diff --git a/0.1 Some Exercises.ipynb b/0.1 Some Exercises.ipynb new file mode 100644 index 0000000..4176a96 --- /dev/null +++ b/0.1 Some Exercises.ipynb @@ -0,0 +1,422 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Numerical Analysis \n", + "\n", + "Autumm Semester 2021 - Sections SIE/GC\n", + "\n", + "prof. Fabio Nobile\n", + "\n", + "## Serie 1: Introduction to `Python`" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "- Before starting to work on this exercise session, you should review the material in the short `Python tutorial` available on Moodle." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Exercise 0 -- Learning Python using Python." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Type the following commands and look at the results. Begin by importing numpy and matplotlib.pyplot as " + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": {}, + "outputs": [], + "source": [ + "import numpy as np\n", + "import matplotlib.pyplot as plt" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "These are the libraries that python relies upon for scientific computing and plotting. " + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "| No. | Code | Instruction |\n", + "| --- | --- | --- |\n", + "| 1 | `a=7` | Define $a$ as a scalar variable which has type `int` |\n", + "| 2 | `a=7.1` | Define $a$ as a scalar variable which has type `float` |\n", + "| 3 | `b=np.array([1,2,3])` | Define $b\\in\\mathbb{R}^{1 \\times 3}$. What happens if you replace `,` by `''` or `;`? |\n", + "| 4 | `g=b[2]` | Return the **third** element of $b$. Note: Indexation starts at 0. |\n", + "| 5 | `E=np.array([[1, 2, 3],[4, 5, 6]])` | Create matrix $E\\in\\mathbb{R}^{2\\times3}$. Note: Python is case sensitive. |\n", + "| 6 | `E[0,:]` | Return first column of $E;$ $(1,4)^T$ |\n", + "| 7 | `h=E[1,2]` | Return the element $E_{2,3}$ of the matrix $E$ ($E_{2,3}=6$). |\n", + "| 8 | `%whos` | Fetch information about the used variables. |\n", + "| 9 | `help(np.sin)` | Fetch help for the command ` np.sin`. Same as `? np.sin` in iPython. |\n", + "| 10 | `A=np.eye(3)` | Return the identity matrix $3\\times 3$ |\n", + "| 11 | `I=np.ones(4)` | Return matrix $4\\times 4$ for which each value is 1 |\n", + "| 12 | `F=np.zeros((2,3))` | Return $F\\in\\mathbb{R}^{2\\times 3}$ for which each value is 0 |\n", + "| 13 | `F.T` | Perform transpose of $F$ |\n", + "| 14 | `x=np.linspace(0,1,3)` | Return a vector of length $3$ whose values are equally distributed between $0$ and $1$ |\n", + "| 15 | `D=np.diag(b)` | Return diagonal matrix with diagonal given by vector b. |\n", + "| 16 | `np.shape(F)` | Return the number of lines and columns of $F$ in a line vector. |\n", + "| 17 | `A@D` | Return product of two matrices |\n", + "| 18 | `E@x` | Return matrix vector product |\n", + "| 19 | `A[-1,1]` | Return element of $A$ which is on the last line, second column |\n", + "| 20 | `H=np.array([[1, 3],[9, 11]])；np.linalg.solve(H,E)` | Compute $H^{-1}E$. Note: Never use `np.linalg.inv(H)@E` |\n", + "| 21 | `np.linalg.det(H)` | Compute determinant of $H$ |" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": {}, + "outputs": [], + "source": [ + "## type here and execute the 21 commands above\"\n", + "#1\n", + "a=7\n", + "print('#1',a),\n", + "#2 continue by yourself" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Exercise 1\n", + "\n", + "Fibonacci's series is defined as follows:\n", + "\\begin{align*}\n", + "F_0&=1\\\\\n", + "F_1&=1\\\\\n", + "F_{n}&=F_{n-1}+F_{n-2}, \\quad \\forall n \\geq 2\n", + "\\end{align*}\n", + "* Complete the `Python` function below which computes the $30^{\\text{th}}$ Fibonacci's number.\n", + "\n", + "* Write a function which takes as an argument a natural number $n$ and returns the $n^{\\text{th}}$ Fibonacci's number.\n" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": { + "jupyter": { + "source_hidden": true + } + }, + "outputs": [], + "source": [ + "#COMPLETE THIS CODE\n", + "def Fibonacci(n):\n", + " # n should be a positive integer\n", + " # Complete this code\n", + " Fn=None #Change this value\n", + " # ...\n", + " return Fn\n", + "\n", + "F30=Fibonacci(30)\n", + "print(F30)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Exercise 2\n", + "Let us consider the matrices \n", + "\\begin{align*}\n", + "A=\n", + "\\begin{bmatrix}\n", + "5 & 3 & 0\\\\\n", + "1 & 1 & -4\\\\\n", + "3 & 0 & 0\n", + "\\end{bmatrix}, \\quad\n", + "B=\n", + "\\begin{bmatrix}\n", + "4 & 3 & 2\\\\\n", + "0 & 1 & 0\\\\\n", + "5 & 0 & 1/2\n", + "\\end{bmatrix}.