text:
\begin{document}

\maketitle
\section{Linear Solvers} \label{sec:solvers}
In this section, we discuss how to solve the linear systems obtained by the TPFA scheme from the main part of the paper. The TPFA scheme solves the unknown $\mathfrak{p}_h$ first, after which the unknowns $\mathfrak{q}_h$ can be recovered.  Therefore, we discuss multigrid methods for the TPFA scheme. In this section, we will solely be dealing with the discrete system, and thus omit the subscripts on $\mathfrak{q}_h$ and $\mathfrak{p}_h$.

The TPFA scheme~\eqref{eqn:TPFA}, following the discussion in Section~\ref{subsec:TPFA},  can be obtained by block Gaussion elimination of the mass lumping mixed-formulation, i.e., its compact matrix form is
\begin{equation} \label{eqn:TPFA-poisson}
\mathfrak{G}^T (\mathfrak{D}_{\bm{\mathfrak{q}}})^{-1} \mathfrak{G} \mathfrak{p} = \mathfrak{r}.
\end{equation}
As we can see, this is a discretization of a Poisson type problem for the unknown $\mathfrak{p}$ in the mixed-dimensional setting.  Since multigrid (MG) methods are designed for solving diffusion type problems efficiently~\cite{TrottenbergOosterleeSchuller2001,XuZikatanov2017}, in our work, we use an algebraic variant of the MG methods to solve the resulting linear system. More precisely, we use the aggregation-based AMG method, which is a suitable choice for solving both Poisson-type problems, see~\cite{Blaheta1986,KimXuZikatanov2003}, and graph Laplacian problems, see~\cite{ColleyLinHuAeron2017,LinCowenHescottHu2018a,HuLinZikatanov2019}.

To validate the efficiency of multi-grid for this problem, we include the solver performance on the three test cases from the manuscript in the tables below.

The performance of the iterative solver for Case 1A is summarized in Table \ref{tab:Case1B-solver}.  The results for Case 1B are essentially identical and are omitted (the deviation is less than $5\%$ in all quantities) . This test case, due to the simplicity of the network, is from a linear algebra perspective almost identical to the 2D homogeneous Poisson problem.  Therefore, the aggregation-based AMG preconditioner performs as expected. Here, the grid complexity (GridComp.) is defined as the ratio between the total size of the matrices on different levels and the size of the matrix on the finest level.  The operator complexity (Oper.Comp.) is defined as the ratio between the total number of nonzeros of the matrices on different levels and the number of nonzeros of the matrix on the finest level.  As we can see, both Grid Complexity and Operator Complexity are bounded by $2$, which means that the aggregation-based AMG method roughly preserves the sparsity on the coarse levels and the overall computational cost of one V-cycle AMG method is optimal (i.e. linearly proportional to the cost of one step of Gauss-Seidel method).  Since we use a simple unsmoothed aggregation AMG method here, the number of iterations grows with respect to the size of the linear system as expected. Overall, the computational cost of the multilevel solver is sub-optimal.  We want to point out that it is possible to obtain optimal computational complexity when a more sophisticated AMG cycle, such as the K-cycle is used~\cite{Vassilevski2008,HuVassilevskiXu2013}. This is a subject of our ongoing and future research since developing special tailored multilevel solvers for mixed-dimensional problems is a challenging and important task itself and beyond the scope of this work.

\begin{table}[]
\scriptsize
\begin{center}
\begin{tabular}{ l  l  l  l  l  l  l  l }
$1/h$ & NLevel & GridComp. & Oper.Comp. & NIter & CPUsetup & CPUSolve & DOF\\
\hline
$16$ &     2 & 1.33 & 1.42 &     19 & 2.14e-03 & 1.54e-02 & 256\\
$32$ &     4 & 1.59 & 1.79 &     30 & 7.50e-03 & 2.32e-02 & 1024\\
$64$ &     5 & 1.66 & 1.90 &     44 & 2.47e-02 & 1.04e-01 & 4096\\
$128$ &     7 & 1.71 & 1.97 &     66 & 9.27e-02 & 3.60e-01 & 16384\\
$256$ &     8 & 1.73 & 2.00 &     95 & 2.96e-01 & 1.66e+00 & 65536\\
$512$ &     9 & 1.74 & 2.02 &    140 & 1.32e+00 & 9.26e+00 & 262144\\
\hline
\end{tabular}
\end{center}
\caption{Solver performance on Case 1A. (NLevel: number of levels; GridComp.: grid complexity; Oper. Comp.: operator complexity; NIter: number of iterations; CPUsetup: CPU time for setup AMG; CPU solve: CPU time for solving using AMG; DOE: degrees of freedoms.) }\label{tab:Case1B-solver}
\end{table}

The solver results for case 2 are presented in Table \ref{tab:Case2-solver}, including the performance on the reference solution, which has more than 33 million degrees of freedom.  Again, we observe that both grid complexity and operator complexity are bounded while the number of iterations grows sublinearly with respect to the size of the linear systems.  This means the overall computational complexity is sub-optimal, which is expected since we are using a simple unsmoothed aggregation AMG preconditioner.

\begin{table}[]
\scriptsize
\begin{center}
\begin{tabular}{ l  l  l  l  l  l  l }
$1/h$ & NLevel & GridComplexity & OperatorComplexity & NIter & CPUAMGsetup & CPUSolve\\
\hline
    $16$ &     06 & 1.49e+00 & 1.76e+00 &     24 & 4.53e-02 & 6.35e-02\\
    $32$ &     08 & 1.58e+00 & 1.93e+00 &     33 & 3.93e-01 & 5.39e-01\\
    $64$ &     10 & 1.63e+00 & 2.02e+00 &     43 & 3.96e+00 & 5.02e+00\\
    $128$ &     12 & 1.65e+00 & 2.07e+00 &     61 & 4.04e+01 & 8.55e+01\\
    $256$ &     14 & 1.66e+00 & 2.09e+00 &     83 & 3.72e+02 & 1.05e+03\\
\hline
\end{tabular}
\end{center}
\caption{Solver performance on Case 2. See \ref{tab:Case1B-solver} for complete caption information. } \label{tab:Case2-solver}
\end{table}

