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-{"id": "algo_chunk_0", "text": "\\documentclass[a4paper]{book}\n\n\\usepackage{style}\n\n\\pagestyle{fancy}\n\\fancyhf{}\\fancyfoot[LE,RO]{\\thepage}\n\n\n\n\\fancyhead[RE]{\\textit{\\leftmark}}\n\\fancyhead[LO]{\\textit{\\rightmark}}\n\\fancyhead[LE]{Algorithms}\n\\fancyhead[RO]{Notes by Joachim Favre}\n\\fancypagestyle{plain}{\\fancyhf{}\\fancyfoot[LE,RO]{\\thepage}}\n\n\\title{Algorithms\\\\ Prof. Mikhail Kapralov --- EPFL}\n\\author{Notes by Joachim Favre}\n\\date{Computer science bachelor --- Semester 3 \\\\ Autumn 2022}\n\n\\begin{document}\n\\maketitle\n\n\\clearemptydoublepage\n\\thispagestyle{empty}\n\n\n\n\n\n\n\\vspace*{\\fill}\nI made this document for my own use, but I thought that typed notes might be of interest to others. There are mistakes, it is impossible not to make any. If you find some, please feel free to share them with me (grammatical and vocabulary errors are of course also welcome). You can contact me at the following e-mail address:\n\\begin{center}\n \\texttt{joachim.favre@epfl.ch}\n\\end{center}\n\nIf you did not get this document through my GitHub repository, then you may be interested by the fact that I have one on which I put those typed notes and their LaTeX code. Here is the link (make sure to read the README to understand how to download the files you're interested in):\n\\begin{center}\n \\url{https://github.com/JoachimFavre/EPFLNotesIN}\n\\end{center}\n\nPlease note that the content does not belong to me. I have made some structural changes, reworded some parts, and added some personal notes; but the wording and explanations come mainly from the Professor, and from the book on which they based their course.", "source": "algo", "format": "latex"}
-{"id": "algo_chunk_1", "text": "Please note that the content does not belong to me. I have made some structural changes, reworded some parts, and added some personal notes; but the wording and explanations come mainly from the Professor, and from the book on which they based their course.\n\nI think it is worth mentioning that in order to get these notes typed up, I took my notes in \\LaTeX during the course, and then made some corrections. I do not think typing handwritten notes is doable in terms of the amount of work. To take notes in \\LaTeX, I took my inspiration from the following link, written by Gilles Castel. If you want more details, feel free to contact me at my e-mail address, mentioned hereinabove.\n\\begin{center}\n \\url{https://castel.dev/post/lecture-notes-1/}\n\\end{center}\n\nI would also like to specify that the words ``trivial'' and ``simple'' do not have, in this course, the definition you find in a dictionary. We are at EPFL, nothing we do is trivial. Something trivial is something that a random person in the street would be able to do. In our context, understand these words more as ``simpler than the rest''. Also, it is okay if you take a while to understand something that is said to be trivial (especially as I love using this word everywhere hihi).\n\nSince you are reading this, I will give you a little advice. Sleep is a much more powerful tool than you may imagine, so do not neglect a good night of sleep in favour of studying (especially the night before an exam). I wish you to have fun during your exams.\n\n\\vspace*{\\fill}\n\\vspace*{\\fill}\n\\begin{center}\n\\initcurrdate\n\\def\\setdateformat{Y--m--d}\n\\textit{Version \\printdate}\n\\end{center}\n\\vspace*{\\fill}\n\n\\clearemptydoublepage\n\n\n\n\\thispagestyle{empty}\n\\vspace*{\\fill}\n\\begin{flushright}\n\\begin{minipage}{7cm}\n{\\itshape \nTo Gilles Castel, whose work has \\\\\ninspired me this note taking method.\n\n\\vspace{1em}", "source": "algo", "format": "latex"}
-{"id": "algo_chunk_2", "text": "\\thispagestyle{empty}\n\\vspace*{\\fill}\n\\begin{flushright}\n\\begin{minipage}{7cm}\n{\\itshape \nTo Gilles Castel, whose work has \\\\\ninspired me this note taking method.\n\n\\vspace{1em}\n\nRest in peace, nobody \\\\\ndeserves to go so young.\n}\n\\end{minipage}\n\\end{flushright}\n\\vspace*{\\fill}\n\\vspace*{\\fill}\n\n\\clearemptydoublepage\n\n\n\n\\tableofcontents\n\n\\cleardoublepage\n\n\n\n\\renewcommand{\\cftchapleader}{\\cftdotfill{\\cftdotsep}}\n\\renewcommand*{\\cftchapfont}{}\n\n\\listoflectures\n\n\n\n\\chapter{Summary by lecture}\n\n\\lecturetitlesummary{1}{2022-09-23}{I'm rambling a bit}\n\\vspace{0.5em}\n\\begin{itemize}[left=0pt]\n \\item Definition of algorithm and instance.\n \\item Recall on asymptotics (I ramble a bit on this subject, but I find it very interesting).\n \\item Definition of the sorting problem, and of loop invariants.\n \\item Explanation of the insertion sort algorithm.\n\\end{itemize}\n\\vspace{2em}\n\n\\lecturetitlesummary{2}{2022-09-26}{Teile und herrsche}\n\\vspace{0.5em}\n\\begin{itemize}[left=0pt]\n \\item Proof that insertion sort works, and analysis of its complexity.\n \\item Definition of divide-and-conquer algorithms.\n \\item Explanation of merge sort, and analysis of its complexity.\n\\end{itemize}\n\\vspace{2em}\n\n\\lecturetitlesummary{3}{2022-09-30}{Trees which grow in the wrong direction}\n\\vspace{0.5em}\n\\begin{itemize}[left=0pt]\n \\item Proof of correctness of Merge-Sort.\n \\item Analysis of the complexity of merge-sort, through the substitution method and through trees.\n\\end{itemize}\n\\vspace{2em}", "source": "algo", "format": "latex"}
-{"id": "algo_chunk_3", "text": "\\lecturetitlesummary{4}{2022-10-03}{Master theorem}\n\\vspace{0.5em}\n\\begin{itemize}[left=0pt]\n \\item Explanation of the master theorem.\n \\item Explanation of how to count the number of inversions in an array.\n\\end{itemize}\n\\vspace{2em}\n\n\\lecturetitlesummary{5}{2022-10-07}{Fast matrix multiplication}\n\\vspace{0.5em}\n\\begin{itemize}[left=0pt]\n \\item Explanation of a solution to the maximum subarray problem.\n \\item Explanation of a divide-and-conquer algorithm for number multiplication.\n \\item Explanation of Strassen's algorithm, a divide-and-conquer algorithm for matrix multiplication.\n\\end{itemize}\n\\vspace{2em}\n\n\\lecturetitlesummary{6}{2022-10-10}{Heap sort}\n\\vspace{0.5em}\n\\begin{itemize}[left=0pt]\n \\item Definition of max-heap.\n \\item Explanation on how to store a heap.\n \\item Explanation of the \\texttt{MaxHeapify} procedure.\n \\item Explanation on how to make a heap out of a random array.\n \\item Explanation on how to use heaps to make heapsort.\n\\end{itemize}\n\\vspace{2em}\n\n\\lecturetitlesummary{7}{2022-10-14}{Queues, stacks and linked list}\n\\vspace{0.5em}\n\\begin{itemize}[left=0pt]\n \\item Explanation on how to implement a priority queue through a heap.\n \\item Explanation on how to implement a stack.\n \\item Explanation on how to implement a queue.\n \\item Explanation on how to implement a linked list.\n\\end{itemize}\n\\vspace{2em}", "source": "algo", "format": "latex"}
-{"id": "algo_chunk_4", "text": "\\lecturetitlesummary{8}{2022-10-17}{More trees growing in the wrong direction}\n\\vspace{0.5em}\n\\begin{itemize}[left=0pt]\n \\item Definition of binary search trees.\n \\item Explanation on how to search, find the extrema, find the successor and predecessor of a given element, how to print and how to insert an element in a binary search tree.\n\\end{itemize}\n\\vspace{2em}\n\n\\lecturetitlesummary{9}{2022-10-21}{Dynamic cannot be a pejorative word}\n\\vspace{0.5em}\n\\begin{itemize}[left=0pt]\n \\item Explanation on how to delete a node from a binary search tree.\n \\item Explanation of top-down with memoisation and bottom-up algorithms in Dynamic Programming, through the example of the Fibonacci numbers.\n\\end{itemize}\n\\vspace{2em}\n\n\\lecturetitlesummary{10}{2022-10-24}{''There are 3 types of mathematicians: the ones who can count, and the ones who cannot'' (Prof. Kapralov) (what do you mean by ``this title is too long''?)}\n\\vspace{0.5em}\n\\begin{itemize}[left=0pt]\n \\item Application of dynamic programming to the rod-cutting, change-making and matrix-multiplication problems.\n\\end{itemize}\n\\vspace{2em}\n\n\\lecturetitlesummary{11}{2022-10-28}{LCS but not LoL's one}\n\\vspace{0.5em}\n\\begin{itemize}[left=0pt]\n \\item Application of dynamic programming to the longest common subsequence problem.\n\\end{itemize}\n\\vspace{2em}\n\n\\lecturetitlesummary{12}{2022-10-31}{More binary search trees}\n\\vspace{0.5em}\n\\begin{itemize}[left=0pt]\n \\item Explanation on how to use dynamic programming in order to find the optimal binary search tree given a sorted sequence and a list of probabilities.\n\\end{itemize}\n\\vspace{2em}", "source": "algo", "format": "latex"}
-{"id": "algo_chunk_5", "text": "\\lecturetitlesummary{13}{2022-11-04}{An empty course.}\n\\vspace{0.5em}\n\\begin{itemize}[left=0pt]\n \\item No really, we only did some revisions for the midterm.\n\\end{itemize}\n\\vspace{2em}\n\n\\lecturetitlesummary{14}{2022-11-07}{I love XKCD}\n\\vspace{0.5em}\n\\begin{itemize}[left=0pt]\n \\item Definition of directed and undirected graphs, and explanation on how to store them in memory.\n \\item Explanation of BFS.\n \\item Explanation of DFS, and of the depth-first forest and edge classification it implies.\n \\item Explanation of the parenthesis theorem.\n \\item Explanation of the white-path theorem.\n \\item Definition of directed acyclic graphs.\n \\item Proof that a DAG is acyclic if and only if it does not have any back edge.\n \\item Definition of topological sort, and explanation of an algorithm to compute it.\n\\end{itemize}\n\\vspace{2em}\n\n\\lecturetitlesummary{15}{2022-11-14}{I definitely really like this date}\n\\vspace{0.5em}\n\\begin{itemize}[left=0pt]\n \\item Definition of SCCs, and proof of their existence and unicity.\n \\item Definition of component graphs.\n \\item Explanation of Kosarju's algorithm for finding component graphs.\n\\end{itemize}\n\\vspace{2em}\n\n\\lecturetitlesummary{16}{2022-11-18}{This date is really nice too, though}\n\\vspace{0.5em}\n\\begin{itemize}[left=0pt]\n \\item Definition of flow network and flow.\n \\item Definition of residual capacity and residual networks.\n \\item Explanation of the Ford-Fulkerson greed algorithm for finding the maximum in a flow network.\n \\item Definition of the cut of a flow network, and its flow and capacity.\n\\end{itemize}\n\\vspace{2em}", "source": "algo", "format": "latex"}
-{"id": "algo_chunk_6", "text": "\\lecturetitlesummary{17}{2022-11-21}{The algorithm may stop, or may not}\n\\vspace{0.5em}\n\\begin{itemize}[left=0pt]\n \\item Explanation and proof of the max-flow min-cut theorem.\n \\item Complexity analysis of the Ford-Fulkerson method.\n \\item Application of the Ford-Fulkerson method to the Bipartite matching problem and the Edge-disjoint paths problem.\n\\end{itemize}\n\\vspace{2em}\n\n\\lecturetitlesummary{18}{2022-11-25}{Either Levi or Mikasa made this function}\n\\vspace{0.5em}\n\\begin{itemize}[left=0pt]\n \\item There exists no other Ackerman in the world, and when they wrote the term ``Inverse Ackermann function'', they definitely made a mistake while writing the word ``Ackerman''.\n \\item Definition of the disjoint-set data structures.\n \\item Explanation of how to implement a disjoint-set data structure though linked lists.\n \\item Explanation of how to implement a disjoint-set data structure though a forest of trees.\n \\item Definition of spanning trees.\n \\item Explanation of the minimum spanning tree problem.\n\\end{itemize}\n\\vspace{2em}\n\n\\lecturetitlesummary{19}{2022-11-28}{Finding the optimal MST}\n\\vspace{0.5em}\n\\begin{itemize}[left=0pt]\n \\item Explanation and proof of Prim's algorithm for finding a MST.\n \\item Explanation and proof of Kruskal's algorithm for finding a MST.\n \\item Definition of the shortest path problem.\n\\end{itemize}\n\\vspace{2em}", "source": "algo", "format": "latex"}
-{"id": "algo_chunk_7", "text": "\\lecturetitlesummary{20}{2022-12-02}{I like the structure of maths courses}\n\\vspace{0.5em}\n\\begin{itemize}[left=0pt]\n \\item Explanation of the Bellman-Ford's algorithm for finding shortest paths and detecting negative cycles.\n \\item Proof of optimality of the Bellmand-Ford algorithm.\n \\item Explanation of Dijkstra's algorithm for finding a shortest path in a weighted graph, and proof of its optimality.\n\\end{itemize}\n\\vspace{2em}\n\n\\lecturetitlesummary{24}{2022-12-16}{Doing fun stuff with matrices (really)}\n\\vspace{0.5em}\n\\begin{itemize}[left=0pt]\n \\item Applying dynamic programming to solve the all-pairs shortest paths problem.\n \\item Translating our dynamic algorithm to matrix usage, in order to use fast exponentiation.\n \\item Explanation of Floyd-Warshall's algorithm for solving the all-pairs shortest paths problem.\n \\item Explanation of Johnson's algorithm for solving the all-pairs shortest paths problem.\n\\end{itemize}\n\\vspace{2em}\n\n\\lecturetitlesummary{21}{2022-12-05}{Stochastic probabilistic randomness}\n\\vspace{0.5em}\n\\begin{itemize}[left=0pt]\n \\item Introduction to probabilistic analysis.\n \\item Definition of indicator random variables.\n\\end{itemize}\n\\vspace{2em}\n\n\\lecturetitlesummary{22}{2022-12-09}{Hachis Parmentier}\n\\vspace{0.5em}\n\\begin{itemize}[left=0pt]\n \\item Explanation of the birthday paradox, and proof of the birthday lemma.\n \\item Definition of hash function.\n \\item Explanation of hash tables.\n \\item Proof of an upper bound on the runtime complexity of unsuccessful search in hash tables.\n\\end{itemize}\n\\vspace{2em}", "source": "algo", "format": "latex"}
-{"id": "algo_chunk_8", "text": "\\lecturetitlesummary{23}{2022-12-12}{Quantum bogosort is a comparison sort in $\\Theta\\left(n\\right)$}\n\\vspace{0.5em}\n\\begin{itemize}[left=0pt]\n \\item Proof of an upper bound on the runtime complexity of successful search in hash tables.\n \\item Explanation of quicksort.\n \\item Proof and analysis of naive quick sort.\n \\item Analysis of randomised quick sort.\n \\item Proof of the $\\Omega\\left(n\\log\\left(n\\right)\\right)$ lower bound for comparison sorts.\n\\end{itemize}\n\\vspace{2em}\n\n\n\n\\cleardoublepage\n\n\\input{Lecture01/lecture01.tex}\n\\input{Lecture02/lecture02.tex}\n\\input{Lecture03/lecture03.tex}\n\\input{Lecture04/lecture04.tex}\n\\input{Lecture05/lecture05.tex}\n\\input{Lecture06/lecture06.tex}\n\\input{Lecture07/lecture07.tex}\n\\input{Lecture08/lecture08.tex}\n\\input{Lecture09/lecture09.tex}\n\\input{Lecture10/lecture10.tex}\n\\input{Lecture11/lecture11.tex}\n\\input{Lecture12/lecture12.tex}\n\\input{Lecture13/lecture13.tex}\n\\input{Lecture14/lecture14.tex}\n\\input{Lecture15/lecture15.tex}\n\\input{Lecture16/lecture16.tex}\n\\input{Lecture17/lecture17.tex}\n\\input{Lecture18/lecture18.tex}\n\\input{Lecture19/lecture19.tex}\n\\input{Lecture20/lecture20.tex}\n\\input{Lecture24/lecture24.tex}\n\\input{Lecture21/lecture21.tex}\n\\input{Lecture22/lecture22.tex}\n\\input{Lecture23/lecture23.tex}\n\n\n\n\n\\clearemptydoublepage\n\n\\end{document}", "source": "algo", "format": "latex"}
-{"id": "intro_to_ML_chunk_0", "text": "\\documentclass[a4paper]{article}\n\n\\usepackage{style}\n\n\\title{Introduction to machine learning --- BA3\\\\ Detailed summary}\n\\author{Joachim Favre\\\\ Course by Prof. Mathieu Salzmann }\n\\date{Autumn semester 2022}\n\n\\begin{document}\n\\maketitle\n\n\\cftsetindents{paragraph}{1.5em}{1em}\n\\setcounter{tocdepth}{5}\n\n\n\\tableofcontents\n\n\\initcurrdate\n\\def\\setdateformat{Y--m--d}\n\\vspace*{\\fill}\n\\vspace*{\\fill}\n\\vspace*{\\fill}\n\\vspace*{\\fill}\n\\vspace*{\\fill}\n\\vspace*{\\fill}\n\\vspace*{\\fill}\n\\vspace*{\\fill}\n\\begin{center}\n\\textit{Version \\printdate}\n\\end{center}\n\\vspace*{\\fill}\n\n\\newpage\n\n\\begin{parag}{Supervised and unsupervised learning}\n In \\important{supervised learning}, we are given data and its groundtruth labels. The goal is, given new data, we want to predict new labels, by doing regression or classification.\n\n In \\important{unsupervised learning}, we are only given data (without any label), and we want to output some information about it, by doing dimensionality reduction or clustering.\n\\end{parag}\n\n\\begin{parag}{Regression and classification}\n The goal of \\important{regression} is to predict a continuous value for a given sample. The goal of \\important{classification} is to output a discrete label (typically encoded in one-hot encoding with 0s and 1s or -1s and 1s).\n\n The main difference is that there is the notion of closeness in regression (when predicting a date, outputting 1970 when it should be 1980 is better than outputting 2100), which is not in classification (when predicting what is on a picture, outputting a car when it should be a cat is not better or worse than outputting an elephant).\n\\end{parag}\n\n\\begin{parag}{Dimensionality reduction}\n Dimensionality reduction has two main advantages.", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_1", "text": "\\begin{parag}{Dimensionality reduction}\n Dimensionality reduction has two main advantages. \n\n The first one is that it allows to decrease the dimension of our data, which typically yield a tremendous speed-up while preserving a lot of the precision.\n\n The second one is that, depending on the model, we can also map data back from the lower dimensional space to the higher one. This can be very interesting since it allows to create new samples. For instance, applying dimensionality reduction on a set of human faces, we could create new points $\\bvec{y}$ randomly thanks to the distribution of our data, and map it back to high dimension. That way, we created a new random face. Another use for this is to denoise the data: mapping a sample to a lower dimensional space and back to high dimensions may result to a lot less noise.\n\\end{parag}\n\n\\begin{parag}{Notations}\n We consider the following notation throughout this course, with some slight exceptions when specified otherwise. $N$ is the number of samples we have, $D$ is the dimensionality (the number of components) of any input, and $C$ is the dimensionality of any output.", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_2", "text": "Without specified otherwise, the input $\\bvec{x}_i \\in \\mathbb{R}^{D+1}$ begins with a constant $1$ to account for a bias, followed by the input data. We let $\\bvec{y}_i \\in \\mathbb{R}^C$ to be the $i$\\Th output. $x_i^{\\left(k\\right)}$ is the $k$\\Th component of the $i$\\Th input, and similarly for $y_i^{\\left(k\\right)}$. To sum up, we have:\n \\[\\bvec{x}_i = \\begin{pmatrix} 1 \\\\ x_i^{\\left(1\\right)} \\\\ \\vdots \\\\ x_i^{\\left(D\\right)} \\end{pmatrix} \\in \\mathbb{R}^{D+1}, \\mathspace \\bvec{y}_i = \\begin{pmatrix} y_i^{\\left(1\\right)} \\\\ \\vdots \\\\ y_i^{\\left(C\\right)} \\end{pmatrix} \\in \\mathbb{R}^{C}\\]", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_3", "text": "Any value output by our model will be represented by a $\\hat{\\bvec{y}} \\in \\mathbb{R}^{C}$, to make a difference with the groundtruth $\\bvec{y} \\in \\mathbb{R}^C$.\n \n We will also need weights, which represent the parameters of our model. $\\bvec{w}_{\\left(i\\right)}$ represents the weight to convert any $\\bvec{x}$ to the $i$\\Th component of the output $y_i$. We need their size to compute dot products with $\\bvec{x}$, i.e. $\\bvec{x}_i^T \\bvec{w}_{\\left(i\\right)}$ must make sense. Thus, without specified otherwise, we use $\\bvec{w}_{\\left(i\\right)} \\in \\mathbb{R}^{\\left(D+1\\right)}$.", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_4", "text": "To simplify the notation and the computations, we will stack our values in matrices. Thus, we let: \n \\[X = \\begin{pmatrix} \\bvec{x}_1^T \\\\ \\vdots \\\\ \\bvec{x}_N^T \\end{pmatrix} = \\begin{pmatrix} 1 & \\cdots & x_1^{\\left(D\\right)} \\\\ \\vdots & \\ddots & \\vdots \\\\ 1 & \\cdots & x_N^{\\left(D\\right)} \\end{pmatrix} \\in \\mathbb{R}^{N \\times \\left(D+1\\right)}\\]\n \\[Y = \\begin{pmatrix} \\bvec{y}_1^T \\\\ \\vdots \\\\ \\bvec{y}_N^T \\end{pmatrix} = \\begin{pmatrix} y_1^{\\left(1\\right)} & \\cdots & y_1^{\\left(C\\right)} \\\\ \\vdots & \\ddots & \\vdots \\\\ y_N^{\\left(1\\right)} & \\cdots & y_n^{\\left(C\\right)} \\end{pmatrix} \\in \\mathbb{R}^{N \\times C}\\]\n \\[W = \\begin{pmatrix} \\bvec{w}_{\\left(1\\right)} & \\hdots & \\bvec{w}_{\\left(C\\right)} \\end{pmatrix} \\in \\mathbb{R}^{\\left(D+1\\right)\\times C}\\]\n\\end{parag}", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_5", "text": "\\begin{parag}{Feature expansion}\n Increasing the amount of dimensions from $D$ to $F$ of our input data may help our models (using non-linear functions, since they would be of no help). Thus, we may let the following function: \n \\[\\phi\\left(\\bvec{x}\\right) = \\begin{pmatrix} 1 & x^{\\left(1\\right)} & \\cdots & x^{\\left(D\\right)} & \\left(x^{\\left(1\\right)}\\right)^2 & \\cdots & \\left(x^{\\left(D\\right)}\\right)^2 & \\cdots \\end{pmatrix}^T \\in \\mathbb{R}^{F}\\]\n \n We also put it in a matrix so simplify notation: \n \\[\\Phi = \\begin{pmatrix} \\phi\\left(\\bvec{x}_1\\right)^T \\\\ \\vdots \\\\ \\phi\\left(\\bvec{x}_N\\right)^T \\end{pmatrix} \\in \\mathbb{R}^{N \\times F} \\]\n \n We can replace $\\bvec{x}$ by $\\phi\\left(\\bvec{x}\\right)$ and $X$ by $\\Phi$ in mostly every model, especially the ones which will be kernalised. Note that, in this case, we need $\\bvec{w} \\in \\mathbb{R}^F$, and thus $W = \\mathbb{R}^{F \\times C}$.\n\n\\begin{subparag}{Remark}\n The 1 we added to the input data to account for the bias is some kind of feature expansion.\n \\end{subparag}\n\n\\begin{subparag}{Cover's Theorem}\n Cover's theorem states (more or less) that doing non-linear feature expansion, then it is more likely for our data to be linearly separable.\n \\end{subparag}\n\\end{parag}", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_6", "text": "\\begin{subparag}{Cover's Theorem}\n Cover's theorem states (more or less) that doing non-linear feature expansion, then it is more likely for our data to be linearly separable.\n \\end{subparag}\n\\end{parag}\n\n\\begin{parag}{Kernel}\n We can notice that defining our $\\phi$ functions for feature expansion can be really tedious. However, since most of our methods depend on a dot product of $\\phi\\left(\\bvec{x}_i\\right)^T \\phi\\left(\\bvec{x}_j\\right)$, which gives some kind of measure of similarity between $\\bvec{x}_i$ and $\\bvec{x}_j$ (since it is proportional to the cosine of their angle), we can define a similarity function named a \\important{kernel}, such that: \n \\[k\\left(\\bvec{x}_i, \\bvec{x}_j\\right) = \\phi\\left(\\bvec{x}_i\\right)^T \\phi\\left(\\bvec{x}_j\\right)\\]\n\n As usual, we put everything in vectors and matrices to simplify our notation. We first have a way to measure the similarity between a sample $\\bvec{x}_i$ and all the other samples:\n \\[k\\left(X, \\bvec{x}_i\\right) = \\begin{pmatrix} k\\left(\\bvec{x}_1, \\bvec{x}_i\\right) \\\\ \\vdots \\\\ k\\left(\\bvec{x}_n, \\bvec{x}_i\\right) \\end{pmatrix} \\in \\mathbb{R}^N\\]", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_7", "text": "And we can then stack all of them in a matrix to define similarity between all pairs of samples: \n \\[K = \\begin{pmatrix} k\\left(\\bvec{x}_1, \\bvec{x}_1\\right) & k\\left(\\bvec{x}_1, \\bvec{x}_2\\right) & \\cdots & k\\left(\\bvec{x}_1, \\bvec{x}_N\\right) \\\\ k\\left(\\bvec{x}_2, \\bvec{x}_1\\right) & k\\left(\\bvec{x}_2, \\bvec{x}_2\\right) & \\cdots & k\\left(\\bvec{x}_2, \\bvec{x}_N\\right) \\\\ \\vdots & \\vdots & \\ddots & \\vdots \\\\ k\\left(\\bvec{x}_N, \\bvec{x}_1\\right) & k\\left(\\bvec{x}_N, \\bvec{x}_2\\right) & \\cdots & k\\left(\\bvec{x}_N, \\bvec{x}_N\\right)\\end{pmatrix} \\in \\mathbb{R}^{N \\times N}\\]\n \n Note that, since $k\\left(\\bvec{x}_i, \\bvec{x}_j\\right) = k\\left(\\bvec{x}_j, \\bvec{x}_i\\right)$ by the commutativity of the dot product, $K$ is symmetric ($K^T = K$).\n\n\\begin{subparag}{Remark}\n The main advantage of a kernel is that we don't need to know what function $\\phi$ is linked to it.\n \\end{subparag}", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_8", "text": "\\begin{subparag}{Remark}\n The main advantage of a kernel is that we don't need to know what function $\\phi$ is linked to it.\n \\end{subparag}\n\n\\begin{subparag}{Examples}\n We can for instance use the \\important{polynomial kernel}: \n \\[k\\left(\\bvec{x}_i, \\bvec{x}_j\\right) = \\left(\\bvec{x}_i^T \\bvec{x}_j + c\\right)^d\\]\n \n $c$ is often set to $1$ and $d$ to 2. For this kernel, the corresponding mapping $\\phi$ is known. This is, except for multiplicative constants, all the possible monomials of degree less than or equal to $d$ composed of the components of $\\bvec{x}_i$ and $\\bvec{x}_j$.\n\n We can also use the \\important{Gaussian kernel} (or radial basis function (RBF)): \n \\[k\\left(\\bvec{x}_i, \\bvec{x}_j\\right) = \\exp\\left(- \\frac{\\left\\|\\bvec{x}_i - \\bvec{x}_j\\right\\|^2}{2 \\sigma ^2}\\right)\\]\n \n $\\sigma$ is typically chosen relatively to the data.\n \\end{subparag}\n\\end{parag}\n\n\\begin{parag}{Representer theorem}\n The minimizer of a regularized empirical risk function can be represented as a linear combination of expanded features. In other words, we can write: \n \\[\\bvec{w}^* = \\sum_{i=1}^{N} \\alpha_i^* \\phi\\left(\\bvec{x}_i\\right) = \\Phi^T \\bvec{\\alpha}^*\\]\n where $\\bvec{\\alpha} \\in \\mathbb{R}^N$.", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_9", "text": "\\begin{subparag}{Remark}\n This theorem is really important to kernalise algorithms. When using it, the goal is to get rid of the $\\Phi$ since we do not know the mapping $\\phi$. Switching our view onto variables $\\bvec{\\alpha}$ instead of variables $\\bvec{w}$ is typically a way to do so.\n \\end{subparag}\n\\end{parag}\n\n\\begin{parag}{Loss function}\n The \\important{loss function} $\\ell \\left(\\hat{\\bvec{y}}_i, \\bvec{y}_i\\right)$ computes an error value between the prediction and the true value.\n\n This is a measure of the error for any given prediction.\n\\end{parag}\n\n\\begin{parag}{Empirical risk}\n Given $N$ training samples $\\left\\{\\left(\\hat{\\bvec{x}}_i, \\hat{\\bvec{y}}_i\\right)\\right\\}$, the \\important{empirical risk} is defined as: \n \\[R\\left(\\left\\{\\hat{\\bvec{x}}_i\\right\\}, \\left\\{\\hat{\\bvec{y}}_i\\right\\}, W\\right) = \\frac{1}{N} \\sum_{i=1}^{N} \\ell \\left(\\hat{\\bvec{y}}_i, \\bvec{y}_i\\right)\\]\n\n This represents the global error on the training samples, this is what we try to minimise: \n \\[W^* = \\argmin_{W} R\\left(\\left\\{\\hat{\\bvec{x}}_i\\right\\}, \\left\\{\\hat{\\bvec{y}}_i\\right\\}, W\\right)\\]", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_10", "text": "\\begin{subparag}{Regularised}\n Sometimes, we want to regularise our objective function, so that we prevent weights to become to large and make a lot of overfitting. We then instead seek to minimise. \n \\[E\\left(W\\right) = R\\left(W\\right) + \\lambda E_W\\left(W\\right)\\]\n where $\\lambda$ is an hyperparameter and $E_W\\left(W\\right)$ is the regulariser.\n\n A regulariser that can be used is Tikhonov regularisation, and it will be used in ridge regressionb: \n \\[E_W\\left(W\\right) = \\left\\|W\\right\\|^2_F\\]\n where $\\left\\|W\\right\\|_F^2$ is the squared Frobenius norm of $W$, meaning the sum of its values squared.\n \\end{subparag}\n\\end{parag}\n\n\\begin{parag}{Gradient descent}\n The goal of \\important{gradient descent} is to minimise a function (an empirical risk $R\\left(W\\right)$ in this context). The idea is to begin with an estimate $W_0$ (typically completely random), and then to update it iteratively, by following the direction of greatest decrease (the opposite of the gradient):\n \\[W_k = W_{k-1} - \\eta \\nabla_{W} R\\left(W_{k-1}\\right)\\]\n where $\\eta > 0$ is the learning rate. \n\n This algorithm can then be stopped after the change in the function reaches a threshold $\\left|R\\left(W_{k}\\right) - R\\left(W_{k-1}\\right)\\right| < \\delta_R$, the change in parameter value is less than a threshold $\\left|W_{k} - W_{k-1}\\right| < \\delta_w$, or if a maximum number of iterations (also known as epochs) is reached (even though this gives no guarantee on a potential convergence).", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_11", "text": "\\begin{subparag}{Remark}\n This algorithm does not always converge and, when it does, not necessarily to a minimum nor to the global minimum.\n \\end{subparag}\n\n\\begin{parag}{Evaluation metrics}\n Once a supervised machine learning model is trained, we want to be able to understand how well it performs on unseen test data (which must absolutely be separated from the train data). \n\n We could use the loss function, but we may also use a different one.\n\n For regression, we typically use \\important{Mean Squared Error} (MSE): \n \\[\\text{MSE} = \\frac{1}{N_t} \\sum_{i=1}^{N_t} \\left\\|\\hat{\\bvec{y}}_i - \\bvec{y}_i\\right\\|^2\\]\n where $N_t$ is the number of test sample.\n\n For classification, this is typically more complicated. Defining $TP$ to be the number true positive predictions (samples correctly predicted positive), $FN$ to be the number of false negative predictions (samples incorrectly predicted negative), and similarly for $TN$ and $FP$, we can define the \\important{accuracy} (the percentage of correctly classified samples), the \\important{precision} (the percentage of samples classified positives, which are truly positives) and the \\important{recall} (the percentage of positive samples that are correctly classified as positives): \n \\[\\text{accuracy} = \\frac{TP + TN}{N_t}, \\mathspace \\text{precision} = \\frac{TP}{TP + FP}, \\mathspace \\text{recall} = \\frac{TP}{P}\\]\n \n We can then combine the two last to make a \\important{F1 score}: \n \\[\\text{F1} = 2 \\frac{\\text{precision} \\cdot \\text{recall}}{\\text{precision} + \\text{recall}}\\]\n \n We can typically then use accuracy and the F1 score to see how well our classification model did.", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_12", "text": "\\begin{subparag}{Remark}\n There are many more metrics for regression and classification. For the former, we could quote RMSE (root mean squared error), MAE (mean absolute error) or the MAPE (mean absolute percentage error). For the latter, making a confusion matrix or computing the AUC (area under the ROC curve) can give good insights too.\n \\end{subparag}\n\\end{parag}\n\n\\begin{parag}{Decision boundary}\n A classifier leads to a decision boundary. This is an object of dimension $D-1$ (a line if our data lies on a plane for instance), which splits the space into two regions: one where samples are considered positive (the predicted value is closer to the value of positive samples), and one where they are considered to be negative (the predicted value is closer to the value of negative samples).\n\n Since it is the set of points which are equivalently distant to both labels, the boundary is parametrised by the following equation: \n \\[\\hat{\\bvec{y}}\\left(\\bvec{x}\\right) = \\frac{y_{pos} + y_{neg}}{2}\\]\n where $y_{pos}$ is the value for positive samples, and $y_{neg}$ is the value for negative samples.\n\n\\begin{subparag}{Remark}\n A classifier is said to be linear if its decision boundary is an hyperplane (a straight line if the data lies on a plane for instance).\n \\end{subparag}\n\\end{parag}\n\n\\begin{parag}{Margin}\n If $C = 1$, the orthogonal distance between the decision boundary and the nearest sample is called the margin.\n \\imagehere[0.5]{margin.png}\n\\end{parag}\n\n\\begin{parag}{Overfitting}\n When we increase the complexity of the model (by changing the hyperparameters) we get better and better result for both training and test data. However, there is a point at which increasing the complexity keeps decreasing error on training data but increases the error on test data. This is a very general phenomenon, named \\important{overfitting}.\n\\end{parag}", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_13", "text": "\\begin{parag}{Cross-validation}\n Cross-validation is a way to find good hyperparameters that prevent overfitting. We test different models (in a predefined set), assign to each of them an error value, and pick the one yielding the smallest error.\n\n The idea of \\important{$k$-fold cross-validation} is, to evaluate the error of a model, to first randomly split the dataset into $k$ partitions (where $k$ is predefined). Then, we do $k$ iterations: at iteration $i$ we drop the $i$\\Th partition, train the model on the $\\left(k-1\\right)$ other folds merged, and use the $i$\\Th partition to compute the error. At the end of all the iterations, we average all the errors.\n\n Note that we never test the model on data we used to train it, allowing to avoid overfitting. Also, we use all of our training data to get a complete insight over it (doing only one iteration, would give less information about the model). It is important to notice that the larger the $k$, the less data we waste but the more expensive this method becomes.\n\n\\begin{subparag}{Remark}\n Note that leaving $k = N$ (where $N$ is the number of training samples) is also sometimes named \\important{leave-one-out cross-validation}. This is really expensive but wastes (almost) no data.\n\n Another way to do cross-validation, which is much cheaper, is to split our training data into training and validation data, using \\important{validation-set cross-validation}. This is like like doing only one iteration of $k$-fold cross-validation, meaning that it is very cheap but wastes a lot of data.\n \\end{subparag}\n\\end{parag}\n\n\\begin{parag}{Data representation}\n All the models we will see only work for fixed size data. If we want to handle data of varying size (such as text or pictures), a good way is to consider \\important{bag of words}: consider the number of times each word from a dictionary appears in the text and put this as a big vector.", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_14", "text": "Note that we don't need to consider the whole English dictionary, only picking the set of words appearing in the training data is enough. Also, it is often interesting to make a histogram out of our vector: divide it by the number of words in the sample, so that we instead have a repartition and give less importance to the length of the text.\n\n Also, we can apply this to images. To do so, we need to extract ``words'', meaning fixed-size picture elements.\n\\end{parag}\n\n\\begin{parag}{Pre-processing}\n The training data might have problems: it might have noise (because of measurement errors), incorrect values, and so on. To fix those, a good idea is to do pre-processing.\n\n\\begin{subparag}{Normalisation}\n To begin with, a good idea is to scale each individual feature dimension to fall within a specified range (so that we don't give more impact to a dimension ranging from 1000 to 10000 than to another dimension ranging from 0 to 1). This can typically be done by using \\important{normalisation}, such as $z$-score normalisation: \n \\[\\widetilde{x}_i^{\\left(d\\right)} = \\frac{x_i^{\\left(d\\right)} - \\mu^{\\left(d\\right)}}{\\sigma^{\\left(d\\right)}}\\]\n where $\\mu^{\\left(d\\right)}$ is the mean of the $d$\\Th dimension, and $\\sigma^{\\left(d\\right)}$ is its standard deviation.", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_15", "text": "Note that there are many other ways to do normalisation, such as min-max normalisation (computing $\\widetilde{x}_i = \\frac{x_i - x_{min}}{x_{max} - x_{min}}$), max-normalisation (computing $\\widetilde{x}_i = \\frac{x_i}{x_{max}}$ where $x_{max}$ is the maximum in absolute value) or decimal-scaling (compute $\\widetilde{x}_i = \\frac{x_i}{10^k}$ where $k$ is the smallest integer $k$ such that $\\left|\\widetilde{x}_i\\right| \\leq 1$ for the largest $\\widetilde{x}_i$).\n \\end{subparag}\n\n\\begin{subparag}{Imbalanced data}\n Another important thing to consider is \\important{imbalanced data}. There might be 10 times as much data in one class as in another (between-class imbalance), or data inside a class might have a lot of representative at some point in space and much less at other points (within-class imbalance).\n\n To fix those problems, we can either work on the data, or on the cost function.\n\n To work on the data, the first method is to decrease the data set by \\important{undersampling}: we remove samples randomly, or more intelligently by removing samples considered redondant. The second method is to increase the data set by \\important{oversampling}: we replicate exactly some of the samples (which might lead to overfitting), or do interpolation (which might create a lot of noise).\n\n To work on the cost function, we can give more impact to some samples: \n \\[R\\left(\\left\\{\\bvec{x}_i\\right\\}, \\left\\{\\bvec{y}_i\\right\\}, W\\right) = \\frac{1}{N} \\sum_{i=1}^{N} \\beta_i \\ell \\left(\\hat{\\bvec{y}}_i, \\bvec{y}_i\\right)\\]", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_16", "text": "We can for instance use weights inversely proportional to the class frequency, such as $\\beta_i = 1 - \\frac{N_k}{N}$.\n \\end{subparag}\n\\end{parag}\n\n\\begin{parag}{Ridge regression}\n The output of \\important{ridge regression} is computed by a simple dot product: \n \\[\\hat{\\bvec{y}}_i = W^T \\phi\\left(\\bvec{x}_i\\right)\\]\n\n The training objective function we want to minimise is a squared Euclidean distance regularised by the sum of squares of the weights:\n \\[E\\left(W\\right) = R\\left(W\\right) + \\lambda E_W \\left(W\\right) = \\sum_{i=1}^{N} \\left\\|\\hat{\\bvec{y}}_i - \\bvec{y}_i\\right\\|^2 + \\lambda \\left\\|W\\right\\|^2_F\\]\n where $\\left\\|W\\right\\|^2_F$ is the Frobenius norm of $W$, meaning the sum of the square of all its values, and $\\lambda \\geq 0$ is a hyperparameter.\n\n This can be solved explicitly: \n \\[W^* = \\left(\\Phi^T \\Phi + \\lambda I_F\\right)^{-1} \\Phi^T Y\\]\n where $I_F$ is the $F \\times F$ identity matrix.\n\n\\begin{subparag}{Linear regression}\n Leaving $\\lambda = 0$, we get the special case of \\important{linear regression}. Then, the closed-form formula can be rephrased as: \n \\[W^* = \\left(\\Phi^T \\Phi\\right)^{-1} \\Phi^T \\bvec{y} = \\Phi^{\\dagger} Y\\]\n where $\\Phi^{\\dagger}$ is the Moore-Penrose pseudo-inverse.\n \\end{subparag}", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_17", "text": "\\begin{subparag}{Kernelisation}\n Using the representer theorem, we can find that: \n \\[A^* = \\left(K + \\lambda I_N\\right)^{-1} Y\\]\n \n This is not of much use on its own, but we can use this result to find how we predict a value $\\hat{\\bvec{y}}$ for a new $\\bvec{x}$: \n \\[\\hat{\\bvec{y}} = Y^T \\left(K + \\lambda I_N\\right)^{-1} k\\left(X, \\bvec{x}\\right)\\]\n \n Note that the value $Y^T \\left(K + \\lambda I_N\\right)^{-1}$ can be computed once during training, and then be reused at every evaluation of the model.\n \\end{subparag}\n\n\\begin{subparag}{Classification}\n Ridge regression can be used for classification tasks, by inputting the result into a softmax function (defined right after), but this does not work very well because we are not encoding this in the objective function.\n\n This is named a \\important{least-square classifier}.\n \\end{subparag}\n\\end{parag}\n\n\\begin{parag}{Logistic regression}\n In \\important{logistic regression} (which is a linear classification algorithm), we consider negative samples to be $y_i = 0$.\n\n The output of logistic regression is computed by using the \\important{softmax function}: \n \\[\\hat{y}^{\\left(k\\right)} = \\frac{\\exp\\left(\\bvec{w}^T_{\\left(k\\right)} \\bvec{x}\\right)}{\\sum_{j=1}^{C} \\exp\\left(\\bvec{w}^T_{\\left(j\\right)} \\bvec{x}\\right)} \\in \\left[0, 1\\right]\\]\n where $\\hat{y}^{\\left(k\\right)}$ represents the probability that the sample $\\bvec{x}$ is in class $k$.", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_18", "text": "The empirical risk we try to minimise is the \\important{cross entropy}: \n \\[R\\left(W\\right) = - \\sum_{i=1}^{N} \\sum_{k=1}^{C} y_i^{\\left(k\\right)} \\ln\\left(\\hat{y}_i^{\\left(k\\right)}\\right)\\]\n \n Note that $y_i^{\\left(k\\right)}$ is non-zero only for a single sample $i$.\n\n There is no closed-form formula, so we need the gradient of the empirical risk in order to do gradient descent (the solution is unique since the function is convex):\n \\[\\nabla_{W} R\\left(W\\right) = \\sum_{i=1}^{N} \\bvec{x}_i \\left(\\hat{\\bvec{y}}_i - \\bvec{y}_i\\right)^T\\]\n\n\\begin{subparag}{One dimension}\n Let's consider $C = 1$. This special case of the softmax function is named the \\important{logistic function}:\n \\[\\hat{y}\\left(\\bvec{x}\\right) = \\sigma\\left(\\bvec{w}^T \\bvec{x}\\right) = \\frac{1}{1 + \\exp\\left(- \\bvec{w}^T \\bvec{x}\\right)}\\]\n\n One-dimensional cross-entropy can be rewritten as: \n \\[R\\left(\\bvec{w}\\right) = -\\sum_{i \\in P}^{} \\ln\\left(\\hat{y}_i\\left(\\bvec{w}\\right)\\right) - \\sum_{i \\in N}^{} \\ln\\left(1 - \\hat{y}_i\\left(\\bvec{w}\\right)\\right)\\]\n where $P$ is the set of positive samples and $N$ the set of negative samples.", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_19", "text": "The gradient of one-dimensional cross-entropy is: \n \\[\\nabla_{\\bvec{w}} R\\left(\\bvec{w}\\right) = \\sum_{i=1}^{N} \\left(\\hat{y}_i - y_i\\right)\\bvec{x}_i\\]\n \\end{subparag}\n\n\\begin{subparag}{Kernelisation}\n This algorithm can be kernalised even though this is not very common.\n \\end{subparag}\n\\end{parag}\n\n\\begin{parag}{Support vector machine}\n In \\important{support vector machine} (SVM) classification (which is also a linear classifier), we consider negative samples to be $y_i = -1$. Also, we leave $\\widetilde{\\bvec{w}}$ to be the vector of parameters without $w^{\\left(0\\right)}$, and $\\bvec{x} \\in \\mathbb{R}^D$ to not have an added 1. Note that we only consider $C = 1$ for now.\n\n The idea of SVM is to maximise the size of the margin. A prediction is given by whether $w^{\\left(0\\right)} + \\widetilde{\\bvec{w}}^T \\bvec{x}_i$ is closest to $-1$ or $1$ (as always for linear classifiers), and the decision boundary is given by $w^{\\left(0\\right)} + \\widetilde{\\bvec{w}}^T \\bvec{x} = \\frac{-1 + 1}{2} = 0$.", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_20", "text": "We show in one of the following paragraphs that the problem can be formulated as:\n \\[\\argmin_{\\bvec{w}, \\left\\{\\xi_i\\right\\}} \\frac{1}{2} \\left\\|\\widetilde{\\bvec{w}}\\right\\|^2 + C \\sum_{i=1}^{N} \\xi_i, \\]\n subject to $y_i \\left(w^{\\left(0\\right)} + \\widetilde{\\bvec{w}}^T \\bvec{x}_i\\right) \\geq 1 - \\xi_i$ and $\\xi_i \\geq 0$, for all $i$. Note that $C$ is an hyperparameter, and the $\\xi_i$ are \\important{slack variables} we added that we also minimise. Those slack variables allow for data which is not linearly separable data to also be used.\n\n When $\\xi_i = 0$, the point is on the correct side of the margin, this is how everything should work. If $0 < \\xi_i < 1$, then the point $i$ are on the wrong side of the margin, but correctly classified. If $\\xi_i = 1$, the point is on the decision boundary (and thus misclassified). If $\\xi_i > 1$, then the point is on the wrong side of the decision boundary, and thus misclassified.\n\n This representation of the problem is known as the \\important{primal problem}.\n\n\\begin{subparag}{Support vectors}\n We notice that, for the margin to be maximised, there must be at least a point from each class lying on it. Such points are named \\important{support vectors}. \n \\end{subparag}", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_21", "text": "\\begin{subparag}{Support vectors}\n We notice that, for the margin to be maximised, there must be at least a point from each class lying on it. Such points are named \\important{support vectors}. \n \\end{subparag}\n\n\\begin{subparag}{Hinge loss}\n By rewriting the constraints, we get: \n \\[\\xi_i \\geq 1 - y_i\\left(w^{\\left(0\\right)} + \\widetilde{\\bvec{w}}^T \\bvec{x}_i\\right)\\]\n \n For samples $i$ that satisfy the support vector machine problem (they are on not in the margin nor misclassified), we have $\\xi_i = 0$ (since they are forced to be non-negative). For samples $i$ which don't, this inequality is held. This allows us to rewrite our SVM primal problem as: \n \\[\\argmin_{\\bvec{w}, \\left\\{\\xi_i\\right\\}} \\frac{1}{2C} \\left\\|\\widetilde{\\bvec{w}}\\right\\|^2 + \\sum_{i=1}^{N} \\max\\left(0, 1 - y_i\\left(w^{\\left(0\\right)} + \\widetilde{\\bvec{w}}^T \\bvec{x}_i\\right)\\right),\\]\n subject to the same conditions\n \n This new term is called the \\important{hinge loss}: \n \\[\\max\\left(0, 1 - y_i \\left(w^{\\left(0\\right)} + \\widetilde{\\bvec{w}}^T \\bvec{x}_i\\right)\\right)\\]\n \\end{subparag}", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_22", "text": "\\begin{subparag}{Dual problem}\n We can reformulate our problem by letting one variable per training sample (meaning that we have $N$ variables instead of $\\left(D+1\\right)$): \n \\[\\argmax_{\\left\\{\\alpha_i\\right\\}} \\left(\\sum_{i=1}^{N} \\alpha_i - \\frac{1}{2} \\sum_{i=1}^{N} \\sum_{j=1}^{N} \\alpha_i \\alpha_j y_i y_j \\bvec{x}_i^T \\bvec{x}_j\\right),\\]\n subject to $\\sum_{i=1}^{N} \\alpha_i y_i = 0$ and $0 \\leq \\alpha_i \\leq C$ for all $i$.\n \n The solution is equivalent to the primal problem: \n \\[\\widetilde{\\bvec{w}}^* = \\sum_{i=1}^{N} \\alpha_i^* y_i \\bvec{x}_i \\implies \\hat{y}\\left(\\bvec{x}\\right) = w^{\\left(0\\right)*} + \\sum_{i = 1}^{N} \\alpha_i y_i \\bvec{x}_i^T \\bvec{x}\\]\n \n Note that $w^{\\left(0\\right)}$ can also be found thanks to those $\\alpha_i^*$.\n\n We can show that, at the solution, we have: \n \\[\\alpha_i^* \\left(y_i \\left(w^{\\left(0\\right)*} + \\widetilde{\\bvec{w}}^{*T} \\bvec{x}_i\\right) - 1 + \\xi_i^*\\right) = 0\\]", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_23", "text": "In other words, for every sample, either of those terms is equal to 0. The samples for which $\\alpha_i^* \\neq 0$ are the support vectors. However, most samples are not support vectors, so we can decrease the computations by leaving $\\mathcal{S}$ to be the set of support vectors: \n \\[\\hat{y}\\left(\\bvec{x}\\right) = w^{\\left(0\\right)*} + \\sum_{i \\in\\mathcal{S}}^{} \\alpha_i^* y_i \\bvec{x}_i^T \\bvec{x}_i\\]\n \\end{subparag}\n\n\\begin{subparag}{Kernelisation}\n We need the dual problem to kernelise the SVM: \n \\[\\argmax_{\\left\\{\\alpha_i\\right\\}} \\left(\\sum_{i=1}^{N} \\alpha_i - \\frac{1}{2} \\sum_{i=1}^{N} \\sum_{j=1}^{N} \\alpha_i \\alpha_j y_i y_j k\\left(\\bvec{x}_i, \\bvec{x}_j\\right)\\right),\\]\n subject to $\\sum_{i=1}^{N} \\alpha_i y_i = 0$ and $0 \\leq \\alpha_i \\leq C$ for all $i$.\n\n The prediction is also similar: \n \\[\\hat{y}\\left(\\bvec{x}\\right) = w^{\\left(0\\right)*} + \\sum_{i \\in \\mathcal{S}}^{} \\alpha_i^* y_i k\\left(\\bvec{x}_i, \\bvec{x}\\right)\\]\n\n Note that we still have $\\alpha_i^* = 0$ for all samples that are not support vectors.\n \\end{subparag}", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_24", "text": "Note that we still have $\\alpha_i^* = 0$ for all samples that are not support vectors.\n \\end{subparag}\n\n\\begin{subparag}{Multi-class SVM}\n To generalise our algorithm to multiple class, we can use multiple ways. The idea is always to use several binary classifiers.\n\n One way is to use \\important{one-vs-rest}: we train classifiers stating if the component is in class $i$ or not. Another way is to use \\important{one-vs-one}: we train classifiers to know if the component is closer to class $i$ or $j$, and then pick the best one. However, in both cases, there are some samples which will give ambiguous answers (belonging to multiple classes or to none).\n \\end{subparag}\n\n\\begin{subparag}{Primal derivation}\n Let's consider how we got the formula for the primal model, since it may typically help to remember and understand it.\n\n First, we know that any two points on the decision boundary have the same prediction (which is 0), which yields that: \n \\[0 = \\hat{y}_1 - \\hat{y}_2 = \\left(w^{\\left(0\\right)} + \\widetilde{\\bvec{w}}^T \\bvec{x}_1\\right) - \\left(w^{\\left(0\\right)} + \\widetilde{\\bvec{w}}^T \\bvec{x}_2\\right) = \\widetilde{\\bvec{w}}^T \\left(\\bvec{x}_1 - \\bvec{x}_2\\right)\\]\n\n This is a dot product, and it thus means that $\\widetilde{\\bvec{w}}$ is orthogonal to the decision boundary.", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_25", "text": "This is a dot product, and it thus means that $\\widetilde{\\bvec{w}}$ is orthogonal to the decision boundary. \n\n Second, we use this information to split any point into a component colinear to $\\widetilde{\\bvec{w}}$, and one orthogonal to it (meaning colinear to the decision boundary): \n \\[\\bvec{x} = \\bvec{x}_{\\perp} + r \\frac{\\widetilde{\\bvec{w}}}{\\left\\|\\widetilde{\\bvec{w}}\\right\\|}\\]\n where $r$ is the signed orthogonal distance of any point to the decision boundary.\n\n Now, looking at the prediction yielded by this point, we get: \n \\[\\hat{y}\\left(\\bvec{x}\\right) = w^{\\left(0\\right)} + \\widetilde{\\bvec{w}}^T \\bvec{x} = w^{\\left(0\\right)} + \\widetilde{\\bvec{w}}^T \\bvec{x}_{\\perp} + r \\frac{\\widetilde{\\bvec{w}}^T \\widetilde{\\bvec{w}}}{\\left\\|\\widetilde{\\bvec{w}}\\right\\|} = y\\left(\\bvec{x}_{\\perp}\\right) + r\\left\\|\\widetilde{\\bvec{w}}\\right\\|\\]", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_26", "text": "However, we know that $y\\left(\\bvec{x}_{\\perp}\\right) = 0$ since it is on the decision boundary, meaning that: \n \\[\\hat{y}\\left(\\bvec{x}\\right) = r\\left\\|\\widetilde{\\bvec{w}}\\right\\| \\iff r = \\frac{\\hat{y}\\left(\\bvec{x}\\right)}{\\left\\|\\widetilde{\\bvec{w}}\\right\\|}\\]\n which is, to recall, the signed orthogonal distance of $\\bvec{x}$ to the decision boundary.\n\n Note that we can then make use of the ground truth label being $-1$ or 1 to make an unsigned distance: \n \\[\\widetilde{r}_i = y_i r_i = \\frac{y_i\\left(w^{\\left(0\\right)} + \\widetilde{\\bvec{w}}^T \\bvec{x}_i\\right)}{\\left\\|\\widetilde{\\bvec{w}}\\right\\|}\\]\n\n This is what we would like to maximise, but it is hard. We thus need to turn it to an equivalent problem. To do so, we notice that there is an infinite number of solutions that matter, since we only want the direction of $\\bvec{w}$ and that its magnitude does not matters. This can be proven mathematically by multiplying our weights by a $\\lambda$, and seeing that we get the same $\\widetilde{r}_i$ above (the $\\lambda$ cancel out in the fraction).", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_27", "text": "So, we may as well require that the margin has size $\\frac{1}{\\left\\|\\widetilde{\\bvec{w}}\\right\\|}$, meaning that: \n \\[r_i \\geq \\frac{1}{\\left\\|\\widetilde{\\bvec{w}}\\right\\|} \\iff r_i \\left\\|\\widetilde{\\bvec{w}}\\right\\| \\geq 1 \\iff y_i \\left(w^{\\left(0\\right)} + \\widetilde{\\bvec{w}}^T \\bvec{x}_i\\right) \\geq 1\\]\n\n Now, maximising the margin means maximising $\\frac{1}{\\left\\|\\widetilde{\\bvec{w}}\\right\\|}$ which can be shown to be equivalent to minimising $\\left\\|\\widetilde{\\bvec{w}}\\right\\|^2$. Our problem has thus become: \n \\[\\argmin_{\\bvec{w}} \\frac{1}{2} \\left\\|\\widetilde{\\bvec{w}}\\right\\|^2, \\mathspace \\text{subject to } y_i \\left(w^{\\left(0\\right)} + \\widetilde{\\bvec{w}}^T \\bvec{x}_i\\right) \\geq 1, \\ \\forall i\\]\n where the factor $\\frac{1}{2}$ was added for convenience.\n\n However, if the data is not linearly separable, we need a way to let some samples violate this rule. This is done by adding \\important{slack variables} $\\xi_i \\geq 0$ for each sample: \n \\[y_i \\left(w^{\\left(0\\right)} + \\widetilde{\\bvec{w}} \\bvec{x}_i\\right) \\geq 1 - \\xi_i\\]", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_28", "text": "In other words, if $0 < \\xi_i \\leq 1$, the sample $i$ lies inside in the margin but is still correctly classified. If $\\xi_i \\geq 1$, then the sample $i$ is misclassified. \n\n We minimise those variables jointly with our original problem, giving us our final formulation:\n \\[\\argmin_{\\bvec{w}, \\left\\{\\xi_i\\right\\}} \\frac{1}{2} \\left\\|\\widetilde{\\bvec{w}}\\right\\|^2 + C \\sum_{i=1}^{N} \\xi_i, \\]\n subject to $y_i \\left(w^{\\left(0\\right)} + \\widetilde{\\bvec{w}}^T \\bvec{x}_i\\right) \\geq 1 - \\xi_i$ and $\\xi_i \\geq 0$, for all $i$.\n \\end{subparag}\n\\end{parag}\n\n\\begin{parag}{$K$-nearest neighbours}\n The idea of \\important{$k$-nearest neighbours} (kNN) is to compute the distance between the test sample $\\bvec{x}$ and all training samples $\\left\\{\\bvec{x}_i\\right\\}$ and find the $k$ samples with minimum distances. Then, we can do classification by finding the most common label amongst these $k$ nearest neighbours, or do regression by computing $\\hat{\\bvec{y}}$ as a function of those neighbours and their distance to $\\bvec{x}$.\n\n To compute close points efficiently we can use datastructures such as $k$-d trees.", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_29", "text": "To compute close points efficiently we can use datastructures such as $k$-d trees.\n\n\\begin{subparag}{Remark}\n The result of this model depends on the choice of the distance function. One can take the \\important{Euclidean distance}: \n \\[d\\left(\\bvec{x}_i, \\bvec{x}\\right) = \\sqrt{\\sum_{d=1}^{D} \\left(x_i^{\\left(d\\right)} - x^{\\left(d\\right)}\\right)^2}\\]\n \n However, for some other structures such as histograms (each data point only has component between 0 and 1, and the sum of all of the components of a data point is equal to 1; like for a probability distribution), then we instead tend to use a Chi-square distance: \n \\[d\\left(\\bvec{x}_i, \\bvec{x}\\right) = \\sqrt{\\chi^2 \\left(\\bvec{x}_i, \\bvec{x}\\right)} = \\sqrt{\\sum_{d=1}^{D} \\frac{\\left(x_i^{\\left(d\\right)} - x^{\\left(d\\right)}\\right)^2}{x_i^{\\left(d\\right)} + x^{\\left(d\\right)}}} \\]\n \n The important thing to remember from this paragraph is that the choice of the distance function is important.\n \\end{subparag}\n\n\\begin{subparag}{Curse of dimensionality}\n Because of a principle named the \\important{curse of dimensionality}, we need exponentially more points to cover a space as the number of dimensions increases. Using dimensionality reduction is a good idea with this algorithm.\n \\end{subparag}\n\n\\begin{subparag}{Complexity}\n Unlike most of the other models, increasing the hyperparameter of this model (the $k$) leads to decreased complexity: the higher the $k$, the less complex the decision boundary is and thus the less overfit we have.\n \\end{subparag}\n\\end{parag}", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_30", "text": "\\begin{parag}{Neural networks}\n Neural networks can do both classification and regression (depending on the output representation and the empirical risk used, typically square loss for regression and cross-entropy for classification), and their main advantage is that they learn a good model during the training.\n\n This method is named \\important{neural network}, \\important{multi-layer perceptron} (MLP), or \\important{deep learning} (as long as there are at least two hidden layers).\n\n The idea is to have layers composed of neurons. Every neuron of a layer is connected to every neurons of the previous layer, meaning that its value is computed by a weighted sum over them, plus a bias. More than that, the important thing for our model not to be just a big linear regression, is that each neuron is passed through a non-linear activation function. Mathematically speaking, this is given by: \n \\[\\bvec{z}_{\\left(l \\right)} = f_{\\left(l \\right)} \\left(W_{\\left(l \\right)}^T \\bvec{z}_{\\left(l - 1\\right)}\\right)\\]\n where $\\bvec{z}_{\\left(0\\right)} = \\bvec{x}$ is the input layer, $\\bvec{z}_{\\left(L+1\\right)} = \\hat{\\bvec{y}}$ is the output layer, $L$ is the number of hidden layers (layers which are neither input nor output), and $f_{\\left(l\\right)} \\left(x\\right)$ is applied to every component of the vector. Note that each $\\bvec{z}_{\\ell }$ has a ``bias term'' just like the input data (meaning a 1 appended at the beginning).", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_31", "text": "This is trained to optimality using stochastic gradient descent, by focusing on a single loss term $\\ell \\left(\\hat{\\bvec{y}}_i, \\bvec{y}_i\\right)$ at a time. To do so, we need to compute the gradients $\\frac{\\partial \\ell_i}{\\partial W_l^{\\left(k, j\\right)}}$ (where we are considering the loss of the $i$\\Th sample, and the weight at position $\\left(k, j\\right)$ of the weight matrix from layer $l$). This is done by an algorithm named \\important{backpropagation}.\n\n We can notice that, by the chain rule (and abusing slightly of the notation of the derivative): \n \\[\\frac{\\partial \\ell _i}{\\partial W_l} = \\frac{\\partial z_{\\left(l\\right)}}{\\partial W_{l}} \\frac{\\partial z_{\\left(l+1\\right)}}{\\partial z_l} \\cdots \\frac{\\partial z_{\\left(L\\right)}}{\\partial z_{\\left(L-1\\right)}} \\frac{\\partial \\ell _i}{\\partial z_{\\left(L\\right)}} \\]\n \n We can compute $\\frac{\\partial z_{\\left(l\\right)}}{\\partial z_{\\left(l-1\\right)}} $ and $\\frac{\\partial z_{\\left(l\\right)}}{\\partial W_l} $ rather easily if we store the values of each layer during the forward pass. We can then propagate backwards those value, by computing the gradients from the end and updating the weights.\n\n\\begin{subparag}{Activation functions}\n There are multiple choice for activation functions. The important thing is that they are non-linear.", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_32", "text": "\\begin{subparag}{Activation functions}\n There are multiple choice for activation functions. The important thing is that they are non-linear.\n\n We can for instance take the ReLU (Rectified Linear Unit) activation function: \n \\begin{functionbypart}{f\\left(a\\right)}\n a, & \\text{if } a > 0 \\\\\n 0, & \\text{otherwise}\n \\end{functionbypart}\n \n Another choice could be the sigmoid: \n \\[f\\left(a\\right) = \\frac{1}{1 + \\exp\\left(-a\\right)}\\]\n \\end{subparag}\n\n\\begin{subparag}{Convolutional network}\n When working with pictures, just vectorising them may give a lot of elements while removing the fact that the picture is inherently two-dimensional. A way to circumvent this problem is using convolutions.\n\n To make a convolution, we need a small matrix of elements (plus a bias). We center this matrix at an element, compute the weighted sum resulting from it, add the bias, and use this as our result. We can then shift it to center it on each element, yielding a new picture. This allows to extract some features of our original picture, such as the edges.\n\n We can also use multiple filters to create multiple channels, increasing the amount of data. If at some point we have 3 channels (for instance) and want to do a convolution with a $5\\times5$ filter, then we will use matrices of size $5\\times 5 \\times 3$ (going three-dimensional over our channels). Note that we can also use some other parameters, such as strides (skip one pixels over two for instance) or padding (add some pixels on the edges).\n\n We can also use pooling layers, splitting the pixels into $k \\times k$ squares and extracting only one value per partition (by taking the maximum value or the average for instance). This allows to decrease the size of our pictures.", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_33", "text": "We can also use pooling layers, splitting the pixels into $k \\times k$ squares and extracting only one value per partition (by taking the maximum value or the average for instance). This allows to decrease the size of our pictures.\n\n The main interest then comes from stacking those operations. For instance, beginning with a $28 \\times 28$ input, we may apply convolutions to get a $3 \\times 24 \\times 24$ layer (we loose some pixels because we may require the filter to ignore pixels where it has to be partly outside of the picture) and then a pooling layer to get a $3 \\times 12 \\times12$ layer.\n\n The idea of a \\important{convolutional neural network} (CNN) is to make some convolutions (typically before the input layers, but also in-between some layers). The main idea is that we also optimise the values inside the filters of our convolutions. To compute the gradient for them, we also use back-propagation.\n \\end{subparag}\n\\end{parag}\n\n\\begin{parag}{Principle component analysis}\n Sometimes, we realise that data is given in many dimensions, but actually lies in many less dimensions. The idea of \\important{principle component analysis} (PCA) is to project the data on some orthogonal axis (of lower dimension), in a way to maximise the kept variance.\n\n Leaving $\\bar{\\bvec{x}} = \\frac{1}{N}\\sum_{i=1}^{N} \\bvec{x}_i$ to be the mean, we have: \n \\[\\bvec{y}_i = W^T \\left(\\bvec{x}_i - \\bar{\\bvec{x}}\\right)\\]", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_34", "text": "To find this matrix $W \\in \\mathbb{R}^{D \\times d}$ (where $d$ is the number of dimensions after the projection), we first need to consider the data covariance matrix: \n \\[C = \\frac{1}{N} \\sum_{i=1}^{N} \\left(\\bvec{x}_i - \\bar{\\bvec{x}}\\right) \\left(\\bvec{x}_i - \\bar{\\bvec{x}}\\right)^T\\]\n \n Then, picking the $d$ eigenvectors which highest eigenvalues of $C$, we get our matrix: \n \\[W = \\begin{pmatrix} & & \\\\ \\bvec{w}_{\\left(1\\right)} & \\cdots & \\bvec{w}_{\\left(d\\right)} \\\\ & & \\end{pmatrix} \\in \\mathbb{R}^{D \\times d}\\]\n\n The explained variance yielded by our projection can be found by computing: \n \\[\\text{exval} = \\frac{\\sum_{j=1}^{d} \\lambda_j}{\\sum_{j=1}^{D} \\lambda_j}\\]\n\n Using PCA can make the computations of the models much faster without losing much precision, or getting some insight of the data.\n\n\\begin{subparag}{Remark}\n Since the axis on which we project the data are orthogonal, we have: \n \\[W^T W = I_d\\]\n\n To make sure of this, we need to take the eigenvector so that they are orthogonal. This can always be done because $C$ is symmetric, thanks to the spectral theorem (this theorem also allows to know that we can compare eigenvalues, since they are real).\n \\end{subparag}", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_35", "text": "\\begin{subparag}{Mapping}\n From our computation, we can notice that, for any point $\\bvec{y} \\in\\mathbb{R}^d$, we can move it to the high-dimensional space: \n \\[\\hat{\\bvec{x}} = \\bar{\\bvec{x}} + W \\bvec{y}\\]\n\n This yields all the advantages presented in the first section of this document.\n \\end{subparag}\n\n\\begin{subparag}{Kerenelisation}\n PCA can be kernalised in a non-trivial fashion.\n\n First, we need to account for the fact that our data may not be centered in input-space (this was done above by considering the mean of our input values), letting: \n \\[\\widetilde{K} = K - 1_N K - K 1_N + 1_N K 1_N\\]\n where $1_N$ is an $N \\times N$ matrix, with every element equal to $\\frac{1}{N}$.\n\n Leaving $\\bvec{a}$ to be the vector of unknowns given by the representer theorem, we can find that it follows the following equation: \n \\[\\widetilde{K} \\bvec{a} = \\lambda N \\bvec{a}\\]\n \n This is an eigenvalue problem, which could be solved to find a solution $\\bvec{a}$. From there, we can project our data: \n \\[y_i = \\sum_{j=1}^{N} a_j k\\left(\\bvec{x}_i, \\bvec{x}_j\\right)\\]\n supposing that we want $d = 1$.\n\n If we want $d > 1$, we can again take the $d$ eigenvectors with greatest eigenvalues, and compute each component of $\\bvec{y}_i$ thanks to a different eigenvector.", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_36", "text": "If we want $d > 1$, we can again take the $d$ eigenvectors with greatest eigenvalues, and compute each component of $\\bvec{y}_i$ thanks to a different eigenvector.\n\n Note that we can no longer map data from $d$ dimensions back to $D$ dimensions with the kernalised method (since it would require us to know the feature expansion mapping $\\phi\\left(\\bvec{x}\\right)$).\n \\end{subparag}\n\\end{parag}\n\n\\begin{parag}{Autoencoder}\n Another way to do dimensionality reduction is through a neural network.\n\n The idea is to have a double funnel shaped neural network: an encoder decreasing the dimension, a layer with $d$ neurons, and a decoder increasing back the number of dimensions.\n \\imagehere[0.7]{Autoencoder.png}\n\n We can train it to output the same data as what we input it, using a least square empirical risk. That way, it must learn an intelligent code.\n\n\\begin{subparag}{Remark}\n This can also be used to both do dimensionality reduction and mapping data back from the low-dimensional space.\n \\end{subparag}\n\n\\begin{subparag}{Convolutional autoencoder}\n We can use convolutional neural networks for autoencoders. To do so, we use the inverse functions of those convolutions.\n \\end{subparag}\n\\end{parag}\n\n\\begin{parag}{Fisher linear discriminant analysis}\n Even though \\important{Fisher linear discriminant analysis} (LDA) is a dimensionality reduction algorithm, it is a supervised learning one. Its goal is to project data on lower space, while keeping classes (hence the supervision) clustered. It considers that clustering should be done relatively to compactness, meaning distance between points within a cluster should be small, whereas distances across clusters should be large.", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_37", "text": "The goal is to minimise the distances within a class, meaning the distance of the elements of a class to their class center $E_W\\left(\\bvec{w}\\right)$, while maximising the distance of cluster centers (weighted by the number of elements in the class) $E_B\\left(\\bvec{w}\\right)$. This is better expressed thanks to the within-class scatter matrix $S_W$ and the between-class scatter matrix $S_B$: \n \\[S_W = \\sum_{c=1}^{C} \\sum_{i \\in c}^{} \\left(\\bvec{x}_i - \\bvec{\\mu}_c\\right) \\left(\\bvec{x}_i - \\bvec{\\mu}_c\\right)^T, \\mathspace S_B = \\sum_{c=1}^{C} N_c \\left(\\bvec{\\mu}_c - \\bar{\\bvec{x}}\\right)\\left(\\bvec{\\mu}_c - \\bar{\\bvec{x}}\\right)^T\\]\n where $\\bar{\\bvec{x}}$ si the mean of all the samples, and $\\bvec{\\mu}_c$ the mean of the values in class $c$.\n\n Our problem can be specified as the following generalised eigenvector problem:\n \\[S_B \\bvec{w}_{\\left(1\\right)} = \\lambda_1 S_w \\bvec{w}_{\\left(1\\right)}\\]\n \n As for PCA we are in fact looking for the $d$ eigenvectors with largest eigenvalues. \n\\end{parag}\n\n\\begin{parag}{$K$-means clustering}\n The idea of \\important{$k$-means clustering} is to consider that clustering should also be done relatively to compactness.", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_38", "text": "\\begin{parag}{$K$-means clustering}\n The idea of \\important{$k$-means clustering} is to consider that clustering should also be done relatively to compactness.\n\n To do so, we consider $K$ (an hyperparameter) cluster centers $\\left\\{\\bvec{\\mu}_1, \\ldots, \\bvec{\\mu}_K\\right\\}$. While we are not converged, we assign each data point $\\bvec{x}_i$ to the nearest center $\\bvec{\\mu}_k$, and then move $\\bvec{\\mu}_k$ to the mean of the points that were assigned to it.\n\n This algorithm is guaranteed to converge, even though it may not be to the expected solution: some clusters may end up completely empty. A good way to fix this problem is to run it multiple times with different initialisation, and pick the solution minimising the sum of the distance between each point and their assigned cluster center.\n\n\\begin{subparag}{Hyperparameter}\n To choose the $K$, a good way is to draw a graph of the average within-cluster sum of distances with respect to the number of cluster, and pick a point at its ``elbow'' (where the drop in the $y$-axis becomes less significant).\n \\end{subparag}\n\\end{parag}\n\n\\begin{parag}{Spectral clustering}\n The idea of \\important{spectral clustering} is to consider that clustering should be done relatively to connectivity instead: we group the points based on edges in a graph, and remove some of the edges with longest length. Let us first consider the case where we only want to make 2 clusters, meaning that we only want one cut.", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_39", "text": "To create the graph, we need a way to give weights to edges in order to represent their affinity, a way to do so is: \n \\[w\\left(i, j\\right) = \\exp\\left(\\frac{-\\left\\|\\bvec{x}_i - \\bvec{x}_j\\right\\|^2}{\\sigma ^2}\\right)\\]\n where $\\sigma$ is an hyper-parameter. Note that this weight decreases as the distance between $\\bvec{x}_i$ and $\\bvec{x}_j$ increases. Also, considering this full graph may be expensive, so we can also restrict to the $k$ nearest neighbours of each points.\n\n The goal is now to find a partition $\\left\\{A, B\\right\\}$, minimising the sum of weights of the edges we have to remove. We thus let this value to be named the cut: \n \\[\\text{cut}\\left(A, B\\right) = \\sum_{i \\in A}^{} \\sum_{j \\in B}^{} w\\left(i, j\\right)\\]\n \n Just minimising the cut might favour imbalanced partitions, so we also define the degree of a node in the graph $d_i$ and the volume of a partition to be given by: \n \\[d_i = \\sum_{j}^{} w\\left(i, j\\right), \\mathspace \\text{vol}\\left(A\\right) = \\sum_{i \\in A}^{} d_i\\]\n \n Our goal is now to minimise a normalised cut: \n \\[\\text{ncut}\\left(A, B\\right) = \\frac{\\text{cut}\\left(A, B\\right)}{\\text{vol}\\left(A\\right)} + \\frac{\\text{cut}\\left(A, B\\right)}{\\text{vol}\\left(B\\right)}\\]", "source": "intro_to_ML", "format": "latex"}
-{"id": "intro_to_ML_chunk_40", "text": "This problem is NP-hard, but it can be relaxed as the following eigenvector problem: \n \\[\\left(D - W\\right)\\bvec{y} = \\lambda D \\bvec{y}\\]\n where $D$ is the diagonal degree matrix ($D_{i, i} = d_i$) and $W$ is the affinity matrix of the graph ($W_{i, j} = w\\left(i, j\\right)$).\n \n The eigenvector with smallest eigenvalue can be shown to be a vector of all ones with eigenvalue 0. We are thus looking for the eigenvector with second smallest eigenvalue. A positive value in this vector indicates that the corresponding point belongs to one partition, and a negative value to the other.\n\n\\begin{subparag}{Remark}\n Since we had to relax the problem, the solution is not always optimal.\n \\end{subparag}\n\n\\begin{subparag}{$K$-way partitions}\n To obtain more than two clusters, we have two choices.\n\n The first one is to recursively apply the two-way partitioning algorithm. This is inefficient and unstable.\n\n The second one is to find $K$ eigenvectors. This leads to each point being represented by a $K$-dimensional vector. We can then apply $K$-means clustering to those resulting vectors.\n \\end{subparag}\n\\end{parag}\n\n\\end{document}", "source": "intro_to_ML", "format": "latex"}
-{"id": "analysis_1_chunk_0", "text": "MATH 101 (en)\u2013 Analysis I (English)\nNotes for the course given in Fall 2021\nTeacher:Roberto Svaldi\nHead Assistant: Stefano Filipazzi\nNotes written by Zsolt Patakfalvi & Roberto Svaldi\nThursday 12th October, 2023\nThis work is licensed under a Creative Commons \u201cAttribution-\nNonCommercial-NoDerivatives 4.0 International\u201d license.\n1", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_1", "text": "CONTENTS\n1\nProofs\n3\n2\nBasic notions\n8\n2.1\nSets\n. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n8\n2.2\nNumber sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n9\n2.2.1\nHalf lines, intervals, balls\n. . . . . . . . . . . . . . . . . . . . . . . . . .\n10\n2.2.2\nExtended real numbers\n. . . . . . . . . . . . . . . . . . . . . . . . . . .\n10\n2.3\nBounds\n. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n11\n2.3.1\nBasic de\ufb01nitions, properties, and results. . . . . . . . . . . . . . . . . . .\n11\n2.3.2\nArchimedean property of R . . . . . . . . . . . . . . . . . . . . . . . . .\n18\n2.3.3\nAn alternative de\ufb01nition for in\ufb01mum/supremum . . . . . . . . . . . . .\n18\n2.3.4\nIn\ufb01mum and supremum for subsets of Z . . . . . . . . . . . . . . . . . .\n20\n2.4\nRational numbers vs real numbers\n. . . . . . . . . . . . . . . . . . . . . . . . .\n21\n2.4.1\n\u221a\n3 is a real number\n. . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n21\n2.4.2\nIntegral part\n. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n22\n2.4.3\nRational numbers are dense in R . . . . . . . . . . . . . . . . . . . . . .\n23", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_2", "text": "22\n2.4.3\nRational numbers are dense in R . . . . . . . . . . . . . . . . . . . . . .\n23\n2.4.4\nIrrational numbers are dense in R\n. . . . . . . . . . . . . . . . . . . . .\n25\n2.5\nAbsolute value\n. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n25\n2.5.1\nProperties of the absolute value . . . . . . . . . . . . . . . . . . . . . . .\n25\n2.5.2\nTriangular inequality . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n26\n3\nComplex numbers\n28\n3.1\nDe\ufb01nition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n29\n3.2\nOperations between complex numbers\n. . . . . . . . . . . . . . . . . . . . . . .\n30\n3.3\nPolar form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n33\n3.4\nEuler formula . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n35\n3.5\nFinding solutions of equations with complex coe\ufb03cients\n. . . . . . . . . . . . .\n36\n3.5.1\nSolving complex equations . . . . . . . . . . . . . . . . . . . . . . . . . .\n38\n4\nSequences\n40\n4.1\nRecursive sequences\n. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n42\n4.2\nInduction\n. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n43\n4.3", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_3", "text": "42\n4.2\nInduction\n. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n43\n4.3\nBernoulli inequality and (non-)boundedness of geometric sequences . . . . . . .\n45\n4.4\nLimit of a sequence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n47\n4.4.1\nLimits and algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n49\n4.5\nSqueeze theorem\n. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n52\n4.5.1\nLimits of recursive sequences\n. . . . . . . . . . . . . . . . . . . . . . . .\n55\n4.5.2\nUnbounded sets and in\ufb01nite limits . . . . . . . . . . . . . . . . . . . . .\n59\n4.6\nMore convergence criteria . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n61\n4.7\nMonotone sequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n63\n4.8\nLiminf, limsup\n. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n65\n4.9\nSubsequences . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n66\n4.10 Cauchy convergence\n. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n68\n2", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_4", "text": "1\nPROOFS\nThe means to explore analysis from a mathematical viewpoint within this course will be\nmathematical proofs. Part of the goal of the course will be for you to learn how to prove\nmathematical statements via mathematical proofs.\nThere are two main types of proof that we will encounter:\n\u25e6Constructive proof: an argument in which, starting from certain hypotheses/assumptions,\none tries to explicitly construct a mathematical object or to explicitly show that a certain\nmathematical property hold for a mathematical object;\n\u25e6Proof by contradiction: an argument in which we assume that the conclusion that we\nare trying to reach does not hold and we show that such assumption, together with our\nstarting hypotheses leads to a contradiction.\nYou have probably already encountered many constructive proofs; on the other hand, the\nreader may be encountering proofs by contradiction for the \ufb01rst time. So, let us start by giving\na classical example of proof by contradiction.\nBefore we explain our \ufb01rst example, let us recall that the set of rational numbers is the set\nof numbers of the form a\nb, with a, b integers, b \u0338= 0, where the following identi\ufb01cation between\ndi\ufb00erent fractions holds: for any non-zero integer c,\na\nb = a \u00b7 c\nb \u00b7 c .\nWe shall start by showing a classical argument by contradiction. For the time being we shall\nassume that we know how to construct the real numbers, and that we know that\n\u221a\n3, that\nis, the positive solution to the equation X2 \u22123 = 0, is a real number. For a more detailed\ndiscussion about the real numbers, we refer the reader to Section 2.\nProposition 1.1. The real number\n\u221a\n3 is not a rational number.\nWe are going to use a proof by contradiction; that is, we are going to assume that\n\u221a\n3 is\nrational and we are going to derive, by means of logical implications, a contradiction to some\nother fact that we already know or to some other fact that is implied by the assumed rationality\nof\n\u221a\n3.\nLet us recall here that a natural number p is prime if and only if the only natural numbers\nthat divide p are 1 and p itself.\nExercise 1.2. Prove that the following two properties for a natural number p are equivalent:\n\u25e6p is prime;\n\u25e6if a, b are natural numbers and p divides ab, then either p divides a or p divides b.\nProof of Proposition 1.1. Assume that\n\u221a\n3 is rational. Thus, we may write\n\u221a\n3 = a\nb", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_5", "text": "\u25e6p is prime;\n\u25e6if a, b are natural numbers and p divides ab, then either p divides a or p divides b.\nProof of Proposition 1.1. Assume that\n\u221a\n3 is rational. Thus, we may write\n\u221a\n3 = a\nb\n(1.2.a)\nfor some integers a and b \u0338= 0. As\n\u221a\n3 > 0, a and b should have the same sign. If they are\nboth negative, by multiplying both by \u22121 we may assume that they are positive. Hence, we\nwill assume that a, b are both positive integers.\nFurthermore, by dividing both a, b by their greatest commond divisor gcd(a, b)1, we may assume\n1Let us recall here the Fundamental Theorem of Arithmetic: any natural number n can be written uniquely\nas a product of powers of the prime numbers: namely, n = pk1\n1 \u00b7 pk2\n2 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 pkn\nn , where p1, . . . , pk are distinct prime\nnumbers and k1, . . . , kn are natural numbers > 0. For example, 36 = 4 \u00b7 9 = 22 \u00b7 32. In view of this, given two\nnatural numbers a, b, then gcd(a, b) is de\ufb01ned by writing it as a product gcd(a, b) = qj1\n1 \u00b7 qj2\n2 \u00b7 . . . qjn\nn\nwhere the\nqi are primes that divide both a and b and ji is the maximal natural number such that qj1\ni\ndivides both a and b.\n3", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_6", "text": "that a and b are relatively prime, that is, they do not share any prime factors. Multiplying\nboth sides of (1.2.a) by b, then, since b \u0338= 0,\nb\n\u221a\n3 = a.\n(1.2.b)\nSquaring both sides of (1.2.b) yields\nb2 \u00b7 3 = a2.\n(1.2.c)\nHence, as 3 divides the left hand side of (1.2.c), 3 must divide the right hand side, too. Thus,\na = 3r.\n(1.2.d)\nSubstituting the relation (1.2.d) into equation (1.2.c), we obtain that\nb2 \u00b7 3 = (3r)2 = 9r2\nHence, b2 = 3r2, which implies that 3|(b2). We write x|y, with x, y integers to mean that x\ndivides y. Again, as 3 is prime, then, since 3|b2,\n3|b,\n(1.2.e)\nBut, (1.2.d)-(1.2.e) together contradict the relatively prime assumption on a and b. Thus, we\nobtained a contradiction with our original assumption, so that\n\u221a\n3 is not a rational number.\nRemark 1.3. The proof of Proposition 1.1 is a nice example of a proof by contradiction. On\nthe other hand, it does not tell us much about the nature of\n\u221a\n3.\nWhat is\n\u221a\n3? Is it a real number? How can we de\ufb01ne real numbers? What notable properties\ndo those have? We will get back to these questions in Section 2.2-2.4.\nWe can generalize the above proof to any prime number p \u2208N.\nExercise 1.4. Imitate the proof of Proposition 1.1, to show that for every prime number p \u2208N,\n\u221ap is not rational.\nIn particular, Exercise 1.4 implies that also\n\u221a\n2 \u0338\u2208Q.\nAs easy as it may seem at a \ufb01rst glance to \ufb01nd and write mathematical proofs, one ought\nto be extremely careful: it is indeed very easy to write wrong proofs! This is often do to that\nthe fact that one may assume something wrong in the course of a proof: if the premise of an\nimplication is false, then anything can follow from it.", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_7", "text": "to be extremely careful: it is indeed very easy to write wrong proofs! This is often do to that\nthe fact that one may assume something wrong in the course of a proof: if the premise of an\nimplication is false, then anything can follow from it.\nExample 1.5. Here is an example of an (incorrect) proof showing that 1 is the largest natural\nnumber, a fact that is clearly false, since 2 > 1 and 2 \u2208N.\nClaim. 1 is the largest integer.\nWRONG PROOF. Let l be the largest integer.\nThen l \u2265l2, so that l \u2212l2 = l(1 \u2212l) \u22650. Hence, there are two possibilities for l(1 \u2212l) \u22650:\na) either l < 0 and 1 \u2212l \u22640; or,\nb) l \u22650 and 1 \u2212l \u22650.\nAs 0 is an integer, we must be in case b), so that l \u22650 and l \u22641. Hence l = 1.\n4", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_8", "text": "This claim cannot possibly be true: in fact, 2 is de\ufb01nitely an integer and 2 > 1. Even better,\nthe set of integeral numbers is not bounded from above2, that is, there is no real number C such\nthat z \u2264C for all z \u2208Z.\nWhat went wrong in the above proof? All the algebraic manipulations that we made following\nthe \ufb01rst line of the proof appear to be correct. [Go back and check that!!] Thus, the issue must\nbe contained in the (absurd) assumption we made in the \ufb01rst sentence:\nLet l be the largest integer.\nIn fact, as we just explained, there cannot be a largest element in the set of integers: in fact,\ngiven an integer l, then l +1 is also an integer and l +1 > l, which clearly shows that the above\nassumption was unreasonable.\nAnalysis is mostly focused on the study of real and complex numbers3 and their properties.\nEven more generally, analysis is concerned with studying (or analyzing, hence the name Anal-\nysis) functions de\ufb01ned over the real (alternatively, over the complex numbers) with values in\nthe real numbers (alternatively, over the complex numbers) and their important properties4.\nIn order to carry out such analysis, we will often need to deal with in\ufb01nity. Roughly speaking,\nwe will often be interested in understanding numbers/functions from the point of view of an\nin\ufb01nitely small or at an in\ufb01nitely large viewpoint. Our main goal will be to provide a frame-\nwork to be able to treat in a formal mathematical way all the di\ufb00erent aspects of in\ufb01nity in\nthe realm of real/complex numbers. To make a slightly better sense of this statement, you may\ntry to think (and formalize) of how to de\ufb01ne the speed of a particle moving linearly on a rod,\nat a given time t.\nHow should we de\ufb01ne the real numbers? Even more importantly, how can we represent\nthem numerically? Intuitively, we have been taught that real numbers are those numbers that\nwe can represent numerically by writing down a decimal expansion, for example,\n\u221a\n2 =1.414213562373095048801688724209698078569671875376948073176679737990", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_9", "text": "we can represent numerically by writing down a decimal expansion, for example,\n\u221a\n2 =1.414213562373095048801688724209698078569671875376948073176679737990\n7324784621070388503875343276415727350138462309122970249248360 . . . .\nAs it suggested from this example, it may be the case that when we try to represent certain\nreal numbers, we have to account for an in\ufb01nite decimal part5 of the expansion, that is, there\nis an in\ufb01nite sequence of digits to the right of the decimal dot \u201c.\u201d. Hence, we may at \ufb01rst\ntempted to adopt the following de\ufb01nition of the set of real numbers:\nThe real numbers are all those numbers that we can represent with a decimal expansion whose\nintegral part (the digits to the left of \u201c.\u201d) can be written using a \ufb01nite number of digits (chosen\nin the set {0, 1, 2, . . . , 9}), whereas its decimal part (the digits to the right of \u201c.\u201d) is any in\ufb01nite\nsequence of digits (as above, chosen in the set {0, 1, 2, . . . , 9}).\nWhile this may seem, at \ufb01rst,\nas an intuitively \ufb01ne de\ufb01nition for the real numbers, it actually hides some subtleties.\nHere we illustrate one of the main subtleties of this de\ufb01nition: namely, we show that,\nin the above de\ufb01nition, we certainly have to be careful. We show that there is non unique\ncorrespondence between a real number and its decimal expansion. An example is given by the\nfollowing proposition, which also provides a great basic example of how we deal with in\ufb01nity\nin Analysis.\n2We will give a formal de\ufb01nition of what being bounded from above means later, cf. De\ufb01nition 2.8.\n3See Section 3 for the de\ufb01nition and basic properties of complex numbers.\n4Some of the most important classes of functions that we will encounter are those of continuous, di\ufb00erentiable,\nintegrable, analytic functions, but there are many more other possible classes of functions that are heavily studied\nin analysis", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_10", "text": "3See Section 3 for the de\ufb01nition and basic properties of complex numbers.\n4Some of the most important classes of functions that we will encounter are those of continuous, di\ufb00erentiable,\nintegrable, analytic functions, but there are many more other possible classes of functions that are heavily studied\nin analysis\n5The decimal part of the expansion is that part of the expansion that lays on the right hand side of the point\n\u201c.\u201d. For example, the decimal part of the expansion of 41369.57693 is the sequence 57693. The integral part of\nthe decimal expansion is instead that part of the expansion that lays on the left hand side of the point \u201c.\u201d. The\nintegral part of 41369.57693 is 41369. The integral part always has \ufb01nite length, that is, it can be written using\na \ufb01nite number of digits.\n5", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_11", "text": "Proposition 1.6. 0.\u00af9 = 1\nBy 0.\u00af9 we denote the real number whose decimal representation is given by an in\ufb01nite\nsequence of 9 in the decimal part, 0.999999 . . ..\nProof. We give two proofs none of which is completely correct, at least as far as our current\nde\ufb01nition and knowledge of the real numbers go. Nevertheless, we carefully explain what the\nissues are in each case; we also explain how these issues will be clari\ufb01ed and taken care of\nduring this course.\n(1) First an elementary proof:\n9 \u00b7 0.\u00af9 = (10 \u22121) \u00b7 0.\u00af9 = 10 \u00b7 0.\u00af9 \u22121 \u00b7 0.\u00af9 = 9.\u00af9 \u22120.\u00af9 = 9\nSo, 0.\u00af9 is a solution of the equation 9X \u22129 = 0; the only solution to this equation is\nclearly X = 1, thus, 0.\u00af9 = 1.\nAt \ufb01rst sight, this proof is de\ufb01nitely a reasonable one from the point of view of the\nalgebraic manipulations that we carried out. However, we assumed that we know what\n0.\u00af9 is. Moreover, we also assumed that we can algebraically manipulate 0.\u00af9 as usual,\ndespite the fact that it has an in\ufb01nite decimal expansion. None of these facts are that\nclear if you think about it, as we have not really de\ufb01ned what the properties of numbers\nlike 0.\u00af9 are.\nSo, what kind of number is 0.\u00af9? What are its properties? For example, what algebraic\nmanipulations are we allowed to make with it?\n(2) Analysis provides us with a precise de\ufb01nition of 0.\u00af9\n0.\u00af9 :=\n\u221e\nX\ni=1\n9\n10i .\nOn the hand, what kind of mathematical object is P\u221e\ni=1\n9\n10i ? This is a series and we will\nstudy series in detail in Section 4. By de\ufb01nition,\n\u221e\nX\ni=1\n9\n10i := lim\nn\u2192\u221e\n n\nX\ni=1\n9\n10i\n!\n.\nWe have yet to learn a precise de\ufb01nition of lim, thus, we cannot quite continue in a precise\nway from here, nevertheless we continue the argument for completeness. If you are not\ncomfortable with it now, it is completely OK, just skip this part of the proof.\nHowever, before we proceed, we need to show an identity for the sum of elements in a\ngeometric series6.", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_12", "text": "way from here, nevertheless we continue the argument for completeness. If you are not\ncomfortable with it now, it is completely OK, just skip this part of the proof.\nHowever, before we proceed, we need to show an identity for the sum of elements in a\ngeometric series6.\nClaim. Let a \u2208R, a \u0338= 1. Then,\na + a2 + \u00b7 \u00b7 \u00b7 + an = a \u2212an+1\n1 \u2212a\n.\n(1.6.f)\nProof of the Claim. To prove this equality, we just multiply the left side by 1 \u2212a to\nobtain:\n(a + a2 + \u00b7 \u00b7 \u00b7 + an)(1 \u2212a) = a \u2212a \u00b7 a + a2 \u2212a2 \u00b7 a + a3 \u2212. . .\n\u2212an\u22121 \u00b7 a + an \u2212an \u00b7 a = a \u2212an+1\nThis shows that (1.6.f) indeed holds, since to obtain the form of the equation in the\nstatement of the claim , it su\ufb03eces to .\n6A geometric series is a series whose elements are of the form caq, for c, a \u2208R and q \u2208N. This will be\nexplicitly de\ufb01ned when we introduce series, later; hence, do not worry about this de\ufb01nition for now.\n6", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_13", "text": "And then we can proceed showing the statement:\n\u221e\nX\ni=1\n9\n10i = 9 \u00b7\n\u221e\nX\ni=1\n1\n10i = 9 \u00b7 lim\nn\u2192\u221e\n n\nX\ni=1\n1\n10i\n!\n=\n9 \u00b7 lim\nn\u2192\u221e\n 1\n10 \u2212\n1\n10n+1\n1 \u22121\n10\n!\n= 9 \u00b7\n1\n10 \u2212lim\nn\u2192\u221e\n1\n10n+1\n1 \u22121\n10\n=\n9 \u00b7\n1\n10\n1 \u22121\n10\n= 91\n9 = 1.\nIn Section 2 and in the following one, we will introduce all the necessary tools, de\ufb01nitions,\nnotations and conventions to answer all of the questions that were raised in these \ufb01rst few\npages.\n7", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_14", "text": "2\nBASIC NOTIONS\n2.1\nSets\nA set S is a collection of objects called elements. If a is an element of S, we say that a\nbelongs to S or that S contains a, and we write a \u2208S. If an element a is not in S, we then\nwrite a \u0338\u2208S. If the elements a, b, c, d, . . . form the set S, we write S = {a, b, c, d, . . . }. We can\nalso de\ufb01ne a set simply by specifying that its elements are given by some condition, and we\nwrite\nS := {s | s satis\ufb01es some condition}.\nNotation 2.1. The symbol := indicates that we are identifying the object on the LHS (left\nhand side) of \u201c:=\u201d with the object on the RHS (right hand side) of \u201c:=\u201d. You can read it as\n\u201cde\ufb01ned as\u201d.\nExample 2.2. The set S = {0, 1, 2, 3, 4, 5} of natural numbers that are at most 5 can be\nde\ufb01ned as follows\nS := {n | n is a natural number and n \u22645}.\nA set T is said to be a subset of a set S if any element of T is also an element of S. If T is\na subset of S, we denote it by writing T \u2286S. Given a set S, one can always de\ufb01ne a subset\nT \u2282S, T := {s \u2208S|\u201c condition\u201d}, that is, S\u2032 is the set formed by those elements of S that\nsatisfy the given condition.\nExample 2.3. The subset 2N of N of even natural numbers can be de\ufb01ned as\n2N := {n \u2208N | 2 divides n}.\nIf T \u2286S, it may happen that there are elements of S which are not contained in T. In this\ncase we say that T is a strict subset of S, or that T is strictly included/contained in S. When\nwe want to stress that we know that a subset T of a set S is strictly included in S we shall\nwrite T \u228aS.\nExample 2.4. 2N \u228aN since 1 \u0338\u22082N.\nIf we just write T \u2286S, we mean that T is a subset of S that may be equal to S, but we are", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_15", "text": "write T \u228aS.\nExample 2.4. 2N \u228aN since 1 \u0338\u22082N.\nIf we just write T \u2286S, we mean that T is a subset of S that may be equal to S, but we are\nnot making any particular statement about whether or not T is a strict subset of S. Hence, in\nthe previous Example 2.4, we may have also used the notation 2N \u2286N and that would have\nbeen correct. To write that a set T is not a subset of a set S, we write T \u0338\u2286S.\nWe will consider the standard operations between sets, such as intersection, union, taking\nthe complent. More precisely, given two subsets U, V , we de\ufb01ne:\nIntersection:\nU \u2229V := {x | x \u2208U and x \u2208V };\nUnion:\nU \u222aV := {x | x \u2208U or x \u2208V };\nComplement:\nU \\ V := {x | x \u2208U and x \u0338\u2208V }.\nExercise 2.5. Given sets E, F and D prove that the following relations hold:\nCommutativity: E \u2229F = F \u2229E and E \u222aF = F \u222aE;\nAssociativity: D \u2229(E \u2229F) = (D \u2229E) \u2229F and D \u222a(E \u222aF) = (D \u222aE) \u222aF;\nDistributivity: D \u2229(E \u222aF) = (D \u2229E) \u222a(D \u2229F) and D \u222a(E \u2229F) = (D \u222aE) \u2229(D \u222aF);\nDe Morgan laws: (E \u2229F)c = Ec \u222aF c and(E \u222aF)c = Ec \u2229F c.\n8", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_16", "text": "2.2\nNumber sets\nThere are a few important sets that we are going to work with all along this course:\n(1) \u2205: the empty set; it is the set which has no elements, \u2205:= { }.\nExercise 2.6. Show that for any set S, \u2205\u2286S.\n(2) N : the set of natural numbers, N := {0, 1, 2, 3, 4, 5, 6, . . . }.\nN is well ordered, that is, all its subsets contain a smallest element. We will prove that\nlater in Proposition 2.34.\n(3) Z : the set of integral numbers7, Z := {. . . , \u22121, 0, 1, . . . } .\n(4) Q : the set of rational numbers, Q := {a\nb | a \u2208Z and b \u2208Z \\ {0}}, where we impose the\nfollowing identi\ufb01cation between fractions\na\nb = a \u00b7 c\nb \u00b7 c ,\nfor c \u2208Z \\ {0}.\n(5) R : the set of real numbers. It is not easy to actually construct it and there are some\nsubtleties in trying to de\ufb01ne real numbers by means of their decimal representation, as\nwe have already understood from Proposition 1.6.\nRemark 2.7. In this course, we will not attempt to provide a rigorous construction of the set of\nreal numbers R, although there are many equivalent constructions. If you are curious, you can\nclick here to \ufb01nd out more about these constructions. Instead of going through the construction\nof R in the course, we proceed to list here certain properties that uniquely de\ufb01ne R [we also do\nnot prove such uniqueness, but, please, believe it] and we will assume them going forward:\n(1) Q \u2286R;\n(2) R is an ordered \ufb01eld\n\u25e6the word \ufb01eld refers to the fact that addition, substraction, multiplication are all\nwell-de\ufb01ned operation within R; moreover, these operations respect commutativity,\nassociativity and distributivity properties and for all x \u2208R, x \u0338= 0 it is possible to\nde\ufb01ned a multiplicative inverse x\u22121 such that x \u00b7 x\u22121 = 1;\n\u25e6the world ordered refers to the fact that given two elements x, y \u2208R we can always\ndecide whether x < y, or x > y, or x = y; moreover, this comparison is also", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_17", "text": "de\ufb01ned a multiplicative inverse x\u22121 such that x \u00b7 x\u22121 = 1;\n\u25e6the world ordered refers to the fact that given two elements x, y \u2208R we can always\ndecide whether x < y, or x > y, or x = y; moreover, this comparison is also\ncompatible with the operations that make R into a \ufb01eld.\n(3) R satis\ufb01es the In\ufb01mum Axiom 2.22, that will be introduced in next section.\nThe following inclusions hold among the sets just de\ufb01ned:\n\u2205\u228aN \u228aZ \u228aQ \u228aR.\nTo justify these inclusions:\n\u25e6\u2205\u228aN : N is non-empty. For example, 0 \u2208N.\n\u25e6N \u228aZ : an integral number can also be negative, for example, \u22121 \u2208Z, while natural\nnumber are always non-negative; thus Z \u220b\u22121 \u0338\u2208N.\n\u25e6Z \u228aQ : 1\n2 \u2208Q, but 1\n2 \u0338\u2208Z.\n\u25e6Q \u228aR : we saw in Proposition 2.38 that\n\u221a\n3 \u0338\u2208Q; we will prove formally in Section 2.4.1\nthat\n\u221a\n3 \u2208R.\n7We will often call an integral number an \u201cinteger\u201d.\n9", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_18", "text": "2.2.1\nHalf lines, intervals, balls\nWe introduce here further notation regarding the real numbers and some special classes of\nsubsets that we will be using all throughout the course.\n(1) Invertible real numbers: R\u2217:= {x \u2208R | x \u0338= 0}.\n(2) Closed half lines: R+ := {x \u2208R | x \u22650}, R\u2212:= {x \u2208R | x \u22640}.\nAt times, these are also denoted by R\u22650 and R\u22640, respectively.\n(3) Open half lines: R\u2217\n+ := {x \u2208R | x > 0}, R\u2217\n\u2212:= {x \u2208R | x < 0}.\nAt times, these are also denoted by R>0 and R<0, respectively.\nWe use the analogous de\ufb01nitions also for the sets\nN\u2217, Z\u2217, Q\u2217,\nN+, Q+, Z+,\nN\u2212, Q\u2212, Z\u2212,\nN\u2217\n+, Q\u2217\n+, Z\u2217\n+,\nN\u2217\n\u2212, Q\u2217\n\u2212, Z\u2217\n\u2212.\n(4) Bounded intervals: if a < b are real numbers, we de\ufb01ne\nOpen bounded interval:\n]a, b[ := {x \u2208R | a < x < b}.\nClosed bounded interval:\n[a, b] := {x \u2208R | a \u2264x \u2264b}.\nHalf-open bounded interval:\n(\n]a, b] := {x \u2208R | a < x \u2264b}.\n[a, b[ := {x \u2208R | a \u2264x < b}.\nIf a = b, then [a, b] = [a, a] = {a}. When we say that a subset I is a bounded interval of\nR of extreme a < b, we mean that I may be either one of\n[a, b], [a, b[, ]a, b], ]a, b[.\n(5) Open balls: let a, \u03b4 \u2208R, \u03b4 > 0; we de\ufb01ne the open ball B(a, \u03b4) \u2286R of radius \u03b4 and center\na as\nB(a, \u03b4) :=]a \u2212\u03b4, a + \u03b4[.\n(6) Closed balls: let a, \u03b4 \u2208R, \u03b4 \u22650; we de\ufb01ne the closed ball B(a, \u03b4) \u2286R of radius \u03b4 and\ncenter a as", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_19", "text": "a as\nB(a, \u03b4) :=]a \u2212\u03b4, a + \u03b4[.\n(6) Closed balls: let a, \u03b4 \u2208R, \u03b4 \u22650; we de\ufb01ne the closed ball B(a, \u03b4) \u2286R of radius \u03b4 and\ncenter a as\nB(a, \u03b4) := [a \u2212\u03b4, a + \u03b4].\nWhen \u03b4 = 0, then B(a, 0) = {a}.\n2.2.2\nExtended real numbers\nThe extended real line is the set\nR := {\u2212\u221e, +\u221e} \u222aR.\nThe symbol +\u221e(resp. \u2212\u221e) is called \u201cplus in\ufb01nity\u201d (resp. \u201cminus in\ufb01nity\u201d). In this course\n\u00b1\u221eshall not be treated as numbers: they are just symbols indicating two elements of the\nextended real line \u00afR. That means that we will not try to make sense of algebraic operations\n10", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_20", "text": "involving \u00b1\u221e; thus, be very careful not to treat those as numbers. If you think carefully a bit,\nyou can see that it is hard to coherently de\ufb01ne for example the result of the addition\n+\u221e+ (\u2212\u221e).\nLater in the course we will use extensively these symbols. For the time being, we just want\nto use them to de\ufb01ne the following subsets of R. Let a \u2208R, then\nOpen unbounded intervals:\n]a, +\u221e[:= {x \u2208R|x > a}, ] \u2212\u221e, a[:= {x \u2208R|x < a}.\nClosed unbounded intervals:\n[a, +\u221e[:= {x \u2208R|x \u2265a}, ] \u2212\u221e, a] := {x \u2208R|x \u2264a}.\nFinally\n] \u2212\u221e, +\u221e[:= R.\nThese sets are also called open/closed half lines, or open/closed unbounded intervals, or\nopen/closed extended intervals, where open/closed is determined by whether or not a belongs\nto the set.\nSo, from now on, when we say that a subset I of R is an interval, we will mean that I has\none of the following forms:\n\u25e6[a, b], ]a, b[, ]a, b], [a, b[,\na, b \u2208R,\na < b;\n\u25e6[a, +\u221e[, ]a, +\u221e[, ] \u2212\u221e, a], ] \u2212\u221e, a[,\na \u2208R;\n\u25e6] \u2212\u221e, +\u221e[= R.\n2.3\nBounds\nWe now start entering the realm of modern (and rigorous) analysis.\nWe start by de\ufb01ning some important properties of subset of R.\n2.3.1\nBasic de\ufb01nitions, properties, and results.\nDe\ufb01nition 2.8. Let S \u2286be a non-empty subset of R.\n(1) A real number a \u2208R is an upper (resp. lower) bound for S if s \u2264a (resp. s \u2265a) holds\nfor all s \u2208S.\n(2) If S has an upper (resp. a lower) bound then S is said to be bounded from above (resp.\nbounded from below).\n(3) The set S is said to be bounded if it is bounded both from above and below.", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_21", "text": "for all s \u2208S.\n(2) If S has an upper (resp. a lower) bound then S is said to be bounded from above (resp.\nbounded from below).\n(3) The set S is said to be bounded if it is bounded both from above and below.\nFor a set S \u2286R in general upper and lower bounds are not unique.\nExample 2.9.\n(1) The set N \u2282R is bounded from below, since \u2200n \u2208N, n \u22650; in particular,\n0 is a lower bound. In fact, any negative real number is also a lower bound for N.\nOn the other hand, N is not bounded. While this fact may appear intuitively clear, it\nis not immediately clear how to prove it formally. Can you \ufb01nd a proof using only the\nconcepts and tools that we have introduced so far in the course? The answer is no, at\nthis time of the course. For a formal proof of the unboundedness of N, we shall need\nArchimedes\u2019 property for R, see Proposition 2.30.\n(2) Z is neither bounded from above nor from below. In fact, it cannot be bounded from\nabove since N \u2286Z. It is also not bounded from below: if a lower bound l \u2208R existed for\nZ, then \u2212l would be an upper bound for N, which we saw above does not hold. [Prove\nthis assertion in detail!].\n11", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_22", "text": "(3) The set S := {n2|n \u2208Z} is bounded from below: in fact, \u2200n \u2208N, n2 \u22650, thus 0 is\na lower bound. On the other hand, it is not bounded. In fact, assume for the sake of\ncontradiction that S were bounded from above, i.e., that there exists u \u2208R and u \u2265s,\n\u2200s \u2208S. Since for any n \u2208N, n2 \u2265n, then it would follow that u > n, for all n \u2208N, but\nthis contradicts part (1).\n(4) The set S := {n3|n \u2208Z} is neither bounded from above nor from below. [Prove it! The\nproof is similar to that in part (2).]\n(5) The set S := {sin(n2)|n \u2208Z} is bounded since for all x \u2208R, \u22121 \u2264sin x \u22641. Examples\nof possible lower bounds are \u22125 and \u221213; example of possible upper bounds are 1 and\n27. As sin x \u2208[\u22121, 1], then it is certainly true that\n\u25e6any real number y such that y \u22651 is an upper bound for S, while\n\u25e6any real number y such that y \u2264\u22121 is a lower bound for S.\n(6) Let S := [3, 5[= {x \u2208R | 3 \u2264x < 5}. Then, 5 is an upper bound for S since for any\nelement x of S, x < 5. Moreover, if c is a real number and c > 5, then c is also an upper\nbound for S, since c > 5 > x for all x \u2208S.\nThe same reasoning shows that 3 is a lower bound for S and that for any real number d\nsuch that d < 3, then d is a lower bound for S as well.\n(It is left to you to prove that in this example you will obtain the exact same con-\nclusions if instead of considering the interval [3, 5[, you considered any of the intervals\n[3, 5], ]3, 5], ]3, 5[.)\nUsing the discussion of the above examples, we summarize here some of the main properties\nof upper and lower bounds.\nProposition 2.10. Let S \u2282R be a non-empty set. Let c \u2208R.", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_23", "text": "[3, 5], ]3, 5], ]3, 5[.)\nUsing the discussion of the above examples, we summarize here some of the main properties\nof upper and lower bounds.\nProposition 2.10. Let S \u2282R be a non-empty set. Let c \u2208R.\n(1) If u is an upper bound for S, then for any d \u2265u, d is also an upper bound for S.\n(2) If l is a lower bound for S, then for any e \u2264l, e is also a lower bound for S.\n(3) If T \u2286S is a non-empty subset and c is a lower (resp. an upper) bound for S, then c is\nalso a lower (resp. an upper) bound for T.\n(4) If T \u2286S is a non-empty subset and T is not bounded from above (resp. from below), then\nalso S is is not bounded from above (resp. from below).\n(5) If S is a bounded interval of extremes a < b, then the set of lower bounds (resp. of upper\nbounds) of S is given by\n] \u2212\u221e, a]\n(resp. [b, +\u221e]).\n(6) If S := [b, +\u221e[ or S :=]b, +\u221e[, b \u2208R, then the set of lower bounds of S is given by\n] \u2212\u221e, b].\n(7) If S :=] \u2212\u221e, a] or S :=] \u2212\u221e, a[, a \u2208R, then the set of upper bounds of S is given by\n[a, +\u221e].\nProof.\n(1) Let u be an upper bound for S. Then \u2200s \u2208S, u \u2265s. If d \u2265u, then \u2200s \u2208S,\nd \u2265u \u2265s, in particular, d \u2265s, which shows the desired property.\n(2) Analogous to (1) and left as an exercise (see the sheet from Week 2).\n12", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_24", "text": "(3) If c is a lower bound for S, then c \u2264s for all element s \u2208S. Since T \u2286S, this means\nthat any element t \u2208T is also an element of S. Hence, a fortiori, the inequality c \u2264s,\n\u2200s \u2208S implies also that c \u2264t, \u2200t \u2208T.\nThe case of an upper bound is analogous, it su\ufb03ces to change the verse of the inequalities.\n(4) Since T is not bounded from above, this means that \u2200u \u2208R, there exists an element\nxu \u2208T (which will depend in general from the real number u we \ufb01x) such that xu > u.\nAs T \u2286S, then xu \u2208S, hence \u2200u \u2208R, there exists an element xu \u2208S such that xu > u\nand u cannot be an upper bound for S. As this holds \u2200u \u2208R, then also S is not bounded\nfrom above.\nThe case of T not bounded from below is analogous, it su\ufb03ces to change the verse of the\ninequalities.\n(5) Let us assume that S :=]a, b] = {x \u2208R | a < x \u2264b}. The other cases are similar \u2013 it\nis left to you to prove that in you will obtain the exact same conclusions if instead of\nconsidering the interval ]a, b], you considered any of the intervals [a, b], [a, b[, ]a, b].\nThen, a is a lower bound for S, since for all s \u2208S, a < s. Also for any real number\nd < a, d is also a lower bound for S, since d < a < s, for all s \u2208S. Similarly, b is an\nupper bound for S, since \u2200s \u2208S, s \u2264b, by de\ufb01nition. Thus, for any real number c > b,\nthen c > b \u2265s, \u2200x \u2208S and c is an upper bound for S. Then, part (1) implies that any\nelement of the half line [b, +\u221e[ (resp. ] \u2212\u221e, a]) is an upper bound (resp. lower bound)\nfor S. To conclude we need to show that no real number c > a (resp. d < b) is a lower\nbound (resp. an upper bound) of S. To show this, it su\ufb03ces to show that there exists an", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_25", "text": "for S. To conclude we need to show that no real number c > a (resp. d < b) is a lower\nbound (resp. an upper bound) of S. To show this, it su\ufb03ces to show that there exists an\nelement m \u2208S such that m < c. Since c > a, then a < a + c\u2212a\n2\n< c. If a + c\u2212a\n2\n\u2208S, it\nsu\ufb03ces to take m := a + c\u2212a\n2 . If a + c\u2212a\n2\n\u0338\u2208S, then a + c\u2212a\n2\n> b then c > b, and it su\ufb03ces\nto take m := b.\n(6) Analogous to the proof of (5).\nWe have just seen that upper/lower bounds of a set S are never unique, when some exist.\nMoreover, if S is an interval of extremes a < b, then a is a lower bound and b is an upper\nbound. We may be tempted to ask whether in general there exists upper lower bounds of a set\nS \u2286R that are element of S itself and what we can say in that case. In general, this is not\nalways true but nonetheless upper/lower bounds of S that are in S are very special elements\nof S.\nDe\ufb01nition 2.11. Let S \u2286R be a non-empty set.\n(1) The maximum of S is a real number M \u2208S which is also an upper bound for S.\n(2) The minimum of S is a real number m \u2208S which is also a lower bound for S.\nIn De\ufb01nition 2.11, we used the determinative article \u201cthe\u201d to intriduce maximum and\nminimum of a set of real numbers. This suggests that they should both be uniquely determined.\nThis is indeed the content of the next exercise.\nProposition 2.12. Let S be a non-empty subset of R. If max S (resp. min S) exists, then it\nis unique.\nNotation 2.13. For S \u2286R, we denote the maximum (resp. the minimum) of S by max S\n(resp. min S).\nProof. Suppose, for the sake of contradiction, that a maximum of S exists and it is not unique.\nThen there are at least two distinct numbers n, n\u2032 \u2208R which are both a maximum for S. As\n13", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_26", "text": "n, n\u2032 are distinct, i.e., n \u0338= n\u2032, we can assume that n < n\u2032. As n\u2032 is a maximum, then n\u2032 \u2208S.\nBut as n is also a maximum, in particular, n is also an upper bound, i.e., n \u2265s, \u2200s \u2208S; hence,\nalso n \u2265n\u2032, which is in contradiction with our assumption above that n\u2032 > n.\nYou can apply a similar argument for the uniqueness of the minimum.\nExample 2.14.\n(1) Let us de\ufb01ne S :=]1, 2[= {x \u2208R | 1 < x < 2}. Then S does not have\nminimum or maximum.\nIn fact, if u \u2208R is an upper bound for S, then, by de\ufb01nition, u \u2265x, \u2200x \u2208]1, 2[, which\nimplies that u \u22652. Hence u \u0338\u2208]1, 2[.\nAnalogously, if l \u2208R is a lower bound for S, then, by de\ufb01nition, l \u2264x, \u2200x \u2208]1, 2[, which\nimplies that l \u22641. Hence l \u0338\u2208]1, 2[.\n(2) S := [1, 2] has both a minimum and a maximum.\nmin S = 1, since 1 \u2208S and 1 \u2264s, \u2200s \u2208S, so that 1 is also a lower bound for S.\nmax S = 2, since 2 \u2208S and 2 \u2265s, \u2200s \u2208S, so that 2 is also an upper bound for S.\n(3) Let a < b be real numbers. S :=]a, b] has maximum but no minimum.\nmax S = b, since b \u2208S and b \u2265s, \u2200s \u2208S, so that b is also an upper bound for S.\nmin S, since any lower bound for S is \u2264a, hence there is no lower bound that is contained\nin S.\nThe above examples suggest that it should not be hard to understand when an interval\nS admits a maximum or a minimum. Indeed, the following characterization is an immediate\nconsequence of De\ufb01nition 2.11 and of Proposition 2.10\nProposition 2.15. Let S \u2286R be a bounded interval of extremes a < b.\n(1) The maximum of S exists if and only if b \u2208S. In this case, max S = b.", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_27", "text": "consequence of De\ufb01nition 2.11 and of Proposition 2.10\nProposition 2.15. Let S \u2286R be a bounded interval of extremes a < b.\n(1) The maximum of S exists if and only if b \u2208S. In this case, max S = b.\n(2) The minimum of S exists if and only if a \u2208S. In this case, min S = b.\nWhen S is not an interval, it may be more complicated to understand whether a maxi-\nmum/minimum exists.\nExample 2.16.\n(1) Take S :=\n\b n\u22121\nn\n| n \u2208Z\u2217\n+\n\t\n. Then S has a minimum but it does not\nhave a maximum.\nIndeed, min S = 0, since 0 = 1\u22121\n1\n\u2208S and n\u22121\nn\n\u22650, \u2200n \u2208Z\u2217\n+, so that 0 is a lower bound\nwhich belongs to S. However, S does not have a maximum. To see this, let l \u2208R, then:\n(i) assume that l < 1.\nThen a natural number n satis\ufb01es n >\n1\n1\u2212l if and only if\n1 \u22121\nn = n\u22121\nn\n> 1 \u2212(1 \u2212l) = l. then 1 \u22121\nn = n\u22121\nn\n> 1 \u2212(1 \u2212l) = 1; Thus, l cannot\nbe an upper bound for S, hence a fortiori it cannot be a maximum either.\n(ii) on the other hand, if a \u22651, then l \u0338\u2208S, so no such l can be a maximum for S.\nOne can actually show that the upper bounds of S are exactly the real numbers \u22651;\nindeed, it is easy to show that any l \u22651 is an upper bound for S, since 1 \u22121\nn \u22641 \u2264a,\nfor all n \u2208Z\u2217\n+. On the other hand (i) above shows that no real number l < 1 can be an\nupper bound for S. Hence, 1 is the least of all possible upper bounds for S.\nExample 2.16.3 above, suggests that we might need a new notion generalizing the concept\nof maximum/minimum.\nIn that example, 1 is very close to being the maximum of S :=\n\b n\u22121\nn\n| n \u2208Z\u2217\n+\n\t\n, as it is the least of all possible upper bounds. On the other hand, 1 cannot\nbe the maximum of S as 1 \u0338\u2208S. This phenomenon motivates the next de\ufb01nition.", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_28", "text": "\b n\u22121\nn\n| n \u2208Z\u2217\n+\n\t\n, as it is the least of all possible upper bounds. On the other hand, 1 cannot\nbe the maximum of S as 1 \u0338\u2208S. This phenomenon motivates the next de\ufb01nition.\nDe\ufb01nition 2.17. Let S \u2286R be a non-empty subset.\n14", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_29", "text": "(1) If the set U of all upper bounds of S is non-empty and U admits a minimum u \u2208U, then\nwe call u the supremum of S.\n(2) If the set L of all lower bounds of S is non-empty and L admits a maximum l \u2208L, then\nwe call l the supremum of S.\nRemark 2.18. Let S \u2286R be a non-empty subset.\nIf the set U of all upper bounds of S is empty, then S is not bounded from above, cf. De\ufb01ni-\ntion 2.8. In this case, then the supremum of S does not exist, by the above de\ufb01nition.\nSimilarly, if the set L of all lower bounds of S is empty, then S is not bounded from below,\ncf. De\ufb01nition 2.8. In this case, then the in\ufb01mum of S does not exist, by the above de\ufb01nition.\nAs in the case of maximum/minimum, the use of the determinative article in De\ufb01nition 2.17\nsuggests that, when they exist, the supremum/in\ufb01mum of a non-empty subset of R should be\nunique.\nProposition 2.19. Let S be a non-empty subset of R. If sup S (resp. inf S) exists, then it is\nunique.\nNotation 2.20. For S \u2286R, we denote the supremum (resp. the in\ufb01mum) of S by sup S (resp.\ninf S), when those exist as real number.\nIf S is not bounded from above, we write sup S = +\u221e. If S is not bounded from below, we\nwrite inf S = \u2212\u221e.\nProof. By de\ufb01nition, if the supremum of S exists, it is the minimum of the set\nU := {u \u2208R | u is an upper bound for S} .\nAs the maximum of a set is unique when it exists, cf. 2.12, then the conclusion follows at once.\nYou can apply a similar argument for the uniqueness of the minimum.\nExample 2.21.\n(1) Let S :=\n\b n\u22121\nn\n\f\f n \u2208Z\u2217\n+\n\t\n. Then, sup S = 1, cf. Example 2.16.3.", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_30", "text": "You can apply a similar argument for the uniqueness of the minimum.\nExample 2.21.\n(1) Let S :=\n\b n\u22121\nn\n\f\f n \u2208Z\u2217\n+\n\t\n. Then, sup S = 1, cf. Example 2.16.3.\n(2) Take S := {n3|n \u2208Z}. Then, S is unbounded. Thus, inf S, sup S do not exist.\n(3) If S is a bounded interval of extremes a < b, then\nsup S = b,\ninf S = a.\nIndeed, we saw in Proposition 2.10 that the set of lower (resp. upper) bounds of S is\n] \u2212\u221e, a] (resp. [b, +\u221e[).\n(4) Similarly, if S := [a, +\u221e[ or S := [a, +\u221e, a \u2208R then inf S = a, while sup S does not exit\nsince S is not bounded from below.\n(5) If S :=] \u2212\u221e, b] or S :=] \u2212\u221e, b[, b \u2208R, then inf S = a, while sup S does not exit since S\nis not bounded from below.\nHow do we know whether the supremum or in\ufb01mum of a non-empty subset S \u2286R exist\nas real numbers? We saw in Remark 2.18 that a necessary condition for the existence of the\nsupremum (resp. in\ufb01mum) of S is that S be bounded from above (resp. below).\nOn the other hand, if, for example, S is bounded from above (resp. below), then we know\nthat the set U (resp. L) of all upper (resp. lower) bounds of S is non-empty. Hence, it is\nlegitimate to ask if U (resp. L), when non-empty, admits a least (resp. largest) element.\nThe existence of the largest of all possible lower bounds (resp. of the least of all possible\nupper bounds) is one of the features of the construction of the real numbers. As we have already\nmentioned that we are not going to explain the construction of R, we will assume the existence\nof such elements. Indeed, it su\ufb03ces to assume the following axiom, which then implies the full\nexistence of in\ufb01ma and suprema, cf. Corollary 2.26.\n15", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_31", "text": "Axiom 2.22. [Infimum axiom] Each non-empty subset S of R\u2217\n+ admits an in\ufb01mum (which\nis a real number).\nRemark 2.23. In Mathematics, an axiom is a statement that we are going to assume to be true,\nwithout requiring for it a formal proof. When we introduce an axiom, we are free to use the\nproperties stated in the axiom, without requiring a proof for them, and we can use those to\nderive other mathematical properties of the objects that we are studying.\nThe property stated in the In\ufb01mum Axiom is a very important one. In a sense, which we\nwill try to make more precise when we introduce sequences of real numbers, this property says\nthat R does not contain any gaps. While at this time, this is a rather nebulous statement,\nlet us at least show that this axiom does not necessarily hold for all the number sets that we\nhave introduced so far, cf. Section 2.2: indeed, it is possible to show that the in\ufb01mum axioms\ndoes not necessarily hold for Q, for example, cf. Example 2.24 below. Hence, the In\ufb01mum\nAxiom is indeed an axiom stating a (very relevant) property that is indeed peculiar to the real\nnumbers and, as such, in this course we actually utilize it to characterize the real numbers,\nagain, cf. Remark 2.7.\nExample 2.24. Let S :=]\n\u221a\n3, 5[ \u2229Q.8 Then S \u2286R\u2217\n+ and the In\ufb01mum Axiom implies that\ninf S exists in the real numbers. We will show in Example 2.46 that inf S =\n\u221a\n3. In particular,\nthe set of lower bounds of S coincides with the real numbers \u2264\n\u221a\n3.\nSince S, by its very de\ufb01nition, is also a subset of Q, we may wonder whether it possible to \ufb01nd\na largest rational number l among the rational numbers which are lower bounds for S. Such\nl \u2208Q would then be an in\ufb01mum for S among the rational numbers. By the above observation,\nwe know that if such l existed, then l <\n\u221a\n3, since\n\u221a\n3 \u0338\u2208Q, cf Proposition 2.38, and l is certainly\na lower bound for l. But then, Proposition 2.44 shows that there exists a rational number m\nsuch that l < m <\n\u221a\n3. As m <\n\u221a", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_32", "text": "we know that if such l existed, then l <\n\u221a\n3, since\n\u221a\n3 \u0338\u2208Q, cf Proposition 2.38, and l is certainly\na lower bound for l. But then, Proposition 2.44 shows that there exists a rational number m\nsuch that l < m <\n\u221a\n3. As m <\n\u221a\n3, then we know that m is also a lower bound for S. This is\nclearly a contradiction, as m \u2208Q nad is a lower bound for S, while we had assumed that l was\nthe largest of all lower bounds of S that are rational. Hence, the in\ufb01mum of S cannot exist in\nQ.\nAxiom 2.22 requires that we work with subsets of R\u2217\n+ to be guaranteeed to \ufb01nd their\nin\ufb01mum. But, in general, we can \ufb01nd the in\ufb01mum also for sets that are not necessarily contained\nin R\u2217\n+, as long as we have some lower bounds.\nExample 2.25. The in\ufb01mum of a set S \u2286R can exist even when S \u0338\u2286R\u2217\n+. For example,\nlet S := {x \u2208R | x > \u2212\n\u221a\n17}. As S contains \u22121, for example, then S \u0338\u2286R\u2217\n+. On the other\nhand, by Proposition 2.10.6, the set of lower bounds of S is given by ] \u2212\u221e, \u2212\n\u221a\n17]. Hence,\ninf S = \u2212\n\u221a\n17.\nUsing the In\ufb01mum Axiom 2.22, we can actually prove that the in\ufb01mum (resp. the supre-\nmum) exists for any subset S \u2286R which is bounded from below (resp. from above).\nCorollary 2.26. Let S \u2286R be a non-empty set.\n(1) If S is bounded from below, then S admits an in\ufb01mum.\n(2) If S is bounded from above, then S admits a supremum.\nProof.\n(1) As S is bounded from below, there exists a lower bound l \u2208R for S, that is, l \u2264s,\nfor all s \u2208S. We can rewrite the previous inequality as\ns \u2212l \u22650,\n\u2200s \u2208S.\n(2.26.a)\n8See Section 2.4.1 for a formal proof that\n\u221a\n3 is actually a real number.\n16", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_33", "text": "Let W \u2286R be the subset obtained by translating the elements of S by \u2212l + 1,\nW := {s \u2212l + 1 | s \u2208S}.\nWhy did we choose to translate the elements of S by \u2212l+1? The reason is that W \u2286R\u2217\n+:\nin fact, by (2.26.a), s\u2212l+1 \u22651 > 0, for all s \u2208S.9 As W \u2286R\u2217\n+, the In\ufb01mum Axiom 2.22\nimplies that inf W exists, call it a := inf W. Then a is the largest lower bound for the\nset W.\nHow can we use a to compute inf S? To construct W, we translated all elements of S\nby \u2212l + 1. If we translate the elements of W back by l \u22121, then we undo what we\ndid before and we recover S. So, what happens if we translate a by l \u22121 as well? The\nnumber we obtain by this translation should be the largest lower bound for S, as addition\nis compatible with the order relation. Let us verify this.\nLet a\u2032 := a + l \u22121. Then a\u2032 \u2264w + l \u22121 for any element w \u2208W. As any w \u2208W is of\nthe form w = s \u2212l + 1 for some s \u2208S, then w + l \u22121 = s. Hence, a\u2032 \u2264s for all s \u2208S\nand a\u2032 is a lower bound for S. If a\u2032 is not the largest lower bound for S, then there is a\nreal number b\u2032 > a\u2032 which is a lower bound for S. But then b\u2032 \u2212l + 1 > a = a\u2032 \u2212l + 1\nand b\u2032 \u2212l + 1 would be a lower bound for W [prove it!]. But this is a contradiction, since\na = inf W.\n(2) The details are left to the reader. Here is a sketch.\nLet S\u2032 \u2286R be the set constructed by \ufb02ipping the sign of the elements of S,\nS\u2032 := {\u2212x | x \u2208S}.\nSince S is bounded from above, then S\u2032 is bounded from below. [Prove this!] Then by\npart (1), inf S\u2032 exists. It is left to you to show that sup S = \u2212inf S\u2032.", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_34", "text": "S\u2032 := {\u2212x | x \u2208S}.\nSince S is bounded from above, then S\u2032 is bounded from below. [Prove this!] Then by\npart (1), inf S\u2032 exists. It is left to you to show that sup S = \u2212inf S\u2032.\nWe have seen the de\ufb01nition of in\ufb01mum/supremum and minimum/maximum.\nBoth the\nin\ufb01mum (resp. supremum) and minimum (resp. maxima) of a set S, provided that they exist,\nare lower bounds (resp. upper bounds) for S. Can we be more precise about what is the\nrelationship among these notions?\nExample 2.27. Let S := [3, 5[ \u2286R. Then, min S = 3 = inf S. On the other hand, max S\ndoes not exist as sup S = 5 is the least upper bound and 5 \u0338\u2208S; hence no upper bound of S is\ncontained in S, as any element of S is < 5.\nThe example above seems to suggest that, at least for intervals, if the minimum (resp.\nmaximum) of an interval exists, then it should coincide with the in\ufb01mum (resp. the supremum)\nof the interval. This property actually holds for any non-empty subset S \u2282R, as long as the\nminimum (resp. maximum) of S exists.\nProposition 2.28. Let S \u2286R a non-empty set.\n(1) If min S exists, then min S = inf S.\n(2) If max S exists, then max S = sup S.\nProof. We prove (1), whereas (2) is left as an exercise. As min S exists, then S is bounded\nfrom below, since min S is in particular a lower bound, cf. De\ufb01nition 2.11. Hence, inf S exists,\nby Corollary 2.26. Then inf S \u2265min S since inf S is the largest of all lower bounds. On the\nother hand, min S \u2208S, and inf S \u2264s, for all s \u2208S. In particular, inf S \u2264min S. Thus,\ninf S \u2264min S and inf S \u2265min S, which implies that inf S = min S.", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_35", "text": "other hand, min S \u2208S, and inf S \u2264s, for all s \u2208S. In particular, inf S \u2264min S. Thus,\ninf S \u2264min S and inf S \u2265min S, which implies that inf S = min S.\n9We could have choosen to translate by \u2212l + c, for any c > 0. Hence the choice of c = 1 was arbitrary, but I\nneeded to choose something explicit, so I went for 1.\n17", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_36", "text": "2.3.2\nArchimedean property of R\nAs we have already mentioned, given any two real numbers x, y we can always compare them,\nthat is, we can decide whether either x = y, or x < y or x > y. On the other hand, whenever\nit makes sense, for example, if x, y are both non-negative real numbers with x < y, we may\nask a more general question: namely, we may ask whether, by taking multiples of x, we can\neventually construct a real number nx > y.\nExample 2.29. Let y = \u03c020 and let x = 1. We want to \ufb01nd a natural number n such that\nnx = n \u00b7 1 = n is > \u03c020. If we write \u03c020 in its decimal representation,\n\u03c020 =8769956796.082699474752255593703897066064114447195437243420984260\n51841239043547990990234985186673598315695604864892372705666 . . .\n.\n10\nThen if we take n = 8769956797, that is, n is equal to the integral part of the decimal repre-\nsentation of \u03c020 + 1, then n = n \u00b7 1 = nx > \u03c020 = y.\nWhen we discussed real numbers at the start of the course, we saw that perhaps it is not\nan ideal method that of relying on their decimal representation. After all, it is not even clear\nthat we can compute e\ufb00ectively the decimal representation of any real number. (Have you\never thought about how computers are able to calculate decimal representations of irrational\nnumbers? If you are curious about that, you may want to take a look here). We said that in\nthis course, we should rather try to prove properties of the real numbers by relying on their\nintrinsic mathematical properties, and by using mathematical proofs as the tools to connect\nproperties and discover new one.\nThe interesting fact, is that we can actually prove that the conclusion of Example 2.29\nholds, in full generality, for any pair of positive numbers x, y.\nProposition 2.30 (Archimedeand property of R). Let x, y be real numbers, with x > 0, y \u22650.", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_37", "text": "holds, in full generality, for any pair of positive numbers x, y.\nProposition 2.30 (Archimedeand property of R). Let x, y be real numbers, with x > 0, y \u22650.\nThen there exists n \u2208N\u2217such that nx > y.\nProof. If y = 0, then take n = 1. Then nx = 1 \u00b7 x = x > 0 and we are done.\nLet us now assume that y > 0. We make a proof by contradiction. Let us assume that\n\u2200n \u2208N,\nnx \u2264y.\n(2.30.b)\nLet S \u2286R be the set\nS := {nx | n \u2208N}.\nThen S is non-empty as x \u2208S, and S is bounded from above, as y is an upper bound by (2.30.b).\nHence, by Corollary 2.26 sup S exists and (n+1)x \u2264sup S for all n \u2208N. Thus, nx \u2264sup S \u2212x\nfor all n \u2208N, that is, s \u2264sup S \u2212x, for all s \u2208S. But this implies that sup S \u2212x is an upper\nbound for S, too. As sup S \u2212x < sup S, since x > 0, this gives a contradiction to the fact that\nsup S is the supremum of S, i.e., the smallest upper bound for S.\nCorollary 2.31. Let y \u2208R+. Then there exists n \u2208N\u2217such that n > y.\nProof. It is enough to apply Proposition 2.30 to y, taking x = 1.\n2.3.3\nAn alternative de\ufb01nition for in\ufb01mum/supremum\nLet S \u2282R be a non-empty set. We have seen in Section 2.3.1 that the in\ufb01mum and supremum\nof S are unique, when they exist. Moreover, as the in\ufb01mum (resp. supremum) of S is the\nlargest (resp. the smallest) lower bound (resp. upper bound) of S, then whenever we take\na number c larger than inf S (resp. smaller than sup S), we must be able to \ufb01nd an element", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_38", "text": "largest (resp. the smallest) lower bound (resp. upper bound) of S, then whenever we take\na number c larger than inf S (resp. smaller than sup S), we must be able to \ufb01nd an element\ns \u2208S contained between inf S and c (resp. between c and sup S), that is, inf S \u2264s < c (resp.\n18", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_39", "text": "c < d \u2264sup S).\nUsing this reasoning, we can characterize the in\ufb01mum (resp. supremum) of S in the following\nalternative way.\nProposition 2.32. Let S \u2282R be a non-empty set.\n(1) A real number u is the supremum of S if and only if\n(i) u is an upper bound for S, and\n(ii) for all \u03b5 > 0, there is s\u03b5 \u2208S, such that s\u03b5 > u \u2212\u03b5.\n(2) A real number l is the in\ufb01mum of S if and only if\n(i\u2019) l is an lower bound for S, and\n(ii\u2019) for all \u03b5 > 0, there is s\u2032\n\u03b5 \u2208S, such that s\u2032\n\u03b5 < l + \u03b5.\nThe criterion just introduced is very useful in practice when trying to prove that a certain\nreal number is the in\ufb01mum/supremum of a given subset of the real numbers.\nExample 2.33. Let S := {1 \u22121\nn | n \u2208Z\u2217\n+}. We show that sup S = 1 using Proposition 2.32.1.\nTo this end, we must verify that 1 satis\ufb01es both properties:\n(i) 1 is an upper bound for S, and\n(ii) for all \u03b5 > 0, there is s\u03b5 \u2208S, such that s\u03b5 > 1 \u2212\u03b5.\nSince 1 \u22651 \u22121\nn, for all n \u2208N\u2217, then, by de\ufb01nition, of upper bound, 1 is an upper bound for\nS; thus, property (i) is satis\ufb01ed.\nTo verify (ii), let, for example, \u03f5 = 3\n17; then we have to show that there exists an element s\u03f5 of\nS such that\n1 \u22123\n17 < s\u03f5 < 1.\n(The second inequality comes for free from the fact that 1 is an upper bound for S). If we take\ns\u03f5 = 1 \u22121\n17, then s\u03f5 \u2208S, and since\n1\n17 < 3\n17\n1 \u22123\n17 < 1 \u22121\n17 < 1\nwhich is what we wanted.\nTo make the proof more general, we have to \ufb01x a positive real number \u03f5 (this could be any\npositive real number, but we are thinking that we have \ufb01xed one speci\ufb01c value for \u03f5). Again,\nwe have to \ufb01nd an element s\u03f5 \u2208S (this element that we construct will depend on the initial", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_40", "text": "positive real number, but we are thinking that we have \ufb01xed one speci\ufb01c value for \u03f5). Again,\nwe have to \ufb01nd an element s\u03f5 \u2208S (this element that we construct will depend on the initial\nchoice of \u03f5, that is why we denote it as s\u03f5, to remind ourselves about the dependence from \u03f5)\nsuch that 1 \u2212\u03f5 < s\u03f5.\nIf \u03f5 > 1 then 1 \u2212\u03f5 < 0, hence we can just take s\u03f5 = 0 = 1 \u22121\n1 \u2208S. If \u03f5 \u22641, then 1 \u2212\u03f5 \u2208[0, 1[.\nHow can \ufb01nd \ufb01nd n \u2208N such that 1 \u2212\u03f5 < 1 \u22121\nn? The inequality 1 \u2212\u03f5 < 1 \u22121\nn is equivalent\nto the inequality n > 1\n\u03f5 [Check that!]. As \u03f5 > 0, also 1\n\u03f5 > 0. Hence, by Corollary 2.31 we can\n\ufb01nd a natural number k such that k > 1\n\u03f5. Then 1 \u2212\u03f5 < 1 \u22121\nk, so that we can take s\u03f5 := 1 \u22121\nk.\nProof of Proposition 2.32. We show part (1). The proof of part (2) is completely analogous\nand is left as an exercise for the reader.\nWe \ufb01rst prove the implication\nl = inf S =\u21d2l satis\ufb01es properties (i) and (ii) in Proposition 2.32.\nLet l = inf S.\nAs inf S is the largest of all lower bounds for S, by Proposition 2.32.1, in\nparticular l is a lower bound for S; thus, l satisfy said property. As inf S is the largest lower\n19", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_41", "text": "bound, by itw de\ufb01nition, then if we take any \u03f5 > 0, l + \u03f5 cannot be a lower bound for S.\nThis means that there exists an element of S (which will depend on the choice of \u03f5 in general,\ncf. Example 2.33), call it s\u03f5, such that s\u03f5 < l + \u03f5, which shows that l satis\ufb01es also property (ii)\nof Proposition 2.32.\nWe then prove the other implication:\nl satis\ufb01es properties (i) and (ii) in Proposition 2.32 =\u21d2l = inf S.\nLet us assume, by contradiction, that l \u0338= inf S. Since by property (i) l is a lower bound, the\nassumption that l \u0338= inf S means that l is not the largest lower bound. Hence, there exists\nl\u2032 \u2208R, l\u2032 > l and l\u2032 is a lower bound for S. In particular,\nfor all s \u2208S, s \u2265l\u2032.\n(2.33.c)\nTake \u03f5 := l\u2032 \u2212l > 0 =\u21d2l + \u03f5 = l\u2032. Then (2.33.c) implies that no element of S is < l + \u03f5.\nBut, this is in contradiction with property (ii) of Proposition 2.32 which we assumed to begin\nwith.\n2.3.4\nIn\ufb01mum and supremum for subsets of Z\nWhen we de\ufb01ned the natural numbers in Section 2.2 we mentioned that any subset of N has a\nminimum. We have now all the tools to prove this statement, which will be one of our standard\ntools for the duration of the course.\nProposition 2.34. Let S \u2286R be a non-empty set of natural numbers. Then, inf S = min S.\nWhat is the important information contained in the statement of the above proposition?\nAs S \u2286N, S is bounded from below.\nHence, the In\ufb01mum Axiom 2.22 implies that inf S\nexists. On the other hand, we know from Proposition 2.28 that if the minimum of S exists,\nthen it must always coincide with inf S. Hence, the important bit of information contained\nin Proposition 2.34 is that the minimum of any set S \u2286N indeed exists, a property that we\nhad already mentioned in Section 2.2.\nExample 2.35. Let", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_42", "text": "then it must always coincide with inf S. Hence, the important bit of information contained\nin Proposition 2.34 is that the minimum of any set S \u2286N indeed exists, a property that we\nhad already mentioned in Section 2.2.\nExample 2.35. Let\nS := {x \u2208R | x \u2208N\u2217and x is divisible by at least 5 distinct prime numbers} .\nThen, by de\ufb01nition, S is a set of natural numbers and certainly 1, 2, 3, 5 are not elements of S;\neven better, no prime number p \u2208N is an element of S. On the other hand, Proposition 2.34\nimplies that S has a minimum.