Ich versuche, mein Dokument zu kompilieren, erhalte jedoch wiederholt den folgenden Fehler:
Package listofitems Error: Empty list ignored, nothing to do. \end{align}
Die Zeile, die diesem Fehler entspricht, lautet:
\begin{align}
\mathbf{A}^{n+1}=\underbrace{\left[\begin{array}{c}\mathbf{\bar{Z}} \\ \hline \mathbf{0}\end{array}\right]}_{n}
\setstackgap{L}{1.2 \normalbaselineskip}
\vcenter{\hbox{\stackunder[0.1pt]{\left.{\Centerstack{}}\right\}\scriptstyle n}{\left.{\Centerstack{}}\right\}\scriptstyle m}}}
\end{align}
Ich weiß nicht, was daran falsch ist. Die benötigte Ausgabe ist wie folgt:
Die gleiche Datei lief auf meinem anderen Computer einwandfrei, deshalb habe ich Latex deinstalliert und neu installiert, aber dieser Fehler tritt immer wieder auf. Die Präambel meiner Hauptdatei sieht folgendermaßen aus:
\documentclass[twoside]{iitbreport}
%% Default spacing: 1.5
%% Default font size: 12pt
%% Default font: txfonts (similar to times new roman)
%% Selectively comment out sections that you want to be left out but
%% maintaining the page numbers and other \ref
\includeonly{%
intro/introduction_v2,
lit/literature,
chapter/chapter1v1,
chapter/chapter2v1,
chapter/chapter3v1,
chapter/chapter4v1,
chapter/conclusion,
rnd/results,
dec,abs,pub,ack
}
%%% Some commonly used packages (make sure your LaTeX installation
%%% contains these packages, if not ask your senior to help installing
%%% the packages)
\usepackage{booktabs}
\graphicspath{{chapter/}}
\usepackage{chemformula} % Formula subscripts using \ch{}
\usepackage[T1]{fontenc} % Use modern font encodings
%\usepackage{amsthm}
\usepackage{blindtext}
\usepackage{graphicx}
\usepackage{chemist}
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{epstopdf}
\epstopdfsetup{update} % only regenerate pdf files when eps file is newer
\usepackage{amsfonts}
\usepackage{float}
\usepackage{mathtools}
\usepackage{enumerate}
\usepackage{color}
\usepackage{subfig}
\usepackage{algorithm}
\usepackage{algpseudocode}
\usepackage{pifont}
\usepackage{mathrsfs}
\usepackage{arydshln}
\usepackage{amsbsy}
%\usepackage{booktabs}
\usepackage{dsfont}
\usepackage{graphicx}
\usepackage{url}
\usepackage{xcolor}
\usepackage{wasysym}
%\usepackage{geometry}
\usepackage{easybmat}
\usepackage{multirow,bigdelim}
\usepackage[usestackEOL]{stackengine}
\usepackage{listofitems}
\usepackage{tabstackengine}
\stackMath
\usepackage[utf8]{inputenc}
\allowdisplaybreaks
%%% Macro definitions for Commonly used symbols
\newcommand{\Rey}{\ensuremath{\mathrm{Re}}}
\newcommand{\avg}[1]{\ensuremath{\overline{#1}}}
\newcommand{\tenpow}[1]{\ensuremath{\times 10^{#1}}}
\newcommand{\pder}[2]{\ensuremath{\frac{\partial#1}{\partial#2}}}
\newtheorem{lemma}{Lemma}
\numberwithin{lemma}{section}
\newtheorem{remark}{Remark}
\numberwithin{remark}{section}
\newtheorem{theorem}{Theorem}
\numberwithin{theorem}{section}
\newtheorem{proof}{Proof}
\numberwithin{proof}{section}
%\theoremstyle{plain}
\newtheorem{assumption}{Assumption}
\numberwithin{assumption}{section}
% Referencing macros
\newcommand{\Eqref}[1]{Equation~\eqref{#1}}
\newcommand{\Tabref}[1]{Table~\ref{#1}}
\newcommand{\Figref}[1]{Figure~\ref{#1}}
\newcommand{\Appref}[1]{Appendix~\ref{#1}}
Die Zeilen über denen, die dem Fehler entsprechen, lauten wie folgt:
\section{Householder Transformation for Square root filtering}
\label{ap:householder}
In this section we briefly discuss the Householder transformation technique to implement square root filtering in EGPOF (Section \ref{sc:online_algo_EGPOF}) for improved numerical precision. Detailed discussion for the same can be found in \cite{Maybeck}.
