Fix hboxes in vol3

Signed-off-by: zeramorphic <[email protected]>

ia/analysis/02_series.tex

@@ -21,7 +21,7 @@ \subsection{Definition}
 			      s_N            & = \sum_{j=1}^N (\lambda a_j + \mu b_j)            \\
 			                     & = \sum_{j=1}^N \lambda a_j + \sum_{j=1}^N \mu b_j \\
 			                     & = \lambda c_N + \mu d_N                           \\
-			      \therefore s_N & \to \lambda c + \mu d
+			      \therefore\ s_N & \to \lambda c + \mu d
 		      \end{align*}
 		\item For any \(n \geq N\), we have
 		      \begin{align*}

ia/analysis/07_differentiability.tex

@@ -242,7 +242,7 @@ \subsection{Inverse function theorem}
 	\begin{align*}
 		\frac{g(y + k) - g(y)}{k}                           & = \frac{x + h - x}{f(x+h) - y}           \\
 		                                                    & = \frac{h}{f(x+h) - f(x)}                \\
-		\therefore \lim_{k \to 0} \frac{g(y + k) - g(y)}{k} & = \lim_{h \to 0} \frac{h}{f(x+h) - f(x)} \\
+		\therefore\ \lim_{k \to 0} \frac{g(y + k) - g(y)}{k} & = \lim_{h \to 0} \frac{h}{f(x+h) - f(x)} \\
 		                                                    & = \frac{1}{f'(x)}
 	\end{align*}
 	as required.

ia/analysis/09_power_series.tex

@@ -205,7 +205,7 @@ \subsection{Infinite differentiability}
 	\begin{align*}
 		(z + h)^n - z^n - nhz^{n-1}                  & = \left( \sum_{r=0}^n \binom{n}{r} z^{n-r} h^r \right)  - z^n - nhz^{n-1}                                                               \\
 		                                             & = \sum_{r=2}^n \binom{n}{r} z^{n-r} h^r                                                                                                 \\
-		\therefore \abs{(z + h)^n - z^n - nhz^{n-1}} & = \abs{\sum_{r=2}^n \binom{n}{r} z^{n-r} h^r}                                                                                           \\
+		\therefore\ \abs{(z + h)^n - z^n - nhz^{n-1}} & = \abs{\sum_{r=2}^n \binom{n}{r} z^{n-r} h^r}                                                                                           \\
 		                                             & \leq \sum_{r=2}^n \abs{\binom{n}{r} z^{n-r} h^r}                                                                                        \\
 		                                             & = \sum_{r=2}^n \binom{n}{r} \abs{z}^{n-r} \abs{h}^r                                                                                     \\
 		                                             & \leq n(n-1) \underbrace{\left[ \sum_{r=2}^n \binom{n-2}{r-2} \abs{z}^{n-r} \abs{h}^{r-2} \right]}_{(\abs{z} + \abs{h})^{n-2}} \abs{h}^2 \\

ia/de/01_differentiation.tex

@@ -190,7 +190,7 @@ \subsection{Equation of a tangent}
 \begin{align*}
 	\eval{\frac{\dd{f}}{\dd{x}}}_{x=x_0} & = \frac{f(x_0 + h) - f(x_0)}{h}                          \\
 	                                     & = \frac{f(x_0 + h) - f(x_0)}{h} + \frac{o(h)}{h}         \\
-	\therefore f(x_0 + h)                & = f(x_0) + \eval{\frac{\dd{f}}{\dd{x}}}_{x=x_0} h + o(h)
+	\therefore\ f(x_0 + h)                & = f(x_0) + \eval{\frac{\dd{f}}{\dd{x}}}_{x=x_0} h + o(h)
 \end{align*}
 If we now take \(x=x_0+h;\,y=f(x);\,y_0=f(x_0)\), we have
 \[

