edit unicode conversion (#280)

* edit latex missing \

* basic edits of unicode typos and missing space

* edit unicode and add commas

* add + for some functions

lectures/tools_and_techniques/geom_series.md

@@ -499,7 +499,7 @@ in a project with gross one period nominal rate of return accumulates
   project
 - thus, $1$ dollar invested at time $0$ pays interest
   $r$ dollars after one period, so we have $r+1 = R$
-  dollars at time$1$
+  dollars at time $1$
 - at time $1$ we reinvest $1+r =R$ dollars and receive interest
   of $r R$ dollars at time $2$ plus the *principal*
   $R$ dollars, so we receive $r R + R = (1+r)R = R^2$
@@ -551,7 +551,7 @@ The **present value** of the lease is
 
 $$
 \begin{aligned} 
-p_0  & = x_0 + x_1/R + x_2/(R^2) + \ddots \\
+p_0  & = x_0 + x_1/R + x_2/(R^2) + \cdots \\
 & = x_0 (1 + G R^{-1} + G^2 R^{-2} + \cdots ) \\
 & = x_0 \frac{1}{1 - G R^{-1}} 
 \end{aligned}
@@ -704,7 +704,7 @@ plot!(plt, T, y_3, label = L"$T$-period Lease First-order Approx. adj.")
 
 Evidently our approximations perform well for small values of $T$.
 
-However, holding $g$ and r fixed, our approximations deteriorate as $T$ increases.
+However, holding $g$ and $r$ fixed, our approximations deteriorate as $T$ increases.
 
 Next we compare the infinite and finite duration lease present values
 over different lease lengths $T$.

lectures/tools_and_techniques/iterative_methods_sparsity.md

@@ -457,7 +457,7 @@ equation through methods such as value-function iteration.
 The condition we will examine here is called [**diagonal dominance**](https://en.wikipedia.org/wiki/Diagonally_dominant_matrix).
 
 $$
-|A_{ii}| \geq \sum_{j\neq i} |A_{ij}| \quad\text{for all } i = 1\ldots N
+|A_{ii}| \geq \sum_{j\neq i} |A_{ij}| \quad\text{for all } i = 1, \ldots N
 $$
 
 That is, in every row, the diagonal element is weakly greater in absolute value than the sum of all of the other elements in the row.  In cases
@@ -466,7 +466,7 @@ where it is strictly greater, we say that the matrix is strictly diagonally domi
 With our example, given that $Q$ is the infinitesimal generator of a Markov chain, we know that each row sums to 0, and hence
 it is weakly diagonally dominant.
 
-However, notice that when $\rho > 0$, and since the diagonal of $Q$ is negative,  $A = rho I - Q$ makes the matrix strictly diagonally dominant.
+However, notice that when $\rho > 0$, and since the diagonal of $Q$ is negative,  $A = \rho I - Q$ makes the matrix strictly diagonally dominant.
 
 ### Jacobi Iteration
 
@@ -1187,7 +1187,7 @@ $$
 If $Q$ is a matrix, we could just take its transpose to find the adoint.  However, with matrix-free methods, we need to implement the
 adjoint-vector product directly.
 
-The logic for the adjoint is that for a given $n = (n_1,\ldots, n_m, \ldots n_M)$, the $Q^T$ product for that row has terms enter when
+The logic for the adjoint is that for a given $n = (n_1,\ldots, n_m, \ldots, n_M)$, the $Q^T$ product for that row has terms enter when
 
 1. $1 < n_m \leq N$, entering into the identical $n$ except with one less customer in the $m$ position
 1. $1 \leq n_m < N$, entering into the identical $n$ except with one more customer in the $m$ position

lectures/tools_and_techniques/linear_algebra.md

@@ -402,7 +402,7 @@ $m > n$ vectors in $\mathbb R ^n$ must be linearly dependent.
 The following statements are equivalent to linear independence of $A := \{a_1, \ldots, a_k\} \subset \mathbb R ^n$.
 
 1. No vector in $A$ can be formed as a linear combination of the other elements.
-1. If $\beta_1 a_1 + \cdots \beta_k a_k = 0$ for scalars $\beta_1, \ldots, \beta_k$, then $\beta_1 = \cdots = \beta_k = 0$.
+1. If $\beta_1 a_1 + \cdots + \beta_k a_k = 0$ for scalars $\beta_1, \ldots, \beta_k$, then $\beta_1 = \cdots = \beta_k = 0$.
 
 (The zero in the first expression is the origin of $\mathbb R ^n$)
 
@@ -415,13 +415,13 @@ In other words, if $A := \{a_1, \ldots, a_k\} \subset \mathbb R ^n$ is
 linearly independent and
 
 $$
-y = \beta_1 a_1 + \cdots \beta_k a_k
+y = \beta_1 a_1 + \cdots + \beta_k a_k
 $$
 
 then no other coefficient sequence $\gamma_1, \ldots, \gamma_k$ will produce
 the same vector $y$.
 
-Indeed, if we also have $y = \gamma_1 a_1 + \cdots \gamma_k a_k$,
+Indeed, if we also have $y = \gamma_1 a_1 + \cdots + \gamma_k a_k$,
 then
 
 $$

lectures/tools_and_techniques/numerical_linear_algebra.md

@@ -626,7 +626,7 @@ Q = Tridiagonal(fill(alpha, N - 1), [-alpha; fill(-2alpha, N - 2); -alpha],
 
 Here we can use `Tridiagonal` to exploit the structure of the problem.
 
-Consider a simple payoff vector $r$ associated with each state, and a discount rate $rho$.  Then we can solve for
+Consider a simple payoff vector $r$ associated with each state, and a discount rate $\rho$.  Then we can solve for
 the expected present discounted value in a way similar to the discrete-time case.
 
 $$
@@ -655,23 +655,23 @@ linear problem.
 v = A \ r
 ```
 
-The $Q$ is also used to calculate the evolution of the Markov chain, in direct analogy to the $psi_{t+k} = psi_t P^k$ evolution with the transition matrix $P$ of the discrete case.
+The $Q$ is also used to calculate the evolution of the Markov chain, in direct analogy to the $\psi_{t+k} = \psi_t P^k$ evolution with the transition matrix $P$ of the discrete case.
 
 In the continuous case, this becomes the system of linear differential equations
 
 $$
-\dot{psi}(t) = Q(t)^T psi(t)
+\dot{\psi}(t) = Q(t)^T \psi(t)
 $$
 
 given the initial condition $\psi(0)$ and where the $Q(t)$ intensity matrix is allowed to vary with time.  In the simplest case of a constant $Q$ matrix, this is a simple constant-coefficient system of linear ODEs with coefficients $Q^T$.
 
-If a stationary equilibrium exists, note that $\dot{psi}(t) = 0$, and the stationary solution $psi^{*}$ needs to satisfy
+If a stationary equilibrium exists, note that $\dot{\psi}(t) = 0$, and the stationary solution $\psi^{*}$ needs to satisfy
 
 $$
-0 = Q^T psi^{*}
+0 = Q^T \psi^{*}
 $$
 
-Notice that this is of the form $0 psi^{*} = Q^T psi^{*}$ and hence is equivalent to finding the eigenvector associated with the $\lambda = 0$ eigenvalue of $Q^T$.
+Notice that this is of the form $0 \psi^{*} = Q^T \psi^{*}$ and hence is equivalent to finding the eigenvector associated with the $\lambda = 0$ eigenvalue of $Q^T$.
 
 With our example, we can calculate all of the eigenvalues and eigenvectors