The computational performance on Case 3, which consists of real data, is summarized in Table \ref{tab:solver-full-brain}. All results are reported relative to the discretization level of the data, which we label $h_0$. The results confirm that even for this large data-set, with data-driven and physuiologically realistic parameters, the solver performs efficiently.  The number of iteration grows sublinearly and the CPU time also indicates that the current solver is sub-optimal.  However, we do not see the degradation of the AMG solver comparing with the previous two cases with simpler geometry, which demonstrates that our discretization is stable and solver-friendly. Moreover, the efficiency of the discretization and solver combinations proposed above allow for analysis of full-resolution medical data.

\begin{table}[]
\scriptsize
\begin{center}
\begin{tabular}{ l l l l l l l l }
$h_0/h$ & NLevel & GComp & OComp & NIter & CPUsetup & CPUSolve & DOF\\
\hline
0.064 &     05 & 1.27 & 1.13 &     68 & 4.14e-02 & 2.64e-01 &   2206\\
0.13 &     06 & 1.12 & 1.02 &    108 & 2.20e-01 & 2.74 &  17179\\
0.25 &     08 & 1.15 & 1.04 &    113 & 2.37 & 22.8 & 137378\\
0.51 &     10 & 1.27 & 1.09 &    122 & 23.4 & 200 & 1103295\\
1 &     12 & 1.53 & 1.36 &    138 & 317 & 2.74e+03 & 8792747\\
\hline
\end{tabular}
\end{center}
\caption{Solver report whole brain simulation. Note that we set $h_0$ as the resolution of the \textit{finest} grid, since this is the resolution of the data. Also, the DOF column reflects that due to the nontrivial shape of the brain, the number of active cells in the computation does not scale exactly with the cube of the grid resolution. }
\label{tab:solver-full-brain}
\end{table}

\section{Derivation of reference solution for section 5.1}

The solution is valid for any domain $\Omega$ containing the origin and with zero Neumann boundary conditions, and for a tree consisting of two nodes: Node 0 is a Dirichlet node and node 1 is a terminal node. Please refer to Figure \ref{fig:case1_illustration} for a visual illustration.

\subsection*{Setup}

Consider polar coordinates $(r,\theta)$ around the origin, included in $\Omega$. In addition to Neumann boundary conditions on $\Omega$, the boundary pressure $p^N_0$ in the Dirichlet node is taken as a given, and we consider for simplicity zero source terms in the terminal node, $r^N_1=0$.

Then we construct up to four concentric domains, bounded by $0 < r_0 \leq r_1 \leq r_2 < r_3 \leq \mathrm{diam}\ (\Omega)$. The inner domain will be a disc, the remaining domains will be doughnuts. We consider homogeneous $k^D(\bm{x})=k^D$, while the remaing parameters in $\Omega$ are strictly functions of $r$. In particular, we choose a transfer permeability which is non-zero only in the two inner-most regions

\begin{equation}\label{eqn:kT-form}
k^T (r) = k^T_0
\begin{cases}
1 \quad \mathrm{for}\quad  r\leq r_0 , \\
a_0^2 \frac{r_1^2 -r^2}{r^2} \quad \mathrm{for}\quad  r\leq r_1, \\
0 \quad \mathrm{for}\quad  r > r_1.
\end{cases}
\end{equation}

Continuity of the transfer function at $r_0$ is obtained by setting
\begin{equation}
a_0^2 = \frac{r_0^2}{r_1^2-r_0^2}
\end{equation}

The source is chosen as non-zero only between $r_2$ and $r_3$,

\begin{equation}
r^D (r) = r^D_0 (r-r_2)^+ (r_3-r)^+,
\end{equation}

where $(a)^+\equiv \max (0,a)$ indicates that only positive values are considered, .

The governing equations in the domain $\Omega$ can be stated in radial coordinates as (our parameters are radially symmetric, as will our solution be, and we drop $\theta$ in the continuation):

\begin{equation} \label{eqn:laplace-rad}
\frac{1}{r}\frac{d}{dr}\left(- r k^D \frac{dp^D}{dr} \right) - q^T = r^D
\end{equation}

where we recall that

\begin{equation}\label{eqn:qT-rad}
q^T = -k^T (p^D-p^N_1)
\end{equation}

\subsection{Solutions}

First, we note that by volume balance, we have that
\begin{equation}
q^N \equiv q^N_{0,1} = -2\pi r_0^D \int_{r_2}^{r_3}r (r-r_2) (r_3-r)= -2\pi  r^D_0\left(\frac{r_3^4-r_2^4}{12} - r_2r_3\frac{r_3^2-r_2^2}{6}\right).
\end{equation}

It then follows that the pressure $p^N_1$ at the terminal node is given by

\begin{equation}
p^N_1 = p^N_0 - k^N q^N
\end{equation}

The combination of equations \eqref{eqn:laplace-rad} and \eqref{eqn:qT-rad} have analytical solutions within each region of the domain. These are as follows (these are reported in numerous text books, we will use Chapter 9 of Abramovich and Stegun as a concrete reference for the use and manipulation of Bessel functions \cite{abramowitz1948handbook}).

\emph{Solution S1}: For $r\leq r_0$:  By substitution of equation \eqref{eqn:qT-rad} into equation \eqref{eqn:laplace-rad}, we obtain
\begin{equation}
r\frac{d}{dr}\left(r\frac{dp^D}{dr} \right) - \frac{k^T_0}{k^D} r^2 (p^D-p^N_1) = 0,
\end{equation}

the solution of which is the modified Bessel function of the first kind $I_0$, since we easily determine that there is no singularity at the origin:

\begin{equation}
p^D(r) - p^N_1 = c_I I_0 (\kappa_0 r)
\end{equation}

where

\begin{equation}
\kappa_0 = \sqrt{\frac{k^T_0}{k^D}}
\end{equation}
The constant $c_I$ will be determined later.  The derivative of presssure is obtained according to according to the differential rules for Bessel functions of integer order (see equation 9.6.27 in \cite{abramowitz1948handbook1}):

\begin{equation}
\left. I_0'\right|_{\kappa_0 r_0}  = I_1(\kappa_0 r_0)
\end{equation}

