equations revamped

javascript/latexContent.js

@@ -121,8 +121,8 @@ export const preamble = `\\documentclass[7x10]{../mitpress/mit}
  \\par%
  \\vspace{12pt}%
  \\noindent%
- \\begin{minipage}{1.0\\linewidth}\\vspace*{-0.85\\baselineskip}\\normalcodesize\\begin{eqnarray*}}{%
- \\end{eqnarray*}%
+ \\begin{minipage}{1.0\\linewidth}\\vspace*{-0.85\\baselineskip}\\normalcodesize\\[\\begin{array}{rcl}}{%
+ \\end{array}\\]%
  \\end{minipage}%
  \\par%
  \\vspace{12pt}%

xml/chapter1/section1/subsection6.xml

@@ -959,7 +959,9 @@ a === 4
  </SPLITINLINE>
  <LATEX>
  \[
- \frac{5+4+\left(2-\left(3-(6+\frac{4}{5})\right)\right)}{3 (6-2) (2-7)}
+ \begin{array}{l}
+ \dfrac{5+4+\left(2-\left(3-(6+\frac{4}{5})\right)\right)}{3 (6-2) (2-7)}
+ \end{array}
  \]
  </LATEX>
  <SOLUTION> 

xml/chapter1/section1/subsection7.xml

@@ -740,7 +740,11 @@ sqrt(3);
  approximation to the cube root of <LATEXINLINE>$x$</LATEXINLINE>, then a better approximation is
  given by the value
  <LATEX>
- \[ \frac{x/y^{2}+2y} {3} \]
+ \[
+ \begin{array}{lll}
+ \dfrac{x/y^{2}+2y} {3}
+ \end{array}
+ \]
  </LATEX>
  Use this formula to implement a cube-root
  <SPLITINLINE>

xml/chapter1/section2/subsection1.xml

@@ -10,14 +10,22 @@
  <INDEX>factorial</INDEX>
  factorial function, defined by
  <LATEX>
- \[ n!=n\cdot(n-1)\cdot(n-2)\cdots3\cdot2\cdot1 \]
+ \[
+ \begin{array}{lll}
+ n! &amp;=&amp; n\cdot(n-1)\cdot(n-2)\cdots3\cdot2\cdot1
+ \end{array}
+ \]
  </LATEX>
  There are many ways to compute factorials. One way is to make use of
  the observation that <LATEXINLINE>$n!$</LATEXINLINE> is equal to 
  <LATEXINLINE>$n$</LATEXINLINE> times <LATEXINLINE>$(n-1)!$</LATEXINLINE> for
  any positive integer<SPACE/><LATEXINLINE>$n$</LATEXINLINE>:
  <LATEX>
- \[ n!=n\cdot\left[(n-1)\cdot(n-2)\cdots3\cdot2\cdot1\right]=n\cdot(n-1)! \]
+ \[
+ \begin{array}{lll}
+ n! &amp;=&amp; n\cdot\left[(n-1)\cdot(n-2)\cdots3\cdot2\cdot1\right] \quad = \quad n \cdot(n-1)!
+ \end{array}
+ \]
  </LATEX>
  Thus, we can compute <LATEXINLINE>$n!$</LATEXINLINE> by computing 
  <LATEXINLINE>$(n-1)!$</LATEXINLINE> and multiplying the
@@ -97,10 +105,12 @@ factorial(5);
  the computation by saying that the counter and the product simultaneously
  change from one step to the next according to the rule
  <LATEX>
- \begin{align*}
- \textrm{product} &amp; \leftarrow \textrm{counter} \cdot \textrm{product}\\
- \textrm{counter} &amp; \leftarrow \textrm{counter} + 1
- \end{align*}
+ \[
+ \begin{array}{lll}
+ \textrm{product} &amp; \leftarrow &amp; \textrm{counter} \cdot \textrm{product}\\
+ \textrm{counter} &amp; \leftarrow &amp; \textrm{counter} + 1
+ \end{array}
+ \]
  </LATEX>
 and stipulating that <LATEXINLINE>$n!$</LATEXINLINE> is the value of the
  product when the counter exceeds <LATEXINLINE>$n$</LATEXINLINE>.

