diff --git a/examples/Demonstration-Physics.ipynb b/examples/Demonstration-Physics.ipynb
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+++ b/examples/Demonstration-Physics.ipynb
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+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ ""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# EXPERIENCE DE LA BILLE SUR GLISSIÈRE"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": 1,
+ "metadata": {},
+ "outputs": [
+ {
+ "data": {
+ "text/html": [
+ "\n",
+ " \n",
+ "
\n",
+ "
Loading BokehJS ...\n",
+ "
\\n\"+\n",
+ " \"
\\n\"+\n",
+ " \"BokehJS does not appear to have successfully loaded. If loading BokehJS from CDN, this \\n\"+\n",
+ " \"may be due to a slow or bad network connection. Possible fixes:\\n\"+\n",
+ " \"
\\n\"+\n",
+ " \"
\\n\"+\n",
+ " \"- re-rerun `output_notebook()` to attempt to load from CDN again, or
\\n\"+\n",
+ " \"- use INLINE resources instead, as so:
\\n\"+\n",
+ " \"
\\n\"+\n",
+ " \"
\\n\"+\n",
+ " \"from bokeh.resources import INLINE\\n\"+\n",
+ " \"output_notebook(resources=INLINE)\\n\"+\n",
+ " \"\\n\"+\n",
+ " \"
\\n\"+\n \"
\\n\"+\n \"BokehJS does not appear to have successfully loaded. If loading BokehJS from CDN, this \\n\"+\n \"may be due to a slow or bad network connection. Possible fixes:\\n\"+\n \"
\\n\"+\n \"
\\n\"+\n \"- re-rerun `output_notebook()` to attempt to load from CDN again, or
\\n\"+\n \"- use INLINE resources instead, as so:
\\n\"+\n \"
\\n\"+\n \"
\\n\"+\n \"from bokeh.resources import INLINE\\n\"+\n \"output_notebook(resources=INLINE)\\n\"+\n \"\\n\"+\n \"
![](\"resources/doubletSketch1.png\")
"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "The stream-function associated to the source-sink pair, evaluated at point $\\text{P}\\left(x,y\\right)$, is\n",
+ "\n",
+ "$$\\psi\\left(x,y\\right) = \\frac{\\sigma}{2\\pi}\\left(\\theta_1-\\theta_2\\right) = -\\frac{\\sigma}{2\\pi}\\Delta\\theta$$\n",
+ "\n",
+ "Let the distance $l$ between the two singularities approach zero while the strength magnitude is increasing so that the product $\\sigma l$ remains constant. In the limit, this flow pattern is a *doublet* and we define its strength by $\\kappa = \\sigma l$.\n",
+ "\n",
+ "The stream-function of a doublet, evaluated at point $\\text{P}\\left(x,y\\right)$, is given by\n",
+ "\n",
+ "$$\\psi\\left(x,y\\right) = \\lim \\limits_{l \\to 0} \\left(-\\frac{\\sigma}{2\\pi}d\\theta\\right) \\quad \\text{and} \\quad \\sigma l = \\text{constant}$$\n",
+ "\n",
+ "![](\"resources/doubletSketch2.png\")
"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Considering the case where $d\\theta$ is infinitesimal, we deduce from the figure above that\n",
+ "\n",
+ "$$a = l\\sin\\theta$$\n",
+ "\n",
+ "$$b = r-l\\cos\\theta$$\n",
+ "\n",
+ "$$d\\theta = \\frac{a}{b} = \\frac{l\\sin\\theta}{r-l\\cos\\theta}$$\n",
+ "\n",
+ "so the stream function becomes\n",
+ "\n",
+ "$$\\psi\\left(r,\\theta\\right) = \\lim \\limits_{l \\to 0} \\left(-\\frac{\\sigma l}{2\\pi}\\frac{\\sin\\theta}{r-l\\cos\\theta}\\right) \\quad \\text{and} \\quad \\sigma l = \\text{constant}$$\n",
+ "\n",
+ "i.e.\n",
+ "\n",
+ "$$\\psi\\left(r,\\theta\\right) = -\\frac{\\kappa}{2\\pi}\\frac{\\sin\\theta}{r}$$\n",
+ "\n",
+ "In Cartesian coordinates, a doublet located at the origin has the stream function\n",
+ "\n",
+ "$$\\psi\\left(x,y\\right) = -\\frac{\\kappa}{2\\pi}\\frac{y}{x^2+y^2}$$\n",
+ "\n",
+ "from which we can derive the velocity components\n",
+ "\n",
+ "$$u\\left(x,y\\right) = \\frac{\\partial\\psi}{\\partial y} = -\\frac{\\kappa}{2\\pi}\\frac{x^2-y^2}{\\left(x^2+y^2\\right)^2}$$\n",
+ "\n",
