Added section on polynomial evaluation to the notebook

docs/.envrc

@@ -0,0 +1 @@
+eval "$(lorri direnv)"

docs/motion-control.ipynb

@@ -466,7 +466,7 @@
   },
   {
    "cell_type": "code",
-   "execution_count": null,
+   "execution_count": 20,
    "id": "dab40dd2-2b62-4445-9a7a-4b02ae87fef0",
    "metadata": {},
    "outputs": [],
@@ -488,7 +488,7 @@
     "    MP_FRAC = 8 # 8 fractional bits\n",
     "    MP_TICK_PER_SEC = 5000\n",
     "    MP_MM_PER_STEP = 0.000\n",
-    "    MP_AMAX = int(\n",
+    "    MP_AMAX = 1\n",
     "        "
    ]
   },
@@ -515,6 +515,53 @@
     "Thus, our fundamental distance unit is that such that $j_{max} = 6$"
    ]
   },
+  {
+   "cell_type": "markdown",
+   "id": "8cca2b8a-42a6-41c7-b0c6-eed4fd33342b",
+   "metadata": {},
+   "source": [
+    "## Polynomial evaluation\n",
+    "\n",
+    "As our position equation is a polynomial, we can evaluate it quickly at successive positions using the method of finite differences.\n",
+    "A semi-naiive evaluation of the formula would take 6 multiplications and 4 additions, whereas FD results in only the four additions.\n",
+    "\n",
+    "The differences can be calculated as follows:"
+   ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": 21,
+   "id": "f8c41a01-fe9c-495c-8644-600ce88ee2e8",
+   "metadata": {},
+   "outputs": [
+    {
+     "data": {
+      "text/latex": [
+       "$\\displaystyle \\left\\{ FD_{0} : \\frac{a_{0}}{2} + \\frac{j}{6} + v_{0}, \\  FD_{1} : a_{0} + j, \\  FD_{2} : j\\right\\}$"
+      ],
+      "text/plain": [
+       "⎧     a₀   j                          ⎫\n",
+       "⎨FD₀: ── + ─ + v₀, FD₁: a₀ + j, FD₂: j⎬\n",
+       "⎩     2    6                          ⎭"
+      ]
+     },
+     "execution_count": 21,
+     "metadata": {},
+     "output_type": "execute_result"
+    }
+   ],
+   "source": [
+    "eq1 = seg_d()['dp']\n",
+    "def fd(eqn, var=t):\n",
+    "    return (eqn.subs(var, var+1) - eqn).simplify()\n",
+    "    \n",
+    "fd1 = fd(eq1)\n",
+    "fd2 = fd(fd1)\n",
+    "fd3 = fd(fd2)\n",
+    "\n",
+    "{sympy.Symbol(f\"FD_{i}\"): e.subs(t,0).simplify() for i, e in  enumerate([fd1, fd2, fd3])}"
+   ]
+  },
   {
    "cell_type": "markdown",
    "id": "d8b4e4c3-f76f-4de7-a570-c28a462f0283",
@@ -550,7 +597,7 @@
    "name": "python",
    "nbconvert_exporter": "python",
    "pygments_lexer": "ipython3",
-   "version": "3.9.11"
+   "version": "3.9.10"
   }
  },
  "nbformat": 4,

docs/shell.nix

@@ -0,0 +1,22 @@
+{ pkgs ? import <nixpkgs> {} }:
+
+let 
+  pythonPackages = pkgs.python3.withPackages (p: with p; [
+    ipython
+    jupyterlab
+    scipy
+    sympy
+    bokeh
+    bkcharts
+  ]);
+in
+
+pkgs.mkShell {
+
+  buildInputs = [
+    pythonPackages
+
+    # keep this line if you use bash
+    pkgs.bashInteractive
+  ];
+}