xml/chapter1/section2/subsection2.xml

@@ -180,8 +180,8 @@ fib(6);
  <LATEXINLINE>$\textrm{Fib}(n)$</LATEXINLINE> is the closest integer to
  <LATEXINLINE>$\phi^{n} /\sqrt{5}$</LATEXINLINE>, where
  <LATEX>
- \[\begin{array}{lll}
- \phi&amp;=&amp;(1+\sqrt{5})/2\approx 1.6180
+ \[\begin{array}{lllll}
+ \phi&amp;=&amp;(1+\sqrt{5})/2 &amp; \approx &amp; 1.6180
  \end{array}\]
  </LATEX>
  is the

xml/chapter1/section2/subsection3.xml

@@ -49,7 +49,11 @@
  <LATEXINLINE>$k_2$</LATEXINLINE> independent of
  <LATEXINLINE>$n$</LATEXINLINE> such that
  <LATEX>
- \[k_1f(n) \leq R(n) \leq k_2f(n)\]
+ \[
+ \begin{array}{lllll}
+ k_1\,f(n) &amp; \leq &amp; R(n) &amp; \leq &amp; k_2\,f(n)
+ \end{array}
+ \]
  </LATEX>
  for any sufficiently large value of <LATEXINLINE>$n$</LATEXINLINE>.
  (In other words, for large <LATEXINLINE>$n$</LATEXINLINE>, 
@@ -281,7 +285,11 @@
  if <LATEXINLINE>$x$</LATEXINLINE> is sufficiently small, and the
  trigonometric identity 
  <LATEX>
- \[\sin x=3\sin {\frac{x}{3}}-4\sin^3{\frac{x}{3}}\]
+ \[
+ \begin{array}{lll}
+ \sin x &amp;=&amp; 3\sin {\dfrac{x}{3}}-4\sin^3{\dfrac{x}{3}}
+ \end{array}
+ \]
  </LATEX>
  to reduce the size of the argument of <LATEXINLINE>$\sin$</LATEXINLINE>.
  (For purposes of this exercise an angle is considered <QUOTE>sufficiently

xml/chapter1/section3/subsection1.xml

@@ -179,7 +179,7 @@ function $name$(a, b) {
  notation,</QUOTE> for example
  <LATEX>
  \[\begin{array}{lll}
- \sum_{n=a}^{b}\ f(n)&amp;=&amp;f(a)+\cdots+f(b)
+ \displaystyle\sum_{n=a}^{b}\ f(n)&amp;=&amp;f(a)+\cdots+f(b)
  \end{array}\]
  </LATEX>
  to express this concept. The power of sigma notation is that it allows
@@ -477,10 +477,12 @@ function pi_sum(a, b) {
  be approximated numerically using the formula
  <LATEX>
  \[
- \int_{a}^{b}f =
- \left[ f\left( a+\frac{dx}{2} \right) + f \left(a+dx+\frac{dx}{2}
- \right) + f \left( a+2dx+\frac{dx}{2} \right)+\cdots
+ \begin{array}{lll}
+ \displaystyle\int_{a}^{b}f &amp; = &amp;
+ \left[\,f\!\left( a+\dfrac{dx}{2} \right)\,+\,f\!\left(a+dx+\dfrac{dx}{2}
+ \right)\,+\,f\!\left( a+2dx+\dfrac{dx}{2}\right)\,+\,\cdots
  \right] dx
+ \end{array}
  \]
  </LATEX>
  for small values of <LATEXINLINE>$dx$</LATEXINLINE>. We can express this
@@ -760,8 +762,12 @@ sum_cubes(1, 10);
 <INDEX>Wallis, John</INDEX>
 John Wallis.</FOOTNOTE>
 <LATEX>
- \[\frac{\pi}{4}\ =\ \frac{2\cdot 4\cdot 4\cdot 6\cdot 6\cdot 8\cdots}{3\cdot
- 3\cdot 5\cdot 5\cdot 7\cdot 7\cdots}\]
+ \[
+ \begin{array}{lll}
+ \dfrac{\pi}{4} &amp; = &amp; \dfrac{2 \cdot 4\cdot 4\cdot 6\cdot 6\cdot 8\cdots}{3\cdot
+ 3\cdot 5\cdot 5\cdot 7\cdot 7\cdots}
+ \end{array}
+ \]
 </LATEX>
  </LI>
  <LI>

xml/chapter1/section3/subsection3.xml

@@ -328,7 +328,11 @@ half_interval_method(x => x * x * x - 2 * x - 3, 1, 2);
  beginning with an initial guess and applying <LATEXINLINE>$f$</LATEXINLINE>
  repeatedly,
  <LATEX>
- \[ f(x), f(f(x)), f(f(f(x))), \ldots \]
+ \[
+ \begin{array}{l}
+ f(x), \ f(f(x)), \ f(f(f(x))), \ \ldots
+ \end{array}
+ \]
  </LATEX>
  until the value does not change very much. Using this idea, we can devise a
  <SPLITINLINE>
@@ -737,9 +741,11 @@ fixed_point_with_average_dampening(x => math_log(1000) / math_log(x), 2.0);
  <EM>continued fraction</EM> is an expression of the form
  <LATEX>
  \[
- f={\dfrac{N_1}{D_1+
+ \begin{array}{lll}
+ f &amp; = &amp; {\dfrac{N_1}{D_1+
  \dfrac{N_2}{D_2+
  \dfrac{N_3}{D_3+\cdots }}}}
+ \end{array}
  \]
  </LATEX>
  As an example, one can show that the infinite continued fraction
@@ -920,10 +926,12 @@ cont_frac(i => 1, i => 1, 20);
  J.H. Lambert:
  <LATEX>
  \[
- \tan x={\dfrac{x}{1-
+ \begin{array}{lll}
+ \tan x &amp; = &amp; {\dfrac{x}{1-
  \dfrac{x^2}{3-
  \dfrac{x^2}{5-
  \dfrac{x^2}{ \ddots }}}}}
+ \end{array}
  \]
  </LATEX>
  where <LATEXINLINE>$x$</LATEXINLINE> is in radians.

