TITLE: A helpful proof in contracting the Christoffel symbol?
QUESTION [9 upvotes]: Out of all of my time learning General relativity, this is the one identity that I cannot get around.
  $$ \Gamma_{\alpha \beta}^{\alpha} = \partial_{\beta}\ln\sqrt{-g} \tag{1}$$
  where $g$ is the determinant of the metric tensor $g_{\alpha \beta}$.

With the Christoffel symbol, we start by contracting
$$ \begin{align}
 \Gamma_{\alpha \beta}^{\alpha} &= \frac{1}{2} g^{\alpha\gamma} (\partial_{\alpha} g_{\beta\gamma}  + \partial_{\beta} g_{\alpha\gamma}  - \partial_{\gamma} g_{\alpha\beta} ) \\
&= \frac{1}{2} g^{\alpha\alpha} ( \partial_{\beta} g_{\alpha\alpha}) \\
&= \frac{1}{2g_{\alpha\alpha}} ( \partial_{\beta} g_{\alpha\alpha})
\end{align}\tag{2}$$
where I took $\gamma \rightarrow \alpha$ and $g^{\alpha\alpha} = 1/g_{\alpha\alpha}$.
The next steps to take now, I have no clue. MTW gives a hint by saying to use the results from some exercise, which are,
$$\det A = \det||A^{\lambda}_{\ \ \rho}|| = \tilde{\epsilon}^{\alpha\beta\gamma\delta}A^{0}_{\ \ \alpha}A^{1}_{\ \ \beta}A^{2}_{\ \ \gamma}A^{3}_{\ \ \delta} $$
$$(A^{-1})^{\mu}_{\ \ \alpha}(\det A) = \frac{1}{3!}\delta_{\alpha\beta\gamma\delta}^{\mu\nu\rho\sigma} A^{\beta}_{\ \ \nu} A^{\gamma}_{\ \ \rho}A^{\delta}_{\ \ \sigma} $$
$$ \mathbf{d}\ln|\det A| = \mathrm{trace}(A^{-1}\mathbf{d}A) ,\tag{3}$$
where $\mathbf{d}A$ is the matrix $||\mathbf{d}A^{\alpha}_{\ \ \mu}||$ whose entries are one-forms.
I fail to reason why the metric turns into the determinant from what I have done and then becomes the result at the top.