\nHow can we compute min S? That is, what is the minimum natural number that is divisible\nby 5 distinct prime numbers? As any natural number can be written essentially uniquely as a\nproduct of prime numbers, then min S is the product of the 5 smallest prime numbers. The\n\ufb01rst few prime numbers are\n2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, . . . .\nHence, min S = 2 \u00b7 3 \u00b7 5 \u00b7 7 \u00b7 11 = 2310.\nProof of Proposition 2.34. Let d := inf S, which exists by Corollary 2.26, since S is bounded\nfrom below. We have to show that d \u2208S.\nAssume by contradiction that d \u0338\u2208S. Then, as inf S is the largest lower bound of S, for each\n\u03b5 > 0, d + \u03b5 is not a lower bound. Hence:\nfor all \u03b5 > 0, there is s\u03b5 \u2208S, such that s\u03b5 < d + \u03b5.\n(2.35.d)\n20", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_43", "text": "Apply (2.35.d) with \u03b5\u2032 := 1\n2. This yields an element s\u03b5\u2032 of S such that\nd < s\u03b5\u2032 < d + \u03b5\u2032 = d + 1\n2\nApply then again the above property of S, but now for \u03b5\u2032\u2032 := s\u03b5\u2032 \u2212d > 0. Then, we can \ufb01nd\ns\u03b5\u2032\u2032 \u2208S such that\nd < s\u03b5\u2032\u2032 < d + \u03b5\u2032\u2032 = s\u03b5\u2032 < d + \u03b5\u2032 = d + 1\n2.\nIn particular, 0 < s\u03b5\u2032 \u2212s\u03b5\u2032\u2032 < d + 1\n2 \u2212d = 1\n2. This gives a contradiction, since s\u03b5\u2032, s\u03b5\u2032\u2032 \u2208N and\nthe distance between two di\ufb00erent natural numbers is always at least 1 one from. Hence, our\ninitial assumption that d \u0338\u2208S must be false, so that d \u2208S.\nExercise 2.36. Let S \u2286R a subset of the integers.\n(1) If S is bounded from below, then min S = inf S.\n(2) If S is bounded from above, then max S = sup S.\n[Hint: for (1), let a be a lower bound for S; then a > [a] \u22121 is an integer > a. Consider the\nset S\u2032 := {s \u2212[a] + 1 | s \u2208S} \u2286N and try to imitate the proof of Corollary 2.26.\nFor (2), de\ufb01ne the set S\u2032\u2032 := {\u2212x | x \u2208S} and then imitate the proof of Corollary 2.26 and use\n(1) to prove (2).]\n2.4\nRational numbers vs real numbers\n2.4.1\n\u221a\n3 is a real number\nWe have seen that\n\u221a\n3 is not a rational number, cf. Proposition 1.1.\nQuestion 2.37. Why is\n\u221a\n3 a real number?\nWe are going to show that using the Axiom 2.22, we can formally show that there exists\na positive real number x satisfying the equation x2 = 3. By, its own de\ufb01nition, then x =\n\u221a\n3.\nTo this end, let us consider S := {x \u2208R | x2 \u22643}. First of all, S is a non-empty subset of R,\nsince 1 \u2208S. Moreover, S is bounded: in fact, 3 is an upper bound and \u22123 is a lower bound for", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_44", "text": "3.\nTo this end, let us consider S := {x \u2208R | x2 \u22643}. First of all, S is a non-empty subset of R,\nsince 1 \u2208S. Moreover, S is bounded: in fact, 3 is an upper bound and \u22123 is a lower bound for\nS. [Prove it! Remember that for real numbers x > y > 0, then x2 > y2 > 0.] As S is bounded\nthen by Corollary 2.26 both the in\ufb01mum and the supremum of S exists. As 1 \u2208S, then the\nsupremum of S is \u22651, in particular it is > 0. We will show that sup S =\n\u221a\n3.\nProposition 2.38. Let S \u2286R be the subset\nS := {x \u2208R | x2 \u22643}.\nThen inf S < 0 < sup S and (sup S)2 = 3 = (inf S)2. Thus, sup S =\n\u221a\n3, inf S = \u2212\n\u221a\n3.\nProof. We have already shown above that inf S and sup S exist. Moreover, as \u00b11 \u2208S, then it\nfollows at once that inf S < \u22121 < 0 < 1 < sup S. Hence, if (sup S)2 = 3 = (inf S)2, then the\nabove chain of inequalities implies that sup S =\n\u221a\n3, inf S = \u2212\n\u221a\n3.\nWe now show that (sup S)2 = 3. The veri\ufb01cation for inf S is analogous.\nLet us assume, for the sake of contradiction, that (sup S)2 \u0338= 3 and let us show that we obtain\na contradiction. We have 2 possible cases:\n(\n(sup S)2 > 3,\n(sup S)2 < 3\n.\nCase 1: Assume (sup S)2 > 3.\nWe shall show that there exists a su\ufb03ciently large n \u2208N such that sup S \u22121\nn is an upper bound\n21", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_45", "text": "for S. This immediately yields the desired contradiction, since sup S \u22121\nn < sup S and sup S is\nby de\ufb01nition the least of all upper bounds.\nAs sup S > 1, then sup S \u22121\nn > 0 for all n \u2208N\u2217. Hence to show that for some n \u2208N\u2217, sup S \u22121\nn\nis an upper bound for S, it su\ufb03ces to show that (sup S \u22121\nn)2 > 3, since for x > 0, x < sup S \u22121\nn\nif and only if x2 < (sup S \u22121\nn)2. But\n(sup S \u22121\nn)2 = (sup S)2 + 1\nn2 \u22122 sup S\nn\n> (sup S)2 \u22122 sup S\nn\n.\nHence, it su\ufb03ces to show that we can \ufb01nd n \u2208N large enough such that (sup S)2 \u22122 sup S\nn\n> 3.\nLet us denote by d := (sup S)2 \u22123 which is a positive real number. But then, \ufb01nding n \u2208N\u2217\nsuch that (sup S)2 \u22122 sup S\nn\n> 3 is equivalent to \ufb01nding n \u2208N\u2217such that 2 sup S\nn\n< d, and the\nlast inequality is equivalent to n >\nd\n2 sup S , since sup S > 0. The existence of n \u2208N\u2217such that\nn >\nd\n2 sup S is guaranteed by the archimidean property, Proposition 2.30. This concludes the\nproof in Case 1.\nCase 2: Assume (sup S)2 < 3.\nWe shall show that there exists n \u2208N\u2217such that (sup S + 1\nn)2 < 3. As sup S + 1\nn > sup S > 0,\nthis implies that sup S + 1\nn \u2208S which will yield the desired contradiction, since sup S must be\nan upper bound of S. Let d\u2032 be the positive real number d\u2032 := 3 \u2212(sup S)2. Then since\n(sup S + 1\nn)2 = (sup S)2 + 1\nn2 + 2 sup S\nn\n< (sup S)2 + 1\nn + 2 sup S\nn\n,\nit su\ufb03ces to show that there exists n \u2208N\u2217such that (sup S)2 + 1\nn + 2 sup S", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_46", "text": "n2 + 2 sup S\nn\n< (sup S)2 + 1\nn + 2 sup S\nn\n,\nit su\ufb03ces to show that there exists n \u2208N\u2217such that (sup S)2 + 1\nn + 2 sup S\nn\n< 3. But this is\nequivalent to \ufb01nding n \u2208N\u2217such that n >\nd\n1+2 sup S . The existence of one such n \u2208N\u2217is again\nguaranteed by the archimidean property of R, cf. Proposition 2.30.\n2.4.2\nIntegral part\nLet x be a real number. According to Exercise 2.36, the set S := {n \u2208N | n \u2264x} has a\nmaximum, since it is bounded from above. Call m := max S. Then m + 1 is not in S, as m is\nthe largest element of S. We call the integer m the integral part of x and we denote it by [x].\nDe\ufb01nition 2.39. Let x \u2208R.\n(1) The round-down \u230ax\u230bof x is the largest integer that is \u2264x.\n(2) The round-up \u2308x\u2309of x is the least integer that is \u2265x.\n(3) The integral part [x] of x is de\ufb01ned as\n[x] :=\n(\n\u230ax\u230b\nfor x \u22650,\n\u2308x\u2309\nfor x < 0.\nWe can also de\ufb01ne the fractional part of x.\nDe\ufb01nition 2.40. Let x be a real number. Then the fractional part {x} of x is de\ufb01ned as\n{x} := x \u2212[x].\nExercise 2.41. For all x \u2208R,\n(1) \u230ax\u230b\u2264x < \u230ax\u230b+ 1;\n(2) \u2308x\u2309\u22121 < x \u2264\u2308x\u2309;\n22", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_47", "text": "(3) [x] = \u2212[\u2212x];\n(4) {x} \u2208] \u22121, 1[ and {x} = \u2212{\u2212x}\n(5) x = [x] + {x};\n(6) x \u2208Z if and only if x = \u230ax\u230b= \u2308x\u2309= [x].\nExample 2.42.\n(1) [\u22124] = \u22124 and hence {\u22124} = 0. In general, if z \u2208Z, then [z] = z,\n{z} = 0.\n(2) Considering the number x = \u03c02 + \u03c0,\n\u03c02 + \u03c0 =13.0111970546791518572971343831556540195108688066158964473882939\n68527861228705414241629808229060669299806174000287305450724866192 . . .\nHence, [\u03c02 + \u03c0] = 13, and {\u03c02 + \u03c0} = \u03c02 + \u03c0 \u221213 \u2013 not a number that we can fully write\ndown with decimals.\n(3) For rational numbers, things are a bit easier. For example,\n\u0002\n\u22123\n2\n\u0003\n= \u22121 and\n\b\n\u22123\n2\n\t\n= \u22121\n2.\n(4) Roughly speaking, when we write a real number x by means of its decimal representation,\nthen the integral part [x] (as its name suggests) stands for the integral number whose\ndigits are left of the \u201c.\u201d dividing integral and decimal part, while {x} stands for the real\nnumber in ] \u22121, 1[ whose digits are right of the \u201c.\u201d dividing integral and decimal part:\nfor example, [7.8324123] = 7, {7.8324123} = 0.8324123.\n2.4.3\nRational numbers are dense in R\nWe have already observed that Q \u228aR. It would be nice to have some more information about\nhow rational numbers are distributed among real numbers. For example, we may ask if we can\n\ufb01nd rational numbers between two arbitrary real numbers.\nExample 2.43. For example, is there a rational number c, such that 0 < c < \u03c0? The left\ninequality, that is, 0 < c, is an easy one to guarantee. It su\ufb03ces to choose c to be a positive", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_48", "text": "\ufb01nd rational numbers between two arbitrary real numbers.\nExample 2.43. For example, is there a rational number c, such that 0 < c < \u03c0? The left\ninequality, that is, 0 < c, is an easy one to guarantee. It su\ufb03ces to choose c to be a positive\nrational number. But, how do we guarantee that the inequality on right holds as well? Well,\nas soon as c is positive, c < \u03c0 is equivalent to 1\nc > 1\n\u03c0. So, if one chooses 1\nc to be any integer\nthat is larger than 1\n\u03c0 we are \ufb01ne. For example, we can choose\n1\nc =\n\u0014 1\n\u03c0\n\u0015\n+ 1\nthat is,\nc =\n\u0012\u0014 1\n\u03c0\n\u0015\n+ 1\n\u0013\u22121\n.\nIt is not too hard to show that the above example can be extended in more generality to\nany two real numbers.\nProposition 2.44. If a < b are real numbers, then there is a rational number c, such that\na < c < b.\nWe can summarize the property stated in Proposition 2.44 by saying that \u201crational numbers\nare arbitrarily close to any real number\u201d. In more precise mathematical terms, we refer to the\nproperty stated in Proposition 2.44 above by saying that Q is dense in R.\nExample 2.45. Let us consider\n\u221a\n2 =1.414213562373095048801688724209698078569671875376948073176679737990\n7324784621070388503875343276415727350138462309122970249248360 . . . .\n23", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_49", "text": "We know that\n\u221a\n2 is not a rational number. Then, how can we show that rational numbers are\narbitrarily close to\n\u221a\n2? We could try to approximate\n\u221a\n2 by means of rational numbers.\nSo, for example, what is a rational number that is close within\n1\n10 of\n\u221a\n2?\nProposition 2.44\ntells us that such approximation certainly exists, as it guarantees that we can \ufb01nd a rational\nnumber c such that\n\u221a\n2 \u22121\n10 < c <\n\u221a\n2. But, in practice, how can we \ufb01nd such c? Using the\ndecimal expansion of\n\u221a\n2 above, we can immediately notice that\n\u221a\n2 \u22121.4 =0.014213562373095048801688724209698078569671875376948073176679737990\n7324784621070388503875343276415727350138462309122970249248360 . . . .\nHence,\n\u221a\n2 \u22121\n10 < 1.4 <\n\u221a\n2.\nIn the same way, if we want to approximate\n\u221a\n2 up to\n1\n10000 with rational, we can search for a\nrational number c\u2032 such\n\u221a\n2 \u2212\n1\n10000 < c\u2032 <\n\u221a\n2. As before, by taking c\u2032 = 1.41421 we obtain\nthat\n\u221a\n2 \u22121.41421 =0.000003562373095048801688724209698078569671875376948073176679\n7379907324784621070388503875343276415727350138462309 \u00b7 \u00b7 \u00b7 <\n1\n10000.\nIn the same way, if we want to approximate\n\u221a\n2 within\n1\n10n , then it is enough to take the\nrational number whose decimal representation is given by taking that of\n\u221a\n2 and truncating it\nafter the n-th decimal digit.\nProof. Let us start with a simple case of our proof.\nEasy case; we assume a = 0:\nWe have\n\u0004 1\nb\n\u0005\n+ 1 > 1\nb and\n\u0004 1\nb\n\u0005\n+ 1 is a positive integer. We conclude that\n0 <\n1\n\u0004 1\nb\n\u0005\n+ 1 < b ,\nso we can take c =\n1\n\u230a1\nb\u230b+1.\nGeneral case:", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_50", "text": "Easy case; we assume a = 0:\nWe have\n\u0004 1\nb\n\u0005\n+ 1 > 1\nb and\n\u0004 1\nb\n\u0005\n+ 1 is a positive integer. We conclude that\n0 <\n1\n\u0004 1\nb\n\u0005\n+ 1 < b ,\nso we can take c =\n1\n\u230a1\nb\u230b+1.\nGeneral case:\nLet us de\ufb01ne the number n :=\nj\n1\nb\u2212a\nk\n+ 1. Then,\nn =\n\u0016\n1\nb \u2212a\n\u0017\n+ 1 \u21d2n >\n1\nb \u2212a \u21d21\nn < b \u2212a\na = an\nn < \u230aan\u230b+ 1\nn\n\u2264an + 1\nn\n= a + 1\nn < a + b \u2212a = b\nFurthermore, \u230aan\u230b+1\nn\nis a rational number. Hence, to conclude it su\ufb03ces to take c = \u230aan\u230b+1\nn\n.\n[This is not the unique rational number between a and b, it is just one example of a rational\nnumber between a and b.]\nExample 2.46. This is a continuation of Example 2.24. We can \ufb01nally prove that for S :=\n]\n\u221a\n3, 5[ \u2229Q then the in\ufb01mum of S in R is\n\u221a\n3.\nBy de\ufb01nition of S, any element of S is >\n\u221a\n3 =\u21d2\n\u221a\n3 is a lower bound.\nLet us assume by contradiction that that\n\u221a\n3 is not the in\ufb01mum of S =\u21d2\n\u221a\n3 < inf S < 5\nand by Proposition 2.44, there exists a rational number c such that\n\u221a\n3 < c < inf S < 5. But\nthen c \u2208S since c \u2208Q and c \u2208]\n\u221a\n3, 5[, and c < inf S, which provides a contradiction. Hence\ninf S =\n\u221a\n3.\n24", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_51", "text": "2.4.4\nIrrational numbers are dense in R\nThe same property of density in R that we showed holds for Q, in the previous section, holds\nalso for the complement R \\ Q of Q in R. The set R \\ Q is called the set of irrational numbers.\nProposition 2.47. If a < b are real numbers, then there is c \u2208R \\ Q, such that a < c < b.\nThe set R \\ Q of real numbers which are not rational is called the set of irrational numbers.\nRemark 2.48. Let us recall that if f \u2208Q\u2217and g \u2208R\u2217\\ Q, then fg \u2208R\u2217\\ Q.\nProof. Apply Proposition 2.44 to\na\n\u221a\n3 <\nb\n\u221a\n3. This yields a rational number d such that\na\n\u221a\n3 <\nd <\nb\n\u221a\n3. Additionally we can assume that d \u0338= 0: indeed, if d = 0 then it su\ufb03ces to replace d\nby the rational number that one can obtain by applying Proposition 2.44 to 0 and\nb\n\u221a\n3. Hence,\na <\n\u221a\n3d < b and d \u0338= 0.\nIt remains to show that\n\u221a\n3d is irrationalbut this follows at once from Remark 2.48.\n2.5\nAbsolute value\nDe\ufb01nition 2.49. If x \u2208R, then the absolute value |x| of x is de\ufb01ned as follows:\n|x| =\n\u001a\nx\nif x \u22650\n\u2212x\nif x \u22640.\nExample 2.50. |3| = 3, | \u22125| = 5, | \u2212\u03c0| = \u03c0, |0| = 0 and |5| = 5.\nIt is useful to remember the graph of the absolute value function,see Figure 1.\n../Images/graph_abs_function.png\nFigure 1: The graph of f(x) = |x|\nAnother way to de\ufb01ne the absolute value |x| of x \u2208R is to de\ufb01ne it as the distance between\nx and 0 on the real line.\n2.5.1\nProperties of the absolute value\nHow does the absolute value |x| of a real number x compare to x itself, in relation to the usual\nordering on R?\nExample 2.51. \u2212\n\f\f\u2212\n\u221a\n3\n\f\f \u2264\u2212\n\u221a\n3 and | \u2212\n\u221a\n3| \u2265\u2212\n\u221a\n3.\nThe inequalities in the above example hold for any real number: that is, for x \u2208R\n\u2212|x| \u2264x \u2264|x|.", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_52", "text": "ordering on R?\nExample 2.51. \u2212\n\f\f\u2212\n\u221a\n3\n\f\f \u2264\u2212\n\u221a\n3 and | \u2212\n\u221a\n3| \u2265\u2212\n\u221a\n3.\nThe inequalities in the above example hold for any real number: that is, for x \u2208R\n\u2212|x| \u2264x \u2264|x|.\n(2.51.a)\nThe absolute value behaves well with respect to the multiplication.\n25", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_53", "text": "Example 2.52. |5 \u00b7 (\u22123)| = | \u221215| = 15 = 5 \u00b7 3 = |5| \u00b7 | \u22123|. Similarly,\n\f\f\f(\u2212\n\u221a\n2) \u00b7 (\u22124)\n\f\f\f =\n\f\f\f4\n\u221a\n2\n\f\f\f = 4\n\u221a\n2 =\n\u221a\n2 \u00b7 4 =\n\f\f\f\u2212\n\u221a\n2\n\f\f\f \u00b7 | \u22124|.\nWe can generalize Example 2.52: indeed, for all x, y \u2208R\n|x| \u00b7 |y| = |x \u00b7 y|.\nTo prove this, you can just list all possible combinations for the signs of x, y (that is,\n\u201cpositive\u201d\u2013\u201cpositive\u201d; \u201cpositive\u201d\u2013 \u201cnegative\u201d; \u201cnegative\u201d\u2013\u201cnegative\u201d) and prove the equality\nin each case. Analogously, in the case of division, for x, y \u2208R, y \u0338= 0, we have that\n\f\f\f\f\nx\ny\n\f\f\f\f = |x|\n|y|.\nExample 2.53.\n\f\f\f 5\n\u22124\n\f\f\f =\n|5|\n|\u22124|.\nThe absolute value is also needed to relate powers and roots.\nExample 2.54.\np\n(\u22123)2 =\n\u221a\n9 = 3 = | \u22123| and\np\n(7)2 =\n\u221a\n49 = 7 = |7|.\nIn general, for x \u2208R,\n\u221a\nx2 = |x|. This can be generalized to any n-th root of the n-th\npower of a real number when n is an even natural number.\n2.5.2\nTriangular inequality\nWhile we have seen that the absolute value is compatible with multiplication,that is, the abso-\nlute value of a product of two terms is equal to the product of the absolute values of the terms,\nthe same does not hold for addition.\nExample 2.55. |(\u22123) + 2| = | \u22121| = 1 \u0338= | \u22123| + |5| = 8. To be more precise, |(\u22123) + 2| =\n1 < 8 = | \u22123| + |5|.\nSo, while it is clear from the above example the the absolute value of a sum of two real\nnumbers is not necessarily equal to the sum of their absolute values, perhaps we may hope to\nstill be able to say something. What the second part of the example suggests is that Is this a\ngeneral property of the absolute value over the real numbers?\nIndeed, it is. A deep property of the absolute value is the so-called triangle inequality,", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_54", "text": "still be able to say something. What the second part of the example suggests is that Is this a\ngeneral property of the absolute value over the real numbers?\nIndeed, it is. A deep property of the absolute value is the so-called triangle inequality,\nwhose name is rooted in geometric considerations that we already clear at the times of Euclid.\nQuestion 2.56. Can you draw a triangle with sides of length 1, 4, and 600?\nI do not think so. On the other hand, it is possible to draw a triangle whose sides have\nlength 3,4 and 6 (give it a try, you might need a compass).\nWhat kind of constraints should we place on The reason is that for every triangle, the sum\nof the length of two edges is always bigger then the length of the third edge. This implies a\ntriangle inequality for the absolute value, we will understand better the relation with triangles\nwhen dealing with complex numbers, let us give a couple of examples now:\n|3 + (\u22127)| \u2264|3| + | \u22127|\nand\n|(\u22125) + (\u22124)| \u2264| \u22125| + | \u22124|\nIn general, we can prove the following.\n26", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_55", "text": "Proposition 2.57 (Triangle inequality). For all x, y \u2208R\n|x + y| \u2264|x| + |y|.\nProof. Recall that x \u2264|x| and y \u2264|y|. So, if x + y \u22650, then |x + y| = x + y \u2264|x| + |y|.\nSimilarly, x \u2265\u2212|x| and y \u2265\u2212|y|. So, if x + y \u22640, then |x + y| = \u2212x \u2212y \u2264|x| + |y|.\nExercise 2.58. Prove that for any x, y \u2208R\n|x \u2212y| \u2265||x| \u2212|y||\n27", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_56", "text": "3\nCOMPLEX NUMBERS\nWhen we work over the real numbers, we will be often working with functions of the form\nf : R \u2192R. We will be intersted in understanding and studying the properties (e.g., deriva-\ntives, integrals, monotonicity) of certain classes of functions (e.g., continuous, di\ufb00ernetiable,\nintegrable functions). Oftentimes, we will also be interested in understanding if and when a\nfunction f : R \u2192R attains a speci\ufb01c value c \u2208bR. Let us give an example.\nExample 3.1. Imagine that we are observing a particle moving along a linear rod. We can\nmodel the linear rod with the real line. We would like to keep track of how the particle moves\nas a function of time. Hence, we can think of the position of the particle as a function p: R \u2192R\nde\ufb01ned as follows\np(t) := position of the particle along the line at time t.\nWe can assume that at time t = 0 (the starting time of our observation) the particle is placed\nat the origin. Let us assume that we also know that at time t = 0 the particle is moving\nwith velocity v11. If no outer forces act on the particle, then the velocity of the particle stays\nconstant and the position can be easily written in terms of time in the form p(t) = v \u00b7 t.\nLet us assume instead that we know that there is there is a force acting on the particle and that\nforce applies a (constant) deceleration to the particle of magnitude a directed in the opposite\nverse than that of the velocity. In this casem then the position of the particle is given by\np(t) = \u22121\n2at2 + vt. Hence, if we wanted to know whether at a certain point in time the particle\npasses at a \ufb01xed point c \u2208R on the rod, we have to solve the equation\np(t) = c\nwhich we can rewrite as\n\u22121\n2at2 + vt \u2212c = 0\nor, equivalently\nat2 \u22122vt + 2c = 0,\nwhere the second equality follows from the \ufb01rst by \ufb02ipping the signs and multiplying the \ufb01rst\nequation by 2. We rewrite the above equation in the form\naT 2 \u22122vT + 2c = 0,\n(3.1.a)", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_57", "text": "where the second equality follows from the \ufb01rst by \ufb02ipping the signs and multiplying the \ufb01rst\nequation by 2. We rewrite the above equation in the form\naT 2 \u22122vT + 2c = 0,\n(3.1.a)\nwhere a, v, c are \ufb01xed real numbers, while the unknown that we are trying to compute is given\nby T. As you probably already know, the above equation admits the following two real solutions\nt1 = 2v +\n\u221a\n4v2 \u22128ac\n2a\n,\nt2 = 2v \u2212\n\u221a\n4v2 \u22128ac\n2a\n,\nprovided that the quantity 4v2 \u22128ac \u22650 (since the square root of a real number is well de\ufb01ned\nonly for non-negative real numbers). If 4v2 \u22128ac < 0, then we cannot possibly \ufb01nd any real\nsolution to (3.1.a)\nHow do we remedy the lack of solutions for polynomial equations in the real numbers?\nPolynomials are a big and relatively simple class of functions that appear rather naturally in\nmany contexts. Hence, it would be nice to know that we can always \ufb01nd solutions to polynomial\nequations. On the other hand, the above example tells us that this is not possible, if we just\nwork with real numbers. The solution to this problem is a classic piece of mathematical wisdom.\nWhen you are lacking a tool, why not invent it yourself? This is the idea behind the de\ufb01nition\nof the complex numbers that we now proceed to explain.\n11Here v could have both positive or negative sign, meaning that the particle is moving in the direction of the\npositive real numbers or in the direction of the negative ones, along the linear rod.\n28", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_58", "text": "3.1\nDe\ufb01nition\nAs discussed in the previous section, one obstruction to \ufb01nding real solutions already for\nquadratic equations is the lack, within the real numbers, of the square root for negative real\nnumbers.\nTo de\ufb01ne the complex numbers, we introduce a new number i called the imaginary unit.\nThe number i is the square root of \u22121, that is, i satis\ufb01es the property\ni2 = \u22121.\n(3.1.a)\nor, equivalently, i is a solution of the equation T 2 + 1 = 0 in the indeterminate T.\nThe introduction of the imaginary unit i can be compared, in terms of the philosophical\nleap that it required for its ideation, to the introduction of 0, or of the negative numbers. It\nis remarkable that the equation x2 = \u22121 has no solution in the set of real numbers, but two\ndistinct solutions in the set of complex numbers, namely i and \u2212i.\nThe complex numbers can be intuitively de\ufb01ned as all those numbers that can be created\nby using the real numbers and the usual operations (+, \u2212, \u00b7, /), together with i, keeping in mind\nthe relation in (3.1.a). Let us give a more formal de\ufb01nition of the complex numbers.\nDe\ufb01nition 3.2.\n(1) A complex number is an expression of the form x + yi, where x, y are\nreal numbers, and i is the imaginary unit de\ufb01ned above.\n(2) The set of complex numbers is denoted by C.\nThus,\nC := {x + yi | x, y \u2208R, i2 = \u22121}.\nOften elements of C are denoted with the letter z.\nDe\ufb01nition 3.3. Let z = x + iy be a complex number.\n(1) The real part Re(z) of z is the real number x.\n(2) The imaginary part Im(z) of z is the real number y.\nWe will write z = x + yi when we want to remind ourselves the real and imaginary part of\nz.\nRemark 3.4. When we write a complex number z whether we write it in the form x + yi,\nx, y \u2208R, or in the form x+iy,both representations stand for the same complex number, as the\nimaginary unit i commutes with all real numbers; that is,\ns \u00b7 i = i \u00b7 s,\n\u2200s \u2208R.", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_59", "text": "x, y \u2208R, or in the form x+iy,both representations stand for the same complex number, as the\nimaginary unit i commutes with all real numbers; that is,\ns \u00b7 i = i \u00b7 s,\n\u2200s \u2208R.\nConsidering the notation for complex numbers introduced in De\ufb01nition 3.2, in the form\nx + yi, taking y = 0 and letting x vary in R, we immediately obtain that R \u2286C. As i \u0338\u2208R, by\nthe de\ufb01nition of i, cf. (3.1.a), then we can be even more precise and write R \u228aC.\nExample 3.5.\n(1) The real numbers 0, 3, and \u2212\u03c0 are complex numbers.\n(2) Other examples of complex numbers are 5 \u2212i, 3i, \u22122i and 1\n2 +\n\u221a\n2i.\n(3) Re(5 + 3i) = 5, Im(5 + 3i) = 3; Re(\u22123i) = 0, Im(\u22123i) = \u22123.\nComplex numbers are not ordered: it makes no sense to ask if a complex number is bigger\nthan another; in particular, it does not make sense to ask if a complex number is positive or\nnegative\n29", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_60", "text": "3.2\nOperations between complex numbers\nWe can add and multiply complex numbers using the standard formal properties of addition\nand multiplication, always remembering that i2 = \u22121.\nExample 3.6.\n(1) (5 + 3i) + (2 \u2212i) = (2 + 5) + (3 \u22121)i = 7 + 2i. In general:\n(x1 + y1i) + (x2 + y2i) = (x1 + x2) + (y1 + y2)i.\n(2) (1 \u22122i)(3 + 4i) = 3 \u22126i + 4i \u22128i2 = 3 \u22126i + 4i + 8 = 11 \u22122i. In general:\n(x1 + y1i) \u00b7 (x2 + y2i) = (x1x2 \u2212y1y2) + (x1y2 + x2y1)i.\nIn the previous section we de\ufb01ned complex numbers as those numbers that we can write in\nthe form x + yi, with x, y \u2208R. In particular, it follows from our de\ufb01nition that any complex\nnumber z \u2208C is completely determined by its real and imaginary part. Hence, we could think\nof parametrizing all complex numbers by means of their real and imaginary part. This is indeed\npossible, as shown in Figure 2. We identify the set of complex numbers with the points in the\nCartesian plane, which we will in this case rename the complex plane.\n../Images/complex-plane1.png\nFigure 2: The complex plane.\nThus, thus for a complex number of the form z = x + yi, we will use the real part x (resp.\nthe imaginary part y) as the cartesian coordinates of z in the complex plane. Then, the line\n{y = 0} in the complex plane is automatically identi\ufb01ed with the set of real numbers within\nthe complex numbers. For this reason, this line is called the real axis. The line {x = 0} in the\ncomplex plane identi\ufb01es instead with the set of complex numbers whose real part is 0. Numbers\nof this form are called purely imaginary numbers. For this reason, the line {x = 0} is called\nthe imaginary axis.\nUsing this representation complex numbers become vectors, and the sum of complex num-", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_61", "text": "complex plane identi\ufb01es instead with the set of complex numbers whose real part is 0. Numbers\nof this form are called purely imaginary numbers. For this reason, the line {x = 0} is called\nthe imaginary axis.\nUsing this representation complex numbers become vectors, and the sum of complex num-\nbers is equal to the sum of vectors, as in Figure 3. Moreover, multiplication of a complex\nnumber z by a positive real number r > 0 corresponds to scaling the length of the vector\nrepresenting z by the factor r.\nDe\ufb01nition 3.7. The modulus (or, absolute value) |z| of a complex number z \u2208C is its distance\nfrom the origin in the complex plane. It is computed using the Pythagorean Theorem in terms\nof the real and imaginary part of z:\n|z| =\np\nRe(z)2 + Im(z)2.\n30", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_62", "text": "Images/sum_complex.png\nFigure 3: Sum of complex numbers as vectors\nExample 3.8.\n(1) |3 \u22122i| =\n\u221a\n32 + 22 =\n\u221a\n25 = 5,\n(2) | \u22123i| =\n\u221a\n32 + 02 = 3\nUsing the representation of a complex number z \u2208C as z = x + yi, then the formula for\nthe modulus |z| of z can be written as\n|z| = |x + yi| =\np\nx2 + y2.\nAs we can represent the addition of complex numbers as addition of the corresponding vectors,\nwe can derive from this the classical triangle inequality\n\u2200z, w \u2208C,\n|z + w| \u2264|z| + |w|\n(3.8.a)\ncf. Figure 4.\nWith reference to the picture, we can compose a triangle using the vector connecting the origin\nto z1 (corresponding to the side of the triangle in the picture of length C), the vector connecting\nthe origin to z1+z2 (corresponding to the side of length AC in the picture), and the translation\nof the vector connecting the origin to z2, where we have moved the starting point of the vector\nto z1 (this corresponds to the side of length B in the picture). The classical triangle inequality\ntell us that A \u2264B+C. But given the way we constructed the triangle, this inequality translates\nto\n|z + w| \u2264|z| + |w|.\n(3.8.b)\nDe\ufb01nition 3.9. The conjugate z of a complex number z = x + yi is de\ufb01ned as the complex\nnumber z = x + iy := x \u2212iy.\nHence, the conjugate of z is simply obtained by changing the sign of the imaginary part of\nz. It is important to understand that geometrically in the complex plane this corresponds to\nre\ufb02ection across the real line.\nExample 3.10. 3 \u22124i = 3 + 4i, 3i = \u22123i, 1 = 1.\n31", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_63", "text": "Images/tr_ineq.png\nFigure 4: Triangle inequality.\nConjugation is compatible with all operations by explicit computation: namely,for z1, z2 \u2208\nC, z3 \u2208C\u2217,\nz1 + z2 = z1 + z2,\nz1 \u00b7 z2 = z1 \u00b7 z2,\n\u0012z1\nz3\n\u0013\n= z1\nz3\n.\nTo verify the formulas above, it su\ufb03ces to write all numbers involved as x + iy and expand all\nthe expressions obtained.\nSimilarly, we can use conjugation also to compute the modulus of a complex number:\nzz = (x + iy)(x + iy) = (x + iy)(x \u2212iy) = x2 + ixy \u2212ixy \u2212i2y2 = x2 + y2 = |z|2\nHence, if z \u0338= 0, we can use the formula above to show that any such z \u2208C has a multiplicative\ninverse, that is, z\u22121 exists12 and moreover it can be computed as\nz\u22121 =\nz\n|z|2 = x \u2212iy\nx2 + y2 .\n(3.10.c)\n12By z\u22121 we denote the (unique) complex number that z \u00b7 z\u22121 = 1 = z\u22121 \u00b7 z.\n32", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_64", "text": "../Images/Complex_conjugate_picture_svg.png\nFigure 5: Conjugate of a complex number.\nWe can use the above formula to better understand division between two complex numbers.\nGiven two complex numbers z and w, with w \u0338= 0, we would like explicitly write z\nw in the form\nx + yi.\nExample 3.11. We can try to turn the denominator of the fraction into a real number by\nmultiplying with the conjugate of w, both above and below.\n2 \u22123i\n5 + i = (2 \u22123i)(5 \u2212i)\n(5 + i)(5 \u2212i) = 7 \u221217i\n26\n= 7\n26 \u221217\n26i\nIn fact, we can write down a general formula using (3.10.c):\nz\nw =\nzw\nw \u00b7 w = zw\n|w|2 .\nExample 3.12. Here is another example.\n1\n3 \u2212\n\u221a\n3i = 3 +\n\u221a\n3i\n12\n= 1\n4 +\n\u221a\n3\n4 i, or\ni\n1 \u2212i = i(1 \u2212i)\n2\n= 1\n2 + 1\n2i\nWe also have the following relation between conjugation, real part and imaginary part\nRe(z) = 1\n2(z + z)\nand\nIm(z) = 1\n2i(z \u2212z).\n3.3\nPolar form\nWe can associate to every non-zero complex number z \u2208C an angle, called the argument or\nthe phase of z, and denoted arg z, in the following way. In the complex plane, we take the the\nangle formed by the half line R+ of the non-negative real numbers and the half-line Lz spanned\nby the vector connecting the origin to z. For example, in Figure 6, the angle arg z has been\ndenoted with \u03c6. The argument arg z is then de\ufb01ned as the angle between R+ and Lz, moving\nin the anti-clockwise direction.\nExample 3.13. arg 3 = 0; arg i = \u03c0\n2 ; arg\n\u221a\n2\n2 (1 + i) = \u03c0\n4 .\nTake now a non-zero complex number z, we have seen that its distance from the origin is\n|z|. Let \u03c6 be its argument. The number\nz\n|z| has distance 1 from the origin, so it lies on the\ntrigonometric (or, unit) circle", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_65", "text": "4 .\nTake now a non-zero complex number z, we have seen that its distance from the origin is\n|z|. Let \u03c6 be its argument. The number\nz\n|z| has distance 1 from the origin, so it lies on the\ntrigonometric (or, unit) circle\nS1 := {z \u2208C | |z| = 1} = {x + yi \u2208C | x, y \u2208R, and x2 + y2 = 1}.\n33", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_66", "text": "Hence, the real part of\nz\n|z| (resp. the imaginary part of z) is just cos(\u03c6) (resp. sin(\u03c6)), where\n\u03c6 is the angle (measured in radiants) formed by the positive part of the real axis and the half\nline passing through the origin and the point z on the complex plane, cf. Figure 6.\n../Images/complex_angle.png\nFigure 6\nThus, under this assumptions, we conclude that\nz = |z|(cos(\u03c6) + sin(\u03c6)i).\n(3.13.a)\nThe expression of a complex number z \u2208C given in (3.13.a) is called the polar form of z. It is a\nvery important and useful way to represent complex numbers, as we will see below. Conversely,\nwhen we write a complex number z in the form x + iy, we say that we are using the Cartesian\nform, or Cartesian representation. Let us note that because of the presence of cos and sin, one\ncan add any multiple of 2\u03c0 to the argument on the right hand side.\nExample 3.14. The polar form of 1 + i is\n\u221a\n2\n\u0000cos( \u03c0\n4 ) + sin(\u03c0\n4 )i\n\u0001\nThe multiplication of complex numbers becomes simple if we use the polar form and we use\nsome well-known trigonometric identities.\nExample 3.15. Let \u03c6 and \u03c8 be two numbers. Then\n(5(cos(\u03c6) + sin(\u03c6)i))(3(cos(\u03c8) + sin(\u03c8)i)) =\n=15(cos(\u03c6) cos(\u03c8) \u2212sin(\u03c6) sin(\u03c8)) + (cos(\u03c6) sin(\u03c8) + sin(\u03c6) cos(\u03c8))i =\n=15(cos(\u03c6 + \u03c8) + sin(\u03c6 + \u03c8)i),\nwhere we have used the addition formulas for sine and cosine\ncos(\u03c6 + \u03c8) = cos(\u03c6) cos(\u03c8) \u2212sin(\u03c6) sin(\u03c8),\n(3.15.b)\nsin(\u03c6 + \u03c8) = cos(\u03c6) sin(\u03c8) + sin(\u03c6) cos(\u03c8).\nThus, the example above can be immediately generalized to show that for two non-zero", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_67", "text": "(3.15.b)\nsin(\u03c6 + \u03c8) = cos(\u03c6) sin(\u03c8) + sin(\u03c6) cos(\u03c8).\nThus, the example above can be immediately generalized to show that for two non-zero\ncomplex numbers z1, z2 \u2208C, then arg z1 \u00b7 z2 = arg z1 + arg z2, while we already saw that\n|z1 \u00b7 z2| = |z1| \u00b7 |z2|,\nz1 \u00b7 z2 = |z1| \u00b7 |z2| \u00b7 (cos(arg z1 + arg z2) + sin(arg z1 + arg z2)i).\n(3.15.c)\nThus, when we multiply two non-zero complex numbers, the modulus of the product is the\nproduct of the moduli and the argument of the product is the sum of the arguments!\nExample 3.16.\n\f\f\f 1\n2 +\n\u221a\n3\n2 i\n\f\f\f = 1, and arg\n\u0010\n1\n2 +\n\u221a\n3\n2 i\n\u0011\n= \u03c0\n3 . Thus,\n \n1\n2 +\n\u221a\n3\n2 i\n!2017\n=\n \n1\n2 +\n\u221a\n3\n2 i\n!\n.\nbecause 12017 = 1, so the absolute values does not change; then 2017 = 336\u00b76+1, so 2017\u00b7 \u03c0\n3 =\n336 \u00b7 2\u03c0 + \u03c0\n3 , so also the argument does not change.\n34", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_68", "text": "The above example shows that the polar form is really useful to compute, for example,\npowers of complex numbers.\nWe can also use the polar form to divide complex numbers. As with multiplication when\nthe moduli (plural of the modulus) multiplied and the arguments added up, with division, we\nhave to do the inverse. That is, the absolute value of a fraction is the fraction of the absolute\nvalues and its argument is just the di\ufb00erence of the arguments:\nz\nw = |z|(cos(\u03c6) + sin(\u03c6)i)\n|w|(cos(\u03c8) + sin(\u03c6)i) = |z|\n|w|(cos(\u03c6 \u2212\u03c8) + sin(\u03c6 \u2212\u03c8)i).\nExample 3.17. Let z \u2208C be given in polar form by\nz := 3\n\u0012\ncos\n\u00122\u03c0\n7\n\u0013\n+ i sin\n\u00122\u03c0\n7\n\u0013\u0013\nThen the inverse of z is\nz\u22121 =1\n3\n\u0012\ncos\n\u0012\n\u22122\u03c0\n7\n\u0013\n+ i sin\n\u0012\n\u22122\u03c0\n7\n\u0013\u0013\n=1\n3\n\u0012\ncos\n\u0012\n2\u03c0 \u22122\u03c0\n7\n\u0013\n+ i sin\n\u0012\n2\u03c0 \u22122\u03c0\n7\n\u0013\u0013\n=1\n3\n\u0012\ncos\n\u001212\u03c0\n7\n\u0013\n+ i sin\n\u001212\u03c0\n7\n\u0013\u0013\n.\n3.4\nEuler formula\nWe can write the polar form of a non-zero complex number z in an even more compact\nform.\nDe\ufb01nition 3.18 (Euler\u2019s formula). Let \u03c6 be a real number. We de\ufb01ne\nei\u03c6 := cos(\u03c6) + i sin(\u03c6)\n(3.18.a)\nImages/eulers_formula.png\nFigure 7: Euler\u2019s formula\nWe will treat the Euler formula above as a formal de\ufb01nition, a shorten notation to describe\nthe points on the unitary circle. At this point, we have not developed the tools to actually\ndiscuss the mathematics behind this formula, as we have not de\ufb01ned exponentiation for complex\nnumbers. So, for now, just think about it as a shortcut for the part of the polar form depending\non the argument.\nAs an immediate consequence of De\ufb01nition 3.18, we have the following properties.\n35", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_69", "text": "Proposition 3.19. Let \u03c6, \u03c8 \u2208R and k \u2208Z. Then,\n(1) ei\u03c6 \u00b7 ei\u03c8 = ei(\u03c6+\u03c8);\n(2) ei\u03c6+2k\u03c0 = ei\u03c6.\nProof.\n(1) Use the trigonometric formulas in (3.15.b).\n(2) As we measure angles in radiants, this is a simple consequence of the 2\u03c0-periodicity of\nthe sine and cosine functions.\nWe have mentioned above that we can use Euler\u2019s formula to write the polar form of z in a\nmore compact form than the one introduce in (3.13.a). Indeed, in view of De\ufb01nition 3.18, we\ncan rewrite the polar form of z as\nz = |z|(cos(\u03c6) + sin(\u03c6)i) = |z|ei\u03c6.\nExample 3.20. Let z = 1 + i. Then\nz =\n\u221a\n2\n\u0012 1\n\u221a\n2 +\ni\n\u221a\n2\n\u0013\n=\n\u221a\n2ei \u03c0\n4 .\nWe can rewrite the formula for multiplication of polar forms, (3.15.c), as\nzw =\n\u0010\n|z|ei\u03c6\u0011 \u0010\n|w|ei\u03c8\u0011\n= |z||w|ei(\u03c6+\u03c8).\n(3.20.b)\n3.5\nFinding solutions of equations with complex coef\ufb01cients\nExample 3.21. The usual formula to compute solutions of quadratic equations works in the\ncomplex numbers exactly as it worked in the real numbers. [The proof is the same as usual,\nverify that!]\nFor example, the solutions of the equation\nT 2 + 2T + 3 = 0\nin the indeterminate T are given by\n\u22122 \u00b1\n\u221a\n22 \u22124 \u00b7 3\n2\n= \u22122 \u00b1 \u221a\u22128\n2\n= \u22121 \u00b1\n\u221a\n\u22122 = \u22121 \u00b1 i\n\u221a\n2\nMore in general, given a quadratic equation of the form\naT 2 + bT + c = 0,\na, b, c \u2208C,\na \u0338= 0,\nin the indeterminate T, then this equation always admits exactly two complex solutions (that\nmay possibly coincide) The two solutions are computed as\nt1 = b +\n\u221a\nb2 \u22124ac\n2a\n,\nt2 = b \u2212\n\u221a\nb2 \u22124ac\n2a\n.\nIn the above expression, what do we mean when we write\n\u221a\nb2 \u22124ac, with a, b, c \u2208C? By", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_70", "text": "t1 = b +\n\u221a\nb2 \u22124ac\n2a\n,\nt2 = b \u2212\n\u221a\nb2 \u22124ac\n2a\n.\nIn the above expression, what do we mean when we write\n\u221a\nb2 \u22124ac, with a, b, c \u2208C? By\nde\ufb01nition of square root as the inverse operation to taking the second power of a given number,\n\u221a\nb2 \u22124ac denotes those complex numbers t \u2208C satisfying t2 = b2 \u22124ac. There are indeed\ntwo distinct such numbers in C that satisfy this equation, when b2 \u22124ac \u0338= 0: given a complex\n36", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_71", "text": "number t is a root of the equation W 2 \u2212(b2 \u22124ac) = 0 (in the indeterminate W), also \u2212t will\nbe a root of that same equation. Hence we can rewrite the formulas for the two solutions as\nt1 = b + t\n2a ,\nt2 = b \u2212t\n2a ,\nwhere t2 = b2 \u22124ac.\nAs we work over the complex numbers, it is not hard to see \u2013 and we shall see it below \u2013 that\nwe can always solve the equation t2 = b2 \u22124ac, where t is the unknown and a, b, c \u2208C are \ufb01xed\nconstants. This is not true in R, where the equation T 2 = \u22121 does not have solutions since for\nany real number t, t2 \u22650.\nThe main importance of complex numbers is that any polynomial equation has a solution\namong the complex numbers. This result is called the fundamental theorem of algebra.\nLet us recall that given a polynomial p(T) in the indeterminate T, a root of p(T) is number\nc such that p(c) = 0.\nTheorem 3.22 (Fundamental Theorem of Algebra). Let p(T) be a polynomial in the indeter-\nminate T with complex coe\ufb03cients. Then, there exists a root t \u2208C of p(T).\nLet us also recall that if t1 is a root of a polynomial p(T), then we can factor p(T) as a\nproduct\np(T) = (T \u2212t1) \u00b7 q(T)\nwhere q(T) is another polynomial with complex coe\ufb03cients and the degree deg q(T) of q(T)\nsatis\ufb01es deg q(T) = deg p(T) \u22121.\nBut then, in turn, Theorem 3.22 implies that also q(T)\npossesses a complex root t2, q(t2) = 0. In turn, then\nq(T) = (T \u2212t2) \u00b7 r(T)\nwhere r(T) is another polynomial with complex coe\ufb03cients and deg r(T) = deg q(T) \u22121.\nMoreover\np(T) = (T \u2212t1) \u00b7 (T \u2212t2) \u00b7 r(T).\nIterating, we obtain that we can factor any degree d polynomial p(T) into d linear factors, that", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_72", "text": "Moreover\np(T) = (T \u2212t1) \u00b7 (T \u2212t2) \u00b7 r(T).\nIterating, we obtain that we can factor any degree d polynomial p(T) into d linear factors, that\nis, there d complex numbers t1, . . . , td (some of them may happen to coincide) such that\np(T) = (T \u2212t1)(T \u2212t2)(T \u2212t3) . . . (T \u2212td\u22121)(T \u2212td).\n(3.22.a)\nThe multiplicity of a root t of p(T) is equal to the number of elements of the set of roots\n{t1, . . . , td} that are equal to t.\nExample 3.23. Let p(T) = T 3 + (\u22121 + 2i)T 2 + (\u22121 \u22122i)T + 1. Then p(1) = 1 + (\u22121 + 2i) \u2212\n(1 + 2i) + 1 = 0 and\np(T) = (T \u22121)(T 2 + 2iT \u22121) = (T \u22121)(T \u2212i)2.\nHence, the roots of p(T) are 1 and i, where the multiplicity of i is 2.\nThe Fundamental Theorem of Algebra, Theorem 3.22, is equivalent to the statement that\nany polynomial p(T) with complex coe\ufb03cients admits a factorization into linear polynomials\nas in (3.22.a).\nWe have already discussed at length how the equation T 2 = \u22121 has no solutions in R,\nwhereas it has two distinct solutions in the complex numbers, z = i and z = \u2212i. Thus, this\nimplies that the Fundamental Theorem of Algebra cannot possibly work over R. Nonetheless,\nwe can still give derive some nice properties.\nTheorem 3.24 (Fundamental Theorem of Algebra over R). Let p(T) be a polynomial in the\nindeterminate T with real coe\ufb03cients.\n(1) If the degree deg p(T) of p(T) is odd, then there exists a real root t \u2208R of p(T).\n(2) In general, if t \u2208C is a complex root of p(T), then also t \u2208C is a complex root. Moreover,\nthe multiplicity of t and t coincide as roots of p(T).\n37", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_73", "text": "3.5.1\nSolving complex equations\nWe are going to learn how to solve some slightly more general equations, thanks to the polar\nform.\nProblem 3.25. For a \u2208C\u2217and n \u2208N\u2217, how many solutions does the equation T n = a in the\nindeterminate T has? Can we compute them all?\nUsing what we discussed in the \ufb01rst part of Section 3.5, it is immediate to see that the\nequations T n = a has exactly n solutions in the complex numbers.\nExample 3.26. When an equation is simple, we can try to solve it directly, by computing the\nreal and imaginary parts of a solutions separately.\nFor example, let us consider the equation in the indeterminate T\nT 2 = i\nWe are searching for a complex number of the form X + iY , where X, Y are indeterminates\nwith real values, such that (X + iY )2 \u2212i = 0.\nWe can rewrite this equation in the form\n(X2 \u2212Y 2) + i(2XY ) \u2212i = 0 which yields two distinct equations with real coe\ufb03cients, by\nseparating real and imaginary part:\n(\nX2 \u2212Y 2 = 0\n2XY = 1\n.\nFrom the \ufb01rst equation we deduce that X = \u00b1Y . We can substitute the two relations X = Y\nand X = \u2212Y into the second equation above to obtain:\nX = Y =\u21d2X \u00b7 (X) = 1\n2 =\u21d2X = \u00b1\n\u221a\n2\n2\nX = \u2212Y =\u21d2X \u00b7 (\u2212X) = 1\n2 =\u21d2X2 = \u22121\n2\n,\nand this last equation cannot be solved in R. Thus, the roots of the equation T 2 \u2212i = 0 are\n\u221a\n2\n2 + i\n\u221a\n2\n2\nand \u2212\n\u221a\n2\n2 \u2212i\n\u221a\n2\n2 .\nTo check that indeed those are solution of the equations, it su\ufb03ces to compute\n\u0010 \u221a\n2\n2 + i\n\u221a\n2\n2\n\u00112\n\u2212i\nand\n\u0010\n\u2212\n\u221a\n2\n2 \u2212i\n\u221a\n2\n2\n\u00112\n\u2212i.\nThe method that we have just seen is a rather ine\ufb03cient one for solving Problem 3.25 as\nit requires us to resolve two equations instead of one, and these two equations will have real\ncoe\ufb03cients and we have to \ufb01nd real solutions for those (which, a priori, we do not know if it is\nalways possible).\nInstead, we are going to take a slightly di\ufb00erent approach, which is more natural given that", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_74", "text": "coe\ufb03cients and we have to \ufb01nd real solutions for those (which, a priori, we do not know if it is\nalways possible).\nInstead, we are going to take a slightly di\ufb00erent approach, which is more natural given that\nwe are working in the \ufb01eld of complex numbers: the main idea is to write a in exponential form\nand do some reasoning with respect to both its modulus and its argument.\nHence, writing a \u2208C in its polar form a = Rei\u03c6, where R is a positive real number, and \u03c6\nis an angle expressed in radiants, we can instead try to \ufb01nd all solutions to the equation\nT n = Rei\u03c6\n(3.26.b)\nwhere T is the unknown and n is a \ufb01xed positive integer. In particular, we shall assume that\n\u03c6 \u2208[0, 2\u03c0). All solutions of (3.26.b) are of the form\nt =\nn\u221a\nRei\u03c8,\n38", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_75", "text": "where \u03c8 is an angle such that ein\u03c8 = ei\u03c6. Since R is a positive real number, then\nn\u221a\nR is well\nde\ufb01ned and it is in turn positive and real, which means that\nn\u221a\nRei\u03c8 is the polar form of a\ncomplex number (uniquely determined by R and \u03c8).\nHow do we compute all possible choices that we have for \u03c6? Since ein\u03c8 = ei\u03c6, we must have\nthat\nn\u03c8 = \u03c6 + 2k\u03c0,\nfor some k \u2208Z,\nby Proposition 3.19. Thus, \u03c8 = \u03c6\nn + 2k\u03c0\nn , where k is an integer between 0 and n \u22121, so that\nthe above equation has always exactly n distinct solutions.\nExample 3.27. The equation\nz2 = 3ei \u03c0\n5\nhas two solutions:\nz =\n\u221a\n3ei \u03c0\n10\nand\nz =\n\u221a\n3ei( \u03c0\n10 +\u03c0)\nThe equation\nz3 = 27ei \u03c0\n7\nhas three solutions:\nz = 3ei \u03c0\n21 ,\nz = 3ei( \u03c0\n21 + 2\u03c0\n3 ),\nand\nz = 3ei( \u03c0\n21 + 4\u03c0\n3 ).\n39", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_76", "text": "4\nSEQUENCES\nDe\ufb01nition 4.1. A sequence is a function x : N \u2192R.\nTraditionally, we denote the value of the function x at n \u2208N by xn, that is, xn := x(n).\nWe denote instead by (xn) the whole sequence.\nLet us start by looking at a few simple examples of sequences.\nExample 4.2.\n(1) Let us \ufb01x a real number C \u2208R. Then the constant sequence of value C\nis the sequence (xn) de\ufb01ned as follows\nxn := C\n\u2200n \u2208N.\n(2) Arithmetic progression: let a, b be real numbers; we de\ufb01ne sequence (xn) by\nx0 := a, x1 := a + b, x2 := a + 2b, . . . , xn := a + nb, . . . .\nWe call the type of sequence just constructed an arithmetic progression. For example,\nthe arithmetic progression given by a = 1 and b = 2 is x0 = 1, x1 = 3, x2 = 5, . . . ; this\nparticular arithmetic progression takes up as values all the positive odd numbers.\n(3) Geometric progression: let a, q be real numbers; we de\ufb01ne a sequence (xn) by\nx0 := a, x1 := aq, x2 := aq2, . . . , xn := aqn, . . . .\nWe call the type of sequence just constructed an geometric progression. For example, the\ngeometric progression given by a = 2 and b = 4\n5 is\nx0 = 2, x1 = 2 \u00b7 4\n5 = 8\n5, x2 = 2 \u00b7\n\u00124\n5\n\u00132\n= 32\n25, x3 = 2 \u00b7\n\u00124\n5\n\u00133\n= 128\n125, . . . .\n(4) Let (xn) be the sequence de\ufb01ned by xn := (\u22121)n. Then the sequence only takes two\nvalues:\nxn =\n(\n\u22121\nif n is odd,\n1\nif n is even.\nNotation 4.3. At times, it may happen that the terms of a sequence (xn) are not de\ufb01ned for\nall natural number values of the index n. For example, the sequence (xn) de\ufb01ned as\nxn := 1\nn\nis only well-de\ufb01ned when n \u0338= 0.", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_77", "text": "all natural number values of the index n. For example, the sequence (xn) de\ufb01ned as\nxn := 1\nn\nis only well-de\ufb01ned when n \u0338= 0.\nIn discussing sequences (and their limits, or lack thereof), we will mostly be concerned with\nproperties of a sequence which are eventually true. That means that we will look for properties\nof a sequence (xn) that hold starting from a certain index l \u2208N and then holds also for all the\nindices > l. Hence, what will matter for us is that all terms of a sequence (xn) are de\ufb01ned for\nall values of n greater or equal of a given natural number l \u2208N.\nHence, when we want to highlight that a sequence the terms of a sequence (xn) are de\ufb01ned\nfor all n \u2265l \u2208N, we will write\n(xn)n\u2265l\nWhen we can take l = 0, we will also write (xn)n\u2208N. When we omit the subscript n \u2265l, i.e.,\nwhen we write (xn), we simply are not specifying what is the initial index starting from which\nthe sequence is de\ufb01ned.\n40", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_78", "text": "Similarly to what we did for the case of subset of R, we would like to de\ufb01ne the concept\nof boundedness, boundedness from above/below also in the case of sequences. To this end, it\nsu\ufb03ces to notice that given a sequence (xn)n\u2265l, then it uniquely de\ufb01nes a subset S \u2282R given\nby all the values that the sequence takes,\nS := {xn | n \u2208N, x \u2265l}.\n(4.3.a)\nWe can then use S to make sense of the concept of boundedness for a sequence, as follows.\nDe\ufb01nition 4.4. Let (xn)n\u2265l be a sequence. We say that (xn)n\u2265l is\n\uf8f1\n\uf8f4\n\uf8f2\n\uf8f4\n\uf8f3\nbounded from above,\nbounded from below,\nbounded,\nif the set {xn | n \u2208N, x \u2265l} of values of the sequence is\n\uf8f1\n\uf8f4\n\uf8f2\n\uf8f4\n\uf8f3\nbounded from above,\nbounded from below,\nbounded,\nrespectively.\nIt is an immediate consequence of De\ufb01nition 2.8 that a sequence (xn)n\u2265l is bounded if and\nonly if it is both bounded from above and below.\nRemark 4.5. Let (xn)n\u2265l be a sequence. Then (xn)n\u2265l is bounded if and only if there exists a\npositive real number C such that the set of values of the sequence is a subset of the interval\n[\u2212C, C]. In particular, (xn)n\u2265l is bounded if and only the sequence (yn)n\u2265l, de\ufb01ned by yn :=\n|xn| is bounded, too.\nExample 4.6.\n(1) Let (xn)n\u2208N be the constant sequence of value C.\nThe set of value of this sequence is the singleton set {C} \u2282R.\n(2) Let (xn)n\u2208N be the sequence de\ufb01ned by xn := (\u22121)n, cf. Example 4.2.4. Then the set of\nvalues of this sequence is {xn \u2208R | n \u2208N} and it coincides with set {\u22121, 1} \u2282R. As\nS is a \ufb01nite subset of R, it follows that it is bounded and possesses both maximum and\nminimum, 1 and \u22121, respectively.\n(3) Let (xn)n\u2208N be an arithmetic progression with a = 0, b = 2. Then", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_79", "text": "S is a \ufb01nite subset of R, it follows that it is bounded and possesses both maximum and\nminimum, 1 and \u22121, respectively.\n(3) Let (xn)n\u2208N be an arithmetic progression with a = 0, b = 2. Then\n{xn | n \u2208N} = {2n | n \u2208N}\nwhere the latter is the set of even numbers. In particular, (xn) is not bounded.\nWe have also the following de\ufb01nitions focusing on the behavior of a sequence (xn) in the\nterms both of ordering of the indices of the sequence, which vary in N, and of the ordering of\nthe values of the sequence, which instead vary in R.\nDe\ufb01nition 4.7. Let (xn)n\u2265l be a sequence.\n(1) We say that\n(xn)n\u2265l is\n\uf8f1\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f2\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f3\nincreasing\nstrictly increasing\ndecreasing\nstrictly decreasing\nif for each n \u2208N, n \u2265l,\n\uf8f1\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f2\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f3\nxn \u2264xn+1\nxn < xn+1\nxn \u2265xn+1\nxn > xn+1\n.\n(2) We say that\n(xn)n\u2265l is\n(\nmonotone,\nstrictly monotone,\nif (xn) is\n(\nincreasing or decreasing\nstrictly increasing or strictly decreasing\n.\n41", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_80", "text": "Example 4.8.\n(1) Let C \u2208R and let (xn)n\u2208N be constant sequence of value C.\nThen\n{xn | n \u2208N} = {C}. Hence (xn)n\u2208N is bounded.\n(2) Let (xn)n\u2208N be an arithmetic progression of the form xn := a + nb, a, b \u2208R. Then,\n(i) the sequence is constant sequence of value a if and only if b = 0;\n(ii) the sequence is increasing if and only if b \u22650: indeed, xn+1 = xn + b. The sequence\nis strictly increasing if and only if b > 0;\n(iii) analogously, the sequence is decreasing if and only if b \u22640. It strictly decreasing if\nand only if b < 0;\n(iv) the sequence is bounded from below if and only if b \u22650: indeed, in that case, we\nalready know that xn+1 \u2265xn, \u2200n \u2208N, thus, xn \u2265x0 \u2200n \u2208N and x0 is a lower\nbound for the set of values of the sequence;\n(v) the sequence is bounded from above if and only if b \u22640: indeed, in that case, we\nalready know that xn+1 \u2264xn, \u2200n \u2208N, thus, xn \u2264x0 \u2200n \u2208N and x0 is an upper\nbound for the set of values of the sequence;\n(vi) the sequence is bounded if and only if b = 0: indeed, (xn)n\u2208N is bounded if and only\nif it is both bounded from above and below. But that is possible if and only if b = 0.\n(3) Let (xn)n\u2208N be an arithmetic progression of the form xn := aqn, a, q \u2208R. Then,\n(i) if a = 0 or q = 0, xn = 0, for all n \u2208N;\n(ii) if q = 1, xn = a, for all n \u2208N;\nHence, in both these cases, (xn)n\u2208N is a constant sequence. We will assume that a \u0338= 0,\nq \u0338= 1.\n(iii) q = \u22121, then xn = (\u22121)na. This sequence is bounded but not monotone;\n(iv) if q = 1\n2, then xn =\na", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_81", "text": "q \u0338= 1.\n(iii) q = \u22121, then xn = (\u22121)na. This sequence is bounded but not monotone;\n(iv) if q = 1\n2, then xn =\na\n2n . This sequence is strictly decreasing and bounded;\n(v) if q = \u22121\n2, then xn = (\u22121)na\n(2)n . This sequence is bounded but not monotone;\n(vi) if q = 2, then xn = 2n. This sequence is strictly increasing and bounded from below;\n(vii) q = \u22122, then xn = (\u22122)n. This sequence is neither bounded nor monotone.\nWe will analyze in general for what values of a and q the sequence (xn) is bounded in\nExamples 4.14 and 4.20.\n(4) The sequence (xn)n\u22651 de\ufb01ned by xn := 5\u22121\nn is strictly increasing. In fact, for all n \u2208N\u2217,\n1\nn >\n1\nn+1, hence xn+1 > xn.\n4.1\nRecursive sequences\nWe say that a sequence (xn) is recursive if the n-th term of the sequence xn is de\ufb01ned by a\nformula f(xn\u22121, . . . , xn\u2212j) which depends on the previous terms xn\u22121, . . . , xn\u2212j of the sequence,\nfor some \ufb01xed integer j > 0 \u2013 here, j does not depend on n. We also require that the formula\nf(. . . ) is \ufb01xed, i.e. it does not depend on n. For this de\ufb01nition to make sense, we will also have\nto \ufb01x the values of x0, x1, . . . , xj\u22121 as those cannot be established using the formula otherwise.\n42", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_82", "text": "Notation 4.9. We will denote a recursive sequence (xn) de\ufb01ned by xn := f(xn\u22121, . . . , xn\u2212j)\nand with assighned initial values c0, c1, c2, . . . , cj\u22121 with the following notation\n\uf8f1\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f2\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f3\nxn = f(xn\u22121, . . . , xn\u2212j)\nx0 = c0\nx1 = c1\nx2 = c2\n...\nxj\u22121 = cj\u22121\nExample 4.10. Let us recall the Fibonacci sequence (xn)n\u2208N:\n(\nxn = xn\u22121 + xn\u22122\nx0 = 1 = x1.\nThen, x2 = 2, x3 = 3, x4 = 5, x5 = 8, . . . and (xn) is strictly increasing as xk > 0, \u2200k \u2208N.\n[Prove this claim!]\nExample 4.11. We can de\ufb01ne arithmetic and geometric progressions as recursive sequences.\n(1) An arithemetic sequence (xn)n\u2208N, xn := a + bn, a, b \u2208R, can be de\ufb01ned recursively as\n(\nxn = xn\u22121 + b\nx0 = a.\n(2) A geometric sequence (xn)n\u2208N, xn := aqn, a, q \u2208R, can be de\ufb01ned recursively as\n(\nxn = qxn\u22121\nx0 = a.\nExample 4.12. Let us consider the following recursive sequence (xn)n\u2208N\n(\nxn = xn\u22121 + (\u22121)nn2,\nx0 = 0.\n(4.12.a)\nEquivalently, xn = Pn\ni=0(\u22121)ii2. What can we say about this sequence? For example, is it\nbounded (resp. bounded from above or from below)? The answer to the above question can\nbe given using induction which we will now introduce.\n4.2\nInduction\nInduction is a method of proving a property P(k) which depends on a parameter k which\nvaries among the natural numbers that are greater or equal than a \ufb01xed natural number C \u2208N.\nMore precisely, we want to be able to prove in\ufb01nitely many di\ufb00erent statements \u2013 all the versions", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_83", "text": "Induction\nInduction is a method of proving a property P(k) which depends on a parameter k which\nvaries among the natural numbers that are greater or equal than a \ufb01xed natural number C \u2208N.\nMore precisely, we want to be able to prove in\ufb01nitely many di\ufb00erent statements \u2013 all the versions\nof property P(k), when k \u2265C in N; hence, we want to \ufb01nd a method that allows us to prove\nall of these statements at once, without having to do in\ufb01nitely many veri\ufb01cations (one for each\nvalue of k).\nTo prove that property P(k) holds when k \u2265C \u2208N and k \u2208N, we can try to use the\nfollowing 2-step recipe, known as a proof by induction:\n(1) we \ufb01rst show that P(k) holds for k = C \u2013 this is called the starting step of a proof by\ninduction;\n43", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_84", "text": "(2) we then proceed to show that P(k) holds for a given value k = n \u2208N (where n here is to\nbe treated as an unspeci\ufb01ed number), under the assumption that we already know that\nP(k) holds all choices of k starting from C and up to n \u22121. This second step is called\nthe inductive step of a proof by induction. The assumption that P(k) statement holds\nfor k = C, C + 1, C + 2, . . . , n \u22122, n \u22121 is called the inductive hypothesis.\nHence, we can think of\nExample 4.13. We continue to work with the sequence (xn)n\u2208N de\ufb01ned in Example 4.12. We\nwill prove by induction the following claim related to this sequence.\nClaim. For the recursive sequence (xn)n\u2208N de\ufb01ned in (4.12.a), the even elements of the\nsequence satisfy the following equality:\nx2k = (2k + 1)k,\n\u2200k \u2208N.\nHence, the property P(k) that we want to prove by induction is the following\nP(k) : \u201cx2k = (2k + 1)k\u201d\nand k is any natural number, i.e., we have to prove that P(k) holds for all values of k \u2208N.\nProof of the Claim. We prove that P(k) holds by induction on k \u22650.\n\u25e6Starting Step: we need to show that P(0) holds.\nTat means that we need to show that the equality x2k = (2k + 1)k holds when we take\nk = 0. But, x0 = 0 and (2 \u00b7 0 + 1) \u00b7 0 = 0, hence, indeed, x2k = (2k + 1)k.\nThe starting step is proven.\n\u25e6Inductive Step: We will now assume that property P(k) holds for all values 0 \u2264k < n,\nthat is, we assume that we know already that for all 0 \u2264k < n, x2k = (2k + 1)k and we\nwill show that P(k) holds for k = n, i.e., we will show that x2n = (2n + 1)n. Thus,\nx2n =\nx2n\u22121 + (2n)2\n|\n{z\n}", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_85", "text": "will show that P(k) holds for k = n, i.e., we will show that x2n = (2n + 1)n. Thus,\nx2n =\nx2n\u22121 + (2n)2\n|\n{z\n}\nrecursive formula applied to x2n\n=\nx2(n\u22121) \u2212(2n \u22121)2\n|\n{z\n}\nrecursive formula applied to x2n\u22121\n+(2n)2\n= x2(n\u22121) \u2212((2n)2 \u22124m + 1)\n|\n{z\n}\n=(2n\u22121)2\n+(2n)2 = x2(n\u22121) + 4n \u22121\n=\n(2n \u22121)(n \u22121)\n|\n{z\n}\ninductive hypothesis: x2(n\u22121) = (2(n \u22121) + 1)(n \u22121)\n+4n \u22121 = 2n2 \u22123n + 1 + 4 \u22121\n= 2n2 + n = (2n + 1)n.\nHence we have shown that P(k) holds for k = n, which concludes the proof of the\ninductive step and, thus, the whole proof by induction of our claim.\nThe claim implies that, as x2k = (2k + 1)k, then (xn)n\u2208N is not bounded from above: in\nfact, for any real number b, we can \ufb01nd kb \u2208N such that\n(2kb + 1)kb = 2(kb)2 + kb \u2265(kb)2 > b.\nIn fact,\n(kb)2 > b\nif and only if\nkb >\np\n|b|,\n(4.13.a)\n44", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_86", "text": "and the Archimedean property Corollary 2.31 shows that the second inequality in (4.13.a) is\nindeed satis\ufb01ed for some kb \u2208N \u2013 it su\ufb03ces to take kb = [\np\n|b|] + 1. Thus, any given b \u2208R\ncannot be an upper bound for the set of values of the sequence, since for m > kb, x2m > x2kb > b.\nWe can also prove that the sequence is not bounded from below, one can also show that [prove\nit, by induction again!]\nx2k+1 = x2k \u2212(2k + 1)2 = (2k + 1)k \u2212(2k + 1)2 = \u2212(2k + 1)(k + 1).\nHence, one can use a similar argument as before to show that (xn)n\u2208N is also not bounded from\nbelow.\n4.3\nBernoulli inequality and (non-)boundedness of geometric sequences\nExample 4.14. Let (xn)n\u2208N be a geometric progression, that is, xn := aqn for some real\nnumbers a and q. If either a = 0 or |q| \u22641, then the sequence is bounded: more precisely,\n(1) for a = 0 or q = 0, 1, the sequence is a constant sequence, cf. Example 4.8.3;\n(2) if instead a \u0338= 0 and |q| \u22641 then |xn| \u2264|a|, for all n \u2208N.\nWe show that (2) holds for xn by induction on n \u2208N. Indeed:\n\u25e6Starting Step: for n = 0, x0 = aq0 = a, hence |x0| = |a|.\n\u25e6Inductive Step: assuming that |xj| \u2264|a|, for all natural numbers j < n then we need to\nprove that also xn \u2264|a|. But then,\n|xn| = |aqn| = |aqn\u22121||q| = |xn\u22121||q| \u2264a \u00b7 1 = a.\nWhat can we say in regards to the boundedness of a geometric progression xn = aqn, when\na \u0338= 0 and |q| > 1? In this section we will show that, when a \u0338= 0 and |q| > 1, then the sequence", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_87", "text": "What can we say in regards to the boundedness of a geometric progression xn = aqn, when\na \u0338= 0 and |q| > 1? In this section we will show that, when a \u0338= 0 and |q| > 1, then the sequence\nis unbounded. In order to do that, we need to show that\n\u2200C \u2208R, \u2203nC \u2208R, such that |xnC| \u2265C,\nwhich is to say that there are no upper or lower bounds for the set of values of the sequence\n(xn). Equivalently, we need to show that\n\u2200C \u2208R, \u2203nC \u2208R, such that |qnC| \u2265C\n|a|.\nWe saw in Example 4.11 that we can de\ufb01ne a geometric sequence recursively. In view of that\nand of the fact that we are assuming |q| > 1, then, as xn = qxn\u22121 it immediately follows that\n|xn| > |xn\u22121|. Even better, we can inductively compute that |xn+l| > ql|xn|, for any l \u2208N.\nHence the absolute value of xn is increasing inde\ufb01nitely with n. Is this enough to prove the\nunboundedness of a geometric sequence with |q| > 1? We will answer this question in the\ncourse of this section.\nExample 4.15. While one may be tempted to think that an increasing sequence must even-\ntually be unbounded, let us show that this is not always the case.\nLet (xn)n\u22651 be the sequence de\ufb01ned as xn := 5 \u22121\nn. We have already seen in Example 4.2\nthat xn is strictly increasing.\nOn the other hand, 0 < xn < 5 which implies that (xn) is\nbounded. Hence, being strictly monotone does not su\ufb03ce to imply boundedness of a sequence\nas this example very simply illustrates.\nBefore we continue in our analysis of geometric sequences, we introduce the following result\nthat will be useful in proving that aqn is unbounded when a \u0338= 0, |q| > 1.\n45", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_88", "text": "Proposition 4.16 (Bernoulli\u2019s inequality). Let q be a positive real number satisfying q > 1.\nThen qn \u22651 + n(q \u22121).\nTo prove Proposition 4.16, we \ufb01rst need to introduce a few new mathematical tools and\nresults. The \ufb01rst is the concept of binomial coe\ufb03cient.\nDe\ufb01nition 4.17. If 0 \u2264k \u2264n are natural numbers, then\n\u0000n\nk\n\u0001\nis de\ufb01ned as\n\u0012n\nk\n\u0013\n:=\nn!\nk!(n \u2212k)! = n \u00b7 (n \u22121) \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 (n \u2212k + 1)\nk \u00b7 (k \u22121) \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 1\n.\nHere the symbol n! for n \u2208N is the factorial notation, that is, n! is the product of the \ufb01rst\nn natural numbers (starting from 1):\nn! = 1 \u00b7 2 \u00b7 3 \u00b7 3 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 (n \u22122) \u00b7 (n \u22121) \u00b7 n.\nWe also de\ufb01ne 0! := 1. The number n! can be recursively de\ufb01ned, for n \u22651, by the recurrence\n(\n(n + 1)! = (n + 1) \u00b7 n!\n0! = 1.\nRemark 4.18. Given natural numbers 0 \u2264k \u2264n, then the natural number\n\u0000n\nk\n\u0001\nis equal to the\nnumber of possible ways one can choose a subset of unordered13 k elements from a set of n\nelements. You can \ufb01nd an explanation of this fact here.\nOne can show, using induction, the following properties of binomial coe\ufb03cients.\nProposition 4.19. Let n, k be natural numbers and let x, y be real numbers. Then,\n(1) For 0 \u2264k \u2264n,\n\u0012n\nk\n\u0013\n=\n\u0012\nn\nn \u2212k\n\u0013\n.\n(2) For 1 \u2264k \u2264n \u22121,\n\u0012n \u22121\nk\n\u0013\n+\n\u0012n \u22121\nk \u22121\n\u0013\n=\n\u0012n\nk\n\u0013\n.\n(3) (Binomial formula) For any x, y \u2208R,\n(x + y)n =\nn\nX\ni=0\n\u0012n\ni\n\u0013\nxiyn\u2212i.\nProof. This is an exercise in the exercise sheet for Week 4.\nWe are now ready to fully prove Bernoulli\u2019s inequality.\nProof of Proposition 4.16. The inequality is an actual equality when n = 0, 1. Then, we can", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_89", "text": "n\nX\ni=0\n\u0012n\ni\n\u0013\nxiyn\u2212i.\nProof. This is an exercise in the exercise sheet for Week 4.\nWe are now ready to fully prove Bernoulli\u2019s inequality.\nProof of Proposition 4.16. The inequality is an actual equality when n = 0, 1. Then, we can\nassume that n \u22652. Let us apply the binomial formula, Proposition 4.19; then,\nqn = (1 + (q \u22121))n =\nn\nX\ni=0\n\u0012n\ni\n\u0013\n(q \u22121)i \u00b7 1n\u2212i\n=\n\u0012n\n0\n\u0013\n(q \u22121)0 +\n\u0012n\n1\n\u0013\n(q \u22121)1 +\nn\nX\ni=2\n\u0012n\ni\n\u0013\n(q \u22121)i \u00b7 1n\u2212i\n=1 + n(q \u22121) +\nn\nX\ni=2\n\u0012n\ni\n\u0013\n(q \u22121)i \u00b7 1n\u2212i.\n13By unordered we mean that we do not distinguish the order in which the k elements are chosen.\n46", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_90", "text": "As q > 1, then (q \u22121)i > 0, for all i \u2208N\u2217. Thus, Pn\ni=2\n\u0000n\ni\n\u0001\n(q \u22121)i \u00b7 1n\u2212i > 0 and\nqn = (1 + (q \u22121))n =\nn\nX\ni=0\n\u0012n\ni\n\u0013\n(q \u22121)i \u00b7 1n\u2212i\n=1 + n(q \u22121) +\nn\nX\ni=2\n\u0012n\ni\n\u0013\n(q \u22121)i \u00b7 1n\u2212i > 1 + n(q \u22121).\nExample 4.20. Let (xn)n\u2208N be a geometric progression for some real numbers a and q, xn :=\naqn. Assume that |q| > 1 and a \u0338= 0.\nUnder these assumptions, Bernoulli\u2019s inequality, Proposition 4.16, implies that (xn) is not\nbounded. In fact,\n|aqn| = |a||q|n \u2265|a|(1 + n(|q| \u22121))\n|\n{z\n}\nBernoulli\u2019s inequality\nWe can turn the latter expression into a sequence (yn), that is, yn := |a|(1 + n(|q| \u22121)). The\nsequence (yn) is not bounded since, for a \ufb01xed positive real number b \u2208R+,\n|a|(1 + n(|q| \u22121)) \u2264b\n\u21d0\u21d2\nn \u2264\nb\n|a| \u22121\n|q| \u22121,\nwhich does not hold for n \u2265\n\u0014\nb\n|a|\u22121\n|q|\u22121\n\u0015\n+ 1. So, no b can be an upper bound for |xn|.\nOne can show similarly:\n(1) (xn)n\u2208N is bounded if and only if |q| \u22641 or a = 0;\n(2) (xn)n\u2208N is increasing if and only if\n\u001a q \u22651 and a \u22650, or\n0 \u2264q \u22641 and a \u22640. ;\n(3) (xn)n\u2208N is strictly increasing if and only if\n\u001a q > 1 and a > 0, or\n0 < q < 1 and a < 0 .\n(4) (xn)n\u2208N is decreasing if and only if\n\u001a 0 \u2264q \u22641 and a \u22650, or\nq \u22651 and a \u22640.\n(5) (xn)n\u2208N is strictly decreasing if and only if", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_91", "text": "0 < q < 1 and a < 0 .\n(4) (xn)n\u2208N is decreasing if and only if\n\u001a 0 \u2264q \u22641 and a \u22650, or\nq \u22651 and a \u22640.\n(5) (xn)n\u2208N is strictly decreasing if and only if\n\u001a 0 < q < 1 and a > 0, or\nq > 1 and a < 0 .\n(6) (xn)n\u2208N is bounded from above if and only if |q| \u22641 or q > 1 and a \u22640;\n(7) (xn)n\u2208N is bounded from below if and only if |q| \u22641 or q > 1 and a \u22650.\n4.4\nLimit of a sequence\nDe\ufb01nition 4.21. Let (xn)n\u2265l be a sequence.\n(1) We say that (xn)n\u2265l converges (or is convergent) to a number y \u2208R, if for each \u03b5 \u2208R\u2217\n+,\nthere exists n\u03b5 \u2208N such that\n\u2200n \u2208N such that n \u2265n\u03b5,\nthen |xn \u2212y| \u2264\u03b5.\n(2) If (xn)n\u2265l does not converge to any y \u2208R then we say that it is not convergent.\n47", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_92", "text": "If the number y \u2208R de\ufb01ned above exists then y is called the limit of the sequence (xn).\nOnce again, as we are talking about the limit of a sequence, this is only possible if the limit is\nunique, when it exists. That is indeed the case.\nProposition 4.22. If a sequence (xn)n\u2265l converges, then its limit is unique.\nProof. Let us assume by contradiction that (xn)n\u2265l admits two distinct limits t1 \u0338= t2 \u2208R.\nThen, for each 0 < \u03b5 \u2208R there are n\u03b5, n\u2032\n\u03b5 \u2208N such that for all n \u2265n\u03b5, then\n|t1 \u2212xn| \u2264\u03b5,\nand for all n \u2265n\u2032\n\u03b5, then\n|t2 \u2212xn| \u2264\u03b5.\nSo, if we take n\u03b5 := max{n\u03b5, n\u2032\n\u03b5}, then both of the above inequalities hold for all integers\nn \u2265n\u03b5. In particular, for such n, we have\n|t1 \u2212t2| \u2264|t1 \u2212xn| + |xn \u2212t2|\n|\n{z\n}\ntriangle inequality\n\u2264\u03b5 + \u03b5 = 2\u03b5\nSince, this holds for all 0 < \u03b5 \u2208R, we obtain that t1 = t2.\nNotation 4.23. When the limit y \u2208R or a sequence (xn)n\u2265l exists, we denote that by\nlim\nn\u2192\u221exn = y. Alternatively, we also write xn\nn\u2192\u221e/ y .\nExample 4.24. The sequence (xn)n\u22651, de\ufb01ned as xn := 1 \u2212\n1\n\u221an, is convergent.\nIndeed, lim\nn\u2192\u221e\n\u0012\n1 \u22121\n\u221an\n\u0013\n= 1. To verify this claim, for any \ufb01xed \u03b5 \u2208R\u2217\n+ we have to \ufb01nd an\nindex n\u03b5 \u2208N such that\n\u2200n \u2265n\u03b5,\n\f\f\f\f1 \u22121\n\u221an \u22121\n\f\f\f\f \u2264\u03b5.\nOn the other hand,\n\f\f\f\f1 \u22121\n\u221an \u22121\n\f\f\f\f =\n\f\f\f\f\n1\n\u221an\n\f\f\f\f =\n1\n\u221an.\nHence, it su\ufb03ces to show that there exists an index n\u03b5 \u2208N such that \u2200n \u2265n\u03b5, then\n1\n\u221an < \u03b5.\nThe latter inequality is equivalent to the inequality \u221an > 1\n\u03b5, which in turn is equivalent to the\ninequality n > 1\n\u03b52 . Hence, for any \ufb01xed \u03b5 \u2208R\u2217\n+, we have to \ufb01nd an index n\u03f5 \u2208N such that", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_93", "text": "1\n\u221an < \u03b5.\nThe latter inequality is equivalent to the inequality \u221an > 1\n\u03b5, which in turn is equivalent to the\ninequality n > 1\n\u03b52 . Hence, for any \ufb01xed \u03b5 \u2208R\u2217\n+, we have to \ufb01nd an index n\u03f5 \u2208N such that\n\u2200n \u2265n\u03b5,\nn > 1\n\u03b52 .\nThus, for a \ufb01xed \u03b5 > 0, it su\ufb03ces to take n\u03b5 :=\n\u0002 1\n\u03b52\n\u0003\n+ 1.\nExample 4.25. Let us introduce an example of a non-converging sequence.\nLet us consider the sequence (xn)n\u2208N de\ufb01ned by xn := (\u22121)n. Indeed, if (xn) was convergent\nwith limit y, then we could apply De\ufb01nition 4.21 with \u03b5 := 1\n2 and \ufb01nd n 1\n2 \u2208N such that for\nall integers n \u2265n 1\n2 , |xn \u2212y| < 1\n2. In particular, if n\u2032 \u2265n 1\n2 is any other integer, then we would\nhave:\n|xn\u2032+1 \u2212xn\u2032| = |xn \u2212y + y \u2212xn\u2032| \u2264|xn \u2212y| + |y \u2212xn\u2032|\n|\n{z\n}\ntriangle inequality\n< 1\n2 + 1\n2 = 1\nHowever, in our sequence |xn\u2032+1 \u2212xn\u2032| = |1 \u2212(\u22121)| = 2 > 1 which prompts a contradiction.\nThus, this sequence cannot converge to any limit y \u2208R.\n48", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_94", "text": "Remark 4.26. In fact, the argument used in Example 4.25 shows that if (xn)n\u2265l is a convergent\nsequence, then for all 0 < \u03b5 \u2208R there is an n\u03b5 \u2208N such that for all n, n\u2032 \u2265n\u03b5, |xn \u2212xn\u2032| < 2\u03b5.\n[Verify this fact using again the triangle inequality!] We will see that this observation can be\nformalized into the notion of Cauchy sequence, see Section 4.10.\nAlso, using a similar argument as in Example 4.25 above, we can show the following result.\nProposition 4.27. Let (xn)n\u2265l be a sequence. If (xn)n\u2265l is convergent, then it is bounded.\nProof. Set y := lim\nn\u2192\u221exn. Applying De\ufb01nition 4.21 with \u03b5 := 1, then there exists n1 \u2208N, such\nthat for all integers n \u2265n1, |xn \u2212y| \u22641. That is, for all integers n \u2265n1,\n\u22121 + y < xn < 1 + y,\nand\n|xn| < max(| \u22121 + y|, |1 + y|).\n(4.27.a)\nLet us de\ufb01ne\nR := max{|xl|, |xl+1|, |xl+2| . . . , |xn1\u22123|, |xn1\u22122|, |xn1\u22121|, |y + 1|, |y \u22121|}.\nWe claim that R is an upper bound and \u2212R is a lower bound for the set of values of the sequence.\nIndeed, R (resp. \u2212R) is an upper bound (resp. a lower bound) for the set {xl, xl+1 . . . , xn1\u22121}\njust because R is \u2265than the absolute values of all these elements of the sequence, by the\nvery de\ufb01nition of R above. Furthermore, R (resp. \u2212R) is an upper bound (resp. a lower\nbound) for the other elements of the sequence, because these elements are lying in the interval\nI = [y \u22121, y + 1], R (resp. \u2212R) is an upper bound (resp. a lower bound) for I, again, by\nde\ufb01nition of R.", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_95", "text": "bound) for the other elements of the sequence, because these elements are lying in the interval\nI = [y \u22121, y + 1], R (resp. \u2212R) is an upper bound (resp. a lower bound) for I, again, by\nde\ufb01nition of R.\nExample 4.28. The sequence (xn)n\u2208N de\ufb01ned by xn :=\n\u221a\nn3 cannot be convergent as it is not\nbounded.\nRemark 4.29. The viceversa of the above proposition is not true: that is, if a sequence (xn)n\u2265l\nis bounded, then it is not necessary convergent. An example of that is given by the sequence\n(xn)n\u2208N de\ufb01ned by xn := (\u22121)n, see Example 4.25.\nIn Section 4.7 we shall see that a monotone bounded sequence (xn)n\u2265l is always convergent.\nOf course, the sequence (xn) de\ufb01ned by xn := (\u22121)n is not monotone.\n4.4.1\nLimits and algebra\nIn this section we show that (\ufb01nite) limits of sequences respect the standard operations.\nProposition 4.30. Let (xn) and (yn) be two convergent sequences and let x := lim\nn\u2192\u221exn and\ny := lim\nn\u2192\u221eyn be their limits. Then:\n(1) the sequence (xn + yn) is also convergent, and lim\nn\u2192\u221e(xn + yn) = x + y,\n(2) the sequence (xn \u00b7 yn) is also convergent, and lim\nn\u2192\u221e(xn \u00b7 yn) = x \u00b7 y,\n(3) if y \u0338= 0, then the sequence\n\u0010\nxn\nyn\n\u0011\nis also convergent, and lim\nn\u2192\u221e\n\u0010\nxn\nyn\n\u0011\n= x\ny, and\n(4) if there is an n0 \u2208N, such that xn \u2264yn for each integer n \u2265n0, then x \u2264y.\nRemark 4.31. Let us note that, since y \u0338= 0, there exists n0 \u2208N such that yn \u0338= 0 for n \u2265n0.\nHence dividing the quotient xn\nyn makes sense for n \u2265n0, provided that xn is de\ufb01ned for such\nchoice of index.\n49", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_96", "text": "Proof. We prove only (1). We refer to 2.3.3 and 2.3.6 in the book for the proofs of the others.\nFix 0 < \u03b5 \u2208R. Let us try to explain how the proof should intuitively go. We need to show\nthat for big enough an index n \u2208N, |(xn + yn) \u2212(x + y)| is smaller than \u03b5. However, as (xn),\n(yn) are both convergent with limit x, y, respectively, we know that |xn \u2212x| and |yn \u2212y| are\nsmall for big n; moreover,\n|(xn + yn) \u2212(x + y)| = |(xn \u2212x) + (yn \u2212y)| \u2264|xn \u2212x| + |yn \u2212y|\n|\n{z\n}\ntriangle inequality\n.\n(4.31.b)\nSo, to make |(xn + yn) \u2212(x + y)| smaller than \u03b5, it su\ufb03ces to make the sum |xn \u2212x| + |yn \u2212y|\nsmaller than \u03b5. That we can attain for example if we make both |xn \u2212x| and |yn \u2212y| smaller\nthan\n\u03b5\n2.\nThe choice of\n\u03b5\n2 is rather arbitrary: the proof would work with any two positive\nnumbers that add up to \u03b5, for example with \u03b5\n3 and 2\u03b5\n3 , but for simplicity, we shall stick with \u03b5\n2.\nAfter this initial discussion, we proceed to the formal proof.\nWe work with \u03b5 > 0 \ufb01xed above. Thus, there exist integers n\u2032\u03b5\n2 and n\u2032\u2032\u03b5\n2 , such that\n\u2200n \u2265n\u2032\u03b5\n2 ,\n|x \u2212xn| \u2264\u03b5\n2,\nand\n\u2200n \u2265n\u2032\u2032\u03b5\n2 ,\n|y \u2212yn| \u2264\u03b5\n2.\nLet us de\ufb01ne n\u03b5 := max\nn\nn\u2032\u03b5\n2 , n\u2032\u2032\u03b5\n2\no\n. Then, (4.31.b) implies that for every n \u2265n\u03b5,\n|(xn + yn) \u2212(x + y)| \u2264|xn \u2212x| + |yn \u2212y| \u2264\u03b5\n2 + \u03b5\n2 = \u03b5.\nThis shows that (xn + yn) satis\ufb01es De\ufb01nition 4.21 for convergence with respect to the \ufb01nite\nlimit x + y.", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_97", "text": "2 + \u03b5\n2 = \u03b5.\nThis shows that (xn + yn) satis\ufb01es De\ufb01nition 4.21 for convergence with respect to the \ufb01nite\nlimit x + y.\nProperty (4) in Proposition 4.30 implies the following immediate corollary.\nCorollary 4.32. Let (xn)n\u2265l be a converging sequence and let x := lim\nn\u2192\u221exn be its limit. If\nthere exists n0 \u2208N such that xn \u22650, \u2200n \u2265n0, then x \u22650\nExample 4.33. In Corollary 4.32, it may well happen that x = 0 even if xn > 0, \u2200n \u2208N as\nshown by the sequence xn = 1\nn.\nExample 4.34. With the above machinery we can already compute the limits of series that\nare de\ufb01ned as fractions of polynomials; a fracion whose numerator and denominator are both\npolynomials is called a rational function.\n(1) xn := n2+2n+3\n4n2+5n+6. Then\nlim\nn\u2192\u221exn = lim\nn\u2192\u221e\nn2 + 2n + 3\n4n2 + 5n + 6 =\nlim\nn\u2192\u221e\n1 + 2\nn + 3\nn2\n4 + 5\nn + 6\nn2\n|\n{z\n}\ndividing\nboth\nthe\nnumerator\nand the denominator by n\n=\nlim\nn\u2192\u221e\n\u00001 + 2\nn + 3\nn2\n\u0001\nlim\nn\u2192\u221e(4 + 5\nn + 6\nn2 )\n|\n{z\n}\nusing\nProposition\n4.30.3\nas\nboth\nthe\nnumerator\nand\nde-\nnominator have \ufb01nite limit\n=\nlim\nn\u2192\u221e1 + lim\nn\u2192\u221e\n2\nn + lim\nn\u2192\u221e\n3\nn2\nlim\nn\u2192\u221e4 + lim\nn\u2192\u221e\n5\nn + lim\nn\u2192\u221e\n6\nn2\n|\n{z\n}\naddition rule for \ufb01nite limits\n=\n1 + lim\nn\u2192\u221e\n2\nn + 3 \u00b7\n\u0010\nlim\nn\u2192\u221e\n1\nn\n\u00112\n4 + lim\nn\u2192\u221e\n5\nn + 6 \u00b7\n\u0010\nlim\nn\u2192\u221e\n1\nn\n\u00112\n|\n{z\n}\nproduct rule for \ufb01nite limits &\nlimits of constant sequences\n= 1 + 0 + 0\n4 + 0 + 0 = 1\n4\nHere are a few comments on the manipulation we just performed:\n50", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_98", "text": "(i) dividing the numerator and the denominator by n is an operation that one cannot\nperform for n = 0. So, after the second equality sign the expression that we wrote\ndoes not make sense for n = 0. But this is not an issues, for when we study a\nsequence for the purpose of understanding its convergence, we are only interested\nin the values of the index for big enough values of the index n. So you are free to\nsubstitute the 0-th term with any real number, e.g., 0, after the second equality sign.\nThe same issue will show up many other times in this sectio, for example, when we\nare computing limits of sequences of the form 2\nn, or\n3\nn2 . Hence, from now onwards,\nwhenever we work with sequences to discuss their convergence, we will not worry too\nmuch about what may happen to a \ufb01nite number of values of the sequence, whenever\nwe perform some algebraic manipulations, or we show that certain estimates holds,\netc.\n(ii) for any number c \u2208R, lim\nn\u2192\u221e\nc\nn = 0: infact, given a \ufb01xed \u03b5 \u2208R\u2217\n+, we may choose\nn\u03b5 :=\n\u0002 c\n\u03b5\n\u0003\n+ 1, and for this choice we have for each integer n \u2265n\u03b5:\n\f\f\f c\nn\n\f\f\f < c\nc\n\u03b5\n= \u03b5.\n(iii) in the step where we use that limits behave well with respect to fractions, we should\ncheck \ufb01rst that the limit of the denominator is not 0. However, following our argu-\nment, we see that this limit is 4, so we are \ufb01ne.\n(2) xn =\nn+2\n3n2+4n+5. Here we will not give the above explanations again (as they are the\nsame):\nlim\nn\u2192\u221exn = lim\nn\u2192\u221e\nn + 2\n3n2 + 4n + 5 = lim\nn\u2192\u221e\n1\nn + 2\nn2\n3 + 4\nn + 5\nn2\n=\n0 + 0\n3 + 0 + 0 = 0\n(3) xn = n2+2n+3\n4n+5\n. For n \u22651, we have 0 \u22643\nn and 1 \u22655\nn. Hence, for n \u22651:\nxn = n2 + 2n + 3\n4n + 5\n= n + 2 + 3\nn\n4 + 5\nn\n\u2265n + 2\n5", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_99", "text": "4n+5\n. For n \u22651, we have 0 \u22643\nn and 1 \u22655\nn. Hence, for n \u22651:\nxn = n2 + 2n + 3\n4n + 5\n= n + 2 + 3\nn\n4 + 5\nn\n\u2265n + 2\n5\nThis shows that (xn) is not bounded and hence cannot be convergent by Proposition 4.27.\nUsing the method of the above exercise one can show the following result on limits of\nsequences de\ufb01ned by means of rational functions.\nProposition 4.35. If (xn) and (yn) are sequences given by polynomials\nxn := P(n),\nP(X) = a0 + a1X + \u00b7 \u00b7 \u00b7 + apXp, with ap \u0338= 0, and\nyn := Q(n),\nQ(X) = b0 + b1X + \u00b7 \u00b7 \u00b7 + bqXq, with bq \u0338= 0,\nthen\n(1) if p \u2264q, then\n\u0010\nxn\nyn\n\u0011\nis convergent, and\n(i) if p = q, then lim\nn\u2192\u221e\nxn\nyn = ap\nbq ,\n(ii) if p < q, then lim\nn\u2192\u221e\nxn\nyn = 0,\n(2) if p > q, then\n\u0010\nxn\nyn\n\u0011\nis not bounded and thus it does not converge.\nProof. See page 22 of the book for a precise proof. The book contains states an unnecessary\nassumption: it is requested that yn \u0338= 0 for all n \u2208N, but in fact it is enough if yn \u0338= 0 for some\nn \u2208N, as, in that case, yn is a given by evaluating a non-zero polynomial at natural numbers,\nand a non-zero polynomial has at most as many roots as its degree.\n51", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_100", "text": "4.5\nSqueeze theorem\nTheorem 4.36 (Squeeze Theorem). Let (xn), (yn), and (zn) be three sequences. Assume\nthat:\n(1) the sequences (xn), (zn) are convergent, and limn\u2192\u221exn = a = limn\u2192\u221ezn; and\n(2) there exists n0 \u2208N such that for all integers n \u2265n0,\nxn \u2264yn \u2264zn.\nThen the sequence (yn) is convergent, and\nlim\nn\u2192\u221eyn = a.\nProof. For each \u03b5 > 0, there are natural numbers n\u2032\n\u03b5 and n\u2032\u2032\n\u03b5, such that\n\u2200n \u2265n\u2032\n\u03b5,\na \u2212\u03b5 < xn,\nand,\n\u2200n \u2265n\u2032\u2032\n\u03b5,\na + \u03b5 > zn.\nSet n\u03b5 := max{n\u2032\n\u03b5, n\u2032\u2032\n\u03b5, n0}. Then, for each integer n \u2265n\u03b5\na \u2212\u03b5 < xn \u2264yn \u2264zn < a + \u03b5,\nwhich in particular implies that |yn \u2212a| < \u03b5.\nExample 4.37. Let (xn)n\u22651 be the sequence de\ufb01ned by xn :=\n1\nn +\n1\n\u221an.\nWe show that\nlim\nn\u2192\u221e\n\u0010\n1\nn +\n1\n\u221an\n\u0011\n= 0.\nIn fact, we may squeeze xn as follows\n0 \u22641\nn + 1\n\u221an \u2264\n2\n\u221an,\n\u2200n \u22651\nIndeed:\n(1) 0 \u22641\nn +\n1\n\u221an holds for every integer n \u22651.\n(2) For every integer n \u22651 we also have:\n1\nn + 1\n\u221an \u2264\n1\n\u221an + 1\n\u221an\n|\n{z\n}\nn\u2264n2\u21d4\u221an\u2264n\n=\n2\n\u221an;\nOn the other hand,\n(i) the limit of the constant sequence of value 0 is 0;\n(ii)\nlim\nn\u2192\u221e\n1\n\u221an = 0 by the computation of Example 4.24.\nHence, we can apply the Squeeze Theorem 4.36 to conclude that (xn)n\u2208N converges and its\nlimit is 0.\nExample 4.38. In general, we can show that a geometric sequence (xn)n\u2208N, xn := aqn is\nconvergent if and only if a = 0 or \u22121 < q \u22641.\nIndeed:\n52", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_101", "text": "(1) When a = 0 or q = 0, 1, then the sequence is constant. If q = \u22121, then xn = (\u22121)na and\nthe sequence does not converge.\n(2) If |q| > 1, we have already shown in Example 4.14 that the sequence is not bounded.\nHence, Proposition 4.27 implies that the sequence is also non-convergent.\n(3) If |q| < 1, we show that the sequence converges and that lim\nn\u2192\u221eaqn = 0.\nTo prove that, we should understand when |aq|n < \u03b5 for a given 0 < \u03b5 \u2208N. But,\n|aq|n < \u03b5\n\u21d0\u21d2\n|a|\n\u03b5 <\n\u0012 1\n|q|\n\u0013n\n(4.38.a)\nAs |q| < 1, then\n\f\f\f 1\nq\n\f\f\f > 1 and we can apply Bernoulli\u2019s inequality, Proposition 4.16,\nshowing that\n\u0012 1\n|q|\n\u0013n\n\u22651 + n\n\u0012 1\n|q| \u22121\n\u0013\n> n\n\u0012 1\n|q| \u22121\n\u0013\n.\n(4.38.b)\nPutting (4.38.a), (4.38.b) together, then the inequality |aq|n < \u03b5 holds as long as\n|a|\n\u03b5 < n\n\u0012 1\n|q| \u22121\n\u0013\n.\n(4.38.c)\nSince the inequality in (4.38.c) is satis\ufb01ed for all integer n \u2265n\u03f5, where\nn\u03b5 =\n\uf8ee\n\uf8f0\n|a|\n\u03b5\n\u0010\n1\n|q| \u22121\n\u0011\n\uf8f9\n\uf8fb+ 1,\nwe can conclude that for all n \u2265n\u03b5, |aqn| < \u03b5.\nExample 4.39. Let (yn)n\u2208N be the sequence de\ufb01ned by yn := 2n\nn! .\nWe claim that lim\nn\u2192\u221e\n2n\nn! = 0. Indeed, we have for all integers n \u22653:\n0\n|{z}\nxn\n\u22642n\nn!\n|{z}\nyn\n\u2264\n2n\n2 \u00b7 3n\u22122 =\n32 \u00b7 2n\n2 \u00b7 32 \u00b7 3n\u22122 = 9\n2 \u00b7\n\u00122\n3\n\u0013n\n|\n{z\n}\nzn\nFurthermore lim\nn\u2192\u221exn = lim\nn\u2192\u221e0 = 0 and lim\nn\u2192\u221ezn = lim\nn\u2192\u221e\n9\n2 \u00b7\n\u0000 2\n3\n\u0001n = 9\n2 \u00b7 lim\nn\u2192\u221e\n\u0000 2\n3\n\u0001n = 9\n2 \u00b7 0 = 0\nby Example 4.38. So, the Squeeze Theorem 4.36 concludes our claim.", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_102", "text": "n\u2192\u221e0 = 0 and lim\nn\u2192\u221ezn = lim\nn\u2192\u221e\n9\n2 \u00b7\n\u0000 2\n3\n\u0001n = 9\n2 \u00b7 lim\nn\u2192\u221e\n\u0000 2\n3\n\u0001n = 9\n2 \u00b7 0 = 0\nby Example 4.38. So, the Squeeze Theorem 4.36 concludes our claim.\nExample 4.40. Let (yn)n\u2208N be the sequence de\ufb01ned by yn :=\nn\u221an. we show that lim\nn\u2192\u221exn = 1.\nTo prove the above claim, we show that we can squeeze the sequence (yn) as follows:\n1\n|{z}\nxn\n\u2264yn \u2264zn := 1 + 1\n\u221an,\n\u2200n \u226b1.\nAs the limit of both sides is 1, and xn is not smaller than 1, it is enough to prove the second\ninequality, for high enough values of n, that is, we shall prove that there exists n0 \u2208N such\nthat the above inequality holds \u2200n \u2265n0. To do that, consider the following equivalence of\ninequalities:\nn\u221an \u22641 + 1\n\u221an\n\u21d0\u21d2\nn \u2264\n\u0012\n1 + 1\n\u221an\n\u0013n\n=\nn\nX\ni=0\n\u0012n\ni\n\u0013\n1\n(\u221an)i\n53", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_103", "text": "Note that the sum on the right hand side for i = 4 is\nn(n \u22121)(n \u22122)(n \u22123)\n24\n1\n(\u221an)4 = n(n \u22121)(n \u22122)(n \u22123)\n24n2\n.\nSo, we know the desired inequality (i.e., that\nn\u221an \u22641 +\n1\n\u221an) as soon as n \u22654 and n \u2264\nn(n\u22121)(n\u22122)(n\u22123)\n24n2\n. The latter is equivalent to\n24n2\n(n \u22121)(n \u22122)(n \u22123) \u22641.\nHowever, we have just learned that\nlim\nn\u2192\u221e\n24n2\n(n \u22121)(n \u22122)(n \u22123) = 0,\nso there is an integer n1, such that for each n \u2265n1,\n\f\f\f\f\n24n2\n(n \u22121)(n \u22122)(n \u22123)\n\f\f\f\f \u22641.\n[There is a di\ufb00erent proof in the book, on page 24: check that out, too!].\nCorollary 4.41. Let (xn) be a convergent sequence, and let (yn) be a bounded sequence. If\nlim\nn\u2192\u221exn = 0, then the sequence (xnyn) is convergent and lim\nn\u2192\u221exnyn = 0.\nRemark 4.42. Let us recall, see also the exercises, that for a sequence (xn), lim\nn\u2192\u221exn = 0 if and\nonly if lim\nn\u2192\u221e|xn| = 0. In fact, if we assume that lim\nn\u2192\u221exn = 0, then we can use a similar argument\nto show that lim\nn\u2192\u221e|xn| = 0.\nProof of Corollary 4.41. Let us note that showing that lim\nn\u2192\u221exnyn = 0 is equivalent to showing\nthat lim\nn\u2192\u221e|xnyn| = 0. This follows immediately from Remark 4.42.\nAs yn is bounded, there is an integer M > 0 such that |yn| \u2264M for all n \u2208N. Hence, we may\nsqueeze |xnyn|:\n0 \u2264|xnyn| \u2264M \u00b7 |xn|,\nSince lim\nn\u2192\u221eM \u00b7 |xn| = M \u00b7 lim\nn\u2192\u221e|xn| = 0, then also lim", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_104", "text": "squeeze |xnyn|:\n0 \u2264|xnyn| \u2264M \u00b7 |xn|,\nSince lim\nn\u2192\u221eM \u00b7 |xn| = M \u00b7 lim\nn\u2192\u221e|xn| = 0, then also lim\nn\u2192\u221e|xnyn| = 0.\nExample 4.43. Let (xn)n\u22651 be the sequence de\ufb01ned by xn :=\n1\nn2 sin(n).\nWe show that\nlim\nn\u2192\u221e\n1\nn2 sin(n) = 0.\nLet us note that we do not know whether sin(n) does or does not converge in itself \u2013 it is possible\nto prove that indeed it does not converge. So, we may not apply the previous multiplication\nrule of limits. However, we may apply the previous corollary, as lim\nn\u2192\u221e\n1\nn2 = 0, and sin(n) is\nbounded (by \u22121 and 1).\nExample 4.44. Let us de\ufb01ne the sequence (xn)n\u2208N recursively as\n(\nxn+1 = sin(xn)\n2\n,\nx0 = 1.\nThen,\n|xn+1|\n|xn|\n=\n| sin(xn)|\n2\n|xn|\n\u22641\n2,\n54", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_105", "text": "where the last inequality follows from the fact that | sin(x)|\n|x|\n\u22641 for all x \u2208R. To show this,\njust notice that |x| measures the length of the circle segment of angle (measured in radiants)\nx, where we count multiple revolutions too, and | sin(x)| gives the absolute value of the y-\ncoordinate of the endpoint of the circle segment, as shown in the \ufb01gure below.\nIn particular,\nImages/SqueezedSin.jpg\nFigure 8: | sin(\u03b8)| \u2264|\u03b8|\n|xn+1| = |xn+1|\n|xn| |xn| \u22641\n2|xn|.\n(4.44.d)\nIterating the observation in (4.44.d), we obtain\n|xn+1| \u22641\n2|xn| \u22641\n22 |xn\u22121| \u22641\n23 |xn\u22122| \u2264\u00b7 \u00b7 \u00b7 \u22641\n2n |x1| \u2264\n1\n2n+1 .\nSo, we may use the Squeeze Theorem 4.36 to show that lim\nn\u2192\u221exn = 0, squeezing since\n0 \u2264xn \u22641\n2n ,\n\u2200n \u2208N.\n4.5.1\nLimits of recursive sequences\nExample 4.45. The Fibonacci sequence (xn)n\u2208N is de\ufb01ned by\n(\nxn+1 = xn + xn\u22121\nx0 = x1 = 1.\nIf we de\ufb01ne the sequence (yn)n\u2208N by yn := xn+1\nxn , then the sequence (yn) admits a recursive\nde\ufb01nition as follows\n(\nyn+1 = 1 + 1\nyn\ny0 = 1.\n(4.45.e)\nWe call (yn)n\u2208N the sequence of Fibonacci quotients.\nProposition 4.46. If (yn) is the sequence of Fibonacci quotients, then \u2200n \u2208N, 1 \u2264yn \u22642.\nProof. We prove the above statements by induction on n \u2208N.\n\u25e6Starting step: the n = 0 case; by de\ufb01nition we have 2 \u2265y0 = 1.\n55", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_106", "text": "\u25e6Inductive step: we can assume that we know that statement for n and then we prove it\nfor n + 1 below:\nyn+1 = 1 + 1\nyn\n\u22651 + 1\n2 \u22651,\nand\nyn+1 = 1 + 1\nyn\n\u22641 + 1\n1 = 2,\nExample 4.47. Let us continue with Example 4.45. As we know that the sequence of Fibonacci\nquotients is bounded, we can ask whether it converges or not.\nIf a limit exists, can we use the recursive relation in (4.45.e) to \ufb01nd what that limit is? Let us\ntry! Let us assume now that (yn) is convergent, and lim\nn\u2192\u221eyn = y. Then, as yn \u22651, it follows\nthat y \u22651. Furthermore, by (4.45.e),\ny = lim\nn\u2192\u221eyn = lim\nn\u2192\u221eyn+1 = lim\nn\u2192\u221e\n\u0012\n1 + 1\nyn\n\u0013\n|\n{z\n}\nrecursive relation in (4.45.e)\n= 1 +\n1\nlim\nn\u2192\u221eyn\n= 1 + 1\ny\n|\n{z\n}\nalgebraic rules of limit\n.\nThis yields that the limit y satis\ufb01es the equation y = 1+ 1\ny which we can rewrite as y2\u2212y\u22121 = 0\n(since we know that y \u0338= 0) and whose solutions are\ny = 1 \u00b1 \u221a1 + 4\n2\n= 1 \u00b1\n\u221a\n5\n2\nAs we have seen that 1 \u2264y \u22642, then this forces the equality y = 1+\n\u221a\n5\n2\n. Thus, if the limit of\n(yn) exists, then y = 1+\n\u221a\n5\n2\nHowever, we have not proven yet that (yn) converges. As we have\n\ufb01gured out that if (yn) converges the only possible limit is 1+\n\u221a\n5\n2\n, we may show that that the\nsequence zn :=\n\f\f\fyn \u22121+\n\u221a\n5\n2\n\f\f\f converges to 0. But then,\nzn+1 =\n\f\f\f\f\fyn+1 \u22121 +\n\u221a\n5\n2\n\f\f\f\f\f =\n\f\f\f\f\f1 + 1\nyn\n\u22121 \u2212\n1\n1+\n\u221a\n5\n2\n\f\f\f\f\f =\n\f\f\f\f\f\n1\nyn\n\u2212\n1\n1+\n\u221a\n5\n2\n\f\f\f\f\f\n|\n{z\n}\nwe apply the de\ufb01nition of the sequence to yn+1, and\nthen as we found 1+\n\u221a\n5\n2\nas the solution of y = 1 + 1\ny\nwe may replace 1+\n\u221a\n5\n2\nby 1 +\n1", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_107", "text": "2\n\f\f\f\f\f =\n\f\f\f\f\f\n1\nyn\n\u2212\n1\n1+\n\u221a\n5\n2\n\f\f\f\f\f\n|\n{z\n}\nwe apply the de\ufb01nition of the sequence to yn+1, and\nthen as we found 1+\n\u221a\n5\n2\nas the solution of y = 1 + 1\ny\nwe may replace 1+\n\u221a\n5\n2\nby 1 +\n1\n1+\n\u221a\n5\n2\n=\n\f\f\fyn \u22121+\n\u221a\n5\n2\n\f\f\f\nyn 1+\n\u221a\n5\n2\n\u2264|zn|\n1+\n\u221a\n5\n2\nIterating this reasoning, we get that\nzn+1 \u2264|zn|\n1+\n\u221a\n5\n2\n\u2264\n|zn\u22121|\n(1+\n\u221a\n5\n2\n)2 \u2264\n|zn\u22122|\n(1+\n\u221a\n5\n2\n)3 \u2264\u00b7 \u00b7 \u00b7 \u2264\n|zn\u2212k|\n(1+\n\u221a\n5\n2\n)k+1 .\nand thus,\n0 \u2264zn \u2264\n|z0|\n\u0010\n1+\n\u221a\n5\n2\n\u0011n ,\nwhere\nlim\nn\u2192\u221e\n|z0|\n\u0010\n1+\n\u221a\n5\n2\n\u0011n = 0.\n56", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_108", "text": "So, the Squeeze Theorem (Theorem 4.36) shows that lim\nn\u2192\u221ezn = 0. This in turn implies, by the\nde\ufb01nition of zn that lim\nn\u2192\u221eyn = 1+\n\u221a\n5\n2\n. Summarizing, we showed that for the Fibonacci sequence\n(xn)\nlim\nn\u2192\u221e\nxn+1\nxn\n= 1 +\n\u221a\n5\n2\nThe number 1+\n\u221a\n5\n2\nis also known as the Golden ratio.\nThe general approach to \ufb01nding the limit of a recursive sequence (xn),\n\uf8f1\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f2\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f3\nxi = f(xi\u22121, xi\u22122, . . . , xi\u2212n)\nx0 = c0\nx1 = c1\nx2 = c2\n...\nxn\u22121 = cn\u22121\nis similar to the one we just explained in Example 4.47.\nIt can be summarized in the following 3-step recipe:\n(1) assuming that there exists a \ufb01nite limit for (xn), lim\nn\u2192\u221exn = x, then \ufb01nd the solutions of\nthe equation\nx = f(x, x, x, . . . , x).\n(4.47.f)\nIn setting up such equation, one has to be careful as to whether the equation itself and its\nsolutions are well-de\ufb01ned \u2013 e.g., one has to be careful when x appear in the denominator\nof a fraction: for which values of x is f(x, x, x, . . . , x) makes sense? Can we make sure that\nthose values of x for which f(x, x, x, . . . , x) is not well-de\ufb01ned are values which cannot\nbe attained by lim\nn\u2192\u221exn?\nIf the above equation does not admit any solutions, then the sequence (xn) cannot admit\nlimit;\n(2) we try to exclude all but one of the possibilities for among the values of x obtained in\nthe previous point by using some argument coming from the explicit de\ufb01nition of (xn);\n(3) if we found a unique solution \u00afx of(4.47.f), we can try to make a direct veri\ufb01cation hat\nlim\nn\u2192\u221exn = \u00afx by showing that the \u00afx satis\ufb01es the de\ufb01nition of limit for (xn).\nExample 4.48. This method of \ufb01nding the limit does not always work.", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_109", "text": "lim\nn\u2192\u221exn = \u00afx by showing that the \u00afx satis\ufb01es the de\ufb01nition of limit for (xn).\nExample 4.48. This method of \ufb01nding the limit does not always work.\nFor example, consider the recursive sequence (xn)n\u2208N de\ufb01ned by\n(\nxn+1 = 1\n2(xn + xn\u22121)\nx0 = C.\nThen applying Step 1 in the above recipe gives the equation x = 1\n2(x + x). Of course, this\nequation is satis\ufb01ed for any value of x \u2208R. Hence, we cannot use it to restrict th epossible\nvalues of the limit of (xn)n\u2208N.\n57", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_110", "text": "Example 4.49. Here is an example of a recursive sequence where our recipe does not work.\nLet a, b \u2208(0, +\u221e) and let (xn)n\u2208N be the recursive sequence de\ufb01ned by the recurrence relation\n(\nxn+1 = ax2\nn\nx0 = b.\nClaim 1. For all n \u2208N, xn = a2n\u22121b2n.\nProof. We prove the claim by induction on n \u2208N.\n\u25e6Starting step: For n = 0, x0 = a20\u22121b20 = b.\n\u25e6Inductive step: Assuming that xk = a2k\u22121b2k for all 0 \u2264k < n, then\nxn =ax2\nn\u22121\n=a \u00b7\n\u0010\na(2n\u22121\u22121) \u00b7 b2n\u22121\u00112\n=a \u00b7 a(2\u00b72n\u22121\u22122) \u00b7 b2\u00b72n\u22121\n=a(2n\u22121)b2n.\nNow, applying the \ufb01rst step of our recipe, we assume that lim\nn\u2192\u221exn = x and we solve the\nequation\nx = ax2.\nSolutions are x = 0 and x = 1\na \u2013 the latter is well de\ufb01ned since a \u0338= 0.\nWe can actually compute lim\nn\u2192\u221exn directly: we have to distinguish 3 di\ufb00erent cases:\n(1) if ab = 1 then\nlim\nn\u2192+\u221exn = lim\nn\u2192\u221e\n(ab)2n\na\n= 1\na.\nHence, in this case the limit of (xn)n\u2208N corresponds to one of the solutions that we found\nabove;\n(2) if ab < 1 then\nlim\nn\u2192+\u221exn = lim\nn\u2192\u221e\n(ab)2n\na\n= 0.\nAlso in this case the limit of (xn)n\u2208N corresponds to one of the solutions that we found\nabove;\n(3) if ab > 1 then xn = (ab)2n\na\nwhich is not bounded; even better,\nlim\nn\u2192+\u221exn = (ab)2n\na\n= +\u221e.\nIn this case, our algorithm could not possibly work since (xn)n\u2208N being unbounded cannot\npossibly converge to a \ufb01nite limit.\n58", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_111", "text": "4.5.2\nUnbounded sets and in\ufb01nite limits\nDe\ufb01nition 4.50. Let (xn) be a sequence.\n(1) We say that (xn) approaches +\u221eif for all C \u2208R there is an index nC \u2208N such that for\nall integers n \u2265nC, xn \u2265C.\n(2) We say that (xn) approaches \u2212\u221eif for all C \u2208R there is an index nC \u2208N such that for\nall integers n \u2265nC, xn \u2264C.\nNotation 4.51. If a sequence (xn) approaches +\u221e(resp. \u2212\u221e), we write\nlim\nn\u2192\u221exn = +\u221e\n(resp. lim\nn\u2192\u221exn = \u2212\u221e).\nIf a sequence (xn) satis\ufb01es lim\nn\u2192\u221exn = \u00b1\u221e, then it cannot possibly converge to a \ufb01nite limit,\nas De\ufb01nition 4.50 implies that (xn) is unbounded.\nExample 4.52. Let (xn)n\u2208N be a geometric sequence, xn := aqn.\n(1) The Bernoulli inequality, see Proposition 4.16, implies that for every geometric progres-\nsion (xn)n\u2208N, xn := aqn, with a > 0 and q > 1 lim xn = +\u221e. An example is the sequence\n(xn)n\u2208N de\ufb01ned by xn := 3 \u00b7 2n. In fact, xn \u22653(1 + n(2 \u22121)) = 3 + 3n, and by the\nArchimidean property, cf. Proposition 2.30, given C \u2208R, then\nif C \u22640, then \u2200n \u2208N,\n3n + 3 > 0 \u2265C,\nif C > 0, then \u2200n \u2265\n\u0014C\n3\n\u0015\n,\n3n + 3 > 3C\n3 = C.\n(2) Similarly, lim xn = \u2212\u221efor every geometric progression xn = aqn with a < 0 and q > 1.\nAn example is the sequence de\ufb01ned by xn := \u22123 \u00b7 2n.\n(3) On the other hand, if a \u0338= 0 and q < 0, then (xn) is unbounded but it neither approaches", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_112", "text": "An example is the sequence de\ufb01ned by xn := \u22123 \u00b7 2n.\n(3) On the other hand, if a \u0338= 0 and q < 0, then (xn) is unbounded but it neither approaches\n+\u221enot it approeaches \u2212\u221e. For example xn = (\u22122)n is non-convergent but it also does\nnot admit limit equal to +\u221eor \u2212\u221e.\nThe in\ufb01nite limits satisfy some algebraic rules, and do not satisfy others. Check out page\n29 and 30 of the book for full list.\nProposition 4.53. Let (xn), (yn) be two sequences.\n(1) Assume that lim\nn\u2192\u221exn = +\u221eand that (yn) is bounded from below. Then,\n(i)\nlim\nn\u2192\u221exn + yn = +\u221e;\n(ii) if there exists A \u2208R\u2217\n+ and n0 \u2208N such that \u2200n \u2265n0, yn \u2265A, then lim\nn\u2192\u221exn\u00b7yn = +\u221e;\n(iii) if (yn) is bounded, then lim\nn\u2192\u221e\nyn\nxn = 0.\n(2) Assume that lim\nn\u2192\u221exn = \u2212\u221eand that (yn) is bounded from above. Then,\n(i)\nlim\nn\u2192\u221exn + yn = \u2212\u221e;\n(ii) if there exists A \u2208R\u2217\n+ and n0 \u2208N such that \u2200n \u2265n0, yn \u2265A, then lim\nn\u2192\u221exn\u00b7yn = \u2212\u221e;\n(iii) if (yn) is bounded, then lim\nn\u2192\u221e\nyn\nxn = 0.\n59", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_113", "text": "Example 4.54. Let (xn)n\u2208N be the sequence de\ufb01ned by xn := 2n + sin(n). Then\nlim\nn\u2192\u221exn = lim\nn\u2192\u221e(2n + sin(n)) = +\u221e,\nbecause lim\nn\u2192\u221e2n = +\u221eand sin(n) \u2265\u22121.\nRemark 4.55. Part (3) for both of the above propositions claims that if lim\nn\u2192\u221e|xn| = +\u221eand (yn)\nis bounded then lim\nn\u2192\u221e\nyn\nxn = 0. It is important to remark that one cannot drop the assumptions\non the boundedness of (yn). That is, if we do not assume that (yn) is bounded, then we cannot\nconclude anything about lim\nn\u2192\u221e\nyn\nxn , as shown by the next examples. In fact, taking\n(1) xn := n, yn := n, then lim\nn\u2192\u221e\nyn\nxn = lim\nn\u2192\u221e1 = 1;\n(2) xn := n, yn := n2, then lim\nn\u2192\u221e\nyn\nxn = lim\nn\u2192\u221en = +\u221e;\n(3) xn := n, yn := \u221an, then lim\nn\u2192\u221e\nyn\nxn = lim\nn\u2192\u221e\n1\n\u221an = 0;\n(4) xn := (\u22121)nn, yn := n, then yn\nxn = (\u22121)n, thus, ( yn\nxn ) does not converge.\nExample 4.56. Here we show examples of sequences (xn) and (yn), for which lim\nn\u2192\u221exn = +\u221e,\nlim\nn\u2192\u221eyn = \u2212\u221eand for which the sequence (xn + yn) displays all possible behaviors in terms of\nits convergence (or lack thereof). In fact, taking\n(1) xn := n, yn := \u2212n lim\nn\u2192\u221e(xn + yn) = 0;\n(2) xn := 2n, yn := \u2212n lim\nn\u2192\u221e(xn + yn) = +\u221e;\n(3) xn := n, yn := \u22122n lim\nn\u2192\u221e(xn + yn) = \u2212\u221e;", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_114", "text": "(2) xn := 2n, yn := \u2212n lim\nn\u2192\u221e(xn + yn) = +\u221e;\n(3) xn := n, yn := \u22122n lim\nn\u2192\u221e(xn + yn) = \u2212\u221e;\n(4) xn := 2n, yn := (\u22121)nn, then\nxn + yn =\n(\nn\nfor n odd,\n3n\nfor n even.\nHence, xn + yn is unbounded, thus, non-converging, and its limit cannot be \u00b1\u221e.\nIt is a homework to cook up similar examples for multiplication and division. For example,\na famous case where \u201danything can happen\u201d for multiplication is that of sequence (xn), (yn)\nsuch that lim\nn\u2192\u221exn = +\u221eand lim\nn\u2192\u221eyn = 0.\nSimilarly to the argument for \ufb01nite limits, we can prove squeeze theorems for in\ufb01nite limits:\nTheorem 4.57 (Squeeze Theorem for sequences approaching in\ufb01nities). Let (xn) and (yn) be\ntwo sequences.\n(1) Assume that there exists n0 \u2208N such that\n\u2200n \u2265n0,\nxn \u2264yn.\n(i) If lim\nn\u2192\u221exn = +\u221e, then lim\nn\u2192\u221eyn = +\u221e.\n(ii) If lim\nn\u2192\u221eyn = \u2212\u221e, then lim\nn\u2192\u221exn = \u2212\u221e.\n(2) Assume that lim\nn\u2192\u221e\nxn\nyn = q.\n60", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_115", "text": "(i\u2019) If lim\nn\u2192\u221exn = +\u221eand q \u2208R\u2217\n+, then lim\nn\u2192\u221eyn = +\u221e.\n(ii\u2019) If lim\nn\u2192\u221eyn = \u2212\u221eand q \u2208R\u2217\n+ \u222a{+\u221e}, then lim\nn\u2192\u221exn = \u2212\u221e.\nProof.\n(1) Let us prove (i). The other case is proven analogously.\nFix C \u2208R. As lim\nn\u2192\u221exn = +\u221e, there exists nC \u2208N such that \u2200n \u2265nC, xn \u2265C. Taking\nn\u2032\nC := max n0, nC, then \u2200n \u2265n\u2032\nC, yn \u2265xn \u2265C. Hence, lim\nn\u2192\u221eyn = +\u221e.\n(2) Let us prove (i\u2032) when q \u2208R\u2217\n+. All the other cases are proven analogously.\nHence, we assume that lim\nn\u2192\u221e\nxn\nyn = q > 0 and lim\nn\u2192\u221exn = +\u221e. In particular, the latter\nimplies that there exists n0 such that \u2200n \u2265n0, xn > 0.\nLet us take \u03f5 = q\n2. Hence, there exists n q\n2 \u2208N such that \u2200n \u2265n q\n2\n\u2212q\n2 \u2264q \u2212xn\nyn\n\u2264q\n2.\nHence, \u2200n \u2265n q\n2 ,\nq\n2 \u2264xn\nyn\n\u22643q\n2.\nThus, \u2200n \u2265max n0, n q\n2 ,\n(\nyn \u2265\nq\n2xn > 0,\nsince n \u2265n0\nxn \u22643q\n2 yn, ,\nsince yn > 0 and n \u2265n q\n2 .\nHence, by part (1) of the theorem, lim\nn\u2192\u221e\n3q\n2 \u00b7 yn = 3q\n2 \u00b7 lim\nn\u2192\u221eyn = 3q\n2 yn = +\u221e.\nExample 4.58.\n(1) Let (xn)n\u2208N be the sequence de\ufb01ne by xn := n!\n2n We compute lim\nn\u2192\u221e\nn!\n2n .\nWe have n!\n2n \u22652\u00b73\u00b73\u00b7\u00b7\u00b73\u00b73\n2n\n= 1\n2\n\u0000 3\n2\n\u0001n\u22122, and lim\nn\u2192\u221e\n1\n2\n\u0000 3\n2\n\u0001n\u22122 = +\u221eaccording to Example 4.52.\nHence, Theorem 4.57 part (1.i) yields lim\nn\u2192\u221e\nn!\n2n = +\u221e.", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_116", "text": "2n\n= 1\n2\n\u0000 3\n2\n\u0001n\u22122, and lim\nn\u2192\u221e\n1\n2\n\u0000 3\n2\n\u0001n\u22122 = +\u221eaccording to Example 4.52.\nHence, Theorem 4.57 part (1.i) yields lim\nn\u2192\u221e\nn!\n2n = +\u221e.\n(2) Similarly, but using part (1.ii) of Theorem 4.57, then lim\nn\u2192\u221e\u2212n!\n2n = \u2212\u221e.\n4.6\nMore convergence criteria\nWe can apply the Squeeze Theorem 4.36 to obtain more convergence criteria.\nCorollary 4.59 (Quotient criterion). Let (xn) be a sequence. Assume that\nlim\nn\u2192\u221e\n|xn+1|\n|xn|\n= q \u2208R+ \u222a{+\u221e}.\n(1) If q < 1, then both (xn) and (|xn|) converge and the limit is 0 for both sequences.\n(2) If q > 1 or q = +\u221e, then (xn) and (|xn|) both are non-converging sequences. Moreover,\nlim\nn\u2192\u221e|xn| = +\u221e.\nRemark 4.60. As in the statement of the Corollary we are assuming that the sequence yn :=\n|xn+1|\n|xn|\nconverges, then since yn \u22650, \u2200n \u226b1, then the limit q of yn is automatically a non-\nnegative real number, cf. Corollary 4.32. Thus q \u22650.\n61", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_117", "text": "Proof. We show here only the 0 \u2264q < 1 case; the other case is similar, and is left as a\nhomework.\nFix \u03b5 := 1\u2212q\n2 . In particular \u03b5 > 0 and q + \u03b5 < 1. There is an index n\u03b5 \u2208N, such that\n\u2200n \u2265n\u03b5,\n\f\f\f\f\n|xn+1|\n|xn|\n\u2212q\n\f\f\f\f < \u03f5,\nor equivalently,\n\u2200n \u2265n\u03b5,\nq \u2212\u03b5 < |xn+1|\n|xn|\n< q + \u03b5,\nthus, |xn+1| < (q + \u03b5)|xn|. Denoting q := q + \u03b5, then\nq < 1\nand\n\u2200i \u2208N, |xn\u03b5+i| \u2264|xn\u03b5| qi.\nHence, as\n\u2200n \u2265n\u03b5,\n0 \u2264|xn| \u2264|xn\u03b5|qn\u2212n\u03b5,\nwe may apply the Squeeze Theorem 4.36 to |xn| since\nlim\nn\u2192\u221e|xn\u03b5|qn\u2212n\u03b5 = |xn\u03b5|\nqn\u03b5\nlim\nn\u2192\u221eqn = |xn\u03b5|\nqn\u03b5 \u00b7 0\n|\n{z\n}\n|q|<1\u21d2limn\u2192\u221e|q|n=0\n= 0.\nExample 4.61. We present some examples showing that if in Corollary 4.59 q = 1, then we\ncannot conclude anything about the behavior of the sequence (xn).\n(1) If xn := n, then (xn) is non-convergent and limn\u2192\u221exn = +\u221e, while\nlim\nn\u2192\u221e\n|xn+1|\n|xn|\n= lim\nn\u2192\u221e\nn + 1\nn\n= 1.\n(2) If xn := (\u22121)nn, then (xn) is not bounded and its limit does not exist in R, while\nlim\nn\u2192\u221e\n|xn+1|\n|xn|\n= lim\nn\u2192\u221e\nn + 1\nn\n= 1.\n(3) If xn := n+1\nn , then (xn) is convergent to 1, while\nlim\nn\u2192\u221e\n|xn+1|\n|xn|\n= lim\nn\u2192\u221e\nn+2\nn+1\nn+1\nn\n= lim\nn\u2192\u221e\n(n + 2)n\n(n + 1)2 = 1.", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_118", "text": "n , then (xn) is convergent to 1, while\nlim\nn\u2192\u221e\n|xn+1|\n|xn|\n= lim\nn\u2192\u221e\nn+2\nn+1\nn+1\nn\n= lim\nn\u2192\u221e\n(n + 2)n\n(n + 1)2 = 1.\n(4) If xn := (\u22121)n, then (xn) is bounded but it does not admit a \ufb01nite limit, while\nlim\nn\u2192\u221e\n|xn+1|\n|xn|\n= lim\nn\u2192\u221e1 = 1.\nHence, all possible behaviors of a sequence, in terms of its convergence or lack thereof, can\nappear when q = 1 in Corollary 4.59.\nAnother consequence\nCorollary 4.62 (Root criterion). Let (xn) be a sequence. Assume that\nlim\nn\u2192\u221e\nnp\n|xn| = q \u2208R+ \u222a{+\u221e}.\n62", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_119", "text": "(1) If q < 1, then both (xn) and (|xn|) converge and their limit is 0.\n(2) If q > 1 or q = +\u221e, then (xn) and (|xn|) both are non-converging sequences. Moreover,\nlim\nn\u2192\u221e|xn| = +\u221e.\nProof. We prove part (2). The proof of the case is analogous.\nWe start assuming that q \u2208(1, +\u221e).\nIf q = +\u221e, instead, then there exists n2 \u2208N such that \u2200n \u2265n2,\nnp\n|xn| \u22652 or, equivalently,\n|xn| \u22652n. Since lim\nn\u2192\u221e2n = +\u221e, by Theorem 4.57, then lim\nn\u2192\u221e|xn| = +\u221e.\n4.7\nMonotone sequences\nLet us recall that we say that a sequence (xn) is monotone if it is increasing or decreasing,\ncf. De\ufb01nition 4.7.\nTheorem 4.63. Let (xn)n\u2265l be a monotone sequence.\n(1) If (xn)n\u2265l is bounded and increasing (resp. decreasing), then (xn)n\u2265l is convergent and\nlim\nn\u2192\u221exn = sup{xn | n \u2208N, n \u2265l}\n(resp. lim\nn\u2192\u221exn = inf{xn | n \u2208N, n \u2265l}).\n(2) If (xn)n\u2265l is unbounded and increasing (resp.\ndecreasing) then lim\nn\u2192\u221exn = +\u221e(resp.\nlim\nn\u2192\u221exn = \u2212\u221e).\nProof. We prove only the increasing case. We leave as a homework to change the words in it\nto obtain a proof for the decreasing case.\nSet S := sup{xn | n \u2208N, n \u2265l} and let 0 < \u03b5 \u2208R be arbitrary. By de\ufb01nition, S is the\nsmallest upper bound, so S \u2212\u03b5 is not an upper bound. Hence, there exists n\u03b5 \u2208N such that\nS \u2212\u03b5 < xn\u03b5. In particular, for any integer n \u2265n\u03b5:\nS \u2212\u03b5\n< xn\u03b5\n| {z }\nde\ufb01nition of n\u03b5\n\u2264xn\n| {z }\n(xn) is monotone\n\u2264S", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_120", "text": "S \u2212\u03b5 < xn\u03b5. In particular, for any integer n \u2265n\u03b5:\nS \u2212\u03b5\n< xn\u03b5\n| {z }\nde\ufb01nition of n\u03b5\n\u2264xn\n| {z }\n(xn) is monotone\n\u2264S\n| {z }\nS is the supremum\n< S + \u03b5.\nExample 4.64 (Nepero\u2019s number e). Let us consider the sequence (xn)n\u22651 de\ufb01ned by\nx :=\n\u0012\n1 + 1\nn\n\u0013n\n, n \u2208N\u2217.\nClaim. The sequence (xn)n\u22651 is strictly increasing.\nProof. We need to show that\n\u00001 + 1\nn\n\u0001n <\n\u0010\n1 +\n1\nn+1\n\u0011n+1\n, \u2200n \u2208N\u2217. Indeed,\n\u0012\n1 + 1\nn\n\u0013n\n=\nn\nX\ni=0\n\u0012n\ni\n\u0013 1\nni =\nn\nX\ni=0\nn!\ni!(n \u2212i)!\n1\nni =\nn\nX\ni=0\n1\ni!\nn(n \u22121) . . . (n \u2212(i \u22121))\nni\n=\n1\n|{z}\n=(n\n0) 1\nn0\n+\n1\n|{z}\n=(n\n1) 1\nn= n\nn\n+\nn\nX\ni=2\n1\ni!\nn\nn\n|{z}\n=1\ni\u22121 terms\nz\n}|\n{\n(n \u22121)\nn\n(n \u22122)\nn\n(n \u2212(i \u22121))\nn\n(4.64.a)\n= 1 + 1 +\nn\nX\ni=2\n1\ni!\n\u0012\n1 \u22121\nn\n\u0013\n. . .\n\u0012\n1 \u2212i \u22121\nn\n\u0013\n|\n{z\n}\ni\u22121 terms\n63", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_121", "text": "Similarly,\n\u0012\n1 +\n1\nn + 1\n\u0013n+1\n=\nn+1\nX\ni=0\n1\ni!\n\u0012\n1 \u2212\n1\nn + 1\n\u0013\n. . .\n\u0012\n1 \u2212i \u22121\nn + 1\n\u0013\n(4.64.b)\n= 1 + 1 +\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\nn\nX\ni=2\n1\ni!\n\u0012\n1 \u2212\n1\nn + 1\n\u0013\n|\n{z\n}\n> (1\u22121\nn)\n\u0012\n1 \u2212\n2\nn + 1\n\u0013\n|\n{z\n}\n> (1\u22122\nn)\n. . .\n\u0012\n1 \u2212i \u22121\nn + 1\n\u0013\n|\n{z\n}\n> (1\u2212i\u22121\nn )\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n+\n\u0012\n1\nn + 1\n\u0013n+1\n> 2 +\nn\nX\ni=2\n1\ni!\n\u0012\n1 \u2212\n1\nn + 1\n\u0013\n. . .\n\u0012\n1 \u2212i \u22121\nn + 1\n\u0013\nHaving proved our claim, then (xn)n\u22651 is a monotone increasing sequence. Is it bounded? Yes,\nit is: indeed,\n\u0012\n1 + 1\nn\n\u0013n\n=2 +\nn\nX\ni=2\n1\ni!\n\u0012\n1 \u22121\nn\n\u0013\n. . .\n\u0012\n1 \u2212i \u22121\nn\n\u0013\n\u2264\nn\nX\ni=0\n1\ni! \u22641 +\nn\nX\ni=1\n1\n2i\u22121 = 1 + 1 \u22121\n2n\n1\n2\n= 3 \u22121\n2n \u22643,\nwhere, for evaluating the sum, we used the formula that we proved in Proposition 1.6\n(1 + \u00b7 \u00b7 \u00b7 + an\u22121) = 1 \u2212an\n1 \u2212a ,\nfor a = 1\n2. Hence, (xn)n\u22651 is not only increasing, but also bounded above by 3. Thus, lim\nn\u2192\u221exn\nexists, according to Theorem 4.63.\nDe\ufb01nition 4.65. We de\ufb01ne e := lim\nn\u2192\u221e\n\u00001 + 1\nn\n\u0001n.\nTheorem 4.63 also gives another method for showing the existence of limits for recursive\nsequences:\nExample 4.66. We consider the recursive sequence (xn)n\u2208N de\ufb01ned as\n(\nxn+1 = 1\n2\n\u0010\nxn +\n1\nxn\n\u0011\nx0 = 2.\nFirst we claim that xn > 0 for all integers n \u2208N. This is certainly true for n = 0, and if we\nassume it for n \u22121, then the recursive formula gives it to us also for n. Hence, by induction,\n\u2200n \u2208N, xn > 0. In particular, the division in the de\ufb01nition does make sense.", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_122", "text": "assume it for n \u22121, then the recursive formula gives it to us also for n. Hence, by induction,\n\u2200n \u2208N, xn > 0. In particular, the division in the de\ufb01nition does make sense.\nNext, we claim that xn \u22651 for all integers n \u22651. Indeed, a similar induction shows that this\nclaim: indeed, for n = 0, we have x0 = 2 \u22651. Furthermore,\nxn+1 = 1\n2\n\u0012\nxn + 1\nxn\n\u0013\n\u22651 \u21d4xn + 1\nxn\n\u22652 \u21d4x2\nn + 1 \u22652xn \u21d4(xn \u22121)2 \u22650,\n(4.66.c)\nwhere we used that we already know that xn > 0, when we multiplied by xn. So, by (4.66.c),\nthe induction step works too. That is, assuming xn \u22651, we obtain that xn+1 \u22651 holds as well.\nNext, we claim that the sequence is decreasing indeed,\nxn \u2212xn+1 = xn \u22121\n2\n\u0012\nxn + 1\nxn\n\u0013\n= 1\n2\n\u0012\nxn \u22121\nxn\n\u0013\n\u22650,\n64", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_123", "text": "where we obtained the last inequality using that xn \u22651 \u2265\n1\nxn .\nSo, (xn) is decreasing (hence bounded from above) and also bounded from below by 1. In\nparticular, xn is convergent, and lim\nn\u2192\u221exn \u22651. Hence to \ufb01nd the actual limit we may just apply\nlimit to the recursive equation to obtain that if y is the limit, then\ny = 1\n2\n\u0012\ny + 1\ny\n\u0013\n\u21d4y\n2 = 1\n2y \u21d4y2 = 1\nAs we also know that y \u22651, y = 1 has to hold. So, lim\nn\u2192\u221exn = 1.\n4.8\nLiminf, limsup\nLet (xn)n\u2265l be a bounded sequence. We de\ufb01ne two new sequences\n(yn)n\u2265l := sup{xk|n \u2264k \u2208N},\n(zn)n\u2265l := inf{xk|n \u2264k \u2208N}.\nThe sequence (yn)n\u2265l (resp. (zn)n\u2265l) is a decreasing (increasing) sequence, as sup (inf) is taken\nover smaller and smaller sets or real numbers as n increases. Moreover, if C \u2208R is an upper\nbound (resp. lower bound) for (xn), then it is also an upper bound (lower bound) for (yn) (for\n(zn)). Hence, Theorem 4.63 implies that the sequence (yn) (resp. (zn)) is convergent since it\nis monotone and bounded. This observation justi\ufb01es the following de\ufb01nition.\nDe\ufb01nition 4.67. With the notation just introduced,\n(1) we call the limit of the sequence (yn) the limsup of the sequence (xn) and we denote it\nby\nlim\nn\u2192\u221esup xn;\n(2) we call the limit of the sequence (zn) the liminf of the sequence (xn) and we denote it by\nlim\nn\u2192\u221einf xn.\nExample 4.68. Let (xn)n\u2208N be the sequence de\ufb01ned as xn = (\u22121)n.\nThen,\nlim\nn\u2192\u221esup{xk|n \u2264k \u2208N} = lim\nn\u2192\u221esup{\u22121, 1} = lim\nn\u2192\u221e1 = 1,\nand\nlim", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_124", "text": "Then,\nlim\nn\u2192\u221esup{xk|n \u2264k \u2208N} = lim\nn\u2192\u221esup{\u22121, 1} = lim\nn\u2192\u221e1 = 1,\nand\nlim\nn\u2192\u221einf{xk|n \u2264k \u2208N} = lim\nn\u2192\u221esup{\u22121, 1} = lim\nn\u2192\u221e\u22121 = \u22121.\nHence,\nlim\nn\u2192\u221esup xn = lim\nn\u2192\u221eyn = 1,\nand\nlim\nn\u2192\u221einf xn = lim\nn\u2192\u221ezn = \u22121.\nIn general, it is not always easy to compute the liminf and limsup of a sequence. Nonetheless,\nwhen a sequence is converging, this task becomes signi\ufb01cantly simpler.\nProposition 4.69. Let (xn)n\u2265l be a converging sequence with lim\nn\u2192\u221exn = l \u2208R. Then,\nlim\nn\u2192\u221einf xn = l = lim\nn\u2192\u221esup xn.\n65", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_125", "text": "4.9\nSubsequences\nDe\ufb01nition 4.70. Let (xn)n\u2265l be a sequence. A subsequence (yk)k\u2208N of (xn)n\u2265l is a sequence\nsequence de\ufb01ned by yk := xnk where nk \u2208N is de\ufb01ned by a function\nf : N \u2192{n \u2208N | n \u2265l}\nk 7\u2192f(k) =: nk\nwhich is a strictly increasing function of k.\nTo say that f is strictly increasing simply means that \u2200k \u2208N, f(k) < f(k + 1).\nThus, a subsequence of (xn)n\u2265l is a new sequence (yk)k\u2208N constructed taking the values of\n(xn)n\u2265l along a subset of the indices of (xn)n\u2265l, where we remember the order in which those\nvalues appear.\nExample 4.71.\n(1) for the sequence (xn)n\u2208N de\ufb01ned by xn := (\u22121)n, then both the constant\n1 sequence and the constant \u22121 sequences are subsequences.\nIn fact for\n(i) for f(k) := 2k, then yk := xnk = x2k = (\u22121)2k = 1; and\n(ii) for f(k) := 2k = 1, then yk := xnk = x2k+1(\u22121)2k+1 = \u22121.\n(2) for the sequence (xn)n\u2208N de\ufb01ned by xn := n2, then yk := xnk = k6 is the subsequence\nobtained by setting nk := k3.\n(3) for the sequence (xn)n\u22651 de\ufb01ned by xn =\n\u00001 + 2\nn\n\u0001n and nk := 2k, then\nlim\nk\u2192\u221exk = lim\nk\u2192\u221e\n\u0012\n1 + 2\n2k\n\u00132k\n= lim\nk\u2192\u221e\n \u0012\n1 + 1\nk\n\u0013k!2\n=\n \nlim\nk\u2192\u221e\n\u0012\n1 + 1\nk\n\u0013k!2\n= e2.\nWe can ask whether (xn)n\u22651 converges and, if so, what its limit is? Is lim\nn\u2192\u221exn = e2?\nThe next proposition illustrates the (simple) connection between the convergence of a se-", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_126", "text": "k\u2192\u221e\n\u0012\n1 + 1\nk\n\u0013k!2\n= e2.\nWe can ask whether (xn)n\u22651 converges and, if so, what its limit is? Is lim\nn\u2192\u221exn = e2?\nThe next proposition illustrates the (simple) connection between the convergence of a se-\nquence and that of a subsequence.\nProposition 4.72. Let (xn) be a sequence.\nIf lim\nn\u2192\u221exn = a \u2208R, then for any subsequence (yk), yk := xnk, lim\nk\u2192\u221eyk = a.\nLet us recall that a \u2208R means that either a is a real number or a = \u00b1\u221e.\nThe proof of Proposition 4.72 is just about invoking the de\ufb01nition of limit, cf. De\ufb01nition 4.21\nand 4.50, thus we do not spell out the details here.\nExample 4.73. Let (xn)n\u22651 be the sequence de\ufb01ned as xn := (\u22121)n \u00001 + 1\nn\n\u0001n.\n(1) If nk = 2k, then the subsequence (yk)k\u22651 de\ufb01ned by yk := x2k =\n\u00001 + 1\n2k\n\u00012k and lim\nk\u2192\u221eyk =\ne;\n(2) if nk = 2k + 1, then the subsequence (yk)k\u22651 de\ufb01ned by yk := x2k+1 = \u2212\n\u0010\n1 +\n1\n2k+1\n\u00112k+1\nand lim\nk\u2192\u221eyk = \u2212e.\nHence, the sequence (xn)k\u22651 cannot converge.\nWe just saw an example of a sequence which does not converge, but which admits converging\nsubsequences \u2013 which converge to di\ufb00erent limits. Given a sequence (xn), does it always admit a\nconverging subsequence? The answer, for a general sequence (xn) is no. In fact, Proposition 4.72\nshows that if lim\nn\u2192\u221exn = \u00b1\u221e, then any subsequence will have the same limit, thus, (xn) will not\nadmit any converging subsequence.\n66", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_127", "text": "Remark 4.74. It actually follows from the de\ufb01nition, that if a sequence (xn) is unbounded then\nit admits a subsequence (yk), yk := xnk such that either lim\nk\u2192\u221eyk = +\u221eor lim\nk\u2192\u221eyk = \u2212\u221e. [Try\nto prove this claim!]\nHence, in view of the claim, we can ask whether for a bounded sequence (xn), there always\nexists a convergent subsequence (yk), yk := xnk. Indeed, we can always answer this question\na\ufb03rmatively, as shown by the following celebrated result.\nTheorem 4.75 (Bolzano-Weierstrass). Let (xn) be a bounded sequence. Then (xn) contains a\nconvergent subsequence.\nProof. We de\ufb01ne nk by induction k \u2208N. We set n0 = 0 - this is the starting step of the\ninduction. So, let us assume nk\u22121 is de\ufb01ned. Let us then de\ufb01ne sk := sup{xn|n > nk\u22121}. Then\nthere is a integer nk > nk\u22121 such that\nxnk > sk \u22121\nk.\nWe claim that (xnk) is convergent. Indeed, this follows from the squeeze principle, as we have\nsk \u22121\nk < xnk < sk,\nif we can prove that (sk) converges. As sk := sup xn | n > nk\u22121, then sk+1 \u2264sk, as the subset\nof R of which we are taking the supremum gets smaller with k. Hence, (sk) is decreasing.\nMoreover, (sk) is bounded, since inf{xn | n \u2208N} \u2264sk \u2264sup{xn | n \u2208N}. Hence, lim\nk\u2192\u221esk =\nl \u2208R, and\nl = lim\nk\u2192\u221esk = lim\nk\u2192\u221esk \u2212lim\nk\u2192\u221e\n1\nk = lim\nk\u2192\u221e(sk \u22121\nk),\nso that also lim\nk\u2192\u221exnk = l.\nExample 4.76. Sometimes, given a sequence (xn), it is possible to write down explicitly some\nconvergent subsequences.\nFor example, de\ufb01ning xn := sin\n\u0000 n\u03c0\n4\n\u0001 \u00001 + 1\nn\n\u0001n, then setting\n(1) nk := 8k + 1, lim\nk\u2192\u221eyk = lim", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_128", "text": "convergent subsequences.\nFor example, de\ufb01ning xn := sin\n\u0000 n\u03c0\n4\n\u0001 \u00001 + 1\nn\n\u0001n, then setting\n(1) nk := 8k + 1, lim\nk\u2192\u221eyk = lim\nk\u2192\u221exnk =\n1\n\u221a\n2e;\n(2) nk := 8k + 2, then lim\nk\u2192\u221eyk = lim\nk\u2192\u221exnk = e,\n(3) nk = 8k + 5, then lim\nk\u2192\u221eyk = lim\nk\u2192\u221exnk = \u22121\n\u221a\n2e.\nExample 4.77. Other times, given a sequence (xn), it is not quite possible to write down\nexplicitly converging subsequences.\nOne example where this is not immediate is given for\nexample by the sequence xn := sin n \u2013 you can read here a discussion of how to obtain a\nconverging subsequence, and how \u201cdi\ufb03cult\u201d that should be.\nIn general, the Bolzano-Weierstrass Theorem 4.75 implies that some convergent subsequence\nexists but it does not a priori indicate how to explicitly obtain one. [Try to write down a\nconverging subsequence of the sequence xn := sin (n)\n\u00001 + 1\nn\n\u0001n.]\nExample 4.78. Let a > 0 be an integer.\nThen de\ufb01ning the sequence (xn)n\u22651 by xn =:\n\u00001 + a\nn\n\u0001n, we can consider the subsequence (yk)k\u22651 de\ufb01ned by yk := xak, to obtain:\nlim\nk\u2192\u221eyk = lim\nk\u2192\u221exak = lim\nk\u2192\u221e\n\u0010\n1 + a\nak\n\u0011ak\n= lim\nk\u2192\u221e\n\u0012\n1 + 1\nk\n\u0013ak\n= lim\nk\u2192\u221e\n \u0012\n1 + 1\nk\n\u0013k!a\n= ea.\n67", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_129", "text": "It is not hard to show that xn is increasing and bounded for a > 0 \u2013 the proof is similar\nto the case where a = 1, using binomial expansion. In particular, xn is convergent, as it is\nbounded \u2013 again the proof of this is similar to the case a = 1. However, if (xn) is convergent\nwe may compute the limit lim\nn\u2192\u221exn by computing the limit of any of subsequence of (xn). Thus,\nlim\nn\u2192\u221exn = ea.\n4.10\nCauchy convergence\nDe\ufb01nition 4.79. A sequence (xn) is a Cauchy sequence if for every \u03b5 \u2208R\u2217\n+ there exists n\u03b5 \u2208N\nsuch that for every integer n, m \u2265n\u03b5, |xn \u2212xm| \u2264\u03b5.\nLet us start with a few examples of Cauchy sequences.\nExample 4.80. Let (xn)n\u22651 be the sequence de\ufb01ned by xn := 1 \u22121\nn, then for all n, m \u2265\n\u00022\n\u03b5\n\u0003\n,\n|xn \u2212xm| =\n\f\f\f\f1 \u22121\nn \u22121 + 1\nm\n\f\f\f\f =\n\f\f\f\f\n1\nm \u22121\nn\n\f\f\f\f \u22641\nm + 1\nn < \u03b5\n2 + \u03b5\n2\n|\n{z\n}\nn,m\u2265[ 2\n\u03b5]\u21d21\nn, 1\nm< \u03b5\n2\n= \u03b5.\nHence, (xn)n\u22651 is a Cauchy sequence. It is easy to compute that the sequence converge and it\nhas limit 1.\nCauchy sequences naturally appear when we try to approximate the decimal representation\nof a real number, by means of rational numbers.\nExample 4.81. Let x \u2208R be a real number.\nLet us think of x by means of a decimal\nrepresentation. We can de\ufb01ne a sequence (xn)n\u2208N, in the following way:\n\u25e6x0 = [x];\n\u25e6for n \u22651, xn is de\ufb01ned as the truncation of the decimal representation of x at the n-th\ndecimal digit.\nWith this de\ufb01nition, we can verify that the sequence (xn)n\u2208N is Cauchy.\nIn fact, for any\nn, m \u2208N, n < m, then\n|xm \u2212xn| < 10\u2212n.\nThus, for a given \u03f5 > 0, it su\ufb03ces to take n\u03f5 \u2208N such that 10\u2212n\u03f5 < \u03f5 \u2013 this is always possible\nsince lim", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_130", "text": "In fact, for any\nn, m \u2208N, n < m, then\n|xm \u2212xn| < 10\u2212n.\nThus, for a given \u03f5 > 0, it su\ufb03ces to take n\u03f5 \u2208N such that 10\u2212n\u03f5 < \u03f5 \u2013 this is always possible\nsince lim\nn\u2192\u221e10\u2212n = 0 \u2013 and thus\n\u2200n, m \u2265n\u03f5,\n|xn \u2212xm| < 10\u2212n\u03f5 < \u03f5.\nThe important fact about Cauchy sequences is that they are always convergent.\nTheorem 4.82. Let (xn) be a sequence. Then, the following two properties are equivalent:\n(1) (xn) is convergent;\n(2) (xn) is a Cauchy sequence.\nIn view of this theorem, we will indicate that a sequence (xn) is a Cauchy sequence (or,\nsimply, Cauchy) by saying that it is Cauchy convergent. Of course, by the above statement, all\nconverging sequences are Cauchy convergent, and viceversa.\n68", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_131", "text": "Proof. (1) =\u21d2(2). First we assume that (xn) is convergent, and then we show that it is Cauchy\nconvergent. Let x := lim\nn\u2192\u221exn and 0 < \u03b5 \u2208R arbitrary. Then there is an n \u03b5\n2 \u2208N such that for\nall integers n \u2265n \u03b5\n2 , we have |xn \u2212x| \u2264\u03b5\n2. Then, for any integers n, m \u2265n \u03b5\n2 we have\n|xn \u2212xm| = |(xn \u2212x) + (x \u2212xm)| \u2264\u03b5\n2 + \u03b5\n2 = \u03b5\n(2) =\u21d2(1). Let us assume that (xn) is Cauchy convergent. We divide this part of the proof\ninto three steps:\n(1) We \ufb01rst claim that then (xn) is bounded. Indeed, there is an n1 \u2208N such that for all\nintegers n \u2265n1, |xn \u2212xm| \u22641. Then, an upper bound for |xn| is\nmax{|x0|, . . . , |xn1\u22121|, |xn1| + 1}.\n(2) As (xn) is bounded, then by Bolzano-Weierstrass, it contains a convergent subsequence\nxnk converging to x \u2208R.\n(3) We show that limn\u2192\u221exn = x.\nFix then a 0 < \u03b5 \u2208R. As (xn) is Cauchy, there is an n \u03b5\n2 \u2208N such that for all integers\nn, m \u2265n \u03b5\n2 ,\n|xn \u2212xm| < \u03b5\n2.\nNow, there is a k such that nk \u2265n \u03b5\n2 and |xnk \u2212x| \u2264\u03b5\n2. For this value of k and any\ninteger n \u2265n \u03b5\n2 we have:\n|xn \u2212x| \u2264|(xn \u2212xnk) + (xnk \u2212x)| \u2264|xn \u2212xnk| + |xnk \u2212x| \u2264\u03b5\n2 + \u03b5\n2 = \u03b5.\n69", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_132", "text": "5\nSERIES\nLet us start this section with the following motivating example.\nExample 5.1 (Zeno\u2019s paradox). Achilles races a tortoise. Achilles runs at 10 m/s, while the\ntortoise moves at 0.1 m/s. Achilles gives the tortoise a head start of 100m.\nStep 1 Achilles runs to the tortoise\u2019s starting point, in 10s, while, at the same time, the tortoise\nhas walked 1m forward.\nStep 2 Achilles advances to where the tortoise was at the end of Step 1, in 0.1s, while the tortoise\ngoes yet 0.001m further.\nStep 3 Achilles advances to where the tortoise was at the end of Step 2, in 0.001s, while the\ntortoise goes yet 0.00001m further.\nStep n Achilles advances to where the tortoise was at the end of Step n \u22121, in\n10\n100n\u22121 s, while\nthe tortoise goes yet\n1\n100n\u22121 m further.\nThe philosopher Zeno doubted that Achilles could ever overtake the tortoise, since however\nmany steps Achilles would ever complete, the tortoise would remain ahead of him.\nIt should be intuitively clear, though, that the more steps Achilles and the tortoise take, the\ncloser they get. So, if they could run for in\ufb01nitely many steps of the above observations of the\nrun, Achilles would reach the tortoise.\nSo, the question is whether by taking in\ufb01nitely many steps of the above observations the time\nthat has passed since the start of the run is going to in\ufb01nity or it is bounded.\nAfter the n-th step, the amount of time sn that has passed since the start of the race is\n(10 + 0.1 + 0.001 + 0.00001 + \u00b7 \u00b7 \u00b7 +\n10\n100n\u22121 )s. We can rewrite this as\nsn =\nn\u22121\nX\ni=0\n10\n100i .\nHence, to understand whether Achilles ever reaches the tortoise, we need to understand the\nconvergence of the sequence (sn).\nTo understand how to solve the problem above, we now introduce the concept of series.\nDe\ufb01nition 5.2. Let (xn)n\u2265l be a sequence. The series associated to (xn)n\u2265l is the sequence\n(sn)n\u2265l de\ufb01ned by the formula", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_133", "text": "To understand how to solve the problem above, we now introduce the concept of series.\nDe\ufb01nition 5.2. Let (xn)n\u2265l be a sequence. The series associated to (xn)n\u2265l is the sequence\n(sn)n\u2265l de\ufb01ned by the formula\nsn :=\nn\nX\ni=l\nxi.\nGiven a sequence (xn) and the associated series (sn) de\ufb01ned above, we will refer to the\nsequence (sn) as the sequence of the truncated sums of (xn). We will also use the symbol\n\u221e\nX\ni=0\nxi\nto refer to the sequence (sn). Depending on the context, we will also use the symbol\n\u221e\nX\ni=0\nxi to\ndenote the limit of the series, that is,\n\u221e\nX\ni=0\nxi := lim\nn\u2192\u221esn, provided that such limit exists.\nExample 5.3. The following are a few examples of sequences (xn) and of their sequences of\ntruncated sums (sn).\n70", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_134", "text": "(1) (Geometric series) Taking xk :=\n1\n2k , then sn =\nn\nX\nk=0\n1\n2k = 1 \u2212\n1\n2n+1\n1 \u22121\n2\n= 2\n\u0012\n1 \u2212\n1\n2n+1\n\u0013\n; in\ngeneral, for q \u2208R, we can de\ufb01ne xn = qn then sn =\nn\nX\nk=0\nqk.\n(2) (Harmonic series) Taking xk := 1\nk, then sn =\nn\nX\nk=1\n1\nk;\n(3) Taking xk := (\u22121)k 1\nk, then sn =\nn\nX\nk=1\n(\u22121)k 1\nk;\n(4) Taking xk :=\n1\nk2 , then sn =\nn\nX\nk=1\n1\nk2 ;\n(5) Taking xk :=\n1\nks , for a \ufb01xed s \u2208Q\u2217\n+, then sn =\nn\nX\nk=1\n1\nks , see. ??;\n(6) (Another de\ufb01nition of e) Taking xk := 1\nk!, then sn =\nn\nX\nk=0\n1\nk!. We shall show in ??, that\nP\u221e\nk=0\n1\nk! = e.