\n", + "\\end{align*}\n", + "Compute (without using any loops), the matrix $C=AB$ (matrix product) and the matrix $D$ which\n", + "has as elements $D_{ij}=A_{ij}B_{ij}$ (element-wise product)." + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "#### Solution:\n", + "\n", + "We can use the Python commands `@` for matrix multiplication and `*` for element-wise multiplication, as shown below:\n" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": {}, + "outputs": [], + "source": [ + "A=np.array([ [5,3,0],\n", + " [1,1,-4],\n", + " [3,0,0]])\n", + "B=np.array([ [4,3,2],\n", + " [0,1,0],\n", + " [5,0,1/2]])\n", + "\n", + "\n", + "# Complete this code\n", + "C=None #Change this value\n", + "D=None #Change this value\n", + "\n", + "\n", + "print('C matrix')\n", + "print(C)\n", + "\n", + "print('D matrix')\n", + "print(D)\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Exercise 3\n", + "\n", + "Define (without using any loops) the\n", + "bidiagonal matrix of size $n=5$ whose main diagonal is a vector\n", + "of equally distributed points between $3$ and $6$, i.e. $$D=(3, 3.75, 4.5, 5.25, 6)^T,$$\n", + "\n", + "and the sub-diagonal is the vector of equally distributed points between $2$ and $3.5$, i.e. $$S=(2, 2.5, 3, 3.5)^T.$$\n", + "\n", + "\n", + "Tip: See ``https://numpy.org/doc/stable/reference/generated/numpy.diag.html``\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "#### Solution\n", + "\n", + "Following the reference for the`diag` method of numpy we obtain:" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": {}, + "outputs": [], + "source": [ + "# Complete this code\n", + "D=None #Change this value\n", + "S=None #Change this value\n", + "M=None #Change this value\n", + "\n", + "print('The matrix M is')\n", + "print(M)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "Remember that you can also obtain help from a python function using the command \n", + "`help(NAME OF FUNCTION)`. In this case: `help(np.diag)`" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": {}, + "outputs": [], + "source": [ + "help(np.diag)" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Exercise 4\n", + "Let us consider the vectors $x=(1,4,7,2,1,2)$ and $y=(0,9,1,4,3,0)$. \n", + "Compute (without using any loops for points a and b):\n", + "1. the product, component by component, between two vectors $x$ and $y$ (tip: use the operator `*`)\n", + "2. the scalar product between the same vectors $x$ and $y$ (tip: use the operator `@`)\n", + "3. a vector whose elements are defined by:\n", + "$v_1=x_1\\,y_n, \\quad v_2=x_2\\, y_{n-1},\\quad \\dots, \\quad v_{n-1}=x_{n-1}\\, y_2,\\quad v_n=x_n\\,y_1$.\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "#### Solution" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": {}, + "outputs": [], + "source": [ + "x=np.array([1,4,7,2,1,2])\n", + "y=np.array([0,9,1,4,3,0])\n", + "\n", + "# Complete this code\n", + "V1=None #Change this value\n", + "V2=None #Change this value\n", + "V3=None #Change this value\n", + "\n", + "print('v1 = '+str(v1))\n", + "print('v2 = '+str(v2))\n", + "print('v3 = '+str(v3))" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Exercise 5\n", + "\n", + "Plot the functions\n", + "\\begin{align*}\n", + "f(x)=x^3 , \\quad x \\in [1,10]\n", + "\\end{align*}\n", + "\\begin{align*}\n", + "g(x)=\\exp(4x) , \\quad x \\in [1,10]\n", + "\\end{align*}\n", + "\n", + "in linear scale (i.e. using the function `plt.plot()` and logarithmic scale (i.e. using the function `plt.semilogy()` and `plt.loglog()`. We will use the definition interval to plot these graphs considering 200 equally distributed points.\n", + "\n", + "\n", + "\n", + "\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "#### Solution \n", + "\n" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": {}, + "outputs": [], + "source": [ + "# Complete this code\n", + "f=None #Change this value to define inline function f\n", + "g=None #Change this value to define inline function g \n", + "\n", + "x=np.linspace(1,10,200)\n", + "\n", + "\n", + "#plot ins plot plot \n", + "plt.title(' Normal scale')\n", + "plt.plot(x,f(x),label=r'$f(x)$')\n", + "# Complete this code\n", + "# add a plot here for g\n", + "plt.legend()\n", + "plt.show()\n", + "\n", + "#plot in semilogy \n", + "\n", + "\n", + "# Complete this code\n", + "\n", + "\n", + "\n", + "\n", + "#plot in log-log\n", + "\n", + "# Complete this code\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "### Exercise 6\n", + "\n", + "1. Given \n", + "\\begin{align*}\n", + "f(x)=\\frac{x^2}{2}\\sin(x) , \\quad x \\in [1,20]\n", + "\\end{align*}\n", + "plot the function $f$ using 10, 20 and 100 equidistant points in the given interval.\n", + "Plot the three graphs on the same figure with three\n", + "different colors. Which one gives the best representation of $f$?\n", + "\n", + "2. Do the same for the functions:\n", + "\\begin{align*}\n", + "g(x) &= \\frac{x^3}{6}\\cos(\\sin(x))\\exp(-x)+\\left(\\frac{1}{1+x}\\right)^2 , \\quad x \\in [1,20] \\\\\n", + "h(x) &= x(1-x)+\\frac{\\sin(x)\\cos(x)}{x^3}, \\quad x \\in [1,20].\n", + "\\end{align*}\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "#### solution" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": {}, + "outputs": [], + "source": [ + "# complete the code here for question 1" + ] + }, + { + "cell_type": "code", + "execution_count": null, + "metadata": {}, + "outputs": [], + "source": [ + "# complete the code here for question 2\n" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 3 (ipykernel)", + "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": "3.7.7" + } + }, + "nbformat": 4, + "nbformat_minor": 4 +}