REPLY [4 votes]: The original and the most general definition of determinant is given by Gauss
. For the determinant of metric tensor we write
\begin{eqnarray}
 g&:=& \frac{1}{4!}\varepsilon^{{\alpha\beta}{\gamma\delta}}\varepsilon^{{\mu\nu}{\rho\sigma}}g_{\alpha\mu}g_{\beta\nu}g_{\gamma\rho}g_{\delta\sigma}.\\
 \therefore \delta g &=& \frac{1}{3!}\varepsilon^{{\alpha\beta}{\gamma\delta}}\varepsilon^{{\mu\nu}{\rho\sigma}}g_{\alpha\mu}g_{\beta\nu}g_{\gamma\rho}\delta g_{\delta\sigma},\\
 &=&\frac{-g}{3!}\epsilon^{{\alpha\beta}{\gamma\delta}}\epsilon^{{\mu\nu}{\rho\sigma}} g_{\alpha\mu}g_{\beta\nu}g_{\gamma\rho}\delta g_{\delta\sigma},\\
 &=&\frac{-g}{3!}\epsilon^{{\alpha\beta}{\gamma\delta}}\epsilon_{{\alpha\beta}\gamma}{}^\sigma \delta g_{\delta\sigma},\\
 &=&g \, g^{\delta\sigma}\delta g_{\delta\sigma}.
\end{eqnarray}
(Equivalently, $$ \delta(\ln |\det g|)=Tr (g^{-1}\delta g)= Tr ( \delta \ln g )$$)
By using this result we the have
$$ \frac{1}{g} \partial_\beta g =g^{\delta\sigma}\partial_\beta g_{\delta\sigma} $$
Note: You may need some basic manipulation of these quantities, $\varepsilon$ is Levi-Civita symbol, $\epsilon$ is Levi-Civita tensor.
$$\epsilon^{{\alpha\beta}{\gamma\delta}} = -\frac{1}{\sqrt{-g}}\varepsilon^{{\alpha\beta}{\gamma\delta}}$$
$$\epsilon_{{\alpha\beta}{\gamma\delta}} = \sqrt{-g}\varepsilon_{{\alpha\beta}{\gamma\delta}}$$
$$\varepsilon_{{\alpha\beta}{\gamma\delta}}=\varepsilon^{{\alpha\beta}{\gamma\delta}}= \delta^{[\alpha}_0 \delta^\beta_1 \delta^\gamma_2 \delta^{\delta]}_3$$
$$\varepsilon^{{\alpha\beta}{\gamma\delta}}\varepsilon_{{\mu\nu}{\rho\sigma}} =-\epsilon^{{\alpha\beta}{\gamma\delta}}\epsilon_{{\mu\nu}{\rho\sigma}}= 4! \delta^{[\alpha}_\mu \delta^\beta_\nu \delta^\gamma_\rho \delta^{\delta]}_\sigma \equiv \delta^{\alpha\beta\gamma\delta}_{\mu\nu\rho\sigma}\equiv \left| \begin{matrix}
\delta^\alpha_\mu  & \delta^\alpha_\nu  & \delta^\alpha_\rho  & \delta^\alpha_\sigma  \\
\delta^\beta_\mu  & \delta^\beta_\nu  & \delta^\beta_\rho  & \delta^\beta_\sigma  \\
\delta^\gamma_\mu  & \delta^\gamma_\nu  & \delta^\gamma_\rho  & \delta^\gamma_\sigma  \\
\delta^\delta_\mu  & \delta^\delta_\nu  & \delta^\delta_\rho  & \delta^\delta_\sigma
\end{matrix} \right|$$
$$\epsilon^{{\alpha\beta}{\gamma\delta}}\epsilon_{{\alpha\nu}{\rho\sigma}} = -\left| \begin{matrix}
\delta^\alpha_\alpha  & \delta^\alpha_\nu  & \delta^\alpha_\rho  & \delta^\alpha_\sigma  \\
\delta^\beta_\alpha  & \delta^\beta_\nu  & \delta^\beta_\rho  & \delta^\beta_\sigma  \\
\delta^\gamma_\alpha  & \delta^\gamma_\nu  & \delta^\gamma_\rho  & \delta^\gamma_\sigma  \\
\delta^\delta_\alpha  & \delta^\delta_\nu  & \delta^\delta_\rho  & \delta^\delta_\sigma
\end{matrix} \right| =-\left| \begin{matrix}
1  & 0  & 0  & 0  \\
0  & \delta^\beta_\nu  & \delta^\beta_\rho  & \delta^\beta_\sigma  \\
0  & \delta^\gamma_\nu  & \delta^\gamma_\rho  & \delta^\gamma_\sigma  \\
0  & \delta^\delta_\nu  & \delta^\delta_\rho  & \delta^\delta_\sigma
\end{matrix} \right| =-\left| \begin{matrix}
 \delta^\beta_\nu  & \delta^\beta_\rho  & \delta^\beta_\sigma  \\
 \delta^\gamma_\nu  & \delta^\gamma_\rho  & \delta^\gamma_\sigma  \\
 \delta^\delta_\nu  & \delta^\delta_\rho  & \delta^\delta_\sigma
\end{matrix} \right|$$
$$\epsilon^{{\alpha\beta}{\gamma\delta}}\epsilon_{{\alpha\beta}{\rho\sigma}} = -\left| \begin{matrix}
 \delta^\beta_\beta  & \delta^\beta_\rho  & \delta^\beta_\sigma  \\
 \delta^\gamma_\beta  & \delta^\gamma_\rho  & \delta^\gamma_\sigma  \\
 \delta^\delta_\beta  & \delta^\delta_\rho  & \delta^\delta_\sigma
\end{matrix} \right|=-\left| \begin{matrix}
 2  & 0  & 0  \\
 0  & \delta^\gamma_\rho  & \delta^\gamma_\sigma  \\
 0  & \delta^\delta_\rho  & \delta^\delta_\sigma
\end{matrix} \right|=-2 \left| \begin{matrix}
 \delta^\gamma_\rho  & \delta^\gamma_\sigma  \\
  \delta^\delta_\rho  & \delta^\delta_\sigma
\end{matrix} \right| $$
$$\epsilon^{{\alpha\beta}{\gamma\delta}}\epsilon_{{\alpha\beta}{\gamma\sigma}} =-2 \left| \begin{matrix}
 \delta^\gamma_\gamma  & \delta^\gamma_\sigma  \\
  \delta^\delta_\gamma  & \delta^\delta_\sigma
\end{matrix} \right| =-2 \left| \begin{matrix}
 3  & 0  \\
  0  & \delta^\delta_\sigma
\end{matrix} \right| = -3! \delta^\delta_\sigma  $$
$$\epsilon^{{\alpha\beta}{\gamma\delta}}\epsilon_{{\alpha\beta}{\gamma\delta}} =-3! \delta^\delta_\delta = -4!$$
Note2:
\begin{eqnarray}
 g^{-1}&:=& \frac{1}{4!}\varepsilon_{{\alpha\beta}{\gamma\delta}}\varepsilon_{{\mu\nu}{\rho\sigma}}g^{\alpha\mu}g^{\beta\nu}g^{\gamma\rho}g^{\delta\sigma}.\\
 \therefore \delta g^{-1} &=& \frac{1}{3!}\varepsilon_{{\alpha\beta}{\gamma\delta}}\varepsilon_{{\mu\nu}{\rho\sigma}}g^{\alpha\mu}g^{\beta\nu}g^{\gamma\rho}\delta g^{\delta\sigma},\\
 &=&\frac{-g^{-1}}{3!}\epsilon_{{\alpha\beta}{\gamma\delta}}\epsilon_{{\mu\nu}{\rho\sigma}} g^{\alpha\mu}g^{\beta\nu}g^{\gamma\rho}\delta g^{\delta\sigma},\\
 &=&\frac{-g^{-1}}{3!}\epsilon_{{\alpha\beta}{\gamma\delta}}\epsilon^{{\alpha\beta}\gamma}{}_\sigma \delta g^{\delta\sigma},\\
 &=&g^{-1} \, g_{\delta\sigma}\delta g^{\delta\sigma}.\\
 \therefore -g^{-2}\delta g &=&g^{-1} \, g_{\delta\sigma}\delta g^{\delta\sigma}.\\
 \delta g &=& -g\,g_{\delta\sigma}\delta g^{\delta\sigma}.
\end{eqnarray}
By comparing this $\delta g$ to the first result, we have
$$\delta g_{\alpha\beta}=- g_{\alpha\mu}g_{\beta\nu} \delta g^{\mu\nu}$$