\nIn the case of the \ufb01rst example one has an explicit expression for sn without involving sums.\nHowever, in the other cases, we are not able to provide such formulas. So, one just has to take\nit as it is, so as a sequence obtained by adding the \ufb01rst n elements of the given other sequence.\nWe can de\ufb01ne a notion of convergence for series, using the notion of convergence already\nintroduced for sequences.\nDe\ufb01nition 5.4. Let (xn)n\u2265l be a sequence.\n(1) The series (sn)n\u2265l, sn := Pn\nk=l xk associated to (xn) is convergent if (sn)n\u2265l converges to\na \ufb01nite limit.\n(2) The series (sn)n\u2265l, sn := Pn\nk=l xk associated to (xn) approaches +\u221e(resp.\n\u2212\u221e) if\nlim\nn\u2192\u221esn = +\u221e(resp. lim\nn\u2192\u221esn = \u2212\u221e).", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_135", "text": "k=l xk associated to (xn) approaches +\u221e(resp.\n\u2212\u221e) if\nlim\nn\u2192\u221esn = +\u221e(resp. lim\nn\u2192\u221esn = \u2212\u221e).\nNotation 5.5. Given a sequence (xn)n\u2265l, such that the series (sn)n\u2265l associated to (xn)n\u2265l is\nconvergent with lim\nn\u2192\u221esn = y, we will write\n\u221e\nX\nk=0\nxk = y.\nto denote .\nIn the course of this section we will discover several techniques to determine when a series\nconverges (or not).\nA \ufb01rst natural condition from convergence stems from the following simple observation:\nwhen a sequence (xn) has values in the positive real numbers R+, then the series (sn) is\nincreasing, hence it converges if and only if it is bounded.\n71", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_136", "text": "Example 5.6. Let (xn)n\u2208N be the sequence de\ufb01ned by xn :=\n1\n2n . Then\nsn :=\nn\nX\nk=0\n1\n2k = 2\n\u0012\n1 \u2212\n1\n2n+1\n\u0013\n.\nThis identity implies that\nlim\nn\u2192\u221esn = lim\nn\u2192\u221e2\n\u0012\n1 \u2212\n1\n2n+1\n\u0013\n= 2 =:\n\u221e\nX\nk=0\n1\n2k .\nSimilarly, taking xn := qn, for q \u2208R, then\nsn :=\nn\nX\nk=0\nqk = 1 \u2212qn+1\n1 \u2212q\n,\ncf. (1.6.f). If |q| < 1, then we showed already that lim\nn\u2192\u221eqn = 0. Thus,\nlim\nn\u2192\u221esn =\n1\n1 \u2212q =:\n\u221e\nX\nk=0\nqk\nThe above observation can be naturally extended to yield the following proposition.\nProposition 5.7. Let (xn)n\u2265l be a sequence.\nAssume that there exists n0 \u2208N such that\n\u2200n \u2265n0, xn \u22650. Then,\n\u221e\nX\nk=l\nxk =\n(\ny \u2208R\nif and only if (sn)n\u2265l, sn := Pn\nk=l xk is a bounded sequence\n+\u221e\nif and only if (sn)n\u2265l is not bounded.\nProof. As \u2200n \u2265n0, xn \u22650, then (sn)n\u2265n0 is increasing starting from n0. Thus, we can conclude\nby Theorem 4.63.\nUsing Cauchy\u2019s convergence criterion for sequences, see Theorem 4.82, we have the following\nbasic convergence criterion for series.\nProposition 5.8. Let (xn)n\u2265l be a sequence. Then, the following conditions are equivalent:\n(1)\n\u221e\nX\nk=l\nxk is convergent;\n(2) (sn)n\u2265l is a Cauchy sequence;\n(3) for every \u03b5 \u2208R, \u03b5 > 0, there is an n\u03b5 \u2208N such that for all integers m, n \u2265n\u03b5, with\nm > n,\n\f\f\f\f\f\nm\nX\nk=n+1\nxk\n\f\f\f\f\f < \u03b5.\nExample 5.9. Let (xn)n\u22651 be sequence de\ufb01ned by xn := 1\nn. We show that\n\u221e\nX\nk=1\n1\nk = +\u221e.\nSince \u2200k \u22651, 1", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_137", "text": "m > n,\n\f\f\f\f\f\nm\nX\nk=n+1\nxk\n\f\f\f\f\f < \u03b5.\nExample 5.9. Let (xn)n\u22651 be sequence de\ufb01ned by xn := 1\nn. We show that\n\u221e\nX\nk=1\n1\nk = +\u221e.\nSince \u2200k \u22651, 1\nk > 0, then we know that either\n\u221e\nX\nk=1\n1\nk either converges to a \ufb01nite limit y \u2208R\nor\n\u221e\nX\nk=1\n1\nk = +\u221e. Thus, let us assume, by contradiction, that\n\u221e\nX\nk=1\n1\nk = y \u2208R. Hence, by ??, for\n72", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_138", "text": "\u03b5 = 1\n4, Cauchy\u2019s condition for the convergence of series is satis\ufb01ed. That is, there exists some\nindex n 1\n4 \u2208N such that for all n, m \u2265n 1\n4 , with m > n, then\n\f\f\f\f\f\nm\nX\ni=n+1\n1\ni\n\f\f\f\f\f < 1\n4.\nIn particular, the above inequality must hold for n := n 1\n4 and m = 2n, in which case,\n1\n4 >\n\f\f\f\f\f\n2n\nX\nk=n+1\n1\nk\n\f\f\f\f\f =\n2n\nX\nk=n+1\n1\nk \u2265\n2n\nX\nk=n+1\n1\n2n\n|\n{z\n}\nk\u22642n\u21d21\nk \u22651\n2n\n= 1\n2\nwhich provides the sought contradiction.\nAn immediate consequence of ?? is the following necessary condition for convergence of a\nseries.\nProposition 5.10. Let (xn)n\u2265l be a sequence. If\n\u221e\nX\nn=l\nxn is convergent, then lim\nn\u2192\u221exn = 0.\nProof. Indeed, by ??, for every 0 < \u03b5 \u2208R, there is an n\u03b5 \u2208N such that for all integers m, n \u2265n\u03b5\nwith m > n,\n\f\f\f\f\f\nm\nX\nk=n+1\nxk\n\f\f\f\f\f \u2264\u03b5.\nIn particular, if we choose n := m \u22121, then we obtain that \u2200m \u2265n\u03f5 + 1,\n\u03b5 \u2265\n\f\f\f\f\f\nm\nX\nk=m\u22121\nxk\n\f\f\f\f\f = |xm|.\nThis implies that lim\nn\u2192\u221exn = 0.\nExample 5.11. The series\n\u221e\nX\nn=0\ncos(n) is not convergent.\nBy ??, it su\ufb03ces to show that\nxn := cos(n) does not converge to 0. Let us assume by contradiction that instead it does.\nThen, so do all its subsequences. However, consider the subsequence given by nk := \u230a2k\u03c0\u230b.\nThus,\nxnk = cos(\u230a2k\u03c0\u230b) \u2265cos(2k\u03c0 \u22121) = cos(\u22121) > 0,\nwhere the inequality follows from the fact that cos(x) is an increasing function in the interval\n2k\u03c0 \u2212\u03c0\n2 \u2264x \u22642k\u03c0, and moreover, as \u03c0\n2 > 1, 2k\u03c0 \u22121 is in this interval.\nAs xnk \u2265cos(\u22121) > 0, \u2200nk, then (xnk) cannot converge to 0; but this is in contradiction with", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_139", "text": "2k\u03c0 \u2212\u03c0\n2 \u2264x \u22642k\u03c0, and moreover, as \u03c0\n2 > 1, 2k\u03c0 \u22121 is in this interval.\nAs xnk \u2265cos(\u22121) > 0, \u2200nk, then (xnk) cannot converge to 0; but this is in contradiction with\nthe the assumption that xn converge to 0.\nOne can use ?? to give a version of the Squeeze Theorem for series.\nTheorem 5.12 (Squeeze theorem for series). Let (xn)n\u2265l, (yn)n\u2265l be sequences. Assume there\nexists n0 \u2208N such that for every integer n \u2265n0, 0 \u2264xn \u2264yn.\n(1) If\n\u221e\nX\nk=l\nyk is convergent, then\n\u221e\nX\nk=l\nxk is also convergent.\n73", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_140", "text": "Images/graphcos.png\nFigure 9: To check that cos(x) is increasing, by using periodicity, it su\ufb03ces to check that the\nsame holds over the inverval [\u2212\u03c0\n2 , 0].\n(2) If\n\u221e\nX\nk=l\nxk = +\u221e, then also\n\u221e\nX\nk=l\nyk = +\u221e.\nProof. For every n, m \u2265n0 with m > n\n0 \u2264\n\f\f\f\f\f\nm\nX\nk=n+1\nxk\n\f\f\f\f\f =\nm\nX\nk=n+1\nxk \u2264\nm\nX\nk=n+1\nyk =\n\f\f\f\f\f\nm\nX\nk=n+1\nyk\n\f\f\f\f\f .\nSo, if the property in ??.3 is veri\ufb01ed for yk then it must also holds for xk. On the other hand,\nif the property in ??.3 is not satis\ufb01ed for the sequence of truncated sums of (xn)n\u2265l, then it\nmust also fail for the sequence of truncated sums of (yn)n\u2265l.\nDe\ufb01nition 5.13. If 0 < s is a rational number, say s = a\nb then we de\ufb01ne ns :=\nb\u221a\nna for all\nn \u2208N.\nExample 5.14. 2\n2\n3 =\n3\u221a\n4 and this is the only positive real solution to the equation X3\u22124 = 0.\nRemark 5.15. The above de\ufb01nition does not depend on the representation of s as a\nb. That is,\nif we replace a\nb by ca\ncb (where c \u2208N), then:\ncb\u221a\nnca =\nbq\nc\u221a\nnca =\nb\u221a\nna.\nMoreover, any x, y \u2208R+, and s, t \u2208Q, then:\n(1) x0 = 1;\n(2) if x > y, s > 0, then xs > ys;\n(3) if x > y, s < 0, then xs < ys;\n(4) if x > 1 and s > t, then xs > xt;\n(5) if x < 1 and s > t, then xs < xt.\n74", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_141", "text": "Example 5.16. If 0 < s = a\nb < 1 is a rational number, then P\u221e\nk=1\n1\nks is divergent.\nIn fact, with the assumption 0 < s \u22641, we can use the Squeeze ??: indeed, for each n \u22651,\n1\nns =\n1\n( b\u221an)a \u2265\n1\n( b\u221an)b\n|\n{z\n}\nb>a and\nb\u221an\u22651\u21d2( b\u221an)a<( b\u221an)b\n= 1\nn,\nand since\n\u221e\nX\nk=1\n1\nk = +\u221e, then the Squeeze Theorem for series implies that for all 0 < s < 1, s \u2208Q\nalso\n\u221e\nX\nk=1\n1\nks = +\u221e.\nExample 5.17. If s > 1 be a rational number, then P\u221e\nk=1\n1\nks is convergent.\nIndeed, when s > 1,\nsn :=\nn\nX\nk=1\n1\nks \u2264\n2n+1\nX\nk=1\n1\nks = 1 +\nn\nX\nk=1\n1\n(2k)s +\nn\nX\nk=1\n1\n(2k + 1)s\n|\n{z\n}\n2k+1>2k\n\u22641 +\nn\nX\nk=1\n1\n(2k)s\n|\n{z\n}\n= 1\n2s sn\n+\nn\nX\nk=1\n1\n(2k)s = 1 + 2\n2s sn = 1 +\n1\n2s\u22121 sn\nBy taking the two ends of this chain of inequalities,\nsn \u22641 + 21\u2212ssn\nor, equivalently, sn \u2264\n1\n1 \u221221\u2212s .\nHence, sn is bounded from above. As it is also increasing, since we are summing positive terms,\nthen (sn) is convergent by Theorem 4.63.\nRemark 5.18. We will show later on in the course that\n(1) \u2200s \u2208(0, 1],\n\u221e\nX\nk=1\n1\nks = +\u221e; and,\n(2) \u2200s \u2208(1, +\u221e),\n\u221e\nX\nk=1\n1\nks converges.\nExample 5.19. The series\n\u221e\nX\ni=1\nsin\n\u00121\ni\n\u0013\ndoes not converge.\nWe have seen in the exercises that\nlim\nn\u2192\u221e\nsin\n\u0000 1\nn\n\u0001\n1\nn\n= 1.\nUsing De\ufb01nition 4.21, taking, for example, \u03f5 := 1\n2, then there exists n 1\n2 \u2208N such that \u2200n \u2265n 1\n2 ,\n\f\f\f\f\nsin( 1", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_142", "text": "\u00121\ni\n\u0013\ndoes not converge.\nWe have seen in the exercises that\nlim\nn\u2192\u221e\nsin\n\u0000 1\nn\n\u0001\n1\nn\n= 1.\nUsing De\ufb01nition 4.21, taking, for example, \u03f5 := 1\n2, then there exists n 1\n2 \u2208N such that \u2200n \u2265n 1\n2 ,\n\f\f\f\f\nsin( 1\nn)\n1\nn\n\u22121\n\f\f\f\f < 1\n2, or equivalently,\n\u22121\n2 < sin\n\u0000 1\nn\n\u0001\n1\nn\n\u22121 < 1\n2\nIn particular,\nsin( 1\nn)\n1\nn\n> 1\n2 from which it follows that 1\n2\n1\nn < sin\n\u0000 1\nn\n\u0001\n. Hence, the Squeeze Theorem\nand ?? imply that\n\u221e\nX\ni=1\nsin\n\u00121\ni\n\u0013\ndoes not converge.\n75", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_143", "text": "The strategy adopted in ?? can be generalized to prove the following convergence criterion.\nProposition 5.20. Let (xn), (yn) be sequences. Assume that xn, yn > 0, \u2200n \u2265n0. Let (zn) be\nthe sequence de\ufb01ned as zn := xn\nyn .\n(1) If (zn) is bounded and\n\u221e\nX\nk=0\nyk converges then\n\u221e\nX\nk=0\nxk converges.\n(2) If lim\nn\u2192\u221e\nxn\nyn = \u03c1 \u0338= 0 and\n\u221e\nX\nk=0\nyk diverges then\n\u221e\nX\nk=0\nxk diverges.\nRemark 5.21. One cannot drop the assumption that \u03c1 \u0338= 0 in part (2) of ??. Indeed, taking\nxn := 2\u2212n, yn := 1\nn, then\nlim\nn\u2192\u221e\nxn\nyn\n= lim\nn\u2192\u221e\nn\n2n = 0,\nbut we know that\n\u221e\nX\nk=0\nxk converges, while\n\u221e\nX\nk=0\nyk does not.\nExample 5.22. The series\n\u221e\nX\nk=1\n(\u22121)k 1\nk is convergent.\nFirst let us look at the odd terms of sn, that is, the subsequence s\u2032\nm := s2m+1 of sn. Then,\ns\u2032\nm = s2m+1 := \u22121 +\n2m+1\nX\nk=2\n(\u22121)k 1\nk = \u22121 +\nm\nX\nk=1\n\u0012\n(\u22121)2k 1\n2k + (\u22121)2k+1\n1\n2k + 1\n\u0013\n= \u22121 +\nm\nX\nk=1\n\u0012 1\n2k \u2212\n1\n2k + 1\n\u0013\n= \u22121 +\nm\nX\nk=1\n1\n2k(2k + 1)\n\u2264\u22121 +\nm\nX\nk=1\n1\n2k \u00b7 2k = \u22121 + 1\n4\nm\nX\nk=1\n1\nk2 \u2264\u22121 +\n1\n1 \u221221\u22122\n|\n{z\n}\nby ??\n= 1\nHence, s2m+1 is an increasing bounded sequence. Theorem 4.63 implies that (s\u2032\nm) is convergent\naccording. Let us de\ufb01ne S := lim\nm\u2192\u221es2m+1.\nWhat happens when we consider also the even terms of (sn)? By the de\ufb01nition, we have\ns2m = s2m\u22121 + 1\n2m.", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_144", "text": "m) is convergent\naccording. Let us de\ufb01ne S := lim\nm\u2192\u221es2m+1.\nWhat happens when we consider also the even terms of (sn)? By the de\ufb01nition, we have\ns2m = s2m\u22121 + 1\n2m.\nIn particular, we may write the sequence sn as the sum sn = xn + yn of two other sequence\nde\ufb01ned by\nxn :=\n(\nsn\nif n is odd ,\nsn\u22121\nif n is even ,\nand\nyn :=\n(\n0\nif n is odd ,\n1\nn\nif n is even .\nClaim. lim\nn\u2192\u221exn = S.\nProof of the claim. xn is increasing and its values are the same as those of s2m+1. Hence, any\nupper bound for s2m+1 is automatically an upper bound for xn too. In particular, xn is not\nonly increasing but also bounded. Therefore, xn is convergent by Theorem 4.63 and then it\nmust converge to the same limit as its subsequence s2m+1, see Proposition 4.72.\n76", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_145", "text": "By the de\ufb01nition of yn we also have lim\nn\u2192\u221eyn = 0 [prove this!]. By the algebraic rules for \ufb01nite\nlimits, cf. Proposition 4.30, sn is convergent and\n|\n{z\n}\nsn=xn+yn\nlim\nn\u2192\u221esn =\nProposition 4.30\nz\n}|\n{\nlim\nn\u2192\u221e(xn + yn) = lim\nn\u2192\u221exn + lim\nn\u2192\u221eyn = S + 0 = S.\nThe previous example can be generalized to the following criterion for the convergence of\nseries.\nProposition 5.23 (Leibniz criterion). Let (xn) be a decreasing sequence. Assume that lim\nn\u2192\u221exk =\n0. Then,\n\u221e\nX\nk=0\n(\u22121)kxk is convergent.\nProof. We refer to the book for the proof, but many of the main ideas are already presented\nin ??.\nExample 5.24. Let s \u2208Q\u2217\n+. Then ?? implies that P\u221e\nk=1(\u22121)k 1\nks is convergent, since \u2200k \u2208N\u2217,\n1\nks >\n1\n(k+1)s and lim\nk\u2192\u221e\n1\nks = 0.\nExample 5.25. The series\n\u221e\nX\ni=1\n(\u22121)i sin\n\u00121\ni\n\u0013\nconverges. By Leibnitz criterion, it su\ufb03ces to\ncheck that the sequence (xn)n\u22651, xn := sin\n\u0000 1\nn\n\u0001\nis decreasing, but this follows from the de\ufb01nition\nof the function sin(x), cf.\nas we already know that lim\nn\u2192\u221esin\n\u0000 1\nn\n\u0001\n= 0, for example, since\nsin\n\u0000 1\nn\n\u0001\n\u22641\nn.\nDe\ufb01nition 5.26. Let (xn) be a sequence. We say that P\u221e\nk=0 xn is absolutely convergent if the\nseries P\u221e\nk=0 |xn| converges.\nExample 5.27. In ?? we showed that\n\u221e\nX\nk=1\n(\u22121)k 1\nk is convergent. On the other hand, ?? implies\nthat\n\u221e\nX\nk=1\n(\u22121)k 1\nk is not absolute convergent. Hence, ?? is a non-trivial criterion to establish the\nconvergence of series whose terms are not all positive or all negative.\nWe now show that it is not possible for the viceversa of this to happen: that is, an abolutely", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_146", "text": "k is not absolute convergent. Hence, ?? is a non-trivial criterion to establish the\nconvergence of series whose terms are not all positive or all negative.\nWe now show that it is not possible for the viceversa of this to happen: that is, an abolutely\nconvergent series is convergent.\nProposition 5.28. If\n\u221e\nX\nn=0\nxn is absolute convergent, then it is convergent.\nProof. As we are assuming that P\u221e\nk=0 |xk| converges, then ?? implies that for any \u03f5 > 0 there\nexists n\u03f5 \u2208N such that \u2200m > n \u2265n\u2032\n\u03f5\nm\nX\nk=n+1\n|xk| < \u03b5.\n(5.28.a)\nLet us \ufb01x \u03f5 > 0. To prove that P\u221e\nk=0 xk converges as well, it su\ufb03ce to \ufb01nd an index n\u03f5 \u2208N\nsuch that \u2200m > n \u2265n\u03f5,\n\f\f\f\f\f\nm\nX\nk=n+1\nxk\n\f\f\f\f\f < \u03b5.\n(5.28.b)\n77", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_147", "text": "By the triangle equality, Proposition 2.57,\n\f\f\f\f\f\nm\nX\nk=n+1\nxk\n\f\f\f\f\f \u2264\nm\nX\nk=n+1\n|xk|.\nHence, taking n\u03f5 := n\u2032\n\u03f5, then \u2200m > n \u2265n\u03f5,\n\f\f\f\f\f\nm\nX\nk=n+1\nxk\n\f\f\f\f\f \u2264\nm\nX\nk=n+1\n|xk| < \u03f5,\nthat is, ?? implies ??.\nExample 5.29. We show that P\u221e\nk=0\n1\nk! = e. This is one of the rare occasions when we will be\nactually able to compute the value of the limit of an in\ufb01nite sum.\nFirst, the sequence (sn), sn :=\nn\nX\nk=0\n1\nk! is a strictly increasing sequence, since\n1\nk! > 0, \u2200k \u2208N.\nFurthermore, (sn) is bounded, because\nn\nX\nk=0\n1\nk! =1 +\nn\nX\nk=1\n1\nk! = 1 +\nn\nX\nk=1\n1\n1 \u00b7 2 \u00b7 3 \u00b7 \u00b7 \u00b7 k = 1 +\nn\nX\nk=1\n1\n1\n|{z}\n=1\n\u00b7\ni\u22121 terms\nz\n}|\n{\n1\n2 \u00b7\n1\n3\n|{z}\n< 1\n2\n\u00b7 \u00b7 \u00b7\n1\nk\n|{z}\n< 1\n2\n\u22641 +\nn\nX\nk=1\n1\n2k\u22121 = 1 +\nn\nX\nk=0\n1\n2k \u22641 +\n\u221e\nX\nk=0\n1\n2k = 3.\nThus,\n\u221e\nX\nk=0\n1\nk! is convergent.\nWe showed earlier, cf. Example 4.64, that\n\u0012\n1 + 1\nn\n\u0013n\n= 2 +\nn\nX\ni=2\n1\ni!\n\u0012\n1 \u22121\nn\n\u0013\n\u00b7\n\u0012\n1 \u22122\nn\n\u0013\n\u00b7 \u00b7 \u00b7 \u00b7 \u00b7\n\u0012\n1 \u2212i \u22121\nn\n\u0013\nBut, for any n \u2208N\u2217, and for any j \u2208{1, 2, 3, . . . , i \u22121} then 1 \u2212i\u22121\nn \u22641 \u2212j\nn < 1, so that\n2 +\nn\nX\ni=2\n1\ni!\n\u0012\n1 \u2212i \u22121\nn\n\u0013i\u22121\n\u2264\n\u0012\n1 + 1\nn\n\u0013n\n\u22642 +\nn\nX\ni=2\n1\ni! =\nn\nX\ni=0\n1\ni!\nUsing ??, then\n\u00001 \u2212i\u22121\nn\n\u0001i\u22121 \u2265\n\u00001 \u2212(i \u22121)i\u22121\nn\n\u0001\nso that we can rewrite the above chain of\ninequality as\n2 +\nn\nX\ni=2\n1\ni!\n\u0012\n1 \u2212(i \u22121)2\nn\n\u0013\n\u2264\n\u0012\n1 + 1\nn\n\u0013n\n\u22642 +", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_148", "text": "Using ??, then\n\u00001 \u2212i\u22121\nn\n\u0001i\u22121 \u2265\n\u00001 \u2212(i \u22121)i\u22121\nn\n\u0001\nso that we can rewrite the above chain of\ninequality as\n2 +\nn\nX\ni=2\n1\ni!\n\u0012\n1 \u2212(i \u22121)2\nn\n\u0013\n\u2264\n\u0012\n1 + 1\nn\n\u0013n\n\u22642 +\nn\nX\ni=2\n1\ni! =\nn\nX\ni=0\n1\ni!\nor, equivalently,\nn\nX\ni=0\n1\ni! \u22121\nn\nn\nX\ni=2\n1\ni!\n\u0012\n1 \u2212(i \u22121)2\nn\n\u0013\n\u2264\n\u0012\n1 + 1\nn\n\u0013n\n\u2264\nn\nX\ni=0\n1\ni!\nSubstituting sn := Pn\ni=0\n1\ni!, zn := Pn\ni=2\n(i\u22121)2\ni!\n, then\nsn + zn\nn \u2264\n\u0012\n1 + 1\nn\n\u0013n\n\u2264sn.\n78", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_149", "text": "If we can prove that (zn) is bounded, then lim\nn\u2192\u221e\nzn\nn = 0 and the Squeeze Theorem 4.36 implies\nthat lim\nn\u2192\u221esn = lim\nn\u2192\u221e\n\u00001 + 1\nn\n\u0001n = e. Why should we think that (zn) is bounded? Well zn =\nPn\ni=2\n(i\u22121)2\ni!\nwhich means that (zn) is the series associated to the sequence de\ufb01ned by xi =\n(i\u22121)2\ni!\n. As xi \u22650, for any i \u2208N, then boundedness of (zn) is equivalent to the convergence of\nn\nX\ni=2\n(i \u22121)2\ni!\n. This will be proven in ??.\nIn the course of the above example. we have used this other version of Proposition 4.16.\nProposition 5.30 (Bernoulli inequality (negative case)). Fix n \u2208N. Given \u22121 < x \u22640, then\n(1 + x)n \u22651 + nx.\nProof. We prove the statement by induction on n \u2208N.\n\u25e6STARTING STEP: for n = 0, then (1 + x)0 = 1 = 1 + 0 \u00b7 x.\n\u25e6INDUCTIVE STEP: let us assume that we know the statement of the proposition holds\nfor any j \u2264n; we need to prove that the statement holds also for n + 1. But then,\n(1 + x)n+1 = (1 + x)(1 + x)n \u2265(1 + x)(1 + nx) = (1 + (n + 1)x + nx2\n|{z}\n\u22650\n) \u22651 + (n + 1)x.\nThere are two further criteria for the convergence of series that will be very useful in the\ncourse.\nProposition 5.31 (Cauchy\u2019s criterion). Let (xn) be a sequence.\n(1) If\n\u0010\nnp\n|xn|\n\u0011\nis bounded and lim\nn\u2192\u221e\nnp\n|xn| = \u03c1 < 1, then\n\u221e\nX\nn=0\nxn is absolutely convergent.\n(2) If\n\u0010\nnp\n|xn|\n\u0011\nis bounded and lim\nn\u2192\u221e\nnp\n|xn| = \u03c1 > 1 or if\n\u0010\nnp\n|xn|\n\u0011\nis not bounded, then\n\u221e\nX\nn=0\nxn is divergent.\nRemark 5.32. Let (xn) be a sequence such that lim\nn\u2192\u221e", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_150", "text": "\u0010\nnp\n|xn|\n\u0011\nis bounded and lim\nn\u2192\u221e\nnp\n|xn| = \u03c1 > 1 or if\n\u0010\nnp\n|xn|\n\u0011\nis not bounded, then\n\u221e\nX\nn=0\nxn is divergent.\nRemark 5.32. Let (xn) be a sequence such that lim\nn\u2192\u221e\nn\u221axn = 1. Then, it is not possible to\npredict the behavior of\n\u221e\nX\nn=0\nxn, as the following examples illustrate.\n(1) xn := (\u22121)n =\u21d2\n\u221e\nX\nn=0\nxn diverges but (sn) is bounded;\n(2) xn := n =\u21d2\n\u221e\nX\nn=0\nxn diverges and (sn) is unbounded;\n(3) xn := 1\nn =\u21d2\n\u221e\nX\nn=0\nxn diverges, (sn) is unbounded and lim\nn\u2192\u221exn = 0;\n(4) xn :=\n1\nn2 =\u21d2\n\u221e\nX\nn=0\nxn converges absolutely, cf. ??;\n79", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_151", "text": "(5) xn := (\u22121)n\nn\n=\u21d2\n\u221e\nX\nn=0\nxn converges, but it does not converges absolutely, cf. ??-??.\nProof of Cauchy\u2019s criterion.\n(1) For the convergence statements, we try to apply the Squeeze\ntheorem for series. It su\ufb03ces to show that there exists n0 \u2208N and 0 < q < 1 such that\nfor any n \u2265n0, 0 \u2264|xn| \u2264qn. The only question is how to choose q.\nAs 0 \u2264\u03c1 < 1, let \u03f5 :=\n1\u2212\u03c1\n2\n> 0.\nThen, there exists n\u03f5 \u2208N, such that \u2200n \u2265n\u03f5,\n| np\n|xn| \u2212\u03c1| < \u03f5. In particular, this implies that 0 \u2264\nnp\n|xn| < \u03c1 + \u03f5 < 1, or equivalently,\nthat 0 \u2264|xn| < (\u03c1 + \u03f5)n, as desired.\n(2) For the divergence statements one just shows that |xn| does not converge to 0. There are\nin\ufb01nitely many elements with\nnp\n|xn| \u22651; this is equivalent to |xn| \u22651.\nExample 5.33. Take xn := qn\nn! , q \u2208R. Then,\nnp\n|xn| =\nn\ns\f\f\f\f\nqn\nn!\n\f\f\f\f = |q|\nn\u221a\nn!\n.\nClaim lim\nn\u2192\u221e\nn\u221a\nn! = +\u221e.\nProof. This is an exercise in Week 7\u2019s exercise sheet.\nThen lim\nn\u2192\u221e\n1\nn\u221a\nn! = 0, thus, lim\nn\u2192\u221e\nn\u221axn = 0, and Cauchy\u2019s criterion implies that\n\u221e\nX\nk=0\nqk\nk! con-\nverges absolutely for any q \u2208R.\nProposition 5.34 (D\u2019Alembert\u2019s criterion). Let (xn) be a sequence such that\n\u0010\n|xn+1|\n|xn|\n\u0011\nis con-\nvergent and let the limit be \u03c1.\n(1) If lim\nn\u2192\u221e\n\u0010\n|xn+1|\n|xn|\n\u0011\n= \u03c1 < 1, then\n\u221e\nX\nk=0\nxn is absolutely convergent.\n(2) If lim\nn\u2192\u221e\n\u0010\n|xn+1|\n|xn|\n\u0011\n= \u03c1 > 1, then\n\u221e\nX\nk=0\nxn is divergent.\nThe ideas behind the proofs.\n(1) For the convergence statements, we try to apply the Squeeze", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_152", "text": "xn is absolutely convergent.\n(2) If lim\nn\u2192\u221e\n\u0010\n|xn+1|\n|xn|\n\u0011\n= \u03c1 > 1, then\n\u221e\nX\nk=0\nxn is divergent.\nThe ideas behind the proofs.\n(1) For the convergence statements, we try to apply the Squeeze\ntheorem for series by showing for any n \u2265of some index n0, 0 \u2264|xn| \u2264qn for some\n0 < q < 1. The only question is what q should one choose. Take xn+1\nxn\n\u2264\u03c1 + \u03b5 after\n\ufb01nitely many steps, say after n \u2265n\u03b5. So, we have |xn| \u2264qn\u2212n\u03b5 |xn\u03b5| if we set q = \u03c1 + \u03b5\nhere too.\n(2) There exists n1 \u2208N, such that \u2200n \u2265n1, |xn+1|\n|xn|\n> 1. Hence, \u2200n \u2265n1, |xn+1| > |xn|, i.e.,\nstarting from n1 the sequence (|xn|) is strictly increasing. As xn \u0338= 0 for all n \u226b1, then\nlim\nn\u2192\u221exn if it exists cannot be 0.\nRemark 5.35. A remark completely analogous to ?? holds also for D\u2019Alembert\u2019s criterion. By\nthis, we mean that if lim\nn\u2192\u221e\n|xn+1|\n|xn|\n= 1, then anything can happen for\n\u221e\nP\nn=0\nxn. That is, it is\npossible to \ufb01nd sequences (xn) such that:\n(1) (xn) is unbounded;\n80", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_153", "text": "(2) (xn) is bounded and\n\u221e\nP\nn=0\nxn diverges;\n(3) (xn) is bounded and\n\u221e\nP\nn=0\nxn converges absolutely;\n(4) (xn) is bounded and\n\u221e\nP\nn=0\nxn converges but not absolutely.\n[This is an exercise in Week 7\u2019s exercise sheet].\nExample 5.36. The series\n\u221e\nX\nk=0\nk\n2k is convergent. This can be proved using both criteria in ??\nor ??. In fact,\nlim\nk\u2192\u221e\nkp\n|xk| = lim\nk\u2192\u221e\nk\nr\nk\n2k = lim\nk\u2192\u221e\nk\u221a\nk\n2\n=\nlim\nk\u2192\u221e\nk\u221a\nk\n2\n= 1\n2,\nlim\nk\u2192\u221e\n|xk+1|\n|xk|\n= lim\nk\u2192\u221e\nk+1\n2k+1\nk\n2k\n= lim\nk\u2192\u221e\nk + 1\n2k\n= 1\n2.\nExample 5.37. Let (xn) xn = nn\nn! . Does the series P\u221e\ni=1 xi converge? For any n \u2208N\u2217, nn \u2265n!.\nHence, \u2200n \u2208N\u2217, xn \u22651. Thus, the series cannot converge, by ??.\nLet us take instead yn := n!\nnn . Does the series P\u221e\ni=1 yi converge instead? In this case,\n|yn+1|\n|yn|\n=\n(n + 1)!\n(n + 1)n+1 \u00b7 nn\n(n)! =\n(n + 1)(n!)\n(n + 1)(n + 1)n \u00b7 nn\n(n)! =\nnn\n(n + 1)n =\n\u0012\n1 + 1\nn\n\u0013\u2212n\n.\nHence, lim\nn\u2192\u221e\n|yn+1|\n|yn|\n= e\u22121 < 1, so that the series\n\u221e\nX\ni=1\nyi converges (absolutely).\nExample 5.38. The series\n\u221e\nX\ni=1\n(i \u22121)2\ni!\nconverges.\nIt su\ufb03ces to apply D\u2019Alembert\u2019s criterion to xn := (n\u22121)2\nn!\n. In fact,\n|xn+1|\n|xn|\n= (n + 1)2\n(n + 1)! \u00b7 n!\nn2 =\n\u0012\n1 + 1\nn\n\u00132\n\u00b7\nn!\n(n + 1)!.\nHence,\nlim\nn\u2192\u221e", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_154", "text": "n!\n. In fact,\n|xn+1|\n|xn|\n= (n + 1)2\n(n + 1)! \u00b7 n!\nn2 =\n\u0012\n1 + 1\nn\n\u00132\n\u00b7\nn!\n(n + 1)!.\nHence,\nlim\nn\u2192\u221e\n|xn+1|\n|xn|\n= lim\nn\u2192\u221e\n\u0012\n1 + 1\nn\n\u00132\n\u00b7\nn!\n(n + 1)! = lim\nn\u2192\u221e\n\u0012\n1 + 1\nn\n\u00132\n\u00b7 lim\nn\u2192\u221e\n1\nn + 1 = 1 \u00b7 0 = 0,\nso that the series\n\u221e\nX\ni=1\nxi converges (absolutely).\n81", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_155", "text": "6\nREAL FUNCTIONS OF ONE VARIABLE\nIn this section, we are going to consider functions f : E \u2192R where E is a subset of R and\nstudy their properties. We \ufb01rst start by recalling general basic properties of functions.\n6.1\nLimits of functions and continuity\nIn this section we will de\ufb01ne and discuss the notion of limit of a function at a given point.\nThe notion of limit aims to give a mathematically precise measure of what the local behavior\nof a function is around a given point.\nExample 6.1. The starting point of our investigation is the function f(x) := sin(x)\nx\nnear x = 0.\nAt \ufb01rst sight, f(0) would seem not to be de\ufb01ned, as x appears in the denominator of sin(x)\nx\n.\nOn the other hand, looking at the graph of the function in ??, it would appear that the closer\nImages/sinx_sux_gr.png\nFigure 10: f(x) = sin(x)\nx\n.\nx is to 0, the closer sin(x)\nx\nis to 1.\nWe can be even more precise if, for example, we consider the sequence (yn), yn := 1\nn, n \u2208N,\nthen lim\nn\u2192\u221eyn = 0 and we can actually show that also the limit lim\nn\u2192\u221ef(yn) exists. Indeed, you\nproved in the exercise sheets that\nlim\nn\u2192\u221e\nsin\n\u0000 1\nn\n\u0001\n1\nn\n= 1.\nSo, even though sin(x)\nx\nis not de\ufb01ned at x = 0, if we set\nf(x) = sin(x)\nx\n, for x \u0338= 0,\nf(0) = 1,\nthen it would appear that f(x) becomes a nice \u201ccontinuous\u201d function around at x = 0, meaning\nthat we could draw the graph of f with just one continuous stroke of the pen.\nThe goal of this section is for us to turn the ideas contained in the previous example into\nsome precise mathematical concepts and de\ufb01nitions and derive further consequences starting\nfrom those. In particular, we will de\ufb01ne precisely why, in the previous example, f(0) = 1 makes\nf(x) \u201ccontinuous\u201d.\nFirst, we need to make of how to unsure that a function f is de\ufb01ned around a point x0 \u2208R,\nso that it makes sense for us to talk about \u201cthe behavior of f around x0\u201d.\n82", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_156", "text": "De\ufb01nition 6.2. A function f : E \u2192R is de\ufb01ned on a punctured neighborhood of x0 \u2208R if for\nsome positive real number \u03b4 \u2208R\u2217\n+, E contains a set of the form (x0 \u2212\u03b4, x0 + \u03b4) \\ {x0}.\nRemark 6.3. Equivalently, we can restate the above de\ufb01nition in the following way:\nA function f : E \u2192R is de\ufb01ned on a punctured neighborhood of x0 \u2208R if for some positive\nreal number \u03b4 \u2208R\u2217\n+, an interval of the form (x0 \u2212\u03b4, x0 + \u03b4) is contained in E \u222a{x0}.\nExample 6.4. The function f(x) := sin(x)\nx\nis de\ufb01ned on any pointed neighborhood of 0. Indeed,\nf is de\ufb01ned on E := R \\ {0} so that\n(\u2212\u03b4, +\u03b4) \\ {0} \u2286E,\n\u2200\u03b4 \u2208R\u2217\n+.\nWe are then ready to give the formal de\ufb01nition of limit.\nDe\ufb01nition 6.5. Let f : E \u2192R and l \u2208R. Assume that E contains a punctured neighborhood\nof x0 \u2208R. Then, lim\nx\u2192x0 f(x) = l if one of the following two equivalent conditions holds:\n(1) For every 0 < \u03b5 \u2208R there exists \u03b4\u03b5 \u2208R\u2217\n+ such that\n\u2200x \u2208E, 0 < |x \u2212x0| < \u03b4\u03b5 \u21d2|f(x) \u2212l| < \u03b5.\n(2) For every sequence (xn) \u2286E \\ {x0} for which lim\nn\u2192\u221exn = x0, we have lim\nn\u2192\u221ef(xn) = l.\nRemark 6.6. Roughly speaking the two de\ufb01nitions mean the following:\n(1) whenever f is de\ufb01ned at x and x is close to x0, then f(x) is close to l. More precisely:\nfor every \u03b5 > 0 there is a \u03b4 > 0 such that if x is closer to x0 than \u03b4 (and f is de\ufb01ned at\nx), then f(x) is closer to l than \u03b5.\n(2) whenever a sequence (yn) is contained in E\\{x0} and it converges to x0, then the sequence", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_157", "text": "x), then f(x) is closer to l than \u03b5.\n(2) whenever a sequence (yn) is contained in E\\{x0} and it converges to x0, then the sequence\n(f(yn)), given by the values of the function f along the xn, converges to l.\nRemark 6.7. We explain why the two conditions in ?? are equivalent.\n?? \u21d2??: Let us \ufb01x a sequence (yn) which is contained in E \\ {x0} and for which\nlim\nn\u2192\u221exn = x0. We have to show that lim\nn\u2192\u221ef(xn) = l. Let us \ufb01x \u03b5 > 0. Then, this yields a\n\u03b4 > 0 as in de\ufb01nition (i). For this \u03b4, there is an n\u03b4 such that |x0 \u2212xn| < \u03b4 for n \u2265n\u03b4,\nand hence for all such n, |l \u2212f(xn)| < \u03b5.\nNOT ?? \u21d2NOT ??: The negation of (i) is that there is an \u03b5 > 0 such that for each\n\u03b4 > 0 we can \ufb01nd y\u03b4 \u2208(x0 \u2212\u03b4, x0 + \u03b4[\\{x0} such that |f(y\u03b4) \u2212l| \u2265\u03b5. De\ufb01ning yn := y 1\nn ,\nthen the sequence (yn) converges to x0, but all values f(yn) have distance at least \u03b5 from\nl, so the sequence (f(xn)) cannot converge to l.\nWe will work more often with the use de\ufb01nition ?? more as it is simpler. Luckily, it is almost\nalways enough for proving that a limit does not exist. We usually use de\ufb01nition ?? only when\n?? does not work.\nExample 6.8. We show that lim\nx\u21922 x2 = 4 using point ?? of ??. In order to do this, we proceed\nas follows: let us \ufb01x \u03b5 > 0; at this point, we need to \ufb01nd \u03b4 > 0 such that\nif 0 < |x \u22122| < \u03b4 then, |x2 \u22124| < \u03b5.\nLet us note that\nx2 \u22124 = (x \u22122)(x + 2).\n83", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_158", "text": "Furthermore, if 0 < |x \u22122| < 1, then 3 < x + 2 < 3; thus, if 0 < |x \u22122| < 1, then\n|x2 \u22124| = |x \u22122||x + 2| < 5|x \u22122|.\n(6.8.a)\nTherefore, taking \u03b4 = min\n\b\n1, \u03b5\n5\n\t\n, we conclude that, if 0 < |x \u22122| < \u03b4 then\n|x \u22122| < 1 and |x \u22122| < \u03b5\n5.\nHence, it follows that\n|x2 \u22124|\n<\n5|x \u22122|\n|\n{z\n}\nUsing ??, since |x \u22122| < 1\n<\n5\u03b5\n5 = \u03b5.\nExample 6.9. We can repeat the same computation as in the previous example, also using ??\nof ??. In general, proving the existence (and \ufb01niteness of the limit using sequence) can be\nrather tricky: you can try for example to use that de\ufb01nition to compute the limit of sin(x)\nx\nat\nx = 0. Let (xn) be a sequence such that lim\nn\u2192\u221exn = 2. Then, by the algebraic properties of the\nlimit, that is, by Proposition 4.30, we know that lim\nn\u2192\u221ex2\nn =\n\u0010\nlim\nn\u2192\u221exn\n\u00112\n= 22 = 4.\nExample 6.10. We can generalize the arguments from the previous examples to show that\nfor any x0 \u2208R, lim\nx\u2192x0 xn = xn\n0 for any n \u2208N.\nUsing the notion of limit, we can also de\ufb01ne the notion of continuity of a function f at a\npoint in D(f).\nDe\ufb01nition 6.11. Let f : E \u2192R be a function, E \u2282R. The function f is continuous at a\npoint x0 \u2208E, if lim\nx\u2192x0 f(x) exists, it is \ufb01nite and lim\nx\u2192x0 f(x) = f(x0).\nRemark 6.12. Implicit in ?? is the fact that the limit of f(x) at x0 exists.\nIn particular,\nthe subset E \u2286R on which f is assumed to be de\ufb01ned must contain not only x0 but also a\npunctured neighborhood of x0.", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_159", "text": "Remark 6.12. Implicit in ?? is the fact that the limit of f(x) at x0 exists.\nIn particular,\nthe subset E \u2286R on which f is assumed to be de\ufb01ned must contain not only x0 but also a\npunctured neighborhood of x0.\nRemark 6.13. The main di\ufb00erence between ?? and ?? is the fact that, while in ?? we do not\nrequire the function f to be de\ufb01ned at the point x0 at which we are trying to compute the limit\nand when taking the limit we only look at the value of f on points of D(f) \\ {x0}, in the case\nof ??, instead, we very much want to allow the value f(x0) to play a role. More precisely, we\nhave the following characterization of continuity at a point, via conditions analogous to those\nin ??.\nProposition 6.14. Let f : E \u2192R be a function and let x0 \u2208E. Assume that there exists a\nopen interval of the form (x0 \u2212\u03b4, x0 + \u03b4[, \u03b4 > 0 contained in E. Then, f is continuous at x0 if\nand only if one of the following two equivalent de\ufb01nitions hold:\n(1) For every \u03b5 \u2208R\u2217\n+ there is a \u03b4\u03b5 \u2208R\u2217\n+ such that\n\u2200x \u2208E such that |x \u2212x0| < \u03b4 \u21d2|f(x) \u2212f(x0)| < \u03b5.\n(2) For every sequence (xn) \u2286E for which lim\nn\u2192\u221exn = x0, then lim\nn\u2192\u221ef(xn) = f(x0).\nIn view of the above proposition, ?? yields the following immediate corollary.\nCorollary 6.15. Fix n \u2208N. Then f(x) = xn is continuous at every x0 \u2208R.\nUsing the next de\ufb01nition, we can rephrase the previous corollary by saying that, for a \ufb01xed\nn \u2208N, the function f : R \u2192R is continuous.\n84", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_160", "text": "De\ufb01nition 6.16. Let f : E \u2192R. Assume that \u2200x0 \u2208E, E contains an open ball centered at\nx0. Then we say that f is continuous if it is continuous at every x0 \u2208E.\nExample 6.17.\n(1) Let us de\ufb01ne the function f : R \u2192R,\nf(x) :=\n(\n0\nx \u0338= 0,\n1\nx = 0.\nThen, f is not continuous at x = 0.\nIn fact, lim\nx\u21920 f(x) = 0, as in the de\ufb01nition we assumed 0 < |x \u2212x0| \u2264\u03b4, so the function\nvalue 1 for x0 = 0 does not cause any problem.\n(2) Let us de\ufb01ne the function f : R \u2192R,\nf(x) :=\n(\n0\nx \u2208Q,\n1\nx \u2208R \\ Q.\nThen, the set of points of R at which f is continuous is empty.\nFor example, let us consider the point 0 \u2208R and let us de\ufb01ne the sequences (y\u2032\nn), (y\u2032\u2032\nn) \u2282\nR \\ {0}, to be\ny\u2032\nn = 1\nn,\ny\u2032\u2032\nn =\n1\n\u221a\nn2 + 1\n,\nn \u22651.\nAs \u2200n \u2208N\u2217,\n1\nn \u2208Q,\n1\n\u221a\nn2+1 \u2208R \\ Q, then \u2200n \u2208N\u2217, f(y\u2032\nn) = 0, while f(y\u2032\u2032\nn) = 1.\nHence, limn\u2192\u221ey\u2032\nn = 0 = limn\u2192\u221ey\u2032\u2032\nn, while limn\u2192\u221ef(y\u2032\nn) = 0, limn\u2192\u221ef(y\u2032\u2032\nn) = 1, hence\nthe limit limx\u21920 f(x) does not exist, and moreover, f is not continuous at 0.\nOne can repeat the same reasoning at any point x0 \u2208R, by taking y\u2032\nn to be a sequence of\nrational numbers converging to x0 (for example, y\u2032\nn = x0 + 1\nn, if x0 is rational, or y\u2032\nn to be\nthe truncation of the decimal form of x0 at the n-th decimal digit, if x0 is irrational) and y\u2032\u2032\nn\nto be a sequence of irrational numbers converging to x0 (for example, y\u2032\u2032\nn = x0 +\n1\n\u221a\nn2+1, if", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_161", "text": "the truncation of the decimal form of x0 at the n-th decimal digit, if x0 is irrational) and y\u2032\u2032\nn\nto be a sequence of irrational numbers converging to x0 (for example, y\u2032\u2032\nn = x0 +\n1\n\u221a\nn2+1, if\nx0 is rational, or y\u2032\u2032\nn = x0+ 1\nn if x0 is irrational). Then, \u2200n \u2208N\u2217, f(y\u2032\nn) = 0, while f(y\u2032\u2032\nn) =\n1, and limn\u2192\u221ey\u2032\nn = x0 = limn\u2192\u221ey\u2032\u2032\nn, while limn\u2192\u221ef(y\u2032\nn) = 0, limn\u2192\u221ef(y\u2032\u2032\nn) = 1, hence\nthe limit limx\u2192x0 f(x) does not exist, and moreover, f is not continuous at x0.\nThis implies that limx\u2192x0 f(x) does not exist at any point x0 \u2208R, in particular, f is not\ncontinuous at any point of R.\nExample 6.18. We claim that lim\nx\u21920 cos(x) = 1.\nIndeed, let (xn) be a sequence converging to 0. Then,\n0 \u2264| cos(xn) \u22121| =\n\f\f\f2 sin2 \u0010xn\n2\n\u0011\f\f\f \u22642x2\nn\n4 = x2\nn\n2 ,\nusing the inequality | sin(x)| \u2264|x|. So, squeeze theorem tells us that lim\nn\u2192\u221e| cos(xn) \u22121| = 0.\nExample 6.19. The limit lim\nx\u21920 sin\n\u00121\nx\n\u0013\ndoes not exist.\nIndeed, consider the sequences xn :=\n1\n\u03c0(2n+ 1\n2) and x\u2032\nn :=\n1\n\u03c0(2n+ 3\n2). Then, \ufb01rst lim\nn\u2192\u221exn = 0 and\nlim\nn\u2192\u221ex\u2032\nn = 0. However,\nlim\nn\u2192\u221esin\n\u0012 1\nxn\n\u0013\n= sin\n\uf8eb\n\uf8ed\n1\n1\n\u03c0(2n+ 1\n2)\n\uf8f6\n\uf8f8= sin\n\u0012\n\u03c0\n\u0012\n2n + 1\n2\n\u0013\u0013\n= 1,\n85", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_162", "text": "but\nlim\nn\u2192\u221esin\n\u0012 1\nx\u2032n\n\u0013\n= sin\n\uf8eb\n\uf8ed\n1\n1\n\u03c0(2n+ 3\n2)\n\uf8f6\n\uf8f8= sin\n\u0012\n\u03c0\n\u0012\n2n + 3\n2\n\u0013\u0013\n= \u22121.\nSo, point ?? of ?? is not satis\ufb01ed, and hence the limit does not exist\n6.1.1\nLimits and algebra\n?? allows us to translate all the statements about limits of sequences to limits of functions.\nIndeed, let us say we are have functions f, g de\ufb01ned around a point x0 \u2208R \u2013 but we are\nnot necessarily assuming that f, g are de\ufb01ned at x0 \u2013 and we want to prove that if l and k\nare the limits of f(x) and g(x) (at x0), then l + k is the limit of (f + g)(x). Let us take\na sequence (yn) converging to x0. We know that lim\nn\u2192\u221ef(yn) = l and lim\nn\u2192\u221eg(yn) = k. But\nthen, Proposition 4.30 implies that lim\nn\u2192\u221ef(xn)+f\u2032(xn) = l+k, that is, lim\nn\u2192\u221e(f +f\u2032)(xn) = l+k,\nwhich is exactly generalizing the statement about limit of sequences and addition, to the case\nof limit of functions.\nUnsurprisingly, at this point, we can do the same with all other properties that we proved\nfor limits of sequences. We collect all the statements one can show along the same arguments:\nProposition 6.20. Let f and g be two functions such that a punctured neighborhood of x0 is\nin the domain of both f and g. Assume that the limits of f and g at x0 exist and they are l\nand k, respectively. Then,\n(1) the limit of f + g exists at x0 and lim\nx\u2192x0(f + g)(x) = l + k\n(2) the limit of f \u00b7 g exists at x0 and lim\nx\u2192x0(f \u00b7 g)(x) = l \u00b7 k\n(3) if k \u0338= 0, then the limit of f\ng exists at x0 and lim\nx\u2192x0\n\u0012f\ng\n\u0013\n(x) = l\nk\n(4) if f(x) \u2264g(x) for any x in a punctured neighborhood of x0, then l \u2264k.", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_163", "text": "(3) if k \u0338= 0, then the limit of f\ng exists at x0 and lim\nx\u2192x0\n\u0012f\ng\n\u0013\n(x) = l\nk\n(4) if f(x) \u2264g(x) for any x in a punctured neighborhood of x0, then l \u2264k.\n(5) Squeeze Theorem: if there is a third function h(x) such that there is also a punctured\nneighborhood of x0 in the domain of h, and:\n(i) on some punctured neighborhood of x0 we have f(x) \u2264h(x) \u2264g(x), and\n(ii) l = k,\nthen lim\nx\u2192x0 h(x) = l.\nExample 6.21. The main example for using point ?? of ?? is that lim\nx\u21920\nsin(x)\nx\n= 1. Indeed,\nwe have already seen, cf. Figure 8, that\n0 \u2264sin(x) \u2264x \u2264tan(x),\nfor x \u2208[0, \u03c0\n2 ],\ntan(x) \u2264x \u2264sin(x) \u22640,\nfor x \u2208[\u2212\u03c0\n2 , 0];\nwhich implies that\n|cos(x)| \u2264\n\f\f\f\f\nsin(x)\nx\n\f\f\f\f \u22641,\nfor x \u2208[\u2212\u03c0\n2 , \u03c0\n2 ].\n(6.21.b)\n86", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_164", "text": "Since \u2200x \u2208[\u2212\u03c0\n2 , \u03c0\n2 ], cos(x) \u22650, whereas sin(x) and x are odd function, so that also \u2200x \u2208\n[\u2212\u03c0\n2 , \u03c0\n2 ] \\ {0}, sin(x)\nx\n\u22650, the chain of inequalities in ?? holds also once we remove the absolute\nvalues. So, by the Squeeze Theorem for limits we can conclude that\nlim\nx\u21920\nsin(x)\nx\n= 1,\nsince lim\nx\u21920 cos(x) = 1, see ??.\nExample 6.22. For any k \u2208N\u2217, lim\nx\u21920 xk sin\n\u00121\nx\n\u0013\n= 0. Indeed, as | sin\n\u0000 1\nx\n\u0001\n| = 1, \u2200x \u2208R\u2217, then\n\u2212xk \u2264xk sin\n\u00121\nx\n\u0013\n\u2264xk,\n\u2200x \u2208R\u2217.\nand the conclusion follows from the Squeeze Theorem.\nThe above proposition has all the nice consequences about continuity.\nProposition 6.23. If f, g: E \u2192R are continuous functions at x0 \u2208E. Then the following\nfunction are also continuous at x0:\n(1) \u03b1f + \u03b2g for any \u03b1, \u03b2 \u2208R,\n(2) f \u00b7 g, and\n(3)\nf\ng , if g|E is nowhere zero (meaning that for all x \u2208E : g(x) \u0338= 0), then.\nExample 6.24. We collect here some example of continuous functions, on their respective\ndomains, that is, each of these functions is continuous at any point where they are de\ufb01ned:\n\u25e6p(x) = a0 + a1x + a2x2 + \u00b7 \u00b7 \u00b7 + arxr, that is, p(x) is a polynomial in one variable x.\n\u25e6f(x) := 1\nx is continuous on R \\ 0.\n\u25e6\nx\nx2\u22123x+1 is continuous on R \\\nn\n3\u00b1\n\u221a\n5\n2\no\n,\n\u25e6In general, if p(x) and q(x) are two polynomials, then p(x)\nq(x) is continuous on {x \u2208R|q(x) \u0338=\n0} (which is the whole real line minus \ufb01nitely many points).\n6.1.2\nLimit and composition\nLet us recall the following de\ufb01nition of composition of functions.\nDe\ufb01nition 6.25. If f : E \u2192R and g : G \u2192R are functions such that R(f) \u2286G then we may", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_165", "text": "6.1.2\nLimit and composition\nLet us recall the following de\ufb01nition of composition of functions.\nDe\ufb01nition 6.25. If f : E \u2192R and g : G \u2192R are functions such that R(f) \u2286G then we may\nde\ufb01ne the composition g \u25e6f (order matters!!) of f with g by\n(g \u25e6f)(x) = g(f(x)).\nExample 6.26. Let us take f : R \u2192R given by f(x) = x2 + 1, and g : R \u2192R given by\ng(y) = y3 + y + 1. Then we have\n(g \u25e6f)(x) = (x2 + 1)3 + (x2 + 1) + 1 = x6 + 3x4 + 4x2 + 3.\n(6.26.c)\nLet us look at an example about whether composition of continuous functions is continuous\nor not.\n87", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_166", "text": "Example 6.27. Consider the functions de\ufb01ned in ??. We want to show that g\u25e6f is continuous\nat x = 0. As (g \u25e6f)(0) = 3 by ??, we then need to show that lim\nx\u21920(g \u25e6f)(x) = 3. and this can\nbe immediately deduced from the last part of the equation in ??. So, indeed g \u25e6f is continuous\nat x = 0.\nIn general the situation is just as nice as in ??.\nProposition 6.28. Let f : E \u2192R and g : G \u2192R be two functions. Assume that:\n(1) f(E) \u2286G,\n(2) f is continuous at x0,\n(3) g is continuous at y0 := f(x0).\nThen g \u25e6f is continuous at x0.\nProof. We verify condition ?? of ??. Let (zn) \u2286E be a sequence such that\nlim\nn\u2192\u221ezn = x0.\n(6.28.d)\nAccording to ?? and our assumption ??, then\nlim\nn\u2192\u221ef(zn) = y0.\n(6.28.e)\nHence\n|\n{z\n}\n??\nlim\nn\u2192\u221e(g \u25e6f)(zn) =\n?? and condition ??\nz\n}|\n{\nlim\nn\u2192\u221eg(f(zn)) = g(y0) .\nRemark 6.29. Let us examine a bit further ??. In the proof of ?? we showed that if\nlim\nx\u2192x0f(x) = y0\nand\nlim\ny\u2192y0 g(y) = l,\n(6.29.f)\nthen lim\nx\u2192x0g \u25e6f(x) = l holds under the assumption that f and g are continuous. We may be\ntempted to think that an analogous statement to ?? should also for the limit of a composition\nof functions, just assuming the condition in ??. However, as we will see in ??, this is not true.\nThe reason is that in ??, contrary to ??, there is nothing said about the behavior at x0 and\ny0. So, we have to assume that f(x) avoids y0 in a punctured neighborhood of x0.\nThe precise statement about composition of functions, in regards to limits, is as follows.", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_167", "text": "y0. So, we have to assume that f(x) avoids y0 in a punctured neighborhood of x0.\nThe precise statement about composition of functions, in regards to limits, is as follows.\nProposition 6.30. Let f : E \u2192R and g : G \u2192R be functions and let x0 \u2208E be a point such\nthat\n(1) f(E) \u2286G,\n(2)\nlim\nx\u2192x0 f(x0) = y0,\n(3) lim\ny\u2192y0 g(y0) = l\n(4) there is a punctured neighborhood (x0 \u2212\u03b4, x0 + \u03b4) \\ {x0} \u2286E such that for every x in this\nneighborhood, f(x) \u0338= y0.\n88", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_168", "text": "Then, lim\nx\u2192x0(g \u25e6f)(x) = l\nProof. We use part ?? of ??. Thus, let us \ufb01x a sequence (zn) \u2286E \\ {x0} such that\nlim\nn\u2192\u221ezn = x0.\n(6.30.g)\nIn particular, by throwing away \ufb01nitely many elements of the sequence, we may assume that\n(zn) \u2286(x0 \u2212\u03b4, x0 + \u03b4) \\ {x0} \u2286E.\n(6.30.h)\nBy the assumption ?? in the statement of the proposition, and by ??, it follows that\nlim\nn\u2192\u221ef(zn) = y0.\n(6.30.i)\nLastly, by our assumption ?? and ?? we have\n(f(zn)) \u2286G \\ {y0}.\n(6.30.j)\nHence, by our assumption ?? and by ??, we have\nlim\nn\u2192\u221e(g \u25e6f)(xn) = lim\nn\u2192\u221eg(f(xn)) = l.\nThe following example shows that condition ?? of ?? is necessary. That is, if we drop\ncondition ??, the statement of ?? would not hold.\nExample 6.31. Consider:\ng(x) =\n(\n0,\nfor x \u0338= 0,\n1,\nfor x = 0,\nand\nf(x) =\n(\nx sin\n\u0000 1\nx\n\u0001\n,\nfor x \u0338= 0\n0,\nfor x = 0.\nThen, lim\nx\u21920 f(x) = 0 and lim\nx\u21920 g(x) = 0. However, lim\nx\u21920(g \u25e6f)(x) \u0338= 0, because the following two\nsequences induce function value sequences with di\ufb00erent limits:\nxn := 1\n\u03c0n\nand\nyn :=\n1\n\u03c0n + \u03c0\n2\n,\nas\nlim\nn\u2192\u221e(g \u25e6f)(xn) = lim\nn\u2192\u221e1 = 1\nand\nlim\nn\u2192\u221e(g \u25e6f)(yn) = lim\nn\u2192\u221e0 = 0.\nAlso, let us note that condition ?? of ?? below is not satis\ufb01ed in this example, as f(x) = 0\nfor x =\n1\n\u03c0n, so there is no punctured neighborhood of 0 on which the function of f avoids the\nvalue 0.\nExample 6.32. A positive example for applying ?? is during the argument of showing that\nlim\nx\u21920", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_169", "text": "for x =\n1\n\u03c0n, so there is no punctured neighborhood of 0 on which the function of f avoids the\nvalue 0.\nExample 6.32. A positive example for applying ?? is during the argument of showing that\nlim\nx\u21920\nsin(x2)\nx2\n= 1. Indeed, if we set g(x) := sin(x)\nx\n, and f(x) = x2, then condition ?? of ?? is also\nsatis\ufb01ed, as f(x) \u0338= 0 for x \u0338= 0.\n89", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_170", "text": "6.1.3\nIn\ufb01nite limits\nDe\ufb01nition 6.33. A neighborhood of +\u221e(resp. \u2212\u221e) is an unbounded interval of the form\n(a, +\u221e) (resp. (\u2212\u221e, a)).\nWe extend the de\ufb01nition of limit to comprise the case where we allow ourselves to work\nwith the extended real line R.\nDe\ufb01nition 6.34. Let x0, l \u2208R, and let f : E \u2192R be a function, E \u2282R. Assume that E\ncontains a punctured neighborhood of x0. We say that the limit of f(x) at x0 is l, if for any\nsequence (yn) \u2286E \\ {x0}14, whenever lim\nn\u2192\u221eyn = x0, then lim\nn\u2192\u221ef(yn) = l.\nExample 6.35. We show that lim\nx\u21920\n1\nx2 = +\u221e. Indeed, if (xn) \u2282R\u2217is a sequence satisfying\nlim\nn\u2192\u221exn = 0, then lim\nn\u2192\u221e\n1\nx2n = +\u221eby algebraic properties of limits of sequences.\nOn the other hand, the limit lim\nx\u21920\n1\nx does not exist. In fact, considering the sequence xn := 1\nn,\nthen lim\nn\u2192\u221e\n1\nxn = lim\nn\u2192\u221en = +\u221e, while for yn = \u22121\nn , then lim\nn\u2192\u221e\n1\nyn = lim\nn\u2192\u221e\u2212n = \u2212\u221e.\nProposition 6.36. Let x0 \u2208R, and let f, g: E \u2192R be functions.\n(1) Addition rule. Assume that the following conditions are satis\ufb01ed:\n\u25e6\nlim\nx\u2192x0 f(x) = +\u221e(resp. \u2212\u221e), and\n\u25e6g(x) is bounded from below (resp. from above)\nthen lim\nx\u2192x0(f + g)(x) = +\u221e(resp. \u2212\u221e).\n(2) Product rule. Assume that the following conditions are satis\ufb01ed:\n\u25e6\nlim\nx\u2192x0|f(x)| = +\u221e,\n\u25e6there exists \u03b4 > 0 such that |g(x)| \u2265\u03b4 for all x \u2208E, and\n\u25e6f(x)g(x) > 0 (resp. < 0) for all x \u2208E,\nthen lim", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_171", "text": "\u25e6\nlim\nx\u2192x0|f(x)| = +\u221e,\n\u25e6there exists \u03b4 > 0 such that |g(x)| \u2265\u03b4 for all x \u2208E, and\n\u25e6f(x)g(x) > 0 (resp. < 0) for all x \u2208E,\nthen lim\nx\u2192x0f(x)g(x) = +\u221e(resp. \u2212\u221e).\n(3) First division rule. If\n\u25e6f(x) is bounded,\n\u25e6g(x) is nowhere zero, and\n\u25e6\nlim\nx\u2192x0|g(x)| = +\u221e.\nThen lim\nx\u2192x0\nf(x)\ng(x) = 0\n(4) Second division rule. If\n\u25e6\nlim\nx\u2192x0g(x) = 0,\n\u25e6there is a \u03b4 > 0 such that |f(x)| \u2265\u03b4 for all x \u2208E, and\n\u25e6f(x)/g(x) > 0 (resp. < 0) for all x \u2208E,\nthen lim\nx\u2192x0\nf(x)\ng(x) = +\u221e(resp. \u2212\u221e).\n14When x0 = \u00b1\u221e, then the condition that x0 does not belong to E is automatically satis\ufb01ed, since E \u2286R.\n90", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_172", "text": "(5) Squeeze. If f(x) \u2264g(x), and\n\u25e6if lim\nx\u2192x0f(x) = +\u221e, then lim\nx\u2192x0g(x) = +\u221e\n\u25e6if lim\nx\u2192x0g(x) = \u2212\u221e, then lim\nx\u2192x0f(x) = \u2212\u221e\nExample 6.37. Here are a few examples.\n\u25e6lim\nx\u21920\n1\nx2\n|{z}\n\u2192\u221e\n+ cos(x)\n| {z }\nbounded\n= +\u221e\n\u25e6\nlim\nx\u2192+\u221ecos(x)\n| {z }\nbounded\n\u00b7 (\u2212x2 + x3)\n|\n{z\n}\n\u2192+\u221e\n= \u2212\u221e, since \u2212x2 + x3 = 0 only for x = 0, 1.\n\u25e6\nlim\nx\u2192+\u221e\nbounded\nz\n}|\n{\narctan(x)\n(\u2212x)\n| {z }\n\u2192\u2212\u221e\n= 0.\nRemark 6.38.\n(1) The assumptions stated in part ?? of ?? are important, as otherwise we can\nhave all di\ufb00erent kinds of limits. We give examples of this using the functions f(x) = x3,\ng(x) = x2 and h(x) = x3 + 1. We have\n\u25e6\nlim\nx\u2192+\u221ef(x) =\nlim\nx\u2192+\u221ex3 = +\u221e,\nlim\nx\u2192+\u221e\u2212f(x) =\nlim\nx\u2192+\u221e\u2212x3 = \u2212\u221e\n\u25e6\nlim\nx\u2192+\u221eh(x) =\nlim\nx\u2192+\u221e(x3 + 1) = +\u221e,\nlim\nx\u2192+\u221e\u2212h(x) =\nlim\nx\u2192+\u221e\u2212(x3 + 1) = \u2212\u221e, and\n\u25e6\nlim\nx\u2192+\u221eg(x) =\nlim\nx\u2192+\u221ex2 = +\u221e, lim\nx\u2192+\u221e\u2212g(x) =\nlim\nx\u2192+\u221e\u2212x2 = \u2212\u221e.\nOn the other hand:\n\u25e6\nlim\nx\u2192+\u221ef(x) \u2212g(x) =\nlim\nx\u2192+\u221ex3 \u2212x2 =\nlim\nx\u2192+\u221ex2(x \u22121) = +\u221e,\n\u25e6\nlim\nx\u2192+\u221eg(x) \u2212f(x) = \u2212\u221e, and\n\u25e6\nlim\nx\u2192+\u221ef(x) \u2212h(x) = \u22121.\nIn particular, never use addition law for limits of the type (+\u221e) + (\u2212\u221e).", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_173", "text": "\u25e6\nlim\nx\u2192+\u221eg(x) \u2212f(x) = \u2212\u221e, and\n\u25e6\nlim\nx\u2192+\u221ef(x) \u2212h(x) = \u22121.\nIn particular, never use addition law for limits of the type (+\u221e) + (\u2212\u221e).\n(2) The above assumptions for point ?? of ?? are also important. We give examples of this\nusing the functions f(x) = x, g(x) = cos(x)\nx\nand h(x) = (\u22121)[x]. We have\n\u25e6\nlim\nx\u2192+\u221e|f(x)| = +\u221e,\n\u25e6|g(x)| is not bounded from below, and\n\u25e6|h(x)| is bounded from below, but f(x)h(x) \u0338> 0.\nThen:\n\u25e6\nlim\nx\u2192+\u221ef(x)g(x) =\nlim\nx\u2192+\u221ecos(x) does not exist, and\n\u25e6\nlim\nx\u2192+\u221ef(x)h(x) =\nlim\nx\u2192+\u221ex(\u22121)[x] does not exist, on the other hand\n\u25e6the product law applies to (f(x)h(x))(h(x)) and yields\nlim\nx\u2192+\u221e(f(x)h(x))(h(x)) =\n+\u221e\nNever try to use product rule to limits of the type 0 \u00b7 \u221e.\n91", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_174", "text": "(3) The assumptions of the \ufb01rst division rule are also important. One can can show that in\nthe \u00b1\u221e\n\u00b1\u221ecase anything can happen for example using 1\nx,\n1\nx2 ,\n1\nx3 ,(\u22121)[ 1\nx] 1\nx with limit at 0:\n\u25e6lim\nx\u21920\n1\nx\n1\nx2 = lim\nx\u21920x = 0,\n\u25e6lim\nx\u21920\n(\u22121)[ 1\nx] 1\nx\n1\nx\n= lim\nx\u21920(\u22121)[ 1\nx] does not exist and bounded,\n\u25e6lim\nx\u21920\n1\nx2\n1\nx = lim\nx\u21920\n1\nx does not exist and unbounded, and\n\u25e6lim\nx\u21920\n1\nx3\n1\nx = lim\nx\u21920\n1\nx2 = +\u221e.\nSimilar examples show that the assumptions are important for the second division rule.\nNever try to use division rules to limits of the form \u00b1\u221e\n\u00b1\u221eand 0\n0.\n6.1.4\nOne sided limits\nThe main question is how to make sense of limits such as at 0 of \u221ax, as here the domain does\nnot contain a punctured neighborhood of 0. The solution for this is the introduction of the\nnotions of left and right limits.\nDe\ufb01nition 6.39. A function f : E \u2192R is de\ufb01ned to the left (resp. to the right) of x0 \u2208R, if\nE contains an interval of the form (x0 \u2212\u03b4, x0[ (resp. (x0, x0 + \u03b4[).\nDe\ufb01nition 6.40. Let f : E \u2192R be a function and x0 \u2208R. Assume that f is de\ufb01ned to the\nleft (resp. right) of x0. Let l \u2208R.\n(1) We say that the limit of f for x that goes to x0 from the left is l if for all sequences\n(xn) \u2286{x \u2208E|x < x0}, whenever lim\nn\u2192\u221exn = x0 then lim\nn\u2192\u221ef(xn) = l. When this condition\nis satis\ufb01ed, we write lim\nx\u2192x\u2212\n0\nf(x) = l.\n(2) We say that the limit of f for x that goes to x0 from the right is l if for all sequences\n(xn) \u2286{x \u2208E|x > x0}, whenever lim\nn\u2192\u221exn = x0 then lim", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_175", "text": "0\nf(x) = l.\n(2) We say that the limit of f for x that goes to x0 from the right is l if for all sequences\n(xn) \u2286{x \u2208E|x > x0}, whenever lim\nn\u2192\u221exn = x0 then lim\nn\u2192\u221ef(xn) = l. When this condition\nis satis\ufb01ed, we write lim\nx\u2192x+\n0\nf(x) = l.\nExample 6.41. Consider the function f : R+ \u2192R de\ufb01ned as f(x) := \u221ax. We claim that\nlim\nx\u21920+\n\u221ax = 0.\nIndeed, \ufb01x a sequence (xn) \u2286R\u2217\n+ such that lim\nn\u2192\u221exn = 0. We have to show that then lim\nn\u2192\u221e\n\u221axn =\n0 too. So, we need to show that for each \u03b5 > 0, there is an n0 such that for every integer n \u2265n0,\n\u221axn \u2264\u03b5. However, we know that lim\nn\u2192\u221exn = 0. So, we know that there is an n0 such that\n|xn| < \u03b52 for all n \u2265n0. But then, for any such n we also have \u221axn < \u03b5.\nProposition 6.42. Let f : E \u2192R be a function such that there is a punctured neighborhood\nof x0 contained in E, and both\nl1 := lim\nx\u2192x\u2212\n0\nf(x)\nand\nl2 := lim\nx\u2192x+\n0\nf(x)\nexists. Then\nlim\nx\u2192x0f(x) = l\n\u21d0\u21d2\nl1 = l2 = l.\n92", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_176", "text": "Example 6.43. Consider the function f(x) = {x} =. Both left and right limits exist at all\npoints, and furthermore:\nlim\nx\u2192x\u2212\n0\n{x} =\n\u001a {x}\nif x /\u2208Z\n1\nif x \u2208Z\nand\nlim\nx\u2192x+\n0\n{x} =\n\u001a {x}\nif x /\u2208Z\n0\nif x \u2208Z\nHence, according to ??,\nlim\nx\u2192x0{x} exists \u21d4x /\u2208Z.\nExample 6.44. ?? together with ?? show that the function f(x) =\np\n|x| is continuous in 0.\nIt is not hard to show that f is actually continuous everywhere in R.\n6.1.5\nMonotone functions\nFor a monotone function f, the left (resp. the right) limits always exists at any point x0 at\nwhich the function is de\ufb01ned at the left (resp. the right) of x0.\nProposition 6.45. Let f : E \u2192R be a monotone function. Then, at each point x0 \u2208E:\n(1) if f is de\ufb01ned on the left of x0, lim\nx\u2192x\u2212\n0\nf(x) exists,\n(2) if f is de\ufb01ned on the right of x0, lim\nx\u2192x+\n0\nf(x) exists, and\n(3) if f is de\ufb01ned in a neighborhood of \u00b1\u221e, then\nlim\nx\u2192\u00b1\u221ef(x) exists.\nProof. We treat only the increasing case, as the decreasing one follows from that by regarding\n\u2212f instead of f. Also, we treat only the \ufb01rst case as the others are similar. Set:\nl := sup{f(x)|x \u2208E, x < x0}.\n(6.45.k)\nLet\n(xn) \u2286{x \u2208E|x < x0}\nsuch that\nlim\nn\u2192\u221exn = x0.\n(6.45.l)\nWe have to show that lim\nn\u2192\u221ef(xn) = l. Fix a \u03b5 > 0. Then, by the de\ufb01nition of l, there is an\nx\u2032 \u2208{x \u2208E|x < x0}, such that\nf(x\u2032) > l \u2212\u03b5.\n(6.45.m)\nAccording to ??, there is an n0 \u2208N such that for all integers n \u2265n0 we have", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_177", "text": "x\u2032 \u2208{x \u2208E|x < x0}, such that\nf(x\u2032) > l \u2212\u03b5.\n(6.45.m)\nAccording to ??, there is an n0 \u2208N such that for all integers n \u2265n0 we have\nx\u2032 < xn < x0.\n(6.45.n)\nHowever, then for all integers n \u2265n0 we have:\nl \u2265f(xn)\n| {z }\n?? and ??\n\u2265\nf(x\u2032)\n| {z }\nf is increasing and ??\n\u2265l \u2212\u03b5.\n| {z }\n??\nThis shows that lim\nn\u2192\u221ef(xn) = l indeed.\nExample 6.46.\n(1) Let\nf(x) := sgn(x) =\n\uf8f1\n\uf8f4\n\uf8f2\n\uf8f4\n\uf8f3\n1\nif x > 0\n0\nif x = 0\n\u22121\nif x < 0.\n93", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_178", "text": "Images/sgn_gr.png\nFigure 11: f(x) = sgn(x).\nThen\nlim\nx\u2192x\u2212\n0\nsgn(x) = \u22121\nand\nlim\nx\u2192x+\n0\nsgn(x) = 1\nNote that these limits exist and neither of them agree with f(0) = 0.\n(2) Let f(x) := \u230ax\u230b. Then:\nlim\nx\u2192x\u2212\n0\nf(x) =\n(\n\u230ax\u230b\nx /\u2208Z\nx \u22121\nx \u2208Z\nand\nlim\nx\u2192x+\n0\nf(x) =\n(\n\u230ax\u230b\nx /\u2208Z\nx\nx \u2208Z.\nSo, the left and right limits exist, despite having di\ufb00erent values whenever x \u2208Z.\nImages/floor_gr.png\nFigure 12: f(x) = \u230ax\u230b.\n94", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_179", "text": "6.1.6\nMore on continuity\nFirst, we note that there are more algebraic rules of continuity (we already discussed addition,\nmultiplication and division in ??):\nProposition 6.47. If f, g: E \u2192R are functions that are continuous at x0 \u2208E, then so are:\n(1) |f|,\n(2) max{f, g}, where\nmax{f, g}(x) := max{f(x), g(x)}\n(3) min{f, g} (de\ufb01ned similarly),\n(4) f+ := max{f, 0},\n(5) f\u2212:= min{f, 0}.\nExample 6.48. We can use for example the continuity of the absolute value for squeezing.\nFor example, let\ng(x) :=\n(\n1\nfor x \u2208Q\nx\nfor x \u2208R \\ Q\nWe claim that g(x) is continuous at x0 = 1. The main idea is that we can try to apply the\nSqueeze Theorem for the limit of functions using the following chain of inequalities:\n\u2212|x \u22121| \u2264g(x) \u22121 \u2264|x \u22121|.\nAccording to point ?? of ??, the function |x \u22121| is continuous everywhere over R; thus,\nlim\nx\u21921 \u2212|x \u22121| = lim\nx\u21921 |x \u22121| = 0,\nso that by ????, it follows that lim\nx\u21921 f(x) \u22121 = 0.\nHence, lim\nx\u21921 f(x) = 1 = f(1) and f is\ncontinuous at x0 = 1.\n6.1.7\nUniform continuity and Lipschitzianity\nWe introduce a stronger version of continuity.\nDe\ufb01nition 6.49. A function f : E \u2192R is said to be uniformly continuous if for every \u03b5 > 0\nthere is a \u03b4\u03b5 > 0 such that for all x, y \u2208E then\nif |x \u2212y| < \u03b4\u03b5 \u21d2|f(x) \u2212f(y)| < \u03b5.\nRemark 6.50. The notion of uniform continuity de\ufb01ned above is much stronger than that\nof continuity, cf. ??. More precisely, using the characterization of continuity for a function\nf : E \u2192R given in ??, for any x0 \u2208E and any \u03b5 > 0 there exists \u03b4 > 0, which depends on \u03b5", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_180", "text": "of continuity, cf. ??. More precisely, using the characterization of continuity for a function\nf : E \u2192R given in ??, for any x0 \u2208E and any \u03b5 > 0 there exists \u03b4 > 0, which depends on \u03b5\nand x0, such that for any x \u2208E\nif |x \u2212x0| < \u03b4 =\u21d2|f(x) \u2212f(x0)| < \u03b5.\nIn ??, the for a \ufb01xed \u03b5 > 0, the existence of \u03b4 > 0 is no longer dependent on the choice of a\nbase point x0 \u2208E; instead, at this point such choice can be made independently (or rather,\nuniformly) from the points of E: it just depends on the choice of \u03b5.\n95", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_181", "text": "In view of the observation of the previous remark, we immediately have the following\nproposing showing that uniform continuity is a stornger property than continuity.\nProposition 6.51. If f : E \u2192R is uniformly continuous then it is continuous.\nExample 6.52. The function f(x) := x2 : R \u2192R is not uniformly continuous. On the other\nhand, we have already seen that it is continuous as it is a polynomial.\nIndeed, for any x, y \u2208R,\n|x2 \u2212y2| = |x + y| \u00b7 |x \u2212y|.\nSo, for any \u03b5 > 0 and \u03b4 > 0, we may chose x, y \u2208R such that |x + y| > 2\u03b5\n\u03b4 , and |x \u2212y| = \u03b4\n2 \u2013 to\ndo that, it su\ufb03ces to choose two real numbers x, y that are very large but very close to each\nother. Thus, it follows that\n|x \u2212y| < \u03b4\nand\n|x2 \u2212y2| > 2\u03b5\n\u03b4\n\u03b4\n2 = \u03b5.\nWe will see in the next section that if we consider a continuous function f over a closed bounded\ninterval [a, b] \u2013 rather than on an unbounded domain as in this case, where x2 is considered\nover R \u2013 then f is absolutely continuous.\nExample 6.53. We show that cos(x) : R \u2192R is uniformly continuous and hence continuous.\nIndeed,\n| cos(x) \u2212cos(y)| = 2\n\f\f\f\fsin\n\u0012x + y\n2\n\u0013\f\f\f\f\n\f\f\f\fsin\n\u0012x \u2212y\n2\n\u0013\f\f\f\f \u22642\n\f\f\f\fsin\n\u0012x \u2212y\n2\n\u0013\f\f\f\f \u22642|x \u2212y|.\nSo, if we set \u03b4 = \u03b5\n2, then we have\n|x \u2212y| \u2264\u03b4 \u21d2| cos(x) \u2212cos(y)| \u22642|x \u2212y| \u22642\u03b4 = 2\u03b5\n2 = \u03b5.\nThis result, together with ??, implies that also functions such as cos(x2), cos2(x), etc. are\ncontinuous.\nWe introduce now a property that makes it particularly easy to show that a function is\nuniformly continuous.\nDe\ufb01nition 6.54. A function f : E \u2192R is said to be Lipschitz if there exists a positive real\nnumber C such that for every x, y \u2208E, |f(x) \u2212f(y)| \u2264C|x \u2212y|.", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_182", "text": "uniformly continuous.\nDe\ufb01nition 6.54. A function f : E \u2192R is said to be Lipschitz if there exists a positive real\nnumber C such that for every x, y \u2208E, |f(x) \u2212f(y)| \u2264C|x \u2212y|.\nWhen the conditions of ?? are satis\ufb01ed we say that C is a Lipschitz constant for the Lipschitz\nfunction f.\nProposition 6.55. Let f : E \u2192R be a function which is Lipschitz with Lipschitz constant C,\nwhere E is an open interval. Then f is uniformly continuous on E; hence f is also continuous\non E.\nProof. For a \ufb01xed positive real number \u03b5 in ??, it su\ufb03ces to take \u03b4 := \u03b5\nC .\nExample 6.56. Let f : [0, 1] \u2192R be the function f(x) = x2. Then for any x, y \u2208[0, 1],\n|x2 \u2212y2| = |x \u2212y||x + y| \u2264C|x \u2212y|,\n(6.56.o)\nwhere C := sup{|x + y| | x, y \u2208[0, 1]. By de\ufb01nition, C \u22642 \u2013 it not hard to show that actually\nC = 2 \u2013 hence we can rewrite ?? as\n|x2 \u2212y2| = |x \u2212y||x + y| \u22642|x \u2212y|.\nHence, f is Lipschitz and thus uniformly continuous. Let us notice that\nWe will see in ?? that there exist functions that are uniformly continuous but not Lipschitz.\n96", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_183", "text": "6.1.8\nLeft and right continuity\nLastly, we introduce left and right continuity, and we use this to de\ufb01ne continuity on a closed\ninterval.\nDe\ufb01nition 6.57. Let f : E \u2192R be a function, and x0 \u2208E.\n(1) f is left continuous at x0, if lim\nx\u2192x\u2212\n0\nf(x) = f(x0).\n(2) f is right continuous at x0, if lim\nx\u2192x+\n0\nf(x) = f(x0).\nIn ?? we de\ufb01ned what it means to be continuous on an open interval. For functions the\ndomains of which are closed intervals the de\ufb01nition has to use left and right limits as well at\nthe two endpoints:\nDe\ufb01nition 6.58. A function f : [a, b] \u2192R is continuous if:\n(1) f is continuous at any point contained in (a, b);\n(2) f is left continuous at b; and,\n(3) f is right continuous at a.\nExample 6.59. The function f : [\u22121, 1] \u2192R de\ufb01ned as f(x) :=\n\u221a\n1 \u2212x2 is continuous. Indeed\nthis is true by the following (where we use that g(y) = \u221ay is continuous on R\u2217\n+, which will be\na consequence of our general theorem about the continuity of the inverse. Indeed, by applying\nthe statement of ?? to f(x) = x2 one obtains that f\u22121 = g is continuous on R\u2217\n+):\n(1) if \u22121 < c < 1, then\n\u221a\n1 \u2212x2 at c is continuous because\n\u221a\n1 \u2212x2 is the composition of \u221ay\nand 1 \u2212x2, and the latter is continuous at c and the former is continuous at 1 \u2212c2 (as\n1 \u2212c2 > 0).\n(2)\n\u221a\n1 \u2212x2 is left continuous at 1, because for all (xn) converging to 1 from the left we have\nlim\nn\u2192\u221e\np\n1 \u2212x2n = 0, as lim\nn\u2192\u221e1 \u2212x2\nn = 0, and lim\nx\u21920+\n\u221ay = 0 according to ??.\n(3)\n\u221a\n1 \u2212x2 is right continuous at \u22121 by almost verbatim the same argument as the previous\npoint, one only needs to take lim", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_184", "text": "n\u2192\u221e1 \u2212x2\nn = 0, and lim\nx\u21920+\n\u221ay = 0 according to ??.\n(3)\n\u221a\n1 \u2212x2 is right continuous at \u22121 by almost verbatim the same argument as the previous\npoint, one only needs to take lim\nn\u2192\u221exn = \u22121 instead of 1.\n6.1.9\nConsequences of Bolzano-Weierstrass\nIn this subsection we shall show how continuous functions de\ufb01ned over bounded closed intervals\nbehave nicely.\nThe proofs of all the results illustrated in this subsection heavily relies on\nBolzano-Weierstrass Theorem 4.75.\nAs we will be assuming, throughout this section, that\nthe domain D(f) of a function f is closed bounded interval, given a sequence (xn) \u2286D(f),\nby Theorem 4.75 we will always be able to assume that we can pass to a converging subsequence\n(xnk) \u2286(xn) whose limit belongs to D(f), since we are assuming D(f) is closed.\nWe start by showing that a continuous function de\ufb01ned over a closed bounded interval is\nalways uniformly continuous.\nTheorem 6.60. Let a, b \u2208R. If f : [a, b] \u2192R is continuous, then f is uniformly continuous.\nProof. Assume that f is not uniformly continuous.\nThen there is a \u03b5 > 0 such that for\nevery 1\nn there are xn and yn \u2208[a, b] such that |xn \u2212yn| \u22641\nn and |f(xn) \u2212f(yn)| > \u03b5. By\nBolzano-Weierstrass (Theorem 4.75) we may assume that lim\nn\u2192\u221exn = x0 \u2208[a, b]. However, then\nthe condition |xn \u2212yn| \u22641\nn yields that we have also lim\nn\u2192\u221eyn = x0. Using again |xn \u2212yn| \u22641\nn\ntogether with the continuity of f we obtain that |f(x0)\u2212f(x0)| \u2265\u03b5. This is a contradiction.\n97", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_185", "text": "Remark 6.61. For ?? to hold true, it is very important that the domain of f is a closed bounded\ninterval [a, b] for a, b \u2208R. We have already seen that the function f : R \u2192R, f(x) = x2 is not\nuniformly continuous on any interval of the form (a, +\u221e[, a \u2208R \u222a{\u2212\u221e}.\nExample 6.62. The following example shows that there exists uniformly continuous functions\nthat are not Lipschitz.\nLet us consider f : R+ \u2192R, f(x) = \u221ax. Let us \ufb01x a real number a > 0, and de\ufb01ne g: [0, a] \u2192R\nby g = f|[0,a], g(x) = \u221ax. ?? shows that g is uniformly continuous, as we are taking the domain\nof de\ufb01nition of g to be a closed bounded interval. To show that g is not Lipschitz, it su\ufb03ces to\nshow that for any C \u2208R\u2217\n+0 there exists s, t \u2208[0, a], s \u0338= t such that |g(s) \u2212g(t)| > C|s \u2212t|, or,\nequivalently, |g(s)\u2212g(t)|\n|s\u2212t|\n> C. Let us \ufb01x C > 0. Then, there exists t \u2208]0, 1[ such that\n1\n\u221a\nt > C,\nsince limx\u21920+\n1\n\u221ax = +\u221e. Taking s = 0, then |g(s)\u2212g(t)|\n|s\u2212t|\n=\n\u221a\nt\nt\n=\n1\n\u221a\nt > C, which is what we\nwanted to prove. It is not hard to show, that actually even f(x) = \u221ax is uniformly continuous\nbut not Lipschitz. The proof is left as an exercise.\nTheorem 6.63. If f : [a, b] \u2192R is continuous for some a, b \u2208R, then there are c, d \u2208[a, b]\nsuch that\nM := sup\nx\u2208[a,b]\nf(x) = max\nx\u2208[a,b] f(x) = f(c),\nm :=\ninf\nx\u2208[a,b] f(x) = min\nx\u2208[a,b] f(x) = f(d).", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_186", "text": "f(x) = max\nx\u2208[a,b] f(x) = f(c),\nm :=\ninf\nx\u2208[a,b] f(x) = min\nx\u2208[a,b] f(x) = f(d).\nRemark 6.64. The above theorem can be restated by saying that for a given function f : [a, b] \u2192\nR, if f is continuous then the range R(f) of f is a closed and bounded interval, R(f) = [c, d].\nProof. We only prove the existence of maxx\u2208[a,b] f(x), the case of minx\u2208[a,b] f(x) follows simi-\nlarly.\nFirst we prove that f is bounded from above, so that supx\u2208[a,b] f(x) must exist. Assume, by con-\ntradiction, that f is not bounded from above. That means that for each integer n > 0 there is\nxn \u2208[a, b] such that f(xn) \u2265n. As (xn) \u2286[a, b] is a bounded sequence, by Bolzano-Weierstrass\nTheorem 4.75, there exists a convergent subsequence (xnk) \u2286(xn). Set c := lim\nk\u2192\u221exnk. Then\nc \u2208[a, b], and the following chain of equalities yields a contradiction:\nR \u220bf(c) =\nlim\nk\u2192\u221ef(xnk)\n|\n{z\n}\nf : [a, b] \u2192R is continuous\n=\n+\u221e\n|{z}\nf(xnk)\u2265nk\u2265k\n.\nThis concludes the statement that f is bounded from above.\nHaving proved that f is bounded from above, sup\nx\u2208[a,b]\nf(x) makes sense. Thus, we must prove that\nsup\nx\u2208[a,b]\nf(x) = max\nx\u2208[a,b] f(x). By de\ufb01nition of supremum, there exists a sequence (yn) \u2286[a, b] such\nthat f(yn) \u2265M \u22121\nn. In particular, lim\nn\u2192\u221ef(yn) = M. By Bolzano-Weierstrass Theorem 4.75,", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_187", "text": "that f(yn) \u2265M \u22121\nn. In particular, lim\nn\u2192\u221ef(yn) = M. By Bolzano-Weierstrass Theorem 4.75,\nthere exists a convergent subsequence (ynk) \u2286(yn). Set c := lim\nk\u2192\u221eynk. Then c \u2208[a, b], and\nf(c) =\nlim\nk\u2192\u221ef(ynk)\n|\n{z\n}\nf : [a, b] \u2192R is continuous\n=\nlim\nn\u2192\u221ef(yn)\n|\n{z\n}\nProposition 4.72\n= M.\nRemark 6.65. The conclusion of the above theorem does not hold, if we do not assume that\nthe domain of f is a closed bounded interval [a, b], a, b \u2208R. For example, take f : R \u2192R,\nf(x) :=\n1\nx2+1. Then f does not attain its minimum as R(f) =]0, 1]: in fact, f(x) > 0, \u2200x \u2208R\nand f converges to 0 as x goes to \u00b1\u221e.\n98", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_188", "text": "Theorem 6.66 (Intermediate value theorem). Let a, b \u2208R. If f : [a, b] \u2192R is continuous,\nthen it takes each value between M := max\nx\u2208[a,b] f(x) and m := min\nx\u2208[a,b] f(x) at least once. More\nprecisely, for each c \u2208[m, M], there exists d \u2208[a, b] such that f(d) = c.\nIdea. We give only the idea and we refer to the precise proof to page 81-82 of the book.\nWe know by the above theorem that there are a\u2032, b\u2032 \u2208[a, b] such that m = f(a\u2032) and\nM = f(b\u2032). Hence, by replacing a with a\u2032 and b with b\u2032 (and some algebraic manipulation in\nthe case when b\u2032 < a\u2032), we may assume that f(a) = m, f(b) = M and m < c < M. Then, the\nidea is to consider\nS := {x \u2208[a, b]|f(x) < c}\nSet d := sup S. By the de\ufb01nition of sup, there is a sequence (xn) \u2286S converging to d from the\nleft and let yn be any sequence converging to d from the right. Applying continuity to the \ufb01rst\nsequence shows that f(d) \u2264c, and by applying it to the second one shows that f(d) \u2265c. So,\nf(d) = c.\nExample 6.67. In other words, ?? says that R(f) = [m, M]. Hence, for example, the image of\nan interval [a, b] via a continuous function f (whose domain contains [a, b]) cannot be [c, d]\u222a[e, d],\nc < d < e < f \u2013 that is, it cannot be the union of two disjoint intervals.\nExample 6.68. If f : R \u2192R is a continuous function, such that f(0) = 1, f(1) = 3, f(2) = \u22121,\nthen f attains the value 2 at least two times. Indeed, our assumptions say that the maximum", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_189", "text": "Example 6.68. If f : R \u2192R is a continuous function, such that f(0) = 1, f(1) = 3, f(2) = \u22121,\nthen f attains the value 2 at least two times. Indeed, our assumptions say that the maximum\nof f|[0,1] is at least 3 and the minimum of f|[0,1] is at most 1. Hence, ?? applied to f|[0,1] yields\nthat there is at least one c \u2208[0, 1] such that f(c) = 2. Similarly, ?? applied to f|[0,1] yields that\nthere is at least one d \u2208[1, 2] such that f(d) = 2. Furthermore, c \u0338= d, because c = d can only\nhappen if c = d = 1. However, f(c) = 3 \u0338= 2. Hence, c and d are two distinct real numbers at\nwhich f takes the value 2.\nWe will apply the above theoretical result to \ufb01nd solutions of equations of the form f(x) = x.\nFor example one can ask, if there is a solution of cos(x) = x for some x \u2208\n\u0002\n0, \u03c0\n2\n\u0003\n. ?? lets us\nanswer this question.\nCorollary 6.69 (Banach \ufb01xed point theorem for closed intervals). Let a, b \u2208R. If f : [a, b] \u2192\n[a, b] is a continuous function, then there exists x \u2208[a, b] such that f(x) = x.\nGiven a set S and function f : S \u2192S, an element s \u2208S such that f(s) = s is called a \ufb01xed\npoint.\nProof. Set g(x) := f(x) \u2212x. Then g(a) = f(a) \u2212a \u22650 and g(b) = f(b) \u2212b \u22640. So, by\nthe intermediate value theorem, there is a real number c \u2208[a, b] such that 0 = g(c). This is\nequivalent to f(c) = c.\nExample 6.70. The function cos(x)|[0, \u03c0\n2 ] :\n\u0002\n0, \u03c0\n2\n\u0003\n\u2192R can be regarded as cos(x)|[0, \u03c0\n2 ] :\n\u0002\n0, \u03c0\n2\n\u0003\n\u2192\n\u0002\n0, \u03c0\n2\n\u0003", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_190", "text": "equivalent to f(c) = c.\nExample 6.70. The function cos(x)|[0, \u03c0\n2 ] :\n\u0002\n0, \u03c0\n2\n\u0003\n\u2192R can be regarded as cos(x)|[0, \u03c0\n2 ] :\n\u0002\n0, \u03c0\n2\n\u0003\n\u2192\n\u0002\n0, \u03c0\n2\n\u0003\n, since R(cos(x)|[0, \u03c0\n2 ]) = [0, 1] \u2282\n\u0002\n0, \u03c0\n2\n\u0003\n. Then the above theorem says that there is a \ufb01xed\npoint x for which cos(x) = x.\n6.2\nMonotonicity and invertibility of continuous functions\nLet us recall the following de\ufb01nition.\nDe\ufb01nition 6.71. Let f : E \u2192R be a function, E \u2282R.\n(1) f is strictly increasing if f(x) < f(y) for all x < y in E.\n99", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_191", "text": "(2) f is strictly decreasing if f(y) > f(x) for all x < y in E.\n(3) f : E \u2192R is strictly monotone, if it is strictly increasing or strictly decreasing.\nCorollary 6.72. Let a, b \u2208R. If f : (a, b) \u2192R is strictly monotone and continuous, then the\nrange R(f) is an open interval.\nProof. Set\nS := sup{ f(x) | x \u2208(a, b) },\nI := inf{ f(x) | x \u2208(a, b) }.\nFirst, we show that S, I \u0338\u2208R(f). We only prove statement about S since the statement about\nI can be proven analogously.\nSo, let us assume by contradiction that S = f(c) for some\nc \u2208(a, b). Choose c < d \u2208(a, b) \u2013 here we are using that the interval is open!. Then, as f is\nstrictly increasing f(d) > f(c) = S, which is a contradiction with the de\ufb01nition of S.\nWe now show that R(f) = (I, S). Let us \ufb01x p \u2208(I, S). By the de\ufb01nition of S and I, there exist\nc, d \u2208(a, b) such that f(c) < p < f(d). Then the Intermediate Value ?? implies that p \u2208R(f),\nsince p \u2208[f(c), f(d)] \u2282R(f).\nTheorem 6.73. Let f : E \u2192F be a continuous function on an interval E. Then, f is strictly\nmonotone if and only if it is injective.\nProof. We do not prove this in class, read the proof from page 84-85 of the book.\nTheorem 6.74. If f : E \u2192F is continuous, strictly monotone and surjective function between\nintervals E, F. Then f\u22121 is also continuous.\nLet us recall that in the hypotheses of ??, the inverse function f\u22121 exists by ??.\nProof. We only show the case when E is an open interval (a, b), for some a, b \u2208R. In this case,", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_192", "text": "Let us recall that in the hypotheses of ??, the inverse function f\u22121 exists by ??.\nProof. We only show the case when E is an open interval (a, b), for some a, b \u2208R. In this case,\nF is also an open interval according to ??. Fix 0 < \u03b5 \u2208R and y0 \u2208F. Set x0 := f\u22121(y0).\nAccording to ??, there exist c, d \u2208R such that\nR(f|(x0\u2212\u03b5,x0+\u03b5)) = (c, d)\n(6.74.a)\nIn particular, there exists \u03b4 > 0 such that for every y \u2208F\nif |y \u2212y0| \u2264\u03b4 \u21d2y \u2208(c, d).\n(6.74.b)\nFor example, it su\ufb03ces to take \u03b4 := min{|c\u2212y0|,|d\u2212y0|}\n2\n: that is a choice of \u03b4 for which the above\ncondition is satis\ufb01ed.\nWe show that with the above choice of \u03b4 the de\ufb01nition of the continuity of f\u22121 at y0 is\nsatis\ufb01ed. That is, for every y \u2208F,\n|y \u2212y0| \u2264\u03b4 \u21d2y \u2208(c, d)\n|\n{z\n}\n??\n\u21d2|f\u22121(y) \u2212x0| \u2264\u03b5.\n|\n{z\n}\n??\nExample 6.75. Neither of the functions sin(x), cos(x), tan(x) and cotan(x) are invertible if\nconsidered as functions R \u2192R, as they are not injective in view of their periodicity. However, if\nwe restrict their domains adequately they become strictly montone, and then, according to ??,\ntheir inverses are continuous too:\n(1) arcsin(x) is the inverse of sin(x)|[\u2212\u03c0\n2 , \u03c0\n2 ]. For example, arcsin\n\u0000\u22121\n2\n\u0001\n= \u2212\u03c0\n6 , and arcsin\n\u0000\u22121\n2\n\u0001\n\u0338=\n7\u03c0\n6 , despite having sin\n\u0000 7\u03c0\n6\n\u0001\n= \u22121\n2 too.\n(2) arccos(x) is the inverse of cos(x)|[0,\u03c0].\n(3) arctan(x) is the inverse of tan(x)|[\u2212\u03c0\n2 , \u03c0\n2 ].\n100", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_193", "text": "Images/arcsin_gr.png\nFigure 13: f(x) = arcsin(x).\nImages/arccos_gr.png\nFigure 14: f(x) = arccos(x).\n7\nDIFFERENTIATION\nLet f : E \u2192R be a real valued one variable function. We would like to approximate it with\na linear one. That is, we would like to write\nf(x) = f(x0) + a(x \u2212x0) + r(x),\n(7.0.a)\nwhere a is a real number, and the error function r(x) is small in a neighborhood of x0. The\nquestion is: how small would we like r(x) to be so that we obtain a \u201cgood\u201d approximation?\nWhat kind of function do then realize formula ?? with our chosen conditions on r(x)?\nWell, if we want our approximation to at least compute the right value of f at x0, since\nlim\nx\u2192x0x \u2212x0 = 0,\nwe need to impose that lim\nx\u2192x0r(x) = 0. Even better, we would like r(x) to be smaller than a\nlinear function, otherwise the linear approximation in ?? will not be very precise. But what does\n101", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_194", "text": "Images/arctan_gr.png\nFigure 15: f(x) = arctan(x).\nit precisely mean that r(x) should be smaller than a linear function? The precise mathematical\nwording is the following:\nlim\nx\u2192x0\nr(x)\nx \u2212x0\n= 0.\n(7.0.b)\nEven better, taking r1(x) :=\nr(x)\nx\u2212x0 , we can rewrite the above condition as\nr(x) = (x \u2212x0)r1(x),\nand\nr1(x)\nx \u2212x0\n= 0.\n(7.0.c)\nThe graph of the function g(x) := f(x0)+a(x\u2212x0) is a line in the cartesian plane. Considering\nImages/etox_tang.png\nFigure 16: A di\ufb00erentiable function and the tangent line to the graph.\nthe graph of f(x), then if we can show that that for f the error function r(x) is smaller than\nlinear, that is, if r(x) satis\ufb01es the condition of ??, then the line representing the graph of g will\nbe tangent to the graph of f at the point (x0, f(x0)).\nAt this point the central question is: for what functions f do a and r(x) exists satisfy-\ning ??, ??, respectively?\n102", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_195", "text": "If both ?? and ?? hold, then\nr(x)\nx \u2212x0\n+ a = f(x) \u2212f(x0)\nx \u2212x0\n,\nx \u0338= x0,\n(7.0.d)\nand, moreover, by taking the limit for x \u2192x0 on both sides of this equation, using ??, it\nfollows that\na = lim\nx\u2192x0\nr(x)\nx \u2212x0\n+ a\n|\n{z\n}\nby ??\n= lim\nx\u2192x0\nf(x) \u2212f(x0)\nx \u2212x0\n|\n{z\n}\nby ??\n.\n(7.0.e)\nSo, the existence of the real number a together with the sub-linear15 behavior of the error\nterm described in ?? imply that the limit on the right of ?? exists and it is \ufb01nite. This discussion\nmotivates the following de\ufb01nition.\nDe\ufb01nition 7.1. Let f : E \u2192R be a function and let x0 \u2208E.\n(1) The function f is di\ufb00erentiable at x0, if the limit\nlim\nx\u2192x0\nf(x) \u2212f(x0)\nx \u2212x0\n(7.1.f)\nexists and it is \ufb01nite. We call the value of the limit in ?? the derivative of f at x0 and\nwe denote it by f\u2032(x0).\n(2) We say that f : E \u2192R is di\ufb00erentiable if it is di\ufb00erentiable at all points x0 \u2208E.\n(3) The function\nf\u2032 : {x \u2208E | f is di\ufb00erentiable at x} \u2192R,\nx 7\u2192f(x)\u2032\nis called the derivative function of f. the domain of f\u2032 is composed of all points of E\nwhere the above limit exists.\nRemark 7.2.\n(1) The derivative f\u2032(x0) of f at x0 can be also de\ufb01ned to be the unique real\nnumber c satisfying\nf(x) = f(x0) + c \u00b7 (x \u2212x0) + r(x),\n(7.2.g)\nwhere the function r(x) satis\ufb01es lim\nx\u2192x0\nr(x)\nx\u2212x0 = 0. As above, we can write r(x) = (x \u2212\nx0)r1(x) and lim", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_196", "text": "(7.2.g)\nwhere the function r(x) satis\ufb01es lim\nx\u2192x0\nr(x)\nx\u2212x0 = 0. As above, we can write r(x) = (x \u2212\nx0)r1(x) and lim\nx\u21920r1(x) = 0. In the reminder of this section, we will also use the notation\n\u03b51(x) to denote the function r1(x).\n(2) The de\ufb01nition of the derivative f\u2032(x0) in ?? can be summarized from a geometrical\nviewpoint by saying that the derivative is the limit (when it exists) for x \u2192x0 of the\nslope of the unique line passing through (x0, f(x0)) and the point (x, f(x)) corresponding\nto x on the graph.\nExample 7.3. Constant functions are di\ufb00erentiable everywhere. In fact, if f : R \u2192R, f(x) =\nC, \u2200x \u2208R, C \u2208R, then for x0 \u2208R\nlim\nx\u2192x0\nf(x) \u2212f(x0)\nx \u2212x0\n= lim\nx\u2192x0\nC \u2212C\nx \u2212x0\n= 0.\n15Sublinear stands for \u201cless than linear\u201d, that is, the condition de\ufb01ned in ??\n103", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_197", "text": "Images/secants.jpg\nExample 7.4. We show that (x2)\u2032 = 2x.\nFor any x0 \u2208R, we need to compute the limit lim\nx\u2192x0\nx2\u2212x2\n0\nx\u2212x0 . Thus,\nlim\nx\u2192x0\nx2 \u2212x2\n0\nx \u2212x0\n= lim\nx\u2192x0\n(x \u2212x0)(x + x0)\nx \u2212x0\n= lim\nx\u2192x0x + x0 = 2x0.\nExample 7.5. Similarly, if a \u2208Z+, then (xa)\u2032 = axa\u22121.\nIndeed,\nlim\nx\u2192x0\nxa \u2212xa\n0\nx \u2212x0\n= lim\nx\u2192x0\n(x \u2212x0)(xa\u22121 + xa\u22122x0 + xa\u22123x2\n0 + \u00b7 \u00b7 \u00b7 + x1xa\u22122\n0\n+ xa\u22121\n0\n)\nx \u2212x0\n= lim\nx\u2192x0xa\u22121 + xa\u22122x0 + xa\u22123x2\n0 + \u00b7 \u00b7 \u00b7 + xxa\u22122\n0\n+ xa\u22121\n0\n= lim\nx\u2192x0xa\u22121 + lim\nx\u2192x0xa\u22122x0 + lim\nx\u2192x0xa\u22123x2\n0 + \u00b7 \u00b7 \u00b7 + lim\nx\u2192x0xxa\u22122\n0\n+ lim\nx\u2192x0xa\u22121\n0\n|\n{z\n}\nby the addition rule for \ufb01nite limits and the fact that \u2200c \u2208N,\nlim\nx\u2192x0xc = xc\n0\n=axa\u22121\n0\n.\nExample 7.6. We show that sin(x)\u2032 = cos(x).\nlim\nx\u2192x0\nsin(x) \u2212sin(x0)\nx \u2212x0\n= lim\nx\u2192x0\n2 cos\n\u0000 x+x0\n2\n\u0001\nsin\n\u0000 x\u2212x0\n2\n\u0001\nx \u2212x0\n= lim\nx\u2192x0 cos\n\u0012x + x0\n2\n\u0013\n\u00b7 lim\nx\u2192x0\nsin\n\u0000 x\u2212x0\n2\n\u0001\nx\u2212x0\n2\n|\n{z\n}\nlimt\u21920\nsin(t)\nt\n=1\n= cos(x0),\nwhere we could break up the limit in the multiplication thanks to ??.\nExample 7.7. Similarly, cos(x)\u2032 = \u2212sin(x).\nlim\nx\u2192x0\ncos(x) \u2212cos(x0)\nx \u2212x0\n= lim\nx\u2192x0\n\u22122 sin", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_198", "text": "where we could break up the limit in the multiplication thanks to ??.\nExample 7.7. Similarly, cos(x)\u2032 = \u2212sin(x).\nlim\nx\u2192x0\ncos(x) \u2212cos(x0)\nx \u2212x0\n= lim\nx\u2192x0\n\u22122 sin\n\u0000 x+x0\n2\n\u0001\nsin\n\u0000 x\u2212x0\n2\n\u0001\nx \u2212x0\n= lim\nx\u2192x0 \u2212sin\n\u0012x + x0\n2\n\u0013\n\u00b7 lim\nx\u2192x0\nsin\n\u0000 x\u2212x0\n2\n\u0001\nx\u2212x0\n2\n|\n{z\n}\nlimt\u21920\nsin(t)\nt\n=1\n= \u2212sin(x0),\n104", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_199", "text": "where we could break up the limit in the multiplication thanks to ??.\nDi\ufb00erentiability is a stronger condition than continuity, as the following proposition readily\nshows.\nProposition 7.8. If f : E \u2192R is di\ufb00erentiable at x0, then it is continuous at x0.\nProof. This is a consequence of the following computation:\nlim\nx\u2192x0f(x) = lim\nx\u2192x0 f(x0) + (x \u2212x0)f(x0)\u2032 + r(x)\n|\n{z\n}\nby ??\n=\nf(x0) + lim\nx\u2192x0r(x)\n|\n{z\n}\n?? and limx\u2192x0 x\u2212x0=0\n=f(x0) + lim\nx\u2192x0\nr(x)\nx \u2212x0\n|\n{z\n}\n=0by ??\n\u00b7 lim\nx\u2192x0(x \u2212x0) = f(x0).\nExample 7.9. The viceversa of ?? is not true. That is, if f is continuous at x0, it does not\nnecessarily have to be di\ufb00erentiable.\nFor example, let us consider the function f(x) := |x|. The function f is continuous on R, in\nparticular, it is continuous at x0 = 0. On the other hand, f is not di\ufb00erentiable at 0, because\nthat would imply that lim\nx\u21920\n|x|\nx exists. However, since\nImages/abs_val_gr.png\nFigure 17: f(x) = |x|.\nlim\nx\u21920\u2212\n|x|\nx = lim\nx\u21920\u2212\n\u2212x\nx = \u22121 \u0338= 1 = lim\nx\u21920+\nx\nx = lim\nx\u21920+\n|x|\nx .\n?? implies that f is not di\ufb00erentiable at 0.\nExample 7.10.\n(1) The function f(x) :=\n(\n1\nx = 0\n0\nx \u0338= 0\nis not di\ufb00erentiable at 0, since it is\nnot continuous at 0. On the other hand, outside 0, f is di\ufb00erentiable, since over R\u2217, f is\nconstant.\n(2) The function f(x) :=\n(\n1\nx \u2208Q\n0\nx \u2208R \\ Q\nis not di\ufb00erentiable at any point of R since it is\nnot continuous at any point of R.\n105", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_200", "text": "7.1\nComputing derivatives\nIn this section we show how to compute derivatives. We \ufb01rst start by studying how deriva-\ntives behave with respect to the usual algebraic operations on R, and then continue by studying\nhow to compute derivatives with respect to composition and taking the inverse.\n7.1.1\nAddition\nProposition 7.11. If f, g: E \u2192R are di\ufb00erentiable at x0, then so is \u03b1f +\u03b2g for any \u03b1, \u03b2 \u2208R,\nand furthermore\n(\u03b1f + \u03b2g)\u2032(x0) = \u03b1f\u2032(x0) + \u03b2g\u2032(x0).\nProof.\n(\u03b1f + \u03b2g)\u2032(x0) = lim\nx\u2192x0\n(\u03b1f + \u03b2g)(x) \u2212(\u03b1f + \u03b2g)(x0)\nx \u2212x0\n=\u03b1 lim\nx\u2192x0\nf(x) \u2212f(x0)\nx \u2212x0\n|\n{z\n}\n=f\u2032(x0)\n+\u03b2 lim\nx\u2192x0\ng(x) \u2212g(x0)\nx \u2212x0\n|\n{z\n}\n=g\u2032(x0)\n= \u03b1f\u2032(x0) + \u03b2g\u2032(x0)\nwhere we could split the limit of the sum into the sum of the limits by the assumption on the\ndi\ufb00erentiability of f, g at x0, using ??.\nExample 7.12.\n\u00005x3 + 6x2\u0001\u2032 =\n\u00005x3\u0001\u2032 +\n\u00006x2\u0001\u2032 = 15x2 + 12x\n7.1.2\nMultiplication\nProposition 7.13. If f, g: E \u2192R are di\ufb00erentiable at x0, then so is f \u00b7 g, and furthermore\n(f \u00b7 g)\u2032(x0) = (fg\u2032 + f\u2032g)\nProof.\n(f \u00b7 g)\u2032(x0) = lim\nx\u2192x0\nf(x)g(x) \u2212f(x0)g(x0)\nx \u2212x0\n= lim\nx\u2192x0\nf(x)g(x) \u2212f(x)g(x0) + f(x)g(x0) \u2212f(x0)g(x0)\nx \u2212x0\n= lim\nx\u2192x0\nf(x)(g(x) \u2212g(x0)) + (f(x) \u2212f(x0))g(x0)\nx \u2212x0\n=\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8edlim", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_201", "text": "x \u2212x0\n= lim\nx\u2192x0\nf(x)(g(x) \u2212g(x0)) + (f(x) \u2212f(x0))g(x0)\nx \u2212x0\n=\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8edlim\nx\u2192x0f(x)\n|\n{z\n}\n=f(x0)\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\u00b7\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8edlim\nx\u2192x0\ng(x) \u2212g(x0)\nx \u2212x0\n|\n{z\n}\n=g\u2032(x0)\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8+ g(x0)\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8edlim\nx\u2192x0\nf(x) \u2212f(x0)\nx \u2212x0\n|\n{z\n}\n=f\u2032(x0)\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n=f(x0)g\u2032(x0) + g(x0)f\u2032(x0)\nwhere the fact that lim\nx\u2192x0f(x) = f(x0) follows from ?? and we could split the limits of sum and\nmultiplications using the di\ufb00erentiability of f, g at x0 and ??.\nExample 7.14.\n\u0000x2 cos(x)\n\u0001\u2032 =\n\u0000x2\u0001\u2032 cos(x) + x2 (cos(x))\u2032 = 2x cos(x) + x2(\u2212sin(x)) = x(2 cos(x) \u2212x sin(x))\n106", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_202", "text": "7.1.3\nDivision\nProposition 7.15. If f, g: E \u2192R are di\ufb00erentiable at x0, and g(x0) \u0338= 0, then f\ng is also\ndi\ufb00erentiable at x0, and furthermore\n\u0012f\ng\n\u0013\u2032\n(x0) =\n\u0012gf\u2032 \u2212fg\u2032\ng2\n\u0013\n(x0)\nIn particular,\n\u00121\ng\n\u0013\u2032\n(x0) =\n\u0012\u2212g\u2032\ng2\n\u0013\n(x0)\nProof. We compute the now familiar limit\nlim\nx\u2192x0\nf\ng (x) \u2212f\ng (x0)\nx \u2212x0\n= lim\nx\u2192x0\nf(x)g(x0) \u2212f(x0)g(x)\ng(x)g(x0)(x \u2212x0)\n= lim\nx\u2192x0\nf(x)g(x0) \u2212f(x0)g(x0) + f(x0)g(x0) \u2212f(x0)g(x)\ng(x)g(x0)(x \u2212x0)\n.\nGrouping together, in the denominator of the last member of the previous equation, those\nterms that depend on g(x0) and f(x0), respectively, we obtain,\nlim\nx\u2192x0\nf(x)g(x0) \u2212f(x0)g(x0) + f(x0)g(x0) \u2212f(x0)g(x)\ng(x)g(x0)(x \u2212x0)\n=\ng(x0)\ng(x0) \u00b7 lim\nx\u2192x0g(x)\n|\n{z\n}\n=g(x0)\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8edlim\nx\u2192x0\nf(x) \u2212f(x0)\nx \u2212x0\n|\n{z\n}\n=f\u2032(x0)\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n\u2212\nf(x0)\ng(x0) \u00b7 lim\nx\u2192x0g(x)\n|\n{z\n}\n=g(x0)\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8edlim\nx\u2192x0\ng(x) \u2212g(x0)\nx \u2212x0\n|\n{z\n}\n=g\u2032(x0)\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n=\n\u0012gf\u2032 \u2212fg\u2032\ng2\n\u0013\n(x0),\nwhere the fact that lim\nx\u2192x0g(x) = g(x0) follows from ?? and we could split the limits of sum and\nmultiplications using the di\ufb00erentiability of f, g at x0 and ??.", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_203", "text": "\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n=\n\u0012gf\u2032 \u2212fg\u2032\ng2\n\u0013\n(x0),\nwhere the fact that lim\nx\u2192x0g(x) = g(x0) follows from ?? and we could split the limits of sum and\nmultiplications using the di\ufb00erentiability of f, g at x0 and ??.\nExample 7.16. If b > 0 is an integer and x \u0338= 0, then:\n\u0012 1\nxb\n\u0013\u2032\n= \u2212\n\u0000xb\u0001\u2032\nx2b\n= \u2212bxb\u22121\nx2b\n= \u2212b\nxb+1 .\nThat is, by setting a = \u2212b we obtain (xa)\u2032 = axa\u22121. In particular, this shows that\n(xa)\u2032 = axa\u22121\nholds for all integer a, not just the non-negative ones.\nExample 7.17. If x \u0338= k\u03c0 + \u03c0\n2 for any k \u2208Z, or, equivalently, if cos(x) \u0338= 0, then\ntan(x)\u2032 =\n\u0012 sin(x)\ncos(x)\n\u0013\u2032\n=cos(x)(sin(x))\u2032 \u2212sin(x)(cos(x))\u2032\ncos(x)2\n=cos(x) cos(x) \u2212sin(x)(\u2212sin(x))\ncos(x)2\n=\n1\ncos2(x).\n107", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_204", "text": "7.1.4\nComposition of functions and derivatives\nProposition 7.18. Let f : E \u2192R, g: G \u2192R be functions such that that f(E) \u2286G. Assume\nthat f is di\ufb00erentiable at x0 \u2208E, and g is di\ufb00erentiable at f(x0) then g \u25e6f : E \u2192R is\ndi\ufb00erentiable at x0, and\n(g \u25e6f)\u2032(x0) = g\u2032(f(x)) \u00b7 f\u2032(x0).\nIdea of the proof.\nlim\nx\u2192x0\ng(f(x)) \u2212g(f(x0))\nx \u2212x0\n= lim\nx\u2192x0\ng(f(x)) \u2212g(f(x0))\nf(x) \u2212f(x0)\nf(x) \u2212f(x0)\nx \u2212x0\n= lim\nx\u2192x0\ng(f(x)) \u2212g(f(x0))\nf(x) \u2212f(x0)\n\u00b7 lim\nx\u2192x0\nf(x) \u2212f(x0)\nx \u2212x0\n= g\u2032(f(x0)) \u00b7 f\u2032(x0)\nExample 7.19. Let f(x) = x2 and g(y) = cos(y). Then f\u2032(x) = 2x and g\u2032(y) = \u2212sin(y). In\nparticular,\ncos(x2)\u2032 = (g \u25e6f)\u2032(x) = (g\u2032 \u25e6f)(x) \u00b7 f\u2032(x) = \u2212sin(x2)2x.\nExample 7.20. Let a \u2208Z, b \u2208Z+, f(x) = xa and g(y) = y\n1\nb . Then according to ?? and ??,\nf(x)\u2032 = axa\u22121 and g(y)\u2032 = 1\nby\n1\nb \u22121. Hence,\n\u0010\nx\na\nb\n\u0011\u2032\n=\n\u0010\n(xa)\n1\nb\n\u0011\u2032\n= (g \u25e6f)\u2032(x) = (g\u2032 \u25e6f)(x) \u00b7 f\u2032(x) = 1\nb (xa)\n1\nb \u22121 axa\u22121 = a\nb x\na\nb \u2212a+a\u22121 = a\nb x\na\nb \u22121\nSo, the formula (xr)\u2032 = rxr\u22121 holds also when r is any rational number (as it did for r \u2208Z in\n??).\n7.1.5", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_205", "text": "b x\na\nb \u2212a+a\u22121 = a\nb x\na\nb \u22121\nSo, the formula (xr)\u2032 = rxr\u22121 holds also when r is any rational number (as it did for r \u2208Z in\n??).\n7.1.5\nInversion of functions and derivatives\nProposition 7.21. Let f : E = (a, b) \u2192F be a bijective continuous function (so f is strictly\nmonotone, and f\u22121 exists and is continuous by ??), and let x0 \u2208E be such that f\u2032(x0) \u0338= 0.\nThen f\u22121 is di\ufb00erentiable at y0 := f(x0), and we have\n\u0000f\u22121\u0001\u2032 (y0) =\n1\nf\u2032(x0) =\n1\nf\u2032(f\u22121(y0))\nProof. The idea behind the proof of the proposition is that if we set y = f(x) and y0 = f(x0,\nwe have\nf\u22121(y) \u2212f\u22121(y0)\ny \u2212y0\n=\n1\ny\u2212y0\nf\u22121(y)\u2212f\u22121(y0)\n=\n1\nf(f\u22121(y))\u2212f(f\u22121(y0))\nf\u22121(y)\u2212f\u22121(y0)\n.\nCheck page 109 for the precise proof.\nExample 7.22. If f(x) = xb for some integer b \u22651, then f\u22121(y) =\nb\u221ay = y\n1\nb . So, f\u2032(x) =\nbxb\u22121, and\n\u0010\ny\n1\nb\n\u0011\u2032\n=\n\u0000f\u22121\u0001\u2032 (y) =\n1\nf\u2032(f\u22121(y)) =\n1\nb\n\u0010\ny\n1\nb\n\u0011b\u22121 = 1\nby\u2212b\u22121\nb\n= 1\nby\n1\nb \u22121.\nSo, for c = 1\nb (where b \u2208Z\u2217\n+), the formula for (yc)\u2032 = cyc\u22121. That is the formula is the same\nas in the case of c being an integer.\nExample 7.23. Let f(x) = sin(x)|[\u2212\u03c0\n2 , \u03c0\n2 ] :\n\u0002\n\u2212\u03c0\n2 , \u03c0\n2\n\u0003\n\u2192[\u22121, 1].\nThen f is invertible and\nf\u22121(y) = arcsin(y). Also, f\u2032(x) = cos(x)|[\u2212\u03c0\n2 , \u03c0\n2 ], thus, for any y \u2208] \u22121, 1[,", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_206", "text": "2\n\u0003\n\u2192[\u22121, 1].\nThen f is invertible and\nf\u22121(y) = arcsin(y). Also, f\u2032(x) = cos(x)|[\u2212\u03c0\n2 , \u03c0\n2 ], thus, for any y \u2208] \u22121, 1[,\narcsin\u2032(y) =\n1\ncos(arcsin(y)) =\n1\np\n1 \u2212sin2(arcsin(y))\n=\n1\np\n1 \u2212y2 .\n108", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_207", "text": "7.1.6\nThe exponential function\nFor our last example in this section, we will discuss in details the exponential and logarithmic\nfunctions. Let us remind the reader that we de\ufb01ned\nex := lim\nn\u2192\u221e\n\u0010\n1 + x\nn\n\u0011n\n.\nDe\ufb01nition 7.24. For x \u2208R, we de\ufb01ne\nex :=\n\u221e\nX\nk=0\nxk\nk! .\nRemark 7.25. Applying ??, to x = 0 yields e0 = 1. Furthermore, according to ??, e1 = e.\nProposition 7.26. For any x, y \u2208R, ex+y = ex \u00b7 ey.\nProof. This is an exercise in Week 10 exercise sheet.\nCorollary 7.27. For any x \u2208R, e\u2212x =\n1\nex .\nProof.\nex \u00b7 e\u2212x = ex+(\u2212x)\n| {z }\n??\n= e0 =\n1\n|{z}\n??\n.\nDividing by ex yields the statement (ex cannot be 0, since then ex\u00b7e\u2212x = 1 could not hold).\nCorollary 7.28. For every x \u2208R, ex > 0.\nProof. For x \u22650, then all the terms in the in\ufb01nite sum in ?? is at least zero, and the \ufb01rst term\nis 1. This implies the statement for x \u22650.\nSo, we may assume from now that x < 0. We have ex =\n1\ne\u2212x by ??. However, as now\n\u2212x > 0 holds, the previous paragraph tells us that e\u2212x > 0, and hence also\n1\ne\u2212x > 0.\nProposition 7.29. (ex)\u2032 = ex\nProof. We need to show that\nlim\nx\u2192x0\nex \u2212ex0\nx \u2212x0\n= ex0.\nThis is equivalent to showing that\n0 = lim\nx\u2192x0\nex \u2212ex0\nx \u2212x0\n\u2212ex0 =\n\u0012\nlim\nx\u2192x0\nex\u2212x0 \u22121\nx \u2212x0\n\u22121\n\u0013\nex0.\nBy setting y = x \u2212x0, what we need to show is that\nlim\ny\u21920\ney \u22121\ny\n\u22121 = 0.\n(7.29.a)\nHowever, for 0 < |y| \u22641:\n0 \u2264\n\f\f\f\f\ney \u22121\ny\n\u22121\n\f\f\f\f =\n\f\f\f\f\f\f\f\f\f\f\n\u221e\nX\nk=0\nyk\nk! \u22121\ny\n\u22121\n\f\f\f\f\f\f\f\f\f\f\n=\n\f\f\f\f\f\n\u221e\nX\nk=2\nyk\u22121\nk!\n\f\f\f\f\f \u2264\n\u221e\nX\nk=2", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_208", "text": "\u22121 = 0.\n(7.29.a)\nHowever, for 0 < |y| \u22641:\n0 \u2264\n\f\f\f\f\ney \u22121\ny\n\u22121\n\f\f\f\f =\n\f\f\f\f\f\f\f\f\f\f\n\u221e\nX\nk=0\nyk\nk! \u22121\ny\n\u22121\n\f\f\f\f\f\f\f\f\f\f\n=\n\f\f\f\f\f\n\u221e\nX\nk=2\nyk\u22121\nk!\n\f\f\f\f\f \u2264\n\u221e\nX\nk=2\n|y|k\u22121\nk!\n\u2264|y|\n\u221e\nX\nk=2\n|y|k\u22122\n(k \u22122)!\n= |y|\n\u221e\nX\nk=0\n|y|k\nk! \u2264|y|e\nHence, by Squeeze Theorem ????, it follows that ?? holds indeed.\n109", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_209", "text": "Proposition 7.30. We have\nlim\nx\u2192+\u221eex = +\u221e, and\nlim\nx\u2192\u2212\u221eex = 0.\nProof. According to ??, for all x > 0, ex \u22651 + x. As\nlim\nx\u2192+\u221e1 + x = +\u221e, squeeze (point ?? of\n??) shows that\nlim\nx\u2192+\u221eex = +\u221e. Then ??, ?? and point ?? of ?? show that\nlim\nx\u2192\u2212\u221eex = 0.\nProposition 7.31. The function ex : R \u2192R is strictly increasing.\nProof. Choose y > x \u2208R. We have to show that ey > ex. This is shown by the following\ncomputation:\ney \u2212ex = (ey\u2212x \u22121)\n|\n{z\n}\n> 1 by ??, us-\ning y > x\n\u00b7\nex\n|{z}\n>\n0 by\n??\n> 0\nCorollary 7.32. The range R(f) of f := ex : R \u2192R is (0, +\u221e) = R\u2217\n+.\nProof. Follows immediately from ?? and ??.\nDe\ufb01nition 7.33.\n(1) We de\ufb01ne the (natural) logarithm function log(x): R\u2217\n+ \u2192R to be the inverse of the\nexponential function f(x) = ex.\n(2) For any a \u2208R\u2217\n+, the a-based exponential functions ax is de\ufb01ned as\nax := ex\u00b7log(a).\nThe logarithm in base a of x is the inverse function of the a-based exponential function\nax,\nloga(x) := log(x)\nlog(a).\n(3) For any a \u2208R, the a-th power functions is de\ufb01ned as\nxa := ea\u00b7log(x).\nRemark 7.34. In the special cases where the functions of ?? have been already de\ufb01ned (so xa\nwhen a \u2208Q, and ax when a = e), they agree with the previously de\ufb01ned functions. This will\nbe an exercise on the exercise sheet.\nExample 7.35.\n(1) If f(x) = ex, then f\u22121(x) = log(x) and f\u2032(x) = ex (??). Hence:\n(log(x))\u2032 =\n1\nelog(x) = 1\nx.\n(2) Let h: R\u2217", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_210", "text": "(1) If f(x) = ex, then f\u22121(x) = log(x) and f\u2032(x) = ex (??). Hence:\n(log(x))\u2032 =\n1\nelog(x) = 1\nx.\n(2) Let h: R\u2217\n+ \u2192R, h(x) := xx. Then, de\ufb01nining f(x) = x log(x), g(y) := ey,\nh(x) = (f \u25e6g)(x).\nThus,\nh\u2032(x) = (f \u25e6g)\u2032(x) = f\u2032(g(x))g\u2032(x) = xx(log(x) + 1).\n110", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_211", "text": "De\ufb01nition 7.36.\n(1) The hyperbolic trigonometric functions are de\ufb01ned below, and they are\ncalled hyperbolic sine/cosine/tangent/cotangent:\nsinh(x) := ex \u2212e\u2212x\n2\ncosh(x) := ex + e\u2212x\n2\ntanh(x) := sinh(x)\ncosh(x)\ncoth(x) := cosh(x)\nsinh(x)\nThe domains of all the above functions is R, except for coth which is de\ufb01ned over R\u2217, since\nsinh(0) = 0.\nProposition 7.37. We have:\n(1) sinh(x)\u2032 = cosh(x)\n(2) cosh(x)\u2032 = sinh(x)\n(3) tanh(x)\u2032 =\n1\ncosh(x)2\n(4) coth(x)\u2032 =\n\u22121\nsinh(x)2\n(5) (xa)\u2032 = axa\u22121\n(6) (ax)\u2032 = log(a) \u00b7 ax\n(7) loga(x)\u2032 =\n1\nlog(a)\u00b7x\nThe proof is left as an exercise.\n7.2\nOne sided derivatives\nDe\ufb01nition 7.38. If f : E \u2192R is a function and x0 \u2208E for which the then we say that the\nleft (resp. right) derivative of f exists at x0 if the function\nf(x) \u2212f(x0)\nx \u2212x0\n: E \\ {x0} \u2192R\nadmits a left (resp. right limit). The value of this limit is then the left (resp. right) derivative.\nExample 7.39. For f(x) = |x| at x = 0 the left derivative is \u22121 and the right derivative is 1.\nIn fact,\nlim\nx\u21920+\nf(x) \u2212f(0)\nx \u22120\n= x\nx = 1,\nlim\nx\u21920\u2212\nf(x) \u2212f(0)\nx \u22120\n= \u2212x\nx = \u22121.\nAs in the case of left and right limits, we can use left and right derivatives to decide when\na function is di\ufb00erentiable at a given point.\nProposition 7.40. Let f : E \u2192R be a function and x0 \u2208E a real number.\nThen f is", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_212", "text": "= \u2212x\nx = \u22121.\nAs in the case of left and right limits, we can use left and right derivatives to decide when\na function is di\ufb00erentiable at a given point.\nProposition 7.40. Let f : E \u2192R be a function and x0 \u2208E a real number.\nThen f is\ndi\ufb00erentiable at a point x0 if and only if both its left and right derivatives exist and they agree.\nFurthermore, then the value of the derivative is the same as the common value of the left and\nthe right derivatives.\nProof. This is an immediate consequence of ??, ?? and ??.\n111", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_213", "text": "Example 7.41.\n(1) Let us consider the function f : R \u2192R de\ufb01ned by\nf(x) :=\n(\nx2,\nx \u22650\nx3,\nx < 0 .\nThe function f is di\ufb00erentiable at 0 with derivative f\u2032(0) = 0. Indeed,\nlim\nh\u21920+\nf(0 + h) \u2212f(0)\nh\n= lim\nh\u21920+\nh2 \u22120\nh\n= lim\nh\u21920+ h = 0.\nSimilarly,\nlim\nh\u21920\u2212\nf(0 + h) \u2212f(0)\nh\n= lim\nh\u21920\u2212\nh3 \u22120\nh\n= lim\nh\u21920\u2212h2 = 0.\nSince the left derivative and the right derivative exist and agree at x0 = 0, we can conclude\nthat\nlim\nh\u21920\nf(0 + h) \u2212f(0)\nh\n= 0.\nThus, f is di\ufb00erentiable at 0 with f\u2032(0) = 0.\n(2) Let us consider the function f : R \u2192R de\ufb01ned by\nf(x) :=\n(\nx + 1\nfor x \u22650\nx\nfor x < 0 .\nThe function f is not continuous at x0 = 0, as lim\nx\u21920\u2212f(x) = 0 whereas lim\nx\u21920+f(x) = 1. As a\ndi\ufb00erentiable function is continuous, then f is not di\ufb00erentiable at 0. On the other hand,\nf is di\ufb00erentiable outside of 0, since on a su\ufb03ciently small neighborhood of any point\nx0 \u0338= 0, f is given by a linear function and we know that linear functions are di\ufb00erentiable.\n7.3\nHigher derivatives\nGiven a function f, we may try to iterate inductively the process of taking the derivative\nof f, thus obtaining what we will call the second derivative of f, the third, derivative of f, etc.\nDe\ufb01nition 7.42. Let f : E \u2192R be a function.\n(1) The second derivative f\u2032\u2032 of f is the function\nf\u2032\u2032 : {x \u2208E | f\u2032 is di\ufb00erentiable at x} \u2192R\nx 7\u2192f\u2032\u2032(x) := (f\u2032)\u2032(x).\n(2) Assume that the n-th derivative f(n) of f has been de\ufb01ned. Then the (n+1)-st derivative\nf(n+1) of f is the function\nf(n+1) : {x \u2208E | f(n) is di\ufb00erentiable at x} \u2192R", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_214", "text": "f(n+1) of f is the function\nf(n+1) : {x \u2208E | f(n) is di\ufb00erentiable at x} \u2192R\nx 7\u2192f(n+1)(x) := (f(n))\u2032(x).\nThe n-th derivative of f at x \u2208E is denoted by f(n)(x). For the \ufb01rst, second and third\nderivative of f, we will adopt the notation f\u2032, f\u2032\u2032, f\u2032\u2032\u2032 rather than f(1), f(2), f(3).\nExample 7.43.\n(1) The second derivative of f(x) = arctan(x) is f\u2032\u2032(x): R \u2192R,\nf\u2032\u2032(x) = (f\u2032)\u2032(x) = (\n1\n1 + x2 )\u2032 =\n\u22122x\n(1 + x2)2 .\n112", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_215", "text": "(2) Let us consider the function f : R\u2217\u2192R de\ufb01ned by f(x) = e\n1\nx . Then,\nf\u2032\u2032(x) = (f\u2032)\u2032(x) = (e\n1\nx \u00b7 (\u22121\nx2 ))\u2032\n= (e\n1\nx )\u2032 \u00b7 (\u22121\nx2 ) + e\n1\nx \u00b7 (\u22121\nx2 )\u2032 = e\n1\nx ( 1\nx4 + \u00b7 2\nx3 ).\nExample 7.44. Here we show an example of a function f(x) such that f(x) is di\ufb00erentiable\ntwo times, but not three times. That is, f\u2032(x) and f\u2032\u2032(x) exist for every x \u2208R, but f\u2032\u2032\u2032(0) does\nnot exist.\nLet us consider f(x) := |x3|. Then, f\u2032(x) exists for all x \u2208R, and:\nf\u2032(x) =\n(\n3x2\nx \u22650\n\u22123x2\nx < 0.\nThis is immediate at x \u0338= 0 from the formula\nf(x) =\n\u001a x3\nfor x \u22650\n\u2212x3\nfor x \u22640\nTo conclude the above \ufb01rst claim we just have to compute the left and the right derivatives of\nf(x) at x = 0, and show that both are 0. Indeed:\nlim\nx\u21920+\nf(x) \u2212f(0)\nx \u22120\n= lim\nx\u21920+\nx3 \u22120\nx\n= lim\nx\u21920+ x2 = 0,\nand\nlim\nx\u21920\u2212\nf(x) \u2212f(0)\nx \u22120\n= lim\nx\u21920\u2212\n\u2212x3 \u22120\nx\n= lim\nx\u21920\u2212\u2212x2 = 0.\nThis concludes our \ufb01rst claim.\nSimilarly, one can prove that f\u2032\u2032(x) exists for all x \u2208R and\nf\u2032\u2032(x) =\n\u001a 6x\nfor x \u22650\n\u22126x\nfor x < 0\nWith other words, f\u2032\u2032(x) = 6|x|. However, as |x| is not di\ufb00erentiable at x = 0, we obtain that\nf\u2032\u2032\u2032(0) does not exist.\nDe\ufb01nition 7.45. f : E \u2192R is called a function of class Cn if its \ufb01rst n derivatives f\u2032, f\u2032\u2032, . . . , f(n)\nexists and are all continuous at all points x0 \u2208E.", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_216", "text": "De\ufb01nition 7.45. f : E \u2192R is called a function of class Cn if its \ufb01rst n derivatives f\u2032, f\u2032\u2032, . . . , f(n)\nexists and are all continuous at all points x0 \u2208E.\nNotation 7.46. To denote that a function f : E \u2192R is a Cn function, we will use the notation\nf \u2208Cn(E, R). We will write f \u2208C\u221e(E, R) if f \u2208Cn(E, R), \u2200n \u2208N, and we will say that f is\na C\u221efunction.\nExample 7.47.\n(1) According to ??, x, x2, etc.\nare Cn for all n, that is they are C\u221e\nfunction. More precisely, for a \u2208N, de\ufb01ning f(x) = xa then\nf(n)(x) =\n(\n0\nif n > a\na \u00b7 (a \u22121) \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 (a \u2212n + 1)xa\u2212n\nfor n \u2264a\n(2) We can repeat the same computation for f : [0, +\u221e) \u2192R, f(x) := x\u03b1 = elog(x)\u03b1, \u03b1 \u2208\nR \\ N. Then f(x) = \u03b1 \u00b7 (\u03b1 \u22121) \u00b7 (\u03b1 \u22122) \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 (\u03b1 \u2212n + 1)x\u03b1\u2212n, x > 0.\n(3) |x| : R \u2192R is not C1, cf. ??.\n(4) |x3| : R \u2192R is C2 but not C3, cf. ??.\n113", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_217", "text": "7.4\nLocal and global extrema\nDe\ufb01nition 7.48. Let f : E \u2192R be a function and let x0 \u2208E.\n(1) The function f admits a point of local maximum at x0 if there is a real number \u03b4 > 0\nsuch that ]x0 \u2212\u03b4, x0 + \u03b4[\u2282E and for every x \u2208E if |x \u2212x0| < \u03b4 then f(x) \u2264f(x0).\n(2) The function f has a point of local minimum at x0 if there is a real number \u03b4 > 0 such\nthat ]x0 \u2212\u03b4, x0 + \u03b4[\u2282E and for every x \u2208E if |x \u2212x0| < \u03b4 then f(x) \u2265f(x0).\n(3) We say that x0 \u2208E is a point of local extremum for f if it is either a point of local\nminimum or of local maximum.\n(4) The function f has a point of global maximum at x0 if f(x0) \u2265f(x), for all x \u2208E.\n(5) The function f has a point of global minimum at x0 if f(x0) \u2264f(x), for all x \u2208E.\nRemark 7.49. We shall also say that f admits a local maximum (resp. local minimum, local\nextremum, global maximum, global minimum) at x0 to indicate that property (1) (resp. (2),\n(3), (4), (5)) de\ufb01ned above is satis\ufb01ed.\nRemark 7.50. Let f : E \u2192R be a function and x0 \u2208E. If x0 is a point of global maximum (resp.\nglobal minimum) for f and E contains a neighborhood of x0 of the form ]x0 \u2212\u03b4, x0 + \u03b4[, \u03b4 > 0,\nthen x0 is also a point of local maximum (resp. local minimum) for f.\nExample 7.51. Let us consider the function f\nThe following proposition shows that any point of local extremum for a function f coincides\nwith a zero of the derivative f\u2032.\nProposition 7.52. If f : E \u2192R is di\ufb00erentiable at x0, and f admits a local extremum at x0,\nthen f\u2032(x0) = 0.", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_218", "text": "with a zero of the derivative f\u2032.\nProposition 7.52. If f : E \u2192R is di\ufb00erentiable at x0, and f admits a local extremum at x0,\nthen f\u2032(x0) = 0.\nProof. We present the local maximum case, as one just need to reverse a few signs, to modify\nthe proof to obtain from it the case of local minimum.\nHence, let us assume that x0 \u2208E is a point of local by ??, there is a real number \u03b4 > 0 such\nthat\n|x \u2212x0| \u2264\u03b4 \u21d2f(x) \u2264f(x0).\n(7.52.a)\nHowever, then\nlim\nx\u2192x+\n0\nf(x) \u2212f(x0)\n(x \u2212x0)\n\u2264lim\nx\u2192x+\n0\n0\n(x \u2212x0)\n|\n{z\n}\nx>x0, and ??\n= 0,\n(7.52.b)\nand\nlim\nx\u2192x\u2212\n0\nf(x) \u2212f(x0)\n(x \u2212x0)\n\u2265lim\nx\u2192x\u2212\n0\n0\n(x \u2212x0)\n|\n{z\n}\nx<x0, and ??\n= 0.\n(7.52.c)\nAs f(x) is di\ufb00erentiable, at x0 the two above limits agree (??). Hence the following stream of\ninequalities have to be all equalities, which conclude our proof:\n0 \u2264lim\nx\u2192x\u2212\n0\nf(x) \u2212f(x0)\n(x \u2212x0)\n|\n{z\n}\n??\n= f\u2032(x0) = lim\nx\u2192x+\n0\nf(x) \u2212f(x0)\n(x \u2212x0)\n\u22640.\n|\n{z\n}\n??\n114", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_219", "text": "Images/x2_gr.png\nFigure 18: f(x) = x2 has a global minimum at x = 0.\nExample 7.53.\n(1) For f(x) = x2, we have f\u2032(x) = 2x. Hence, f\u2032(x) = 0 if and only if\nx = 0. So, x = 0 is the only option for stationary point of f, thus also for a point of local\nextremum, and, indeed, f admits a local (and global) minimum at x = 0.\n(2) For f(x) = x3, we have f\u2032(x) = 3x2. So f\u2032(x) = 0 if and only if x = 0 (as in the previous\ncase). However, f(0) = 0 is not a local extremum. This underlines that ?? yields only a\nnecessary, but not a su\ufb03cient condition for having a local extremum.\nImages/x3_gr.png\nFigure 19: f(x) = x3, f\u2032(0) = 0, but 0 is not a point of local extremum for f.\nDe\ufb01nition 7.54. Let f : E \u2192R be a function and let x0 \u2208E. Assume that f is di\ufb00erentiable\nat x0. Then, we say that x0 a stationary point (for f) if f\u2032(x0) = 0.\nWe have just seen that when f\u2032(x0) = 0, we cannot necessarily conclude that x0 is a point\nof local extremum for f. On the other hand, if the domain of f is a closed bounded interval\n[a, b], then ?? implies that f admits both a global maximum and a global minimum in [a, b].\nTherefore, using ??, a point of global maximum for f can only be:\n\u25e6either a stationary point, x \u2208(a, b) such that f\u2032(x) = 0, or\n115", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_220", "text": "\u25e6x = a, or x = b.\nSimilarly, a point of global minimum for f can only be:\n\u25e6either a stationary point, x \u2208(a, b) such that f\u2032(x) = 0, or\n\u25e6x = a, or x = b.\nHence, to \ufb01nd the value of the global maximum and the global minimum of f over the interval\n[a, b], it su\ufb03ces to compute:\nmax\nx\u2208[a,b] f = sup{f(x) | x = a or x = b or f\u2032(x) = 0},\nmin\nx\u2208[a,b] f = inf{f(x) | x = a or x = b or f\u2032(x) = 0}.\nThis procedure guves an algorithmic approach to \ufb01nding the values of global extrema.\nExample 7.55. We compute the global minimum and the global maximum of\nf(x) = 4\n3x3 + 3\n2x2 \u2212x + 2\non the closed bounded interval\n\u0002\n\u22122, 1\n2\n\u0003\n. By the discussion in the paragraph before the example\nwe have to compute:\nf\u2032(x) = 4x2 + 3x \u22121,\nand then \ufb01nd the solutions of the equation f\u2032(x) = 0. These are:\nx = \u22123 \u00b1\n\u221a\n25\n8\n= \u22123 \u00b1 5\n8\n= \u22121, and x = 1\n4.\nThen, we have to compute the function values at these two points, and at the endpoints of our\ninterval. The point, where the function value is the maximal yields the maximum and where\nthe function value is minimal yields the minimum of f(x) on\n\u0002\n\u22122, 1\n2\n\u0003\n:\nvalue of x\nf(x)\n\u22122\n\u22124\n38 + 3\n24 + 2 + 2 = 36\u221264\n6\n+ 4 = 4 \u221228\n6 = \u22124\n6\n\u22121\n\u22124\n3 + 3\n2 + 1 + 2 = 9\u22128\n6\n+ 3 = 3 + 1\n6\n1\n4\n4\n3\u00b764 +\n3\n2\u00b716 \u22121\n4 + 2 = 4+18\u221248\n192\n+ 2 = \u221226\n192 + 2 = 2 \u221213\n96\n1\n2\n4\n3\u00b78 +\n3\n2\u00b74 \u22121\n2 + 2 = 4+9\u221212\n24\n+ 2 = 2 + 1\n24\nSo, f(x) on\n\u0002\n\u22122, 1\n2\n\u0003\ntakes its minimum at x = \u22122 and its maximum at x = \u22121.\n7.5", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_221", "text": "96\n1\n2\n4\n3\u00b78 +\n3\n2\u00b74 \u22121\n2 + 2 = 4+9\u221212\n24\n+ 2 = 2 + 1\n24\nSo, f(x) on\n\u0002\n\u22122, 1\n2\n\u0003\ntakes its minimum at x = \u22122 and its maximum at x = \u22121.\n7.5\nRolle\u2019s and Mean Value theorem\nWe are ready to state and prove the two main results of this chapter: Rolle\u2019s theorem (??)\nand the Mean value theorem (??).\nTheorem 7.56 (Rolle\u2019s Theorem). Let f : [a, b] \u2192R is a continuous function, for a, b \u2208R.\nAssume that f is di\ufb00erentiable on (a, b), and that f(a) = f(b). Then there exists c \u2208(a, b)\nsuch that f\u2032(c) = 0.\nProof. If f is constant on [a, b], then f\u2032(x) = 0, \u2200x \u2208(a, b) and so we are done.\nHence, we can assume that f is not constant. Then, according to ??, f has both a maximum\nand a minimum on [a, b]. However, as f is non-constant, one of these values have to be not equal\nto f(a) = f(b). Formally, this means that there is a c \u2208[a, b], such that f(c) \u0338= f(a) = f(b).\nIn particular, we must have a < c < b. Then, f is di\ufb00erentiable at c, and as f has a (local)\nextremum at c, we have f\u2032(c) = 0 according to ??.\n116", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_222", "text": "Remark 7.57. The di\ufb00erentiability assumption is needed for ?? to hold. For example, consider-\ning f(x) := |x| on [\u22121, 1], then f(\u22121) = f(1), but there is no point in [\u22121, 1] at the derivative\nof f is 0.\nTheorem 7.58 (Mean value theorem). Let f : [a, b] \u2192R be a continuous function, a, b \u2208R.\nAssume that f is di\ufb00erentiable on (a, b). Then there exists c \u2208(a, b) such that f\u2032(c)(b \u2212a) =\nf(b) \u2212f(a).\nProof. Apply ?? to g(x) = f(x) \u2212f(a) \u2212f(b)\u2212f(a)\nb\u2212a\n(x \u2212a).\nExample 7.59. Let f be a C1 function (see ??) on [\u22121, 1] such that f(\u22121) = 2, f(0) = 4 and\nf(1) = 3. We show using ?? that there is a c \u2208] \u22121, 1[ such that f\u2032(c) = 1:\n(1) Applying ?? to f|[\u22121,0] we obtain that there is an a \u2208]\u22121, 0[ such that f\u2032(a) =\n4\u22122\n0\u2212(\u22121) = 2.\n(2) Applying ?? to f|[0,1] we obtain that there is a b \u2208]0, 1[ such that f\u2032(b) = 3\u22124\n1\u22120 = \u22121.\n(3) As f is C1, f\u2032 is continuous. Hence, ?? implies that there is a c \u2208(a, b) \u2286] \u22121, 1[ such\nthat f\u2032(c) = 1.\nCorollary 7.60. Let f, g: [a, b] \u2192R be continuous functions, a, b \u2208R. Assume that f, g are\ndi\ufb00erentiable over (a, b) and that f\u2032(x) = g\u2032(x) for each x \u2208(a, b). Then there exists a real\nnumber C such that f(x) = g(x) + C.", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_223", "text": "di\ufb00erentiable over (a, b) and that f\u2032(x) = g\u2032(x) for each x \u2208(a, b). Then there exists a real\nnumber C such that f(x) = g(x) + C.\nProof. By taking h(x) := f(x) \u2212g(x), it su\ufb03ce to apply ??.\nLemma 7.61. Let h: [a, b] \u2192R a continuous function which is di\ufb00erentiable on (a, b). Assume\nthat h\u2032(x) = 0, for all x \u2208(a, b). Then, h is a constant function.\nRecall that by saying that h is a constant function we simply mean that \u2200x \u2208[a, b], h(x) = C\nfor some \ufb01xed real number C \u2208R (indipendent of x).\nProof. Assume that h is not constant. Then, there exists c, d \u2208[a, b], c < d such that h(c) \u0338=\nh(d). Then, the Mean Value ?? implies that there exists e \u2208(c, d), such that h\u2032(e) = h(d)\u2212h(c)\nd\u2212c\n\u0338=\n0; nonetheless, this contradicts our assumption that h\u2032().\n7.5.1\nMonotone functions and di\ufb00erentials\nWe can apply the Mean Value Theorem to characterize the derivative of monotone (di\ufb00eren-\ntiable) functions.\nCorollary 7.62. Let f : [a, b] \u2192R be a continuous function, a, b \u2208R.\nAssume that f is\ndi\ufb00erentiable on (a, b). Then,\n(1) f is increasing (resp. decreasing) if and only if f\u2032(x) \u22650 (resp. \u22640);\n(2) if f\u2032(x) > 0 (resp. < 0) for all x \u2208(a, b), then f is strictly increasing (resp. strictly\ndecreasing).\nProof. We only prove the increasing case of ??, as the others are similar.\n\u25e6First we assume that f is increasing. Then,\nx \u2265x0 =\u21d2f(x) \u2265f(x0) =\u21d2f(x) \u2212f(x0)\nx \u2212x0\n\u22650,", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_224", "text": "Proof. We only prove the increasing case of ??, as the others are similar.\n\u25e6First we assume that f is increasing. Then,\nx \u2265x0 =\u21d2f(x) \u2265f(x0) =\u21d2f(x) \u2212f(x0)\nx \u2212x0\n\u22650,\nx \u2264x0 =\u21d2f(x) \u2264f(x0) =\u21d2f(x) \u2212f(x0)\nx \u2212x0\n\u22650.\nThus,\nf\u2032(x0) = lim\nx\u2192x0\nf(x) \u2212f(x0)\nx \u2212x0\n\u22650\n117", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_225", "text": "Images/x3_gr.png\nFigure 20: f(x) = x3\n\u25e6Second, let us assume by contradiction that f\u2032(x) \u22650 \u2200x \u2208(a, b) and that f is not\nincreasing. Hence, there are a \u2264c < d \u2264b, such that f(c) > f(d). However, then ??\ntells us that then there esists e \u2208R, c < e < d such that f\u2032(e) = f(d)\u2212f(c)\nd\u2212c\n< 0.\nExample 7.63. Let f : E \u2192R be a strictly increasing (resp. strictly decreasing) function.\nThen, it does not necessarily follow that f\u2032(x) > 0 (resp. f\u2032(x) < 0). For example, f(x) = x3\nis strictly increasing, but f\u2032(0) = 3 \u00b7 02 = 0.\nExample 7.64. Let us consider the function fa : R \u2192R, f(x) := sin(x)+ax, where a \u2208R is a\n\ufb01xed real number. Let us compute for what value of a fa is monotone. As fa(x) is di\ufb00erentiable\non R, then fa is monotone if and only if either f\u2032(x) \u22650, \u2200x \u2208R or f\u2032(x) \u22640, \u2200x \u2208R. Thus,\nlet us compute f\u2032(x):\nf\u2032(x) = cos(x) + a.\nThus,\n\u25e6f is increasing if and only if a \u22651;\n\u25e6f is decreasing if and only if a \u2264\u22121.\nExample 7.65. Using ?? and ?? we obtain that the all the functions of ?? are either monotone,\n118", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_226", "text": "or become monotone when restricted to R\u2217\n+ or to R\u2217\n\u2212.\nf(x)\nD(f)\nf\u2032\nmonotonicity\nsinh(x)\nR\ncosh(x)\nincreasing over R\ncosh(x)\nR\nsinh(x)\ndecreasing over R\u2217\n\u2212and increasing over R\u2217\n+\ntanh(x)\nR\n1\ncosh(x)2\nincreasing over R\ncoth(x)\nR\u2217\n\u22121\nsinh(x)2\ndecreasing over R\u2217\n\u2212and over R\u2217\n+\nxa, a > 0\nR+\naxa\u22121, x \u0338= 0\nincreasing over R\u2217\n+\nxa, a < 0\nR+\naxa\u22121, x \u0338= 0\ndecreasing over R\u2217\n+\nax, a > 1\nR\nlog(a) \u00b7 ax\nincreasing over R\nax, 0 < a < 1\nR\nlog(a) \u00b7 ax\ndecreasing over R\nloga(x), a > 1\nR\u2217\n+\n1\nlog(a)\u00b7x\nincreasing over R\u2217\n+\nloga(x), 0 < a < 1\nR\u2217\n+\n1\nlog(a)\u00b7x\ndecreasing over R\u2217\n+\n7.5.2\nL\u2019H\u02c6opital\u2019s rule\nL\u2019H\u02c6opital rule gives a method to compute limits of fractions of function which are in the\nindeterminate forms\n0\n0, \u221e\n\u221e,\nthat is, either both values of the limit of the denominator and of the limit of the numerator\napproach 0, or they both approach \u2212\u221eor +\u221e\u2013 in the latter case, the sign of \u221edoes not\nreally matter.\nExample 7.66. How can we compute\nlim\nx\u2192+\u221e\nex\nx ? In this case,\nlim\nx\u2192+\u221eex = +\u221e=\nlim\nx\u2192+\u221ex.\nIn this example, we cannot answer using the algebraic rules of ??.\nLuckily, the following theorem provides us with new tools to carry out this kind of compu-\ntations.\nTheorem 7.67 (L\u2019H\u02c6opital rule). Let f, g: (a, b) \u2192R be di\ufb00erentiable functions, and let a, b \u2208\nR. Assume that the following conditions hold:\n(1) (exactly) one the following conditions hold for x0:\n(i) x0 \u2208(a, b);\n(ii) x0 = a \u2208R;\n(iii) x0 = b \u2208R;\n(iv) x0 = a = \u2212\u221e;\n(v) x0 = b = +\u221e;", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_227", "text": "(i) x0 \u2208(a, b);\n(ii) x0 = a \u2208R;\n(iii) x0 = b \u2208R;\n(iv) x0 = a = \u2212\u221e;\n(v) x0 = b = +\u221e;\n(2) g(x) \u0338= 0 and g\u2032(x) \u0338= 0 for all x \u2208(a, b) \\ {x0};\n119", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_228", "text": "(3)\nlim\nx\u2192x0f(x) = lim\nx\u2192x0g(x) = \u03b1 for \u03b1 = 0 or \u03b1 = \u00b1\u221e.\nThen, in the respective cases we have the following implications for any \u00b5 \u2208R:\nCases ??, ?? and ??\nif\nlim\nx\u2192x0\nf\u2032(x)\ng\u2032(x) = \u00b5\n\u21d2\nlim\nx\u2192x0\nf(x)\ng(x) = \u00b5\nCase ??\nif\nlim\nx\u2192x+\n0\nf\u2032(x)\ng\u2032(x) = \u00b5\n\u21d2\nlim\nx\u2192x+\n0\nf(x)\ng(x) = \u00b5\nCase ??\nif\nlim\nx\u2192x\u2212\n0\nf\u2032(x)\ng\u2032(x) = \u00b5\n\u21d2\nlim\nx\u2192x\u2212\n0\nf(x)\ng(x) = \u00b5\nProof. We prove only the \u03b1 = 0 and x0 \u2208(a, b) case, and we refer to page 121-122 of the book\nfor the rest. As f and g are di\ufb00erentiable at x0 they are also continuous there, and hence\nf(x0) = lim\nx\u2192x0f(x) = \u03b1 = 0\nand\ng(x0) = lim\nx\u2192x0g(x) = \u03b1 = 0.\n(7.67.a)\nSo, by the mean value theorem for derivatives, there is a real number c(x) between x and x0\nsuch that\nf\u2032(c(x)) = f(x) \u2212f(x0)\nx \u2212x0\n.\n(7.67.b)\nIn particular, c(x): E \\ x0 \u2192I \\ x0 is a function such that lim\nx\u2192x0c(x) = x0. Then:\n\u00b5 = lim\nx\u2192x0\nf\u2032(x)\ng\u2032(x)\n|\n{z\n}\nde\ufb01nition of \u00b5\n= lim\nx\u2192x0\nf\u2032(c(x))\ng\u2032(c(x))\n|\n{z\n}\n??\n= lim\nx\u2192x0\nf(x)\u2212f(x0)\nx\u2212x0\ng(x)\u2212g(x0)\nx\u2212x0\n|\n{z\n}\n??\n= lim\nx\u2192x0\nf(x) \u2212f(x0)\ng(x) \u2212g(x0) = lim\nx\u2192x0\nf(x)\ng(x)\n|\n{z\n}\n??", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_229", "text": "x\u2212x0\n|\n{z\n}\n??\n= lim\nx\u2192x0\nf(x) \u2212f(x0)\ng(x) \u2212g(x0) = lim\nx\u2192x0\nf(x)\ng(x)\n|\n{z\n}\n??\nRemark 7.68. We show that the property (3) in the statement of ?? is a necessary one. Indeed,\nwe show that if the limit lim\nx\u2192x0\nf\u2032(x)\ng\u2032(x) does not exist, then we cannot conclude anything about the\nlimit lim\nx\u2192x0\nf(x)\ng(x).\n(1) Let us take f(x) = x + sin(x), g(x) = x. Then f, g are di\ufb00erentiable over R,\nlim\nx\u2192+\u221ef(x) = +\u221e= lim\nx\u21920g(x),\ng(x) \u0338= 0 \u0338= g\u2032(x), \u2200x \u2208R\u2217.\nMoreover,\nlim\nx\u2192+\u221e\nf(x)\ng(x) =\nlim\nx\u2192+\u221e1 + sin(x)\nx\n= 1.\nOn the other hand,\nlim\nx\u2192+\u221e\nf\u2032(x)\ng\u2032(x) =\nlim\nx\u2192+\u221e\n1 + cos(x)\n1\nwhich is not de\ufb01ned since the limit\nlim\nx\u2192+\u221ecos(x) does not exist. Hence, since the limit of\nthe quotient of the derivatives of f, g does not exist, a priori, we cannot conclude anything\nabout the limit of the quotient of f, g. Nonetheless, in this case we are lucky and we can\nstill carry out the computation.\n120", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_230", "text": "(2) Consider f(x) = \u221ax + sin(x), g(x) = x. Then f, g are di\ufb00erentiable over R\u2217\n+,\nlim\nx\u2192+\u221ef(x) = +\u221e= lim\nx\u21920g(x),\ng(x) \u0338= 0 \u0338= g\u2032(x), \u2200x \u2208R\u2217.\nMoreover,\nlim\nx\u2192+\u221e\nf(x)\ng(x) =\nlim\nx\u2192+\u221e\n1\n\u221ax + sin(x)\nx\n= 0.\nOn the other hand,\nlim\nx\u2192+\u221e\nf\u2032(x)\ng\u2032(x) =\nlim\nx\u2192+\u221e\n1\n2\u221ax + cos(x)\n1\nwhich is not de\ufb01ned since the limit\nlim\nx\u2192+\u221ecos(x) does not exist.\n(3) Consider f(x) = x + sin(x), g(x) = x. Then f, g are di\ufb00erentiable over R\u2217\n+,\nlim\nx\u2192+\u221ef(x) = +\u221e= lim\nx\u21920g(x),\ng(x) \u0338= 0 \u0338= g\u2032(x), \u2200x \u2208R\u2217.\nMoreover,\nlim\nx\u2192+\u221e\nf(x)\ng(x) =\nlim\nx\u2192+\u221e\n1\n\u221ax + sin(x)\nx\n= 0.\nOn the other hand,\nlim\nx\u2192+\u221e\nf\u2032(x)\ng\u2032(x) =\nlim\nx\u2192+\u221e\n1\n2\u221ax + cos(x)\n1\nwhich is not de\ufb01ned since the limit\nlim\nx\u2192+\u221ecos(x) does not exist.\nHence, if the limit of the quotient of the derivatives of f, g does not exist, a priori, we cannot\nconclude anything about the limit of the quotient of f, g. Nonetheless, in some cases, such as\n() here, we are lucky and we can still carry out the computation.\nExample 7.69. Let us consider the limit lim\nx\u21920\narcsin(x)\nsin(x) . Then, lim\nx\u21920 arcsin(x) = 0 = lim\nx\u21920 sin(x)\nand sin(x)\u2032 = cos(x), arcsin(x)\u2032 =\n1\n\u221a\n1\u2212x2 .\nMoreover, both sin(x) and cos(x) are non-zero over the pointed neighborhood ] \u2212\u03c0\n2 , \u03c0\n2 [\\{0} of", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_231", "text": "and sin(x)\u2032 = cos(x), arcsin(x)\u2032 =\n1\n\u221a\n1\u2212x2 .\nMoreover, both sin(x) and cos(x) are non-zero over the pointed neighborhood ] \u2212\u03c0\n2 , \u03c0\n2 [\\{0} of\n0. Hence, we can apply ?? to get\nlim\nx\u21920\narcsin(x)\nsin(x)\n= lim\nx\u21920\n1\n\u221a\n1\u2212x2\ncos(x) = 1\nExample 7.70. Let us consider the limit\nlim\nx\u2192+\u221e\nex\nxn . Then, f(x) = ex, g(x) = xn, n \u2208N. Let\nus start with the case n = 1. Then,\nlim\nx\u2192+\u221e\nex\nx =\nlim\nx\u2192+\u221e\nex\n1 = +\u221e.\nFor n = 2,\nlim\nx\u2192+\u221e\nex\nx2 =\nlim\nx\u2192+\u221e\nex\n2x =\nlim\nx\u2192+\u221e\nex\n2 = +\u221e.\nHence, inductively, one can prove that\nlim\nx\u2192+\u221e\nex\nxn =\nlim\nx\u2192+\u221e\nex\nnxn\u22122 =\nlim\nx\u2192+\u221e\nex\nn(n \u22121)xn\u22122 = ... =\nlim\nx\u2192+\u221e\nex\nn! = +\u221e.\nHence, the exponential function ex goes to +\u221e\u2013 as x goes to +\u221e\u2013 faster than any monomial\nxn; a similar argument shows that it goes faster than any polynomial.\n121", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_232", "text": "Images/arcsin_gr.png\nFigure 21: f(x) = arcsin(x)\nExample 7.71. Similarly to the previous example,\nlim\nx\u21920+ xn log(x) = lim\nx\u21920+\nlog(x)\n1\nxn\n= lim\nx\u21920+\n1\nx\n\u2212n\nxn+1\n= lim\nx\u21920+ \u2212xn\nn = 0,\nwhile,\nlim\nx\u2192+\u221e\nlog(x)\nxn\n=\nlim\nx\u2192+\u221e\n1\nx\nnxn\u22121 =\nlim\nx\u2192+\u221e\n1\nnxn 0\nSo, log goes to \u2212\u221eas x goes to 0 and to +\u221eas x goes to +\u221eslower than 1\nx and x, respectively.\n7.5.3\nTaylor expansion\nDe\ufb01nition 7.72. Let f : E \u2192R be a function and let x0 \u2208E Assume that there is a neighbor-\nhood of a \u2208E which is contained in the domain (so in E). We say that f admits an expansion\nto the n-th order x0 if there is an equality of the form\nf(x) = a0 + a1(x \u2212a) + a2(x \u2212a)2 + \u00b7 \u00b7 \u00b7 + an(x \u2212a)n + (x \u2212a)n\u03f5(x),\n(7.72.c)\nwhere ai are real number, and \u03f5n(x): E \u2192R satis\ufb01es lim\nx\u2192x0 \u03f5n(x) = 0.\nProposition 7.73. In the hypotheses of ??, if a function f admits an n-th order expansion\naround a point x0 \u2208D(f), then the coe\ufb03cients ai in ?? are uniquely determined.\nProof. Let\nf(x) = a0 + a1(x \u2212x0) + a2(x \u2212x0)2 + \u00b7 \u00b7 \u00b7 + an(x \u2212x0)n + (x \u2212x0)n\u03f5n(x),\nf(x) = a\u2032\n0 + a\u2032\n1(x \u2212x0) + a\u2032\n2(x \u2212x0)2 + \u00b7 \u00b7 \u00b7 + a\u2032\nn(x \u2212x0)n + (x \u2212x0)n\u03f5\u2032\nn(x),\nbe two di\ufb00erent expansions to order n of f around x0. We show by induction on i that ai = a\u2032\ni.", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_233", "text": "2(x \u2212x0)2 + \u00b7 \u00b7 \u00b7 + a\u2032\nn(x \u2212x0)n + (x \u2212x0)n\u03f5\u2032\nn(x),\nbe two di\ufb00erent expansions to order n of f around x0. We show by induction on i that ai = a\u2032\ni.\nFor i = 0 this is given by passing to the limit as x \u2192x0 of the two expansion:\na0 = lim\nx\u2192x0 a0 + a1(x \u2212x0) + a2(x \u2212x0)2 + \u00b7 \u00b7 \u00b7 + an(x \u2212x0)n + (x \u2212x0)n\u03f5n(x)\n= lim\nx\u2192x0 f(x)\n= lim\nx\u2192x0 a\u2032\n0 + a\u2032\n1(x \u2212x0) + a\u2032\n2(x \u2212x0)2 + \u00b7 \u00b7 \u00b7 + a\u2032\nn(x \u2212x0)n + (x \u2212x0)n\u03f5\u2032\nn(x) = a\u2032\n0\n122", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_234", "text": "Let us prove the induction step. Thus, let us assume that we know that aj = a\u2032\nj for j =\n0, . . . , i \u22121. Then,\nf(x) =a0 + a1(x \u2212x0) + a2(x \u2212x0)2 + \u00b7 \u00b7 \u00b7 + an(x \u2212x0)n + (x \u2212x0)n\u03f5n(x)\n=a\u2032\n0 + a\u2032\n1(x \u2212x0) + a\u2032\n2(x \u2212x0)2 + \u00b7 \u00b7 \u00b7 + a\u2032\nn(x \u2212x0)n + (x \u2212x0)n\u03f5\u2032\nn(x)\n=a0 + a1(x \u2212x0) + \u00b7 \u00b7 \u00b7 + ai\u22121(x \u2212x0)i\u22121 + a\u2032\ni(x \u2212x0)i + \u00b7 \u00b7 \u00b7 + a\u2032\nn(x \u2212x0)n + (x \u2212x0)n\u03f5\u2032\nn(x)\nHence, taking the expansions on the 1st and the 3rd line of the previous chain of equalities,\nand subtracting from both expansions a0 + a1(x \u2212a) + \u00b7 \u00b7 \u00b7 + ai\u22121(x \u2212a)i\u22121 and then dividing\nboth by (x \u2212a)i, we obtain\nai + ai+1(x \u2212x0) + \u00b7 \u00b7 \u00b7 + an(x \u2212x0)n\u2212i + (x \u2212x0)n\u2212i\u03f5n(x)\n= a\u2032\ni + a\u2032\ni+1(x \u2212x0) + \u00b7 \u00b7 \u00b7 + a\u2032\nn(x \u2212x0)n\u2212i + (x \u2212x0)n\u2212i\u03f5\u2032\nn(x).\nTaking limit of this equality as x \u2192x0 yields that ai = a\u2032\ni, which concludes the induction step.\nHence, ai = a\u2032\ni for each i. In particular, it also follows that \u03f5(x) = \u03f5\u2032(x) for each x \u2208E.\nWhen is it that we can \ufb01nd an expansion to order n for a function f around a point\nx0 \u2208D(f)? The following theorem provides a \ufb01rst answer.\nTheorem 7.74. Let n \u22650 be an integer. Let f : E \u2192R be a function de\ufb01ned on an open\ninterval E, and let x0 \u2208E. Assume that f is n + 1 times di\ufb00erentiable over E. Then, for each", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_235", "text": "Theorem 7.74. Let n \u22650 be an integer. Let f : E \u2192R be a function de\ufb01ned on an open\ninterval E, and let x0 \u2208E. Assume that f is n + 1 times di\ufb00erentiable over E. Then, for each\nx \u2208E there exists x\u2032 \u2208]x0, x[, if x > x0 (resp. x\u2032 \u2208]x, x0[, if x < x0) and such that\nf(x) =\n n\nX\ni=0\nf(i)(a)\ni!\n(x \u2212a)i\n!\n+ f(n+1)(x\u2032)(x \u2212a)n+1\n(n + 1)! .\nRemark 7.75.\n?? not only tells us that, under the hypotheses posed in its statement, it is\npossible to \ufb01nd an order n expansion for a function f around a point x0 but also that, when\nthat is the case, we have a recipe to compute the coe\ufb03cients which are given by the formula\naj = f(j)(x0)\nj!\n.\nMoreover, we can also compute the error term in the\nProof. To understand the proof, note that the statement for n = 0 is just the Mean Value\nTheorem, cf. ??. Indeed, that results implies that there exists x\u2032 \u2208]x0, x[, if x > x0 (resp.\nx\u2032 \u2208]x, x0[, if x < x0) such that f\u2032(x\u2032) = f(x)\u2212f(x0)\nx\u2212x0\n. Multiply by x \u2212x0, then\nf(x) = f(x0) + f\u2032(x0 + \u03b8x,x0(x \u2212x0))(x \u2212x0),\nwhere \u03b8x,x0(x \u2212x0) \u2208[0, 1] and x\u2032 = x0 + \u03b8x,x0(x \u2212x0) \u2013 which is possible exactly because x\u2032\nis contained between x and x0. Let us recall that the proof of ?? was an application of Rolle\u2019s\ntheorem to the function g(y) := f(y) \u2212f(x0) \u2212f(x)\u2212f(x0)\nx\u2212x0\n(y \u2212x0). Furthermore, this techique", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_236", "text": "theorem to the function g(y) := f(y) \u2212f(x0) \u2212f(x)\u2212f(x0)\nx\u2212x0\n(y \u2212x0). Furthermore, this techique\nwas working since g(x0) = g(x), and g\u2032(y) = f\u2032(y) \u2212f(x)\u2212f(x0)\nx\u2212x0\n, so that g\u2032(y) being 0 yielded\nexactly the above equation.\nLet us now de\ufb01ne\nPn(x) :=\nn\nX\ni=0\nf(i)(x0)\ni!\n(x \u2212x0)i,\nand let us consider\ng(y) = f(y) \u2212Pn(y) + Pn(x) \u2212f(x)\n(x \u2212x0)n+1 (y \u2212x0)n+1.