whoops, fix 4b

psets/pset4.ipynb

@@ -118,7 +118,7 @@
     "\n",
     "**(a)** If $sz = [x_1, x_2/\\kappa]$ exactly, then on the next step we would have the new $x \\leftarrow x - sz = [0, (1-\\frac{1}{\\kappa}) x_2)]$.   However, explain why a more careful calculation shows that the new $x - sz \\approx [O(1/\\kappa^2), x_2-\\frac{1}{\\kappa} x_2 + O(1/\\kappa^3)]$, i.e. the first component is proportional to $1/\\kappa^2$ to leading order in $1/\\kappa$.\n",
     "\n",
-    "**(b)** If you start with an $x = [\\#/\\kappa^2, x_2]$, i.e. where the first component is proportional to $1/\\kappa^2$ and $\\#$ is some number of the same order of magnitude as $x_2$, show that after one steepest-descent step (for $\\kappa \\gg 1$) the $x_1$ component is *still* roughly order $1/\\kappa^2$ but of the opposite sign, and $x_2$ again subtracts a term roughly proportional to $x_2/\\kappa$.\n",
+    "**(b)** If you start with an $x = [\\#/\\kappa, x_2]$, i.e. where the first component is proportional to $1/\\kappa$ and $\\#$ is some number of the same order of magnitude as $x_2$, show that after one steepest-descent step (for $\\kappa \\gg 1$) the $x_1$ component is *still* roughly order $1/\\kappa$ but of the opposite sign, and $x_2$ again subtracts a term roughly proportional to $x_2/\\kappa$.\n",
     "\n",
     "**(c)** Implement this steepest-descent process numerically for $\\kappa = 100$ and a starting point $x_1 = 0.01234567$ and $x_2 = 0.8910$.  Plot $100x_1$ and $x_2$ for 100 iterations of steepest descent.  A more careful analysis would show a convergence proportional to $\\left( \\frac{\\kappa - 1}{\\kappa + 1} \\right)^k$, where $k$ is the iteration number, following equation (4) in the Strang book — include this function for comparison on your plot."
    ]