Thus the radial component of flux is given by

\begin{equation}
q_r^D = - k^D c_I \kappa_0 I_1(\kappa_0 r_0)
\end{equation}

\emph{Solution S2}: For $r_0< r\leq r_1$:  By substitution of equation \eqref{eqn:qT-rad} into equation \eqref{eqn:laplace-rad}, we obtain in this region
\begin{equation}
r \frac{d}{dr}\left(r\frac{dp^D}{dr} \right) + \frac{k^T_0}{k^D}a_0^2 (r^2-r_1^2) (p^D-p^N_1) = 0,
\end{equation}

the solution of which are the (unmodified) Bessel functions of the first and second kind $J$ and $Y$, of fractional order:

\begin{equation}
p^D(r) - p^N_1 = c_J J_{\kappa_1 r_1} (\kappa_1 r) + c_Y Y_{\kappa_1 r_1} (\kappa_1 r) = \sum_{\mcC=J,Y}c_\mcC \mcC_{\kappa_1 r_1} (\kappa_1 r)
\end{equation}

where $\mcC$ indexes the Bessel functions $J$ and $Y$, and

\begin{equation}
\kappa_1 = a_0 \sqrt{\frac{k^T_0}{k^D}}
\end{equation}
The constants $c_J$ and $c_Y$ will be determined later. From the pressure, we determine the derivative according to
(see equation 9.1.30 in \cite{abramowitz1948handbook})

\begin{equation}
\mcC_\nu' (z) = \mcC_{\nu-1}(z) - \nu z^{-1} \mcC_{\nu}(z)
\end{equation}
thus the radial component of flux is given by

\begin{equation}
q_r^D = - k^D \kappa_1 \sum_{\mcC=J,Y}c_\mcC  \left(\mcC_{\kappa_1 r_1-1} (\kappa_1 r) - \frac{r_1}{r} \mcC_{\kappa_1 r_1} (\kappa_1 r)\right)
\end{equation}

\emph{Solution S3}: For $r_1< r\leq r_2$:  By equation \eqref{eqn:laplace-rad}
\begin{equation}
\frac{d}{dr}\left(r\frac{dp^D}{dr} \right) = 0,
\end{equation}

the solution of which is the logarithm, and since $k^T(r)=0$ for $r>r_1$, we know that the total flux around any perimeter enclosing $r_1$ is given by $q^N$. This determines the constant for the logarithmic term, and we obtain:

\begin{equation}
p^D(r) - p^D(r_1) = - \frac{q^N}{2\pi k^D} \ln \left(\frac{r}{r_1}\right)
\end{equation}

The radial component of flux is given by
\begin{equation}
q_r^D =  \frac{q^N }{2\pi } \frac{1}{r}
\end{equation}

\emph{Solution S4}: For $r_2< r\leq r_3$:  By equation \eqref{eqn:laplace-rad}
\begin{equation}
\frac{1}{r}\frac{d}{dr}\left(r\frac{dp^D}{dr} \right) = - \frac{r^D_0}{k^D} (r-r_2) (r_3-r),
\end{equation}

the solution of which is a fourth-order polynomial plus a logarithmic term.

\begin{equation}
p^D(r) - p^D(r_2) = c_4 \ln \left(\frac{r}{r_2}\right) + \frac{r^D_0}{k^D} \left(\frac{r^4-r_2^4}{16}-(r_3+r_2)\frac{r^3-r_2^3}{9} +r_2r_3\frac{r^2-r_2^2}{4}\right)
\end{equation}

The radial component of flux is given by
\begin{equation}
q_r^D = - k^D\frac{c_4}{r} -
r^D_0 \left(\frac{r^3}{4}-(r_3+r_2)\frac{r^2}{3} +r_2r_3\frac{r}{2}\right)
\end{equation}

However, since the outer boundary condition of $\Omega$ is zero Neumann, and we have continuity of flux with respect to the solution S4 at $r_3$, this implies that we can determine the constant multiplier of the logarithmic term by setting $q_r^D(r_3)=0$, which yields:

\begin{equation}
c_4 = \frac{r^D_0}{k^D} \left(\frac{r_3^4}{12}-\frac{r_2r_3^3}{6}\right)
\end{equation}

\emph{Solution S5}: For $r_3 < r$:  By equation \eqref{eqn:laplace-rad}
\begin{equation}
\frac{d}{dr}\left(r\frac{dp^D}{dr} \right) = 0,
\end{equation}

the solution of which is again the logarithm as in S3. However, in accordance with the zero Neumann boundary condition, the flux is zero, and all that remains is a constant.
\begin{equation}
p^D(r) - p^D(r_3) = 0
\end{equation}

and

\begin{equation}
q_r^D = 0
\end{equation}

\subsection*{Constants}
It remains to determine the constants $c_I$, $c_J$ and $c_Y$ from solutions S1 and S2. The solutions S1 and S2 must satisfy three criteria: 1) The pressure must be continuous at $r_0$, 2) the flux must be continuous at $r_0$, and 3) The flux must be continuous at $r_1$. This leads to three linear constraints on the constants.

These criteria can be made precise using the solutions derived above as follows.

\emph{2) Pressure continuity}  Continuity of pressure at $r_0$ requires that
\begin{equation}\label{eqn-const2}
c_I I_0 (\kappa_0 r_0)  = \sum_{\mcC=J,Y}c_\mcC \mcC_{\kappa_1 r_1}(\kappa_1 r_0)
\end{equation}

\emph{3) Flux continuity at $r_0$}  Continuity of flux at $r_0$ requires that

\begin{equation}\label{eqn-const3}
c_I \kappa_0 I_1(\kappa_0 r_0)  = \kappa_1 \sum_{\mcC=J,Y}c_\mcC  \left(\mcC_{\kappa_1 r_1-1} (\kappa_1 r_0) - \frac{\kappa_1 r_1}{\kappa_1 r_0} \mcC_{\kappa_1 r_1} (\kappa_1 r_0)\right)
\end{equation}

\emph{3) Flux continuity at $r_1$} Continuity of flux at $r_0$ requires that
\begin{equation}\label{eqn-const1}
\frac{q^N }{2\pi} \frac{1}{r_1}
= - k^D\kappa_1 \sum_{\mcC=J,Y}c_\mcC  \left(\mcC_{\kappa_1 r_1-1} (\kappa_1 r_1) - \mcC_{\kappa_1 r_1} (\kappa_1 r_1)\right)
\end{equation}

The three equations \eqref{eqn-const1}, \eqref{eqn-const2} and \eqref{eqn-const3} provide the three linear constraints for the unknown constants.

