IN_PROGRESS: Updating optimization documentation.

doc/algebraic_equations.dox

@@ -31,7 +31,7 @@
 \page algebraic_equations_page Algebraic Equations
 
 This page describes methods for the rigorous numerical solution of algebraic equations.
-For details on how this is implemented in Ariadne, see the \ref AlgebraicEquationSubModule documentation
+For details on how this is implemented in %Ariadne, see the \ref AlgebraicEquationSubModule documentation
 
 Consider the system of nonlinear algebraic equations
 \f[ f(x) = 0; \quad x\in D \f]

doc/logic.dox

@@ -74,7 +74,7 @@ On computing yet more digits we might find \f$r_1=3.141592653589793\,238\cdots\f
 Of course, if \f$r_1\neq r_2\f$, by computing enough digits, we will \e eventually be able to decide which of \f$r_1<r_2\f$ or \f$r_2>r_1\f$ is true, but if \f$r_1=r_2\f$, then no matter how many digits we compute, we will still never know which if \f$r_1 \lesseqgtr r_2\f$ holds.
 In this case again the computation would run forever, and we would consider the value \a indeterminate.
 
-The resulting logic is <em>%Kleenean</em> logic \f$\K=\{T,F,I\}\f$, with three values \a true \f$\top,\mathsc{T}\f$, \a false \f$\bot,\mathsc{F}\f$ and \a indeterminate \f$\uparrow,\mathsc{I}\f$.
+The resulting logic is <em>%Kleenean</em> logic \f$\K=\{T,F,I\}\f$, with three values \a true \f$\top,\mathsf{T}\f$, \a false \f$\bot,\mathsf{F}\f$ and \a indeterminate \f$\uparrow,\mathsf{I}\f$.
 The %Kleenean type represents the result of <em>quasidecidable</em> predicates.
 
 A subtype of the %Kleeneans is given by the <em>%Sierpinskian</em> type \f$\S=\{\tru,\indt\}\f$ with open sets \f$\{\},\{\tru\},\{\tru,\indt\}\f$. The %Sierpinsian type represents the result of <em>verifiable</em> predicates.
@@ -147,7 +147,7 @@ One may also think of \f$\S\f$ as "positive" %Kleeneans \f$\K^+\f$.
 \section logical_operations Logical Operations
 
 The standard logical operators \f$\neg,\wedge,\vee,\rightarrow,\leftrightarrow\f$ are all defined on \f$\K\f$.
-They can be extracted from their %Boolean counterparts by considering a set-valued interpretation with \f$\mathsc{I}=\{\mathsc{T},\mathsc{F}\}\f$. Explicitly, the operators are given by
+They can be extracted from their %Boolean counterparts by considering a set-valued interpretation with \f$\mathsf{I}=\{\mathsf{T},\mathsf{F}\}\f$. Explicitly, the operators are given by
 \f[ \begin{array}[t]{|c|c|}\hline p&\!\neg{p}\!\\\hline \fls&\tru\\\indt&\indt\\\tru&\fls\\\hline \end{array} \qquad
     \begin{array}[t]{|c|ccc|}\hline \!p \wedge q\!&\fls&\indt&\tru\\\hline \fls&\fls&\fls&\fls\\\indt&\fls&\indt&\indt\\\tru&\fls&\indt&\tru\\\hline\end{array} \quad
     \begin{array}[t]{|c|ccc|}\hline \!p \vee q\!&\fls&\indt&\tru\\\hline \fls&\fls&\indt&\tru\\\indt&\indt&\indt&\tru\\\tru&\tru&\tru&\tru\\\hline\end{array} \quad

doc/macros.js

@@ -17,23 +17,24 @@ MathJax.Hub.Config({
       Y: "{\\mathbb{Y}}",
       A: "{\\mathbb{A}}",
       seq: ["{\\vec{#1}}",1],
+      dt: ["{\\dot{#1}}",1],
       fto: "{\\longrightarrow}",
+      pfto: "{\\dashrightarrow}",
+      psfto: "{\\dashrightarrow}",
       interval: ["{[#1]}",1],
       ivl: ["{[#1]}",1],
-      dt: ["{\\dot{#1}}",1],
       unl: ["{\\underline{#1}}",1],
       ovl: ["{\\overline{#1}}",1],
       der: ["{\\dot{#1}}",1],
-      dag: "{\\dagger}",
 
       hatR: "{\\,\\widehat{\\!R}}",
       hatX: "{\\,\\widehat{\\!X}}",
       hatY: "{\\widehat{Y}}",
 