+ "$$v\\left(x,y\\right) = -\\frac{\\partial\\psi}{\\partial x} = -\\frac{\\kappa}{2\\pi}\\frac{2xy}{\\left(x^2+y^2\\right)^2}$$\n",
+ "\n",
+ "Now we have done the math, it is time to code and visualize what the streamlines look like. We start by creating a mesh grid."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {},
+ "outputs": [],
+ "source": [
+ "N = 50 # Number of points in each direction\n",
+ "x_start, x_end = -2.0, 2.0 # x-direction boundaries\n",
+ "y_start, y_end = -1.0, 1.0 # y-direction boundaries\n",
+ "x = numpy.linspace(x_start, x_end, N) # creates a 1D-array for x\n",
+ "y = numpy.linspace(y_start, y_end, N) # creates a 1D-array for y\n",
+ "X, Y = numpy.meshgrid(x, y) # generates a mesh grid"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "We consider a doublet of strength $\\kappa=1.0$ located at the origin."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {},
+ "outputs": [],
+ "source": [
+ "kappa = 1.0 # strength of the doublet\n",
+ "x_doublet, y_doublet = 0.0, 0.0 # location of the doublet"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "As seen in the previous notebook, we play smart by defining functions to calculate the stream function and the velocity components that could be re-used if we decide to insert more than one doublet in our domain."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {},
+ "outputs": [],
+ "source": [
+ "def get_velocity_doublet(strength, xd, yd, X, Y):\n",
+ " \"\"\"\n",
+ " Returns the velocity field generated by a doublet.\n",
+ " \n",
+ " Parameters\n",
+ " ----------\n",
+ " strength: float\n",
+ " Strength of the doublet.\n",
+ " xd: float\n",
+ " x-coordinate of the doublet.\n",
+ " yd: float\n",
+ " y-coordinate of the doublet.\n",
+ " X: 2D Numpy array of floats\n",
+ " x-coordinate of the mesh points.\n",
+ " Y: 2D Numpy array of floats\n",
+ " y-coordinate of the mesh points.\n",
+ " \n",
+ " Returns\n",
+ " -------\n",
+ " u: 2D Numpy array of floats\n",
+ " x-component of the velocity vector field.\n",
+ " v: 2D Numpy array of floats\n",
+ " y-component of the velocity vector field.\n",
+ " \"\"\"\n",
+ " u = (- strength / (2 * math.pi) *\n",
+ " ((X - xd)**2 - (Y - yd)**2) /\n",
+ " ((X - xd)**2 + (Y - yd)**2)**2)\n",
+ " v = (- strength / (2 * math.pi) *\n",
+ " 2 * (X - xd) * (Y - yd) /\n",
+ " ((X - xd)**2 + (Y - yd)**2)**2)\n",
+ " \n",
+ " return u, v\n",
+ "\n",
+ "def get_stream_function_doublet(strength, xd, yd, X, Y):\n",
+ " \"\"\"\n",
+ " Returns the stream-function generated by a doublet.\n",
+ " \n",
+ " Parameters\n",
+ " ----------\n",
+ " strength: float\n",
+ " Strength of the doublet.\n",
+ " xd: float\n",
+ " x-coordinate of the doublet.\n",
+ " yd: float\n",
+ " y-coordinate of the doublet.\n",
+ " X: 2D Numpy array of floats\n",
+ " x-coordinate of the mesh points.\n",
+ " Y: 2D Numpy array of floats\n",
+ " y-coordinate of the mesh points.\n",
+ " \n",
+ " Returns\n",
+ " -------\n",
+ " psi: 2D Numpy array of floats\n",
+ " The stream-function.\n",
+ " \"\"\"\n",
+ " psi = - strength / (2 * math.pi) * (Y - yd) / ((X - xd)**2 + (Y - yd)**2)\n",
+ " \n",
+ " return psi"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Once the functions have been defined, we call them using the parameters of the doublet: its strength `kappa` and its location `x_doublet`, `y_doublet`."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {},
+ "outputs": [],
+ "source": [
+ "# compute the velocity field on the mesh grid\n",
+ "u_doublet, v_doublet = get_velocity_doublet(kappa, x_doublet, y_doublet, X, Y)\n",
+ "\n",
+ "# compute the stream-function on the mesh grid\n",