ia/de/04_linear_ordinary_differential_equations.tex

@@ -58,19 +58,19 @@ \subsection{Eigenfunction forcing}
 \begin{align*}
 	\dot a                    & = -k_a a \implies a = a_0 e^{-k_a t} \\
 	\dot b                    & = k_a a - k_b b                      \\
-	\therefore \dot b + k_b b & = k_a a_0 e^{-k_a t}
+	\therefore\ \dot b + k_b b & = k_a a_0 e^{-k_a t}
 \end{align*}
 So we have a linear first order ODE with an eigenfunction as the forcing function.
 We can guess that the particular integral is of the form \(b_p = \lambda e^{-k_a t}\).
 \begin{align*}
 	-k_a\lambda e^{-k_a t} + k_b \lambda e^{-k_a t} & = k_a a_0 e^{-k_a t}        \\
 	\lambda(k_b-k_a)                                & = k_a a_0                   \\
-	\therefore \lambda                              & = \frac{k_a}{k_b - k_a} a_0
+	\therefore\ \lambda                              & = \frac{k_a}{k_b - k_a} a_0
 \end{align*}
 We can form the complementary function by solving:
 \begin{align*}
 	\dot{b_c} + k_b b_c & = 0           \\
-	\therefore b_c      & = Ae^{-k_b t}
+	\therefore\ b_c      & = Ae^{-k_b t}
 \end{align*}
 So combining everything, we have
 \[

ia/de/06_isoclines_and_solution_curves.tex

@@ -175,14 +175,14 @@ \subsection{Fixed points and perturbation analysis}
 	\item (\(y = 1\)) \begin{align*}
 		      \dot \varepsilon                           & = -2(1)t \varepsilon     \\
 		                                                 & = -2t\varepsilon         \\
-		      \therefore \varepsilon                     & = \varepsilon_0 e^{-t^2} \\
+		      \therefore\ \varepsilon                     & = \varepsilon_0 e^{-t^2} \\
 		      \lim_{t \to \infty} \varepsilon_0 e^{-t^2} & = 0
 	      \end{align*}
 	      so this point is stable.
 	\item (\(y = 1\)) \begin{align*}
 		      \dot \varepsilon                          & = -2(-1)t \varepsilon   \\
 		                                                & = 2t\varepsilon         \\
-		      \therefore \varepsilon                    & = \varepsilon_0 e^{t^2} \\
+		      \therefore\ \varepsilon                    & = \varepsilon_0 e^{t^2} \\
 		      \lim_{t \to \infty} \varepsilon_0 e^{t^2} & = \pm\infty
 	      \end{align*}
 	      so this point is unstable.
@@ -194,7 +194,7 @@ \subsection{Autonomous differential equations}
 Therefore, near a fixed point \(y=a\), we have:
 \begin{align*}
 	y                         & = a + \varepsilon(t)                                   \\
-	\therefore\dot\varepsilon & = \varepsilon \frac{\dd{f}}{\dd{y}}(a) = \varepsilon k
+	\therefore\\dot\varepsilon & = \varepsilon \frac{\dd{f}}{\dd{y}}(a) = \varepsilon k
 \end{align*}
 where \(k\) is the constant value \(\frac{\dd{f}}{\dd{y}}(a)\).
 Note that we can use normal derivatives in place of partial derivatives because \(f\) depends only on \(y\).

ia/de/07_phase_portraits.tex

@@ -145,7 +145,7 @@ \subsection{Phase portraits}
 Let \(\alpha y\) denote the birth rate, and \(\beta y\) be the death rate.
 Then, we can model this using a linear model by:
 \[
-	\frac{\dd{y}}{\dd{t}} = \alpha y - \beta y \quad \therefore y = y_0 e^{(\alpha - \beta) t}
+	\frac{\dd{y}}{\dd{t}} = \alpha y - \beta y \quad \therefore\ y = y_0 e^{(\alpha - \beta) t}
 \]
 If \(\alpha > \beta\) then we have exponential growth; if \(\alpha < \beta\) then we have exponential decay.
 This is an unrealistic model, so we can use a nonlinear model to increase accuracy.