-      tru: "{\\mathsc{T}}",
-      fls: "{\\mathsc{F}}",
-      indt: "{\\mathsc{I}}",
-      unkn: "{\\mathsc{U}}",
+      tru: "{\\mathsf{T}}",
+      fls: "{\\mathsf{F}}",
+      indt: "{\\mathsf{I}}",
+      unkn: "{\\mathsf{U}}",
 
       precless: "\\prec\\!<",
       precprec: "\\prec\\!\\!\\!\\prec",
@@ -42,24 +43,24 @@ MathJax.Hub.Config({
       succsucc: "\\succ\\!\\!\\!\\succ",
       gtrsucc: ">\\!\\succ",
 
-      and: "\\wedge",
-      or: "\\vee",
+//      and: "\\wedge",
+//      or: "\\vee",
+
+      dom: "{\\mathrm{dom}}",
 
       tand: "{\\text{ and }}",
       tor: "{\\text{ or }}",
       timplies: "{\\text{ implies }}",
 
-      twoheaddownarrow: "{\\!\\downarrow\\!\\!\\downarrow}",
-      twoheaduparrow: "{\\!\\uparrow\\!\\!\\uparrow}",
 
-      pfto: "{\\dashrightarrow}",
-      psfto: "{\\dashrightarrow}",
+      dag: "{\\dagger}",
 
-      dom: "{\\mathrm{dom}}",
+      twoheaddownarrow: "{\\!\\downarrow\\!\\!\\downarrow}",
+      twoheaduparrow: "{\\!\\uparrow\\!\\!\\uparrow}",
 
       cline: ["",1],
-      textsc: ["{\\mathsc{#1}}",1],
       mathsc: ["\\text{#1}",1],
+      textsc: ["{\\mathsc{#1}}",1],
     }
   }
 });

doc/macros.sty

@@ -1,12 +1,13 @@
-\usepackage[top=25mm,left=25mm,bottom=25mm,right=25mm]{geometry}
+%\usepackage[top=25mm,left=25mm,bottom=25mm,right=25mm]{geometry}
+
+\usepackage{amssymb}
 
-\newcommand{\interval}[1]{{[#1]}}
-\newcommand{\ivl}[1]{{[#1]}}
 \newcommand{\B}{\mathbb{N}}
 \newcommand{\K}{\mathbb{K}}
-\newcommand{\S}{\mathbb{S}}
+\renewcommand{\S}{\mathbb{S}}
 \newcommand{\N}{\mathbb{N}}
 \newcommand{\Z}{\mathbb{Z}}
+\newcommand{\Q}{\mathbb{Q}}
 \newcommand{\R}{\mathbb{R}}
 \newcommand{\F}{\mathbb{F}}
 \newcommand{\I}{\mathbb{I}}
@@ -18,13 +19,21 @@
 \newcommand{\dt}[1]{\dot{#1}}
 \newcommand{\fto}{\longrightarrow}
 \newcommand{\pfto}{\dashrightarrow}
+\newcommand{\psfto}{\dashrightarrow}
+\newcommand{\interval}[1]{{[#1]}}
+\newcommand{\ivl}[1]{{[#1]}}
+\newcommand{\unl}[1]{{\underline{#1}}}
+\newcommand{\ovl}[1]{{\overline{#1}}}
+\newcommand{\der}[1]{\dot{#1}}
 
 \newcommand{\hatR}{\widehat{R}}
 \newcommand{\hatX}{\widehat{X}}
 \newcommand{\hatY}{\widehat{Y}}
 