+ "psi_doublet = get_stream_function_doublet(kappa, x_doublet, y_doublet, X, Y)"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "We are ready to do a nice visualization."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {},
+ "outputs": [],
+ "source": [
+ "# plot the streamlines\n",
+ "width = 10\n",
+ "height = (y_end - y_start) / (x_end - x_start) * width\n",
+ "pyplot.figure(figsize=(width, height))\n",
+ "pyplot.xlabel('x', fontsize=16)\n",
+ "pyplot.ylabel('y', fontsize=16)\n",
+ "pyplot.xlim(x_start, x_end)\n",
+ "pyplot.ylim(y_start, y_end)\n",
+ "pyplot.streamplot(X, Y, u_doublet, v_doublet,\n",
+ " density=2, linewidth=1, arrowsize=1, arrowstyle='->')\n",
+ "pyplot.scatter(x_doublet, y_doublet, color='#CD2305', s=80, marker='o');"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Just like we imagined that the streamlines of a source-sink pair would look from very far away. What is this good for, you might ask? It does not look like any streamline pattern that has a practical use in aerodynamics. If that is what you think, you would be wrong!"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Uniform flow past a doublet"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "A doublet alone does not give so much information about how it can be used to represent a practical flow pattern in aerodynamics. But let's use our superposition powers: our doublet in a uniform flow turns out to be a very interesting flow pattern. Let's first define a uniform horizontal flow."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {},
+ "outputs": [],
+ "source": [
+ "u_inf = 1.0 # freestream speed"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Remember from our previous lessons that the Cartesian velocity components of a uniform flow in the $x$-direction are given by $u=U_\\infty$ and $v=0$. Integrating, we find the stream-function, $\\psi = U_\\infty y$.\n",
+ "\n",
+ "So let's calculate velocities and stream function values for all points in our grid. And as we now know, we can calculate them all together with one line of code per array."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {},
+ "outputs": [],
+ "source": [
+ "u_freestream = u_inf * numpy.ones((N, N), dtype=float)\n",
+ "v_freestream = numpy.zeros((N, N), dtype=float)\n",
+ "\n",
+ "psi_freestream = u_inf * Y"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Below, the stream function of the flow created by superposition of a doublet in a free stream is obtained by simple addition. Like we did before in the [Source & Sink in a Freestream](02_Lesson02_sourceSinkFreestream.ipynb) notebook, we find the *dividing streamline* and plot it separately in red. \n",
+ "\n",
+ "The plot shows that this pattern can represent the flow around a cylinder with center at the location of the doublet. All the streamlines remaining outside the cylinder originated from the uniform flow. All the streamlines inside the cylinder can be ignored and this area assumed to be a solid object. This will turn out to be more useful than you may think."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {},
+ "outputs": [],
+ "source": [
+ "# superposition of the doublet on the freestream flow\n",
+ "u = u_freestream + u_doublet\n",
+ "v = v_freestream + v_doublet\n",
+ "psi = psi_freestream + psi_doublet\n",
+ "\n",
+ "# plot the streamlines\n",
+ "width = 10\n",
+ "height = (y_end - y_start) / (x_end - x_start) * width\n",
+ "pyplot.figure(figsize=(width, height))\n",
+ "pyplot.xlabel('x', fontsize=16)\n",
+ "pyplot.ylabel('y', fontsize=16)\n",
+ "pyplot.xlim(x_start, x_end)\n",
+ "pyplot.ylim(y_start, y_end)\n",
+ "pyplot.streamplot(X, Y, u, v,\n",
+ " density=2, linewidth=1, arrowsize=1, arrowstyle='->')\n",
