opalt.asciidoc

@@ -791,7 +791,7 @@ we obtain the recurrence relation
 
 [latexmath]
 ++++
-C_{n,k}(z) = - \frac{1}{4k(n+k)}   \frac{d^2 C_{n,k-1}} {dz^2}, k = 1,2, \dots
+C_{n,k}(z) = - \frac{1}{4k(n+k)} \frac{d^2 C_{n,k-1}} {dz^2}, k = 1,2, \ldots
 ++++
 
 The solution of the recursion relation becomes
@@ -859,7 +859,7 @@ profile on the central axis is the k-parameter Enge function
 ++++
 \begin{aligned}
         C_{n,0}(z) & = \frac{G_0}{1+exp[P(d(z))]}, \quad G_0 = \frac{B_0}{r_0^{n-1}} \\
-        P(d) & = C_0 + C_1 \left( \frac{d}{\lambda} \right) + C_2 \left( \frac{d}{\lambda} \right)^2 + \dots + C_{k-1} \left( \frac{d}{\lambda} \right)^{k-1} 
+        P(d) & = C_0 + C_1 \left( \frac{d}{\lambda} \right) + C_2 \left( \frac{d}{\lambda} \right)^2 + \ldots + C_{k-1} \left( \frac{d}{\lambda} \right)^{k-1}
     \end{aligned}
 ++++
 
@@ -913,17 +913,19 @@ latexmath:[ \nabla \cdot \mathbf{B} = 0 ] and
 latexmath:[ \nabla \times \mathbf{B} = 0 ] in the above coordinates
 yield
 
+[#eq-21]
 [latexmath]
 ++++
-\frac{\partial}{\partial x} \left( (1+x/ \rho) B_x \right) + \frac{\partial B_s}{\partial s} + (1+x/ \rho) \frac{\partial B_z}{\partial z} = 0 \label{eq:21}
+\frac{\partial}{\partial x} \left( (1+x/ \rho) B_x \right) + \frac{\partial B_s}{\partial s} + (1+x/ \rho) \frac{\partial B_z}{\partial z} = 0
 ++++
 
+[#eq-22-23-24]
 [latexmath]
 ++++
 \begin{aligned}
-            \frac{\partial B_z}{\partial s} & = (1+x/ \rho) \frac{\partial B_s}{\partial z} \label{eq:22}  \\
-            \frac{\partial B_x}{\partial z} & = \frac{\partial B_z}{\partial s} \label{eq:23}  \\
-             \frac{\partial B_x}{\partial s} & = \frac{\partial}{\partial x} \left( (1+x/ \rho) B_s \right) \label{eq:24}
+            \frac{\partial B_z}{\partial s} & = (1+x/ \rho) \frac{\partial B_s}{\partial z}  \\
+            \frac{\partial B_x}{\partial z} & = \frac{\partial B_z}{\partial s}   \\
+            \frac{\partial B_x}{\partial s} & = \frac{\partial}{\partial x} \left( (1+x/ \rho) B_s \right)
         \end{aligned}
 ++++
 
@@ -938,17 +940,20 @@ recursion relations. ([eq:23]) gives
 
 Equating the powers in latexmath:[x^i z^{2k}]
 
+[#eq-11]
 [latexmath]
 ++++
-a_{i,k} = \frac{i+1}{2k+1} b_{i+1,k} \label{eq:11}
+a_{i,k} = \frac{i+1}{2k+1} b_{i+1,k}
 ++++
 
 A similar result is obtained from ([eq:24])
+
+[#eq-12]
 [latexmath]
 ++++
 \begin{aligned}
         \sum_{i,k=0}^{\infty} \partial_s b_{i,k} x^i z^{2k} & = \left( 1+ \frac{x}{\rho} \right) \sum_{i,k=0}^{\infty} c_{i,k} (2k+1) x^i z^{2k} \\
-        \Rightarrow c_{i,k}  & + \frac{1}{\rho} c_{i-1,k} = \frac{1}{2k+1} \partial_s b_{i,k} \label{eq:12} 
+        \Rightarrow c_{i,k}  & + \frac{1}{\rho} c_{i-1,k} = \frac{1}{2k+1} \partial_s b_{i,k}
     \end{aligned}
 ++++
 
@@ -993,7 +998,7 @@ field latexmath:[B_z] can be measured at the mid-plane in the form
 
 [latexmath]
 ++++
-B_z(z=0) = B_{0,0} + B_{1,0}x + B_{2,0} x^2 + B_{3,0} x^3 + \dots
+B_z(z=0) = B_{0,0} + B_{1,0}x + B_{2,0} x^2 + B_{3,0} x^3 + \ldots
 ++++
 
 where latexmath:[B_{i,0}] are functions of latexmath:[s] and they
@@ -1045,7 +1050,7 @@ written as
 
 [latexmath]
 ++++
-\psi = z f_0(x,s) + \frac{z^3}{3!} f_1(x,s) + \frac{z^5}{5!} f_3(x,s) + \dots
+\psi = z f_0(x,s) + \frac{z^3}{3!} f_1(x,s) + \frac{z^5}{5!} f_3(x,s) + \ldots
 ++++
 
 The given transverse profile requires that
@@ -1113,11 +1118,12 @@ latexmath:[\rho = \rho (s)]. Laplace’s equation is
 
 The last step is again the substitution to get
 
+[#eq-40]
 [latexmath]
 ++++
 \begin{aligned}
         f_{n+1}(x,s) & = - \left[ \frac{\partial_x}{\rho h_s} + \partial_x^2 + \partial_z^2 + \frac{1}{h_s^2}\partial_s^2 + \frac{x}{\rho^2 h_s^3} (\partial_s \rho) \partial_s \right] f_n(x,s) \\
-        f_{n}(x,s) & = (-1)^n \left[ \frac{\partial_x}{\rho h_s} + \partial_x^2 + \partial_z^2 + \frac{\partial_s^2}{h_s^2} + \frac{x}{\rho^2 h_s^3} (\partial_s \rho) \partial_s \right]^n f_0(x,s) \label{eq:40}
+        f_{n}(x,s) & = (-1)^n \left[ \frac{\partial_x}{\rho h_s} + \partial_x^2 + \partial_z^2 + \frac{\partial_s^2}{h_s^2} + \frac{x}{\rho^2 h_s^3} (\partial_s \rho) \partial_s \right]^n f_0(x,s)
     \end{aligned}
 ++++