-\newcommand{\unl}[1]{{\underline{#1}}}
-\newcommand{\ovl}[1]{{\overline{#1}}}
+\newcommand{\tru}{\mathsf{T}}
+\newcommand{\fls}{\mathsf{F}}
+\newcommand{\indt}{\mathsf{I}}
+\newcommand{\unkn}{\mathsf{U}}
 
 \newcommand{\precless}{\prec\!<}
 \newcommand{\precprec}{\prec\!\!\prec}
@@ -33,19 +42,11 @@
 \newcommand{\succsucc}{\succ\!\!\succ}
 \newcommand{\gtrsucc}{>\!\!\!\succ}
 
-\newcommand{\and}{\wedge}
-\newcommand{\or}{\vee}
+%\newcommand{\and}{\wedge}
+%\newcommand{\or}{\vee}
 
+\newcommand{\dom}{\mathrm{dom}}
 
 \newcommand{\tand}{\text{ and }}
 \newcommand{\tor}{\text{ or }}
 \newcommand{\timplies}{\text{ implies }}
-
-\newcommand{\dom}{\mathrm{dom}}
-
-\newcommand{\der}[1]{\dot{#1}}
-
-\newcommand{\tru}{\mathsf{T}}
-\newcommand{\fls}{\mathsf{F}}
-\newcommand{\indt}{\mathsf{I}}
-\newcommand{\unkn}{\mathsf{U}}

doc/modules.dox

@@ -506,7 +506,7 @@ namespace Ariadne {
  * \ingroup SolverModule
  * \brief Classes and functions for solving nonlinear algebraic equations.
  *
- * \details For the mathematical theory of algebraic equations, including implicit function problems, see the \ref algebraic_equations_page Page.
+ * \details <em>For the mathematical theory of algebraic equations, including implicit function problems, see the \ref algebraic_equations_page Page.</em>
  *
  * A \a Solver class provides functionality for solving algebraic equations, as defined in the \ref SolverInterface.
  * %Two basic kinds of problem are considered.
@@ -529,9 +529,12 @@ namespace Ariadne {
  * \ingroup SolverModule
  * \brief Classes and functions for solving linear and nonlinear programming problems.
  *
- * The LinearProgram class supports construction and solution of linear
- * programming problems. Tests for feasibility only are also possible. The
- * computations are performed by a BLAS/LAPACK style lpslv() routing.
+ * \details <em>For the mathematical theory of linear programming, see the \ref linear_programming_page Page, and for the theory of nonlinear programming, see the \ref nonlinear_programming_page Page.</em>
+ *
+ * The \a LinearProgram class supports construction and solution of linear
+ * programming problems. Tests for feasibility only are also possible.
+ *
+ *
  */
 
 

doc/nonlinear_programming.dox

@@ -30,37 +30,44 @@
 
 \page nonlinear_programming_page NonLinear Programming
 
+This page describes the theory of nonlinear programming and algorithms for the rigorous numerical solution of nonlinear optimisation problems.
+For details on how this is implemented in %Ariadne, see the \ref OptimisationSubModule documentation
+
 
 \section nonlinear_optimisation Nonlinear Constrained Optimisation Problems
 
+We first give the theory of nonlinear programming, including first- and second-order conditions for a local optimum.
+
 \subsection standard_nonlinear_optimisation Standard optimisation problem
 