ia/de/09_forced_second_order_odes.tex

@@ -262,7 +262,7 @@ \subsection{Resonance in undamped systems}
 	C(-\omega^2 + \omega_0^2) = 1
 \]
 \[
-	\therefore y_p = \frac{1}{\omega_0^2 - \omega^2}\sin\omega t
+	\therefore\ y_p = \frac{1}{\omega_0^2 - \omega^2}\sin\omega t
 \]
 As the system is linear in \(y\) and its derivatives, we can freely add some multiple of the complementary function and it will remain a solution.
 \[

ia/de/11_discrete_equations_and_the_method_of_frobenius.tex

@@ -168,7 +168,7 @@ \subsection{Fuch's theorem}
 		a_n            & = \frac{n-3}{n-1}a_{n-2}                                         \\
 		a_n            & = \frac{n-3}{n-1}\frac{n-5}{n-3}a_{n-4} = \frac{n-5}{n-1}a_{n-4} \\
 		a_n            & = \frac{n-5}{n-1}\frac{n-7}{n-5}a_{n-6} = \frac{n-7}{n-1}a_{n-6} \\
-		\therefore a_n & = \frac{-1}{n-1}a_0
+		\therefore\ a_n & = \frac{-1}{n-1}a_0
 	\end{align*}
 	Therefore
 	\[

ia/de/12_multivariate_calculus.tex

@@ -199,7 +199,7 @@ \subsection{Contours near stationary points}
 	f = \text{constant (since \(f\) is a contour)} \approx f(\vb x_s) = \frac{1}{2}\delta \vb x \cdot H \cdot \delta \vb x^\transpose
 \]
 \begin{equation}\label{contourhessian}
-	\therefore \lambda_1 \xi^2 + \lambda_2 \eta^2 \approx \text{constant}
+	\therefore\ \lambda_1 \xi^2 + \lambda_2 \eta^2 \approx \text{constant}
 \end{equation}
 Near a minimum or maximum point, \(\lambda_1\) and \(\lambda_2\) have the same sign.
 \eqref{contourhessian} implies that the contours of \(f\) are elliptical.

ia/dr/06_angular_motion_and_orbits.tex

@@ -68,7 +68,7 @@ \subsection{Polar coordinates in the plane}
 We can compute expressions for velocity and acceleration in terms of these new coordinates.
 \begin{align*}
 	\vb r                  & = r \vb e_r                                    \\
-	\therefore \dot{\vb r} & = \dot r \vb e_r + r \frac{\dd}{\dd{t}}\vb e_r \\
+	\therefore\ \dot{\vb r} & = \dot r \vb e_r + r \frac{\dd}{\dd{t}}\vb e_r \\
 	                       & = \dot r \vb e_r + r \dot\theta \vb e_\theta
 \end{align*}
 So \(\dot r\) is the radial component of the velocity, and \(r\dot\theta\) is the angular component of the velocity.

ia/dr/16_relativistic_physics.tex

@@ -23,7 +23,7 @@ \subsection{Proper time}
 	                        & = \frac{1}{c}\sqrt{c^2\dd{t}^2 - \abs{\dd{\vb x}}^2}     \\
 	                        & = \frac{1}{c}\sqrt{c^2\dd{t}^2 - \abs{\vb u}^2 \dd{t}^2} \\
 	                        & = \qty(1 - \frac{\vb u^2}{c^2})^{\frac{1}{2}}\dd{t}      \\
-	\therefore \dv{t}{\tau} & = \gamma_{\vb u}
+	\therefore\ \dv{t}{\tau} & = \gamma_{\vb u}
 \end{align*}
 where \(\gamma_{\vb u} = \qty(1 - \frac{\vb u^2}{c^2})^{\frac{1}{2}}\).
 Now, the total time observed by a particle moving along its world line is

ia/probability/08_combinations_of_random_variables.tex

@@ -156,7 +156,7 @@ \subsection{Properties of conditional expectation}
 We can see by the standard properties of the expectation that
 \begin{align*}
 	\expect{X \mid Y}                     & = \sum_y 1(Y = y) \expect{X \mid Y = y}                              \\
-	\therefore \expect{\expect{X \mid Y}} & = \sum_y \expect{1(Y = y)} \expect{X \mid Y = y}                     \\
+	\therefore\ \expect{\expect{X \mid Y}} & = \sum_y \expect{1(Y = y)} \expect{X \mid Y = y}                     \\
 	                                      & = \sum_y \prob{Y = y} \expect{X \mid Y = y}                          \\
 	                                      & = \sum_y \prob{Y = y} \frac{\expect{X \cdot 1(Y = y)}}{\prob{Y = y}} \\
 	                                      & = \sum_y \expect{X \cdot 1(Y = y)}                                   \\

