minor updates

Author: HumphreyYang

lectures/newton_method.md

@@ -28,10 +28,6 @@ kernelspec:
 :depth: 2
 ```
 
-```{seealso}
-A version of this lecture using [JAX](https://github.com/jax-ml/jax) is {doc}`available here <jax:newtons_method>`
-```
-
 ## Overview
 
 Many economic problems involve finding [fixed
@@ -69,10 +65,10 @@ problem where we know the solution and solve it using both successive
 approximation and Newton's method.
 
 Then we apply Newton's method to multidimensional settings to solve
-market for equilibria with multiple goods.
+for market equilibria with multiple goods.
 
 At the end of the lecture, we leverage the power of automatic
-differentiation in [`jax`](https://docs.jax.dev/en/latest/_autosummary/jax.grad.html) to solve a very high-dimensional equilibrium problem
+differentiation in [`jax`](https://docs.jax.dev/en/latest/_autosummary/jax.grad.html) to solve a very high-dimensional equilibrium problem.
 
 
 We use the following imports in this lecture
@@ -112,7 +108,7 @@ zero population growth, the law of motion for capital is
 Here
 
 - $k_t$ is capital stock per worker,
-- $A, \alpha>0$ are production parameters, $\alpha < 1$
+- $A, \alpha>0$ are production parameters with $\alpha < 1$
 - $s>0$ is a savings rate, and
 - $\delta \in(0,1)$ is a rate of depreciation
 
@@ -121,7 +117,7 @@ of $g$, the law of motion for capital.
 
 In other words, we seek a $k^* > 0$ such that $g(k^*) = k^*$.
 
-* such a $k^*$ is called a [steady state](https://en.wikipedia.org/wiki/Steady_state),
+* Such a $k^*$ is called a [steady state](https://en.wikipedia.org/wiki/Steady_state),
   since $k_t = k^*$ implies $k_{t+1} = k^*$.
 
 Using pencil and paper to solve $g(k) = k$, you will be able to confirm that
@@ -228,7 +224,7 @@ Here's a time series from a particular choice of $k_0$.
 
 ```{code-cell} ipython3
 def compute_iterates(k_0, f, params, n=25):
-    """Compute time series of length n generated by arbitrary function f."""
+    """Compute time series of length n generated by function f."""
     k = k_0
     k_iterates = []
     for t in range(n):
@@ -489,13 +485,13 @@ Let's start by computing the market equilibrium of a two-good problem.
 We consider a market for two related products, good 0 and good 1, with
 price vector $p = (p_0, p_1)$
 
-Supply of good $i$ at price $p$,
+Supply of good $i$ at price $p$ is
 
 $$ 
 q^s_i (p) = b_i \sqrt{p_i} 
 $$
 
-Demand of good $i$ at price $p$ is,
+Demand of good $i$ at price $p$ is
 
 $$ 
 q^d_i (p) = \exp(-(a_{i0} p_0 + a_{i1} p_1)) + c_i
@@ -505,7 +501,7 @@ Here $c_i$, $b_i$ and $a_{ij}$ are parameters.
 
 For example, the two goods might be computer components that are typically used together, in which case they are complements. Hence demand depends on the price of both components.
 
-The excess demand function is,
+The excess demand function is
 
 $$
 e_i(p) = q^d_i(p) - q^s_i(p), \quad i = 0, 1
@@ -818,7 +814,7 @@ The excess demand function is essentially the same, but now the matrix $A$ is $3
 dim = 3000
 
 # Create JAX random key
-key = jax.random.PRNGKey(123)
+key = jax.random.PRNGKey(0)
 
 # Create a random matrix A and normalize the columns to sum to one
 A = jax.random.uniform(key, (dim, dim))