 Consider the nonlinear programming problem
-\f[ \boxed{ \max f(x) \text{ s.t. } g_j(x)\geq 0,\ j=1,\ldots,l; \ h_k(x) = 0,\ k=1,\ldots,m . } \f]
-Applying a penalty function yields the unconstrained maximisation of
-\f[ f(x) + \mu  \sum_{j=1}^{l} \log g_j(x) - \frac{1}{2\nu} \sum_{k=1}^{m} (h_k(x) )^2 . \f]
+\f[ \boxed{ \min f(x) \text{ s.t. } g_j(x)\geq 0,\ j=1,\ldots,l; \ h_k(x) = 0,\ k=1,\ldots,m . } \f]
+Applying a barrier function for the inequality constraints and a penalty function for the equality constraints yields the unconstrained minimisation of
+\f[ f(x) - \mu  \sum_{j=1}^{l} \log(g_j(x)) + \frac{1}{2\nu} \sum_{k=1}^{m} (h_k(x) )^2 . \f]
 Differentiating with respect to \f$x\f$ yields
 \f[ \nabla_{\!i\,} f(x) + \mu  \sum_{j=1}^{l} \frac{\nabla_{\!i\,}g_j(x)}{g_j(x)} - \frac{1}{\nu} \sum_{k=1}^{m} h_k(x)\nabla_{\!i\,}h_k(x) = 0. \f]
 Setting \f$\lambda_j = \mu/g_j(x)\f$ and \f$\kappa_k = - h_k(x)/\nu\f$ yields
 \f[ \nabla_{\!i\,} f(x) +  \sum_{j=1}^{l} \lambda_j \nabla_{\!i\,}g_j(x) + \sum_{k=1}^{m} \kappa_k\,\nabla_{\!i\,}h_k(x) . \f]
 Combining these equations yields
 \f[ \begin{gathered} \nabla f(x) + \sum_{j=1}^{l} \lambda_j \nabla g_j(x) + \sum_{k=1}^{m} \kappa_k \nabla h_k(x) = 0; \\ \lambda_j g_j(x) - \mu = 0; \\ h_k(x) + \nu \kappa_k = 0; \\ \lambda_j \geq 0; \ g_j(x) \geq 0 . \end{gathered} \f]
-Setting \f$\nu=0\f$ these equations yields the <em>central path</em> for the problem, which is a relaxation of the optimality conditions
-\f[ \boxed{ \begin{gathered} \nabla f(x) + \sum_{j=1}^{l} \lambda_j \nabla g_j(x) + \sum_{k=1}^{m} \kappa_k \nabla h_k(x) = 0; \\ \lambda_j g_j(x) = \mu; \\ h_k(x) = 0; \\ \lambda_j \geq 0; \ g_j(x) \geq 0 . \end{gathered} } \f]
-Taking \f$\mu\to0\f$ yields the standard Karush-Kuhn-Tucker conditions for optimality
+Taking \f$\nu\to0\f$ in these equations yields the <em>central path</em> for the problem, which is a relaxation of the optimality conditions
+\f[ \boxed{ \begin{gathered} \nabla f(x) + \sum_{j=1}^{l} \lambda_j \nabla g_j(x) + \sum_{k=1}^{m} \kappa_k \nabla h_k(x) = 0; \\ \lambda_j g_j(x) = \mu; \\ h_k(x) = 0; \\ \lambda_j \gt 0; \ g_j(x) \gt 0 . \end{gathered} } \f]
+Taking \f$\mu\to0\f$ yields the standard Karush-Kuhn-Tucker (KKT) conditions for optimality
 \f[ \boxed{ \begin{gathered} \nabla f(x) + \sum_{j=1}^{l} \lambda_j \nabla g_j(x) + \sum_{k=1}^{m} \kappa_k \nabla h_k(x) = 0; \\ \lambda_j g_j(x) = 0; \\ h_k(x) = 0; \\ \lambda_j \geq 0; \ g_j(x) \geq 0 . \end{gathered} } \f]
+The Karush-Kuhn-Tucker conditions are necessary conditions for a <em>regular</em> local optimum i.e. one for which that gradients \f$\nabla h_k(x)\f$ of the equality constraints, and \f$\nabla g_j(x)\f$ for the <em>active</em> inequality constraints, are a linearly independent set of vectors.
 
 A Lagrangian for this problem is
 \f[ \boxed{\displaystyle L(x,\lambda,\kappa) = f(x) + \sum_{j=1}^{l} \lambda_j g_j(x) + \sum_{k=1}^{m} \kappa_k h_k(x) . } \f]
 The standard Karush-Kuhn-Tucker conditions for optimality are obtained by setting the partial derivatives of \f$L\f$ to zero, with the inequality constraints \f$\lambda_j,g_j(x)\geq0\f$ allowing for the relaxation \f$\lambda_j=0 \vee g_j(x)=0\f$.
 