For the special case where $r_0 = r_1$, two of the constants, $c_J$ and $c_Y$ are superfluous, and the remaining constant is obtained by combining equations \eqref{eqn-const2} and \eqref{eqn-const3} as:

\begin{equation}
c_I k^D \kappa_0 I_1(\kappa_0 r_1)  = -\frac{q^N }{2\pi} \frac{1}{r_1}
\end{equation}

\section{Full-size figures for section 5.2}

This section contains enlarged versions of the subfigures in Figure 5 of the article, which contains the colorbars for the figures in this section.

\begin{figure}[H]\label{fig:SM1}
    \centering
    \includegraphics[width=12cm]{pN-venous-tree.png}

    \includegraphics[width=12cm]{pN-arterial-tree.png}
    \caption{Simulation results for Case 3, showing the pressure solution for venous (upper) and arterial (lower) trees, $p^N$}
\end{figure}

\begin{figure}[H]\label{fig:SM2}
    \centering
    \includegraphics[width=12cm]{pD-venous-domain.png}

    \includegraphics[width=12cm]{pD-arterial-domain.png}
    \caption{Simulation results for Case 3, showing the pressure solution for venous (upper) and arterial (lower) parts of the domain, $p^D$}
\end{figure}

\begin{figure}[H]
    \centering
    \includegraphics[width=9cm]{qT-venous-network.png}

    \includegraphics[width=9cm]{qT-arterial-network.png}
    \includegraphics[width=9cm]{qP.png}
    \caption{Simulation results for Case 3, showing the fluxes associated with the venous (upper) and arterial (middle right) unresolved trees, $q^T$, as well as perfusion $q^P$ (lower).}
\end{figure}

\bibliographystyle{siamplain}
\bibliography{BloodFlowBrainSI.bib}

\end{document}


TITLE: Examples for when the quotient ring is necessarily / not necessarily an extension of the residue field.
QUESTION [1 upvotes]: Let $R$ be an integral domain with unique maximal ideal $\mathfrak{m}$. Let $F:=\operatorname{Frac}(R)$ be the field of fraction of $R$.
I know that $F/k(B)$ where $k(B):=R/\mathfrak{m}$ is a residue field is not always true. For instance, take the integers $\mathbb{Z}$ and localize at $p\mathbb{Z}$. The residue field is $\mathbb{Z}_p/p\mathbb{Z}_p$ while the quotient field of $\mathbb{Z}_p$ has no zero-divisors.
However, I was wondering if there were contexts in which this could be true. For instance, if $B$ were a finitely generated algebra over some field (or maybe even if $B$ is the localization of some such algebra). Something along this line is mentioned here and this is the context I would love for it to be true in.
A disproof/proof of this last statement would be greatly helpful (I'll need to keep thinking about it while this post is up). I will even accept a reference.

REPLY [1 votes]: I can at least explain the situation you linked to.
Let $R$ be a finitely generated algebra over an algebraically closed field $k$. Let $\mathfrak{m}$ be a maximal ideal of $R$, and let $B = R_{\mathfrak{m}}$.
Now the quotient $B/\mathfrak{m}B$ is a field which is also a finitely generated $k$-algebra, so by Zariski's Lemma $B/\mathfrak{m}B$ is a finite extension of $k$. But now since $k$ is algebraically closed, this forces $B/\mathfrak{m}B$ to just be $k$. So, $B/\mathfrak{m}B$ embeds in $B$, and thus also in $\operatorname{Frac}(B)$.


TITLE: $N = Poisson(\lambda)$, $\{U_i\}$ iid $\implies (N_1, N_2) = Po(\lambda p_1)$x $Po(\lambda p_2)$
QUESTION [0 upvotes]: Let $\{N\}\cup\{U_i\}$ be independent random variables.
$N = $ Poisson$(\lambda)$
$\{U_i\}$ iid, taking values in $\{1,2\}$, $\mathbb{P}[U_i = 1] = p_1$ and  $\mathbb{P}[U_i = 2] = p_2$, $p_1 + p_2 =1$.
Define $N_j =  \sum_{i=1}^N\textbf{1}_{[U_i =j]}$, $j \in \{1,2\}$.
Then $(N_1, N_2) = Po(\lambda p_1)$x $Po(\lambda p_2)$, that is,
\begin{align*}
\mathbb{P}[N_1 = k_1, N_2 = k_2] = \frac{e^{\lambda p_1}(\lambda p_1)^{k_1}}{k_1!}\frac{e^{\lambda p_2}(\lambda p_2)^{k_2}}{k_2!}
\end{align*}
My attempt:
\begin{align*}
&[N_1 = k_1, N_2 = k_2] = \big[\sum_{i=1}^{k_1 + k_2} \textbf{1}_{[U_i =1]} = k_1, \sum_{i=1}^{k_1 + k_2} \textbf{1}_{[U_i = 2]} = k_2 \big]\\  &= \big[ \exists    i^{1}_{1}, \dots, i^{1}_{k_1}, i^{2}_{1}, \dots, i^{2}_{k_2} \in \{1, \dots, k_1 + k_2\} : U_{i^1_l} =1, l \in \{1, \dots, k_1\}, U_{i^2_m} = 2, m \in \{1, \dots, k_2\} \big] \\ &= \cup_{i^{1}_{1}, \dots, i^{1}_{k_1}, i^{2}_{1}, \dots, i^{2}_{k_2}} \big[ [U_{i^1_1} = 1]\cap\dots\cap[U_{i^1_{k_1}} = 1]\cap[U_{i^2_1} = 2]\cap\dots\cap[U_{i^2_{k_2}} = 2]\big]]
\end{align*}
Then,
$\mathbb{P}[N_1 = k_1, N_2 = k_2]$ = $\mathbb{P}\Big[\cup_{i^{1}_{1}, \dots, i^{1}_{k_1}, i^{2}_{1}, \dots, i^{2}_{k_2}} \big[ [U_{i^1_1} = 1]\cap\dots\cap[U_{i^1_{k_1}} = 1]\cap[U_{i^2_1} = 2]\cap\dots\cap[U_{i^2_{k_2}} = 2]\big] \Big]$ = $\sum_{i^{1}_{1}, \dots, i^{1}_{k_1}, i^{2}_{1}, \dots, i^{2}_{k_2}} \mathbb{P}\Big[[U_{i^1_1} = 1]\cap\dots\cap[U_{i^1_{k_1}} = 1]\cap[U_{i^2_1} = 2]\cap\dots\cap[U_{i^2_{k_2}} = 2]\Big]$ = $\sum_{i^{1}_{1}, \dots, i^{1}_{k_1}, i^{2}_{1}, \dots, i^{2}_{k_2}} \prod_{l=1}^{k_1}\mathbb{P}\big[U_{i^1_l} = 1]\prod_{m=1}^{k_2}\mathbb{P}\big[U_{i^2_m} = 1]$ =  $\sum_{i^{1}_{1}, \dots, i^{1}_{k_1}, i^{2}_{1}, \dots, i^{2}_{k_2}}\prod_{l=1}^{k_1}p_1\prod_{m=1}^{k_2} p_2 $