\n123", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_237", "text": "Then,\n0 = g(x) = g(a) = g\u2032(a) = \u00b7 \u00b7 \u00b7 = g(n)(a),\nwhich means that there exists y1 between x0 and x such that g\u2032(y1) = 0 by Rolle\u2019s theorem. But\nthen applying Rolle\u2019s theorem again we obtain a y2 between x0 and y1 such that g(2)(y2) = 0.\nIterating this process we obtain a point yn+1 between x0 and x such that g(n+1)(yn+1) = 0. In\nparticular, by setting x\u2032 := yn+1, then\n0 = g(n+1)(x\u2032) = f(x\u2032) + Pn(x) \u2212f(x)\n(x \u2212x0)n+1 (n + 1)!.\nReorganizing the latter equation yields exactly the statement of the theorem.\nCorollary 7.76. Let n \u22650 be a real number. Let f : E \u2192R be a function de\ufb01ned on an open\ninterval E. Assume that f \u2208Cn(E, R), and let x0 \u2208E. Then, the n-th order expansion of f\naround x0 exists and is given by the formula\nf(x) =\nn\nX\nj=0\nf(j)(x0)\nj!\n(x \u2212x0)j + (x \u2212x0)n\u03f5n(x).\nThe idea behind the proof of the corollary is that by the previous theorem the error term\nis f(n+1)(x\u2032) \u2212f(n+1)(x), which converges to zero as x goes to a as x\u2032 is between a and x, and\nf(n) is continuous. For the precise proof we refer to page 126 of the book.\nExample 7.77. Applying ?? to f(x) =\n1\n1\u2212x and x0 = 0 yields that the order n expansion\ntakes the form\n1\n1 \u2212x = 1 + x + x2 + \u00b7 \u00b7 \u00b7 + xn + xn\u03f5n(x),\nsince\nf(i)(x) =\ni!\n(1 \u2212x)i+1\n\u21d2\nf(i)(0) = i!\n\u21d2\nf(i)(0)\ni!\n= 1.", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_238", "text": "since\nf(i)(x) =\ni!\n(1 \u2212x)i+1\n\u21d2\nf(i)(0) = i!\n\u21d2\nf(i)(0)\ni!\n= 1.\nExample 7.78. Applying ?? to f(x) = ex and x0 = 0 yields that the order n expansion takes\nthe form\nex =\nn\nX\ni=0\nxi\ni! + xn\u03f5n(x),\nsince\nf(i)(x) = ex\n\u21d2\nf(i)(0) = 1\n\u21d2\nf(i)(0)\ni!\n= 1\ni!.\nExample 7.79. Similarly, the (2n + 1)-st order expansion of cos(x) around x = 0 is\ncos(x) =\nn\nX\nj=0\n(\u22121)n x2j\n(2j)! + x2n+1, \u03f5(x)\nwhile the (2n + 2)-nd order expansion of sin(x) around x = 0 is\nsin(x) =\nn\nX\nj=0\n(\u22121)n\nx2j+1\n(2j + 1)! + x2n+2\u03f52n+2(x).\n124", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_239", "text": "Example 7.80. One can also \ufb01gure out expansions of products, sums, compositions, etc.\nFor example the 3-rd order expansion of sin(cos(x)) is as follows:\ncos(sin(x)) = cos\n\u0012\nx \u2212x3\n6 + x3\u03f53(x)\n\u0013\n=1 \u2212\n\u0010\nx \u2212x3\n6 + x3\u03f53(x)\n\u00112\n2\n+\n\u0012\nx \u2212x3\n6 + x3\u03f53(x)\n\u00133\n\u03b73(sin(x)) = 1 \u2212x2\n2 + x3\u03c43(x),\nwhere x3\u03c43(x) is the sum of all terms of the form x3h(x), where lim\nx\u21920h(x) = 0. In particular,\nlim\nx\u21920\u03c43(x) = 0 and hence the above is indeed the 3-rd order expansion.\nIn general, we can compute the expansion to order n of a composition (f \u25e6g)(x) around a\npoint f(x0) by substituting the expension to order n of g around x0 into the expansion of f\naround f(x0) to order n and then re-ordering all the terms thus obtained up to order n. Let\nus highlight how one should be careful that the base-point of the expansion of the function f\nshould be the value g(x0) of the function g at the base-point x0. So, for example, sin(cos(x))\nat 0 is not easy to compute this way, because one would need the expansion of sin around\ncos(0) = 1, for which there is no nice formula.\nAnother example is by taking h(x) =\n1\n1\u2212(ex\u22121), x0 = 0. Then we can rewrite h as the\ncomposition h = f \u25e6g of f(y) =\n1\n1\u2212y and g(x) = ex \u22121. Then,\n1\n1 \u2212(ex \u22121) =\n1\n1 \u2212\n\u0010\nx + x2\n2 + x3\n6 + x3\u03f53(x)\n\u0011\n=1 +\n\u0012\nx + x2\n2 + x3\n6 + x3\u03f53(x)\n\u0013\n+\n\u0012\nx + x2\n2 + x3\n6 + x3\u03f53(x)\n\u00132\n+\n\u0012\nx + x2\n2 + x3\n6 + x3\u03f53(x)\n\u00133\n\u03b73(x + x2\n2 + x3\n6 + x3\u03f53(x))\n=1 + x +\n\u00121\n2 + 1\n\u0013\nx2 +\n\u00121\n6 + 2 \u00b7 1 \u00b7 1", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_240", "text": "6 + x3\u03f53(x)\n\u00132\n+\n\u0012\nx + x2\n2 + x3\n6 + x3\u03f53(x)\n\u00133\n\u03b73(x + x2\n2 + x3\n6 + x3\u03f53(x))\n=1 + x +\n\u00121\n2 + 1\n\u0013\nx2 +\n\u00121\n6 + 2 \u00b7 1 \u00b7 1\n2 + 1\n\u0013\nx3 + x3\u03c43(x)\n=1 + x + 3\n2x2 + 13\n6 x3 + x3\u03c43(x).\nSimilarly, one can write the order 3 expansion of\n1\n1\u2212x \u00b7 ex around 0 as\n1\n1 \u2212x \u00b7 ex =(1 + x + x2 + x3 + x3\u03f53(x))\n\u0012\n1 + x + x2\n2 + x3\n6 + x3\u03b73(x)\n\u0013\n=1 + 2x + 5\n2x2 + 8\n3x2 + x3\u03c43(x)\nYou can \ufb01nd more examples in the book, pages 127-131.\nExample 7.81. One can use expansions also to avoid using ??. For example, to compute\nlim\nx\u21920\n(ex \u22121 \u2212x) + x sin(x)\ncos(x) \u22121\n,\nthen we can try to compute the 2-nd order expansions \ufb01rst:\n(ex \u22121 \u2212x) + x sin(x) =\n\u0012\n1 + x + x2\n2 + x2\u03f52(x) \u22121 \u2212x\n\u0013\n+ x(x + x2\u03b72(x))\n=x2\n2 + x2 + x2(\u03b72(x) + \u03f52(x)) = 3\n2x2 + x2\n\u03b32(x)\n| {z }\n\u03b32(x):=\u03b72(x)+\u03f52(x)\n),\ncos(x) \u22121 =1 \u2212x2\n2 + x2\u03c42(x) \u22121 = \u2212x2\n2 + x2\u03c44(x)\n125", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_241", "text": "Then,\nlim\nx\u21920\n(ex \u22121 \u2212x) + x sin(x)\ncos(x) \u22121\n= lim\nx\u21920\n3\n2x2 + x2\u03f53(x)\n\u2212x2\n2 + x2\u03f54(x)\n= lim\nx\u21920\n3\n2 + \u03f53(x)\n\u22121\n2 + \u03f54(x) = \u22123.\n7.5.4\nApplication of Taylor expansion to local extrema and in\ufb02ection points\nWe have proven that if f has a point of local extremum at x0 \u2208D(f), then f\u2032(x0) = 0, cf. ??.\nHowever, we have also shown that the converse implication does not hold, cf. ??. Nevertheless,\nwe would like to know whether, for example, by imposing suitable conditions on the higher\nderivatives of a function f, we can still characterize when a stationary point is a point of local\nextremum for a function.\nLet f : E \u2192R be a function, and let x0 \u2208E be a stationary point for f. Moreover, let us\nassume that for some even natural number n, the \ufb01rst n \u22121 derivatives of f vanish at x0\nf\u2032(x0) = 0 = f\u2032\u2032(x0) = \u00b7 \u00b7 \u00b7 = f(n\u22121)(x0),\nwhile the n-th derivative of f is non-zero and f(n)(x0) > 0. Then, writing the n-th order\nexpansion of f around x0,\nf(x) = f(x0) + f(n)(x0)\nn!\n(x \u2212x0)n + (x \u2212x0)n\u03f5n(x).\nThus, for x su\ufb03ciently close to x0 it holds that |\u03b5n(x)| < 1\n2 \u00b7 f(n)(x0)\nn!\nholds. In particular, for\nsuch values of x, then\nf(x0) < f(x0) + 1\n2 \u00b7 f(n)(x0)\nn!\n(x \u2212x0)n \u2264f(x)\nThis shows that x0 is a point of local minimum for f. One can imitate this argument for the\ncase where f(n)(x0) < 0 to yield that x0 is a point of local maximum for f. Thus, we can\nsummarize the results obtained so far in the following theorem.", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_242", "text": "case where f(n)(x0) < 0 to yield that x0 is a point of local maximum for f. Thus, we can\nsummarize the results obtained so far in the following theorem.\nTheorem 7.82. Let n \u22652 be an even integer.\nLet f : E \u2192R be a function on an open\ninterval E and let x0 \u2208E. Assume that f is di\ufb00erentiable n times on E and that f(i)(x0) = 0,\n\u2200i = 1, 2, 3, . . . , n \u22121.\n(1) If f(n) > 0, then f has a point of local minimum at x0.\n(2) If f(n) < 0, then f has a point of local maximum at x0.\nExample 7.83. Consider the function f(x) = sin(x) + 1\n2x over the interval [0, 2\u03c0], cf. ??.\nThen, f\u2032(x) = 0 if and only if cos(x) = \u22121\n2, which is equivalent to x = 2\u03c0\n3 or 4\u03c0\n3 . Whether or\nnot we have a maximum or minimum at these points is decided by the sign of f\u2032\u2032(x) = \u2212sin(x).\n\u25e6At x = 2\u03c0\n3 , f(x)\u2032\u2032 < 0, so f(x) has a local maximum, and\n\u25e6At x = 4\u03c0\n3 , f(x)\u2032\u2032 > 0, so f(x) has a local minimum.\nQuestion 7.84. What happens if we assume that n is an odd natural number in the statement\nof ???\nIn that case, the expansion to order n for f around x0 takes the same expression as before\nf(x) = f(x0) + f(n)(x0)\nn!\n(x \u2212x0)n + (x \u2212x0)n\u03f5n(x),\n126", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_243", "text": "Images/sinxplushalfx_gr.png\nFigure 22: f(x) = sin(x) + 1\n2x over the interval [0, 2\u03c0]\nImages/hor_flex.png\nFigure 23: f(x) = x3 has a stationary point at x = 0 which is a \ufb02ex, as the graph goes through\nthe tangent line y = 0.\nbut this time the \ufb01rst non-leading term will be of the form (x \u2212x0)3, (x \u2212x0)5, or (x \u2212x0)7,\netc., depending on the precise value of n. But then for x > x0, (x \u2212x0)n > 0, while x < x0,\n(x \u2212x0)n < 0. This type of behavior characterizes what is called an in\ufb02ection. That is to say,\nthat for a stationary point x0 to be an in\ufb02ection point for f, we require that, on one side of x0\nthe graph of the function is above the tangent line to the graph of f through (x0, f(x0)), and\non the other side it is below it, cf. ??\nWe can actually generalize this tentative de\ufb01nition, as follows, to comprise not just the case\nof stationary points.\nDe\ufb01nition 7.85. Let f : E \u2192R be a function. Assume that f is di\ufb00erentiable at x0 \u2208E.\nWe say that f has an in\ufb02ection point at x0 if there exists \u03b4 > 0 such that either one of the\nfollowing two conditions is satis\ufb01ed:\n(1) {x \u2208E|a < x < a + \u03b4} \u21d2f(x) \u2212f(a) \u2212f\u2032(a)(x \u2212a) > 0, and\n{x \u2208E|a \u2212\u03b4 < x < a} \u21d2f(x) \u2212f(a) \u2212f\u2032(a)(x \u2212a) < 0; or,\n127", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_244", "text": "(2) {x \u2208E|a < x < a + \u03b4} \u21d2f(x) \u2212f(a) \u2212f\u2032(a)(x \u2212a) < 0, and\n{x \u2208E|a \u2212\u03b4 < x < a} \u21d2f(x) \u2212f(a) \u2212f\u2032(a)(x \u2212a) > 0.\nThe reasoning contained in the paragraph before ?? immediately yields the following result.\nTheorem 7.86. Let n \u22653 be an odd integer. Let f : E \u2192R be a function de\ufb01ned over an\nopen interval E and let x0 \u2208E. Assume that f is di\ufb00erentiable n times on E and that\nf\u2032\u2032(x0) = \u00b7 \u00b7 \u00b7 = f(n\u22121)(x0) = 0,\nwhile f(n)(x0) \u0338= 0. Then, f has an in\ufb02ection point at x0.\nExample 7.87. Let us consider the function f(x) = 2 sin(x) \u2212x. Then f\u2032(x) = 2 cos(x) \u22121,\nf\u2032\u2032(x) = \u22122 sin(x) and f\u2032\u2032\u2032(x) = \u22122 cos(x). Hence, f\u2032(0) = f\u2032\u2032(0) = 0, and f\u2032\u2032\u2032(0) \u0338= 0. Hence\nf(x) has an in\ufb02ection point at x = 0 according to ??.\nImages/obl_flex.png\nFigure 24: The function f(x) = 2 sin(x) \u2212x has a \ufb02ex at the point x = 0, which is non-\nstationary, as f\u2032(0) = 1, through the tangent line y = x to the graph of f at the point (0, 0).\n7.5.5\nConvex and concave functions\nDe\ufb01nition 7.88. Let f : E \u2192R be a function de\ufb01ned over an open interval E. We say that\nf is convex (resp. concave) if for every a, b \u2208E and every \u03bb \u2208[0, 1] we have:\nf(\u03bba + (1 \u2212\u03bb)b) \u2264\u03bbf(a) + (1 \u2212\u03bb)f(b).", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_245", "text": "f is convex (resp. concave) if for every a, b \u2208E and every \u03bb \u2208[0, 1] we have:\nf(\u03bba + (1 \u2212\u03bb)b) \u2264\u03bbf(a) + (1 \u2212\u03bb)f(b).\n(resp. f(\u03bba + (1 \u2212\u03bb)b) \u2265\u03bbf(a) + (1 \u2212\u03bb)f(b) ).\nRemark 7.89. Let us maintain the same notation as in the above de\ufb01nition. We may assume\nthat a < b. Then, for \u03bb \u2208[0, 1], x := \u03bba + (1 \u2212\u03bb)b is a point between a and b. Geometrically,\nthe above de\ufb01nition means the following:\n(1) f is convex, if for any choice of a, b \u2208E, then between a and b, the graph of f lies\ncompletely below the line segment connecting (a, f(a)) and (b, f(b));\n(2) if f is concave, then between a and b, the graph of f lies completely above the line\nsegment connecting (a, f(a)) and (b, f(b)).\nWe can characterize convexity and concavity of a function which is di\ufb00erentiable by means\nof the monotonicity of its \ufb01rst derivative.\n128", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_246", "text": "Images/convex.png\nFigure 25: The graph of the function f(x) = (x+1)2 \u22123 lies below the line segment connecting\nthe points (\u22124, f(\u22124)) = \u22124, 6 and (1, f(1)) = (1, 1).\nImages/concave.png\nFigure 26: The graph of the function f(x) = \u2212(x\u22121)2+5 lies below the line segment connecting\nthe points (\u22122, f(\u22122)) = (\u22122, 4) and (1, f(1)) = (1, 1).\nTheorem 7.90. Let f : E \u2192R be a function de\ufb01ned on an open interval E. Assume that f\nis di\ufb00erentiable. Then f is convex (resp. concave) if and only if f\u2032 : E \u2192R is an increasing\n(resp. decreasing) function.\nProof. We prove only the statements about convexity, as f is convex if and only if \u2212f is\nconcave.\n(1) First, let us assume that f is convex. Let a < b be points of I. We want to prove that\nf\u2032(a) \u2264f\u2032(b). By the above characterization of convexity we have\nf(b) \u2212f(\u03bba + (1 \u2212\u03bb)b)\nb \u2212(\u03bba + (1 \u2212\u03bb)b)\n\u2265f(b) \u2212f(a)\nb \u2212a\n, and f(\u03bba + (1 \u2212\u03bb)b) \u2212f(a)\n(\u03bba + (1 \u2212\u03bb)b) \u2212a\n\u2264f(b) \u2212f(a)\nb \u2212a\n.\nNow, as \u03bb goes to 0, the left side of the \ufb01rst inequality converges to f\u2032(b), and as \u03bb goes\n129", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_247", "text": "to 1 the left side of the second inequality converges to f\u2032(a). This yields:\nf\u2032(b) \u2265f(b) \u2212f(a)\nb \u2212a\n\u2265f\u2032(a)\n(2) For, the other direction let us assume that f\u2032 is increasing. Fix a < b \u2208E. Set x :=\n\u03bba + (1 \u2212\u03bb)b for any \u03bb \u2208]0, 1[ (for \u03bb = 0 and 1 the convexity inequality is automatic).\nThen, the mean value theorem tells us that there are a < x1 < x < x2 < b such that\nf(x)\u2212f(a)\nx\u2212a\n= f\u2032(x1) and f(b)\u2212f(x)\nb\u2212x\n= f\u2032(x2). In particular, by our assumption that the\nderivative is increasing it follows that f(x)\u2212f(a)\nx\u2212a\n\u2264f(b)\u2212f(x)\nb\u2212x\n. But this shows that f is\nconvex by the above characterization of convexity in terms of slopes.\nIf f\u2032 is di\ufb00erentiable \u2013 or, equivalently, f is twice di\ufb00erentiable \u2013 then f\u2032 being increasing\n(resp. decreasing) is equivalent to f\u2032\u2032 \u22650 (resp. f\u2032\u2032 \u22640). Thus, we can characterize convexity\nand concavity of a function which is twice di\ufb00erentiable by means of the sign of its second\nderivative.\nCorollary 7.91. Let f : E \u2192R be a two times di\ufb00erentiable function on an open interval.\nThen f is convex (resp. concave) if and only if f\u2032\u2032(x) \u22650 (resp. f\u2032\u2032(x) \u22640) for all x \u2208E.\nExample 7.92.\n(1) Let us consider f(x) = ex. Then, f\u2032\u2032(x) = ex, thus f : R \u2192R is convex,\nsince the second derivative ex > 0, \u2200x \u2208R.\n(2) Let us consider f(x) = log(x). Then, f\u2032\u2032(x) =\n\u0000 1\nx\n\u0001\u2032 = \u22121\nx2 , so the function log(x), which\nis de\ufb01ned over R\u2217\n+ is concave over its entire domain.\nExample 7.93. Here we explain why the graph of a di\ufb00erentiable convex function f must lie", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_248", "text": "\u0000 1\nx\n\u0001\u2032 = \u22121\nx2 , so the function log(x), which\nis de\ufb01ned over R\u2217\n+ is concave over its entire domain.\nExample 7.93. Here we explain why the graph of a di\ufb00erentiable convex function f must lie\nabove the tangent line to the graph through a point (a, f(a)). That is to say, we show that for\na di\ufb00erentiable convex function f,\nf(x) \u2265f(a) + f\u2032(a)(x \u2212a).\nWe take x > a, and leave the case x < a to the reader. Thus, assuming that x > a, we wish to\nshow that f(x) \u2265f(a) + f\u2032(a)(x \u2212a), or equivalently that\nf(x) \u2212f(a)\nx \u2212a\n\u2265f\u2032(a).\n(7.93.d)\nIndeed, by the Mean value theorem (??) there exists a x\u2032 \u2208R, a < x\u2032 < x such that\nf(x) \u2212f(a)\nx \u2212a\n= f\u2032(x\u2032).\n(7.93.e)\nAs f is convex, f\u2032 is increasing by ??, hence, f\u2032(x\u2032) \u2265f\u2032(a). Adding this observation the\nequality in ??, we have shown that ?? must hold.\n7.6\nAsymptotes\nDe\ufb01nition 7.94.\n(1) If for some c \u2208R, lim\nx\u2192c\u2212f(x) = \u00b1\u221eor lim\nx\u2192c+ f(x) = \u00b1\u221e, then we say\nthat the function f has a vertical asymptote at x = c.\n(2) If for some c \u2208R,\nlim\nx\u2192+\u221ef(x) = c (resp.\nlim\nx\u2192\u2212\u221ef(x) = c), then we say that the function f\nhas a horizontal asymptote at +\u221e(resp. \u2212\u221e) at y = c.\n130", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_249", "text": "(3) If for some a \u0338= 0, b \u2208R,\nlim\nx\u2192+\u221ef(x) \u2212ax = b (resp.\nlim\nx\u2192\u2212\u221ef(x) \u2212ax = b), then we say\nthat f has an oblique (or slant) asymptote at +\u221e(resp. \u2212\u221e) along the line y = ax + b.\nExample 7.95. Here we give a few examples of the di\ufb00erent notions introduce in the above\nde\ufb01nition.\n(1) Vertical asymptote: f(x) =\n1\n1\u2212x has a vertical asymptote at at x = 1, cf. ??;\n(2) Horizontal asymptote: the function f(x) = 2 \u2212e\u2212x has a horizontal asymptote at +\u221eof\nvalue y = 2, cf. ??;\n(3) Slant asymptote: the function f(x) = 2+3x+ 1\nx2 has a slant asymptote both at +\u221eand\n\u2212\u221ealong the line y = 3x + 2, cf. ??.\nImages/1suxmeno1.png\nFigure 27: The function f(x) =\n1\nx\u22121, and the line x = 1.\n131", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_250", "text": "Images/2-emenox.png\nFigure 28: The function f(x) = 2 \u2212e\u2212x, and the line y = 2.\n8\nINTEGRATION\n8.1\nDe\ufb01nition\nThe idea behind integration is that the integral\nZ b\na\nf(x)dx of a bounded function f on a\nclosed interval [a, b] should be the area under the graph of f. However, it is not that easy to\nsay what this area means and when it is computable at all. If it is computable, we say that\nthe function is integrable (??), and the value of this area is then called the integral\nZ b\na\nf(x)dx\nof f.\nNow, the idea of trying to de\ufb01ne the area under the graph of f is simple. We start with\nthe only area that we can compute trustably, that is of rectangles, and then we try to ap-\nproximate the area under the graph of f from above and from below using rectangles. These\napproximations are called upper and lower Darboux sums (??). We say that the area under\nthe graph of f is computable, which as above means that the function is integrable, if these\ntwo approximations meet in the limit. This is spelled out in precise mathematical terms below.\nDe\ufb01nition 8.1.\n(1) A partition \u03c3 = (xi) of a bounded interval [a, b] is an ordered collection\na = x0 < x1 < \u00b7 \u00b7 \u00b7 < xn\u22121 < xn = b of points of [a, b].\n(2) The norm or mesh of \u03c3 is\nmax{xi \u2212xi\u22121|1 \u2264i \u2264n}.\n(3) A re\ufb01nement \u03c3\u2032 = (x\u2032\ni) of \u03c3 is a partition such that each value of xi shows up amongst\nx\u2032\ni. we indicate that \u03c3\u2032 is a re\ufb01nement of \u03c3 by writing \u03c3 \u2ab0\u03c3.\n(4) The regular partition of length n is xi := a + ib\u2212a\nn , i = 0, 1, 2, . . . , n.\nProposition 8.2. Given a bounded interval [a, b], any two partitions \u03c3, \u03c3\u2032 have a common\nre\ufb01nement \u03c3\u2032\u2032. Moreover, each partition can be re\ufb01ned to a new one with arbitrarily small\nnorm.\n132", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_251", "text": "Images/2piu3xpiu1sux2.png\nFigure 29: The function f(x) = 2 + 3x + 1\nx2 , and the line y = 3x + 2.\nDe\ufb01nition 8.3. Let f : [a, b] \u2192R be a bounded function and \u03c3 = (xi) a partition of [a, b].\nThen, the upper Darboux sum of f with respect to \u03c3 is\nS\u03c3 =\nn\nX\ni=1\nMi(xi \u2212xi\u22121),\nwhere Mi := supx\u2208[xi\u22121,xi] f(x). The lower Darboux sum of f with respect to \u03c3 is\nS\u03c3 =\nn\nX\ni=1\nmi(xi \u2212xi\u22121),\nwhere mi := infx\u2208[xi\u22121,xi] f(x).\nExample 8.4. Let us consider a constant function f(x) = c, c \u2208R over a closed bounded\n133", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_252", "text": "interval [a, b]. Then for any partition \u03c3 of [a, b],\nS\u03c3 = S\u03c3 =\nn\nX\ni=1\nc(xi \u2212xi\u22121) = cxn \u2212cx0\n|\n{z\n}\ntelescopic sum\n= c(b \u2212a).\nExample 8.5. Let us consider the function f(x) = x over a closed bounded interval [a, b], and\nlet \u03c3n =\n\u0010\na + i(b\u2212a)\nn\n\u0011\nbe the regular partition of length n. Then,\nS\u03c3n =\nn\nX\ni=1\n\u0012\na + ib \u2212a\nn\n\u0013 b \u2212a\nn\n= a(b \u2212a) + n(n + 1)\n2\n(b \u2212a)2\nn2\nand\nS\u03c3n =\nn\nX\ni=1\n\u0012\na + (i \u22121)b \u2212a\nn\n\u0013 b \u2212a\nn\n= a(b \u2212a) + (n \u22121)n\n2\n(b \u2212a)2\nn2\n,\nwhere in both cases we used the following identity\nn\nX\ni=1\ni = (n + 1)n\n2\n.\nNote that lim\nn\u2192\u221eS\u03c3n = lim\nn\u2192\u221eS\u03c3n = a(b \u2212a) + (b\u2212a)2\n2\n= b2\n2 \u2212a2\n2 .\nProposition 8.6. Let f : [a, b] \u2192R be a function (not necessarily a continuous one). Assume\nthat f admits an upper bound M (resp. a lower bound m) for its range R(f). Then, for any\npartition \u03c3 of [a, b], m(b \u2212a) \u2264S\u03c3, S\u03c3 \u2264M(b \u2212a). In particular, the sets\n{S\u03c3|\u03c3 is a partition of [a, b]}\nand\n{S\u03c3|\u03c3 is a partition of [a, b]}\nare bounded.\nProof. This follows immediately from the de\ufb01nitions, since for any interval [xi, xi+1] \u2282[a, b]\nthen\nm \u2264\ninf\n[xi,xi+1] f \u2264\nsup\n[xi,xi+1]\nf \u2264M.\nHence, for a partition \u03c3 = {xi} of [a, b],\nm(b \u2212a) =m\nn\u22121\nX\ni=0\nm(xi+1 \u2212xi) \u2264\nn\u22121\nX\ni=0\n\u0012\ninf\n[xi,xi+1] f\n\u0013", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_253", "text": "f \u2264M.\nHence, for a partition \u03c3 = {xi} of [a, b],\nm(b \u2212a) =m\nn\u22121\nX\ni=0\nm(xi+1 \u2212xi) \u2264\nn\u22121\nX\ni=0\n\u0012\ninf\n[xi,xi+1] f\n\u0013\n(xi+1 \u2212xi) = S\u03c3\n\u2264\nn\u22121\nX\ni=0\n \nsup\n[xi,xi+1]\nf\n!\n(xi+1 \u2212xi) = S\u03c3 \u2264\nn\u22121\nX\ni=0\nM(xi+1 \u2212xi) = M(b \u2212a)\nDe\ufb01nition 8.7. Let f : [a, b] \u2192R be a bounded function.\n(1) The upper Darboux integral of f on [a, b] is de\ufb01ned as\nS := inf{S\u03c3|\u03c3 is a partition of [a, b]}.\n(2) The lower Darboux integral of f on [a, b] is de\ufb01ned as\nS := sup{S\u03c3|\u03c3 is a partition of [a, b]}\n134", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_254", "text": "Example 8.8. Using the above computation for the constant function, cf. ??, we see that if f\nis the constant function on [a, b], then S = S = (b \u2212a)c.\nProposition 8.9. Let f : [a, b] \u2192R be a bounded function.\n(1) If \u03c3 is a partition of [a, b] and \u03c3\u2032 is a re\ufb01nement of \u03c3, then:\nS\u03c3 \u2264S\u03c3\u2032, and S\u03c3 \u2265S\u03c3\u2032.\n(2) If \u03c3 is a partition of [a, b], then:\nS\u03c3 \u2264S\u03c3.\nCorollary 8.10. If f : [a, b] \u2192R is a bounded function, then S \u2264S.\nProof. It is enough to prove that S\u03c31 \u2264S\u03c32 for any partitions \u03c31 and \u03c32 of [a, b]. However,\nthis follows straight from ??. Indeed, if \u03c3 is a common re\ufb01nement of \u03c31 and \u03c32, then ?? yields\nthat\nS\u03c31 \u2264S\u03c3\n|{z}\n??.??\n\u2264S\u03c3\n|{z}\n??.??\n\u2264S\u03c32\n|{z}\n??.??\n.\nDe\ufb01nition 8.11. Let f : [a, b] \u2192R be a bounded function. We say that f is integrable, if\nS = S, in which case this common value is called the integral of f between a and b, and it is\ndenoted by\nZ b\na\nf(x)dx.\nRemark 8.12. Using ??, f is integrable over a closed bounded interval [a, b] if one exhibits a\nsequence (\u03c3n) of partitions such that lim\nn\u2192\u221eS\u03c3n = lim\nn\u2192\u221eS\u03c3n. Indeed, this follows immediately by\nthe following chain of inequalities\nlim\nn\u2192\u221eS\u03c3n \u2264S \u2264S \u2264lim\nn\u2192\u221eS\u03c3n,\n(8.12.a)\npassing to the limit for n \u2192\u221e.\nExample 8.13. Using ??, the constant functions are integrable on [a, b], and\nZ b\na\nc dx = (b \u2212a)c\nExample 8.14. Using ?? and the computation of ?? for f(x) := x over a closed bounded", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_255", "text": "Example 8.13. Using ??, the constant functions are integrable on [a, b], and\nZ b\na\nc dx = (b \u2212a)c\nExample 8.14. Using ?? and the computation of ?? for f(x) := x over a closed bounded\ninterval [a, b] then f is integrable, and\nZ b\na\nxdx = b2\n2 \u2212a2\n2\nExample 8.15. Consider the function [0, 2] \u2192R given by\nf(x) =\n\u001a0\nif x \u2208Q\n3\nif x \u0338\u2208Q\nThen, for all partition \u03c3, S\u03c3 = 6, and S\u03c3 = 0. So, S = 6, S = 0, and hence f is not integrable.\n135", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_256", "text": "Proposition 8.16. If f : [a, b] \u2192R is continuous then it is integrable.\nProof. As f is continuous over a closed bounded interval [a, b], ?? implies the uniform continuity\nof f. Let us \ufb01x \u03b5 > 0. Let \u03b4 > 0 be the constant in the de\ufb01nition of uniform continuity\nassociated to\n\u03b5\nb\u2212a \u2013 that is,\n|x \u2212y| \u2264\u03b4 \u21d2|f(x) \u2212f(y)| \u2264\n\u03b5\nb \u2212a.\nClaim Let \u03c3 be a partition of [a, b] with norm at most \u03b4. Then S\u03c3 \u2212S\u03c3 \u2264\u03b5.\nProof. In fact,\nS\u03c3 \u2212S\u03c3 =\nn\nX\ni=1\n( max\n[xi,xi+1] f \u2212\nmin\n[xi,xi+1] f)(xi \u2212xi\u22121)\n\u2264\nn\nX\ni=1\n\u03b5\nb \u2212a(xi \u2212xi\u22121) =\n\u03b5\nb \u2212a\n(b \u2212a)\n| {z }\nPn\ni=1(xi\u2212xi\u22121)=b\u2212a\n= \u03b5\n8.2\nBasic properties\nProposition 8.17. Let f, g: [a, b] \u2192R be integrable functions. Then,\n(1) If f extends over [b, c] for some c \u2208R, c > b and f is also integrable over [b, c], then it is\nintegrable over [a, c], and\nZ b\na\nf(x)dx +\nZ c\nb\nf(x)dx =\nZ c\na\nf(x)dx.\n(2) Given \u03b1, \u03b2 \u2208R, \u03b1f + \u03b2g is integrable on [a, b], and\nZ b\na\n(\u03b1f + \u03b2g)(x)dx = \u03b1\nZ b\na\nf(x)dx + \u03b2\nZ b\na\ng(x)dx\n(3) If f \u2264g, then\nZ b\na\nf(x)dx \u2264\nZ b\na\ng(x)dx\n(4) The function |f| is integrable on [a, b], and\nZ b\na\n|f(x)|dx \u2265\n\f\f\f\f\nZ b\na\nf(x)dx\n\f\f\f\f\nProof. The proofs of all these statements follow the same pattern: one writes up the inequalities", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_257", "text": "(4) The function |f| is integrable on [a, b], and\nZ b\na\n|f(x)|dx \u2265\n\f\f\f\f\nZ b\na\nf(x)dx\n\f\f\f\f\nProof. The proofs of all these statements follow the same pattern: one writes up the inequalities\nfor lower (resp. upper) Darboux sums for a \ufb01xed partition \u03c3. Then these inequalities remain\nvalid when taking sup (resp. inf) of the lower (resp. upper) Darboux sums along all possible\npartitions of an interval [a, b]. This gives inequalities in both direction, which then implies\nequalities.\nFor example, let us show how this strategy works in the case of point ?? \u2013 we leave the other\n136", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_258", "text": "cases to the reader. Let \u03c3, and \u03c4 be partitions for [a, b] and [b, c], respectively. Then the union\nof \u03c3 and \u03c4 gives a partition \u03c1 for [a, c] and by de\ufb01nition we have\nS\u03c1 = S\u03c3 + S\u03c4,\n(8.17.a)\nS\u03c1 = S\u03c3 + S\u03c4.\n(8.17.b)\nAs both these equalities are true for all choice of partitions \u03c3 of [a, b] and \u03c4 of [b, c], by taking the\ninf (resp. the sup) of ?? (resp. of ??) along all possible choices of partitions of [a, b], [b, c], [a, c],\nthen\nS\n[a,c] \u2264S\n[a,b] + S\n[b,c], and S[a,c] \u2265S[a,b] + S[b,c],\n(8.17.c)\nwhere S\n[a,c] (resp. S[a,c]) denotes the upper (resp. lower) Darboux integral of partitions of\n[a, c], and similarly for the other cases. As f is integrable both on [a, b] and on [b, c], we have\nR b\na f(x)dx = S\n[a,b] = S[a,b] and\nR c\nb f(x)dx = S\n[b,c] = S[b,c]. Thus, ?? yields\nZ b\na\nf(x)dx +\nZ c\nb\nf(x)dx \u2264S[a,c] \u2264S\n[a,c] \u2264\nZ b\na\nf(x)dx +\nZ c\nb\nf(x)dx.\n(8.17.d)\nAs the two ends of ?? are the same, we have everywhere equalities. This concludes both the\nintegrability of f over [a, c] as well as the statement of ??.\nExample 8.18.\nZ b\na\n(1 + x)dx =\nZ b\na\n1dx +\nZ b\na\nxdx\n|\n{z\n}\npoint ?? of ??\n= (b \u2212a)\n| {z }\n??\n+ b2 \u2212a2\n2\n| {z }\n??\n.\n8.3", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_259", "text": "Z b\na\n(1 + x)dx =\nZ b\na\n1dx +\nZ b\na\nxdx\n|\n{z\n}\npoint ?? of ??\n= (b \u2212a)\n| {z }\n??\n+ b2 \u2212a2\n2\n| {z }\n??\n.\n8.3\nFundamental theorem of calculus\nIn this section, we learn how to compute integrals using the anti-derivative, cf. ??. We start\nby giving the de\ufb01nition of an anti-derivative.\nDe\ufb01nition 8.19. Let f : [a, b] \u2192R be a continuous function. A function G : [a, b] \u2192R is\ncalled an anti-derivative of f if\n(1) G is continuous on [a, b],\n(2) G is di\ufb00erentiable on ]a, b[, and\n(3) G\u2032(x) = f(x) for all x \u2208]a, b[.\nRemark 8.20. Given a continuous function f : [a, b] \u2192R admitting an anti-derivative G: [a, b] \u2192\nR, then for any C \u2208R, also GC : [a, b] \u2192R, GC(x) := G(x) + C is an anti-derivative for f.\nAccording to ??, the vice versa is also true: namely, if G, H are anti-derivatives of f, then\nthere exists C \u2208R such that G(x) = H(x) + C, \u2200x \u2208R.\nNotation 8.21. The anti-derivatives of f are at times denoted by\nR\nf(x)dx+c, where c \u2208R is\na constant that is free to vary in R. Also, sometimes the expressions\nR\nf(x)dx + c is also called\nthe inde\ufb01nite integral, while what we de\ufb01ned as the integral of f over [a, b],\nR b\na f(x)dx is then\ncalled the de\ufb01nite integral \u2013 where the de\ufb01nitiveness comes from the fact that we computed\nthe integral over the closed bounded integral [a, b]. We use the integral/anti-derivative naming\nin this course.\nExample 8.22. We collect here a few important functions together with their anti-dervatives.\nfunction\nex\ncos(x)\nsin(x)\n1\nx\nxn, n \u2208N\n. . .\nanti-derivative\nex", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_260", "text": "in this course.\nExample 8.22. We collect here a few important functions together with their anti-dervatives.\nfunction\nex\ncos(x)\nsin(x)\n1\nx\nxn, n \u2208N\n. . .\nanti-derivative\nex\nsin(x)\n\u2212cos(x)\nlog |x|\nxn+1\nn+1\n. . .\n137", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_261", "text": "Now that we have de\ufb01ned the notion of anti-derivative of a continuous function, there are\ntwo questions that arise spontaneously:\n\u25e6Does an anti-derivative for a continuous function f : [a, b] \u2192R over a closed bounded\ninterval [a, b] always exist?\n\u25e6If an anti-derivative exists for a continous function f : [a, b] \u2192R, does it help us in any\nway in computing the value of the integral\nR b\na f(x)dx?\nThese two questions have some simple but very powerful answers provided by the following\ntwo theorems, that are usually called the \ufb01rst and second fundamental theorems of calculus.\nTheorem 8.23. Fundamental theorem of calculus I\nLet f : [a, b] \u2192R be continuous. Then,\nF(x) :=\nZ x\na\nf(t)dt\nis an anti-derivative of f.\nTheorem 8.24. Fundamental theorem of calculus II\nLet f : [a, b] \u2192R be continuous and let G be an anti-derivative of f. Then,\nZ b\na\nf(x)dx = G(b) \u2212G(a).\nNotation 8.25. In order for the statement of ?? to make full sense, we need to introduce some\nfurther notation: so far we de\ufb01ned\nR b\na f(x)dx only for a < b. If a = b, then we de\ufb01ne\nZ a\na\nf(x)dx := 0.\nIf a > b, then we also de\ufb01ne\nZ b\na\nf(x) := \u2212\nZ a\nb\nf(x)dx.\nWith these notations our previously proven rules give that if f : [a, b] \u2192R is continuous, and\nc, d \u2208[a, b] are any points, then\nZ c\na\nf(x)dx +\nZ d\nc\nf(x)dx =\nZ d\na\nf(x)dx.\nThe next statement is not too interesting in itself but it is needed in the proof of ??.\nTheorem 8.26. Mean Value Theorem for Integrals\nIf f : [a, b] \u2192R is continuous, then there is a c \u2208[a, b], such that\nZ b\na\nf(x)dx = f(c)(b \u2212a).\nProof. As [a, b] is closed and f is continuous, by ??, f admits global maximum and minimum", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_262", "text": "Z b\na\nf(x)dx = f(c)(b \u2212a).\nProof. As [a, b] is closed and f is continuous, by ??, f admits global maximum and minimum\nover [a, b]. Set M := max\nx\u2208[a,b] f(x) and m := min\nx\u2208[a,b] f(x). By ??, f takes all values in [m, M].\nHowever, by ??, then\nm \u2264\nR b\na f(x)dx\nb \u2212a\n\u2264M,\nso there is a c \u2208[a, b] such that f(c) equals the above fraction, which is exactly the statement\nof the theorem.\n138", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_263", "text": "Let us now give the proofs of the two fundamental theorems of calculus.\nProof of ??. Fix x0 \u2208]a, b[. Then, for any x0 \u0338= x \u2208]a, b[:\nF(x) \u2212F(x0)\nx \u2212x0\n=\n1\nx \u2212x0\nZ x\nx0\nf(t)dt = f(c(x))\n| {z }\n??\n,\nfor a real number c(x) between x and x0. Hence:\nlim\nx\u2192x0\nF(x) \u2212F(x0)\nx \u2212x0\n= lim\nx\u2192x0f(c(x)) =\nlim\nx\u2192x0f(x)\n|\n{z\n}\nlim\nx\u2192x0\nc(x)=x0\n=\nf(x0)\n| {z }\nf is continuous\n.\nProof of ??. We have already shown in ?? that F(x) =\nR x\na f(t)dt is an anti-derivative of f. As\nboth F and G are anti-derivatives, they di\ufb00er by a constant C \u2208R, that is, (F \u2212G)(x) = C,\n\u2200x \u2208[a, b]. Then:\nG(b) \u2212G(a) =(G(b) + c) \u2212(G(a) + c) = F(b) \u2212F(a)\n=\nZ b\na\nf(x)dx \u2212\nZ a\na\nf(x)dx =\nZ b\na\nf(x)dx.\nNotation 8.27. The expression G(b)\u2212G(a) appearing in the statement of ?? is usually denoted\nby\nG(x)|b\na\nor\nG(x)|x=b\nx=a.\nExample 8.28.\nZ \u22121\n\u22125\n1\nx =\n\u0000log |x|)\n\f\fx=\u22121\nx=\u22125 = log 1 \u2212log 5 = \u2212log 5\n8.4\nSubstitution\nTheorem 8.29. Let f : [a, b] \u2192R be a continuous function, and let \u03c6 : [\u03b1, \u03b2] \u2192[a, b] be a C1\nfunction. Then,\nZ \u03c6(\u03b2)\n\u03c6(\u03b1)\nf(x)dx =\nZ \u03b2\n\u03b1\nf(\u03c6(t))\u03c6\u2032(t)dt.\n(8.29.a)\nProof. De\ufb01ne G(x) :=\nR x", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_264", "text": "function. Then,\nZ \u03c6(\u03b2)\n\u03c6(\u03b1)\nf(x)dx =\nZ \u03b2\n\u03b1\nf(\u03c6(t))\u03c6\u2032(t)dt.\n(8.29.a)\nProof. De\ufb01ne G(x) :=\nR x\na f(u)du. By ??, G is an anti-derivative of f, so that ?? tells us\nZ \u03c6(\u03b2)\n\u03c6(\u03b1)\nf(x)dx = G(\u03c6(\u03b2)) \u2212G(\u03c6(\u03b1)).\nSo, it is enough to show that the value of the right side of ?? is the same. To show that, let us\njust note that by the chain rule G(\u03c6(t))\u2032 = G\u2032(\u03c6(t))\u03c6\u2032(t) = f(\u03c6(t))\u03c6\u2032(t). Then applying ?? to\nthe integral\nR \u03b2\n\u03b1 f(\u03c6(t))\u03c6\u2032(t)dt implies that\nZ \u03b2\n\u03b1\nf(\u03c6(t))\u03c6\u2032(t)dt = G(\u03c6(\u03b2)) \u2212G(\u03c6(\u03b1)),\nwhich concludes the proof.\n139", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_265", "text": "Example 8.30. In this example, we go from the right side of ?? to the left side.\nZ 1\n0\n\u221a\nexexdx =\nZ e\n1\n\u221audu\n|\n{z\n}\nu=ex\n(ex)\u2032=ex\n= u\n3\n2\n3\n2\n\f\f\f\f\f\nu=e\nu=1\n= 2\n3\n\u0010\ne\n3\n2 \u22121\n\u0011\nExample 8.31. Let us consider the function f : [0, 1] \u2192R, f(x) :=\n\u221a\n1 \u2212x2.\nWe want to compute the integral\nR 1\n0 f(x)dx. This integral computes the area of a quarter of\nImages/sqrt_1minusx2_gr.png\nFigure 30: f(x) :=\n\u221a\n1 \u2212x2.\na circle of radius 1, as shown by ?? so the result should be \u03c0\n4 . Indeed, the above computation\nshows that are train of thought is correct. Note that, opposite to the previous example, in this\nargument at our \ufb01rst substitution we go from the left side of ?? to the right side.\nZ 1\n0\np\n1 \u2212x2dx =\nZ\n\u03c0\n2\n0\np\n1 \u2212(sin(t))2 cos(t)dt\n|\n{z\n}\nx=sin(t)\nsin(t)\u2032=cos(t)\n=\nZ\n\u03c0\n2\n0\np\ncos(t)2 cos(t)dt\n=\nZ\n\u03c0\n2\n0\n| cos(t)| cos(t)dt =\nZ\n\u03c0\n2\n0\ncos(t) cos(t)dt\n|\n{z\n}\nt\u2208[0, \u03c0\n2 ] \u21d2cos(t)\u22650 \u21d2| cos(t)|=cos(t)\n=\nZ\n\u03c0\n2\n0\ncos(2t) + 1\n2\ndt =\nZ \u03c0\n0\n\u0012cos(u) + 1\n2\n\u0013 1\n2du\n|\n{z\n}\nt= u\n2\n=1\n4\nZ \u03c0\n0\n(cos(u) + 1)du = 1\n4 (sin(u) + u) |u=\u03c0\nu=0\n=1\n4(sin(\u03c0) + \u03c0 \u2212sin(0) \u22120) = \u03c0\n4\nExample 8.32. Recall that sinh(x) : R \u2192R is an odd function and it is strictly increas-", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_266", "text": "u=0\n=1\n4(sin(\u03c0) + \u03c0 \u2212sin(0) \u22120) = \u03c0\n4\nExample 8.32. Recall that sinh(x) : R \u2192R is an odd function and it is strictly increas-\ning (indeed, sinh(x)\u2032 = cosh(x) > 0). In particular, it has an inverse, which we denote by\n140", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_267", "text": "sinh\u22121(x): R \u2192R. With this we may compute similarly :\nZ 1\n0\np\n1 + x2dx =\nZ sinh\u22121(1)\n0\np\n1 + (sinh(t))2 cosh(t)dt\n|\n{z\n}\nx=sinh(t)\nsinh(t)\u2032=cosh(t)\n=\nZ sinh\u22121(1)\n0\np\ncosh(t)2 cosh(t)dt\n=\nZ sinh\u22121(1)\n0\n| cosh(t)2| cosh(t)dt =\nZ sinh\u22121(1)\n0\ncosh(t) cosh(t)dt\n|\n{z\n}\ncosh(t)>0 \u21d2| cosh(t)|=cosh(t)\n=\nZ sinh\u22121(1)\n0\ncosh(2t) + 1\n2\ndt\n=\nZ 2 sinh\u22121(1)\n0\ncosh(u) + 1\n2\n1\n2du\n|\n{z\n}\nt= u\n2\n= 1\n4\nZ 2 sinh\u22121(1)\n0\ncosh(u) + 1du = 1\n4 (sinh(u) + u) |u=2 sinh\u22121(1)\nu=0\n=sinh(2 sinh\u22121(1)) + 2 sinh\u22121(1)\n4\n= 2 sinh(sinh\u22121(1)) cosh(sinh\u22121(1)) + 2 sinh\u22121(1)\n4\n=\n2 sinh(sinh\u22121(1))\nq\n1 + sinh(sinh\u22121(1))2 + 2 sinh\u22121(1)\n4\n= 2 \u00b7 1 \u00b7\n\u221a\n1 + 12 + 2 sinh\u22121(1)\n4\n=2\n\u221a\n2 + 2 sinh\u22121(1)\n4\nExample 8.33. Substitution can be used the generally integrate cos(x)n and sin(x)n when n\nis a positive integer.\n(1) The simplest case is when n odd. Here is an example of that:\nZ\n\u03c0\n2\n0\ncos(x)5dx =\nZ\n\u03c0\n2\n0\ncos(x)(1 \u2212sin(x)2)2dx =\nZ 1\n0\n(1 \u2212u2)2du\n|\n{z\n}\nu(x)=sin(x)\nu(x)\u2032=cos(x)\n=\nZ 1\n0\n(1 \u22122u2 + u4)du =\n\u0012", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_268", "text": "Z 1\n0\n(1 \u2212u2)2du\n|\n{z\n}\nu(x)=sin(x)\nu(x)\u2032=cos(x)\n=\nZ 1\n0\n(1 \u22122u2 + u4)du =\n\u0012\nu \u22122u3\n3\n+ u5\n5\n\u0013\f\f\f\f\nu=1\nu=0\n=1 \u22122\n3 + 1\n5 = 6\n15 = 2\n5.\n(2) On the other hand, when n is even, by reverse-engineering duplication formulas for sine\nand cosine, we can reduce again to the case of an odd power:\nZ\n\u03c0\n2\n0\nsin4(x)dx =\nZ\n\u03c0\n2\n0\n\u00121 \u2212cos(2x)\n2\n\u00132\ndx =\nZ\n\u03c0\n2\n0\n1\n4 \u2212cos(2x)\n2\n+ cos(2x)2\n4\ndx\n=\nZ\n\u03c0\n2\n0\n1\n4dx \u2212\nZ\n\u03c0\n2\n0\ncos(2x)\n2\ndx +\nZ\n\u03c0\n2\n0\ncos(2x)2\n4\ndx\n(8.33.b)\n=\n\u0012\u03c0\n8 \u2212sin(2x)\n4\n\u0013\f\f\f\f\nx= \u03c0\n2\nx=0\n+\nZ\n\u03c0\n2\n0\ncos(4x) + 1\n8\ndx\n=\n\u0012\u03c0\n8 + \u03c0\n16 + sin(4x)\n32\n\u0013\f\f\f\f\nx= \u03c0\n2\nx=0\n= \u03c0\n8 + \u03c0\n16 = 3\u03c0\n16\n141", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_269", "text": "8.5\nIntegration by parts\nTheorem 8.34. If f, g: : E \u2192R be two C1 functions on an open interval E, and let a < b be\nelements of E. Then,\nZ b\na\nf(x)g\u2032(x)dx = f(x)g(x)|b\na \u2212\nZ b\na\nf\u2032(x)g(x)dx,\n(8.34.a)\nProof. It is enough to show that\nZ b\na\n(f(x)g\u2032(x) + f\u2032(x)g(x))dx = f(x)g(x)|b\na .\nThis follows immediately from the Leibniz formula, ??,\n(f(x)g(x))\u2032 = f(x)g\u2032(x) + f\u2032(x)g(x).\nGenerally integration by parts are useful for products. The main question in applying ?? is\nhow one chooses f and g. There is a rule which works in most cases (but not always!). Using\nthe list below, when you encounter a product of two functions both of which belong to one of\nthe categories in the list, the idea is that you should assign g\u2032 in the formula ?? to the function\nthat belongs to the category appearing earlier in the list.\n\u25e6E(xponential)\n\u25e6T(rigonometric)\n\u25e6A(lgebraic, that is, polynomial)\n\u25e6L(ogarithm)\n\u25e6I(nverse trigonometric).\nWe present now a few examples.\nExample 8.35.\nZ b\na\nxexdx\n|\n{z\n}\ng\u2032(x)=ex\ng(x)=ex\nf(x)=x\nf\u2032(x)=1\n= (xex)|b\nx=a \u2212\nZ b\na\nexdx\n= (xex \u2212ex)|b\nx=a = ((x \u22121)ex)|b\nx=a\nExample 8.36.\nZ b\na\nsin(x)exdx\n|\n{z\n}\ng\u2032(x)=ex,g(x)=ex,f(x)=sin(x),f\u2032(x)=cos(x)\n= (sin(x)ex)|b\nx=a \u2212\nZ b\na\ncos(x)exdx\n|\n{z\n}", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_270", "text": "= (sin(x)ex)|b\nx=a \u2212\nZ b\na\ncos(x)exdx\n|\n{z\n}\ng\u2032(x)=ex,g(x)=ex,f(x)=cos(x),f\u2032(x)=\u2212sin(x)\n= (sin(x)ex \u2212cos(x)ex)|b\nx=a +\nZ b\na\n(\u2212sin(x))exdx\nThus, we can rewrite the equality between the LHS of the \ufb01rst line and the second line as\nZ b\na\nsin(x)exdx = (ex(sin(x) \u2212cos(x)))|b\nx=a\n2\n.\n142", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_271", "text": "Example 8.37.\nZ b\na\nlog(x)dx\n|\n{z\n}\nf(x)=log(x),f\u2032(x)= 1\nx ,g\u2032(x)=1,g(x)=x\n= (x log(x))|b\nx=a \u2212\nZ b\na\n1dx = (x log(x) \u2212x)|b\nx=a\nExample 8.38.\nZ b\na\narctan(x)dx\n|\n{z\n}\nf(x)=arctan(x),f\u2032(x)=\n1\n1+x2 ,g\u2032(x)=1,g(x)=x\n= (x arctan(x))|b\nx=a \u2212\nZ b\na\nx\n1 + x2 dx\n= (x arctan(x))|b\nx=a \u2212\n1\n2\nZ 1\nudu\n|\n{z\n}\nu(x)=1+x2,u(x)\u2032=2x\n=\n\u0012\nx arctan(x) \u22121\n2 log |u|\n\u0013\f\f\f\f\nb\nx=a\n=\n\u0012\nx arctan(x) \u22121\n2 log |1 + x2|\n\u0013\f\f\f\f\nb\nx=a\n8.6\nIntegrating rational functions\nA rational function, is a function of the form P(x)\nQ(x), where P(x) and Q(x) are polynomials\nwith real coe\ufb03cients.\nWe start by recalling how polynomials with real coe\ufb03cients can be factorized.\nTheorem 8.39 (Fundamental theorem of algebra over R). Let Q(x) be a polynomial with real\ncoe\ufb03cients. Then, Q(x) can be factored as\nQ(x) = (x \u2212a1)k1 . . . (x \u2212an)kn(x2 + 2b1x + c1)l1 . . . (x2 + 2bmx + cm)lm,\n(8.39.a)\nwhere the ai, bi and ci are real numbers, ki, li > 0 are positive integers, and the quadratic\npolynomials x2 + 2bix + ci, 1 \u2264i \u2264m are irreducible, that is, there is no real number x0 such\nthat x2\n0 + 2bix0 + ci = 0, 1 \u2264i \u2264m.\nRemark 8.40. The Fundamental Theorem of Algebra is originally for polynomials R(x) with\ncomplex coe\ufb03cients. For those polynomial, the statement is even better: namely, polynomials", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_272", "text": "that x2\n0 + 2bix0 + ci = 0, 1 \u2264i \u2264m.\nRemark 8.40. The Fundamental Theorem of Algebra is originally for polynomials R(x) with\ncomplex coe\ufb03cients. For those polynomial, the statement is even better: namely, polynomials\nwith complex coe\ufb03cients can be factored into linear terms. That is,\nR(x) = (x \u2212d1)s1 . . .\n\u0000x \u2212d\u2032\nn\n\u0001sn ,\ndi \u2208C, si \u2208N\u2217.\n(8.40.b)\nThis does work also for R(x) := Q(x) a real polynomial,as R \u2282C, however it may happen\nthat some of the di that are complex and not real. Then, the expression cannot be used for\nintegration because we did not learn integration of complex valued functions. Hence, the idea\nis to collect the di that are real numbers. These numbers provide the ai in ??. Furthermore,\nas we are working with a polynomial R(x) with real coe\ufb03cients,then\n?? is invariant under\nconjugation, since R(x) is. Hence, whenever di is not real, then also the conjugate of (x \u2212di)\nhas to show up in ?? with the same power. That is, we have a factor of the right hand side of\n?? of the form:\n(x \u2212di)si \u0000x \u2212di\n\u0001si =\n\u0000(x \u2212di)\n\u0000x \u2212di\n\u0001\u0001si\n=\n\u0000x \u22122\n\u0000di + di\n\u0001\n+ didi\n\u0001si =\n\u0000x \u22122 Re(di) + |di|2\u0001si\nThen, setting bj := \u2212Re(di), lj := si and cj := |bi|2, we obtain one of the terms of the form\n(x2 + 2bjx + cj)lj in ??.\n143", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_273", "text": "Example 8.41. Take Q(x) = x3 + x2 \u22122, and consider the factorization as in ??. As the\ndegree of Q is three, there must be a linear term (the product of the non-linear terms has even\ndegree). This correspond to a real root of Q(x), so let us search for it.\n(1) Finding the real root.\n1st try: Q(0) = \u22122. As\nlim\nx\u2192+\u221eQ(x) = +\u221e, the Intermediate value theorem, ??, implies\nthat Q has a root greater than 0.\n2nd try: Q(1) = 0. We found the root, great.\n(2) Factoring out the linear term.\nx3\n+\nx2\n\u2212\n2\nx \u22121\nx2 + 2x + 2\nx3\n\u2212\nx2\n2x2\n\u2212\n2\n2x2\n\u2212\n2x\n2x\n\u2212\n2\n2x\n\u2212\n2\n0\nHence, we have\n(x2 + 2x + 2)(x \u22121) = x3 + x2 \u22122\nRemark 8.42. Unfortunately, for polynomials of degree \u22655 there is no algorithm for \ufb01nding\nthe roots; one just has to try to use the Intermediate Value theorem, hoping that the given\npolynomial yields a nice root.\nUsing ??, we have the following nice factorization of rational functions.\nProposition 8.43. Any rational function P(x)\nQ(x) can be written as\nP(x)\nQ(x) = \u03b11R1(x) + \u00b7 \u00b7 \u00b7 + \u03b1tRt(x),\nwhere the \u03b1i are real numbers, and Ri(x) are of the form:\n(1) polynomial, or\n(2)\n1\n(x\u2212r)p , or\n(3)\nx+c\n(x2+2rx+s)p .\nInstead of giving a proof, we explain the idea behind ?? in the following example.\nExample 8.44. Given the rational function 4x3+9x2+11x+8\n(x2+x+1)2\n, let us try to factorize it as follows:\n4x3 + 9x2 + 11x + 8\n(x2 + x + 1)2\n=\nAx + B\n(x2 + x + 1)2 +\nCx + D\nx2 + x + 1\n=Ax + B + (Cx + D)(x2 + x + 1)", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_274", "text": "4x3 + 9x2 + 11x + 8\n(x2 + x + 1)2\n=\nAx + B\n(x2 + x + 1)2 +\nCx + D\nx2 + x + 1\n=Ax + B + (Cx + D)(x2 + x + 1)\n(x2 + x + 1)2\n=Cx3 + (C + D)x2 + (A + C + D)x + (B + D)\n(x2 + x + 1)2\n,\n144", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_275", "text": "which yields the following linear system\n\uf8f1\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f2\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f3\nC\n= 4\nC + D\n= 9\nA + C + D\n= 11\nB + D\n= 8\nfor which the solutions are\nC = 4 \u21d24 + D = 9 \u21d2D = 5\n\u21d2A + 4 + 5 = 11; B + 5 = 8\n\u21d2A = 2; B = 3\nThus,\n4x3 + 9x2 + 11x + 8\n(x2 + x + 1)2\n=\n2x + 3\n(x2 + x + 1)2 +\n4x + 5\nx2 + x + 1.\nHaving the decomposition stated in ??, the question is how we integrate these terms sepa-\nrately.\nExample 8.45.\n\u25e6For p > 1, then\nZ\n1\n(x \u2212r)p dx = (x \u2212r)1\u2212p\n1 \u2212p\n.\n\u25e6For p = 1, then\nZ\n1\n(x \u2212r)dx = log |x \u2212r|.\nExample 8.46.\nZ\nx + c\n(x2 + 2rx + s)p =1\n2\nZ\n2(x + r)\n(x2 + 2rx + s)p dx +\nZ\nc \u2212r\n(x2 + 2rx + s)p dx\n=1\n2\nZ\n2(x + r)\n(x2 + 2rx + s)p dx + (c \u2212r)\nZ\n1\n(x2 + 2rx + s)p dx.\nHence, we need to compute the two integrals\nZ\n2(x + r)\n(x2 + 2rx + s)p dx,\nZ\n1\n(x2 + 2rx + s)p dx,\nindividually.\n\u25e6Using the substitution u = x2 + 2rx + s, then\nZ\n2(x + r)\n(x2 + 2rx + s)p dx =\n\uf8f1\n\uf8f2\n\uf8f3\nlog |x2 + 2rx + s|\nif p = 1\n(x2+2rx+s)1\u2212p\n1\u2212p\nif p > 1\n\u25e6Hence, we now know how to integrate all the terms in ??, except for the integral\nZ\n1\n(x2 + 2rx + s)p dx,\np > 0.\n145", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_276", "text": "Hence,\nZ\n1\n(x2 + 2rx + s)p dx =\nZ\n1\n((x + r)2 + (s \u2212r2))p dx\n=\n1\n(s \u2212r2)p\nZ\n1\n\u0012\u0010\nx+r\n\u221a\ns\u2212r2\n\u00112\n+ 1\n\u0013p dx\n|\n{z\n}\ns\u2212r2>0, as x2+2rx+s has no real roots\n=\n1\n(s \u2212r2)p\u22121\n2\nZ\n1\n(u2 + 1)p du\n|\n{z\n}\nu=\nx+r\n\u221a\ns\u2212r2\n,\nand we are left to compute the integral\nZ\n1\n(u2 + 1)p du,\nfor p > 0.\nSetting\nIp :=\nZ\n1\n(u2 + 1)p du,\nthen I1 := arctan(u) and furthermore, if p \u22651, then we obtain a recursive formula as\nfollows:\nIp =\nZ\n1\n(u2 + 1)p du =\nu\n(u2 + 1)p \u2212\nZ\n(\u2212p)u \u00b7 2u\n(u2 + 1)p+1 du\n|\n{z\n}\nintegrating by parts with f(u)=\n1\n(u2+1)p ,g\u2032(u)=1\n=\nu\n(u2 + 1)p + 2p\nZ\nu2 + 1 \u22121\n(u2 + 1)p+1 du =\nu\n(u2 + 1)p + 2pIp \u22122pIp+1\nSo, by looking at the two ends of the equation, we obtain the recursive equality:\nIp+1 =\nu\n(u2+1)p + (2p \u22121)Ip\n2p\n.\nRemark 8.47. Be careful, there is an error on page 201 of the book where they prove the\nformulas above: intead of 2p \u22121, they wrote 2(p \u22121)!!\nExample 8.48. Let us compute I2 for example:\nI2 =\nu\nu2+1 + I1\n2\n= 1\n2\n\u0012\nu\nu2 + 1 + arctan(u)\n\u0013\n146", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_277", "text": "Example 8.49. Let us get back to ??:\nZ\n4x3 + 9x2 + 11x + 8\n(x2 + x + 1)2\ndx =\nZ\n2x + 3\n(x2 + x + 1)2 dx +\nZ\n4x + 5\nx2 + x + 1dx\n|\n{z\n}\nby ??\n=\nZ\n2x + 3\n\u0010\u0000x + 1\n2\n\u00012 + 3\n4\n\u00112 dx +\nZ\n4x + 5\n\u0000x + 1\n2\n\u00012 + 3\n4\ndx\n|\n{z\n}\ncompleting the square in the denominators\n=\nZ\n(2x + 1) + 2\n\u0010\u0000x + 1\n2\n\u00012 + 3\n4\n\u00112 dx +\nZ\n(4x + 2) + 3\n\u0000x + 1\n2\n\u00012 + 3\n4\ndx\n|\n{z\n}\nwriting the numerators in terms of a multiple of x + 1\n2\n=\nZ\n\u0012 2\n\u221a\n3\n\u00133\n2\n\u221a\n3(2x + 1) +\n4\n\u221a\n3\n\u0012\u0010\n2\n\u221a\n3\n\u0000x + 1\n2\n\u0001\u00112\n+ 1\n\u00132 dx +\nZ\n2\n\u221a\n3\n2\n\u221a\n3(4x + 2) +\n6\n\u221a\n3\n\u0010\n2\n\u221a\n3\n\u0000x + 1\n2\n\u0001\u00112\n+ 1\ndx\n|\n{z\n}\nWe multiply the numerators and the denominators by adequate multiples of\n2\n\u221a\n3 ,\nto make them of the form u2 + 1 or (u2 + 1)2\n= 4\n3\nZ\n2u +\n4\n\u221a\n3\n(u2 + 1)2 du +\nZ\n4u +\n6\n\u221a\n3\nu2 + 1 du\n|\n{z\n}\nu= 2\n\u221a\n3(x+ 1\n2)\n\u21d2\nx=\n\u221a\n3\n2 u\u22121\n2\n\u21d2\nx(u)\u2032=\n\u221a\n3\n2\n=4\n3\nZ\n2u\n(u2 + 1)2 du + 16\n3\n\u221a\n3\nZ\n1\n(u2 + 1)2 du\n+2\nZ\n2u\nu2 + 1dx + 6\n\u221a\n3\nZ\n1\nu2 + 1du\n=4\n3\n\u22121\nu2 + 1 +\n8\n3\n\u221a\n3\n\u0012\nu\nu2 + 1 + arctan(u)\n\u0013\n+ 2 log |u2 + 1| + 6\n\u221a\n3 arctan(u)\n=4\n3\n\u22121\n\u0010\n2\n\u221a\n3\n\u0000x + 1\n2\n\u0001\u00112\n+ 1\n+\n8\n3\n\u221a\n3\n\uf8eb\n\uf8ec\n\uf8ed\n2\n\u221a\n3\n\u0000x + 1\n2\n\u0001\n\u0010\n2\n\u221a\n3\n\u0000x + 1\n2\n\u0001\u00112\n+ 1\n+ arctan\n\u0012 2\n\u221a\n3\n\u0012\nx + 1\n2\n\u0013\u0013\n\uf8f6\n\uf8f7\n\uf8f8\n+2 log\n\f\f\f\f\f\n\u0012 2\n\u221a\n3\n\u0012\nx + 1\n2\n\u0013\u00132\n+ 1\n\f\f\f\f\f + 6\n\u221a\n3 arctan\n\u0012 2\n\u221a\n3\n\u0012\nx + 1\n2\n\u0013\u0013\n147", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_278", "text": "8.6.1\nRational functions in exponentials\nThere is a method of integrating functions obtained by plugging in ex into a rational function.\nWe explain it via the next example:\nExample 8.50.\nZ\n1\nex + 1dx =\nZ\n1\n(ex + 1)ex exdx =\nZ\n1\n(t + 1)tdt\n|\n{z\n}\nt(x)=ex\nt(x)\u2032=ex\n=\nZ \u00121\nt \u2212\n1\nt + 1\n\u0013\ndt = log |t| \u2212log |t + 1|\n= log |ex| \u2212log |ex + 1| = x \u2212log |ex + 1|\n8.6.2\nRational functions in roots\nThere is a method of integrating functions obtained by plugging in \u221ax into a rational function.\nWe explain it via the next example.\nExample 8.51.\nZ\n1\n\u221ax + 1dx =\nZ \u0012\n1\n\u221ax + 12\u221ax\n\u0013 1\n2\n1\n\u221axdx\n=\nZ\n2t\nt + 1dt\n|\n{z\n}\nt(x)=\u221ax\nt(x)\u2032= 1\n2\n1\n\u221ax\n=\nZ\n2dt \u2212\nZ\n2\nt + 1dt\n=2t \u22122 log |t + 1| = 2\u221ax \u22122 log |\u221ax + 1|.\n8.7\nImproper integrals\nSo far we have studied integrals of functions (mostly continuous ones) that are de\ufb01ned over\na closed bounded interval.\nBut how can we make sense of integrating a function over an unbounded interval, for\nexample,\nR +\u221e\n1\n1\nx2 dx? Or more generally, we have a real valued function f that is continuous\non an interval I of the form [a, b[, ]a, b] or ]a, b[, where a, b \u2208R, but either f does not extend\ncontinuously to [a, b], or the interval [a, b] does not exist at all \u2013 such as in the case when a or\nb are \u00b1\u221e.\nDe\ufb01nition 8.52. Let f : I \u2192R be a continuous function.\n(1) If I = [a, b[, a \u2208R, and either b \u2208R or b = +\u221e, then we de\ufb01ne the improper integral of\nf on I to be the limit\nZ b\u2212\na\nf(t)dt := lim\nx\u2192b\u2212\n\u0012Z x\na\nf(t)dt\n\u0013\n,\nprovided that the above limits exist.", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_279", "text": "f on I to be the limit\nZ b\u2212\na\nf(t)dt := lim\nx\u2192b\u2212\n\u0012Z x\na\nf(t)dt\n\u0013\n,\nprovided that the above limits exist.\n(2) If I =]a, b], b \u2208R, and either a \u2208R or a = \u2212\u221e\u221e, then we de\ufb01ne the improper integral\nof f on I to be the limit\nZ b\na+ f(t)dt := lim\nx\u2192a+\n\u0012Z b\nx\nf(t)dt\n\u0013\n,\nprovided that the above limits exist.\n148", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_280", "text": "(3) If I =]a, b[, a, b \u2208R, then we de\ufb01ne the improper integral of f on I to be the limit\nZ b\u2212\na+ f(t)dt :=\nZ c\na+ f(t)dt +\nZ b\u2212\nc\nf(t)dt\nfor any chosen c \u2208I, provided that the improper integrals\nZ c\na+ f(t)dt,\nZ b\u2212\nc\nf(t)dt\nexist.\nIf the limits above exist and are \ufb01nite, then we say that the improper integrals converge. If the\nabove limits diverge, we say that the corresponding improper integral is divergent.\nRemark 8.53.\n(1) It is an easy exercise to verify that part (3) of the above the de\ufb01nition does\nnot depend on the choice of c \u2208I.\n(2) By abuse of notation many times the + and the \u2212is forgotten from the lower and upper\nlimits.\nExample 8.54.\nZ 1\n0+\n1\n\u221a\ntdt = lim\nx\u21920+ 2t\n1\n2\n\f\f\f\nt=1\nt=x = lim\nx\u21920+ 2 \u22122\u221ax = 2\nExample 8.55.\nZ 1\n0+\n1\nt dt = lim\nx\u21920+ log(t)|t=1\nt=x = lim\nx\u21920+ \u2212log(x) = +\u221e\nSo,\nR 1\n0+ 1\nt dt is divergent.\nExample 8.56.\nZ 1\n0+ log(t)dt = lim\nx\u21920+ (log(t)t \u2212t)|t=1\nt=x = \u22121 \u2212lim\nx\u21920+ (log(x)x \u2212x) = \u22121 \u2212lim\nx\u21920+(log(x)x)\nHere, we may compute limx\u21920+(log(x)x) using L\u2019Hospital\u2019s rule:\nlim\nx\u21920+(log(x)x) = lim\nx\u21920+\nlog(x)\n1\nx\n= lim\nx\u21920+\n1\nx\n\u22121\nx2\n= lim\nx\u21920+ \u2212x = 0.\nHence,\nR 1\n0+ log(t)dt = \u22121.\nImproper integrals enjoy many of the basic features of standard integrals.\nProposition 8.57. Let f, g: I \u2192R be continuous functions de\ufb01ned over an interval I, where\nI is either one of the following intervals", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_281", "text": "Hence,\nR 1\n0+ log(t)dt = \u22121.\nImproper integrals enjoy many of the basic features of standard integrals.\nProposition 8.57. Let f, g: I \u2192R be continuous functions de\ufb01ned over an interval I, where\nI is either one of the following intervals\n[a, b\u2032[, ]a\u2032, b], ]a\u2032, b\u2032[,\na, b \u2208R,\na\u2032, b\u2032 \u2208R.\nThen,\n(1) If I = [a, b[ (resp. I =]a, b], I =]a, b[), a, b \u2208R and f extends to a continuous function\nde\ufb01ned over the interval [a, b], then the improper interval\nZ b\u2212\na\nf(x)dx (resp.\nZ b\na+\nf(x)dx,\nZ b\u2212\na+\nf(x)dx).\nconverges and it is equal to\nZ b\na\nf(x)dx.\n149", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_282", "text": "(2) If c \u2208I and c \u0338= sup I, inf I then if the improper interval of f on I converges, we have\nthat\nZ b\na\nf(x)dx +\nZ c\na\nf(x)dx =\nZ b\nc\nf(x)dx.\n(3) Given \u03b1, \u03b2 \u2208R, if the improper integral of f, g over I converge, then also \u03b1f + \u03b2g is\nintegrable on I, and\nZ b\na\n(\u03b1f + \u03b2g)(x)dx = \u03b1\nZ b\na\nf(x)dx + \u03b2\nZ b\na\ng(x)dx\n(4) If 0 \u2264f \u2264g, then\n(i) if the improper integral\nR b\na g(x)dx converges then also the improper integral\nR b\na f(x)dx\ndoes;\n(ii) if the improper integral\nR b\na f(x)dx diverges then also the improper integral\nR b\na g(x)dx\ndoes.\nZ b\na\nf(x)dx \u2264\nZ b\na\ng(x)dx\nRemark 8.58. In part (2-4) of the previous proposition, we have used the simpli\ufb01ed notation\nfor improper integrals that was introduced in ??(2).\nDe\ufb01nition 8.59. In the hypotheses of ??, we say that an improper integral is absolutely\nconvergent if the improper integral de\ufb01ned by integrating the function |f| instead of f is\nconvergent.\nThe following is an an analogue, for improper integrals, of ??.\nProposition 8.60. If an improper integral is absolutely convergent, then it is also convergent.\nExample 8.61. The backwards implication of ?? does not hold, as shown by the next example.\nZ +\u221e\n\u03c0\n4\nsin(t)\nt\ndt =\nlim\nx\u2192+\u221e\n\u2212cos(x)\nx\n\u2212\u2212cos\n\u0000 \u03c0\n4\n\u0001\n\u03c0\n4\n\u2212\nZ +\u221e\n\u03c0\n4\n\u2212cos(t)\n\u2212t2\ndt\n|\n{z\n}\ng\u2032=sin(t)\nf= 1\nt\ng=\u2212cos(t)\nf\u2032=\n1\n\u2212t2\n=\n\u221a\n2\n\u03c0 \u2212\nZ +\u221e\n\u03c0\n4\ncos(t)\nt2\ndt.\n(8.61.a)\nSo,\nZ +\u221e\n\u03c0\n4\nsin(t)\nt\ndt is convergent if so is\nZ +\u221e\n\u03c0\n4\ncos(t)", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_283", "text": "f\u2032=\n1\n\u2212t2\n=\n\u221a\n2\n\u03c0 \u2212\nZ +\u221e\n\u03c0\n4\ncos(t)\nt2\ndt.\n(8.61.a)\nSo,\nZ +\u221e\n\u03c0\n4\nsin(t)\nt\ndt is convergent if so is\nZ +\u221e\n\u03c0\n4\ncos(t)\nt2\ndt. However, the latter is convergent because\nit is absolute convergent:\nZ +\u221e\n\u03c0\n4\n\f\f\f\f\ncos(t)\nt2\n\f\f\f\f dt \u2264\nZ +\u221e\n\u03c0\n4\n1\nt2 dt =\nlim\nx\u2192+\u221e\n \n\u22121\nt\n\f\f\f\f\nx\n\u03c0\n4\n!\n= 4\n\u03c0 +\nlim\nx\u2192+\u221e\n1\nx = 4\n\u03c0.\nThis yields that\nZ +\u221e\n\u03c0\n4\nsin(t)\nt\ndt is convergent.\n150", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1_chunk_284", "text": "However, be careful, the fact that\nZ +\u221e\n\u03c0\n4\ncos(t)\nt2\ndt is absolute convergent, does not mean that\nso is\nZ +\u221e\n\u03c0\n4\nsin(t)\nt\ndt. That is, equation ?? does not work for sin(t)\nt\nreplaced by\n\f\f\f sin(t)\nt\n\f\f\f. And, in\nfact,\nZ +\u221e\n\u03c0\n4\nsin(t)\nt\ndt is not absolute convergent, because\nZ n\u03c0\n\u03c0\n4\n\f\f\f\f\nsin(t)\nt\n\f\f\f\f dt \u2265\nn\nX\nk=1\nZ k\u03c0\u2212\u03c0\n4\nk\u03c0\u22123\u03c0\n4\n\f\f\f\f\nsin(t)\nt\n\f\f\f\f dt \u2265\nn\nX\nk=1\n\u03c0\n2\n \nmin\nt\u2208[k\u03c0\u22123\u03c0\n4 ,k\u03c0\u2212\u03c0\n4 ]\n| sin(t)|\nt\n!\n\u2265\nn\nX\nk=1\n\u03c0\n2\n\uf8eb\n\uf8ec\n\uf8ed\nmin\nt\u2208[k\u03c0\u22123\u03c0\n4 ,k\u03c0\u2212\u03c0\n4 ]\n| sin(t)|\nmax\nt\u2208[k\u03c0\u22123\u03c0\n4 ,k\u03c0\u2212\u03c0\n4 ]\nt\n\uf8f6\n\uf8f7\n\uf8f8=\nn\nX\nk=1\n\u03c0\n2\n1\n\u221a\n2\nk \u2212\u03c0\n4\n=\n\u03c0\n2\n\u221a\n2\nn\nX\nk=1\n1\nk.\nAs\n\u221e\nX\nk=1\n1\nk is divergent, lim\nn\u2192\u221e\nZ n\u03c0\n\u03c0\n4\n\f\f\f\f\nsin(t)\nt\n\f\f\f\f dt does not exist. Therefore,\nZ \u221e\n\u03c0\n4\n\f\f\f\f\nsin(t)\nt\n\f\f\f\f dt is diver-\ngent.\nExample 8.62. A typical application of improper integral is to give an upper bound on in\ufb01nite\nsums. For example\n\u221e\nX\nk=10\n1\nk2 \u2264\nZ +\u221e\n9\n1\nx2 dx = \u22121\nx\n\f\f\f\f\nx\u2192+\u221e\nx=9\n=\n\u0012\nlim\nn\u2192\u221ex\u2192+\u221e\n\u22121\nx\n\u0013\n\u2212\u22121\n9 = 1\n9\n151", "source": "analysis_1", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_0", "text": "MATH 101 (en)\u2013 Analysis I (English)\nNotes for the course given in Fall 2021\nTeacher:Roberto Svaldi\nHead Assistant: Stefano Filipazzi\nNotes written by Zsolt Patakfalvi & Roberto Svaldi\nThursday 30th November, 2023\nThis work is licensed under a Creative Commons \u201cAttribution-\nNonCommercial-NoDerivatives 4.0 International\u201d license.\n1", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_1", "text": "CONTENTS\n1\nReal functions of one variable\n3\n1.1\nLimits of functions and continuity\n. . . . . . . . . . . . . . . . . . . . . . . . .\n3\n1.1.1\nLimits and algebra . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n7\n1.1.2\nLimit and composition . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n8\n1.1.3\nIn\ufb01nite limits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n11\n1.1.4\nOne sided limits\n. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n13\n1.1.5\nMonotone functions\n. . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n14\n1.1.6\nMore on continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n15\n1.1.7\nUniform continuity and Lipschitzianity . . . . . . . . . . . . . . . . . . .\n16\n1.1.8\nLeft and right continuity . . . . . . . . . . . . . . . . . . . . . . . . . . .\n17\n1.1.9\nConsequences of Bolzano-Weierstrass . . . . . . . . . . . . . . . . . . . .\n18\n1.2\nMonotonicity and invertibility of continuous functions\n. . . . . . . . . . . . . .\n20\n2\nDi\ufb00erentiation\n22\n2.1\nComputing derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n25\n2.1.1\nAddition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n26\n2.1.2", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_2", "text": "25\n2.1.1\nAddition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n26\n2.1.2\nMultiplication . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n26\n2.1.3\nDivision . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n26\n2.1.4\nComposition of functions and derivatives\n. . . . . . . . . . . . . . . . .\n27\n2.1.5\nInversion of functions and derivatives . . . . . . . . . . . . . . . . . . . .\n28\n2.1.6\nThe exponential function\n. . . . . . . . . . . . . . . . . . . . . . . . . .\n29\n2.2\nOne sided derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n31\n2.3\nHigher derivatives\n. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n32\n2.4\nLocal and global extrema\n. . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n34\n2.5\nRolle\u2019s and Mean Value theorem\n. . . . . . . . . . . . . . . . . . . . . . . . . .\n36\n2.5.1\nMonotone functions and di\ufb00erentials . . . . . . . . . . . . . . . . . . . .\n37\n2.5.2\nL\u2019H\u02c6opital\u2019s rule . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n39\n2.5.3\nTaylor expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n42\n2.5.4", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_3", "text": "39\n2.5.3\nTaylor expansion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n42\n2.5.4\nApplication of Taylor expansion to local extrema and in\ufb02ection points .\n46\n2.5.5\nConvex and concave functions . . . . . . . . . . . . . . . . . . . . . . . .\n48\n2.6\nAsymptotes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n51\n3\nIntegration\n52\n3.1\nDe\ufb01nition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n52\n3.2\nBasic properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n56\n3.3\nFundamental theorem of calculus . . . . . . . . . . . . . . . . . . . . . . . . . .\n57\n3.4\nSubstitution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n59\n3.5\nIntegration by parts\n. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n62\n3.6\nIntegrating rational functions . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n63\n3.6.1\nRational functions in exponentials\n. . . . . . . . . . . . . . . . . . . . .\n68\n3.6.2\nRational functions in roots\n. . . . . . . . . . . . . . . . . . . . . . . . .\n68\n3.7\nImproper integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n69\n4\nPower series\n73\n4.1", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_4", "text": "68\n3.7\nImproper integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n69\n4\nPower series\n73\n4.1\nTaylor series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .\n78\n2", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_5", "text": "1\nREAL FUNCTIONS OF ONE VARIABLE\nIn this section, we are going to consider functions f : E \u2192R where E is a subset of R and\nstudy their properties. We \ufb01rst start by recalling general basic properties of functions.\n1.1\nLimits of functions and continuity\nIn this section we will de\ufb01ne and discuss the notion of limit of a function at a given point.\nThe notion of limit aims to give a mathematically precise measure of what the local behavior\nof a function is around a given point.\nExample 1.1. The starting point of our investigation is the function f(x) := sin(x)\nx\nnear x = 0.\nAt \ufb01rst sight, f(0) would seem not to be de\ufb01ned, as x appears in the denominator of sin(x)\nx\n.\nOn the other hand, looking at the graph of the function in Figure 1, it would appear that the\nImages/sinx_sux_gr.png\nFigure 1: f(x) = sin(x)\nx\n.\ncloser x is to 0, the closer sin(x)\nx\nis to 1.\nWe can be even more precise if, for example, we consider the sequence (yn), yn := 1\nn, n \u2208N,\nthen lim\nn\u2192\u221eyn = 0 and we can actually show that also the limit lim\nn\u2192\u221ef(yn) exists. Indeed, you\nproved in the exercise sheets that\nlim\nn\u2192\u221e\nsin\n\u0000 1\nn\n\u0001\n1\nn\n= 1.\nSo, even though sin(x)\nx\nis not de\ufb01ned at x = 0, if we set\nf(x) = sin(x)\nx\n, for x \u0338= 0,\nf(0) = 1,\nthen it would appear that f(x) becomes a nice \u201ccontinuous\u201d function around at x = 0, meaning\nthat we could draw the graph of f with just one continuous stroke of the pen.\nThe goal of this section is for us to turn the ideas contained in the previous example into\nsome precise mathematical concepts and de\ufb01nitions and derive further consequences starting\nfrom those. In particular, we will de\ufb01ne precisely why, in the previous example, f(0) = 1 makes\nf(x) \u201ccontinuous\u201d.\nFirst, we need to make of how to unsure that a function f is de\ufb01ned around a point x0 \u2208R,\nso that it makes sense for us to talk about \u201cthe behavior of f around x0\u201d.\n3", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_6", "text": "De\ufb01nition 1.2. A function f : E \u2192R is de\ufb01ned on a punctured neighborhood of x0 \u2208R if for\nsome positive real number \u03b4 \u2208R\u2217\n+, E contains a set of the form (x0 \u2212\u03b4, x0 + \u03b4) \\ {x0}.\nRemark 1.3. Equivalently, we can restate the above de\ufb01nition in the following way:\nA function f : E \u2192R is de\ufb01ned on a punctured neighborhood of x0 \u2208R if for some positive\nreal number \u03b4 \u2208R\u2217\n+, an interval of the form (x0 \u2212\u03b4, x0 + \u03b4) is contained in E \u222a{x0}.\nExample 1.4. The function f(x) := sin(x)\nx\nis de\ufb01ned on any pointed neighborhood of 0. Indeed,\nf is de\ufb01ned on E := R \\ {0} so that\n(\u2212\u03b4, +\u03b4) \\ {0} \u2286E,\n\u2200\u03b4 \u2208R\u2217\n+.\nWe are then ready to give the formal de\ufb01nition of limit.\nDe\ufb01nition 1.5. Let f : E \u2192R and l \u2208R. Assume that E contains a punctured neighborhood\nof x0 \u2208R. Then, lim\nx\u2192x0 f(x) = l if one of the following two equivalent conditions holds:\n(1) For every 0 < \u03b5 \u2208R there exists \u03b4\u03b5 \u2208R\u2217\n+ such that\n\u2200x \u2208E, 0 < |x \u2212x0| < \u03b4\u03b5 \u21d2|f(x) \u2212l| < \u03b5.\n(2) For every sequence (xn) \u2286E \\ {x0} for which lim\nn\u2192\u221exn = x0, we have lim\nn\u2192\u221ef(xn) = l.\nRemark 1.6. Roughly speaking the two de\ufb01nitions mean the following:\n(1) whenever f is de\ufb01ned at x and x is close to x0, then f(x) is close to l. More precisely:\nfor every \u03b5 > 0 there is a \u03b4 > 0 such that if x is closer to x0 than \u03b4 (and f is de\ufb01ned at\nx), then f(x) is closer to l than \u03b5.\n(2) whenever a sequence (yn) is contained in E\\{x0} and it converges to x0, then the sequence", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_7", "text": "x), then f(x) is closer to l than \u03b5.\n(2) whenever a sequence (yn) is contained in E\\{x0} and it converges to x0, then the sequence\n(f(yn)), given by the values of the function f along the xn, converges to l.\nRemark 1.7. We explain why the two conditions in De\ufb01nition 1.5 are equivalent.\n(1) \u21d2(2): Let us \ufb01x a sequence (yn) which is contained in E \\ {x0} and for which\nlim\nn\u2192\u221exn = x0. We have to show that lim\nn\u2192\u221ef(xn) = l. Let us \ufb01x \u03b5 > 0. Then, this yields a\n\u03b4 > 0 as in de\ufb01nition (i). For this \u03b4, there is an n\u03b4 such that |x0 \u2212xn| < \u03b4 for n \u2265n\u03b4,\nand hence for all such n, |l \u2212f(xn)| < \u03b5.\nNOT (1) \u21d2NOT (2): The negation of (i) is that there is an \u03b5 > 0 such that for each\n\u03b4 > 0 we can \ufb01nd y\u03b4 \u2208(x0 \u2212\u03b4, x0 + \u03b4[\\{x0} such that |f(y\u03b4) \u2212l| \u2265\u03b5. De\ufb01ning yn := y 1\nn ,\nthen the sequence (yn) converges to x0, but all values f(yn) have distance at least \u03b5 from\nl, so the sequence (f(xn)) cannot converge to l.\nWe will work more often with the use de\ufb01nition (2) more as it is simpler. Luckily, it is almost\nalways enough for proving that a limit does not exist. We usually use de\ufb01nition (1) only when\n(2) does not work.\nExample 1.8. We show that lim\nx\u21922 x2 = 4 using point (1) of De\ufb01nition 1.5. In order to do this,\nwe proceed as follows: let us \ufb01x \u03b5 > 0; at this point, we need to \ufb01nd \u03b4 > 0 such that\nif 0 < |x \u22122| < \u03b4 then, |x2 \u22124| < \u03b5.\nLet us note that\nx2 \u22124 = (x \u22122)(x + 2).\n4", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_8", "text": "Furthermore, if 0 < |x \u22122| < 1, then 3 < x + 2 < 3; thus, if 0 < |x \u22122| < 1, then\n|x2 \u22124| = |x \u22122||x + 2| < 5|x \u22122|.\n(1.8.a)\nTherefore, taking \u03b4 = min\n\b\n1, \u03b5\n5\n\t\n, we conclude that, if 0 < |x \u22122| < \u03b4 then\n|x \u22122| < 1 and |x \u22122| < \u03b5\n5.\nHence, it follows that\n|x2 \u22124|\n<\n5|x \u22122|\n|\n{z\n}\nUsing (1.8.a), since |x \u22122| < 1\n<\n5\u03b5\n5 = \u03b5.\nExample 1.9. We can repeat the same computation as in the previous example, also using (2)\nof De\ufb01nition 1.5. In general, proving the existence (and \ufb01niteness of the limit using sequence)\ncan be rather tricky: you can try for example to use that de\ufb01nition to compute the limit of\nsin(x)\nx\nat x = 0. Let (xn) be a sequence such that lim\nn\u2192\u221exn = 2. Then, by the algebraic properties\nof the limit, that is, by ??, we know that lim\nn\u2192\u221ex2\nn =\n\u0010\nlim\nn\u2192\u221exn\n\u00112\n= 22 = 4.\nExample 1.10. We can generalize the arguments from the previous examples to show that\nfor any x0 \u2208R, lim\nx\u2192x0 xn = xn\n0 for any n \u2208N.\nUsing the notion of limit, we can also de\ufb01ne the notion of continuity of a function f at a\npoint in D(f).\nDe\ufb01nition 1.11. Let f : E \u2192R be a function, E \u2282R. The function f is continuous at a\npoint x0 \u2208E, if lim\nx\u2192x0 f(x) exists, it is \ufb01nite and lim\nx\u2192x0 f(x) = f(x0).\nRemark 1.12. Implicit in De\ufb01nition 1.11 is the fact that the limit of f(x) at x0 exists. In\nparticular, the subset E \u2286R on which f is assumed to be de\ufb01ned must contain not only x0\nbut also a punctured neighborhood of x0.", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_9", "text": "particular, the subset E \u2286R on which f is assumed to be de\ufb01ned must contain not only x0\nbut also a punctured neighborhood of x0.\nRemark 1.13. The main di\ufb00erence between De\ufb01nition 1.5 and De\ufb01nition 1.11 is the fact that,\nwhile in De\ufb01nition 1.5 we do not require the function f to be de\ufb01ned at the point x0 at which\nwe are trying to compute the limit and when taking the limit we only look at the value of f on\npoints of D(f) \\ {x0}, in the case of De\ufb01nition 1.11, instead, we very much want to allow the\nvalue f(x0) to play a role. More precisely, we have the following characterization of continuity\nat a point, via conditions analogous to those in De\ufb01nition 1.5.\nProposition 1.14. Let f : E \u2192R be a function and let x0 \u2208E. Assume that there exists a\nopen interval of the form (x0 \u2212\u03b4, x0 + \u03b4[, \u03b4 > 0 contained in E. Then, f is continuous at x0 if\nand only if one of the following two equivalent de\ufb01nitions hold:\n(1) For every \u03b5 \u2208R\u2217\n+ there is a \u03b4\u03b5 \u2208R\u2217\n+ such that\n\u2200x \u2208E such that |x \u2212x0| < \u03b4 \u21d2|f(x) \u2212f(x0)| < \u03b5.\n(2) For every sequence (xn) \u2286E for which lim\nn\u2192\u221exn = x0, then lim\nn\u2192\u221ef(xn) = f(x0).\nIn view of the above proposition, Example 1.10 yields the following immediate corollary.\nCorollary 1.15. Fix n \u2208N. Then f(x) = xn is continuous at every x0 \u2208R.\nUsing the next de\ufb01nition, we can rephrase the previous corollary by saying that, for a \ufb01xed\nn \u2208N, the function f : R \u2192R is continuous.\n5", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_10", "text": "De\ufb01nition 1.16. Let f : E \u2192R. Assume that \u2200x0 \u2208E, E contains an open ball centered at\nx0. Then we say that f is continuous if it is continuous at every x0 \u2208E.\nExample 1.17.\n(1) Let us de\ufb01ne the function f : R \u2192R,\nf(x) :=\n(\n0\nx \u0338= 0,\n1\nx = 0.\nThen, f is not continuous at x = 0.\nIn fact, lim\nx\u21920 f(x) = 0, as in the de\ufb01nition we assumed 0 < |x \u2212x0| \u2264\u03b4, so the function\nvalue 1 for x0 = 0 does not cause any problem.\n(2) Let us de\ufb01ne the function f : R \u2192R,\nf(x) :=\n(\n0\nx \u2208Q,\n1\nx \u2208R \\ Q.\nThen, the set of points of R at which f is continuous is empty.\nFor example, let us consider the point 0 \u2208R and let us de\ufb01ne the sequences (y\u2032\nn), (y\u2032\u2032\nn) \u2282\nR \\ {0}, to be\ny\u2032\nn = 1\nn,\ny\u2032\u2032\nn =\n1\n\u221a\nn2 + 1\n,\nn \u22651.\nAs \u2200n \u2208N\u2217,\n1\nn \u2208Q,\n1\n\u221a\nn2+1 \u2208R \\ Q, then \u2200n \u2208N\u2217, f(y\u2032\nn) = 0, while f(y\u2032\u2032\nn) = 1.\nHence, limn\u2192\u221ey\u2032\nn = 0 = limn\u2192\u221ey\u2032\u2032\nn, while limn\u2192\u221ef(y\u2032\nn) = 0, limn\u2192\u221ef(y\u2032\u2032\nn) = 1, hence\nthe limit limx\u21920 f(x) does not exist, and moreover, f is not continuous at 0.\nOne can repeat the same reasoning at any point x0 \u2208R, by taking y\u2032\nn to be a sequence of\nrational numbers converging to x0 (for example, y\u2032\nn = x0 + 1\nn, if x0 is rational, or y\u2032\nn to be\nthe truncation of the decimal form of x0 at the n-th decimal digit, if x0 is irrational) and y\u2032\u2032\nn\nto be a sequence of irrational numbers converging to x0 (for example, y\u2032\u2032\nn = x0 +\n1\n\u221a\nn2+1, if", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_11", "text": "the truncation of the decimal form of x0 at the n-th decimal digit, if x0 is irrational) and y\u2032\u2032\nn\nto be a sequence of irrational numbers converging to x0 (for example, y\u2032\u2032\nn = x0 +\n1\n\u221a\nn2+1, if\nx0 is rational, or y\u2032\u2032\nn = x0+ 1\nn if x0 is irrational). Then, \u2200n \u2208N\u2217, f(y\u2032\nn) = 0, while f(y\u2032\u2032\nn) =\n1, and limn\u2192\u221ey\u2032\nn = x0 = limn\u2192\u221ey\u2032\u2032\nn, while limn\u2192\u221ef(y\u2032\nn) = 0, limn\u2192\u221ef(y\u2032\u2032\nn) = 1, hence\nthe limit limx\u2192x0 f(x) does not exist, and moreover, f is not continuous at x0.\nThis implies that limx\u2192x0 f(x) does not exist at any point x0 \u2208R, in particular, f is not\ncontinuous at any point of R.\nExample 1.18. We claim that lim\nx\u21920 cos(x) = 1.\nIndeed, let (xn) be a sequence converging to 0. Then,\n0 \u2264| cos(xn) \u22121| =\n\f\f\f2 sin2 \u0010xn\n2\n\u0011\f\f\f \u22642x2\nn\n4 = x2\nn\n2 ,\nusing the inequality | sin(x)| \u2264|x|. So, squeeze theorem tells us that lim\nn\u2192\u221e| cos(xn) \u22121| = 0.\nExample 1.19. The limit lim\nx\u21920 sin\n\u00121\nx\n\u0013\ndoes not exist.\nIndeed, consider the sequences xn :=\n1\n\u03c0(2n+ 1\n2) and x\u2032\nn :=\n1\n\u03c0(2n+ 3\n2). Then, \ufb01rst lim\nn\u2192\u221exn = 0 and\nlim\nn\u2192\u221ex\u2032\nn = 0. However,\nlim\nn\u2192\u221esin\n\u0012 1\nxn\n\u0013\n= sin\n\uf8eb\n\uf8ed\n1\n1\n\u03c0(2n+ 1\n2)\n\uf8f6\n\uf8f8= sin\n\u0012\n\u03c0\n\u0012\n2n + 1\n2\n\u0013\u0013\n= 1,\n6", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_12", "text": "but\nlim\nn\u2192\u221esin\n\u0012 1\nx\u2032n\n\u0013\n= sin\n\uf8eb\n\uf8ed\n1\n1\n\u03c0(2n+ 3\n2)\n\uf8f6\n\uf8f8= sin\n\u0012\n\u03c0\n\u0012\n2n + 3\n2\n\u0013\u0013\n= \u22121.\nSo, point (2) of De\ufb01nition 1.5 is not satis\ufb01ed, and hence the limit does not exist\n1.1.1\nLimits and algebra\nDe\ufb01nition 1.5 allows us to translate all the statements about limits of sequences to limits of\nfunctions. Indeed, let us say we are have functions f, g de\ufb01ned around a point x0 \u2208R \u2013 but\nwe are not necessarily assuming that f, g are de\ufb01ned at x0 \u2013 and we want to prove that if l\nand k are the limits of f(x) and g(x) (at x0), then l + k is the limit of (f + g)(x). Let us\ntake a sequence (yn) converging to x0. We know that lim\nn\u2192\u221ef(yn) = l and lim\nn\u2192\u221eg(yn) = k. But\nthen, ?? implies that lim\nn\u2192\u221ef(xn) + f\u2032(xn) = l + k, that is, lim\nn\u2192\u221e(f + f\u2032)(xn) = l + k, which is\nexactly generalizing the statement about limit of sequences and addition, to the case of limit\nof functions.\nUnsurprisingly, at this point, we can do the same with all other properties that we proved\nfor limits of sequences. We collect all the statements one can show along the same arguments:\nProposition 1.20. Let f and g be two functions such that a punctured neighborhood of x0 is\nin the domain of both f and g. Assume that the limits of f and g at x0 exist and they are l\nand k, respectively. Then,\n(1) the limit of f + g exists at x0 and lim\nx\u2192x0(f + g)(x) = l + k\n(2) the limit of f \u00b7 g exists at x0 and lim\nx\u2192x0(f \u00b7 g)(x) = l \u00b7 k\n(3) if k \u0338= 0, then the limit of f\ng exists at x0 and lim\nx\u2192x0\n\u0012f\ng\n\u0013\n(x) = l\nk", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_13", "text": "(2) the limit of f \u00b7 g exists at x0 and lim\nx\u2192x0(f \u00b7 g)(x) = l \u00b7 k\n(3) if k \u0338= 0, then the limit of f\ng exists at x0 and lim\nx\u2192x0\n\u0012f\ng\n\u0013\n(x) = l\nk\n(4) if f(x) \u2264g(x) for any x in a punctured neighborhood of x0, then l \u2264k.\n(5) Squeeze Theorem: if there is a third function h(x) such that there is also a punctured\nneighborhood of x0 in the domain of h, and:\n(i) on some punctured neighborhood of x0 we have f(x) \u2264h(x) \u2264g(x), and\n(ii) l = k,\nthen lim\nx\u2192x0 h(x) = l.\nExample 1.21. The main example for using point (5) of Proposition 1.20 is that lim\nx\u21920\nsin(x)\nx\n=\n1. Indeed, we have already seen, cf. ??, that\n0 \u2264sin(x) \u2264x \u2264tan(x),\nfor x \u2208[0, \u03c0\n2 ],\ntan(x) \u2264x \u2264sin(x) \u22640,\nfor x \u2208[\u2212\u03c0\n2 , 0];\nwhich implies that\n|cos(x)| \u2264\n\f\f\f\f\nsin(x)\nx\n\f\f\f\f \u22641,\nfor x \u2208[\u2212\u03c0\n2 , \u03c0\n2 ].\n(1.21.b)\n7", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_14", "text": "Since \u2200x \u2208[\u2212\u03c0\n2 , \u03c0\n2 ], cos(x) \u22650, whereas sin(x) and x are odd function, so that also \u2200x \u2208\n[\u2212\u03c0\n2 , \u03c0\n2 ] \\ {0}, sin(x)\nx\n\u22650, the chain of inequalities in (1.21.b) holds also once we remove the\nabsolute values. So, by the Squeeze Theorem for limits we can conclude that\nlim\nx\u21920\nsin(x)\nx\n= 1,\nsince lim\nx\u21920 cos(x) = 1, see Example 1.18.\nExample 1.22. For any k \u2208N\u2217, lim\nx\u21920 xk sin\n\u00121\nx\n\u0013\n= 0. Indeed, as | sin\n\u0000 1\nx\n\u0001\n| = 1, \u2200x \u2208R\u2217, then\n\u2212xk \u2264xk sin\n\u00121\nx\n\u0013\n\u2264xk,\n\u2200x \u2208R\u2217.\nand the conclusion follows from the Squeeze Theorem.\nThe above proposition has all the nice consequences about continuity.\nProposition 1.23. If f, g: E \u2192R are continuous functions at x0 \u2208E. Then the following\nfunction are also continuous at x0:\n(1) \u03b1f + \u03b2g for any \u03b1, \u03b2 \u2208R,\n(2) f \u00b7 g, and\n(3)\nf\ng , if g|E is nowhere zero (meaning that for all x \u2208E : g(x) \u0338= 0), then.\nExample 1.24. We collect here some example of continuous functions, on their respective\ndomains, that is, each of these functions is continuous at any point where they are de\ufb01ned:\n\u25e6p(x) = a0 + a1x + a2x2 + \u00b7 \u00b7 \u00b7 + arxr, that is, p(x) is a polynomial in one variable x.\n\u25e6f(x) := 1\nx is continuous on R \\ 0.\n\u25e6\nx\nx2\u22123x+1 is continuous on R \\\nn\n3\u00b1\n\u221a\n5\n2\no\n,\n\u25e6In general, if p(x) and q(x) are two polynomials, then p(x)\nq(x) is continuous on {x \u2208R|q(x) \u0338=\n0} (which is the whole real line minus \ufb01nitely many points).\n1.1.2\nLimit and composition\nLet us recall the following de\ufb01nition of composition of functions.", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_15", "text": "q(x) is continuous on {x \u2208R|q(x) \u0338=\n0} (which is the whole real line minus \ufb01nitely many points).\n1.1.2\nLimit and composition\nLet us recall the following de\ufb01nition of composition of functions.\nDe\ufb01nition 1.25. If f : E \u2192R and g : G \u2192R are functions such that R(f) \u2286G then we may\nde\ufb01ne the composition g \u25e6f (order matters!!) of f with g by\n(g \u25e6f)(x) = g(f(x)).\nExample 1.26. Let us take f : R \u2192R given by f(x) = x2 + 1, and g : R \u2192R given by\ng(y) = y3 + y + 1. Then we have\n(g \u25e6f)(x) = (x2 + 1)3 + (x2 + 1) + 1 = x6 + 3x4 + 4x2 + 3.\n(1.26.c)\nLet us look at an example about whether composition of continuous functions is continuous\nor not.\n8", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_16", "text": "Example 1.27. Consider the functions de\ufb01ned in Example 1.26. We want to show that g\u25e6f is\ncontinuous at x = 0. As (g\u25e6f)(0) = 3 by (1.26.c), we then need to show that lim\nx\u21920(g\u25e6f)(x) = 3.\nand this can be immediately deduced from the last part of the equation in (1.26.c). So, indeed\ng \u25e6f is continuous at x = 0.\nIn general the situation is just as nice as in Example 1.27.\nProposition 1.28. Let f : E \u2192R and g : G \u2192R be two functions. Assume that:\n(1) f(E) \u2286G,\n(2) f is continuous at x0,\n(3) g is continuous at y0 := f(x0).\nThen g \u25e6f is continuous at x0.\nProof. We verify condition (2) of Proposition 1.14. Let (zn) \u2286E be a sequence such that\nlim\nn\u2192\u221ezn = x0.\n(1.28.d)\nAccording to (1.28.d) and our assumption (2), then\nlim\nn\u2192\u221ef(zn) = y0.\n(1.28.e)\nHence\n|\n{z\n}\nDe\ufb01nition 1.25\nlim\nn\u2192\u221e(g \u25e6f)(zn) =\n(1.28.e) and condition (3)\nz\n}|\n{\nlim\nn\u2192\u221eg(f(zn)) = g(y0) .\nRemark 1.29. Let us examine a bit further Example 1.27. In the proof of Proposition 1.28 we\nshowed that if\nlim\nx\u2192x0f(x) = y0\nand\nlim\ny\u2192y0 g(y) = l,\n(1.29.f)\nthen lim\nx\u2192x0g \u25e6f(x) = l holds under the assumption that f and g are continuous.\nWe may\nbe tempted to think that an analogous statement to Proposition 1.28 should also for the\nlimit of a composition of functions, just assuming the condition in (1.29.f). However, as we\nwill see in Example 1.31, this is not true.\nThe reason is that in De\ufb01nition 1.5, contrary\nto Proposition 1.14, there is nothing said about the behavior at x0 and y0. So, we have to", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_17", "text": "will see in Example 1.31, this is not true.\nThe reason is that in De\ufb01nition 1.5, contrary\nto Proposition 1.14, there is nothing said about the behavior at x0 and y0. So, we have to\nassume that f(x) avoids y0 in a punctured neighborhood of x0.\nThe precise statement about composition of functions, in regards to limits, is as follows.\nProposition 1.30. Let f : E \u2192R and g : G \u2192R be functions and let x0 \u2208E be a point such\nthat\n(1) f(E) \u2286G,\n(2)\nlim\nx\u2192x0 f(x0) = y0,\n(3) lim\ny\u2192y0 g(y0) = l\n(4) there is a punctured neighborhood (x0 \u2212\u03b4, x0 + \u03b4) \\ {x0} \u2286E such that for every x in this\nneighborhood, f(x) \u0338= y0.\n9", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_18", "text": "Then, lim\nx\u2192x0(g \u25e6f)(x) = l\nProof. We use part (2) of De\ufb01nition 1.5. Thus, let us \ufb01x a sequence (zn) \u2286E \\ {x0} such that\nlim\nn\u2192\u221ezn = x0.\n(1.30.g)\nIn particular, by throwing away \ufb01nitely many elements of the sequence, we may assume that\n(zn) \u2286(x0 \u2212\u03b4, x0 + \u03b4) \\ {x0} \u2286E.\n(1.30.h)\nBy the assumption (2) in the statement of the proposition, and by (1.30.g), it follows that\nlim\nn\u2192\u221ef(zn) = y0.\n(1.30.i)\nLastly, by our assumption (4) and (1.30.h) we have\n(f(zn)) \u2286G \\ {y0}.\n(1.30.j)\nHence, by our assumption (3) and by (1.30.j), we have\nlim\nn\u2192\u221e(g \u25e6f)(xn) = lim\nn\u2192\u221eg(f(xn)) = l.\nThe following example shows that condition (4) of Proposition 1.30 is necessary. That is,\nif we drop condition (4), the statement of Proposition 1.30 would not hold.\nExample 1.31. Consider:\ng(x) =\n(\n0,\nfor x \u0338= 0,\n1,\nfor x = 0,\nand\nf(x) =\n(\nx sin\n\u0000 1\nx\n\u0001\n,\nfor x \u0338= 0\n0,\nfor x = 0.\nThen, lim\nx\u21920 f(x) = 0 and lim\nx\u21920 g(x) = 0. However, lim\nx\u21920(g \u25e6f)(x) \u0338= 0, because the following two\nsequences induce function value sequences with di\ufb00erent limits:\nxn := 1\n\u03c0n\nand\nyn :=\n1\n\u03c0n + \u03c0\n2\n,\nas\nlim\nn\u2192\u221e(g \u25e6f)(xn) = lim\nn\u2192\u221e1 = 1\nand\nlim\nn\u2192\u221e(g \u25e6f)(yn) = lim\nn\u2192\u221e0 = 0.\nAlso, let us note that condition (4) of Proposition 1.30 below is not satis\ufb01ed in this example,\nas f(x) = 0 for x =\n1", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_19", "text": "and\nlim\nn\u2192\u221e(g \u25e6f)(yn) = lim\nn\u2192\u221e0 = 0.\nAlso, let us note that condition (4) of Proposition 1.30 below is not satis\ufb01ed in this example,\nas f(x) = 0 for x =\n1\n\u03c0n, so there is no punctured neighborhood of 0 on which the function of f\navoids the value 0.\nExample 1.32. A positive example for applying Proposition 1.30 is during the argument of\nshowing that lim\nx\u21920\nsin(x2)\nx2\n= 1. Indeed, if we set g(x) := sin(x)\nx\n, and f(x) = x2, then condition\n(4) of Proposition 1.30 is also satis\ufb01ed, as f(x) \u0338= 0 for x \u0338= 0.\n10", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_20", "text": "1.1.3\nIn\ufb01nite limits\nDe\ufb01nition 1.33. A neighborhood of +\u221e(resp. \u2212\u221e) is an unbounded interval of the form\n(a, +\u221e) (resp. (\u2212\u221e, a)).\nWe extend the de\ufb01nition of limit to comprise the case where we allow ourselves to work\nwith the extended real line R.\nDe\ufb01nition 1.34. Let x0, l \u2208R, and let f : E \u2192R be a function, E \u2282R. Assume that E\ncontains a punctured neighborhood of x0. We say that the limit of f(x) at x0 is l, if for any\nsequence (yn) \u2286E \\ {x0}1, whenever lim\nn\u2192\u221eyn = x0, then lim\nn\u2192\u221ef(yn) = l.\nExample 1.35. We show that lim\nx\u21920\n1\nx2 = +\u221e. Indeed, if (xn) \u2282R\u2217is a sequence satisfying\nlim\nn\u2192\u221exn = 0, then lim\nn\u2192\u221e\n1\nx2n = +\u221eby algebraic properties of limits of sequences.\nOn the other hand, the limit lim\nx\u21920\n1\nx does not exist. In fact, considering the sequence xn := 1\nn,\nthen lim\nn\u2192\u221e\n1\nxn = lim\nn\u2192\u221en = +\u221e, while for yn = \u22121\nn , then lim\nn\u2192\u221e\n1\nyn = lim\nn\u2192\u221e\u2212n = \u2212\u221e.\nProposition 1.36. Let x0 \u2208R, and let f, g: E \u2192R be functions.\n(1) Addition rule. Assume that the following conditions are satis\ufb01ed:\n\u25e6\nlim\nx\u2192x0 f(x) = +\u221e(resp. \u2212\u221e), and\n\u25e6g(x) is bounded from below (resp. from above)\nthen lim\nx\u2192x0(f + g)(x) = +\u221e(resp. \u2212\u221e).\n(2) Product rule. Assume that the following conditions are satis\ufb01ed:\n\u25e6\nlim\nx\u2192x0|f(x)| = +\u221e,\n\u25e6there exists \u03b4 > 0 such that |g(x)| \u2265\u03b4 for all x \u2208E, and\n\u25e6f(x)g(x) > 0 (resp. < 0) for all x \u2208E,\nthen lim", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_21", "text": "\u25e6\nlim\nx\u2192x0|f(x)| = +\u221e,\n\u25e6there exists \u03b4 > 0 such that |g(x)| \u2265\u03b4 for all x \u2208E, and\n\u25e6f(x)g(x) > 0 (resp. < 0) for all x \u2208E,\nthen lim\nx\u2192x0f(x)g(x) = +\u221e(resp. \u2212\u221e).\n(3) First division rule. If\n\u25e6f(x) is bounded,\n\u25e6g(x) is nowhere zero, and\n\u25e6\nlim\nx\u2192x0|g(x)| = +\u221e.\nThen lim\nx\u2192x0\nf(x)\ng(x) = 0\n(4) Second division rule. If\n\u25e6\nlim\nx\u2192x0g(x) = 0,\n\u25e6there is a \u03b4 > 0 such that |f(x)| \u2265\u03b4 for all x \u2208E, and\n\u25e6f(x)/g(x) > 0 (resp. < 0) for all x \u2208E,\nthen lim\nx\u2192x0\nf(x)\ng(x) = +\u221e(resp. \u2212\u221e).\n1When x0 = \u00b1\u221e, then the condition that x0 does not belong to E is automatically satis\ufb01ed, since E \u2286R.\n11", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_22", "text": "(5) Squeeze. If f(x) \u2264g(x), and\n\u25e6if lim\nx\u2192x0f(x) = +\u221e, then lim\nx\u2192x0g(x) = +\u221e\n\u25e6if lim\nx\u2192x0g(x) = \u2212\u221e, then lim\nx\u2192x0f(x) = \u2212\u221e\nExample 1.37. Here are a few examples.\n\u25e6lim\nx\u21920\n1\nx2\n|{z}\n\u2192\u221e\n+ cos(x)\n| {z }\nbounded\n= +\u221e\n\u25e6\nlim\nx\u2192+\u221ecos(x)\n| {z }\nbounded\n\u00b7 (\u2212x2 + x3)\n|\n{z\n}\n\u2192+\u221e\n= \u2212\u221e, since \u2212x2 + x3 = 0 only for x = 0, 1.\n\u25e6\nlim\nx\u2192+\u221e\nbounded\nz\n}|\n{\narctan(x)\n(\u2212x)\n| {z }\n\u2192\u2212\u221e\n= 0.\nRemark 1.38.\n(1) The assumptions stated in part (1) of Proposition 1.36 are important, as\notherwise we can have all di\ufb00erent kinds of limits. We give examples of this using the\nfunctions f(x) = x3, g(x) = x2 and h(x) = x3 + 1. We have\n\u25e6\nlim\nx\u2192+\u221ef(x) =\nlim\nx\u2192+\u221ex3 = +\u221e,\nlim\nx\u2192+\u221e\u2212f(x) =\nlim\nx\u2192+\u221e\u2212x3 = \u2212\u221e\n\u25e6\nlim\nx\u2192+\u221eh(x) =\nlim\nx\u2192+\u221e(x3 + 1) = +\u221e,\nlim\nx\u2192+\u221e\u2212h(x) =\nlim\nx\u2192+\u221e\u2212(x3 + 1) = \u2212\u221e, and\n\u25e6\nlim\nx\u2192+\u221eg(x) =\nlim\nx\u2192+\u221ex2 = +\u221e, lim\nx\u2192+\u221e\u2212g(x) =\nlim\nx\u2192+\u221e\u2212x2 = \u2212\u221e.\nOn the other hand:\n\u25e6\nlim\nx\u2192+\u221ef(x) \u2212g(x) =\nlim\nx\u2192+\u221ex3 \u2212x2 =\nlim\nx\u2192+\u221ex2(x \u22121) = +\u221e,\n\u25e6\nlim\nx\u2192+\u221eg(x) \u2212f(x) = \u2212\u221e, and\n\u25e6\nlim\nx\u2192+\u221ef(x) \u2212h(x) = \u22121.", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_23", "text": "lim\nx\u2192+\u221ex2(x \u22121) = +\u221e,\n\u25e6\nlim\nx\u2192+\u221eg(x) \u2212f(x) = \u2212\u221e, and\n\u25e6\nlim\nx\u2192+\u221ef(x) \u2212h(x) = \u22121.\nIn particular, never use addition law for limits of the type (+\u221e) + (\u2212\u221e).\n(2) The above assumptions for point (2) of Proposition 1.36 are also important. We give\nexamples of this using the functions f(x) = x, g(x) = cos(x)\nx\nand h(x) = (\u22121)[x]. We have\n\u25e6\nlim\nx\u2192+\u221e|f(x)| = +\u221e,\n\u25e6|g(x)| is not bounded from below, and\n\u25e6|h(x)| is bounded from below, but f(x)h(x) \u0338> 0.\nThen:\n\u25e6\nlim\nx\u2192+\u221ef(x)g(x) =\nlim\nx\u2192+\u221ecos(x) does not exist, and\n\u25e6\nlim\nx\u2192+\u221ef(x)h(x) =\nlim\nx\u2192+\u221ex(\u22121)[x] does not exist, on the other hand\n\u25e6the product law applies to (f(x)h(x))(h(x)) and yields\nlim\nx\u2192+\u221e(f(x)h(x))(h(x)) =\n+\u221e\nNever try to use product rule to limits of the type 0 \u00b7 \u221e.\n12", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_24", "text": "(3) The assumptions of the \ufb01rst division rule are also important. One can can show that in\nthe \u00b1\u221e\n\u00b1\u221ecase anything can happen for example using 1\nx,\n1\nx2 ,\n1\nx3 ,(\u22121)[ 1\nx] 1\nx with limit at 0:\n\u25e6lim\nx\u21920\n1\nx\n1\nx2 = lim\nx\u21920x = 0,\n\u25e6lim\nx\u21920\n(\u22121)[ 1\nx] 1\nx\n1\nx\n= lim\nx\u21920(\u22121)[ 1\nx] does not exist and bounded,\n\u25e6lim\nx\u21920\n1\nx2\n1\nx = lim\nx\u21920\n1\nx does not exist and unbounded, and\n\u25e6lim\nx\u21920\n1\nx3\n1\nx = lim\nx\u21920\n1\nx2 = +\u221e.\nSimilar examples show that the assumptions are important for the second division rule.\nNever try to use division rules to limits of the form \u00b1\u221e\n\u00b1\u221eand 0\n0.\n1.1.4\nOne sided limits\nThe main question is how to make sense of limits such as at 0 of \u221ax, as here the domain does\nnot contain a punctured neighborhood of 0. The solution for this is the introduction of the\nnotions of left and right limits.\nDe\ufb01nition 1.39. A function f : E \u2192R is de\ufb01ned to the left (resp. to the right) of x0 \u2208R, if\nE contains an interval of the form (x0 \u2212\u03b4, x0[ (resp. (x0, x0 + \u03b4[).\nDe\ufb01nition 1.40. Let f : E \u2192R be a function and x0 \u2208R. Assume that f is de\ufb01ned to the\nleft (resp. right) of x0. Let l \u2208R.\n(1) We say that the limit of f for x that goes to x0 from the left is l if for all sequences\n(xn) \u2286{x \u2208E|x < x0}, whenever lim\nn\u2192\u221exn = x0 then lim\nn\u2192\u221ef(xn) = l. When this condition\nis satis\ufb01ed, we write lim\nx\u2192x\u2212\n0\nf(x) = l.\n(2) We say that the limit of f for x that goes to x0 from the right is l if for all sequences\n(xn) \u2286{x \u2208E|x > x0}, whenever lim\nn\u2192\u221exn = x0 then lim", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_25", "text": "0\nf(x) = l.\n(2) We say that the limit of f for x that goes to x0 from the right is l if for all sequences\n(xn) \u2286{x \u2208E|x > x0}, whenever lim\nn\u2192\u221exn = x0 then lim\nn\u2192\u221ef(xn) = l. When this condition\nis satis\ufb01ed, we write lim\nx\u2192x+\n0\nf(x) = l.\nExample 1.41. Consider the function f : R+ \u2192R de\ufb01ned as f(x) := \u221ax. We claim that\nlim\nx\u21920+\n\u221ax = 0.\nIndeed, \ufb01x a sequence (xn) \u2286R\u2217\n+ such that lim\nn\u2192\u221exn = 0. We have to show that then lim\nn\u2192\u221e\n\u221axn =\n0 too. So, we need to show that for each \u03b5 > 0, there is an n0 such that for every integer n \u2265n0,\n\u221axn \u2264\u03b5. However, we know that lim\nn\u2192\u221exn = 0. So, we know that there is an n0 such that\n|xn| < \u03b52 for all n \u2265n0. But then, for any such n we also have \u221axn < \u03b5.\nProposition 1.42. Let f : E \u2192R be a function such that there is a punctured neighborhood\nof x0 contained in E, and both\nl1 := lim\nx\u2192x\u2212\n0\nf(x)\nand\nl2 := lim\nx\u2192x+\n0\nf(x)\nexists. Then\nlim\nx\u2192x0f(x) = l\n\u21d0\u21d2\nl1 = l2 = l.\n13", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_26", "text": "Example 1.43. Consider the function f(x) = {x} =. Both left and right limits exist at all\npoints, and furthermore:\nlim\nx\u2192x\u2212\n0\n{x} =\n\u001a {x}\nif x /\u2208Z\n1\nif x \u2208Z\nand\nlim\nx\u2192x+\n0\n{x} =\n\u001a {x}\nif x /\u2208Z\n0\nif x \u2208Z\nHence, according to Proposition 1.42,\nlim\nx\u2192x0{x} exists \u21d4x /\u2208Z.\nExample 1.44. Example 1.41 together with Proposition 1.42 show that the function f(x) =\np\n|x| is continuous in 0. It is not hard to show that f is actually continuous everywhere in R.\n1.1.5\nMonotone functions\nFor a monotone function f, the left (resp. the right) limits always exists at any point x0 at\nwhich the function is de\ufb01ned at the left (resp. the right) of x0.\nProposition 1.45. Let f : E \u2192R be a monotone function. Then, at each point x0 \u2208E:\n(1) if f is de\ufb01ned on the left of x0, lim\nx\u2192x\u2212\n0\nf(x) exists,\n(2) if f is de\ufb01ned on the right of x0, lim\nx\u2192x+\n0\nf(x) exists, and\n(3) if f is de\ufb01ned in a neighborhood of \u00b1\u221e, then\nlim\nx\u2192\u00b1\u221ef(x) exists.\nProof. We treat only the increasing case, as the decreasing one follows from that by regarding\n\u2212f instead of f. Also, we treat only the \ufb01rst case as the others are similar. Set:\nl := sup{f(x)|x \u2208E, x < x0}.\n(1.45.k)\nLet\n(xn) \u2286{x \u2208E|x < x0}\nsuch that\nlim\nn\u2192\u221exn = x0.\n(1.45.l)\nWe have to show that lim\nn\u2192\u221ef(xn) = l. Fix a \u03b5 > 0. Then, by the de\ufb01nition of l, there is an\nx\u2032 \u2208{x \u2208E|x < x0}, such that\nf(x\u2032) > l \u2212\u03b5.\n(1.45.m)", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_27", "text": "n\u2192\u221ef(xn) = l. Fix a \u03b5 > 0. Then, by the de\ufb01nition of l, there is an\nx\u2032 \u2208{x \u2208E|x < x0}, such that\nf(x\u2032) > l \u2212\u03b5.\n(1.45.m)\nAccording to (1.45.l), there is an n0 \u2208N such that for all integers n \u2265n0 we have\nx\u2032 < xn < x0.\n(1.45.n)\nHowever, then for all integers n \u2265n0 we have:\nl \u2265\nf(xn)\n| {z }\n(1.45.k) and (1.45.n)\n\u2265\nf(x\u2032)\n| {z }\nf is increasing and (1.45.n)\n\u2265l \u2212\u03b5.\n| {z }\n(1.45.m)\nThis shows that lim\nn\u2192\u221ef(xn) = l indeed.\nExample 1.46.\n(1) Let\nf(x) := sgn(x) =\n\uf8f1\n\uf8f4\n\uf8f2\n\uf8f4\n\uf8f3\n1\nif x > 0\n0\nif x = 0\n\u22121\nif x < 0.\n14", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_28", "text": "Images/sgn_gr.png\nFigure 2: f(x) = sgn(x).\nThen\nlim\nx\u2192x\u2212\n0\nsgn(x) = \u22121\nand\nlim\nx\u2192x+\n0\nsgn(x) = 1\nNote that these limits exist and neither of them agree with f(0) = 0.\n(2) Let f(x) := \u230ax\u230b. Then:\nlim\nx\u2192x\u2212\n0\nf(x) =\n(\n\u230ax\u230b\nx /\u2208Z\nx \u22121\nx \u2208Z\nand\nlim\nx\u2192x+\n0\nf(x) =\n(\n\u230ax\u230b\nx /\u2208Z\nx\nx \u2208Z.\nSo, the left and right limits exist, despite having di\ufb00erent values whenever x \u2208Z.\n1.1.6\nMore on continuity\nFirst, we note that there are more algebraic rules of continuity (we already discussed addition,\nmultiplication and division in Proposition 1.23):\nProposition 1.47. If f, g: E \u2192R are functions that are continuous at x0 \u2208E, then so are:\n(1) |f|,\n(2) max{f, g}, where\nmax{f, g}(x) := max{f(x), g(x)}\n(3) min{f, g} (de\ufb01ned similarly),\n(4) f+ := max{f, 0},\n(5) f\u2212:= min{f, 0}.\nExample 1.48. We can use for example the continuity of the absolute value for squeezing.\nFor example, let\ng(x) :=\n(\n1\nfor x \u2208Q\nx\nfor x \u2208R \\ Q\n15", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_29", "text": "We claim that g(x) is continuous at x0 = 1. The main idea is that we can try to apply the\nSqueeze Theorem for the limit of functions using the following chain of inequalities:\n\u2212|x \u22121| \u2264g(x) \u22121 \u2264|x \u22121|.\nAccording to point (1) of Proposition 1.47, the function |x \u22121| is continuous everywhere over\nR; thus,\nlim\nx\u21921 \u2212|x \u22121| = lim\nx\u21921 |x \u22121| = 0,\nso that by Proposition 1.20(5), it follows that lim\nx\u21921 f(x) \u22121 = 0. Hence, lim\nx\u21921 f(x) = 1 = f(1)\nand f is continuous at x0 = 1.\n1.1.7\nUniform continuity and Lipschitzianity\nWe introduce a stronger version of continuity.\nDe\ufb01nition 1.49. A function f : E \u2192R is said to be uniformly continuous if for every \u03b5 > 0\nthere is a \u03b4\u03b5 > 0 such that for all x, y \u2208E then\nif |x \u2212y| < \u03b4\u03b5 \u21d2|f(x) \u2212f(y)| < \u03b5.\nRemark 1.50. The notion of uniform continuity de\ufb01ned above is much stronger than that of\ncontinuity, cf. De\ufb01nition 1.11. More precisely, using the characterization of continuity for a\nfunction f : E \u2192R given in Proposition 1.14, for any x0 \u2208E and any \u03b5 > 0 there exists \u03b4 > 0,\nwhich depends on \u03b5 and x0, such that for any x \u2208E\nif |x \u2212x0| < \u03b4 =\u21d2|f(x) \u2212f(x0)| < \u03b5.\nIn De\ufb01nition 1.49, the for a \ufb01xed \u03b5 > 0, the existence of \u03b4 > 0 is no longer dependent on the\nchoice of a base point x0 \u2208E; instead, at this point such choice can be made independently\n(or rather, uniformly) from the points of E: it just depends on the choice of \u03b5.\nIn view of the observation of the previous remark, we immediately have the following\nproposing showing that uniform continuity is a stornger property than continuity.\nProposition 1.51. If f : E \u2192R is uniformly continuous then it is continuous.", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_30", "text": "In view of the observation of the previous remark, we immediately have the following\nproposing showing that uniform continuity is a stornger property than continuity.\nProposition 1.51. If f : E \u2192R is uniformly continuous then it is continuous.\nExample 1.52. The function f(x) := x2 : R \u2192R is not uniformly continuous. On the other\nhand, we have already seen that it is continuous as it is a polynomial.\nIndeed, for any x, y \u2208R,\n|x2 \u2212y2| = |x + y| \u00b7 |x \u2212y|.\nSo, for any \u03b5 > 0 and \u03b4 > 0, we may chose x, y \u2208R such that |x + y| > 2\u03b5\n\u03b4 , and |x \u2212y| = \u03b4\n2 \u2013 to\ndo that, it su\ufb03ces to choose two real numbers x, y that are very large but very close to each\nother. Thus, it follows that\n|x \u2212y| < \u03b4\nand\n|x2 \u2212y2| > 2\u03b5\n\u03b4\n\u03b4\n2 = \u03b5.\nWe will see in the next section that if we consider a continuous function f over a closed bounded\ninterval [a, b] \u2013 rather than on an unbounded domain as in this case, where x2 is considered\nover R \u2013 then f is absolutely continuous.\n16", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_31", "text": "Example 1.53. We show that cos(x) : R \u2192R is uniformly continuous and hence continuous.\nIndeed,\n| cos(x) \u2212cos(y)| = 2\n\f\f\f\fsin\n\u0012x + y\n2\n\u0013\f\f\f\f\n\f\f\f\fsin\n\u0012x \u2212y\n2\n\u0013\f\f\f\f \u22642\n\f\f\f\fsin\n\u0012x \u2212y\n2\n\u0013\f\f\f\f \u22642|x \u2212y|.\nSo, if we set \u03b4 = \u03b5\n2, then we have\n|x \u2212y| \u2264\u03b4 \u21d2| cos(x) \u2212cos(y)| \u22642|x \u2212y| \u22642\u03b4 = 2\u03b5\n2 = \u03b5.\nThis result, together with Proposition 1.28, implies that also functions such as cos(x2), cos2(x),\netc. are continuous.\nWe introduce now a property that makes it particularly easy to show that a function is\nuniformly continuous.\nDe\ufb01nition 1.54. A function f : E \u2192R is said to be Lipschitz if there exists a positive real\nnumber C such that for every x, y \u2208E, |f(x) \u2212f(y)| \u2264C|x \u2212y|.\nWhen the conditions of De\ufb01nition 1.54 are satis\ufb01ed we say that C is a Lipschitz constant\nfor the Lipschitz function f.\nProposition 1.55. Let f : E \u2192R be a function which is Lipschitz with Lipschitz constant C,\nwhere E is an open interval. Then f is uniformly continuous on E; hence f is also continuous\non E.\nProof. For a \ufb01xed positive real number \u03b5 in De\ufb01nition 1.49, it su\ufb03ces to take \u03b4 := \u03b5\nC .\nExample 1.56. Let f : [0, 1] \u2192R be the function f(x) = x2. Then for any x, y \u2208[0, 1],\n|x2 \u2212y2| = |x \u2212y||x + y| \u2264C|x \u2212y|,\n(1.56.o)\nwhere C := sup{|x + y| | x, y \u2208[0, 1]. By de\ufb01nition, C \u22642 \u2013 it not hard to show that actually\nC = 2 \u2013 hence we can rewrite (1.56.o) as\n|x2 \u2212y2| = |x \u2212y||x + y| \u22642|x \u2212y|.\nHence, f is Lipschitz and thus uniformly continuous. Let us notice that", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_32", "text": "C = 2 \u2013 hence we can rewrite (1.56.o) as\n|x2 \u2212y2| = |x \u2212y||x + y| \u22642|x \u2212y|.\nHence, f is Lipschitz and thus uniformly continuous. Let us notice that\nWe will see in Example 1.62 that there exist functions that are uniformly continuous but\nnot Lipschitz.\n1.1.8\nLeft and right continuity\nLastly, we introduce left and right continuity, and we use this to de\ufb01ne continuity on a closed\ninterval.\nDe\ufb01nition 1.57. Let f : E \u2192R be a function, and x0 \u2208E.\n(1) f is left continuous at x0, if lim\nx\u2192x\u2212\n0\nf(x) = f(x0).\n(2) f is right continuous at x0, if lim\nx\u2192x+\n0\nf(x) = f(x0).\nIn De\ufb01nition 1.16 we de\ufb01ned what it means to be continuous on an open interval. For\nfunctions the domains of which are closed intervals the de\ufb01nition has to use left and right\nlimits as well at the two endpoints:\n17", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_33", "text": "De\ufb01nition 1.58. A function f : [a, b] \u2192R is continuous if:\n(1) f is continuous at any point contained in (a, b);\n(2) f is left continuous at b; and,\n(3) f is right continuous at a.\nExample 1.59. The function f : [\u22121, 1] \u2192R de\ufb01ned as f(x) :=\n\u221a\n1 \u2212x2 is continuous. Indeed\nthis is true by the following (where we use that g(y) = \u221ay is continuous on R\u2217\n+, which will be\na consequence of our general theorem about the continuity of the inverse. Indeed, by applying\nthe statement of Theorem 1.74 to f(x) = x2 one obtains that f\u22121 = g is continuous on R\u2217\n+):\n(1) if \u22121 < c < 1, then\n\u221a\n1 \u2212x2 at c is continuous because\n\u221a\n1 \u2212x2 is the composition of \u221ay\nand 1 \u2212x2, and the latter is continuous at c and the former is continuous at 1 \u2212c2 (as\n1 \u2212c2 > 0).\n(2)\n\u221a\n1 \u2212x2 is left continuous at 1, because for all (xn) converging to 1 from the left we have\nlim\nn\u2192\u221e\np\n1 \u2212x2n = 0, as lim\nn\u2192\u221e1 \u2212x2\nn = 0, and lim\nx\u21920+\n\u221ay = 0 according to Example 1.41.\n(3)\n\u221a\n1 \u2212x2 is right continuous at \u22121 by almost verbatim the same argument as the previous\npoint, one only needs to take lim\nn\u2192\u221exn = \u22121 instead of 1.\n1.1.9\nConsequences of Bolzano-Weierstrass\nIn this subsection we shall show how continuous functions de\ufb01ned over bounded closed intervals\nbehave nicely.\nThe proofs of all the results illustrated in this subsection heavily relies on\nBolzano-Weierstrass ??. As we will be assuming, throughout this section, that the domain\nD(f) of a function f is closed bounded interval, given a sequence (xn) \u2286D(f), by ?? we will\nalways be able to assume that we can pass to a converging subsequence (xnk) \u2286(xn) whose\nlimit belongs to D(f), since we are assuming D(f) is closed.", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_34", "text": "always be able to assume that we can pass to a converging subsequence (xnk) \u2286(xn) whose\nlimit belongs to D(f), since we are assuming D(f) is closed.\nWe start by showing that a continuous function de\ufb01ned over a closed bounded interval is\nalways uniformly continuous.\nTheorem 1.60. Let a, b \u2208R. If f : [a, b] \u2192R is continuous, then f is uniformly continuous.\nProof. Assume that f is not uniformly continuous. Then there is a \u03b5 > 0 such that for every\n1\nn there are xn and yn \u2208[a, b] such that |xn \u2212yn| \u22641\nn and |f(xn) \u2212f(yn)| > \u03b5. By Bolzano-\nWeierstrass (??) we may assume that lim\nn\u2192\u221exn = x0 \u2208[a, b].\nHowever, then the condition\n|xn \u2212yn| \u22641\nn yields that we have also lim\nn\u2192\u221eyn = x0. Using again |xn \u2212yn| \u22641\nn together with\nthe continuity of f we obtain that |f(x0) \u2212f(x0)| \u2265\u03b5. This is a contradiction.\nRemark 1.61. For Theorem 1.60 to hold true, it is very important that the domain of f is a\nclosed bounded interval [a, b] for a, b \u2208R. We have already seen that the function f : R \u2192R,\nf(x) = x2 is not uniformly continuous on any interval of the form (a, +\u221e[, a \u2208R \u222a{\u2212\u221e}.\nExample 1.62. The following example shows that there exists uniformly continuous functions\nthat are not Lipschitz.\nLet us consider f : R+ \u2192R, f(x) = \u221ax. Let us \ufb01x a real number a > 0, and de\ufb01ne g: [0, a] \u2192R\nby g = f|[0,a], g(x) = \u221ax. Theorem 1.60 shows that g is uniformly continuous, as we are taking\nthe domain of de\ufb01nition of g to be a closed bounded interval. To show that g is not Lipschitz,\nit su\ufb03ces to show that for any C \u2208R\u2217", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_35", "text": "the domain of de\ufb01nition of g to be a closed bounded interval. To show that g is not Lipschitz,\nit su\ufb03ces to show that for any C \u2208R\u2217\n+0 there exists s, t \u2208[0, a], s \u0338= t such that |g(s) \u2212g(t)| >\nC|s \u2212t|, or, equivalently, |g(s)\u2212g(t)|\n|s\u2212t|\n> C. Let us \ufb01x C > 0. Then, there exists t \u2208]0, 1[ such\nthat\n1\n\u221a\nt > C, since limx\u21920+\n1\n\u221ax = +\u221e. Taking s = 0, then |g(s)\u2212g(t)|\n|s\u2212t|\n=\n\u221a\nt\nt =\n1\n\u221a\nt > C, which is\nwhat we wanted to prove. It is not hard to show, that actually even f(x) = \u221ax is uniformly\ncontinuous but not Lipschitz. The proof is left as an exercise.\n18", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_36", "text": "Theorem 1.63. If f : [a, b] \u2192R is continuous for some a, b \u2208R, then there are c, d \u2208[a, b]\nsuch that\nM := sup\nx\u2208[a,b]\nf(x) = max\nx\u2208[a,b] f(x) = f(c),\nm :=\ninf\nx\u2208[a,b] f(x) = min\nx\u2208[a,b] f(x) = f(d).\nRemark 1.64. The above theorem can be restated by saying that for a given function f : [a, b] \u2192\nR, if f is continuous then the range R(f) of f is a closed and bounded interval, R(f) = [c, d].\nProof. We only prove the existence of maxx\u2208[a,b] f(x), the case of minx\u2208[a,b] f(x) follows simi-\nlarly.\nFirst we prove that f is bounded from above, so that supx\u2208[a,b] f(x) must exist. Assume, by\ncontradiction, that f is not bounded from above. That means that for each integer n > 0\nthere is xn \u2208[a, b] such that f(xn) \u2265n. As (xn) \u2286[a, b] is a bounded sequence, by Bolzano-\nWeierstrass ??, there exists a convergent subsequence (xnk) \u2286(xn). Set c := lim\nk\u2192\u221exnk. Then\nc \u2208[a, b], and the following chain of equalities yields a contradiction:\nR \u220bf(c) =\nlim\nk\u2192\u221ef(xnk)\n|\n{z\n}\nf : [a, b] \u2192R is continuous\n=\n+\u221e\n|{z}\nf(xnk)\u2265nk\u2265k\n.\nThis concludes the statement that f is bounded from above.\nHaving proved that f is bounded from above,\nsup\nx\u2208[a,b]\nf(x) makes sense. Thus, we must prove\nthat sup\nx\u2208[a,b]\nf(x) = max\nx\u2208[a,b] f(x). By de\ufb01nition of supremum, there exists a sequence (yn) \u2286[a, b]\nsuch that f(yn) \u2265M \u22121", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_37", "text": "that sup\nx\u2208[a,b]\nf(x) = max\nx\u2208[a,b] f(x). By de\ufb01nition of supremum, there exists a sequence (yn) \u2286[a, b]\nsuch that f(yn) \u2265M \u22121\nn. In particular, lim\nn\u2192\u221ef(yn) = M. By Bolzano-Weierstrass ??, there\nexists a convergent subsequence (ynk) \u2286(yn). Set c := lim\nk\u2192\u221eynk. Then c \u2208[a, b], and\nf(c) =\nlim\nk\u2192\u221ef(ynk)\n|\n{z\n}\nf : [a, b] \u2192R is continuous\n= lim\nn\u2192\u221ef(yn)\n|\n{z\n}\n??\n= M.\nRemark 1.65. The conclusion of the above theorem does not hold, if we do not assume that\nthe domain of f is a closed bounded interval [a, b], a, b \u2208R. For example, take f : R \u2192R,\nf(x) :=\n1\nx2+1. Then f does not attain its minimum as R(f) =]0, 1]: in fact, f(x) > 0, \u2200x \u2208R\nand f converges to 0 as x goes to \u00b1\u221e.\nTheorem 1.66 (Intermediate value theorem). Let a, b \u2208R. If f : [a, b] \u2192R is continuous,\nthen it takes each value between M := max\nx\u2208[a,b] f(x) and m := min\nx\u2208[a,b] f(x) at least once. More\nprecisely, for each c \u2208[m, M], there exists d \u2208[a, b] such that f(d) = c.\nIdea. We give only the idea and we refer to the precise proof to page 81-82 of the book.\nWe know by the above theorem that there are a\u2032, b\u2032 \u2208[a, b] such that m = f(a\u2032) and\nM = f(b\u2032). Hence, by replacing a with a\u2032 and b with b\u2032 (and some algebraic manipulation in\nthe case when b\u2032 < a\u2032), we may assume that f(a) = m, f(b) = M and m < c < M. Then, the\nidea is to consider", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_38", "text": "the case when b\u2032 < a\u2032), we may assume that f(a) = m, f(b) = M and m < c < M. Then, the\nidea is to consider\nS := {x \u2208[a, b]|f(x) < c}\nSet d := sup S. By the de\ufb01nition of sup, there is a sequence (xn) \u2286S converging to d from the\nleft and let yn be any sequence converging to d from the right. Applying continuity to the \ufb01rst\n19", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_39", "text": "sequence shows that f(d) \u2264c, and by applying it to the second one shows that f(d) \u2265c. So,\nf(d) = c.\nExample 1.67. In other words, Theorem 1.66 says that R(f) = [m, M]. Hence, for example,\nthe image of an interval [a, b] via a continuous function f (whose domain contains [a, b]) cannot\nbe [c, d] \u222a[e, d], c < d < e < f \u2013 that is, it cannot be the union of two disjoint intervals.\nExample 1.68. If f : R \u2192R is a continuous function, such that f(0) = 1, f(1) = 3, f(2) = \u22121,\nthen f attains the value 2 at least two times. Indeed, our assumptions say that the maximum\nof f|[0,1] is at least 3 and the minimum of f|[0,1] is at most 1. Hence, Theorem 1.66 applied to\nf|[0,1] yields that there is at least one c \u2208[0, 1] such that f(c) = 2. Similarly, Theorem 1.66\napplied to f|[0,1] yields that there is at least one d \u2208[1, 2] such that f(d) = 2. Furthermore,\nc \u0338= d, because c = d can only happen if c = d = 1. However, f(c) = 3 \u0338= 2. Hence, c and d are\ntwo distinct real numbers at which f takes the value 2.\nWe will apply the above theoretical result to \ufb01nd solutions of equations of the form f(x) = x.\nFor example one can ask, if there is a solution of cos(x) = x for some x \u2208\n\u0002\n0, \u03c0\n2\n\u0003\n. Corollary 1.69\nlets us answer this question.\nCorollary 1.69 (Banach \ufb01xed point theorem for closed intervals). Let a, b \u2208R. If f : [a, b] \u2192\n[a, b] is a continuous function, then there exists x \u2208[a, b] such that f(x) = x.\nGiven a set S and function f : S \u2192S, an element s \u2208S such that f(s) = s is called a \ufb01xed\npoint.", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_40", "text": "[a, b] is a continuous function, then there exists x \u2208[a, b] such that f(x) = x.\nGiven a set S and function f : S \u2192S, an element s \u2208S such that f(s) = s is called a \ufb01xed\npoint.\nProof. Set g(x) := f(x) \u2212x. Then g(a) = f(a) \u2212a \u22650 and g(b) = f(b) \u2212b \u22640. So, by\nthe intermediate value theorem, there is a real number c \u2208[a, b] such that 0 = g(c). This is\nequivalent to f(c) = c.\nExample 1.70. The function cos(x)|[0, \u03c0\n2 ] :\n\u0002\n0, \u03c0\n2\n\u0003\n\u2192R can be regarded as cos(x)|[0, \u03c0\n2 ] :\n\u0002\n0, \u03c0\n2\n\u0003\n\u2192\n\u0002\n0, \u03c0\n2\n\u0003\n, since R(cos(x)|[0, \u03c0\n2 ]) = [0, 1] \u2282\n\u0002\n0, \u03c0\n2\n\u0003\n. Then the above theorem says that there is a \ufb01xed\npoint x for which cos(x) = x.\n1.2\nMonotonicity and invertibility of continuous functions\nLet us recall the following de\ufb01nition.\nDe\ufb01nition 1.71. Let f : E \u2192R be a function, E \u2282R.\n(1) f is strictly increasing if f(x) < f(y) for all x < y in E.\n(2) f is strictly decreasing if f(y) > f(x) for all x < y in E.\n(3) f : E \u2192R is strictly monotone, if it is strictly increasing or strictly decreasing.\nCorollary 1.72. Let a, b \u2208R. If f : (a, b) \u2192R is strictly monotone and continuous, then the\nrange R(f) is an open interval.\nProof. Set\nS := sup{ f(x) | x \u2208(a, b) },\nI := inf{ f(x) | x \u2208(a, b) }.\nFirst, we show that S, I \u0338\u2208R(f). We only prove statement about S since the statement about\nI can be proven analogously.\nSo, let us assume by contradiction that S = f(c) for some", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_41", "text": "First, we show that S, I \u0338\u2208R(f). We only prove statement about S since the statement about\nI can be proven analogously.\nSo, let us assume by contradiction that S = f(c) for some\nc \u2208(a, b). Choose c < d \u2208(a, b) \u2013 here we are using that the interval is open!. Then, as f is\n20", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_42", "text": "strictly increasing f(d) > f(c) = S, which is a contradiction with the de\ufb01nition of S.\nWe now show that R(f) = (I, S). Let us \ufb01x p \u2208(I, S). By the de\ufb01nition of S and I, there exist\nc, d \u2208(a, b) such that f(c) < p < f(d). Then the Intermediate Value Theorem 1.66 implies\nthat p \u2208R(f), since p \u2208[f(c), f(d)] \u2282R(f).\nTheorem 1.73. Let f : E \u2192F be a continuous function on an interval E. Then, f is strictly\nmonotone if and only if it is injective.\nProof. We do not prove this in class, read the proof from page 84-85 of the book.\nTheorem 1.74. If f : E \u2192F is continuous, strictly monotone and surjective function between\nintervals E, F. Then f\u22121 is also continuous.\nLet us recall that in the hypotheses of Theorem 1.74, the inverse function f\u22121 exists by\nTheorem 1.73.\nProof. We only show the case when E is an open interval (a, b), for some a, b \u2208R. In this\ncase, F is also an open interval according to Corollary 1.72. Fix 0 < \u03b5 \u2208R and y0 \u2208F. Set\nx0 := f\u22121(y0). According to Corollary 1.72, there exist c, d \u2208R such that\nR(f|(x0\u2212\u03b5,x0+\u03b5)) = (c, d)\n(1.74.a)\nIn particular, there exists \u03b4 > 0 such that for every y \u2208F\nif |y \u2212y0| \u2264\u03b4 \u21d2y \u2208(c, d).\n(1.74.b)\nFor example, it su\ufb03ces to take \u03b4 := min{|c\u2212y0|,|d\u2212y0|}\n2\n: that is a choice of \u03b4 for which the above\ncondition is satis\ufb01ed.\nWe show that with the above choice of \u03b4 the de\ufb01nition of the continuity of f\u22121 at y0 is\nsatis\ufb01ed. That is, for every y \u2208F,\n|y \u2212y0| \u2264\u03b4 \u21d2y \u2208(c, d)\n|\n{z\n}", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_43", "text": "We show that with the above choice of \u03b4 the de\ufb01nition of the continuity of f\u22121 at y0 is\nsatis\ufb01ed. That is, for every y \u2208F,\n|y \u2212y0| \u2264\u03b4 \u21d2y \u2208(c, d)\n|\n{z\n}\n(1.74.b)\n\u21d2|f\u22121(y) \u2212x0| \u2264\u03b5.\n|\n{z\n}\n(1.74.a)\nExample 1.75. Neither of the functions sin(x), cos(x), tan(x) and cotan(x) are invertible if\nconsidered as functions R \u2192R, as they are not injective in view of their periodicity. However,\nif we restrict their domains adequately they become strictly montone, and then, according\nto Theorem 1.74, their inverses are continuous too:\n(1) arcsin(x) is the inverse of sin(x)|[\u2212\u03c0\n2 , \u03c0\n2 ]. For example, arcsin\n\u0000\u22121\n2\n\u0001\n= \u2212\u03c0\n6 , and arcsin\n\u0000\u22121\n2\n\u0001\n\u0338=\n7\u03c0\n6 , despite having sin\n\u0000 7\u03c0\n6\n\u0001\n= \u22121\n2 too.\n(2) arccos(x) is the inverse of cos(x)|[0,\u03c0].\n(3) arctan(x) is the inverse of tan(x)|[\u2212\u03c0\n2 , \u03c0\n2 ].\n21", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_44", "text": "2\nDIFFERENTIATION\nLet f : E \u2192R be a real valued one variable function. We would like to approximate it with\na linear one. That is, we would like to write\nf(x) = f(x0) + a(x \u2212x0) + r(x),\n(2.0.a)\nwhere a is a real number, and the error function r(x) is small in a neighborhood of x0. The\nquestion is: how small would we like r(x) to be so that we obtain a \u201cgood\u201d approximation?\nWhat kind of function do then realize formula (2.0.a) with our chosen conditions on r(x)?\nWell, if we want our approximation to at least compute the right value of f at x0, since\nlim\nx\u2192x0x \u2212x0 = 0,\nwe need to impose that lim\nx\u2192x0r(x) = 0. Even better, we would like r(x) to be smaller than\na linear function, otherwise the linear approximation in (2.0.a) will not be very precise. But\nwhat does it precisely mean that r(x) should be smaller than a linear function? The precise\nmathematical wording is the following:\nlim\nx\u2192x0\nr(x)\nx \u2212x0\n= 0.\n(2.0.b)\nEven better, taking r1(x) :=\nr(x)\nx\u2212x0 , we can rewrite the above condition as\nr(x) = (x \u2212x0)r1(x),\nand\nr1(x)\nx \u2212x0\n= 0.\n(2.0.c)\nThe graph of the function g(x) := f(x0)+a(x\u2212x0) is a line in the cartesian plane. Considering\nImages/etox_tang.png\nFigure 3: A di\ufb00erentiable function and the tangent line to the graph.\nthe graph of f(x), then if we can show that that for f the error function r(x) is smaller than\nlinear, that is, if r(x) satis\ufb01es the condition of (2.0.b), then the line representing the graph of\ng will be tangent to the graph of f at the point (x0, f(x0)).\nAt this point the central question is: for what functions f do a and r(x) exists satisfy-", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_45", "text": "g will be tangent to the graph of f at the point (x0, f(x0)).\nAt this point the central question is: for what functions f do a and r(x) exists satisfy-\ning (2.0.a), (2.0.b), respectively?\nIf both (2.0.a) and (2.0.b) hold, then\nr(x)\nx \u2212x0\n+ a = f(x) \u2212f(x0)\nx \u2212x0\n,\nx \u0338= x0,\n(2.0.d)\n22", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_46", "text": "and, moreover, by taking the limit for x \u2192x0 on both sides of this equation, using (2.0.b), it\nfollows that\na = lim\nx\u2192x0\nr(x)\nx \u2212x0\n+ a\n|\n{z\n}\nby (2.0.b)\n= lim\nx\u2192x0\nf(x) \u2212f(x0)\nx \u2212x0\n|\n{z\n}\nby (2.0.d)\n.\n(2.0.e)\nSo, the existence of the real number a together with the sub-linear2 behavior of the error\nterm described in (2.0.b) imply that the limit on the right of (2.0.e) exists and it is \ufb01nite. This\ndiscussion motivates the following de\ufb01nition.\nDe\ufb01nition 2.1. Let f : E \u2192R be a function and let x0 \u2208E.\n(1) The function f is di\ufb00erentiable at x0, if the limit\nlim\nx\u2192x0\nf(x) \u2212f(x0)\nx \u2212x0\n(2.1.f)\nexists and it is \ufb01nite. We call the value of the limit in (2.1.f) the derivative of f at x0\nand we denote it by f\u2032(x0).\n(2) We say that f : E \u2192R is di\ufb00erentiable if it is di\ufb00erentiable at all points x0 \u2208E.\n(3) The function\nf\u2032 : {x \u2208E | f is di\ufb00erentiable at x} \u2192R,\nx 7\u2192f(x)\u2032\nis called the derivative function of f. the domain of f\u2032 is composed of all points of E\nwhere the above limit exists.\nRemark 2.2.\n(1) The derivative f\u2032(x0) of f at x0 can be also de\ufb01ned to be the unique real\nnumber c satisfying\nf(x) = f(x0) + c \u00b7 (x \u2212x0) + r(x),\n(2.2.g)\nwhere the function r(x) satis\ufb01es lim\nx\u2192x0\nr(x)\nx\u2212x0 = 0. As above, we can write r(x) = (x \u2212\nx0)r1(x) and lim\nx\u21920r1(x) = 0. In the reminder of this section, we will also use the notation\n\u03b51(x) to denote the function r1(x).", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_47", "text": "x0)r1(x) and lim\nx\u21920r1(x) = 0. In the reminder of this section, we will also use the notation\n\u03b51(x) to denote the function r1(x).\n(2) The de\ufb01nition of the derivative f\u2032(x0) in De\ufb01nition 2.1 can be summarized from a geomet-\nrical viewpoint by saying that the derivative is the limit (when it exists) for x \u2192x0 of the\nslope of the unique line passing through (x0, f(x0)) and the point (x, f(x)) corresponding\nto x on the graph.\nExample 2.3. Constant functions are di\ufb00erentiable everywhere. In fact, if f : R \u2192R, f(x) =\nC, \u2200x \u2208R, C \u2208R, then for x0 \u2208R\nlim\nx\u2192x0\nf(x) \u2212f(x0)\nx \u2212x0\n= lim\nx\u2192x0\nC \u2212C\nx \u2212x0\n= 0.\nExample 2.4. We show that (x2)\u2032 = 2x.\nFor any x0 \u2208R, we need to compute the limit lim\nx\u2192x0\nx2\u2212x2\n0\nx\u2212x0 . Thus,\nlim\nx\u2192x0\nx2 \u2212x2\n0\nx \u2212x0\n= lim\nx\u2192x0\n(x \u2212x0)(x + x0)\nx \u2212x0\n= lim\nx\u2192x0x + x0 = 2x0.\n2Sublinear stands for \u201cless than linear\u201d, that is, the condition de\ufb01ned in (2.0.b)\n23", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_48", "text": "Images/secants.jpg\nExample 2.5. Similarly, if a \u2208Z+, then (xa)\u2032 = axa\u22121.\nIndeed,\nlim\nx\u2192x0\nxa \u2212xa\n0\nx \u2212x0\n= lim\nx\u2192x0\n(x \u2212x0)(xa\u22121 + xa\u22122x0 + xa\u22123x2\n0 + \u00b7 \u00b7 \u00b7 + x1xa\u22122\n0\n+ xa\u22121\n0\n)\nx \u2212x0\n= lim\nx\u2192x0xa\u22121 + xa\u22122x0 + xa\u22123x2\n0 + \u00b7 \u00b7 \u00b7 + xxa\u22122\n0\n+ xa\u22121\n0\n= lim\nx\u2192x0xa\u22121 + lim\nx\u2192x0xa\u22122x0 + lim\nx\u2192x0xa\u22123x2\n0 + \u00b7 \u00b7 \u00b7 + lim\nx\u2192x0xxa\u22122\n0\n+ lim\nx\u2192x0xa\u22121\n0\n|\n{z\n}\nby the addition rule for \ufb01nite limits and the fact that \u2200c \u2208N,\nlim\nx\u2192x0xc = xc\n0\n=axa\u22121\n0\n.\nExample 2.6. We show that sin(x)\u2032 = cos(x).\nlim\nx\u2192x0\nsin(x) \u2212sin(x0)\nx \u2212x0\n= lim\nx\u2192x0\n2 cos\n\u0000 x+x0\n2\n\u0001\nsin\n\u0000 x\u2212x0\n2\n\u0001\nx \u2212x0\n= lim\nx\u2192x0 cos\n\u0012x + x0\n2\n\u0013\n\u00b7 lim\nx\u2192x0\nsin\n\u0000 x\u2212x0\n2\n\u0001\nx\u2212x0\n2\n|\n{z\n}\nlimt\u21920\nsin(t)\nt\n=1\n= cos(x0),\nwhere we could break up the limit in the multiplication thanks to Proposition 1.20.\nExample 2.7. Similarly, cos(x)\u2032 = \u2212sin(x).\nlim\nx\u2192x0\ncos(x) \u2212cos(x0)\nx \u2212x0\n= lim\nx\u2192x0\n\u22122 sin\n\u0000 x+x0\n2\n\u0001\nsin\n\u0000 x\u2212x0\n2\n\u0001\nx \u2212x0\n= lim\nx\u2192x0 \u2212sin\n\u0012x + x0\n2\n\u0013\n\u00b7 lim\nx\u2192x0\nsin\n\u0000 x\u2212x0\n2\n\u0001\nx\u2212x0\n2\n|\n{z\n}\nlimt\u21920\nsin(t)\nt\n=1\n= \u2212sin(x0),\nwhere we could break up the limit in the multiplication thanks to Proposition 1.20.\nDi\ufb00erentiability is a stronger condition than continuity, as the following proposition readily\nshows.", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_49", "text": "2\n\u0001\nx\u2212x0\n2\n|\n{z\n}\nlimt\u21920\nsin(t)\nt\n=1\n= \u2212sin(x0),\nwhere we could break up the limit in the multiplication thanks to Proposition 1.20.\nDi\ufb00erentiability is a stronger condition than continuity, as the following proposition readily\nshows.\n24", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_50", "text": "Proposition 2.8. If f : E \u2192R is di\ufb00erentiable at x0, then it is continuous at x0.\nProof. This is a consequence of the following computation:\nlim\nx\u2192x0f(x) = lim\nx\u2192x0 f(x0) + (x \u2212x0)f(x0)\u2032 + r(x)\n|\n{z\n}\nby (2.2.g)\n=\nf(x0) + lim\nx\u2192x0r(x)\n|\n{z\n}\nProposition 1.20 and limx\u2192x0 x\u2212x0=0\n=f(x0) +\nlim\nx\u2192x0\nr(x)\nx \u2212x0\n|\n{z\n}\n=0by Remark 2.2\n\u00b7 lim\nx\u2192x0(x \u2212x0) = f(x0).\nExample 2.9. The viceversa of Proposition 2.8 is not true. That is, if f is continuous at x0,\nit does not necessarily have to be di\ufb00erentiable.\nFor example, let us consider the function f(x) := |x|. The function f is continuous on R, in\nparticular, it is continuous at x0 = 0. On the other hand, f is not di\ufb00erentiable at 0, because\nthat would imply that lim\nx\u21920\n|x|\nx exists. However, since\nImages/abs_val_gr.png\nFigure 4: f(x) = |x|.\nlim\nx\u21920\u2212\n|x|\nx = lim\nx\u21920\u2212\n\u2212x\nx = \u22121 \u0338= 1 = lim\nx\u21920+\nx\nx = lim\nx\u21920+\n|x|\nx .\nProposition 1.42 implies that f is not di\ufb00erentiable at 0.\nExample 2.10.\n(1) The function f(x) :=\n(\n1\nx = 0\n0\nx \u0338= 0\nis not di\ufb00erentiable at 0, since it is\nnot continuous at 0. On the other hand, outside 0, f is di\ufb00erentiable, since over R\u2217, f is\nconstant.\n(2) The function f(x) :=\n(\n1\nx \u2208Q\n0\nx \u2208R \\ Q\nis not di\ufb00erentiable at any point of R since it is\nnot continuous at any point of R.\n2.1\nComputing derivatives\nIn this section we show how to compute derivatives. We \ufb01rst start by studying how deriva-\ntives behave with respect to the usual algebraic operations on R, and then continue by studying\nhow to compute derivatives with respect to composition and taking the inverse.", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_51", "text": "not continuous at any point of R.\n2.1\nComputing derivatives\nIn this section we show how to compute derivatives. We \ufb01rst start by studying how deriva-\ntives behave with respect to the usual algebraic operations on R, and then continue by studying\nhow to compute derivatives with respect to composition and taking the inverse.\n25", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_52", "text": "2.1.1\nAddition\nProposition 2.11. If f, g: E \u2192R are di\ufb00erentiable at x0, then so is \u03b1f +\u03b2g for any \u03b1, \u03b2 \u2208R,\nand furthermore\n(\u03b1f + \u03b2g)\u2032(x0) = \u03b1f\u2032(x0) + \u03b2g\u2032(x0).\nProof.\n(\u03b1f + \u03b2g)\u2032(x0) = lim\nx\u2192x0\n(\u03b1f + \u03b2g)(x) \u2212(\u03b1f + \u03b2g)(x0)\nx \u2212x0\n=\u03b1 lim\nx\u2192x0\nf(x) \u2212f(x0)\nx \u2212x0\n|\n{z\n}\n=f\u2032(x0)\n+\u03b2 lim\nx\u2192x0\ng(x) \u2212g(x0)\nx \u2212x0\n|\n{z\n}\n=g\u2032(x0)\n= \u03b1f\u2032(x0) + \u03b2g\u2032(x0)\nwhere we could split the limit of the sum into the sum of the limits by the assumption on the\ndi\ufb00erentiability of f, g at x0, using Proposition 1.20.\nExample 2.12.\n\u00005x3 + 6x2\u0001\u2032 =\n\u00005x3\u0001\u2032 +\n\u00006x2\u0001\u2032 = 15x2 + 12x\n2.1.2\nMultiplication\nProposition 2.13. If f, g: E \u2192R are di\ufb00erentiable at x0, then so is f \u00b7 g, and furthermore\n(f \u00b7 g)\u2032(x0) = (fg\u2032 + f\u2032g)\nProof.\n(f \u00b7 g)\u2032(x0) = lim\nx\u2192x0\nf(x)g(x) \u2212f(x0)g(x0)\nx \u2212x0\n= lim\nx\u2192x0\nf(x)g(x) \u2212f(x)g(x0) + f(x)g(x0) \u2212f(x0)g(x0)\nx \u2212x0\n= lim\nx\u2192x0\nf(x)(g(x) \u2212g(x0)) + (f(x) \u2212f(x0))g(x0)\nx \u2212x0\n=\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8edlim\nx\u2192x0f(x)\n|\n{z\n}\n=f(x0)\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\u00b7\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8edlim\nx\u2192x0\ng(x) \u2212g(x0)\nx \u2212x0\n|\n{z\n}\n=g\u2032(x0)\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8+ g(x0)\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8edlim", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_53", "text": "x \u2212x0\n=\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8edlim\nx\u2192x0f(x)\n|\n{z\n}\n=f(x0)\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\u00b7\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8edlim\nx\u2192x0\ng(x) \u2212g(x0)\nx \u2212x0\n|\n{z\n}\n=g\u2032(x0)\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8+ g(x0)\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8edlim\nx\u2192x0\nf(x) \u2212f(x0)\nx \u2212x0\n|\n{z\n}\n=f\u2032(x0)\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n=f(x0)g\u2032(x0) + g(x0)f\u2032(x0)\nwhere the fact that lim\nx\u2192x0f(x) = f(x0) follows from Proposition 2.8 and we could split the limits\nof sum and multiplications using the di\ufb00erentiability of f, g at x0 and Proposition 1.20.\nExample 2.14.\n\u0000x2 cos(x)\n\u0001\u2032 =\n\u0000x2\u0001\u2032 cos(x) + x2 (cos(x))\u2032 = 2x cos(x) + x2(\u2212sin(x)) = x(2 cos(x) \u2212x sin(x))\n2.1.3\nDivision\nProposition 2.15. If f, g: E \u2192R are di\ufb00erentiable at x0, and g(x0) \u0338= 0, then f\ng is also\ndi\ufb00erentiable at x0, and furthermore\n\u0012f\ng\n\u0013\u2032\n(x0) =\n\u0012gf\u2032 \u2212fg\u2032\ng2\n\u0013\n(x0)\nIn particular,\n\u00121\ng\n\u0013\u2032\n(x0) =\n\u0012\u2212g\u2032\ng2\n\u0013\n(x0)\n26", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_54", "text": "Proof. We compute the now familiar limit\nlim\nx\u2192x0\nf\ng (x) \u2212f\ng (x0)\nx \u2212x0\n= lim\nx\u2192x0\nf(x)g(x0) \u2212f(x0)g(x)\ng(x)g(x0)(x \u2212x0)\n= lim\nx\u2192x0\nf(x)g(x0) \u2212f(x0)g(x0) + f(x0)g(x0) \u2212f(x0)g(x)\ng(x)g(x0)(x \u2212x0)\n.\nGrouping together, in the denominator of the last member of the previous equation, those\nterms that depend on g(x0) and f(x0), respectively, we obtain,\nlim\nx\u2192x0\nf(x)g(x0) \u2212f(x0)g(x0) + f(x0)g(x0) \u2212f(x0)g(x)\ng(x)g(x0)(x \u2212x0)\n=\ng(x0)\ng(x0) \u00b7 lim\nx\u2192x0g(x)\n|\n{z\n}\n=g(x0)\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8edlim\nx\u2192x0\nf(x) \u2212f(x0)\nx \u2212x0\n|\n{z\n}\n=f\u2032(x0)\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n\u2212\nf(x0)\ng(x0) \u00b7 lim\nx\u2192x0g(x)\n|\n{z\n}\n=g(x0)\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8edlim\nx\u2192x0\ng(x) \u2212g(x0)\nx \u2212x0\n|\n{z\n}\n=g\u2032(x0)\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n=\n\u0012gf\u2032 \u2212fg\u2032\ng2\n\u0013\n(x0),\nwhere the fact that lim\nx\u2192x0g(x) = g(x0) follows from Proposition 2.8 and we could split the limits\nof sum and multiplications using the di\ufb00erentiability of f, g at x0 and Proposition 1.20.\nExample 2.16. If b > 0 is an integer and x \u0338= 0, then:\n\u0012 1\nxb\n\u0013\u2032\n= \u2212\n\u0000xb\u0001\u2032\nx2b\n= \u2212bxb\u22121\nx2b\n= \u2212b\nxb+1 .\nThat is, by setting a = \u2212b we obtain (xa)\u2032 = axa\u22121. In particular, this shows that\n(xa)\u2032 = axa\u22121", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_55", "text": "xb\n\u0013\u2032\n= \u2212\n\u0000xb\u0001\u2032\nx2b\n= \u2212bxb\u22121\nx2b\n= \u2212b\nxb+1 .\nThat is, by setting a = \u2212b we obtain (xa)\u2032 = axa\u22121. In particular, this shows that\n(xa)\u2032 = axa\u22121\nholds for all integer a, not just the non-negative ones.\nExample 2.17. If x \u0338= k\u03c0 + \u03c0\n2 for any k \u2208Z, or, equivalently, if cos(x) \u0338= 0, then\ntan(x)\u2032 =\n\u0012 sin(x)\ncos(x)\n\u0013\u2032\n=cos(x)(sin(x))\u2032 \u2212sin(x)(cos(x))\u2032\ncos(x)2\n=cos(x) cos(x) \u2212sin(x)(\u2212sin(x))\ncos(x)2\n=\n1\ncos2(x).\n2.1.4\nComposition of functions and derivatives\nProposition 2.18. Let f : E \u2192R, g: G \u2192R be functions such that that f(E) \u2286G. Assume\nthat f is di\ufb00erentiable at x0 \u2208E, and g is di\ufb00erentiable at f(x0) then g \u25e6f : E \u2192R is\ndi\ufb00erentiable at x0, and\n(g \u25e6f)\u2032(x0) = g\u2032(f(x)) \u00b7 f\u2032(x0).\n27", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_56", "text": "Idea of the proof.\nlim\nx\u2192x0\ng(f(x)) \u2212g(f(x0))\nx \u2212x0\n= lim\nx\u2192x0\ng(f(x)) \u2212g(f(x0))\nf(x) \u2212f(x0)\nf(x) \u2212f(x0)\nx \u2212x0\n= lim\nx\u2192x0\ng(f(x)) \u2212g(f(x0))\nf(x) \u2212f(x0)\n\u00b7 lim\nx\u2192x0\nf(x) \u2212f(x0)\nx \u2212x0\n= g\u2032(f(x0)) \u00b7 f\u2032(x0)\nExample 2.19. Let f(x) = x2 and g(y) = cos(y). Then f\u2032(x) = 2x and g\u2032(y) = \u2212sin(y). In\nparticular,\ncos(x2)\u2032 = (g \u25e6f)\u2032(x) = (g\u2032 \u25e6f)(x) \u00b7 f\u2032(x) = \u2212sin(x2)2x.\nExample 2.20. Let a \u2208Z, b \u2208Z+, f(x) = xa and g(y) = y\n1\nb . Then according to Example 2.16\nand Example 2.22, f(x)\u2032 = axa\u22121 and g(y)\u2032 = 1\nby\n1\nb \u22121. Hence,\n\u0010\nx\na\nb\n\u0011\u2032\n=\n\u0010\n(xa)\n1\nb\n\u0011\u2032\n= (g \u25e6f)\u2032(x) = (g\u2032 \u25e6f)(x) \u00b7 f\u2032(x) = 1\nb (xa)\n1\nb \u22121 axa\u22121 = a\nb x\na\nb \u2212a+a\u22121 = a\nb x\na\nb \u22121\nSo, the formula (xr)\u2032 = rxr\u22121 holds also when r is any rational number (as it did for r \u2208Z in\nExample 2.16).\n2.1.5\nInversion of functions and derivatives\nProposition 2.21. Let f : E = (a, b) \u2192F be a bijective continuous function (so f is strictly\nmonotone, and f\u22121 exists and is continuous by Theorem 1.74), and let x0 \u2208E be such that\nf\u2032(x0) \u0338= 0. Then f\u22121 is di\ufb00erentiable at y0 := f(x0), and we have", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_57", "text": "monotone, and f\u22121 exists and is continuous by Theorem 1.74), and let x0 \u2208E be such that\nf\u2032(x0) \u0338= 0. Then f\u22121 is di\ufb00erentiable at y0 := f(x0), and we have\n\u0000f\u22121\u0001\u2032 (y0) =\n1\nf\u2032(x0) =\n1\nf\u2032(f\u22121(y0))\nProof. The idea behind the proof of the proposition is that if we set y = f(x) and y0 = f(x0,\nwe have\nf\u22121(y) \u2212f\u22121(y0)\ny \u2212y0\n=\n1\ny\u2212y0\nf\u22121(y)\u2212f\u22121(y0)\n=\n1\nf(f\u22121(y))\u2212f(f\u22121(y0))\nf\u22121(y)\u2212f\u22121(y0)\n.