tutorials/tutorial3_orthogonalization.ipynb

@@ -94,7 +94,7 @@
     "q_1 = \\frac{a_1}{\\|a_1\\|}\n",
     "$$\n",
     "$$\n",
-    "q_k = \\frac{a_k - \\sum_{j=1}^{k-1} (q_j \\cdot a_k) q_j}{\\left\\|a_k - \\sum_{j=1}^{k-1} (q_j \\cdot a_k) q_j\\right\\|}\n",
+    "q_k = \\frac{a_k - \\sum_{j=1}^{k-1} (q_j \\cdot a_k) q_j}{\\left\\|a_k - \\sum_{j=1}^{k-1} (q_j \\cdot a_k) q_j\\right\\|}, \\qquad k = 2, \\ldots, n\n",
     "$$\n",
     "\n",
     "Matrix form:\n",
@@ -236,17 +236,17 @@
     "\n",
     "We want to solve\n",
     "\n",
-    "$$ \\min_x \\|Ax - b\\|_2. $$\n",
+    "$$ \\min_x \\Vert Ax - b \\Vert_2^2. $$\n",
     "\n",
     "If $A = QR$, then\n",
     "\n",
-    "$$ \\min_x \\|Ax - b\\|_2 = \\min_x \\|QRx - b\\|_2. $$\n",
+    "$$ \\min_x \\Vert Ax - b \\Vert_2^2 = \\min_x \\Vert QRx - b \\Vert_2^2. $$\n",
     "\n",
-    "Since $Q$ has orthonormal columns:\n",
+    "Since $Q$ has orthonormal columns, and the normal equations boils down to\n",
     "\n",
-    "$$ R x = Q^T b. $$\n",
+    "$$ R x = Q^T b, $$\n",
     "\n",
-    "So we can solve using back-substitution.\n"
+    "we can therefore solve for $x$ by using back-substitution.\n"
    ]
   },
   {

tutorials/tutorial4_root_finding.ipynb

@@ -55,7 +55,7 @@
    "source": [
     "## 2. Bisection Method\n",
     "\n",
-    "**Assumption (Intermediate Value Theorem):** If f is continuous on ([a,b]) and (f(a),f(b) < 0),\n",
+    "**Assumption (Intermediate Value Theorem):** If f is continuous on $[a,b]$ and $f(a),f(b) < 0$,\n",
     "then there exists $x^\\star$ in (a,b) with $f(x^\\star)=0$.\n",
     "\n",
     "- Assumes $f$ is continuous on $[a,b]$ with $f(a)f(b)<0$.\n",