TITLE: How do I verify the Coq proof of Feit-Thompson?
QUESTION [79 upvotes]: I probably don't have the appropriate background to even ask this question.  I know next to nothing about formal or computer-aided proof, and very little even about group theory.  And this question is more "tech support" than math.
But: after reading that Georges Gonthier and collaborators had formalized a proof of the Feit-Thompson theorem in Coq, I thought it would be fun to try to verify it on my own computer.
So I installed Coq (8.4pl2), OCaml, etc, on my Ubuntu box, and then went here and downloaded feit_thompson.tar.gz (v2.0).  I unpacked it and ran make.  It ran various invocations of coqc for about two hours, and then finished with no error reported.  A bunch of .vo files were produced, including one for theories/stripped_odd_order_theorem.v, which even the casual observer can see contains what appears to be a statement of the Feit-Thompson theorem.
Is that it?  Does that mean that Coq agrees with the correctness of the proof?  Or are there more steps?
I was particularly curious because I saw that Makefile.coq has a validate target (which isn't run by just doing make).  So I ran make -f Makefile.coq validate.  This started a coqchk command which ran for about five minutes and then failed with:
 User error: Signature components for label fun_of_fin do not match
 make: *** [validate] Error 1

Was I not supposed to do that?  Or did I just find a flaw in their proof? :-)
Again, I freely admit that I do not really know what I'm doing.  In particular, it's not clear to me whether compilation of a Coq file actually verifies the theorems and proofs within it, or merely prepares it for later validation.  If an expert answers "Go away until you understand Coq" I'll meekly accept it.  But it does seem like it might be helpful for even a newbie to be able to understand how to properly run Coq on a proof created by someone else, and I couldn't find documentation that clearly explained it.

REPLY [26 votes]: Following Enrico Tassi's answer, I compiled the current snapshot of Coq 8.4pl4 from https://github.com/coq/coq.  With it, I was able to successfully compile and validate the proof.
 Modules were successfully checked

What an impressive piece of work!


TITLE: Log-Likelihood ratio derivation
QUESTION [0 upvotes]: I am struggling to understand the derivation of the log likelihood in the proof of Lemma 1 in Kauffman14 (https://arxiv.org/pdf/1407.4443.pdf).
I will give a bit of context: the lemma is about a lower bound on the sample complexity for multi-armed bandit through a change of measure argument. The log-likelihood in the derivation after $t$ rounds where the agent chooses arm $A_t$ and observe reward $Z_t$ is stated as
$$L_{t}=L_{t}\left(A_{1}, \ldots, A_{t}, Z_{1}, \ldots, Z_{t}\right):=\sum_{a=1}^{K} \sum_{s=1}^{t} \mathbb{1}_{\left(A_{s}=a\right)} \log \left(\frac{f_{a}\left(Z_{s}\right)}{f_{a}^{\prime}\left(Z_{s}\right)}\right)$$
My question is concerns how to write this from what I know to be the log likelihood ration
$$L_{t}=L_{t}\left(A_{1}, \ldots, A_{t}, Z_{1}, \ldots, Z_{t}\right):= \log \left(\frac{f_{a}\left(A_1,Z_1,\dots,A_t,Z_t\right)}{f_{a}^{\prime}\left(A_1,Z_1,\dots,A_t,Z_t\right)}\right)$$
Do they apply some sort of conditioning or some martingale properties on the expectation of the sigma algebra generated by the observations? It would be very helpful to see the complete derivation. Thanks a lot!

REPLY [1 votes]: They may have used three steps

$f_{a}\left(A_1,Z_1,\dots,A_t,Z_t\right) = \prod\limits_{s=1}^t   f_{a}\left(A_s,Z_s\right)$ assuming independence, and similarly for $f'$
$\log\left( \frac{\prod\limits_{s=1}^t f_{a}\left(A_s,Z_s\right)}{\prod\limits_{s=1}^t f'_{a}\left(A_s,Z_s\right)}\right)=\sum\limits_{s=1}^t \log\left( \frac{f_{a}\left(A_s,Z_s\right)}{f'_{a}\left(A_s,Z_s\right)}\right) $ using properties of logarithms
$\log\left(\frac{f_{a}\left(A_s,Z_s\right)}{f'_{a}\left(A_s,Z_s\right)}\right) = \sum\limits_{a=1}^K \mathbf 1_{(A_s=a)} \log\left(\frac{f_{a}\left(Z_s\right)}{f'_{a}\left(Z_s\right)}\right)$ since each $A_s$ only takes one value


TITLE: Matrix order in Dirac equations
QUESTION [3 upvotes]: The trace of matrix is always sum of its eigen values , which can be seen if $\hat{U}$ transforms the matrix $\alpha_i$ into it's diagonal form .
$$ \begin{pmatrix}
  A_1 & 0 & \cdots & 0 \\
  0 & A_2 & \cdots & 0 \\
  \vdots  & \vdots  & \ddots & \vdots  \\
 0 & 0 & \cdots & A_N
 \end{pmatrix}= \hat{U}\alpha_i\hat{U}^{-1} =tr\alpha_i\hat{U}\hat{U}^{-1} =tr\alpha_i
$$
Therefore we can say that order of matrices have to be even.
We also have following requirements from dirac equations
$$(\alpha_i \alpha_j+ \alpha_j \alpha_i)= 2\delta_{i\,j} I$$
$$\left\{ \alpha_i , \alpha_i \right\}= 2\delta_{i\,j} I$$
$$(\alpha_i \beta+ \beta \alpha_i) =0 $$
Now how can we prove that $\alpha$ and $\beta$ are 4*4 matrices? Please, don't skip too much.