ia/probability/11_branching_processes.tex

@@ -30,8 +30,8 @@ \subsection{Expectation of generation size}
 		                                     & = \expect{Y_{1,n} + \cdots + Y_{m,n} \mid X_n = m}   \\
 		                                     & = m \expect{Y_{1,n}}                                 \\
 		                                     & = m \expect{X_1}                                     \\
-		\therefore \expect{X_{n+1} \mid X_n} & = X_n \cdot \expect{X_1}                             \\
-		\therefore \expect{X_{n+1}}          & = \expect{X_n \cdot \expect{X_1}}                    \\
+		\therefore\ \expect{X_{n+1} \mid X_n} & = X_n \cdot \expect{X_1}                             \\
+		\therefore\ \expect{X_{n+1}}          & = \expect{X_n \cdot \expect{X_1}}                    \\
 		                                     & = \expect{X_n} \cdot \expect{X_1}
 	\end{align*}
 \end{proof}
@@ -51,7 +51,7 @@ \subsection{Probability generating functions}
 		\expect{z^{X_{n+1}} \mid X_n = m}                 & = \expect{z^{Y_{1, n} + \dots + Y_{m,n}} \mid X_n = m} \\
 		                                                  & = \expect{z^{X_1}}^m                                   \\
 		                                                  & = G(z)^m                                               \\
-		\therefore \expect{\expect{z^{X_{n+1}} \mid X_n}} & = \expect{G(z)^{X_n}}                                  \\
+		\therefore\ \expect{\expect{z^{X_{n+1}} \mid X_n}} & = \expect{G(z)^{X_n}}                                  \\
 		                                                  & = G_n(G(z))
 	\end{align*}
 \end{proof}

ia/probability/13_multivariate_density_functions.tex

@@ -241,7 +241,7 @@ \subsection{Order statistics of a random sample}
 	\cdot \prob{X_1 \leq x_1, \dots, X_n \leq x_n, X_1 < \dots < X_n}                                               \\
 	                                                & = n!
 	\int_{-\infty}^{x_1} \int_{u_1}^{x_2} \cdots \int_{u_{n-1}}^{x_n} f(u_1) \cdots f(u_n) \dd{u_1} \cdots \dd{u_n} \\
-	\therefore f_{Y_1, \dots, Y_n}(x_1, \dots, x_n) & = n!
+	\therefore\ f_{Y_1, \dots, Y_n}(x_1, \dots, x_n) & = n!
 	f(x_1) \cdots f(x_n)
 \end{align*}
 when \(x_1 < x_2 < \dots < x_n\), and the joint density is zero otherwise.

ia/probability/17_simulation_of_random_variables.tex

@@ -84,5 +84,5 @@ \subsection{Rejection sampling}
 	                               & = \frac{1}{\lambda} \int_B f(x) \dd{x}                                                                                         \\
 	\abs{A}                        & = \frac{1}{\lambda} \int_{[0,1]^{d-1}} f(x) \dd{x}                                                                             \\
 	                               & = \frac{1}{\lambda}                                                                                                            \\
-	\therefore \prob{X \in B}      & = \int_B f(x) \dd{x}
+	\therefore\ \prob{X \in B}      & = \int_B f(x) \dd{x}
 \end{align*}

ia/vc/08_maxwell_s_equations.tex

@@ -74,7 +74,7 @@ \subsection{Electromagnetic waves}
 	                 & = \frac{1}{c^2} \pdv[2]{\vb E}{t}
 \end{align*}
 \[
-	\therefore \laplacian \vb E - \frac{1}{c^2} \pdv[2]{\vb E}{t} = \vb 0
+	\therefore\ \laplacian \vb E - \frac{1}{c^2} \pdv[2]{\vb E}{t} = \vb 0
 \]
 which is the wave equation for waves travelling at speed \(c\).
 Hence, in a vacuum, the electric field propagates at speed \(c\).
@@ -87,7 +87,7 @@ \subsection{Electromagnetic waves}
 	                 & = \frac{1}{c^2} \pdv[2]{\vb B}{t}
 \end{align*}
 \[
-	\therefore \laplacian \vb B - \frac{1}{c^2} \pdv[2]{\vb B}{t} = \vb 0
+	\therefore\ \laplacian \vb B - \frac{1}{c^2} \pdv[2]{\vb B}{t} = \vb 0
 \]
 Hence the magnetic field also propagates at speed \(c\).
 So in general, we can say that electromagnetic waves always travel at speed \(c\) in a vacuum.

