TITLE: How do you find the sum of series?
QUESTION [0 upvotes]: I was doing revision on sequences and series when I found I was unable to answer this particular question.

I have gone completely blank and would appreciate any help.
Thanks,

REPLY [2 votes]: \begin{align*}
\sum_{i=1}^{n}i(8i^{2}+1)&=\sum_{i=1}^{n}\left ( 8i^{3}+i \right )\\
&=\sum_{i=1}^{n}8i^{3}+\sum_{i=1}^{n}i\\
&=8\sum_{i=1}^{n}i^{3}+\sum_{i=1}^{n}i\\
&=8\left ( \frac{n(n+1)}{2} \right )^{2}+\frac{n(n+1)}{2}.
\end{align*}


TITLE: Simple trigonometrical equations
QUESTION [0 upvotes]: I'm having difficulties in solving the simultaneous equations
$$
\begin{cases}
\sin(x+y)=\frac{1}{\sqrt{2}}\\
\cos(2x+y)=\frac12
\end{cases}
$$
for $0^{\circ}\le x,y\le 90^{\circ}$.
The answer is $x=15^{\circ}$ degrees and $y= 30^{\circ}$.

REPLY [0 votes]: $$\sin(x+y)=\dfrac1{\sqrt2}=\sin45^\circ$$
$x+y=n180^\circ+(-1)^n45^\circ$ where $n$ is any integer
As $0^\circ<x,y<90^\circ,0^\circ<x+y<180^\circ\implies x+y=45^\circ\  \ \  \ (1)$ or $x+y=135^\circ\  \ \  \ (2)$
Similarly, $2x+y=360^\circ m\pm 60^\circ$ where $m$ is any integer
As $0^\circ<2x+y<270^\circ,2x+y=60^\circ\  \ \  \ (3)$
Solve $(1),(3)$  and $(2),(3)$


TITLE: Determine amount of congruent numbers
QUESTION [0 upvotes]: I found the claim in a paper that there are at max 8 integers mod $2^{130}-5$ congruent to one integer mod $2^{128}$.
$$u \pmod {2^{130}-5} \equiv g  \pmod {2^{128}} \quad\text{ with }u \in U \quad \#U \le 8 $$
So $2^{130}-5$ is almost 4 times larger than $2^{128}$ and therefore $g , g + 2^{128}, g +2^{129}$ should be congruent integers to g. What are the other 5 integers?

REPLY [1 votes]: This is not what the paper says!  The paper refers to integers in the interval $[-2^{130}+6, 2^{130}-6]$, not integers modulo $2^{130}-5$.  There are about twice as many of the former than the latter.  You're missing all the negative values.


TITLE: A subtle time-dependent inequality
QUESTION [0 upvotes]: Let $x(t)\in\mathbb{R}$ be a positive function of $t\in\mathbb{R}$, $t\ge 0$ and suppose the following inequality holds for all $t\ge 0$
$$\tag{$\star$}\label{ineq}
x(t)\le c e^{-at + b x(t) t^2},
$$
where $a,b,c$ are positive scalar.

My question: Do there exist conditions on the coefficients $a$, $b$ and $c$ guaranteeing that
  $$
\lim_{t\to \infty}x(t)=0\ ?
$$

What makes the problem hard (at least to me) is the term $b x(t) t^2$ on the exponential in \eqref{ineq}. I tried to take the logarithm on both sides of \eqref{ineq}, but this does not seem to help much. So any suggestion/comment is very appreciated, thanks!

REPLY [1 votes]: Take $x(t)=1$ and consider the set of $a,b,c$ that satisfy $x(t)\leq ce^{-at+bx(t)t^2}$. This reduces to $1\leq ce^{-at+bt^2}$, or $0\leq \ln c-at+bt^2$, which implies $bt^2-at-\ln c\geq 0$. The minimum of the left side occurs at $t=\frac{a}{2b}$, giving $\frac{a^2}{4b}-\frac{a^2}{2b}-\ln c = \frac{a^2(1-2b)}{2b}-\ln(c).$ There are plenty of solutions $(a,b,c)$ that satisfy this, for example if you fix a $c$, just ensure $a$ is large and $b=1/2$. But none of these solutions imply $x(t)\rightarrow 0$.