xml/chapter1/section3/subsection4.xml

@@ -304,7 +304,9 @@ cube_root(27);
  fixed point of the function <LATEXINLINE>$x\mapsto f(x)$</LATEXINLINE> where
  <LATEX>
  \[
- f(x)\ =\ x - \frac{g(x)}{Dg(x)}
+ \begin{array}{lll}
+ f(x) &amp; = &amp; x - \dfrac{g(x)}{Dg(x)}
+ \end{array}
  \]
  </LATEX>
  and <LATEXINLINE>$Dg(x)$</LATEXINLINE> is the derivative of
@@ -351,7 +353,11 @@ cube_root(27);
  function whose value at any number <LATEXINLINE>$x$</LATEXINLINE> is given
  (in the limit of small <LATEXINLINE>$dx$</LATEXINLINE>) by
  <LATEX>
- \[Dg(x)\ =\ \frac {g(x+dx) - g(x)}{dx} \]
+ \[
+ \begin{array}{lll}
+ Dg(x) &amp; = &amp; \dfrac {g(x+dx) - g(x)}{dx}
+ \end{array}
+ \]
  </LATEX>
  Thus, we can express the idea of derivative (taking
  <LATEXINLINE>$dx$</LATEXINLINE> to be, say, 0.00001) as the

xml/chapter2/section1/subsection1.xml

@@ -83,36 +83,18 @@
 
  <LATEX>
  \[
- \frac{n_{1}}{d_{1}}+\frac{n_{2}}{d_{2}}
- =\frac{n_{1}d_{2}+n_{2}d_{1}}{d_{1}d_{2}}
- \]
- </LATEX>
-
- <LATEX>
- \[
- \frac{n_{1}}{d_{1}}-\frac{n_{2}}{d_{2}}
- =\frac{n_{1}d_{2}-n_{2}d_{1}}{d_{1}d_{2}}
- \]
- </LATEX>
-
- <LATEX>
- \[
- \frac{n_{1}}{d_{1}}\cdot\frac{n_{2}}{d_{2}}
- =\frac{n_{1}n_{2}}{d_{1}d_{2}}
- \]
- </LATEX>
-
- <LATEX>
- \[
- \frac{n_{1}/d_{1}}{n_{2}/d_{2}}
- =\frac{n_{1}d_{2}}{d_{1}n_{2}}
- \]
- </LATEX>
-
- <LATEX>
- \[ 
- \frac{n_{1}}{d_{1}}
- =\frac{n_{2}}{d_{2}}\ \textrm{if and only if} \ n_{1}d_{2}=n_{2}d_{1}
+ \begin{array}{rll}
+ \dfrac{n_{1}}{d_{1}}+\dfrac{n_{2}}{d_{2}}
+ &amp;=&amp;\dfrac{n_{1}d_{2}+n_{2}d_{1}}{d_{1}d_{2}}\\[15pt]
+ \dfrac{n_{1}}{d_{1}}-\dfrac{n_{2}}{d_{2}}
+ &amp;=&amp;\dfrac{n_{1}d_{2}-n_{2}d_{1}}{d_{1}d_{2}}\\[15pt]
+ \dfrac{n_{1}}{d_{1}}\cdot\dfrac{n_{2}}{d_{2}}
+ &amp;=&amp;\dfrac{n_{1}n_{2}}{d_{1}d_{2}}\\[15pt]
+ \dfrac{n_{1}/d_{1}}{n_{2}/d_{2}}
+ &amp;=&amp;\dfrac{n_{1}d_{2}}{d_{1}n_{2}}\\[15pt]
+ \dfrac{n_{1}}{d_{1}}
+ &amp;=&amp;\dfrac{n_{2}}{d_{2}}\ \quad \textrm{if and only if}\ \ \ n_{1}d_{2}\ =\ n_{2}d_{1}
+ \end{array}
  \]
  </LATEX>
  </TEXT>

