Recurrence relations

src/applications.xml

@@ -14,5 +14,6 @@
 
     <xi:include href="./least-squares.xml" />
     <xi:include href="./curve-fitting.xml" />
+    <xi:include href="./recurrence-relations.xml" />
 
 </chapter>

src/bookinfo.xml

@@ -12,6 +12,20 @@
     <!-- For title into HTML head, then tabs, bookmarks, etc -->
     <initialism>SCLA</initialism>
 
+
+    <!-- For LaTeX-based images; inserted both into tex preamble, -->
+    <!-- and into standalone preambles for svg production         -->
+
+    <latex-image-preamble>
+    \usepackage{tikz}
+    \usetikzlibrary{backgrounds}
+    \usetikzlibrary{arrows}
+    \usetikzlibrary{matrix}
+    \usepackage{forest}
+    \usepackage{wasysym}
+    </latex-image-preamble>
+
+
     <!-- These subsequent macro definitions allow for a consistent look for   -->
     <!-- mathematical structures.  It is relatively safe to play around with  -->
     <!-- variants if you desire other notation or looks.                      -->

src/cholesky.xml

@@ -56,7 +56,7 @@
             <mrow>&amp;=\adjoint{U_1}A_1U_1</mrow>
         </md>  The only obstacle to this computation is the square root of the entry in the top left corner of <m>A</m>, and the result should be positive.  If we apply the positive definite condition, with <m>\vect{x}=\vect{e}_1</m> (the first column of the identity matrix) then we have <me>a=\innerproduct{\vect{e}_1}{A\vect{e}_1}\gt 0</me>.</p>
 
-        <p>Can we repeat this decomposition on the <m>(n-1)\times(n-1)</m> matrix <m>B-\frac{1}{a}\vect{y}\adjoint{\vect{y}}</m>?  As before we just need a strictly positive entry in the upper left corner of this slightly smaller matrix.  Similar to before, employ the positive definite condition for <m>A</m> using <m>\vect{x}=\inverse{U_1}\vect{e}_2</m> and employ the version of <m>A</m> defining <m>A_1</m> (see Exercise<nbsp /><xref ref="exercise-compute-cholesky-positive-definite" />).  What is the result after <m>n</m> iterations?  <me>A=\adjoint{U_1}\adjoint{U_2}\dots\adjoint{U_n}IU_n\dotsU_2 U_1=\adjoint{U}U</me>  Here we have used the observation that a product of upper triangular matrices is again upper triangular, and you should notice the appearance of the positive diagonal entries.   So we have our desired factorization.</p>
+        <p>Can we repeat this decomposition on the <m>(n-1)\times(n-1)</m> matrix <m>B-\frac{1}{a}\vect{y}\adjoint{\vect{y}}</m>?  As before we just need a strictly positive entry in the upper left corner of this slightly smaller matrix.  Similar to before, employ the positive definite condition for <m>A</m> using <m>\vect{x}=\inverse{U_1}\vect{e}_2</m> and employ the version of <m>A</m> defining <m>A_1</m> (see Exercise<nbsp /><xref ref="exercise-compute-cholesky-positive-definite" />).  What is the result after <m>n</m> iterations?  <me>A=\adjoint{U_1}\adjoint{U_2}\dots\adjoint{U_n}IU_n\dots U_2 U_1=\adjoint{U}U</me>  Here we have used the observation that a product of upper triangular matrices is again upper triangular, and you should notice the appearance of the positive diagonal entries.   So we have our desired factorization.</p>
 
         <exercise xml:id="exercise-compute-cholesky-positive-definite">
             <statement>