TITLE: Help with Baby Rudin Theorem 3.29
QUESTION [4 upvotes]: Theorem 3.29 If $p >1$ then $\sum_{n=2}^{\infty} \frac{1}{n(\log n)^p}$ converges; if $p \leq 1$, the series diverges.
Proof:
The monotonicity of the logarithmic function implies that $\{\log n\}$ increases. Hence $\frac{1}{n\log n}$ decreases, and we can apply theorem 3.27; this leads us to the series
$\sum_{k=1}^{\infty} 2^k \frac{1}{2^k(\log 2^k)^p}= \sum_{k=1}^{\infty} \frac{1}{(k\log 2)^p}= \frac{1}{(\log 2)^p} \sum_{k=1}^{\infty} \frac{1}{k^p}$ and the conclusion follows from Theorem 3.28.
I understand the proof thus far, but Rudin goes on to say that this procedure may evidently be continued. For instance,
$\sum_{k=3}^{\infty} \frac{1}{n\log n(\log\log n)}$ diverges, whereas
$\sum_{k=3}^{\infty} \frac{1}{n\log n(\log\log n)^2}$ converges.
I want to continue the procedure to show for which $p$ does $\sum_{k=3}^{\infty} \frac{1}{n\log n(\log\log n)^p}$ converge.
Here are the relevant theorems:
Theorem 3.27 Suppose $a_1 \geq a_2 \geq ... \geq 0$. Then the series $\sum_{k=1}^{\infty} a_n$ converges if and only if $\sum_{k=1}^{\infty} 2^n a_{2^n}$ converges.
Theorem 3.28 $\sum \frac{1}{n^p}$ converges if $p>1$ and diverges if $p \leq 1$.
Here is what I have so far:
By Theorem 3.27 $\sum_{k=3}^{\infty} \frac{1}{n\log n(\log\log n)^p}$ converges if and only if $\sum_{k=2}^{\infty}2^n \frac{1}{2^n\log 2^n(\log\log 2^n)^p} = \frac{1}{\log 2} \sum_{k=2}^{\infty} \frac{1}{n(\log\log 2^n)^p}$
Now I'm stuck. How may the procedure be continued? Please note it is my goal to show a result similar to Theorem 3.29. It is not my goal to simply show that $\sum_{k=3}^{\infty} \frac{1}{n\log n(\log\log n)^p}$ converges for some $p$. Any hint would be greatly appreciated. Thank you.

REPLY [1 votes]: I think I figured it out.
By Theorem 3.29 if $p>1$, then $\Sigma_{n=2}^{\infty}\frac{1}{n(logn)^p}$ converges.
By Theorem 3.27 $\Sigma_{n=3}^{\infty} \frac{1}{nlogn(loglogn)^p}$ converges if and only if $\frac{1}{log2} \Sigma_{n=2}^{\infty} \frac{1}{n(loglog2^n)^p}$ converges.
But $\frac{1}{n(loglog2^n)^p} = \frac{1}{n(logn+loglog2)^p} \leq \frac{1}{n(logn)^p}$
Hence $\frac{1}{log2} \Sigma_{n=2}^{\infty} \frac{1}{n(loglog2^n)^p}$ converges for $p>1$, and therefore $\Sigma_{n=3}^{\infty} \frac{1}{nlogn(loglogn)^p}$ converges for $p>1$.


text:
\begin{document}

\section{INTRODUCTION}

The boundary element method (BEM) has offered an alternative to the finite element method and has been attractive for certain types of problems, such as those involving an infinite or semi-inifinite domain \cite{BeerSmithDuenser2008b}.

The isogeometric approach \cite{Hughes2005a} has led to renewed interest in the method since it only requires a surface discretization and a direct link can be established with geometric modeling technology, without the need to generate a mesh.
Using NURBS instead of the traditional Serendipity or Lagrange functions  for describing the variation of boundary values, additional benefits are gained because of their higher continuity and efficient refinement strategies \cite{Beer2013}.

Trimmed NURBS patches have been successfully applied to problems where two solids intersect.  Such models can be created quickly in a CAD program and data exported in IGES format. The exported IGES data contain the description of the boundary of objects via NURBS patches and trimming curves defined in the local coordinate system of a patch. This information can then be used to trim the NURBS patch, i.e. to remove part of the patch surface.

In this paper we present a simple but effective approach to the analysis of trimmed surfaces. To the best of our knowledge the proposed method of double mapping has not been previously published.
Furthermore we propose to generalize the isogeometric concept by approximating the boundary values (tractions, displacements) with functions that are different from the ones used to describe the geometry. Our motivation for this comes from the fact that the data obtained from CAD programs describing the boundary, may not be suitable for describing the boundary values.

\section{GEOMETRY DEFINITION WITH TRIMMED NURBS PATCHES}
\subsection{NURBS patches}

In CAD programs the geometry is described by NURBS patches which are mapped from a unit square with coordinates $u,v$ to the global coordinates \textbf{x} (x,y,z) by

\begin{equation}
\mathbf{x}=  \sum_{b=1}^{B}\sum_{a=1} ^{A}R_{a,b}^{p,q} (u,v)\mathbf{x}_{a,b}
\label{Gmap}
 \end{equation}

where $\mathbf{x}_{a,b}$ are the coordinates of the control points, $p$ and $q$ are the function orders in $u$ and $v$ direction, $A$ and $B$ are the number of control points in $u$ and $v$ direction and $R_{a,b}^{p,q}$ are tensor products of NURBS functions:

\begin{equation}
R_{a,b}^{p,q}=\frac{N_{a,p}(u)N_{b,q}(v)w_{a,b}}{\sum_{\bar{b}=1}^{B}\sum_{\bar{a}=1}^{A}N_{\bar{a},p}(u)N_{\bar{b},q}(v)w_{\bar{a},\bar{b}}}
 \end{equation}

