Updating week 5

Author: bjmorgan

CH40208/geometry_optimisation/Lennard_Jones_optimisation.ipynb

@@ -7,7 +7,7 @@
    "source": [
     "### Exercise: Geometry Optimisation of a Lennard-Jones Potential\n",
     "\n",
-    "The potential energy surface for a pair of &lqduo;Lennard–Jones” is given by\n",
+    "The potential energy surface for a pair of “Lennard–Jones” is given by\n",
     "\n",
     "$$U_\\mathrm{LJ} = \\frac{A}{r^{12}} - \\frac{B}{r^6}$$\n",
     "\n",
@@ -34,6 +34,14 @@
     "\n",
     "Investigate how the Newton-Raphson method performs with starting points of $r$ = 3.2 Å, $r$ = 4.4 Å, and $r$ = 6.0 Å."
    ]
+  },
+  {
+   "cell_type": "code",
+   "execution_count": null,
+   "id": "10ceb78f-ff2f-4f67-ae24-851d7765668d",
+   "metadata": {},
+   "outputs": [],
+   "source": []
   }
  ],
  "metadata": {
@@ -52,7 +60,7 @@
    "name": "python",
    "nbconvert_exporter": "python",
    "pygments_lexer": "ipython3",
-   "version": "3.11.10"
+   "version": "3.12.3"
   }
  },
  "nbformat": 4,