REPLY [1 votes]: Note that for fixed $n$, $X_n := \sum_{i = 1}^n I(U_i = 1) \sim \text{Bin}(n, p_1)$, and $\sum_{i = 1}^n I(U_i = 2) = n - X_n$.
Thus for $k_1, k_2 \in \{0, 1, \ldots\}$, by law of total probability, we have
\begin{align}
& P[N_1 = k_1, N_2 = k_2] \\
= & \sum_{n = 0}^\infty P[N_1 = k_1, N_2 = k_2 \mid N = n]P[N = n] \\
= & \sum_{n = 0}^\infty P\left[\sum_{i = 1}^n I(U_i = 1)= k_1, \sum_{i = 1}^n I(U_i = 2) = k_2 \mid N = n\right]P[N = n] \\
= & \sum_{n = 0}^\infty P\left[X_n = k_1, n - X_n = k_2\right]P[N = n] \quad\text{ by independence}\\
= & P[X_{k_1 + k_2} = k_1]P[N = k_1 + k_2] \quad\text{ the summand is not zero if and only if $n = k_1 + k_2$.}\\
= & \binom{k_1 + k_2}{k_1}p_2^{k_2}p_1^{k_1}e^{-\lambda}\frac{\lambda^{k_1 + k_2}}{(k_1 + k_2)!} \\
= & \frac{e^{-\lambda p_1}(\lambda p_1)^{k_1}}{k_1!}\times \frac{e^{-\lambda p_2}(\lambda p_2)^{k_2}}{k_2!}.
\end{align}


TITLE: How to prove that $n! \mid 2^n \prod_{i=1}^{n}(2i-1)$?
QUESTION [3 upvotes]: i would like to prove that
$$
n! \mid 2^n \prod_{i=1}^{n}(2i-1)
$$
Thanks!

REPLY [5 votes]: HINT: First,
$$(2n)!=\left(\prod_{k=1}^n2k\right)\left(\prod_{k=1}^n(2k-1)\right)=2^nn!\prod_{k=1}^n(2k-1)\;.$$
Now combine this with the fact that $\dbinom{2n}n$ is an integer.

REPLY [3 votes]: You can notice that $$2^n \prod\limits_{i=1}^n (2i-1)= 2^n \prod\limits_{i=1}^n \frac{2i(2i-1)}{2i}= \frac{1}{n!} \prod\limits_{i=1}^n 2i(2i-1)= \frac{(2n)!}{n!}=n!C_{2n}^n$$


TITLE: Inverse relation graphical shape
QUESTION [1 upvotes]: I have a few points among two quantities which are inversely related. The points are $(0,20), (1,19)$ and $(5,15)$. I have drawn these points in Matlab but it gives me an inverted triangle which isn't present in the inverse relationship among the two quantities.  Clearly stated, if I have two quantities and they are inversely related, what shape should it show me if I were given only three points?  Please help me.

REPLY [0 votes]: Any three points when plotted on a plane will appear as a triangle.
Inversely related quanties will appear somewhat like the curve y = 1/x.
Points should appear more or less along such a curve.  It is up to you to decide, given three points, how the curve may lie.  The points will appear as a broken line.  Which side of the inverted triangle to remove will require some thought.
Expecting a quick answer from a computer is often a waste of time.


TITLE: Are there universal homological functors?
QUESTION [5 upvotes]: There is a bifunctor $H: Stab^{op} \times Ab \to Top$ where $H(C,A)$ is the space of homological functors $C \to A$. Is this bifunctor left or right representable?
That is, for each small abelian category $A$ does there exist a small stable $\infty$-category $\underline A$ and a homological functor $\underline A \to A$ such that every homological functor $C \to A$ (where $C$ is a small stable $\infty$-category) factors uniquely through $\underline A \to A$ via an exact $\infty$-functor $C \to \underline A$?
On the other hand, for each small stable $\infty$ category $C$, does there exist an abelian category $\overline C$ and a homological functor $C \to \overline C$ such that every homological functor $C \to A$ (where $A$ is a small abelian category) factors uniquely through $C \to \overline C$ via an exact functor $\overline C \to A$?

REPLY [8 votes]: Here is half an answer (i.e. an answer to the last question). Given a small stable $\infty$-category $C$ and a small abelian category $A$, any functor $C\to A$ factors through the homotopy category $ho(C)$ of $C$. Hence the category of homological functors from $C$ to $A$ is equivalent to the category of homological functors from the triangulated category $ho(C)$ to $A$. There is a universal homological functor from $ho(C)\to A(C)$: this is documented in Verdier's thesis (see Théorème 3.2.1 in Astérisque 239, an electronic copy being available here), although Verdier thanks Freyd for sharing the construction with him. A modern reference in English is Theorem 5.1.18 in Neeman's book.


TITLE: For $n \in \Bbb N$ define $f(x) = \frac {x^n} {1+x^n}$, then $f(x)$ converges on $[c, \infty]$ for $c>1$
QUESTION [1 upvotes]: For $n \in \Bbb N$ define $f(x) = \frac {x^n} {1+x^n}$, I am trying to show that on the interval $[c, \infty]$ where $c > 1$ that $f(x)$ converges uniformly.
I have to show that for $\epsilon >0$ that there is an $N$ such that for all $n>N$
$|\frac {x^n} {1+x^n} - 1| < \epsilon$, but what confuses me is that it would need to hold for all $x$ in the interval.