$N_{a,p}(u)$ and $N_{b,q}(v)$ are B-spline functions of local
coordinates $u$ or $v$ of order $p$ or $q$ ($0$ constant , $1$ linear
, $2$ quadratic etc.) and $w_{a,b}$ are weights. The B-spline
functions are defined by a Knot vector with non-decreasing values of
the local coordinate and a recursive computation which starts at order
0. For example for $N_{a,p}(u)$ with a Knot vector $\Xi_{u}=\left(\begin{array}{cccc}u_{1} & u_{2} & \cdots & u_{A+p+1}\end{array}\right)$ we have for $p=0$
\begin{eqnarray}
N_{a,p}(u)&=&1  \quad  for  \quad  u_{a}\leqslant u < u_{a+1}\\
N_{a,p}(u)&=&0 \quad otherwise
\end{eqnarray}

and for $p$=1,2,3$\cdots$

\begin{equation}
\label{ }
N_{a,p}(u)=\frac{u-u_{p}}{u_{a+p}-u_{a}} \cdot N_{a,p-1} + \frac{u_{a+p+1}-u}{u_{a+p+1}-u_{a+1}} \cdot N_{a+1,p-1} \end{equation}

For a more detailed description of NURBS the reader is referred to \cite{Piegl1997b}.

As an example we show in Figure \ref{Cyl} the geometrical representation of a quarter cylinder.
\begin{figure}[t]
\begin{center}
\includegraphics[scale=0.8]{Cyl1.pdf}
\caption{Quarter cylinder with control points and local u,v coordinate system}
\label{Cyl}
\end{center}
\end{figure}

For this example the knot vectors in u,v direction are
\begin{eqnarray}
\Xi_{u} & = & \left(\begin{array}{cccccc}0 & 0 & 0 & 1 & 1 & 1\end{array}\right) \\
\Xi_{v} & = & \left(\begin{array}{cccccc}0 & 0  & 1 & 1\end{array}\right)
\end{eqnarray}

and the weights are given by

\begin{equation}
\label{ }
\textbf{w}=\left(\begin{array}{ccc}1 & 0.7 & 1 \\1 & 0.7 & 1\end{array}\right)
\end{equation}

\subsection{Analysis with trimmed NURBS patches}

If there is an intersection of NURBS patches the CAD program provides trimming information.
The trimming information comprises one or more trimming curves, which are B-splines and are defined in the local coordinate of the NURBS patch to be trimmed. A method for performing an analysis on trimmed surfaces has already been presented in \cite{Schmidt2012} and involves finding the intersection of the trimming curve with the underlying NURBS surface and a reconstruction of the knot spans and control points.

\begin{figure}
\begin{center}
\includegraphics[scale=1.0]{Map.pdf}
\caption{Trimming of a quarter cylinder using the double mapping algorithm}
\label{TrimC}
\end{center}
\end{figure}

Here we present a novel approach that appears to be simpler to implement and more efficient, since all it involves is a mapping.
For explaining the proposed trimming method we use a simple example and assume that the cylinder in Figure \ref{Cyl} is trimmed by 2 trimming curves obtained from the CAD program, marked I and II in Figure \ref{TrimC}. Trimming curve I is a B-spline of order $p=1$ and has 2 control points and  trimming curve II is of order $p=3$ and has 6 control points. The idea is to map the trimmed area from the $u,v$ coordinate system to an $\bar{u}$ ,$\bar{v}$ coordinate system that represents a unit square. The trimming curves map as straight lines along $\bar{v}$ at $\bar{u}=0$ and $\bar{u}=1$ in this system.

The mapping from the $\bar{u}$ ,$\bar{v}$ to the $u,v$ system is given by

\begin{eqnarray}
u & = & N_{1}(\bar{u})u_{I} (\bar{v}) + N_{2}(\bar{u})u_{II} (\bar{v})\\
v & = & N_{1}(\bar{u})v_{I} (\bar{v}) + N_{2}(\bar{u})v_{II} (\bar{v})
\end{eqnarray}

where
\begin{eqnarray}
N_{1}(\bar{u})& = &1-\bar{u} \\
N_{2}(\bar{u})& = & \bar{u}
\end{eqnarray}

For trimming curve I we compute the points along $\bar{u}=0$ as

\begin{eqnarray}
u_{I} (\bar{v})& = & \sum_{b=1}^{B} N_{b,p}^{I} (\bar{v})u_{b,I}\\
v_{I} (\bar{v})& = & \sum_{b=1}^{B} N_{b,p}^{I} (\bar{v})v_{b,I}
\end{eqnarray}

where $N_{b,p}^{I} (\bar{v})$ are the B-spline functions defining the trimming curve and $u_{b,I}$, $v_{b,I}$ are the local coordinates of the control points.