\nCheck page 109 for the precise proof.\nExample 2.22. If f(x) = xb for some integer b \u22651, then f\u22121(y) =\nb\u221ay = y\n1\nb . So, f\u2032(x) =\nbxb\u22121, and\n\u0010\ny\n1\nb\n\u0011\u2032\n=\n\u0000f\u22121\u0001\u2032 (y) =\n1\nf\u2032(f\u22121(y)) =\n1\nb\n\u0010\ny\n1\nb\n\u0011b\u22121 = 1\nby\u2212b\u22121\nb\n= 1\nby\n1\nb \u22121.\nSo, for c = 1\nb (where b \u2208Z\u2217\n+), the formula for (yc)\u2032 = cyc\u22121. That is the formula is the same\nas in the case of c being an integer.\nExample 2.23. Let f(x) = sin(x)|[\u2212\u03c0\n2 , \u03c0\n2 ] :\n\u0002\n\u2212\u03c0\n2 , \u03c0\n2\n\u0003\n\u2192[\u22121, 1].\nThen f is invertible and\nf\u22121(y) = arcsin(y). Also, f\u2032(x) = cos(x)|[\u2212\u03c0\n2 , \u03c0\n2 ], thus, for any y \u2208] \u22121, 1[,\narcsin\u2032(y) =\n1\ncos(arcsin(y)) =\n1\np\n1 \u2212sin2(arcsin(y))\n=\n1\np\n1 \u2212y2 .\n28", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_58", "text": "2.1.6\nThe exponential function\nFor our last example in this section, we will discuss in details the exponential and logarithmic\nfunctions. Let us remind the reader that we de\ufb01ned\nex := lim\nn\u2192\u221e\n\u0010\n1 + x\nn\n\u0011n\n.\nDe\ufb01nition 2.24. For x \u2208R, we de\ufb01ne\nex :=\n\u221e\nX\nk=0\nxk\nk! .\nRemark 2.25. Applying De\ufb01nition 2.24, to x = 0 yields e0 = 1. Furthermore, according to ??,\ne1 = e.\nProposition 2.26. For any x, y \u2208R, ex+y = ex \u00b7 ey.\nProof. This is an exercise in Week 10 exercise sheet.\nCorollary 2.27. For any x \u2208R, e\u2212x =\n1\nex .\nProof.\nex \u00b7 e\u2212x =\nex+(\u2212x)\n| {z }\nProposition 2.26\n= e0 =\n1\n|{z}\nRemark 2.25\n.\nDividing by ex yields the statement (ex cannot be 0, since then ex\u00b7e\u2212x = 1 could not hold).\nCorollary 2.28. For every x \u2208R, ex > 0.\nProof. For x \u22650, then all the terms in the in\ufb01nite sum in De\ufb01nition 2.24 is at least zero, and\nthe \ufb01rst term is 1. This implies the statement for x \u22650.\nSo, we may assume from now that x < 0. We have ex =\n1\ne\u2212x by Corollary 2.27. However, as\nnow \u2212x > 0 holds, the previous paragraph tells us that e\u2212x > 0, and hence also\n1\ne\u2212x > 0.\nProposition 2.29. (ex)\u2032 = ex\nProof. We need to show that\nlim\nx\u2192x0\nex \u2212ex0\nx \u2212x0\n= ex0.\nThis is equivalent to showing that\n0 = lim\nx\u2192x0\nex \u2212ex0\nx \u2212x0\n\u2212ex0 =\n\u0012\nlim\nx\u2192x0\nex\u2212x0 \u22121\nx \u2212x0\n\u22121\n\u0013\nex0.\nBy setting y = x \u2212x0, what we need to show is that\nlim\ny\u21920\ney \u22121\ny\n\u22121 = 0.\n(2.29.a)\nHowever, for 0 < |y| \u22641:\n0 \u2264\n\f\f\f\f\ney \u22121\ny\n\u22121\n\f\f\f\f =\n\f\f\f\f\f\f\f\f\f\f\n\u221e\nX\nk=0\nyk\nk! \u22121\ny\n\u22121", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_59", "text": "lim\ny\u21920\ney \u22121\ny\n\u22121 = 0.\n(2.29.a)\nHowever, for 0 < |y| \u22641:\n0 \u2264\n\f\f\f\f\ney \u22121\ny\n\u22121\n\f\f\f\f =\n\f\f\f\f\f\f\f\f\f\f\n\u221e\nX\nk=0\nyk\nk! \u22121\ny\n\u22121\n\f\f\f\f\f\f\f\f\f\f\n=\n\f\f\f\f\f\n\u221e\nX\nk=2\nyk\u22121\nk!\n\f\f\f\f\f \u2264\n\u221e\nX\nk=2\n|y|k\u22121\nk!\n\u2264|y|\n\u221e\nX\nk=2\n|y|k\u22122\n(k \u22122)!\n= |y|\n\u221e\nX\nk=0\n|y|k\nk! \u2264|y|e\nHence, by Squeeze Theorem Proposition 1.20(5), it follows that (2.29.a) holds indeed.\n29", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_60", "text": "Proposition 2.30. We have\nlim\nx\u2192+\u221eex = +\u221e, and\nlim\nx\u2192\u2212\u221eex = 0.\nProof. According to De\ufb01nition 2.24, for all x > 0, ex \u22651 + x. As\nlim\nx\u2192+\u221e1 + x = +\u221e, squeeze\n(point (5) of Proposition 1.36) shows that\nlim\nx\u2192+\u221eex = +\u221e. Then Corollary 2.27, Corollary 2.28\nand point (3) of Proposition 1.36 show that\nlim\nx\u2192\u2212\u221eex = 0.\nProposition 2.31. The function ex : R \u2192R is strictly increasing.\nProof. Choose y > x \u2208R. We have to show that ey > ex. This is shown by the following\ncomputation:\ney \u2212ex = (ey\u2212x \u22121)\n|\n{z\n}\n> 1 by De\ufb01ni-\ntion 2.24, using\ny > x\n\u00b7\nex\n|{z}\n>\n0 by\nCorol-\nlary 2.28\n> 0\nCorollary 2.32. The range R(f) of f := ex : R \u2192R is (0, +\u221e) = R\u2217\n+.\nProof. Follows immediately from Proposition 2.30 and Proposition 2.31.\nDe\ufb01nition 2.33.\n(1) We de\ufb01ne the (natural) logarithm function log(x): R\u2217\n+ \u2192R to be the inverse of the\nexponential function f(x) = ex.\n(2) For any a \u2208R\u2217\n+, the a-based exponential functions ax is de\ufb01ned as\nax := ex\u00b7log(a).\nThe logarithm in base a of x is the inverse function of the a-based exponential function\nax,\nloga(x) := log(x)\nlog(a).\n(3) For any a \u2208R, the a-th power functions is de\ufb01ned as\nxa := ea\u00b7log(x).\nRemark 2.34. In the special cases where the functions of De\ufb01nition 2.36 have been already\nde\ufb01ned (so xa when a \u2208Q, and ax when a = e), they agree with the previously de\ufb01ned\nfunctions. This will be an exercise on the exercise sheet.\nExample 2.35.", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_61", "text": "de\ufb01ned (so xa when a \u2208Q, and ax when a = e), they agree with the previously de\ufb01ned\nfunctions. This will be an exercise on the exercise sheet.\nExample 2.35.\n(1) If f(x) = ex, then f\u22121(x) = log(x) and f\u2032(x) = ex (Proposition 2.29).\nHence:\n(log(x))\u2032 =\n1\nelog(x) = 1\nx.\n(2) Let h: R\u2217\n+ \u2192R, h(x) := xx. Then, de\ufb01nining f(x) = x log(x), g(y) := ey,\nh(x) = (f \u25e6g)(x).\nThus,\nh\u2032(x) = (f \u25e6g)\u2032(x) = f\u2032(g(x))g\u2032(x) = xx(log(x) + 1).\n30", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_62", "text": "De\ufb01nition 2.36.\n(1) The hyperbolic trigonometric functions are de\ufb01ned below, and they are\ncalled hyperbolic sine/cosine/tangent/cotangent:\nsinh(x) := ex \u2212e\u2212x\n2\ncosh(x) := ex + e\u2212x\n2\ntanh(x) := sinh(x)\ncosh(x)\ncoth(x) := cosh(x)\nsinh(x)\nThe domains of all the above functions is R, except for coth which is de\ufb01ned over R\u2217, since\nsinh(0) = 0.\nProposition 2.37. We have:\n(1) sinh(x)\u2032 = cosh(x)\n(2) cosh(x)\u2032 = sinh(x)\n(3) tanh(x)\u2032 =\n1\ncosh(x)2\n(4) coth(x)\u2032 =\n\u22121\nsinh(x)2\n(5) (xa)\u2032 = axa\u22121\n(6) (ax)\u2032 = log(a) \u00b7 ax\n(7) loga(x)\u2032 =\n1\nlog(a)\u00b7x\nThe proof is left as an exercise.\n2.2\nOne sided derivatives\nDe\ufb01nition 2.38. If f : E \u2192R is a function and x0 \u2208E for which the then we say that the\nleft (resp. right) derivative of f exists at x0 if the function\nf(x) \u2212f(x0)\nx \u2212x0\n: E \\ {x0} \u2192R\nadmits a left (resp. right limit). The value of this limit is then the left (resp. right) derivative.\nExample 2.39. For f(x) = |x| at x = 0 the left derivative is \u22121 and the right derivative is 1.\nIn fact,\nlim\nx\u21920+\nf(x) \u2212f(0)\nx \u22120\n= x\nx = 1,\nlim\nx\u21920\u2212\nf(x) \u2212f(0)\nx \u22120\n= \u2212x\nx = \u22121.\nAs in the case of left and right limits, we can use left and right derivatives to decide when\na function is di\ufb00erentiable at a given point.\nProposition 2.40. Let f : E \u2192R be a function and x0 \u2208E a real number.\nThen f is", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_63", "text": "= \u2212x\nx = \u22121.\nAs in the case of left and right limits, we can use left and right derivatives to decide when\na function is di\ufb00erentiable at a given point.\nProposition 2.40. Let f : E \u2192R be a function and x0 \u2208E a real number.\nThen f is\ndi\ufb00erentiable at a point x0 if and only if both its left and right derivatives exist and they agree.\nFurthermore, then the value of the derivative is the same as the common value of the left and\nthe right derivatives.\nProof. This is an immediate consequence of Proposition 1.42, De\ufb01nition 2.1 and De\ufb01nition 2.38.\n31", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_64", "text": "Example 2.41.\n(1) Let us consider the function f : R \u2192R de\ufb01ned by\nf(x) :=\n(\nx2,\nx \u22650\nx3,\nx < 0 .\nThe function f is di\ufb00erentiable at 0 with derivative f\u2032(0) = 0. Indeed,\nlim\nh\u21920+\nf(0 + h) \u2212f(0)\nh\n= lim\nh\u21920+\nh2 \u22120\nh\n= lim\nh\u21920+ h = 0.\nSimilarly,\nlim\nh\u21920\u2212\nf(0 + h) \u2212f(0)\nh\n= lim\nh\u21920\u2212\nh3 \u22120\nh\n= lim\nh\u21920\u2212h2 = 0.\nSince the left derivative and the right derivative exist and agree at x0 = 0, we can conclude\nthat\nlim\nh\u21920\nf(0 + h) \u2212f(0)\nh\n= 0.\nThus, f is di\ufb00erentiable at 0 with f\u2032(0) = 0.\n(2) Let us consider the function f : R \u2192R de\ufb01ned by\nf(x) :=\n(\nx + 1\nfor x \u22650\nx\nfor x < 0 .\nThe function f is not continuous at x0 = 0, as lim\nx\u21920\u2212f(x) = 0 whereas lim\nx\u21920+f(x) = 1. As a\ndi\ufb00erentiable function is continuous, then f is not di\ufb00erentiable at 0. On the other hand,\nf is di\ufb00erentiable outside of 0, since on a su\ufb03ciently small neighborhood of any point\nx0 \u0338= 0, f is given by a linear function and we know that linear functions are di\ufb00erentiable.\n2.3\nHigher derivatives\nGiven a function f, we may try to iterate inductively the process of taking the derivative\nof f, thus obtaining what we will call the second derivative of f, the third, derivative of f, etc.\nDe\ufb01nition 2.42. Let f : E \u2192R be a function.\n(1) The second derivative f\u2032\u2032 of f is the function\nf\u2032\u2032 : {x \u2208E | f\u2032 is di\ufb00erentiable at x} \u2192R\nx 7\u2192f\u2032\u2032(x) := (f\u2032)\u2032(x).\n(2) Assume that the n-th derivative f(n) of f has been de\ufb01ned. Then the (n+1)-st derivative\nf(n+1) of f is the function\nf(n+1) : {x \u2208E | f(n) is di\ufb00erentiable at x} \u2192R", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_65", "text": "f(n+1) of f is the function\nf(n+1) : {x \u2208E | f(n) is di\ufb00erentiable at x} \u2192R\nx 7\u2192f(n+1)(x) := (f(n))\u2032(x).\nThe n-th derivative of f at x \u2208E is denoted by f(n)(x). For the \ufb01rst, second and third\nderivative of f, we will adopt the notation f\u2032, f\u2032\u2032, f\u2032\u2032\u2032 rather than f(1), f(2), f(3).\nExample 2.43.\n(1) The second derivative of f(x) = arctan(x) is f\u2032\u2032(x): R \u2192R,\nf\u2032\u2032(x) = (f\u2032)\u2032(x) = (\n1\n1 + x2 )\u2032 =\n\u22122x\n(1 + x2)2 .\n32", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_66", "text": "(2) Let us consider the function f : R\u2217\u2192R de\ufb01ned by f(x) = e\n1\nx . Then,\nf\u2032\u2032(x) = (f\u2032)\u2032(x) = (e\n1\nx \u00b7 (\u22121\nx2 ))\u2032\n= (e\n1\nx )\u2032 \u00b7 (\u22121\nx2 ) + e\n1\nx \u00b7 (\u22121\nx2 )\u2032 = e\n1\nx ( 1\nx4 + \u00b7 2\nx3 ).\nExample 2.44. Here we show an example of a function f(x) such that f(x) is di\ufb00erentiable\ntwo times, but not three times. That is, f\u2032(x) and f\u2032\u2032(x) exist for every x \u2208R, but f\u2032\u2032\u2032(0) does\nnot exist.\nLet us consider f(x) := |x3|. Then, f\u2032(x) exists for all x \u2208R, and:\nf\u2032(x) =\n(\n3x2\nx \u22650\n\u22123x2\nx < 0.\nThis is immediate at x \u0338= 0 from the formula\nf(x) =\n\u001a x3\nfor x \u22650\n\u2212x3\nfor x \u22640\nTo conclude the above \ufb01rst claim we just have to compute the left and the right derivatives of\nf(x) at x = 0, and show that both are 0. Indeed:\nlim\nx\u21920+\nf(x) \u2212f(0)\nx \u22120\n= lim\nx\u21920+\nx3 \u22120\nx\n= lim\nx\u21920+ x2 = 0,\nand\nlim\nx\u21920\u2212\nf(x) \u2212f(0)\nx \u22120\n= lim\nx\u21920\u2212\n\u2212x3 \u22120\nx\n= lim\nx\u21920\u2212\u2212x2 = 0.\nThis concludes our \ufb01rst claim.\nSimilarly, one can prove that f\u2032\u2032(x) exists for all x \u2208R and\nf\u2032\u2032(x) =\n\u001a 6x\nfor x \u22650\n\u22126x\nfor x < 0\nWith other words, f\u2032\u2032(x) = 6|x|. However, as |x| is not di\ufb00erentiable at x = 0, we obtain that\nf\u2032\u2032\u2032(0) does not exist.\nDe\ufb01nition 2.45. f : E \u2192R is called a function of class Cn if its \ufb01rst n derivatives f\u2032, f\u2032\u2032, . . . , f(n)\nexists and are all continuous at all points x0 \u2208E.", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_67", "text": "De\ufb01nition 2.45. f : E \u2192R is called a function of class Cn if its \ufb01rst n derivatives f\u2032, f\u2032\u2032, . . . , f(n)\nexists and are all continuous at all points x0 \u2208E.\nNotation 2.46. To denote that a function f : E \u2192R is a Cn function, we will use the notation\nf \u2208Cn(E, R). We will write f \u2208C\u221e(E, R) if f \u2208Cn(E, R), \u2200n \u2208N, and we will say that f is\na C\u221efunction.\nExample 2.47.\n(1) According to Example 2.5, x, x2, etc. are Cn for all n, that is they are\nC\u221efunction. More precisely, for a \u2208N, de\ufb01ning f(x) = xa then\nf(n)(x) =\n(\n0\nif n > a\na \u00b7 (a \u22121) \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 (a \u2212n + 1)xa\u2212n\nfor n \u2264a\n(2) We can repeat the same computation for f : [0, +\u221e) \u2192R, f(x) := x\u03b1 = elog(x)\u03b1, \u03b1 \u2208\nR \\ N. Then f(x) = \u03b1 \u00b7 (\u03b1 \u22121) \u00b7 (\u03b1 \u22122) \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 (\u03b1 \u2212n + 1)x\u03b1\u2212n, x > 0.\n(3) |x| : R \u2192R is not C1, cf. Example 2.39.\n(4) |x3| : R \u2192R is C2 but not C3, cf. Example 2.44.\n33", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_68", "text": "2.4\nLocal and global extrema\nDe\ufb01nition 2.48. Let f : E \u2192R be a function and let x0 \u2208E.\n(1) The function f admits a point of local maximum at x0 if there is a real number \u03b4 > 0\nsuch that ]x0 \u2212\u03b4, x0 + \u03b4[\u2282E and for every x \u2208E if |x \u2212x0| < \u03b4 then f(x) \u2264f(x0).\n(2) The function f has a point of local minimum at x0 if there is a real number \u03b4 > 0 such\nthat ]x0 \u2212\u03b4, x0 + \u03b4[\u2282E and for every x \u2208E if |x \u2212x0| < \u03b4 then f(x) \u2265f(x0).\n(3) We say that x0 \u2208E is a point of local extremum for f if it is either a point of local\nminimum or of local maximum.\n(4) The function f has a point of global maximum at x0 if f(x0) \u2265f(x), for all x \u2208E.\n(5) The function f has a point of global minimum at x0 if f(x0) \u2264f(x), for all x \u2208E.\nRemark 2.49. We shall also say that f admits a local maximum (resp. local minimum, local\nextremum, global maximum, global minimum) at x0 to indicate that property (1) (resp. (2),\n(3), (4), (5)) de\ufb01ned above is satis\ufb01ed.\nRemark 2.50. Let f : E \u2192R be a function and x0 \u2208E. If x0 is a point of global maximum (resp.\nglobal minimum) for f and E contains a neighborhood of x0 of the form ]x0 \u2212\u03b4, x0 + \u03b4[, \u03b4 > 0,\nthen x0 is also a point of local maximum (resp. local minimum) for f.\nExample 2.51. Let us consider the function f\nThe following proposition shows that any point of local extremum for a function f coincides\nwith a zero of the derivative f\u2032.\nProposition 2.52. If f : E \u2192R is di\ufb00erentiable at x0, and f admits a local extremum at x0,\nthen f\u2032(x0) = 0.", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_69", "text": "with a zero of the derivative f\u2032.\nProposition 2.52. If f : E \u2192R is di\ufb00erentiable at x0, and f admits a local extremum at x0,\nthen f\u2032(x0) = 0.\nProof. We present the local maximum case, as one just need to reverse a few signs, to modify\nthe proof to obtain from it the case of local minimum.\nHence, let us assume that x0 \u2208E is a point of local by De\ufb01nition 2.48, there is a real number\n\u03b4 > 0 such that\n|x \u2212x0| \u2264\u03b4 \u21d2f(x) \u2264f(x0).\n(2.52.a)\nHowever, then\nlim\nx\u2192x+\n0\nf(x) \u2212f(x0)\n(x \u2212x0)\n\u2264\nlim\nx\u2192x+\n0\n0\n(x \u2212x0)\n|\n{z\n}\nx>x0, and (2.52.a)\n= 0,\n(2.52.b)\nand\nlim\nx\u2192x\u2212\n0\nf(x) \u2212f(x0)\n(x \u2212x0)\n\u2265\nlim\nx\u2192x\u2212\n0\n0\n(x \u2212x0)\n|\n{z\n}\nx<x0, and (2.52.a)\n= 0.\n(2.52.c)\nAs f(x) is di\ufb00erentiable, at x0 the two above limits agree (Proposition 2.40).\nHence the\nfollowing stream of inequalities have to be all equalities, which conclude our proof:\n0 \u2264lim\nx\u2192x\u2212\n0\nf(x) \u2212f(x0)\n(x \u2212x0)\n|\n{z\n}\n(2.52.c)\n= f\u2032(x0) = lim\nx\u2192x+\n0\nf(x) \u2212f(x0)\n(x \u2212x0)\n\u22640.\n|\n{z\n}\n(2.52.b)\n34", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_70", "text": "Images/x2_gr.png\nFigure 5: f(x) = x2 has a global minimum at x = 0.\nExample 2.53.\n(1) For f(x) = x2, we have f\u2032(x) = 2x. Hence, f\u2032(x) = 0 if and only if\nx = 0. So, x = 0 is the only option for stationary point of f, thus also for a point of local\nextremum, and, indeed, f admits a local (and global) minimum at x = 0.\n(2) For f(x) = x3, we have f\u2032(x) = 3x2. So f\u2032(x) = 0 if and only if x = 0 (as in the previous\ncase). However, f(0) = 0 is not a local extremum. This underlines that Proposition 2.52\nyields only a necessary, but not a su\ufb03cient condition for having a local extremum.\nImages/x3_gr.png\nFigure 6: f(x) = x3, f\u2032(0) = 0, but 0 is not a point of local extremum for f.\nDe\ufb01nition 2.54. Let f : E \u2192R be a function and let x0 \u2208E. Assume that f is di\ufb00erentiable\nat x0. Then, we say that x0 a stationary point (for f) if f\u2032(x0) = 0.\nWe have just seen that when f\u2032(x0) = 0, we cannot necessarily conclude that x0 is a point\nof local extremum for f. On the other hand, if the domain of f is a closed bounded interval\n[a, b], then Theorem 1.63 implies that f admits both a global maximum and a global minimum\nin [a, b]. Therefore, using Proposition 2.52, a point of global maximum for f can only be:\n\u25e6either a stationary point, x \u2208(a, b) such that f\u2032(x) = 0, or\n35", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_71", "text": "\u25e6x = a, or x = b.\nSimilarly, a point of global minimum for f can only be:\n\u25e6either a stationary point, x \u2208(a, b) such that f\u2032(x) = 0, or\n\u25e6x = a, or x = b.\nHence, to \ufb01nd the value of the global maximum and the global minimum of f over the interval\n[a, b], it su\ufb03ces to compute:\nmax\nx\u2208[a,b] f = sup{f(x) | x = a or x = b or f\u2032(x) = 0},\nmin\nx\u2208[a,b] f = inf{f(x) | x = a or x = b or f\u2032(x) = 0}.\nThis procedure guves an algorithmic approach to \ufb01nding the values of global extrema.\nExample 2.55. We compute the global minimum and the global maximum of\nf(x) = 4\n3x3 + 3\n2x2 \u2212x + 2\non the closed bounded interval\n\u0002\n\u22122, 1\n2\n\u0003\n. By the discussion in the paragraph before the example\nwe have to compute:\nf\u2032(x) = 4x2 + 3x \u22121,\nand then \ufb01nd the solutions of the equation f\u2032(x) = 0. These are:\nx = \u22123 \u00b1\n\u221a\n25\n8\n= \u22123 \u00b1 5\n8\n= \u22121, and x = 1\n4.\nThen, we have to compute the function values at these two points, and at the endpoints of our\ninterval. The point, where the function value is the maximal yields the maximum and where\nthe function value is minimal yields the minimum of f(x) on\n\u0002\n\u22122, 1\n2\n\u0003\n:\nvalue of x\nf(x)\n\u22122\n\u22124\n38 + 3\n24 + 2 + 2 = 36\u221264\n6\n+ 4 = 4 \u221228\n6 = \u22124\n6\n\u22121\n\u22124\n3 + 3\n2 + 1 + 2 = 9\u22128\n6\n+ 3 = 3 + 1\n6\n1\n4\n4\n3\u00b764 +\n3\n2\u00b716 \u22121\n4 + 2 = 4+18\u221248\n192\n+ 2 = \u221226\n192 + 2 = 2 \u221213\n96\n1\n2\n4\n3\u00b78 +\n3\n2\u00b74 \u22121\n2 + 2 = 4+9\u221212\n24\n+ 2 = 2 + 1\n24\nSo, f(x) on\n\u0002\n\u22122, 1\n2\n\u0003\ntakes its minimum at x = \u22122 and its maximum at x = \u22121.\n2.5", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_72", "text": "96\n1\n2\n4\n3\u00b78 +\n3\n2\u00b74 \u22121\n2 + 2 = 4+9\u221212\n24\n+ 2 = 2 + 1\n24\nSo, f(x) on\n\u0002\n\u22122, 1\n2\n\u0003\ntakes its minimum at x = \u22122 and its maximum at x = \u22121.\n2.5\nRolle\u2019s and Mean Value theorem\nWe are ready to state and prove the two main results of this chapter: Rolle\u2019s theorem\n(Theorem 2.56) and the Mean value theorem (Theorem 2.58).\nTheorem 2.56 (Rolle\u2019s Theorem). Let f : [a, b] \u2192R is a continuous function, for a, b \u2208R.\nAssume that f is di\ufb00erentiable on (a, b), and that f(a) = f(b). Then there exists c \u2208(a, b)\nsuch that f\u2032(c) = 0.\nProof. If f is constant on [a, b], then f\u2032(x) = 0, \u2200x \u2208(a, b) and so we are done.\nHence, we can assume that f is not constant. Then, according to Theorem 1.63, f has both\na maximum and a minimum on [a, b]. However, as f is non-constant, one of these values have\nto be not equal to f(a) = f(b). Formally, this means that there is a c \u2208[a, b], such that\nf(c) \u0338= f(a) = f(b). In particular, we must have a < c < b. Then, f is di\ufb00erentiable at c, and\nas f has a (local) extremum at c, we have f\u2032(c) = 0 according to Proposition 2.52.\n36", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_73", "text": "Remark 2.57. The di\ufb00erentiability assumption is needed for Theorem 2.56 to hold. For example,\nconsidering f(x) := |x| on [\u22121, 1], then f(\u22121) = f(1), but there is no point in [\u22121, 1] at the\nderivative of f is 0.\nTheorem 2.58 (Mean value theorem). Let f : [a, b] \u2192R be a continuous function, a, b \u2208R.\nAssume that f is di\ufb00erentiable on (a, b). Then there exists c \u2208(a, b) such that f\u2032(c)(b \u2212a) =\nf(b) \u2212f(a).\nProof. Apply Theorem 2.56 to g(x) = f(x) \u2212f(a) \u2212f(b)\u2212f(a)\nb\u2212a\n(x \u2212a).\nExample 2.59. Let f be a C1 function (see De\ufb01nition 2.45) on [\u22121, 1] such that f(\u22121) = 2,\nf(0) = 4 and f(1) = 3. We show using Theorem 2.58 that there is a c \u2208] \u22121, 1[ such that\nf\u2032(c) = 1:\n(1) Applying Theorem 2.58 to f|[\u22121,0] we obtain that there is an a \u2208] \u22121, 0[ such that\nf\u2032(a) =\n4\u22122\n0\u2212(\u22121) = 2.\n(2) Applying Theorem 2.58 to f|[0,1] we obtain that there is a b \u2208]0, 1[ such that f\u2032(b) =\n3\u22124\n1\u22120 = \u22121.\n(3) As f is C1, f\u2032 is continuous. Hence, Theorem 1.66 implies that there is a c \u2208(a, b) \u2286]\u22121, 1[\nsuch that f\u2032(c) = 1.\nCorollary 2.60. Let f, g: [a, b] \u2192R be continuous functions, a, b \u2208R. Assume that f, g are\ndi\ufb00erentiable over (a, b) and that f\u2032(x) = g\u2032(x) for each x \u2208(a, b). Then there exists a real\nnumber C such that f(x) = g(x) + C.", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_74", "text": "di\ufb00erentiable over (a, b) and that f\u2032(x) = g\u2032(x) for each x \u2208(a, b). Then there exists a real\nnumber C such that f(x) = g(x) + C.\nProof. By taking h(x) := f(x) \u2212g(x), it su\ufb03ce to apply Lemma 2.61.\nLemma 2.61. Let h: [a, b] \u2192R a continuous function which is di\ufb00erentiable on (a, b). Assume\nthat h\u2032(x) = 0, for all x \u2208(a, b). Then, h is a constant function.\nRecall that by saying that h is a constant function we simply mean that \u2200x \u2208[a, b], h(x) = C\nfor some \ufb01xed real number C \u2208R (indipendent of x).\nProof. Assume that h is not constant. Then, there exists c, d \u2208[a, b], c < d such that h(c) \u0338=\nh(d).\nThen, the Mean Value Theorem 2.58 implies that there exists e \u2208(c, d), such that\nh\u2032(e) = h(d)\u2212h(c)\nd\u2212c\n\u0338= 0; nonetheless, this contradicts our assumption that h\u2032().\n2.5.1\nMonotone functions and di\ufb00erentials\nWe can apply the Mean Value Theorem to characterize the derivative of monotone (di\ufb00eren-\ntiable) functions.\nCorollary 2.62. Let f : [a, b] \u2192R be a continuous function, a, b \u2208R.\nAssume that f is\ndi\ufb00erentiable on (a, b). Then,\n(1) f is increasing (resp. decreasing) if and only if f\u2032(x) \u22650 (resp. \u22640);\n(2) if f\u2032(x) > 0 (resp. < 0) for all x \u2208(a, b), then f is strictly increasing (resp. strictly\ndecreasing).\nProof. We only prove the increasing case of (1), as the others are similar.\n37", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_75", "text": "Images/x3_gr.png\nFigure 7: f(x) = x3\n\u25e6First we assume that f is increasing. Then,\nx \u2265x0 =\u21d2f(x) \u2265f(x0) =\u21d2f(x) \u2212f(x0)\nx \u2212x0\n\u22650,\nx \u2264x0 =\u21d2f(x) \u2264f(x0) =\u21d2f(x) \u2212f(x0)\nx \u2212x0\n\u22650.\nThus,\nf\u2032(x0) = lim\nx\u2192x0\nf(x) \u2212f(x0)\nx \u2212x0\n\u22650\n\u25e6Second, let us assume by contradiction that f\u2032(x) \u22650 \u2200x \u2208(a, b) and that f is not increas-\ning. Hence, there are a \u2264c < d \u2264b, such that f(c) > f(d). However, then Theorem 2.58\ntells us that then there esists e \u2208R, c < e < d such that f\u2032(e) = f(d)\u2212f(c)\nd\u2212c\n< 0.\nExample 2.63. Let f : E \u2192R be a strictly increasing (resp. strictly decreasing) function.\nThen, it does not necessarily follow that f\u2032(x) > 0 (resp. f\u2032(x) < 0). For example, f(x) = x3\nis strictly increasing, but f\u2032(0) = 3 \u00b7 02 = 0.\nExample 2.64. Let us consider the function fa : R \u2192R, f(x) := sin(x)+ax, where a \u2208R is a\n\ufb01xed real number. Let us compute for what value of a fa is monotone. As fa(x) is di\ufb00erentiable\non R, then fa is monotone if and only if either f\u2032(x) \u22650, \u2200x \u2208R or f\u2032(x) \u22640, \u2200x \u2208R. Thus,\nlet us compute f\u2032(x):\nf\u2032(x) = cos(x) + a.\nThus,\n\u25e6f is increasing if and only if a \u22651;\n\u25e6f is decreasing if and only if a \u2264\u22121.\n38", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_76", "text": "Example 2.65. Using Corollary 2.62 and Proposition 2.37 we obtain that the all the functions\nof De\ufb01nition 2.36 are either monotone, or become monotone when restricted to R\u2217\n+ or to R\u2217\n\u2212.\nf(x)\nD(f)\nf\u2032\nmonotonicity\nsinh(x)\nR\ncosh(x)\nincreasing over R\ncosh(x)\nR\nsinh(x)\ndecreasing over R\u2217\n\u2212and increasing over R\u2217\n+\ntanh(x)\nR\n1\ncosh(x)2\nincreasing over R\ncoth(x)\nR\u2217\n\u22121\nsinh(x)2\ndecreasing over R\u2217\n\u2212and over R\u2217\n+\nxa, a > 0\nR+\naxa\u22121, x \u0338= 0\nincreasing over R\u2217\n+\nxa, a < 0\nR+\naxa\u22121, x \u0338= 0\ndecreasing over R\u2217\n+\nax, a > 1\nR\nlog(a) \u00b7 ax\nincreasing over R\nax, 0 < a < 1\nR\nlog(a) \u00b7 ax\ndecreasing over R\nloga(x), a > 1\nR\u2217\n+\n1\nlog(a)\u00b7x\nincreasing over R\u2217\n+\nloga(x), 0 < a < 1\nR\u2217\n+\n1\nlog(a)\u00b7x\ndecreasing over R\u2217\n+\n2.5.2\nL\u2019H\u02c6opital\u2019s rule\nL\u2019H\u02c6opital rule gives a method to compute limits of fractions of function which are in the\nindeterminate forms\n0\n0, \u221e\n\u221e,\nthat is, either both values of the limit of the denominator and of the limit of the numerator\napproach 0, or they both approach \u2212\u221eor +\u221e\u2013 in the latter case, the sign of \u221edoes not\nreally matter.\nExample 2.66. How can we compute\nlim\nx\u2192+\u221e\nex\nx ? In this case,\nlim\nx\u2192+\u221eex = +\u221e=\nlim\nx\u2192+\u221ex.\nIn this example, we cannot answer using the algebraic rules of Proposition 1.20.\nLuckily, the following theorem provides us with new tools to carry out this kind of compu-\ntations.\nTheorem 2.67 (L\u2019H\u02c6opital rule). Let f, g: (a, b) \u2192R be di\ufb00erentiable functions, and let a, b \u2208\nR. Assume that the following conditions hold:\n(1) (exactly) one the following conditions hold for x0:\n(i) x0 \u2208(a, b);\n(ii) x0 = a \u2208R;", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_77", "text": "R. Assume that the following conditions hold:\n(1) (exactly) one the following conditions hold for x0:\n(i) x0 \u2208(a, b);\n(ii) x0 = a \u2208R;\n(iii) x0 = b \u2208R;\n(iv) x0 = a = \u2212\u221e;\n(v) x0 = b = +\u221e;\n39", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_78", "text": "(2) g(x) \u0338= 0 and g\u2032(x) \u0338= 0 for all x \u2208(a, b) \\ {x0};\n(3)\nlim\nx\u2192x0f(x) = lim\nx\u2192x0g(x) = \u03b1 for \u03b1 = 0 or \u03b1 = \u00b1\u221e.\nThen, in the respective cases we have the following implications for any \u00b5 \u2208R:\nCases (1|i), (1|iv) and (1|v)\nif\nlim\nx\u2192x0\nf\u2032(x)\ng\u2032(x) = \u00b5\n\u21d2\nlim\nx\u2192x0\nf(x)\ng(x) = \u00b5\nCase (1|ii)\nif\nlim\nx\u2192x+\n0\nf\u2032(x)\ng\u2032(x) = \u00b5\n\u21d2\nlim\nx\u2192x+\n0\nf(x)\ng(x) = \u00b5\nCase (1|iii)\nif\nlim\nx\u2192x\u2212\n0\nf\u2032(x)\ng\u2032(x) = \u00b5\n\u21d2\nlim\nx\u2192x\u2212\n0\nf(x)\ng(x) = \u00b5\nProof. We prove only the \u03b1 = 0 and x0 \u2208(a, b) case, and we refer to page 121-122 of the book\nfor the rest. As f and g are di\ufb00erentiable at x0 they are also continuous there, and hence\nf(x0) = lim\nx\u2192x0f(x) = \u03b1 = 0\nand\ng(x0) = lim\nx\u2192x0g(x) = \u03b1 = 0.\n(2.67.a)\nSo, by the mean value theorem for derivatives, there is a real number c(x) between x and x0\nsuch that\nf\u2032(c(x)) = f(x) \u2212f(x0)\nx \u2212x0\n.\n(2.67.b)\nIn particular, c(x): E \\ x0 \u2192I \\ x0 is a function such that lim\nx\u2192x0c(x) = x0. Then:\n\u00b5 = lim\nx\u2192x0\nf\u2032(x)\ng\u2032(x)\n|\n{z\n}\nde\ufb01nition of \u00b5\n= lim\nx\u2192x0\nf\u2032(c(x))\ng\u2032(c(x))\n|\n{z\n}\nProposition 1.30\n= lim\nx\u2192x0\nf(x)\u2212f(x0)\nx\u2212x0\ng(x)\u2212g(x0)\nx\u2212x0\n|\n{z\n}", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_79", "text": "f\u2032(c(x))\ng\u2032(c(x))\n|\n{z\n}\nProposition 1.30\n= lim\nx\u2192x0\nf(x)\u2212f(x0)\nx\u2212x0\ng(x)\u2212g(x0)\nx\u2212x0\n|\n{z\n}\n(2.67.b)\n= lim\nx\u2192x0\nf(x) \u2212f(x0)\ng(x) \u2212g(x0) = lim\nx\u2192x0\nf(x)\ng(x)\n|\n{z\n}\n(2.67.a)\nRemark 2.68. We show that the property (3) in the statement of Theorem 2.67 is a necessary\none.\nIndeed, we show that if the limit lim\nx\u2192x0\nf\u2032(x)\ng\u2032(x) does not exist, then we cannot conclude\nanything about the limit lim\nx\u2192x0\nf(x)\ng(x).\n(1) Let us take f(x) = x + sin(x), g(x) = x. Then f, g are di\ufb00erentiable over R,\nlim\nx\u2192+\u221ef(x) = +\u221e= lim\nx\u21920g(x),\ng(x) \u0338= 0 \u0338= g\u2032(x), \u2200x \u2208R\u2217.\nMoreover,\nlim\nx\u2192+\u221e\nf(x)\ng(x) =\nlim\nx\u2192+\u221e1 + sin(x)\nx\n= 1.\nOn the other hand,\nlim\nx\u2192+\u221e\nf\u2032(x)\ng\u2032(x) =\nlim\nx\u2192+\u221e\n1 + cos(x)\n1\nwhich is not de\ufb01ned since the limit\nlim\nx\u2192+\u221ecos(x) does not exist. Hence, since the limit of\nthe quotient of the derivatives of f, g does not exist, a priori, we cannot conclude anything\nabout the limit of the quotient of f, g. Nonetheless, in this case we are lucky and we can\nstill carry out the computation.\n40", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_80", "text": "(2) Consider f(x) = \u221ax + sin(x), g(x) = x. Then f, g are di\ufb00erentiable over R\u2217\n+,\nlim\nx\u2192+\u221ef(x) = +\u221e= lim\nx\u21920g(x),\ng(x) \u0338= 0 \u0338= g\u2032(x), \u2200x \u2208R\u2217.\nMoreover,\nlim\nx\u2192+\u221e\nf(x)\ng(x) =\nlim\nx\u2192+\u221e\n1\n\u221ax + sin(x)\nx\n= 0.\nOn the other hand,\nlim\nx\u2192+\u221e\nf\u2032(x)\ng\u2032(x) =\nlim\nx\u2192+\u221e\n1\n2\u221ax + cos(x)\n1\nwhich is not de\ufb01ned since the limit\nlim\nx\u2192+\u221ecos(x) does not exist.\n(3) Consider f(x) = x + sin(x), g(x) = x. Then f, g are di\ufb00erentiable over R\u2217\n+,\nlim\nx\u2192+\u221ef(x) = +\u221e= lim\nx\u21920g(x),\ng(x) \u0338= 0 \u0338= g\u2032(x), \u2200x \u2208R\u2217.\nMoreover,\nlim\nx\u2192+\u221e\nf(x)\ng(x) =\nlim\nx\u2192+\u221e\n1\n\u221ax + sin(x)\nx\n= 0.\nOn the other hand,\nlim\nx\u2192+\u221e\nf\u2032(x)\ng\u2032(x) =\nlim\nx\u2192+\u221e\n1\n2\u221ax + cos(x)\n1\nwhich is not de\ufb01ned since the limit\nlim\nx\u2192+\u221ecos(x) does not exist.\nHence, if the limit of the quotient of the derivatives of f, g does not exist, a priori, we cannot\nconclude anything about the limit of the quotient of f, g. Nonetheless, in some cases, such as\n() here, we are lucky and we can still carry out the computation.\nExample 2.69. Let us consider the limit lim\nx\u21920\narcsin(x)\nsin(x) . Then, lim\nx\u21920 arcsin(x) = 0 = lim\nx\u21920 sin(x)\nand sin(x)\u2032 = cos(x), arcsin(x)\u2032 =\n1\n\u221a\n1\u2212x2 .\nMoreover, both sin(x) and cos(x) are non-zero over the pointed neighborhood ] \u2212\u03c0\n2 , \u03c0\n2 [\\{0} of", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_81", "text": "and sin(x)\u2032 = cos(x), arcsin(x)\u2032 =\n1\n\u221a\n1\u2212x2 .\nMoreover, both sin(x) and cos(x) are non-zero over the pointed neighborhood ] \u2212\u03c0\n2 , \u03c0\n2 [\\{0} of\n0. Hence, we can apply Theorem 2.67 to get\nlim\nx\u21920\narcsin(x)\nsin(x)\n= lim\nx\u21920\n1\n\u221a\n1\u2212x2\ncos(x) = 1\nExample 2.70. Let us consider the limit\nlim\nx\u2192+\u221e\nex\nxn . Then, f(x) = ex, g(x) = xn, n \u2208N. Let\nus start with the case n = 1. Then,\nlim\nx\u2192+\u221e\nex\nx =\nlim\nx\u2192+\u221e\nex\n1 = +\u221e.\nFor n = 2,\nlim\nx\u2192+\u221e\nex\nx2 =\nlim\nx\u2192+\u221e\nex\n2x =\nlim\nx\u2192+\u221e\nex\n2 = +\u221e.\nHence, inductively, one can prove that\nlim\nx\u2192+\u221e\nex\nxn =\nlim\nx\u2192+\u221e\nex\nnxn\u22122 =\nlim\nx\u2192+\u221e\nex\nn(n \u22121)xn\u22122 = ... =\nlim\nx\u2192+\u221e\nex\nn! = +\u221e.\nHence, the exponential function ex goes to +\u221e\u2013 as x goes to +\u221e\u2013 faster than any monomial\nxn; a similar argument shows that it goes faster than any polynomial.\n41", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_82", "text": "Images/arcsin_gr.png\nFigure 8: f(x) = arcsin(x)\nExample 2.71. Similarly to the previous example,\nlim\nx\u21920+ xn log(x) = lim\nx\u21920+\nlog(x)\n1\nxn\n= lim\nx\u21920+\n1\nx\n\u2212n\nxn+1\n= lim\nx\u21920+ \u2212xn\nn = 0,\nwhile,\nlim\nx\u2192+\u221e\nlog(x)\nxn\n=\nlim\nx\u2192+\u221e\n1\nx\nnxn\u22121 =\nlim\nx\u2192+\u221e\n1\nnxn 0\nSo, log goes to \u2212\u221eas x goes to 0 and to +\u221eas x goes to +\u221eslower than 1\nx and x, respectively.\n2.5.3\nTaylor expansion\nDe\ufb01nition 2.72. Let f : E \u2192R be a function and let x0 \u2208E Assume that there is a neighbor-\nhood of a \u2208E which is contained in the domain (so in E). We say that f admits an expansion\nto the n-th order x0 if there is an equality of the form\nf(x) = a0 + a1(x \u2212a) + a2(x \u2212a)2 + \u00b7 \u00b7 \u00b7 + an(x \u2212a)n + (x \u2212a)n\u03f5(x),\n(2.72.c)\nwhere ai are real number, and \u03f5n(x): E \u2192R satis\ufb01es lim\nx\u2192x0 \u03f5n(x) = 0.\nProposition 2.73. In the hypotheses of De\ufb01nition 2.72, if a function f admits an n-th order\nexpansion around a point x0 \u2208D(f), then the coe\ufb03cients ai in (2.72.c) are uniquely deter-\nmined.\nProof. Let\nf(x) = a0 + a1(x \u2212x0) + a2(x \u2212x0)2 + \u00b7 \u00b7 \u00b7 + an(x \u2212x0)n + (x \u2212x0)n\u03f5n(x),\nf(x) = a\u2032\n0 + a\u2032\n1(x \u2212x0) + a\u2032\n2(x \u2212x0)2 + \u00b7 \u00b7 \u00b7 + a\u2032\nn(x \u2212x0)n + (x \u2212x0)n\u03f5\u2032\nn(x),\nbe two di\ufb00erent expansions to order n of f around x0. We show by induction on i that ai = a\u2032\ni.", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_83", "text": "2(x \u2212x0)2 + \u00b7 \u00b7 \u00b7 + a\u2032\nn(x \u2212x0)n + (x \u2212x0)n\u03f5\u2032\nn(x),\nbe two di\ufb00erent expansions to order n of f around x0. We show by induction on i that ai = a\u2032\ni.\nFor i = 0 this is given by passing to the limit as x \u2192x0 of the two expansion:\na0 = lim\nx\u2192x0 a0 + a1(x \u2212x0) + a2(x \u2212x0)2 + \u00b7 \u00b7 \u00b7 + an(x \u2212x0)n + (x \u2212x0)n\u03f5n(x)\n= lim\nx\u2192x0 f(x)\n= lim\nx\u2192x0 a\u2032\n0 + a\u2032\n1(x \u2212x0) + a\u2032\n2(x \u2212x0)2 + \u00b7 \u00b7 \u00b7 + a\u2032\nn(x \u2212x0)n + (x \u2212x0)n\u03f5\u2032\nn(x) = a\u2032\n0\n42", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_84", "text": "Let us prove the induction step. Thus, let us assume that we know that aj = a\u2032\nj for j =\n0, . . . , i \u22121. Then,\nf(x) =a0 + a1(x \u2212x0) + a2(x \u2212x0)2 + \u00b7 \u00b7 \u00b7 + an(x \u2212x0)n + (x \u2212x0)n\u03f5n(x)\n=a\u2032\n0 + a\u2032\n1(x \u2212x0) + a\u2032\n2(x \u2212x0)2 + \u00b7 \u00b7 \u00b7 + a\u2032\nn(x \u2212x0)n + (x \u2212x0)n\u03f5\u2032\nn(x)\n=a0 + a1(x \u2212x0) + \u00b7 \u00b7 \u00b7 + ai\u22121(x \u2212x0)i\u22121 + a\u2032\ni(x \u2212x0)i + \u00b7 \u00b7 \u00b7 + a\u2032\nn(x \u2212x0)n + (x \u2212x0)n\u03f5\u2032\nn(x)\nHence, taking the expansions on the 1st and the 3rd line of the previous chain of equalities,\nand subtracting from both expansions a0 + a1(x \u2212a) + \u00b7 \u00b7 \u00b7 + ai\u22121(x \u2212a)i\u22121 and then dividing\nboth by (x \u2212a)i, we obtain\nai + ai+1(x \u2212x0) + \u00b7 \u00b7 \u00b7 + an(x \u2212x0)n\u2212i + (x \u2212x0)n\u2212i\u03f5n(x)\n= a\u2032\ni + a\u2032\ni+1(x \u2212x0) + \u00b7 \u00b7 \u00b7 + a\u2032\nn(x \u2212x0)n\u2212i + (x \u2212x0)n\u2212i\u03f5\u2032\nn(x).\nTaking limit of this equality as x \u2192x0 yields that ai = a\u2032\ni, which concludes the induction step.\nHence, ai = a\u2032\ni for each i. In particular, it also follows that \u03f5(x) = \u03f5\u2032(x) for each x \u2208E.\nWhen is it that we can \ufb01nd an expansion to order n for a function f around a point\nx0 \u2208D(f)? The following theorem provides a \ufb01rst answer.\nTheorem 2.74. Let n \u22650 be an integer. Let f : E \u2192R be a function de\ufb01ned on an open\ninterval E, and let x0 \u2208E. Assume that f is n + 1 times di\ufb00erentiable over E. Then, for each", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_85", "text": "Theorem 2.74. Let n \u22650 be an integer. Let f : E \u2192R be a function de\ufb01ned on an open\ninterval E, and let x0 \u2208E. Assume that f is n + 1 times di\ufb00erentiable over E. Then, for each\nx \u2208E there exists x\u2032 \u2208]x0, x[, if x > x0 (resp. x\u2032 \u2208]x, x0[, if x < x0) and such that\nf(x) =\n n\nX\ni=0\nf(i)(a)\ni!\n(x \u2212a)i\n!\n+ f(n+1)(x\u2032)(x \u2212a)n+1\n(n + 1)! .\nRemark 2.75. Theorem 2.74 not only tells us that, under the hypotheses posed in its statement,\nit is possible to \ufb01nd an order n expansion for a function f around a point x0 but also that,\nwhen that is the case, we have a recipe to compute the coe\ufb03cients which are given by the\nformula\naj = f(j)(x0)\nj!\n.\nMoreover, we can also compute the error term in the\nProof. To understand the proof, note that the statement for n = 0 is just the Mean Value\nTheorem, cf. Theorem 2.58. Indeed, that results implies that there exists x\u2032 \u2208]x0, x[, if x > x0\n(resp. x\u2032 \u2208]x, x0[, if x < x0) such that f\u2032(x\u2032) = f(x)\u2212f(x0)\nx\u2212x0\n. Multiply by x \u2212x0, then\nf(x) = f(x0) + f\u2032(x0 + \u03b8x,x0(x \u2212x0))(x \u2212x0),\nwhere \u03b8x,x0(x \u2212x0) \u2208[0, 1] and x\u2032 = x0 + \u03b8x,x0(x \u2212x0) \u2013 which is possible exactly because x\u2032 is\ncontained between x and x0. Let us recall that the proof of Theorem 2.58 was an application\nof Rolle\u2019s theorem to the function g(y) := f(y) \u2212f(x0) \u2212f(x)\u2212f(x0)\nx\u2212x0\n(y \u2212x0). Furthermore, this", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_86", "text": "of Rolle\u2019s theorem to the function g(y) := f(y) \u2212f(x0) \u2212f(x)\u2212f(x0)\nx\u2212x0\n(y \u2212x0). Furthermore, this\ntechique was working since g(x0) = g(x), and g\u2032(y) = f\u2032(y) \u2212f(x)\u2212f(x0)\nx\u2212x0\n, so that g\u2032(y) being 0\nyielded exactly the above equation.\nLet us now de\ufb01ne\nPn(x) :=\nn\nX\ni=0\nf(i)(x0)\ni!\n(x \u2212x0)i,\n43", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_87", "text": "and let us consider\ng(y) = f(y) \u2212Pn(y) + Pn(x) \u2212f(x)\n(x \u2212x0)n+1 (y \u2212x0)n+1.\nThen,\n0 = g(x) = g(a) = g\u2032(a) = \u00b7 \u00b7 \u00b7 = g(n)(a),\nwhich means that there exists y1 between x0 and x such that g\u2032(y1) = 0 by Rolle\u2019s theorem. But\nthen applying Rolle\u2019s theorem again we obtain a y2 between x0 and y1 such that g(2)(y2) = 0.\nIterating this process we obtain a point yn+1 between x0 and x such that g(n+1)(yn+1) = 0. In\nparticular, by setting x\u2032 := yn+1, then\n0 = g(n+1)(x\u2032) = f(x\u2032) + Pn(x) \u2212f(x)\n(x \u2212x0)n+1 (n + 1)!.\nReorganizing the latter equation yields exactly the statement of the theorem.\nCorollary 2.76. Let n \u22650 be a real number. Let f : E \u2192R be a function de\ufb01ned on an open\ninterval E. Assume that f \u2208Cn(E, R), and let x0 \u2208E. Then, the n-th order expansion of f\naround x0 exists and is given by the formula\nf(x) =\nn\nX\nj=0\nf(j)(x0)\nj!\n(x \u2212x0)j + (x \u2212x0)n\u03f5n(x).\nThe idea behind the proof of the corollary is that by the previous theorem the error term\nis f(n+1)(x\u2032) \u2212f(n+1)(x), which converges to zero as x goes to a as x\u2032 is between a and x, and\nf(n) is continuous. For the precise proof we refer to page 126 of the book.\nExample 2.77. Applying Corollary 2.76 to f(x) =\n1\n1\u2212x and x0 = 0 yields that the order n\nexpansion takes the form\n1\n1 \u2212x = 1 + x + x2 + \u00b7 \u00b7 \u00b7 + xn + xn\u03f5n(x),\nsince", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_88", "text": "Example 2.77. Applying Corollary 2.76 to f(x) =\n1\n1\u2212x and x0 = 0 yields that the order n\nexpansion takes the form\n1\n1 \u2212x = 1 + x + x2 + \u00b7 \u00b7 \u00b7 + xn + xn\u03f5n(x),\nsince\nf(i)(x) =\ni!\n(1 \u2212x)i+1\n\u21d2\nf(i)(0) = i!\n\u21d2\nf(i)(0)\ni!\n= 1.\nExample 2.78. Applying Corollary 2.76 to f(x) = ex and x0 = 0 yields that the order n\nexpansion takes the form\nex =\nn\nX\ni=0\nxi\ni! + xn\u03f5n(x),\nsince\nf(i)(x) = ex\n\u21d2\nf(i)(0) = 1\n\u21d2\nf(i)(0)\ni!\n= 1\ni!.\nExample 2.79. Similarly, the (2n + 1)-st order expansion of cos(x) around x = 0 is\ncos(x) =\nn\nX\nj=0\n(\u22121)n x2j\n(2j)! + x2n+1, \u03f5(x)\nwhile the (2n + 2)-nd order expansion of sin(x) around x = 0 is\nsin(x) =\nn\nX\nj=0\n(\u22121)n\nx2j+1\n(2j + 1)! + x2n+2\u03f52n+2(x).\n44", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_89", "text": "Example 2.80. One can also \ufb01gure out expansions of products, sums, compositions, etc.\nFor example the 3-rd order expansion of sin(cos(x)) is as follows:\ncos(sin(x)) = cos\n\u0012\nx \u2212x3\n6 + x3\u03f53(x)\n\u0013\n=1 \u2212\n\u0010\nx \u2212x3\n6 + x3\u03f53(x)\n\u00112\n2\n+\n\u0012\nx \u2212x3\n6 + x3\u03f53(x)\n\u00133\n\u03b73(sin(x)) = 1 \u2212x2\n2 + x3\u03c43(x),\nwhere x3\u03c43(x) is the sum of all terms of the form x3h(x), where lim\nx\u21920h(x) = 0. In particular,\nlim\nx\u21920\u03c43(x) = 0 and hence the above is indeed the 3-rd order expansion.\nIn general, we can compute the expansion to order n of a composition (f \u25e6g)(x) around a\npoint f(x0) by substituting the expension to order n of g around x0 into the expansion of f\naround f(x0) to order n and then re-ordering all the terms thus obtained up to order n. Let\nus highlight how one should be careful that the base-point of the expansion of the function f\nshould be the value g(x0) of the function g at the base-point x0. So, for example, sin(cos(x))\nat 0 is not easy to compute this way, because one would need the expansion of sin around\ncos(0) = 1, for which there is no nice formula.\nAnother example is by taking h(x) =\n1\n1\u2212(ex\u22121), x0 = 0. Then we can rewrite h as the\ncomposition h = f \u25e6g of f(y) =\n1\n1\u2212y and g(x) = ex \u22121. Then,\n1\n1 \u2212(ex \u22121) =\n1\n1 \u2212\n\u0010\nx + x2\n2 + x3\n6 + x3\u03f53(x)\n\u0011\n=1 +\n\u0012\nx + x2\n2 + x3\n6 + x3\u03f53(x)\n\u0013\n+\n\u0012\nx + x2\n2 + x3\n6 + x3\u03f53(x)\n\u00132\n+\n\u0012\nx + x2\n2 + x3\n6 + x3\u03f53(x)\n\u00133\n\u03b73(x + x2\n2 + x3\n6 + x3\u03f53(x))\n=1 + x +\n\u00121\n2 + 1\n\u0013\nx2 +\n\u00121\n6 + 2 \u00b7 1 \u00b7 1", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_90", "text": "6 + x3\u03f53(x)\n\u00132\n+\n\u0012\nx + x2\n2 + x3\n6 + x3\u03f53(x)\n\u00133\n\u03b73(x + x2\n2 + x3\n6 + x3\u03f53(x))\n=1 + x +\n\u00121\n2 + 1\n\u0013\nx2 +\n\u00121\n6 + 2 \u00b7 1 \u00b7 1\n2 + 1\n\u0013\nx3 + x3\u03c43(x)\n=1 + x + 3\n2x2 + 13\n6 x3 + x3\u03c43(x).\nSimilarly, one can write the order 3 expansion of\n1\n1\u2212x \u00b7 ex around 0 as\n1\n1 \u2212x \u00b7 ex =(1 + x + x2 + x3 + x3\u03f53(x))\n\u0012\n1 + x + x2\n2 + x3\n6 + x3\u03b73(x)\n\u0013\n=1 + 2x + 5\n2x2 + 8\n3x2 + x3\u03c43(x)\nYou can \ufb01nd more examples in the book, pages 127-131.\nExample 2.81. One can use expansions also to avoid using Theorem 2.67. For example, to\ncompute\nlim\nx\u21920\n(ex \u22121 \u2212x) + x sin(x)\ncos(x) \u22121\n,\n45", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_91", "text": "then we can try to compute the 2-nd order expansions \ufb01rst:\n(ex \u22121 \u2212x) + x sin(x) =\n\u0012\n1 + x + x2\n2 + x2\u03f52(x) \u22121 \u2212x\n\u0013\n+ x(x + x2\u03b72(x))\n=x2\n2 + x2 + x2(\u03b72(x) + \u03f52(x)) = 3\n2x2 + x2\n\u03b32(x)\n| {z }\n\u03b32(x):=\u03b72(x)+\u03f52(x)\n),\ncos(x) \u22121 =1 \u2212x2\n2 + x2\u03c42(x) \u22121 = \u2212x2\n2 + x2\u03c44(x)\nThen,\nlim\nx\u21920\n(ex \u22121 \u2212x) + x sin(x)\ncos(x) \u22121\n= lim\nx\u21920\n3\n2x2 + x2\u03f53(x)\n\u2212x2\n2 + x2\u03f54(x)\n= lim\nx\u21920\n3\n2 + \u03f53(x)\n\u22121\n2 + \u03f54(x) = \u22123.\n2.5.4\nApplication of Taylor expansion to local extrema and in\ufb02ection points\nWe have proven that if f has a point of local extremum at x0 \u2208D(f), then f\u2032(x0) = 0,\ncf. Proposition 2.52. However, we have also shown that the converse implication does not hold,\ncf. Example 2.53. Nevertheless, we would like to know whether, for example, by imposing\nsuitable conditions on the higher derivatives of a function f, we can still characterize when a\nstationary point is a point of local extremum for a function.\nLet f : E \u2192R be a function, and let x0 \u2208E be a stationary point for f. Moreover, let us\nassume that for some even natural number n, the \ufb01rst n \u22121 derivatives of f vanish at x0\nf\u2032(x0) = 0 = f\u2032\u2032(x0) = \u00b7 \u00b7 \u00b7 = f(n\u22121)(x0),\nwhile the n-th derivative of f is non-zero and f(n)(x0) > 0. Then, writing the n-th order\nexpansion of f around x0,\nf(x) = f(x0) + f(n)(x0)\nn!\n(x \u2212x0)n + (x \u2212x0)n\u03f5n(x).", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_92", "text": "expansion of f around x0,\nf(x) = f(x0) + f(n)(x0)\nn!\n(x \u2212x0)n + (x \u2212x0)n\u03f5n(x).\nThus, for x su\ufb03ciently close to x0 it holds that |\u03b5n(x)| < 1\n2 \u00b7 f(n)(x0)\nn!\nholds. In particular, for\nsuch values of x, then\nf(x0) < f(x0) + 1\n2 \u00b7 f(n)(x0)\nn!\n(x \u2212x0)n \u2264f(x)\nThis shows that x0 is a point of local minimum for f. One can imitate this argument for the\ncase where f(n)(x0) < 0 to yield that x0 is a point of local maximum for f. Thus, we can\nsummarize the results obtained so far in the following theorem.\nTheorem 2.82. Let n \u22652 be an even integer.\nLet f : E \u2192R be a function on an open\ninterval E and let x0 \u2208E. Assume that f is di\ufb00erentiable n times on E and that f(i)(x0) = 0,\n\u2200i = 1, 2, 3, . . . , n \u22121.\n(1) If f(n) > 0, then f has a point of local minimum at x0.\n(2) If f(n) < 0, then f has a point of local maximum at x0.\nExample 2.83. Consider the function f(x) = sin(x)+ 1\n2x over the interval [0, 2\u03c0], cf. Figure 9.\nThen, f\u2032(x) = 0 if and only if cos(x) = \u22121\n2, which is equivalent to x = 2\u03c0\n3 or 4\u03c0\n3 . Whether or\nnot we have a maximum or minimum at these points is decided by the sign of f\u2032\u2032(x) = \u2212sin(x).\n\u25e6At x = 2\u03c0\n3 , f(x)\u2032\u2032 < 0, so f(x) has a local maximum, and\n46", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_93", "text": "Images/sinxplushalfx_gr.png\nFigure 9: f(x) = sin(x) + 1\n2x over the interval [0, 2\u03c0]\n\u25e6At x = 4\u03c0\n3 , f(x)\u2032\u2032 > 0, so f(x) has a local minimum.\nQuestion 2.84. What happens if we assume that n is an odd natural number in the statement\nof Theorem 2.82?\nIn that case, the expansion to order n for f around x0 takes the same expression as before\nf(x) = f(x0) + f(n)(x0)\nn!\n(x \u2212x0)n + (x \u2212x0)n\u03f5n(x),\nbut this time the \ufb01rst non-leading term will be of the form (x \u2212x0)3, (x \u2212x0)5, or (x \u2212x0)7,\netc., depending on the precise value of n. But then for x > x0, (x \u2212x0)n > 0, while x < x0,\n(x \u2212x0)n < 0. This type of behavior characterizes what is called an in\ufb02ection. That is to say,\nthat for a stationary point x0 to be an in\ufb02ection point for f, we require that, on one side of x0\nthe graph of the function is above the tangent line to the graph of f through (x0, f(x0)), and\non the other side it is below it, cf. Figure 10\nImages/hor_flex.png\nFigure 10: f(x) = x3 has a stationary point at x = 0 which is a \ufb02ex, as the graph goes through\nthe tangent line y = 0.\n47", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_94", "text": "We can actually generalize this tentative de\ufb01nition, as follows, to comprise not just the case\nof stationary points.\nDe\ufb01nition 2.85. Let f : E \u2192R be a function. Assume that f is di\ufb00erentiable at x0 \u2208E.\nWe say that f has an in\ufb02ection point at x0 if there exists \u03b4 > 0 such that either one of the\nfollowing two conditions is satis\ufb01ed:\n(1) {x \u2208E|a < x < a + \u03b4} \u21d2f(x) \u2212f(a) \u2212f\u2032(a)(x \u2212a) > 0, and\n{x \u2208E|a \u2212\u03b4 < x < a} \u21d2f(x) \u2212f(a) \u2212f\u2032(a)(x \u2212a) < 0; or,\n(2) {x \u2208E|a < x < a + \u03b4} \u21d2f(x) \u2212f(a) \u2212f\u2032(a)(x \u2212a) < 0, and\n{x \u2208E|a \u2212\u03b4 < x < a} \u21d2f(x) \u2212f(a) \u2212f\u2032(a)(x \u2212a) > 0.\nThe reasoning contained in the paragraph before De\ufb01nition 2.85 immediately yields the\nfollowing result.\nTheorem 2.86. Let n \u22653 be an odd integer. Let f : E \u2192R be a function de\ufb01ned over an\nopen interval E and let x0 \u2208E. Assume that f is di\ufb00erentiable n times on E and that\nf\u2032\u2032(x0) = \u00b7 \u00b7 \u00b7 = f(n\u22121)(x0) = 0,\nwhile f(n)(x0) \u0338= 0. Then, f has an in\ufb02ection point at x0.\nExample 2.87. Let us consider the function f(x) = 2 sin(x) \u2212x. Then f\u2032(x) = 2 cos(x) \u22121,\nf\u2032\u2032(x) = \u22122 sin(x) and f\u2032\u2032\u2032(x) = \u22122 cos(x). Hence, f\u2032(0) = f\u2032\u2032(0) = 0, and f\u2032\u2032\u2032(0) \u0338= 0. Hence\nf(x) has an in\ufb02ection point at x = 0 according to Theorem 2.86.\nImages/obl_flex.png", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_95", "text": "f(x) has an in\ufb02ection point at x = 0 according to Theorem 2.86.\nImages/obl_flex.png\nFigure 11: The function f(x) = 2 sin(x) \u2212x has a \ufb02ex at the point x = 0, which is non-\nstationary, as f\u2032(0) = 1, through the tangent line y = x to the graph of f at the point (0, 0).\n2.5.5\nConvex and concave functions\nDe\ufb01nition 2.88. Let f : E \u2192R be a function de\ufb01ned over an open interval E. We say that\nf is convex (resp. concave) if for every a, b \u2208E and every \u03bb \u2208[0, 1] we have:\nf(\u03bba + (1 \u2212\u03bb)b) \u2264\u03bbf(a) + (1 \u2212\u03bb)f(b).\n(resp. f(\u03bba + (1 \u2212\u03bb)b) \u2265\u03bbf(a) + (1 \u2212\u03bb)f(b) ).\n48", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_96", "text": "Remark 2.89. Let us maintain the same notation as in the above de\ufb01nition. We may assume\nthat a < b. Then, for \u03bb \u2208[0, 1], x := \u03bba + (1 \u2212\u03bb)b is a point between a and b. Geometrically,\nthe above de\ufb01nition means the following:\n(1) f is convex, if for any choice of a, b \u2208E, then between a and b, the graph of f lies\ncompletely below the line segment connecting (a, f(a)) and (b, f(b));\nImages/convex.png\nFigure 12: The graph of the function f(x) = (x+1)2 \u22123 lies below the line segment connecting\nthe points (\u22124, f(\u22124)) = \u22124, 6 and (1, f(1)) = (1, 1).\n(2) if f is concave, then between a and b, the graph of f lies completely above the line\nsegment connecting (a, f(a)) and (b, f(b)).\nImages/concave.png\nFigure 13: The graph of the function f(x) = \u2212(x\u22121)2+5 lies below the line segment connecting\nthe points (\u22122, f(\u22122)) = (\u22122, 4) and (1, f(1)) = (1, 1).\nWe can characterize convexity and concavity of a function which is di\ufb00erentiable by means\nof the monotonicity of its \ufb01rst derivative.\n49", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_97", "text": "Theorem 2.90. Let f : E \u2192R be a function de\ufb01ned on an open interval E. Assume that f\nis di\ufb00erentiable. Then f is convex (resp. concave) if and only if f\u2032 : E \u2192R is an increasing\n(resp. decreasing) function.\nProof. We prove only the statements about convexity, as f is convex if and only if \u2212f is\nconcave.\n(1) First, let us assume that f is convex. Let a < b be points of I. We want to prove that\nf\u2032(a) \u2264f\u2032(b). By the above characterization of convexity we have\nf(b) \u2212f(\u03bba + (1 \u2212\u03bb)b)\nb \u2212(\u03bba + (1 \u2212\u03bb)b)\n\u2265f(b) \u2212f(a)\nb \u2212a\n, and f(\u03bba + (1 \u2212\u03bb)b) \u2212f(a)\n(\u03bba + (1 \u2212\u03bb)b) \u2212a\n\u2264f(b) \u2212f(a)\nb \u2212a\n.\nNow, as \u03bb goes to 0, the left side of the \ufb01rst inequality converges to f\u2032(b), and as \u03bb goes\nto 1 the left side of the second inequality converges to f\u2032(a). This yields:\nf\u2032(b) \u2265f(b) \u2212f(a)\nb \u2212a\n\u2265f\u2032(a)\n(2) For, the other direction let us assume that f\u2032 is increasing. Fix a < b \u2208E. Set x :=\n\u03bba + (1 \u2212\u03bb)b for any \u03bb \u2208]0, 1[ (for \u03bb = 0 and 1 the convexity inequality is automatic).\nThen, the mean value theorem tells us that there are a < x1 < x < x2 < b such that\nf(x)\u2212f(a)\nx\u2212a\n= f\u2032(x1) and f(b)\u2212f(x)\nb\u2212x\n= f\u2032(x2). In particular, by our assumption that the\nderivative is increasing it follows that f(x)\u2212f(a)\nx\u2212a\n\u2264f(b)\u2212f(x)\nb\u2212x\n. But this shows that f is\nconvex by the above characterization of convexity in terms of slopes.\nIf f\u2032 is di\ufb00erentiable \u2013 or, equivalently, f is twice di\ufb00erentiable \u2013 then f\u2032 being increasing", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_98", "text": "x\u2212a\n\u2264f(b)\u2212f(x)\nb\u2212x\n. But this shows that f is\nconvex by the above characterization of convexity in terms of slopes.\nIf f\u2032 is di\ufb00erentiable \u2013 or, equivalently, f is twice di\ufb00erentiable \u2013 then f\u2032 being increasing\n(resp. decreasing) is equivalent to f\u2032\u2032 \u22650 (resp. f\u2032\u2032 \u22640). Thus, we can characterize convexity\nand concavity of a function which is twice di\ufb00erentiable by means of the sign of its second\nderivative.\nCorollary 2.91. Let f : E \u2192R be a two times di\ufb00erentiable function on an open interval.\nThen f is convex (resp. concave) if and only if f\u2032\u2032(x) \u22650 (resp. f\u2032\u2032(x) \u22640) for all x \u2208E.\nExample 2.92.\n(1) Let us consider f(x) = ex. Then, f\u2032\u2032(x) = ex, thus f : R \u2192R is convex,\nsince the second derivative ex > 0, \u2200x \u2208R.\n(2) Let us consider f(x) = log(x). Then, f\u2032\u2032(x) =\n\u0000 1\nx\n\u0001\u2032 = \u22121\nx2 , so the function log(x), which\nis de\ufb01ned over R\u2217\n+ is concave over its entire domain.\nExample 2.93. Here we explain why the graph of a di\ufb00erentiable convex function f must lie\nabove the tangent line to the graph through a point (a, f(a)). That is to say, we show that for\na di\ufb00erentiable convex function f,\nf(x) \u2265f(a) + f\u2032(a)(x \u2212a).\nWe take x > a, and leave the case x < a to the reader. Thus, assuming that x > a, we wish to\nshow that f(x) \u2265f(a) + f\u2032(a)(x \u2212a), or equivalently that\nf(x) \u2212f(a)\nx \u2212a\n\u2265f\u2032(a).\n(2.93.d)\nIndeed, by the Mean value theorem (Theorem 2.58) there exists a x\u2032 \u2208R, a < x\u2032 < x such that\nf(x) \u2212f(a)\nx \u2212a\n= f\u2032(x\u2032).\n(2.93.e)", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_99", "text": "(2.93.d)\nIndeed, by the Mean value theorem (Theorem 2.58) there exists a x\u2032 \u2208R, a < x\u2032 < x such that\nf(x) \u2212f(a)\nx \u2212a\n= f\u2032(x\u2032).\n(2.93.e)\nAs f is convex, f\u2032 is increasing by Theorem 2.90, hence, f\u2032(x\u2032) \u2265f\u2032(a). Adding this observation\nthe equality in (2.93.e), we have shown that (2.93.d) must hold.\n50", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_100", "text": "2.6\nAsymptotes\nDe\ufb01nition 2.94.\n(1) If for some c \u2208R, lim\nx\u2192c\u2212f(x) = \u00b1\u221eor lim\nx\u2192c+ f(x) = \u00b1\u221e, then we say\nthat the function f has a vertical asymptote at x = c.\n(2) If for some c \u2208R,\nlim\nx\u2192+\u221ef(x) = c (resp.\nlim\nx\u2192\u2212\u221ef(x) = c), then we say that the function f\nhas a horizontal asymptote at +\u221e(resp. \u2212\u221e) at y = c.\n(3) If for some a \u0338= 0, b \u2208R,\nlim\nx\u2192+\u221ef(x) \u2212ax = b (resp.\nlim\nx\u2192\u2212\u221ef(x) \u2212ax = b), then we say\nthat f has an oblique (or slant) asymptote at +\u221e(resp. \u2212\u221e) along the line y = ax + b.\nExample 2.95. Here we give a few examples of the di\ufb00erent notions introduce in the above\nde\ufb01nition.\n(1) Vertical asymptote: f(x) =\n1\n1\u2212x has a vertical asymptote at at x = 1, cf. Figure 14;\n(2) Horizontal asymptote: the function f(x) = 2 \u2212e\u2212x has a horizontal asymptote at +\u221eof\nvalue y = 2, cf. Figure 15;\n(3) Slant asymptote: the function f(x) = 2+3x+ 1\nx2 has a slant asymptote both at +\u221eand\n\u2212\u221ealong the line y = 3x + 2, cf. Figure 16.\nImages/1suxmeno1.png\nFigure 14: The function f(x) =\n1\nx\u22121, and the line x = 1.\n51", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_101", "text": "Images/2-emenox.png\nFigure 15: The function f(x) = 2 \u2212e\u2212x, and the line y = 2.\n3\nINTEGRATION\n3.1\nDe\ufb01nition\nThe idea behind integration is that the integral\nZ b\na\nf(x)dx of a bounded function f on a\nclosed interval [a, b] should be the area under the graph of f. However, it is not that easy to\nsay what this area means and when it is computable at all. If it is computable, we say that\nthe function is integrable (De\ufb01nition 3.11), and the value of this area is then called the integral\nZ b\na\nf(x)dx of f.\nNow, the idea of trying to de\ufb01ne the area under the graph of f is simple. We start with the\nonly area that we can compute trustably, that is of rectangles, and then we try to approximate\nthe area under the graph of f from above and from below using rectangles. These approxima-\ntions are called upper and lower Darboux sums (De\ufb01nition 3.3). We say that the area under\nthe graph of f is computable, which as above means that the function is integrable, if these\ntwo approximations meet in the limit. This is spelled out in precise mathematical terms below.\nDe\ufb01nition 3.1.\n(1) A partition \u03c3 = (xi) of a bounded interval [a, b] is an ordered collection\na = x0 < x1 < \u00b7 \u00b7 \u00b7 < xn\u22121 < xn = b of points of [a, b].\n(2) The norm or mesh of \u03c3 is\nmax{xi \u2212xi\u22121|1 \u2264i \u2264n}.\n(3) A re\ufb01nement \u03c3\u2032 = (x\u2032\ni) of \u03c3 is a partition such that each value of xi shows up amongst\nx\u2032\ni. we indicate that \u03c3\u2032 is a re\ufb01nement of \u03c3 by writing \u03c3 \u2ab0\u03c3.\n(4) The regular partition of length n is xi := a + ib\u2212a\nn , i = 0, 1, 2, . . . , n.\nProposition 3.2. Given a bounded interval [a, b], any two partitions \u03c3, \u03c3\u2032 have a common", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_102", "text": "(4) The regular partition of length n is xi := a + ib\u2212a\nn , i = 0, 1, 2, . . . , n.\nProposition 3.2. Given a bounded interval [a, b], any two partitions \u03c3, \u03c3\u2032 have a common\nre\ufb01nement \u03c3\u2032\u2032. Moreover, each partition can be re\ufb01ned to a new one with arbitrarily small\nnorm.\n52", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_103", "text": "Images/2piu3xpiu1sux2.png\nFigure 16: The function f(x) = 2 + 3x + 1\nx2 , and the line y = 3x + 2.\nDe\ufb01nition 3.3. Let f : [a, b] \u2192R be a bounded function and \u03c3 = (xi) a partition of [a, b].\nThen, the upper Darboux sum of f with respect to \u03c3 is\nS\u03c3 =\nn\nX\ni=1\nMi(xi \u2212xi\u22121),\nwhere Mi := supx\u2208[xi\u22121,xi] f(x). The lower Darboux sum of f with respect to \u03c3 is\nS\u03c3 =\nn\nX\ni=1\nmi(xi \u2212xi\u22121),\nwhere mi := infx\u2208[xi\u22121,xi] f(x).\nExample 3.4. Let us consider a constant function f(x) = c, c \u2208R over a closed bounded\n53", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_104", "text": "interval [a, b]. Then for any partition \u03c3 of [a, b],\nS\u03c3 = S\u03c3 =\nn\nX\ni=1\nc(xi \u2212xi\u22121) = cxn \u2212cx0\n|\n{z\n}\ntelescopic sum\n= c(b \u2212a).\nExample 3.5. Let us consider the function f(x) = x over a closed bounded interval [a, b], and\nlet \u03c3n =\n\u0010\na + i(b\u2212a)\nn\n\u0011\nbe the regular partition of length n. Then,\nS\u03c3n =\nn\nX\ni=1\n\u0012\na + ib \u2212a\nn\n\u0013 b \u2212a\nn\n= a(b \u2212a) + n(n + 1)\n2\n(b \u2212a)2\nn2\nand\nS\u03c3n =\nn\nX\ni=1\n\u0012\na + (i \u22121)b \u2212a\nn\n\u0013 b \u2212a\nn\n= a(b \u2212a) + (n \u22121)n\n2\n(b \u2212a)2\nn2\n,\nwhere in both cases we used the following identity\nn\nX\ni=1\ni = (n + 1)n\n2\n.\nNote that lim\nn\u2192\u221eS\u03c3n = lim\nn\u2192\u221eS\u03c3n = a(b \u2212a) + (b\u2212a)2\n2\n= b2\n2 \u2212a2\n2 .\nProposition 3.6. Let f : [a, b] \u2192R be a function (not necessarily a continuous one). Assume\nthat f admits an upper bound M (resp. a lower bound m) for its range R(f). Then, for any\npartition \u03c3 of [a, b], m(b \u2212a) \u2264S\u03c3, S\u03c3 \u2264M(b \u2212a). In particular, the sets\n{S\u03c3|\u03c3 is a partition of [a, b]}\nand\n{S\u03c3|\u03c3 is a partition of [a, b]}\nare bounded.\nProof. This follows immediately from the de\ufb01nitions, since for any interval [xi, xi+1] \u2282[a, b]\nthen\nm \u2264\ninf\n[xi,xi+1] f \u2264\nsup\n[xi,xi+1]\nf \u2264M.\nHence, for a partition \u03c3 = {xi} of [a, b],\nm(b \u2212a) =m\nn\u22121\nX\ni=0\nm(xi+1 \u2212xi) \u2264\nn\u22121\nX\ni=0\n\u0012\ninf\n[xi,xi+1] f\n\u0013", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_105", "text": "f \u2264M.\nHence, for a partition \u03c3 = {xi} of [a, b],\nm(b \u2212a) =m\nn\u22121\nX\ni=0\nm(xi+1 \u2212xi) \u2264\nn\u22121\nX\ni=0\n\u0012\ninf\n[xi,xi+1] f\n\u0013\n(xi+1 \u2212xi) = S\u03c3\n\u2264\nn\u22121\nX\ni=0\n \nsup\n[xi,xi+1]\nf\n!\n(xi+1 \u2212xi) = S\u03c3 \u2264\nn\u22121\nX\ni=0\nM(xi+1 \u2212xi) = M(b \u2212a)\nDe\ufb01nition 3.7. Let f : [a, b] \u2192R be a bounded function.\n(1) The upper Darboux integral of f on [a, b] is de\ufb01ned as\nS := inf{S\u03c3|\u03c3 is a partition of [a, b]}.\n(2) The lower Darboux integral of f on [a, b] is de\ufb01ned as\nS := sup{S\u03c3|\u03c3 is a partition of [a, b]}\n54", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_106", "text": "Example 3.8. Using the above computation for the constant function, cf. Example 3.4, we\nsee that if f is the constant function on [a, b], then S = S = (b \u2212a)c.\nProposition 3.9. Let f : [a, b] \u2192R be a bounded function.\n(1) If \u03c3 is a partition of [a, b] and \u03c3\u2032 is a re\ufb01nement of \u03c3, then:\nS\u03c3 \u2264S\u03c3\u2032, and S\u03c3 \u2265S\u03c3\u2032.\n(2) If \u03c3 is a partition of [a, b], then:\nS\u03c3 \u2264S\u03c3.\nCorollary 3.10. If f : [a, b] \u2192R is a bounded function, then S \u2264S.\nProof. It is enough to prove that S\u03c31 \u2264S\u03c32 for any partitions \u03c31 and \u03c32 of [a, b]. However,\nthis follows straight from Proposition 3.9. Indeed, if \u03c3 is a common re\ufb01nement of \u03c31 and \u03c32,\nthen Proposition 3.9 yields that\nS\u03c31 \u2264\nS\u03c3\n|{z}\nProposition 3.9.(1)\n\u2264\nS\u03c3\n|{z}\nProposition 3.9.(2)\n\u2264\nS\u03c32\n|{z}\nProposition 3.9.(1)\n.\nDe\ufb01nition 3.11. Let f : [a, b] \u2192R be a bounded function. We say that f is integrable, if\nS = S, in which case this common value is called the integral of f between a and b, and it is\ndenoted by\nZ b\na\nf(x)dx.\nRemark 3.12. Using Corollary 3.10, f is integrable over a closed bounded interval [a, b] if one\nexhibits a sequence (\u03c3n) of partitions such that lim\nn\u2192\u221eS\u03c3n =\nlim\nn\u2192\u221eS\u03c3n.\nIndeed, this follows\nimmediately by the following chain of inequalities\nlim\nn\u2192\u221eS\u03c3n \u2264S \u2264S \u2264lim\nn\u2192\u221eS\u03c3n,\n(3.12.a)\npassing to the limit for n \u2192\u221e.\nExample 3.13. Using Example 3.4, the constant functions are integrable on [a, b], and\nZ b\na\nc dx = (b \u2212a)c", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_107", "text": "n\u2192\u221eS\u03c3n,\n(3.12.a)\npassing to the limit for n \u2192\u221e.\nExample 3.13. Using Example 3.4, the constant functions are integrable on [a, b], and\nZ b\na\nc dx = (b \u2212a)c\nExample 3.14. Using Remark 3.12 and the computation of Example 3.5 for f(x) := x over a\nclosed bounded interval [a, b] then f is integrable, and\nZ b\na\nxdx = b2\n2 \u2212a2\n2\nExample 3.15. Consider the function [0, 2] \u2192R given by\nf(x) =\n\u001a0\nif x \u2208Q\n3\nif x \u0338\u2208Q\nThen, for all partition \u03c3, S\u03c3 = 6, and S\u03c3 = 0. So, S = 6, S = 0, and hence f is not integrable.\n55", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_108", "text": "Proposition 3.16. If f : [a, b] \u2192R is continuous then it is integrable.\nProof. As f is continuous over a closed bounded interval [a, b], Theorem 1.60 implies the uniform\ncontinuity of f. Let us \ufb01x \u03b5 > 0. Let \u03b4 > 0 be the constant in the de\ufb01nition of uniform\ncontinuity associated to\n\u03b5\nb\u2212a \u2013 that is,\n|x \u2212y| \u2264\u03b4 \u21d2|f(x) \u2212f(y)| \u2264\n\u03b5\nb \u2212a.\nClaim Let \u03c3 be a partition of [a, b] with norm at most \u03b4. Then S\u03c3 \u2212S\u03c3 \u2264\u03b5.\nProof. In fact,\nS\u03c3 \u2212S\u03c3 =\nn\nX\ni=1\n( max\n[xi,xi+1] f \u2212\nmin\n[xi,xi+1] f)(xi \u2212xi\u22121)\n\u2264\nn\nX\ni=1\n\u03b5\nb \u2212a(xi \u2212xi\u22121) =\n\u03b5\nb \u2212a\n(b \u2212a)\n| {z }\nPn\ni=1(xi\u2212xi\u22121)=b\u2212a\n= \u03b5\n3.2\nBasic properties\nProposition 3.17. Let f, g: [a, b] \u2192R be integrable functions. Then,\n(1) If f extends over [b, c] for some c \u2208R, c > b and f is also integrable over [b, c], then it is\nintegrable over [a, c], and\nZ b\na\nf(x)dx +\nZ c\nb\nf(x)dx =\nZ c\na\nf(x)dx.\n(2) Given \u03b1, \u03b2 \u2208R, \u03b1f + \u03b2g is integrable on [a, b], and\nZ b\na\n(\u03b1f + \u03b2g)(x)dx = \u03b1\nZ b\na\nf(x)dx + \u03b2\nZ b\na\ng(x)dx\n(3) If f \u2264g, then\nZ b\na\nf(x)dx \u2264\nZ b\na\ng(x)dx\n(4) The function |f| is integrable on [a, b], and\nZ b\na\n|f(x)|dx \u2265\n\f\f\f\f\nZ b\na\nf(x)dx\n\f\f\f\f\nProof. The proofs of all these statements follow the same pattern: one writes up the inequalities", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_109", "text": "(4) The function |f| is integrable on [a, b], and\nZ b\na\n|f(x)|dx \u2265\n\f\f\f\f\nZ b\na\nf(x)dx\n\f\f\f\f\nProof. The proofs of all these statements follow the same pattern: one writes up the inequalities\nfor lower (resp. upper) Darboux sums for a \ufb01xed partition \u03c3. Then these inequalities remain\nvalid when taking sup (resp. inf) of the lower (resp. upper) Darboux sums along all possible\npartitions of an interval [a, b]. This gives inequalities in both direction, which then implies\nequalities.\nFor example, let us show how this strategy works in the case of point (1) \u2013 we leave the other\n56", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_110", "text": "cases to the reader. Let \u03c3, and \u03c4 be partitions for [a, b] and [b, c], respectively. Then the union\nof \u03c3 and \u03c4 gives a partition \u03c1 for [a, c] and by de\ufb01nition we have\nS\u03c1 = S\u03c3 + S\u03c4,\n(3.17.a)\nS\u03c1 = S\u03c3 + S\u03c4.\n(3.17.b)\nAs both these equalities are true for all choice of partitions \u03c3 of [a, b] and \u03c4 of [b, c], by taking\nthe inf (resp. the sup) of (3.17.a) (resp. of (3.17.b)) along all possible choices of partitions of\n[a, b], [b, c], [a, c], then\nS\n[a,c] \u2264S\n[a,b] + S\n[b,c], and S[a,c] \u2265S[a,b] + S[b,c],\n(3.17.c)\nwhere S\n[a,c] (resp. S[a,c]) denotes the upper (resp. lower) Darboux integral of partitions of\n[a, c], and similarly for the other cases. As f is integrable both on [a, b] and on [b, c], we have\nR b\na f(x)dx = S\n[a,b] = S[a,b] and\nR c\nb f(x)dx = S\n[b,c] = S[b,c]. Thus, (3.17.c) yields\nZ b\na\nf(x)dx +\nZ c\nb\nf(x)dx \u2264S[a,c] \u2264S\n[a,c] \u2264\nZ b\na\nf(x)dx +\nZ c\nb\nf(x)dx.\n(3.17.d)\nAs the two ends of (3.17.d) are the same, we have everywhere equalities. This concludes both\nthe integrability of f over [a, c] as well as the statement of (1).\nExample 3.18.\nZ b\na\n(1 + x)dx =\nZ b\na\n1dx +\nZ b\na\nxdx\n|\n{z\n}\npoint (2) of Proposition 3.17\n=\n(b \u2212a)", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_111", "text": "Example 3.18.\nZ b\na\n(1 + x)dx =\nZ b\na\n1dx +\nZ b\na\nxdx\n|\n{z\n}\npoint (2) of Proposition 3.17\n=\n(b \u2212a)\n| {z }\nExample 3.13\n+\nb2 \u2212a2\n2\n| {z }\nExample 3.14\n.\n3.3\nFundamental theorem of calculus\nIn this section, we learn how to compute integrals using the anti-derivative, cf. Theorem 3.24.\nWe start by giving the de\ufb01nition of an anti-derivative.\nDe\ufb01nition 3.19. Let f : [a, b] \u2192R be a continuous function. A function G : [a, b] \u2192R is\ncalled an anti-derivative of f if\n(1) G is continuous on [a, b],\n(2) G is di\ufb00erentiable on ]a, b[, and\n(3) G\u2032(x) = f(x) for all x \u2208]a, b[.\nRemark 3.20. Given a continuous function f : [a, b] \u2192R admitting an anti-derivative G: [a, b] \u2192\nR, then for any C \u2208R, also GC : [a, b] \u2192R, GC(x) := G(x) + C is an anti-derivative for f.\nAccording to Corollary 2.60, the vice versa is also true: namely, if G, H are anti-derivatives of\nf, then there exists C \u2208R such that G(x) = H(x) + C, \u2200x \u2208R.\nNotation 3.21. The anti-derivatives of f are at times denoted by\nR\nf(x)dx+c, where c \u2208R is\na constant that is free to vary in R. Also, sometimes the expressions\nR\nf(x)dx + c is also called\nthe inde\ufb01nite integral, while what we de\ufb01ned as the integral of f over [a, b],\nR b\na f(x)dx is then\ncalled the de\ufb01nite integral \u2013 where the de\ufb01nitiveness comes from the fact that we computed\nthe integral over the closed bounded integral [a, b]. We use the integral/anti-derivative naming\nin this course.\n57", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_112", "text": "Example 3.22. We collect here a few important functions together with their anti-dervatives.\nfunction\nex\ncos(x)\nsin(x)\n1\nx\nxn, n \u2208N\n. . .\nanti-derivative\nex\nsin(x)\n\u2212cos(x)\nlog |x|\nxn+1\nn+1\n. . .\nNow that we have de\ufb01ned the notion of anti-derivative of a continuous function, there are\ntwo questions that arise spontaneously:\n\u25e6Does an anti-derivative for a continuous function f : [a, b] \u2192R over a closed bounded\ninterval [a, b] always exist?\n\u25e6If an anti-derivative exists for a continous function f : [a, b] \u2192R, does it help us in any\nway in computing the value of the integral\nR b\na f(x)dx?\nThese two questions have some simple but very powerful answers provided by the following\ntwo theorems, that are usually called the \ufb01rst and second fundamental theorems of calculus.\nTheorem 3.23. Fundamental theorem of calculus I\nLet f : [a, b] \u2192R be continuous. Then,\nF(x) :=\nZ x\na\nf(t)dt\nis an anti-derivative of f.\nTheorem 3.24. Fundamental theorem of calculus II\nLet f : [a, b] \u2192R be continuous and let G be an anti-derivative of f. Then,\nZ b\na\nf(x)dx = G(b) \u2212G(a).\nNotation 3.25. In order for the statement of Theorem 3.23 to make full sense, we need to\nintroduce some further notation: so far we de\ufb01ned\nR b\na f(x)dx only for a < b. If a = b, then we\nde\ufb01ne\nZ a\na\nf(x)dx := 0.\nIf a > b, then we also de\ufb01ne\nZ b\na\nf(x) := \u2212\nZ a\nb\nf(x)dx.\nWith these notations our previously proven rules give that if f : [a, b] \u2192R is continuous, and\nc, d \u2208[a, b] are any points, then\nZ c\na\nf(x)dx +\nZ d\nc\nf(x)dx =\nZ d\na\nf(x)dx.\nThe next statement is not too interesting in itself but it is needed in the proof of Theo-\nrem 3.24.", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_113", "text": "Z c\na\nf(x)dx +\nZ d\nc\nf(x)dx =\nZ d\na\nf(x)dx.\nThe next statement is not too interesting in itself but it is needed in the proof of Theo-\nrem 3.24.\nTheorem 3.26. Mean Value Theorem for Integrals\nIf f : [a, b] \u2192R is continuous, then there is a c \u2208[a, b], such that\nZ b\na\nf(x)dx = f(c)(b \u2212a).\n58", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_114", "text": "Proof. As [a, b] is closed and f is continuous, by Theorem 1.63, f admits global maximum and\nminimum over [a, b]. Set M := max\nx\u2208[a,b] f(x) and m := min\nx\u2208[a,b] f(x). By Theorem 1.66, f takes all\nvalues in [m, M]. However, by Proposition 3.6, then\nm \u2264\nR b\na f(x)dx\nb \u2212a\n\u2264M,\nso there is a c \u2208[a, b] such that f(c) equals the above fraction, which is exactly the statement\nof the theorem.\nLet us now give the proofs of the two fundamental theorems of calculus.\nProof of Theorem 3.23. Fix x0 \u2208]a, b[. Then, for any x0 \u0338= x \u2208]a, b[:\nF(x) \u2212F(x0)\nx \u2212x0\n=\n1\nx \u2212x0\nZ x\nx0\nf(t)dt =\nf(c(x))\n| {z }\nTheorem 3.26\n,\nfor a real number c(x) between x and x0. Hence:\nlim\nx\u2192x0\nF(x) \u2212F(x0)\nx \u2212x0\n= lim\nx\u2192x0f(c(x)) =\nlim\nx\u2192x0f(x)\n|\n{z\n}\nlim\nx\u2192x0\nc(x)=x0\n=\nf(x0)\n| {z }\nf is continuous\n.\nProof of Theorem 3.24. We have already shown in Theorem 3.23 that F(x) =\nR x\na f(t)dt is an\nanti-derivative of f. As both F and G are anti-derivatives, they di\ufb00er by a constant C \u2208R,\nthat is, (F \u2212G)(x) = C, \u2200x \u2208[a, b]. Then:\nG(b) \u2212G(a) =(G(b) + c) \u2212(G(a) + c) = F(b) \u2212F(a)\n=\nZ b\na\nf(x)dx \u2212\nZ a\na\nf(x)dx =\nZ b\na\nf(x)dx.\nNotation 3.27. The expression G(b) \u2212G(a) appearing in the statement of Theorem 3.24 is\nusually denoted by\nG(x)|b\na\nor", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_115", "text": "f(x)dx \u2212\nZ a\na\nf(x)dx =\nZ b\na\nf(x)dx.\nNotation 3.27. The expression G(b) \u2212G(a) appearing in the statement of Theorem 3.24 is\nusually denoted by\nG(x)|b\na\nor\nG(x)|x=b\nx=a.\nExample 3.28.\nZ \u22121\n\u22125\n1\nx =\n\u0000log |x|)\n\f\fx=\u22121\nx=\u22125 = log 1 \u2212log 5 = \u2212log 5\n3.4\nSubstitution\nTheorem 3.29. Let f : [a, b] \u2192R be a continuous function, and let \u03c6 : [\u03b1, \u03b2] \u2192[a, b] be a C1\nfunction. Then,\nZ \u03c6(\u03b2)\n\u03c6(\u03b1)\nf(x)dx =\nZ \u03b2\n\u03b1\nf(\u03c6(t))\u03c6\u2032(t)dt.\n(3.29.a)\n59", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_116", "text": "Proof. De\ufb01ne G(x) :=\nR x\na f(u)du. By Theorem 3.23, G is an anti-derivative of f, so that The-\norem 3.24 tells us\nZ \u03c6(\u03b2)\n\u03c6(\u03b1)\nf(x)dx = G(\u03c6(\u03b2)) \u2212G(\u03c6(\u03b1)).\nSo, it is enough to show that the value of the right side of (3.29.a) is the same. To show\nthat, let us just note that by the chain rule G(\u03c6(t))\u2032 = G\u2032(\u03c6(t))\u03c6\u2032(t) = f(\u03c6(t))\u03c6\u2032(t). Then\napplying Theorem 3.24 to the integral\nR \u03b2\n\u03b1 f(\u03c6(t))\u03c6\u2032(t)dt implies that\nZ \u03b2\n\u03b1\nf(\u03c6(t))\u03c6\u2032(t)dt = G(\u03c6(\u03b2)) \u2212G(\u03c6(\u03b1)),\nwhich concludes the proof.\nExample 3.30. In this example, we go from the right side of (3.29.a) to the left side.\nZ 1\n0\n\u221a\nexexdx =\nZ e\n1\n\u221audu\n|\n{z\n}\nu=ex\n(ex)\u2032=ex\n= u\n3\n2\n3\n2\n\f\f\f\f\f\nu=e\nu=1\n= 2\n3\n\u0010\ne\n3\n2 \u22121\n\u0011\nExample 3.31. Let us consider the function f : [0, 1] \u2192R, f(x) :=\n\u221a\n1 \u2212x2.\nWe want to compute the integral\nR 1\n0 f(x)dx. This integral computes the area of a quarter\nImages/sqrt_1minusx2_gr.png\nFigure 17: f(x) :=\n\u221a\n1 \u2212x2.\nof a circle of radius 1, as shown by Figure 17 so the result should be \u03c0\n4 . Indeed, the above\ncomputation shows that are train of thought is correct. Note that, opposite to the previous\nexample, in this argument at our \ufb01rst substitution we go from the left side of (3.29.a) to the\n60", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_117", "text": "right side.\nZ 1\n0\np\n1 \u2212x2dx =\nZ\n\u03c0\n2\n0\np\n1 \u2212(sin(t))2 cos(t)dt\n|\n{z\n}\nx=sin(t)\nsin(t)\u2032=cos(t)\n=\nZ\n\u03c0\n2\n0\np\ncos(t)2 cos(t)dt\n=\nZ\n\u03c0\n2\n0\n| cos(t)| cos(t)dt =\nZ\n\u03c0\n2\n0\ncos(t) cos(t)dt\n|\n{z\n}\nt\u2208[0, \u03c0\n2 ] \u21d2cos(t)\u22650 \u21d2| cos(t)|=cos(t)\n=\nZ\n\u03c0\n2\n0\ncos(2t) + 1\n2\ndt =\nZ \u03c0\n0\n\u0012cos(u) + 1\n2\n\u0013 1\n2du\n|\n{z\n}\nt= u\n2\n=1\n4\nZ \u03c0\n0\n(cos(u) + 1)du = 1\n4 (sin(u) + u) |u=\u03c0\nu=0\n=1\n4(sin(\u03c0) + \u03c0 \u2212sin(0) \u22120) = \u03c0\n4\nExample 3.32. Recall that sinh(x) : R \u2192R is an odd function and it is strictly increas-\ning (indeed, sinh(x)\u2032 = cosh(x) > 0). In particular, it has an inverse, which we denote by\nsinh\u22121(x): R \u2192R. With this we may compute similarly :\nZ 1\n0\np\n1 + x2dx =\nZ sinh\u22121(1)\n0\np\n1 + (sinh(t))2 cosh(t)dt\n|\n{z\n}\nx=sinh(t)\nsinh(t)\u2032=cosh(t)\n=\nZ sinh\u22121(1)\n0\np\ncosh(t)2 cosh(t)dt\n=\nZ sinh\u22121(1)\n0\n| cosh(t)2| cosh(t)dt =\nZ sinh\u22121(1)\n0\ncosh(t) cosh(t)dt\n|\n{z\n}\ncosh(t)>0 \u21d2| cosh(t)|=cosh(t)\n=\nZ sinh\u22121(1)\n0\ncosh(2t) + 1\n2\ndt\n=\nZ 2 sinh\u22121(1)\n0\ncosh(u) + 1\n2\n1\n2du\n|\n{z\n}\nt= u\n2\n= 1\n4", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_118", "text": "=\nZ sinh\u22121(1)\n0\ncosh(2t) + 1\n2\ndt\n=\nZ 2 sinh\u22121(1)\n0\ncosh(u) + 1\n2\n1\n2du\n|\n{z\n}\nt= u\n2\n= 1\n4\nZ 2 sinh\u22121(1)\n0\ncosh(u) + 1du = 1\n4 (sinh(u) + u) |u=2 sinh\u22121(1)\nu=0\n=sinh(2 sinh\u22121(1)) + 2 sinh\u22121(1)\n4\n= 2 sinh(sinh\u22121(1)) cosh(sinh\u22121(1)) + 2 sinh\u22121(1)\n4\n=\n2 sinh(sinh\u22121(1))\nq\n1 + sinh(sinh\u22121(1))2 + 2 sinh\u22121(1)\n4\n= 2 \u00b7 1 \u00b7\n\u221a\n1 + 12 + 2 sinh\u22121(1)\n4\n=2\n\u221a\n2 + 2 sinh\u22121(1)\n4\nExample 3.33. Substitution can be used the generally integrate cos(x)n and sin(x)n when n\nis a positive integer.\n61", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_119", "text": "(1) The simplest case is when n odd. Here is an example of that:\nZ\n\u03c0\n2\n0\ncos(x)5dx =\nZ\n\u03c0\n2\n0\ncos(x)(1 \u2212sin(x)2)2dx =\nZ 1\n0\n(1 \u2212u2)2du\n|\n{z\n}\nu(x)=sin(x)\nu(x)\u2032=cos(x)\n=\nZ 1\n0\n(1 \u22122u2 + u4)du =\n\u0012\nu \u22122u3\n3\n+ u5\n5\n\u0013\f\f\f\f\nu=1\nu=0\n=1 \u22122\n3 + 1\n5 = 6\n15 = 2\n5.\n(2) On the other hand, when n is even, by reverse-engineering duplication formulas for sine\nand cosine, we can reduce again to the case of an odd power:\nZ\n\u03c0\n2\n0\nsin4(x)dx =\nZ\n\u03c0\n2\n0\n\u00121 \u2212cos(2x)\n2\n\u00132\ndx =\nZ\n\u03c0\n2\n0\n1\n4 \u2212cos(2x)\n2\n+ cos(2x)2\n4\ndx\n=\nZ\n\u03c0\n2\n0\n1\n4dx \u2212\nZ\n\u03c0\n2\n0\ncos(2x)\n2\ndx +\nZ\n\u03c0\n2\n0\ncos(2x)2\n4\ndx\n(3.33.b)\n=\n\u0012\u03c0\n8 \u2212sin(2x)\n4\n\u0013\f\f\f\f\nx= \u03c0\n2\nx=0\n+\nZ\n\u03c0\n2\n0\ncos(4x) + 1\n8\ndx\n=\n\u0012\u03c0\n8 + \u03c0\n16 + sin(4x)\n32\n\u0013\f\f\f\f\nx= \u03c0\n2\nx=0\n= \u03c0\n8 + \u03c0\n16 = 3\u03c0\n16\n3.5\nIntegration by parts\nTheorem 3.34. If f, g: : E \u2192R be two C1 functions on an open interval E, and let a < b be\nelements of E. Then,\nZ b\na\nf(x)g\u2032(x)dx = f(x)g(x)|b\na \u2212\nZ b\na\nf\u2032(x)g(x)dx,\n(3.34.a)\nProof. It is enough to show that\nZ b\na\n(f(x)g\u2032(x) + f\u2032(x)g(x))dx = f(x)g(x)|b\na .\nThis follows immediately from the Leibniz formula, Theorem 3.24,", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_120", "text": "Proof. It is enough to show that\nZ b\na\n(f(x)g\u2032(x) + f\u2032(x)g(x))dx = f(x)g(x)|b\na .\nThis follows immediately from the Leibniz formula, Theorem 3.24,\n(f(x)g(x))\u2032 = f(x)g\u2032(x) + f\u2032(x)g(x).\nGenerally integration by parts are useful for products.\nThe main question in apply-\ning (3.34.a) is how one chooses f and g.\nThere is a rule which works in most cases (but\nnot always!). Using the list below, when you encounter a product of two functions both of\nwhich belong to one of the categories in the list, the idea is that you should assign g\u2032 in the\nformula (3.34.a) to the function that belongs to the category appearing earlier in the list.\n\u25e6E(xponential)\n\u25e6T(rigonometric)\n\u25e6A(lgebraic, that is, polynomial)\n\u25e6L(ogarithm)\n62", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_121", "text": "\u25e6I(nverse trigonometric).\nWe present now a few examples.\nExample 3.35.\nZ b\na\nxexdx\n|\n{z\n}\ng\u2032(x)=ex\ng(x)=ex\nf(x)=x\nf\u2032(x)=1\n= (xex)|b\nx=a \u2212\nZ b\na\nexdx\n= (xex \u2212ex)|b\nx=a = ((x \u22121)ex)|b\nx=a\nExample 3.36.\nZ b\na\nsin(x)exdx\n|\n{z\n}\ng\u2032(x)=ex,g(x)=ex,f(x)=sin(x),f\u2032(x)=cos(x)\n= (sin(x)ex)|b\nx=a \u2212\nZ b\na\ncos(x)exdx\n|\n{z\n}\ng\u2032(x)=ex,g(x)=ex,f(x)=cos(x),f\u2032(x)=\u2212sin(x)\n= (sin(x)ex \u2212cos(x)ex)|b\nx=a +\nZ b\na\n(\u2212sin(x))exdx\nThus, we can rewrite the equality between the LHS of the \ufb01rst line and the second line as\nZ b\na\nsin(x)exdx = (ex(sin(x) \u2212cos(x)))|b\nx=a\n2\n.\nExample 3.37.\nZ b\na\nlog(x)dx\n|\n{z\n}\nf(x)=log(x),f\u2032(x)= 1\nx ,g\u2032(x)=1,g(x)=x\n= (x log(x))|b\nx=a \u2212\nZ b\na\n1dx = (x log(x) \u2212x)|b\nx=a\nExample 3.38.\nZ b\na\narctan(x)dx\n|\n{z\n}\nf(x)=arctan(x),f\u2032(x)=\n1\n1+x2 ,g\u2032(x)=1,g(x)=x\n= (x arctan(x))|b\nx=a \u2212\nZ b\na\nx\n1 + x2 dx\n= (x arctan(x))|b\nx=a \u2212\n1\n2\nZ 1\nudu\n|\n{z\n}", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_122", "text": "= (x arctan(x))|b\nx=a \u2212\nZ b\na\nx\n1 + x2 dx\n= (x arctan(x))|b\nx=a \u2212\n1\n2\nZ 1\nudu\n|\n{z\n}\nu(x)=1+x2,u(x)\u2032=2x\n=\n\u0012\nx arctan(x) \u22121\n2 log |u|\n\u0013\f\f\f\f\nb\nx=a\n=\n\u0012\nx arctan(x) \u22121\n2 log |1 + x2|\n\u0013\f\f\f\f\nb\nx=a\n3.6\nIntegrating rational functions\nA rational function, is a function of the form P(x)\nQ(x), where P(x) and Q(x) are polynomials\nwith real coe\ufb03cients.\nWe start by recalling how polynomials with real coe\ufb03cients can be factorized.\nTheorem 3.39 (Fundamental theorem of algebra over R). Let Q(x) be a polynomial with real\ncoe\ufb03cients. Then, Q(x) can be factored as\nQ(x) = (x \u2212a1)k1 . . . (x \u2212an)kn(x2 + 2b1x + c1)l1 . . . (x2 + 2bmx + cm)lm,\n(3.39.a)\nwhere the ai, bi and ci are real numbers, ki, li > 0 are positive integers, and the quadratic\npolynomials x2 + 2bix + ci, 1 \u2264i \u2264m are irreducible, that is, there is no real number x0 such\nthat x2\n0 + 2bix0 + ci = 0, 1 \u2264i \u2264m.\n63", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_123", "text": "Remark 3.40. The Fundamental Theorem of Algebra is originally for polynomials R(x) with\ncomplex coe\ufb03cients. For those polynomial, the statement is even better: namely, polynomials\nwith complex coe\ufb03cients can be factored into linear terms. That is,\nR(x) = (x \u2212d1)s1 . . .\n\u0000x \u2212d\u2032\nn\n\u0001sn ,\ndi \u2208C, si \u2208N\u2217.\n(3.40.b)\nThis does work also for R(x) := Q(x) a real polynomial,as R \u2282C, however it may happen\nthat some of the di that are complex and not real. Then, the expression cannot be used for\nintegration because we did not learn integration of complex valued functions. Hence, the idea is\nto collect the di that are real numbers. These numbers provide the ai in (3.39.a). Furthermore,\nas we are working with a polynomial R(x) with real coe\ufb03cients,then (3.40.b) is invariant under\nconjugation, since R(x) is. Hence, whenever di is not real, then also the conjugate of (x \u2212di)\nhas to show up in (3.40.b) with the same power. That is, we have a factor of the right hand\nside of (3.40.b) of the form:\n(x \u2212di)si \u0000x \u2212di\n\u0001si =\n\u0000(x \u2212di)\n\u0000x \u2212di\n\u0001\u0001si\n=\n\u0000x \u22122\n\u0000di + di\n\u0001\n+ didi\n\u0001si =\n\u0000x \u22122 Re(di) + |di|2\u0001si\nThen, setting bj := \u2212Re(di), lj := si and cj := |bi|2, we obtain one of the terms of the form\n(x2 + 2bjx + cj)lj in (3.39.a).\nExample 3.41. Take Q(x) = x3 + x2 \u22122, and consider the factorization as in (3.39.a). As the\ndegree of Q is three, there must be a linear term (the product of the non-linear terms has even\ndegree). This correspond to a real root of Q(x), so let us search for it.\n(1) Finding the real root.\n1st try: Q(0) = \u22122.\nAs\nlim\nx\u2192+\u221eQ(x) = +\u221e, the Intermediate value theorem, Theo-", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_124", "text": "degree). This correspond to a real root of Q(x), so let us search for it.\n(1) Finding the real root.\n1st try: Q(0) = \u22122.\nAs\nlim\nx\u2192+\u221eQ(x) = +\u221e, the Intermediate value theorem, Theo-\nrem 1.66, implies that Q has a root greater than 0.\n2nd try: Q(1) = 0. We found the root, great.\n(2) Factoring out the linear term.\nx3\n+\nx2\n\u2212\n2\nx \u22121\nx2 + 2x + 2\nx3\n\u2212\nx2\n2x2\n\u2212\n2\n2x2\n\u2212\n2x\n2x\n\u2212\n2\n2x\n\u2212\n2\n0\nHence, we have\n(x2 + 2x + 2)(x \u22121) = x3 + x2 \u22122\nRemark 3.42. Unfortunately, for polynomials of degree \u22655 there is no algorithm for \ufb01nding\nthe roots; one just has to try to use the Intermediate Value theorem, hoping that the given\npolynomial yields a nice root.\nUsing (3.39.a), we have the following nice factorization of rational functions.\nProposition 3.43. Any rational function P(x)\nQ(x) can be written as\nP(x)\nQ(x) = \u03b11R1(x) + \u00b7 \u00b7 \u00b7 + \u03b1tRt(x),\nwhere the \u03b1i are real numbers, and Ri(x) are of the form:\n64", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_125", "text": "(1) polynomial, or\n(2)\n1\n(x\u2212r)p , or\n(3)\nx+c\n(x2+2rx+s)p .\nInstead of giving a proof, we explain the idea behind Proposition 3.43 in the following\nexample.\nExample 3.44. Given the rational function 4x3+9x2+11x+8\n(x2+x+1)2\n, let us try to factorize it as follows:\n4x3 + 9x2 + 11x + 8\n(x2 + x + 1)2\n=\nAx + B\n(x2 + x + 1)2 +\nCx + D\nx2 + x + 1\n=Ax + B + (Cx + D)(x2 + x + 1)\n(x2 + x + 1)2\n=Cx3 + (C + D)x2 + (A + C + D)x + (B + D)\n(x2 + x + 1)2\n,\nwhich yields the following linear system\n\uf8f1\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f2\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f3\nC\n= 4\nC + D\n= 9\nA + C + D\n= 11\nB + D\n= 8\nfor which the solutions are\nC = 4 \u21d24 + D = 9 \u21d2D = 5\n\u21d2A + 4 + 5 = 11; B + 5 = 8\n\u21d2A = 2; B = 3\nThus,\n4x3 + 9x2 + 11x + 8\n(x2 + x + 1)2\n=\n2x + 3\n(x2 + x + 1)2 +\n4x + 5\nx2 + x + 1.\nHaving the decomposition stated in Proposition 3.43, the question is how we integrate these\nterms separately.\nExample 3.45.\n\u25e6For p > 1, then\nZ\n1\n(x \u2212r)p dx = (x \u2212r)1\u2212p\n1 \u2212p\n.\n\u25e6For p = 1, then\nZ\n1\n(x \u2212r)dx = log |x \u2212r|.\nExample 3.46.\nZ\nx + c\n(x2 + 2rx + s)p =1\n2\nZ\n2(x + r)\n(x2 + 2rx + s)p dx +\nZ\nc \u2212r\n(x2 + 2rx + s)p dx\n=1\n2\nZ\n2(x + r)\n(x2 + 2rx + s)p dx + (c \u2212r)\nZ\n1\n(x2 + 2rx + s)p dx.\n65", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_126", "text": "Hence, we need to compute the two integrals\nZ\n2(x + r)\n(x2 + 2rx + s)p dx,\nZ\n1\n(x2 + 2rx + s)p dx,\nindividually.\n\u25e6Using the substitution u = x2 + 2rx + s, then\nZ\n2(x + r)\n(x2 + 2rx + s)p dx =\n\uf8f1\n\uf8f2\n\uf8f3\nlog |x2 + 2rx + s|\nif p = 1\n(x2+2rx+s)1\u2212p\n1\u2212p\nif p > 1\n\u25e6Hence, we now know how to integrate all the terms in Proposition 3.43, except for the\nintegral\nZ\n1\n(x2 + 2rx + s)p dx,\np > 0.\nHence,\nZ\n1\n(x2 + 2rx + s)p dx =\nZ\n1\n((x + r)2 + (s \u2212r2))p dx\n=\n1\n(s \u2212r2)p\nZ\n1\n\u0012\u0010\nx+r\n\u221a\ns\u2212r2\n\u00112\n+ 1\n\u0013p dx\n|\n{z\n}\ns\u2212r2>0, as x2+2rx+s has no real roots\n=\n1\n(s \u2212r2)p\u22121\n2\nZ\n1\n(u2 + 1)p du\n|\n{z\n}\nu=\nx+r\n\u221a\ns\u2212r2\n,\nand we are left to compute the integral\nZ\n1\n(u2 + 1)p du,\nfor p > 0.\nSetting\nIp :=\nZ\n1\n(u2 + 1)p du,\nthen I1 := arctan(u) and furthermore, if p \u22651, then we obtain a recursive formula as\nfollows:\nIp =\nZ\n1\n(u2 + 1)p du =\nu\n(u2 + 1)p \u2212\nZ\n(\u2212p)u \u00b7 2u\n(u2 + 1)p+1 du\n|\n{z\n}\nintegrating by parts with f(u)=\n1\n(u2+1)p ,g\u2032(u)=1\n=\nu\n(u2 + 1)p + 2p\nZ\nu2 + 1 \u22121\n(u2 + 1)p+1 du =\nu\n(u2 + 1)p + 2pIp \u22122pIp+1\nSo, by looking at the two ends of the equation, we obtain the recursive equality:\nIp+1 =\nu\n(u2+1)p + (2p \u22121)Ip\n2p\n.\n66", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_127", "text": "Remark 3.47. Be careful, there is an error on page 201 of the book where they prove the\nformulas above: intead of 2p \u22121, they wrote 2(p \u22121)!!\nExample 3.48. Let us compute I2 for example:\nI2 =\nu\nu2+1 + I1\n2\n= 1\n2\n\u0012\nu\nu2 + 1 + arctan(u)\n\u0013\nExample 3.49. Let us get back to Example 3.44:\nZ\n4x3 + 9x2 + 11x + 8\n(x2 + x + 1)2\ndx =\nZ\n2x + 3\n(x2 + x + 1)2 dx +\nZ\n4x + 5\nx2 + x + 1dx\n|\n{z\n}\nby Example 3.44\n=\nZ\n2x + 3\n\u0010\u0000x + 1\n2\n\u00012 + 3\n4\n\u00112 dx +\nZ\n4x + 5\n\u0000x + 1\n2\n\u00012 + 3\n4\ndx\n|\n{z\n}\ncompleting the square in the denominators\n=\nZ\n(2x + 1) + 2\n\u0010\u0000x + 1\n2\n\u00012 + 3\n4\n\u00112 dx +\nZ\n(4x + 2) + 3\n\u0000x + 1\n2\n\u00012 + 3\n4\ndx\n|\n{z\n}\nwriting the numerators in terms of a multiple of x + 1\n2\n67", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_128", "text": "=\nZ\n\u0012 2\n\u221a\n3\n\u00133\n2\n\u221a\n3(2x + 1) +\n4\n\u221a\n3\n\u0012\u0010\n2\n\u221a\n3\n\u0000x + 1\n2\n\u0001\u00112\n+ 1\n\u00132 dx +\nZ\n2\n\u221a\n3\n2\n\u221a\n3(4x + 2) +\n6\n\u221a\n3\n\u0010\n2\n\u221a\n3\n\u0000x + 1\n2\n\u0001\u00112\n+ 1\ndx\n|\n{z\n}\nWe multiply the numerators and the denominators by adequate multiples of\n2\n\u221a\n3 ,\nto make them of the form u2 + 1 or (u2 + 1)2\n= 4\n3\nZ\n2u +\n4\n\u221a\n3\n(u2 + 1)2 du +\nZ\n4u +\n6\n\u221a\n3\nu2 + 1 du\n|\n{z\n}\nu= 2\n\u221a\n3(x+ 1\n2)\n\u21d2\nx=\n\u221a\n3\n2 u\u22121\n2\n\u21d2\nx(u)\u2032=\n\u221a\n3\n2\n=4\n3\nZ\n2u\n(u2 + 1)2 du + 16\n3\n\u221a\n3\nZ\n1\n(u2 + 1)2 du\n+2\nZ\n2u\nu2 + 1dx + 6\n\u221a\n3\nZ\n1\nu2 + 1du\n=4\n3\n\u22121\nu2 + 1 +\n8\n3\n\u221a\n3\n\u0012\nu\nu2 + 1 + arctan(u)\n\u0013\n+ 2 log |u2 + 1| + 6\n\u221a\n3 arctan(u)\n=4\n3\n\u22121\n\u0010\n2\n\u221a\n3\n\u0000x + 1\n2\n\u0001\u00112\n+ 1\n+\n8\n3\n\u221a\n3\n\uf8eb\n\uf8ec\n\uf8ed\n2\n\u221a\n3\n\u0000x + 1\n2\n\u0001\n\u0010\n2\n\u221a\n3\n\u0000x + 1\n2\n\u0001\u00112\n+ 1\n+ arctan\n\u0012 2\n\u221a\n3\n\u0012\nx + 1\n2\n\u0013\u0013\n\uf8f6\n\uf8f7\n\uf8f8\n+2 log\n\f\f\f\f\f\n\u0012 2\n\u221a\n3\n\u0012\nx + 1\n2\n\u0013\u00132\n+ 1\n\f\f\f\f\f + 6\n\u221a\n3 arctan\n\u0012 2\n\u221a\n3\n\u0012\nx + 1\n2\n\u0013\u0013\n3.6.1\nRational functions in exponentials\nThere is a method of integrating functions obtained by plugging in ex into a rational function.\nWe explain it via the next example:\nExample 3.50.\nZ\n1\nex + 1dx =\nZ\n1\n(ex + 1)ex exdx =\nZ\n1\n(t + 1)tdt\n|\n{z\n}\nt(x)=ex\nt(x)\u2032=ex\n=\nZ \u00121\nt \u2212\n1\nt + 1\n\u0013\ndt = log |t| \u2212log |t + 1|\n= log |ex| \u2212log |ex + 1| = x \u2212log |ex + 1|\n3.6.2\nRational functions in roots\nThere is a method of integrating functions obtained by plugging in \u221ax into a rational function.\nWe explain it via the next example.\n68", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_129", "text": "Example 3.51.\nZ\n1\n\u221ax + 1dx =\nZ \u0012\n1\n\u221ax + 12\u221ax\n\u0013 1\n2\n1\n\u221axdx\n=\nZ\n2t\nt + 1dt\n|\n{z\n}\nt(x)=\u221ax\nt(x)\u2032= 1\n2\n1\n\u221ax\n=\nZ\n2dt \u2212\nZ\n2\nt + 1dt\n=2t \u22122 log |t + 1| = 2\u221ax \u22122 log |\u221ax + 1|.\n3.7\nImproper integrals\nSo far we have studied integrals of functions (mostly continuous ones) that are de\ufb01ned over\na closed bounded interval.\nBut how can we make sense of integrating a function over an unbounded interval, for\nexample,\nR +\u221e\n1\n1\nx2 dx? Or more generally, we have a real valued function f that is continuous\non an interval I of the form [a, b[, ]a, b] or ]a, b[, where a, b \u2208R, but either f does not extend\ncontinuously to [a, b], or the interval [a, b] does not exist at all \u2013 such as in the case when a or\nb are \u00b1\u221e.\nDe\ufb01nition 3.52. Let f : I \u2192R be a continuous function.\n(1) If I = [a, b[, a \u2208R, and either b \u2208R or b = +\u221e, then we de\ufb01ne the improper integral of\nf on I to be the limit\nZ b\u2212\na\nf(t)dt := lim\nx\u2192b\u2212\n\u0012Z x\na\nf(t)dt\n\u0013\n,\nprovided that the above limits exist.\n(2) If I =]a, b], b \u2208R, and either a \u2208R or a = \u2212\u221e\u221e, then we de\ufb01ne the improper integral\nof f on I to be the limit\nZ b\na+ f(t)dt := lim\nx\u2192a+\n\u0012Z b\nx\nf(t)dt\n\u0013\n,\nprovided that the above limits exist.\n(3) If I =]a, b[, a, b \u2208R, then we de\ufb01ne the improper integral of f on I to be the limit\nZ b\u2212\na+ f(t)dt :=\nZ c\na+ f(t)dt +\nZ b\u2212\nc\nf(t)dt\nfor any chosen c \u2208I, provided that the improper integrals\nZ c\na+ f(t)dt,\nZ b\u2212\nc\nf(t)dt\nexist.", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_130", "text": "a+ f(t)dt :=\nZ c\na+ f(t)dt +\nZ b\u2212\nc\nf(t)dt\nfor any chosen c \u2208I, provided that the improper integrals\nZ c\na+ f(t)dt,\nZ b\u2212\nc\nf(t)dt\nexist.\nIf the limits above exist and are \ufb01nite, then we say that the improper integrals converge. If the\nabove limits diverge, we say that the corresponding improper integral is divergent.\nRemark 3.53.\n(1) It is an easy exercise to verify that part (3) of the above the de\ufb01nition does\nnot depend on the choice of c \u2208I.\n(2) By abuse of notation many times the + and the \u2212is forgotten from the lower and upper\nlimits.\n69", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_131", "text": "Example 3.54.\nZ 1\n0+\n1\n\u221a\ntdt = lim\nx\u21920+ 2t\n1\n2\n\f\f\f\nt=1\nt=x = lim\nx\u21920+ 2 \u22122\u221ax = 2\nExample 3.55.\nZ 1\n0+\n1\nt dt = lim\nx\u21920+ log(t)|t=1\nt=x = lim\nx\u21920+ \u2212log(x) = +\u221e\nSo,\nR 1\n0+ 1\nt dt is divergent.\nExample 3.56.\nZ 1\n0+ log(t)dt = lim\nx\u21920+ (log(t)t \u2212t)|t=1\nt=x = \u22121 \u2212lim\nx\u21920+ (log(x)x \u2212x) = \u22121 \u2212lim\nx\u21920+(log(x)x)\nHere, we may compute limx\u21920+(log(x)x) using L\u2019Hospital\u2019s rule:\nlim\nx\u21920+(log(x)x) = lim\nx\u21920+\nlog(x)\n1\nx\n= lim\nx\u21920+\n1\nx\n\u22121\nx2\n= lim\nx\u21920+ \u2212x = 0.\nHence,\nR 1\n0+ log(t)dt = \u22121.\nImproper integrals enjoy many of the basic features of standard integrals.\nProposition 3.57. Let f, g: I \u2192R be continuous functions de\ufb01ned over an interval I, where\nI is either one of the following intervals\n[a, b\u2032[, ]a\u2032, b], ]a\u2032, b\u2032[,\na, b \u2208R,\na\u2032, b\u2032 \u2208R.\nThen,\n(1) If I = [a, b[ (resp. I =]a, b], I =]a, b[), a, b \u2208R and f extends to a continuous function\nde\ufb01ned over the interval [a, b], then the improper interval\nZ b\u2212\na\nf(x)dx (resp.\nZ b\na+\nf(x)dx,\nZ b\u2212\na+\nf(x)dx).\nconverges and it is equal to\nZ b\na\nf(x)dx.\n(2) If c \u2208I and c \u0338= sup I, inf I then if the improper interval of f on I converges, we have\nthat\nZ b\na\nf(x)dx +\nZ c\na\nf(x)dx =\nZ b\nc", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_132", "text": "a\nf(x)dx.\n(2) If c \u2208I and c \u0338= sup I, inf I then if the improper interval of f on I converges, we have\nthat\nZ b\na\nf(x)dx +\nZ c\na\nf(x)dx =\nZ b\nc\nf(x)dx.\n(3) Given \u03b1, \u03b2 \u2208R, if the improper integral of f, g over I converge, then also \u03b1f + \u03b2g is\nintegrable on I, and\nZ b\na\n(\u03b1f + \u03b2g)(x)dx = \u03b1\nZ b\na\nf(x)dx + \u03b2\nZ b\na\ng(x)dx\n(4) If 0 \u2264f \u2264g, then\n70", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_133", "text": "(i) if the improper integral\nR b\na g(x)dx converges then also the improper integral\nR b\na f(x)dx\ndoes;\n(ii) if the improper integral\nR b\na f(x)dx diverges then also the improper integral\nR b\na g(x)dx\ndoes.\nZ b\na\nf(x)dx \u2264\nZ b\na\ng(x)dx\nRemark 3.58. In part (2-4) of the previous proposition, we have used the simpli\ufb01ed notation\nfor improper integrals that was introduced in Remark 3.53(2).\nDe\ufb01nition 3.59. In the hypotheses of De\ufb01nition 3.52, we say that an improper integral is\nabsolutely convergent if the improper integral de\ufb01ned by integrating the function |f| instead\nof f is convergent.\nThe following is an an analogue, for improper integrals, of ??.\nProposition 3.60. If an improper integral is absolutely convergent, then it is also convergent.\nExample 3.61. The backwards implication of Proposition 3.60 does not hold, as shown by\nthe next example.\nZ +\u221e\n\u03c0\n4\nsin(t)\nt\ndt =\nlim\nx\u2192+\u221e\n\u2212cos(x)\nx\n\u2212\u2212cos\n\u0000 \u03c0\n4\n\u0001\n\u03c0\n4\n\u2212\nZ +\u221e\n\u03c0\n4\n\u2212cos(t)\n\u2212t2\ndt\n|\n{z\n}\ng\u2032=sin(t)\nf= 1\nt\ng=\u2212cos(t)\nf\u2032=\n1\n\u2212t2\n=\n\u221a\n2\n\u03c0 \u2212\nZ +\u221e\n\u03c0\n4\ncos(t)\nt2\ndt.\n(3.61.a)\nSo,\nZ +\u221e\n\u03c0\n4\nsin(t)\nt\ndt is convergent if so is\nZ +\u221e\n\u03c0\n4\ncos(t)\nt2\ndt. However, the latter is convergent because\nit is absolute convergent:\nZ +\u221e\n\u03c0\n4\n\f\f\f\f\ncos(t)\nt2\n\f\f\f\f dt \u2264\nZ +\u221e\n\u03c0\n4\n1\nt2 dt =\nlim\nx\u2192+\u221e\n \n\u22121\nt\n\f\f\f\f\nx\n\u03c0\n4\n!\n= 4\n\u03c0 +\nlim\nx\u2192+\u221e\n1\nx = 4\n\u03c0.\nThis yields that\nZ +\u221e\n\u03c0\n4\nsin(t)\nt\ndt is convergent.\nHowever, be careful, the fact that\nZ +\u221e\n\u03c0\n4\ncos(t)\nt2\ndt is absolute convergent, does not mean that\nso is\nZ +\u221e\n\u03c0\n4\nsin(t)\nt", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_134", "text": "\u03c0.\nThis yields that\nZ +\u221e\n\u03c0\n4\nsin(t)\nt\ndt is convergent.\nHowever, be careful, the fact that\nZ +\u221e\n\u03c0\n4\ncos(t)\nt2\ndt is absolute convergent, does not mean that\nso is\nZ +\u221e\n\u03c0\n4\nsin(t)\nt\ndt. That is, equation (3.61.a) does not work for sin(t)\nt\nreplaced by\n\f\f\f sin(t)\nt\n\f\f\f. And,\nin fact,\nZ +\u221e\n\u03c0\n4\nsin(t)\nt\ndt is not absolute convergent, because\nZ n\u03c0\n\u03c0\n4\n\f\f\f\f\nsin(t)\nt\n\f\f\f\f dt \u2265\nn\nX\nk=1\nZ k\u03c0\u2212\u03c0\n4\nk\u03c0\u22123\u03c0\n4\n\f\f\f\f\nsin(t)\nt\n\f\f\f\f dt \u2265\nn\nX\nk=1\n\u03c0\n2\n \nmin\nt\u2208[k\u03c0\u22123\u03c0\n4 ,k\u03c0\u2212\u03c0\n4 ]\n| sin(t)|\nt\n!\n\u2265\nn\nX\nk=1\n\u03c0\n2\n\uf8eb\n\uf8ec\n\uf8ed\nmin\nt\u2208[k\u03c0\u22123\u03c0\n4 ,k\u03c0\u2212\u03c0\n4 ]\n| sin(t)|\nmax\nt\u2208[k\u03c0\u22123\u03c0\n4 ,k\u03c0\u2212\u03c0\n4 ]\nt\n\uf8f6\n\uf8f7\n\uf8f8=\nn\nX\nk=1\n\u03c0\n2\n1\n\u221a\n2\nk \u2212\u03c0\n4\n=\n\u03c0\n2\n\u221a\n2\nn\nX\nk=1\n1\nk.\nAs\n\u221e\nX\nk=1\n1\nk is divergent, lim\nn\u2192\u221e\nZ n\u03c0\n\u03c0\n4\n\f\f\f\f\nsin(t)\nt\n\f\f\f\f dt does not exist. Therefore,\nZ \u221e\n\u03c0\n4\n\f\f\f\f\nsin(t)\nt\n\f\f\f\f dt is diver-\ngent.\n71", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_135", "text": "Example 3.62. A typical application of improper integral is to give an upper bound on in\ufb01nite\nsums. For example\n\u221e\nX\nk=10\n1\nk2 \u2264\nZ +\u221e\n9\n1\nx2 dx = \u22121\nx\n\f\f\f\f\nx\u2192+\u221e\nx=9\n=\n\u0012\nlim\nn\u2192\u221ex\u2192+\u221e\n\u22121\nx\n\u0013\n\u2212\u22121\n9 = 1\n9\n72", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_136", "text": "4\nPOWER SERIES\nDe\ufb01nition 4.1. Let (ak) \u2282R be a sequence of real numbers and let x0 \u2208R be a \ufb01xed real\nnumber.\nA power series centered at x0 is an expression of the form\n\u221e\nX\nk=0\nak(x \u2212x0)k.\nRemark 4.2. In the expression above, the coe\ufb03cients ak and the center of the power series x0\nare \ufb01xed real numbers (the ones we \ufb01xed at the start of the de\ufb01nition), while the variable x is\nfree to vary.\nAs the value of the series\n\u221e\nX\nk=0\nak(x \u2212x0)k varies when x \u2208R varies, in particular, the series\nwill converge for certain values of x and diverge for others. For example, if we take x = x0, then\nthe value of the series above is a0. For those values of x for which\n\u221e\nX\nk=0\nak(x\u2212x0)k converges, we\ncan then discuss the value of the series, and we will see below that we can construct a function\nout of it.\nHence, in view of this observation, we are compelled to introduce the following\nde\ufb01nition.\nDe\ufb01nition 4.3. The domain of convergence D \u2282R of the power series\n\u221e\nX\nk=0\nak(x \u2212x0)k is the\nset\nD :=\n(\nx \u2208R\n\f\f\f\f\f\n\u221e\nX\nk=0\nak(x \u2212x0)k is convergent\n)\nAs already noted above, for a power series\n\u221e\nX\nk=0\nak(x \u2212x0)k centered at x0 \u2208R, then x0 \u2208D\nalways, as the only non-zero term in\n\u221e\nX\nk=0\nak(x0 \u2212x0)k is a0. The following theorem illustrates\nthat the domain of convergence of a power series\n\u221e\nX\nk=0\nak(x0 \u2212x0)k has a relatively simple form:\nnamely, it is an interval.\nTheorem 4.4. Let\n\u221e\nX\nk=0\nak(x0 \u2212x0)k be a power series centered at x0 \u2208R. Then, there is a real\nnumber R \u2208R+ such that if\n\uf8f1\n\uf8f4\n\uf8f4\n\uf8f2\n\uf8f4\n\uf8f4\n\uf8f3\n|x \u2212x0| < R,\n|x \u2212x0| > R,\nthen\n\uf8f1\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f2\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f3\n\u221e\nX\nk=0\nak(x \u2212x0)k is convergent\n\u221e\nX\nk=0\nak(x \u2212x0)k is divergent\n.", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_137", "text": "number R \u2208R+ such that if\n\uf8f1\n\uf8f4\n\uf8f4\n\uf8f2\n\uf8f4\n\uf8f4\n\uf8f3\n|x \u2212x0| < R,\n|x \u2212x0| > R,\nthen\n\uf8f1\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f2\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f3\n\u221e\nX\nk=0\nak(x \u2212x0)k is convergent\n\u221e\nX\nk=0\nak(x \u2212x0)k is divergent\n.\nThe number R is called the radius of convergence of the power series\n\u221e\nX\nk=0\nak(x \u2212x0)k.\nProof. Let D be the domain of convergence of\n\u221e\nX\nk=0\nak(x\u2212x0)k. We have to show that if x\u2032 \u2208D,\nthen for all x with 0 < |x \u2212x0| < |x\u2032 \u2212x0|, x \u2208D.\n73", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_138", "text": "So, let us assume that for a given x\u2032 \u0338= x0,\n\u221e\nX\nk=0\nak(x\u2032 \u2212x0)k is convergent.\nIn particular,\nlim\nk\u2192\u221eak(x\u2032 \u2212x0)k \u21920, and hence the sequence bk := ak(x\u2032 \u2212x0)k is bounded, that is, there\nexists a non-negative real number B, such that |bk| \u2264B. However, setting y :=\nx\u2212x0\nx\u2032\u2212x0 , then\n|y| < 1, by our assumption, and\n0 \u2264\n\f\f\fak(x \u2212x0)k\f\f\f =\n\f\f\fbkyk\f\f\f \u2264B|y|k.\nSince\n\u221e\nX\nk=0\nB|y|k is a geometric series and |y| < 1, then\n\u221e\nX\nk=0\nB|y|k is convergent. The Squeeze\nTheorem for series, (??), then implies that\n\u221e\nX\nk=0\nak(x \u2212x0)k is absolutely convergent, and hence\nalso convergent. We then de\ufb01ne\nR := sup{r \u2208R | \u2200x \u2208]x0 \u2212r, x0 + r[ ,\n\u221e\nX\nk=0\nak(x \u2212x0)k converges absolutely}.\nA similar argument to the one used in the \ufb01rst part of the the proof can be used to show\nthe other part of the statement about the non-convergence of the power series for x \u2208R,\n|x \u2212x0| > R.\nRemark 4.5. Let\n\u221e\nX\nk=0\nak(x \u2212x0)k be a power series and let R be its radius of convergence. If\nR = +\u221e, then the domain of convergence D is R. If R = 0, then D = {x0}. If R > 0, R \u0338= +\u221e,\nthen Theorem 4.4 implies that D \u2287]x0 \u2212R, x0 + R[ and for any x \u2208R such that |x \u2212x0| > R\nthen x \u0338\u2208D. Thus, D\\ ]x0 \u2212R, x0 + R[ \u2286{x0 \u2212R, x0 + R}. One can prove that it cannot\nhappen that both x0 \u2212R, x0 +R belong to the domain of convergence, cf. Example 4.7 for some\nexamples.\nNow that we know that the radius of convergence exists, the question is how we can compute", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_139", "text": "happen that both x0 \u2212R, x0 +R belong to the domain of convergence, cf. Example 4.7 for some\nexamples.\nNow that we know that the radius of convergence exists, the question is how we can compute\nit. In ?? we saw several criteria to prove the convergence of a series. In particular, it is not\ntoo hard to adapt Cauchy\u2019s and D\u2019Alembert criteria, ?? and ??, to compute the radius of\nconvergence of a series.\nTheorem 4.6. Let\n\u221e\nX\nk=0\nak(x \u2212x0)k be a power series, with radius of convergence R.\n(1) If l1 := lim\nn\u2192\u221e\n\f\f\f an+1\nan\n\f\f\f exists for some l1 \u2208R, then R = 1\nl1 .\n(2) If l2 := lim\nn\u2192\u221e\nnp\n|an| exists for some l2 \u2208R, then R = 1\nl2 .\nThe above formulas are to be understood with the following notation when li = 0, +\u221e:\nif\n(\nli = 0\nli = +\u221e\nthen\n(\nR = +\u221e\nR = 0\n.\nProof. Let x \u0338= x0.\nWe want to understand when\n\u221e\nX\nk=0\nak(x \u2212x0)k is convergent using the\nexistence of the limits in the hypotheses of parts (1)-(2) of the statement.\n74", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_140", "text": "(1) Let us use the quotient criterion:\nlim\nn\u2192\u221e\n\f\f\f\f\nan+1(x \u2212x0)n+1\nan(x \u2212x0)n\n\f\f\f\f = lim\nn\u2192\u221e\n\f\f\f\f\nan+1(x \u2212x0)n+1\nan(x \u2212x0)n\n\f\f\f\f = |x \u2212x0| lim\nn\u2192\u221e\n\f\f\f\f\nan+1\nan\n\f\f\f\f = |x \u2212x0|l\nSo, by the quotient criterion, that is, ??,\n\u221e\nX\nk=0\nak(x \u2212x0)k is convergent if |x \u2212x0| < 1\nl and\nit is divergent if |x \u2212x0| > 1\nl .\n(2) We use the Alembert\u2019s criterion:\nlim\nn\u2192\u221e\nnp\n|an(x \u2212x0)n| = lim\nn\u2192\u221e| n\u221aan| |x \u2212x0| = L|x \u2212x0|\nSo, by Alembert\u2019s criterion, that is, ??,\n\u221e\nX\nk=0\nak(x \u2212x0)k is convergent if |x \u2212x0| < 1\nL and\nit is divergent if |x \u2212x0| > 1\nL.\nExample 4.7.\n(1) We have seen earlier that\nex =\n\u221e\nX\nn=0\nxn\nn!\nis convergent for all x \u2208R, which we can now verify at once:\nlim\nn\u2192\u221e\nan+1\nan\n= lim\nn\u2192\u221e\n1\n(n+1)!\n1\nn!\n= lim\nn\u2192\u221e\n1\nn + 1 = 0\nSo, indeed, the radius of convergence of the above series is \u221eand the domain of converge\nis R.\n(2) According to Theorem 4.6, the radius of convergence of\n\u221e\nX\nn=0\nnxn\nis 1\nl , where\nl = lim\nn\u2192\u221e\nn + 1\nn\n= 1.\nSo, the radius of convergence is 1. In particular the domain of convergence D of the series\nsatis\ufb01es\n] \u22121, 1[\u2286D \u2286[\u22121, 1].\nTo determine D fully, we need to understand whether 1, \u22121 belong to D. For x = 1, the\npower series becomes\n\u221e\nX\nk=0\nn\nwhich is clearly divergent as the sequence cn := n does not converge to 0; while, for\nx = \u22121, the power series becomes\n\u221e\nX\nk=0\n(\u22121)nn\n75", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_141", "text": "which is divergent as the sequence of the partial sums sj := Pj\nk=0(\u22121)n satis\ufb01es\nsj =\n(\n\u22121\nfor n odd,\n1\nfor n even,\nthus the series does not converge. Thus, the domain of convergence is D =] \u22121, 1[.\n(3) According to Theorem 4.6, the radius of convergence of\n\u221e\nX\nn=1\n1\nnxn\nis 1\nl , where\nl = lim\nn\u2192\u221e\nn\nr\n1\nn = 1.\nSo, the radius of convergence is 1. In particular the domain of convergence D of the series\nsatis\ufb01es\n] \u22121, 1[\u2286D \u2286[\u22121, 1].\nTo determine D fully, we need to understand whether 1, \u22121 belong to D. For x = 1, the\npower series becomes\n\u221e\nX\nk=1\n1\nn\nwhich is clearly divergent, being the harmonic series, cf. ??; while, for x = \u22121, the power\nseries becomes\n\u221e\nX\nk=0\n(\u22121)n 1\nn\nwhich is convergent by Leibniz\u2019s criterion ??. Thus, the domain of convergence is D =\n[\u22121, 1[.\n(4) According to Theorem 4.6, the radius of convergence of the series\n\u221e\nX\nn=0\nen+1xn,\ncentered at 0, is 1\nl , where\nl = lim\nn\u2192\u221e\nen+1\nen\n= lim\nn\u2192\u221ee = e.\nSo, the radius of convergence is 1\ne. In particular the domain of convergence D of the\nseries satis\ufb01es\n] \u22121\ne, 1\ne[\u2286D \u2286[\u22121\ne, 1\ne].\nTo determine D fully, we need to understand whether 1\ne, \u22121\ne belong to D. For x = 1\ne, the\npower series becomes\n\u221e\nX\nk=0\nen+1\n\u00121\ne\n\u0013n\n=\n\u221e\nX\nk=0\ne\n76", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_142", "text": "which is clearly divergent as the constant sequence cn := e does not converge to 0. For\nx = \u22121\ne, the power series becomes\n\u221e\nX\nk=0\n(\u22121)ne\nwhich is divergent for the same reason as above. Thus, the domain of convergence is\nD =] \u22121\ne, 1\ne[.\nLet\n\u221e\nX\nk=0\nak(x \u2212x0)k be a power series and assume that the radius of convergence R > 0.\nThis means that the domain of convergence of the power series contains the open interval\n]x0 \u2212R, x0 + R[, hence we can de\ufb01ne a function\nf : ]x0 \u2212R, x0 + R[ \u2192R,\nx 7\u2192f(x) :=\n\u221e\nX\nk=0\nak(x \u2212x0)k.\nWe are interested in understanding what kind of regularity properties the function f possesses.\nIs it continuous? Is it di\ufb00erentiable? How many derivatives does it have? Are those continuous\nover the domain of convergence of the series? All these questions are answered by the following\nresult.\nTheorem 4.8. Let\n\u221e\nX\nk=0\nak(x \u2212x0)k be a power series with radius of convergence R > 0. Let\nf(x):\n]x0 \u2212R, x0 + R[ \u2192R the function f(x) :=\n\u221e\nX\nk=0\nak(x \u2212x0)k.\nThen, for every x \u2208\n]x0 \u2212R, x0 + R[,\nf\u2032(x) =\n\u221e\nX\nk=1\nkak(x \u2212x0)k\u22121,\n(4.8.a)\nZ x\nx0\nf(t)dt =\n\u221e\nX\nk=0\nak\nk + 1(x \u2212x0)k+1.\n(4.8.b)\nRemark 4.9. Given a power series\n\u221e\nX\nk=0\nak(x \u2212x0)k with radius of convergence R > 0, then one\ncan show, using the same argument as in the proof of Theorem 4.6, that both power series\n\u221e\nX\nk=1\nkak(x \u2212x0)k\u22121, and\n\u221e\nX\nk=0\nak\nk + 1(x \u2212x0)k+1\nhave the same radius of convergence R.\nIn particular, this implies that we can iteratively\nuse Theorem 4.8 to compute the n-th derivative of f and this will be given by the power series", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_143", "text": "\u221e\nX\nk=0\nak\nk + 1(x \u2212x0)k+1\nhave the same radius of convergence R.\nIn particular, this implies that we can iteratively\nuse Theorem 4.8 to compute the n-th derivative of f and this will be given by the power series\nf(n)(x) =\n\u221e\nX\nk=n\nk(k \u22121)(k \u22122) \u00b7 \u00b7 \u00b7 (k \u2212n + 2)(k \u2212n + 1)ak(x \u2212x0)k\u2212n.\nIn particular, f \u2208C\u221e(]x0 \u2212R, x0 + R, R).\nThe proof of Theorem 4.8 is omitted.\n77", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_144", "text": "Example 4.10. We consider the power series\n\u221e\nX\nk=1\n(\u22121)k+1\nk\nxk. Then, using either one of the\ncriteria from Theorem 4.6, it is not hard to see that the radius of convergence of the series is\n1. Thus, we have a well-de\ufb01ned associated function f : ] \u22121, 1[\u2192R, f(x) =\n\u221e\nX\nk=1\n(\u22121)k+1\nk\nxk and\nusing the above theorem,\nf\u2032(x) =\n\u221e\nX\nk=1\n(\u22121)k+1xk\u22121,\nZ x\n0\nf(t)dt =\n\u221e\nX\nk=1\n(\u22121)k+1\nk(k + 1)(x)k+1.\n4.1\nTaylor series\nIn Section 2.5.3, we saw how to approximate a function f : E \u2192R around a point x0 \u2208E\nby means of the values of f and its derivatives at x0. When the function f \u2208C\u221e(E, R), then\nwe may be tempted to take expansions to larger and larger order of f around x0 and wonder\nhow they may compare with each other when we let n \u2192\u221e. To this end, we introduce the\nfollowing de\ufb01nition.\nDe\ufb01nition 4.11. Let f : E \u2192R be a function de\ufb01ned on an open interval E and let x0 \u2208E.\nAssume that f \u2208C\u221e(E, R). Then the Taylor series of f is\nTf,x0(x) :=\n\u221e\nX\nk=0\nf(k)(x0)\nk!\n(x \u2212x0)k.\nRemark 4.12. The terms of the Taylor series Tf,x0(x) of f centered at x0 are the same as those\nof the Taylor expansions of order n, for any n \u2208N, of f centered at x0, cf. Theorem 2.74.\nHence, given a function f : E \u2192R, f \u2208C\u221e(E, R), and x0 \u2208E, we can consider the Taylor\nseries Tf,x0(x) of f centered at x0\nTf,x0(x) :=\n\u221e\nX\nk=0\nf(k)(x0)\nk!\n(x \u2212x0)k.", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_145", "text": "series Tf,x0(x) of f centered at x0\nTf,x0(x) :=\n\u221e\nX\nk=0\nf(k)(x0)\nk!\n(x \u2212x0)k.\nAt this point, we may be tempted to wonder whether f(x) = Tf,x0(x), for x su\ufb03ciently close\nto x0. Let us remind the reader that in Theorem 2.74 we saw how if f is di\ufb00erentiable n times\non E, then\nf(x) =\n n\u22121\nX\ni=0\nf(i)(x0)\ni!\n(x \u2212x0)i\n!\n+ f(n)(c)(x \u2212x0)n\nn!\n,\n(4.12.a)\nfor some c \u2208]x0, x[ (resp. c \u2208]x, x0[). Hence, the Taylor series Tf,x0(x) equals f(x) if and only\nif the error term f(n)(c)(x\u2212x0)n\nn!\nincluded in (4.12.a) converges to 0 as n \u2192\u221e. This yields the\nfollowing proposition.\nProposition 4.13. In the situation of De\ufb01nition 4.11, if f \u2208C\u221e(E, R), then f(x) = Tf(x)\nfor a value x \u2208E, x > x0 (resp. x < x0) if\nlim\nn\u2192\u221e\n \nsup\ny\u2208]x0,x[\n\f\f\ff(n)(y)\n\f\f\f\n!\n|x \u2212x0|n\nn!\n= 0\n(resp. lim\nn\u2192\u221e\n \nsup\ny\u2208]x,x0[\n\f\f\ff(n)(y)\n\f\f\f\n!\n|x \u2212x0|n\nn!\n= 0).\n78", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_146", "text": "Remark 4.14. Sometimes the Taylor series Tf,x0(x) of a a function f equals the original function\nf only in the center point x0. A famous example is the following: let f : R \u2192R,\nf(x) =\n(\ne\u22121\nx2\nfor x \u0338= 0\n0\nx = 0\nWe claim that f \u2208C\u221e(R, R), and f(n)(0) = 0, \u2200n \u2208N. We only give the idea of how to prove\nthis claim: one proves by induction that outside 0, f(n) is a function which is a sum of terms of\nthe form e\n\u22121\nx2\nxj , for some j \u22650, and furthermore f(n)(0) = 0. In view of this claim, the Taylor\nseries of f around 0 is the constant 0 function. So, the radius of convergence is +\u221e, but apart\nfrom 0 there is no positive x at which the Taylor series Tf,0(x) = f(x).\nExample 4.15.\n(1) The Taylor series of f(x) = ex centered at x = 0 is given by\nTf,0(x) :=\n\u221e\nX\nk=0\n1\nk!xk.\nIn fact, f(0) = e0 = 1,\nf(n)(0) = (ex)(0) = 1\nWe have already seen that the radius of convergence of Tf,0(x) is +\u221e. In particular,\naccording to Theorem 4.6, Tf,0(x) \u2200x \u2208R. Hence, we have to understand whether for\nx \u2208R\nex = Tf,0(x).\n(4.15.b)\nLet us use Proposition 4.13. For x < 0, then\nsup\ny\u2208]x,0[\n|f(n)(y)| = sup\ny\u2208]x,0[\n|ey| = 1,\nand\nlim\nn\u2192\u221e\n \nsup\ny\u2208]x,0[\n\f\f\ff(n)(y)\n\f\f\f\n!\n|x|n\nn! = lim\nn\u2192\u221e\n|x|n\nn! = 0,\nsince x belongs to the domain of convergence of the power series. Similarly, for x > 0,\nthen\nsup\ny\u2208]0,x[\n|f(n)(y)| = sup\ny\u2208]x,0[\n|ey| = ex,\nand\nlim\nn\u2192\u221e\n \nsup", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_147", "text": "since x belongs to the domain of convergence of the power series. Similarly, for x > 0,\nthen\nsup\ny\u2208]0,x[\n|f(n)(y)| = sup\ny\u2208]x,0[\n|ey| = ex,\nand\nlim\nn\u2192\u221e\n \nsup\ny\u2208]0,x[\n\f\f\ff(n)(y)\n\f\f\f\n!\n|x|n\nn! = lim\nn\u2192\u221e\nex|x|n\nn!\n= 0,\nby the same argument as above. Then, for x \u2208R,\nex =\n\u221e\nX\nk=0\n1\nk!xk.\n(2) The Taylor series of f(x) = log(1 + x) centered at x = 0 is given by\nTf,0(x) :=\n\u221e\nX\nk=1\n(\u22121)k\u22121\nk\nxk.\nIn fact, f(0) = log(1) = 0,\nf\u2032(0) = (\n1\n1 + x)(0) =\n1\n1 + 0 = 1,\nf(n)(0) =\n\u0012(\u22121)n\u22121(n \u22121)!\n(1 + x)n\n\u0013\n(0) = (\u22121)n\u22121(n \u22121)!.\n79", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_148", "text": "Let us determine the radius of convergence of Tf,0(x):\nlim\nn\u2192\u221e\n\f\f\f\f\f\n(\u22121)n\nn+1\n(\u22121)n\u22121\nn\n\f\f\f\f\f = 1,\nso the radius of convergence of Tf,0(x) is 1. In particular, according to Theorem 4.6,\n\u221e\nX\nk=1\n(\u22121)k+1\nk\nxk converges whenever |x| < 1 and diverges whenever |x| > 1. Hence, we\nhave to understand whether for x \u2208] \u22121, 1[, then\nlog(1 + x) = Tf,0(x).\n(4.15.c)\nLet us use Proposition 4.13. For \u22121 < x < 0, then\nsup\ny\u2208]x,0[\n|f(n)(y)| = sup\ny\u2208]x,0[\n| (n \u22121)!\n(1 + y)n | = (n \u22121)!\n(1 + x)n ,\nand\nlim\nn\u2192\u221e\n \nsup\ny\u2208]x,0[\n\f\f\ff(n)(y)\n\f\f\f\n!\n|x|n\nn! = lim\nn\u2192\u221e\n|x|n\nn(1 + x)n = 0,\nsince lim\nn\u2192\u221e|x|n = lim\nn\u2192\u221e(1 + x)n = 0. Similarly, for 0 < x < 1, then\nsup\ny\u2208]0,x[\n|f(n)(y)| = sup\ny\u2208]x,0[\n| (n \u22121)!\n(1 + y)n | = (n \u22121)!,\nand\nlim\nn\u2192\u221e\n \nsup\ny\u2208]0,x[\n\f\f\ff(n)(y)\n\f\f\f\n!\n|x|n\nn! = lim\nn\u2192\u221e\n|x|n\nn\n= 0,\nsince |x| < 1. Then, for x \u2208] \u22121, 1[,\nlog(1 + x) =\n\u221e\nX\nk=1\n(\u22121)k\u22121\nk\nxk.\nExample 4.16. Here we collect a list of important functions with their Taylor series centered\nat 0.\nf(x)\nTf,0(x)\nradius of convergence of Tf,0(x)\nf(x) = Tf,0(x) over the interval\n1\n1\u2212x\nP\u221e\nk=0 xk\n1\n] \u22121, 1[\n1\n1+x\nP\u221e", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_149", "text": "at 0.\nf(x)\nTf,0(x)\nradius of convergence of Tf,0(x)\nf(x) = Tf,0(x) over the interval\n1\n1\u2212x\nP\u221e\nk=0 xk\n1\n] \u22121, 1[\n1\n1+x\nP\u221e\nk=0(\u22121)kxk\n1\n] \u22121, 1[\nex\nP\u221e\nk=0\n1\nk!xk\n+\u221e\nR\nlog(1 + x)\nP\u221e\nk=1\n(\u22121)k+1\nk\nxk\n1\n] \u22121, 1[\ncos(x)\nP\u221e\nk=0\n(\u22121)k\n(2k)! x2k\n+\u221e\nR\nsin(x)\nP\u221e\nk=0\n(\u22121)k\n(2k+1)!x2k+1\n+\u221e\nR\narctan(x)\nP\u221e\nk=0\n(\u22121)k\n2k+1 x2k+1\n1\n] \u22121, 1[\ncosh(x)\nP\u221e\nk=0\n1\n(2k)!x2k\n+\u221e\nR\nsinh(x)\nP\u221e\nk=0\n1\n(2k+1)!x2k+1\n+\u221e\nR\n80", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "analysis_1(2)_chunk_150", "text": "Example 4.17. Similar example as above is the Taylor series for\n1\nx+1.\nUsing the Taylor\nexpansion, computed in Example 2.77, the Taylor series is\n\u221e\nX\nk=0\n(\u22121)kxk\nAs in Example 4.15, using Theorem 4.6, the radius of convergence is 1. Furthermore, using\nProposition 4.13 as in Example 4.15, for x \u2208] \u22121, 1[,\nf(x) =\n1\nx + 1 =\n\u221e\nX\nk=0\n(\u22121)kxk.\n(4.17.d)\nExample 4.18. If we want to compute the derivative and integral of f(x) = log(x + 1), then,\nover ] \u22121, 1[, we can do so using Theorem 4.8 and deriving/integrating the power series term\nby term. So, for x \u2208] \u22121, 1[:\nf\u2032(x) =\n\u221e\nX\nk=1\n(\u22121)k+1k\nk\n(x)k\u22121 =\n\u221e\nX\nk=1\n(\u22121)k\u22121(x)k\u22121\n=\n\u221e\nX\nk=0\n(\u22121)k(x)k =\n1\n1 + x,\nand\nZ x\n0\nf(t)dt =\n\u221e\nX\nk=1\n(\u22121)k+1\nk(k + 1)(x)k+1 =\n\u221e\nX\nk=1\n(\u22121)k+1\n\u00121\nk \u2212\n1\nk + 1\n\u0013\n(x \u22121)k+1\n=(x \u22121)\n\u221e\nX\nk=1\n(\u22121)k+1 1\nkxk +\n\u221e\nX\nk=1\n(\u22121)k+2\n1\nk + 1xk+1\n=(x \u22121)\n\u221e\nX\nk=1\n(\u22121)k+1 1\nkxk +\n\u221e\nX\nk=2\n(\u22121)k+1 1\nkxk\n=x log(1 + x) + log(x + 1) \u2212x = (x + 1) log(x + 1) \u2212x\nwhere the result of the second computation is exactly what we obtained in Example 3.37\n81", "source": "analysis_1(2)", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_0", "text": "LINEAR ALGEBRA (MATH-111(EN)) FALL 2022\nKALAN WALMSLEY\nAcknowledgments. These lecture notes were written by Kalan Walmsley based on the course \u2019Linear\nAlgebra (En)\u2019 given at EPFL in fall 2022. They are an abridged version and do not cover all examples or\nadditional explanation given during the courses. If you find mistakes or have other suggestions, you can\nsend an email to matthias.ruf@epfl.ch.\nMotivation\nWe start with an easy example:\n(\nx + 2y = 1\n\u2212x + y = 0\nThe second equation tells us that x = y. Inserting this information into the first equation: 3y = 1 and\ntherefore x = y = 1/3\nGeneral system of equations:\n(\u2217)\n\uf8f1\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f2\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f3\na1,1x1 + a1,2x2 + \u00b7 \u00b7 \u00b7 + a1,nxn = b1\na2,1x1 + a2,2x2 + \u00b7 \u00b7 \u00b7 + a2,nxn = b2\n...\nam,1x1 + am,2x2 + \u00b7 \u00b7 \u00b7 + am,nxn = bm\nx1, \u00b7 \u00b7 \u00b7 , xn \u2208R\nKnown:\nai,j \u2208R,\n1 \u2264i \u2264m, 1 \u2264j \u2264n\nbi \u2208R,\n1 \u2264i \u2264m\nUnknowns (variables): x1, . . . , xn \u2208R\nQuestion: Existence of solutions? Uniqueness of solutions? How to compute them efficiently?\n1. Linear equations in linear algebra\n1.1. Sytems of linear equations. Each equation in (\u2217) is called a linear equation. (\u2217) is called a\nsystem of linear equations, the solutions to which are a list {s1, \u00b7 \u00b7 \u00b7 , sn} such that (\u2217) is true for the\nchoice (x1, \u00b7 \u00b7 \u00b7 , xn) = (s1, \u00b7 \u00b7 \u00b7 , sn)\nMatrix notation. A central goal when solving linear systems is to replace them by simpler linear systems\nthat have the same solutions. Before we do that, we introduce a useful notion that reduces notation.\nDefinition. A matrix is a rectangular array which contains numbers. We say that it is an m \u00d7 n matrix\nif it has m rows and n columns. Typically, the entries are real numbers, but later we will also consider\nso-called complex numbers.\nFor instance, the matrix\n\u00121\n2\n0\n0\n\u03c0\n1\n\u0013\nis a 2 \u00d7 3-matrix.\n1", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_1", "text": "2\nKALAN WALMSLEY\nTo each linear system (\u2217), we can associate its so-called coefficient matrix\nA =\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\na1,1\na1,2\n\u00b7 \u00b7 \u00b7\na1,n\na2,1\na2,2\n\u00b7 \u00b7 \u00b7\na2,n\n...\nam,1\nam,2\n\u00b7 \u00b7 \u00b7\nam,n\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\nand the so-called augmented matrix\n(A | b) =\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\na1,1\na1,2\n\u00b7 \u00b7 \u00b7\na1,n\nb1\na2,1\na2,2\n\u00b7 \u00b7 \u00b7\na2,n\nb2\n...\nam,1\nam,2\n\u00b7 \u00b7 \u00b7\nam,n\nbm\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8.\nNote that A is an m \u00d7 n-matrix, while (A|b) is an m \u00d7 (n + 1)-matrix.\nSolving linear systems. Three elementary operations suffice to solve a linear system. In what follows\non the left-hand side (LHS) we describe the operation on the linear system, on the right-hand side (RHS)\nthe effect on the (augmented) matrix:\n(1) Swap two equations\nSwap two rows\n(2) Multiply an equation by \u03bb \u0338= 0\nMultiply the elements of a row by \u03bb \u0338= 0\n(3) Add to an equation a multiple of another equation\nAdd to a row a multiple of another row\nIt is clear that operations of type (1) do not change the set of solutions. Moreover, if \u03bb \u0338= 0, then looking\nat the ith equation we have the equivalence\nai,1x1 + . . . + ai,nxn = bi \u21d0\u21d2\u03bbai,1x1 + . . . + \u03bbai,nxn = \u03bbbi,\nso that also the second operation does not change the set of solutions. Finally, for the third operation we\nconsider the case that we add \u03bb times the ith equation to the jth equation. Then we have the equivalence\n(\nai,1x1 + . . . + ai,nxn = bi\naj,1x1 + . . . aj,nxn = bj\n\u21d0\u21d2\n(\nai,1x1 + . . . + ai,nxn = bi\n(aj,1 + \u03bbai,1)x1 + . . . + (aj,n + \u03bbai,n)xn = bj + \u03bbbi,\nwhile the other equations are not affected by this operation.\nExample. Demonstrate elementary operations on the augmented matrix\n\uf8eb\n\uf8ed\n1\n\u22122\n1\n0\n0\n2\n\u22128\n8\n5\n0\n\u22125\n10\n\uf8f6\n\uf8f8and the\ncorresponding linear system.", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_2", "text": "while the other equations are not affected by this operation.\nExample. Demonstrate elementary operations on the augmented matrix\n\uf8eb\n\uf8ed\n1\n\u22122\n1\n0\n0\n2\n\u22128\n8\n5\n0\n\u22125\n10\n\uf8f6\n\uf8f8and the\ncorresponding linear system.\n1.2. Row reduction and echelon form. Here we discuss how we can use the above operations to turn\na given augmented matrix into a matrix with simpler structure. Row reduction is an algorithmic way to\napply the elementary operations to (A | b) to solve (\u2217)\nDefinition. An m \u00d7 n matrix is in echelon form if\na) All rows containing only zeros (also called \u2019zero rows\u2019) are at the bottom\nb) If a row is non-zero, its leftmost entry (called leading entry or pivot) is in a column to the right\nof the leading entry of the row above\nIt is in reduced echelon form if additionally\nc) Each leading entry is equal to 1\nd) Each leading entry is the only non-zero entry in its column\nTheorem 1.1. Every m \u00d7 n matrix is row equivalent to a unique m \u00d7 n matrix in reduced echelon form,\nthat means it can be transformed by elementary operators of type 1, 2, 3 into a matrix in reduced echelon\nform and this matrix is unique.", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_3", "text": "LECTURE NOTES\n3\nThe row reduction algorithm (Gaussian elimination). Let A =\n\uf8eb\n\uf8ec\n\uf8ed\na1,1\n\u00b7 \u00b7 \u00b7\na1,n\n...\nam,1\n\u00b7 \u00b7 \u00b7\nam,n\n\uf8f6\n\uf8f7\n\uf8f8be a general\nmatrix (not necessarily an augmented matrix of a linear system) and let Aj be the jth column of A (so\nj \u2208{1, . . . , n} and A = (A1 \u00b7 \u00b7 \u00b7 An)).\nThe following algorithm produces an echelon form of A:\nStep 1 : For j = 1, . . . , n:\nIf Aj =\n\uf8eb\n\uf8ec\n\uf8ed\n0\n...\n0\n\uf8f6\n\uf8f7\n\uf8f8, go on with Aj+1\nIf Aj \u0338=\n\uf8eb\n\uf8ec\n\uf8ed\n0\n...\n0\n\uf8f6\n\uf8f7\n\uf8f8, choose an index i with ai,j \u0338= 0 (preferably with a simple value like 1) and swap\nwith the 1st row. Then use transformations of type 3 to turn all elements below the new top\nelement to zero.\nStep 2 : Repeat Step 1 with the matrix you obtain by removing the columns 1, \u00b7 \u00b7 \u00b7 , j and the first row\n(so the parameter n gets smaller)\nThe above procedure terminates after at most n runs. The following operations yield the reduced\nechelon form:\nStep 3: Normalize all leading entries (pivots) to 1 (type 2 operation)\nStep 4: Use type 3 operations to put all elements above leading entries to zero\nExample. Perform row reduction on the matrices\n\uf8eb\n\uf8ed\n0\n0\n2\n1\n0\n0\n2\n1\n0\n1\n4\n2\n\uf8f6\n\uf8f8and\n\uf8eb\n\uf8ed\n1\n1\n1\n2\n1\n0\n4\n\u22128\n3\n\uf8f6\n\uf8f8.\nFree variables & solutions in parametric form. Let (A | b) be an augmented matrix in reduced\nechelon form. The last equation will be of form\n\u00000\n\u00b7 \u00b7 \u00b7\n0\n1\nam,k+1\n\u00b7 \u00b7 \u00b7\nam,n\nbm\n\u0001\n(the leading\nentry is in the kth column). So we have\nxk = bm \u2212am,k+1xk+1 \u2212\u00b7 \u00b7 \u00b7 \u2212am,nxn\nwhere xk+1, \u00b7 \u00b7 \u00b7 , xn are free variables.\nThe penultimate equation will be of form\n\u00000\n\u00b7 \u00b7 \u00b7\n0\n1\n\u22c6\n0\n\u00b7 \u00b7 \u00b7\nbm\u22121\n\u0001\n(the leading entry is in\nthe jth column, the zero after the star is the kth column, j < k). So we have xj = bm\u22121 \u2212\u00b7 \u00b7 \u00b7 not\ndepending on xk.\nContinuing this procedure we obtain an equation for all the variables corresponding to leading entries", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_4", "text": "bm\u22121\n\u0001\n(the leading entry is in\nthe jth column, the zero after the star is the kth column, j < k). So we have xj = bm\u22121 \u2212\u00b7 \u00b7 \u00b7 not\ndepending on xk.\nContinuing this procedure we obtain an equation for all the variables corresponding to leading entries\nand the equations do not depend on the other variables corresponding to leading entries. The set of these\nequations is called the parametric form of the solution.\nIn other words:\n\u2022 free variables = variables corresponding to columns with no leading entry\n\u2022 basic variables = variables corresponding to columns with leading entry\n\u2022 solutions in parametric form = equations for all basic variables in terms of free variables\nExample. Find solution in parametric vector form for the augmented matrix (A|b) =\n\uf8eb\n\uf8ed\n1\n1\n0\n2\n0\n0\n1\n4\n0\n0\n0\n0\n\uf8f6\n\uf8f8.\nExistence of uniqueness of solutions.\nTheorem 1.2.\n(i) A linear system has at least one solution if and only if (iff) any echelon form of\nthe augmented matrix has no row of the form\n\u00000\n\u00b7 \u00b7 \u00b7\n0\nb\u0001\nwith b \u0338= 0\n(ii) If there exists a solution, then it is unique iff there are no free variables", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_5", "text": "4\nKALAN WALMSLEY\nProof.\n(i) Two directions to show: first assume that the linear system has at least one solution. If in\nan echelon form of the augmented matrix there exists a row of the form\n\u00000\n\u00b7 \u00b7 \u00b7\n0\nb\u0001\n, then the\ncorresponding equation reads 0x1 + . . . + 0xn = b, so existence of a solution requires that b = 0.\nNext we assume that in any echelon form of the augmented matrix there is no row of the form\n\u00000\n\u00b7 \u00b7 \u00b7\n0\nb\n\u0001\nwith b \u0338= 0. Then any row is either zero or it has a leading entry strictly before\nthe last column. Turn the matrix into the reduced echelon form (which has the same property).\nThen we can solve all equations since each variable corresponding to leading entries only appears\nin a single equation.\n(ii) If there are free variables, then for any value of the free variables we have a solution, so uniqueness\ncannot hold. If there are no free variables, then we have a leading entry in each column (except\nmaybe the last one which corresponds to the RHS in the linear system). Therefore every variable\nis a basic variable and since there are no free variables, the equation for solutions in parametric\nform uniquely determines the basic variables. Hence the solution is unique.\n\u25a1\n1.3. Vector equations.\nDefinition. Vectors in Rn are ordered n-tuples of real numbers\nx =\n\uf8eb\n\uf8ec\n\uf8ed\nx1\n...\nxn\n\uf8f6\n\uf8f7\n\uf8f8.\nGiven x, y \u2208Rn, x =\n\uf8eb\n\uf8ec\n\uf8ed\nx1\n...\nxn\n\uf8f6\n\uf8f7\n\uf8f8, y =\n\uf8eb\n\uf8ec\n\uf8ed\ny1\n...\nyn\n\uf8f6\n\uf8f7\n\uf8f8, \u03bb \u2208R (scalar), we define\nx + y =\n\uf8eb\n\uf8ec\n\uf8ed\nx1 + y1\n...\nxn + yn\n\uf8f6\n\uf8f7\n\uf8f8,\n\u03bbx =\n\uf8eb\n\uf8ec\n\uf8ed\n\u03bbx1\n...\n\u03bbxn\n\uf8f6\n\uf8f7\n\uf8f8.\nWe write \u2212x = (\u22121)x and if no confusion occurs we write 0 =\n\uf8eb\n\uf8ec\n\uf8ed\n0\n...\n0\n\uf8f6\n\uf8f7\n\uf8f8for the zero vector.\nRemark (Rules for vector calculus). Let u, v, w \u2208Rn and \u03bb, \u00b5 \u2208R, then\ni) u + v = v + u\nii) (u + v) + w = u + (v + w)\niii) u + 0 = u\niv) u + (\u2212u) = 0\nv) \u03bb(u + v) = \u03bbu + \u03bbv\nvi) (\u03bb + \u00b5)u = \u03bbu + \u00b5u\nvii) \u03bb(\u00b5u) = (\u03bb\u00b5)u\nviii) 1 \u00b7 u = u", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_6", "text": "iv) u + (\u2212u) = 0\nv) \u03bb(u + v) = \u03bbu + \u03bbv\nvi) (\u03bb + \u00b5)u = \u03bbu + \u00b5u\nvii) \u03bb(\u00b5u) = (\u03bb\u00b5)u\nviii) 1 \u00b7 u = u\nDefinition. Given u1, \u00b7 \u00b7 \u00b7 , uk \u2208Rn and \u03bb1, \u00b7 \u00b7 \u00b7 , \u03bbk \u2208R, then vector\nu = \u03bb1u1 + \u00b7 \u00b7 \u00b7 + \u03bbkuk =\nk\nX\ni=1\n\u03bbiui\nis called a linear combination of the vectors u1, \u00b7 \u00b7 \u00b7 , uk.\nThe set of all possible linear combinations of u1, \u00b7 \u00b7 \u00b7 , uk is denoted by span{u1, \u00b7 \u00b7 \u00b7 , uk} = {\u03bb1u1 +\n\u00b7 \u00b7 \u00b7 + \u03bbkuk | \u03bb1, \u00b7 \u00b7 \u00b7 , \u03bbk \u2208R} \u2286Rn\nGiven b \u2208Rn, how to determine if it belongs to span{u1, \u00b7 \u00b7 \u00b7 , uk} ?\nThis is equivalent to solving \u03bb1u1 + \u00b7 \u00b7 \u00b7 + \u03bbkuk = b, where \u03bb1, . . . , \u03bbk are the variables, so we can write", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_7", "text": "LECTURE NOTES\n5\nthe problem as\n\u03bb1\n\uf8eb\n\uf8ec\n\uf8ed\nu1,1\n...\nu1,n\n\uf8f6\n\uf8f7\n\uf8f8+ \u00b7 \u00b7 \u00b7 + \u03bbk\n\uf8eb\n\uf8ec\n\uf8ed\nuk,1\n...\nuk,n\n\uf8f6\n\uf8f7\n\uf8f8=\n\uf8eb\n\uf8ec\n\uf8ed\nb1\n...\nbn\n\uf8f6\n\uf8f7\n\uf8f8.\nThis is a linear system so we can use row reduction for its augmented matrix\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n...\n...\n...\nb1\nu1\nu2\n\u00b7 \u00b7 \u00b7\nuk\n...\n...\n...\n...\nbn\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\nExample (Geometric interpretation). Give visual interpretation of addition, multiplication by scalars of\nvectors, and span of one or two vectors in R2.\n1.4. More matrix notation.\nDefinition. Let A be an m \u00d7 n matrix with columns A1, \u00b7 \u00b7 \u00b7 , An \u2208Rm and x =\n\uf8eb\n\uf8ec\n\uf8ed\nx1\n...\nxn\n\uf8f6\n\uf8f7\n\uf8f8\u2208Rn. The\n(matrix-vector) product Ax is defined by Ax = x1A1 + x2A2 + \u00b7 \u00b7 \u00b7 + xnAn \u2208Rm. It is only defined if the\nnumber of columns of A is equal to the number of entries of x. Writing out all components, we have\nA =\n\uf8eb\n\uf8ec\n\uf8ed\na1,1\na1,2\n\u00b7 \u00b7 \u00b7\na1,n\n...\nam,1\nam,2\n\u00b7 \u00b7 \u00b7\nam,n\n\uf8f6\n\uf8f7\n\uf8f8\nx =\n\uf8eb\n\uf8ec\n\uf8ed\nx1\n...\nxn\n\uf8f6\n\uf8f7\n\uf8f8\nAx = x1A1 + \u00b7 \u00b7 \u00b7 + xnAn =\n\uf8eb\n\uf8ec\n\uf8ed\nx1a1,1 + x2a1,2 + \u00b7 \u00b7 \u00b7 + xna1,n\n...\nx1am,1 + x2am,2 + \u00b7 \u00b7 \u00b7 + xnam,n\n\uf8f6\n\uf8f7\n\uf8f8\nBelow we collect some further properties of the matrix-vector product that will allow us to understand\nlinear systems a bit better.\nProposition 1.3. Let A \u2208Rm\u00d7n, x, y \u2208Rn, \u03bb \u2208R. Then\ni) A(x + y) = Ax + Ay,\nii) A(\u03bbx) = \u03bbAx.\nProof. i) By definition and the calculus rules for vectors we have\nA(x + y) = (x1 + y1)A1 + . . . + (xn + yn)An = x1A1 + y1A1 + . . . + xnAn + ynAn\n= x1A1 + . . . + xnAn + y1A1 + . . . + ynAn = Ax + Ay;\nA(\u03bbx) = \u03bbx1A1 + . . . + \u03bbxnAn = \u03bb(x1A1 + . . . + xnAn)) = \u03bbAx.\n\u25a1", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_8", "text": "A(\u03bbx) = \u03bbx1A1 + . . . + \u03bbxnAn = \u03bb(x1A1 + . . . + xnAn)) = \u03bbAx.\n\u25a1\nDefinition. We denote by In the n \u00d7 n matrix written\nIn =\n\uf8eb\n\uf8ec\n\uf8ed\n1\n0\n...\n0\n1\n\uf8f6\n\uf8f7\n\uf8f8", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_9", "text": "6\nKALAN WALMSLEY\nIt is called the identity matrix and has the interesting property\nInx = x1\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n1\n0\n...\n...\n0\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n+ x2\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n0\n1\n0\n...\n0\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n+ \u00b7 \u00b7 \u00b7 + xn\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n0\n...\n...\n0\n1\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n=\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\nx1\n...\n...\n...\nxn\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n= x\nWhen the parameter n is clear from the context, we also write I instead of In.\nThe following results connectes the matrix-vector product to linear systems.