REPLY [2 votes]: Your idea of using a $\delta-\epsilon$ proof works.  Note that for $0<\epsilon<1$
$$\begin{align}
\left|\frac{x^n}{1+x^n}-1\right|&=\frac{1}{1+x^n}\\\\
&\le \frac{1}{1+c^n}\\\\
&<\epsilon \tag 1
\end{align}$$
whenever $n>\frac{\log(1-\epsilon)-\log(\epsilon)}{\log(c)}$.  And for $\epsilon\ge 1$, the inequality in $(1)$ holds trivially.

Note that the convergence is not uniform on $(1,\infty)$ since for $\epsilon=2/3$, we can take $x=(1/2)^{1/n}$.  Then, we have
$$\frac{1}{1+x^n}=\frac{1}{1+1/2}=2/3=\epsilon$$


TITLE: Connection between two notions of undecidability
QUESTION [1 upvotes]: As far as I understand, there are two common notions of "undecidability" that I will call

"Mathematical undecidability":="the statement can neither be proved nor disproved from the axioms", equivalently, it is true in some model and false in some other model, and
"Algorithmic undecidability":="there is no algorithm that, for a given input, corretly decides yes/no". More formally, the possible inputs are ordered and the set of indices of true statements is not a recursive set.

If I'm (at least partially) right in the above, I would like to understand the link between those two notions. On wikipedia, I read

The connection between these two is that if a decision problem is
  undecidable (in the recursion theoretical sense) then there is no
  consistent, effective formal system which proves for every question A
  in the problem either "the answer to A is yes" or "the answer to A is
  no".

This is quite vaguely formulated.
Question:

Is it true that for every algorithmically undecidable problem, for
  some particular instance of the input, the statement is mathematically
  undecidable (using Peano arithmetic)?

REPLY [1 votes]: Here is an answer which is a bit closer to the quote from Wikipedia. If you have some property $\phi(n)$ in the language of PA and if, for every natural number constant $\overline{n}$, either PA proves $\phi(\overline{n})$ or PA disproves $\phi(\overline{n})$, then $\phi(n)$ is algorithmically decidable: given an input $\overline{n}$, enumerate proofs in PA until you find one that either proves $\phi(\overline{n})$ or disproves it. This is guaranteed to terminate by our assumption on $\phi$ and when it terminates we know whether $\phi(\overline{n})$ is true or not (because PA is sound).


TITLE: Two block system with strings
QUESTION [2 upvotes]: Cant figure it out, can anyone tell me how to do it plus the general methodology for solving 2 or 3 block system questions.
Equations: (ū is coefficient of friction)
P-f1-f2=M2a
f1-Tcos45=Ma
f1=(M2g-Tcos45)ū
f2=(M1g+M2g-Tcos45)ū
4 equations but 5 variables?
I took friction between M2 and ground as f2 and M1 and M2 as f1.

REPLY [2 votes]: Because the blocks "just slide", what you have here is a situation where the forces are (almost) in balance. Forget about the acceleration; get rid of acceleration (you are not trying to solve for that) and make one of the expressions an inequality.  This should allow you to solve the problem.


TITLE: Closed form of integral $\int_a^b e^{-ix^2} dx$
QUESTION [0 upvotes]: Does any know how to find the closed form of integral  $\int_a^b e^{-ix^2} dx$ for any real $a$ and $b$. It seems that I need to use the fresnel integrals.

REPLY [1 votes]: $$\int_a^b e^{-ix^2} dx = \int_a^b \cos x^2 dx - i\int_a^b \sin x^2 dx$$


TITLE: Given convex differentiable function $h: R^m \to R$, what is geometric interpretation of $h(x)/\|\nabla h(x)\|$ vis à vis the level-set $h=0$?
QUESTION [1 upvotes]: Let $h:\mathbb R^m \to \mathbb R$ be a convex diifferentiable function and define $C_h := \{x \in \mathbb R^m \mid h(x) = 0\}$, assumed to be non-empty.

Question. Given $x \in \mathbb R^m$ such that $\nabla h(x) \ne 0$, is there any geometric interpretation of the quantity $\omega_h(x) := (h(x))_+/\|\nabla h(x)\|$ vis à vis the set $C_h$ ?

Motivation. In the linear case where $h(x) = v^Tx$ for some unit-vector $v \in \mathbb R^m$, one has

$C_h = v^\perp$, the hyperplane perpendicular to $v$.
$\|\nabla h(x)\| = \|v\| = 1$ for all $x \in \mathbb R^m$, and so $\omega_h(x) = (h(x))_+ = (v^Tx)_+$, which is the distance of $x$ from $C_h$.

REPLY [1 votes]: Given a unit-vector $v \in \mathbb S_{m-1} := \{v \in \mathbb R^m \mid \|v\| = 1\}$, let $\mathscr L_v := \{x + tv \mid
t \ge 0\} \subseteq  \mathbb R^m$ be the ray generated by $v$ and starting at $x$. The
distance of $x$ from $C_h$ can be written as
$$
r_h(x) := d(x,C_h) := \inf_{x' \in C_h}\|x'-x\| = \inf_{v
  \in \mathbb S_{m-1}}\inf_{x' \in
  \mathscr L_v \cap C_h}\|x'-x\|.
$$
Now, by definition, $x' \in \mathscr L_v \cap C_h$ if and only if
there exists a time $t \in [0,\infty)$ such that $x' = x + tv$ and $h(x') = 0$.
Thus,
\begin{eqnarray}
  r_h(x) = \inf\{t \mid (v,t) \in \mathbb S_{m-1} \times [0,\infty),\; h(x + tv) = 0\}.
  \tag{1}
\end{eqnarray}
The convex case
If $h$ is convex and differentiable, then for all $(v,t) \in \mathbb S_{m-1} \times
[0,\infty)$ one has
\begin{eqnarray*}
  \begin{split}
    h(x+tv) &\ge h(x) + t\nabla h(x)^Tv\ge h(x) - t\|\nabla h(x)\|,\\
    -h(x+tv) &\le -h(x) - t\nabla h(x)^Tv \le -h(x) + t\|\nabla h(x)\|,
\end{split}
\end{eqnarray*}
where we have used Cauchy-Schwarz to reach the final term of each chain of
inequalities. Combining the above inequalities and (1) one
computes
\begin{eqnarray*}
  \begin{split}
r_h(x) &\ge \inf\{t \ge 0 \mid h(x) -
t\|\nabla h(x)\|\le 0\} = \frac{\max(h(x),0)}{\|\nabla h(x)\|},\\
r_h(x) &\ge \inf\{t \ge 0 \mid -h(x) +
t\|\nabla h(x)\|\ge 0\} = \frac{\max(-h(x), 0)}{\|\nabla h(x)\|},
\end{split}
\end{eqnarray*}
We have thus proved