For triming curve II we compute the points at $\bar{u}=1$ as

\begin{equation}
u_{II} (\bar{v}) =  \sum_{b=1}^{B} N_{b,p}^{II} (\bar{v})u_{b,II}
\end{equation}

\begin{equation}
v_{II} (\bar{v}) =  \sum_{b=1}^{B} N_{b,p}^{II} (\bar{v})v_{b,II}
\end{equation}

where $N_{b,p}^{II} (\bar{v})$ are the B-spline functions defining the trimming curve and  $u_{b,II}$, $v_{b,II}$ are the local coordinates of the control points. This represents a linear interpolation between the trimming curves.
The evaluation of the integrals and the definition of the basis functions is carried out in the $\bar{u}$,$\bar{v}$ coordinate system and then mapped onto the $u,v$ and then the $x,y,z$ coordinate system (the mapping involves two Jacobians). An extension of the method to more than 2 trimming curves is possible.
The proposed mapping however would not work for the case where the trimming curve is a closed contour inside the $u,v$ domain. In this case the NURBS patch may be split into two or more patches.

\subsection{Geometry definition}

To explain the definition of the problem geometry we use the example of a branched tunnel.
Figure \ref{Tun} depicts the CAD model of the tunnel with a branch at 90$^{\circ}$.

\begin{figure}
\begin{center}
\includegraphics[scale=1.0]{Geomtun.pdf}
\caption{CAD model of tunnel branch and resulting 1/4 simulation  model depicting NURBS patches and control points}
\label{Tun}
\end{center}
\end{figure}

For the simulation, symmetry conditions were applied which meant that only 1/4 of the problem needed to be considered. The geometry definition with 6 NURBS patches (2 of them trimmed) and 3 infinite plane strain NURBS patches \cite{BeerBordas2014} is shown on the right of Figure \ref{Tun}. This geometry description is as accurate as the CAD description and needs no further refinement.

\section{SIMULATION }

An ideal companion to CAD is the boundary element method, as both rely on a description of the problem by surfaces. Therefore this method will be used for the simulation. In the following we only present a brief description of the method. Details of the implementation of the isogeometric BEM can be found in \cite{Scott2013197} and \cite{BeerBordas2014}.

\subsection{Boundary Element Method}
The boundary integral equation for an elastic continuum without body forces can be written as:

\begin{eqnarray}
\label{eq:bie}
	\textbf{c}\left( P \right) \mathbf{u}\left( P \right) & = &
 \int_{S} \mathbf{U}\left( P,Q \right) \mathbf{t}\left( Q \right) dS  -   \int_{S}
\mathbf{T}\left( P,Q \right) \mathbf{u}\left( Q \right) dS
\end{eqnarray}

The coefficient $ \mathbf{c}\left( P \right) $ is a free term
related to the boundary geometry.

$ \mathbf{u}\left( Q \right) $ and $ \mathbf{t}\left( Q
\right) $ are the displacements and tractions on the boundary and
$\mathbf{U}\left( P,Q \right) $ and $ \mathbf{T}\left( P,Q \right) $  are matrices containing Kelvin's fundamental solutions (\textit{Kernels}) for the displacements and
tractions respectively.  $P$ is the source point and $Q$ is the field point.

For the purpose of explaining the proposed simulation procedure we use the tunnel problem where tractions due to excavation are known and the displacements are unknown.

The integral equation can be discretized by using an interpolation of the displacements:

 \begin{equation}
\mathbf{u}=  \sum_{b=1}^{B}\sum_{a=1} ^{A}R_{a,b}^{p_{d},q_{d}} (\bar{u},\bar{v})\mathbf{d}_{a,b}^{e}
 \end{equation}

where  $\mathbf{d}_{a,b}^{e}$ denote the parameters for $\mathbf{u}$ at points $a,b$. The subscript $d$ of $p$ and $q$ indicates that the functions differ form the ones used for the description of the geometry. Therefore we use the terminology \textit{generalized IGA} as in the classical IGA reported in the literature the same functions are used.