+ "pyplot.contour(X, Y, psi,\n",
+ " levels=[0.], colors='#CD2305', linewidths=2, linestyles='solid')\n",
+ "pyplot.scatter(x_doublet, y_doublet, color='#CD2305', s=80, marker='o')\n",
+ "\n",
+ "# calculate the stagnation points\n",
+ "x_stagn1, y_stagn1 = +math.sqrt(kappa / (2 * math.pi * u_inf)), 0.0\n",
+ "x_stagn2, y_stagn2 = -math.sqrt(kappa / (2 * math.pi * u_inf)), 0.0\n",
+ "\n",
+ "# display the stagnation points\n",
+ "pyplot.scatter([x_stagn1, x_stagn2], [y_stagn1, y_stagn2],\n",
+ " color='g', s=80, marker='o');"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##### Challenge question"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "What is the radius of the circular cylinder created when a doublet of strength $\\kappa$ is added to a uniform flow $U_\\infty$ in the $x$-direction?"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##### Challenge task"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "You have the streamfunction of the doublet in cylindrical coordinates above. Add the streamfunction of the free stream in those coordinates, and study it. You will see that $\\psi=0$ at $r=a$ for all values of $\\theta$. The line $\\psi=0$ represents the circular cylinder of radius $a$. Now write the velocity components in cylindrical coordinates, find the speed of the flow at the surface. What does this tell you?"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "### Bernoulli's equation and the pressure coefficient"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "A very useful measurement of a flow around a body is the *coefficient of pressure* $C_p$. To evaluate the pressure coefficient, we apply *Bernoulli's equation* for ideal flow, which says that along a streamline we can apply the following between two points:\n",
+ "\n",
+ "$$p_\\infty + \\frac{1}{2}\\rho U_\\infty^2 = p + \\frac{1}{2}\\rho U^2$$\n",
+ "\n",
+ "We define the pressure coefficient as the ratio between the pressure difference with the free stream, and the dynamic pressure:\n",
+ "\n",
+ "$$C_p = \\frac{p-p_\\infty}{\\frac{1}{2}\\rho U_\\infty^2}$$\n",
+ "\n",
+ "i.e.,\n",
+ "\n",
+ "$$C_p = 1 - \\left(\\frac{U}{U_\\infty}\\right)^2$$\n",
+ "\n",
+ "In an incompressible flow, $C_p=1$ at a stagnation point. Let's plot the pressure coefficient in the whole domain."
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {},
+ "outputs": [],
+ "source": [
+ "# compute the pressure coefficient field\n",
+ "cp = 1.0 - (u**2 + v**2) / u_inf**2\n",
+ "\n",
+ "# plot the pressure coefficient field\n",
+ "width = 10\n",
+ "height = (y_end - y_start) / (x_end - x_start) * width\n",
+ "pyplot.figure(figsize=(1.1 * width, height))\n",
+ "pyplot.xlabel('x', fontsize=16)\n",
+ "pyplot.ylabel('y', fontsize=16)\n",
+ "pyplot.xlim(x_start, x_end)\n",
+ "pyplot.ylim(y_start, y_end)\n",
+ "contf = pyplot.contourf(X, Y, cp,\n",
+ " levels=numpy.linspace(-2.0, 1.0, 100), extend='both')\n",
+ "cbar = pyplot.colorbar(contf)\n",
+ "cbar.set_label('$C_p$', fontsize=16)\n",
+ "cbar.set_ticks([-2.0, -1.0, 0.0, 1.0])\n",
+ "pyplot.scatter(x_doublet, y_doublet,\n",
+ " color='#CD2305', s=80, marker='o')\n",
+ "pyplot.contour(X,Y,psi,\n",
+ " levels=[0.], colors='#CD2305', linewidths=2, linestyles='solid')\n",
+ "pyplot.scatter([x_stagn1, x_stagn2], [y_stagn1, y_stagn2],\n",
+ " color='g', s=80, marker='o');"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##### Challenge task"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Show that the pressure coefficient distribution on the surface of the circular cylinder is given by\n",
+ "\n",
+ "$$C_p = 1-4\\sin^2\\theta$$\n",
+ "\n",
+ "and plot the coefficient of pressure versus the angle."