The a prior covariance matrix for EGPOF is $\mathbf{P}(k+1|k)$, and its square root is $\mathbf{P_{sq}}(k+1|k)$ such that $\mathbf{P}(k+1|k)=\mathbf{P_{sq}}(k+1|k)[\mathbf{P_{sq}}(k+1|k)]^T$. The aposterior error covariance matrix is $\mathbf{P}(k+1|k+1)$, and its square root is $\mathbf{P_{sq}}(k+1|k+1)$. Also, square root of $\left[\mathbf{\bar{Q}}-\mathbf{S}[\mathbf{R}]^{-1}[\mathbf{S}]^T\right]$ is given by $\mathbf{Q_{sq}}$ such that $\left[\mathbf{\bar{Q}}-\mathbf{S}[\mathbf{R}]^{-1}[\mathbf{S}]^T\right]=\mathbf{Q_{sq}}[\mathbf{Q_{sq}}]^T$. Then using Eq. \eqref{eq:pred_cov_lin}, the predicted covariance can be written as:
\begin{align}
\label{eq:pred_cov_household}
\mathbf{P}(k+1|k) &=\boldsymbol{\Phi}(k)\mathbf{P_{sq}}(k|k)[\mathbf{P_{sq}}(k|k)]^T\boldsymbol{\Phi}(k)^T+\boldsymbol{\Gamma}(k)\mathbf{Q_{sq}}[\mathbf{Q_{sq}}]^T[\boldsymbol{\Gamma}(k)]^T
\end{align}
Now it is desired to find the propagation equation for the square root of $\mathbf{P}(k+1|k)$; to find the relation to yield $\mathbf{P_{sq}}(k+1|k)$ such that $\mathbf{P_{sq}}(k+1|k)[\mathbf{P_{sq}}(k+1|k)]^T$ is equal to the right hand side of Eq. \eqref{eq:pred_cov_household}. Suppose that we can find an orthogonal $(n+m)\times (n+m)$ matrix $\mathbf{\tilde{T}}$ i.e. $\mathbf{\tilde{T}}[\mathbf{\tilde{T}}]^T=\mathbf{I}$ such that
\begin{align}
\left[\begin{array}{c:c}
\boldsymbol{\Phi}(k)\mathbf{P_{sq}}(k|k) & \boldsymbol{\Gamma}(k)\mathbf{Q_{sq}}
\end{array}\right]\mathbf{\tilde{T}}&=\left. \left[\begin{array}{c:c}\underbrace{\mathbf{P_{sq}}(k+1|k)}_{\text{n columns}} & \underbrace{\mathbf{0}}_{\text{m columns}} \end{array}\right]\right\}\text{n rows}
\end{align}
then in fact an $n \times n$ square root matrix $\mathbf{P_{sq}}(k+1|k)$ will have been found which satisfies the desired relationship. Further, note that the same procedure could also be applied to:
\begin{align}
\mathbf{P}(k+1|k+1)=\big[\mathbf{I}-\mathbf{L}(k+1)\mathbf{C}(k+1)\big] \mathbf{P}(k+1|k)
\big[\mathbf{I}-\mathbf{L}(k+1)\mathbf{C}(k+1)\big]^T + \mathbf{L}(k+1) \mathbf{R} [\mathbf{L}(k+1)]^T
\end{align}
where, we find an orthogonal $(n+r)\times (n+r)$ matrix $\mathbf{\tilde{T}}$ i.e. $\mathbf{\tilde{T}}[\mathbf{\tilde{T}}]^T=\mathbf{I}$ such that
\begin{align}
\left[\begin{array}{c:c}
\big[\mathbf{I}-\mathbf{L}(k+1)\mathbf{C}(k+1)\big]\mathbf{P_{sq}}(k+1|k) & \mathbf{L}(k+1) [\mathbf{R}]^{1/2}