REPLY [4 votes]: You can't prove it, because it isn't true. Given any 4 by 4 matrices that satisfy the Dirac algebra, you can make 8 by 8 matrices $\gamma_i$ which are equal to the four by four matrices in their upper diagonal, and also equal to the same thing in the lower diagonal. The goal is to look for the lowest dimensional representation, the smallest possible matrices. In this stupid trick, the upper 4 components of the spinor transforms exactly the same as the lower 4 components, meaning that you can reduce the representation by setting the top and bottom components to be equal. You are looking for an irreducible representation, which just means you can't do that.
To understand the size of the lowest dimensional representation Dirac matrices, there is a nice trick described in an article of Scherk's which does this for arbitrary dimensions. In Euclidean signature, make complex dimensions out of pairs of real dimensions
$$ z_1 = x_1 + i x_2 $$
$$ z_2 = x_3 + i x_4 $$
and then linearly combine the Dirac matrices as indicated by this coordinate change:
$$ \gamma'_1 = \gamma_1 + i \gamma_2 $$
$$ \gamma'_2 = \gamma_3 + i \gamma_4 $$
Then the $\gamma$ algebra in terms of the new $\gamma'$ matrices and their conjugates turns into the commuting Fermionic raising and lowering operators. These have a minimal representation which begins by defining the state $|0>$ which is annihilated by all lowering operators, and then acting the raising operators at most once, to produce states. This produces $2^n$ different states, where n is the dimension divided by 2, and this is the basic starting point, ignoring three annoying complications.
These states produced by raising and lowering give you the dimension of the spinor (ignoring the two complications) This starting point tells you that the Dirac algebra should be represented by 4 by 4 matrices in 4d, by 8 by 8 matrices in 6d, by 16 by 16 in 8d, 32 by 32 in 10d. Two to the power of half the dimensions.
The three annoying complications are: odd dimensions, Weyl Fermions, and Majorana Fermions, which each are a separate not so long discussion, but you can see how big the Dirac matrices are supposed to be from the argument above, at least approximately, and you can figure out the useful forms in 2d,3d, and 4d from just piddling around, as Pauli, Dirac, Weyl and Majorana did. The generalization to higher dimensions for string theory is the only time you need to get systematic about it, so it isn't discussed well outside of string theory literature.


TITLE: What is the main point in defining electric potential?
QUESTION [1 upvotes]: What I mean is that as the electric field is a necessity for us to define it because it has practical applications and similarly we need electric potential energy because it is the energy we require to move a unit charge from one place to another. So, when we have everything, why do we need to define electric potential?
I am not understanding its practical use.

REPLY [1 votes]: To elaborate on Mr Galalgher's answer,
In electro statics, two of maxwell's equations will suffice:
$\nabla\cdot\vec{E}=\rho/\epsilon_0$, equivalent to Coulomb's law, and, since we are only worried about static charge distributions there is no time varying Magnetic field:
$\nabla \times \vec{E}=0$
By a vector identity theorem, the second equation allows us to assert that $\vec{E}=-\nabla V$ for some scalar V.
Plug this into the first equation, we get:
$\nabla^2V=-\rho/\epsilon_0$.
This is a famous equation: Poisson's equation
There are several well known solutions depending on a system's symmetry.
Once you solve the differential equation for $V$, you can take its negative gradient to get the electric field.
Solution methods are available even if there is insufficient geometry to permit use of a Gaussian Surface, making it a more general procedure. There are also associated linear algebra techniques that make numerical solutions faster than a straight forward integration.

REPLY [0 votes]: In quantum mechanics and classical mechanics the potential is the relevant quantity. Also on electronics the potential is used and never the electric field. Thinking in terms of energy - and momentum - is more convenient than thinking in terms of force.


TITLE: $x_n \to a \iff x_n\in U$ where $U$ is an open neighborhood
QUESTION [1 upvotes]: Let there be a metric space $M$ and let $\{x_n\}_{n\in \mathbb{N}}$ be a sequence in $M$ and $a\in M$ then: $x_n\to a\iff$ for all open neighborhood $U$ of $a$ there is an $N$ s.t for all $N<n$ we have $x_n\in U$

$x_n\to a$ then, by definition for all $\varepsilon>0$ there is $N$ such that for all  $N<n$ we have $d(x_n,a)<\varepsilon$ $\iff$
$\iff$ for all $\varepsilon>0$ there is $N$ such that for all  $N<n$ we have $x_n\in B(a,\varepsilon)$ and every open ball is an open set or in every open set there is a ball contained in it (by definition of an open set) $\iff$
$\iff$ So $x_n\in B(a,\varepsilon)$ is and open set and $a\in B(a,\varepsilon)$ because $a\in B(a,\varepsilon)\to d(a,a)=0<\varepsilon$ and therefore $x_n\in B(a,\varepsilon)$ is an open neighborhood.
Is my reasoning correct?