ia/vm/01_complex_numbers.tex

@@ -43,7 +43,7 @@ \subsection{Definition and basic theorems}
 \begin{align*}
 	\abs{z_2 - z_1}            & \geq \abs{z_2} - \abs{z_1}       \\
 	\text{or } \abs{z_2 - z_1} & \geq \abs{z_1} - \abs{z_2}       \\
-	\therefore \abs{z_2 - z_1} & \geq \abs{\abs{z_2} - \abs{z_1}} \\
+	\therefore\ \abs{z_2 - z_1} & \geq \abs{\abs{z_2} - \abs{z_1}} \\
 \end{align*}
 
 De Moivre's Theorem states that

ia/vm/05_vectors_in_real_euclidean_space.tex

@@ -103,7 +103,7 @@ \subsection{Inner product spaces}
 Then by the Cauchy--Schwarz inequality, we have
 \begin{align*}
 	\abs{\langle f, g \rangle}               & \leq \norm{f} \cdot \norm{g}                                    \\
-	\therefore \abs{\int_0^1 f(x)g(x)\dd{x}} & \leq \sqrt{\int_0^1 f(x)^2 \dd{x}}\sqrt{\int_0^1 g(x)^2 \dd{x}}
+	\therefore\ \abs{\int_0^1 f(x)g(x)\dd{x}} & \leq \sqrt{\int_0^1 f(x)^2 \dd{x}}\sqrt{\int_0^1 g(x)^2 \dd{x}}
 \end{align*}
 
 \begin{lemma}
@@ -114,7 +114,7 @@ \subsection{Inner product spaces}
 	\begin{align*}
 		\left\langle \vb v_j, \sum_i \alpha_i \vb v_i \right\rangle     & = 0 \\
 		\intertext{And because each vector that is not \(\vb v_j\) is orthogonal to it, those terms cancel, leaving}
-		\therefore \left\langle \vb v_j, \alpha_j \vb v_j \right\rangle & = 0 \\
+		\therefore\ \left\langle \vb v_j, \alpha_j \vb v_j \right\rangle & = 0 \\
 		\alpha_j \left\langle \vb v_j, \vb v_j \right\rangle            & = 0 \\
 		\alpha_j = 0
 	\end{align*}

ia/vm/08_linear_maps.tex

@@ -337,7 +337,7 @@ \subsection{Matrices}
 	      Therefore:
 	      \begin{align*}
 		      x_i'              & = x_i + \lambda b_j x_j a_i = S_{ij}x_j \\
-		      \therefore S_{ij} & = \delta_{ij} + \lambda a_i b_j
+		      \therefore\ S_{ij} & = \delta_{ij} + \lambda a_i b_j
 	      \end{align*}
 	      For example in \(\mathbb R^2\) with \(\vb a = \begin{pmatrix}
 		      1 \\ 0
@@ -353,7 +353,7 @@ \subsection{Matrices}
 	      \begin{align*}
 		      \vb x'            & = R\vb x = (\cos \theta)\vb x + (1 - \cos \theta)(\nhat \cdot \vb x)\nhat + (\sin \theta)(\nhat \times \vb x) \\
 		      x_i'              & = (\cos \theta)x_i + (1 - \cos \theta)n_j x_j n_i - (\sin \theta) \varepsilon_{ijk}x_j n_k = R_{ij} x_j       \\
-		      \therefore R_{ij} & = \delta_{ij}(\cos \theta) - (1 - \cos \theta)n_i n_j - (\sin \theta)\varepsilon_{ijk} n_k
+		      \therefore\ R_{ij} & = \delta_{ij}(\cos \theta) - (1 - \cos \theta)n_i n_j - (\sin \theta)\varepsilon_{ijk} n_k
 	      \end{align*}
 \end{enumerate}
 \end{example}