\nTheorem 1.4. Let A \u2208Rm\u00d7n with columns A1, \u00b7 \u00b7 \u00b7 , An and b \u2208Rm. The equation Ax = b is equivalent\nto the vector equation\nx1A1 + \u00b7 \u00b7 \u00b7 + xnAn = b,\nwhich is in turn equivalent to the linear system with augmented matrix (A | b).\nProof. The claim follows from the definition of the matrix-vector product and the componentwise-\ninterpretation.\n\u25a1\nMore important is the following observation: Ax = b has a solution iff b \u2208span{A1, \u00b7 \u00b7 \u00b7 , An}. This will\nbe a recurrent fact in this course. For instance, it allows us to answer the question whether we can solve\nAx = b for all b \u2208Rm instead of just a given single vector b.\nTheorem 1.5. Let A \u2208Rm\u00d7n, then the following statements are equivalent:\ni) for all b \u2208Rm the equation Ax = b has a solution\nii) span{A1, \u00b7 \u00b7 \u00b7 , An} = Rm\niii) Any echelon form of A has a leading entry in every row\nProof. We show that i) =\u21d2ii) =\u21d2iii) =\u21d2i), which allows us to close the circle and then implies all\nequivalences.\ni) \u21d2ii) Let b \u2208Rm. We can assume i), so Ax = b has a solution. Thus \u2203x1, \u00b7 \u00b7 \u00b7 , xn \u2208R s.t. b =\nx1A1 + . . . + xnAn\n|\n{z\n} \u2208span{A1, \u00b7 \u00b7 \u00b7 , An}. Since b was arbitrary, this implies span{A1, \u00b7 \u00b7 \u00b7 , An} =\nRm.\nii) \u21d2iii) Let \u02dcA be an echelon form of A and assume by contradiction that \u02dcA has a row with no pivot\nelement =\u21d2\n\u02dcA has a zero row at the bottom =\u21d2( \u02dcA | em), em =\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n0\n...\n0\n1\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8has no solution. Using", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_10", "text": "Rm.\nii) \u21d2iii) Let \u02dcA be an echelon form of A and assume by contradiction that \u02dcA has a row with no pivot\nelement =\u21d2\n\u02dcA has a zero row at the bottom =\u21d2( \u02dcA | em), em =\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n0\n...\n0\n1\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8has no solution. Using\nelementary transformations of type 1, 2, 3, we can transform this augmented matrix back to a\nsystem of the form (Ax | b) for some b \u2208Rm (its \"original\" state). Since the solution does not\nchange, Ax = b has no solution =\u21d2b /\u2208span{A1, \u00b7 \u00b7 \u00b7 , An}\nE\niii) \u21d2i) Statement iii) implies that any echelon form of A has no zero row. Consider Ax = b and solve\nit by row reduction =\u21d2( \u02dcA | \u02dcb) has no row of the form\n\u00000\n\u00b7 \u00b7 \u00b7\n0\nb\u0001\nwith b \u0338= 0. Theorem 1.2\nthus implies that solutions exist\n\u25a1\nExample. Discuss existence and uniqueness for the matrix A =\n\uf8eb\n\uf8ed\n1\n1\n1\n0\n0\n1\n1\n1\n1\n0\n1\n1\n\uf8f6\n\uf8f8, using the above theorem\nand an argument on the number of free variables (which does not depend on the RHS).", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_11", "text": "LECTURE NOTES\n7\nRemark. For a matrix Am\u00d7n, we can make the following elementary observations:\n\u2022 If n > m, there are always free variables since we have at most one pivot element per row. Thus\nthe solution of Ax = b can never be unique\n\u2022 If n < m, since pivot elements are strictly moving to the right starting from the top of the echelon\nform, there are at most n pivot elements and therefore there exists a row without pivot element.\nHence we cannot solve Ax = b for all b \u2208Rm (but surely for some).\n\u2022 n < m does not necessarily imply uniqueness (take A =\n\uf8eb\n\uf8ed\n1\n1\n2\n2\n4\n4\n\uf8f6\n\uf8f8and b =\n\uf8eb\n\uf8ed\n0\n0\n0\n\uf8f6\n\uf8f8) and n > m does\nnot necessarily imply existence for all b (take A =\n\u0012\n1\n2\n4\n1\n2\n4\n\u0013\nand b =\n\u0012\n1\n0\n\u0013\n).\nRemark. To verify the existence of solutions for all b, computing an echelon form and counting pivot\nelements suffices. To find the solution(s) we need to consider the augmented matrix instead.\n1.5. Structure of solutions of linear systems. Idea: using vectors, we can describe the solutions\nof linear systems more geometrically than with the parametric form. Moreover, we shall see that it is\nimportant to understand the equation Ax = 0, the so-called homogeneous equation.\nRemark. Adding solutions and scaling solutions of the homogeneous equation produces again solutions.\n\u2022 If Ax = 0 and Ay = 0, A(x + y) = Ax + Ay = 0 + 0 = 0\n\u2022 If Ax = 0 and \u03bb \u2208R, A(\u03bbx) = \u03bbAx = \u03bb \u00b7 0 = 0\nWe know that Ax = 0 always has the solution x = 0 \u2208Rn Thm 1.2\n=\u21d2\n0 is the only solution iff there are\nno free variables in any echelon form of A. The following result describes the structure of solutions of the\ngeneral equation Ax = b, provided that there exists at least one solution.\nTheorem 1.6. Let A \u2208Rm\u00d7n and b \u2208Rm. Assume \u2203sb \u2208Rn s.t. Asb = b. Then the set of solutions of\nthe equation Ax = b is given by S = {sb + sh | Ash = 0}, i.e. for each solution sh of Ax = 0 the vector\nsb + sh solves Ax = b and conversely every solution s of Ax = b is of the form s = sb + sh for some sh", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_12", "text": "the equation Ax = b is given by S = {sb + sh | Ash = 0}, i.e. for each solution sh of Ax = 0 the vector\nsb + sh solves Ax = b and conversely every solution s of Ax = b is of the form s = sb + sh for some sh\nsolving Ax = 0\nProof. There are two statements to show:\ni) Let sh be s.t. Ash = 0. Then A(sb + sh) = Asb\n|{z}\nb\n+ Ash\n|{z}\n0\n= b + 0 = b.\nii) Let s solve As = b We need to find sh s.t. Ash = 0 and s = sb + sh We try sh = s \u2212sb =\u21d2\nA(s \u2212sb) = As\n|{z}\nb\n\u2212Asb\n|{z}\nb\n= b \u2212b = 0.\n\u25a1\nThe parametric vector form. Goal: represent the solution set of a linear system Ax = b in the form\nsb + span({some vectors}).\nExample.\na) Find the parametric vector form for A =\n\uf8eb\n\uf8ed\n3\n5\n\u22124\n\u22123\n\u22122\n4\n6\n1\n\u22128\n\uf8f6\n\uf8f8and b =\n\uf8eb\n\uf8ed\n7\n\u22121\n\u22124\n\uf8f6\n\uf8f8.\nb) Find the parametric vector form for the linear equation 2x1 + 4x2 \u2212x3 = 8.\nHere the general strategy:\n(1) Find the reduced echelon form of (A | b)\n(2) Express basic variables in terms of free variables\n(3) Write x =\n\uf8eb\n\uf8ec\n\uf8ed\nx1\n...\nxn\n\uf8f6\n\uf8f7\n\uf8f8and insert all equations for the basic variables\n(4) Decompose x into an x-independent part and a linear combination, where the free variables act\nas weights/parameters", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_13", "text": "8\nKALAN WALMSLEY\n1.6. Linear independence.\nRemark. v1 \u2208Rn, v2 = \u03bbv1, \u03bb \u2208R\n=\u21d2\nspan{v1, v2} = span{v1} Indeed, let \u00b51v1 + \u00b52v2 \u2208\nspan{v1, v2}, \u00b51, \u00b52 \u2208R, then\n\u00b51v1 + \u00b52v2 = \u00b51v1 + \u00b52\u03bbv1 = (\u00b51 + \u00b52\u03bb\n|\n{z\n}\n\u2208R\n)v1\ndef\n\u2208span{v1}\nOn the other hand, if \u00b51v1 \u2208span{v1} =\u21d2\u00b51v1 + 0 \u00b7 v2 \u2208span{v1, v2}\n=\u21d2span{v1, v2} = span{v1}\nIn the previous remark v2 was a multiple of v1. The following definition generalizes this concept to\nmore than 2 elements.\nDefinition. Let {v1, \u00b7 \u00b7 \u00b7 , vk} \u2282Rn be a set of vectors. The set is called linearly independent if the\nequation\n\u03bb1v1 + \u00b7 \u00b7 \u00b7 + \u03bbkvk = 0\n(\u2208Rn)\nhas only the solution \u03bb1 = \u03bb2 = \u00b7 \u00b7 \u00b7 = \u03bbk = 0 Otherwise, it is called linearly dependent.\nFor sets only containing one vector v \u2208Rn, {v} is linearly independent iff v \u0338= 0. Indeed,\n\u03bbv =\n\uf8eb\n\uf8ec\n\uf8ed\n\u03bbv1\n...\n\u03bbvn\n\uf8f6\n\uf8f7\n\uf8f8\u21d0\u21d2\n(\neither \u03bb = 0\nor vj = 0 \u2200j\nFor sets containing several vectors we have the following characterization:\nTheorem 1.7. Let k \u22652 and {v1, \u00b7 \u00b7 \u00b7 , vk} \u2282Rn. This set is linearly dependent iff \u2203j \u2208{1, \u00b7 \u00b7 \u00b7 , k} s.t. vj\nis a linear combination of the other vectors.\nProof. (Note that an \u2019iff\u2019 statement always involves two claims to show)\n\u21d2Let {v1, \u00b7 \u00b7 \u00b7 , vk} be linearly dependent \u21d4\u2203\u03bb1, \u00b7 \u00b7 \u00b7 , \u03bbk \u2208R, \u03bbj \u0338= 0 for at least one j s.t. 0 =\n\u03bb1v1 + \u00b7 \u00b7 \u00b7\n\u2191\n\u03bbjvj\n+ \u03bbkvk", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_14", "text": "\u21d2Let {v1, \u00b7 \u00b7 \u00b7 , vk} be linearly dependent \u21d4\u2203\u03bb1, \u00b7 \u00b7 \u00b7 , \u03bbk \u2208R, \u03bbj \u0338= 0 for at least one j s.t. 0 =\n\u03bb1v1 + \u00b7 \u00b7 \u00b7\n\u2191\n\u03bbjvj\n+ \u03bbkvk\n=\u21d2\u03bbjvj = \u2212\u03bb1v1 \u2212\u00b7 \u00b7 \u00b7 \u2212\u03bbj\u22121vj\u22121 \u2212\u03bbj+1vj+1 \u2212\u00b7 \u00b7 \u00b7 \u2212\u03bbkvk\n=\u21d2vj = \u2212\u03bb1\n\u03bbj v1 \u2212\u00b7 \u00b7 \u00b7 \u2212\u03bbj\u22121\n\u03bbj vj\u22121 \u2212\u03bbj+1\n\u03bbj vj+1 \u2212\u00b7 \u00b7 \u00b7 \u2212\u03bbk\n\u03bbj vk\n=\u21d2vj is a linear combination of the other vectors\n\u21d0Assume \u2203j \u2208{1, \u00b7 \u00b7 \u00b7 , k}, \u03bb1, \u00b7 \u00b7 \u00b7 , \u03bbj\u22121, \u03bbj+1, \u03bbk \u2208R s.t.\nvj = \u03bb1v1 + \u00b7 \u00b7 \u00b7 + \u03bbj\u22121vj\u22121 + \u03bbj+1vj+1 + \u00b7 \u00b7 \u00b7 + \u03bbkvk\n0 = \u03bb1v1 + \u00b7 \u00b7 \u00b7 + \u03bbj\u22121vj\u22121 \u2212vj + \u03bbj+1vj+1 + \u00b7 \u00b7 \u00b7 + \u03bbkvk\n\u2212vj\n=\u21d2\u03bbj = \u22121 \u0338= 0 =\u21d2{v1, \u00b7 \u00b7 \u00b7 , vk} is linearly dependent.\n\u25a1\nPractical check for linear dependence.\n\u03bb1v1 + \u00b7 \u00b7 \u00b7 + \u03bbkvk = 0, v1, \u00b7 \u00b7 \u00b7 , vk \u2208Rn\n\u21d0\u21d2\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ed\n...\n...\nv1\n\u00b7 \u00b7 \u00b7\nvk\n...\n...\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f8\n\uf8eb\n\uf8ec\n\uf8ed\n\u03bb1\n...\n\u03bbk\n\uf8f6\n\uf8f7\n\uf8f8=\n\uf8eb\n\uf8ec\n\uf8ed\n0\n...\n0\n\uf8f6\n\uf8f7\n\uf8f8\n0 is the only solution of this linear system \u21d4no free variables in any echelon form of the matrix \u21d4\nlinearly independent.\nTheorem 1.8. If k > n, then any set {v1, \u00b7 \u00b7 \u00b7 , vk} \u2282Rn is linearly dependent.", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_15", "text": "LECTURE NOTES\n9\nProof. The matrix\n\u0000v1\n\u00b7 \u00b7 \u00b7\nvk\n\u0001\n\u2208Rn\u00d7k has more columns than rows =\u21d2there are free variables.\n\u25a1\nTheorem 1.9. If vj = 0 for some j \u2208{1, \u00b7 \u00b7 \u00b7 , k}, then {v1, \u00b7 \u00b7 \u00b7 , vk} is linearly dependent.\nProof. 0 = 0 \u00b7 v1 + \u00b7 \u00b7 \u00b7 + 0 \u00b7 vj\u22121 + 1 \u00b7 vj + 0 \u00b7 vj+1 + . . . + 0 \u00b7 vk\n\u25a1\nImportant properties.\n\u2022 If the set {v1, \u00b7 \u00b7 \u00b7 , vk} is linearly dependent, then adding vectors keeps the set linearly dependent.\n(Proof: If there exists a non-zero solution, then this solution also solves \u03bb1v1 + \u00b7 \u00b7 \u00b7 + \u03bbkvk + 0 \u00b7\nvk+1 + \u00b7 \u00b7 \u00b7 + 0 \u00b7 vm = 0)\n\u2022 If the set {v1, \u00b7 \u00b7 \u00b7 , vk} is linearly independent, then removing vectors keeps the set linearly inde-\npendent.\n(Proof: If {v1, \u00b7 \u00b7 \u00b7 , vk} is linearly independent and assume that {v1, \u00b7 \u00b7 \u00b7 , vk\u22121} is linearly depen-\ndent, then by the above proof, we would obtain that the full set is also linearly dependent. This\nis a contradiction.)\nCorollary 1.10. Let A \u2208Rm\u00d7n. Then\ni) \u2200b \u2208Rm the equation Ax = b has at most one solution iff the columns {A1, \u00b7 \u00b7 \u00b7 , An} are linearly\nindependent\nii) \u2200b \u2208Rm, Ax = b has at least one solution iff span{A1, \u00b7 \u00b7 \u00b7 , An} = Rm\niii) Ax = b has a unique solution \u2200b \u2208Rm iff span{A1, \u00b7 \u00b7 \u00b7 , An} = Rm and {A1, \u00b7 \u00b7 \u00b7 , An} is linearly\nindependent\nProof. For the proof we just need to combine previous results, for details see the exercises.\n\u25a1\n1.7. Matrices and linear maps. So far Ax = b corresponds to a linear system. Now a new point of\nview: functions x 7\u2192Ax.\nDefinition. Let X, Y be two non-empty sets. A function (map) is a rule that assigns to each element", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_16", "text": "\u25a1\n1.7. Matrices and linear maps. So far Ax = b corresponds to a linear system. Now a new point of\nview: functions x 7\u2192Ax.\nDefinition. Let X, Y be two non-empty sets. A function (map) is a rule that assigns to each element\nx \u2208X one (and only one) element f(x) \u2208Y . We write f : X \u2192Y . X is called the domain of f and Y\nthe codomain of f. The range of f is the set {f(x) | x \u2208X}.\nDefinition. A map f : Rn \u2192Rm is called linear if \u2200x, y \u2208Rn, \u03bb \u2208R we have\ni) f(x + y) = f(x) + f(y);\nii) f(\u03bbx) = \u03bbf(x).\nRemark (Quick check for linearity). Any linear map satisfies f(0) = 0\n(Indeed, f( 0\n\u2191\nRn\n) = f(0\n\u2191\nR\n\u00b7 0\n\u2191\nRn\n) = 0\n\u2191\nR\n\u00b7 f( 0\n\u2191\nRn\n) = 0.)\nHowever, this is just a necessary condition. It does not imply linearity.\nOn the other hand, linearity is equivalent to f(\u03bbx + y) = \u03bbf(x) + f(y) for all x, y \u2208Rn and \u03bb \u2208R;\nsee exercises.\nImportant properties of linear maps: Let \u03bb1v1 + \u03bb2v2 + \u00b7 \u00b7 \u00b7 + \u03bbkvk \u2208Rn be a linear combination of\nv1, \u00b7 \u00b7 \u00b7 , vk \u2208Rn, then\nf(\u03bb1v1 + \u00b7 \u00b7 \u00b7 + \u03bbkvk) = \u03bb1f(v1) + \u00b7 \u00b7 \u00b7 + \u03bbkf(vk).\nThis follows from iterating the linearity property.\nTheorem 1.11. Functions writable as matrices are always linear. The columns of a matrix are the\nimages of the basis vectors.\ni) For every matrix A \u2208Rm\u00d7n, the function fA : Rn \u2192Rm s.t. fA(x) = Ax is linear\nii) If the function fA : Rn \u2192Rm is linear, then there exists a unique matrix A \u2208Rm\u00d7n s.t. f(x) =\nAx, \u2200x \u2208Rn\nThe matrix A is given by A =\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ed\n...\n...\nf(e1)\n\u00b7 \u00b7 \u00b7\nf(en)\n...\n...\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f8", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_17", "text": "10\nKALAN WALMSLEY\nProof. i)) follows from Proposition 1.3.\nii)) Let f be linear and x \u2208Rn, so we have x = x1e1 + x2e2 + \u00b7 \u00b7 \u00b7 + xnen\n=\u21d2f(x) = f(x1e1 + x2e2 + \u00b7 \u00b7 \u00b7 + xnen) = x1f(e1) + \u00b7 \u00b7 \u00b7 + xnf(en)\n=\n\u0000f(e1)\n\u00b7 \u00b7 \u00b7\nf(en)\u0001\n|\n{z\n}\nA\u2208Rm\u00d7n\nx\nNow consider another matrix B \u2208Rm\u00d7n s.t. f(x) = Bx, \u2200x \u2208Rn. Choose x = ei =\u21d2f(ei) = Bei =\nBi =\u21d2ith column of B equals the ith column of A for all i =\u21d2A = B.\n\u25a1\nLemma 1.12. Let f, g : Rn \u2192Rm, \u03bb \u2208R. Define f + g : Rn \u2192Rm by (f + g)(x) = f(x) + g(x) and\n\u03bbf : Rn \u2192Rm by (\u03bbf)(x) = \u03bbf(x). Then f + g and \u03bbf are linear functions.\nProof. For all x, y \u2208Rn and \u03b1 \u2208R we have\n(f + g)(\u03b1x + y) = f(\u03b1x + y) + g(\u03b1x + y)\nf, g\nlinear\n\u2193=\n\u03b1f(x) + f(y) + \u03b1g(x) + g(y)\n= \u03b1(f(x) + g(x)) + f(y) + g(y) = \u03b1(f + g)(x) + (f + g)(y).\n\u25a1\nDefinition. Let f : X \u2192Y be a function. We say that f is\ni) Injective if f(x1) = f(x2) =\u21d2x1 = x2 (related to uniqueness)\nii) Surjective if \u2200y \u2208Y, \u2203x \u2208X s.t. f(x) = y (related to existence)\niii) Bijective if f is injective and surjective\nTheorem 1.13. Let f : Rn \u2192Rm be linear and let A \u2208Rm\u00d7n be such that f(x) = Ax, \u2200x \u2208Rn, then", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_18", "text": "iii) Bijective if f is injective and surjective\nTheorem 1.13. Let f : Rn \u2192Rm be linear and let A \u2208Rm\u00d7n be such that f(x) = Ax, \u2200x \u2208Rn, then\ni) f is injective \u21d0\u21d2f(x) = 0 implies x = 0 \u21d0\u21d2the columns of A are linearly independent\nii) f is surjective \u21d0\u21d2span{A1, \u00b7 \u00b7 \u00b7 , An} = Rm\nProof.\ni)\n\u21d2If f is injective, then f(x) = 0 has at most one solution. f is linear =\u21d2x = 0 is a\nsolution =\u21d2f(x) = 0 implies x = 0\n\u21d0Suppose f(x) = 0 implies x = 0 and assume f(x1) = f(x2) =\u21d20 = f(x1) \u2212f(x2)\n=\n\u2191\nf linear\nf(x1 \u2212x2) =\u21d2x1 \u2212x2 = 0 =\u21d2x1 = x2\nNext, f(x) = 0 \u21d0\u21d2Ax = 0 \u21d0\u21d2x1A1 + \u00b7 \u00b7 \u00b7 + xnAn = 0\n=\u21d2\nf(x) = 0 implies x = 0 is\nequivalent to {A1, \u00b7 \u00b7 \u00b7 , An} being linearly independent.\nii) f is surjective \u21d0\u21d2\u2200b \u2208Rm f(x) = b has a solution \u21d0\u21d2\u2200b \u2208Rm Ax = b has a solution\nThm 1.5\n\u21d0\u21d2\nspan{A1, \u00b7 \u00b7 \u00b7 , An} = Rm\n\u25a1\nCorollary 1.14. Let f : Rn \u2192Rm be linear, then f is\ni) not surjective if m > n\nii) not injective if m < n\niii) not bijective if m \u0338= n\nProof.\ni) If m > n, the matrix A s.t. f(x) = Ax has more rows than columns\n=\u21d2\nan echelon\nform of A can have at most n pivot elements\n=\u21d2\nnot every row contains a pivot element\nthm 1.5\n=\u21d2\nspan{A1, \u00b7 \u00b7 \u00b7 , An} \u0338= Rm thm 1.13\n=\u21d2\nf is not surjective\nii) If m < n, then by theorem 1.8 the columns of A are linearly dependent\nthm 1.13\n=\u21d2\nf is not injective\niii) Consequence of the two previous statements\n\u25a1", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_19", "text": "=\u21d2\nf is not surjective\nii) If m < n, then by theorem 1.8 the columns of A are linearly dependent\nthm 1.13\n=\u21d2\nf is not injective\niii) Consequence of the two previous statements\n\u25a1\nTheorem 1.15. Let f : Rn \u2192Rm be linear, A \u2208Rm\u00d7n s.t. f(x) = Ax,\n\u2200x \u2208Rn. Let \u02dcA be an echelon form of A, then\ni) f is injective \u21d0\u21d2\u02dcA has n pivot elements", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_20", "text": "LECTURE NOTES\n11\nii) f is surjective \u21d0\u21d2\u02dcA has m pivot elements\niii) f is bijective \u21d0\u21d2n = m and \u02dcA has n pivot elements\nProof.\ni) Theorem 1.13 states f is injective \u21d0\u21d2x = 0 is the only solution of Ax = 0\nthm 1.2\n\u21d0\u21d2there\nare no free variables \u21d0\u21d2each column has a pivot element \u21d0\u21d2\u02dcA has n pivot elements\nii) Theorem 1.13 also states f is surjective \u21d0\u21d2span{A1, \u00b7 \u00b7 \u00b7 , An} = Rm\nthm 1.5\n\u21d0\u21d2\n\u02dcA has a pivot element in every row \u21d0\u21d2\u02dcA has m pivot elements\niii) f is bijective\ncorol 1.14\n=\u21d2\nn = m pivot elements =\u21d2f is bijective\n\u25a1\nSummary of Chapter 1.\n\u2022 Linear systems:\nA \u2208Rm\u00d7n, Ax = b \u21d0\u21d2f : Rn \u2192Rm linear , f(x) = b\n\u2022 Existence of solutions for given b:\n\u2013 \u02dcA | \u02dcb echelon form of augmented matrix has no row of the form\n\u00000\n\u00b7 \u00b7 \u00b7\n0\nk\u0001\nwith k \u0338= 0\n\u2013 b \u2208span{A1, \u00b7 \u00b7 \u00b7 , An}\n\u2022 Uniqueness of solutions for given b: (assume \u2203a solution)\n\u2013 \u02dcA has no free variables \u21d0\u21d2n pivot elements\n\u2013 A1, \u00b7 \u00b7 \u00b7 , An are linearly independent\nThese don\u2019t depend on b !\n\u2022 Existence of solutions \u2200b:\n\u2013 \u02dcA has m pivot elements (m \u2264n necessary)\n\u2013 span{A1, \u00b7 \u00b7 \u00b7 , An} = Rm\n\u2022 Uniqueness of solutions \u2200b:\n\u2013 Same as for a single b\n\u2022 Computing the solution: row reduction and parametric vector form\n\u2022 Linear independence of v1, \u00b7 \u00b7 \u00b7 , vk \u2208Rm\n\u2013 Solve\n\u0000v1\n\u00b7 \u00b7 \u00b7\nvk\n|\n0\u0001\n, if the solution is unique, v1, \u00b7 \u00b7 \u00b7 , vk are linearly independent", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_21", "text": "12\nKALAN WALMSLEY\n2. Matrix algebra\n2.1. Matrix operations. Recall: A \u2208Rm\u00d7n, ai,j \u2208R is the entry for the ith row and the jth column\n(sometimes (A)i,j).\nA =\n\uf8eb\n\uf8ec\n\uf8ed\na1,1\n\u00b7 \u00b7 \u00b7\na1,n\n...\nam,1\n\u00b7 \u00b7 \u00b7\nam,n\n\uf8f6\n\uf8f7\n\uf8f8=\n\u0000A1\n\u00b7 \u00b7 \u00b7\nAn\n\u0001\nDefinition. Given A, B \u2208Rm\u00d7n and \u03bb \u2208R, we define A + B \u2208R as the matrix with coefficients\n(A + B)i,j = ai,j+bi,j (componentwise addition) and \u03bbA \u2208Rm\u00d7n as the matrix with coefficients (\u03bbA)i,j =\n\u03bbai,j.\nNotation.\n\u2022 0 \u2208Rm\u00d7n is the matrix with coefficients ai,j = 0\n\u2200i, j\n\u2022 \u2212A \u2208Rm\u00d7n is the matrix with coefficients \u2212(A)i,j = \u2212ai,j\n\u2200i, j\nRemark. Let f, g : Rn \u2192Rm be linear with A =\n\u0000f(e1)\n\u00b7 \u00b7 \u00b7\nf(en)\u0001\n, and B =\n\u0000g(e1)\n\u00b7 \u00b7 \u00b7\ng(en)\u0001\n.\nThen \u03bbf has matrix \u03bbA and f + g has matrix A + B since for instance\nA + B =\n\u0000f(e1) + g(e1)\n\u00b7 \u00b7 \u00b7\nf(en) + g(en)\u0001\n=\n\u0000f(e1)\n\u00b7 \u00b7 \u00b7\nf(en)\u0001\n+\n\u0000g(e1)\n\u00b7 \u00b7 \u00b7\ng(en)\u0001\n.\nThe argument for \u03bbf is similar.\nFor the sum of matrices and the multiplication by scalars we have the following rules for computation:\nTheorem 2.1. Let A, B, C \u2208Rm\u00d7n, \u03bb, \u00b5, \u2208R. Then\ni) A + B = B + A\nii) (A + B) + C = A + (B + C)\niii) A + 0 = A\niv) \u03bb(A + B) = \u03bbA + \u03bbB\nv) (\u03bb + \u00b5)A = \u03bbA + \u00b5A\nvi) \u03bb(\u00b5A) = (\u03bb\u00b5)A\nProof. All operations are componentwise and the properties hold for real numbers.\n\u25a1\nMatrix product. In order to motivate the matrix product, we first make the following observation: let", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_22", "text": "v) (\u03bb + \u00b5)A = \u03bbA + \u00b5A\nvi) \u03bb(\u00b5A) = (\u03bb\u00b5)A\nProof. All operations are componentwise and the properties hold for real numbers.\n\u25a1\nMatrix product. In order to motivate the matrix product, we first make the following observation: let\nf : Rn \u2192Rm, g : Rp \u2192Rn both be linear. Then their composition f \u25e6g : Rp \u2192Rm, (f \u25e6g)(x) = f (g (x))\nis well defined and linear. Indeed, due to the linearity of f and g, for all x, y \u2208Rp and \u03bb \u2208R we have\nthat\n(f \u25e6g)(\u03bbx + y) = f(g(\u03bbx + y))\ng linear\n=\nf(\u03bbg(x) + g(y))\nf linear\n=\n\u03bbf(g(x)) + f(g(y))\n= \u03bb(f \u25e6g)(x) + (f \u25e6g)(y).\nThis shows the linearity of the composition.\nNext, we know that f(x) = Ax and g(x) = Bx, A \u2208Rm\u00d7n, B \u2208Rn\u00d7p, so what is the matrix f \u25e6g ?\nDefinition. Let A \u2208Rm\u00d7n, B \u2208Rn\u00d7p, B =\n\u0000B1 \u00b7 \u00b7 \u00b7 Bp\n\u0001\n. We define A \u00b7 B \u2208Rm\u00d7p to be the matrix with\ncolumns AB1, \u00b7 \u00b7 \u00b7 , ABp.\nAB = A\n\u0000B1 \u00b7 \u00b7 \u00b7 Bp\n\u0001\n=\n\u0000AB1 \u00b7 \u00b7 \u00b7 ABp\n\u0001\nRemark. There is a more practical way of computing AB\n(AB)i,j = (ABj)i\n=\nith row of A \u2022 jth column of B\n=\nn\nX\nl=1\nai,lbl,j", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_23", "text": "LECTURE NOTES\n13\nGraphically\nA = i\n\uf8eb\n\uf8ed\u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7\n\uf8f6\n\uf8f8\nB =\nj\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n.........\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\nTheorem 2.2. Let A \u2208Rm\u00d7n and B, C be matrices s.t. the following formulas are well defined. Then\ni) A(BC) = (AB)C\nB \u2208Rn\u00d7p, C \u2208Rp\u00d7l\nii) A(B + C) = AB + AC\nB, C \u2208Rn\u00d7p\niii) (B + C)A = BA + CA\nB, C \u2208Rn\u00d7p\niv) \u03bb(AB) = (\u03bbA)B = A(\u03bbB)\n\u2200\u03bb \u2208R, B \u2208Rn\u00d7p\nv) ImA = A = AIn\nProof. Let f(x) = Ax, g(x) = Bx, h(x) = Cx, then\ni) \u2200x\n(A(BC))x = (f \u25e6(g \u25e6h))(x)\n= f((g \u25e6h)(x))\n= f(g(h(x)))\n= (f \u25e6g)(h(x))\n= ((f \u25e6g) \u25e6h)(x) = ((AB)C)x\n=\u21d2choose x = ei, \u2200i =\u21d2A(BC) = (AB)C\nii) \u2200x\nA(B + C) = (f \u25e6(g + h))(x)\n= f((g + h)(x))\n= f(g(x) + h(x))\nf linear \u2192= f(g(x)) + f(h(x))\n= (f \u25e6g)(x) + (f \u25e6h)(x) = ABx + ACx\niii) Similar to previous.\niv) \u2200x\n\u03bb(AB)x = (\u03bb(f \u25e6g))(x)\n= \u03bb(f \u25e6g)(x)\n= \u03bbf(g(x))\n= f(\u03bbg(x)) = A(\u03bbB)x\nv) In : Rn \u2192Rn, Inx = x =\u21d2Im \u25e6f = f = f \u25e6In\n\u25a1\nRemark.\ni) AB \u0338= BA in general\nii) If AB = AC, then in general B \u0338= C\niii) If AB = 0, then not necessarily A = 0 or B = 0", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_24", "text": "v) In : Rn \u2192Rn, Inx = x =\u21d2Im \u25e6f = f = f \u25e6In\n\u25a1\nRemark.\ni) AB \u0338= BA in general\nii) If AB = AC, then in general B \u0338= C\niii) If AB = 0, then not necessarily A = 0 or B = 0\nMatrix equations. Given A \u2208Rm\u00d7n and C \u2208Rm\u00d7p, we look for X \u2208Rn\u00d7p s.t. AX = C. So we write\nX =\n\u0000X1\n\u00b7 \u00b7 \u00b7\nXp\n\u0001\nwith columns in Rn.\nAX =\n\u0000AX1\n\u00b7 \u00b7 \u00b7\nAXp\n\u0001\n=\n\u0000C1\n\u00b7 \u00b7 \u00b7\nCp\n\u0001\n\u21d0\u21d2\nAXj = Cj\n|\n{z\n}\np-independent linear systems\n\u2200j \u2208[[1, p]]", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_25", "text": "14\nKALAN WALMSLEY\nWe can solve all of these systems at once by using the augmented matrix\n\u0000A\n|\nC1\n\u00b7 \u00b7 \u00b7\nCp\n\u0001\nand by\nperforming row reduction to find\n\u0000 \u02dcA\n|\n\u02dcC1\n\u00b7 \u00b7 \u00b7\n\u02dcCp\n\u0001\nwith \u02dcA the (reduced) echelon form. Then solve\nsystems \u02dcAXj = \u02dcCj.\nDefinition. Let A \u2208Rn\u00d7n. We define A0 = In and for k \u2208N\nAk := A \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 A\n|\n{z\n}\nk times\n\u2208Rn\u00d7n\nDefinition. Let A \u2208Rn\u00d7n be a square matrix. We say that A is a diagonal matrix if ai,j = 0 whenever\ni \u0338= j.\nA =\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\na1,1\n0\n\u00b7 \u00b7 \u00b7\n0\n0\n...\n...\n...\n...\n0\n0\n\u00b7 \u00b7 \u00b7\n0\nan,n\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\nWe say that A is a lower (resp. upper) triangular matrix if ai,j = 0 whenever j > i (resp. j < i).\nA =\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\na1,1\n0\n\u00b7 \u00b7 \u00b7\n0\n\u22c6\n...\n...\n...\n...\n0\n\u22c6\n\u00b7 \u00b7 \u00b7\n\u22c6\nan,n\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\nlower triangular\nA =\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\na1,1\n\u22c6\n\u00b7 \u00b7 \u00b7\n\u22c6\n0\n...\n...\n...\n...\n\u22c6\n0\n\u00b7 \u00b7 \u00b7\n0\nan,n\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\nupper triangular\nLemma 2.3.\ni) Let A \u2208Rn\u00d7n be a diagonal matrix. Then Ak is also a diagonal matrix with\nAk =\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\nak\n1,1\n0\n\u00b7 \u00b7 \u00b7\n0\n0\n...\n...\n...\n...\n0\n0\n\u00b7 \u00b7 \u00b7\n0\nak\nn,n\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\nii) If A, B \u2208Rn\u00d7n are both lower (resp. upper) triangular matrices, then A \u00b7 B is also a lower (resp.\nupper) triangular matrix. In particular, Ak is always a lower (resp. upper) triangular matrix\nwhenever A is.\nProof.\ni) We use induction. Start with k = 1, A1 = A is diagonal by assumption. k =\u21d2k + 1\nAk+1 = AkA =\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\nak\n1,1\n0\n\u00b7 \u00b7 \u00b7\n0\n0\n...\n...\n...\n...\n0\n0\n\u00b7 \u00b7 \u00b7\n0\nak\nn,n\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\na1,1\n0\n\u00b7 \u00b7 \u00b7\n0\n0\n...\n...\n...\n...\n0\n0\n\u00b7 \u00b7 \u00b7\n0\nan,n\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n=\n\u0000Aka1,1e1\n|\n{z\n}\na1,1\u00b71st\ncolumn of Ak\n\u00b7 \u00b7 \u00b7\nAkan,nen\n|\n{z\n}", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_26", "text": "n,n\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\na1,1\n0\n\u00b7 \u00b7 \u00b7\n0\n0\n...\n...\n...\n...\n0\n0\n\u00b7 \u00b7 \u00b7\n0\nan,n\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n=\n\u0000Aka1,1e1\n|\n{z\n}\na1,1\u00b71st\ncolumn of Ak\n\u00b7 \u00b7 \u00b7\nAkan,nen\n|\n{z\n}\nan,n\u00b7nth\ncolumn of Ak\n\u0001\n=\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\nak+1\n1,1\n0\n\u00b7 \u00b7 \u00b7\n0\n0\n...\n...\n...\n...\n0\n0\n\u00b7 \u00b7 \u00b7\n0\nak+1\nn,n\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n= Ak+1", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_27", "text": "LECTURE NOTES\n15\nii) Analyse the entries of (A \u00b7 B) for A, B lower triangular matrices. We only have to observe the\nentries verifying j > i (supposed to be zero).\n(A \u00b7 B)i,j =\nn\nX\np=1\nai,pbp,j =\ni\nX\np=1\nai,p\nbp,j\n|{z}\n0\nbecause\np\u2264i<j\nand\nlower\ntriangular\n+\nn\nX\np=i+1\nai,p\n|{z}\n0\nbecause\ni<p\nand\nlower\ntriangular\nbp,j = 0\nSimilar for upper triangular matrices.\n\u25a1\nThe transpose of a matrix.\nDefinition. Let A \u2208Rm\u00d7n. The transpose of A is the n \u00d7 m matrix, denoted by AT , whose columns\nformed by the corresponding rows of A.\n\u0000AT \u0001\ni,j = aj,i\nTheorem 2.4. Let A \u2208Rm\u00d7n, B \u2208Rm\u00d7n, C \u2208Rn\u00d7p. Then\ni)\n\u0000AT \u0001T = A\nii) (A + B)T = AT + BT\niii) (\u03bbA)T = \u03bbAT\n\u2200\u03bb \u2208R\niv) (AC)T = CT AT\nProof.\ni)\n\u0000AT \u0001T\ni,j = AT\nj,i = Ai,j = A\nii) (A + B)T\ni,j = (A + B)j,i = (A)j,i + (B)j,i = AT + BT\niii) (\u03bbA)T\ni,j = (\u03bbA)j,i = \u03bb (A)j,i = \u03bb\n\u0000AT \u0001\ni,j = \u03bbAT\niv)\n(AC)T\ni,j = (AC)j,i =\nn\nX\nk=1\naj,kbk,i =\nn\nX\nk=1\n\u0000AT \u0001\nk,j\n\u0000CT \u0001\ni,k\n=\nn\nX\nk=1\n\u0000CT \u0001\ni,k\n\u0000AT \u0001\nk,j =\n\u0000CT AT \u0001\ni,j = CT AT\n=\u21d2(AC)T = CT AT\n\u25a1\n2.2. The inverse of a matrix. Motivation: given a matrix A can we find another matrix A\u22121 such that\nAx = b =\u21d2x\n?= A\u22121b.\nDefinition. Let A \u2208Rn\u00d7n. We say that A is invertible if \u2203C \u2208Rn\u00d7n s.t. AC = CA = In. We write\nC = A\u22121 for the inverse (inverse matrix) of A.", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_28", "text": "Ax = b =\u21d2x\n?= A\u22121b.\nDefinition. Let A \u2208Rn\u00d7n. We say that A is invertible if \u2203C \u2208Rn\u00d7n s.t. AC = CA = In. We write\nC = A\u22121 for the inverse (inverse matrix) of A.\nRemark. The inverse is unique (if exists). Indeed, suppose that B is another inverse. Then B = BIn =\nB(AC) = (BA)C = InC = C =\u21d2B = C.\nRemark. The formula for the inverse of a 2 \u00d7 2 matrix is\nA =\n\u0012a\nb\nc\nd\n\u0013\n=\u21d2A\u22121 =\n1\nad \u2212bc\n\u0012 d\n\u2212b\n\u2212c\na\n\u0013\n.\nTheorem 2.5.\ni) Let A \u2208Rn\u00d7n be invertible. Then A\u22121 is invertible with\n\u0000A\u22121\u0001\u22121 = A.\nii) A, B \u2208Rn\u00d7n invertible \u21d2A \u00b7 B invertible with (AB)\u22121 = B\u22121A\u22121\niii) A \u2208Rn\u00d7n invertible \u21d2AT invertible with\n\u0000AT \u0001\u22121 =\n\u0000A\u22121\u0001T", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_29", "text": "16\nKALAN WALMSLEY\nProof.\ni) We need to find C \u2208Rn\u00d7n s.t. A\u22121C = CA\u22121 = In. Try C = A.\nii)\n\u21d2ABB\u22121A\u22121 = A(\nIn\nz }| {\nBB\u22121)A\u22121 = AInA\u22121 = AA\u22121 = In\n\u21d0B\u22121A\u22121AB = B\u22121(A\u22121A\n| {z }\nIn\n)B = B\u22121InB = B\u22121B = In\niii)\n\u21d2AT \u0000A\u22121\u0001T = (\nIn\nz }| {\nA\u22121A)T = (In)T = In\n\u21d0\n\u0000A\u22121\u0001T AT =\n\u0000AA\u22121\n| {z }\nIn\n\u0001T = (In)T = In\n\u25a1\nRemark. More generally (A1 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 Ak)\u22121 = A\u22121\nk\n\u00b7 \u00b7 \u00b7 \u00b7 \u00b7 A\u22121\n1\nHow can we compute A\u22121 ?\nConsider AC = In =\u21d2n linear equations =\u21d2row reduction. But how can we assure that CA = In ?\nElementary matrices.\nI) Multiplying a row i by a scalar \u03bb \u0338= 0. Realizable by multiplication from the left with a matrix\nof the form (identity but one entry on the diagonal replaced by a scalar \u03bb \u0338= 0)\ni\ni\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n1\n0\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n0\n0\n...\n...\n...\n...\n1\n...\n...\n...\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u03bb\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n...\n...\n...\n1\n...\n...\n...\n...\n0\n0\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n0\n1\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\nand invertible with inverse matrix of the form\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n1\n0\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n0\n0\n...\n...\n...\n1\n...\n...\n1\n\u03bb\n...\n...\n1\n...\n...\n...\n0\n0\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n0\n1\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_30", "text": "LECTURE NOTES\n17\nII) Exchanging two rows i, k with i < k. Realizable by multiplication from the left with a matrix of\nthe form (identity except two rows interchanged)\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n\u00b7 \u00b7 \u00b7\ne1\n\u00b7 \u00b7 \u00b7\n...\n\u00b7 \u00b7 \u00b7\nek\n\u00b7 \u00b7 \u00b7\n...\n\u00b7 \u00b7 \u00b7\nei\n\u00b7 \u00b7 \u00b7\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\nek \u2194ei\n\uf8ee\n\uf8ef\uf8ef\uf8ef\uf8f0\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n1\n...\n1\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n0\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n1\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n1\n...\n1\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n1\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n0\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n1\n...\n1\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\nand invertible with inverse matrix equal to the matrix itself.\nIII) Adding a multiple of a row i to the kth row. Realizable by multiplication from the left with a\nmatrix of the form (identity except one non-zero entry off the diagonal)\ni\nk\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n1\n...\n...\n...\n...\n\u00b7 \u00b7 \u00b7\n\u03bb\n\u00b7 \u00b7 \u00b7\n...\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n...\n...\n...\n1\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\nand invertible with inverse matrix of the form\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n1\n...\n...\n\u2212\u03bb\n...\n...\n1\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\nDefinition. An elementary matrix is a matrix that realizes an elementary operation for row reduction\nby multiplication from the left. Such matrices are invertible. One can show that all elementary operations\ncan be realized by elementary matrices.\nTheorem 2.6. A \u2208Rn\u00d7n is invertible iff A can be transformed by elementary operations into In. In this\ncase, the same elementary operations transform In into A\u22121.", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_31", "text": "18\nKALAN WALMSLEY\nProof.\n\u21d2Assume A is invertible. Then \u2200b \u2208Rn, Ax = b has the solution x = A\u22121b\nthm 1.4\n=\u21d2\n\u02dcA\n(reduced echelon form of A) has n pivot elements. \u02dcA has n columns, so \u02dcA has the form\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n1\n0\n\u00b7 \u00b7 \u00b7\n0\n0\n...\n...\n...\n...\n0\n0\n\u00b7 \u00b7 \u00b7\n0\n1\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n= In\n\u21d0Now assume that A can be transformed by elementary operations into In. Let E1, \u00b7 \u00b7 \u00b7 , Ek denote\nthe corresponding elementary matrices.\nEk \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 E1A = In\n(Ek \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 E1\n|\n{z\n}\ninvertible since\neach Ej\nis invertible\n) A = In\n\u21d0\u21d2\nA =\nIn\nz\n}|\n{\n(Ek \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 E1)\u22121 (Ek \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 E1) A\n|\n{z\n}\nIn\n= (Ek \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 E1)\u22121\nthm 2.5\n=\u21d2\nA is invertible and A\u22121 = Ek \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 E1 = Ek \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 E1In\n\u25a1\nRemark. When computing (A | In) to get\n\u0000In | A\u22121\u0001\n, always start row reduction with, if exists, a row\nstaring with 1.\nRemark. Solving Ax = ej yields the jth column of A\u22121. This saves time if one is interested in a single\ncolumn of the inverse.\n2.3. Characterization of invertible matrices.\nTheorem 2.7. Let A \u2208Rn\u00d7n. Then the following statements are equivalent\na) A is invertible\nb) A can be transformed by elementary operations into In\nc) The reduced echelon form has n pivot elements ( \u02dcA = In)\nd) The equation Ax = 0 has only the trivial solution x = 0\ne) The columns of A are linearly independent\nf) The linear transformation f(x) = Ax is injective\ng) The equation Ax = b has at least one solution \u2200b \u2208Rn\nh) The columns of A span Rn\ni) f(x) = Ax is surjective\nj) \u2203C \u2208Rn\u00d7n s.t. CA = In\nk) \u2203D \u2208Rn\u00d7n s.t. AD = In\nl) AT is invertible\nThe matrices in j) and k) satisfy C = D = A\u22121.\nProof.\na \u21d2b) Theorem 2.6", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_32", "text": "j) \u2203C \u2208Rn\u00d7n s.t. CA = In\nk) \u2203D \u2208Rn\u00d7n s.t. AD = In\nl) AT is invertible\nThe matrices in j) and k) satisfy C = D = A\u22121.\nProof.\na \u21d2b) Theorem 2.6\nb \u21d2c) In has n pivot elements\nc \u21d2d) Theorem 1.15 and 1.13\nd \u21d2e) Theorem 1.13\ne \u21d2f) Theorem 1.13\nf \u21d2g) Theorem 1.15 with n = m\ng \u21d2h) Theorem 1.4 with n = m\nh \u21d2i) Theorem 1.13", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_33", "text": "LECTURE NOTES\n19\ni \u21d2j) Theorem 1.15 implies that \u02dcA has n = m pivot elements =\u21d2\n\u02dcA = In\nthm 2.6\n=\u21d2\nA is invertible\n=\u21d2C = A\u22121 works\nj \u21d2k) Assume that Ax = 0 =\u21d2CAx\n| {z }\n0\n= x\nthm 1.13\n=\u21d2\nf(x) = Ax is injective\nthm 1.15\n=\u21d2\n\u02dcA has n pivot\nelements =\u21d2\n\u02dcA = In\nthm 2.6\n=\u21d2\nA is invertible =\u21d2D = A\u22121 works\nk \u21d2l) Statement k) implies that f(x) = Ax is surjective because Db solves Ax = b \u2200b =\u21d2as\nabove, it follows that A is invertible\nthm 2.5\n=\u21d2\nAT is invertible\nl \u21d2a) Theorem 2.5\nThat proves the quivalence. Finally, if\n(\nCA = In\nAD = In\nC = CAA\u22121 = InA\u22121 = A\u22121\nD = A\u22121AD = A\u22121In = A\u22121\n)\nD = C = A\u22121\n\u25a1\nRemark. To check whether A is invertible, it suffices to compute any echelon form and count the number\nof pivot elements. To find A\u22121, we solve Ax = In by row reduction.\nInvertible linear transformations.\nRecall. A \u2208Rn\u00d7n \u2190\u2192f(x) = Ax linear, A =\n\u0000f(e1)\n\u00b7 \u00b7 \u00b7\nf(en)\u0001\nWe already know from the invertible\nmatrix theorem (2.7) that A invertible is equivalent to f bijective.\nFunction perspective: f : X \u2192Y bijective \u21d0\u21d2\u2203f \u22121 : Y \u2192X\n(f \u25e6f \u22121)(y) = y, \u2200y \u2208Y\n(f \u22121 \u25e6f)(x) = x, \u2200x \u2208X\nTheorem 2.8. Let f : Rn \u2192Rn be linear and A =\n\u0000f(e1)\n\u00b7 \u00b7 \u00b7\nf(en)\n\u0001\n, A \u2208Rn\u00d7n. Then f is bijective\niff A is invertible. Moreover, the linear map f \u22121(x) = A\u22121x is the unique map s.t.\n(f \u22121 \u25e6f)(x) = x = (f \u25e6f \u22121)(x), \u2200x \u2208Rn", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_34", "text": "iff A is invertible. Moreover, the linear map f \u22121(x) = A\u22121x is the unique map s.t.\n(f \u22121 \u25e6f)(x) = x = (f \u25e6f \u22121)(x), \u2200x \u2208Rn\nProof. f bijective \u21d0\u21d2A invertible follows from theorem 2.7. The map f \u22121(x) = A\u22121x is linear since\nA\u22121 \u2208Rn\u00d7n. It holds that\n(f \u22121 \u25e6f)(x) = A(A\u22121x) = x\n(f \u25e6f \u22121)(x) = A\u22121(Ax) = x\nTo show uniqueness, observe that whenever g : Rn \u2192Rn such that g \u25e6f = f \u25e6g = In, then f \u22121 =\nf \u22121 \u25e6(f \u25e6g) = (f \u22121 \u25e6f) \u25e6g = In \u25e6g = g.\n\u25a1", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_35", "text": "20\nKALAN WALMSLEY\n3. Determinants\n3.1. Introduction to determinants. We already know that\nA =\n\u0012a\nb\nc\nd\n\u0013\n: A invertible \u21d0\u21d2ad \u2212bc \u0338= 0\nDeterminants are a generalization of ad \u2212bc to n \u00d7 n matrices.\nDefinition. Let A \u2208Rn\u00d7n. Given 1 \u2264i, j \u2264n, we denote by Aij the (n \u22121) \u00d7 (n \u22121) matrix obtained\nfrom A by removing the ith row and the jth column. The determinant of A is defined by det(A) = a1,1 if\nn = 1 and\ndet(a) = a1,1 det\n\u0000A1,1\n\u0001\n\u2212a1,2 det\n\u0000A1,2\n\u0001\n+ a1,3 det\n\u0000A1,3\n\u0001\n\u2212\u00b7 \u00b7 \u00b7 + (\u22121)n+1a1,n det\n\u0000A1,n\n\u0001\n3.2. Properties of the determinant.\nTheorem 3.1. Let A \u2208Rn\u00d7n, A =\n\u0000A1\n\u00b7 \u00b7 \u00b7\nAn\n\u0001\n, then\ni) det(In) = 1\nii) Fixing A1, \u00b7 \u00b7 \u00b7 , Aj\u22121, Aj+1, \u00b7 \u00b7 \u00b7 , An, the function\nx 7\u2192det\n\u0000A1\nAj\u22121\nx\nAj+1\nAn\n\u0001\nis a linear map from Rn to R (det is not linear but if you fix all but one column, it becomes\nlinear).\niii) Let A1 be the matrix obtained from A by interchanging two columns, then the determinant changes\nsign\ndet(A1) = \u2212det(A)\nProof.\ni) By induction\ni) n = 1 : det(1)\nii) n \u22121 =\u21d2n : det(In) = 1 \u00b7 det\n\u0010\n(In)1,1\n\u0011\n= 1 \u00b7 det (In\u22121) = 1\nii) Let C =\n\u0000A1\n\u00b7 \u00b7 \u00b7\nAj\u22121\nAj + \u03bbBj\nAj+1\n\u00b7 \u00b7 \u00b7\nAn\n\u0001\n. We want to show that\ndet(C) = det(A) + \u03bb det\n\u0000A1\n\u00b7 \u00b7 \u00b7\nAj\u22121\n\u03bbBj\nAj+1\n\u00b7 \u00b7 \u00b7\nAn\n\u0001\n|\n{z\n}\nB\ni) n = 1 : det(a + \u03bbb) = a + \u03bbb = det(a) + \u03bb det(b)\nii) n \u22121 =\u21d2n :\ndet(C) =\nn\nX\nk=1\nk\u0338=j", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_36", "text": "Aj+1\n\u00b7 \u00b7 \u00b7\nAn\n\u0001\n|\n{z\n}\nB\ni) n = 1 : det(a + \u03bbb) = a + \u03bbb = det(a) + \u03bb det(b)\nii) n \u22121 =\u21d2n :\ndet(C) =\nn\nX\nk=1\nk\u0338=j\n(\u22121)k+1c1,k det(C1,k) + (\u22121)j+1c1,j det(C1,j)\nC =\n\u0000A1\n\u00b7 \u00b7 \u00b7\nAj\u22121\nAj + \u03bbBj\nAj+1\n\u00b7 \u00b7 \u00b7\nAn\n\u0001\n|\n{z\n}\nstill a linear combination\nin one column\nA linear combination appears in every term of det so we can separate it into A1,k + \u03bbB1,k\n(by induction hypothesis).\n=\u21d2det (C1,k) = det (A1,k) + \u03bb det (B1,k)\n=\nn\nX\nk=1\nk\u0338=j\n(\u22121)k+1 (a1,k det (A1,k) + \u03bbb1,k det (B1,k))\n+ (\u22121)j+1c1,j \u2212det (C1,j)", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_37", "text": "LECTURE NOTES\n21\nWe have c1,j = a1,j + \u03bbb1,j.\nC =\n\u0000A1\n\u00b7 \u00b7 \u00b7\nAj\u22121\nAj + \u03bbBj\nAj+1\n\u00b7 \u00b7 \u00b7\nAn\n\u0001\n=\u21d2C1,j = A1,j, C1,j = B1,j\n=\u21d2det (C1,j) = det (A1,j) = det (B1,j)\n=\nn\nX\nk=1\nk\u0338=j\n(\u22121)k+1 (a1,k det (A1,k) + \u03bbb1,k det (B1,k))\n+ (\u22121)j+1 (a1,j det (A1,j) + \u03bbb1,j det (B1,j))\n= det(A) + \u03bb det(B)\n=\u21d2det(C) = det(A) + \u03bb det(B)\niii) Omitted\n\u25a1\nLemma 3.2. Let A =\n\u0000A1\n\u00b7 \u00b7 \u00b7\nAn\n\u0001\n\u2208Rn\u00d7n\ni) If Ai = Aj for some i \u0338= j then det(A) = 0.\nii) If \u2203Aj that is a linear combination of the other columns, then det(A) = 0. In particular, if A is\nnot invertible, then det(A) = 0.\nProof.\ni) If Ai = Aj, the exchanging of Ai and Aj does not change A. But theorem 3.1 implies that\ndet(A) = \u2212det(A) =\u21d2det(A) = 0\nii) Write Aj =\nnP\nk=1\nk\u0338=j\n\u03bbkak. By linearity in the jth column, we have\ndet\n \nA1\n\u00b7 \u00b7 \u00b7\nAj\u22121\nnP\nk=1\nk\u0338=j\n\u03bbkAk\nAj+1\n\u00b7 \u00b7 \u00b7\nAn\n!\n=\nn\nX\nk=1\nk\u0338=j\n\u03bbk det\n\u0000A1\n\u00b7 \u00b7 \u00b7\nAj\u22121\nAk\nAj+1\n\u00b7 \u00b7 \u00b7\nAn\n\u0001\n|\n{z\n}\n0 by i)\n= 0\nIf A is not invertible, we know that the columns of A are linearly dependent =\u21d2det(A) = 0 by\nthe above property (if n \u22652, otherwise obvious).\n\u25a1\nThe determinant of elementary matrices. Notation:\n\u2022 Pi,j \u2208Rm\u00d7m interchanges row i and row j.\n\u2022 Di(\u03bb) \u2208Rm\u00d7m multiplies the ith row by \u03bb\n\u2022 Li,j(\u03bb) \u2208Rm\u00d7m add \u03bb \u00d7 ith row to the jth row\nLemma 3.3.", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_38", "text": "\u2022 Pi,j \u2208Rm\u00d7m interchanges row i and row j.\n\u2022 Di(\u03bb) \u2208Rm\u00d7m multiplies the ith row by \u03bb\n\u2022 Li,j(\u03bb) \u2208Rm\u00d7m add \u03bb \u00d7 ith row to the jth row\nLemma 3.3.\ni) det (Pi,j) = \u22121\nii) det (Di(\u03bb)) = \u03bb\niii) det (Li,j(\u03bb)) = 1\nProof.\ni) det (Pi,j) = det (Pi,jIm)\nthm 3.1\n=\n\u2212det(Im) = \u22121 (Pi,j exchanges two rows of Im, which is\nthe same as switching two of its columns)\nii) det (Di(\u03bb)) = det (Di(\u03bbIm)) = \u03bb det(Im\u22121) = \u03bb (Di(\u03bb) multiplies the ith row of Im by \u03bb, which\nis the same as multiplying its ith column)\niii) det (Li,j(\u03bb))\nlinearity\n=\ndet\n\u0000e1\n\u00b7 \u00b7 \u00b7\nen\n\u0001\n|\n{z\n}\nIn\n+\u03bb det\n\u00000\n\u00b7 \u00b7 \u00b7 0\niej\n0\n\u00b7 \u00b7 \u00b7 0\u0001\n|\n{z\n}\n0\n= 1 (Li,j(\u03bb) adds \u03bb \u00d7\nith row of In to the jth row of In, which is the same as adding \u03bb \u00d7 ith column to the jth\ncolumn of In)", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_39", "text": "22\nKALAN WALMSLEY\n\u25a1\nTheorem 3.4. Let A, B \u2208Rn\u00d7n, then det(AB) = det(A) det(B)\nProof. Case i) If B is not invertible, then AB is not invertible =\u21d2det(B) = 0 = det(AB)\nCase ii) If B is invertible, B = Ek \u00b7 \u00b7 \u00b7 E1, with elementary matrices E1, \u00b7 \u00b7 \u00b7 , Ek. So\ndet(AB) = det(AEk \u00b7 \u00b7 \u00b7 E1)\n=\n\uf8f1\n\uf8f4\n\uf8f2\n\uf8f4\n\uf8f3\n\u2212det(AEk \u00b7 \u00b7 \u00b7 E2) if E1 swaps rows\n\u03bb det(AEk \u00b7 \u00b7 \u00b7 E2) if E1 multiplies a row by \u03bb\ndet(AEk \u00b7 \u00b7 \u00b7 E2) if E1 is of type III)\nlem 3.3\n=\ndet(AEk \u00b7 \u00b7 \u00b7 E2) det(E1)\n= \u00b7 \u00b7 \u00b7\n= det(A) det(Ek) \u00b7 \u00b7 \u00b7 det(E1)\n|\n{z\n}\ndet(Ek\u00b7\u00b7\u00b7E1)\n= det(A) det(B)\n=\u21d2det(AB) = det(A) det(B)\n\u25a1\nRemark. Elementary matrices applied from the right act on columns\n\u0010\n(CE)T \u0011T\n=\n\u0000CT ET \u0001T (ET is\nacting on the rows of the transpose).\nRemark. det(A + B) \u0338= det(A) + det(B)\nTheorem 3.5. Let A \u2208Rn\u00d7n, then A is invertible iff det(A) \u0338= 0. In this case, det\n\u0000A\u22121\u0001\n=\u21d2\ndet\n\u0000A\u22121\u0001\n=\n1\ndet(A)\nProof.\n\u21d2Assume A is invertible, then\n1 = det(In) = det\n\u0000AA\u22121\u0001 thm 3.4\n=\ndet(A) det\n\u0000A\u22121\u0001\n=\u21d2det(A) \u0338= 0\ndet\n\u0000A\u22121\u0001\n=\n1\ndet(A)\n\u21d0Lemma 3.2\n\u25a1\nTheorem 3.6. Let A \u2208Rn\u00d7n, then det\n\u0000AT \u0001\n= det(A)\nRemark. Transpose of elementary matrices\ni) P T\ni,j = Pi,j =\u21d2det\n\u0000P T\ni,j\n\u0001\n= det (Pi,j)\nii) Di(\u03bb)T = Di(\u03bb) =\u21d2det\n\u0000Di(\u03bb)T \u0001\n= det (Di(\u03bb))", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_40", "text": "Remark. Transpose of elementary matrices\ni) P T\ni,j = Pi,j =\u21d2det\n\u0000P T\ni,j\n\u0001\n= det (Pi,j)\nii) Di(\u03bb)T = Di(\u03bb) =\u21d2det\n\u0000Di(\u03bb)T \u0001\n= det (Di(\u03bb))\niii) Li,j(\u03bb)T = Lj,i(\u03bb) =\u21d2det\n\u0000Li,j(\u03bb)T \u0001 lem 3.3\n=\n1 = det (Li,j(\u03bb))\nProof. Case I A not invertible =\u21d2AT not invertible =\u21d2det(A) = det\n\u0000AT \u0001\n= 0\nCase II A invertible\n=\u21d2A = Ek \u00b7 \u00b7 \u00b7 E1 =\u21d2AT = (Ek \u00b7 \u00b7 \u00b7 E1)T = ET\n1 \u00b7 \u00b7 \u00b7 ET\nk\n=\u21d2det(A) = det (Ek) \u00b7 \u00b7 \u00b7 det (E1)\n=\u21d2det\n\u0000AT \u0001\n= det\n\u0000ET\n1\n\u0001\n\u00b7 \u00b7 \u00b7 det\n\u0000ET\nk\n\u0001\n= det\n\u0000ET\nk\n\u0001\n\u00b7 \u00b7 \u00b7 det\n\u0000ET\n1\n\u0001\n= det (Ek) \u00b7 \u00b7 \u00b7 det (E1) = det(A)\nSo in all cases det\n\u0000AT \u0001\n= det(A)\n\u25a1", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_41", "text": "LECTURE NOTES\n23\n3.3. Computation of determinants. Recall that Ai,j is obtained from A by removing the ith row and\nthe jth column.\nDefinition. Given A \u2208Rn\u00d7n, we call ci,j = (\u22121)i+j det (Ai,j) the (i, j)-cofactor matrix of A. Note that\nby definition det(A) =\nnP\nk=1\na1,kc1,k (cofactor expansion accross the first row).\nTheorem 3.7. Let A \u2208Rn\u00d7n, then\ni) det(A) = ai,1ci,1 + ai,2ci,2 + \u00b7 \u00b7 \u00b7 + ai,nci,n (cofactor expansion accross the ith row)\nii) det(A) = a1,jc1,j + a2,jc2,j + \u00b7 \u00b7 \u00b7 + an,jcn,j (cofactor expansion down the jth column)\nProof.\ni) Column expansion down the ith column of A is the row expansion accross the ith row of\nAT .\nn\nX\nk=1\n(\u22121)i+k aT\ni,k det\n\u0010\u0000AT \u0001\ni,k\n\u0011\n|\n{z\n}\nrow expansion of AT\n=\nn\nX\nk=1\n(\u22121)i+k ak,i det\n\u0010\n(Ak,i)T \u0011\n=\nn\nX\nk=1\n(\u22121)i+k ak,i det (Ak,i)\n|\n{z\n}\ncolumn expansion of A\nSince det\n\u0000AT \u0001\n= det(A), column expansion down the first column holds.\nii) Show that column expansion holds down any column. Theorem 3.1 implies that\n\u22121\n\u21b7\n\u21b7\n\u21b7\n\u22121\n\u21b7\ndet\n\u0000Ai\nA1\n\u00b7 \u00b7 \u00b7\nAi\u22121\nAi+1\n\u00b7 \u00b7 \u00b7\nAn\n\u0001\n= (\u22121)i\u22121 det\n\u0000A1\n\u00b7 \u00b7 \u00b7\nAn\n\u0001\n= (\u22121)i\u22121 det(A)\ndown 1st column\n=\nn\nX\nk=1\n(\u22121)k+1 ak,i det (Ak,i) =\nn\nX\nk=1\nak,ick,i\n=\u21d2det(A) =\nn\nX\nk=1\n(\u22121)k+1 ak,i det (Ak,i) =\nn\nX\nk=1\nak,ick,i\niii) Show that row expansion holds accross any row. Apply the previous two claims together with\ndet\n\u0000AT \u0001\n= det(A)\n\u25a1", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_42", "text": "n\nX\nk=1\n(\u22121)k+1 ak,i det (Ak,i) =\nn\nX\nk=1\nak,ick,i\niii) Show that row expansion holds accross any row. Apply the previous two claims together with\ndet\n\u0000AT \u0001\n= det(A)\n\u25a1\nTheorem 3.8. Let A \u2208Rn\u00d7n and assume that B is obtained by applying an elementary transformation\non the rows (respectively the columns) of A, then\ndet(A) =\n\uf8f1\n\uf8f4\n\uf8f2\n\uf8f4\n\uf8f3\n\u2212det(B) if of type I\n\u03bb\u22121 det(B) if of type II\ndet(B) if of type III\nProof.\ni) E type I det(B) = det(EA) = det(E) det(A) = \u22121 \u00b7 det(A) =\u21d2det(A) = \u2212det(B)\nii) E type II det(B) = det(EA) = det(E) det(A) = \u03bb \u00b7 det(A) =\u21d2det(A) = \u03bb\u22121 det(B)\niii) E type III det(B) = det(EA) = det(E) det(A) = 1 \u00b7 det(A)\n=\u21d2\ndet(A) = det(B) (E is an\nupper/lower triangular matrix)\n\u25a1\nTheorem 3.9. Let A \u2208Rn\u00d7n be an upper or lower triangular matrix, then\ndet(A) = a1,1a2,2 \u00b7 \u00b7 \u00b7 an,n =\nn\nY\ni=1\nai,i", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_43", "text": "24\nKALAN WALMSLEY\nIn particular, such a matrix is invertible iff all elements on the diagonal are non-zero.\nProof.\ni) For a lower triangular matrix\ndet(A) = a1,1 det(A1,1) = a1,1\n\u0010\na2,2 det\n\u0010\n(A1,1)1,1\n\u0011\u0011\n= \u00b7 \u00b7 \u00b7 = a1,1 \u00b7 \u00b7 \u00b7 an,n =\nn\nY\ni=1\nai,i.\nii) For an upper triangular matrix det(A) = det\n\u0000AT\n|{z}\nlower\ntriangular\n\u0001\n=\nnQ\ni=1\nai,i.\n\u25a1\nDeterminants and volume.\nTheorem 3.10.\ni) Let v1, v2 \u2208R2 determine a parallelogram in R2. Then the area of that parallel-\nogram is given by |det (v1, v2)|\nii) Let w1, w2, w3 \u2208R3 determine a parallelepiped in R3. Then its volume is given by |det (w1, w2, w3)|\nRemark. Given v1, \u00b7 \u00b7 \u00b7 , vn \u2208Rn, they are linearly independent if and only if det(v1, \u00b7 \u00b7 \u00b7 , vn) \u0338= 0\nThe final theorem in this chapter is not relevant for the exam.\nTheorem 3.11. Let A \u2208Rn\u00d7n be invertible. Then\n\u0000A\u22121\u0001\ni,j =\n1\ndet(A)ci,j, where ci,j = (\u22121)i+j det (Ai,)", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_44", "text": "LECTURE NOTES\n25\n4. Vector spaces\nVector spaces are a generalization of Rn (adding vectors, multiplying by a scalar).\n4.1. Vector spaces and subspaces.\nDefinition. A (real) vector space V is a set of objects, called \"vectors\", s.t. \u2203two operations, called\n\"addition\" and \"multiplication by scalars\", satisfying the following properties \u2200u, v, w \u2208V, \u2200\u03bb, \u00b5 \u2208R\ni) u + v \u2208V\nii) u + v = v + u\niii) (u + v) + w = u + (v + w)\niv) \u2203zero vector 0 \u2208V or 0V s.t. u + 0 = u\nv) \u2200u \u2208V\n\u2203! \u2212u \u2208V s.t. u + (\u2212u) = 0V\nvi) \u03bbu \u2208V\nvii) (\u03bb + \u00b5)v = \u03bbv + \u00b5v\nviii) \u03bb(u + v) = \u03bbu + \u03bbv\nix) (\u03bb\u00b5)v = \u03bb(\u00b5v)\nx) 1 \u00b7 v = v\nExample. V = Rn\nExample. V = {f : R \u2192R} the set of all function from R to R\nExample. Let A be a non-empty set and V1 be a vector space, then V = {f : A \u2192V1}, the set of all\nfunctions from A to V1, is a vector space.\n(f + g)(x) = f(x) + g(x)\n(\u03bbf)(x) = \u03bbf(x)\n(The domain does not have to be a vector space but the codomain does.)\nExample. The set V = {A \u2208Rm\u00d7n} of m \u00d7 n matrices.\nLemma 4.1. Let V be a vector space. Then the element 0V is unique. Moreover, the element \u2212u is\nunique \u2200u \u2208V . Finally\nI) 0 \u00b7 u = 0V\nII) \u03bb0V = 0V\nIII) \u2212u = (\u22121)u\nProof. Let \u02dc0V be another zero vector, then\n\u02dc0V\niv)\n= \u02dc0V + 0V\nii)= 0V + \u02dc0V\niv)\n= 0V\nNext, assume that \u2212\u02dcu satisfies u + \u2212\u02dcu = 0V , then\n\u2212\u02dcu\niv)\n= \u2212\u02dcu + 0V\nv)= \u2212\u02dcu + (u + (\u2212u))\niii)\n= (\u2212\u02dcu + u) + (\u2212u)\nii)=\n(u + \u2212\u02dcu", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_45", "text": "iv)\n= 0V\nNext, assume that \u2212\u02dcu satisfies u + \u2212\u02dcu = 0V , then\n\u2212\u02dcu\niv)\n= \u2212\u02dcu + 0V\nv)= \u2212\u02dcu + (u + (\u2212u))\niii)\n= (\u2212\u02dcu + u) + (\u2212u)\nii)=\n(u + \u2212\u02dcu\n| {z }\n0V\n) + (\u2212u)\nv)= 0V + (\u2212u) = (\u2212u) + 0V = \u2212u\nFinally\ni)\n0 \u00b7 u = (0 + 0)u = 0 \u00b7 u + 0 \u00b7 u\n=\u21d20 \u00b7 u + \u2212(0 \u00b7 u)\n|\n{z\n}\n0V\n= 0 \u00b7 u + (0 \u00b7 u + \u2212(0 \u00b7 u)) = 0 \u00b7 u + 0V = 0 \u00b7 u\nii) \u03bb0V = \u03bb(0V + 0V ) = \u03bb0V + \u03bb0V Apply \u2212(\u03bb0V ) to both sides =\u21d20V = \u03bb0V\niii) u + (\u22121)u\nx)= 1 \u00b7 u + (\u22121) \u00b7 u = (1 + (\u22121))u = 0 \u00b7 u\ni)= 0V By uniqueness, \u2212u = \u22121 \u00b7 u\n\u25a1\nDefinition. Let V be a vector space and H \u2286V . We say that H is a subspace of V if\ni) 0V \u2208H\nii) u, w \u2208H =\u21d2u + w \u2208H\niii) \u03bb \u2208R, u \u2208H =\u21d2\u03bbu \u2208H", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_46", "text": "26\nKALAN WALMSLEY\nLemma 4.2. Let V be a vector space and H \u2286V be a subspace. Then H is itself a vector space with the\noperations inherited from V .\nProof. Properties ii), iii), vi), vii), viii), ix), and x) hold for all elements in V , so also for all elements in\nH.\ni)) Is true by item ii) of the definition.\niv)) 0V \u2208H by part i) and 0 + u = u\u2200u \u2208H since H \u2286V .\nv)) By lemma 4.1, \u2212u = (\u22121)u. If u \u2208H, then (\u22121)u \u2208H by part iii).\n\u25a1\nExample. Let P = {p : R \u2192R, p a polynomial}. This is a subspace of V = {f : R \u2192R}. Indeed, f(x) =\n0 is a polynomial. Moreover, the sum of two polynomials is again a polynomial, also if you multiply a\npolynomial by any \u03bb \u2208R.\nSubspace spanned by a set.\nDefinition. Let V be a vector space and {v1, \u00b7 \u00b7 \u00b7 , vk} \u2286V . We define\nspan {v1, \u00b7 \u00b7 \u00b7 , vk} =\n( k\nX\nn=1\n\u03bbnvn | \u03bb1, \u00b7 \u00b7 \u00b7 , \u03bbk \u2208R\n)\nWe call span {v1, \u00b7 \u00b7 \u00b7 , vk} the subspace spanned by v1, \u00b7 \u00b7 \u00b7 , vk.\nRemark. One can also define span W where W \u2286V contains infinitely many elements. In thes setting,\nlinear combinations are elements of the form\nm\nP\nn=1\n\u03bbnvn, \u03bb \u2208R, v1, \u00b7 \u00b7 \u00b7 , vn \u2208W, m \u2208N i.e. the number\nof vectors has to be finite but can vary.\nTheorem 4.3. Let V be a vector space and v1, \u00b7 \u00b7 \u00b7 , vk \u2208V . Then span {v1, \u00b7 \u00b7 \u00b7 , vk} is a subspace of V .\nIt is the smallest subspace that contains v1, \u00b7 \u00b7 \u00b7 , vk.\nProof.\ni) 0V =\nkP\nn=1\n0 \u00b7 vn \u2208span {v1, \u00b7 \u00b7 \u00b7 , vk}\nii) Let u =\nkP\nn=1\n\u03bbnvn, w =\nkP\nn=1", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_47", "text": "Proof.\ni) 0V =\nkP\nn=1\n0 \u00b7 vn \u2208span {v1, \u00b7 \u00b7 \u00b7 , vk}\nii) Let u =\nkP\nn=1\n\u03bbnvn, w =\nkP\nn=1\n\u00b5nvn. Then u + w\ndef\n=\nkP\nn=1\n(\u03bbn + \u00b5n) vn \u2208span {v1, \u00b7 \u00b7 \u00b7 , vk}\niii) \u03bbu =\nkP\nn=1\n(\u03bb\u03bbn\n|{z}\n\u2208R\n)vn \u2208span {v1, \u00b7 \u00b7 \u00b7 , vk}\nNow let W \u2286V be any subspace that contains v1, \u00b7 \u00b7 \u00b7 , vk. By part iii) \u03bb1v1, \u00b7 \u00b7 \u00b7 , \u03bbkvk \u2208W. By part\nii) \u03bb1v1 + \u03bb2v2 \u2208W\niterate\n=\u21d2\u03bb1v1 + \u00b7 \u00b7 \u00b7 + \u03bbkvk \u2208W. This holds \u2200\u03bb1, \u00b7 \u00b7 \u00b7 , \u03bbk \u2208R =\u21d2span {v1, \u00b7 \u00b7 \u00b7 , vk} \u2286\nW.\n\u25a1\nDefinition. Given a subspace H \u2286V , a spanning (or generating) set for H is any set {v1, \u00b7 \u00b7 \u00b7 , vk} s.t.\nspan {v1, \u00b7 \u00b7 \u00b7 , vk} = H.\nRemark. A spanning set is in general not unique. Take as an example span {e1} = span {\u2212e1} in R2 or\nspan {e1, e2} = span\n\u001a\u00121\n1\n\u0013\n,\n\u0012 1\n\u22121\n\u0013\u001b\n.\nExample. Let V = P be the vector space of all polynomials and let monomials p0(x) = 1, p1(x) =\nx, p2(x) = x2, \u00b7 \u00b7 \u00b7 , pn(x) = xn. Then span {p0, \u00b7 \u00b7 \u00b7 , pn} = {p \u2208P | p of deg at most n}.\nExample. Let V = R3 and H =\n\uf8f1\n\uf8f2\n\uf8f3x =\n\uf8eb\n\uf8ed\na \u2212b\n2b\na + b\n\uf8f6\n\uf8f8| a, b \u2208R\n\uf8fc\n\uf8fd\n\uf8fe. Then we have\n\uf8eb\n\uf8ed\na \u2212b\n2b\na + b\n\uf8f6\n\uf8f8= a\n\uf8eb\n\uf8ed\n1\n0\n1\n\uf8f6\n\uf8f8+\nb\n\uf8eb\n\uf8ed\n\u22121\n2\n1\n\uf8f6\n\uf8f8=\u21d2H = span\n\uf8f1\n\uf8f2\n\uf8f3\n\uf8eb\n\uf8ed\n1\n0\n1\n\uf8f6\n\uf8f8,\n\uf8eb\n\uf8ed\n\u22121\n2\n1\n\uf8f6\n\uf8f8\n\uf8fc\n\uf8fd\n\uf8fe=\u21d2H is a subspace of R3", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_48", "text": "LECTURE NOTES\n27\nLemma 4.4. Let V be a vector space and S, T \u2286V be two subspaces of V , then\na) S \u2229T = {x \u2208V | x \u2208S \u2227x \u2208T} is a subspace of V\nb) S + T = {s + t | s \u2208S, t \u2208T} is a subspace of V\nProof.\na)\ni) Since the zero element 0V is in both S and T, it is also in S \u2229T\nii) If u and v are in S \u2229T, they are both in S and T, so u + v \u2208S \u2229T\niii) If u is in S \u2229T and \u03bb \u2208R, then \u03bb \u00b7 u is in both S and T, so it is in S \u2229T\nb)\ni) Since S and T are both subspaces, they both contain the zero vector 0V , hence S+T contains\n0V + 0V = 0V .\nii) If s1 + t1 and s2 + t2 are both elements of S + T, then we have that (s1 + t1) + (s2 + s2) =\n(s1 + s2) + (t1 + t2) is an element of S + T because s1 + s2 \u2208S and t1 + t2 \u2208T.\niii) If s + t is an element of S + T and \u03bb \u2208R, then we have that \u03bb \u00b7 (s + t) = (\u03bb \u00b7 s) + (\u03bb \u00b7 t) is\nan element of S + T because \u03bb \u00b7 s \u2208S and \u03bb \u00b7 t \u2208T.\n\u25a1\n4.2. Null spaces, column spaces, linear transformations.\nDefinition. The null space (kernel) of a matrix A \u2208Rm\u00d7n is defined as Nul (A) = Ker (A) = {x \u2208Rn | Ax = 0}.\nTheorem 4.5. Let A \u2208Rm\u00d7n. Then Nul (A) is a subspace of Rn.\nProof.\n\u2022 0 \u2208Nul (A) since A \u00b7 0\n\u2191\n\u2208Rn\n=\n\u2208Rm\n\u2193\n0.\n\u2022 Let x, y \u2208Rn be s.t. Ax = 0 and Ay = 0, then A(x + y) = Ax + Ay = 0 + 0 = 0 =\u21d2x + y \u2208\nNul (A).", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_49", "text": "\u2191\n\u2208Rn\n=\n\u2208Rm\n\u2193\n0.\n\u2022 Let x, y \u2208Rn be s.t. Ax = 0 and Ay = 0, then A(x + y) = Ax + Ay = 0 + 0 = 0 =\u21d2x + y \u2208\nNul (A).\n\u2022 Let \u03bb \u2208R, then x \u2208Nul (A) =\u21d2A(\u03bbx) = \u03bbAx = \u03bb \u00b7 0 = 0 =\u21d2\u03bbx \u2208Nul (A).\n\u25a1\nDefinition. The column space (range) of A \u2208Rm\u00d7n is defined as Col (A) = Ran (A) = span {A1, \u00b7 \u00b7 \u00b7 , An},\nwhere A1, \u00b7 \u00b7 \u00b7 , An are the columns of A.\nTheorem 4.6. Let A \u2208Rm\u00d7n, then Col (A) is a subspace of Rm.\nProof. See theorem 4.3.\n\u25a1\nRemark. Clearly Col (A) = {b \u2208Rm | b = Ax for some x \u2208Rn}\nKernel and range of linear transformations.\nDefinition. Let V, W be vector spaces. A function f : V \u2192W is called linear if\ni) f(u\nin V\n\u2193\n+ v) = f(u)\nin W\n\u2193\n+ f(v)\n\u2200u, v \u2208V\nii) f(\u03bb\nin V\n\u2193\u00b7 u) = \u03bb\nin W\n\u2193\u00b7 f(u)\n\u2200\u03bb \u2208R, \u2200u \u2208V\nDefinition. The kernel of a linear transformation f is defined by\nKer (f) = {v \u2208V | f(v) = 0W }\nDefinition. The range of a linear transformation f is defined by\nRan (f) = {w \u2208W | w = f(v) for some v \u2208V }\nTheorem 4.7. Let V, W be vector spaces and f : V \u2192W be linear. Then Ker (f) is a subspace of V and\nRan (f) is a subspace of W.\nProof. Kernel\n\u2022 0V \u2208Ker (f) since f(0V ) = f(0 \u00b7 0V )\nf linear\n=\n0 \u00b7 f(0V ) = 0W .\n\u2022 Let x, y \u2208Ker (f), then f(x + y)\nf linear\n=\nf(x) + f(y) = 0W + 0W = 0W =\u21d2x + y \u2208Ker (f).", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_50", "text": "28\nKALAN WALMSLEY\n\u2022 Let \u03bb \u2208R, x \u2208Ker (f), then f(\u03bbx)\nf linear\n=\n\u03bbf(x) = \u03bb0W = 0W =\u21d2\u03bbx \u2208Ker (f).\nRange\n\u2022 0W \u2208Ran (f) since f(0V ) = f(0 \u00b7 0V )\nf linear\n=\n0 \u00b7 f(0V ) = 0W .\n\u2022 If w1, w2 \u2208Ran (f), then \u2203v1, v2 \u2208V s.t. w1 = f(v1) and w2 = f(v2). So f(v1 + v2)\nf linear\n=\nf(v1) + f(v2) = w1 + w2 =\u21d2w1 + w2 \u2208Ran (f).\n\u2022 For \u03bb \u2208R, f(\u03bbv)\nf linear\n=\n\u03bbf(v) = \u03bbw =\u21d2\u03bbw \u2208Ran (f).\n\u25a1\nExample. V = W = P, the space of all polynomials. Define T : P \u2192P, T(p)(x) = p\u2032(x). We must verify\nthat T(p1 + p2) = T(p1) + T(p2) and that T(\u03bbp) = \u03bbT(p).\nLet p1 =\nnP\ni=0\naixi, p2 =\nm\nP\ni=0\nbixi and wlog n = m (some ai, bi can be zero).\n=\u21d2(p1 + p2)(x) =\nn\nX\ni=0\n(ai + bi)xi =\u21d2T(p1 + p2) =\nn\nX\ni=0\n(ai + bi)ixi\u22121\n=\nn\nX\ni=0\naiixi\u22121 +\nn\nX\ni=0\nbiixi\u22121 = T(p1)(x) + T(p2)(x)\nSimilarly T(\u03bbp1)(x) =\nnP\ni=0\n\u03bbaiixi\u22121 = \u03bb\nnP\ni=0\naiixi\u22121 = \u03bbT(p1)(x) =\u21d2T is linear.\nKer (T) = {p \u2208P | p\u2032(x) = 0\n\u2200x \u2208R} = {p \u2208P | p consant}\nRan (T) = P\nThis last statement because given p \u2208P, the function q(x) =\nxR\n0", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_51", "text": "Ker (T) = {p \u2208P | p\u2032(x) = 0\n\u2200x \u2208R} = {p \u2208P | p consant}\nRan (T) = P\nThis last statement because given p \u2208P, the function q(x) =\nxR\n0\np(t) dt is a polynomial and q\u2032(x) = p(x).\nRemark. T : P \u2192P is surjective, but not injective (non-trivial kernel), which is very different from the\ncase f : Rn \u2192Rn linear.\nLemma 4.8. Let f : V \u2192W be linear. Then f is injective if and only if Ker (f) = {0V }.\nProof.\n\u21d2Assume that f is injective. If f(x) = 0, then f(x)\nf linear\n=\nf(0) =\u21d2x\nf injective\n=\n0.\n\u21d0Assume that Ker (f) = {0} and that f(x) = f(y), then we have 0 = f(x) \u2212f(y)\nf linear\n=\nf(x \u2212\ny) =\u21d2x \u2212y = 0 =\u21d2x = y.\n\u25a1\n4.3. Linearly independent sets and bases.\nDefinition. Let V be a vector space and v1, \u00b7 \u00b7 \u00b7 , vk \u2208V . Then family {v1, \u00b7 \u00b7 \u00b7 , vk} is called linearly\nindependent if the equation \u03bb1v1 + \u00b7 \u00b7 \u00b7 + \u03bbkvk has only the solution \u03bb1 = \u00b7 \u00b7 \u00b7 = \u03bbk = 0. Otherwise they\nare called linearly dependent.\nRemark. If S \u2286V contains infinitely many elements, one calls S linearly independent if any finite\nsubset is linearly independent.\nDefinition. Let V be a vector space and B \u2286V . We call B a basis for V if\ni) B is linearly independent\nii) span (B) = V\nRemark. We will focus mostly on the case where B = {b1, \u00b7 \u00b7 \u00b7 , bk} is a finite set.\nExample. Let P be the space of all polynomials, then the set of all monomials,\n\b\np0(x) = 1, p1(x) = x, p2(x) = x2, \u00b7 \u00b7 \u00b7\n\t\n,\nis a basis for P. To prove this, we must prove linear independence for this set and that it spans V .", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_52", "text": "LECTURE NOTES\n29\n\u2022 It suffices to prove\nnP\ni=0\n\u03bbixi = 0 \u2200x \u2208R =\u21d2\u03bb0 = \u03bb1 = \u00b7 \u00b7 \u00b7 = \u03bbn = 0. Try x = 0 =\u21d2\u03bb0 = 0 and\n\u03bb1x + \u00b7 \u00b7 \u00b7 + \u03bbnxn = 0\n\u2200x \u2208R. Then, take the 1st derivative: \u03bb1 + 2\u03bb2x + 3\u03bb3x2 + \u00b7 \u00b7 \u00b7 n\u03bbnxn\u22121 =\n0\n\u2200x \u2208R. Try again x = 0 =\u21d2\u03bb1 = 0. Iterate to get \u03bb0 = \u00b7 \u00b7 \u00b7 = \u03bbn = 0.\nExample. Let {v1, \u00b7 \u00b7 \u00b7 , vn} \u2286Rn, then {v1, \u00b7 \u00b7 \u00b7 , vn} is a basis if and only if det(v1, \u00b7 \u00b7 \u00b7 , vn) \u0338= 0. This\nfollows from the invertible matrix theorem. In particular {e1, \u00b7 \u00b7 \u00b7 , en}\n|\n{z\n}\nstandard basis\nis a basis for Rn.\nTheorem 4.9. Let V be a vector space and {v1, \u00b7 \u00b7 \u00b7 , vk} \u2286V with k \u22652. Then this set is linearly\ndependent iff \u2203j \u2208[[1, k]] s.t. vj is a linear combination of the other vectors.\nProof. Same as in Rn.\n\u25a1\nRemark. A set containing one vector is linearly independent iff this vector is not 0V . Indeed, if \u03bbv = 0V\nand \u03bb \u0338= 0, then we have that v =\n\u0000\u03bb\u22121\u03bb\n\u0001\nv = \u03bb\u22121 (\u03bbv) = \u03bb\u221210V = 0V .\nThe spanning set theorem.\nTheorem 4.10. Let V be a vector space, V = span {v1, \u00b7 \u00b7 \u00b7 , vk}, then\ni) If \u2203vj that is a linear combination of the other vectors, then V = span {v1, \u00b7 \u00b7 \u00b7 , vj\u22121, vj+1, \u00b7 \u00b7 \u00b7 , vk}\nii) If V \u0338= {0}, then some subset of {v1, \u00b7 \u00b7 \u00b7 , vk} forms a basis for V .\nProof.\ni) Clearly span {v1, \u00b7 \u00b7 \u00b7 , vj\u22121, vj+1, \u00b7 \u00b7 \u00b7 , vk} \u2286V . Now let v \u2208V , v =\nkP\nn=1\n\u03bbnvn, vj =\nkP\nn=1\nn\u0338=j", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_53", "text": "Proof.\ni) Clearly span {v1, \u00b7 \u00b7 \u00b7 , vj\u22121, vj+1, \u00b7 \u00b7 \u00b7 , vk} \u2286V . Now let v \u2208V , v =\nkP\nn=1\n\u03bbnvn, vj =\nkP\nn=1\nn\u0338=j\n\u00b5nvn\n=\u21d2v =\nk\nX\nn=1\nn\u0338=j\n\u03bbnvn + \u03bbj\nk\nX\nn=1\nn\u0338=j\n\u00b5nvn \u2208span {v1, \u00b7 \u00b7 \u00b7 , vj\u22121, vj+1, \u00b7 \u00b7 \u00b7 , vk}\nii) If {v1, \u00b7 \u00b7 \u00b7 , vk} is linearly independent, we are done. If not, by theorem 4.9 and part I, we can\nremove one of the vectors, say vj, and still have a generating set. If this set is linearly independent,\nwe are done. Otherwise, iterate. This way, we either find a basis or we end with a single vector that\ngenerates V . The vector cannot be zero since V \u0338= {0} =\u21d2this vector is linearly independent.\n\u25a1\nBases for Nul (A) and Col (A). Using the parametric vector form, for some vectors v1, \u00b7 \u00b7 \u00b7 , vk \u2208Rn we\ncan write Nul (A) = span {v1, \u00b7 \u00b7 \u00b7 , vk} (with k the number of free variables). Are these vectors linearly\nindependent ? Let xi1, \u00b7 \u00b7 \u00b7 , xik be the free variables.\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\nx1\n...\nxi\n...\nxn\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n= xi1\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n...\n...\nv1\n...\n...\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n+ xi2\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n...\n...\nv2\n...\n...\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n+ \u00b7 \u00b7 \u00b7 + xik\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n...\n...\nvk\n...\n...\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n\u2022 Check the i1th coordinate: v1 has entry 1 and v2, \u00b7 \u00b7 \u00b7 , vk have entry 0.\n\u2022 Check the i2th coordinate: v2 has entry 1 and v1, v3, \u00b7 \u00b7 \u00b7 , vk have entry 0.\n\u2022 In total we have k such coordinates\n=\u21d2if \u03bb1v1 + \u00b7 \u00b7 \u00b7 + \u03bbkvk = 0, we get \u03bb1 = \u00b7 \u00b7 \u00b7 = \u03bbk = 0.", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_54", "text": "30\nKALAN WALMSLEY\nA basis for Nul (A) is given by the vectors appearing in the parametric vector form.\nExample.\nA =\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ed\n1\n2\n0\n4\n0\n0\n0\n0\n1\n2\n0\n0\n0\n0\n0\n0\n1\n0\n0\n0\n0\n0\n0\n1\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f8\n=\u21d2\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\nx1\nx2\nx3\nx4\nx5\nx6\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n=\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n\u22122x2 \u22124x4\nx2\n\u22122x4\nx4\n0\n0\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n= x2\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n\u22122\n1\n0\n0\n0\n0\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n+ x4\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n\u22124\n0\n\u22122\n1\n0\n0\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n=\u21d2B =\n\uf8f1\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f2\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f3\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n\u22122\n1\n0\n0\n0\n0\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n,\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n\u22124\n0\n\u22122\n1\n0\n0\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n\uf8fc\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8fd\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8fe\nis a basis for Nul (A)\nNow let us look at Col (A): we need to find a column that is a linear combination of the other columns\nand remove it, then iterate. Note that row reduction does not change the set of solutions of Ax = 0 (it\ndoes for Ax = b). =\u21d2If x1A1 + \u00b7 \u00b7 \u00b7 + xnAn = 0 with some xi \u0338= 0, then also x1 \u02dcA1 + \u00b7 \u00b7 \u00b7 + xn \u02dcAn = 0\nwith same xi \u0338= 0. So Ai or \u02dcAi is a linear combination of the other columns of A or \u02dcA.\nLet \u02dcA be the reduced echelon form of A. The pivot columns (the columns corresponding to the basic\nvariables) are linearly independent. Moreover, any non-pivot column of \u02dcA is a linear combination of the\npivot columns (e1, \u00b7 \u00b7 \u00b7 , ek) because the all belong to Rk \u00d7 {0}m\u2212k (add m \u2212k zeros).\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n0\n1\n\u22c6\n0\n\u22c6\n...\n...\n0\n1\n\u22c6\n0\n\u22c6\n...\n...\n0\n1\n\u22c6\n0\n\u22c6\n...\n...\n0\n1\n\u22c6\n\u22c6\n...\n...\n...\n0\n...\n...\n...\n...\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\nTheorem 4.11. The pivot columns of A (not its reduced echelon form) form a basis for the column space\nCol (A).\nProof. Above picture.\n\u25a1\nLinear dependence for polynomials and functions. Given f1, \u00b7 \u00b7 \u00b7 , fn : R \u2192R, linear dependence\nboils down to studying\n\u03bb1f1(x) + \u00b7 \u00b7 \u00b7 + \u03bbnfn(x) = 0\n\u2200x \u2208R\nPossible strategies\n\u2022 Insert some well-chosen x and obtain a linear system", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_55", "text": "LECTURE NOTES\n31\nExample. \u03bb1 sin (x) + \u03bb2 cos (x) = 0\n\u2200x \u2208R\n\u2013 x = 0: \u03bb1 sin (0) + \u03bb2 cos (0) = 0 =\u21d2\u03bb2 = 0\n\u2013 x = \u03c0\n2 : \u03bb1 sin ( \u03c0\n2 ) + \u03bb2 cos ( \u03c0\n2 ) = 0 =\u21d2\u03bb1 = 0\n\u2022 Take derivatives (see example of polynomials)\n\u2022 Use tricks in the following examples\nExample. f1(x) = e2xx, f2(x) = e2xx2, f3(x) = e2xx3\n\u03bb1f1(x) + \u03bb2f2(x) + \u03bb3f3(x) = e2x \u0000\u03bb1x + \u03bb2x2 + \u03bb3x3\u0001\n= 0\n\u2200x \u2208R\ne2x\u0338=0 \u2200x\n=\u21d2\n\u03bb1x + \u03bb2x2 + \u03bb3x3 = 0\n\u2200x \u2208R\nmonomials linearly\nindependent\n=\u21d2\n\u03bb1 = \u03bb2 = \u03bb3 = 0\nExample. p1(x) = 1 + x, p2(x) = x + x2, p3(x) = 1 \u2212x3\n0 = \u03bb1p1(x) + \u03bb2p2(x) + \u03bb3p3(x) = 1 \u00b7 (\u03bb1 + \u03bb3) + x \u00b7 (\u03bb1 + \u03bb2) + x2 \u00b7 \u03bb2 + x3 \u00b7 (\u2212\u03bb3)\nmonomials linearly\nindependent\n=\u21d2\n\uf8f1\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f2\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f3\n\u03bb1 + \u03bb3 = 0\n\u03bb1 + \u03bb2 = 0\n\u03bb2 = 0\n\u2212\u03bb3 = 0\n=\u21d2\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ed\n1\n0\n1\n0\n1\n1\n0\n0\n0\n1\n0\n0\n0\n0\n1\n0\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f8=\u21d2\u03bb1 = \u03bb2 = \u03bb3 = 0\nA similar method works for testing whether p3 \u2208span {p1, p2} \u21d0\u21d2\u03bb1p1 + \u03bb2p2 = p3.\n4.4. Coordinate systems.\nTheorem 4.12. Let V be a vector space and B = {b1, \u00b7 \u00b7 \u00b7 , bn} be a basis for V . Then \u2200v \u2208V\n\u2203!\u03bb1, \u00b7 \u00b7 \u00b7 , \u03bbn \u2208R s.t. v = \u03bb1b1 + \u00b7 \u00b7 \u00b7 + \u03bbnbn.\nProof.\n\u2022 Existence ? B spans V\n\u2022 Uniqueness ? Assume that also v =\nnP\ni=1\n\u00b5ibi\n=\u21d20 = v \u2212v =\nn\nX\ni=1", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_56", "text": "Proof.\n\u2022 Existence ? B spans V\n\u2022 Uniqueness ? Assume that also v =\nnP\ni=1\n\u00b5ibi\n=\u21d20 = v \u2212v =\nn\nX\ni=1\n\u03bbibi \u2212\nn\nX\ni=1\n\u00b5ibi =\nn\nX\ni=1\n(\u03bbi \u2212\u00b5i)bi\nB linearly independent =\u21d2\u03bbi \u2212\u00b5i = 0\n\u2200i =\u21d2\u03bbi = \u00b5i\n\u25a1\nDefinition. If V is a vector space and B = {b1, \u00b7 \u00b7 \u00b7 , bn} is a basis with a given order (changing the\norder leads to a different basis), the coordinates of v \u2208V relative to B are the scalars \u03bb1, \u00b7 \u00b7 \u00b7 , \u03bbn \u2208R s.t.\nv = \u03bb1b1 + \u00b7 \u00b7 \u00b7 + \u03bbnbn.\nThey form the vector [v]B =\n\uf8eb\n\uf8ec\n\uf8ed\n\u03bb1\n...\n\u03bbn\n\uf8f6\n\uf8f7\n\uf8f8\u2208Rn, the coordinate vector of v relative to B (or B-coordinate\nvector).\nExample. B = {e1, \u00b7 \u00b7 \u00b7 , en} standard basis for Rn. v =\n\uf8eb\n\uf8ec\n\uf8ed\nv1\n...\nvn\n\uf8f6\n\uf8f7\n\uf8f8\u2208Rn =\u21d2[v]B = v since v = v1e1 +\n\u00b7 \u00b7 \u00b7 + vnen.", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_57", "text": "32\nKALAN WALMSLEY\nExample. V = R2, B =\n\u001a\u0012\n1\n0\n\u0013\n,\n\u0012\n1\n1\n\u0013\u001b\nis a basis since det\n\u0012\n1\n1\n0\n1\n\u0013\n\u0338= 0.\nGiven B and x, we can compute [x]B by solving\n\u0000b1\n\u00b7 \u00b7 \u00b7\nbn\n\u0001\n[x]B = x \u21d0\u21d2\n\u0000b1\n\u00b7 \u00b7 \u00b7\nbn\nx\u0001\nExample. x =\n\u0012\n3\n2\n\u0013\n=\u21d2\u03bb1b1 + \u03bb2b2 = x \u21d0\u21d2\n\u0012\n1\n1\n3\n0\n1\n2\n\u0013\n\u2192\n\u0012\n1\n0\n1\n0\n1\n2\n\u0013\n\u21d2\u03bb1 = 1, \u03bb2 = 2\n=\u21d2\n[x]B =\n\u00121\n2\n\u0013\nGiven B and [x]B, we can compute x by the formula\nx = ([x]B)1 b1 + \u00b7 \u00b7 \u00b7 + ([x]B)n bn\nExample. [x]B =\n\u00122\n2\n\u0013\n=\u21d2x = 2b1 + 2b2 = 2\n\u00121\n0\n\u0013\n+ 2\n\u00121\n1\n\u0013\n=\n\u00124\n2\n\u0013\nCoordinates in Rn. Let B be a basis for Rn, B = {b1, \u00b7 \u00b7 \u00b7 , bn} and let x \u2208Rn. To find [x]B, we need to\nsolve \u03bb1b1 + \u00b7 \u00b7 \u00b7 + \u03bbnbn = x \u21d0\u21d2\n\u0000b1\n\u00b7 \u00b7 \u00b7\nbn\n\u0001\n\uf8eb\n\uf8ec\n\uf8ed\n\u03bb1\n...\n\u03bbn\n\uf8f6\n\uf8f7\n\uf8f8= x.\nDefinition. Let B = {b1, \u00b7 \u00b7 \u00b7 , bn} be an orderred basis for Rn. The matrix PB =\n\u0000b1\n\u00b7 \u00b7 \u00b7\nbn\n\u0001\n\u2208Rn\u00d7n is\ncalled the change-of-coordinates matrix from B to the standard basis (it satisfies the system x = PB [x]B).\nSince B is a basis, the columns of PB are linearly independent, which implies that PB is invertible\n=\u21d2[x]B = P \u22121\nB x.\nThe coordinate mapping. x 7\u2192[x]B \"identifies\" the abstract space V with Rn.\nTheorem 4.13. Let V be a vector space and B = {b1, \u00b7 \u00b7 \u00b7 , bn} be a basis for V . Then the map x 7\u2192[x]B\nis a bijective, linear map from V to Rn. Also, the inverse map is linear.\nProof. To prove this theorem, we must prove that the map x 7\u2192[x]B is linear, injective and surjective\n(bijective), and that its inverse is linear.\nLinearity: We must show [\u03bbx + y]B = \u03bb [x]B + [y]B", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_58", "text": "(bijective), and that its inverse is linear.\nLinearity: We must show [\u03bbx + y]B = \u03bb [x]B + [y]B\n\u2200\u03bb \u2208R, x, y \u2208V . Let x = \u03b11b1 + \u00b7 \u00b7 \u00b7 + \u03b1nbn, y =\n\u03b21b1 + \u00b7 \u00b7 \u00b7 + \u03b2nbn, then\n\u03bbx + y = \u03bb\u03b11b1 + \u00b7 \u00b7 \u00b7 + \u03bb\u03b1nbn + y = \u03b21b1 + \u00b7 \u00b7 \u00b7 + \u03b2nbn\ncalculus\nrules in V\n=\n(\u03bb\u03b11 + \u03b21)b1 + \u00b7 \u00b7 \u00b7 + (\u03bb\u03b1n + \u03b2n)bn =\u21d2[\u03bbx + y]B =\n\uf8eb\n\uf8ec\n\uf8ed\n\u03bb\u03b11 + \u03b21\n...\n\u03bb\u03b1n + \u03b2n\n\uf8f6\n\uf8f7\n\uf8f8\n= \u03bb\n\uf8eb\n\uf8ec\n\uf8ed\n\u03b11\n...\n\u03b1n\n\uf8f6\n\uf8f7\n\uf8f8+\n\uf8eb\n\uf8ec\n\uf8ed\n\u03b21\n...\n\u03b2n\n\uf8f6\n\uf8f7\n\uf8f8= \u03bb [x]B + [y]B\nInjectivity: Assume [x]B = [y]B, then\nx = ([x]B)1\n| {z }\n([y]B)1\nb1 + \u00b7 \u00b7 \u00b7 + ([x]B)n\n| {z }\n([y]B)n\nbn = ([y]B)1 b1 + \u00b7 \u00b7 \u00b7 + ([y]B)n bn", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_59", "text": "LECTURE NOTES\n33\nSurjectivity: Let z =\n\uf8eb\n\uf8ec\n\uf8ed\nz1\n...\nzn\n\uf8f6\n\uf8f7\n\uf8f8\u2208Rn. Define x = z1b1 + \u00b7 \u00b7 \u00b7 + znbn \u2208V , then [x]B = z.\nLinearity of the inverse: Let T : V \u2192W be linear and bijective, then T \u22121 : W \u2192V is also linear. Let\n\u03bb \u2208R, w1, w2 \u2208W, then\nT\n\u0000\u03bbT \u22121(w1) + T \u22121(w2)\n\u0001\nT linear\n\u2193= \u03bbT\n\u0000T \u22121(w1)\n\u0001\n+ T\n\u0000T \u22121(w2)\n\u0001\n= \u03bbw1 + w2\n=\u21d2T \u22121(\u03bbw1 + w2) = \u03bbT \u22121(w1) + T \u22121(w2)\n\u25a1\nCorollary 4.14. Let V be a vector space and B = {b1, \u00b7 \u00b7 \u00b7 , bn} be a basis. Let further S = {v1, \u00b7 \u00b7 \u00b7 , vk} \u2286\nV . Then\ni) S is linearly independent iff {[v1]B , \u00b7 \u00b7 \u00b7 , [vk]B} is linearly independent in Rn.\nii) S generates V iff {[v1]B , \u00b7 \u00b7 \u00b7 , [vk]B} generates Rn.\nProof.\ni) Assume that S is linearly independent. Since the coordinate map x 7\u2192[x]B is injective, we\nhave that [v1]B , \u00b7 \u00b7 \u00b7 , [vk]B is also linearly independent.\nThe other implication follows the same way since the inverse of the coordinate map is also\nlinearly independent (theorem 4.13).\nii) Let S generate V and let x \u2208Rn. Let T \u22121 : Rn \u2192V be the inverse coordinate map, then T \u22121(x) \u2208\nV =\u21d2T \u22121(x)\u03bb1v1 + \u00b7 \u00b7 \u00b7 + \u03bbnvn for some \u03bb1, \u00b7 \u00b7 \u00b7 , \u03bbn \u2208R. =\u21d2x =\n\u0002\nT \u22121(x)\n\u0003\nB =\n\u2191\n[\u00b7\u00b7\u00b7 ]B linear\n\u03bb1 [v1]B + \u00b7 \u00b7 \u00b7 +\n\u03bbk [vk]B =\u21d2[v1]B , \u00b7 \u00b7 \u00b7 , [vk]B generate Rn.\nThe other implication follows the same way since T \u22121 is also linear and T \u22121 (T(v)) = v.", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_60", "text": "\u03bbk [vk]B =\u21d2[v1]B , \u00b7 \u00b7 \u00b7 , [vk]B generate Rn.\nThe other implication follows the same way since T \u22121 is also linear and T \u22121 (T(v)) = v.\n(T \u22121 (T(v)) = \u02dcv, where \u02dcv is the unique element s.t. T (\u02dcv) = T(v)\ninjective\n=\u21d2\n\u02dcv = v.)\n\u25a1\nRemark. In particular, S is a basis iff [v1]B , \u00b7 \u00b7 \u00b7 , [vk]B are a basis for Rn.\nExample. V = P2\nB =\n\b\np0(x) = 1, p2(x) = x, p3(x) = x2\t\nIs\n\b\n1 + 2x2, 4 + x + 5x2, 3 + 2x\n\t\na basis for P2 ?\n\u0002\n1 + 2x2\u0003\nB =\n\uf8eb\n\uf8ed\n1\n0\n2\n\uf8f6\n\uf8f8\n\u0002\n4 + x + 5x2\u0003\nB =\n\uf8eb\n\uf8ed\n4\n1\n5\n\uf8f6\n\uf8f8\n[3 + 2x]B =\n\uf8eb\n\uf8ed\n3\n2\n0\n\uf8f6\n\uf8f8\n=\u21d2\n\uf8eb\n\uf8ed\n1\n4\n3\n0\n1\n2\n2\n5\n0\n\uf8f6\n\uf8f8\u2192\n\uf8eb\n\uf8ed\n1\n4\n3\n0\n1\n2\n0\n0\n0\n\uf8f6\n\uf8f8=\u21d2not invertible\n=\u21d2not linearly independent\n4.5. The dimension of a vector space.\nTheorem 4.15. Let V be a vector space with a basis B = {b1, \u00b7 \u00b7 \u00b7 , bn}, then any subset of V that contains\nmore than n elements has to be linearly independent.\nProof. Let S be such a set. In the case where S has infinitely many elements, we replace it by a finite\nsubset that contains more than n elements.\nWrite S = {v1, \u00b7 \u00b7 \u00b7 , vk} with k > n\n=\u21d2\n[v1]B , \u00b7 \u00b7 \u00b7 , [vk]B \u2208Rn has to be linearly dependent since\nk > n (theorem 1.8).\ncorol. 4.14\n=\u21d2\nv1, \u00b7 \u00b7 \u00b7 , vk is linearly dependent.\n\u25a1\nCorollary 4.16. Let V be a vector space and B a basis with n elements. Then any other basis also has\nn elements.\nIf B has infinitely many elements, then any other basis also has infinitely many elements.", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_61", "text": "34\nKALAN WALMSLEY\nProof. Let B1 be another basis for V , then B1 has at most n elements since it is linearly independent.\nIf it had strictly fewer elements than B, then by the same argument, also B would also have strictly\nfewer elements than B.\nE\nIf B has infinitely many elements, then B1 cannot have infinitely many elements, by the first part.\n\u25a1\nDefinition. Let V be a vector space. If V has a basis with finitely many elements, we say that V is finite\ndimensional and the number of elements in this basis is called the dimension of V , dim (V ).\nIf V has no finite basis, it is called infinite dimensional, and finally, dim({0}) = 0.\nExample. dim(Pn) = n + 1, dim(Rm\u00d7n) = m \u00b7 n\nSubspaces of finite dimensional spaces.\nTheorem 4.17. Let V be a vector space, dim (V ) < +\u221e, and H \u2286V be a subspace. Then any linearly\nindependent set S \u2286H can be expanded to a basis for H. Moreover, dim (H) \u2264dim (V ).\nProof. If H = {0V }, then dim (H) = 0 \u2264dim (V ). Since, in this case, there is no linearly independent\nsubset of H, there is nothing left to prove.\nIf H \u0338= {0V }, let S = {v1, \u00b7 \u00b7 \u00b7 , vk} \u2286H be linearly independent. If span (S) = H, then S is a basis for\nH. Otherwise, \u2203vk+1 \u2208H such that vk+1 /\u2208span (S). We then claim that {v1, \u00b7 \u00b7 \u00b7 , vk, vk+1} is linearly\ndependent.\nProof: Assume that \u03bb1v1 + \u00b7 \u00b7 \u00b7 + \u03bbkvk + \u03bbk+1vk+1 = 0. If \u03bbk+1 = 0, then \u03bb1 = \u00b7 \u00b7 \u00b7 = \u03bbk = 0 by linear\nindependent of S. If \u03bbk+1 \u0338= 0, then vk+1 =\n\u22121\n\u03bbk+1 (\u03bb1v1 + \u00b7 \u00b7 \u00b7 + \u03bbkvk) \u2208span (S). By assumption E\nSo we continue adding elements until either the new set is a basis for H or the new set has more than\ndim (V ) elements which is a contradiction.", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_62", "text": "\u22121\n\u03bbk+1 (\u03bb1v1 + \u00b7 \u00b7 \u00b7 + \u03bbkvk) \u2208span (S). By assumption E\nSo we continue adding elements until either the new set is a basis for H or the new set has more than\ndim (V ) elements which is a contradiction.\nIn the second case, we obtain a linearly independent set with too many vectors, contradicting theorem\n4.15. By definition, the cardinality (number of elements in a set) of the new set agrees with the dimension\nof H. To obtain a basis for V , we either can take the same basis or we have to add elements\n=\u21d2\ndim (H) \u2264dim (V ).\n\u25a1\nTheorem 4.18. Let V be a vector space, dim (V ) = n \u22651 and let B = {v1, \u00b7 \u00b7 \u00b7 , vn} \u2286V . Then the\nfollowing are equivalent:\n(i) B is a basis for V\n(ii) B is linearly independent\n(iii) span (B) = V\nProof.\ni) \u21d2ii) Let B be linearly independent. Apply theorem 4.17 to H = V , we know that B can\nbe expanded to a basis B\u2032 for V . B\u2032 contains n elements, same for B =\u21d2B\u2032 = B.\niii) \u21d2i) Assume that span (B) = V . Since V \u0338= {0}, the spanning set theorem (4.10) implies that B\ncontains a basis B\u2032. B\u2032 has n elements, same for B =\u21d2B\u2032 = B =\u21d2is a basis.\n\u25a1\nThe dimension of Nul (A) and Col (A). We already know that a basis for Nul (A) can be obtained\nfrom the parametric vector form and that a basis for Col (A) is given by the pivot columns of A.\nCorollary 4.19. Let A \u2208Rm\u00d7n, then\nI) dim (Nul (A)) = number of free variables = number of non-pivot columns\nII) dim (Col (A)) = number of pivot columns\nIn particular n = dim (Nul (A))\n|\n{z\n}\n\u2286Rn\n+ dim (Col (A))\n|\n{z\n}\n\u2286Rm\n.\nProof. We know a basis, so we know the dimension.\n\u25a1\nExample.\nA =\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ed\n1\n1\n0\n0\n0\n0\n1\n0\n0\n0\n0\n1\n0\n0\n0\n0\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f8=\u21d2\n\uf8f1\n\uf8f2\n\uf8f3", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_63", "text": "|\n{z\n}\n\u2286Rn\n+ dim (Col (A))\n|\n{z\n}\n\u2286Rm\n.\nProof. We know a basis, so we know the dimension.\n\u25a1\nExample.\nA =\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ed\n1\n1\n0\n0\n0\n0\n1\n0\n0\n0\n0\n1\n0\n0\n0\n0\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f8=\u21d2\n\uf8f1\n\uf8f2\n\uf8f3\ndim (Nul (A)) = 1\ndim (Col (A)) = 3", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_64", "text": "LECTURE NOTES\n35\n4.6. The rank of a matrix.\nDefinition. Let A \u2208Rm\u00d7n. The rank of A is defined as\nrank (A) = dim Col (A)\ni.e. the dimension of the column space.\nRemark. dim Col (A) = dim Row (A)\nThe row space of a matrix.\nDefinition. Let A \u2208Rm\u00d7n and interpret its rows as vectors in Rn. Then Row (A), the row space of A,\nis the span of the rows of A in Rn.\nExample.\nA =\n\u00121\n2\n3\n4\n3\n1\n\u0013\n=\u21d2Row (A) = span\n\uf8f1\n\uf8f2\n\uf8f3\n\uf8eb\n\uf8ed\n1\n2\n3\n\uf8f6\n\uf8f8,\n\uf8eb\n\uf8ed\n4\n3\n1\n\uf8f6\n\uf8f8\n\uf8fc\n\uf8fd\n\uf8fe= Col\n\u0000AT \u0001\nTheorem 4.20. If A and B are matrices that are obtained from each-other by elementary row operations,\nthen Row (A) = Row (B). If B is in echelon form, the non-zero rows of B form a basis for Row (B) and\nRow (A).\nProof. It suffices to show that Row (A) \u2286Row (B) (since operations are reversible, the statement is\nsymmetric in A and B).\nAfter any elementary row operation, the rows of A are linear combinations of the rows of B\n=\u21d2\nRow (A) \u2286Row (B). Now assume that B is in echelon form with non-zero rows b1, \u00b7 \u00b7 \u00b7 , bk.\ni1\ni2\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\nik\nb1\nb2\n...\nbk\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n0\n\u22c6\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n0\n0\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n...\n\u22c6\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n...\n0\n\u22c6\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n...\n...\n0\n\u22c6\n\u00b7 \u00b7 \u00b7\n...\n...\n...\n0\n...\n...\n...\n...\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\nLet i1, \u00b7 \u00b7 \u00b7 , ik be the indices of the pivot columns in the rows b1, \u00b7 \u00b7 \u00b7 , bk.\n\u03bb1b1 + \u00b7 \u00b7 \u00b7 + \u03bbkbk = 0\ni1-component: \u03bb1 \u00b7 \u22c6= 0 =\u21d2\u03bb1 = 0\ni2-component: \u03bb2 \u00b7 \u22c6= 0 =\u21d2\u03bb2 = 0\nik-component: \u03bbk \u00b7 \u22c6= 0 =\u21d2\u03bbk = 0", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_65", "text": "i1-component: \u03bb1 \u00b7 \u22c6= 0 =\u21d2\u03bb1 = 0\ni2-component: \u03bb2 \u00b7 \u22c6= 0 =\u21d2\u03bb2 = 0\nik-component: \u03bbk \u00b7 \u22c6= 0 =\u21d2\u03bbk = 0\n=\u21d2linear independence. Of course these rows also span the row space as all other rows are zero.\n\u25a1\nTheorem 4.21 (Rank theorem). Let A \u2208Rm\u00d7n, then dim Row (A) = rank (A) and\nn = rank (A) + dim Nul (A)\nProof. By theorem 4.20, dim Row (A) = number of pivot elements of A\ncorol 4.19\n=\nrank (A). The second\nformula is already stated in corollary 4.19.\n\u25a1\nRemark. A \u2208Rm\u00d7n the image of A is at most m dimensional.", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_66", "text": "36\nKALAN WALMSLEY\nExample. Let A \u2208R2\u00d76, then what is the minimal dimension of the null space of A ?\ndim Nul (A) = n \u2212rank (A) = 6 \u2212rank (A) \u22656 \u22122 = 4\nHere, we use that rank (A) is at most 2 since A is a map from R6 to R2.\nExample. A \u2208Rm\u00d7n\n=\u21d2\nCol (A) \u2286Rm\n=\u21d2\nrank (A) \u2264m and n = rank (A) + dim Nul (A)\n|\n{z\n}\n\u22650\n\u2265\nrank (A)\nTheorem (Continuation of invertible matrix theorem). Let A \u2208Rn\u00d7n. Then A being invertible is equiv-\nalent to each of the following statements:\nxiii) The columns of A form a basis for Rn\nxiv) rank (A) = n\nxv) Nul (A) = {0}\nxvi) dim Nul (A) = 0\n4.7. Change of basis.\nRecall. Let V be a vector space, B a finite basis {b1, \u00b7 \u00b7 \u00b7 , bn}, then [x]b are the B-coordinates satisfying\n([x]B)1 b1 + \u00b7 \u00b7 \u00b7 + ([x]B)n bn = x. Let C = {c1, \u00b7 \u00b7 \u00b7 , cn} be another basis. How to obtain [x]C from [x]B ?\nWrite [x]B =\n\uf8eb\n\uf8ec\n\uf8ed\n\u03bb1\n...\n\u03bbn\n\uf8f6\n\uf8f7\n\uf8f8, [x]C =\n\uf8eb\n\uf8ec\n\uf8ed\n\u00b51\n...\n\u00b5n\n\uf8f6\n\uf8f7\n\uf8f8, then \u03bb1b1 +\u00b7 \u00b7 \u00b7+\u03bbnbn = x = \u00b51c1 +\u00b7 \u00b7 \u00b7+\u00b5ncn. Apply the coordinate\nmap [x] 7\u2192[x]C, then\n[\u03bb1b1 + \u00b7 \u00b7 \u00b7 + \u03bbnbn]C = [\u00b51c1 + \u00b7 \u00b7 \u00b7 + \u00b5ncn]C =\u21d2\n\u03bb1 [b1]C + \u00b7 \u00b7 \u00b7 + \u03bbn [bn]C\n|\n{z\n}\n( [b1]C \u00b7\u00b7\u00b7 [bn]C )\n|\n{z\n}\n:=P\nC\u2190B\n \u03bb1\n...\n\u03bbn\n!\n| {z }\n[x]B\n= \u00b51 [c1]C\n|{z}\ne1\n+ \u00b7 \u00b7 \u00b7 + \u00b5n [cn]C\n|{z}\nen\n=\n\uf8eb\n\uf8ec\n\uf8ed\n\u00b51\n...\n\u00b5n\n\uf8f6\n\uf8f7\n\uf8f8\n=\u21d2[x]C = P", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_67", "text": "C\u2190B\n \u03bb1\n...\n\u03bbn\n!\n| {z }\n[x]B\n= \u00b51 [c1]C\n|{z}\ne1\n+ \u00b7 \u00b7 \u00b7 + \u00b5n [cn]C\n|{z}\nen\n=\n\uf8eb\n\uf8ec\n\uf8ed\n\u00b51\n...\n\u00b5n\n\uf8f6\n\uf8f7\n\uf8f8\n=\u21d2[x]C = P\nC\u2190B [x]B\nTheorem 4.22. Let V be a vector space with basis B = {b1, \u00b7 \u00b7 \u00b7 , bn} and another basis C = {c1, \u00b7 \u00b7 \u00b7 , cn}.\nDefine the matrix\nP\nC\u2190B =\n\u0000[b1]C\n\u00b7 \u00b7 \u00b7\n[bn]C\n\u0001\n\u2208Rn\u00d7n\nthen [x]C = P\nC\u2190B [x]B\n\u2200x \u2208V . The matrix\nP\nC\u2190B is invertible with\n\u0010\nP\nC\u2190B\n\u0011\u22121\n= P\nB\u2190C\nProof. It remains to show that\nP\nC\u2190B is invertible (implying surjectivity). Let v \u2208Rn be s.t.\nP\nC\u2190Bv = 0,\nthen x = v1b1 + \u00b7 \u00b7 \u00b7 + vnbn satisfies [x]C = 0 (since [x]B = v). So x = 0 \u00b7 c1 + \u00b7 \u00b7 \u00b7 + 0 \u00b7 cn = 0\nB a basis\n=\u21d2\nv1 =\n\u00b7 \u00b7 \u00b7 = vn = 0.\n\u25a1\nHow to compute\nP\nC\u2190B ?\nFind the C-coordinates of all b1, \u00b7 \u00b7 \u00b7 , bn, so solve\n\u03bb1c1 + \u00b7 \u00b7 \u00b7 + \u03bbncn = bj\n\u2191\nj th column\nof\nP\nC\u2190B\n\u2200j \u2208[[1, n]]", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_68", "text": "LECTURE NOTES\n37\nExample. V = P2\nB =\n\b\n1, 1 \u2212x, 1 + x + x2\t\nC =\n\b\n1, 2x + x2, 1 \u2212x \u2212x2\t\nWe want to find\nP\nC\u2190B.\n\u0010\nP\nC\u2190B\n\u0011\n1 : \u03bb1 \u00b7 1 + \u03bb2\n\u00002x + x2\u0001\n+ \u03bb3\n\u00001 \u2212x \u2212x2\u0001\n= 1 \u21d2\u03bb1 = 1, \u03bb2 = \u03bb3 = 0\n\u0010\nP\nC\u2190B\n\u0011\n2 : \u03bb1 \u00b71+\u03bb2\n\u00002x + x2\u0001\n+\u03bb3\n\u00001 \u2212x \u2212x2\u0001\n= 1\u2212x \u21d0\u21d2Let E =\n\b\n1, x, x2\t\n, then [c1]E =\n\uf8eb\n\uf8ed\n1\n0\n0\n\uf8f6\n\uf8f8, [c2]E =\n\uf8eb\n\uf8ed\n0\n2\n1\n\uf8f6\n\uf8f8, [c1]E =\n\uf8eb\n\uf8ed\n1\n\u22121\n\u22121\n\uf8f6\n\uf8f8, [b2]E =\n\uf8eb\n\uf8ed\n1\n\u22121\n0\n\uf8f6\n\uf8f8\n=\u21d2\n\uf8eb\n\uf8ed\n1\n0\n1\n1\n0\n2\n\u22121\n\u22121\n0\n1\n\u22121\n0\n\uf8f6\n\uf8f8\u2192\n\uf8f1\n\uf8f4\n\uf8f2\n\uf8f4\n\uf8f3\n\u03bb1 = 2\n\u03bb2 = \u22121\n\u03bb3 = \u22121\n\u0010\nP\nC\u2190B\n\u0011\n3 :\n\uf8eb\n\uf8ed\n1\n0\n1\n1\n0\n2\n\u22121\n1\n0\n1\n\u22121\n1\n\uf8f6\n\uf8f8\u2192\n\uf8f1\n\uf8f4\n\uf8f2\n\uf8f4\n\uf8f3\n\u03bb1 = 2\n\u03bb2 = 0\n\u03bb3 = \u22121\n=\u21d2\nP\nC\u2190B =\n\uf8eb\n\uf8ed\n1\n2\n2\n0\n\u22121\n0\n0\n\u22121\n\u22121\n\uf8f6\n\uf8f8\nChange of basis in Rn. For PC =\n\u0000c1\n\u00b7 \u00b7 \u00b7\ncn\n\u0001\nwe must solve PC\n\uf8eb\n\uf8ec\n\uf8ed\n\u03bb1\n...\n\u03bbn\n\uf8f6\n\uf8f7\n\uf8f8= bj to find the jth column\nof\nP\nC\u2190B. We can do this simultaneously.\n\u0000c1\n\u00b7 \u00b7 \u00b7\ncn\n|\nb1\n\u00b7 \u00b7 \u00b7\nbn\n\u0001\nrow operations\n\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2212\u2192\nsince\nP\nC\u2190Binvertible\n\u0012\nIn\nP\nC\u2190B\n\u0013\nThe matrix of a general linear map. Let V, W be vector spaces. Let B = {b1, \u00b7 \u00b7 \u00b7 , bn} be a basis for\nV and Q = {q1, \u00b7 \u00b7 \u00b7 , qm} a basis for W. We know yous can convert x \u2208V to an element in Rn and an\nelement w \u2208W to an element in Rm. What about linear maps ?\nTheorem 4.23. Let V, W, B, Q as above and T : V \u2192W be a linear map. Then \u2203a unique matrix\nA \u2208Rm\u00d7n s.t. [T(v)]Q = A [v]B. The jth column of the matrix is given by [T(bj)]Q.\nProof. Define A =\n\u0000[T(b1)]Q\n\u00b7 \u00b7 \u00b7", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_69", "text": "A \u2208Rm\u00d7n s.t. [T(v)]Q = A [v]B. The jth column of the matrix is given by [T(bj)]Q.\nProof. Define A =\n\u0000[T(b1)]Q\n\u00b7 \u00b7 \u00b7\n[T(b1)]Q\n\u0001\n\u2208Rm\u00d7n, then \u2200v \u2208V we can write v = \u03bb1b1 + \u00b7 \u00b7 \u00b7 + \u03bbnbn\nand by linearity of T\n[T(v)]Q = [\u03bb1T(b1) + \u00b7 \u00b7 \u00b7 + \u03bbnT(bn)]Q = \u03bb1 [T(b1)]Q + \u00b7 \u00b7 \u00b7 + \u03bbn [T(bn)]Q\n= A\n\uf8eb\n\uf8ec\n\uf8ed\n\u03bb1\n...\n\u03bbn\n\uf8f6\n\uf8f7\n\uf8f8= A [v]B\nNow, to prove uniqueness, assume that C [v]B = [T(v)]Q\n\u2200v \u2208V . We can try v = bj,\nC [bj]B\n=\n[T(bj)]Q =\u21d2C = A\n= Cej = jth column of C", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_70", "text": "38\nKALAN WALMSLEY\n\u25a1\nExample. T : P3 \u2192P2, T(p) = p\u2032, B =\n\b\n1, x, x2, x3\t\n, Q =\n\b\n1, x, x2\t\n=\u21d2\nT(b1) = 0, T(b2) =\n1, T(b3) = 2x, T(b4) = 3x2 =\u21d2A =\n\uf8eb\n\uf8ed\n0\n1\n0\n0\n0\n0\n2\n0\n0\n0\n0\n3\n\uf8f6\n\uf8f8\nCorollary 4.24. Let V, W, B, Q, T, A as above. It then holds that\ni) Ker (T) = {x \u2208V | [x]B \u2208Nul (A)}\nii) Ran (T) =\n\b\nw \u2208W | [w]Q \u2208Col (A)\n\t\nProof.\ni) Let x \u2208Ker (T) \u21d0\u21d2T(x) = 0 \u21d0\u21d2[T(x)]Q\n=\nA [x]B =\u21d2[x]B \u2208Nul (A)\n= 0\nii) w \u2208Ran (T) \u21d0\u21d2\u2203v \u2208V s.t. T(v) = w\nQ-coordinate\n\u21d0========\u21d2\nmap bijective\n\u2203v \u2208V s.t. [T(v)]Q\n=\nA [v]B =\u21d2[w]Q \u2208Col (A)\n= [w]Q\n\u25a1\nExample. Given\n\u2022 T\n\u0012\na1,1\na1,2\na2,1\na2,2\n\u0013\n=\n\u0012\na1,1 \u2212a1,2\na1,1 + a2,1\n\u0013\n\u2022 B =\n\u001a\u00121\n0\n0\n0\n\u0013\n,\n\u00120\n1\n0\n0\n\u0013\n,\n\u00120\n0\n1\n0\n\u0013\n,\n\u00120\n0\n0\n1\n\u0013\u001b\n\u2022 Q = {e1, e2}\nA =\n\u0000[T(b1)]Q\n\u00b7 \u00b7 \u00b7\n[T(b4)]Q\n\u0001\n=\u21d2\n\uf8f1\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f2\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f3\nT(b1) =\n\u0012\n1\n1\n\u0013\n= e1 + e2\nT(b2) =\n\u0012\u22121\n0\n\u0013\n= \u2212e1\nT(b3) =\n\u00120\n1\n\u0013\n= e2\nT(b4) =\n\u00120\n0\n\u0013\n= 0\n=\u21d2A =\n\u00121\n\u22121\n0\n0\n1\n0\n1\n0\n\u0013\n=\u21d2\n\uf8f1\n\uf8f4\n\uf8f4\n\uf8f2\n\uf8f4\n\uf8f4\n\uf8f3\nNul (A) = span\n\uf8f1\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f2\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f3\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n0\n0\n0\n1\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8,\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n1\n1\n\u22121\n0\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n\uf8fc\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8fd\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8fe\nCol (A) = R2\n=\u21d2\n\uf8f1\n\uf8f2\n\uf8f3\nRan (T) = R2\nKer (T) =\n( \na\na\n\u2212a\nb\n!\n| a, b \u2208R\n)", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_71", "text": "LECTURE NOTES\n39\n5. Eigenvalues and eigenvectors\nAs we will see, the concept of eigenvalues involves zeros of certain polynomials. Hence we will also use\ncomplex numbers, which we briefly recall here.\nComplex numbers.\nDefinition. A complex number is an element of the form a+bi with a, b \u2208R and i the so-called imaginary\nunit. We say that two complex numbers a + bi and c + di are equal iff a = c and b = d. We write C for\nthe set of all complex numbers. On C we define the following operations\ni) (a + bi) + (c + di) = (a + c) + (b + d)i \u2208C\nii) (a + bi)(c + di) = (ac \u2212bd) + (bc + ad)i \u2208C\nIf z = a + bi, then a = \u211c(z) = Re(z) is called the real part of z, and b = \u2111(z) = Im(z) is called the\nimaginary part of z. We interpret real numbers as complex numbers with imaginary part zero. Complex\nnumbers have the following properties, which imply that C is a field:\ni) z1 + z2 = z2 + z1\nii) z1 + (z2 + z3) = (z1 + z2) + z3\niii) z1 + 0 = z1\niv) z1 \u2212z1 = 0 where \u2212z1 = (\u22121)z1\nv) z1 \u00b7 z2 = z2 \u00b7 z1\nvi) z1(z2 \u00b7 z3) = (z1 \u00b7 z2)z3\nvii) z1(z2 + z3) = z1z2 + z1z3\nviii) 1 \u00b7 z1 = z1\nix) Moreover \u2200z \u2208C \\ {0} \u22031\nz s.t. z \u00b7 1\nz = 1. Indeed, for z = a + bi define the complex conjugate\nz = a \u2212bi \u2208C, then\n1\nz = z\nzz\nThis is well-defined: z \u00b7 z = (a + bi)(a \u2212bi) = a2 \u2212(\u2212b2) + (ba \u2212ab)i = a2 + b2 > 0 since a \u0338= 0\nor b \u0338= 0 or both. We have\ni2 = (0 + 1 \u00b7 i)(0 + 1 \u00b7 i) = \u22121", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_72", "text": "or b \u0338= 0 or both. We have\ni2 = (0 + 1 \u00b7 i)(0 + 1 \u00b7 i) = \u22121\nWe write z \u2208C =\u21d2|z| =\n\u221a\nz \u00b7 z, which is called the modulus of z.\nz\nw := z \u00b7 1\nw = z w\n|w|2\nProof. We only show that z 1\nz = 1, the other properties are straightforward.\nz = a + bi =\u21d2z \u00b7 z\nzz =\nzz\na2 + b2 = (a + bi)(a \u2212bi)\na2 + b2\n= a2 + b2\na2 + b2 = 1\n\u25a1\nLemma 5.1. Let z, w \u2208C, then\ni) z = z\nii) z + w = z + w\niii) z \u00b7 w = z \u00b7 w\niv) |z \u00b7 w| = |z| |w|\nGeometric interpretation. C \u223c= R2 The product of one complex number by z represents scaling the\nfirst by |z| and rotating by \u03c6 (counterclockwise). More rigorously, for z = |z| (cos (\u03c6) + i sin (\u03c6)) , w =\n|w| (cos (\u03b8) + i sin (\u03b8)),\nzw = |z| |w| (cos (\u03c6) cos (\u03b8) \u2212sin (\u03c6) sin (\u03b8)) + i (cos (\u03c6) sin (\u03b8) + cos (\u03b8) sin (\u03c6))\n= |z| |w| (cos (\u03c6 + \u03b8) + i sin (\u03c6 + \u03b8))\nTheorem (Fundamental theorem of algebra). Every polynomial with complex coefficients and of degree\nn has at least n roots (counted with multiplicity).", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_73", "text": "40\nKALAN WALMSLEY\nRemark. All the results proven so far also hold when we replace R by C (Rm\u00d7n \u2192Cm\u00d7n, real vector\nspaces \u2192imaginary vector spaces).\n5.1. Eigenvalues and eigenvectors. Intuitively, eigenvectors of a matrix are vectors that are just\nscaled by the matrix, i.e for \u03bb \u2208R, Av = \u03bbv. We must exclude v = 0. Moreover, eigenvectors can only\nexist for square matrices.\nDefinition. Let A \u2208Rn\u00d7n. A vector v \u2208Rn \\ {0} is called an eigenvector if A if \u2203\u03bb \u2208R s.t. Av = \u03bbv.\nA scalar \u03bb \u2208R is called an eigenvalue of A if \u2203v \u2208Rn \\ {0} s.t. Av = \u03bbv.\nThe equation Av = \u03bbv has two unknowns, \u03bb and v. Later we will see a method how to eliminate v, so\nthat we look first for eigenvalues and then for eigenvectors in a second step.\nExample.\ni) Test if v is an eigenvector of A:\nA =\n\u00121\n2\n3\n4\n\u0013\n, v =\n\u00121\n0\n\u0013\n=\u21d2Av =\n\u00121\n3\n\u0013\n?= \u03bb\n\u00121\n0\n\u0013\n=\u21d2v is not an eigenvector since 3 \u0338= 0.\nii) Test if \u03bb is an eigenvalue of A\nAv = \u03bbv = \u03bbInv \u21d0\u21d2(Av \u2212\u03bbIn) v = 0\n=\u21d2row reduction to see if \u2203a non-zero solution.\nHow can we find \u03bb and v ? Eigenvalues: \u03bb is an eigenvalue of A \u21d0\u21d2(A \u2212\u03bbIn) v = 0 has a solution\nv \u0338= 0 \u21d0\u21d2A \u2212\u03bbIn is not invertible \u21d0\u21d2det(A \u2212\u03bbIn) = det(\u03bbIn \u2212A) = 0\n5.2. The characteristic polynomial.\nProposition 5.2. Let A \u2208Rn\u00d7n and \u03bb be a scalar. Then the function \u03c7A(\u03bb) := det(\u03bbIn \u2212A) is a\npolynomial of degree n with leading coefficient 1 i.e.\n\u03c7A(\u03bb) = \u03bbn +\nn\u22121\nX\nk=0\nbk\u03bbk\nfor some scalars bk.\nProof.\n\u03bbIn \u2212A =\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n\u03bb \u2212a1,1\n\u2212a1,2\n\u00b7 \u00b7 \u00b7\n\u2212a1,n\n\u2212a2,1\n...\n...\n...\n\u2212an,1", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_74", "text": "n\u22121\nX\nk=0\nbk\u03bbk\nfor some scalars bk.\nProof.\n\u03bbIn \u2212A =\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n\u03bb \u2212a1,1\n\u2212a1,2\n\u00b7 \u00b7 \u00b7\n\u2212a1,n\n\u2212a2,1\n...\n...\n...\n\u2212an,1\n\u03bb \u2212an,n\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\nWe claim if B(\u03bb) \u2208Rn\u00d7n is s.t. bi,j(\u03bb) is a polynomial of degree at most ones, then det(B(\u03bb)) is a\npolynomial of degree at most n. We perform induction on n. n = 1 verifies the property so we now show\nn \u22121 =\u21d2n using Laplace expansion.\ndet(\u03bbIn \u2212A) = (\ndeg=1\nz\n}|\n{\n\u03bb \u2212a1,1) det\n\u0010\ndeg\u2264n\u22121\nz\n}|\n{\n(n\u22121)\u00d7(n\u22121)\nz\n}|\n{\n(\u03bbIn \u2212A)1,1\n\u0011\n+\ndeg=0\nz}|{\na1,2 det\n\u0010\ndeg\u2264n\u22121\nz\n}|\n{\n(\u03bbIn \u2212A)1,2\n\u0011\n= sum of polynomials of degree at most n", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_75", "text": "LECTURE NOTES\n41\n=\u21d2det(\u03bbIn \u2212A) is a polynomial of degree at most n. The leading coefficient is given by\nlim\n\u03bb\u2192+\u221e\n1\n\u03bbn det(\u03bbIn \u2212A) =\nlim\n\u03bb\u2192+\u221edet\n\uf8eb\n\uf8ec\n\uf8ed\n1\n\u03bb\n...\n1\n\u03bb\n\uf8f6\n\uf8f7\n\uf8f8det(\u03bbIn \u2212A)\n=\nlim\n\u03bb\u2192+\u221edet\n\u0012 1\n\u03bbIn (\u03bbIn \u2212A)\n\u0013\n=\nlim\n\u03bb\u2192+\u221edet\n\u0012\nIn \u22121\n\u03bbA\n\u0013\nt= 1\n\u03bb\n= lim\nt\u21920 det (In \u2212tA) =\n\u2191\nt7\u2192det(In\u2212tA)\nis a polynomial\n=\u21d2continuous\ndet(In) = 1\n\u25a1\nDefinition. Given A \u2208Rn\u00d7n, the function \u03c7A(\u03bb) = det(\u03bbIn \u2212A) is called the characteristic polynomial\nof A.\nTheorem 5.3. Let A \u2208Rn\u00d7n, then \u03bb is an eigenvalue of A iff \u03c7A(\u03bb) = 0. In particular\ni) A can have at most n eigenvalues\nii) A invertible \u21d0\u21d20 is not an eigenvalue\niii) If A is a triangular matrix, then its eigenvalues are the elements on the diagonal\nProof. The first statement was already proven in a previous paragraph (How can we find \u03bb and v ?).\ni) \u03c7A can have at most n zeros since it has degree n.\nii) A invertible \u21d0\u21d2det(A) \u0338= 0 \u21d0\u21d2det(\u2212A) \u0338= 0 =\u21d2\u03c7A(0) \u0338= 0 \u21d0\u21d20 is not an eigenvalue.\niii) A triangular =\u21d2\u03bbIn \u2212A also triangular =\u21d2det(\u03bbIn \u2212A) = (\u03bb \u2212a1,1) \u00b7 \u00b7 \u00b7 (\u03bb \u2212an,n) =\u21d2\ndet(\u03bbIn \u2212A) = 0 iff \u03bb = ai,i for some i.\n\u25a1\nDefinition. Given A \u2208Rn\u00d7n and eigenvalue \u03bb, we define\nEA(\u03bb) = Ker (\u03bbIn \u2212A) = {0} \u222a{v \u2208Rn \\ {0} | v eigenvector of A}\nto be the so-called eigenspace of A corresponding to \u03bb. Its dimension is called the geometric multiplicity\nof the eigenvalue \u03bb.\nRemark. A non-zero vector can belong to at most one eigenspace.", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_76", "text": "to be the so-called eigenspace of A corresponding to \u03bb. Its dimension is called the geometric multiplicity\nof the eigenvalue \u03bb.\nRemark. A non-zero vector can belong to at most one eigenspace.\nTheorem 5.4. Let A \u2208Rn\u00d7n. Suppose \u03bb1, \u00b7 \u00b7 \u00b7 , \u03bbj are distinct eigenvalues of A and v1, \u00b7 \u00b7 \u00b7 , vj are the\nvectors corresponding to \u03bb1, \u00b7 \u00b7 \u00b7 , \u03bbj, respectively. Then {v1, \u00b7 \u00b7 \u00b7 , vj} is linearly independent.\nProof. Induction on j\n\u2022 j = 1: v1 \u0338= 0 since it is an eigenvector =\u21d2{v1} linearly independent\n\u2022 j \u22121 7\u2192j: Suppose \u00b51v1 + \u00b7 \u00b7 \u00b7 + \u00b5jvj = 0\nA (\n0\nz\n}|\n{\n\u00b51v1 + \u00b7 \u00b7 \u00b7 + \u00b5jvj) \u2212\u03bbj (\n0\nz\n}|\n{\n\u00b51v1 + \u00b7 \u00b7 \u00b7 + \u00b5jvj) =\n\u00b51Av1 + \u00b7 \u00b7 \u00b7 + \u00b5jAvj \u2212\u03bbj\u00b51v1 \u2212\u00b7 \u00b7 \u00b7 \u2212\u03bbj\u00b5jvj =\n\u00b51(\u03bb1v1 \u2212\u03bbjv1) + \u00b7 \u00b7 \u00b7 + \u00b5j\u22121(\u03bbj\u22121vj\u22121 \u2212\u03bbjvj\u22121) =\n\u00b51(\u03bb1 \u2212\u03bbj)v1 + \u00b7 \u00b7 \u00b7 + \u00b5j\u22121(\u03bbj\u22121 \u2212\u03bbj)vj\u22121\ninduction\n======\u21d2\u00b51 (\n\u0338=0\nz }| {\n\u03bb1 \u2212\u03bbj) = \u00b7 \u00b7 \u00b7 = \u00b5j\u22121 (\n\u0338=0\nz\n}|\n{\n\u03bbj\u22121 \u2212\u03bbj) = 0\n=\u21d2\u00b51 = \u00b7 \u00b7 \u00b7 = \u00b5j\u22121 = 0\n=\u21d2\u00b5jvj = 0\nvj\u0338=0\n===\u21d2\u00b5j = 0", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_77", "text": "42\nKALAN WALMSLEY\n\u25a1\nDefinition. Let A, B \u2208Rn\u00d7n. We say that A and B are similar if \u2203P \u2208Rn\u00d7n invertible s.t.\nA = P \u22121BP\nIn this case also B = Q\u22121AQ with Q = P \u22121.\nProposition 5.5. Let A, B \u2208Rn\u00d7n be similar matrices. Then we have \u03c7A(\u03bb) = \u03c7B(\u03bb)\n\u2200\u03bb and therefore\nA and B have the same eigenvalues.\nProof.\n\u03c7A(\u03bb) = det(\u03bbIn \u2212A) = det\n\u0000\u03bbIn \u2212P \u22121BP\n\u0001\n= det\n\u0000\u03bbP \u22121P \u2212P \u22121BP\n\u0001\n=\ndet\n\u0000P \u22121(\u03bbIn \u2212B)P\n\u0001\n= det\n\u0000P \u22121\u0001\ndet(P)\n|\n{z\n}\n1\ndet(\u03bbIn \u2212B) =\ndet(\u03bbIn \u2212B)\n|\n{z\n}\n\u03c7B(\u03bb)\n= \u03c7B(\u03bb)\n\u25a1\nRemark.\na) Similarity has nothing to do with row equivalence.\nb) Row operations on A change the eigenvalues of the matrix.\nc) Two matrices can have the same characteristic polynomial without being similar.\nd) Similar matrices have the same eigenvalues and the corresponding eigenspaces have the same\ndimension.\nLet us collect some further properties of eigenvalues. The following notion will be useful.\nDefinition. For A \u2208Rn\u00d7n, \u03c3(A) :\u2212{set of all eigenvalues of A}. It is also called the spectrum of A.\nLemma 5.6. Let A \u2208Rn\u00d7n, then\ni) If A is invertible, then \u03bb \u2208\u03c3(A) \u21d0\u21d21\n\u03bb \u2208\u03c3\n\u0000A\u22121\u0001\nii) \u03c3(A) = \u03c3\n\u0000AT \u0001\niii) \u03bb \u2208\u03c3(A) =\u21d2\u03bbk \u2208\u03c3\n\u0000Ak\u0001\n5.3. Diagonalization. For many purposes, diagonal matrices are very simple. For example Dk =\n\uf8eb\n\uf8ed\n\u03bbk\n1\n...\n\u03bbk\n1\n\uf8f6\n\uf8f8.\nFor non-diagonal matrices, a change of basis may help.\nDefinition. Let A \u2208Rn\u00d7n, then A is said to be diagonalizable if \u2203P \u2208Rn\u00d7n invertible s.t. A = PDP \u22121\nfor a diagonal matrix D.\nRemark.\na) If A is diagonalizable, then \u03c7A(\u03bb) = \u03c7D(\u03bb) =\u21d2diagonal entries of D have to be the\neigenvalues of A.", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_78", "text": "for a diagonal matrix D.\nRemark.\na) If A is diagonalizable, then \u03c7A(\u03bb) = \u03c7D(\u03bb) =\u21d2diagonal entries of D have to be the\neigenvalues of A.\nb) For A diagonalizable,\nAk =\n\u0000PDP \u22121\u0001k = PDP \u22121 \u00b7 \u00b7 \u00b7 PDP \u22121\n|\n{z\n}\nk times\n= PDP \u22121P\n| {z }\n=I\nDP \u22121 \u00b7 \u00b7 \u00b7 PDP \u22121P\n| {z }\n=I\nDP \u22121 = PDkP \u22121\nTheorem 5.7. A \u2208Rn\u00d7n is diagonalizable iff \u2203a basis of eigenvectors of A for Rn. In this case,\nA = PDP \u22121, where P =\n\u0000v1\n\u00b7 \u00b7 \u00b7\nvn\n\u0001\nand D =\n\uf8eb\n\uf8ec\n\uf8ed\n\u03bb1\n...\n\u03bbn\n\uf8f6\n\uf8f7\n\uf8f8with Avi = \u03bbivi\n\u22001 \u2264i \u2264n.", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_79", "text": "LECTURE NOTES\n43\nProof.\n\u21d2Assume A diagonalizable, then A = PDP \u22121 with P invertible \u21d2columns of P form a\nbasis for Rn. Write P =\n\u0000v1\n\u00b7 \u00b7 \u00b7\nvn\n\u0001\n. We need to show Avi = \u03bbivi, where D =\n \u03bb1\n...\n\u03bbn\n!\n.\nAvi = PDP \u22121vi\nP ei=vi\u21d0\u21d2ei=P \u22121vi\n\uf8e6\uf8e6y\n= PDei = P\u03bbiei = \u03bbiPei = \u03bbivi\n\u21d0Let {v1, \u00b7 \u00b7 \u00b7 , vn} be a basis of eigenvectors, then we have that P :\u2212\n\u0000v1\n\u00b7 \u00b7 \u00b7\nvn\n\u0001\nis invertible.\nP \u22121APei = P \u22121Avi\nvi eigenvector\n\u2193= P \u22121\u03bbivi = \u03bbiP \u22121vi = \u03bbiei\n=\u21d2P \u22121AP =\n\uf8eb\n\uf8ec\n\uf8ed\n\u03bb1\n...\n\u03bbn\n\uf8f6\n\uf8f7\n\uf8f8\n|\n{z\n}\nD\n=\u21d2PDP \u22121 = A\n\u25a1\nMethod for diagonalizing a matrix.\nStep 1: Find all eigenvectors of A (\u03c7A(\u03bb) = 0)\nStep 2: Find a basis for each eigenspace (Ker (A \u2212\u03bbI) =\u21d2parametric vector form)\nStep 3:\n\u2022 If in total one has n basis vectors, put them in P and write the eigenvalues in D (with same\norder)\n\u2022 If one has less than n vectors, A not diagonalizable\nRemark. A helpful technique for factorising third degree polynomials. Any polynomial with rational\ncoefficients a0, \u00b7 \u00b7 \u00b7 , an \u2208Q possibly has rational roots. If \u2203a rational root, it is of the form a\nb , where a\nis a factor the constant coefficient of a0 and b a factor of the leading coefficient an. Since the leading\ncoefficient of the characteristic is always 1, any rational root (if exists) will be a factor of an. Let us take\nfor example x3\u22123x2\u221226x+56. The possible roots are thus 2, 7, 4, 8 (if the constant coefficient is negative,\nthe possible roots are either positive or negative). We can use Horner\u2019s scheme. If we guess 2\n1\n\u22123\n\u221226\n56\n2\n2\n\u22122\n\u221256\n1\n\u22121\n\u221228\n0\n+\nFrom this we get (x2 \u2212x \u221228)(x \u22122), which we know how to solve.", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_80", "text": "1\n\u22123\n\u221226\n56\n2\n2\n\u22122\n\u221256\n1\n\u22121\n\u221228\n0\n+\nFrom this we get (x2 \u2212x \u221228)(x \u22122), which we know how to solve.\nCorollary 5.8. Let A \u2208Rn\u00d7n. If A has n distinct eigenvalues, then A is diagonalizable.\nProof. Combine theorem 5.4 and theorem 5.7.\n\u25a1\nTheorem 5.9. Let A \u2208Rn\u00d7n and \u03c3(A) = {\u03bb1, \u00b7 \u00b7 \u00b7 , \u03bbk} (k \u2264n), then\ni) \u2200\u03bbj \u2208\u03c3(A) the dimension of the eigenspace E\u03bbj (called geometric multiplicity) is less than or\nequal to the multiplicity of \u03bbj as a root of \u03c7A (called algebraic multiplicity).\nmultgeom\u03bb (A) \u2264multalg\u03bb (A)\nii) A is diagonalizable iff\ni) \u03c7A can be factorised into n polynomials of degree 1 (always possible in C).\nii) In multgeom\u03bb (A) \u2264multalg\u03bb (A) there holds equality. This is in turn equivalent to the fact\nthat the sum of the geometric multiplicities equals n.\niii) If Bj is a basis for E\u03bbj, then the union of these bases (B1, \u00b7 \u00b7 \u00b7 , Bk) is linearly independent.", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_81", "text": "44\nKALAN WALMSLEY\nSketch of the proof.\ni) Let {v1, \u00b7 \u00b7 \u00b7 , vl} be a basis for E\u03bbj. We can complete this family to a basis\n{v1, \u00b7 \u00b7 \u00b7 , vl, vl+1, \u00b7 \u00b7 \u00b7 , vn} for Rn. Define P =\n\u0000v1\n\u00b7 \u00b7 \u00b7\nvn\n\u0001\n, then\nP \u22121AP =\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n\u03bbj\n0\n\u00b7 \u00b7 \u00b7\n0\n\u22c6\n0\n...\n\u22c6\n...\n...\n\u22c6\n0\n\u03bbj\n\u22c6\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n=\u21d2\u03c7A(\u03bb)\nProp 5.5\n=\n\u03c7P \u22121AP (\u03bb)\n= det\n\uf8eb\n\uf8ec\n\uf8ed\n\u03bb \u2212\u03bbj\n\u22c6\n...\n\u22c6\n\u03bb \u2212\u03bbj\n\u22c6\n\uf8f6\n\uf8f7\n\uf8f8= (\u03bb \u2212\u03bbj)l det (\u03bbIn\u2212l \u2212\u22c6)\n=\u21d2multalg\u03bbj (A) \u2265l = multgeom\u03bbj (A)\nii)\ni) \u21d2ii)\nAssume that A = PDP \u22121, then \u03c7A(\u03bb)\nprop 5.5\n=\n\u03c7D(\u03bb) =\nnQ\ni=1\n(\u03bb \u2212di,i) \u21d2\u03c7A can\nbe factorised. Let {\nfor \u03bb1\nz }| {\nv1, \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 \u00b7 , vn\nfor \u03bbk\nz }| {\n} be a basis of eigenvectors (columns of P)\n=\u21d2\nn \u2264\nkP\nj=1\ndim (E\u03bbj) \u2264\nkP\nj=1\nmultalg\u03bbj (A) \u2264n\n=\u21d2\nin every inequality we have equality\n=\u21d2\ndim (E\u03bbj) = multalg\u03bbj (A).\nii) \u21d2iii)\nkP\nj=1\nmultgeom\u03bbj (A)\nii)=\nkP\nj=1\nmultalg\u03bbj (A)\nii)= n\niii) \u21d2i) follows from iii) and theorem 5.7.\niii)\nvectors composing E\u03bb1\nz\n}|\n{\n\u03bb1b1\n1 + \u00b7 \u00b7 \u00b7 + \u03bbjb1\nj\n|\n{z\n}\nv1\u2208E\u03bb1\n+ \u00b7 \u00b7 \u00b7 + \u03bbrbk\n1 + \u00b7 \u00b7 \u00b7 + \u03bbsbk\ni\n|\n{z\n}\nvk\u2208E\u03bbk\n= 0\n\u2022 If vj = 0, then all its coefficients are zero.\n\u2022 If \u2203vj \u0338= 0, then this contradicts theorem 5.4.\n\u25a1\n5.4. Eigenvectors and linear transformations.", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_82", "text": "i\n|\n{z\n}\nvk\u2208E\u03bbk\n= 0\n\u2022 If vj = 0, then all its coefficients are zero.\n\u2022 If \u2203vj \u0338= 0, then this contradicts theorem 5.4.\n\u25a1\n5.4. Eigenvectors and linear transformations.\nRecall. V, W vector spaces, T : V \u2192W linear, B a basis for V , C a basis for W, then [T(v)]C = A [v]B\nfor A =\n\u0000[T(b1)]C\n\u00b7 \u00b7 \u00b7\n[T(bn)]C\n\u0001\n. Special case: V = W, C = B. In this case, A is called the B-matrix of\nT. Now change the basis from B to Q: what is the Q-matrix of T ?\nA [v]B\n=\nA P\nB\u2190Q [v]Q\n= [T(v)]B =\nP\nB\u2190Q [T(v)]Q =\u21d2[T(v)]Q = P \u22121\nB\u2190QA P\nB\u2190Q\n|\n{z\n}\nsimilar to A\n[v]Q\n=\u21d2Diagonalization is a change of basis.\nTheorem 5.10. Let A \u2208Rn\u00d7n and B = {b1, \u00b7 \u00b7 \u00b7 , bn} be a basis for Rn. For P =\n\u0000b1\n\u00b7 \u00b7 \u00b7\nbn\n\u0001\n, the\nB-matrix for the linear map x 7\u2192Ax is given by P \u22121AP.\nProof. E standard basis for Rn \u21d2\nP\nB\u2190E =\n\u0010\nP\nE\u2190B\n\u0011\u22121\n=\n\u0000b1\n\u00b7 \u00b7 \u00b7\nbn\n\u0001\n= P\n\u25a1\n5.5. Complex Eigenvalues. Problem: \u03c7A can have complex roots. In this case, we can consider complex\neigenvalues and eigenvectors (Av cannot be complex if A \u2208Rn\u00d7n, v \u2208Rn).\nDefinition. Cn is the vector space of all n-tuples with complex entries. If x, y \u2208Cn, \u03bb \u2208C, then\nx + y \u2208Cn and \u03bbx \u2208Cn are defined componentwise using the operations of C =\u21d2Cn is a vector space.", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_83", "text": "LECTURE NOTES\n45\nGiven A \u2208Cn\u00d7n, v \u2208Cn, the product Av is defined as usual. Also, determinants (and every other\ntopic discussed), works in the complex case.\n=\u21d2det(\u03bbI \u2212A) = 0 \u21d0\u21d2\u2203v \u2208Cn \\ {0} s.t. Av = \u03bbv,\n\u03bb \u2208C a complex eigenvalue, v \u2208Cn a complex eigenvector.\nFurther operations on Cn.\nv \u2208Cn, v = v1 + iv2, v1, v2 \u2208Rn =\u21d2\n(\nv1 = Re(v)\nv2 = Im(v)\n=\u21d2v = v1 \u2212iv2\nSimilarly, A \u2208Cn\u00d7n =\u21d2A componentwise complex conjugate of A.\nRemark (Complex calculus rules). Let \u03bb \u2208C, v \u2208Rn, A, B \u2208Cn\u00d7n, then\ni) \u03bbv = \u03bb v\nii) Av = A v\niii) \u03bbA = \u03bb A\niv) AB = A B\nProof. We only prove the fourth property, the others are shown in a similar fashion. For the coefficient\nat position (i, j) we have\n(AB)i,j =\nn\nX\nk=1\nai,kbk,j =\nn\nX\nk=1\nai,kbk,j =\nn\nX\nk=1\nai,k bk,j =\n\u0000A B\n\u0001\ni,j .\nHence the two matrices are equal.\n\u25a1\nEigenvalues and eigenvectors of A \u2208Rn\u00d7n acting on Cn.\nTheorem 5.11. Let A \u2208Rn\u00d7n and \u03bb \u2208C be an eigenvalue with eigenvector v \u2208Cn. Then \u03bb \u2208C is also\nan eigenvalue with eigenvector v \u2208Cn.\nProof. Av = \u03bbv =\u21d2Av = \u03bbv = \u03bb v\nA real\n====\u21d2Av = \u03bb v and v \u0338= 0 because v \u0338= 0.\n\u25a1\nComputation of complex eigenvectors. We use row reduction in C. The above theorem halves computation\ntime, since complex eigenvectors come in pairs. It is also nice to remember\n1\ni = \u2212i\nfor computations and\nAP = PD\nas a practical check for correctness. This relationship allows one verify computations without computing\nP \u22121.\nGeometric meaning of complex eigenvalues. Case A \u2208R2\u00d72:", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_84", "text": "1\ni = \u2212i\nfor computations and\nAP = PD\nas a practical check for correctness. This relationship allows one verify computations without computing\nP \u22121.\nGeometric meaning of complex eigenvalues. Case A \u2208R2\u00d72:\nTheorem 5.12. Let A \u2208R2\u00d72 with \u03c3(A) = {a + ib, a \u2212ib} , b \u0338= 0. Let A = PCP \u22121, where P =\n\u0000Re (v)\nIm (v)\u0001\n\u2208R2\u00d72, C =\n\u0012a\n\u2212b\nb\na\n\u0013\n. C is a scaling and rotation matrix.\nProof. We have Av = A\n\u0000Re (v) + i Im (v)\u0001\n= A\n\u0000Re (v)\u0001\n+iA\n\u0000Im (v)\u0001\n. Since A has real-valued coefficients,\nboth the vectors A\n\u0000Re (v)\u0001\nand A\n\u0000Im (v)\u0001\nare real-valued. Hence the above decomposition gives the real\nand imaginary part of Av. We use that\nAv = \u03bb2v = av \u2212ibv = a Re (v) + b Im (v) + i (a Im (v) \u2212b Re (v))", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_85", "text": "46\nKALAN WALMSLEY\n=\u21d2\n\uf8f1\n\uf8f4\n\uf8f2\n\uf8f4\n\uf8f3\nA\n\u0010\nRe (v)\n\u0011\n= a Re (v) + b Im (v)\nA\n\u0010\nIm (v)\n\u0011\n= a Im (v) \u2212b Re (v)\n.\nBy definition of the matrix-matrix product we have that\nAP =\n\u0000A\n\u0000Re (v)\u0001\nA\n\u0000Im (v)\u0001\u0001\n=\n\u0000a Re (v) + b Im (v)\na Im (v) \u2212b Re (v)\n\u0001\n= PC\nWe still need to show that P must be invertible. We know that Im (v) \u0338= 0 (otherwise v cannot be an\neigenvector for a complex eigenvalue). Now assume by contradiction that the columns of P are linearly\ndependent. Then \u00b51 Re (v) + \u00b52 Im (v) = 0 has a non-trivial solution (\u00b51, \u00b51) = (0, 0). Since Im (v) is\nnon-zero, we actually know that \u00b51 \u0338= 0 (otherwise \u00b52 Im (v) = 0, which is only possible for \u00b52 = 0).\nHence Re (v) = \u2212\u00b52\n\u00b51 Im (v) and therefore\nv = Re (v) \u2212i Im (v) =\n\u2212\u00b52\n\u00b51 \u2212i\n\u2212\u00b52\n\u00b51 + i\n\u0000Re (v) + i Im (v)\n\u0001\n=\n\u2212\u00b52\n\u00b51 \u2212i\n\u2212\u00b52\n\u00b51 + i\n|\n{z\n}\n:\u2212k\nv\nThen Av = Akv = kAv = \u03bb2kv = \u03bb2v, but we know that (theorem 5.11) v has to be an eigenvector for\n\u03bb1 \u0338= \u03bb2.\nE\n\u25a1\nRemark. Any rotation matrix in R2\u00d72 has the form\n\u0012cos (\u03c6)\n\u2212sin (\u03c6)\nsin (\u03c6)\ncos (\u03c6)\n\u0013\n.\nCase A \u2208Rn\u00d7n with a pair of complex eigenvalues. \u2203a two dimensional subspace which is inveriant\nunder the action of A and, on this subspace, A acts like a scaling and rotating matrix.", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_86", "text": "LECTURE NOTES\n47\n6. Orthogonality and least-squares\n6.1. Inner product, length, and orthogonality.\nDefinition. Let u, v \u2208Rn. Then the inner product between u and v is defined by\nu \u00b7 v = uT v = u1v1 + \u00b7 \u00b7 \u00b7 + unvn\nCalculus rules for the inner product\ni) u \u00b7 v = v \u00b7 u\nii) (u + v) \u00b7 w = u \u00b7 w + v \u00b7 w\niii) (\u03bbu) \u00b7 v = u \u00b7 (\u03bbv) = \u03bb(u \u00b7 v)\n\u2200\u03bb \u2208R\niv) u \u00b7 u \u22650, u \u00b7 u = u2\n1 + \u00b7 \u00b7 \u00b7 + u2\nn = 0 \u21d0\u21d2u = 0\n=\u21d2(\u03bb1v1 + \u00b7 \u00b7 \u00b7 + \u03bbnvn) \u00b7 u = \u03bb1v1 \u00b7 u + \u00b7 \u00b7 \u00b7 + \u03bbnvn \u00b7 u\nDefinition. The length (or norm) of a vector u \u2208Rn is defined by\n\u2225u\u2225= \u221au \u00b7 u =\nq\nu2\n1 + \u00b7 \u00b7 \u00b7 + u2n =\u21d2\n(\n\u2225u\u22252 = u \u00b7 u\n\u2225\u03bbu\u2225= |\u03bb|\u2225u\u2225\n\u2200\u03bb \u2208R\nGiven u \u2208Rn \\ {0}, we can normalise it by multiplying with\n1\n\u2225u\u2225, i.e.,\n\r\r\r u\n\u2225u\u2225\n\r\r\r = 1\n\u2200u \u2208Rn \\ {0}.\nDefinition. Given u, v \u2208Rn, the distance between u and v is defined by \u2225u \u2212v\u2225.\nDefinition. We say that u, v \u2208Rn are orthogonal, u \u22a5v, if\nu \u00b7 v = 0\nIn particular, 0 \u2208Rn is orthogonal to any vector v \u2208Rn. If u, v are both non-zero, the angle \u03b8 between u\nand v is defined by the relation (Cauchy-Schwarz inequality)\ncos (\u03b8) =\nu \u00b7 v\n\u2225u\u2225\u2225v\u2225\u2208[\u22121, 1]\nTheorem 6.1. u, v \u2208Rn are orthogonal iff \u2225u + v\u22252 = \u2225u\u22252 + \u2225v\u22252.\nProof.\n\u2225u + v\u22252 = (u + v) \u00b7 (u + v) = u \u00b7 u + 2u \u00b7 v + v \u00b7 v = \u2225u\u22252 + \u2225v\u22252 + 2u \u00b7 v\n!= \u2225u\u22252 + \u2225v\u22252 \u21d0\u21d22u \u00b7 v = 0 \u21d0\u21d2u \u22a5v\n\u25a1\nOrthogonal complements.\nDefinition. Let W be a subspace of Rn. A vector u \u2208Rn is called orthogonal to W if u\u00b7w = 0\n\u2200w \u2208W.\nThe orthogonal complement of W is defined by\nW \u22a5= {z \u2208Rn | z \u00b7 w = 0\n\u2200w \u2208W}", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_87", "text": "\u25a1\nOrthogonal complements.\nDefinition. Let W be a subspace of Rn. A vector u \u2208Rn is called orthogonal to W if u\u00b7w = 0\n\u2200w \u2208W.\nThe orthogonal complement of W is defined by\nW \u22a5= {z \u2208Rn | z \u00b7 w = 0\n\u2200w \u2208W}\nLemma 6.2.\ni) W \u22a5is a subspace of Rn, W \u2229W \u22a5= {0}.\nii) If B spans W, then W \u22a5= {z \u2208Rn | z \u00b7 b = 0\n\u2200b \u2208B}.\niii) dim\n\u0000W \u22a5\u0001\n= n \u2212dim (W).\nProof.\ni) Let w \u2208W be an arbitrary element. Then 0 \u00b7 w = 0, so 0 \u2208W \u22a5. If x, y \u2208W \u22a5and \u03bb \u2208R,\nthen x \u00b7 w = 0 and y \u00b7 w = 0, therefore (\u03bbx + y) \u00b7 w = \u03bbx \u00b7 w + y \u00b7 w = 0, so \u03bbx + y \u2208W \u22a5. So W \u22a5\nis a subspace of Rn. As the zero vector belongs to every subspace, it only remains to show that\nif w \u2208W \u2229W \u22a5, then w = 0. For such w, we have 0 = w\n\u2208W\n\u00b7 w\n\u2208W \u22a5\n= \u2225w\u22252 and therefore w = 0.", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_88", "text": "48\nKALAN WALMSLEY\nii) If B spans W, then in particular B \u2286W and therefore we have W \u22a5\u2286{z \u2208Rn | z \u00b7 b = 0\n\u2200b \u2208B}.\nTo prove the reverse inclusion, let w \u2208W and write w =\nkP\ni=1\n\u03bbibi for some bi \u2208B. If z\u00b7b = 0\n\u2200b \u2208\nB, we get that z \u00b7 w =\nkP\ni=1\n\u03bbi z \u00b7 bi\n|{z}\n=0\n= 0, so z \u2208W \u22a5.\niii) Let B = {b1, \u00b7 \u00b7 \u00b7 , bk} be a basis for W and write those vectors as columns in a matrix B.\nThis matrix is of size n \u00d7 k and has rank k. We know from ii) and theorem 6.3 that W \u22a5=\nCol (B)\u22a5= Nul\n\u0000BT \u0001\n. The matrix BT is of size k \u00d7 n and by the rank theorem we know that\nn = rank\n\u0000BT \u0001\n+ dim Nul\n\u0000BT \u0001\n= k + dim\n\u0000W \u22a5\u0001\n. Since we have k = dim (W) this proves the\nclaim.\n\u25a1\nTheorem 6.3. Let A \u2208Rm\u00d7n, then\nRow (A)\u22a5\n=\nCol\n\u0000AT \u0001\u22a5\n= Nul (A)\nCol (A)\u22a5= Nul\n\u0000AT \u0001\nProof. The first statement implies the second one since\n\u0000AT \u0001T = A. We just need to show that Row (A)\u22a5=\nNul (A). Let v \u2208Row (A)\u22a5, then Av =\n\uf8eb\n\uf8ec\n\uf8ed\na1,1\n\u00b7 \u00b7 \u00b7\na1,n\n...\nam,1\n\u00b7 \u00b7 \u00b7\nam,n\n\uf8f6\n\uf8f7\n\uf8f8v =\n\uf8eb\n\uf8ec\n\uf8ed\n0\n...\n0\n\uf8f6\n\uf8f7\n\uf8f8=\u21d2v \u2208Nul (A). Let v \u2208Nul (A),\nthen Av =\n\uf8eb\n\uf8ec\n\uf8ed\n0\n...\n0\n\uf8f6\n\uf8f7\n\uf8f8=\u21d2v \u00b7 Rowi (A) = 0\n\u2200i \u2208[[1, m]]\nlem 6.2\n=====\u21d2v \u2208Row (A)\u22a5.\n\u25a1\n6.2. Orthogonal sets.\nDefinition. A set {u1, \u00b7 \u00b7 \u00b7 , uk} \u2286Rn is called orthogonal if\nui \u00b7 uj = 0\n\u2200i \u0338= j\nTheorem 6.4. If S = {u1, \u00b7 \u00b7 \u00b7 , uk} \u2286Rn is orthogonal and uj \u0338= 0\n\u2200j, then S is linearly independent.\nProof. Assume that 0 = \u03bb1u1 + \u00b7 \u00b7 \u00b7 + \u03bbkuk, then\n0\n\u2208R= 0\n\u2208Rn", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_89", "text": "Theorem 6.4. If S = {u1, \u00b7 \u00b7 \u00b7 , uk} \u2286Rn is orthogonal and uj \u0338= 0\n\u2200j, then S is linearly independent.\nProof. Assume that 0 = \u03bb1u1 + \u00b7 \u00b7 \u00b7 + \u03bbkuk, then\n0\n\u2208R= 0\n\u2208Rn\n\u00b7 uj = (\u03bb1u1 + \u00b7 \u00b7 \u00b7 + \u03bbkuk) \u00b7 uj = \u03bb1 (u1 \u00b7 uj\n| {z }\n=0\nif j\u0338=1\n) + \u00b7 \u00b7 \u00b7 + \u03bbk (uk \u00b7 uj\n| {z }\n=0\nif j\u0338=k\n)\nS orthogonal\n=\n\u03bbjuj \u00b7 uj = \u03bbj \u2225uj\u22252\n| {z }\n\u0338=0\nsince uj\u0338=0\n=\u21d2\u03bbj = 0\n\u2200j\n\u25a1\nDefinition. Let W be a subspace of Rn. An orthogonal basis for W is a basis for W which is also\northogonal.\nTheorem 6.5. Let W be a subspace of Rn and let B = {u1, \u00b7 \u00b7 \u00b7 , uk} be an orthogonal basis for W. Then\n\u2200w \u2208W\nw = \u03bb1u1 + \u00b7 \u00b7 \u00b7 + \u03bbkuk with \u03bbi = wi\u00b7ui\nui\u00b7ui .\nProof. B basis =\u21d2w = \u03bb1u1 + \u00b7 \u00b7 \u00b7 + \u03bbkuk for some \u03bb1, \u00b7 \u00b7 \u00b7 , \u03bbk \u2208R\nas for\n=====\u21d2\nthm 6.4\nw \u00b7 ui = \u03bbiui \u00b7 ui \u21d0\u21d2\u03bbi =\nwi\u00b7ui\nui\u00b7ui .\n\u25a1", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_90", "text": "LECTURE NOTES\n49\nExample. B =\n\u001a\u00121\n2\n\u0013\n,\n\u0012\u22122\n1\n\u0013\u001b\n, x =\n\u00121\n1\n\u0013\n\u21d2[x]B =\n\uf8eb\n\uf8ec\n\uf8ed\nx\u00b7( 1\n2)\n5\nx\u00b7\n\u0010 \u22122\n1\n\u0011\n5\n\uf8f6\n\uf8f7\n\uf8f8=\n\u0012 3\n5\n\u22121\n5\n\u0013\nGeometric interpretation. Consider a non-zero vector u \u2208Rn. Let us try to decompose any x \u2208Rn\ninto a part proportional to u and a part which is orthogonal to u. =\u21d2x = \u03b1u + \u02dcu, \u02dcu \u2208span {u}\u22a5.\nIf this is true, then 0 = (x \u2212\u03b1u) \u00b7 u = x \u00b7 u \u2212\u03b1u \u00b7 u\n=\u21d2\n\u03b1 =\nx\u00b7u\nu\u00b7u, \u02dcu = x \u2212\u03b1u = x \u2212x\u00b7u\nu\u00b7uu,\nso we have that Pu(x) :\u2212x \u00b7 u\nu \u00b7 u\n| {z }\n\u2208R\nu, Pu : Rn \u2192Rn is linear. Replacing u with \u03bbu, \u03bb \u0338= 0, we obtain\nP\u03bbu(x) =\nx\u00b7\u03bbu\n\u03bbu\u00b7\u03bbu\u03bbu = x\u00b7u\nu\u00b7u = Pu(x) =\u21d2Pu only depends on L = span {u}.\nDefinition. Given a one dimensional subspace L of Rn with a basis vector u \u2208L \\ {0}, the linear map\nprojL (x) = x \u00b7 u\nu \u00b7 uu\nis called the orthogonal projection onto L (the vector x \u2212projL (x) is orthogonal to L).\nRemark. So theorem 6.5 decomposes a vector into k orthogonal projections onto the one dimensional\nsubspaces spanned by the single basis vectors.\nOrthonormal sets and matrices.\nDefinition. A set {u1, \u00b7 \u00b7 \u00b7 , uk} \u2282Rn is called orthonormal if it is orthogonal and \u2225uj\u2225= 1\n\u2200j. Simi-\nlarly, an orthonormal basis (ONB) of a subspace W of Rn is an orthogonal basis s.t. all its elements have\nlength one.\nRemark. An orthogonal set (without zero) or a basis can always by turned into an orthonormal set/basis\nby normalising all vectors, i.e. dividing each one by its norm.\nTheorem 6.6. A matrix U \u2208Rm\u00d7n has orthonormal columns iff\nU T U = In\nMoreover, for any such U\ni) \u2225Ux\u2225= \u2225x\u2225\nii) Ux \u00b7 Uy = x \u00b7 y", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_91", "text": "Theorem 6.6. A matrix U \u2208Rm\u00d7n has orthonormal columns iff\nU T U = In\nMoreover, for any such U\ni) \u2225Ux\u2225= \u2225x\u2225\nii) Ux \u00b7 Uy = x \u00b7 y\niii) Ux \u00b7 Uy = 0 \u21d0\u21d2x \u00b7 y = 0\nProof.\n\uf8eb\n\uf8ec\n\uf8ed\n\u00b7 \u00b7 \u00b7\nu1\n\u00b7 \u00b7 \u00b7\n...\n\u00b7 \u00b7 \u00b7\nun\n\u00b7 \u00b7 \u00b7\n\uf8f6\n\uf8f7\n\uf8f8\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ed\n...\n...\nu1\n\u00b7 \u00b7 \u00b7\nun\n...\n...\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f8=\n\uf8eb\n\uf8ec\n\uf8ed\nu1 \u00b7 u1\n\u00b7 \u00b7 \u00b7\nu1 \u00b7 un\n...\nun \u00b7 u1\n\u00b7 \u00b7 \u00b7\nun \u00b7 un\n\uf8f6\n\uf8f7\n\uf8f8\n=\u21d2U T U = In \u21d0\u21d2\n(\n\u2225uj\u22252 = 1\n\u2200j\nui \u00b7 uj = 0\n\u2200i \u0338= j\ni) Ux \u00b7 Uy = (Ux)T Uy = xT (\nIn\nz }| {\nU T U) y = xT y = x \u00b7 y\ni) \u2225Ux\u22252 = Ux \u00b7 Ux\nII)\n= x \u00b7 x = \u2225x\u22252\niii) Ux \u00b7 Uy\nii)= x \u00b7 y\n\u25a1\nDefinition. A matrix U \u2208Rn\u00d7n is called orthogonal if its columns are orthonormal. In this case\nU \u22121 = U T", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_92", "text": "50\nKALAN WALMSLEY\n6.3. Orthogonal projections.\nTheorem 6.7. Let W be a subspace of Rn, then every x \u2208Rn can be uniquely written in the form\nx = w + \u02dcw\nw \u2208W, \u02dcw \u2208W \u22a5\nIf {b1, \u00b7 \u00b7 \u00b7 , bn} is an onb for W, then\nw = x \u00b7 b1\nb1 \u00b7 b1\nb1 + \u00b7 \u00b7 \u00b7 + x \u00b7 bk\nbk \u00b7 bk\nbk\n\u02dcw = x \u2212w\nProof. Uniqueness: assume that x = w1+ \u02dcw1 = w2+ \u02dcw2, w1, w2 \u2208W, \u02dcw1, \u02dcw2 \u2208W \u22a5. So w1 \u2212w2\n|\n{z\n}\n\u2208W\n= \u02dcw1 \u2212\u02dcw2\n|\n{z\n}\n\u2208W \u22a5\n,\nbut W \u2229W \u22a5= {0} =\u21d2w1 = w2, \u02dcw1 = \u02dcw2.\nExistence: clear if W = {0} because {0}\u22a5= Rn. So assume W \u0338= {0}. Let B = {b1, \u00b7 \u00b7 \u00b7 , bk} be a basis\nfor W (not necessarily orthogonal). So by lemma 6.2, and the fact that dim (W) = dim (B) = k, we have\nthat dim\n\u0000W \u22a5\u0001\n= n \u2212k. Let Q = {bk+1, \u00b7 \u00b7 \u00b7 , bn} be a basis for W \u22a5. We show that {b1, \u00b7 \u00b7 \u00b7 , bn} is a basis\nfor Rn.\n\u03bb1b1 + \u00b7 \u00b7 \u00b7 + \u03bbkbk\n|\n{z\n}\nx\u2208W\n+ \u03bbk+1bk+1 + \u00b7 \u00b7 \u00b7 + \u03bbnbn\n|\n{z\n}\ny\u2208W \u22a5\n= 0 =\u21d2x\n\u2208W\n= \u2212y\n\u2208W \u22a5\n=\u21d2x = y = 0\nB basis =\u21d2\u03bb1 = \u00b7 \u00b7 \u00b7 = \u03bbk = 0\nQ basis =\u21d2\u03bbk+1 = \u00b7 \u00b7 \u00b7 = \u03bbn = 0\n\u001b\n=\u21d2linearly independent\nSince B \u222aQ is a bais, we can write\nx = \u00b51b1 + \u00b7 \u00b7 \u00b7 + \u00b5kbk\n|\n{z\n}\n\u2208W\n+ \u00b5k+1bk+1 + \u00b7 \u00b7 \u00b7 + \u00b5nbn\n|\n{z\n}\n\u2208W \u22a5\nwhich implies existence.\nIf B is an orthogonal basis for W, then as previously \u22001 \u2264j \u2264k, x\u00b7bj = (w + \u02dcw)\u00b7bj = w \u00b7bj + \u02dcw \u00b7 bj\n| {z }\n0\n=", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_93", "text": "|\n{z\n}\n\u2208W \u22a5\nwhich implies existence.\nIf B is an orthogonal basis for W, then as previously \u22001 \u2264j \u2264k, x\u00b7bj = (w + \u02dcw)\u00b7bj = w \u00b7bj + \u02dcw \u00b7 bj\n| {z }\n0\n=\nw \u00b7 bj = \u00b5jbj \u00b7 bj =\u21d2\u00b5j = x\u00b7bj\nbj\u00b7bj .\n\u25a1\nRemark. The uniqueness property of theorem 6.7 shows that the formula w = x\u00b7b1\nb1\u00b7b1 b1 +\u00b7 \u00b7 \u00b7+ x\u00b7bk\nbk\u00b7bk bk does\nnot depend on the basis chosen. We define projW (x) = x\u00b7b1\nb1\u00b7b1 b1 + \u00b7 \u00b7 \u00b7 + x\u00b7bk\nbk\u00b7bk bk. This is a linear map from\nRn to itself. By uniqueness, projW (w) = w\n\u2200w \u2208W.\nTheorem 6.8. Let {u1, \u00b7 \u00b7 \u00b7 , uk} be an onb for a subspace W of Rn, then\nprojW (x) = (x \u00b7 u1)u1 + \u00b7 \u00b7 \u00b7 + (x \u00b7 uk)uk\nWriting U =\n\u0000u1\n\u00b7 \u00b7 \u00b7\nuk\n\u0001\n\u2208Rn\u00d7k, then also\nprojW (x) = UU T x\nProof. \u2200j\n\u2225uj\u2225= 1 =\u21d2uj \u00b7 uj = 1, so we only need to show that UU T x = projW (x).\nUU T =\nU\n\uf8eb\n\uf8ec\n\uf8ed\n\u00b7 \u00b7 \u00b7\nu1\n\u00b7 \u00b7 \u00b7\n...\n\u00b7 \u00b7 \u00b7\nuk\n\u00b7 \u00b7 \u00b7\n\uf8f6\n\uf8f7\n\uf8f8x = U\n\uf8eb\n\uf8ec\n\uf8ed\nu1 \u00b7 x\n...\nuk \u00b7 x\n\uf8f6\n\uf8f7\n\uf8f8= (u1 \u00b7 x)u1 + \u00b7 \u00b7 \u00b7 + (uk \u00b7 x)uk\n= projW (x)\n\u25a1", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_94", "text": "LECTURE NOTES\n51\nOrthogonal projection as closest point.\nTheorem 6.9. Let W be a subspace of Rn. For some x \u2208Rn, we write x = w + \u02dcw, w \u2208W \u22a5. Then\n\u2225x \u2212w\u2225< \u2225x \u2212w\u2032\u2225\n\u2200w\u2032 \u2208W \\ {w}\nProof. Let w\u2032 \u2208W \\ {w} =\u21d2w \u2212w\u2032 \u2208W, x \u2212w = \u02dcw \u2208W \u22a5=\u21d2(w \u2212w\u2032) \u00b7 (x \u2212w) = 0, so\n\u2225x \u2212w\u2032\u22252 = \u2225x \u2212w + w \u2212w\u2032\u22252 thm 6.1\n=\n\u2225x \u2212w\u22252 + \u2225w \u2212w\u2032\u22252\n|\n{z\n}\n>0\n> \u2225x \u2212w\u22252\n\u25a1\n6.4. The Gram-Schmidt process.\nTheorem 6.10. Let W be a subspace of Rn and {b1, \u00b7 \u00b7 \u00b7 , bk} be a basis for W, then define\nu1 = b1\nu2 = b2 \u2212b2 \u00b7 u1\nu1 \u00b7 u1\nu1\nu3 = b3 \u2212b3 \u00b7 u1\nu1 \u00b7 u1\nu1 \u2212b3 \u00b7 u2\nu2 \u00b7 u2\nu2\n...\nuk = bk \u2212bk \u00b7 u1\nu1 \u00b7 u1\nu1 \u2212\u00b7 \u00b7 \u00b7 \u2212\nbk \u00b7 uk\u22121\nuk\u22121 \u00b7 uk\u22121\nuk\u22121\n= bk \u2212projspan{u1,\u00b7\u00b7\u00b7 ,uk\u22121} (bk)\nThen family {u1, \u00b7 \u00b7 \u00b7 , uk} is an orthogonal basis for W. Moreover, span {u1, \u00b7 \u00b7 \u00b7 , up} = span {b1, \u00b7 \u00b7 \u00b7 , bp}\n\u22001 \u2264\np \u2264k.\nProof. For 1 \u2264p \u2264k, define Wp = span {b1, \u00b7 \u00b7 \u00b7 , bp}. Now, induction:\np = 1 =\u21d2span {u1} = W1 and {u1} orthogonal.\np \u22121 7\u2192p: Assume that span {u1, \u00b7 \u00b7 \u00b7 , up\u22121} = Wp\u22121 and that family {u1, \u00b7 \u00b7 \u00b7 , up\u22121} is an orthogonal\nbasis for Wp\u22121. Then by definition we have up\nind\n= bp\u2212projWp\u22121 (bp) \u2208W \u22a5\np\u22121 =\u21d2{u1, \u00b7 \u00b7 \u00b7 , up} orthogonal.\nIs up = 0 ? No, otherwise bp = projWp\u22121 (bp) \u2208Wp\u22121\n=\u21d2\nb1, \u00b7 \u00b7 \u00b7 , bp not linearly independent", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_95", "text": "p\u22121 =\u21d2{u1, \u00b7 \u00b7 \u00b7 , up} orthogonal.\nIs up = 0 ? No, otherwise bp = projWp\u22121 (bp) \u2208Wp\u22121\n=\u21d2\nb1, \u00b7 \u00b7 \u00b7 , bp not linearly independent\n=\u21d2up \u0338= 0 =\u21d2{u1, \u00b7 \u00b7 \u00b7 , up} linearly independent.\nFinally, up = bp\n\u2208Wp\n\u2212projWp\u22121 (bp)\n\u2208Wp\u22121\u2286Wp\n\u2208Wp =\u21d2span {u1, \u00b7 \u00b7 \u00b7 , up} = Wp.