To solve the discretized integral equation we use the \textit{Collocation} method, that is we satisfy it only at discrete points on the boundary $P_{n}$

The discretized integral equation can be written as:

\begin{equation}
\label{eq:bie_a}
\begin{split}
\mathbf{c}\left( P_{n} \right) \: \sum_{b=1}^{B}\sum_{a=1} ^{A}R_{a,b}^{p_{d},q_{d}} (\bar{u},\bar{v})\mathbf{d}_{a,b}^{ec}=
\int_{S_{e}}
\mathbf{U}\left( P_{n},Q \right) \: \textbf{t }\: dS \\ - \sum_{e=1}^{E}
 \sum_{b=1}^{B}\sum_{a=1} ^{A}\int_{S_{e}} (\mathbf{T}\left( P_{n},Q \right) \: R_{a,b}^{p_{d},q_{d}} (\bar{u},\bar{v}) dS) \mathbf{d}_{a,b}^{e}\\ \quad for \quad n=1,2,3...N
\end{split}
 \end{equation}

where $ec$ denotes the patch that contains the collocation point and $E$ is the number of patches. The integrals over NURBS patches are computed using Gauss Quadrature. In some cases the patches have to divided into integration regions. For the case where the integrand tends to infinity inside a patch special procedures have to be applied. The reader is referred to \cite{BeerBordas2014} for a detailed discussion on this topic.

\begin{figure}
\begin{center}
\includegraphics[scale=0.6]{DOFplot.pdf}
\caption{Plot showing numbering of patches and parameters for tunnel example}
\label{DOF}
\end{center}
\end{figure}

The collocation points $P_{n} $ are first computed in the local coordinate system and then transferred to the global coordinate system as explained previously.
The local coordinate of the collocation points can be computed using the method proposed by Greville \cite{Greville}:

\begin{equation}
\bar{u}(P_{i})= \frac{\bar{u}_{i+1}+\bar{u}_{i+2} + \dots +\bar{u}_{i+p_{d}}}{p_{d}} \qquad i=0,1, \dots ,I
\end{equation}

\begin{equation}
\bar{v}(P_{j})= \frac{\bar{v}_{j+1}+\bar{v}_{j+2} + \dots +\bar{v}_{j+q_{d}}}{q_{d}} \qquad j=0,1, \dots ,J
\end{equation}

where $i$ and $j$ denote the local (patch) numbering of the collocation point and $I$ and $J$ are the number of parameter points of the patch in $\bar{u}$ and $\bar{v}$ direction. $\bar{u}_{n}$ and $\bar{v}_{n}$ denote the corresponding entries in the Knot vector of the basis functions approximating the unknown displacements.

\textbf{Remark}: The Greville formulae also compute collocation points at the edges of NURBS patches. The coordinates of these points, computed in local coordinates of the different connecting NURBS, have to be the same. For trimmed NURBS this would only be the case if the parameter spaces of the trimming curves match. For the tunnel example presented here this was the case, so the collocation points matched, but this may not be guaranteed for a general application. However, this can be resolved by using discontiuous collocation \cite{Marus}.

The final system of equations to be solved is
\begin{equation}
\label{ }
[\textbf{T}] \{ \textbf{u} \}= \{ \textbf{F} \}
\end{equation}

where $[\textbf{T}]$ and $\{ \textbf{F} \}$ are assembled from patch contributions and $\{ \textbf{u} \}$ contains all displacement values.

\subsection{Solution and refinement strategies}

Because the idea is to completely separate the description of the geometry from the approximation of the unknown  the unknown parameters are numbered independently. The parameter numbering is shown for the tunnel example in Figure \ref{DOF} for the coarsest discretization. This is updated automatically by the program during the refinement process.
The variation of the unknown is defined in the local $\bar{u},\bar{v}$ coordinate system and then mapped into the global coordinate system. For trimmed patches the procedure outlined in 2.2 is used. Figure \ref{Mapf} shows this for patch 2 of the tunnel example and the first basis function. We start the simulation with basis functions of order $p=q=2$ and use the different refinement strategies available in IGA. For the problem of the tunnel intersection it was found that raising the order to $p=q=4$ gave the best results (Figure \ref{Ref}).

\begin{figure}
\begin{center}
\includegraphics[scale=0.8]{Mapfunc.pdf}
\caption{Example of mapping a basis function onto patch 2}
\label{Mapf}
\end{center}
\end{figure}

\begin{figure}
\begin{center}
\includegraphics[scale=0.7]{Refine.pdf}
\caption{Refinement process: Basis functions on edges of NURBS patch 2: a) in global coordinate system for $p=q=2$, b) in local coordinate system  for $p=q=2$ and c) refinement to $p=q=4$}
\label{Ref}
\end{center}
\end{figure}

\section{Example}

For testing the proposed algorithm we analyze the tunnel intersection with the following properties

\begin{itemize}
  \item Elastic domain with E= 1000 MPa, v= 0
  \item Virgin stress: $\sigma_{z}=1 Mpa$ compression, all other components zero
  \item Symmetry about x-z and x-y planes
  \item Single stage excavation
\end{itemize}

For the case of a single stage excavation this is a pure Neumann problem and the tractions are given as:

\begin{equation}
\label{ }
\textbf{t}= \textbf{n} \boldsymbol \sigma
\end{equation}

where $\textbf{n}$ is the outward normal and $\boldsymbol \sigma$ is the pseudo-vector of virgin stress.

