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{
"cells": [
{
"cell_type": "markdown",
"id": "d66e2b4b-047b-4670-897d-7877d27a1597",
"metadata": {},
"source": [
"# Numerical Differentiation\n",
"\n",
"Finding a derivative of tabular data can be done using a finite\n",
"difference. Here we essentially pick two points on a function or a set\n",
"of data points and calculate the slope from there. Let’s imagine a\n",
"domain $x$ as a vector such that $\\vec{x}$ =\n",
"$\\pmatrix{x_0, x_1, x_2, ...}$. Then we can use the following methods to\n",
"approximate derivatives\n",
"\n",
"## Forward Difference\n",
"\n",
"Uses the point at which we want to find the derivative and a point\n",
"forwards on the line. $$\n",
"f'(x_i) = \\frac{f(x_{i+1})-f(x_i)}{x_{i+1}-x_i}\n",
"$$ *Hint: Consider what happens at the last point.*"
]
},
{
"cell_type": "code",
"execution_count": 2,
"id": "d2fe3ad1-8af0-499a-8954-41a88a49b834",
"metadata": {},
"outputs": [
{
"ename": "IndexError",
"evalue": "index 99 is out of bounds for axis 0 with size 99",
"output_type": "error",
"traceback": [
"\u001b[0;31m---------------------------------------------------------------------------\u001b[0m",
"\u001b[0;31mIndexError\u001b[0m Traceback (most recent call last)",
"Cell \u001b[0;32mIn[2], line 10\u001b[0m\n\u001b[1;32m 6\u001b[0m y \u001b[38;5;241m=\u001b[39m \u001b[38;5;241m34\u001b[39m \u001b[38;5;241m*\u001b[39m np\u001b[38;5;241m.\u001b[39mexp(\u001b[38;5;241m3\u001b[39m \u001b[38;5;241m*\u001b[39m x)\n\u001b[1;32m 8\u001b[0m dydx \u001b[38;5;241m=\u001b[39m (y[\u001b[38;5;241m1\u001b[39m:] \u001b[38;5;241m-\u001b[39m y[:\u001b[38;5;241m-\u001b[39m\u001b[38;5;241m1\u001b[39m]) \u001b[38;5;241m/\u001b[39m (x[\u001b[38;5;241m1\u001b[39m:] \u001b[38;5;241m-\u001b[39m x[:\u001b[38;5;241m-\u001b[39m\u001b[38;5;241m1\u001b[39m])\n\u001b[0;32m---> 10\u001b[0m dydx[\u001b[38;5;28mlen\u001b[39m(y)]\u001b[38;5;241m=\u001b[39mdydx[\u001b[38;5;28mlen\u001b[39m(dydx)]\n\u001b[1;32m 12\u001b[0m \u001b[38;5;66;03m# Plot the function\u001b[39;00m\n\u001b[1;32m 13\u001b[0m plt\u001b[38;5;241m.\u001b[39mplot(x, y, label\u001b[38;5;241m=\u001b[39m\u001b[38;5;124mr\u001b[39m\u001b[38;5;124m'\u001b[39m\u001b[38;5;124m$y(x)$\u001b[39m\u001b[38;5;124m'\u001b[39m)\n",
"\u001b[0;31mIndexError\u001b[0m: index 99 is out of bounds for axis 0 with size 99"
]
}
],
"source": [
"import numpy as np\n",
"import matplotlib.pyplot as plt\n",
"\n",
"# Initiate vectors\n",
"x = np.linspace(0, 2, 100) \n",
"y = 34 * np.exp(3 * x)\n",
"\n",
"dydx = (y[1:] - y[:-1]) / (x[1:] - x[:-1])\n",
"\n",
"dydx[len(y)]=dydx[len(dydx)]\n",
"\n",
"# Plot the function\n",
"plt.plot(x, y, label=r'$y(x)$')\n",
"plt.plot(x, dydx, label=r'$\\frac{dy}{dx}$')\n",
"plt.xlabel('x')\n",
"plt.ylabel('y')\n",
"plt.title('Plot of $34e^{3x}$')\n",
"plt.grid(True)\n",
"plt.legend()\n",
"plt.show()"
]
},
{
"cell_type": "markdown",
"id": "a54ad04f-05e5-4589-a633-d6dea2a5acfc",
"metadata": {},
"source": [
"## Backwards Difference\n",
"\n",
"Uses the point at which we want to find $$\n",
"f'(x_i) = \\frac{f(x_{i})-f(x_{i-1})}{x_i - x_{i-1}}\n",
"$$"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "415d0b3c-a70d-48ef-93ad-6e827127b222",
"metadata": {},
"outputs": [],
"source": [
"dydx = (y[1:] - y[:-1]) / (x[1:] - x[:-1])\n",
"\n",
"# Plot the function\n",
"plt.plot(x, y, label=r'$y(x)$')\n",
"plt.plot(x, dydx, label=b'$/frac{dy}{dx}$')\n",
"plt.xlabel('x')\n",
"plt.ylabel('y')\n",
"plt.title('Plot of $34e^{3x}$')\n",
"plt.grid(True)\n",
"plt.legend()\n",
"plt.show()"
]
},
{
"cell_type": "markdown",
"id": "21c87595-add2-4f7a-8d10-2da3b4284e02",
"metadata": {},
"source": [
"## Central Difference\n",
"\n",
"$$\n",
"f'(x_i) = \\frac{f(x_{i+1})-f(x_{i-1})}{x_{i+1}-x_{i-1}}\n",
"$$"
]
},
{
"cell_type": "code",
"execution_count": null,
"id": "6a67e63c-9516-415d-87d6-97d9d986ba33",
"metadata": {},
"outputs": [],
"source": []
}
],
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|