 The standard (Fritz) John conditions for optimality are
-\f[ \boxed{ \begin{gathered} \mu \nabla f(x) + \sum_{j=1}^{l} \lambda_j \nabla g_j(x) + \sum_{k=1}^{m} \kappa_k \nabla h_k(x) = 0; \\ \lambda_j g_j(x)=0; \\ h_k(x)=0; \\ \mu + \sum_j \lambda_j + \sum_k \kappa_k^2 = 0; \\ \lambda_j \geq 0; \ g_j(x) \geq 0 . \end{gathered} } \f]
+\f[ \boxed{ \begin{gathered} \mu \nabla f(x) + \sum_{j=1}^{l} \lambda_j \nabla g_j(x) + \sum_{k=1}^{m} \kappa_k \nabla h_k(x) = 0; \\ \lambda_j g_j(x)=0; \\ h_k(x)=0; \\ \mu + \sum_j \lambda_j + \sum_k \kappa_k^2 = 1; \\ \mu\geq 0; \ \lambda_j \geq 0; \ g_j(x) \geq 0 . \end{gathered} } \f]
+The John conditions are necessary conditions for any local optimum (assuming differentiability of \f$f,g_j,h_k\f$.
 
-\note Taking \f$f(x) = -cx\f$, \f$g(x)=-x\f$ and \f$h(x)=Ax-b\f$ yields the primal linear programming problem \f$\min cx \mid Ax=b,\ x\geq 0\f$.
+\note Taking \f$f(x) = -cx\f$, \f$g(x)=x\f$ and \f$h(x)=Ax-b\f$ yields the primal linear programming problem \f$\min cx \mid Ax=b,\ x\geq 0\f$.
 Taking \f$f(y)=yb\f$, \f$g(y) = yA-c\f$ yields the dual linear programming problem \f$\max yb \mid yA\leq c\f$.
 This means that the standard \em primal nonlinear programming problem with only affine inequality constraints more closely resembles the \em dual linear programming problem.
-Despite this, we shall use \c x for the primal variables of the standard primal nonlinear programming problem.
+Despite this, we shall use \f$x\f$ for the primal variables of the standard primal nonlinear programming problem.
 
 
 
@@ -78,48 +85,46 @@ Taking \f$\mu\to 0\f$, we obtain the standard Karush-Kuhn-Tucker conditions.
 
 Consider the problem
 \f[ \max f(x) \text{ s.t. } h(x) = 0 . \f]
-The central path is defined by
-\f[ \boxed{ \begin{gathered}  \nabla f(x) + \kappa \cdot \nabla h(x) = 0; \\ h(x) + \kappa \mu = 0. \end{gathered} } \f]
 The Karush-Kuhn-Tucker optimality conditions are
 \f[ \boxed{ \begin{gathered}  \nabla f(x) + \kappa \cdot \nabla h(x) = 0; \\ h(x) = 0. \end{gathered} } \f]
 
-The problem can be relaxed by adding the penalty function \f$-g(x)^2/2\mu\f$.
+The problem can be relaxed by adding the penalty function \f$-h(x)^2/2\nu\f$.
 Then the optimality conditions are
-\f[ \nabla f(x) - \frac{g(x)}{\mu} \nabla g(x) = 0\f]
-Taking Lagrange multiplier \f$\lambda = g(x)/\mu\f$, we obtain
-\f[ \boxed{ \begin{gathered} \nabla f(x) -  \lambda \nabla g(x) = 0; \\ g(x) - \lambda \mu = 0 \end{gathered} } \f]
-which relaxes to \f$g(x)=0\f$ as \f$\mu\to0\f$.
+\f[ \nabla f(x) - \frac{h(x)}{\nu} \nabla h(x) = 0\f]
+Taking Lagrange multiplier \f$\kappa = -h(x)/\nu\f$, we obtain
+\f[ \boxed{ \begin{gathered} \nabla f(x) +  \kappa \cdot \nabla h(x) = 0; \\ h(x) + \kappa \nu = 0 \end{gathered} } \f]
+which relaxes to \f$h(x)=0\f$ as \f$\nu\to0\f$.
 