Figure \ref{Mesh} shows the location of collocation points and the subdivision into integration regions for the finest discretization ($p=q=4$, 291 degrees of freedom).  In the program, subdivision lines are generated automatically through collocation points. In addition, the user may define additional subdivision lines. Further subdivisions are automatically made by the program for the case where the source point $P$ is close to the integration region, using a quad tree method (for details see \cite{Beer2015})

\begin{figure}
\begin{center}
\includegraphics[scale=0.4]{Meshsub.pdf}
\caption{Location of collocation points and subdivision into integration regions}
\label{Mesh}
\end{center}
\end{figure}

Figure \ref{Def} shows one result of the analysis namely the deformed shape.

\begin{figure}
\begin{center}
\includegraphics[scale=0.4]{Deformed.pdf}
\caption{Deformed shape}
\label{Def}
\end{center}
\end{figure}

To check the accuracy, the results are compared with a conventional BEM analysis using Serendipity functions for describing the geometry and the variation of the unknowns.

Figure \ref {BEFE} shows the mesh used for the analysis with the simulation program BEFE \cite{BEFE2003}. Two analyses were performed, one with linear and one with quadratic shape functions. The latter had 2895 unknowns.

\begin{figure}
\begin{center}
\includegraphics[scale=0.4]{MeshBEFE.pdf}
\caption{Mesh used for the conventional BEM analysis with isoparametric elements}
\label{BEFE}
\end{center}
\end{figure}

The z-displacement along the trimming curve is shown in Figure \ref{Comp} for the conventional BEM and the new approach.

\begin{figure}
\begin{center}
\includegraphics[scale=0.35]{Comparison.pdf}
\caption{Variation of the vertical displacement along the trimming line, comparison of new method (IGA) with isoparametric BEM (BEFE)}
\label{Comp}
\end{center}
\end{figure}

It can be seen that the conventional BEM results converge towards the results obtained by the method presented here.

\section{SUMMARY AND CONCLUSIONS}

We have presented a novel approach to the simulation with the boundary element method and trimmed NURBS patches. The innovations are in two parts. First, a procedure is presented for analyzing trimmed surfaces, which is much simpler to implement and more efficient than published methods. Secondly, we propose that the approximation of the unknown is completely independent from the definition of the boundary geometry. Our motivation comes from the fact that the boundary geometry is described with the same accuracy as the CAD model and needs no further refinement. Efficient refinement strategies available for NURBS  can then be applied to the description of the unknown only.

Comparison of the results of the analysis of a branched tunnel with a conventional isoparametric analysis shows good agreement. However, the number of degrees of freedom required to achieve the same (if not better) result is an order of magnitude smaller. The reason for this is that the approximation of the geometry is much more accurate and that the functions describing the variation of the unknowns exhibit a much higher continuity. In addition the need for the generation a mesh is completely avoided as data are taken directly from the CAD program.

The algorithm for trimming still requires further testing on more complex cases and may have some limitations.
Future work will concentrate on the implementation of efficient procedures for non-linear analysis into the proposed framework and on methods of reducing storage and run times for large simulations.

\section*{ACKNOWLEDGMENTS}
This work was supported by the Austrian Science Fund (Project "Fast
isogeometric BEM"), Grant Number P24974-N30.

\bibliographystyle{plain}
\bibliography{ifbbib}

\end{document}

config: arxiv


TITLE: Visualizing mathematics and geometry
QUESTION [4 upvotes]: Im writing a paper on the role of visualization in mathematics and specifically geometry.  I was wondering if it is possible to represent any arbitrary system of relations and manipulable objects visually, and how this interpretation imbues a meaning to the mathematical structures.  At this point I am simply trying to gather the opinions and musings of fellow mathematicians and logicians on this topic to see which ideas are the most interesting and worth writing about.  I am particularly interested in the extreme cases of things which are very difficult to interpret visually, as well as the things which are best interpreted visually(fractals and impossible figures).  Let me know what you guys think about this topic, Im sure this community has a wealth of interesting ideas.
Thanks! looking forward to hearing your thoughts
EDIT:
I realize my question is unclear, but Im not looking for an answer per se but more of some general thoughts on the subject.  For example, when you say that there are far more mathematical objects which cannot be easily visualized, what aspects or properties make visualization difficult or impossible?  Im interested in discussing the nature of visualization and its role in mathematics.  Is visualization simply a tool to help us intuitively understand the more abstract concepts and relations?  Or is visualization a process deeply connected to mathematics itself?  If so is this due to our human intuition imbuing its own structural artifacts into algebra and logic?
I realize I am not asking a specific question, so answering in a concrete way is difficult, but as I said I am more interested in discussion and related examples of interesting phenomena.