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "##### Think"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "Don't you find it a bit fishy that the pressure coefficient (and the surface distribution of pressure) is symmetric about the vertical axis?\n",
+ "\n",
+ "That means that the pressure in the front of the cylinder is the same as the pressure in the *back* of the cylinder. In turn, this means that the horizontal components of forces are zero.\n",
+ "\n",
+ "We know that, even at very low Reynolds number (creeping flow), there *is* in fact a drag force. The theory is unable to reflect that experimentally observed fact! This discrepancy is known as *d'Alembert's paradox*. \n",
+ "\n",
+ "Here's how creeping flow around a cylinder *really* looks like:"
+ ]
+ },
+ {
+ "cell_type": "code",
+ "execution_count": null,
+ "metadata": {},
+ "outputs": [],
+ "source": [
+ "from IPython.display import YouTubeVideo\n",
+ "YouTubeVideo('Ekd8czwELOc')"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "If you look carefully, there is a slight asymmetry in the flow pattern. Can you explain it? What is the consequence of that?\n",
+ "\n",
+ "Here's a famous visualization of actual flow around a cylinder at a Reynolds number of 1.54. This image was obtained by S. Taneda and it appears in the \"Album of Fluid Motion\", by Milton Van Dyke. A treasure of a book.\n",
+ "\n",
+ "![](\"resources/Cylinder-Re=1dot54.png\")
"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "---"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 3",
+ "language": "python",
+ "name": "python3"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 3
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython3",
+ "version": "3.6.9"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 4
+}
diff --git a/examples/TextbookC-Stats.ipynb b/examples/TextbookC-Stats.ipynb
new file mode 100644
index 0000000..17b611b
--- /dev/null
+++ b/examples/TextbookC-Stats.ipynb
@@ -0,0 +1,789 @@
+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ ""
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Introduction to hypothesis testing\n",
+ "\n",
+ "In this notebook we look at data on a type of flower called Iris and we analyze it using a programming language called Python."
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Learning goals\n",
+ "\n",
+ "This notebook is designed for you to learn:\n",
+ "* How to distinguish between \"population\" datasets and \"sample\" datasets when dealing with experimental data\n",
+ "* How to compare a sample to a population using a statistical test called the \"t-test\" and interpret its results\n",
+ "* How to evaluate the magnitude of the difference between a sample and a population using a measure of the effect size called \"Cohen's d\" and interpret the results\n",
+ "* How to use Python scripts to make statistical analyses on a dataset"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "---"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "## Introduction\n",
+ "\n",
+ "\n",
+ "
![\"iris](\"figs/iris-picture.jpg\")
\n",
+ "\n",
+ "###### Iris virginica (Credit: Frank Mayfield CC BY-SA 2.0)\n",
+ "\n",
+ "
\n",
+ "
![](\"figs/diagram-samples.png\")
\n",
+ "
![](\"figs/diagram-normalcurveAB.png\")
\n",
+ "
![](\"figs/diagram-normalcurveAlpha.png\")
\n",
+ "
Bibliography
\n",
+ "\n",
+ "[1] E. Anderson (1935). \"The Irises of the Gaspe Peninsula.\" Bulletin of the American Iris Society 59: 2–5.\n",
+ "\n",
+ "[2] R. A. Fisher (1936). \"The use of multiple measurements in taxonomic problems\". Annals of Eugenics. 7 (2): 179–188. doi:10.1111/j.1469-1809.1936.tb02137.x\n",
+ "\n",
+ "More about the Iris Dataset on Wikipedia: https://en.wikipedia.org/wiki/Iris_flower_data_set\n",
+ "\n",
+ "*Please note that the datasets used in this notebook have been generated using a random generator, they do not come from real measurement and cannot be used for any research purpose.*"
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 3",
+ "language": "python",
+ "name": "python3"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 3
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
+ "nbconvert_exporter": "python",
+ "pygments_lexer": "ipython3",
+ "version": "3.6.9"
+ }
+ },
+ "nbformat": 4,
+ "nbformat_minor": 4
+}
diff --git a/examples/figs/Glissiere.png b/examples/figs/Glissiere.png
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index 0000000..7dbf8ce
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diff --git a/examples/figs/iris-picture.jpg b/examples/figs/iris-picture.jpg
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diff --git a/examples/figs/piano.jpg b/examples/figs/piano.jpg
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diff --git a/examples/lib/InterpolationLib.py b/examples/lib/InterpolationLib.py
new file mode 100644
index 0000000..c1c26ad
--- /dev/null
+++ b/examples/lib/InterpolationLib.py
@@ -0,0 +1,165 @@
+#!/usr/bin/env python3
+# -*- coding: utf-8 -*-
+"""
+Created on Thu Jan 30
+
+@author: Simone Deparis
+"""
+
+import numpy as np
+
+# Defining the mxn Vandermonde matrix
+def VandermondeMatrix(x, m=0):
+ # Input
+ # x : +1 array with interpolation nodes
+ # m : degree of the polynomial. If empty, chooses m=size(x)-1
+ # Output
+ # Matrix of Vandermonde of size m x n
+
+ n = x.size - 1
+ if (m == 0):
+ m = n
+
+ # The following is a trick to put toghether the matrix A. Don't need to learn by heart ...