Theorem.
If $h:\mathbb R^m \to \mathbb R$ is convex differentiable, we have
\begin{eqnarray}
  r_h(x) \ge \frac{|h(x)|}{\|\nabla h(x)\|},\;\forall x \in \mathbb R^m
\end{eqnarray}

Lipschitz case
If we instead assume $h$ Lipschitz continuous (but not necessarily convex), then $|h(x + tv) -  h(x)| \le
Lt$ for all $(v,t) \in \S_{m-1} \times [0,\infty)$. Using (1), we may then compute
\begin{eqnarray}
  \begin{split}
r_h(x) &\ge \inf\{t \ge 0\mid h(x) - tL \ge 0\} =
\frac{\max(h(x),0)}{L},\\
r_h(x) &\ge \inf\{t \ge 0\mid h(x) + tL \ge 0\} =
\frac{\max(-h(x),0)}{L}.
\end{split}
\end{eqnarray}
Putting things together, we get

Theorem. If $h:\mathbb R^m \to \mathbb R$ is $L$-Lipschitz, then we have
\begin{eqnarray}
  r_h(x) \ge \frac{|h(x)|}{L},\;\forall x \in \mathbb R^m.
\end{eqnarray}


TITLE: Solar System and the Plane of our Galaxy
QUESTION [0 upvotes]: Why do all of the space probes launched travel within the plane of the solar system?  Can we direct a probe perpendicular to the plane of the solar system and galaxy.

REPLY [2 votes]: In fact we have got a probe flying out of the plane of the Solar System because Voyager 1 is heading out of the plane of the ecliptic, though admittedly not perpendicular to it. This wasn't deliberate, or at least it wasn't planned for. Voyager 1 was diverted to get a close view of Titan and a side effect of this was to send it out of the plane of the Solar System.
As Jerry says, there's nothing interesting out of the plane of the ecliptic until you start reaching nearby stars, which in Voyager's case will take about 40,000 years by which time it will be dead (along with you and me of course).


TITLE: Pointwise and uniform convergence of $f_n(x)=\cos \frac{nx}{1+n^2}$
QUESTION [3 upvotes]: Study the pointwise and uniform convergence of $f_n(x)=\cos \frac{nx}{1+n^2}$ for $x \in \mathbb{R}$.
If $x=0$ it is $f_n(0)=1$, for $x \ne 0$ it is
$$\lim_{n \to \infty} \cos\frac{nx}{1+n^2}=\cos 0=1$$
So $f_n(x)$ converges pointwise to $f(x)=1$ for all $x \in \mathbb{R}$. To study the uniform convergence I must evaluate
$$\lim_{n \to \infty} \sup_{x \in \mathbb{R}}\left|\cos \frac{nx}{1+n^2}-1\right|=\lim_{n \to \infty} \sup_{x \in \mathbb{R}}\left(1-\cos \frac{nx}{1+n^2}\right)$$
Since the derivative of $1-\cos \frac{nx}{1+n^2}$ exists for all $x\in\mathbb{R}$, it is
$$\frac{d}{dx} \left(1-\cos \frac{nx}{1+n^2}\right)=\frac{n}{1+n^2} \sin \frac{nx}{1+n^2} \geq 0$$
$$\iff 2k\pi \leq \frac{nx}{1+n^2} \leq (2k+1)\pi \iff \frac{1+n^2}{n}2k\pi \leq x \leq \frac{1+n^2}{n}(2k+1)\pi$$
For $k\in\mathbb{Z}$, so
$$\lim_{n \to \infty} \sup_{x \in \mathbb{R}}\left(1-\cos \frac{nx}{1+n^2}\right)=\lim_{n \to \infty} \left(1-\cos \frac{n \left(\frac{1+n^2}{n}(2k+1)\pi\right)}{1+n^2}\right)=\lim_{n\to\infty}(1-\cos\pi)=2$$
So $f_n$ is not uniformly convergent in $\mathbb{R}$.
Is this correct? Some more questions:

I've noticed that $f_n(\frac{1+n^2}{n})=\cos 1$, can I conclude that since there exists at least one $\bar{x}=\frac{1+n^2}{n}$ such that $|f_n(\bar{x})-f(\bar{x})|=1-\cos 1 \ne 0$ so $f_n$ cannot be uniform convergent in whole $\mathbb{R}$ because it it will never be $|f_n(x)-f(x)|<\varepsilon$ for all $x \in \mathbb{R}$ because of $\bar{x}$. Is this correct?

It is $\cos \frac{nx}{1+n^2} \approx 1-\frac{n^2 x^2}{2(1+n^2)^2}$ as $n \to \infty$, can this be useful to evaluate the limit of the supremum? Or is it useless because the asymptotic behaviour can't give informations about a supremum?

Is there a way to identify if there are some subsets of $\mathbb{R}$ where $f_n$ can converge uniformly to $f$?

REPLY [1 votes]: For $(1)$, the answer is yes.
Another simple way to see this is that $\text{Im}(f_n) = \text{Im}(\cos) = [-1,1]$ for all $n$, and since $\text{Im}(f) = \{1\}$ convergence cannot be uniform.
For $(2)$, I don't really see how that helps.
If $x$ is fixed and you're letting $n$ grow, all you're getting help with is calculating the pointwise limit.
For $(3)$, notice that as $n\to\infty$, $n/(1+n^2)$ approaches $0$.
I guess the natural approach here would be to consider what happens if $x$ is bounded.
And then you'd have to ask yourself what happens if $x$ is not bounded?


TITLE: K-Theory of Schemes: Monoidal vs. Exact
QUESTION [7 upvotes]: There are several ways for defining the K-Theory of a category depending on which structure it admits. The K-Theory of schemes is commonly defined as the "group completion" of the category of algebraic vector bundles with respect to its exact category structure rather than with respect to its symmetric monoidal category structure.
In general, these two are not the same and the former is a quotient of the latter. In particular, for schemes these two constructions are not equivalent since exact sequences of algebraic vector bundles do not always split (e.g. projective line $\mathbb{P}_R^1$).
My question: why is the "correct" definition to choose the exact category structure? What goes wrong when one just considers the monoidal category structure which seems more natural since that is how for instance K-Theory of rings is defined.