REPLY [1 votes]: I think that "vision" in mathematics is more deep that in common sense. When a matematician "visualises" something it is not just a two-dimensional picture. It is a picture in a much more rich world.
For instance, there are famous mathematicians which don't see (I don't know the word in english). And some of them do... GEOMETRY! (and at least one 4-dimensional geometry)
To visualise is like "tranlsate in a world that you are confortable with".
Visualisation's power is proportional to how many examples and counterexamples you have "seen" in your life.
In this sense, visualising is absolutely fundamental for mathematicians.
Here my favourite "difficult thing to visualise":
Russel paradox: The set of sets that does not contain itlsef, does contain itself?
In general, can you immagine the universe? (the mathematical universe, the class of all the sets) is it bounded in your "vision"?


TITLE: Help me about continuity of graph
QUESTION [0 upvotes]: For the function:
  $$f(x) =
\begin{cases}0 & ~~\text {if}~~ x=-1;\\ -x & ~~\text{if}~~ -1 \lt x \lt 0;\\ x & ~~\text{if } 0 \le x\le 1 \end{cases}$$
  $$f(x+2) = f(x) + 1$$
  make this graph, and watch the graph.

I'm sure that $y=f(x)$ is discontinuous at $x=1$, but about $x \in [0,1]$, $x$ is continuous at $x=1$ ?

REPLY [1 votes]: Let's sketch the graph of this function first over $[-1,1]$.
It looks messy:
$f(x)= \cases{\color{darkgreen}{0}, & $\color{darkgreen}{x=-1}$ \cr\color{maroon}{ -x }& $\color{maroon}{-1\lt x\lt 0} $\cr  \color{darkblue}{x}, & $\color{darkblue}{ {0\le x\lt 1} }$ \cr}$
$f$ s a piecewise defined function. To sketch the graph of $f$, look at each "piece" separately:
When $\color{darkgreen}{x=-1}$, the function has the value $0$; so the point $(-1,0)$ is on the graph. We plot this point in the graph below.
For the range of $x$-values in $\color{maroon}{-1<x<0}$, the rule for $f$ is $\color{maroon}{f(x)=-x}$.  Over the interval $(-1,0)$ sketch the graph of $\color{maroon}{y=-x}$.  Note that the value of $f$ at $-1$ is 0 and is not given by the rule $f(x)=-x$.
The "closed" and "open" dots on the graph below indicate what value $f$ takes at the point $x=0$.
For the range of $x$-values in $\color{darkblue}{0\le x\le1}$, the rule for $f$ is
$\color{darkblue}{f(x)= x}$.  Over the interval $(0,1)$ sketch the graph of $\color{darkblue}{y=-x}$.  Note that the value of $f$ at $ 1$ is 1.

So, that's the graph of $f$ over $[-1,1]$.
What about the rest of the graph? Well, that's obtained from the rule
$$
f(x+2)=f(x)+1.
$$
So, for example: $f(3)=f(1+2)=f(2)+1$ and $f(3.5)=f(1.5+2)=f(1.5)+1$.
To evaluate $f$ at $f(x+2)$, you just take the value of $f$ at $x$, and add 1.
So, to obtain the graph of $f$ over $[1,3]$, you just draw the same shape as the graph over $[1-2, 3-2]=[-1,1]$ but raise it vertically by 1 unit.
To obtain the graph of $f$ over $[3,5]$, you just draw the same shape as the graph over $[3-2, 5-2]=[1,3]$ but raise it by 1 unit. etc...
Below, is the graph of $f$ over three of its "periods":

Now, looking at the graph, you can see that $f$  discontinuous at every odd integer $x\ge -1$ and that $f$ is continuous for any $x>-1$ that is not an odd integer.


TITLE: 2nd solution of $\cos x \cos 2x\cos 3x= \frac 1 4 $
QUESTION [2 upvotes]: $\cos x \cos 2x\cos 3x= \dfrac 1 4 $

Attempt explained:
$(2\cos x \cos 3x)\cos 2x = \frac1 2 $
$(\cos 4x +\cos 2x )\cos 2x = \frac 1 2 \\\cos ^2y + \cos y (2\cos^2y- 1)= \frac1 2 \\ $
(Let, y = 2x)
$\implies 4\cos^3 y+2\cos^2y- 2\cos y-1=0$
I solved this equation using Rational Root Theorem and got $y = \frac 1 2$
$\implies x= m\pi \pm \dfrac\pi3 \forall x\in \mathbb {Z}$
Using Remainder theorem, the other solution is $\cos^2 y = \dfrac 1 2 $
$\implies x = \dfrac n2\pi \pm\dfrac \pi 8 $
But answer key states: $ x= m\pi \pm \dfrac\pi3or x =(2n+1)\dfrac \pi 8$
Why don't I get the second solution correct?