+ # np.tile(x, (n+1, 1)) : repete n+1 times of the array x
+ # .T : transpose
+ # np.linspace(0,n,n+1) : [0,...,n+1]
+ # np.tile : again: repete n+1 times the array [0,...,n+1]
+ # np.power : element by element power funcion
+
+ A = np.power( np.tile(x, (m+1, 1)).T , np.tile(np.linspace(0,m,m+1), (n+1, 1)))
+
+ return A
+
+
+
+
+# Defining the Lagrange basis functions
+def LagrangeBasis(t,k,z):
+ # the input variables are:
+ # t : the interpolatory points
+ # k : which basis function
+ # z : where to evaluate the function
+
+
+ # careful, there are n+1 interpolation points!
+ n = t.size - 1
+
+ # init result to one, of same type and size as z
+ result = np.zeros_like(z) + 1
+
+ # first few checks on k:
+ if (type(k) != int) or (t.size < 1) or (k > n) or (k < 0):
+ raise ValueError('Lagrange basis needs a positive integer k, smaller than the size of t')
+
+ # loop on n to compute the product
+ for j in range(0,n+1) :
+ if (j == k) :
+ continue
+ if (t[k] == t[j]) :
+ raise ValueError('Lagrange basis: all the interpolation points need to be distinct')
+
+ result = result * (z - t[j]) / (t[k] - t[j])
+
+ return result
+
+
+# Piece-wise linear interpolation, equidistributed points
+def PiecewiseLinearInterpolation(a,b,N,f,z):
+ # the input variables are:
+ # a,b : x[0] = a, x[n] = b
+ # f : the corresponding data at the points x
+ # z : where to evaluate the function
+
+ def localLinearInterpolation (a, H, x, y, zk):
+ # first find out in which interval we are. Easy here since equidistributed point
+ i = int( (zk-a)//H )
+
+ # if we are not in the interval, return constant function
+ if x[i] == zk :
+ return y[i]
+
+ # use formula for local linear interpolation
+ return y[i] + (y[i+1] - y[i])/(x[i+1] - x[i])* (zk - x[i])
+
+ # careful, there are N+1 points
+ H = (b-a)/N
+ if np.isclose(H,0) :
+ raise ValueError('PiecewiseLinearInterpolation : a and b are too close')
+
+ x = np.linspace(a,b,N+1)
+
+ # if f is a function, we need to evaluate it at alle the locations
+ from types import FunctionType
+ if isinstance(f, FunctionType) :
+ y = f(x)
+ else :
+ y = f
+
+ # if we want to evaluate the interpolating function in a single point
+ if np.ndim(z) == 0 :
+ return localLinearInterpolation (b, H, x, y, z)
+
+ # if we are here, it means that we need to evaluate the interpolation on many points.
+ # init result to zero, of same type and size as z
+ result = np.zeros_like(z)
+
+ # The following part is not optimized
+ for k in range(z.size) :
+
+ result[k] = localLinearInterpolation (a, H, x, y, z[k])
+
+ return result
+
+# Piece-wise quadratic interpolation, equidistributed points
+def PiecewiseQuadraticInterpolation(a,b,N,f,z):
+ # the input variables are:
+ # a,b : x[0] = a, x[n] = b
+ # f : the corresponding data at the points x
+ # only works if fis a fonction
+ # z : where to evaluate the function
+
+ def localQuadraticInterpolation (a, H, x, f, zk):
+ # first find out in which interval we are. Easy here since equidistributed point
+ i = int( (zk-a)//H )
+
+ # if we are not in the interval, return constant function
+ if x[i] == zk :
+ return f(x[i])
+
+ t = np.array([ x[i], (x[i]+x[i+1])/2, x[i+1]] )
+
+ # use formula for local quadratic interpolation
+ return LagrangeInterpolation(t,f(t),zk)
+
+ # careful, there are N+1 points
+ H = (b-a)/N
+ if np.isclose(H,0) :
+ raise ValueError('PiecewiseQuadraticInterpolation : a and b are too close')
+
+ x = np.linspace(a,b,N+1)
+
+ # if f is a function, we need to evaluate it at alle the locations
+ # from types import FunctionType
+ # if ~isinstance(f, FunctionType) :
+ # raise ValueError('PiecewiseQuadraticInterpolation : works only with a given function')
+
+ # if we want to evaluate the interpolating function in a single point
+ if np.ndim(z) == 0 :
+ return localQuadraticInterpolation (b, H, x, f, z)
+
+ # if we are here, it means that we need to evaluate the interpolation on many points.