\n\u25a1\nCorollary 6.11. Every non-zero subspace of Rn has an onb.\nProof. W has a basis\nthm 6.10\n======\u21d2W has an orthogonal basis\nnormalisation\n=========\u21d2W has an onb.\n\u25a1\nQR-factorisation of matrices.\nTheorem 6.12. Let A \u2208Rm\u00d7n have linearly independent columns, then we can write\nA = QR\nwhere Q \u2208Rm\u00d7n has as columns an ONB of Col (A), and W \u2208Rn\u00d7n is an upper triangular matrix s.t.\nall diagonal entries are positive.\nProof. We apply Gram-Schmidt to the columns of A and normalise the output so that we get an ONB\nof Col (A). Write A =\n\u0000A1\n\u00b7 \u00b7 \u00b7\nAn\n\u0001\nand onb = {q1, \u00b7 \u00b7 \u00b7 , qn}. If Ak \u00b7 qk \u22640, multiply qk by \u22121 =\u21d2\nwe obtain an onb {q1, \u00b7 \u00b7 \u00b7 , qn} s.t.\ni) Ak \u00b7 qk \u22650\nii) span {A1, \u00b7 \u00b7 \u00b7 , An} = span {q1, \u00b7 \u00b7 \u00b7 , qn}", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_96", "text": "52\nKALAN WALMSLEY\nDefine\nQ =\n\u0000q1\n\u00b7 \u00b7 \u00b7\nqn\n\u0001\n\u2208Rm\u00d7n\nWe know that Ak\nii)\n\u2208span {q1, \u00b7 \u00b7 \u00b7 , qk}, so Ak = (Ak \u00b7 q1) q1 + \u00b7 \u00b7 \u00b7 + (Ak \u00b7 qk) qk. Define\nR =\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\nA1 \u00b7 q1\nA2 \u00b7 q1\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\nAn \u00b7 q1\n0\nA2 \u00b7 q2\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n...\n...\n0\n...\n...\n...\n...\n...\n...\n0\n0\nAn \u00b7 qn\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n\u2208Rn\u00d7n\n=\u21d2QR = A\nDiagonal entries of R: Ak\u00b7qk\ni)\n\u22650. Can we have Ak\u00b7qk = 0 ? No because otherwise, Ak \u2208span {q1, \u00b7 \u00b7 \u00b7 , qk\u22121} =\nspan {A1, \u00b7 \u00b7 \u00b7 , Ak\u22121} but the columns of A were assumed to be linearly independent.\n\u25a1\nRemark. The above proof almost gives a formula for Q and R. For R, see above matrix. Let ok be the\nkth output of Gram-Schmidt, then\nqk = sgn (Ak \u00b7 ok)\n\u2225ok\u2225\nok\nExample. A =\n\uf8eb\n\uf8ed\n1\n2\n2\n1\n1\n0\n\uf8f6\n\uf8f8\nu1 =\n\uf8eb\n\uf8ed\n1\n2\n1\n\uf8f6\n\uf8f8\nu2 =\n\uf8eb\n\uf8ed\n2\n1\n0\n\uf8f6\n\uf8f8\u2212\n\u0010 2\n1\n0\n\u0011\n\u00b7\n\u0010 1\n2\n1\n\u0011\n6\n\uf8eb\n\uf8ed\n1\n2\n1\n\uf8f6\n\uf8f8= 1\n3\n\uf8eb\n\uf8ed\n4\n\u22121\n\u22122\n\uf8f6\n\uf8f8\n=\u21d2q1 =\n1\n\u221a\n6\n\uf8eb\n\uf8ed\n1\n2\n1\n\uf8f6\n\uf8f8, A1 \u00b7 q1 > 0\nq2 =\n1\n\u221a\n21\n\uf8eb\n\uf8ed\n4\n\u22121\n\u22122\n\uf8f6\n\uf8f8, A2 \u00b7 q2 > 0\n=\u21d2Q =\n\uf8eb\n\uf8ec\n\uf8ed\n1\n\u221a\n6\n4\n\u221a\n21\n2\n\u221a\n6\n\u22121\n\u221a\n21\n1\n\u221a\n6\n\u22122\n\u221a\n21\n\uf8f6\n\uf8f7\n\uf8f8, R =\n \u221a\n6\n4\n\u221a\n6\n0\n7\n\u221a\n21\n!\n6.5. Least-squares problem. Problem: Ax = b has no solution (usually because too few variables)\n=\u21d2find a good approximation, i.e., minimise \u2225Ax \u2212b\u2225over x \u2208Rn.\nDefinition. Let A \u2208Rm\u00d7n and b \u2208Rm. A least-squares solution of Ax = b is an x0 \u2208Rn s.t.\n\u2225Ax0 \u2212b\u2225\u2264\u2225Ax \u2212b\u2225\n\u2200x \u2208Rn.\nRemark. Ax \u2208Col (A)\n\u2200x \u2208Rn\n=\u21d2we look for the closest point to b in Col (A). A least-squares", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_97", "text": "\u2225Ax0 \u2212b\u2225\u2264\u2225Ax \u2212b\u2225\n\u2200x \u2208Rn.\nRemark. Ax \u2208Col (A)\n\u2200x \u2208Rn\n=\u21d2we look for the closest point to b in Col (A). A least-squares\nsolution satisfies Ax0 = projCol (A) (b) and, vice versa, every solution of Ax0 = projCol (A) (b) is a least-\nsquares solution.\nTheorem 6.13. The set of least-squares solutions of Ax = b coincides with the (non-empty) set of\nsolutions of the equation\nAT Ax = AT b\nProof. We already know x0 least-squares solution of Ax = b\n\u21d0\u21d2\nAx0 = projCol (A) (b). Next,\nb \u2212projCol (A) (b)\nthm 6.7\n\u2208\nCol (A)\u22a5\nthm 6.3\n=\nNul\n\u0000AT \u0001\n\u21d2AT \u0010\nb \u2212projCol (A) (b)\n\u0011\n= 0\n=\u21d2\nAT b =\nAT projCol (A) (b) = AT Ax0.", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_98", "text": "LECTURE NOTES\n53\nNow assume that AT Ax0 = AT b \u21d2Ax0 \u2212b \u2208Nul\n\u0000AT \u0001\n= Col (A)\u22a5\u21d2b = b \u2212Ax0\n|\n{z\n}\n\u2208Col(A)\u22a5\n+ Ax0\n|{z}\n\u2208Col (A)\nthm 6.7\n=====\u21d2\nAx0 = projCol (A) (b) =\u21d2x0 least-squares solution. Existence: projCol (A) (b) \u2208Col (A) =\u21d2\u2203x0 \u2208Rn\nsuch that Ax0 = projCol (A) (b).\n\u25a1\nExample. A =\n\uf8eb\n\uf8ed\n1\n0\n2\n\u22121\n1\n2\n\uf8f6\n\uf8f8\nb =\n\uf8eb\n\uf8ed\n2\n2\n0\n\uf8f6\n\uf8f8\nLeast-squares solution: AT A =\n\u00126\n0\n0\n5\n\u0013\nAT b =\n\u0012 6\n\u22122\n\u0013\nNeed to solve AT Ax = AT b\nrow\n======\u21d2\nreduction\nx0 =\n\u0012 1\n\u22122\n5\n\u0013\nLeast-squares error: \u2225Ax0 \u2212b\u2225\nRemark. AT A does not have to be invertible.\nTheorem 6.14. Let A \u2208Rm\u00d7n, then the following are equivalent:\ni) Ax = b has a unique least-squares solution \u2200b \u2208Rm\nii) The columns of A are linearly independent\niii) AT A invertible\nProof.\ni) \u21d2ii)\nLet b = 0 =\u21d2least-squares solutions of Ax = 0 are exactly the vectors in Nul (A).\nUniqueness of solution =\u21d2Nul (A) = {0} =\u21d2columns of A are linearly independent.\nii) \u21d2iii)\nAT A \u2208Rn\u00d7n is invertible iff Nul\n\u0000AT A\n\u0001\n= {0}. Assume that AT Ax = 0 =\u21d2xT AT Ax =\n0 = (Ax)T Ax = \u2225Ax\u22252 =\u21d2Ax = 0\nii)\n=\u21d2x = 0.\niii) \u21d2i)\nHere, AT Ax = AT b has the unique solution\n\u0000AT A\n\u0001\u22121 AT b.\n\u25a1\nAlternative methods for least-squares solutions. Suppose that A \u2208Rm\u00d7n has orthogonal columns,\nA =\n\u0000A1\n\u00b7 \u00b7 \u00b7\nAn\n\u0001\n, Aj \u0338= 0\n\u2200j, so\nprojCol (A) (b) =\nb \u00b7 A1\nA1 \u00b7 A1\nA1\n+ \u00b7 \u00b7 \u00b7 +\nb \u00b7 An\nAn \u00b7 An\nAn\nAx =\nx1A1\n+ \u00b7 \u00b7 \u00b7 +\nxnAn\n\uf8fc\n\uf8fd\n\uf8fe=\u21d2x =\n\uf8eb\n\uf8ec\n\uf8ed\nb\u00b7A1\nA1\u00b7A1\n...\nb\u00b7An\nAn\u00b7An\n\uf8f6\n\uf8f7\n\uf8f8", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_99", "text": "\u2200j, so\nprojCol (A) (b) =\nb \u00b7 A1\nA1 \u00b7 A1\nA1\n+ \u00b7 \u00b7 \u00b7 +\nb \u00b7 An\nAn \u00b7 An\nAn\nAx =\nx1A1\n+ \u00b7 \u00b7 \u00b7 +\nxnAn\n\uf8fc\n\uf8fd\n\uf8fe=\u21d2x =\n\uf8eb\n\uf8ec\n\uf8ed\nb\u00b7A1\nA1\u00b7A1\n...\nb\u00b7An\nAn\u00b7An\n\uf8f6\n\uf8f7\n\uf8f8\nThis solution is unique, since we assume that A1, \u00b7 \u00b7 \u00b7 , An are linearly independent. In general, we can use\nthe QR-factorisation\nA = QR =\u21d2QT A = QT Q\n| {z }\nIn\nR = R\nTheorem 6.15. Let A \u2208Rm\u00d7n have linearly independent columns and let A = QR be a QR-factorisation\nof A. Then \u2200b \u2208Rm, the unique least-squares solution of Ax = b is given by x = R\u22121QT b.\nProof. Uniqueness follows from theorem 6.14.\nAx = AR\u22121QT b = Q RR\u22121\n| {z }\nIn\nQT b = QQT b =\n\u2191\nthm 6.12\nsince the columns\nof Q are an\nonb for Col (A)\nprojCol (A) (b)\n\u25a1", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_100", "text": "54\nKALAN WALMSLEY\n6.6. Applications to linear models. Framework: for some x1, \u00b7 \u00b7 \u00b7 , xm \u2208R (e.g. timings), we have\nmeasurements y1, \u00b7 \u00b7 \u00b7 , ym \u2208R. Assumption: y depends on x, but measurements can contain noise =\u21d2\nfind best linear model given. So we try to search for f(x) = ax + b (line of regression).\naxi + b\nb= ideal value\nyi\nb= measurement\n)\nminimise |axi + b \u2212yi|\n=\u21d2minimise |ax1 + b \u2212y1|2 + \u00b7 \u00b7 \u00b7 + |axm + b \u2212ym|2\nVariables:\n\u0012a\nb\n\u0013\n\u2208R2\nData: A =\n\uf8eb\n\uf8ec\n\uf8ed\nx1\n1\n...\n...\nxm\n1\n\uf8f6\n\uf8f7\n\uf8f8,\n\uf8eb\n\uf8ec\n\uf8ed\ny1\n...\nym\n\uf8f6\n\uf8f7\n\uf8f8= y\n=\u21d2minimise\n\r\r\r\rA\n\u0012a\nb\n\u0013\n\u2212y\n\r\r\r\r\n2\nLine of regression: f(x) = ax + b, where\n\u0012\na\nb\n\u0013\n\u2208R2 solves\nAT A\n\u0012a\nb\n\u0013\n= AT y\nExample. x =\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ed\n1\n2\n3\n4\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f8, y =\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ed\n0\n3\n3\n2\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f8, A =\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ed\n1\n1\n2\n1\n3\n1\n4\n1\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f8\n=\u21d2AT A =\n\u001230\n10\n10\n4\n\u0013\n, AT y =\n\u001223\n8\n\u0013\n=\u21d2AT A\n\u0012a\nb\n\u0013\n= AT y has the unique solution\n\u0012a\nb\n\u0013\n=\n\u0012 3\n51\n2\n\u0013\n.", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_101", "text": "LECTURE NOTES\n55\n7. Symmetric matrices, SVD and the exponential of a matrix\n7.1. Symmetric matrices.\nDefinition. A matrix A \u2208Rn\u00d7n is called symmetric if\nAT = A\nRemark. The set of symmetric matrices is a subspace of Rn\u00d7n. Its dimension is (n+1)n\n2\n.\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n\u22c6\n\u22c6\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u22c6\n\u22c6\n\u22c6\n\u22c6\n\u22c6\n\u22c6\n\u22c6\n\u22c6\n...\n\u22c6\n...\n...\n...\n...\n...\n...\n\u22c6\n\u22c6\n...\n\u22c6\n\u22c6\n\u22c6\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n=\u21d2\n\uf8f1\n\uf8f2\n\uf8f3\n1 + 2 + 3 + \u00b7 \u00b7 \u00b7 + (n \u22121) + n\n=\nn\nX\ni=1\ni = (n + 1)n\n2\nRemark. A symmetric and invertible =\u21d2A\u22121 symmetric. Indeed,\n\u0000A\u22121\u0001T =\n\u0000AT \u0001\u22121 = A\u22121\nEvery symmetric matrix is diagonalizable and the eigenvectors can be chosen as an onb for Rn.\nThis means A = QDQT , whith Q an orthogonal matrix, D diagonal.\nRemark. Whenever A = QDQT ad above, then we necessarily have AT =\n\u0000QDQT \u0001T =\n\u0000QT \u0001T DT QT =\nQDQT = A =\u21d2the symmetry of A is both necessary and sufficient.\nDefinition. We say that A \u2208Rn\u00d7n is orthogonally diagonalizable if \u2203an orthogonal matrix Q \u2208Rn\u00d7n\n(i.e. Q\u22121 = QT ) s.t. A = QDQT for a diagonal matrix D \u2208Rn\u00d7n.\nRemark. We already know that D has to contain the eigenvalues of A. Moreover, the columns of Q have\nto be linearly independent eigenvectors. Orthogonaly diagonalisation thus requires in addition an onb of\neigenvectors.\nTheorem 7.1. The spectral theorem\nA matrix A \u2208Rn\u00d7n is orthogonally diagonalizable iff AT = A. In particular, if AT = A, then\ni) A has n real eigenvalues (counted whith multiplicity)\nii) multgeomA (\u03bb) = multalgA (\u03bb)\n\u2200\u03bb \u2208\u03c3(A)\niii) If vi \u2208Rn is an eigenvector associated to \u03bbi \u0338= \u03bbj, then vi \u00b7 vj = 0\nProof. i), ii), iii) follow the fact that A is orthogonally diagonalizable. Here is a direct proof for iii)", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_102", "text": "\u2200\u03bb \u2208\u03c3(A)\niii) If vi \u2208Rn is an eigenvector associated to \u03bbi \u0338= \u03bbj, then vi \u00b7 vj = 0\nProof. i), ii), iii) follow the fact that A is orthogonally diagonalizable. Here is a direct proof for iii)\n\u03bbivi \u00b7 vj = Avi \u00b7 vj = (Avi)T vj = vT\ni AT vj = vT\ni Avj = \u03bbjvT\ni vj = \u03bbjvi \u00b7 vj\n=\u21d2(\u03bbi \u2212\u03bbj) vi \u00b7 vj = 0\n\u03bbi\u0338=\u03bbj\n====\u21d2vi \u00b7 vj = 0\nWe still need to prove AT = A =\u21d2\u2203Q \u2208Rn\u00d7n orthogonal, D \u2208Rn\u00d7n diagonal s.t. A = QDQT .", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_103", "text": "56\nKALAN WALMSLEY\nStep 1: eigenvalues are real. Let v \u2208Cn be an eigenvector with eigenvalue \u03bb \u2208\u03c3(A). We need to show\nthat \u03bb = \u03bb.\nAv = \u03bbv =\u21d2vT Av = vT \u03bbv = \u03bbvT v = \u03bb\n\u0000v1\n\u00b7 \u00b7 \u00b7\nvn\n\u0001\n\uf8eb\n\uf8ec\n\uf8ed\nv1\n...\nvn\n\uf8f6\n\uf8f7\n\uf8f8\n= \u03bb (v1v1 + \u00b7 \u00b7 \u00b7 + vnvn) = \u03bb\n\u0010\n|v1|2 + \u00b7 \u00b7 \u00b7 + |vn|2\u0011\n|\n{z\n}\n\u2208R\\{0}\n=\u21d2\u03bbvT v = \u03bb vT v = vT Av = vT Av = vT Av = vT AT v\n= (Av)T v = (\u03bbv)T v = \u03bb vT v\n|{z}\n=vT v\nvT v\u0338=0\n====\u21d2\u03bb = \u03bb\n=\u21d2all eigenvalues are real.\nStep 2: induction on n.\nn = 1 : a1,1 = 1 \u00b7 a1,1 \u00b7 1\nn \u22121 7\u2192n : fundamental theorem of algebra\n=\u21d2\nA has at least one eigenvalue \u03bb1 \u2208\nC\nstep 1\n====\u21d2\u03bb1 \u2208R. Let v1 \u2208Rn be a normalised eigenvector for \u03bb1. Gram-Schmidt\n=\u21d2\n\u2203\nonb {v1, v2, \u00b7 \u00b7 \u00b7 , vn} of Rn. Attention, v2, \u00b7 \u00b7 \u00b7 , vn are not necessarily eigenvectors of A. Write\nP =\n\u0000v1\n\u00b7 \u00b7 \u00b7\nvn\n\u0001\n\u2208Rn\u00d7n =\u21d2P orthogonal. Define B = P T AP\n=\u21d2BT =\n\u0000P T AP\n\u0001T =\nP T AT P = P T AP = B. First column of B :\nBe1 = P T APe1 = P T Av1 = P T \u03bb1v1 = \u03bb1P T v1\nP orthogonal\n=\n\u03bb1P \u22121v1\nP e1=v1\n=\n\u03bb1e1 =\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n\u03bb1\n0\n...\n0\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\nBT =B\n=====\u21d2B =\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n\u03bb1\n0\n\u00b7 \u00b7 \u00b7\n0\n0\nC\n...\n0\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\nwith C \u2208R(n\u22121)\u00d7(n\u22121) symmetric. The induction hypothesis applied to C shows that QT\n1 CQ1 =\nD1, Q1 \u2208R(n\u22121)\u00d7(n\u22121) orthogonal , D1 \u2208R(n\u22121)\u00d7(n\u22121) diagonal.\nDefine U =\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_104", "text": "1 CQ1 =\nD1, Q1 \u2208R(n\u22121)\u00d7(n\u22121) orthogonal , D1 \u2208R(n\u22121)\u00d7(n\u22121) diagonal.\nDefine U =\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n1\n0\n\u00b7 \u00b7 \u00b7\n0\n0\nQ1\n...\n0\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n\u21d2U has orthonormal columns\n=\u21d2\nU orthogonal\n=\u21d2\nQ = PU is also orthogonal. Finally,", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_105", "text": "LECTURE NOTES\n57\nQT AQ = (PU)T APU = U T\nB\nz }| {\nP T AP U = U T BU =\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n1\n0\n\u00b7 \u00b7 \u00b7\n0\n0\nQT\n1\n...\n0\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n\u03bb1\n0\n\u00b7 \u00b7 \u00b7\n0\n0\nC\n...\n0\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n1\n0\n\u00b7 \u00b7 \u00b7\n0\n0\nQ1\n...\n0\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n=\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n\u03bb1\n0\n\u00b7 \u00b7 \u00b7\n0\n0\nQT\n1 CQ1\n...\n0\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n=\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n\u03bb1\n0\n\u00b7 \u00b7 \u00b7\n0\n0\nD1\n...\n0\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\nwhich is diagonal.\n\u25a1\nHow to orthogonally diagonalize a matrix.\n(1) Diagonalize as usual (eigenvalues, eigenvectors, etc.)\n(2) Apply Gram-Schmidt and normalisation in each eigenspace separately\n(3) Put the output as columns in Q\nExample. A =\n\uf8eb\n\uf8ed\n\u22124\n3\n0\n3\n4\n0\n0\n0\n5\n\uf8f6\n\uf8f8\n\u2022 Eigenvalues: \u03c7A(\u03bb) = (\u03bb \u22125)2 (\u03bb + 5) =\u21d2\n(\n\u03bb1 = 5\n\u03bb2 = \u22125\n\u2022 Eigenvectors:\n\u03bb1:\n\uf8eb\n\uf8ed\n\u22129\n3\n0\n3\n\u22121\n0\n0\n0\n0\n\uf8f6\n\uf8f8\u2192\n\uf8eb\n\uf8ed\n\u22129\n3\n0\n0\n0\n0\n0\n0\n0\n\uf8f6\n\uf8f8=\u21d2v1 =\n\uf8eb\n\uf8ed\n0\n0\n1\n\uf8f6\n\uf8f8, v2 =\n\uf8eb\n\uf8ed\n1\n3\n0\n\uf8f6\n\uf8f8\n\u03bb2:\n\uf8eb\n\uf8ed\n1\n3\n0\n3\n9\n0\n0\n0\n10\n\uf8f6\n\uf8f8\u2192\n\uf8eb\n\uf8ed\n1\n3\n0\n0\n0\n0\n0\n0\n1\n\uf8f6\n\uf8f8=\u21d2v3 =\n\uf8eb\n\uf8ed\n3\n\u22121\n0\n\uf8f6\n\uf8f8\n\u2022 Normalisation: Q =\n\uf8eb\n\uf8ed\n0\n1\n\u221a\n10\n3\n\u221a\n10\n0\n3\n\u221a\n10\n\u22121\n\u221a\n10\n1\n0\n0\n\uf8f6\n\uf8f8, D =\n\uf8eb\n\uf8ed\n5\n0\n0\n0\n5\n0\n0\n0\n\u22125\n\uf8f6\n\uf8f8\nRemark.\n\u0012\na\nb\n\u0013\n\u22a5\n\u0012\n\u2212b\na\n\u0013\n\uf8eb\n\uf8ed\na\nb\nc\n\uf8f6\n\uf8f8\u22a5\n\uf8eb\n\uf8ed\n0\n\u2212c\nb\n\uf8f6\n\uf8f8", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_106", "text": "58\nKALAN WALMSLEY\nGeometric interpretation of AT = A. If A = QDQT , Q orthogonal, D diagonal, then\nA =\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ed\n...\n...\nv1\n\u00b7 \u00b7 \u00b7\nvn\n...\n...\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f8\n\uf8eb\n\uf8ec\n\uf8ed\n\u03bb1\n...\n\u03bbn\n\uf8f6\n\uf8f7\n\uf8f8\n\uf8eb\n\uf8ec\n\uf8ed\n\u00b7 \u00b7 \u00b7\nv1\n\u00b7 \u00b7 \u00b7\n...\n\u00b7 \u00b7 \u00b7\nvn\n\u00b7 \u00b7 \u00b7\n\uf8f6\n\uf8f7\n\uf8f8\n=\n\u0000\u03bb1v1\n\u00b7 \u00b7 \u00b7\n\u03bbnvn\n\u0001\n\uf8eb\n\uf8ec\n\uf8ed\n\u00b7 \u00b7 \u00b7\nv1\n\u00b7 \u00b7 \u00b7\n...\n\u00b7 \u00b7 \u00b7\nvn\n\u00b7 \u00b7 \u00b7\n\uf8f6\n\uf8f7\n\uf8f8= \u03bb1v1vT\n1 + \u00b7 \u00b7 \u00b7 + \u03bbnvnvT\nn\nThe matrix vivT\ni is s.t. it projects onto span {vi}. vi \u2208Rn\u00d71, vT\ni \u2208R1\u00d7n =\u21d2vivT\ni \u2208Rn\u00d7n. =\u21d2A is a\nweighted sum of orthogonal projections onto 1D eigenspaces.\nExample.\n\uf8eb\n\uf8ed\na1\na2\na3\n\uf8f6\n\uf8f8\u0000b1\nb2\nb3\n\u0001\n=\n\uf8eb\n\uf8ed\na1b1\na1b2\na1b3\na2b1\na2b2\na2b3\na3b1\na3b2\na3b3\n\uf8f6\n\uf8f8\nFor A = QDQT , Q orthogonal, D diagonal,\nA = \u03bb1v1vT\n1 + \u00b7 \u00b7 \u00b7 + \u03bbnvnvT\nn\nis called the spectral decomposition of A.\nExamples of symmetric matrices.\n(1) Reflection matrices of the form In\u2212UU T are symmetric. Indeed,\n\u0000In \u2212UU T \u0001T = IT\nn \u2212\n\u0000UU T \u0001T =\nIn \u2212\n\u0000U T \u0001T U T = In \u2212UU T .\n(2) Projection matrices of the form aaT are symmetric \u2200a \u2208Rn.\n(3) Given A \u2208Rm\u00d7n, the matrix AT A \u2208Rn\u00d7n is symmetric. Indeed,\n\u0000AT A\n\u0001T = AT \u0000AT \u0001T = AT A.\n7.2. The singular value decomposition. Can we factorise a non-symmetric matrix into orthogonal\nand diagonal matrices ? Yes, in the form A = U\u03a3V T for U, V orthogonal, \u03a3 \"diagonal\". The general\nstructure of a singular value decomposition A \u2208Rm\u00d7n is (the \"diagonal\" of \u03a3 depends on n and m, here\nm > n).\n\uf8eb\n\uf8ec\n\uf8ed\na1,1\n\u00b7 \u00b7 \u00b7\na1,n\n...\nam,1\n\u00b7 \u00b7 \u00b7\nam,n\n\uf8f6\n\uf8f7\n\uf8f8=\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ed\n...\n...\nu1\n\u00b7 \u00b7 \u00b7\num\n...\n...\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f8\n\u2208Rm\u00d7m\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n\u22c6\n0\n\u00b7 \u00b7 \u00b7\n0\n0\n...\n...\n...", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_107", "text": "\uf8eb\n\uf8ec\n\uf8ed\na1,1\n\u00b7 \u00b7 \u00b7\na1,n\n...\nam,1\n\u00b7 \u00b7 \u00b7\nam,n\n\uf8f6\n\uf8f7\n\uf8f8=\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ed\n...\n...\nu1\n\u00b7 \u00b7 \u00b7\num\n...\n...\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f8\n\u2208Rm\u00d7m\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n\u22c6\n0\n\u00b7 \u00b7 \u00b7\n0\n0\n...\n...\n...\n...\n0\n...\n\u22c6\n...\n0\n...\n...\n0\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n0\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n\u2208Rm\u00d7n\n\uf8eb\n\uf8ec\n\uf8ed\n\u00b7 \u00b7 \u00b7\nv1\n\u00b7 \u00b7 \u00b7\n...\n\u00b7 \u00b7 \u00b7\nvn\n\u00b7 \u00b7 \u00b7\n\uf8f6\n\uf8f7\n\uf8f8\n\u2208Rn\u00d7n\nRemark.\n\u2022 V T x has the same length as x (orthogonal). Indeed, V T x \u00b7 V T x = xV V T x = x \u00b7 x.\n\u2022 \u03a3 stretches/contracts and adds/removes dimensions\n\u2022 U preserves the length\nCompute \u2225Ax\u22252 = (Ax)T Ax = xT AT A\n| {z }\nsymmetric \u21d2\u2203onb\nof eigenvectors\n{v1,\u00b7\u00b7\u00b7 ,vn} for AT A\nx \u21d2write x = \u03b11v1 + \u00b7 \u00b7 \u00b7 + \u03b1nvn", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_108", "text": "LECTURE NOTES\n59\n=\u21d2xT AT Ax = xT AT A (\u03b11v1 + \u00b7 \u00b7 \u00b7 + \u03b1nvn)\n= xT \u0000\u03b11AT Av1 + \u00b7 \u00b7 \u00b7 + \u03b1nAT Avn\n\u0001\n= xT (\u03b11\u03bb1v1 + \u00b7 \u00b7 \u00b7 + \u03b1n\u03bbnvn)\n=\n\u0000\u03b11vT\n1 + \u00b7 \u00b7 \u00b7 + \u03b1nvT\nn\n\u0001\n(\u03b11\u03bb1v1 + \u00b7 \u00b7 \u00b7 + \u03b1n\u03bbnvn)\nonb \u21d2vi\u00b7vj=0\n\u2200i\u0338=j\n=\n\u03b12\n1\u03bb1 + \u00b7 \u00b7 \u00b7 + \u03b12\nn\u03bbn\n=\u21d2\u03bb1, \u00b7 \u00b7 \u00b7 , \u03bbn change the length of the vector. Choose x = vi =\u21d2\u2225Avi\u22252 = \u03bbi \u22650.\nDefinition. Let A \u2208Rm\u00d7n. The singular values of A are given by \u03c3i = \u221a\u03bbi, where \u03bb1, \u00b7 \u00b7 \u00b7 , \u03bbn are the\neigenvalues of AT A arranged s.t. \u03bb1 \u2265\u00b7 \u00b7 \u00b7 \u2265\u03bbn \u22650.\nLemma 7.2. Let {v1, \u00b7 \u00b7 \u00b7 , vn} be an onb for Rn consisting of eigenvectors of AT A corresponding to\neigenvalues \u03bb1 \u2265\u00b7 \u00b7 \u00b7 \u2265\u03bbn \u22650. Assume that A has r non-zero singular values. Then {Av1, \u00b7 \u00b7 \u00b7 , Avr} is\nan orthogonal basis for Col (A) and rank (A) = r.\nProof. It suffices to show that {Av1, \u00b7 \u00b7 \u00b7 , Avr} is orthogonal, does not contain the zero vector, and that\nit generates Col (A).\n\u2022 Avi \u00b7 Avj = (Avi)T Avj = vT\ni AT Avj = vT\ni \u03bbjvj = \u03bbj vT\ni vj\n|{z}\n=0\n= 0\n\u2022 \u2225Avj\u22252 = Avj \u00b7 Avj = (Avj)T Avj = \u03bbj vT\nj vj\n|{z}\n=1\n= \u03bbj > 0 if j \u2208[[1, r]]\n\u2022 Let y = Ax \u2208Col (A). Write x = \u03b11v1 + \u00b7 \u00b7 \u00b7 + \u03b1nvn =\u21d2\nAx = \u03b11Av1 + \u00b7 \u00b7 \u00b7 + \u03b1rAvr\n|\n{z\n}\n\u2208span{Av1,\u00b7\u00b7\u00b7 ,Avr}\n+ \u03b1r+1\n\u2225Avr+1\u22252=\u03bbr+1=0\nz }| {", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_109", "text": "Ax = \u03b11Av1 + \u00b7 \u00b7 \u00b7 + \u03b1rAvr\n|\n{z\n}\n\u2208span{Av1,\u00b7\u00b7\u00b7 ,Avr}\n+ \u03b1r+1\n\u2225Avr+1\u22252=\u03bbr+1=0\nz }| {\nAvr+1 + \u00b7 \u00b7 \u00b7 + \u03b1nAvn\n|\n{z\n}\n0\n\u25a1\nTheorem 7.3. Let A \u2208Rm\u00d7n, rank (A) = r. Denote by {\u03c31, \u00b7 \u00b7 \u00b7 , \u03c3r} the first r singular values of A,\narranged in decreasing order. Then \u2203orthogonal matrices U \u2208Rm\u00d7m and V \u2208Rn\u00d7n s.t.\nA = U\u03a3V T\nwhere\n\u03a3 =\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n\u03c31\n0\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n0\n0\n...\n...\n...\n\u03c3r\n...\n...\n0\n...\n...\n...\n...\n0\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n0\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n\u2208Rm\u00d7n\nProof. Let {v1, \u00b7 \u00b7 \u00b7 , vr} be an onb of AT A as in lemma 7.2. Define\nui =\nAvi\n\u2225Avi\u2225= Avi\n\u03c3i\n1 \u2264i \u2264r", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_110", "text": "60\nKALAN WALMSLEY\n=\u21d2\n{u1, \u00b7 \u00b7 \u00b7 , ur} is an onb for Col (A). Extend {u1, \u00b7 \u00b7 \u00b7 , ur} to and onb for Rm\n=\u21d2\nset U =\n\u0000u1\n\u00b7 \u00b7 \u00b7\num\n\u0001\n, V =\n\u0000v1\n\u00b7 \u00b7 \u00b7\nvn\n\u0001\n=\u21d2U, V orthogonal =\u21d2show that AV = U\u03a3.\nAV =\n\u0000Av1\n\u00b7 \u00b7 \u00b7\nAvn\n\u0001\n=\n\u0000\u03c31u1\n\u00b7 \u00b7 \u00b7\n\u03c3rur\n0\n\u00b7 \u00b7 \u00b7\n0\u0001\nU\u03a3 =\n\u0000u1\n\u00b7 \u00b7 \u00b7\nun\n\u0001\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n\u03c31\n...\n\u03c3r\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n=\n\u0000\u03c31u1\n\u00b7 \u00b7 \u00b7\n\u03c3rur\n0\n\u00b7 \u00b7 \u00b7\n0\u0001\n\u25a1\nHow to compute a singular value decomposition.\n\u2022 Orthogonaly diagonalise AT A, then rearrange eigenvalues and eigenvectors\n\u2022 Put onb into V\n\u2022 Compute Av1, \u00b7 \u00b7 \u00b7 , Avr and normalise\n\u2022 Extend new vectors to an onb for Rm (find a basis for Col (A)\u22a5\u21d2linear system, then Gram-\nSchmidt)\n\u2022 Put onb into U\n\u2022 Put singular values into \u03a3\nExample. A =\n\uf8eb\n\uf8ed\n1\n\u22121\n\u22122\n2\n2\n\u22122\n\uf8f6\n\uf8f8\n\u2022 AT A =\n\u0012 9\n\u22129\n\u22129\n9\n\u0013\n, \u03c7AT A(\u03bb) = (\u03bb \u221218)\u03bb \u21d2\u03c3\n\u0000AT A\n\u0001\n= {0, 18} =\u21d2\n(\n\u03c31 =\n\u221a\n18\n\u03c32 = 0\n=\u21d2\u03a3 =\n\uf8eb\n\uf8ed\n\u221a\n18\n0\n0\n0\n0\n0\n\uf8f6\n\uf8f8\n(\u21d2rank (A) = 1)\n\u2022 Eigenvectors for AT A :\nv1 =\n \n1\n\u221a\n2\n\u22121\n\u221a\n2\n!\n, v2 =\n 1\n\u221a\n2\n1\n\u221a\n2\n!\n=\u21d2V =\n \n1\n\u221a\n2\n1\n\u221a\n2\n\u22121\n\u221a\n2\n1\n\u221a\n2\n!\n\u2022 Av1 =\n\uf8eb\n\uf8ec\n\uf8ed\n2\n\u221a\n2\n\u22124\n\u221a\n2\n4\n\u221a\n2\n\uf8f6\n\uf8f7\n\uf8f8=\u21d2normalise u1 = 1\n3\n\uf8eb\n\uf8ed\n1\n\u22122\n2\n\uf8f6\n\uf8f8, Av2 = 0\nNeed to extend {u1} to an onb of R3 :\n\u00001\n\u22122\n2\u0001\nx = 0 =\u21d2linear system. Basis for solution space\nb1 =\n\uf8eb\n\uf8ed\n2\n1\n0\n\uf8f6\n\uf8f8, b2 =\n\uf8eb\n\uf8ed\n0\n1\n1\n\uf8f6\n\uf8f8=\u21d2Gram-Schmidt and normalisation :\nu2 =\n1\n\u221a\n5\n\uf8eb\n\uf8ed\n2\n1\n0\n\uf8f6\n\uf8f8\n\u02dcu3 =\n\uf8eb\n\uf8ed\n0\n1\n1\n\uf8f6\n\uf8f8\u22121\n\u221a\n5\n\uf8eb\n\uf8ed\n2\n1\n0\n\uf8f6\n\uf8f8\u00b7\n\uf8eb\n\uf8ed\n0\n1\n1\n\uf8f6\n\uf8f81\n\u221a\n5\n\uf8eb\n\uf8ed\n2\n1\n0\n\uf8f6\n\uf8f8=\n\uf8eb\n\uf8ed\n\u22122\n5\n4\n5\n1\n\uf8f6\n\uf8f8\nu3 =\n\uf8eb\n\uf8ec\n\uf8ed\n\u2212\n2\n\u221a\n45\n4\n\u221a\n45\n5\n\u221a\n45\n\uf8f6\n\uf8f7\n\uf8f8", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_111", "text": "LECTURE NOTES\n61\n=\u21d2U =\n\uf8eb\n\uf8ec\n\uf8ed\n1\n3\n2\n\u221a\n5\n\u2212\n2\n\u221a\n45\n\u22122\n3\n1\n\u221a\n5\n4\n\u221a\n45\n2\n3\n0\n5\n\u221a\n45\n\uf8f6\n\uf8f7\n\uf8f8\nThe reduced singular value decomposition. Goal: use matrices of less size if rank (A) < max {n, m}.\nWe already know that\nA = U\u03a3V T =\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ed\n...\n...\nu1\n\u00b7 \u00b7 \u00b7\num\n...\n...\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f8\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n\u03c31\n...\n\u03c3r\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n\uf8eb\n\uf8ec\n\uf8ed\n\u00b7 \u00b7 \u00b7\nv1\n\u00b7 \u00b7 \u00b7\n...\n\u00b7 \u00b7 \u00b7\nvn\n\u00b7 \u00b7 \u00b7\n\uf8f6\n\uf8f7\n\uf8f8\n=\n\u0000\u03c31u1\n\u00b7 \u00b7 \u00b7\n\u03c3rur\n0\n\u00b7 \u00b7 \u00b7\n0\u0001\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n\u00b7 \u00b7 \u00b7\nv1\n\u00b7 \u00b7 \u00b7\n...\n\u00b7 \u00b7 \u00b7\nvr\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\nvr+1\n\u00b7 \u00b7 \u00b7\n...\n\u00b7 \u00b7 \u00b7\nvn\n\u00b7 \u00b7 \u00b7\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n=\n\u0000\u03c31u1\n\u00b7 \u00b7 \u00b7\n\u03c3rur\n\u0001\n\uf8eb\n\uf8ec\n\uf8ed\n\u00b7 \u00b7 \u00b7\nv1\n\u00b7 \u00b7 \u00b7\n...\n\u00b7 \u00b7 \u00b7\nvr\n\u00b7 \u00b7 \u00b7\n\uf8f6\n\uf8f7\n\uf8f8\n=\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ed\n...\n...\nu1\n\u00b7 \u00b7 \u00b7\nur\n...\n...\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f8\n\u2208Rm\u00d7r\n\uf8eb\n\uf8ec\n\uf8ed\n\u03c31\n...\n\u03c3r\n\uf8f6\n\uf8f7\n\uf8f8\n\u2208Rr\u00d7r\n\uf8eb\n\uf8ec\n\uf8ed\n\u00b7 \u00b7 \u00b7\nv1\n\u00b7 \u00b7 \u00b7\n...\n\u00b7 \u00b7 \u00b7\nvr\n\u00b7 \u00b7 \u00b7\n\uf8f6\n\uf8f7\n\uf8f8\n\u2208Rr\u00d7n\n|\n{z\n}\nreduced svd\n= Ur\u03a3rV T\nr\nAdvantages: no extension to onb needed\nDisadvantages: U, V are no longer square =\u21d2not invertible\nRemark. When we compute the eigenvalues of AT A, if A \u2208Rm\u00d7n where m < n it is possible to reduce\nthe size of the matrix we use to get the characteristic polynomial by instead computing the eigenvalues\nfor AAT . As is shown below, these eigenvalues correspond with the non-zero eigenvalues of AT A.\nAT A = V \u03a3T\nIn\nz }| {\nU T U \u03a3V T = V\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n\u03c32\n1\n...\n\u03c32\nr\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\nz}|{\n\u03a3T \u03a3 V T\nAAT = U\u03a3 V T V\n| {z }\nIn\n\u03a3T U T = U \u03a3\u03a3T\n|{z}\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n\u03c32\n1\n...\n\u03c32\nr\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\nU T\n\uf8fc\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8fd\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8f4\n\uf8fe\n=\u21d2same eigenvalues \u0338= 0\n7.3. The matrix exp(tA) and ordinary differential equations. Motivation: consider the ordinary\ndifferential equations (ode)\n(\nx\u2032\n1(t) = ax1t + bx2(t), x1(0) = x0\n1 \u2208R\nx\u2032", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_112", "text": "7.3. The matrix exp(tA) and ordinary differential equations. Motivation: consider the ordinary\ndifferential equations (ode)\n(\nx\u2032\n1(t) = ax1t + bx2(t), x1(0) = x0\n1 \u2208R\nx\u2032\n2(t) = cx1t + dx2(t), x2(0) = x0\n2 \u2208R", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_113", "text": "62\nKALAN WALMSLEY\nIntroducing the (time dependent) function x(t) =\n\u0012\nx1(t)\nx2(t)\n\u0013\n, this ode can be written as x\u2032(t) =\n\u0012\na\nb\nc\nd\n\u0013\n|\n{z\n}\nA\nx(t).\nWe can solve the ODE in the simplest case b = c = 0:\n(\nx1(t) = eatx0\n1\nx2(t) = edtx0\n2\nBut what can we do if A is not diagonal?\nAnsatz:\n(\nx1(t) = \u03b11e\u03c91t + \u03b21e\u03c92t\nx2(t) = \u03b13e\u03c93t + \u03b22e\u03c94t\nBut sometimes this does not work =\u21d2assume\n\u03b11, \u03b12, \u03b21, \u03b22\n\u03c91, \u03c92, \u03c93, \u03c94\n\u001b\n\u2208C. We will always consider odes of the\nform\nx\u2032(t) = Ax(t)\nx(0) = x0 \u2208Rn, A \u2208Rn\u00d7n, x : R \u2192Rn\nRemark. Any ode of the form\ny(n)(t) + an\u22121y(n\u22121)(t) + \u00b7 \u00b7 \u00b7 + a1y\u2032(t) + a0y(t) = 0\nwith initial conditions for y(n\u22121)(0), \u00b7 \u00b7 \u00b7 , y(0) can be transform into the above form by setting\nx1(t) = y(t)\nx2(t) = y\u2032(t)\n...\nxn(t) = y(n\u22121)(t)\nthen\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\nx\u2032\n1(t)\n...\n...\nx\u2032\nn(t)\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n=\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\nx2(t)\n...\nxn(t)\ny(n)(t)\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n=\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\nx2(t)\n...\nxn(t)\n\u2212an\u22121xn(t) \u2212\u00b7 \u00b7 \u00b7 \u2212a0x1(t)\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\n=\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n0\n1\n0\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n0\n0\n0\n1\n0\n\u00b7 \u00b7 \u00b7\n0\n...\n...\n...\n...\n...\n0\n0\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n0\n1\n\u2212a0\n\u2212a1\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u2212an\u22121\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\nIdea: define exp(tA) s.t.\nd\ndt exp(tA) = A exp(tA). We know from analysis ex =\n\u221e\nP\nk=0\n1", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_114", "text": "\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n0\n1\n\u2212a0\n\u2212a1\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u00b7 \u00b7 \u00b7\n\u2212an\u22121\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\nIdea: define exp(tA) s.t.\nd\ndt exp(tA) = A exp(tA). We know from analysis ex =\n\u221e\nP\nk=0\n1\nk!xk =\u21d2insert tA.\nDefinition. Given A \u2208Rn\u00d7n, t \u2208R, we define\nexp(tA) :\u2212\n\u221e\nX\nk=0\n1\nk!tkAk \u2208Rn\u00d7n\nRemark.\ni) exp(0) = In\nii) In general, exp(A + B) \u0338= exp(A) \u00b7 exp(B)\niii) exp((t1 + t2)A) = exp(t1A) \u00b7 exp(t2A) =\u21d2exp(tA)\u22121 = exp(\u2212tA)", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_115", "text": "LECTURE NOTES\n63\nNow we will differentiate t 7\u2192exp(tA) :\nd\ndt exp(tA) = d\ndt\n\u221e\nX\nk=0\ntkAk\nk!\n=\n\u221e\nX\nk=0\nd\ndt\ntkAk\nk!\n=\n\u221e\nX\nk=0\nktk\u22121 Ak\nk!\n=\n\u221e\nX\nk=1\ntk\u22121\nAk\n(k \u22121)!\nj=k\u22121\n=\nA\n\u221e\nX\nj=0\ntj Aj\nj! = A exp(tA)\nTheorem 7.4. Let A \u2208Rn\u00d7n, x0 \u2208Rn. Then the function x : R \u2192Rn defined by\nx(t) = exp(tA)x0\nsolves the ode x\u2032(t) = Ax(t), x(0) = x0.\nProof. We have x(0) = exp(0 \u00b7 A)x0 = Inx0 = x0. Moreover\nx\u2032(t) = d\ndt (exp(tA)x0) =\n\u0012 d\ndt exp(tA)\n\u0013\n|\n{z\n}\nA exp(tA)\nx0 + exp(tA) d\ndtx0\n| {z }\n0\n= A exp(tA)x0 = Ax(t)\n\u25a1\nHow to compute exp(tA). We only consider the case where A is diagonalizable, \u21d2A = PDP \u22121,\nP \u2208Cn\u00d7n invertible, D \u2208Cn\u00d7n diagonal.\nexp(tA) =\n\u221e\nX\nk=0\n1\nk!tk \u0000PDP \u22121\u0001k =\n\u221e\nX\nk=0\n1\nk!tkPDkP \u22121\n= P\n \u221e\nX\nk=0\n1\nk!tkDk\n!\nP \u22121 = P\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ed\n\u221e\nX\nk=0\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ed\ntk\u03bbk\n1\nk!\n...\ntk\u03bbk\nn\nk!\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f8\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f8P \u22121\n= P\n\uf8eb\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ec\n\uf8ed\n\u221e\nP\nk=0\ntk\u03bbk\n1\nk!\n...\n\u221e\nP\nk=0\ntk\u03bbk\nn\nk!\n\uf8f6\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f7\n\uf8f8\nP \u22121 = P\n\uf8eb\n\uf8ec\n\uf8ed\net\u03bb1\n...\net\u03bbn\n\uf8f6\n\uf8f7\n\uf8f8P \u22121\nCorollary 7.5. If A \u2208Rn\u00d7n is diagonalizable, i.e. A = PDP \u22121, with D =\n\uf8eb\n\uf8ec\n\uf8ed\n\u03bb1\n...\n\u03bbn\n\uf8f6\n\uf8f7\n\uf8f8, then\nexp(tA) = P\n\uf8eb\n\uf8ec\n\uf8ed\net\u03bb1\n...\net\u03bbn\n\uf8f6\n\uf8f7\n\uf8f8P \u22121", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_116", "text": "\uf8f6\n\uf8f7\n\uf8f8P \u22121\nCorollary 7.5. If A \u2208Rn\u00d7n is diagonalizable, i.e. A = PDP \u22121, with D =\n\uf8eb\n\uf8ec\n\uf8ed\n\u03bb1\n...\n\u03bbn\n\uf8f6\n\uf8f7\n\uf8f8, then\nexp(tA) = P\n\uf8eb\n\uf8ec\n\uf8ed\net\u03bb1\n...\net\u03bbn\n\uf8f6\n\uf8f7\n\uf8f8P \u22121\nRemark. In corollary 7.5, exp(tA)x0 will be a vector which has a linear combination of exponential\nfunctions as coefficients. If A is not diagonalizable, one can use the Jordan normal form to show that in\ngeneral exp(tA) contains linear combinations of exponential functions multiplied by certain polynomials.", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Lecture_notes_Linear_Algebra__en_.Walmsley_chunk_117", "text": "64\nKALAN WALMSLEY\nExample.\n(\nx\u2032\n1(t) = x1(t) + x2(t)\nx1(0) = 2\nx\u2032\n2(t) = x1(t) + x2(t)\nx2(0) = 4\n=\u21d2x\u2032(t) =\n\u00121\n1\n1\n1\n\u0013\nx(t), x0 =\n\u00122\n4\n\u0013\n=\u21d2A =\n\u00121\n1\n1\n1\n\u0013\n=\u21d2\u03c7A(\u03bb) = (\u03bb \u22121)2 \u22121 = \u03bb (\u03bb \u22122) =\u21d2\n(\n\u03bb1 = 2\n\u03bb2 = 0\nv2 =\n\u0012\n1\n\u22121\n\u0013\nv1 =\n\u0012\n1\n1\n\u0013\n=\u21d2P =\n\u0012\n1\n1\n1\n\u22121\n\u0013\nP \u22121 = 1\n2\n\u0012\n1\n1\n1\n\u22121\n\u0013\nHence\nexp(tA)\n=\nP\n\u0012e2t\n0\n0\n1\n\u0013\nP \u22121\n=\n1\n2\n\u00121\n1\n1\n\u22121\n\u0013 \u0012e2t\n0\n0\n1\n\u0013 \u00121\n1\n1\n\u22121\n\u0013\n=\n1\n2\n\u0012e2t + 1\ne2t \u22121\ne2t \u22121\ne2t + 1\n\u0013\n=\u21d2x(t) = exp(tA)\n\u00122\n4\n\u0013\n=\n\u0012e2t + 1\ne2t \u22121\ne2t \u22121\ne2t + 1\n\u0013 \u00121\n2\n\u0013\n=\n\u00123e2t \u22121\n3e2t + 1\n\u0013\n=\n\u0012x1(t)\nx2(t)\n\u0013\nAnd may the odds be ever in your favour !", "source": "Lecture_notes_Linear_Algebra__en_.Walmsley", "format": "pdf"}
-{"id": "Meca_Notes_chunk_0", "text": "Classical Mechanics: MIT 8.01 Course Notes  \nChapter 1 Introduction to Classical Mechanics................................................................  \nChapter 2 Units, Dimensional Analysis, Problem Solving, and Estimation ............  \nChapter 3 Vectors .......................................................................................................................  \nChapter 4 One Dimensional Kinematics.............................................................................  \nChapter 5 Two Dimensional Kinematics ............................................................................  \nChapter 6 Circular Motion .......................................................................................................", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_1", "text": "Chapter 7 Newton\u2019s Laws of Motion.....................................................................................  \nChapter 8 !pplications of Newton\u2019s Second Law.............................................................  \nChapter 9 Circular Motion Dynamics...................................................................................  \nChapter 10 Momentum, System of Particles, and Conservation of Momentum ...  \nChapter 11 Reference Frames................................................................................................  \nChapter 12 Momentum and the Flow of Mass...................................................................", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_2", "text": "Chapter 13 Energy, Kinetic Energy, and Work .................................................................  \nChapter 14 Potential Energy and Conservation of Energy...........................................  \nChapter 15 Collision Theory ...................................................................................................  \nChapter 16 Two Dimensional Rotational Kinematics....................................................  \nChapter 17 Two Dimensional Rotational Dynamics ......................................................  \nChapter 18 Static Equilibrium................................................................................................", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_3", "text": "Chapter 19 Angular Momentum............................................................................................  \nChapter 20 Rigid Body Kinematics About a Fixed Axis .................................................  \nChapter 21 Rigid Body Dynamics About a Fixed Axis ....................................................  \n  \nChapter 23 Simple Harmonic Motion ..................................................................................", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_4", "text": "2\n\n5\n10\n15\n20\n25\nChapter 1 Introduction to Classical Mechanics \nChapter 2 Units, Dimensional Analysis, Problem Solving, and \nEstimation \nChapter 3 Vectors \nChapter 4 One Dimensional Kinematics \nChapter  Two Dimensional Kinematics \nChapter 6 Circular Motion \nChapter 7 Newton\u2019s Laws of Motion \nChapter 8 Applications of Newton\u2019s Second Law \nChapter 9 Circular Motion Dynamics \nChapter \n Momentum, System of Particles, and Conservation \nof Momentum \nChapter 11 Reference Frames \nChapter 12 Momentum and the Flow of Mass \nChapter 13 Energy, Kinetic Energy, and Work \nChapter 14 Potential Energy and Conservation of Energy \nChapter \n Collision Theory \nChapter 16 Two Dimensional Rotational Kinematics \nChapter 17 Two Dimensional Rotational Dynamics \nChapter 18 Static Equilibrium \nChapter 19 Angular Momentum \nChapter \n Rigid Body Kinematics About a Fixed Axis \nChapter 21 Rigid Body Dynamics About a Fixed Axis \nChapter 22 Three Dimensional Rotations and Gyroscopes \nChapter 23 Simple Harmonic Motion \nChapter 24 Physical Pendulum \nChapter \n Celestial Mechanics \n3\n\nChapter 26 Elastic Properties of Materials \nChapter 27 Static Fluids \nChapter 28 Fluid Dynamics \nChapter 29 Kinetic Theory of Gases \nPhysical Constants \nAstronomical Data \n4\n\nFor Dorothea\n\nFundamental Physical Constants \nQuantity \nSymbol \nValue \nAvogadro\u2019s number \nN A \n6.02214129(27) \u00d7 1023 / mol \nBoltzmann\u2019s constant \nkB \n1.3806488(13) \u00d7 10\u221223 J/K \nCoulomb constant \nk e = 1/ 4\u03c0\u03b50 \n8.987551787\uf04c\u00d7109 N \u22c5m2 \u22c5C\u22122 \nElementary charge \ne \n1.602176565(35) \u00d7 10\u221219 C \nElectron mass \nm e \n9.10938215(45) \u00d7 10\u221231 kg \nGravitational constant \nG \n6.67384(80)\uf04c\u00d710\u221211 N \u22c5m2 \u22c5 kg\u22122 \nNeutron mass \nm n \n1.674927351(74) \u00d7 10\u221227 kg \nPermeability of free space \n\u00b50 \n4\u03c0 \u00d7 10\u22127 T \u22c5 m/A \nPermittivity of free space \n\u03b50 = 1/ \u00b50c2 \n8.854187817\uf04c\u00d710\u221212 C2 / N \u22c5m2 \nPlanck\u2019s constant \nh \n6.62606957(29) \u00d710\u221234 J \u22c5s \nProton mass \nm p \n1.672621777(74) \u00d7 10\u221227 kg \nSpeed of light \nc \n2.99792458 \u00d7108 m \u22c5s\u22121", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_5", "text": "Astronomical Data \nEarth \nSolar mass \nEarth mass \nEarth mean radius \nMean solar day \nEarth-Sun Orbit \nAphelion \nPerihelion \nEccentricity \nOrbital Period \nMoon \nMoon mass \nMoon mean radius \nMoon orbital period \n(sidereal month) \nMoon synodic period \nValue \n(1.98855 \u00b1 0.00025) \u00d71030 kg \n5.97219 \u00d7 1024kg \n6.371009 \u00d7106 m \n8.6400 \u00d7104 s \n1.52098232 \u00d71011 km \n1.470098290 \u00d71011 km \n0.01671123 \n3.15581495 \u00d7107 s \n7.3477 \u00d7 1022kg \n1.73710 \u00d7106 m \n2.3605847 \u00d7106 s \n2.5514429 \u00d7106 s\n\nChapter 1 The History and Limitations of Classical Mechanics \nChapter 1.1\t\r \u00a0Introduction ................................................................................................................1", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_6", "text": "Chapter 1 The History and Limitations of Classical Mechanics \nChapter 1.1 Introduction \nClassical mechanics is the mathematical science that studies the displacement of bodies \nunder the action of forces. Gailieo Galilee initiated the modern era of mechanics by using \nmathematics to describe the motion of bodies. His Mechanics, published in 1623, \nintroduced the concepts of force and described the constant accelerated motion of objects \nnear the surface of the Earth. Sixty years later Isaac Newton formulated his Laws of \nMotion, which he published in 1687 under the title, Philosophiae Naturalis Principia \nMathematica (Mathematical Principles of Natural Philosophy). In the third book, \nsubtitled De mundi systemate (On the system of the world), Newton solved the greatest \nscientific problem of his time by applying his Universal Law of Gravitation to determine \nthe motion of planets. Newton established a mathematical approach to the analysis of \nphysical phenomena in which he stated that it was unnecessary to introduce final causes \n(hypothesis) that have no experimental basis, \n\u201cHypotheses non fingo (I frame no \nhypotheses), but that physical models are built from experimental observations and then \nmade general by induction. This led to a great century of applications of the principles of \nNewtonian mechanics to many new problems culminating in the work of Leonhard Euler. \nEuler began a systematic study of the three dimensional motion of rigid bodies, leading to \na set of dynamical equations now known as Euler\u2019s equations of motion. \nAlongside this development and refinement of the concept of force and its \napplication to the description of motion, the concept of energy slowly emerged, \nculminating in the middle of the nineteenth century in the discovery of the principle of \nconservation of energy and its immediate applications to the laws of thermodynamics. \nConservation principles are now central to our study of mechanics; the conservation of \nmomentum, energy, and angular momentum enabled a new reformulation of classical \nmechanics. \nDuring this period, the experimental methodology and mathematical tools of \nNewtonian mechanics were applied to other non-rigid systems of particles leading to the \ndevelopment of continuum mechanics. The theories of fluid mechanics, wave mechanics, \nand electromagnetism emerged leading to the development of the wave theory of light. \nHowever there were many perplexing aspects of the wave theory of light, for example, \ndoes light propagate through a medium, the \u201cether\u201d? A series of optics experiments, \nculminating in the Michelson-Morley experiment in 1887 ruled out the hypothesis of a \nstationary medium. Many attempts were made to reconcile the experimental evidence", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_7", "text": "does light propagate through a medium, the \u201cether\u201d? A series of optics experiments, \nculminating in the Michelson-Morley experiment in 1887 ruled out the hypothesis of a \nstationary medium. Many attempts were made to reconcile the experimental evidence \nwith classical mechanics but the challenges were more fundamental. The basics concepts \nof absolute time and absolute space, which Newton had defined in the Principia, were \nthemselves inadequate to explain a host of experimental observations. Albert Einstein, by \ninsisting on a fundamental rethinking of the concepts of space and time, and the relativity \nof motion, in his special theory of relativity (1905) was able to resolve the apparent \nconflicts between optics and Newtonian mechanics. In particular, special relativity \nprovides the necessary framework for describing the motion of rapidly moving objects \n(speed greater than v > 0.1 c ). \n1-1", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_8", "text": "A second limitation on the validity of Newtonian mechanics appeared at the \nmicroscopic length scale. A new theory, statistical mechanics, was developed relating the \nmicroscopic properties of individual atoms and molecules to the macroscopic or bulk \nthermodynamic properties of materials. Started in the middle of the nineteenth century, \nnew observations at very small scales revealed anomalies in the predicted behavior of \ngases (heat capacity). It became increasingly clear that classical mechanics did not \nadequately explain a wide range of newly discovered phenomena at the atomic and sub-\natomic length scales. An essential realization was that the language of classical \nmechanics was not even adequate to qualitatively describe certain microscopic \nphenomena. By the early part of the twentieth century, quantum mechanics provided a \nmathematical description of microscopic phenomena in complete agreement with our \nempirical knowledge of all non-relativistic phenomena. \nIn the twentieth century, as experimental observations led to a more detailed \nknowledge of the large-scale properties of the universe, Newton\u2019s Universal Law of \nGravitation no longer accurately modeled the observed universe and needed to be \nreplaced by general relativity. By the end of the twentieth century and beginning of the \ntwenty-first century, many new observations, for example the accelerated expansion of \nthe Universe, have required introduction of new concepts like dark energy that may lead \nonce again to a fundamental rethinking of the basic concepts of physics in order to \nexplain observed phenomena. \n1-2", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_9", "text": "Chapter 2 Units, Dimensional Analysis, and Estimation \n2.1 The Speed of light ................................................................................................... 1  \n2.2 International System of Units ................................................................................ 1  \n2.2.1 Standard Mass.................................................................................................... 2  \nExample 2.1 The International Prototype Kilogram................................................... 3  \nExample 2.2 Mass of a Silicon Crystal....................................................................... 4", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_10", "text": "2.2.2 Atomic Clock and the Definition of the Second................................................ 5  \n2.2.3 Meter .................................................................................................................. 6  \nExample 2.3 Light-Year.............................................................................................. 6  \n2.2.4 Radians............................................................................................................... 7", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_11", "text": "Example 2.4 Parsec..................................................................................................... 8  \n2.2.5 Steradians........................................................................................................... 9  \n2.2.6 Radiant Intensity ................................................................................................ 9  \n2.3 Dimensions of Commonly Encountered Quantities............................................. 9  \n2.3.1 Dimensional Analysis ...................................................................................... 10", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_12", "text": "Example 2.5 Period of a Pendulum........................................................................... 11  \n2.4 Order of Magnitude Estimates - Fermi Problems ............................................. 13  \n2.4.1 Methodology for Estimation Problems............................................................ 13  \nExample 2.6 Lining Up Pennies ............................................................................... 14  \nExample 2.7 Estimation of Mass of Water on Earth ................................................ 14", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_13", "text": "Chapter 2 Units, Dimensional Analysis, Problem Solving,  \nEstimation, and Error Analysis  \nBut we must not forget that all things in the world are connected with one another \nand depend on one another, and that we ourselves and all our thoughts are also a \npart of nature. It is utterly beyond our power to measure the changes of things by \ntime. Quite the contrary, time is an abstraction, at which we arrive by means of \nthe change of things; made because we are not restricted to any one definite \nmeasure, all being interconnected. A motion is termed uniform in which equal \nincrements of space described correspond to equal increments of space described \nby some motion with which we form a comparison, as the rotation of the earth. A \nmotion may, with respect to another motion, be uniform. But the question whether \na motion is in itself uniform, is senseless. With just as little justice, also, may we \nspeak of an \u201cabsolute time\u201d --- of a time independent of change. This absolute \ntime can be measured by comparison with no motion; it has therefore neither a \npractical nor a scientific value; and no one is justified in saying that he knows \naught about it. It is an idle metaphysical conception.1 \nErnst Mach \n2.1 The Speed of light \nWhen we observe and measure phenomena in the world, we try to assign numbers to the \nphysical quantities with as much accuracy as we can possibly obtain from our measuring \nequipment. For example, we may want to determine the speed of light, which we can \ncalculate by dividing the distance a known ray of light propagates over its travel time, \ndistance \nspeed of light = \n. \n(2.1.1)\ntime \nIn 1983 the General Conference on Weights and Measures defined the speed of \nlight to be \nc = 299, 792, 458 meters/second . \n(2.1.2) \nThis number was chosen to correspond to the most accurately measured value of \nthe speed of light and is well within the experimental uncertainty. \n2.2 International System of Units \nThe system of units most commonly used throughout science and technology today is the \nSyst\u00e8me International (SI). It consists of seven base quantities and their corresponding \nbase units, shown in Table 2.1. \n1 E. Mach, The Science of Mechanics, translated by Thomas J. McCormack, Open Court\nPublishing Company, La Salle, Illinois, 1960, p. 273. \n2-1", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_14", "text": "Table 2.1 International System of Units \nBase Quantity \nBase Unit \nLength \nmeter (m) \nMass \nkilogram (kg) \nTime \nsecond (s) \nElectric Current \nampere (A) \nTemperature \nkelvin (K) \nAmount of Substance \nmole (mol) \nLuminous Intensity \ncandela (cd) \nWe shall refer to the dimension of the base quantity by the quantity itself, for example \ndim length \u2261 length \u2261 L, dim mass \u2261 mass \u2261 M, dim time \u2261 time \u2261 T. \n(2.2.1) \nMechanics is based on just the first three of these quantities, the MKS or meter-\nkilogram-second system. An alternative metric system, still widely used, is the CGS \nsystem (centimeter-gram-second). \n2.2.1 Standard Mass \nThe unit of mass, the kilogram (kg), remains the only base unit in the \nInternational System of Units (SI) that is still defined in terms of a physical artifact, \nknown as the \u201cInternational Prototype of the Standard Kilogram.\u201d George Matthey (of \nJohnson Matthey) made the prototype in 1879 in the form of a cylinder, 39 mm high and \n39 mm in diameter, consisting of an alloy of 90 % platinum and 10 % iridium. The \ninternational prototype is kept in the Bureau International des Poids et M\u00e8sures (BIPM) \nat Sevres, France, under conditions specified by the 1st Conf\u00e9rence G\u00e9n\u00e9rale des Poids et \nM\u00e8sures (CGPM) in 1889 when it sanctioned the prototype and declared \u201cThis prototype \nshall henceforth be considered to be the unit of mass.\u201d It is stored at atmospheric pressure \nin a specially designed triple bell-jar. The prototype is kept in a vault with six official \ncopies. \nThe 3rd Conf\u00e9rence G\u00e9n\u00e9rale des Poids et M\u00e8sures CGPM (1901), in a declaration \nintended to end the ambiguity in popular usage concerning the word \u201cweight\u201d confirmed \nthat: \nThe kilogram is the unit of mass; it is equal to the mass of the international \nprototype of the kilogram. \nThere is a stainless steel one-kilogram standard that is used for comparisons with \nstandard masses in other laboratories. In practice it is more common to quote a \nconventional mass value (or weight-in-air, as measured with the effect of buoyancy), than \nthe standard mass. Standard mass is normally only used in specialized measurements \nwherever suitable copies of the prototype are stored. \n2-2", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_15", "text": "Example 2.1 The International Prototype Kilogram \nIn order to minimize the effects of corrosion, the platinum-iridium prototype kilogram is \na right cylinder with dimensions chosen to minimize the surface area for a given fixed \nvolume. The standard kilogram is an alloy of 90 % platinum and 10 % iridium. The \ndensity of the alloy is \u03c1 = 21.56 g \u22c5 cm\u22123 . Based on this information, (i) determine the \nradius of the prototype kilogram, and (ii) the ratio of the radius to the height. \nSolution: The volume for a cylinder of radius r and height h is given by \nV = \u03c0r 2h . \n(2.2) \nThe surface area can be expressed as a function of the radius r and the constant volume \nV according to \n2 + 2V\nA = 2\u03c0r 2 + 2\u03c0rh = 2\u03c0r\n. \n(2.3)\nr \nTo find the smallest surface area for a fixed volume, minimize the surface area with \nrespect to the radius by setting \ndA \n2V\n0 =\n= 4\u03c0 r \u2212 \n, \n(2.4)\ndr \nr 2 \nwhich we can solve for the radius \n1/3 \n\u239b V \u239e \nr = \n. \n(2.5)\n\u239d\u239c 2\u03c0 \u23a0\u239f \nBecause we also know that V = \u03c0r 2h , we can rewrite Eq. (2.5) as \n2h\n\u03c0 r \nr3 = \n, \n(2.6)\n2\u03c0 \nwhich implies that ratio of the radius to the height is \nr \n1 \n= \n. \n(2.7)\nh \n2 \nThe standard kilogram is an alloy of 90% platinum and 10% iridium. The density of \nplatinum is 21.45 g\u22c5cm\u22123 and the density of iridium is 22.55 g \u22c5cm\u22123 . Thus the density of \nthe standard kilogram is \n\u03c1 = (0.90)(21.45 g \u22c5cm\u22123) + (0.10)(22.55 g \u22c5cm\u22123) = 21.56 g \u22c5cm\u22123, \n(2.8) \n2-3", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_16", "text": "and its volume is \nV = m / \u03c1 = (1000 g) / (21.56 g \u22c5cm\u22123 ) = 46.38 cm3 . \n(2.9) \nFor the standard mass, the radius is \nr = V \n2\u03c0 \n\u239b \n\u239d\u239c \n\u239e \n\u23a0\u239f \n1 3 \n= 46.38 cm3 \n2\u03c0 \n\u239b \n\u239d\u239c \n\u239e \n\u23a0\u239f \n1 3 \n\u2245 1.95 cm . \n(2.10) \nBecause the prototype kilogram is an artifact, there are some intrinsic problems \nassociated with its use as a standard. It may be damaged, or destroyed. The prototype \ngains atoms due to environment wear and cleaning, at a rate of change of mass \ncorresponding to approximately 1 \u00b5g / year , (1 \u00b5g \u2261 1microgram \u2261 1\u00d7 10-6 g ). \nSeveral new approaches to defining the SI unit of mass [kg] are currently being \nexplored. One possibility is to define the kilogram as a fixed number of atoms of a \nparticular substance, thus relating the kilogram to an atomic mass. Silicon is a good \ncandidate for this approach because it can be grown as a large single crystal, in a very \npure form. \nExample 2.2 Mass of a Silicon Crystal \nA given standard unit cell of silicon has a volume V0 and contains N0 atoms. The \nnumber of molecules in a given mole of substance is given by Avogadro\u2019s constant \n= 6.02214129(27) \u00d71023 mol-1 . The molar mass of silicon is given by Mmol . Find the \nN A \nmass m of a volume V in terms of V0 , N0 , V , Mmol , and N A . \nSolution: The mass m0 of the unit cell is the density \u03c1 of the silicon cell multiplied by \nthe volume of the cell V0 , \n(2.11)\nm0 = \u03c1V0 . \nThe number of moles in the unit cell is the total mass, m0 , of the cell, divided by the \nmolar mass Mmol , \n(2.12)\nn0 = m0 / Mmol = \u03c1V0 / Mmol . \nThe number of atoms in the unit cell is the number of moles times the Avogadro constant, \nN A , \n\u03c1V0 N A\n= \n. \n(2.13)\nN0 = n0 N A  \nMmol  \n2-4", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_17", "text": "The density of the crystal is related to the mass m of the crystal divided by the volume V \nof the crystal, \n\u03c1 = m / V . \n(2.14) \nThe number of atoms in the unit cell can be expressed as \nmV0 N A\nN0 = VMmol \n. \n(2.15) \nThe mass of the crystal is \nm = Mmol \nN A \nV \nV0 \nN0 \n(2.16) \nThe molar mass, unit cell volume and volume of the crystal can all be measured directly. \nNotice that Mmol / N A is the mass of a single atom, and (V / V0 \nis the number of \n)N0 \natoms in the volume. This accuracy of the approach depends on how accurate the \nAvogadro constant can be measured. Currently, the measurement of he Avogadro \nconstant has a relative uncertainty of 1 part in 108, which is equivalent to the uncertainty \nin the present definition of the kilogram. \n2.2.2 Atomic Clock and the Definition of the Second \nIsaac Newton, in the Philosophiae Naturalis Principia Mathematica (\u201cMathematical \nPrinciples of Natural Philosophy\u201d), distinguished between time as duration and an \nabsolute concept of time, \n\u201cAbsolute true and mathematical time, of itself and from its own nature, \nflows equably without relation to anything external, and by another name \nis called duration: relative, apparent, and common time, is some sensible \nand external (whether accurate or unequable) measure of duration by \nmeans of motion, which is commonly used instead of true time; such as an \nhour, a day, a month, a year. \u201d2. \nThe development of clocks based on atomic oscillations allowed measures of \ntiming with accuracy on the order of 1 part in 1014 , corresponding to errors of less than \none microsecond (one millionth of a second) per year. Given the incredible accuracy of \nthis measurement, and clear evidence that the best available timekeepers were atomic in \nnature, the second [s] was redefined in 1967 by the International Committee on Weights \n2 Isaac Newton. Mathematical Principles of Natural Philosophy. Translated by Andrew\nMotte (1729). Revised by Florian Cajori. Berkeley: University of California Press, 1934. \np. 6. \n2-5", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_18", "text": "and Measures as a certain number of cycles of electromagnetic radiation emitted by \ncesium atoms as they make transitions between two designated quantum states: \nThe second is the duration of 9,192,631,770 periods of the radiation \ncorresponding to the transition between the two hyperfine levels of the \nground state of the cesium 133 atom. \n2.2.3 Meter \nThe meter [m] was originally defined as 1/10,000,000 of the arc from the Equator to the \nNorth Pole along the meridian passing through Paris. To aid in calibration and ease of \ncomparison, the meter was redefined in terms of a length scale etched into a platinum bar \npreserved near Paris. Once laser light was engineered, the meter was redefined by the \n17th Conf\u00e9rence G\u00e9n\u00e9rale des Poids et M\u00e8sures (CGPM) in 1983 to be a certain number \nof wavelengths of a particular monochromatic laser beam. \nThe meter is the length of the path traveled by light in vacuum during a \ntime interval of 1/299 792 458 of a second. \nExample 2.3 Light-Year \nAstronomical distances are sometimes described in terms of light-years [ly]. A light-year \nis the distance that light will travel in one year [yr]. How far in meters does light travel in \none year? \nSolution: Using the relationship distance = (speed of light) \u22c5 (time) , one light year \ncorresponds to a distance. Because the speed of light is given in terms of meters per \nsecond, we need to know how many seconds are in a year. We can accomplish this by \nconverting units. We know that \n1 year = 365.25 days, 1 day = 24 hours, 1 hour = 60 minutes, 1 minute = 60 seconds \nPutting this together we find that the number of seconds in a year is \n1 year =(365.25 day) 24 hours \n1 day \n\u239b \n\u239d\u239c \n\u239e \n\u23a0\u239f \n60 min \n1 hour \n\u239b \n\u239d\u239c \n\u239e \n\u23a0\u239f \n60 s \n1 min \n\u239b \n\u239d\u239c \n\u239e \n\u23a0\u239f =31,557,600 s . \n(2.2.17) \nThe distance that light travels in a one year is \n1 ly = 299,792,458 m \n1s \n\u239b \n\u239d\u239c \n\u239e \n\u23a0\u239f \n31,557,600 s \n1 yr \n\u239b \n\u239d\u239c \n\u239e \n\u23a0\u239f (1 yr) = 9.461\u00d71015 m . \n(2.2.18) \nThe distance to the nearest star, a faint red dwarf star, Proxima Centauri, is \n4.24 ly . \n2-6", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_19", "text": "2.2.4 Radians \nConsider the triangle drawn in Figure 2.1. The basic trigonometric functions of an angle \n\u03b8 in a right-angled triangle ONB are sin(\u03b8) = y / r , cos(\u03b8) = x / r , and tan(\u03b8) = y / x . \nO\nr\nB\ny\ns\nA\nN\nx\nX\nP\nFigure 2.1 Trigonometric relations \nIt is very important to become familiar with using the measure of the angle \u03b8 \nitself as expressed in radians [rad]. Let \u03b8 be the angle between two straight lines OX \nand OP . Draw a circle of radius r centered at O . The lines OP and OX cut the circle \nat the points A and B where OA = OB = r . Denote the length of the arc AB by s , \nthen the radian measure of \u03b8 is given by \n\u03b8 = s / r , \n(2.2.19) \nand the ratio is the same for circles of any radii centered at O -- just as the ratios y / r \nand y / x are the same for all right triangles with the angle \u03b8 at O . As \u03b8 approaches \n360\uf06f , s approaches the complete circumference 2\u03c0r of the circle, so that \n360\uf06f = 2\u03c0 rad . \ntan( ) \nsin( ) \n/ 2 \n0 \n1.0 \n/ 2 \nFigure 2.2 Radians compared to trigonometric functions. \nLet\u2019s compare the behavior of sin(\u03b8) , tan(\u03b8) and \u03b8 itself for small angles. One \ncan see from Figure 2.1 that s / r > y / r . It is less obvious that y / x > \u03b8 . It is very \ninstructive to plot sin(\u03b8) , tan(\u03b8) , and \u03b8 as functions of \u03b8 [rad] between 0 and \u03c0 / 2 \n2-7", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_20", "text": "on the same graph (see Figure 2.2). For small \u03b8 , the values of all three functions are \nalmost equal. But how small is \u201csmall\u201d? An acceptable condition is for \u03b8 << 1 in radians. \nWe can show this with a few examples. Recall that 360\uf06f = 2\u03c0 rad , 57.3\uf06f = 1rad , so an \nangle 6\uf06f \u2245 (6\uf06f )(2\u03c0 rad / 360o ) \u2245 0.1 rad when expressed in radians. In Table 2.2 we \ncompare the value of \u03b8 (measured in radians) with sin(\u03b8 ) , tan(\u03b8 ) , (\u03b8 \u2212 sin\u03b8 ) / \u03b8 , and \n(\u03b8 \u2212 tan\u03b8 ) / \u03b8 , for \u03b8 = 0.1 rad , 0.2 rad , 0.5 rad , and 1.0 rad . \nTable 2.2 Small Angle Approximation \n\u03b8 [rad] \n\u03b8 [deg] \nsin(\u03b8 ) \ntan(\u03b8 ) \n(\u03b8 \u2212 sin\u03b8 ) / \u03b8 \n(\u03b8 \u2212 tan\u03b8 ) / \u03b8 \n0.1 \n5.72958 \n0.09983 \n0.10033 \n0.00167 \n-0.00335 \n0.2 \n11.45916 \n0.19867 \n0.20271 \n0.00665 \n-0.01355 \n0.5 \n28.64789 \n0.47943 \n0.54630 \n0.04115 \n-0.09260 \n1.0 \n57.29578 \n0.84147 \n1.55741 \n0.15853 \n-0.55741 \nThe values for (\u03b8 \u2212 sin\u03b8 ) / \u03b8 , and (\u03b8 \u2212 tan\u03b8 ) / \u03b8 , for \u03b8 = 0.2 rad are less than \u00b11.4% . \nProvided that \u03b8 is not too large, the approximation that \nsin(\u03b8) \uf03b tan(\u03b8) \uf03b \u03b8 , \n(2.2.20) \ncalled the small angle approximation, can be used almost interchangeably, within some \nsmall percentage error. This is the basis of many useful approximations in physics \ncalculations. \nExample 2.4 Parsec \nA standard astronomical unit is the parsec . Consider two objects that are separated by a \ndistance of one astronomical unit, 1AU = 1.50 \u00d71011 m , which is the mean distance \nbetween the earth and sun. (One astronomical unit is roughly equivalent to eight light", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_21", "text": "Example 2.4 Parsec \nA standard astronomical unit is the parsec . Consider two objects that are separated by a \ndistance of one astronomical unit, 1AU = 1.50 \u00d71011 m , which is the mean distance \nbetween the earth and sun. (One astronomical unit is roughly equivalent to eight light \nminutes, 1AU = 8.3light-minutes .) One parsec is the distance at which one astronomical \nunit subtends an angle \u03b8 = 1arcsecond = (1/ 3600) degree . Suppose is a spacecraft is \nlocated in a space a distance 1 parsec from the Sun as shown in Figure 2.3. How far is the \nspacecraft in terms of light years and meters? \nEarth \nSun \n1 AU \n1 arcsecond \nspacecraft \n1 parsec \nFigure 2.3 Example 2.4 \n2-8", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_22", "text": "Because one arc second corresponds to a very small angle, one parsec is therefore equal \nto distance divided by angle, hence \n(1 AU) \n\u239b 1.50 \u00d71011 m \u239e\n1pc =\n= (2.06 \u00d7105 AU) \n\u23a0\u239f = 3.09 \u00d71016 m\n(1/3600) \n\u239d\u239c \n1AU \n. (2.2.21)\n\u239b \n1ly \n\u239e \n= (3.09 \u00d71016 m) \n\u23a0\u239f = 3.26 ly \n\u239d\u239c 9.46 \u00d71015 m \n2.2.5 Steradians \nThe steradian [sr] is the unit of solid angle that, having its vertex in the center of a sphere, \ncuts off an area of the surface of the sphere equal to that of a square with sides of length \nequal to the radius of the sphere. The conventional symbol for steradian measure is \u03a9 , \nthe uppercase Greek letter \u201cOmega.\u201d The total solid angle \u2126sph of a sphere is then found \nby dividing the surface area of the sphere by the square of the radius, \n\u03a9sph = 4\u03c0 r2 / r2 = 4\u03c0 \n(2.2.22) \nThis result is independent of the radius of the sphere. \n2.2.6 Radiant Intensity \n\u201cThe SI unit, candela, is the luminous intensity of a source that emits \nmonochromatic radiation of frequency 540 \u00d7 1012 s-1 , in a given direction, \nand that has a radiant intensity in that direction of 1/683 watts per \nsteradian.\u201d \nNote that \"in a given direction\" cannot be taken too literally. The intensity is measured \nper steradian of spread, so if the radiation has no spread of directions, the luminous \nintensity would be infinite. \n2.3 Dimensions of Commonly Encountered Quantities \nMany physical quantities are derived from the base quantities by a set of algebraic \nrelations defining the physical relation between these quantities. The dimension of the \nderived quantity is written as a power of the dimensions of the base quantities. For \nexample velocity is a derived quantity and the dimension is given by the relationship \ndim velocity = (length)/(time) = L \u22c5 T-1 . \n(2.3.1) \nwhere L \u2261 length , T \u2261 time . Force is also a derived quantity and has dimension \n2-9", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_23", "text": "(mass)(dim velocity) \ndim force = \n. \n(2.3.2)\n(time) \nwhere M \u2261 mass . We can also express force in terms of mass, length, and time by the \nrelationship \n(mass)(length) = M \u22c5 L \u22c5 T-2 \ndim force = \n. \n(2.3.3)\n(time)2 \nThe derived dimension of kinetic energy is \ndim kineticenergy = (mass)(dim velocity)2 , \n(2.3.4) \nwhich in terms of mass, length, and time is \n(mass)(length)2 \n= M \u22c5 L2 \u22c5 T-2 \ndim kineticenergy = \n. \n(2.3.5)\n(time)2 \nThe derived dimension of work is \ndim work = (dim force)(length) , \n(2.3.6) \nwhich in terms of our fundamental dimensions is \n(mass)(length)2 \n= M \u22c5 L2 \u22c5 T-2 \ndim work = \n. \n(2.3.7)\n(time)2 \nSo work and kinetic energy have the same dimensions. Power is defined to be the rate of \nchange in time of work so the dimensions are \ndim work (dim force)(length) (mass)(length)2 \n= M \u22c5 L2 \u22c5 T-3 \ndim power = \n= \n= \n.(2.3.8)\ntime \ntime \n(time)3 \nIn Table 2.3 we include the derived dimensions of some common mechanical quantities \nin terms of mass, length, and time. \n2.3.1 Dimensional Analysis \nThere are many phenomena in nature that can be explained by simple relationships \nbetween the observed phenomena. \n2-10", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_24", "text": "Table 2.3 Dimensions of Some Common Mechanical Quantities \nM \u2261 mass , L \u2261 length , T \u2261 time \nQuantity \nDimension \nMKS unit \nAngle \ndimensionless \nDimensionless = radian \nSolid Angle \ndimensionless \nDimensionless = sterradian \nArea \nL2 \nm2 \nVolume \nL3 \nm3 \nFrequency \nT-1 \ns\u22121 = hertz = Hz \nVelocity \nL \u22c5 T-1 \nm \u22c5 s\u22121 \nAcceleration \nL \u22c5 T-2 \nm \u22c5 s\u22122 \nAngular Velocity \nT-1 \nrad \u22c5 s\u22121 \nAngular Acceleration \nT-2 \nrad \u22c5 s\u22122 \nDensity \nM \u22c5 L-3 \nkg \u22c5 m\u22123 \nMomentum \nM \u22c5 L \u22c5 T-1 \nkg \u22c5 m \u22c5 s\u22121 \nAngular Momentum \nM \u22c5 L2 \u22c5 T-1 \nkg \u22c5 m2 \u22c5 s\u22121 \nForce \nM \u22c5 L \u22c5 T-2 \nkg\u22c5m \u22c5s\u22122 = newton = N \nWork, Energy \nM \u22c5 L2 \u22c5 T-2 \nkg \u22c5 m2 \u22c5 s\u22122 = joule = J \nTorque \nM \u22c5 L2 \u22c5 T-2 \nkg \u22c5 m2 \u22c5 s\u22122 \nPower \nM \u22c5 L2 \u22c5 T-3 \nkg \u22c5 m2 \u22c5 s\u22123 = watt = W \nPressure \nM \u22c5 L-1 \u22c5 T-2 \nkg \u22c5 m\u22121 \u22c5 s\u22122 = pascal= Pa \nExample 2.5 Period of a Pendulum \nConsider a simple pendulum consisting of a massive bob suspended from a fixed point by \na string. Let T denote the time interval (period of the pendulum) that it takes the bob to \ncomplete one cycle of oscillation. How does the period of the simple pendulum depend \non the quantities that define the pendulum and the quantities that determine the motion? \nSolution: What possible quantities are involved? The length of the pendulum l , the mass \nof the pendulum bob m , the gravitational acceleration g , and the angular amplitude of \nthe bob \u03b80 are all possible quantities that may enter into a relationship for the period of \nthe swing. Have we included every possible quantity? We can never be sure but let\u2019s first \nwork with this set and if we need more than we will have to think harder! Our problem is \nthen to find a function f such that \n2-11", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_25", "text": "T = f (l,m, g,\u03b80) \n(2.3.9) \nWe first make a list of the dimensions of our quantities as shown in Table 2.4. \nTable 2.4 Dimensions of Quantities Relevant to the Period of Pendulum \nName of Quantity \nSymbol \nDimensional Formula \nTime of swing \nt \nT \nLength of pendulum \nl \nL \nMass of pendulum \nm \nM \nGravitational acceleration \ng \nL \u22c5 T-2 \nAngular amplitude of swing \n\u03b80 \nNo dimension \nOur first observation is that the mass of the bob cannot enter into our relationship, \nas our final quantity has no dimensions of mass and no other quantity has dimensions of \nmass. Let\u2019s focus on the length of the string and the gravitational acceleration. In order to \neliminate length, these quantities must divide each other when appearing in some \nfunctional relation for the period T . If we choose the combination l / g , the dimensions \nare \nlength \ndim[l / g] =\n= (time)2 \n(2.3.10)\nlength/(time)2 \nIt appears that the time of swing may proportional to the square root of this ratio. Thus \nwe have a candidate formula \n1/2 \n\u239b l \u239e\nT \uf03a \n. \n(2.3.11)\n\u239d\u239c g \u23a0\u239f \n(in the above expression, the symbol \u201c \uf03a \u201d represents a proportionality, not an \napproximation). Because the angular amplitude \u03b80 is dimensionless, it may or may not \nappear. We can account for this by introducing some function y(\u03b80) into our relationship, \nwhich is beyond the limits of this type of analysis. The period is then \n1/2 \n\u239b l \u239e\nT = y(\u03b80) \n. \n(2.3.12)\n\u239d\u239c g \u23a0\u239f \nWe shall discover later on that y(\u03b80) is nearly independent of the angular amplitude \u03b80 \nfor very small amplitudes and is equal to y(\u03b80) = 2\u03c0 , \n2-12", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_26", "text": "T = 2\u03c0 l \ng \n\u239b \n\u239d\u239c \n\u239e \n\u23a0\u239f \n1/2 \n(2.3.13) \n2.4 Order of Magnitude Estimates - Fermi Problems \nCounting is the first mathematical skill we learn. We came to use this skill by \ndistinguishing elements into groups of similar objects, but counting becomes problematic \nwhen our desired objects are not easily identified, or there are too many to count. Rather \nthan spending a huge amount of effort to attempt an exact count, we can try to estimate \nthe number of objects. For example, we can try to estimate the total number of grains of \nsand contained in a bucket of sand. Because we can see individual grains of sand, we \nexpect the number to be very large but finite. Sometimes we can try to estimate a number, \nwhich we are fairly sure but not certain is finite, such as the number of particles in the \nuniverse. \nWe can also assign numbers to quantities that carry dimensions, such as mass, \nlength, time, or charge, which may be difficult to measure exactly. We may be interested \nin estimating the mass of the air inside a room, or the length of telephone wire in the \nUnited States, or the amount of time that we have slept in our lives. We choose some set \nof units, such as kilograms, miles, hours, and coulombs, and then we can attempt to \nestimate the number with respect to our standard quantity. \nOften we are interested in estimating quantities such as speed, force, energy, or \npower. We may want to estimate our natural walking speed, or the force of wind acting \nagainst a bicycle rider, or the total energy consumption of a country, or the electrical \npower necessary to operate a university. All of these quantities have no exact, well-\ndefined value; they instead lie within some range of values. \nWhen we make these types of estimates, we should be satisfied if our estimate is \nreasonably close to the middle of the range of possible values. But what does \u201creasonably \nclose\u201d mean? Once again, this depends on what quantities we are estimating. If we are \ndescribing a quantity that has a very large number associated with it, then an estimate \nwithin an order of magnitude should be satisfactory. The number of molecules in a breath \nof air is close to 1022 ; an estimate anywhere between 1021 and 1023 molecules is close \nenough. If we are trying to win a contest by estimating the number of marbles in a glass \ncontainer, we cannot be so imprecise; we must hope that our estimate is within 1% of the", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_27", "text": "of air is close to 1022 ; an estimate anywhere between 1021 and 1023 molecules is close \nenough. If we are trying to win a contest by estimating the number of marbles in a glass \ncontainer, we cannot be so imprecise; we must hope that our estimate is within 1% of the \nreal quantity. These types of estimations are called Fermi problems. The technique is \nnamed after the physicist Enrico Fermi, who was famous for making these sorts of \u201cback \nof the envelope\u201d calculations. \n2.4.1 Methodology for Estimation Problems \nEstimating is a skill that improves with practice. Here are two guiding principles that may \nhelp you get started. \n(1) You must identify a set of quantities that can be estimated or calculated. \n2-13", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_28", "text": "(2) You must establish an approximate or exact relationship between these quantities \nand the quantity to be estimated in the problem. \nEstimations may be characterized by a precise relationship between an estimated quantity \nand the quantity of interest in the problem. When we estimate, we are drawing upon what \nwe know. But different people are more familiar with certain things than others. If you \nare basing your estimate on a fact that you already know, the accuracy of your estimate \nwill depend on the accuracy of your previous knowledge. When there is no precise \nrelationship between estimated quantities and the quantity to be estimated in the problem, \nthen the accuracy of the result will depend on the type of relationships you decide upon. \nThere are often many approaches to an estimation problem leading to a reasonably \naccurate estimate. So use your creativity and imagination! \nExample 2.6 Lining Up Pennies \nSuppose you want to line pennies up, diameter to diameter, until the total length is \n1 kilometer . How many pennies will you need? How accurate is this estimation? \nSolution: The first step is to consider what type of quantity is being estimated. In this \nexample we are estimating a dimensionless scalar quantity, the number of pennies. We \ncan now give a precise relationship for the number of pennies needed to mark off 1 \nkilometer \ntotaldistance \n# of pennies = \n(2.4.1)\ndiameter of penny . \nWe can estimate a penny to be approximately 2 centimeters wide. Therefore the number \nof pennies is \ntotaldistance \n(1km) \n# of pennies= \n= \nlengthof a penny \n(2 cm)(1km /105 cm) \n(2.4.2) \n= 50,000 pennies = 5 \u00d7104 pennies. \nWhen applying numbers to relationships we must be careful to convert units \nwhenever necessary. How accurate is this estimation? If you measure the size of a penny, \nyou will find out that the width is 1.9 cm , so our estimate was accurate to within 5%. \nThis accuracy was fortuitous. Suppose we estimated the length of a penny to be 1 cm. \nThen our estimate for the total number of pennies would be within a factor of 2, a margin \nof error we can live with for this type of problem. \nExample 2.7 Estimation of Mass of Water on Earth \nEstimate the mass of the water on the Earth. \nSolution: In this example we are estimating mass, a quantity that is a fundamental in SI \nunits, and is measured in kg. We start by approximating that the amount of water on \n2-14", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_29", "text": "Earth is approximately equal to the amount of water in all the oceans. Initially we will try \nto estimate two quantities: the density of water and the volume of water contained in the \noceans. Then the relationship we want is \nmass = (density)(volume) . \n(2.4.3) \nOne of the hardest aspects of estimation problems is to decide which relationship applies. \nOne way to check your work is to check dimensions. Density has dimensions of \nmass/volume, so our relationship is correct dimensionally. \nThe density of fresh water is \u03c1 = 1.0 g \u22c5cm\u22123 ; the density of seawater is slightly \nhigher, but the difference won\u2019t matter for this estimate. You could estimate this density \nby estimating how much mass is contained in a one-liter bottle of water. (The density of \nwater is a point of reference for all density problems. Suppose we need to estimate the \ndensity of iron. If we compare iron to water, we might estimate that iron is 5 to 10 times \ndenser than water. The actual density of iron is \u03c1iron = 7.8 g \u22c5 cm-3 ). \nBecause there is no precise relationship, estimating the volume of water in the \noceans is much harder. Let\u2019s model the volume occupied by the oceans as if the water \ncompletely covers the earth, forming a spherical shell of radius RE and thickness d \n(Figure 2.4, which is decidedly not to scale), where RE is the radius of the earth and d is \nthe average depth of the ocean. The volume of that spherical shell is \n2\nvolume \u2245 4\u03c0 Rearth d . \n(2.4.4) \nRE \nd \nFigure 2.4 A model for estimating the mass of the water on Earth. \nWe also estimate that the oceans cover about 75% of the surface of the earth. So we can \nrefine our estimate that the volume of the oceans is \nvolume \u2245 (0.75)(4\u03c0 RE \n2d) . \n(2.4.5) \n2-15", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_30", "text": "We therefore have two more quantities to estimate, the average depth of the ocean, which \nwe can estimate as d \u2245 1km , and the radius of the earth, which is approximately \n\u2245 6 \u00d7103km . (The quantity that you may remember is the circumference of the earth, \nRE \nabout 25,000 miles . Historically the circumference of the earth was defined to be \n4 \u00d7 107 m ). The radius RE and the circumference s are exactly related by \ns = 2\u03c0 RE . \n(2.4.6) \nThus \ns \n(2.5\u00d7104 mi )(1.6 km \u22c5 mi-1 )\n=\n= \n= 6.4 \u00d7103 km \n(2.4.7)\nRE \n2\u03c0 \n2\u03c0 \nWe will use RE \u2245 6 \u00d7103km ; additional accuracy is not necessary for this problem, since \nthe ocean depth estimate is clearly less accurate. In fact, the factor of 75% is not needed, \nbut included more or less from habit. Altogether, our estimate for the mass of the oceans \nis \nmass = (density)(volume) \u2245\u03c1(0.75)(4\u03c0 RE \n2d) \n\u239b 1g \u239e\u239b 1 kg \u239e\u239b (105 cm)3 \u239e \nmass \u2245 \n\u23a0\u239f (0.75)(4\u03c0 )(6 \u00d7103 km)2(1km) \n(2.4.8)\n3\n\u239d\u239c cm \u23a0\u239f \u239d\u239c 103 g\u23a0\u239f \u239d\u239c (1km)3 \nmass \u2245 3\u00d71020 kg \u2245 1020 kg. \n2-16", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_31", "text": "Chapter 3 Vectors \n3.1 Vector Analysis ....................................................................................................... 1  \n3.1.1 Introduction to Vectors ................................................................................... 1  \n3.1.2 Properties of Vectors ....................................................................................... 1  \n3.2 Coordinate Systems ................................................................................................ 6  \n3.2.1 Cartesian Coordinate System ......................................................................... 6", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_32", "text": "3.2.2 Cylindrical Coordinate System....................................................................... 8  \n3.3 Vectors ................................................................................................................... 10  \n3.3.1 The Use of Vectors in Physics ....................................................................... 10  \n3.3.2 Vectors in Cartesian Coordinates................................................................. 11  \nExample 3.1 Vector Addition................................................................................. 14", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_33", "text": "Example 3.2 Sinking Sailboat ................................................................................ 14  \nExample 3.3 Vector Addition................................................................................. 16  \nExample 3.4 Vector Description of a Point on a Line.......................................... 18  \n3.3.2 Transformation of Vectors in Rotated Coordinate Systems...................... 19  \nExample 3.5 Vector Decomposition in Rotated Coordinate Systems................. 21  \n3.4 Vector Product (Cross Product).......................................................................... 23  \n3.4.1 Right-hand Rule for the Direction of Vector Product................................ 23  \n3.4.2 Properties of the Vector Product.................................................................. 25", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_34", "text": "3.4.3 Vector Decomposition and the Vector Product: Cartesian Coordinates.. 25  \n3.4.4 Vector Decomposition and the Vector Product: Cylindrical Coordinates27  \nExample 3.6 Vector Products................................................................................. 27  \nExample 3.7 Law of Sines....................................................................................... 28  \nExample 3.8 Unit Normal....................................................................................... 28  \nExample 3.9 Volume of Parallelepiped ................................................................. 29  \nExample 3.10 Vector Decomposition..................................................................... 29", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_35", "text": "Chapter 3 Vectors \nPhilosophy is written in this grand book, the universe which stands \ncontinually open to our gaze. But the book cannot be understood unless \none first learns to comprehend the language and read the letters in which \nit is composed. It is written in the language of mathematics, and its \ncharacters are triangles, circles and other geometric figures without \nwhich it is humanly impossible to understand a single word of it; without \nthese, one wanders about in a dark labyrinth.1 \nGalileo Galilee \n3.1 Vector Analysis \n3.1.1 Introduction to Vectors \nCertain physical quantities such as mass or the absolute temperature at some point in \nspace only have magnitude. A single number can represent each of these quantities, with \nappropriate units, which are called scalar quantities. There are, however, other physical \nquantities that have both magnitude and direction. Force is an example of a quantity that \nhas both direction and magnitude (strength). Three numbers are needed to represent the \nmagnitude and direction of a vector quantity in a three dimensional space. These \nquantities are called vector quantities. Vector quantities also satisfy two distinct \noperations, vector addition and multiplication of a vector by a scalar. We can add two \nforces together and the sum of the forces must satisfy the rule for vector addition. We can \nmultiply a force by a scalar thus increasing or decreasing its strength. Position, \ndisplacement, velocity, acceleration, force, and momentum are all physical quantities that \ncan be represented mathematically by vectors. The set of vectors and the two operations \nform what is called a vector space. There are many types of vector spaces but we shall \nrestrict our attention to the very familiar type of vector space in three dimensions that \nmost students have encountered in their mathematical courses. We shall begin our \ndiscussion by defining what we mean by a vector in three dimensional space, and the \nrules for the operations of vector addition and multiplication of a vector by a scalar. \n3.1.2 Properties of Vectors \nA vector is a quantity that has both direction and magnitude. Let a vector be denoted by \n\uf072 \n\uf072\n\uf072 \nthe symbol A . The magnitude of A is | A |\u2261 A . We can represent vectors as geometric \nobjects using arrows. The length of the arrow corresponds to the magnitude of the vector. \nThe arrow points in the direction of the vector (Figure 3.1). \n1 Galileo Galilei, The Assayer, tr. Stillman Drake (1957), Discoveries and Opinions of\nGalileo pp. 237-8. \n1", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_36", "text": "Figure 3.1 Vectors as arrows. \nThere are two defining operations for vectors: \n(1) Vector Addition: \n\uf072 \n\uf072 \n\uf072\n\uf072\n\uf072 \nVectors can be added. Let A and B be two vectors. We define a new vector, C = A + B , \n\uf072\n\uf072 \nthe \u201cvector addition\u201d of A and B , by a geometric construction. Draw the arrow that \n\uf072 \n\uf072\n\uf072 \nrepresents A . Place the tail of the arrow that represents B at the tip of the arrow for A \n\uf072\n\uf072 \nas shown in Figure 3.2a. The arrow that starts at the tail of A and goes to the tip of B is \n\uf072 \n\uf072\n\uf072 \ndefined to be the \u201cvector addition\u201d C = A + B . There is an equivalent construction for the \n\uf072\n\uf072 \nlaw of vector addition. The vectors A and B can be drawn with their tails at the same \npoint. The two vectors form the sides of a parallelogram. The diagonal of the \n\uf072\n\uf072\n\uf072 \nparallelogram corresponds to the vector C = A + B , as shown in Figure 3.2b. \nC = A + B\nC = A + B \nB\nB \nA\nA \n(a) head to tail \n(b) parallelogram \nFigure 3.2a \nFigure 3.2b \nVector addition satisfies the following four properties: \n(i) Commutativity: \nThe order of adding vectors does not matter; \n\uf072 \nA +  \n\uf072 \nB =  \n\uf072 \nB +  \n\uf072 \nA .  \n(3.1.1)  \nOur geometric definition for vector addition satisfies the commutative property (3.1.1).  \nWe can understand this geometrically because in the head to tail representation for the  \n2", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_37", "text": "addition of vectors, it doesn\u2019t matter which vector you begin with, the sum is the same \nvector, as seen in Figure 3.3. \nC = B + A \nC = A + B \nB \nA \nB \nA \nFigure 3.3 Commutative property of vector addition. \n(ii) Associativity: \nWhen adding three vectors, it doesn\u2019t matter which two you start with \n\uf072\n\uf072 \n\uf072\n\uf072 \n\uf072\n\uf072 \n(A + B) + C = A + (B + C) . \n(3.1.2) \n\uf072 \nB +  \n\uf072 \nC ) +  \n\uf072 \nA , and use commutativity to get \n\uf072 \nA  \n\uf072 \nB +  \n\uf072 \nC ) . In figure, \nIn Figure 3.4a, we add ( \n+ (  \n\uf072 \nA +  \n\uf072 \nB ) +  \n\uf072 \nC \nwe add ( \nto arrive at the same vector as in Figure 3.4a. \nB + C \nC\nA \nA + (B + C)\n(A + B) + C \nB \nA + B\nB  \nA\nC \nFigure 3.4a Associative law. \n(iii) Identity Element for Vector Addition: \n\uf072 \nThere is a unique vector, 0 , that acts as an identity element for vector addition. For all \n\uf072 \nvectors A , \n\uf072 \n\uf072 \n\uf072 \n\uf072 \n\uf072 \nA + 0 = 0 + A = A . \n(3.1.3) \n(iv) Inverse Element for Vector Addition: \n\uf072\n\uf072 \nFor every vector A , there is a unique inverse vector \u2212A such that \n3", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_38", "text": "\uf072 \nA + (\u2212  \n\uf072 \nA ) =  \n\uf072\n0 . \n(3.1.4)  \n\uf072 \n\uf072\n\uf075\uf072 \n\uf075\uf072 \nThe vector \u2212A has the same magnitude as A , | A | |= \u2212A |= A , but they point in opposite \ndirections (Figure 3.5). \nA \nA \nFigure 3.5 Additive inverse \n(2) Scalar Multiplication of Vectors: \n\uf072 \nVectors can be multiplied by real numbers. Let A be a vector. Let c be a real positive \n\uf072 \nnumber. Then the multiplication of A by c is a new vector, which we denote by the \n\uf072\n\uf072 \n\uf072 \nsymbol c A . The magnitude of c A is c times the magnitude of A (Figure 3.6a), \n\uf072 \n\uf072 \ncA = c A . \n(3.1.5) \n\uf072\n\uf072 \nLet c > 0 , then the direction of c A is the same as the direction of A . However, the \n\uf072\n\uf072 \ndirection of \u2212c A is opposite of A (Figure 3.6). \nA \nc A\nc A \n\uf072 \nFigure 3.6 Multiplication of vector A by c > 0 , and \u2212c < 0 . \nScalar multiplication of vectors satisfies the following properties: \n(i) Associative Law for Scalar Multiplication: \nThe order of multiplying numbers is doesn\u2019t matter. Let b and c be real numbers. Then \n\uf072 \n\uf072\n\uf072\n\uf072 \n( A) = bc)A = (cb A) = c b\n( A) . \n(3.1.6)\nb c \n( \n(ii) Distributive Law for Vector Addition: \nVectors satisfy a distributive law for vector addition. Let c be a real number. Then \n4", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_39", "text": "c( \n\uf072 \nA +  \n\uf072 \nB ) = c \n\uf072 \nA + c \n\uf072 \nB .  \n(3.1.7)  \nFigure 3.7 illustrates this property. \nc(A + B) \nc A + cB \nC = A + B \ncB \nc A \nB \nA \nFigure 3.7 Distributive Law for vector addition. \n(iii) Distributive Law for Scalar Addition: \nVectors also satisfy a distributive law for scalar addition. Let b and c be real numbers. \nThen \n(b + c) \n\uf072 \nA = b \n\uf072 \nA + c \n\uf072 \nA \n(3.1.8)  \nOur geometric definition of vector addition and scalar multiplication satisfies this \ncondition as seen in Figure 3.8. \nA \nbA \nc A \n(b + c) A \nbA + c A \nFigure 3.8 Distributive law for scalar multiplication. \n(iv) Identity Element for Scalar Multiplication: \nThe number 1 acts as an identity element for multiplication, \n\uf072\n\uf072 \n1 A = A . \n(3.1.9) \n5", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_40", "text": "Unit vector: \nDividing a vector by its magnitude results in a vector of unit length which we denote with \na caret symbol \n\u02c6A = A\n\uf072\n\uf072. \n(3.1.10) \nA \n\uf072 \n\uf072\n\u02c6\nNote that A = A / A = 1. \n3.2 Coordinate Systems \nPhysics involve the study of phenomena that we observe in the world. In order to connect \nthe phenomena to mathematics we begin by introducing the concept of a coordinate \nsystem. A coordinate system consists of four basic elements: \n(1) Choice of origin \n(2) Choice of axes \n(3) Choice of positive direction for each axis \n(4) Choice of unit vectors at every point in space \nThere are three commonly used coordinate systems: Cartesian, cylindrical and spherical. \nIn this chapter we will describe a Cartesian coordinate system and a cylindrical \ncoordinate system. \n3.2.1 Cartesian Coordinate System \nCartesian coordinates consist of a set of mutually perpendicular axes, which intersect at a \ncommon point, the origin O . We live in a three-dimensional spatial world; for that \nreason, the most common system we will use has three axes. \n(1) Choice of Origin: Choose an origin O at any point that is most convenient. \n(2) Choice of Axes: The simplest set of axes is known as the Cartesian axes, x -axis, y \u00ad\naxis, and the z -axis, that are at right angles with respect to each other. Then each point \nP in space can be assigned a triplet of values (xP , yP , zP ) , the Cartesian coordinates of \nthe point P . The ranges of these values are: \u2212\u221e < xP < +\u221e , \n\u2212\u221e < yP < +\u221e , \u2212\u221e < zP < +\u221e . \n(3) Choice of Positive Direction: Our third choice is an assignment of positive direction \nfor each coordinate axis. We shall denote this choice by the symbol + along the positive \naxis. In physics problems we are free to choose our axes and positive directions any way \nthat we decide best fits a given problem. Problems that are very difficult using the \n6", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_41", "text": "conventional choices may turn out to be much easier to solve by making a thoughtful \nchoice of axes. \n(4) Choice of Unit Vectors: We now associate to each point P in space, a set of three \n\u02c6 \n\u02c6 \n\u02c6\nunit vectors (\u02c6iP ,\u02c6jP ,k\u02c6 \nP ) . A unit vector has magnitude one: \n= 1, and\n= 1, \n= 1.\niP \njP \nk P \nWe assign the direction of \u02c6iP to point in the direction of the increasing x -coordinate at \nthe point P . We define the directions for \u02c6jP and k\u02c6 \nP in the direction of the increasing \ny -coordinate and z -coordinate respectively, (Figure 3.10). If we choose a different point \nS , and define a similar set of unit vectors (\u02c6iS , \u02c6jS , k\u02c6 \nS ) , the unit vectors at S and P \nsatisfy the equalities \n\u02c6iS = \u02c6iP , \u02c6jS = \u02c6jP , and k\u02c6 \nS = k\u02c6 \nP , \n(3.2.1) \nbecause vectors are equal if they have the same direction and magnitude regardless of \nwhere they are located in space. \nFigure 3.10 Choice of unit vectors at points P and S . \nA Cartesian coordinate system is the only coordinate system in which Eq. (3.2.1) holds \nfor all pair of points. We therefore drop the reference to the point P and use (\u02c6i, \u02c6j, k\u02c6 ) to \nrepresent the unit vectors in a Cartesian coordinate system (Figure 3.11). \n+z \n+ y \n+x \nxP \nyP \nzP . \nP \n0 \n\u02c6k \n\u02c6j\n\u02c6i \nFigure 3.11 Unit vectors in a Cartesian coordinate system \n7", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_42", "text": "3.2.2 Cylindrical Coordinate System \nMany physical objects demonstrate some type of symmetry. For example if you rotate a \nuniform cylinder about the longitudinal axis (symmetry axis), the cylinder appears \nunchanged. The operation of rotating the cylinder is called a symmetry operation, and the \nobject undergoing the operation, the cylinder, is exactly the same as before the operation \nwas performed. This symmetry property of cylinders suggests a coordinate system, called \na cylindrical coordinate system, that makes the symmetrical property under rotations \ntransparent. \nFirst choose an origin O and axis through O , which we call the z -axis. The \ncylindrical coordinates for a point P are the three numbers (r,\u03b8,z) (Figure 3.12). The \nnumber z represents the familiar coordinate of the point P along the z -axis. The \nnonnegative number r represents the distance from the z -axis to the point P . The points \nin space corresponding to a constant positive value of r lie on a circular cylinder. The \nlocus of points corresponding to r = 0 is the z -axis. In the plane z = 0 , define a \nreference ray through O , which we shall refer to as the positive x -axis. Draw a line \nthrough the point P that is parallel to the z -axis. Let D denote the point of intersection \nbetween that line PD and the plane z = 0 . Draw a ray OD from the origin to the point \nD . Let \u03b8 denote the directed angle from the reference ray to the ray OD . The angle \u03b8 is \npositive when measured counterclockwise and negative when measured clockwise. \nr \n(r, , z)\n. \nz \ny \nr \nr \nP \nO \nD . \nx \n+x \n+ y \n+z \n(r, ,0) \nFigure 3.12 Cylindrical Coordinates \nThe coordinates (r,\u03b8) are called polar coordinates. The coordinate transformations \nbetween (r,\u03b8) and the Cartesian coordinates (x, y) are given by \n8", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_43", "text": "x = r cos\u03b8 , \n(3.2.2) \ny = r sin\u03b8 . \n(3.2.3) \nConversely, if we are given the Cartesian coordinates (x, y) , the coordinates (r,\u03b8) can \nbe determined from the coordinate transformations \n2 + y2)1 2 \nr = +(x\n, \n(3.2.4) \n\u03b8 = tan\u22121( y / x) . \n(3.2.5) \nWe choose a set of unit vectors (r\u02c6P ,\u03b8\u02c6 \nP ,k\u02c6 \nP ) at the point P as follows. We choose k\u02c6 \nP to \npoint in the direction of increasing z . We choose r\u02c6P to point in the direction of \n\u02c6\nincreasing r , directed radially away from the z -axis. We choose \u03b8P to point in the \ndirection of increasing \u03b8 . This unit vector points in the counterclockwise direction, \ntangent to the circle (Figure 3.13a). One crucial difference between cylindrical \ncoordinates and Cartesian coordinates involves the choice of unit vectors. Suppose we \n\u02c6\n\u02c6\nconsider a different point S in the plane. The unit vectors (r\u02c6\n) at the point S are\nS ,\u03b8S ,k S \n\u02c6\n\u02c6\nalso shown in Figure 3.13. Note that r\u02c6 \u2260 r\u02c6\nand \u03b8 \u2260\u03b8\nbecause their direction differ.\nP \nS \nP \nS \nWe shall drop the subscripts denoting the points at which the unit vectors are defined at \nand simple refer to the set of unit vectors at a point as (r\u02c6,\u03b8\u02c6,k\u02c6 ) , with the understanding \n\u02c6\nthat the directions of the set (r\u02c6,\u03b8) depend on the location of the point in question. \n. \nP \nO \n+x \n+ y \n+z \n\u02c6 \nP \n\u02c6rP \n\u02c6k P\n. \n\u02c6rS \nS \n\u02c6k S \n\u02c6 \nS \nFigure 3.13a Unit vectors at two different points in cylindrical coordinates. \n9", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_44", "text": "\u02c6r\n\u02c6 \u02c6j \n\u02c6i \n+ x \n+ y \nFigure 3.13b Unit vectors in polar coordinates and Cartesian coordinates. \nThe unit vectors (r\u02c6,\u03b8\u02c6) at the point P also are related to the Cartesian unit vectors (\u02c6i, \u02c6j ) \nby the transformations \nr\u02c6 = cos\u03b8 \u02c6i + sin\u03b8 \u02c6j , \n(3.2.6) \n\u03b8\u02c6 = \u2212sin\u03b8 \u02c6i + cos\u03b8 \u02c6j . \n(3.2.7) \nSimilarly the inverse transformations are given by \n\u02c6i = cos\u03b8 r\u02c6 \u2212 sin\u03b8 \u03b8\u02c6 , \n(3.2.8) \n\u02c6j = sin\u03b8 r\u02c6 + cos\u03b8 \u03b8\u02c6 . \n(3.2.9) \nA cylindrical coordinate system is also a useful choice to describe the motion of an object \nmoving in a circle about a central point. Consider a vertical axis passing perpendicular to \nthe plane of motion passing through that central point. Then any rotation about this \nvertical axis leaves circles unchanged. \n3.3 Vectors \n3.3.1 The Use of Vectors in Physics \nFrom the last section we have three important ideas about vectors, (1) vectors can exist at \nany point P in space, (2) vectors have direction and magnitude, and (3) any two vectors \nthat have the same direction and magnitude are equal no matter where in space they are \nlocated. When we apply vectors to physical quantities it\u2019s nice to keep in the back of our \nminds all these formal properties. However from the physicist\u2019s point of view, we are \ninterested in representing physical quantities such as displacement, velocity, acceleration, \nforce, impulse, and momentum as vectors. We can\u2019t add force to velocity or subtract \nmomentum from force. We must always understand the physical context for the vector \nquantity. Thus, instead of approaching vectors as formal mathematical objects we shall \ninstead consider the following essential properties that enable us to represent physical \nquantities as vectors. \n10", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_45", "text": "3.3.2 Vectors in Cartesian Coordinates \n(1) Vector Decomposition: Choose a coordinate system with an origin, axes, and unit \nvectors. We can decompose a vector into component vectors along each coordinate axis \n(Figure 3.14). \n+z \n+ y \n+x \n. \n0 \nA \nA z \nA y \nA x \nFigure 3.14 Component vectors in Cartesian coordinates. \n\uf072 \nA vector A at P can be decomposed into the vector sum, \n\uf072 \nA =  \n\uf072 \nA +  \n\uf072 \nA +  \n\uf072 \nA ,  \n(3.3.1) \nx\ny\nz \n\uf072 \nwhere A is the x -component vector pointing in the positive or negative x -direction,\nx \n\uf072\n\uf072 \nA y is the y -component vector pointing in the positive or negative y -direction, and A z \nis the z -component vector pointing in the positive or negative z -direction. \n(2) Vector Components: Once we have defined unit vectors (\u02c6i, \u02c6j, k\u02c6 ) , we then define the \ncomponents of a vector. Recall our vector decomposition, \n\uf072 \nA =  \n\uf072 \nA +  \n\uf072 \nA +  \n\uf072 \nA . We define \nx\ny\nz \n\uf072 \nthe x-component vector, Ax , as \n\uf072 \n\u02c6\nAx = Ax i . \n(3.3.2) \nIn this expression the term Ax , (without the arrow above) is called the x -component of \n\uf072 \nthe vector A . The x -component Ax can be positive, zero, or negative. It is not the \n\uf072 \n2 1/ 2 \nmagnitude of A\nwhich is given by (A ) . The x -component A is a scalar quantity \nx\nx \nx \n\uf072 \nand the x -component vector, Ax is a vector. In a similar fashion we define the y \u00ad\n\uf072 \ncomponent, Ay , and the z -component, Az , of the vector A according to \n\uf072\n\uf072\n\u02c6\n\u02c6\nA = A j, \nA = A k . \n(3.3.3)\ny\ny \nz\nz \n11", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_46", "text": "\uf072 \nA vector A is represented by its three components ( A , A , A ) . Thus we need three \nx\ny\nz \n\uf072 \nnumbers to describe a vector in three-dimensional space. We write the vector A as \n\uf072 \n\u02c6\n\u02c6\n\u02c6\nA = A i + A j + A k . \n(3.3.4)\nx\ny\nz \n\uf072 \n(3) Magnitude: Using the Pythagorean theorem, the magnitude of A is, \n2\n2\n2\nA = A + A + A . \n(3.3.5)\nx \ny\nz \n\uf072 \n(4) Direction: Let\u2019s consider a vector A = (Ax , Ay ,0) . Because the z -component is zero, \n\uf072\n\uf072 \nthe vector A lies in the x y plane. Let \u03b8 denote the angle that the vector A makes in \n-\nthe counterclockwise direction with the positive x -axis (Figure 3.15). \n+y \n\u02c6j \n\u02c6i\nA \nA y\nP A x \n+x \nFigure 3.15 Components of a vector in the xy -plane. \nThen the x -component and y -component are \nAx = Acos(\u03b8), Ay = Asin(\u03b8) . \n(3.3.6) \nWe now write a vector in the xy -plane as \n\uf072 \n\u02c6\n\u02c6\n= Acos( ) i + Asin( ) \n\u03b8 j \n(3.3.7)\nA \n\u03b8 \nOnce the components of a vector are known, the tangent of the angle \u03b8 can be \ndetermined by \nAy \nAsin(\u03b8)\n=\n= tan(\u03b8) , \n(3.3.8)\nAx \nAcos(\u03b8) \nand hence the angle \u03b8 is given by \n\u239b Ay \u239e\n\u03b8 = tan\u22121 \u239c\n\u239f . \n(3.3.9)\n\u239d Ax \u23a0 \n12", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_47", "text": "Clearly, the direction of the vector depends on the sign of Ax and Ay . For example, if \nboth A > 0 and A > 0 , then 0 < \u03b8 < \u03c0 / 2 . If A < 0 and A > 0 then \u03c0 / 2 < \u03b8 < \u03c0 .\nx\ny \nx \ny \nIf A < 0 and A < 0 then \u03c0 < \u03b8 < 3\u03c0 / 2 . If A > 0 and A < 0 , then 3\u03c0 / 2 < \u03b8 < 2\u03c0 . \nx\ny \nx\ny \nNote that tan(\u03b8 ) is a double valued function because \n\u2212 A\nA \nA \n\u2212 A \ny\ny \ny\ny\n= \n, and \n= \n. \n(3.3.10)\n\u2212 A\nA \n\u2212 A\nA \nx\nx \nx\nx \n\uf072\n\uf072 \n\u02c6\n\u02c6\n\u02c6 \n\u02c6\n(5) Unit Vectors: Unit vector in the direction of A : Let A = A i + A j + A k . Let A\nx\ny\nz \n\uf072 \ndenote a unit vector in the direction of A . Then \n\uf072 \n\u02c6\nA\nA x \u02c6i + Ay \u02c6j + Az k\u02c6\nA = \uf072 = \n. \n(3.3.11)\n2 + \n2 + \n2)1/ 2 \n( A \nA\nA\nA \nx \ny \nz \n\uf072\n\uf072 \n(6) Vector Addition: Let A and B be two vectors in the x-y plane. Let \u03b8 A and \u03b8B \n\uf072\n\uf072 \ndenote the angles that the vectors A and B make (in the counterclockwise direction) \nwith the positive x -axis. Then \n\uf072 \nA = Acos(\u03b8 A ) \u02c6i + Asin(\u03b8 A ) \u02c6j, \n(3.3.12) \n\uf072 \nB = Bcos(\u03b8B ) \u02c6i + Bsin(\u03b8B ) \u02c6j \n(3.3.13) \n\uf072\n\uf072\n\uf072 \nIn Figure 3.16, the vector addition C = A + B is shown. Let \u03b8C denote the angle that the \n\uf072 \nvector C makes with the positive x-axis. \nC = A + B \nA \nB \nA x \nA y \nB x \nB y \nC x = A x + B x \nC y = A y + B y \n+x \n+ y \nA \nB \nC \nFigure 3.16 Vector addition using components. \n\uf072 \nFrom Figure 3.16, the components of C are \nC = A + B , \nC = A + B . \n(3.3.14)\nx\nx\nx \ny\ny\ny \n13", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_48", "text": "In terms of magnitudes and angles, we have \nCx = C cos(\u03b8C ) = Acos(\u03b8 A )+ Bcos(\u03b8 B ) \n(3.3.15)\nCy = C sin(\u03b8C ) = Asin(\u03b8 A ) + Bsin(\u03b8 B ). \n\uf072 \nWe can write the vector C as \n\uf072 \nC = ( A + B ) \u02c6i + ( A + B ) \u02c6j = C cos(\u03b8C ) \u02c6i + C sin(\u03b8C ) \u02c6j , \n(3.3.16)\nx\nx \ny\ny \nExample 3.1 Vector Addition \n\uf072\n\uf072 \n\uf072 \n\uf072\n\u02c6 \n\u02c6\n\u02c6 \n\u02c6\n\u02c6\n\u02c6\nGiven two vectors, A = 2 i + \u22123 j + 7 k and B = 5i + j + 2k , find: (a) \n; (b) \n; (c) \nA \nB \n\uf072\n\uf072\n\uf072\n\uf072 \n\uf072\n\u02c6\n\u02c6\nA + B ; (d) A \u2212 B ; (e) a unit vector A pointing in the direction of A ; (f) a unit vector B\n\uf072 \npointing in the direction of B . \nSolution: \n\uf072 \n\uf072 \n52 +12 + 22\n(a) A =(22 + (\u22123)2 + 72 )\n1/2 = 62 = 7.87 . (b) B =(\n)\n1/2 = 30 = 5.48. \n\uf072\n\uf072 \nA + B = ( A + B ) \u02c6i + ( A + B ) \u02c6j + ( A + B ) k\u02c6 \nx \nx \ny\ny \nz\nz \n(c) \n= (2 + 5) \u02c6i + (\u22123+1) \u02c6j + (7 + 2) k\u02c6 \n= 7 \u02c6i \u2212 2 \u02c6j + 9 k\u02c6. \n\uf072\n\uf072 \nA \u2212 B = ( A \u2212 B ) \u02c6i + ( A \u2212 B ) \u02c6j + ( A \u2212 B ) k\u02c6 \nx\nx\ny\ny \nz\nz \n(d) \n= (2 \u2212 5) \u02c6i + (\u22123\u22121) \u02c6j + (7 \u2212 2) k\u02c6 \n= \u22123\u02c6i \u2212 4 \u02c6j + 5 k\u02c6. \n\uf072\n\uf072 \n(e) A unit vector A\u02c6 in the direction of A can be found by dividing the vector A by the \n\uf072 \nmagnitude of A . Therefore \n\uf072 \n\uf072 \nA\u02c6 = A / A = (2 \u02c6i + \u22123 \u02c6j + 7 k\u02c6 ) / \n\uf072 \n\uf072 \n(f) In a similar fashion, B\u02c6 = B / B = (5\u02c6i + \u02c6j + 2k\u02c6 ) / \nExample 3.2 Sinking Sailboat \n62 . \n30 . \n14", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_49", "text": "A Coast Guard ship is located 35 km away from a checkpoint in a direction 52\uf06f north of \nwest. A distressed sailboat located in still water 24 km from the same checkpoint in a \ndirection 18\uf06f south of east is about to sink. Draw a diagram indicating the position of \nboth ships. In what direction and how far must the Coast Guard ship travel to reach the \nsailboat? \nSolution: The diagram of the set-up is Figure 3.17. \n52 \n18 \n35 km \n24 km \nsailboat \nN \nW \nE \nS \ncheckpoint \nCoast Guard \nship \nFigure 3.17 Example 3.2 \nN\nCoast Guard \nship \nsailboat \nW \nE \n1 \n\u02c6i \n\u02c6j \n2 \nr1 \nr2 \n+ x \n+ y \nS \nFigure 3.18 Coordinate system for sailboat and ship \nChoose the checkpoint as the origin of a Cartesian coordinate system with the positive x \u00ad\naxis in the East direction and the positive y \u2013axis in the North direction. Choose the \ncorresponding unit vectors \u02c6i and \u02c6j as shown in Figure 3.18. The Coast Guard ship is \nthen a distance r1 = 35 km at an angle \u03b81 = 180\uf06f \u2212 52\uf06f = 128\uf06f from the positive x -axis, \nand the sailboat is at a distance r2 = 24 km at an angle \u03b82 = \u221218\uf06f from the positive x \u00ad\naxis. The position of the Coast Guard ship is then \n15", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_50", "text": "\uf072 r1 = r1(cos\u03b81\u02c6i + sin\u03b81\u02c6j) \nr \uf072 = \u221221.5km \u02c6i + 27.6km \u02c6j \nand the position of the sailboat is \n\uf072 r2 = r2(cos\u03b82 \u02c6i + sin\u03b82 \u02c6j) \nr \uf072 \n2 = 22.8km \u02c6i \u2212 7.4km \u02c6j. \nN\nCoast Guard \nW \nE \n21 \nS \nship \n\u02c6j \nr1 \nr2 \nr1 \nr2 \nsailboat \nFigure 3.19 Relative position vector from ship to sailboat \nThe relative position vector from the Coast Guard ship to the sailboat is (Figure 3.19) \n\u02c6i  \n\uf072\u2212\nr2 \uf072 \n\uf072  \nr1 = (22.8km \u02c6i \u2212 7.4km \u02c6j) \u2212 (\u221221.5km \u02c6i + 27.6km \u02c6j) \n\uf072 \n\u2212\nr2 \nr1 = 44.4km \u02c6i \u2212 35.0km \u02c6j. \nThe distance between the ship and the sailboat is \n\uf072 \u2212 \uf072 \nr\nr = ((44.4km)2 + (\u221235.0km)2 )1/2 = 56.5km \n2\n1 \nThe rescue ship\u2019s heading would be the inverse tangent of the ratio of the y - and x \u00ad\ncomponents of the relative position vector, \n= tan\u22121(\u221235.0km/44.4km) = \u221238.3\uf06f .\n\u03b821 \nor 38.3\uf06f South of East. \nExample 3.3 Vector Addition \n16", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_51", "text": "\uf072\n\uf072 \n\uf072\n\uf072 \nB = 2 A\nTwo vectors A and B , such that \n, have a resultant \n\uf072 \nC =  \n\uf072 \nA +  \n\uf072 \nB of magnitude \n26.5. \n\uf072 \nThe vector C makes an angle \u03b8C = 41\uf06f with respect to vector \n\uf072 \nA . Find the \n\uf072 \n\uf072 \nmagnitude of each vector and the angle between vectors A and B . \n\uf072 \nSolution: We begin by making a sketch of the three vectors, choosing A to point in the \npositive x -direction (Figure 3.20). \nC = A + B \nB \n\u02c6j \n\u02c6i \nA \nC \nFigure 3.20 Choice of coordinates system for Example 3.3 \n\uf072\n\uf072 \n)2 + (C )2\nDenote the magnitude of C by C \u2261 \n= (Cx \ny \n= 26.5. The components of \nC \n\uf072 \nC =  \n\uf072 \nA +  \n\uf072 \nB are given by \nC = A + B = C cos\u03b8C = (26.5)cos(41\uf06f ) = 20 \n(3.3.17)\nx \nx\nx \nCy = By = C sin\u03b8C = (26.5)sin(41\uf06f ) = 17.4 . \n(3.3.18) \n\uf072\n\uf072 \nB = 2 A\nFrom the condition that \n, the square of their magnitudes satisfy \n(B )2 + (B )2 = 4( A )2 . \n(3.3.19)\nx\ny \nx \nUsing Eqs. (3.3.17) and (3.3.18), Eq. (3.3.19)becomes \n)2 + (C )2 \n)2\n(C \u2212 A \n= 4( A \nx \nx\ny \nx \n)2 \n)2\n(C )2 \u2212 2C A + ( A )2 + (C \n= 4( A \n. \nx \nx \nx \nx\ny \nx \nThis is a quadratic equation \n0 = 3( A )2 + 2C A \u2212 C 2 \nx \nx\nx \nwhich we solve for the component Ax : \n17", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_52", "text": "A = \u22122C x \u00b1 (2C x )2 + (4)(3)(C 2 ) = \u22122(20) \u00b1 (40))2 + (4)(3)(26.5)2 \n= 10.0, \nx \n6\n6 \nwhere we choose the positive square root because we originally chose Ax > 0 . The \n\uf072 \ncomponents of B are then given by Eqs. (3.3.17) and (3.3.18): \nB = C \u2212 A = 20.0 \u221210.0 = 10.0 \nx \nx\nx \nBy = 17.4. \n\uf072 \n)2 + (B )2\nThe magnitude of \n= (Bx \ny \n= 20.0 which is equal to two times the magnitude \nB \n\uf072 \n\uf072\n\uf072 \nof A = 10.0 . The angle between A and B is given by \n\uf072 \n\u03b8 = sin\u22121(By / B ) = sin\u22121(17.4 / 20.0 N) = 60\uf06f . \nExample 3.4 Vector Description of a Point on a Line \n. \n. \nP1 \nP2 \n. \nA \na \na\nd \n\u02c6i \n\u02c6j \n+ x \n+ y \n\uf072r\n1 \n2 \n\uf072r\n1\n2 \n\uf072r \n\uf072r\n\uf072r\n1\n2 \n\uf072r \nFigure 3.21 Example 3.4 \nConsider two points, P1 with coordinates (x1, y1) and P2 with coordinates (x1, y1) , that \n\uf072 \nare separated by distance d . Find a vector A from the origin to the point on the line \nconnecting P1 and P2 that is located a distance a from the point P1 (Figure 3.21). \n\uf072\n\uf072\nSolution: Let r1 = x1 \u02c6i + y1 \u02c6j be the position vector of P1 and r2 = x2 \u02c6i + y2 \u02c6j the position \n\uf072 \u2212 \uf072\nvector of P . Let r\nr be the vector from P to P (Figure 3.22a). The unit vector \n2 \n1 \n2 \n2\n1 \n\u02c6r21\npointing from \nP\nP2\n1 is given by \n= (  \u2212  ) /  \u2212  = (  \u2212  ) / d , where\nto \n)2)1/2 \nd = ((x2 \u2212 x1)2 + ( y2 \u2212 y1 \n18", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_53", "text": "P1 . \n. \nA \nP1\nr1 r2 \nA \nr1 \n. s = a(r1 . \n) / d \nr2 \n.\nP2 \nr1 \nP2 \nr2 \nr2 \n\uf072\n\uf072\n\uf072\nThe vector s in Figure 3.22b connects A to the point at r1, points in the direction of r\u02c612 , \n1 \nFigure 3.22\nFigure 3.22b\n: Relative positio\n: Relative position vector\nn vector\na\n\uf072r\n\uf072 \n\uf072 \n\uf072 \n\uf072 \nA \uf072s\n= a\u02c6r21\nand has length a . Therefore \n= a( \u2212  ) / d . The vector \n. Therefore\n+ \ns \nr1 r2 \n=  \n\uf072 \n\uf072 \n\uf072 \n\uf072 \n\uf072  \nr1 \u2212 \uf072  \nr2 \n\uf072r1 + (a / d)\uf072  \nr2\nA =  \nA = (1\u2212 a / d)(x1  \n\u2212  \n\u2212 a( \n) / d = (1\u2212 a / d)\nr1 \ns = r1\n\uf072 \n\u02c6i + y1\u02c6j) + (a / d)(x2\u02c6i + y2 \u02c6j) \n\u239b\n\u2212 x1) \n\u239e\n\u239b \n\u2212 y1) \n\u239e\na(x2 \n+ \na( y2 \n\u239d\u239c \n((x2 \u2212 x1)2 + ( y2 \u2212 y1)2)1/2 \u23a0\u239f \u02c6i + \n\u239d\u239c y1 \n((x2 \u2212 x1)2 + ( y2 \u2212 y1)2)1/2 \u23a0\u239f \u02c6j. \n\uf072  \nA =  x1 + \n3.3.2 Transformation of Vectors in Rotated Coordinate Systems \nConsider two Cartesian coordinate systems S and S\u2032 such that the (x\u2032, y\u2032) coordinate \naxes in S\u2032 are rotated by an angle \u03b8 with respect to the (x, y) coordinate axes in S , \n(Figure 3.23). \n\u02c6j\n\u02c6j \n\u02c6i\n\u02c6i \n\u02c6j\n+ y \nrotation \n+ y' \nby angle \n\u02c6i \n+x ' \n+ x \nFigure 3.23 Rotated coordinate systems  \nThe components of the unit vector \u02c6i\u2032 in the \u02c6i and \u02c6j direction are given by  \n\u02c6i\u2032 \n\u02c6i\u2032\ncos\u03b8 = cos\u03b8 and i\u2032 = \nsin\u03b8 = sin\u03b8 . Therefore\ni\u2032 = \ny \n\u02c6i\u2032 = ix \u2032 \u02c6i + i\u2032 \ny \u02c6j = \u02c6icos\u03b8 + \u02c6jsin\u03b8 . \n(3.3.20) \n19\nx", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_54", "text": "A similar argument holds for the components of the unit vector \u02c6j\u2032 . The components of \u02c6j\u2032 \nin the \u02c6i and \u02c6j direction are given by jx \u2032 = \u2212 \u02c6j\u2032 \n\u02c6j\u2032\nsin\u03b8 = \u2212sin\u03b8 and j\u2032 \ny = \ncos\u03b8 = cos\u03b8 . \nTherefore \n\u02c6j\u2032 = jx \u2032 \u02c6i + j\u2032 \ny \u02c6j = \u02c6jcos\u03b8 \u2212 \u02c6isin\u03b8 . \n(3.3.21) \nConversely, from Figure 3.23 and similar vector decomposition arguments, the \ncomponents of \u02c6i and \u02c6j in S\u2032 are given by \n\u02c6i = \u02c6i\u2032cos\u03b8 \u2212 \u02c6j\u2032sin\u03b8 , \n(3.3.22) \n\u02c6j = \u02c6i\u2032sin\u03b8 + \u02c6j\u2032cos\u03b8 . \n(3.3.23) \n\uf072\nConsider a fixed vector r = x \u02c6i + y\u02c6j with components (x, y) in coordinate system S . In \n\uf072\ncoordinate system S\u2032 , the vector is given by r = x\u2032 \u02c6i\u2032 + y\u2032\u02c6j\u2032 , where (x\u2032, y\u2032) are the \ncomponents in S\u2032 , (Figure 3.24). \nx ' \ny' \n\u02c6i\n\u02c6j\n\u02c6i \n\u02c6j \nrotation \nby angle \nx \ny \nr \nr \nFigure 3.24 Transformation of vector components \nUsing the Eqs. (3.3.20) and (3.3.21), we have that \n\uf072 r = x \u02c6i + y\u02c6j = x(\u02c6i\u2032cos\u03b8 \u2212 \u02c6j\u2032sin\u03b8) + y(\u02c6j\u2032cos\u03b8 + \u02c6i\u2032sin\u03b8 ) \n(3.3.24) \nr \uf072 = (xcos\u03b8 + ysin\u03b8 )\u02c6i\u2032 + (xsin\u03b8 \u2212 ycos\u03b8 )\u02c6j\u2032. \nTherefore the components of the vector transform according to \nx\u2032 = xcos\u03b8 + ysin\u03b8 , \n(3.3.25) \ny\u2032 = x sin\u03b8 \u2212 ycos\u03b8 . \n(3.3.26) \nWe now consider an alternate approach to understanding the transformation laws for the \ncomponents of the position vector of a fixed point in space. In coordinate system S , \n\uf072\n\uf072 \nr\nsuppose the position vector r has length r = \nand makes an angle \u03c6 with respect to \nthe positive x -axis (Figure 3.25). \n20", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_55", "text": "x ' \ny' \n\u02c6i\n\u02c6j\n\u02c6i \n\u02c6j \nrotation \nby angle \nx \ny \nr \nr \nFigure 3.25 Transformation of vector components of the position vector \n\uf072\nThen the components of r in S are given by \nx = r cos\u03c6 , \n(3.3.27) \ny = r sin\u03c6 . \n(3.3.28) \n\uf072\nIn coordinate system S\u2032 , the components of r are given by \nx\u2032 = r cos(\u03c6 \u2212\u03b8 ) , \n(3.3.29) \ny\u2032 = r sin(\u03c6 \u2212\u03b8 ) . \n(3.3.30) \nApply the addition of angle trigonometric identities to Eqs. (3.3.29) and (3.3.30) yielding \nx\u2032 = r cos(\u03c6 \u2212\u03b8 ) = r cos\u03c6 cos\u03b8 + r sin\u03c6 sin\u03b8 = xcos\u03b8 + ysin\u03b8 , \n(3.3.31) \ny\u2032 = r sin(\u03c6 \u2212\u03b8 ) = r sin\u03c6 cos\u03b8 \u2212 r cos\u03c6 sin\u03b8 = ycos\u03b8 \u2212 xsin\u03b8 , \n(3.3.32) \nin agreement with Eqs. (3.3.25) and (3.3.26). \nExample 3.5 Vector Decomposition in Rotated Coordinate Systems \n\uf072 \nWith respect to a given Cartesian coordinate system S , a vector A has components \nA = 5 , A = \u22123, A = 0 . Consider a second coordinate system S\u2032 such that the (x\u2032, y\u2032)\nx\ny \nz \ncoordinate axes in S\u2032 are rotated by an angle \u03b8 = 60\uf06f with respect to the (x, y) \ncoordinate axes in S , (Figure 3.26). (a) What are the components A and A of vector \nx ' \ny ' \n\uf072 \nA in coordinate system S\u2032 ? (b) Calculate the magnitude of the vector using the ( Ax , Ay ) \ncomponents and using the ( A , A ) components. Does your result agree with what you \nx ' \ny ' \nexpect? \n21", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_56", "text": "\u02c6i \n\u02c6j \nA \nA x \nA y \n\u02c6j \n\u02c6i \nrotation \n= 60\nby angle \n= 60 \n\u0003A \nFigure 3.26 Example 3.4 \n\uf072 \nSolution: a) We begin by considering the vector decomposition of A with respect to the \ncoordinate system S , \n\uf072 \nA = Ax \u02c6i + Ay \u02c6j . \n(3.3.33) \nNow we can use our results for the transformation of unit vectors \u02c6i and \u02c6j in terms of \u02c6i\u2032 \n\uf072 \nand \u02c6j\u2032 , (Eqs. (3.3.22) and (3.3.23)) in order decompose the vector A in coordinate \nsystem S\u2032 \n\uf072 \nA = A \u02c6i + A \u02c6j = A (cos\u03b8\u02c6i\u2032 \u2212 sin\u03b8\u02c6j\u2032) + A (sin\u03b8\u02c6i\u2032 + cos\u03b8\u02c6j\u2032)\nx\ny\nx \ny \n= ( Ax cos\u03b8 + Ay sin\u03b8 )\u02c6i\u2032 + (\u2212 Ax sin\u03b8 + Ay cos\u03b8 )\u02c6j\u2032 \n(3.3.34) \n= A \u02c6i + A \u02c6j,\nx\u2032 \ny\u2032 \nwhere \nA = A cos\u03b8 + A sin\u03b8 \n(3.3.35)\nx\u2032 \nx\ny \nA = \u2212 A sin\u03b8 + A cos\u03b8 . \n(3.3.36)\ny\u2032 \nx \ny \nWe now use the given information that Ax = 5 , Ay = \u22123, and \u03b8 = 60\uf06f to solve for the \n\uf072 \ncomponents of A in coordinate system S\u2032 \nA = A cos\u03b8 + A sin\u03b8 = (1/ 2)(5\u2212 3 3) ,\nx\u2032 \nx\ny \nA = \u2212 A sin\u03b8 + A cos\u03b8 = (1/ 2)(\u22125 3 \u2212 3) . \ny\u2032 \nx\ny \nb) The magnitude can be calculated in either coordinate system \n\uf072 \n)2\nA = ( Ax )2 + ( Ay \n= (5)2 + (\u22123)2 = 34 \n\uf072 \nA = ( A )2 + ( A )2 = ((1/ 2)(5\u2212 3 3))\n2 \n+((1/ 2)(\u22125 3 \u2212 3) )\n2 \n= 34 . \nx\u2032 \ny\u2032 \n\uf072 \nThis result agrees with what I expect because the length of vector A is independent of \nthe choice of coordinate system. \n22", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_57", "text": "3.4 Vector Product (Cross Product) \n\uf072\n\uf072 \nLet A and B be two vectors. Because any two non-parallel vectors form a plane, we \n\uf072\n\uf072 \ndenote the angle \u03b8 to be the angle between the vectors A and B as shown in Figure \n\uf072\n\uf072 \n\uf072\n\uf072 \n3.27. The magnitude of the vector product A \u00d7 B of the vectors A and B is defined to \n\uf072\n\uf072 \nbe product of the magnitude of the vectors A and B with the sine of the angle \u03b8 \nbetween the two vectors, \n\uf072 \nA \u00d7  \n\uf072 \nB =  \n\uf072 \nB  \n\uf072 \nA \nsin(\u03b8) . \n(3.3.37) \nThe angle \u03b8 between the vectors is limited to the values 0 \u2264\u03b8 \u2264\u03c0 ensuring that \nsin( ) \u2265 0.\n\u03b8 \nFigure 3.27 Vector product geometry. \n\uf072\n\uf072 \nThe direction of the vector product is defined as follows. The vectors A and B form a \nplane. Consider the direction perpendicular to this plane. There are two possibilities: we \nshall choose one of these two (the one shown in Figure 3.27) for the direction of the \n\uf072\n\uf072 \nvector product A \u00d7 B using a convention that is commonly called the \u201cright-hand rule\u201d. \n3.4.1 Right-hand Rule for the Direction of Vector Product \n\uf072\n\uf072 \nThe first step is to redraw the vectors A and B so that the tails are touching. Then draw \n\uf072\n\uf072 \nan arc starting from the vector A and finishing on the vector B . Curl your right fingers \nthe same way as the arc. Your right thumb points in the direction of the vector product \n\uf072\n\uf072 \nA \u00d7 B (Figure 3.28). \n23", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_58", "text": "C = A B \nA \nB \nFigure 3.28 Right-Hand Rule. \n\uf072 \n\uf072 \nYou should remember that the direction of the vector product \n\uf072 \n\uf072 \nA \u00d7 B is perpendicular to \nthe plane formed by A and B . We can give a geometric interpretation to the magnitude \nof the vector product by writing the magnitude as \n\uf072 \nA \u00d7  \n\uf072 \nB =  \n\uf072 \nA ( \n\uf072 \nB sin\u03b8 ). \n(3.3.38) \n\uf072\n\uf072 \nThe vectors A and B form a parallelogram. The area of the parallelogram is equal to the \nheight times the base, which is the magnitude of the vector product. In Figure 3.29, two \ndifferent representations of the height and base of a parallelogram are illustrated. As \n\uf072 \n\uf072 \ndepicted in Figure 3.29(a), the term B sin\u03b8 is the projection of the vector B in the \n\uf072 \ndirection perpendicular to the vector B . We could also write the magnitude of the vector \nproduct as\n\uf072 \nA \u00d7  \n\uf072 \nB = ( \n\uf072 \nA sin\u03b8 )\n\uf072 \nB .  \n(3.3.39)  \n\uf072 \n\uf072 \nA\nThe term \nsin\u03b8 is the projection of the vector A in the direction perpendicular to the \n\uf072 \nvector B as shown in Figure 3.29(b). The vector product of two vectors that are parallel \n(or anti-parallel) to each other is zero because the angle between the vectors is 0 (or \u03c0 ) \nand sin(0) = 0 (or sin( ) \n\u03c0 = 0 ). Geometrically, two parallel vectors do not have a unique \ncomponent perpendicular to their common direction. \nA sin \nB sin \nB \nB \nA\nA \n(a) \n(b) \n24", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_59", "text": "Figure 3.29 Projection of (a) \n\uf072 \nB \n\uf072\n\uf072  \nperpendicular to A , (b) of A perpendicular to \n\uf072 \nB \n3.4.2 Properties of the Vector Product \n(1)  The vector product is anti-commutative because changing the order of the vectors \nchanges the direction of the vector product by the right hand rule: \n\uf072\n\uf072 \n\uf072\n\uf072 \nA \u00d7 B = \u2212B \u00d7 A .  \n(3.3.40) \n\uf072  \n(2) The vector product between a vector c A where c is a scalar and a vector \n\uf072 \nB is \n\uf072\n\uf072 \n\uf072\n\uf072 \nc A \u00d7 B = c (A \u00d7 B) . \n(3.3.41) \nSimilarly, \n\uf072 \n\uf072 \n\uf072\n\uf072 \nA \u00d7 c B = c (A \u00d7 B) .  \n(3.3.42) \n\uf072\n\uf072  \n\uf072 \n(3) The vector product between the sum of two vectors A and B with a vector C is \n\uf072 \n\uf072 \n\uf072 \n\uf072\n\uf072\n\uf072\n\uf072 \n(A + B)\u00d7C = A \u00d7C + B\u00d7C \n(3.3.43) \nSimilarly, \n\uf072  \n\uf072\n\uf072 \n\uf072\n\uf072\n\uf072\n\uf072 \nA \u00d7 (B + C) = A \u00d7 B + A \u00d7C .  \n(3.3.44) \n3.4.3 Vector Decomposition and the Vector Product: Cartesian Coordinates \nWe first calculate that the magnitude of vector product of the unit vectors \u02c6i and \u02c6j : \n\u02c6 \u02c6 \n\u02c6 \u02c6\n| i \u00d7 j | | || |sin( \n= i\nj \n\u03c0 / 2) = 1,  \n(3.3.45) \n\u02c6\n\u02c6\nbecause the unit vectors have magnitude | | | |\ni = j = 1 and sin( \u03c0 / 2) = 1. By the right hand \nrule, the direction of \u02c6i \u00d7 \u02c6j is in the +k\u02c6 as shown in Figure 3.30. Thus \u02c6i \u00d7 \u02c6j = k\u02c6 . \n\u02c6j\nk\u02c6 = \u02c6i \n\u02c6i \n\u02c6j \nFigure 3.30 Vector product of \u02c6i \u00d7 \u02c6j \n25", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_60", "text": "We note that the same rule applies for the unit vectors in the y and z directions, \n\u02c6\n\u02c6 \n\u02c6\n\u02c6\n\u02c6\n\u02c6\nj\u00d7k = i, k \u00d7 i = j . \n(3.3.46) \nBy the anti-commutatively property (1) of the vector product, \n\u02c6j\u00d7 \u02c6i = \u2212k\u02c6, \u02c6i \u00d7k\u02c6 = \u2212\u02c6j \n(3.3.47) \nThe vector product of the unit vector \u02c6i with itself is zero because the two unit vectors are \nparallel to each other, (sin(0) = 0 ), \n\u02c6 \u02c6 \n\u02c6 \u02c6\n| i \u00d7 i | | || | sin(0) \ni\ni \n= 0 . \n(3.3.48)\n= \nThe vector product of the unit vector \u02c6j with itself and the unit vector k\u02c6 with itself are \nalso zero for the same reason, \n\u02c6 \u02c6 \n\u02c6\n\u02c6 = 0 . \n(3.3.49)\n\u00d7\nj j = 0, k \u00d7k \nWith these properties in mind we can now develop an algebraic expression for the vector \nproduct in terms of components. Let\u2019s choose a Cartesian coordinate system with the \n\uf072 \nvector B pointing along the positive x-axis with positive x-component Bx . Then the \n\uf072\n\uf072 \nvectors A and B can be written as \n\uf072 \n\u02c6\n\u02c6 \n\u02c6\nA = A i + A j + A k \n(3.3.50)\nx\ny\nz \n\uf072 \n\u02c6\nB = Bx i , \n(3.3.51) \nrespectively. The vector product in vector components is \n\uf072\n\uf072 \n\u02c6\n\u02c6\n\u02c6 \n\u02c6\nA \u00d7 B = ( A i + Ay j + A k)\u00d7 Bx i . \n(3.3.52)\nx\nz \nThis becomes, \n\uf072\n\uf072 \nA \u00d7 B = (A \u02c6i \u00d7 B \u02c6i) + (A \u02c6j\u00d7 B \u02c6i) + (A k\u02c6 \u00d7 B \u02c6i)\nx\nx \ny\nx\nz\nx \n= A B (\u02c6i \u00d7 \u02c6i) + A B (\u02c6j\u00d7 \u02c6i) + A B (k\u02c6 \u00d7 \u02c6i) . \n(3.3.53)\nx\nx \ny\nx\nz\nx \n= \u2212A B k\u02c6 + A B \u02c6j\ny\nx \nz\nx \nThe vector component expression for the vector product easily generalizes for arbitrary \nvectors \n\uf072 \nA = A \u02c6i + A \u02c6j+ A k\u02c6 \n(3.3.54)\nx \ny \nz \n\uf072 \n\u02c6\n\u02c6\n\u02c6\nB = B i + B j + B k , \n(3.3.55)\nx\ny\nz \nto yield \n26", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_61", "text": "\uf072\n\uf072 \n\u02c6\n\u02c6 \n\u02c6\nA \u00d7 B = (A B \u2212 A B ) i + (A B \u2212 A B ) j + (A B \u2212 A B ) k . \n(3.3.56)\ny\nz\nz\ny \nz\nx\nx\nz \nx\ny\ny\nx \n3.4.4 Vector Decomposition and the Vector Product: Cylindrical Coordinates \nRecall the cylindrical coordinate system, which we show in Figure 3.31. We have chosen \ntwo directions, radial and tangential in the plane, and a perpendicular direction to the \nplane. \n! \n\u02c6r \n\u02c6\n\u02c6k\nr \nz \n+ x \n+ y \n+ z \nFigure 3.31 Cylindrical coordinates  \nThe unit vectors are at right angles to each other and so using the right hand rule, the  \nvector product of the unit vectors are given by the relations \n\u02c6r \u00d7 \u02c6\u03b8 = \u02c6k \n(3.3.57) \n\u02c6\u03b8 \u00d7 \u02c6k = \u02c6r \n(3.3.58) \n\u02c6k \u00d7 \u02c6r = \u02c6\u03b8 . \n(3.3.59) \nBecause the vector product satisfies \n\uf072 \nA \u00d7  \n\uf072 \nB = \u2212  \n\uf072 \nB \u00d7  \n\uf072 \nA , we also have that \nFinally \n\u02c6\u03b8 \u00d7 \u02c6r = \u2212 \u02c6k \n\u02c6k \u00d7 \u02c6\u03b8 = \u2212 \u02c6r \n\u02c6r \u00d7 \u02c6k = \u2212 \u02c6\u03b8 . \n\u02c6r \u00d7 \u02c6r = \u02c6\u03b8 \u00d7 \u02c6\u03b8 = \u02c6k \u00d7 \u02c6k = \n\uf072 \n0 . \n(3.3.60) \n(3.3.61) \n(3.3.62) \n(3.3.63) \nExample 3.6 Vector Products \nGiven two vectors, \n\u02c6 \n\u02c6 \n\u02c6\n2 \n3 \n7\n= \n+ \u2212 \n+\nA \ni \nj \nk \n\uf072 \nand \n\u02c6 \u02c6 \n\u02c6\n5 \n2\n= \n+ +\nB \ni \nj \nk \n\uf072 \n, find \n\uf072 \nA \u00d7 \n\uf072 \nB . \nSolution: \n27", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_62", "text": "\uf072\n\uf072 \nA \u00d7 B = ( A B \u2212 A B ) \u02c6i + ( A B \u2212 A B ) \u02c6j + ( A B \u2212 A B ) k\u02c6 \ny\nz\nz\ny \nz\nx\nx\nz \nx\ny\ny\nx \n= ((\u22123)(2) \u2212 (7)(1)) \u02c6i + ((7)(5) \u2212 (2)(2)) \u02c6j+ ((2)(1) \u2212 (\u22123)(5)) k\u02c6 \n= \u221213 \u02c6i + 31\u02c6j+17 k\u02c6. \nExample 3.7 Law of Sines \nFor \nthe \ntriangle\n\uf072 \nB \nshown \nin \nFigure \n3.32(a), \nprove \nthe \nlaw \nof \nsines, \n\uf072 \nA / sin\u03b1 = \n/ sin \u03b2 = \n\uf072 \nC / sin\u03b3 , using the vector product. \nFigure 3.32(a) Example 3.6 \nFigure 3.32(b) Vector analysis \n\uf072\n\uf072 \n\uf072 \nSolution: Consider the area of a triangle formed by three vectors A , B , and C , where \n\uf072\n\uf072\n\uf072 \n\uf072\n\uf072\n\uf072 \nA + B + C = 0 \n(Figure \n3.32(b)). \nBecause \nA + B + C = 0 , \nwe \nhave \nthat \n\uf072 \n\uf072\n\uf072\n\uf072 \n\uf072\n\uf072\n\uf072\n\uf072 \n\uf072\n\uf072 \n\uf072\n\uf072 \n\uf072\n\uf072 \n\uf072\n\uf072 \n0 = A \u00d7 (A + B + C) = A \u00d7 B + A \u00d7C . Thus A \u00d7 B = \u2212A \u00d7C or \n. From\nA \u00d7 B = A \u00d7C \n\uf072\n\uf072 \n\uf072 \n\uf072\n\uf072 \n\uf072 \uf072\n\uf072 \nFigure 17.7b we see that \nsin \u03b2 . Therefore\nA \u00d7 B = A B sin \u03b3 and A \u00d7C = A C \n\uf072 \n\uf072 \uf072 \n\uf072\n\uf072 \n\uf072 \nsin \u03b2 , and hence \n/ sin \u03b3 . A similar argument shows that \nA B sin \u03b3 = A C \nB / sin \u03b2 = C \n\uf072\n\uf072 \n/ sin\u03b1 proving the law of sines.\nB / sin \u03b2 = A \nExample 3.8 Unit Normal \nFind a unit vector perpendicular to \n\u02c6 \u02c6 \n\u02c6\n= + \u2212\nA i \nj k \n\uf072 \nand \n\u02c6 \u02c6 \n\u02c6\n2 \n3\n= \u2212 \u2212 +\nB \ni j \nk \n\uf072 \n. \n\uf072 \n\uf072 \n\uf072 \n\uf072 \nSolution: The vector product A \u00d7 B is perpendicular to both A and B . Therefore the \n\uf072\n\uf072\n\uf072\n\uf072 \n\uf072\n\uf072 \nunit vectors n\u02c6 = \u00b1A \u00d7 B / \nare perpendicular to both A and B . We first calculate\nA \u00d7 B \n\uf072\n\uf072 \nA \u00d7 B = ( A B \u2212 A B ) \u02c6i + ( A B \u2212 A B ) \u02c6j+ ( A B \u2212 A B ) k\u02c6 \ny\nz\nz\ny \nz\nx\nx\nz \nx\ny\ny\nx \n= ((1)(3) \u2212 (\u22121)(\u22121)) \u02c6i + ((\u22121)(2) \u2212 (1)(3)) \u02c6j + ((1)(\u22121) \u2212 (1)(2)) k\u02c6", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_63", "text": "y \nz\nx\nx\nz \nx\ny\ny\nx \n= ((1)(3) \u2212 (\u22121)(\u22121)) \u02c6i + ((\u22121)(2) \u2212 (1)(3)) \u02c6j + ((1)(\u22121) \u2212 (1)(2)) k\u02c6 \n= 2 \u02c6i \u2212 5 \u02c6j\u2212 3 k\u02c6. \nWe now calculate the magnitude \n28", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_64", "text": "\uf072\n\uf072 \n= (22 + 52 + 32)1/2 = (38)1/2 \nA \u00d7 B \n. \nTherefore the perpendicular unit vectors are \n\uf072\n\uf072 \n\uf072\n\uf072 \nn\u02c6 = \u00b1A \u00d7 B / A \u00d7 B = \u00b1(2 \u02c6i \u2212 5 \u02c6j\u2212 3 k\u02c6 ) / (38)1/2 . \nExample 3.9 Volume of Parallelepiped \n\uf072\n\uf072 \nShow that the volume of a parallelepiped with edges formed by the vectors A , B , and \n\uf072 \n\uf072\n\uf072\n\uf072 \nC is given by A \u22c5(B\u00d7C) . \nSolution: The volume of a parallelepiped is given by area of the base times height. If the \n\uf072\n\uf072 \nbase is formed by the vectors B and C , then the area of the base is given by the \n\uf072\n\uf072 \n\uf072\n\uf072 \n\uf072\n\uf072 \nmagnitude of B \u00d7C. The vector B \u00d7C = \nn\u02c6 where n\u02c6 is a unit vector perpendicular \nB\u00d7C \nto the base (Figure 3.33). \nFigure 3.33 Example 3.9 \n\uf072 \nThe projection of the vector A along the direction n\u02c6 gives the height of the \n\uf072 \nparallelepiped. This projection is given by taking the dot product of A with a unit vector \n\uf072 \n\u22c5 \u02c6\nand is equal to A n = height . Therefore \n\uf072\n\uf072\n\uf072 \n\uf072 \n\uf072\n\uf072 \n\uf072\n\uf072 \uf072 \nA \u22c5 (B \u00d7 C) = A \u22c5 ( B \u00d7 C )n\u02c6 = ( B \u00d7 C )A \u22c5 n\u02c6 = (area)(height) = (volume) . \nExample 3.10 Vector Decomposition \n\uf072 \nLet A be an arbitrary vector and let n\u02c6 be a unit vector in some fixed direction. Show \n\uf072\n\uf072 \n\uf072 \nthat A = (A \u22c5 n\u02c6)n\u02c6 + (n\u02c6 \u00d7 A) \u00d7 n\u02c6 . \n\uf072\n\uf072 \nSolution: Let A = A\uf050n\u02c6 + A\u22a5e\u02c6 where A\uf050 is the component A in the direction of n\u02c6 , e\u02c6 is \n\uf072 \nthe direction of the projection of A in a plane perpendicular to n\u02c6 , and A\u22a5 is the \n29\n\n\uf072\n\uf072  \n\u22c5 \nA \u22c5 n =\ncomponent of A in the direction of e\u02c6 . Because e n \n\u02c6 \u02c6 = 0 , we have that \n\u02c6 \nA\uf050 . Note \nthat \n\uf072 \nn\u02c6 \u00d7 A = n\u02c6 \u00d7 ( An\u02c6 + A\u22a5 e\u02c6) = n\u02c6 \u00d7 A\u22a5 e\u02c6 = \n(n\u02c6 \u00d7 e\u02c6) .