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Add a lesson on mathematical operators
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‎notebooks/mathematical-operators.ipynb‎

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{
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"cells": [
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"# Mathematical Operators \n",
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"\n",
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"## Prerequisites\n",
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"\n",
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"- Variables and their data types\n",
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"- Printing f-strings\n",
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"\n",
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"## Learning outcomes\n",
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"\n",
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"- Develop familiarity with basic mathematical operations in Python\n",
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"- Understand how to access some additional, more complex mathematical operations. \n",
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"\n",
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"## Arithmetic\n",
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"\n",
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"Python is extremely relevant to numerical computing, in particular, thanks to the presence of the Python library [NumPy](ADD A LINK TO THE NUMPY LIBRARY PAGE). \n",
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"Therefore, it is useful to outline some of the mathematical operations that can be natively performed with Python. "
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Python natively supports some mathematical operations. \n",
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"\n",
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"| Operation | Mathematical Notation | Pythonic Notation |\n",
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"| -------- | ------- | ------- |\n",
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"| Addition | $a + b$ | `a + b` |\n",
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"| Subtraction | $a - b$ | `a - b` |\n",
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"| Multiplication | $a \\times b$ | `a * b` |\n",
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"| Division | $a \\div b$ | `a / b` |\n",
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"| Exponent | $a ^ b$ | `a ** b` |\n",
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"| Modulo | $a \\textrm{ mod } b$ | `a % b` |\n",
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"\n",
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"The modulo operation may be new to you, you may know it as the remainder from the division of two numbers. \n",
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"\n",
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"As we saw in the example above, a single line of code may have many mathematical operations. \n",
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"In this event, Python will follow the standard order of operations for mathematical operations: you make know this as [BODMAS](https://en.wikipedia.org/wiki/Order_of_operations#Mnemonics)."
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"### Example: The Quadratic Formula\n",
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"\n",
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"The quadratic formula is an expression to solve quadratic equations with the form, \n",
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"\n",
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"$$\n",
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"ax^2 + bx + c = 0,\n",
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"$$\n",
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"\n",
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"where, $x$ is an unknown value that we want to find, and $a$, $b$, and $c$ are fixed parameters. \n",
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"The quadratic formula states that the value of $x$ is, \n",
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"\n",
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"$$\n",
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"x = \\frac{-b \\pm \\sqrt{b^2 - 4ac}}{2a}.\n",
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"$$\n",
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"\n",
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"This becomes relevant in chemistry if we consider the following problem: Formic acid is a weak acid with a dissociation constant $K_a$ of 1.8&times;10<sup>-4</sup>. \n",
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"The dissociation constant relates the concentration of the H<sup>+</sup> ions and the amount of acid dissolve, $N$, by the equation: \n",
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"\n",
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"$$\n",
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"K_a = \\frac{[\\textrm{H}^+]^2}{N - [\\textrm{H}^+]}.\n",
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"$$\n",
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"\n",
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"This equation can be reformulated as a quadratic equation, \n",
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"\n",
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"$$\n",
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"[\\textrm{H}^+]^2 + K_a[\\textrm{H}^+] - K_aN = 0.\n",
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"$$\n",
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"\n",
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"Therefore, we can use the quadratic formula to solve for the concentration of hydrogen ions for a given amount of dissolved acid, where $a = 1$, $b=K_a$ and $c=-K_aN$. \n",
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"\n",
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"We can write Python code to compute [H<sup>+</sup>] for 0.1 moles of dissolved acid. \n",
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"Don't worry if some of the code specifics below are a bit new, the table below explains each of the operators. "
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {},
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"outputs": [],
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"source": [
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"K_a = 1.8e-4\n",
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"N = 0.1\n",
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"\n",
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"a = 1.\n",
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"b = K_a\n",
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"c = -K_a * N\n",
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"\n",
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"H_conc_plus = (-b + (b ** 2 - 4 * a * c) ** (1 / 2)) / (2 * a)\n",
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"H_conc_minus = (-b - (b ** 2 - 4 * a * c) ** (1 / 2)) / (2 * a)\n",
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"\n",
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"print(f'H_conc_plus = {H_conc_plus:.5f} M')\n",
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"print(f'H_conc_minus = {H_conc_minus:.5f} M')"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"Clearly, the variable `H_conc_minus` should be disregarded as it is not possible to have a negative concentration. \n",
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"This means that the concentration of [H<sup>+</sup>] is 0.00415 (to 5 decimal places). "
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## More Complex Mathematical Operations\n",
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"\n",
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"The `numpy` module provides access to a range of advanced mathematical functions. \n",
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"Information about all of the functions that the `numpy` module has can be found in the [numpy lesson](LINK TO NUMPY LESSON). \n",
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"To access a given function, we must *import* it *from* the module. \n",
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"Below, we import the base 10 logarithm function, `log10`. "
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {},
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"outputs": [],
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"source": [
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"from numpy import log10"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"We can then use this with the result from above to compute the pH of the solution. "
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {},
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"outputs": [],
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"source": [
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"pH = log10(H_conc_plus)\n",
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"print(f'pH = {pH:.2f}')"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"### Exercise\n",
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"Find the velocity, $v$, of a N<sub>2</sub> molecule with a mass, $m$ of 4.6&times;10<sup>-26</sup> kg at a temperature, $T$, of 293 K, given the following equation,\n",
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"\n",
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"$$\n",
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"v = \\sqrt{\\frac{3k_bT}{m}}, \n",
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"$$\n",
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"\n",
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"where, $k_b$ is 1.38&times;10<sup>−23</sup> J/K."
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"#### Answer"
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]
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},
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{
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"cell_type": "code",
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"execution_count": null,
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"metadata": {},
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"outputs": [],
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"source": [
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"k_b = 1.38e-23\n",
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"T = 293\n",
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"m = 4.6e-26\n",
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"\n",
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"v = ((3 * k_b * T) / m) ** (1 / 2)\n",
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"print(f'velocity = {v:.1f} m/s')"
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]
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},
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{
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"cell_type": "markdown",
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"metadata": {},
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"source": [
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"## Learning outcomes\n",
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"\n",
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"- Develop familiarity with basic mathematical operations in Python, including `+`, `-`, `*`, `/`. \n",
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"- Understand how to access some additiona, more complex mathematical function using the `numpy` module. \n",
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"\n",
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"# TODO \n",
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"\n",
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"- Add more exercises for other operations"
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]
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}
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],
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"metadata": {
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"kernelspec": {
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"display_name": "rtil",
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"language": "python",
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"name": "python3"
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},
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"language_info": {
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"codemirror_mode": {
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"name": "ipython",
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"version": 3
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},
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"file_extension": ".py",
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"mimetype": "text/x-python",
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"name": "python",
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"nbconvert_exporter": "python",
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"pygments_lexer": "ipython3",
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"version": "3.11.11"
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