REPLY [1 votes]: If $a \in U $ and $U$ open then there is some $\epsilon >0 $ such that $B(a, \epsilon )\subset U.$ We can take $n$ large enough so that $x_n $ is in the ball and so lies in $U$. Here we use the fact that $x_n$ converges to $a$.
For the converse it's simply because $B(a, \epsilon)$ is an open neighbourhood of $a$ for each $\epsilon >0$. So by assumption for each $\epsilon >0 $ we can find an $N \in \mathbb{N}$ such that if $n>N$ then $x_n \in B(a, \epsilon ).$ So $d(x_n, a )<\epsilon $.


text:
\section{Strategy for Finding Minimal Obstructions} \label{sec:strategy}

In this section, we describe the general strategy to show that every non-extendible partial
representation contains one of the obstructions described in
Section~\ref{sec:definition_of_obstructions}.

For any two disjoint subtrees $T_i$ and $T_j$ of the MPQ-tree $T$, we write $T_i \wlt T_j$ if and
only if there exist cliques $a \in T_i$ and $b \in T_j$ such that $a \wlt b$. In this situation, the
maximal cliques of $T_i$ are forced to appear on the left of the maximal cliques of $T_j$.

\heading{Testing Extendibility by MPQ-tree Reordering.}
Recall that a MPQ-tree $T$ represents all feasible orderings of the maximal cliques of a given
interval graph $G$. By Lemma~\ref{lem:repext_char}, a partial representation is extendible if and
only if there exists a reordering $T'$ of $T$ such that the frontier of $T'$ extends $\wlt$. This
condition can be tested by the following algorithm (see~\cite{kkosv}).

We process the MPQ-tree $T$ from the bottom to the root. When a P-node is processed, we test whether
there exists a linear extension of $\wlt$ on its subtrees. It exists if and only if $\wlt$ induced
on the subtrees of the node is acyclic. Thus, if there is a cycle, the MPQ-tree cannot be reordered
according to $\wlt$. When a Q-node is processed, there are two possible orderings of its subtrees,
and we check whether any of them is compatible with $\wlt$. The partial representation is extendible
if and only if all nodes can be reordered in this manner. See Fig.~\ref{fig:reordering_example} for
an example.

\begin{figure}[b!]
\centering
\includegraphics{strategy_reordering_example.eps}
\caption[]{This example is from~\cite{kkosv}, and it shows from left to right the way in which the reordering
algorithm works. We depict comparable pairs of maximal cliques by directed edges. The processed
trees are contracted into vertices.

\hskip 1em First, we reorder the highlighted P-node on the left. The subdigraph induced by $a$, $b$
and $c$ is ordered $b \rightarrow a \rightarrow c$. We contract this subtree $T_1$ into a vertex.
Next, we keep the order of the highlighted Q-node and contract its subtree $T_2$ into a vertex. When
we reorder the root P-node, the algorithm finds a two-cycle between $T_1$ and $T_2$, and outputs
``no''.}
\label{fig:reordering_example}
\end{figure}

A node that cannot be reordered is called \emph{obstructed}. A set of maximal cliques
\emph{creates} an obstruction if the ordering of this set in $\wlt$ makes the node
obstructed.

\heading{Strategy.} Suppose that a partial representation $\calR'$ is non-extendible.
From~\cite{kkosv}, we know that there exists an obstructed node in the MPQ-tree. We divide the
argument into three cases, according to the type of this node: \emph{an obstructed leaf}
(Section~\ref{sec:leaves}), \emph{an obstructed P-node} (Section~\ref{sec:p_nodes}), and \emph{an
obstructed Q-node} (Section~\ref{sec:q_nodes}).  Figure~\ref{fig:diagram} shows an overview of the
proof.

\begin{figure}[t!]
\centering
\includegraphics{strategy_diagram.eps}
\caption{Overview of the main steps of the proof, it starts in the middle. The obtained
obstructions are highlighted in gray, and three tools are depicted with highlighted borders. The
most involved case is in Section~\ref{sec:q_nodes_three_subtrees}.}
\label{fig:diagram}
\end{figure}

First, we argue that there exist at most three maximal cliques creating an obstruction.
Then, we consider their positions in the MPQ-tree and their open intervals from the definition of
$\wlt$. We use tools of Sections~\ref{sec:maximal_cliques} and~\ref{sec:interval_orders} to derive
positions of several pre-drawn intervals forming one of the obstructions.

In Section~\ref{sec:q_nodes_k_fat}, we prove a key tool called $k$-FAT Lemma~\ref{lem:k_fat}:
If three non-adjacent vertices $x_k$, $y_k$, and $z_k$ are pre-drawn in an order that is
different from their order in the sections of a Q-node, then they induce a $k$-FAT obstruction.
The proof is done by induction for $k$, and it explains why complicated obstructions are
needed.

config: arxiv


TITLE: Should one talk about THE Quantumfield
QUESTION [0 upvotes]: I studied QED and QCD and I know that there is an vacuum state for each theory $|{0}\rangle_{QED}, |{0}\rangle_{QCD}$ where we describe the particles as excitations of that vacuum state. But can we also say that this state is our quantum field and further is there one single vacuum state for the Standardmodel theory?
In other words: Does it makes sense if we talk about the one and only quantum field and all interactions and particle creations and anihilations are only excitations of this field?