+ # init result to zero, of same type and size as z
+ result = np.zeros_like(z)
+
+ # The following part is not optimized
+ for k in range(z.size) :
+
+ result[k] = localQuadraticInterpolation (a, H, x, f, z[k])
+
+ return result
+
+
+
+
+
diff --git a/examples/resources/Cylinder-Re=1dot54.png b/examples/resources/Cylinder-Re=1dot54.png
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index 0000000..a6ea0de
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diff --git a/examples/resources/background.png b/examples/resources/background.png
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index 0000000..18dcffe
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diff --git a/examples/resources/background2.png b/examples/resources/background2.png
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diff --git a/examples/resources/data-05-04-19.csv b/examples/resources/data-05-04-19.csv
new file mode 100644
index 0000000..731676a
--- /dev/null
+++ b/examples/resources/data-05-04-19.csv
@@ -0,0 +1,74 @@
+"VideoAnalysis: Time (s)","VideoAnalysis: X (cm)","VideoAnalysis: Y (cm)","VideoAnalysis: X Velocity (cm/s)","VideoAnalysis: Y Velocity (cm/s)"
+0,46.6055836842,36.2237701107,-3.17942792473,-2.67511151126
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diff --git a/examples/resources/doubletSketch1.png b/examples/resources/doubletSketch1.png
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index 0000000..3a388e1
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diff --git a/examples/resources/iris-sample1-vullierens.csv b/examples/resources/iris-sample1-vullierens.csv
new file mode 100644
index 0000000..7928165
--- /dev/null
+++ b/examples/resources/iris-sample1-vullierens.csv
@@ -0,0 +1,51 @@
+petal_length
+6.0
+5.9
+5.5
+5.7
+5.4
+5.9
+6.7
+5.8
+6.1
+6.2
+6.8
+5.2
+5.0
+4.4
+4.4
+5.9
+6.4
+4.9
+6.0
+5.3
+5.8
+6.1
+4.6
+6.0
+5.4
+5.7
+5.1
+5.1
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+5.6
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+6.0
+5.8
+6.5
+6.6
+4.8
+4.8
+6.0
+5.5
+5.0
+5.4
diff --git a/examples/resources/iris-sample2-meiji.csv b/examples/resources/iris-sample2-meiji.csv
new file mode 100644
index 0000000..3919d65
--- /dev/null
+++ b/examples/resources/iris-sample2-meiji.csv
@@ -0,0 +1,51 @@
+petal_length
+5.7
+4.5
+5.8
+5.1
+6.2
+6.1
+5.3
+5.7
+6.4
+5.2
+5.7
+6.7
+5.8
+5.6
+5.6
+5.3
+6.1
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+5.6
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+5.7
+6.3
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+5.3
+5.4
+6.2
+5.9
diff --git a/examples/resources/piano.wav b/examples/resources/piano.wav
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index 0000000..b573343
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diff --git a/examples/solution/TextbookC-Stats-solution.ipynb b/examples/solution/TextbookC-Stats-solution.ipynb
new file mode 100644
index 0000000..d499e8f
--- /dev/null
+++ b/examples/solution/TextbookC-Stats-solution.ipynb
@@ -0,0 +1,81 @@
+{
+ "cells": [
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "# Solution of the exercise\n"
+ ]
+ },
+ {
+ "cell_type": "markdown",
+ "metadata": {},
+ "source": [
+ "**A. Getting familiar with the code.**\n",
+ "1. In the code cells above, where is the t-test computed using the predefined Python function? \n",
+ " 🡆 The predefined Python function for computing a one-sample t-test is `stats.ttest_1samp`. You can search for it in the notebook by typing `ctrl + F` or using the top menu `Edit` > `Find`.\n",
+ "\n",
+ "1. What are the two parameters that the t-test function takes as input? \n",
+ " 🡆 The two input parameters of the `stats.ttest_1samp` function are: a) the list of all the sample values, in our case `sample_data[\"petal_length\"]`, and b) the population reference value for comparison, in our case `mu`.\n",
+ "\n",
+ "1. If you wanted to change the population mean to a different value, like $\\mu = 5.4$ cm for instance, in which cell would you change it? \n",