CH40208/geometry_optimisation/newton_raphson_method.ipynb

@@ -9,61 +9,62 @@
     "\n",
     "An alternative method for finding the minimum of a function is the Newton-Raphson Method (also called Newton's Method).\n",
     "\n",
-    "The central idea here is that if we zoom in on any function, then we can approximate the local shape of that function as quadratic, i.e.\n",
+    "The key insight is that **at a minimum, the first derivative equals zero**: $f^\\prime(x) = 0$. So finding a minimum is equivalent to finding where the derivative vanishes.\n",
+    "\n",
+    "The Newton-Raphson method uses the idea that if we zoom in on any function, we can approximate the local shape of that function as quadratic, i.e.\n",
     "\n",
     "$$f(x) \\approx ax^2 + bx + c$$\n",
     "\n",
     "over some small range of $x$.\n",
     "\n",
-    "If $f(x)$ has a minimum, then at that point\n",
+    "If $f(x)$ is approximately quadratic near our current position, then its first derivative is approximately **linear**:\n",
+    "\n",
+    "$$f^\\prime(x) \\approx 2ax + b$$\n",
     "\n",
-    "$$f^{\\prime\\prime}(x) = 0$$\n",
+    "If we know the coefficients $a$ and $b$, we can directly solve for where this linear approximation crosses zero (i.e., where $f^\\prime(x) = 0$):\n",
     "\n",
-    "and, if we know the coefficients $a$ and $b$, then we can directly calculate the value of $x$ that minimises our function.\n",
+    "$$2ax + b = 0$$\n",
     "\n",
-    "$$f^{\\prime\\prime}(x) = 2ax + b = 0$$\n",
+    "$$x = \\frac{-b}{2a}$$\n",
     "\n",
-    "$$x = \\frac{-b}{2a}$$\n"
+    "This gives us our predicted location of the minimum.\n"
    ]
   },
   {
    "cell_type": "markdown",
    "id": "641bab47-12b3-485d-afcb-4d25ad60030f",
    "metadata": {},
    "source": [
-    "To find the value of the coefficients $a$ and $b$, we use the property that any function can be expanded as a **Taylor series**\n",
-    "(see, e.g., Chapter 6 in [Foundations of Science Mathematics](https://bath.primo.exlibrisgroup.com/permalink/44BAT_INST/iac705/alma991004175199902761)).\n",
+    "To find the coefficients $a$ and $b$ of our local quadratic approximation, we use the property that any smooth function can be expanded as a **Taylor series** (see, e.g., Chapter 6 in [Foundations of Science Mathematics](https://bath.primo.exlibrisgroup.com/permalink/44BAT_INST/iac705/alma991004175199902761)).\n",
     "\n",
     "If we are at some point $x_0$, then we can approximate our function locally as a polynomial in $(x-x_0)$:\n",
     "\n",
     "$$f(x) = f(x_0) + f^\\prime(x_0)(x-x_0) + \\frac{1}{2}f^{\\prime\\prime}(x_0)(x-x_0)^2 + \\ldots$$\n",
     "\n",
-    "And, if we remain close to $x_0$, where the higher order terms in $(x-x_0)$ are small, we have the approximation that \n",
+    "If we remain close to $x_0$, where the higher-order terms in $(x-x_0)$ are small, we can truncate this series after the quadratic term:\n",
     "\n",
     "$$f(x) \\approx f(x_0) + f^\\prime(x_0)(x-x_0) + \\frac{1}{2}f^{\\prime\\prime}(x_0)(x-x_0)^2$$\n",
     "\n",
-    "i.e., we approximate $f(x)$ as quadratic.\n",
+    "This is our quadratic approximation. Differentiating this expression with respect to $x$ gives us the approximation for the first derivative:\n",
     "\n",
-    "Differentiating this approximate expression with respect to $x$gives:\n",
+    "$$f^\\prime(x) \\approx f^\\prime(x_0) + f^{\\prime\\prime}(x_0)(x - x_0)$$\n",
     "\n",
-    "$$f^\\prime(x) = f^\\prime + f^{\\prime\\prime}(x - x_0) = 0$$\n",
+    "At a minimum, $f^\\prime(x) = 0$, so we can set this expression to zero and solve for $x$:\n",
     "\n",
-    "and setting this to zero allows us to solve for $x$:\n",
+    "$$f^\\prime(x_0) + f^{\\prime\\prime}(x_0)(x - x_0) = 0$$\n",
     "\n",
-    "$$x = x_0 - \\frac{f^\\prime(x_0)}{f^{\\prime\\prime}(x_0)}$$"
-   ]
-  },
-  {
-   "cell_type": "markdown",
-   "id": "4c58aee1-7acf-4ac2-a237-a8f77cf57138",
-   "metadata": {},
-   "source": [
-    "In the context of finding the minimum of a potential energy surface, this looks like:\n",
+    "$$x = x_0 - \\frac{f^\\prime(x_0)}{f^{\\prime\\prime}(x_0)}$$\n",
+    "\n",
+    "This is the **Newton-Raphson equation**. Starting from our current position $x_0$, it predicts a new position $x$ that should be closer to the minimum.\n",
+    "\n",
+    "In the context of finding the minimum of a potential energy surface, we can rewrite this using our usual notation for energy and bond length:\n",
+    "\n",
+    "1. Starting at an initial guess $r_n$, calculate the first and second derivatives of $U(r_n)$ at that point.\n",
+    "2. Use the Newton-Raphson equation to predict a new position:\n",
+    "\n",
+    "$$r_{n+1} = r_n - \\frac{U^\\prime(r_n)}{U^{\\prime\\prime}(r_n)}$$\n",
     "\n",
-    "1. Starting at an initial guess, $r_n$, calculate the first and second derivatives of $U(r_n)$ at that point.\n",
-    "2. Use the Newton-Raphson equation $r_{n+1} = r_n - U^\\prime(r_n) / U^{\\prime\\prime}(r_n)$ to predict the position of the minimum.\n",
-    " \n",
-    "In reality, our potential energy surface is never exactly quadratic (except for the special case of a harmonic potential), so step 2. only gives us an approximate **guess** for the minimum position. We then repeat steps 1. and 2. as an iterative process, until we (hopefully) converge on the energy minimum."
+    "In reality, our potential energy surface is never exactly quadratic (except for the special case of a harmonic potential), so this equation only gives us an approximate guess for the minimum position. We therefore repeat steps 1 and 2 as an iterative process, recalculating the derivatives at each new position, until we (hopefully) converge on the energy minimum."
    ]
   },
   {
@@ -97,7 +98,7 @@
    "name": "python",
    "nbconvert_exporter": "python",
    "pygments_lexer": "ipython3",
-   "version": "3.11.10"
+   "version": "3.12.3"
   }
  },
  "nbformat": 4,