REPLY [4 votes]: It's $$\cos2x(\cos2x+\cos4x)=\frac{1}{2}.$$
Now, let $\cos2x=t$.
Thus, $$t(t+2t^2-1)=\frac{1}{2}$$ or
$$4t^3+2t^2-2t-1=0$$ or
$$2t^2(2t+1)-(2t+1)=0$$ or
$$(2t^2-1)(2t+1)=0,$$ which gives
$$\cos4x=0$$ or $$\cos2x=-\frac{1}{2}$$ and we can write the answer.


TITLE: Matrix "dimensional analysis" of Lagrangians in QFT
QUESTION [0 upvotes]: Since the important things in the QFT Lagrangian are vectors and matrices, I wanted to do a "matrix dimensional analysis" of each term.
The electromagnetic Lagrangian (ignoring all constants and signs) is :
$\bar{\psi}\gamma^{\mu}\partial_{\mu}\psi + \bar{\psi}\gamma^{\mu}\partial^{\mu}A_{\mu}\psi + \bar{\psi}\psi + F_{\mu\nu}F^{\mu\nu}$
Since $\psi$ is a (4x1) vector, $\bar{\psi}$ is (1x4), and each $\gamma^\mu$ is (4x4), the first term is :
(1x4)(4x4)(4x1) = (1x1) or a scalar.
Since $A_{\mu}$ for a given $\mu$ is a scalar, the second term is :
(1x4)(4x4)(1x1)(4x1) = (1x1), a scalar.
The third term is :
(1x4)(4x1) = (1x1)
And because of the summations in the fourth term, the two (4x4) terms collapse into a scalar as well.
The problem I'm having is with the QCD Lagrangian :
$\bar{\psi}\gamma^{\mu}\partial_{\mu}\psi + \bar{\psi}\gamma^{\mu}\lambda\cdot A_{\mu}\psi + \bar{\psi}\psi + F_{\mu\nu}^aF^{a\mu\nu}$
Now, $\psi$ is primarily a 4-vector spinor, so it's still (4x1) (which it must be in order to be multiplied by the $\gamma$'s).  So the first and third terms don't change at all.
But the $\lambda$'s are 3x3 color matrices, and it was clarified here that each $A_\mu$ is a 3x3 matrix, which makes sense if they're to be multiplied together.  And I'm assuming $\lambda\cdot A_{\mu}$ = $\lambda_aA^a_\mu$ (a=1..8) is still a 3x3 matrix (a "weighted sum", if you will, of each $A_\mu$).
So then the second term looks like :
(1x4)(4x4)(3x3)(3x3)(4x1), which does not work.
Now, I understand that $\psi$ in QCD is a Dirac spinor of the quark field, so some component of $\psi$ must represent the colors $[\psi^r \psi^b \psi^g]^T$.  But even if $\psi$ represents a color vector in the third term, there's still the 4x4 $\gamma$'s.
Question 1 : what is the resolution to this? How can $\psi$ represent both a spinor and color space in the same equation?  And is it only in the third term that $\psi$ represents the color space? (Ok, questions 1-3 :)
Regarding the $F_{\mu\nu}$ term, since it is created from the $A_\mu$ term, is it a 4x4 matrix of 3x3 matrices?  If so, the summations over $\mu$ and $\nu$ (0..3) would still leave a 3x3 matrix.
Question 2 : what is the resolution to this? Is there an implied summation over the color matrices? Or something else?
What I would most appreciate is the QCD Lagrangian, with both spinor and color indices included in every variable.
(edited to correct mistakes pointed out below)

REPLY [3 votes]: In the case of QCD $\psi$ has both a spinor and a color index. If you put all appropriate indices on everything you will see how it works.
Edit: Answer extended after a comment from OP.
Let $\mu, \nu, \cdots=0, \cdots, 3$ be Lorentz indices, $\alpha,\beta, \cdots = 1,\cdots, 4 $ spinor indices, $a,b,\cdots = 1,\cdots ,8$ indices for the generators of the Lie algebra  ($su(3)$ for QCD) and $r,s,\cdots = 1,2,3$ be the indices for the $\bf{3}$ representation of $su(3)$.
We then have $\psi_{\alpha \, r}$ for the spinor,  $(\lambda^a)_{rs}$ is an element a generator, $A_{\mu a}$ is the gauge field. These are now all $1\times 1$ in your notation.
As an example
$$
\bar{\psi} \gamma^\mu \lambda \cdot A_\mu \psi = \psi^\dagger_{\alpha r} (\gamma^0)_{\alpha\beta} (\gamma^\mu)_{\beta\delta}(\lambda^a)_{rs} A_{\mu a} \psi_{\delta s}
$$
As you see all indices are contracted. The $\partial_\mu$ should not be present in that term. I trust you can fill in the rest.