\end{array}\right]\mathbf{\tilde{T}}&=\left. \left[\begin{array}{c:c}\underbrace{\mathbf{P_{sq}}(k+1|k+1)}_{\text{n columns}} & \underbrace{\mathbf{0}}_{\text{r columns}} \end{array}\right]\right\}\text{n rows}
\end{align}
One of the method of generating such a $\mathbf{P_{sq}}(k+1|k)$ and $\mathbf{P_{sq}}(k+1|k+1)$ is Householder transformation technique. Conceptually, it generated $\mathbf{\tilde{T}}$ as
\begin{align}
\mathbf{\tilde{T}}=\mathbf{\tilde{T}}^n\mathbf{\tilde{T}}^{(n-1)}\ldots \mathbf{\tilde{T}}^1
\end{align}
where $\mathbf{\tilde{T}}^k$ is generated recursively as
\begin{align}
\mathbf{\tilde{T}}^k=\mathbf{I}-d^k\mathbf{u}^k[\mathbf{u}^k]^T
\end{align}
with the scalar $d^k$ and the $(n+m)$-vector $\mathbf{u}^k$ defined in the following. However, the computational algorithm never calculates these $\mathbf{\tilde{T}}^k$'s or $\mathbf{\tilde{T}}$ explicitly. The initial condition on the $(n+m)\times n$ $\mathbf{A}^k$ is
\begin{align}
\mathbf{A}^1&=[\mathbf{P_{sq}}(k+1|k)]^T=\left[\begin{array}{c:c}
\boldsymbol{\Phi}(k)\mathbf{P_{sq}}(k|k) & \boldsymbol{\Gamma}(k)\mathbf{Q_{sq}}
\end{array}\right]^T
\end{align}
Now letting $\mathbf{A}^k_j$ represent the $j^{th}$ column of $\mathbf{A}^k$, perform the $n$-step recursion, for $k=1,2,\ldots,n$:
\begin{align}
a^k &=\sqrt{\sum_{j=k}^{n+m}[A^k_{jk}]^2}\text{sign}\{A^k_{kk}\} \nonumber \\
d^k&=\frac{1}{a^k(a^k+A^k_{kk})} \nonumber \\
u_j^k&=\begin{cases}
0& j< k\\
a^k+A^k_{kk}& j=k \\
A^k_{jk} & j=(k+1),\ldots ,(n+m)
\end{cases} \nonumber \\
y^k_j&=\begin{cases}
0 & j<k \\
1 & j=k \\
d^k[\mathbf{u}^{k}]^T\mathbf{A}^k_j & j=(k+1), \ldots , n
\end{cases} \nonumber\\
\mathbf{A}^{k+1}&=\mathbf{A}^k-\mathbf{u}^k [\mathbf{y}^{k}]^T
\label{eq:iter}
\end{align}
At stage $k$, the first $(k-1)$ columns of $\mathbf{A}^k$ are zero below the diagonal of the upper square partition, and $\mathbf{u}^k$ has been chosen so that the sub-diagonal elements of $\mathbf{A}^{k+1}_k$ will be zero. After the $n$ iterations of Eq. \eqref{eq:iter},
\begin{align}
\mathbf{A}^{n+1}=\underbrace{\left[\begin{array}{c}\mathbf{\bar{Z}} \\ \hline \mathbf{0}\end{array}\right]}_{n}
\setstackgap{L}{1.2 \normalbaselineskip}
\vcenter{\hbox{\stackunder[0.1pt]{\left.{\Centerstack{}}\right\}\scriptstyle n}{\left.{\Centerstack{}}\right\}\scriptstyle m}}}
\end{align}
and then $\mathbf{P_{sq}}(k+1|k)$ is generated as
\begin{align}
\mathbf{P_{sq}}(k+1|k)=\left[\mathbf{\bar{Z}}\right]^T
\end{align}