 \subsection nonlinear_bounded_constraints Bounded constraints
 
 Consider the problem
 \f[ \max f(x) \text{ s.t. } |g(x)| \leq \delta \f]
 We can handle the problem in two ways; either by considering the smooth constraint \f$\delta^2-g(x)^2\geq0\f$ or the two constraints \f$\delta + g(x) \geq 0\f$ and \f$\delta - g(x)\geq 0\f$.
 
-In the former case, we obtain the conditions
-\f[ \begin{gathered} \nabla f(x) + 2 \lambda g(x) \nabla g(x) = 0; \\ \lambda (\delta^2 - g(x)^2) = \mu. \end{gathered} \f]
-Replacing the standard Lagrange multiplier \f$\lambda\f$ with \f$-2\lambda g(x)\f$, we obtain
-\f[ \begin{gathered} \nabla f(x) - \lambda \nabla g(x) = 0; \\ -\lambda (\delta^2 - g(x)^2) / 2g(x) = \mu. \end{gathered} \f]
+In the former case considering \f$\delta^2-g(x)^2\geq0\f$, we obtain the central path equations
+\f[ \begin{gathered} \nabla f(x) - 2 \lambda g(x) \nabla g(x) = 0; \\ \lambda (\delta^2 - g(x)^2) = \mu. \end{gathered} \f]
+Replacing the standard Lagrange multiplier \f$\lambda\f$ with \f$ - 2\lambda g(x)\f$, we obtain
+\f[ \begin{gathered} \nabla f(x) + \lambda \nabla g(x) = 0; \\ - \lambda (\delta^2 - g(x)^2) / 2g(x) = \mu. \end{gathered} \f]
 Rearranging the second formula gives
-\f[ \boxed{ \begin{gathered} \nabla f(x) - \lambda \nabla g(x) = 0; \\ \bigl(\delta^2-g(x)^2\bigr) \lambda + 2 g(x) \mu = 0 . \end{gathered} } \f]
+\f[ \boxed{ \begin{gathered} \nabla f(x) + \lambda \cdot \nabla g(x) = 0; \\ \bigl(\delta^2-g(x)^2\bigr) \lambda + 2 g(x) \mu = 0 . \end{gathered} } \f]
 Differentiating gives
-\f[ (\delta^2-g(x)^2) {\Delta\lambda} + (2 \mu - 2 g(x) \lambda) \nabla g(x) {\Delta x} = 0\f]
+\f[ (\delta^2-g(x)^2) {\Delta\lambda} - 2 ( g(x) \lambda - \mu ) \nabla g(x) {\Delta x} = 0\f]
 In order to solve for \f${\Delta x}\f$ in terms of \f${\Delta\lambda}\f$, we require
-\f$ \mu -  g(x) \lambda \neq 0\f$.
-Since the optimal \f$x^*,\lambda^*\f$ satisfy \f$(\delta^2-g(x^*)^2) \lambda^* + 2 g(x^*) \mu = 0\f$, we have \f$\lambda^* g(x^*) \geq 0\f$.
+\f$ g(x) \lambda - \mu \neq 0\f$.
+Since the optimal \f$x^*,\lambda^*\f$ satisfy \f$(\delta^2-g(x^*)^2) \lambda^* + 2 g(x^*) \mu = 0\f$, we have \f$\lambda^* g(x^*) \leq 0\f$.
 
-In the former case, we obtain the conditions
-\f[ \begin{gathered} \nabla f(x) - \lambda^+ \nabla g(x) + \lambda^- \nabla g(x) = 0; \\ \lambda^+ (\delta + g(x)) = \mu; \\ \lambda^- (\delta - g(x)) = \mu \end{gathered} \f]
-Set \f$\lambda = \lambda^+ - \lambda^-\f$. Then we have
-\f[ \lambda = \mu \biggl( \frac{1}{\delta+g(x)} - \frac{1}{\delta-g(x)}\biggr) = \frac{-2\mu g(x)}{\delta^2-g(x)^2} . \f]
+In the latter case considering \f$\delta+g(x)\geq0\f$ and \f$\delta - g(x) \geq0\f$,, we obtain the conditions
+\f[ \begin{gathered} \nabla f(x) + \lambda^+ \nabla g(x) - \lambda^- \nabla g(x) = 0; \\ \lambda^+ (\delta + g(x)) = \mu; \\ \lambda^- (\delta - g(x)) = \mu \end{gathered} \f]
+Set \f$\lambda = \lambda^+ - \lambda^-\f$. Then we have \f$\nabla{f}+\lambda \cdot \nabla{g}(x) = 0\f$ and
+\f[ \lambda = \mu \biggl( \frac{1}{\delta+g(x)} - \frac{1}{\delta-g(x)}\biggr) = \frac{-2 g(x) \mu}{\delta^2-g(x)^2} . \f]
 Rearranging again gives
 \f[ (\delta^2-g(x)^2) \lambda + 2 g(x) \mu = 0\f]
 Note that \f$\lambda^+ + \lambda^- = 2\delta\mu/(\delta^2-g(x)^2) \geq \bigl|\lambda^+-\lambda^-\bigr| \f$
 