REPLY [0 votes]: Quantum field theory just reflects our current understanding of the universe, which continuous to develop as time goes. We already know that the Standard Model is not the ultimate description of our universe. For example, quantum gravity is not included in the Standard Model. So even if we can unify QED and QCD and everything in the Standard Model into a Grand Unified Theory, it is still not the ultimate quantum field that can describe everything (although it can describe pretty much a lot of things we have known). As our knowledge of the universe increases, new quantum fields may be introduced (such as supersymmetry fields) or we may even go beyond quantum field theory (such as the string theory). If you believe that we can never know the ultimate truth of the universe, then we should not be obsessed with the idea of THE quantum field.
The vacuum state of a quantum field theory (QFT) is not unique. Even if we are given a well-formulated QFT, its ground state (or its vacuum state) is not unique. A QFT typically has infinitely many vacuum states related by symmetry or by large gauge transformations. We do not even know which vacuum state we are living in currently. There is no local detection/measurement to tell the different vacuum states apart.


TITLE: $ \sum_{n=1}^\infty ( n ( \sum_{k=n}^\infty \frac{1}{k^2})^2 - \frac1n) = \frac32 - \frac12 \zeta(2) + \frac32\zeta(3)$
QUESTION [1 upvotes]: Prove  $$\sum_{n=1}^\infty \left( n \left( \sum_{k=n}^\infty \frac{1}{k^2}\right)^2 - \frac1n\right) = \frac32 - \frac12 \zeta(2) + \frac32\zeta(3)$$

I decided to write the LHS of the equation as
\begin{align}
\sum_{n=1}^\infty \left( n \left( \sum_{k=n}^\infty \frac{1}{k^2}\right)^2 - \frac1n\right) &= \sum_{n=1}^\infty n \left( \left( \sum_{k=n}^\infty \frac{1}{k^2}\right)^2 - \frac{1}{n^2}\right)\\
&= \sum_{n=1}^\infty n \left(\sum_{k=n}^\infty \frac{1}{k^2} - \frac{1}{n^2}\right)\left( \sum_{k=n}^\infty \frac{1}{k^2} + \frac{1}{n^2}\right)
\end{align}
I'm not sure how to proceed from here. Some hints would be greatly appreciated!

REPLY [3 votes]: One way is to reduce the power of summand by summation by parts.
Take
$$a_n=\bigg(\sum_{k=n}^\infty\frac1{k^2}\bigg)^2-\frac1{n^2},\quad b_n=n,\quad B_n=\sum_{k=1}^nb_n=\frac12n(n+1),$$
we wish to find the closed form of
$$S=\sum_{n=1}^\infty a_nb_n.$$
By summation by parts, with some simplifications,
\begin{align*}
S&=\underbrace{\lim_{n\to\infty}a_nB_n}_{:=L=0}-
\sum_{n=1}^\infty B_n (a_{n+1}-a_n)\\
&=\sum_{n=1}^\infty \frac12\left(\frac1{n^2}+\frac1{n^3}+\frac1{n(n+1)}\right)
+\frac1n\sum_{k=n+1}^\infty\frac1{k^2}+\sum_{k=n+1}^\infty\frac1{k^2}-\frac1n\\
&=\frac12(\zeta(2)+\zeta(3)+1)
+\underbrace{\sum_{n=1}^\infty\sum_{k=n+1}^\infty\frac1{nk^2}}_{:=S'=\zeta(3)}
+\underbrace{\sum_{n=1}^\infty\bigg(\sum_{k=n+1}^\infty\frac1{k^2}-\frac1n\bigg)}_{:=S''=1-\zeta(2)}.
\end{align*}

Proof of $L=0$: By Stolz–Cesàro,
$$\lim_{n\to\infty}\frac{\sum_{k=n}^\infty\frac1{k^2}}{\frac1n}
=\lim_{n\to\infty}\frac{\frac1{n^2}}{\frac1{n(n+1)}}=1\quad
\implies\quad \sum_{k=n}^\infty\frac1{k^2}=\frac1n+o\left(\frac1n\right).$$
Thus,
$$\lim_{n\to\infty}a_nB_n=\lim_{n\to\infty}o\left(\frac1{n^2}\right)\frac{n(n+1)}2=0.$$

Proof of $S'=\zeta(3)$:
\begin{align*}
\sum_{n=1}^\infty\sum_{k=n+1}^\infty\frac1{nk^2}
&=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{n(n+k)^2}\\
&=\frac12\sum_{n=1}^\infty\sum_{k=1}^\infty\left(\frac1n+\frac1k\right)\frac1{(n+k)^2}\\
&=\frac12\sum_{n=1}^\infty\sum_{k=1}^\infty\frac1{nk}\int_0^1 x^{n+k-1}\,\mathrm dx\\
&=\frac12\int_0^1\frac1x\sum_{n=1}^\infty\frac{x^n}n\sum_{k=1}^\infty\frac{x^k}k\,\mathrm dx\\
&=\frac12\int_0^1\frac{\ln^2(1-x)}x\,\mathrm dx\\
&=\zeta(3).
\end{align*}