REPLY [10 votes]: There is no "universally correct" definition, but $K^0(X)$ usually denotes the K-theory of the exact category of vector bundles over $X$. As you say, there are various approaches for (higher) algebraic K-theory (summarized in Weibel's K-book, for instance), and which one one uses depends on the applications, but also on the realizability. Namely, there are lots of exact sequences of vector bunbles on projective varieties which give you relations in the K-theory, but these exact sequences don't split. Although this makes $K^0(X)$ relatively small compared to $K^0_{\oplus}(X)$ (the K-theory of the monoidal category of vector bundles w.r.t. $\oplus$) and perhaps doesn't contain enough information for what you want, it is at least computable! For example, $K^0(\mathbb{P}^r) \cong \mathbb{Z}[x]/(x^{r+1})$ is well-known (SGA 6), but little is known about $K^0_{\oplus}(\mathbb{P}^r)$ (cf. MO/20444): It is the free abelian group on indecomposable vector bundles on $\mathbb{P}^r$ (Atiyah), but these have not been classified yet.
Grothendieck's celebrated and important resolution theorem won't tell you anything about $K^0_{\oplus}(X)$ because of the appearance of exact seqeunces and alike, so there won't be a nice comparision between K-theory and G-theory. Many other theorems for $K^0$, such as Grothendieck's homotopy-invariance $K^0(X) \cong K^0(X \times \mathbb{A}^1)$ will probably fail for $K^0_{\oplus}$. The upshot is: Although $K^0_{\oplus}$ contains more information, it is not flexible enough for computations.
If $X$ is an affine scheme, the canonical map $K^0_{\oplus}(X) \to K^0(X)$ is an isomorphism. Because of that, many authors define $K_0(R)$ of a ring $R$ to be the free abelian group on iso-classes of f.g. proj. $R$-modules modulo the relation $[P \oplus Q] = [P]+[Q]$, but this should not be seen as the correct definition for the general case of a scheme. In particular I don't agree with your sentence "... the monoidal category structure which seems more natural since that is how for instance K-Theory of rings is defined."

REPLY [8 votes]: You don't need to go to non-affine schemes to run into trouble with your proposed definition.
In the simplest case, if $R$ is a regular ring and $f$ a non-zero element, you want a long exact localization sequence relating the $K$-theory of $R$ to the $K$-theory of $R[f^{-1}]$.  The missing term turns out to be the $K$-theory of the category of finitely generated $R/fR$-modules.  In most cases $R/fR$ won't be regular (so you can't just replace this with the category of projective modules), and this category won't have the property that all short exact sequences split.   Therefore it matters which $K$-theory you use, and it's the $K$-theory of exact categories that makes the localization theorem true.


TITLE: what is general theory for those type of problem: To think about the condition must be exist if a decimal with infinite digit in base 10 ...
QUESTION [0 upvotes]: what is general theory for those type of problem: To think about the condition must be exist if a decimal with infinite digit in base 10 also have infinite digit in both base 3, base 4? please prove the theory please.

REPLY [1 votes]: Qiaochu has answered the question in general in the comments. Applying this to the specific cases mentioned in the question, we have the following:
If $x$ is a real irrational, it has a non-terminating expansion in base 10, base 3, base 4, base whatever-$n$-you want (so long as you want an integer 2 or greater).
If $x$ is rational, and has a non-terminating expansion in base 10, then it also has a non-terminating expansion in base 3, unless its denominator is a power of 3; it is guaranteed to have a non-terminating expansion in base 4.


TITLE: How to find $\int_{-1}^1 \frac{\cos x}{a^x+1}\mathrm dx$
QUESTION [11 upvotes]: Evaluate $$\int_{-1}^1 \frac{\cos x}{a^x+1}\mathrm dx$$ where $a$ is a real parameter $a\geq1$.

I can easily find the definite integral for $a=1$. It is $\sin(1)$.
In wolframalpha.com when I put $\displaystyle\int_{-1}^1 \frac{\cos x}{a^x+1}\text{d}x$ it shows me a very complicated formula with complex numbers and functions I didn't study but it says that definite integral is $= 0.841471\ldots$.
How can I find that integral?

REPLY [1 votes]: Let $$I=\int_{-1}^1 \frac{\cos (x)}{a^x+1}\mathrm dx\tag1$$
Using identity $$\int_a^bf(x)\;\mathrm dx=\int_a^bf(a+b-x)\;\mathrm dx$$
we get
$$I=\int_{-1}^1 \frac{\cos (-x)}{a^{-x}+1}\mathrm dx=\int_{-1}^1 \frac{a^{x}\cos (x)}{1+a^{x}}\mathrm dx\tag2$$
Adding $(1)$ and $(2)$, we get
$$2I=\int_{-1}^1\cos x\;\mathrm dx=\sin x\,\Bigg|_{-1}^1=2\sin 1\qquad\implies\qquad I=\bbox[8pt,border:3px #FF69B4 solid]{\color{red}{\large\sin 1}}$$
Note that, $\cos(-x)=\cos(x)$ and $\sin(-x)=-\sin(x)$. You might be interested in seeing even and odd functions.


TITLE: Compactness and disjoint open sets
QUESTION [3 upvotes]: Let $C1$ and $C2$ be two disjoint compact subsets of $R^n$. Prove that there exist disjoint open sets $U_1$ and $U_2$ such that $C_1 \subseteq U_1$ and $C_2 \subseteq U_2$.
Intuitively, this seems to be correct, however, I could not figure out the importance of compactness for this problem - why arbitrary subsets of $R^n$ could hold such property for some open sets $U_1$ and $U_2$? Is it due to the fact of boundedness and closeness of the sets $C1$ and $C2$?

REPLY [4 votes]: Here is a second tack.  The function $(x,y)\mapsto d(x,y)$ is a continuous function on the compact set $C_1\times C_2$.  It therefore has a positive minimum value $v$.  Let
$$U_k = \bigcup_{x\in C_k} B_{v/2}(x), \qquad k = 0, 1.$$
These two sets do the job for you.