invariant. Thus a
  Tachyon will be found to have negative energy in certain inertial frames.

https://iopscience.iop.org/article/10.1088/1742-6596/196/1/012022/pdf
If the tachyon does have negative stress-energy, it could repel gravitationally.
Question:

Could a BH made of tachyons repel gravitationally (in case we agree they have negative mass)?

REPLY [2 votes]: The no-hair theorems tell us that the properties of a black hole are limited to mass, charge, and angular momentum. A black hole has no memory of what went into it, other than these three quantities.


TITLE: Bound on probabilities
QUESTION [2 upvotes]: Let $p_n$ be a set of probabilities so that $\sum_{n=1}^N p_n = 1$.
I believe that the following should hold:
\begin{align}
\left(\sum_{n=1}^N p_n^{\frac{k-2}{k-1}} \right)^{k-1}\leq N
\end{align}
for $k \geq 2$, with equality if $p_n = 1/N$. What is the easiest way to show this?

REPLY [3 votes]: Here is a way, using Jensen's inequality:
$$
\sum_{n=1}^N p_n^{\frac{k-2}{k-1}}
= N \cdot \frac{1}{N}\sum_{n=1}^N p_n^{\frac{k-2}{k-1}}
\leq N \cdot \left(\frac{1}{N}\sum_{n=1}^N p_n \right)^{\frac{k-2}{k-1}}
= N^{\frac{1}{k-1}}
$$
using that the function $x\mapsto x^{\frac{k-2}{k-1}}$ is concave.
In general, convexity/concavity+Jensen, Cauchy-Schwarz (or Hölder), and monotonicity of $\ell_p$ norms will get you a long way.