Proof of $S''=1-\zeta(2)$: We consider the partial sum. Summation by parts gives
$$\sum_{n=1}^\ell\sum_{k=n+1}^\infty\frac1{k^2}
=\ell\sum_{n=\ell+1}^\infty\frac1{k^2}
+\sum_{n=1}^{\ell-1}\frac n{(n+1)^2}.$$
As $\ell\to\infty$,
$$\sum_{n=1}^\infty\bigg(\sum_{k=n+1}^\infty\frac1{k^2}-\frac1n\bigg)
=\underbrace{\lim_{\ell\to\infty}\ell\sum_{n=\ell+1}^\infty\frac1{k^2}}_{=1}
+\underbrace{\sum_{n=1}^{\infty}\bigg(\frac n{(n+1)^2}-\frac1n\bigg)}_{=-\zeta(2)}.$$


TITLE: How to compute confidence intervals and standard error for nonlinear regression with three parameters?
QUESTION [0 upvotes]: I have been working on a personal project trying to emulate the nonlinear regression functionality of Mathematica for three free parameters. I am able to accurately fit functions, yet I am unsure how to report the standard error and confidence intervals like Mathematica.
See here:

Whenever I apply the algorithms in the below tutorials for three free parameters, I get imaginary numbers as my answers:
https://www.youtube.com/watch?v=3IgIToOV2Wk
https://www.youtube.com/watch?v=y6bCBQhwtKQ&list=RDCMUCKVGxWqAcyGibKC2RKD19RQ&start_radio=1&rv=y6bCBQhwtKQ&t=244
Is there a problem with using the above algorithms, which are demonstrated with two parameters, for three parameters? How would I compute the standard error and confidence intervals for three parameters?
If it is helpful, here is the function that has given me imaginary results:
$$c * x^b * e^{(a * log10(x)^2)}. $$

REPLY [0 votes]: May I recommend a Bayesian inversion for estimating confidence intervals on non-linear regression problems.  See Data Analysis, A Bayesian Tutorial, D. S. Silva.  Oxford Press (2006).


TITLE: Equilibrium point of a system described by a Lagrangian
QUESTION [1 upvotes]: I have the Lagrangian $L=\frac{m}{2}\Big(\dot{r}^2+r^2\dot{\theta}^2\Big) - \kappa\big(r^2+a^2-ar(1+\cos\theta)\big)+mgr\sin\theta$ describing a system. In this system a mass is displaced slightly from equilibrium so that we can write $r=R+\epsilon$ and $\theta=\theta_0+\delta$. I am also given that $\epsilon,\delta,\dot{\epsilon},\dot{\delta}$ are all much less than zero. I need to find approximate linear equations of motion for $\epsilon$ and $\delta$.
Firstly I need to find the position $r=R$, $\theta=\theta_0$ for which a particle remains at rest in this system. From the Euler-Lagrange equations and setting $\dot{r}=0, \dot{\theta}=0$ I end up with the $2$ equations:
$-2kr+ka(1+\cos\theta)+mg\sin\theta=0$, from the E-L equation for the coordinate $r$, and
$-kar\sin\theta +mgr\cos\theta=0$, for the E-L equation for $\theta$.
So from the second one I find that $\tan\theta_0=\frac{mg}{ka}$. After making use of $1+\tan^2\theta=\sec^2\theta$ and a bit of tedious algebra I get the pretty ugly expression $R=\frac{1}{2}\Big(\frac{k^2a^2+m^2g^2}{k\sqrt{1+m^2g^2}}+a\Big)$.
Are these correct? If so how do I proceed from here? I have a feeling that I am supposed to substitute in $R$, $\theta_0$ and Taylor expand, disregarding the $O(\epsilon^2), O(\delta^2)$ terms but this makes everything pretty messy so I am a bit stuck.

REPLY [0 votes]: What is $\kappa$ ?
Assuming that $\kappa$ is a Lagrange multiplier to handle the holonomic constraint $r^2+a^2-ar(1+\cos\theta)=0$ the movement equations are
$$
\left\{
\begin{array}{rcl}
 -a \kappa  (\cos (\theta)+1)+g m \sin (\theta)+r \left(m \theta '^2+2 \lambda \right)-m r'' & = & 0\\
 r \left(g m \cos (\theta)+a \kappa \sin (\theta)-2 m r' \theta '-m \theta ''\right) & = & 0\\
 2 r'^2+2 a \sin (\theta (t)) \theta ' r'-a (\cos (\theta)+1) r''(t)+r \left(a \cos (\theta)\theta '^2+2 r''+a \sin (\theta) \theta ''\right) & = & 0\\
\end{array}
\right.
$$
here the last equation is the holonomic constraint derived twice regarding $t$. Now after solving for $(r'',\theta'',\kappa)$ we have
$$
\kappa = \frac{m \left(a g \sin (\theta)+r(t) (a-2 r) \theta '^2-2 g r \sin (\theta )-2 r'^2\right)}{2 a^2
   (\cos (\theta)+1)-4 a r (\cos (\theta)+1)+4 r^2}
$$