 \subsection nonlinear_state_constraints State constraints
 
 Setting \f$g(x)=x\f$ and taking \f$\underline{x}\leq x\leq\overline{x}\f$, we obtain the state constraints
-\f[ \boxed{ \begin{gathered} \nabla f(x) - \lambda \nabla g(x) = 0; \\ (\overline{x}-x)(x-\underline{x})\lambda + (2x-\underline{x}-\overline{x}) \mu = 0 . \end{gathered} } \f]
+\f[ \boxed{ \begin{gathered} \nabla f(x) + \lambda \cdot \nabla g(x) = 0; \\ (\overline{x}-x)(x-\underline{x})\lambda + (2x-\underline{x}-\overline{x}) \mu = 0 . \end{gathered} } \f]
 If \f$\overline{x}=-\underline{x}=\delta\f$, this simplifies to
 \f[ (\delta^2-x^2) \lambda + 2 x \mu = 0 . \f]
 Differentiating yields

doc/topology.dox

@@ -359,10 +359,10 @@ However, taking these as a sub-base for all open sets yields a topology that may
 
 <i>Example</i> We now give an example to show that the lower and upper sets, when taken as closed sets, need not generate an order topology.
 <br/>
-Consider \f$\R\times\R\f$ with the standard product order \f$(x_1,y_1)\leq(x_2,y_2)\iff x_1\leq x_2\and y_1\leq y_2\f$.
-Suppose we take the sets \f$\{(x,y)\mid x\leq a\and y\leq b\}\f$ and \f$\{(x,y)\mid a\leq x\and b\leq y\}\f$ to generate the closed sets of the topology, so a sub-base of open sets is given by \f$\{(x,y)\mid x < a \or y < b\}\f$ and \f$\{(x,y)\mid a < x \or b < y\}\f$.
-Then any open set \f$U\f$ must contain a quadrants of the form \f$x< a \and y>b\f$ and \f$x>a \and y< b\f$, so any two open sets intersect, so the topology cannot yield a partially-ordered space.
-Note that in this case, the sets \f$\{ (x,y)  \mid \underline{a} < x < \overline{a} \and \underline{b} < y < \overline{b}\}\f$ are not open.
+Consider \f$\R\times\R\f$ with the standard product order \f$(x_1,y_1)\leq(x_2,y_2)\iff x_1\leq x_2\wedge y_1\leq y_2\f$.
+Suppose we take the sets \f$\{(x,y)\mid x\leq a\wedge y\leq b\}\f$ and \f$\{(x,y)\mid a\leq x\wedge b\leq y\}\f$ to generate the closed sets of the topology, so a sub-base of open sets is given by \f$\{(x,y)\mid x < a \vee y < b\}\f$ and \f$\{(x,y)\mid a < x \vee b < y\}\f$.
+Then any open set \f$U\f$ must contain a quadrants of the form \f$x< a \wedge y>b\f$ and \f$x>a \wedge y< b\f$, so any two open sets intersect, so the topology cannot yield a partially-ordered space.
+Note that in this case, the sets \f$\{ (x,y)  \mid \underline{a} < x < \overline{a} \wedge \underline{b} < y < \overline{b}\}\f$ are not open.
 
 \subsubsection partialorderedspace Partially ordered space