xml/chapter2/section1/subsection3.xml

@@ -75,18 +75,22 @@
  <JAVASCRIPT>
  <LATEX>
  \[
- \frac{\texttt{numer}(\texttt{x})}{\texttt{denom}(\texttt{x})}
- =
- \frac{\texttt{n}}{\texttt{d}}
+ \begin{array}{lll}
+ \dfrac{\texttt{numer}(\texttt{x})}{\texttt{denom}(\texttt{x})}
+ &amp;=&amp;
+ \dfrac{\texttt{n}}{\texttt{d}}
+ \end{array}
  \]
  </LATEX>
  </JAVASCRIPT>
  <SCHEME>
  <LATEX>
  \[
- \frac{(\texttt{numer}~\texttt{x})}{(\texttt{denom}~\texttt{x})}
- =
- \frac{\texttt{n}}{\texttt{d}}
+ \begin{array}{lll}
+ \dfrac{(\texttt{numer}~\texttt{x})}{(\texttt{denom}~\texttt{x})}
+ &amp;=&amp;
+ \dfrac{\texttt{n}}{\texttt{d}}
+ \end{array}
  \]
  </LATEX>
  </SCHEME>

xml/chapter2/section1/subsection4.xml

@@ -24,7 +24,9 @@
  using the formula
  <LATEX>
  \[
- R_{p}\ =\ \frac{1}{1/R_{1}+1/R_{2}}
+ \begin{array}{lll}
+ R_{p} &amp; = &amp; \dfrac{1}{1/R_{1}+1/R_{2}}
+ \end{array}
  \]
  </LATEX>
  Resistance values are usually known only up to some 
@@ -583,13 +585,13 @@ percent(my_interval);
  algebraically equivalent ways:
  <LATEX>
  \[
- \frac{R_{1}R_{2}}{R_{1}+R_{2}}
+ \dfrac{R_{1}R_{2}}{R_{1}+R_{2}}
  \]
  </LATEX>
  and
  <LATEX>
  \[
- \frac{1}{1/R_{1}+1/R_{2}}
+ \dfrac{1}{1/R_{1}+1/R_{2}}
  \]
  </LATEX>
  He has written the following two programs, each of which computes the

xml/chapter2/section3/subsection2.xml

@@ -83,14 +83,16 @@
  <INDEX>differentiation<SUBINDEX>rules for</SUBINDEX></INDEX>
  reduction rules:
  <LATEX>
- \begin{align*}
- \frac{dc}{dx} &amp; = 
- 0\text{ for $c$ a constant or a variable different from $x$} \\[3mm]
- \frac{dx}{dx} &amp; = 1 \\[3mm]
- \frac{d(u+v)}{dx} &amp; = \frac{du}{dx}+\frac{dv}{dx} \\[3mm]
- \frac{d(uv)}{dx} &amp; = u\left( \frac{dv}{dx}\right)+v\left(
- \frac{du}{dx}\right)
- \end{align*}
+ \[
+ \begin{array}{rll}
+ \dfrac{dc}{dx} &amp; = &amp; 
+ 0\quad \text{for $c$ a constant or a variable different from $x$} \\[12pt]
+ \dfrac{dx}{dx} &amp; = &amp; 1 \\[12pt]
+ \dfrac{d(u+v)}{dx} &amp; = &amp; \dfrac{du}{dx}+\dfrac{dv}{dx} \\[12pt]
+ \dfrac{d(uv)}{dx} &amp; = &amp; u\left( \dfrac{dv}{dx}\right)+v\left(
+ \dfrac{du}{dx}\right)
+ \end{array}
+ \]
  </LATEX>
  Observe that the latter two rules are recursive in nature. That is, to
  obtain the derivative of a sum we first find the derivatives of the terms
@@ -724,7 +726,11 @@ list("+", list("*", list("*", "x", "y"),
  The program produces answers that are correct; however, they are
  unsimplified. It is true that
  <LATEX>
- \[ \frac{d(xy)}{dx} = x\cdot 0+1\cdot y \]
+ \[
+ \begin{array}{lll}
+ \dfrac{d(xy)}{dx} &amp; = &amp; x\cdot 0+1\cdot y
+ \end{array}
+ \]
  </LATEX>
  but we would like the program to know that
  <LATEXINLINE>$x\cdot 0 = 0$</LATEXINLINE>,
@@ -992,7 +998,11 @@ list("+",
  <INDEX>differentiation<SUBINDEX>rules for</SUBINDEX></INDEX>
  For instance, implement the differentiation rule
  <LATEX>
- \[ \frac {d(u^{n})}{dx} = nu^{n-1}\left( \frac{du}{dx}\right) \]
+ \[
+ \begin{array}{lll}
+ \dfrac {d(u^{n})}{dx} &amp; = &amp; nu^{n-1}\left( \dfrac{du}{dx}\right)
+ \end{array}
+ \]
  </LATEX>
  by adding a new clause to the <SCHEMEINLINE>deriv</SCHEMEINLINE> program
  and defining appropriate