TITLE: Sample variance formula vs. Population variance formula usage
QUESTION [1 upvotes]: Just for sake of not having to write this equation out a few times, I will denote the residual sum of squares for a set of data points $X$ to be
$$
RSS= \sum_{x\in X} \left(x - \overline{x}\right)^2
$$
Now I understand that there are two different formulas for variance, with sample variance being defined as
$$
s^2 = \dfrac{RSS}{n-1}
$$
and population variance being
$$
\sigma^2 = \dfrac{RSS}{n}
$$
for $n$ data points.
Now I understand that we use the sample formula, as the population formula is a biased estimator for the true variance.
Now with that I was listening to one of my machine learning lectures and saw that my professor told us to calculate the variance for Naive Baysian classifier using the population formula since it is proven to be the Maximum Likelyhood Estimator for Gaussian Distribution variance. I also understood this proof, but I was confused about why we would select to use the formula for population variance on our sample when we know it is a biased estimator?
Anyone who could provide some insight would be appreciated!

REPLY [1 votes]: For reasonably large n it is not going to make a significant difference.
The "sample variance" is not the best(optimal) estimator for the variance. The optimality criteria being MSE(Mean Square Error). The estimator with lower MSE
is considered to be better. It is more accurate.

Item 2 in the list requires more explanation. Let's denote
$$ \hat{[\sigma}^{*}]^2 = \frac{1}{k-1}\sum_k (x_k - \bar{x})^2 \tag{1} \label{1}$$
$$ \hat{\sigma}^2 = \frac{1}{k}\sum_k (x_k - \bar{x})^2 \tag{2} \label{2} $$
The MSE for $\eqref{1}$ is $e^{*} = E[(\hat{[\sigma}^{*}]^2 - \sigma^2)^2]$
The MSE for $\eqref{2}$ is $e = E[(\hat{\sigma}^2 - \sigma^2)^2]$
Where $\sigma^2$ is true variance.
It can be shown that $e^{*} > e$. Although $\eqref{2}$ is biased it has lower MSE.


TITLE: what kind of E this is?
QUESTION [1 upvotes]: I could not figure out what kind of expectation is C because clearly it is not for a binomial distribution.
Any help would be appreciated a lot!
Thanks,

REPLY [3 votes]: By the binomial theorem, $$
\left(1+{x\over2}\right)^{100}=\sum_{k=0}^{100}{100\choose k}{x^k\over2^k}$$
Differentiate both sides:$$
50\left(1+{x\over2}\right)^{99}=\sum_{k=1}^{100}k{100\choose k}{x^{k-1}\over2^k}$$
Set $x=1$:
$$50\cdot\left(\frac32\right)^{99}=\sum_{k=1}^{100}k{100\choose k}\left(\frac12\right)^k$$


TITLE: Linear Algebra: if matrices $\mathbf{A} + \mathbf{B} = \mathbf{I}$, then do $\mathbf{A}$ and $\mathbf{B}$ have the same eigenvectors?
QUESTION [0 upvotes]: Just had a quick question which I think ought to be simple, but I can't fully convince myself of the result.
Question: If matrices $\mathbf{A} + \mathbf{B} = \mathbf{I}$, ($\mathbf{I}$ is the identity matrix) then do $\mathbf{A}$ and $\mathbf{B}$ have the same eigenvectors? Let us assume that $A$ and $B$ are invertible and that they have real-valued entries.
Context: The context of this question is in graph theory where we have the graph Laplacian $L$, which can be expressed as the difference between the degree matrix $D$ and the adjacency matrix $A$ in: $L = D - A \rightarrow D = L + A$. Then this can be 'normalized' to form symmetric matrices $A_{\text{sym}} = D^{-\frac{1}{2}}AD^{-\frac{1}{2}}$ and $L_{\text{sym}} = D^{-\frac{1}{2}}LD^{-\frac{1}{2}}$ and then the equation becomes:
$$ I = L_{\text{sym}} + A_{\text{sym}} $$
and I have read that $L_{\text{sym}}$ and $A_{\text{sym}}$ can both be diagonalized with the same eigenbasis, so I was wondering why. I suppose the problem simplifies to any two matrices $A$ and $B$, so I just took the context out of the question.
Attempt: From thinking about it, all real vectors $\mathbf{v}$ satisfy $I\mathbf{v} = \mathbf{v}$, so
$$ I\mathbf{v} = \mathbf{v} = (A + B)\mathbf{v} \rightarrow \mathbf{v} = A\mathbf{v} + B\mathbf{v} $$
Then I don't know how to proceed from there - maybe some sort of proof by contradiction (come from an engineering background, some haven't seen these types of proofs since before college/university).
Any help is appreciated. If the question/understanding is wrong, please let me know.

REPLY [5 votes]: You can write $B=I−A$. So eigenvectors of $A$ are also eigenvectors of $B$ and vice versa.


TITLE: Need to find the diameter of graphs
QUESTION [0 upvotes]: I got a question on my assignment and I'm really not sure where to start or even what some of these graphs are so I'm hoping someone will be able to help me.

I need to find the diam of $K_{n}$, $P_{n}$, $C_{n}$ and $B_{3}$.

REPLY [1 votes]: Hint: The distance between two vertices is the length of the shortest path between them. The diameter is the longest distance in the graph.