ia/vm/09_transpose_and_hermitian_conjugate.tex

@@ -160,7 +160,7 @@ \subsection{Orthogonal matrices}
 \]
 because
 \[
-	H_{ij} = \delta_{ij} - 2n_i n_j \therefore H = \begin{pmatrix}
+	H_{ij} = \delta_{ij} - 2n_i n_j \therefore\ H = \begin{pmatrix}
 		1 - 2 \sin^2 \frac{\theta}{2}              & 2\sin\frac{\theta}{2}\cos\frac{\theta}{2} \\
 		2\sin\frac{\theta}{2} \cos\frac{\theta}{2} & 1-2\cos^2\frac{\theta}{2}
 	\end{pmatrix}

ia/vm/20_symmetries_and_transformation_groups.tex

@@ -28,7 +28,7 @@ \subsection{2D Minkowski space}
 	\end{pmatrix}
 \]
 \[
-	\therefore \left( \begin{pmatrix}
+	\therefore\ \left( \begin{pmatrix}
 			x_0 \\ x_1
 		\end{pmatrix}, \begin{pmatrix}
 			y_0 \\ y_1

ib/antop/03_metric_spaces.tex

@@ -326,10 +326,15 @@ \subsection{Continuity}
 \end{proposition}
 \begin{proof}
 	Let \( \varepsilon > 0 \).
-	We want to find \( \delta > 0 \) such that \( \forall x \in M \), \( d(x,a) < \delta \) implies \( d''(g(f(x)), g(f(a))) < \varepsilon \).
-	Since \( g \) is continuous at \( f(a) \), there exists \( \eta > 0 \) such that \( \forall y \in M' \), \( d'(y,f(a)) < \eta \implies d''(g(y), g(f(a))) < \varepsilon \).
-	Now, since \( f \) is continuous at \( a \), for this \( \eta \) there exists \( \delta \) such that for all \( x \in M \), \( d(x,a) < \delta \implies d'(f(x) - f(a)) < \eta \).
-	Then \( d(x,a) < \delta \implies d''(g(f(x)), g(f(a))) < \varepsilon \) as required.
+	We want to find \( \delta > 0 \) such that \( \forall x \in M \),
+	\[ d(x,a) < \delta \implies d''(g(f(x)), g(f(a))) < \varepsilon \]
+	Since \( g \) is continuous at \( f(a) \), there exists \( \eta > 0 \) such that \( \forall y \in M' \),
+	\[ d'(y,f(a)) < \eta \implies d''(g(y), g(f(a))) < \varepsilon \]
+	Now, since \( f \) is continuous at \( a \), for this \( \eta \) there exists \( \delta \) such that for all \( x \in M \),
+	\[ d(x,a) < \delta \implies d'(f(x) - f(a)) < \eta \]
+	Then
+	\[ d(x,a) < \delta \implies d''(g(f(x)), g(f(a))) < \varepsilon \]
+	as required.
 \end{proof}
 
 \begin{example}
@@ -362,9 +367,9 @@ \subsection{Isometric, Lipschitz, and uniformly continuous functions}
 	Let \( f \colon M \to M' \) be a function between metric spaces.
 	Then, \( f \) is
 	\begin{enumerate}
-		\item \textit{isometric}, if \( \forall x,y \in M, d'(f(x),f(y)) = d(x,y) \)
-		\item \textit{Lipschitz}, or \( c \)-Lipschitz, if \( \exists c \in \mathbb R^+, \forall x,y \in M, d'(f(x),f(y)) \leq c\cdot d(x,y) \)
-		\item \textit{uniformly continuous}, if \( \forall \varepsilon > 0, \exists \delta > 0, \forall x,y \in M, d(x,y) < \delta \implies d'(f(x), f(y)) < \varepsilon \)
+		\item \textit{isometric}, if \[ \forall x,y \in M, d'(f(x),f(y)) = d(x,y) \]
+		\item \textit{Lipschitz}, or \( c \)-Lipschitz, if \[ \exists c \in \mathbb R^+, \forall x,y \in M, d'(f(x),f(y)) \leq c\cdot d(x,y) \]
+		\item \textit{uniformly continuous}, if \[ \forall \varepsilon > 0, \exists \delta > 0, \forall x,y \in M, d(x,y) < \delta \implies d'(f(x), f(y)) < \varepsilon \]
 	\end{enumerate}
 \end{definition}
 \begin{remark}