+ "Then what would you need to do to update the result of the analysis in the notebook? \n",
+ " 🡆 You could change it in two different places: in the cell where `mu` is defined, at the top of the notebook, or you could replace `mu` with the value `5.4` in the cell where the function `stats.ttest_1samp` is called. \n",
+ " Then you would have to *re-execute the cells* of the notebook which do the analysis to update the results.\n",
+ "\n",
+ "1. What is the result of the t-test if you compare the Vullierens sample to a population mean of $\\mu = 5.4$ cm? \n",
+ " 🡆 The difference becomes statistically significant with $t=3.025$ and $p=0.004 \\lt \\alpha$, in which case we would reject our null hypothesis and conclude that our sample is probably from a different population. \n",
+ " However, the effect size remains below the medium threshold, $d=0.428$.\n",
+ "\n",
+ "*Change the value of $\\mu$ back to 5.552 before working on the following questions.*\n",
+ "\n",
+ "**B. Analyzing another dataset.**\n",
+ "\n",
+ "A researcher from Tokyo sends you the results of a series of measurements she has done on the Irises of the [Meiji Jingu Imperial Gardens](http://www.meijijingu.or.jp/english/nature/2.html). The dataset can be found in the `iris-sample2-meiji.csv` file. \n",
+ "How similar (or different) is the Meiji sample compared to the Iris virginica population documented by Edgar Anderson? \n",
+ "The following questions are designed to guide you in analyzing this new dataset using this notebook.\n",
+ "\n",
+ "1. Which of the code cells above loads the data from the file containing the Vullierens dataset? Modify it to load the Meiji dataset. \n",
+ " 🡆 The function to read a CSV file in Python is `pan.read_csv`. To load the Meiji dataset, you have to replace its input parameter with the name of the file containing the second dataset, `'iris-sample2-meiji.csv'`.\n",
+ "\n",
+ "1. Do you need to modify anything else in the code to analyze this new dataset? \n",
+ " 🡆 There is nothing else to change in the code to analyse this new dataset since the new sample values are simply stored in the `sample_data` variable, in replacement of the previous values. \n",
+ " However, you have to *re-execute all the code cells* which compute the analysis so that all the calculations are made with the new values. \n",
+ " Note that you can use the top menu `Run` > `Run All Cells` to execute all the code cells of the notebook all at once.\n",
+ "\n",
+ "1. What can you conclude about the Meiji sample from this analysis? \n",
+ " 🡆 The difference in petal length mean between the Meiji sample and the Anderson population is statistically significant at the 5% level with $t=2.352$ and $p=0.023$, but a relatively small effect size $d=0.333$. \n",
+ " In this case we could reject our null hypothesis and conclude that the Meiji sample is probably from a different population.\n",
+ "\n",
+ "**C. Going a bit further in the interpretation of the t-test.**\n",
+ "1. In the code cells above, where is the cut-off point $\\alpha$ defined? Change its value to 0.01 and re-execute the notebook. \n",
+ " 🡆 Search for the variable `alpha` in the notebook and modify its assignment in `alpha = 0.01`. Then re-execute the notebook.\n",
+ "\n",
+ "1. How does this affect the result of the t-test for the Meiji sample? \n",
+ " 🡆 When choosing a significance level of 1% ($\\alpha = .01$), the difference in petal length mean between the Meiji sample and the Anderson population cannot be considered statistically significant. \n",
+ " This means that the evidence we have is not strong enough if we want to be 99% sure. "
+ ]
+ }
+ ],
+ "metadata": {
+ "kernelspec": {
+ "display_name": "Python 3",
+ "language": "python",
+ "name": "python3"
+ },
+ "language_info": {
+ "codemirror_mode": {
+ "name": "ipython",
+ "version": 3
+ },
+ "file_extension": ".py",
+ "mimetype": "text/x-python",
+ "name": "python",
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+}