\n\uf050\nA\u22a5 \nThe unit vector n\u02c6 \u00d7e\u02c6 lies in the plane perpendicular to n\u02c6 and is also perpendicular to e\u02c6 .  \nTherefore (n\u02c6 \u00d7 e\u02c6) \u00d7 n\u02c6 is also a unit vector that is parallel to e\u02c6 (by the right hand rule. So  \n\uf072 \n(n\u02c6 \u00d7 A) \u00d7 n\u02c6 = A\u22a5 e\u02c6 . Thus \n\uf072 \n\uf072\n\uf072 \nA = A\uf050n\u02c6 + A\u22a5e\u02c6 = (A \u22c5n\u02c6)n\u02c6 + (n\u02c6 \u00d7 A) \u00d7 n\u02c6 . \n30", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_65", "text": "Chapter 4 One Dimensional Kinematics \n4.1 Introduction............................................................................................................. 2  \n4.2 Position, Time Interval, Displacement.................................................................. 3  \n4.2.1 Position.............................................................................................................. 3  \n4.2.2 Time Interval.................................................................................................... 3", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_66", "text": "4.2.3 Displacement .................................................................................................... 3  \n4.3 Velocity .................................................................................................................... 4  \n4.3.1 Average Velocity .............................................................................................. 4  \n4.3.3 Instantaneous Velocity..................................................................................... 5  \nExample 4.1 Determining Velocity from Position.................................................. 6", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_67", "text": "Example 4.1 Determining Velocity from Position.................................................. 6  \nExample 4.2 Mean Value Theorem ......................................................................... 7  \n4.4 Acceleration............................................................................................................. 9  \n4.4.1 Average Acceleration....................................................................................... 9  \n4.4.2 Instantaneous Acceleration ............................................................................. 9  \nExample 4.3 Determining Acceleration from Velocity........................................ 10", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_68", "text": "Example 4.3 Determining Acceleration from Velocity........................................ 10  \n4.5 Constant Acceleration .......................................................................................... 11  \n4.5.1 Velocity: Area Under the Acceleration vs. Time Graph ............................ 11  \n4.5.2 Displacement: Area Under the Velocity vs. Time Graph........................... 11  \nExample 4.4 Accelerating Car ............................................................................... 13  \nExample 4.5 Catching a Bus .................................................................................. 15  \n4.6 One Dimensional Kinematics and Integration................................................... 16  \n4.6.1 Change of Velocity as the Indefinite Integral of Acceleration................... 16", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_69", "text": "4.6.1 Change of Velocity as the Indefinite Integral of Acceleration................... 16  \nExample 4.6 Non-constant acceleration................................................................ 17  \n4.6.2 Area as the Indefinite Integral of Acceleration........................................... 17  \n4.6.3 Change of Velocity as the Definite Integral of Acceleration ...................... 19  \n4.6.4 Displacement as the Definite Integral of Velocity....................................... 22  \nExample 4.5 Non-constant Acceleration ............................................................... 22  \nExample 4.6 Bicycle and Car ................................................................................. 24", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_70", "text": "Chapter 4 One Dimensional Kinematics  \nIn the first place, what do we mean by time and space? It turns out that these deep \nphilosophical questions have to be analyzed very carefully in physics, and this is \nnot easy to do. The theory of relativity shows that our ideas of space and time are \nnot as simple as one might imagine at first sight. However, for our present \npurposes, for the accuracy that we need at first, we need not be very careful about \ndefining things precisely. Perhaps you say, \u201cThat\u2019s a terrible thing\u2014I learned \nthat in science we have to define everything precisely.\u201d We cannot define \nanything precisely! If we attempt to, we get into that paralysis of thought that \ncomes to philosophers, who sit opposite each other, one saying to the other, \u201cYou \ndon\u2019t know what you are talking about!\u201d The second one says. \u201cWhat do you \nmean by know? What do you mean by talking? What do you mean by you?\u201d, and \nso on. In order to be able to talk constructively, we just have to agree that we are \ntalking roughly about the same thing. You know as much about time as you need \nfor the present, but remember that there are some subtleties that have to be \ndiscussed; we shall discuss them later.1 \nRichard Feynman \n4.1 Introduction \nKinematics is the mathematical description of motion. The term is derived from the \nGreek word kinema, meaning movement. In order to quantify motion, a mathematical \ncoordinate system, called a reference frame, is used to describe space and time. Once a \nreference frame has been chosen, we shall introduce the physical concepts of position, \nvelocity and acceleration in a mathematically precise manner. Figure 4.1 shows a \nCartesian coordinate system in one dimension with unit vector \u02c6i pointing in the direction \nof increasing x -coordinate. \n\u02c6i \n0 \n+ x \nFigure 4.1 A one-dimensional Cartesian coordinate system. \n1 Richard P. Feynman, Robert B. Leighton, Matthew Sands, The Feynman Lectures on \nPhysics, Addison-Wesley, Reading, Massachusetts, (1963), p. 12-2. \n1", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_71", "text": "4.2 Position, Time Interval, Displacement \n4.2.1 Position \nConsider a point-like object moving in one dimension. We denote the position \ncoordinate of the object with respect to the choice of origin by x(t) . The position \ncoordinate is a function of time and can be positive, zero, or negative, depending on the \nlocation of the object. The position of the object with respect to the origin has both \ndirection and magnitude, and hence is a vector (Figure 4.2), which we shall denote as the \nposition vector (or simply position) and write as \n\uf072 r(t) = x(t) \u02c6i . \n(4.2.1) \nWe denote the position coordinate at t = 0 by the symbol x0 \u2261 x t( = 0) . The SI unit for \nposition is the meter [m]. \n\u02c6i \nr(t) \n+ x\n0 \nx(t) \nFigure 4.2 The position vector, with reference to a chosen origin. \n4.2.2 Time Interval \nConsider a closed interval of time [t1, t2] . We characterize this time interval by the \ndifference in endpoints of the interval, \n\u0394t = t2 \u2212 t1. \n(4.2.2) \nThe SI units for time intervals are seconds [s]. \n4.2.3 Displacement \nThe displacement of a body during a time interval [t1, t2] (Figure 4.3) is defined \nto be the change in the position of the body \n\uf072\n\uf072 \n\uf072\n\u0394r \u2261 r(t2 ) \u2212 r(t1) = (x(t2) \u2212 x(t1)) \u02c6i \u2261\u0394x(t) \u02c6i . \n(4.2.3) \nDisplacement is a vector quantity. \n2", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_72", "text": "\u02c6i \n0 \n+ x \nx(t2 )\nx(t1 ) \nr(t2 ) \nr \nr(t1) \nFigure 4.3 The displacement vector of an object over a time interval is the vector \ndifference between the two position vectors \n4.3 Velocity \nWhen describing the motion of objects, words like \u201cspeed\u201d and \u201cvelocity\u201d are used in \nnatural language; however when introducing a mathematical description of motion, we \nneed to define these terms precisely. Our procedure will be to define average quantities \nfor finite intervals of time and then examine what happens in the limit as the time interval \nbecomes infinitesimally small. This will lead us to the mathematical concept that velocity \nat an instant in time is the derivative of the position with respect to time. \n4.3.1 Average Velocity \nThe x -component of the average velocity, v \n, for a time interval \u0394t is defined\nx,ave \nto be the displacement \u0394x divided by the time interval \u0394t , \n\u0394x \nv \n\u2261 \n. \n(4.3.1)\nx,ave \n\u0394t \nBecause we are describing one-dimensional motion we shall drop the subscript x and \ndenote \nv \n= v \n. \n(4.3.2)\nave \nx,ave \nWhen we introduce two-dimensional motion we will distinguish the components of the \nvelocity by subscripts. The average velocity vector is then \nv \uf072 \n\u2261\u0394x \u02c6i = v \n\u02c6i . \n(4.3.3)\nave \nave \n\u0394t \nThe SI units for average velocity are meters per second \u23a1m s\u22c5\n\u22121 \u23a4\u23a6 . The average velocity is \n\u23a3 \nnot necessarily equal to the distance in the time interval \u0394t traveled divided by the time \ninterval \u0394t . For example, during a time interval, an object moves in the positive x \u00ad\ndirection and then returns to its starting position, the displacement of the object is zero, \nbut the distance traveled is non-zero. \n3", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_73", "text": "4.3.3 Instantaneous Velocity \nConsider a body moving in one direction. During the time interval [ ,t t + \u0394t] , the average \nvelocity corresponds to the slope of the line connecting the points ( ,t x t( )) and \n(t + \u0394t, x(t + \u0394t)) . The slope, the rise over the run, is the change in position divided by \nthe change in time, and is given by \nrise \n\u0394x\nx(t + \u0394t) \u2212 x(t)\nv \n\u2261 \n=\n= \n. \n(4.3.4)\nave \nrun \n\u0394t \n\u0394t \nt \n, the slope of the lines connecting the points (t, x(t)) and (t + \u0394t, x(t + \u0394t)) ,\nAs \u0394 \u2192 0 \napproach slope of the tangent line to the graph of the function x(t) at the time t (Figure \n4.4). \nt \nt + t \nx(t) \nx(t + t) \nt \nx \ntangent line \nat time t \nx(t) \nFigure 4.4 Plot of position vs. time showing the tangent line at time t . \nThe limiting value of this sequence is defined to be the x -component of the \ninstantaneous velocity at the time t . \nThe x -component of instantaneous velocity at time t is given by the \nslope of the tangent line to the graph of the position function at time t : \n\u0394x\nx(t + \u0394t) \u2212 x(t) \ndx \nv(t) \u2261 lim vave = lim \n= lim \n\u2261 \n. \n(4.3.5)\n\u0394t\u21920 \n\u0394t\u21920 \u0394t \n\u0394t\u21920 \n\u0394t \ndt \nThe instantaneous velocity vector is then \n\uf072 v(t) = v(t) \u02c6i . \n(4.3.6) \nThe component of the velocity, v(t) , can be positive, zero, or negative, depending on \nwhether the object is travelling in the positive x -direction, instantaneously at rest, or the \nnegative x -direction. \n4", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_74", "text": "Example 4.1 Determining Velocity from Position \nConsider an object that is moving along the x -coordinate axis with the position function \ngiven by \nx(t) = x0 + 1 bt2 \n(4.3.7)\n2 \nwhere x0 is the initial position of the object at t = 0 . We can explicitly calculate the x \u00ad\ncomponent of instantaneous velocity from Equation (4.3.5) by first calculating the \ndisplacement in the x -direction, \u0394x = x t( + \u0394t) \u2212 x t( ) . We need to calculate the position \nat time t + \u0394t , \nx(t + \u0394t) = x0 + 1 b(t + \u0394t)2 = x0 + 1 b(t2 + 2t\u0394t + \u0394t2) . \n(4.3.8)\n2\n2 \nThen the x -component of instantaneous velocity is \n\u239b \n1 \n\u239e\n\u239b \n1 \n\u239e \n+ b(t2 + 2t \u0394t + \u0394t2) \n+ bt2\nx0 \nx0\nx(t + \u0394t) \u2212 x(t) \n\u239d\u239c \n2 \n\u23a0\u239f \u2212\u239d\u239c \n2 \n\u23a0\u239f \nv(t) = lim \n= lim \n. (4.3.9)\n\u0394t\u21920 \n\u0394t \n\u0394t\u21920 \n\u0394t \nThis expression reduces to \n\u239b \n1 \n\u239e \nv(t) = lim bt + b\u0394t\u23a0\u239f . \n(4.3.10)\n\u0394t\u21920 \u239d\u239c \n2 \nThe first term is independent of the interval \u0394t and the second term vanishes because in \nthe limit as \u0394 \u2192 0 , the term (1/ 2)b\u0394t \u2192 0 is zero. Therefore the x -component of \nt \ninstantaneous velocity at time t is \nv(t) = bt . \n(4.3.11) \nIn Figure 4.5 we plot the instantaneous velocity, v(t) , as a function of time t . \nt \nv(t) \nv(t) = bt \nFigure 4.5 Plot of instantaneous velocity instantaneous velocity as a function of time. \n5", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_75", "text": "Example 4.2 Mean Value Theorem \nConsider an object that is moving along the x -coordinate axis with the position function \ngiven by \nx(t) = x0 + v0t + 1 \n2 bt2 . \n(4.3.12) \nThe graph of x(t) vs. t is shown in Figure 4.6. \nx(t) \nx0 \nslope = v ave \nslope = v(t1) \nt \nx(t) = x0 + v0t + 1 \n2 bt2 \nti \nt f \nx(t f ) \nx(ti ) \nt1 = (t f \nti ) / 2 \nFigure 4.6 Intermediate Value Theorem \nThe x -component of the instantaneous velocity is \ndx(t)\nv(t) =\n+ bt . \n(4.3.13)\n= v0\ndt \nFor the time interval [ti ,t f ], the displacement of the object is \n) = \u0394x = v0\n) + 1 \n2 \u2212 ti \n2) = v0\n) + 1 \n) . (4.3.14)\nx(t f ) \u2212 x(ti \n(t f \u2212 ti \nb(t f \n(t f \u2212 ti \nb(t f \u2212 ti )(t f + ti\n2\n2 \nRecall that the x -component of the average velocity is defined by the condition that \n\u0394x = v \n) . \n(4.3.15)\nave (t f \u2212 ti \nWe can determine the average velocity by substituting Eq. (4.3.15) into Eq. (4.3.14) \nyielding \n6", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_76", "text": "1 \nv \n+ \n) . \n(4.3.16)\nave = v0 \nb(t f + ti\n2 \nThe Mean Value Theorem from calculus states that there exists an instant in time t1, with \n< t f , such that the x -component of the instantaneously velocity, v(t1) , satisfies\nti < t1 \n\u0394x = v(t1)(t f \n) . \n(4.3.17)\n\u2212 ti \nGeometrically this means that the slope of the straight line (blue line in Figure 4.6) \nconnecting the points (ti , x(ti )) to (t f ,x(t f )) is equal to the slope of the tangent line (red \nline in Figure 4.6) to the graph of x(t) vs. t at the point (t1, x(t1)) (Figure 4.6), \nv(t1) = vave . \n(4.3.18) \nWe know from Eq. (4.3.13) that \nv(t1) = v0 + bt1. \n(4.3.19) \nWe can solve for the time t1 by substituting Eqs. (4.3.19) and (4.3.16) into Eq. (4.3.18) \nyielding \n) / 2 \n(4.3.20)\nt1 = (t f + ti \nThis intermediate value v(t1) is also equal to one-half the sum of the initial velocity and \nfinal velocity \n) \n+ bt f )\nv(ti ) + v(t f \n(v0 + bti ) + (v0 \n1 \nv(t1) =\n= \n= v0 + b(t f \n) = v0 \n. (4.3.21)\n+ ti \n+ bt1\n2 \n2\n2 \nFor any time interval, the quantity (v(ti ) + v(t f )) / 2 , is the arithmetic mean of the initial \nvelocity and the final velocity (but unfortunately is also sometimes referred to as the \naverage velocity). The average velocity, which we defined as v \n) / \u0394t , and the \nave = (x f \u2212 xi \narithmetic mean, (v(ti ) + v(t f )) / 2 , are only equal in the special case when the velocity is \na linear function in the variable t as in this example, (Eq. (4.3.13)). We shall only use the", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_77", "text": "arithmetic mean, (v(ti ) + v(t f )) / 2 , are only equal in the special case when the velocity is \na linear function in the variable t as in this example, (Eq. (4.3.13)). We shall only use the \nterm average velocity to mean displacement divided by the time interval. \n7", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_78", "text": "4.4 Acceleration \nWe shall apply the same physical and mathematical procedure for defining acceleration, \nas the rate of change of velocity with respect to time. We first consider how the \ninstantaneous velocity changes over a fixed time interval of time and then take the limit \nas the time interval approaches zero. \n4.4.1 Average Acceleration \nAverage acceleration is the quantity that measures a change in velocity over a particular \ntime interval. Suppose during a time interval \u0394t a body undergoes a change in velocity \n\uf072\n\uf072 \n\uf072\n\u0394v = v(t + \u0394t) \u2212 v( )t . \n(4.3.22) \nThe change in the x -component of the velocity, \u0394v , for the time interval [ ,t t + \u0394t] is \nthen \n\u0394v = v(t + \u0394t) \u2212 v(t) . \n(4.3.23) \nThe x -component of the average acceleration for the time interval \u0394t is defined \nto be \n\uf072 \n\u0394v \n(v(t + \u0394t) \u2212 v(t)) \na \n= a \n\u02c6i \u2261 \n\u02c6i = \n\u02c6i . \n(4.3.24)\nave \nave \n\u0394t \n\u0394t \nThe SI units for average acceleration are meters per second squared, [m\u22c5s\u22122 ]. \n4.4.2 Instantaneous Acceleration \nConsider the graph of the x -component of velocity, v(t) , (Figure 4.7). \nv(t \nslope = \nt\nt \nt + t\na(t) \nv(t) \nv\nt)\n+ \nt \n8", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_79", "text": "Figure 4.7 Graph of velocity vs. time showing the tangent line at time t . \nThe average acceleration for a fixed time interval \u0394t is the slope of the straight line \nconnecting the two points (t, v(t)) and (t + \u0394t, v(t + \u0394t)) . In order to define the x \u00ad\ncomponent of the instantaneous acceleration at time t , we employ the same limiting \nargument as we did when we defined the instantaneous velocity in terms of the slope of \nthe tangent line. \nThe x -component of the instantaneous acceleration at time t is the slope of the \ntangent line at time t of the graph of the x -component of the velocity as a \nfunction of time, \na(t) \u2261 lim \n\u0394t\u21920 \n\u0394v \n\u0394t = lim \n\u0394t\u21920 \n(v(t + \u0394t) \u2212 v(t)) \n\u0394t \n\u2261 dv \ndt . \n(4.3.25) \nThe instantaneous acceleration vector at time t is then \n\uf072 a(t) = a(t) \u02c6i . \n(4.3.26) \nBecause the velocity is the derivative of position with respect to time, the x -component \nof the acceleration is the second derivative of the position function, \nd 2\ndv \nx \na =\n= \n. \n(4.3.27)\ndt \ndt2 \nExample 4.3 Determining Acceleration from Velocity \nLet\u2019s continue Example 4.1, in which the position function for the body is given by \nx = x0 + (1/ 2)bt2 , and the x -component of the velocity is v = bt . The x -component of \nthe instantaneous acceleration is the first derivative (with respect to time) of the x \u00ad\ncomponent of the velocity: \ndv \nv(t + \u0394t) \u2212 v(t) \nbt + b\u0394t \u2212 bt \na =\n= lim \n= lim \n= b. \n(4.3.28)\ndt \n\u0394t\u21920 \n\u0394t \n\u0394t\u21920 \n\u0394t \nNote that in Eq. (4.3.28), the ratio \u0394v / \u0394t is independent of t , consistent with the \nconstant slope as shown in Figure 4.5. \n9", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_80", "text": "4.5 Constant Acceleration \nv(t) \nt \nv(t) = v0 + at \n(a) \na(t) \na \nv0 \nt \na(t) = a \n(b) \nArea = at \nFigure 4.8 Constant acceleration: (a) velocity, (b) acceleration \nWhen the x -component of the velocity is a linear function (Figure 4.8(a)), the average \nacceleration, \u0394v / \u0394t , is a constant and hence is equal to the instantaneous acceleration \n(Figure 4.8(b)). Let\u2019s consider a body undergoing constant acceleration for a time interval \n[0, t] , where \u0394t = t . Denote the x -component of the velocity at time t = 0 by \n\u2261 v(t = 0) . Therefore the x -component of the acceleration is given by\nv0 \n\u0394v\nv(t) \u2212 v0\na(t) =\n= \n. \n(4.4.1)\n\u0394t\nt \nThus the x -component of the velocity is a linear function of time given by \nv(t) = v0 + at . \n(4.4.2) \n4.5.1 Velocity: Area Under the Acceleration vs. Time Graph \nIn Figure 4.8(b), the area under the acceleration vs. time graph, for the time interval \n\u0394t = t \u2212 0 = t , is \nArea(a(t), t) = at . \n(4.4.3) \nFrom Eq. (4.4.2), the area is the change in the x -component of the velocity for the \ninterval [0, t]: \nArea(a(t),t) = at = v(t) \u2212 v0 = \u0394v . \n(4.4.4) \n4.5.2 Displacement: Area Under the Velocity vs. Time Graph \nIn Figure 4.9 shows a graph of the x -component of the velocity vs. time for the case of \nconstant acceleration (Eq. (4.4.2)). \n10", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_81", "text": "t \nv(t) \nv(t) = v0 + at \nv0 \nA1 = v0 t \nA2 = 1 \n2 (v(t) v0 ) \nO \nFigure 4.9 Graph of velocity as a function of time for a constant. \nThe region under the velocity vs. time curve is a trapezoid, formed from a rectangle with \narea \nt , and a triangle with area \n= (1/ 2)(v(t) \u2212 v0) . The total area of the \nA1 = v0 \nA2 \ntrapezoid is given by \n1\nArea(v(t),t) = A1 \nt + (v(t) \u2212 v0) . \n(4.4.5)\n+ A2 = v0 \n2 \nSubstituting for the velocity (Eq. (4.4.2)) yields \nArea(v(t),t) = v0 t + 1 at2 . \n(4.4.6)\n2 \nRecall that from Example 4.2 (setting b = a and \u0394t = t ), \n1 \nv \n+ at = \u0394x / t , \n(4.4.7)\nave = v0 \n2 \ntherefore Eq. (4.4.6) can be rewritten as \n1\nArea(v(t),t) = (v0 + at)t = v\nt = \u0394x \n(4.4.8)\n2 \nave \nThe displacement is equal to the area under the graph of the x -component of the velocity \nvs. time. The position as a function of time can now be found by rewriting Equation \n(4.4.8) as \n1 \nx(t) = x0 \nt + at2 . \n(4.4.9)\n+ v0 \n2 \nFigure 4.10 shows a graph of this equation. Notice that at t = 0 the slope is non-zero, \ncorresponding to the initial velocity component v0 . \n11", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_82", "text": "x(t) )  \nx0 \nO \nt \nslope = v0 \nFigure 4.10 Graph of position vs. time for constant acceleration. \nExample 4.4 Accelerating Car \nA car, starting at rest at t = 0 , accelerates in a straight line for 100 m with an unknown \nconstant acceleration. It reaches a speed of 20 m \u22c5 s\u22121 and then continues at this speed for \nanother 10 s . (a) Write down the equations for position and velocity of the car as a \nfunction of time. (b) How long was the car accelerating? (c) What was the magnitude of \nthe acceleration? (d) Plot speed vs. time, acceleration vs. time, and position vs. time for \nthe entire motion. (e) What was the average velocity for the entire trip? \nSolutions: (a) For the acceleration a , the position x(t) and velocity v(t) as a function of \ntime t for a car starting from rest are \nx(t) = (1/ 2) at2 \n(4.4.10)\nvx (t) = at. \nb) Denote the time interval during which the car accelerated by t1. We know that the \nposition x(t1) = 100m and v(t1) = 20 m \u22c5 s\u22121 . Note that we can eliminate the acceleration \na between the Equations (4.4.10) to obtain \nx(t) = (1/ 2)v(t) t . \n(4.4.11) \nWe can solve this equation for time as a function of the distance and the final speed \ngiving \nt = 2 x(t) \nv(t) . \n(4.4.12) \nWe can now substitute our known values for the position x(t1 ) = 100m and \nv(t1) = 20 m \u22c5 s\u22121 and solve for the time interval that the car has accelerated \n12", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_83", "text": "x(t1) \n100 m \n= 2 \n= 2 \n= 10s . \n(4.4.13)\nt1 \n\u22121\nv(t1) \n20 m \u22c5 s \nc) We can substitute into either of the expressions in Equation (4.4.10); the second is \nslightly easier to use, \nv(t1) \n20 m \u22c5 s\u22121 \n\u22122\na =\n= \n= 2.0m \u22c5 s . \n(4.4.14)\n10s \nt1 \nd) The x -component of acceleration vs. time, x -component of the velocity vs. time, and \nthe position vs. time are piece-wise functions given by \n-2 ;\n\u23a72 m \u22c5s\n0 < t \u2264 10 s \na(t) = \u23a8 \n, \n\u23a90; \n10 s < t < 20 s \n-2 )t;\n\u23a7 \n0 < t \u2264 10 s \n\u23aa(2 m \u22c5s \nv(t) = \u23a8 \n,\n-1;\n\u23aa20 m \u22c5s \n10 s \u2264 t \u2264 20 s \n\u23a9 \n-2 )t2;\n\u23a7 \n0 < t \u2264 10 s \n\u23aa(1/ 2)(2 m \u22c5s \nx(t) = \u23a8 \n. \n\u23a9\u23aa100 m +(20 m \u22c5s-2 )(t \u221210 s); \n10 s \u2264 t \u2264 20 s \nThe graphs of the x -component of acceleration vs. time, x -component of the velocity vs. \ntime, and the position vs. time are shown in Figure 4.11. \n(e) After accelerating, the car travels for an additional ten seconds at constant speed and \nduring this interval the car travels an additional distance \u0394x = v(t1) \u00d7 10s=200m (note \nthat this is twice the distance traveled during the 10s of acceleration), so the total \ndistance traveled is 300m and the total time is 20s , for an average velocity of \n300m \n\u22121\nv \n=\n=15m\u22c5s . \n(4.4.15)\nave \n20s \nx(t)\na(t) \nv(t) \n-2 \n100 m \n20 m s\n2 m s \nt \n-1 \nt \nt\n10 s \n20 s \n10 s \n20 s \n10 s \n20 s \nFigure 4.11 Graphs of the x-components of acceleration, velocity and position as piece\u00ad\nwise functions \n13", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_84", "text": "Example 4.5 Catching a Bus \nAt the instant a traffic light turns green, a car starts from rest with a given constant \nacceleration, 3.0 m \u22c5s-2 . Just as the light turns green, a bus, traveling with a given \nconstant velocity, 1.6 \u00d7 101 m \u22c5 s-1 , passes the car. The car speeds up and passes the bus \nsome time later. How far down the road has the car traveled, when the car passes the bus? \nSolution: \nThere are two moving objects, bus and the car. Each object undergoes one stage of one-\ndimensional motion. We are given the acceleration of the car, the velocity of the bus, and \ninfer that the position of the car and the bus are equal when the bus just passes the car. \nFigure 4.12 shows a qualitative sketch of the position of the car and bus as a function of \ntime. \nx \nx2 (t)\nbus \nx1(t)\ncar \n0 \nt \nta \nFigure 4.12 Position vs. time of the car and bus \nChoose a coordinate system with the origin at the traffic light and the positive x \u00ad\ndirection such that car and bus are travelling in the positive x -direction. Set time t = 0 as \nthe instant the car and bus pass each other at the origin when the light turns green. Figure \n4.13 shows the position of the car and bus at time t . \nx2 (t) \nx1(t) \n0 \n+ x \nFigure 4.13 Coordinate system for car and bus \nLet x1(t) denote the position function of the car, and x2(t) the position function for the \nbus. The initial position and initial velocity of the car are both zero, x1,0 = 0 and v1,0 = 0 , \n14", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_85", "text": "and the acceleration of the car is non-zero a1 \u2260 0 . Therefore the position and velocity \nfunctions of the car are given by \n1 \nx1(t) = a1t2 ,\n2 \nv1(t) = a1t . \nThe initial position of the bus is zero, x2,0 = 0 , the initial velocity of the bus is non-zero, \n\u2260 0 , and the acceleration of the bus is zero, a2 = 0 . Therefore the velocity is constant, \nv2,0 \n, and the position function for the bus is given by x2 \nt .\nv2 (t) = v2,0 \n(t) = v2,0 \nLet t = ta correspond to the time that the car passes the bus. Then at that instant, the \nposition functions of the bus and car are equal, x1(ta ) = x2(ta ) . We can use this condition \nto solve for t : \na \n2\n2v2,0 \n(2)(1.6 \u00d7101 m \u22c5s-1)\n(1/ 2)a1ta = v2,0ta \u21d2 ta =\n= \n-2 ) \n= 1.1\u00d7101s . \n(3.0m \u22c5s\na1 \nTherefore the position of the car at ta is \n1 \n2\n2v2,0 \n2 \n(2)(1.6 \u00d7101 m \u22c5s \nx1(ta ) = a1ta =\n= \n-2 ) \n-1)2 \n= 1.7 \u00d7102m .\n2 \n(3.0 m \u22c5s\na1 \n4.6 One Dimensional Kinematics and Integration \nWhen the acceleration a(t) of an object is a non-constant function of time, we would like \nto determine the time dependence of the position function x(t) and the x -component of \nthe velocity v(t) . Because the acceleration is non-constant we no longer can use Eqs. \n(4.4.2) and (4.4.9). Instead we shall use integration techniques to determine these \nfunctions. \n4.6.1 Change of Velocity as the Indefinite Integral of Acceleration \nConsider a time interval t1 < t < t2 . Recall that by definition the derivative of the velocity \nv(t) is equal to the acceleration a(t) , \ndv(t) = a(t) . \n(4.5.1)\ndt \n15", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_86", "text": "Integration is defined as the inverse operation of differentiation or the \u2018anti-derivative\u2019. \nFor our example, the function v(t) is called the indefinite integral of a(t) with respect \nto t , and is unique up to an additive constant C . We denote this by writing \nv(t) + C = \u222b a(t) dt .  \n(4.5.2) \nThe symbol \u222b ...dt means the \u2018integral, with respect to t , of \u2026\u201d, and is thought of as the \nd\ninverse of the symbol \n.... . Equivalently we can write the differential dv(t) = a(t)dt ,\ndt \ncalled the integrand, and then Eq. (4.5.2) can be written as \nv(t) + C = \u222b dv(t) , \n(4.5.3) \nwhich we interpret by saying that the integral of the differential of function is equal to the \nfunction plus a constant. \nExample 4.6 Non-constant acceleration \nSuppose an object at time t = 0 has initial non-zero velocity \nand acceleration \nv0 \na(t) = bt2 , where b is a constant. Then dv(t) = bt2dt = d(bt3 / 3) . The velocity is then \nv(t) + C = \u222b d(bt3 / 3) = bt3 / 3. At t = 0 , we have that v0 + C = 0. Therefore C = \u2212v0 and \nthe velocity as a function of time is then v(t) = v0 + (bt3 / 3) . \n4.6.2 Area as the Indefinite Integral of Acceleration \nConsider the graph of a positive-valued acceleration function a(t) vs. t for the \ninterval t1 \u2264 t \u2264 t2 , shown in Figure 4.14a. Denote the area under the graph of a(t) over \n2\nthe interval t1 \u2264 t \u2264 t2 by A t\nt . \n1 \na(t) )  \nFigure 4.14a: Area under the graph of \nacceleration over an interval t1 \u2264 t \u2264 t2 \nt1 \nt \na(t1) \na(t2 ) \nt2\nt c \nArea = A t1 \nt2 \n16", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_87", "text": "t1 \nt \na(t1) \na(t2 ) \nt2\nt c \na(t c ) \na(t)  \nFigure \n4.14b: \nIntermediate \nvalue \nTheorem. The shaded regions above and \nbelow the curve have equal areas. \nThe Intermediate Value Theorem states that there is at least one time t such that the \nc \n2\narea A t\nt is equal to\n1 \nA \nt\nt1\n2 = a(tc )(t2 \u2212 t1) . \n(4.5.4) \nIn Figure 4.14b, the shaded regions above and below the curve have equal areas, and \n2\nhence the area A t\nt under the curve is equal to the area of the rectangle given by \n1 \na(t \n\u2212 t1) . \nc )(t2 \na(t) \nt1 \nt \na(t1) \na(t2 ) \nt2\nt t + t \nA t1 \nt \nA t \nt+ t \nFigure 4.15 Area function is additive \nWe shall now show that the derivative of the area function is equal to the acceleration and \nthererfore we can write the area function as an indefinite integral. From Figure 4.15, the \narea function satisfies the condition that \nA t + A t+\u0394t\nt+\u0394t \nt \n= A \n.  \n(4.5.5)\nt1 \nt1 \nA t+\u0394t\u2212 A t\nLet the small increment of area be denoted by \u0394A t =\n= A t+\u0394t . By the\nt1 \nt1 \nt1 \nt \nIntermediate Value Theorem \n17", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_88", "text": "\u0394A t\nt = a(tc )\u0394t , \n(4.5.6)\n1 \nwhere t \u2264 tc \u2264 t + \u0394t . In the limit as \u0394t \u2192 0, \n\u0394A t\ndA t \nt1 \nt1\n= lim \n= lim a(t c ) = a(t) , \n(4.5.7)\ndt \n\u0394t\u21920 \u0394t \nt \u2192t\nc \nwith the initial condition that when t = t1, the area A\nt1 = 0 is zero. Because v(t) is also an \nt1 \nintegral of a(t) , we have that \nA t \nt = \u222b a(t) dt = v(t) + C . \n(4.5.8)\n1 \nWhen t = t1, the area A\nt1 = 0 is zero, therefore v(t1) + C = 0 , and so C = \u2212v(t1) . Therefore\nt1 \nEq. (4.5.8) becomes \nA t\nt = v(t) \u2212 v(t1) = \u222b a(t)dt . \n(4.5.9)\n1 \nWhen we set t = t2 , Eq. (4.5.9) becomes \nA\nt2 = v(t2) \u2212 v(t1) = \u222b a(t) dt . \n(4.5.10)\nt1 \nThe area under the graph of the positive-valued acceleration function for the interval \n\u2264 t \u2264 t2 can be found by integrating a(t) .\nt1 \n4.6.3 Change of Velocity as the Definite Integral of Acceleration \nLet a(t) be the acceleration function over the interval ti \u2264 t \u2264 t f . Recall that the velocity \nv(t) is an integral of a(t) because dv(t) / dt = a(t) . Divide the time interval [ti , t f ] into \nn equal time subintervals \u0394t = (t f \u2212 ti ) / n . For each subinterval [t j ,t j+1] , where the index \nj = 1, 2, ...,n , t1 = ti and tn+1 = t f , let t be a time such that t j \u2264 t \u2264 t j+1 . Let \nc j \nc j \nj=n \nSn = \u2211a(t ) \u0394t . \n(4.5.11)\nc j\nj=1", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_89", "text": "c j \nc j \nj=n \nSn = \u2211a(t ) \u0394t . \n(4.5.11)\nc j\nj=1 \nSn is the sum of the blue rectangle shown in Figure 4.16a for the case n = 4 . The \nFundamental Theorem of Calculus states that in the limit as n \u2192\u221e , the sum is equal \nto the change in the velocity during the interval [ti , t f ] \n18", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_90", "text": "j=n \nlim Sn = lim \u2211a(t ) \u0394t = v(t f ) \u2212 v(ti ) . \n(4.5.12)\nn\u2192\u221e \nn\u2192\u221e \nc j\nj=1 \nt1 \nt2 \nt3 \nt4 \nt5\nt c1 \nt c2 \nt c3 \nt c4 \na( ) \na(t c2 ) \na(t c3 ) \na(t c4 ) \na(t) \nt c1 \nt \nt1 \nt2 \nt3 \nt4 \nt5\nt c1 \nt c2 \nt c3 \nt c4 \na( ) \na(t c2 ) \na(t c3 ) \na(t c4 ) \na(t) \nt c1 \nt \nFigure 4.16a Graph of a(t) vs. t \nFigure 4.16b Graph of a(t) vs. t \nThe limit of the sum in Eq. (4.5.12) is a number, which we denote by the symbol \nt f \nj=n \n\u222b a(t) dt \u2261 lim \u2211a(t ) \u0394t = v(t f ) \u2212 v(ti ) , \n(4.5.13)\nc j\nj=1\nn\u2192\u221e \nti \nand is called the definite integral of a(t) from ti to t f . The times ti and t f are called \nthe limits of integration, ti the lower limit and t f the upper limit. The definite integral is \na linear map that takes a function a(t) defined over the interval [ti , t f ] and gives a \nnumber. The map is linear because \nt f \nt f\nt f \n(a1(t) + a2(t)) dt = a1(t) dt + a2(t) dt , \n(4.5.14)\n\u222b \n\u222b\n\u222b \nti \nti\nti \nSuppose the times t , j = 1,...,n , are selected such that each t satisfies the Intermediate \nc j \nc j \nValue Theorem, \ndv(tc j )\n\u0394v j \u2261 v(t j+1) \u2212 v(t j ) = \ndt \n\u0394t = a(t c j )\u0394t , \n(4.5.15) \n19", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_91", "text": "where a(t ) is the instantaneous acceleration at t , (Figure 4.16b). Then the sum of the \nc j \nc j \nchanges in the velocity for the interval [ti , t f ] is \nj=n \n\u2211\u0394v j = (v(t2) \u2212 v(t1)) + (v(t3) \u2212 v(t2)) + \uf04c + (v(tn+1) \u2212 v(tn )) = v(tn+1) \u2212 v(t1) \nj=1 \n(4.5.16) \n). \n= v(t f ) \u2212 v(ti \nwhere v(t f ) = v(tn+1) and v(ti ) = v(t1) . Substituting Eq. (4.5.15) into Equation (4.5.16) \nyields the exact result that the change in the x -component of the velocity is give by this \nfinite sum. \nj=n\nj=n \nv(t f ) \u2212 v(ti ) = \u2211\u0394v j = \u2211a(tc j ) \u0394t . \n(4.5.17) \nj=1 \nj=1 \nWe do not specifically know the intermediate values a(tc j ) and so Eq. (4.5.17) is not \nuseful as a calculating tool. The statement of the Fundamental Theorem of Calculus is \nthat the limit as n \u2192\u221e of the sum in Eq. (4.5.12) is independent of the choice of the set \nof tc j . Therefore the exact result in Eq. (4.5.17) is the limit of the sum. \nThus we can evaluate the definite integral if we know any indefinite integral of the \nintegrand a(t)dt = dv(t). \nAdditionally, provided the acceleration function has only non-negative values, the limit is \nalso equal to the area under the graph of a(t) vs. t for the time interval, [ti , t f ]: \nt f \nt f =\nA ti \na(t) dt . \n(4.5.18)\n\u222b \nti \nIn Figure 4.14, the red areas are an overestimate and the blue areas are an underestimate. \nAs N \u2192\u221e , the sum of the red areas and the sum of the blue areas both approach zero. If \nthere are intervals in which a(t) has negative values, then the summation is a sum of", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_92", "text": "As N \u2192\u221e , the sum of the red areas and the sum of the blue areas both approach zero. If \nthere are intervals in which a(t) has negative values, then the summation is a sum of \nsigned areas, positive area above the t -axis and negative area below the t -axis. \nWe can determine both the change in velocity for the time interval [ti , t f ] and the area \nunder the graph of a(t) vs. t for [ti , t f ] by integration techniques instead of limiting \narguments. We can turn the linear map into a function of time, instead of just giving a \nnumber, by setting t f = t . In that case, Eq. (4.5.13) becomes \n20", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_93", "text": "t\u2032=t \n) = \na(t\u2032)dt\u2032 . \n(4.5.19)\nv(t) \u2212 v(ti \n\u222b \nt\u2032=ti \nBecause the upper limit of the integral, t f = t , is now treated as a variable, we shall use \nthe symbol t\u2032 as the integration variable instead of t . \n4.6.4 Displacement as the Definite Integral of Velocity \nWe can repeat the same argument for the definite integral of the x -component of the \nvelocity v(t) vs. time t . Because x(t) is an integral of v(t) the definite integral of v(t) \nfor the time interval [ti , t f ] is the displacement \nt\u2032=t f \n) \u2212 x(ti ) = \nv(t\u2032)dt\u2032 . \n(4.5.20)\nx(t f \n\u222b \nt\u2032=ti \nIf we set t f = t , then the definite integral gives us the position as a function of time \nt\u2032=t \nx(t) = x(ti ) + \nv(t\u2032)dt\u2032 . \n(4.5.21)\n\u222b \nt\u2032=ti \nSummarizing the results of these last two sections, for a given acceleration a(t) , we can \nuse integration techniques, to determine the change in velocity and change in position for \nan interval [ti , t] , and given initial conditions (xi ,vi ) , we can determine the position x(t) \nand the x -component of the velocity v(t) as functions of time. \nExample 4.5 Non-constant Acceleration \nLet\u2019s consider a case in which the acceleration, a(t) , is not constant in time, \nt2 . \n(4.5.22)\na(t) = b0 + b1 t + b2 \nThe graph of the x -component of the acceleration vs. time is shown in Figure 4.16 \n21", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_94", "text": "a(t) \nt \na(t) = b0 + b1 t + b2 t2 \nb0 \nslope = b1 + \nFigure 4.16 Non-constant acceleration vs. time graph. \nDenote the initial velocity at t = 0 by v0 . Then, the change in the x -component of the \nvelocity as a function of time can be found by integration: \nt\u2032=t\nt\u2032=t\nt2 \nt3\nb1 \nb2\n= \na(t\u2032) dt\u2032 = \nt\u20322) dt\u2032 \nt +\n+ \n. \n(4.5.23)\nv(t) \u2212 v0 \n\u222b\n\u222b (b0 + b1 t\u2032 + b2 \n= b0\n2 \n3\nt\u2032=0 \nt\u2032=0 \nThe x -component of the velocity as a function in time is then \nt2 \nt3\nb1 \nb2\nv(t) = v0 \nt +\n+ \n. \n(4.5.24)\n+ b0 \n2\n3 \nDenote the initial position at t = 0 by x0 . The displacement as a function of time is \nt\u2032=t \n= \nv(t\u2032) dt\u2032. \n(4.5.25)\nx(t) \u2212 x0 \n\u222b \nt\u2032=0 \nUse Equation (4.5.27) for the \nx-component of the velocity in Equation \n(4.5.24) and then integrate to determine the displacement as a function of time: \nt\u2032=t \nx(t) \u2212 x0 = \nv(t\u2032) dt\u2032\n\u222b \nt\u2032=0 \n(4.5.26)\nt\u2032=t \u239b \nt\u20322 \nt\u20323 \u239e \nt2 \nt3 \nt4\nb1 \nb2 \nb0 \nb1 \nb2\n= \nt\u2032 +\n+ \n\u239f dt\u2032 = v0 t + \n+\n+ \n.\n\u222b\u239c v0 + b0 \n2 \n3 \n2\n6\n12\nt\u2032=0 \u239d\n\u23a0 \nFinally the position as a function of time is then \nt2 \nt3 \nt4\nb0 \nb1 \nb2\n+ v\nt + \n+\n+ \n. \n(4.5.27)\nx(t) = x0 \nx,0 \n2 \n6\n12 \n22", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_95", "text": "Example 4.6 Bicycle and Car \nA car is driving through a green light at t = 0 located at x = 0 with an initial speed \nvc,0 = 12 m \u22c5s-1 . At time t1 = 1s , the car starts braking until it comes to rest at time t2 . The \nacceleration of the car as a function of time is given by the piecewise function \nac (t) = \n\u23a7\u23aa\u23a8\n\u23aa\u23a9  \n0; \n0 < t < t1 = 1s \n, \nb(t \u2212 t1); 1s < t < t2 \n-3 ) .\nwhere b = \u2212(6 m \u22c5s \n(a) Find the x -component of the velocity and the position of the car as a function of time. \n(b) A bicycle rider is riding at a constant speed of vb,0 and at t = 0 is 17 m behind the car. \nThe bicyclist reaches the car when the car just comes to rest. Find the speed of the bicycle. \nSolution: a) In order to apply Eq. (4.5.19), we shall treat each stage separately. For the \ntime interval 0 < t < t1 , the acceleration is zero so the x -component of the velocity is \nconstant. For the second time interval t1 < t < t2 , the definite integral becomes \nt\u2032=t \nv (t) \u2212 v (t1) = \nb(t\u2032 \u2212 t1) dt\u2032 \nc\nc \n\u222b \nt\u2032=t1 \nBecause vc (t1) = vc0 , the x -component of the velocity is then \n\u23a7vc0; \n0 < t \u2264 t1 \nv (t) = \u23aa \nt\u2032=t \n. \nc \n\u23a8 v + \nb(t\u2032 \u2212 t1)dt\u2032; t1 \u2264 t < t2\n\u23aa c0 \u222b \n\u23a9 \nt\u2032=t1 \nIntegrate and substitute the two endpoints of the definite integral, yields \n\u23a7v ;\n0 < t \u2264 t1\n\u23aa c0 \nv (t) = \n. \nc \n\u23a8 \n1 \nv + b(t \u2212 t1)2; t1 \u2264 t < t2\n\u23aa c0\n\u23a9 \n2 \nIn order to use Eq. (4.5.25), we need to separate the definite integral into two integrals \ncorresponding to the two stages of motion, using the correct expression for the velocity \nfor each integral. The position function is then \n23", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_96", "text": "t\u2032=t1\n\u23a7 \n\u23aax + \nv dt\u2032;\n0 < t \u2264 t1\nc0 \u222b \nc0\n\u23aa \nt\u2032=0\nx (t) = \u23a8 \n. \nc\nt\u2032=t \u239b \n1 \n)2 \u239e\n\u23aa\n\u23aax c (t1) + \u222b \u239d\u239c v c0 + 2 b(t\u2032 \u2212 t1\n\u23a0\u239fdt; t1 \u2264 t < t2 \n\u23a9 \nt\u2032=t1 \nUpon integration we have \n\u23a7xc (0) + vc0 t;\n0 < t \u2264 t1 \nt\u2032=t\n\u23aa \nxc (t) = \u23a8\n\u239b \n1 \n)3 \u239e \n. \nx (t1) + v (t\u2032 \u2212 t1) + b(t\u2032 \u2212 t1 \n; t1 \u2264 t < t2\n\u23aa c \n\u239d\u239c c0 \n\u23a0\u239f\n6 \nt\u2032=t1\n\u23aa\u23a9 \nWe chose our coordinate system such that the initial position of the car was at the origin, \nx = 0 , therefore x \n) = v \n. So after substituting in the endpoints of the integration\nc0 \nc (t1 \nc0 t1 \ninterval we have that \n\u23a7vc0t;\n0 < t \u2264 t1\n\u23aa \nxc (t) = \u23a8 \n. \nv \n+ v (t \u2212 t1) + 1 b(t \u2212 t1)3; t1 \u2264 t < t2\n\u23aa c0 t1 \nc0\n\u23a9 \n6 \n(b) We are looking for the instant t2 that the car has come to rest. So we use our \nexpression for the x -component of the velocity the interval t1 \u2264 t < t2 , where we set t = t2 \nand v \n) = 0: \nc (t2 \n1 \n)2\n0 = v (t2) = v + \n\u2212 t1\n. \nc\nc0 \nb(t2\n2 \nSolving for t2 yields \n+\nt2 = t1 \nwhere we have taken the positive square root. Substitute the given values then yields \n\u2212 2v c0 \nb , \nt2 = 1s + \u2212 2(12 m \u22c5s\u22121) \n(\u22126 m \u22c5s\u22123 ) = 3 s . \nThe position of the car at t2 is then given by \n24", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_97", "text": "1 \n)3\nx (t2) = v \n+ v \n) + \nc\nc0 t1 \nc0 (t2 \u2212 t1\nb(t2 \u2212 t1\n6 \n1\n/ b)3/2 \nx (t2) = v \n+ v \n\u22122v / b + b(\u22122v \nc\nc0 t1 \nc0 \nc0 \nc0\n6 \n3/2 )\n2 2(vc0\nxc (t2) = vc0 t1 + 3(\u2212b)1/2 \nwhere we used the condition that t2 \n= \u22122v / b . Substitute the given values then \n\u2212 t1 \nc0 \nyields \n-1 )3/2 \n4 2(vc0  \n4 2((12 m \u22c5s \nxc (t2) = vc0 t1 + 2 3(\u2212b)1/2 \n)3/2 \n= (12 m \u22c5s-1)(1s) + \n\u22123))1/2 \n= 28 m . \n3((6 m \u22c5s \nb)  Because the bicycle is traveling at a constant speed with an initial position \n= \u221217 m , the position of the bicycle is given by xb (t) = \u221217 m + vbt . The bicycle and \nxb0 \ncar intersect at time t2 = 3 s , where xb (t2) = xc (t2) . Therefore \u221217 m + vb (3 s) = 28 m . So \nthe speed of the bicycle is vb = 15 m \u22c5s\u22121 . \n25", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_98", "text": "Chapter 5 Two Dimensional Kinematics \n5.1 Introduction to the Vector Description of Motion in Two Dimensions ............. 1  \n5.2 Projectile Motion..................................................................................................... 2  \nExample 5.1 Time of Flight and Maximum Height of a Projectile ...................... 5  \n5.2.1 Orbit equation .................................................................................................. 8  \nExample 5.2 Hitting the Bucket............................................................................. 10", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_99", "text": "Chapter 5 Two Dimensional Kinematics \nWhere was the chap I saw in the picture somewhere? Ah yes, in the dead \nsea floating on his back, reading a book with a parasol open. Couldn\u2019t \nsink if you tried: so thick with salt. Because the weight of the water, no, \nthe weight of the body in the water is equal to the weight of the what? Or \nis it the volume equal to the weight? It\u2019s a law something like that. Vance \nin High school cracking his fingerjoints, teaching. The college curriculum. \nCracking curriculum. What is weight really when you say weight? \nThirtytwo feet per second per second. Law of falling bodies: per second \nper second. They all fall to the ground. The earth. It\u2019s the force of gravity \nof the earth is the weight. 1 \nJames Joyce \n5.1 Introduction to the Vector Description of Motion in Two Dimensions \nWe have introduced the concepts of position, velocity and acceleration to describe \nmotion in one dimension; however we live in a multidimensional universe. In order to \nexplore and describe motion in more than one dimension, we shall study the motion of a \nprojectile in two-dimension moving under the action of uniform gravitation. \nWe extend our definitions of position, velocity, and acceleration for an object that \nmoves in two dimensions (in a plane) by treating each direction independently, which we \ncan do with vector quantities by resolving each of these quantities into components. For \nexample, our definition of velocity as the derivative of position holds for each component \n!\nseparately. In Cartesian coordinates, the position vector r(t) with respect to some choice \nof origin for the object at time t is given by \n! r(t) = x(t) \u02c6i + y(t) \u02c6j. \n(5.1.1) \n\uf072\nThe velocity vector v( )t at time t is the derivative of the position vector, \n\uf072 \ndx t( ) \ndy t( ) \n\u02c6 \n\u02c6\n\u02c6\n\u02c6\nv( )t = \ni + \nj \u2261 v ( )t i + v ( )t j , \n(5.1.2)\ndt\ndt\nx \ny \nwhere vx ( )t \u2261 dx t( )/ dt and vy ( )t \u2261 dy t( )/ dt denote the x - and y -components of the \nvelocity respectively. \n\uf072\nThe acceleration vector a( )t is defined in a similar fashion as the derivative of the \nvelocity vector, \n1 James Joyce, Ulysses, The Corrected Text edited by Hans Walter Gabler with Wolfhard", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_100", "text": "velocity respectively. \n\uf072\nThe acceleration vector a( )t is defined in a similar fashion as the derivative of the \nvelocity vector, \n1 James Joyce, Ulysses, The Corrected Text edited by Hans Walter Gabler with Wolfhard\nSteppe and Claus Melchior, Random House, New York. \n1", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_101", "text": "\uf072 \ndvx ( )t \ndvy ( )t\n\u02c6 \n\u02c6\n\u02c6\n\u02c6\na( )t = \ni + \nj \u2261 a ( )t i + a ( )t j, \n(5.1.3)\ndt\ndt\nx \ny \nwhere a ( ) \u2261 dv ( )/ \nt dt and a ( ) \u2261 dv ( )/ \nt \nt \nt\ndt denote the x - and y -components of the\nx\nx \ny\ny \nacceleration. \n5.2 Projectile Motion \n!\nConsider the motion of a body that is released at time t = 0 with an initial velocity v0 . \nTwo paths are shown in Figure 5.1. \nparabolic orbit \nactual orbit \nv0 \nFigure 5.1 Actual orbit accounting for air resistance and parabolic orbit of a projectile \nThe dotted path represents a parabolic trajectory and the solid path represents the actual \ntrajectory. The difference between the two paths is due to air resistance acting on the \n! \nFair \nobject, \n= \u2212bv2v\u02c6 , where v\u02c6 is a unit vector in the direction of the velocity. (For the \n!\n2 \u22c5 m-2 \norbits shown in Figure 5.1, b = 0.01 N \u22c5s\n, \n= 30.0 m \u22c5s , the initial launch angle \nv0 \nwith respect to the horizontal \u03b80 = 21! , and the actual horizontal distance traveled is \n71.7% of the projectile orbit.). There are other factors that can influence the path of \nmotion; a rotating body or a special shape can alter the flow of air around the body, \nwhich may induce a curved motion or lift like the flight of a baseball or golf ball. We \nshall begin our analysis by neglecting all interactions except the gravitational interaction. \ny \n. \nx(t) \ny(t) \nO \nx \ny0 \nr(t)\n0 \nv0 \n\u02c6i \n\u02c6j \nFigure 5.2 A coordinate sketch for parabolic motion. \nChoose coordinates with the positive y-axis in the upward vertical direction and the \npositive x-axis in the horizontal direction in the direction that the object is moving \nhorizontally. Choose the origin at the ground immediately below the point the object is \n2", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_102", "text": "!\nreleased. Figure 5.2 shows our coordinate system with the position of the object r(t) at \n!\ntime t , the initial velocity v0 , and the initial angle \u03b80 with respect to the horizontal, and \nthe coordinate functions x(t) and y(t) . \nInitial Conditions: \n. \nO \nx \ny \n0 \nv0 \n\u02c6i \n\u02c6j \nv x,0 \nv y,0 \nFigure 5.3 A vector decomposition of the initial velocity \nDecompose the initial velocity vector into its components: \n! v0 = vx,0 \u02c6i + vy,0 \u02c6j . \n(5.1.4) \nThe vector decomposition for the initial velocity is shown in Figure 5.3. Often the \ndescription of the flight of a projectile includes the statement, \u201ca body is projected with \nan initial speed v0 at an angle \u03b80 with respect to the horizontal.\u201d The components of the \ninitial velocity can be expressed in terms of the initial speed and angle according to \nvx,0 = v0 cos\u03b80 , \n(5.1.5) \nvy,0 = v0 sin\u03b80 . \n(5.1.6) \nBecause the initial speed is the magnitude of the initial velocity, we have that \n2 \n2 )1/2 \nv = (v \n+ v \n. \n(5.1.7)\n0 \nx,0 \ny,0 \nThe angle \u03b80 is related to the components of the initial velocity by \n\u03b80 = tan\u22121(vy,0 / vx,0 ). \n(5.1.8) \nEquation (5.1.8) will give two values for the angle \u03b80 , so care must be taken to choose \nthe correct physical value. The initial position vector generally is given by \n! r0 = x0 \u02c6i + y0 \u02c6j . \n(5.1.9) \nNote that the trajectory in Figure 5.3 has x0 = 0 , but this will not always be the case. \n3", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_103", "text": "Force Diagram: \nWe begin by neglecting all forces other than the gravitational interaction between the \nobject and the earth. This force acts downward with magnitude mg , where m is the \nmass of the object and g = 9.8 m \u22c5s\u22122 . Figure 5.4 shows the force diagram on the object. \ny \n. \nO \nx \n\u02c6i \n\u02c6j \nFg \nFigure 5.4 Free-body force diagram on the object with the action of gravity \nThe vector decomposition of the force is \n\uf072 \nFg = \u2212mg \u02c6j. \n(5.1.10) \nEquations of Motions: \nThe force diagram reminds us that the force is acting in the y -direction. Newton\u2019s \n! \nFtotal \nSecond Law states that the sum of the force, \n, acting on the object is equal to the \n\uf072\nproduct of the mass m and the acceleration vector a , \n\uf072 \nFtotal \n\uf072 \n= ma . \n(5.1.11) \n\uf072\n\uf072 \nFtotal = Fg\nBecause we are modeling the motion with only one force, we have that \n. This is \na vector equation; the components are equated separately: \n\u2212mg = ma y , \n(5.1.12) \n0 = ma x . \n(5.1.13) \nTherefore the y -component of the acceleration is \nay = \u2212g. \n(5.1.14) \nWe see that the acceleration is a constant and is independent of the mass of the object. \nNotice that ay < 0 . This is because we chose our positive y -direction to point upwards. \nThe sign of the y -component of acceleration is determined by how we choose our \n4", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_104", "text": "coordinate system. Because there are no horizontal forces acting on the object, we \nconclude that the acceleration in the horizontal direction is also zero \nax = 0 . \n(5.1.15) \nTherefore the x -component of the velocity remains unchanged throughout the flight of \nthe object. \nThe acceleration in the vertical direction is constant for all bodies near the surface of the \nEarth, independent of the mass of the object, thus confirming Galileo\u2019s Law of Free \nFalling Bodies. Notice that the equation of motion (Equation (5.1.14)) generalizes the \nexperimental observation that objects fall with constant acceleration. Our statement about \nthe acceleration of objects near the surface of Earth depends on our model force law Eq. \n(5.1.10), and if subsequent observations show the acceleration is not constant then we \neither must include additional forces (for example, air resistance), or modify the force \nlaw (for objects that are no longer near the surface of Earth, or consider that Earth is a \nnon-symmetric non-uniform body), or take into account the rotational motion of the Earth. \nWe can now integrate the equation of motions (Eqs. (5.1.14) and (5.1.15)) \nseparately for the x - and y - directions to find expressions for the x - and y -components \nof velocity and position: \nt \u2032=t \nv (t) \u2212 v \n= \na (t\u2032) dt\u2032 = 0 \u21d2 v (t) = v \nx\nx,0 \n\u222b \nx \nx\nx,0 \nt \u2032=0 \nt \u2032=t\nt\u2032=t \nx(t) \u2212 x = \nv (t\u2032) dt\u2032 = \nv \ndt\u2032 = v\nt \u21d2 x(t) = x + v\nt\n0 \n\u222b \nx \n\u222b \nx,0 \nx,0 \n0 \nx,0 \nt \u2032=0 \nt\u2032=0 \nt \u2032=t\nt \u2032=t \nv (t) \u2212 v \n= \na (t\u2032) dt\u2032 = \u2212 \ng dt\u2032 = \u2212gt \u21d2 v (t) = v \n\u2212 gt\ny\ny,0 \n\u222b \ny \n\u222b \ny\ny,0 \nt \u2032=0 \nt \u2032=0 \nt\u2032=t\nt\u2032=t \ny(t) \u2212 y0 = \u222b vy (t\u2032)dt\u2032 = \u222b (v y,0 \u2212 gt) dt\u2032 = vy,0 t \u2212 (1/ 2)gt2 \u21d2 y(t) = y0 + vy,0t \u2212 (1/ 2)gt2. \nt\u2032=0", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_105", "text": "t\u2032=0 \nt\u2032=0 \nThe complete set of vector equations for position and velocity for each independent \ndirection of motion are given by \n\uf072 \n+ v \n+ v\nt + (1/ 2)a t2) \u02c6j , \n(5.1.16)\nr(t) = x(t) \u02c6i + y(t) \u02c6j = (x0 \nx,0 t) \u02c6i + ( y0 \ny,0 \ny \n\uf072 v(t) = v (t) \u02c6i + v (t) \u02c6j = v \n\u02c6i + (v \n+ a t) \u02c6j, \n(5.1.17)\nx\ny \nx,0 \ny,0 \ny \n\uf072 \n\u02c6 \n\u02c6\n\u02c6\na( )t = a ( )t i + a ( )t j = a j . \n(5.1.18)\nx\ny \ny \nExample 5.1 Time of Flight and Maximum Height of a Projectile \n5", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_106", "text": "A person throws a stone at an initial angle \u03b80 = 45! from the horizontal with an initial \nspeed of v0 = 20 m \u22c5 s-1 . The point of release of the stone is at a height d = 2 m above the \nground. You may neglect air resistance. a) How long does it take the stone to reach the \nhighest point of its trajectory? b) What was the maximum vertical displacement of the \nstone? Ignore air resistance. \nSolution: Choose the origin on the ground directly underneath the point where the stone \nis released. We choose the positive y-axis in the upward vertical direction and the \npositive x-axis in the horizontal direction in the direction that the object is moving \nhorizontally. Set t = 0 the instant the stone is released. At t = 0 the initial conditions are \nthen x0 = 0 and y0 = d . The initial x - and y -components of the velocity are given by \nEqs. (5.1.5) and (5.1.6). \nAt time t the stone has coordinates (x(t), y(t)) . These coordinate functions are shown in \nFigure 5.5. \nFigure 5.5: Coordinate functions for stone \nFigure 5.6 Plot of the y-component of the position as a function of time \nThe slope of this graph at any time t yields the instantaneous y-component of the \nvelocity vy (t) at that time t . Figure 5.5 is a plot of y(t) vs. x(t) and Figure 5.6 is a plot \nof y(t) vs. t . There are several important things to notice about Figures 5.5 and 5.6. The \nfirst point is that the abscissa axes are different in both figures. The second thing to notice \nis that at t = 0 , the slope of the graph in Figure 5.5 is equal to \n6", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_107", "text": "v\ndy \n\u239b dy / dt \u239e \n= \n= \ny,0 = tan\u03b80 , \n(5.1.19)\ndx t=0 \n\u239d\u239c dx / dt \u23a0\u239f \nv \nt=0 \nx,0 \nwhile at t = 0 the slope of the graph in Figure 5.6 is equal to \ndy \n= vy,0 . \n(5.1.20)\ndt t=0 \nThe slope of this graph in Figure 5.6 at any time t yields the instantaneous y-component \nof the velocity vy (t) at that time t . Let t = t1 correspond to the instant the stone is at its \nmaximal vertical position, the highest point in the flight. The final thing to notice about \nFigure 5.6 is that at t = t1 the slope is zero or vy (t = t1) = 0 . Therefore \nv \nsin\u03b80 \n= 0 . \n(5.1.21)\ny (t1) = v0 \n\u2212 gt1 \nSolving Eq. (5.1.21) for t1 yields, \nv0 sin\u03b80 \n(20 m \u22c5s-1)sin(45! )\n=\n= \n= 1.44 s . \n(5.1.22)\nt1 \n-2 \ng \n9.8 m \u22c5s\nThe graph in Figure 5.7 shows a plot of vy (t) as a function of time. Notice that at t = 0 \nthe intercept is positive indicting that vy,0 is positive which means that the stone was \nthrown upwards. The y -component of the velocity changes sign at t = t1 indicating that \nthe stone is reversing its direction and starting to move downwards. \n15 \n10 \n5 \n0 \n-5 \n-10 \n-15 \n1 \n2 \n3 \n4 \nt1 \nt \nv y (t) \n[m s 1] \n[s] \nFigure 5.7 y -component of the velocity as a function of time \n7", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_108", "text": "We now substitute the expression for t = ttop (Eq. (5.1.22)) into the y -component of the \nposition in Eq. (5.1.16) to find the maximal height of the stone above the ground \n\u239e \n2 \nv0 sin\u03b80\n1 \u239b v0 sin\u03b80\ny(t = t \nsin\u03b80 \n\u2212 g\ntop ) = d + v0 \ng \n2 \u239d\u239c \ng \n\u23a0\u239f \n, \n(5.1.23) \nv0\n2 sin2 \u03b80 \n(20 m \u22c5s-1)2 sin2(45\uf06f )\n= d +\n= 2 m +\n= 12.2 m \n-2 )\n2g \n2(9.8 m \u22c5s \n5.2.1 Orbit equation \nSo far our description of the motion has emphasized the independence of the spatial \ndimensions, treating all of the kinematic quantities as functions of time. We shall now \neliminate time from our equation and find the orbit equation of the body undergoing \nprojectile motion. We begin with the x -component of the position in Eq. (5.1.16), \nx t( ) = x0 + vx,0 t \n(5.1.24) \nand solve Equation (5.1.24) for time t as a function of x t( ) , \nx t( ) \u2212 x\nt = \n0 . \n(5.1.25)\nvx,0 \nThe y -component of the position in Eq. (5.1.16) is given by \n1\n2\ny t( ) = y0 + vy,0 t \u2212 g t . \n(5.1.26)\n2 \nWe then substitute Eq. (5.1.25) into Eq. (5.1.26) yielding \n\u239b x t( ) \u2212 x0 \u239e 1 \u239b x t( ) \u2212 x0 \u239e\n2 \ny t( ) = y0 + vy,0 \u239c \n\u239f\u2212 g \u239c\n\u239f . \n(5.1.27)\n\u239c \n\u239f\n\u239c \n\u239f\nv \n2 \nv\n\u239d \nx,0 \n\u23a0\n\u239d \nx,0 \n\u23a0 \nA little algebraic simplification yields the equation for a parabola: \n1 g \n2 \n\u239b g x0 \nvy,0 \u239e \nvy,0 \n1 g \n2\ny t( ) = \u2212 \nx t( ) + \u239c \n+ \n\u239f x t( ) \u2212 \nx \u2212 \nx + y . \n(5.1.28)\n2 \n2 \n0\n2\n0\n0\n\u239c\n\u239f\n2 v \nv\nv\nv \n2 v\nx,0 \n\u239d \nx,0 \nx,0 \u23a0 \nx,0", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_109", "text": "1 g \n2\ny t( ) = \u2212 \nx t( ) + \u239c \n+ \n\u239f x t( ) \u2212 \nx \u2212 \nx + y . \n(5.1.28)\n2 \n2 \n0\n2\n0\n0\n\u239c\n\u239f\n2 v \nv\nv\nv \n2 v\nx,0 \n\u239d \nx,0 \nx,0 \u23a0 \nx,0 \nx,0 \nThe graph of y t( ) as a function of x t( ) is shown in Figure 5.8. \n8", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_110", "text": "Figure 5.8 The parabolic orbit \nThe velocity vector is given by \n\uf072 \ndx t( ) \ndy t( ) \nv( )t = \n\u02c6i + \n\u02c6j \u2261 vx ( )t \u02c6i + vy ( )t \u02c6j . \n(5.1.29)\ndt \ndt \nThe direction of the velocity vector at a point (x(t), y(t)) can be determined from the \ncomponents. Let \u03b8 be the angle that the velocity vector forms with respect to the positive \nx -axis. Then \n\u239b vy (t)\u239e\n\u239b dy / dt \u239e\n\u239b dy \u239e\n\u03b8 = tan\u22121 \u239c\n\u239f = tan\u22121 \n\u239d\u239c \n\u23a0\u239f = tan\u22121 \n\u239d\u239c \n\u23a0\u239f . \n(5.1.30)\n\u239d v x (t)\u23a0 \ndx / dt \ndx \nDifferentiating Eq. (5.1.28) with respect to x yields \n\u239b \n\u239e \ndy \ng x0 \nv \n= \u2212 g \n2 \ny,0 \nx,0 \n(5.1.31) \nx + \n+ \n\u239c\u239d  \n\u239f\u23a0  \n. \n2\ndx \nv\nv \nv \nx,0 \nx,0 \nThe direction of the velocity vector at a point (x(t), y(t)) is therefore \n\u239b \n\u239e \n\u239b \n\u239e \ng x0 \nv \n\u2212 g \n2 \ny,0 \nx,0 \n\u03b8 = tan\u22121 \n(5.1.32) \nx + \n+ \n\u239c\n\u239d  \n\u239f\n\u23a0  \n\u239c\u239d  \n\u239f\u23a0  \n. \n2 \nv\nv \nv \nx,0 \nx,0 \nAlthough we can determine the angle of the velocity, we cannot determine how fast the \nbody moves along the parabolic orbit from our graph of y(x) ; the magnitude of the \nvelocity cannot be determined from information about the tangent line. \n9", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_111", "text": "If we choose our origin at the initial position of the body at t = 0 , then x0 = 0 and y0 = 0 . \nOur orbit equation, Equation (5.1.28) can now be simplified to \n1 g x(t)2 + \nvy,0 \ny(t) = \u2212 \n2 \nx(t) .  \n(5.1.33)\n2 v\nv \nx,0 \nx,0 \nExample 5.2 Hitting the Bucket \nA person is holding a pail while standing on a ladder. The person releases the pail from \nrest at a height h1 above the ground. A second person, standing a horizontal distance s \nfrom the pail, aims and throws a ball the instant the pail is released in order to hit the pail. \nThe person releases the ball at a height h2 above the ground, with an initial speed v0 , and \nat an angle \u03b80 with respect to the horizontal. Assume that v0 is large enough so that the \nstone will at least travel a horizontal distance s before it hits the ground. You may ignore \nair resistance. \nh2 \nh1 \ns \nFigure 5.9: Example 5.2 \na)  Find an expression for the angle \u03b80 that the person aims the ball in order to hit the \npail. Does the answer depend on the initial velocity? \nb)  Find an expression for the time of collision as a function of the initial speed of the \nball v0 , and the quantities h1, h2 , and s . \nc)  Find an expression for the height above the ground where the collision occurred \nas a function of the initial speed of the ball v0 , and the quantities h1, h2 , and s . \nSolution: \nThere are two objects involved in this problem. Each object is undergoing free fall, so \nthere is only one stage of motion for each object. The pail is undergoing one-dimensional \n10", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_112", "text": "motion. The ball is undergoing two-dimensional motion. The parameters h1, h2 , v0 , and \ns are unspecified, so our answers will be functions of those quantities. Figure 5.9 shows \na sketch of the motion of all the bodies in this problem. \nChoose an origin on the ground directly underneath the point where the ball is released, \nupwards for the positive y -direction and towards the pail for the positive x -direction. \nChoose position coordinates for the pail as follows. The horizontal coordinate is constant \nand given by x1 = s . The vertical coordinate represents the height above the ground and \nis denoted by y1(t) . The ball has coordinates (x2(t), y2(t)) . We show these coordinates in \nthe Figure 5.10. \nh2 \nh1 \nx2 (t) \ny2 (t) \ny1(t) \nx1 = s \nv0 \n0 \nO \nFigure 5.10: Coordinate System \nThe pail undergoes constant acceleration a1,y = \u2212g in the vertical direction and the ball \nundergoes uniform motion in the horizontal direction and constant acceleration in the \nvertical direction, with a2,x = 0 and a2,y = \u2212g . \nThe initial conditions for the pail are (v1,0 )y = 0 , x1,0 = s , y1,0 = h1 . The equations for \nposition and velocity of the pail simplify to \ny1(t) = h1 \u2212 1 gt2 \n(5.1.34)\n2 \nvy,1(t) = \u2212gt . \n(5.1.35) \nThe initial position is given by x2,0 = 0, y2,0 = h2 . The components of the initial velocity \nare given by (v2,0 )y = v0 sin(\u03b80) and(v2,0 )x = v0 cos(\u03b80) , where v0 is the magnitude of the \ninitial velocity and \u03b80 is the initial angle with respect to the horizontal. The equations for \nthe position and velocity of the ball simplify to \n11", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_113", "text": "x2(t) = v0 cos(\u03b80)t \n(5.1.36) \ncos(\u03b80) \n(5.1.37)\nv2,x (t) = v0 \ny2(t) = h2 + v0 sin(\u03b80)t \u2212 1 gt2 \n(5.1.38)\n2 \nsin(\u03b80) \u2212 gt . \n(5.1.39)\nv2,y (t) = v0 \nNote that the quantities h1, h2 , v0 , and s should be treated as known quantities although \nno numerical values were given. There are six independent equations with 8 as yet \nunspecified quantities y1(t), t , y2(t), x2(t) , v1,y (t) , v2,y (t), v2,x (t) , and \u03b80 . \nSo we need two more conditions, in order to find expressions for the initial angle, \u03b80 , the \ntime of collision, ta , and the spatial location of the collision point specified by y1(ta ) or \n(t ) . At the collision time t = t , the collision occurs when the two balls are located at \ny2 \na \na \nthe same position. Therefore \ny1(t a ) = \n(t a ) \n(5.1.40)\ny2 \n= s . \n(5.1.41)\nx2 (t a ) = x1 \nWe shall now apply these conditions that must be satisfied in order for the ball to hit the \npail. \n1\n2 \n1\n2\n\u2212 gt \nsin \u03b80 \n\u2212 gt \n(5.1.42)\nh1 \na = h2 + v0 \n( )t a \na\n2\n2 \ns = v0 cos(\u03b80)ta . \n(5.1.43) \nEq. (5.1.42) simplifies to \nsin(\u03b80)t \n(5.1.44)\nv0 \na = h1 \u2212 h2 \nDividing Eq. (5.1.44) by Eq. (5.1.43) yields \nsin(\u03b80)t\nv0 \na \nh1 \u2212 h2\n= tan(\u03b80) = \n. \n(5.1.45)\ncos(\u03b80)t\nv0 \na\ns2 \nSo the initial angle \u03b80 is independent of v0 , and is given by \n\u03b80 = tan\u22121((h1 \u2212 h2) / s) .", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_114", "text": "= tan(\u03b80) = \n. \n(5.1.45)\ncos(\u03b80)t\nv0 \na\ns2 \nSo the initial angle \u03b80 is independent of v0 , and is given by \n\u03b80 = tan\u22121((h1 \u2212 h2) / s) . \n(5.1.46) \nFrom the Figure 5.11 we can see that tan(\u03b80) = (h1 \u2212 h2) / s implies that the second person \naims the ball at the initial position of the pail. \n12", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_115", "text": "h2 \nh1 \ns \n0 \nO \nh1 \nh2 \nFigure 5.11: Geometry of collision \nIn order to find the time that the ball collides with the pail, we begin by squaring both \nEqs. (5.1.44) and (5.1.43), then utilize the trigonometric identity sin2(\u03b80) + cos2(\u03b80) = 1. \nOur squared equations become \n2 \n)2\n2 sin2(\u03b80)t \n(5.1.47)\nv0 \na = (h1 \u2212 h2 \n2 \n2\n2\ncos2(\u03b80)t = s . \n(5.1.48)\nv0 \na \nAdding these equations together and using the identity sin2(\u03b80) + cos2(\u03b80) = 1 and taking \nsquare roots yields \n)2)1/2 \nt = (s2 + (h1 \n. \n(5.1.49)\nv0 a \n\u2212 h2 \nWe can solve Eq. (5.1.49) for the time of collision \n1 \n2 + (h1 \n)2)1 2 . \n(5.1.50)\nt = \n(s\na \n\u2212 h2\nv0 \nWe can now use the y -coordinate function of either the ball or the pail at t = ta to find \nthe height that the ball collides with the pail. Because the pail had no initial y \u00ad\ncomponent of the velocity, it\u2019s easier to use the condition for the pail, \ng(s2 + (h1 \u2212 h2)2)\ny1(t ) = h1 \u2212 \n. \n(5.1.51)\na \n2v0\n2 \nComments: \n(1) Eqs. (5.1.49) and (5.1.50) can be arrived at in a very direct way. Suppose we analyze \n! \nthe motion in a reference frame that is accelerating downward with A = \u2212g \u02c6j . In that \n13\n\nreference frame both the pail and the stone are not accelerating; the pail is at rest and the \nstone is travelling with speed v0 , at an angle \u03b80 . Therefore in order to hit the stationary \npail, the stone must be thrown at the angle given by Eq. (5.1.46) and the time that it takes \nto hit the stone is just given by distance traveled divided by speed, Eq. (5.1.50). \n14", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_116", "text": "Chapter 6 Circular Motion \n6.1 Introduction............................................................................................................. 1  \n6.2 Circular Motion: Velocity and Angular Velocity ................................................ 2  \n6.2.1 Geometric Derivation of the Velocity for Circular Motion.......................... 4  \n6.3 Circular Motion: Tangential and Radial Acceleration ....................................... 5  \nExample 6.1 Circular Motion Kinematics .............................................................. 7  \n6.4 Period and Frequency for Uniform Circular Motion.......................................... 7  \n6.4.1 Geometric Interpretation for Radial Acceleration for Uniform Circular", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_117", "text": "6.4.1 Geometric Interpretation for Radial Acceleration for Uniform Circular  \nMotion ........................................................................................................................ 9  \n6.5 Angular Velocity and Angular Acceleration ...................................................... 10  \n6.5.1. Angular Velocity ........................................................................................... 10  \nExample 6.2 Angular Velocity ............................................................................... 12  \nExample 6.3 Integration and Circular Motion Kinematics ................................ 14", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_118", "text": "Example 6.3 Integration and Circular Motion Kinematics ................................ 14  \n6.5 Non-circular Central Motion............................................................................... 15  \nExample 6.4 Spiral Motion..................................................................................... 16", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_119", "text": "Chapter 6 Central Motion \nAnd the seasons they go round and round \nAnd the painted ponies go up and down \nWe're captive on the carousel of time \nWe can't return we can only look \nBehind from where we came \nAnd go round and round and round \nIn the circle game 1 \nJoni Mitchell \n6.1 Introduction \nWe shall now investigate a special class of motions, motion in a plane about a central \npoint, a motion we shall refer to as central motion, the most outstanding case of which is \ncircular motion. Special cases often dominate our study of physics, and circular motion \nabout a central point is certainly no exception. There are many instances of central \nmotion about a point; a bicycle rider on a circular track, a ball spun around by a string, \nand the rotation of a spinning wheel are just a few examples. Various planetary models \ndescribed the motion of planets in circles before any understanding of gravitation. The \nmotion of the moon around the earth is nearly circular. The motions of the planets around \nthe sun are nearly circular. Our sun moves in nearly a circular orbit about the center of \nour galaxy, 50,000 light years from a massive black hole at the center of the galaxy. \nWhen Newton solved the two-body under a gravitational central force, he discovered that \nthe orbits can be circular, elliptical, parabolic or hyperbolic. All of these orbits still \ndisplay central force motion about the center of mass of the two-body system. Another \nexample of central force motion is the scattering of particles by a Coulombic central force, \nfor example Rutherford scattering of an alpha particle (two protons and two neutrons \nbound together into a particle identical to a helium nucleus) against an atomic nucleus \nsuch as a gold nucleus. \nWe shall begin by describing the kinematics of circular motion, the position, velocity, \nand acceleration, as a special case of two-dimensional motion. We will see that unlike \nlinear motion, where velocity and acceleration are directed along the line of motion, in \ncircular motion the direction of velocity is always tangent to the circle. This means that as \nthe object moves in a circle, the direction of the velocity is always changing. When we \nexamine this motion, we shall see that the direction of the change of the velocity is \ntowards the center of the circle. This means that there is a non-zero component of the \nacceleration directed radially inward, which is called the centripetal acceleration. If our \nobject is increasing its speed or slowing down, there is also a non-zero tangential", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_120", "text": "towards the center of the circle. This means that there is a non-zero component of the \nacceleration directed radially inward, which is called the centripetal acceleration. If our \nobject is increasing its speed or slowing down, there is also a non-zero tangential \nacceleration in the direction of motion. But when the object is moving at a constant speed \nin a circle then only the centripetal acceleration is non-zero. \n1 Joni Mitchell, The Circle Game, Siquomb Publishing Company. \n1", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_121", "text": "In 1666, twenty years before Newton published his Principia, he realized that the \nmoon is always \u201cfalling\u201d towards the center of the earth; otherwise, by the First Law, it \nwould continue in some linear trajectory rather than follow a circular orbit. Therefore \nthere must be a centripetal force, a radial force pointing inward, producing this \ncentripetal acceleration. \nIn all of these instances, when an object is constrained to move in a circle, there \n! \nmust exist a force F acting on the object directed towards the center. Because Newton\u2019s \nSecond Law is a vector equality, the radial component of the Second Law is \nFr = ma r . \n(6.1.1) \n6.2 Circular Motion: Velocity and Angular Velocity \nWe begin our description of circular motion by choosing polar coordinates. In Figure 6.1\n\uf072\nwe sketch the position vector r ( )t of the object moving in a circular orbit of radius r . \n\u02c6 \n\u02c6j \n\u02c6i \n+ x \n+ y \n\u02c6r(t)\n(t) \n(t) \nr P . \nFigure 6.1 A circular orbit with unit vectors. \nAt time t , the particle is located at the point P with coordinates (r, \u03b8(t)) and position \nvector given by \n!r(t) = r r\u02c6(t) . \n(6.2.1) \n\u02c6\nAt the point P , consider two sets of unit vectors ( r\u02c6(t) , \u03b8(t)) and ( \u02c6i , \u02c6j), as shown in \nFigure 6.1. The vector decomposition expression for r\u02c6(t) and \u03b8\u02c6(t) in terms of \u02c6i and \u02c6j is \ngiven by \nr\u02c6(t) = cos\u03b8(t) \u02c6i + sin\u03b8(t) \u02c6j , \n(6.2.2) \n\u03b8\u02c6(t) = \u2212sin\u03b8(t) \u02c6i + cos\u03b8(t) \u02c6j . \n(6.2.3) \nBefore we calculate the velocity, we shall calculate the time derivatives of Eqs. (6.2.2) \nand (6.2.3). Let\u2019s first begin with d r\u02c6(t) / dt : \n2", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_122", "text": "d r\u02c6(t) \nd \n= \n(cos\u03b8(t) \u02c6i + sin\u03b8(t) \u02c6j) = (\u2212sin\u03b8(t) d\u03b8(t) \u02c6i + cos\u03b8(t) d\u03b8(t) \u02c6j)\ndt\ndt \ndt \ndt \n; \n(6.2.4)\nd\u03b8(t) \nd\u03b8(t) \u02c6\n= \n(\u2212sin\u03b8(t) \u02c6i + cos\u03b8(t) \u02c6j) = \n\u03b8(t)\ndt \ndt \nwhere we used the chain rule to calculate that \nd cos\u03b8(t) = \u2212sin\u03b8(t) d\u03b8(t) , \n(6.2.5)\ndt \ndt \nd sin\u03b8(t) = cos\u03b8(t) d\u03b8(t) . \n(6.2.6)\ndt \ndt \nThe calculation for d \u03b8\u02c6(t) / dt is similar: \nmagnitude of the velocity by v \u2261 \n, The angular speed is the magnitude of the rate of \ndt \ndt \nd \u03b8\u02c6 (t) \nd \n= \n(\u2212sin\u03b8(t) \u02c6i + cos\u03b8(t \u02c6j) = (\u2212 cos\u03b8(t) d\u03b8(t) \u02c6i \u2212 sin(t) d\u03b8(t) \u02c6j)\ndt \ndt \ndt \ndt \n.\nd\u03b8(t) \nd\u03b8(t)\n= \n(\u2212 cos\u03b8(t) \u02c6i \u2212 sin\u03b8(t) \u02c6j) = \u2212 \nr\u02c6(t) \n(6.2.7) \nThe velocity vector is then \n!v(t) = dr!(t) \ndt = r d r\u02c6 \ndt = r d\u03b8 \ndt \u03b8(t) = v\u03b8 \u03b8(t) , \n\u02c6 \n\u02c6 \n(6.2.8) \nwhere the \u03b8\u02c6 -component of the velocity is given by \nv\u03b8 = r d\u03b8 \ndt , \n(6.2.9) \na quantity we shall refer to as the tangential component of the velocity. Denote the \n! v \nchange of angle with respect to time, which we denote by the Greek letter \u03c9 , \nd\u03b8 . \n(6.2.10)\n\u03c9 \u2261 dt \n3", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_123", "text": "6.2.1 Geometric Derivation of the Velocity for Circular Motion \nConsider a particle undergoing circular motion. At time t , the position of the particle is \n!\n!\nr(t) . During the time interval \u0394t , the particle moves to the position r(t + \u0394t) with a \n!\ndisplacement \u0394r . \nr \nr(t)\nr(t + t) \nr \nFigure 6.2 Displacement vector for circular motion \nThe magnitude of the displacement, \u0394r \uf072 , is represented by the length of the horizontal \n\uf072 \nvector \u0394r joining the heads of the displacement vectors in Figure 6.2 and is given by \n\u0394! r = 2r sin(\u0394\u03b8 / 2) . \n(6.2.11) \nWhen the angle \u0394\u03b8 is small, we can approximate \nsin(\u0394\u03b8 / 2) \u2245 \u0394\u03b8 / 2 . \n(6.2.12) \nThis is called the small angle approximation, where the angle \u0394\u03b8 (and hence \u0394\u03b8 / 2 ) is \nmeasured in radians. This fact follows from an infinite power series expansion for the \nsine function given by \n\u239b \u0394\u03b8 \u239e\nsin \u239c \n\u239f\n\u239d 2 \u23a0 \n= \u0394\u03b8 \n2 \u2212 1 \u239b \u0394\u03b8 \u239e\n3 \n\u239c \n\u239f\n3!\u239d 2 \u23a0 \n+ 1 \u239b \u0394\u03b8 \u239e\n5 \n\u239c \n\u239f\n5!\u239d 2 \u23a0 \n\u2212 \u22c5\u22c5\u22c5 . \n(6.2.13) \nWhen the angle \u0394\u03b8 / 2 is small, only the first term in the infinite series contributes, as \nsuccessive terms in the expansion become much smaller. For example, when \n3 \n\u22124\n\u0394\u03b8 / 2 = \u03c0 /30 \u2245 0.1, corresponding to 6o, (\u0394\u03b8 / 2) /3! \u2245 1.9\u00d710 ; this term in the power \nseries is three orders of magnitude smaller than the first and can be safely ignored for \nsmall angles. \nUsing the small angle approximation, the magnitude of the displacement is \n4", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_124", "text": "\u0394\uf072 r \u2245 r \u0394\u03b8 . \n(6.2.14) \nThis result should not be too surprising since in the limit as \u0394\u03b8 approaches zero, the \nlength of the chord approaches the arc length r \u0394\u03b8 . \nThe magnitude of the velocity, v , is proportional to the rate of change of the \nmagnitude of the angle with respect to time, \n\u0394! r \nr \u0394\u03b8 \n\u0394\u03b8 \nd\u03b8 = r\u03c9 . \n(6.2.15)\nv \u2261 v !(t) = lim \n= lim \n= r lim \n= r \n\u0394t\u21920 \u0394t \n\u0394t\u21920 \u0394t \n\u0394t\u21920 \u0394t \ndt \nThe direction of the velocity can be determined by considering that in the limit as \n\uf072\n\u0394t \u2192 0 (note that \u0394\u03b8 \u2192 0 ), the direction of the displacement \u0394r approaches the \ndirection of the tangent to the circle at the position of the particle at time t (Figure 6.3). \nr \nr(t)\nr(t + t) \nr \nv(t) \n\u02c6r(t)\n\u02c6(t) \ntangent line \nFigure 6.3 Direction of the displacement approaches the direction of the tangent line \n\uf072\n\uf072 \n\uf072\nThus, in the limit \u0394t \u2192 0 , \u0394r \u22a5 r , and so the direction of the velocity v(t) at time t is \n\uf072\nperpendicular to the position vector r(t) and tangent to the circular orbit in the +\u03b8\u02c6 \u00ad\ndirection for the case shown in Figure 6.3. \n6.3 Circular Motion: Tangential and Radial Acceleration \nWhen the motion of an object is described in polar coordinates, the acceleration has two \ncomponents, the tangential component a\u03b8 , and the radial component, ar . We can write \nthe acceleration vector as \n! \n\u02c6\na = ar r\u02c6(t) + a\u03b8 \u03b8(t) . \n(6.3.1) \n5", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_125", "text": "Keep in mind that as the object moves in a circle, the unit vectors r\u02c6(t) and \u03b8\u02c6(t) change \ndirection and hence are not constant in time. \nWe will begin by calculating the tangential component of the acceleration for \ncircular motion. Suppose that the tangential velocity v\u03b8 = r d\u03b8 / dt is changing in \nmagnitude due to the presence of some tangential force; we shall now consider that \nd\u03b8 / dt is changing in time, (the magnitude of the velocity is changing in time). Recall \nthat in polar coordinates the velocity vector Eq. (6.2.8) can be written as \nv!(t) = r d\u03b8 \ndt \u03b8\u02c6(t) . \n(6.3.2) \nWe now use the product rule to determine the acceleration. \n!a(t) = dv!(t) \ndt = r d 2\u03b8(t) \u02c6\u03b8(t) + r\ndt 2 \nd\u03b8(t) d\u03b8\u02c6(t) .\ndt \ndt \n(6.3.3) \nRecall from Eq. (6.2.3) that \u03b8\u02c6(t) = \u2212sin\u03b8(t)\u02c6i + cos\u03b8(t) \u02c6j. So we can rewrite Eq. (6.3.3) as \n! \nd 2\u03b8(t) \u02c6 \nd\u03b8(t) d\na(t) = r \n\u03b8(t) + r \n(\u2212sin\u03b8(t)\u02c6i + cos\u03b8(t) \u02c6j) . \n(6.3.4)\ndt 2 \ndt dt \nWe again use the chain rule (Eqs. (6.2.5) and (6.2.6)) and find that \n! \nd 2\u03b8(t) \u02c6 \nd\u03b8(t) \u239b\na(t) = r \n\u03b8(t) + r \n\u239d\u239c \u2212 cos\u03b8(t) d\u03b8(t) \u02c6i \u2212 sin\u03b8(t) d\u03b8(t) \u02c6j\u239e\n\u23a0\u239f . \n(6.3.5)\ndt 2 \ndt \ndt\ndt \nRecall that \u03c9 \u2261 d\u03b8 / dt , and from Eq. (6.2.2), r\u02c6(t) = cos\u03b8(t) \u02c6i + sin\u03b8(t) \u02c6j, therefore the \nacceleration becomes \n!\nd 2\u03b8(t) \n\u239b d\u03b8(t)\u239e \n2 \na(t) = r \n\u03b8\u02c6(t) \u2212 r \nr\u02c6(t) . \n(6.3.6)\n\u239d\u239c \n\u23a0\u239f\ndt 2 \ndt \nThe tangential component of the acceleration is then", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_126", "text": "acceleration becomes \n!\nd 2\u03b8(t) \n\u239b d\u03b8(t)\u239e \n2 \na(t) = r \n\u03b8\u02c6(t) \u2212 r \nr\u02c6(t) . \n(6.3.6)\n\u239d\u239c \n\u23a0\u239f\ndt 2 \ndt \nThe tangential component of the acceleration is then \nd 2\u03b8(t)\na\u03b8 = r \n. \n(6.3.7)\ndt2 \nThe radial component of the acceleration is given by \n6", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_127", "text": "\u239b d\u03b8(t)\u239e \n2 \nar = \u2212r \n= \u2212r\u03c9 2 < 0 . \n(6.3.8)\n\u239d\u239c dt \u23a0\u239f \n\uf072\nBecause ar < 0 , that radial vector component a r (t) = \u2212r\u03c9 2 r\u02c6(t) is always directed \ntowards the center of the circular orbit. \nExample 6.1 Circular Motion Kinematics \nA particle is moving in a circle of radius R . At t = 0 , it is located on the x -axis. The \nangle the particle makes with the positive x -axis is given by \u03b8(t) = At3 \u2212 Bt , where A \nand B are positive constants. Determine (a) the velocity vector, and (b) the acceleration \nvector. Express your answer in polar coordinates. At what time is the centripetal \nacceleration zero? \nSolution: \nThe derivatives of the angle function \u03b8(t) = At3 \u2212 Bt are d\u03b8 / dt = 3At2 \u2212 B and \nd 2\u03b8 / dt2 = 6 At . Therefore the velocity vector is given by \n!v(t) = R d\u03b8(t) \u03b8\u02c6(t) = R(3At 2 \u2212 Bt)\u03b8\u02c6(t) .\ndt \nThe acceleration is given by \na\uf072(t) = R d 2\u03b8(t) \u03b8\u02c6(t) \u2212 R\u239b\n\u239d\u239c \nd\u03b8(t)\u239e\n\u23a0\u239f \n2 \nr\u02c6(t)\ndt 2 \ndt \n. \n= R(6At) \u03b8\u02c6(t) \u2212 R(3At 2 \u2212 B)\n2 r\u02c6(t) \nThe centripetal acceleration is zero at time t = t1 when \n3At1\n2 \u2212 B = 0 \u21d2 t1 = \n6.4 Period and Frequency for Uniform Circular Motion \nIf the object is constrained to move in a circle and the total tangential force acting on the \ntotal \nobject is zero, F\u03b8 \n= 0 , then (Newton\u2019s Second Law), the tangential acceleration is zero, \na\u03b8 = 0 . \n(6.4.1) \nB / 3A . \n7", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_128", "text": "This means that the magnitude of the velocity (the speed) remains constant. This motion \nis known as uniform circular motion. The acceleration is then given by only the \nacceleration radial component vector \n\uf072a r (t) = \u2212r\u03c9 2 (t) r\u02c6(t) uniform circular motion . \n(6.4.2) \n\u03c9\nBecause the speed v = r \nis constant, the amount of time that the object takes to \ncomplete one circular orbit of radius r is also constant. This time interval, T , is called \nthe period. In one period the object travels a distance s = vT equal to the circumference, \ns = 2\u03c0r ; thus \ns = 2\u03c0r = vT . \n(6.4.3) \nThe period T is then given by \n2\u03c0r \n2\u03c0r \n2\u03c0\nT =\n=\n= \n. \n(6.4.4)\nv\nr\u03c9\n\u03c9 \nThe frequency f is defined to be the reciprocal of the period, \n1 \n\u03c9\nf =\n= \n. \n(6.4.5)\nT \n2\u03c0 \n\u23a1\u22121\nThe SI unit of frequency is the inverse second, which is defined as the hertz, s \u23a4\u2261 [Hz] .\n\u23a3\n\u23a6 \nThe magnitude of the radial component of the acceleration can be expressed in several \nequivalent forms since both the magnitudes of the velocity and angular velocity are \nrelated by v = r\u03c9 . Thus we have several alternative forms for the magnitude of the \ncentripetal acceleration. The first is that in Equation (6.5.3). The second is in terms of the \nradius and the angular velocity, \n= r\u03c9 2\na \n. \n(6.4.6)\nr \nThe third form expresses the magnitude of the centripetal acceleration in terms of the \nspeed and radius, \n2v \na = \n. \n(6.4.7)\nr \nr \nRecall that the magnitude of the angular velocity is related to the frequency by \n\u03c9 = 2\u03c0 f , so we have a fourth alternate expression for the magnitude of the centripetal \nacceleration in terms of the radius and frequency, \na = 4\u03c0 2r f 2 . \n(6.4.8)\nr \n8", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_129", "text": "A fifth form commonly encountered uses the fact that the frequency and period \nare related by f = 1/ T = \u03c9 / 2\u03c0 . Thus we have the fourth expression for the centripetal \nacceleration in terms of radius and period, \n4\u03c0 2r \na = \n. \n(6.4.9)\nr \nT 2 \nOther forms, such as 4\u03c0 2r 2 f / T or 2\u03c0r\u03c9 f , while valid, are uncommon. \nOften we decide which expression to use based on information that describes the \norbit. A convenient measure might be the orbit\u2019s radius. We may also independently \nknow the period, or the frequency, or the angular velocity, or the speed. If we know one, \nwe can calculate the other three but it is important to understand the meaning of each \nquantity. \n6.4.1 Geometric Interpretation for Radial Acceleration for Uniform Circular \nMotion \nAn object traveling in a circular orbit is always accelerating towards the center. Any \nradial inward acceleration is called centripetal acceleration. Recall that the direction of \nthe velocity is always tangent to the circle. Therefore the direction of the velocity is \nconstantly changing because the object is moving in a circle, as can be seen in Figure 6.4. \nBecause the velocity changes direction, the object has a nonzero acceleration. \nFigure 6.5 Change in velocity vector. \nFigure 6.4 Direction of the velocity for \ncircular motion. \n9", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_130", "text": "The calculation of the magnitude and direction of the acceleration is very similar to the \ncalculation for the magnitude and direction of the velocity for circular motion, but the \n\uf072\nchange in velocity vector, \u0394v , is more complicated to visualize. The change in velocity \n\uf072\n\uf072 \n\uf072\n\u0394v = v(t + \u0394t) \u2212 v( )t is depicted in Figure 6.5. The velocity vectors have been given a \n\uf072\ncommon point for the tails, so that the change in velocity, \u0394v , can be visualized. The \nlength \u0394v \uf072 of the vertical vector can be calculated in exactly the same way as the \ndisplacement \u0394r \uf072 . The magnitude of the change in velocity is \n\u0394v \uf072 = 2vsin(\u0394\u03b8 / 2) . \n(6.5.1) \nWe can use the small angle approximation sin (\u0394\u03b8 / 2)\u2245 \u0394\u03b8 / 2 to approximate the \nmagnitude of the change of velocity, \n\u0394v \uf072 \u2245 v \u0394\u03b8 . \n(6.5.2) \nThe magnitude of the radial acceleration is given by \n\u0394\uf072 v \nv \u0394\u03b8 \n\u0394\u03b8 \nd\u03b8 \na = lim \n= lim \n= v lim \n= v \n= v \u03c9 . \n(6.5.3)\nr \n\u0394t\u21920 \u0394t \n\u0394t\u21920 \u0394t \n\u0394t\u21920 \u0394t \ndt \nThe direction of the radial acceleration is determined by the same method as the direction \n\uf072\n\uf072\nof the velocity; in the limit \u0394\u03b8 \u2192 0 , \u0394v \u22a5 v , and so the direction of the acceleration \n\uf072\n\uf072\nradial component vector a r (t) at time t is perpendicular to position vector v( )t and \ndirected inward, in the \u2212r\u02c6 -direction. \n6.5 Angular Velocity and Angular Acceleration \n6.5.1. Angular Velocity \nWe shall always choose a right-handed cylindrical coordinate system. If the positive z \u00ad\naxis points up, then we choose \u03b8 to be increasing in the counterclockwise direction as \nshown in Figures 6.6. \nO \n+x \n+ y \n.\nr \n\u02c6k \n\u02c6 \n\u02c6r \n+ z \nFigure 6.6 Right handed coordinate system \n10", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_131", "text": "For a point object undergoing circular motion about the z -axis, the angular velocity \n! \nvector \u03c9 is directed along the z -axis with z -component equal to the time derivative of \nthe angle \u03b8 , \n\u03c9! = d\u03b8 k\u02c6 = \u03c9 z k\u02c6 . \n(6.5.4)\ndt \nThe SI units of angular velocity are [rad \u22c5 s\u22121]. Note that the angular speed is just the \nmagnitude of the z -component of the angular velocity, \nd\u03b8 . \n(6.5.5)\n\u03c9 \u2261 \u03c9 = \nz \ndt \nIf the velocity of the object is in the +\u03b8\u02c6 -direction, (rotating in the counterclockwise \ndirection in Figure 6.7(a)), then the z -component of the angular velocity is positive, \n\u03c9 z = d\u03b8 / dt > 0 . The angular velocity vector then points in the +k\u02c6 -direction as shown in \nFigure 6.7(a). If the velocity of the object is in the \u2212\u03b8\u02c6 -direction, (rotating in the \nclockwise direction in Figure 6.7(b)), then the z -component of the angular velocity \nangular velocity is negative, \u03c9 z = d\u03b8 / dt < 0 . The angular velocity vector then points in \nthe \u2212k\u02c6 -direction as shown in Figure 6.7(b). \nO \n+x \n+ y \n.\nr \nv \n\u02c6k \n\u02c6 \n\u02c6r \nz = d \ndt > 0\n+ z \nO \n+x \n+ y \n.\nr \nv \n\u02c6k \n\u02c6 \n\u02c6r \n+ z \nz = d \ndt < 0 \nFigure 6.7(a) Angular velocity vector \nFigure 6.7(b) Angular velocity \nvector for motion with d\u03b8 / dt > 0 . \nfor motion with d\u03b8 / dt < 0 . \nThe velocity and angular velocity are related by \n! \n!\n! \nd\u03b8 \nd\u03b8 \nv = \u03c9 \u00d7 r = \nk\u02c6 \u00d7 rr\u02c6 = r \n\u03b8\u02c6 . \n(6.5.6)\ndt \ndt \n11", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_132", "text": "Example 6.2 Angular Velocity \nA particle is moving in a circle of radius R . At t = 0 , it is located on the x -axis. The \nangle the particle makes with the positive x -axis is given by \u03b8(t) = At \u2212 Bt3 , where A \nand B are positive constants. Determine (a) the angular velocity vector, and (b) the \nvelocity vector. Express your answer in polar coordinates. (c) At what time, t = t1 , is the \nangular velocity zero? (d) What is the direction of the angular velocity for (i) t < t1, and \n(ii) t > t1? \nSolution: The derivative of \u03b8(t) = At \u2212 Bt3 is \nd\u03b8(t) = A \u2212 3Bt2 .\ndt \nTherefore the angular velocity vector is given by \n! \nd\u03b8(t)\n\u03c9(t) = dt k\u02c6 = (A \u2212 3Bt 2)k\u02c6 . \nThe velocity is given by \n!v(t) = R d\u03b8(t) \u03b8\u02c6(t) = R(A \u2212 3Bt 2)\u03b8\u02c6(t).\ndt \nThe angular velocity is zero at time t = t1 when \nA \u2212 3Bt1\n2 = 0 \u21d2 t1 = \nd\u03b8(t) \n!\nFor t < t1 , \n= A \u2212 3Bt1\n2 > 0 hence \u03c9(t) points in the positive k\u02c6 -direction.\ndt \nd\u03b8(t) \n!\nFor t > t1 , \n= A \u2212 3Bt1\n2 < 0 hence \u03c9(t) points in the negative k\u02c6 -direction.\ndt \n6.5.2 Angular Acceleration \nIn a similar fashion, for a point object undergoing circular motion about the fixed z -axis, \nthe angular acceleration is defined as \nd 2\u03b8\n\u03b1! = dt2 k\u02c6 = \u03b1 k\u02c6 . \n(6.5.7)\nz \nA / 3B . \n12", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_133", "text": "The SI units of angular acceleration are [rad \u22c5s\u22122]. The magnitude of the angular \nacceleration is denoted by the Greek symbol alpha, \nd 2\u03b8\n!\n\u03b1 \u2261 \u03b1 = \n. \n(6.5.8)\ndt2 \nThere are four special cases to consider for the direction of the angular velocity. Let\u2019s \n!\nfirst consider the two types of motion with \u03b1 pointing in the +k\u02c6 -direction: (i) if the \nobject is rotating counterclockwise and speeding up then both d\u03b8 / dt > 0 and \nd 2\u03b8 / dt2 > 0 (Figure 6.8(a), (ii) if the object is rotating clockwise and slowing down then \nd\u03b8 / dt < 0 but d 2\u03b8 / dt2 > 0 (Figure 6.8(b). There are two corresponding cases in which \n!\u03b1 pointing in the \u2212k\u02c6 -direction: (iii) if the object is rotating counterclockwise and \nslowing down then d\u03b8 / dt > 0 but d 2\u03b8 / dt2 < 0 (Figure 6.9(a), (iv) if the object is \nrotating clockwise and speeding up then both d\u03b8 / dt < 0 and d 2\u03b8 / dt2 < 0 (Figure 6.9(b). \nO \n+x \n+ y \n.\nr \nv \n\u02c6k \n\u02c6 \n\u02c6r \n+ z \nz = d \ndt\nz = d 2 \ndt2 > 0 \n> 0 \nO \n+x \n+ y \n.\nr \nv \n\u02c6k \n\u02c6 \n\u02c6r \n+ z \nz = d \ndt\nz = d 2 \ndt2 > 0 \n0\n< \nFigure 6.8(a) Angular acceleration vector \nFigure 6.8(b) Angular velocity \nvector for motion with d\u03b8 / dt > 0 , \nfor motion with d\u03b8 / dt < 0 , and \nand d 2\u03b8 / dt2 > 0. \nd 2\u03b8 / dt2 > 0. \nO \n+x \n+ y \n.\nr \nv \n\u02c6k \n\u02c6 \n\u02c6r \n+ z \nz = d \ndt\nz = d 2 \ndt2 \n> 0\n< 0 \nO \n+x \n+ y \n.\nr \nv \n\u02c6k \n\u02c6 \n\u02c6r \n+ z \nz = d \ndt < 0 \nz = d 2 \ndt2 < 0 \nFigure 6.9(a) Angular acceleration vector \nFigure 6.9(b) Angular velocity \nvector for motion with d\u03b8 / dt > 0 , \nfor motion with d\u03b8 / dt < 0 , and \nand d 2\u03b8 / dt2 < 0. \nd 2\u03b8 / dt2 < 0. \n13", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_134", "text": "Example 6.3 Integration and Circular Motion Kinematics \nA point-like object is constrained to travel in a circle. The z -component of the angular \nacceleration of the object for the time interval [0, t1] is given by the function \n\u23a7\u239b \nt \u239e \n\u23aab \n\u239d\u239c1\u2212 \n\u23a0\u239f ; 0 \u2264 t \u2264 t1\n\u03b1 z (t) = \u23a8 \nt1\n, \n\u23aa0; t > t1\n\u23a9 \nwhere b is a positive constant with units rad \u22c5s\u22122 . \na) Determine an expression for the angular velocity of the object at t = t1. \nb) Through what angle has the object rotated at time t = t1? \nSolution: \na) The angular velocity at time t = t1 is given by \nt\u2032=t1 \nt\u2032=t1 \u239b \nt\u2032\u239e \n\u239b \n2 \u239e \nbt1\n\u03c9 z (t1) \u2212\u03c9 z (t = 0) = \u222b\u03b1 z (t\u2032) dt\u2032 = \u222b b 1\u2212 \n\u23a0\u239f dt\u2032 = b t1 \u2212 t1 \nt\u2032=0 \nt\u2032=0 \u239d\u239c \nt1 \n\u239d\u239c \n2t1 \u23a0\u239f = 2 \nb) In order to find the angle \u03b8(t1) \u2212\u03b8(t = 0) that the object has rotated through at time \nt = t1 , you first need to find \u03c9 z (t) by integrating the z-component of the angular \nacceleration \nt\u2032=t\nt\u2032=t \nt2 \u239e\n\u239b \nt\u2032\u239e \n\u239b\n\u03c9 z (t) \u2212\u03c9 z (t = 0) = \u222b\u03b1 z (t\u2032) dt\u2032 = \u222b b 1\u2212 \n\u23a0\u239f dt\u2032 = b t \u2212 \n\u23a0\u239f . \n\u239d\u239c \n\u239d\u239c \n2t1\nt\u2032=0 \nt\u2032=0 \nt1 \n\u239b \nBecause it started from rest, \u03c9 (t = 0) = 0 , hence \u03c9 (t) = b t \u2212 t2 \u239e\n\u23a0\u239f ; 0 \u2264 t \u2264 t1 . \nz\nz \n\u239d\u239c \n2t1 \nThen integrate \u03c9 z (t) between t = 0 and t = t1 to find that \nt\u2032=t1 \nt\u2032=t1 \nt\u20322 \u239e \n2\n3 \n2\n\u239b\n\u239e \nbt1\n\u03b8(t1) \u2212\u03b8(t = 0) = \u222b\u03c9 (t\u2032) dt\u2032 = \u222b b t\u2032 \u2212 \n\u23a0\u239f dt\u2032 = b \n\u239b t1 \u2212 t1 \n. \n\u239d\u239c \n\u239d\u239c 2\nt\u2032=0 \nz \nt\u2032=0 \n2t1 \n6t1 \u23a0\u239f = 3 \n14", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_135", "text": "6.5 Non-circular Central Planar Motion \nLet\u2019s now consider central motion in a plane that is non-circular. In Figure 6.10, we show \nthe spiral motion of a moving particle. In polar coordinates, the key point is that the time \nderivative dr / dt of the position function r is no longer zero. The second derivative \nd 2r / dt2 also may or may not be zero. In the following calculation we will drop all \nexplicit references to the time dependence of the various quantities. The position vector is \nstill given by Eq. (6.2.1), which we shall repeat below \n!r = r r\u02c6 . \n(6.5.9) \nBecause dr / dt \u2260 0 , when we differentiate Eq. (6.5.9), we need to use the product rule \nd !\n! \nr \ndr \nd r\u02c6 \nv =\n= \nr\u02c6 + r \n. \n(6.5.10)\ndt\ndt \ndt \nSubstituting Eq. (6.2.4) into Eq. (6.5.10) \n! \nr \ndr \nd\u03b8\nv = d! \n= \nr\u02c6 + r \n\u03b8\u02c6 = vr r\u02c6 + v\u03b8\u03b8\u02c6 . \n(6.5.11)\ndt\ndt \ndt \nThe velocity is no longer tangential but now has a radial component as well \ndr \nvr = dt . \n(6.5.12) \nIn order to determine the acceleration, we now differentiate Eq. (6.5.11), again using the \nproduct rule, which is now a little more involved: \nd ! \n\u02c6\n! \nv \nd 2r \ndr d r\u02c6 \ndr d\u03b8 \nd 2\u03b8 \nd\u03b8 d \u03b8\n\u02c6\n\u02c6\na =\n= \nr\u02c6 +\n+ \n\u03b8 + r \n\u03b8 + r \n. \n(6.5.13)\ndt \ndt 2 \ndt dt \ndt dt \ndt 2 \ndt dt \nNow substitute Eqs. (6.2.4) and (6.2.7) for the time derivatives of the unit vectors in Eq. \n(6.5.13), and after collecting terms yields \n! \u239b d 2r \n\u239b d\u03b8 \u239e \n2 \u239e\n\u239b 2 dr d\u03b8 \nd 2\u03b8 \u239e \u02c6\na = \nr + \n\u03b8 \n\u239d\u239c dt 2 \u2212 r \u239d\u239c dt \u23a0\u239f \u23a0\u239f \u02c6 \n\u239d\u239c dt dt + r dt 2 \u23a0\u239f . \n(6.5.14) \n\u02c6\n= arr\u02c6 + a\u03b8\u03b8 \nThe radial and tangential components of the acceleration are now more complicated than \nthen in the case of circular motion due to the non-zero derivatives of dr / dt and d 2r / dt2 . \nThe radial component is \n15", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_136", "text": "d 2r \n\u239b d\u03b8 \u239e \n2 \nar = \n. \n(6.5.15)\ndt 2 \u2212 r \u239d\u239c dt \u23a0\u239f \nand the tangential component is \n= 2 dr d\u03b8 \nd 2\u03b8 . \n(6.5.16)\na\u03b8 \ndt dt + r dt 2 \nThe firs term in the tangential component of the acceleration, 2(dr / dt)(d\u03b8 / dt) has a \nspecial name, the coriolis acceleration, \n= 2 dr d\u03b8 \nacor \n. \n(6.5.17)\ndt dt \nExample 6.4 Spiral Motion \nA particle moves outward along a spiral starting from the origin at t = 0 . Its trajectory is \ngiven by r = b\u03b8 , where b is a positive constant with units [m \u22c5rad-1] . \u03b8 increases in time \naccording to \u03b8 = ct2 , where c > 0 is a positive constant (with units [rad \u22c5s\u22122] ). \na) Determine the acceleration as a function of time. \nb) Determine the time at which the radial acceleration is zero. \nc) What is the angle when the radial acceleration is zero? \nd) Determine the time at which the radial and tangential accelerations have equal \nmagnitude. \nSolution: \na) The position coordinate as a function of time is given by r = b\u03b8 = bct2 . The \nacceleration is given by Eq. (6.5.14). In order to calculate the acceleration, we need to \ncalculate the four derivatives dr / dt = 2bct , d 2r / dt2 = 2bc , d\u03b8 / dt = 2ct , and \nd 2\u03b8 / dt2 = 2c . The acceleration is then \n!\n\u02c6\n\u02c6\na = (2bc \u2212 4bc3t 4 ) r\u02c6 +(8bc2t 2 + 2bc2t 2 )\u03b8 = (2bc \u2212 4bc3t 4 ) r\u02c6 +10bc2t 2 \u03b8 . \nb) The radial acceleration is zero when \nt1 = \u239b\u239c\u239d  \n1  \n2c2 \n\u239e\u239f\u23a0  \n1/4 \n.  \nc) The angle when the radial acceleration is zero is \n16", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_137", "text": "= ct1\n2 = 2 / 2 .\n\u03b81 \nd)  The radial and tangential accelerations have equal magnitude when after some \nalgebra \n(2bc \u2212 4bc3t 4) = 10bc2t 2 \u21d2 0 = t 4 + (5 / 2c)t 2 \u2212 (1/ 2c2) . \nThis equation has as only positive solution for t 2 : \n\u2212(5 / 2c) \u00b1 ((5 / 2c)2 + 2c2 )1/2 \n33 \u2212 5\nt2\n2 = \n= \n.\n2 \n4c \nTherefore the magnitudes of the two components are equal when \nt2 = \n33 \u2212 5 \n4c \n. \n17", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_138", "text": "Chapter 7 Newton\u2019s Laws of Motion \n7.1 Force and Quantity of Matter................................................................................ 1\nExample 7.1 Vector Decomposition Solution ......................................................... 3\n7.1.1 Mass Calibration.............................................................................................. 4\n7.2 Newton\u2019s First Law................................................................................................. 5\n7.3 Momentum, Newton\u2019s Second Law and Third Law............................................ 6\n7.4 Newton\u2019s Third Law: Action-Reaction Pairs....................................................... 7\n\u00a9 Peter Dourmashkin 2012", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_139", "text": "Chapter 7 Newton\u2019s Laws of Motion \nI have not as yet been able to discover the reason for these properties of \ngravity from phenomena, and I do not feign hypotheses. For whatever is \nnot deduced from the phenomena must be called a hypothesis; and \nhypotheses, whether metaphysical or physical, or based on occult \nqualities, or mechanical, have no place in experimental philosophy. In this \nphilosophy particular propositions are inferred from the phenomena, and \nafterwards rendered general by induction. 1\nIsaac Newton \n7.1 Force and Quantity of Matter \nIn our daily experience, we can cause a body to move by either pushing or pulling that \nbody. Ordinary language use describes this action as the effect of a person\u2019s strength or \nforce. However, bodies placed on inclined planes, or when released at rest and undergo \nfree fall, will move without any push or pull. Galileo referred to a force acting on these \nbodies, a description of which he published in Mechanics in 1623. In 1687, Isaac Newton \npublished his three laws of motion in the Philosophiae Naturalis Principia Mathematica \n(\u201cMathematical Principles of Natural Philosophy\u201d), which extended Galileo\u2019s \nobservations. The First Law expresses the idea that when no force acts on a body, it will \nremain at rest or maintain uniform motion; when a force is applied to a body, it will \nchange its state of motion. \nMany scientists, especially Galileo, recognized the idea that force produces motion \nbefore Newton but Newton extended the concept of force to any circumstance that \nproduces acceleration. When a body is initially at rest, the direction of our push or pull \ncorresponds to the direction of motion of the body. If the body is moving, the direction of \nthe applied force may change both the direction of motion of the body and how fast it is \nmoving. Newton defined the force acting on an object as proportional to the acceleration \nof the object. \nAn impressed force is an action exerted upon a body, in order to change \nits state, either of rest, or of uniform motion in a right line.2 \nIn order to define the magnitude of the force, he introduced a constant of \nproportionality, the inertial mass, which Newton called \u201cquantity of matter\u201d. \n1 Isaac Newton (1726). Philosophiae Naturalis Principia Mathematica, General\nScholium. Third edition, page 943 of I. Bernard Cohen and Anne Whitman's 1999 \ntranslation, University of California Press. \n2 Isaac Newton. Mathematical Principles of Natural Philosophy. Translated by Andrew", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_140", "text": "1 Isaac Newton (1726). Philosophiae Naturalis Principia Mathematica, General\nScholium. Third edition, page 943 of I. Bernard Cohen and Anne Whitman's 1999 \ntranslation, University of California Press. \n2 Isaac Newton. Mathematical Principles of Natural Philosophy. Translated by Andrew\nMotte (1729). Revised by Florian Cajori. Berkeley: University of California Press, 1934. \np. 2.\n7-1", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_141", "text": "The quantity of matter is the measure of the same, arising from its density \nand bulk conjointly. \nThus air of double density, in a double space, is quadruple in quantity; in a \ntriple space, sextuple in quantity. The same thing is to be understood of \nsnow, and fine dust or powders, that are condensed by compression or \nliquefaction, and of all bodies that are by any causes whatever differently \ncondensed. I have no regard in this place to a medium, if any such there is, \nthat freely pervades the interstices between the parts of bodies. It is this \nquantity that I mean hereafter everywhere under the name of body or mass. \nAnd the same is known by the weight of each body, for it is proportional to \nthe weight, as I have found by experiment on pendulums, very accurately \nmade, which shall be shown hereafter.3 \nSuppose we apply a force to a body that is an identical copy of the standard mass, (we \nshall refer to this body as a standard body). The force will induce the standard body to \n\uf072 a\naccelerate with magnitude \nthat can be measured by an accelerometer (any device that \n\uf072 \nF\nmeasures acceleration). The magnitude of the force \nacting on the standard body is \ndefined to be the product of the standard mass m s with the magnitude of the acceleration \n\uf072 a . Force is a vector quantity. The direction of the force on the standard body is defined\nto be the direction of the acceleration of the body. Thus \n\uf072\n\uf072\nF \u2261 m s a \n(7.1.1) \nIn order to justify the statement that force is a vector quantity, we need to apply two\n\uf072\n\uf072 \nforces F1 and F2 simultaneously to our standard body and show that the resultant force \n\uf072 \nFT is the vector sum of the two forces when the forces are applied one at a time. \nFigure 7.1 Acceleration add as vectors \nFigure 7.2 Force adds as vectors. \n\uf072\n\uf072\nWe apply each force separately and measure the accelerations a1 and a2., noting that \n\uf072\n\uf072\nF = m a \n(7.1.2)\n1 \ns \n1 \n3 Ibid. p. 1. \n\u001a\u0010\u0015", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_142", "text": "\uf072\n\uf072 \n= m a2 . \n(7.1.3)\nF2 \ns \n\uf072\nWhen we apply the two forces simultaneously, we measure the acceleration a . The force \nby definition is now \nF\n\uf072T \n\uf072\n\u2261 m s a . \n(7.1.4) \nWe then compare the accelerations. The results of these three measurements, and for that \nmatter any similar experiment, confirms that the accelerations add as vectors (Figure 7.1) \n\uf072a =  \uf072a + \uf072a\n(7.1.5) \n2 .\n1 \nTherefore the forces add as vectors as well (Figure 7.2), \n\uf072 \nF T =  \n\uf072 \nF1 +  \n\uf072 \nF2 .  \n(7.1.6)  \nThis last statement is not a definition but a consequence of the experimental result \ndescribed by Equation (7.1.5) and our definition of force. \nExample 7.1 Vector Decomposition Solution \nTwo horizontal ropes are attached to a post that is stuck in the ground. The ropes pull the \n\uf076\n\uf076 \npost producing the vector forces F1 = 70 N \u02c6i + 20 N \u02c6j and F2 = \u221230 N \u02c6i + 40 N \u02c6j as \nshown in Figure 7.3. Find the direction and magnitude of the horizontal component of a \ncontact force of the ground on the post. \nFigure 7.3 Example 7.1 \nFigure 7.4 Vector sum of the horizontal \nforces \n\u001a\u00103", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_143", "text": "Solution: Because the ropes are pulling the post horizontally, the contact force must have \na horizontal component that is equal to the negative of the sum of the two horizontal \nforces exerted by the rope on the post (Figure 7.4). There is an additional vertical \ncomponent of the contact force that balances the gravitational force exerted on the post \nby the earth. We restrict our attention to the horizontal component of the contact force. \n\uf076\n\uf076 \nLet F3 denote the sum of the forces due to the ropes. Then we can write the vector F3 as \n\uf076F3 = (F1x + F2x ) \u02c6i + (F1y + F2 y ) \u02c6j = (70 N + \u2212 30 N) \u02c6i + (20 N + 40 N) \u02c6j \n= (40 N) \u02c6i + (60 N) \u02c6j \nTherefore the horizontal component of the contact force satisfies the condition that \n\uf072 \n\uf076\n\uf076\n\uf076 \nFhor = \u2212F3 = \u2212(F1 + F2) = (\u221240 N) \u02c6i + (\u221260 N) \u02c6j . \n\uf072 \nThe magnitude is \n= (\u221240 N)2 + (\u221260 N)2 = 72 N . The horizontal component of\nFhor \nthe contact force makes an angle \n\u23a1 60 N \u23a4\n\u03b8 = tan\u22121 \n= 56.3\u00b0 \n\u23a2\n\u23a5\n\u23a3 40 N \u23a6 \nas shown in the figure above. \n7.1.1 Mass Calibration \nSo far, we have only used the standard body to measure force. Instead of performing \nexperiments on the standard body, we can calibrate the masses of all other bodies in \nterms of the standard mass by the following experimental procedure. We shall refer to the \nmass measured in this way as the inertial mass and denote it by min . \nWe apply a force of magnitude F to the standard body and measure the \nmagnitude of the acceleration a s . Then we apply the same force to a second body of \nunknown mass min and measure the magnitude of the acceleration ain . Because the same \nforce is applied to both bodies, \nain = m a \ns \n, \n(1.7)\nF = min \ns \nthe ratio of the inertial mass to the standard mass is equal to the inverse ratio of the \nmagnitudes of the accelerations, \nmin \na s\n= \n. \n(1.8)\nm s \nain \nTherefore the second body has inertial mass equal to \n7-4", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_144", "text": "a \n= m \ns . \n(1.9)\nmin \ns ain \nThis method is justified by the fact that we can repeat the experiment using a different \nforce and still find that the ratios of the acceleration are the same. For simplicity we shall \ndenote the inertial mass by m . \n7.2 Newton\u2019s First Law \nThe First Law of Motion, commonly called the \u201cPrinciple of Inertia,\u201d was first realized \nby Galileo. (Newton did not acknowledge Galileo\u2019s contribution.) Newton was \nparticularly concerned with how to phrase the First Law in Latin, but after many rewrites \nNewton choose the following expression for the First Law (in English translation): \nLaw 1: Every body continues in its state of rest, or of uniform motion in a \nright line, unless it is compelled to change that state by forces impressed \nupon it. \nProjectiles continue in their motions, so far as they are not retarded by the \nresistance of air, or impelled downwards by the force of gravity. A top, \nwhose parts by their cohesion are continually drawn aside from \nrectilinear motions, does not cease its rotation, otherwise than as it is \nretarded by air. The greater bodies of planets and comets, meeting with \nless resistance in freer spaces, preserve their motions both progressive \nand circular for a much longer time.4 \nThe first law is an experimental statement about the motions of bodies. When a \nbody moves with constant velocity, there are either no forces present or the sum of all the \nforces acting on the body is zero. If the body changes its velocity, it has non-zero \nacceleration, and hence the sum of all the forces acting on the body must be non-zero as \nwell. If the velocity of a body changes in time, then either the direction or magnitude \nchanges, or both can change. \nAfter a bus or train starts, the acceleration is often so small we can barely perceive \nit. We are often startled because it seems as if the station is moving in the opposite \ndirection while we seem to be at rest. Newton\u2019s First Law states that there is no physical \nway to distinguish between whether we are moving or the station is moving, because \nthere is nearly zero total force acting on the body. Once we reach a constant velocity, our \nminds dismiss the idea that the ground is moving backwards because we think it is \nimpossible, but there is no actual way for us to distinguish whether the train is moving or \nthe ground is moving. \n4 Ibid. p. 13. \n\u001a\u0010\u0018", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_145", "text": "7.3 Momentum, Newton\u2019s Second Law and Third Law \nNewton began his analysis of the cause of motion by introducing the quantity of motion: \nDefinition: Quantity of Motion \nThe quantity of motion is the measure of the same, arising from the \nvelocity and quantity of matter conjointly. \nThe motion of the whole is the sum of the motion of all its parts; and \ntherefore in a body double in quantity, with equal velocity, the motion \nis double, with twice the velocity, it is quadruple.5 \n\uf072p \nOur modern term for quantity of motion is momentum and it is a vector quantity \n\uf072\n\uf072 \np = mv ,\n(7.3.1) \n\uf072 \nF \n\uf072\nwhere m is the inertial mass and v is the velocity of the body. Newton\u2019s Second Law\nstates that \nLaw II: The change of motion is proportional to the motive force \nimpressed, and is made in the direction of the right line in which that force \nis impressed. \nIf any force generates a motion, a double force will generate double the \nmotion, a triple force triple the motion, whether that force is impressed \naltogether and at once or gradually and successively. And this motion \n(being always directed the same way with the generating force), if the \nbody moved before, is added or subtracted from the former motion, \naccording as they directly conspire with or are directly contrary to each \nother; or obliquely joined, when they are oblique, so as to produce a new \nmotion compounded from the determination of both.6 \nSuppose that a force is applied to a body for a time interval \u0394t . The impressed force or\n\uf072 \nimpulse (a vector quantity I ) produces a change in the momentum of the body, \n\u0394t = \u0394 \n\uf072\nI =  \n(7.3.2) \n.  \nFrom the commentary to the second law, Newton also considered forces that were \napplied continually to a body instead of impulsively. The instantaneous action of the total \n5 Ibid. p. 1.\n6 Ibid. p. 13. \n7-\u0019", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_146", "text": "force acting on a body at a time t is defined by taking the mathematical limit as the time \ninterval \u0394t becomes smaller and smaller, \n\uf072\n\uf072 \n\u0394p \ndp \uf072 \nF = lim \n\u2261 \n. \n(7.3.3)\n\u0394t\u21920 \u0394t \ndt \nWhen the mass remains constant in time, the Second Law can be recast in its more \nfamiliar form, \n\uf072 \nd\uf072 v\nF = m \n. \n(7.3.4)\ndt \nBecause the derivative of velocity is the acceleration, the force is the product of mass and \nacceleration, \n\uf072\n\uf072\nF = ma . \n(7.3.5) \nBecause we defined force in terms of change in motion, the Second Law appears to \nbe a restatement of this definition, and devoid of predictive power since force is only \ndetermined by measuring acceleration. What transforms the Second Law from just a \ndefinition is the additional input that comes from force laws that are based on \nexperimental observations on the interactions between bodies. Throughout this book, we \nshall investigate these force laws and learn to use them in order to determine the forces \nand accelerations acting on a body (left-hand-side of Newton\u2019s Second Law). When a \nphysical body is constrained to move along a surface, or inside a container (for example \ngas molecules in a container), there are constraint forces that are not determined \nbeforehand by any force law but are only determined by their effect on the motion of the \nbody. For any given constrained motion, these constraint forces are unknown and must be \ndetermined by the particular motion of the body that we are studying, for example the \ncontact force of the surface on the body, or the force of the wall on the gas particles. \nThe right-hand-side of Newton\u2019s Second Law is the product of mass with \nacceleration. Acceleration is a mathematical description of how the velocity of a body \nchanges. Knowledge of all the forces acting on the body enables us to predict the \nacceleration. Eq. (7.3.5) is known as the equation of motion. Once we know this equation \nwe may be able to determine the velocity and position of that body at all future times by \nintegration techniques, or computational techniques. For constrained motion, if we know \nthe acceleration of the body, we can also determine the constraint forces acting on the \nbody. \n7.4 Newton\u2019s Third Law: Action-Reaction Pairs \nNewton realized that when two bodies interact via a force, then the force on one body is \nequal in magnitude and opposite in direction to the force acting on the other body. \n7-\u001a", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_147", "text": "Law III: To every action there is always opposed an equal reaction: or, \nthe mutual action of two bodies upon each other are always equal, and \ndirected to contrary parts. \nWhatever draws or presses another is as much drawn or pressed by that \nother. If you press on a stone with your finger, the finger is also pressed \nby the stone.7 \nThe Third Law, commonly known as the \u201caction-reaction\u201d law, is the most \nsurprising of the three laws. Newton\u2019s great discovery was that when two objects interact, \nthey each exert the same magnitude of force on each other but in opposite directions. We \nshall refer to the pair of forces between two interacting bodies as an interaction pair of \nforce, or more briefly as an interaction pair. \nConsider two bodies engaged in a mutual interaction. Label the bodies 1 and 2 \n\uf072\n\uf072 \nrespectively. Let \nbe the force on body 2 due to the interaction with body 1, and \nF1, 2 \nF2,1 \nbe the force on body 1 due to the interaction with body 2. These forces are depicted in \nFigure 7.5. \nFigure 7.5 Interaction pair of forces \nThese two vector forces are equal in magnitude and opposite in direction, \n\uf072\n\uf072 \n= \u2212 \n(7.4.1)\nF1, 2 \nF2,1 . \nWe shall employ these definitions, Newton\u2019s three laws, and force laws to \ndescribe the motion of bodies, a subject known as classical mechanics or Newtonian \nMechanics, and hence explain a vast range of phenomena. Newtonian mechanics has \nimportant limits. It does not satisfactorily explain systems of objects moving at speeds \ncomparable to the speed of light ( v > 0.1 c ) where we need the theory of special \nrelativity, nor does it adequately explain the motion of electrons in atoms, where we need \nquantum mechanics. We also need general relativity and cosmology to explain the large-\nscale structure of the universe. \n7 Ibid p. 13. \n7-\u001b", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_148", "text": "Chapter 8 Applications of Newton\u2019s Second Law \n8.1 Force Laws 2  \n8.1.1 Hooke\u2019s Law ..................................................................................................... 2  \n8.2.2 Principle of Equivalence: ................................................................................ 6  \n8.2.3 Gravitational Force near the Surface of the Earth....................................... 7  \n8.2.4 Electric Charge and Coulomb\u2019s Law............................................................. 7  \nExample 8.1 Coulomb\u2019s Law and the Universal Law of Gravitation .................. 8  \n8.3 Constraint Forces.................................................................................................... 9", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_149", "text": "8.3.1 Contact Forces................................................................................................ 10  \nExample 8.2 Normal Component of the Contact Force and Weight ................. 11  \n8.3.2 Kinetic and Static Friction ............................................................................ 13  \n8.4 Free-body Force Diagram .................................................................................... 16  \n8.4.1 System ............................................................................................................. 16", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_150", "text": "8.4.5 Modeling ......................................................................................................... 16  \n8.5 Tension in a Rope.................................................................................................. 17  \n8.5.1 Definition of Tension in a Rope .................................................................... 17  \nExample 8.3 Tension in a Massive Rope............................................................... 19  \nExample 8.4 Tension in a Suspended Rope .......................................................... 22  \n8.5.2 Continuous Systems and Newton\u2019s Second Law as a Differential", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_151", "text": "8.5.2 Continuous Systems and Newton\u2019s Second Law as a Differential  \nEquations ................................................................................................................. 24  \n8.6 Drag Forces in Fluids ........................................................................................... 26  \nExample 8.5 Drag Force at Low Speeds ............................................................... 28  \nExample 8.6 Drag Forces at High Speeds............................................................. 31  \n8.7 Worked Examples................................................................................................. 32", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_152", "text": "Example 8.7 Staircase............................................................................................. 32  \nExample 8.8 Cart Moving on a Track................................................................... 34  \nExample 8.9 Pulleys and Ropes Constraint Conditions ...................................... 38  \nExample 8.10 Accelerating Wedge ........................................................................ 42  \nExample 8.11: Capstan........................................................................................... 45  \nExample 8.12 Free Fall with Air Drag .................................................................. 48", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_153", "text": "\u001b-1\n\nChapter 8 Applications of Newton\u2019s Second Law \nThose who are in love with practice without knowledge are like the sailor \nwho gets into a ship without rudder or compass and who never can be \ncertain whether he is going. Practice must always be founded on \nsound theory\u20261 \nLeonardo da Vinci \n8.1 Force Laws \nThere are forces that don't change appreciably from one instant to another, which we \nrefer to as constant in time, and forces that don't change appreciably from one point to \nanother, which we refer to as constant in space. The gravitational force on an object near \nthe surface of the earth is an example of a force that is constant in space. \nThere are forces that depend on the configuration of a system. When a mass is \nattached to one end of a spring, the spring force acting on the object increases in strength \nwhether the spring is extended or compressed. \nThere are forces that spread out in space such that their influence becomes less \nwith distance. Common examples are the gravitational and electrical forces. The \ngravitational force between two objects falls off as the inverse square of the distance \nseparating the objects provided the objects are of a small dimension compared to the \ndistance between them. More complicated arrangements of attracting and repelling \ninteractions give rise to forces that fall off with other powers of r : constant, 1/ r , 1/ r 2 , \n1/ r3, \u2026,. \nA force may remain constant in magnitude but change direction; for example the \ngravitational force acting on a planet undergoing circular motion about a star is directed \ntowards the center of the circle. This type of attractive central force is called a centripetal \nforce. \nA force law describes the relationship between the force and some measurable \nproperty of the objects involved. We shall see that some interactions are describable by \nforce laws and other interactions cannot be so simply described. \n8.1.1 Hooke\u2019s Law \nIn order to stretch or compress a spring from its equilibrium length, a force must be \nexerted on the spring. Consider an object of mass m that is lying on a horizontal surface. \nAttach one end of a spring to the object and fix the other end of the spring to a wall. Let \ndenote the equilibrium length of the spring (neither stretched or compressed). Assume\nl0 \n1 Notebooks of Leonardo da Vinci Complete, tr. Jean Paul Richter, 1888, Vol.1. \n\u001b-2", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_154", "text": "that the contact surface is smooth and hence frictionless in order to consider only the \neffect of the spring force. If the object is pulled to stretch the spring or pushed to \ncompress the spring, then by Newton\u2019s Third Law the force of the spring on the object is \nequal and opposite to the force that the object exerts on the spring. We shall refer to the \nforce of the spring on the object as the spring force and experimentally determine a \nrelationship between that force and the amount of stretch or compress of the spring. \nChoose a coordinate system with the origin located at the point of contact of the spring \nand the object when the spring-object system is in the equilibrium configuration. Choose \nthe \u02c6i unit vector to point in the direction the object moves when the spring is being \nstretched. Choose the coordinate function x to denote the position of the object with \nrespect to the origin (Figure 8.1). \nl0 \nx = 0 \n\u02c6i\nfrictionless \nwall \nm \nequilibrium configuration +\nsurface \nx \nl0 \n\u02c6i\nm \nx = 0 \nx \n\u02c6i\nstretched: x > 0 \ncompressed: x < 0 \nx = 0 \nFigure 8.1 Spring attached to a wall and an object \nInitially stretch the spring until the object is at position x . Then release the object \nand measure the acceleration of the object the instant the object is released. The\n\uf072 \n\uf072\nF = m a\nmagnitude of the spring force acting on the object is \n. Now repeat the\nexperiment for a range of stretches (or compressions). Experiments show that for each \nspring, there is a range of maximum values x max > 0 for stretching and minimum values \n< 0 for compressing such that the magnitude of the measured force is proportional to \nxmin \nthe stretched or compressed length and is given by the formula \nm \n! \nF = k x , \n(8.1.1) \n\u001b-3", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_155", "text": "where the spring constant k has units N \u22c5 m\u22121 . The free-body force diagram is shown in \nFigure 8.2. \n\u02c6i \nx \nF = F x \u02c6i = kx\u02c6i \nx = 0 \nFigure 8.2 Spring force acting on object \nThe constant k is equal to the negative of the slope of the graph of the force vs. the \ncompression or stretch (Figure 8.3). \nF \nslope = -k \nx \nx max \nxmin \nx \n. \nFigure 8.3 Plot of x -component of the spring force Fx vs. x \nThe direction of the acceleration is always towards the equilibrium position whether the \nspring is stretched or compressed. This type of force is called a restoring force. Let Fx \ndenote the x -component of the spring force. Then \nFx = \u2212kx . \n(8.1.2) \nNow perform similar experiments on other springs. For a range of stretched \nlengths, each spring exhibits the same proportionality between force and stretched length, \nalthough the spring constant may differ for each spring. \nIt would be extremely impractical to experimentally determine whether this \nproportionality holds for all springs, and because a modest sampling of springs has \nconfirmed the relation, we shall infer that all ideal springs will produce a restoring force, \nwhich is linearly proportional to the stretched (or compressed) length. This experimental \nrelation regarding force and stretched (or compressed) lengths for a finite set of springs \nhas now been inductively generalized into the above mathematical model for ideal springs, \na force law known as a Hooke\u2019s Law. \nThis inductive step, referred to as Newtonian induction, is the critical step that \nmakes physics a predictive science. Suppose a spring, attached to an object of mass m, is \n8-4", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_156", "text": "stretched by an amount \u0394x . Use the force law to predict the magnitude of the force \n\uf072 \nF = k \u0394x\nbetween the rubber band and the object, \n, without having to experimentally \n\u23a7\u23aa\u23a8\n\u23aa\u23a9  \nmeasure the acceleration. Now use Newton\u2019s Second Law to predict the magnitude of the \nacceleration of the object \n\uf072 \nF \nk \u0394x\n\uf072 a =\n= \n. \n(8.1.3)\nm \nm \nCarry out the experiment, and measure the acceleration within some error bounds. \nIf the magnitude of the predicted acceleration disagrees with the measured result, then the \nmodel for the force law needs modification. The ability to adjust, correct or even reject \nmodels based on new experimental results enables a description of forces between objects \nto cover larger and larger experimental domains. \nMany real springs have been wound such that a force of magnitude F0 must be applied \nbefore the spring begins to stretch. The value of F0 is referred to as the pre-tension of the \nspring. Under these circumstances, Hooke\u2019s law must be modified to account for this \npretension, \nF \n\u2212 kx, x > 0 \nx = \u2212F0 \n. \n(8.1.4)\nFx = +F1 \u2212 kx, x < 0 \nNote the value of the pre-tension F0 and F1 may differ for compressing or stretching a \nspring. \n8.2 Fundamental Laws of Nature \nForce laws are mathematical models of physical processes. They arise from observation \nand experimentation, and they have limited ranges of applicability. Does the linear force \nlaw for the spring hold for all springs? Each spring will most likely have a different range \nof linear behavior. So the model for stretching springs still lacks a universal character. As \nsuch, there should be some hesitation to generalize this observation to all springs unless \nsome property of the spring, universal to all springs, is responsible for the force law. \nPerhaps springs are made up of very small components, which when pulled apart \ntend to contract back together. This would suggest that there is some type of force that \ncontracts spring molecules when they are pulled apart. What holds molecules together? \nCan we find some fundamental property of the interaction between atoms that will suffice \nto explain the macroscopic force law? This search for fundamental forces is a central \ntask of physics. \nIn the case of springs, this could lead into an investigation of the composition and \nstructural properties of the atoms that compose the steel in the spring. We would \ninvestigate the geometric properties of the lattice of atoms and determine whether there is \nsome fundamental property of the atoms that create this lattice. Then we ask how stable is \n8-5", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_157", "text": "this lattice under deformations. This may lead to an investigation into the electron \nconfigurations associated with each atom and how they overlap to form bonds between \natoms. These particles carry charges, which obey Coulomb\u2019s Law, but also the Laws of \nQuantum Mechanics. So in order to arrive at a satisfactory explanation of the elastic \nrestoring properties of the spring, we need models that describe the fundamental physics \nthat underline Hooke\u2019s Law. \n8.2.1 Universal Law of Gravitation \nAt points significantly far away from the surface of Earth, the gravitational force is no \nlonger constant with respect to the distance to the center of Earth. Newton\u2019s Universal \nLaw of Gravitation describes the gravitational force between two objects with masses, \nand m2 . This force points along the line connecting the objects, is attractive, and its \nm1 \nmagnitude is proportional to the inverse square of the distance, r1,2 , between the two \npoint-like objects (Figure 8.4a). The force on object 2 due to the gravitational interaction \nbetween the two objects is given by \n\uf072 G\nm1 m2\n= \u2212G \n\u02c6 \n(8.2.1)\nF1,2 \n2 r1,2 , \nr1,2 \n\uf072  \n1,2 \n2\n1 \n\uf072 \nis a unit vector directed from object 1 to object 2 (Figure 8.4b). The\n\u02c6 = \uf072 /\nr1,2 \nr1,2 \nconstant of proportionality in SI units is G = 6.67 \u00d7 10\u221211N \u22c5 m2 \u22c5 kg\u22122 . \n\uf072r  \n\uf072r\nr1,2  \n\uf072r \nwhere \n\u2212  \nis a vector directed from object 1 to object 2, \nand\n=  \n,\nr1,2 = r1,2 \nFigure 8.4 (a) Gravitational force between two point-like objects. Figure 8.4 (b) \nCoordinate system for the two-body problem. \n8.2.2 Principle of Equivalence: \nThe Principle of Equivalence states that the mass that appears in the Universal Law of \nGravity is identical to the inertial mass that is determined with respect to the standard \n8-6", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_158", "text": "kilogram. From this point on, the equivalence of inertial and gravitational mass will be \nassumed and the mass will be denoted by the symbol m. \n8.2.3 Gravitational Force near the Surface of the Earth \nNear the surface of Earth, the gravitational interaction between an object and Earth is \nmutually attractive and has a magnitude of \n\uf072 \nFG \n= mg \n(8.2.2)\nearth,object \nwhere g is a positive constant. \nThe International Committee on Weights and Measures has adopted as a standard \nvalue for the acceleration of an object freely falling in a vacuum g = 9.80665 m \u22c5 s\u22122 . The \nactual value of g varies as a function of elevation and latitude. If \u03c6 is the latitude and h \nthe elevation in meters then the acceleration of gravity in SI units is \ng = (9.80616\u2212 0.025928cos(2\u03c6) + 0.000069cos2(2\u03c6) \u2212 3.086 \u00d710\u22124 h) m \u22c5s\u22122 . \n(8.2.3) \nThis is known as Helmert\u2019s equation. The strength of the gravitational force on the \nstandard kilogram at 42\uf06f latitude is 9.80345 N \u22c5 kg\u22121, and the acceleration due to gravity \nat sea level is therefore g = 9.80345 m \u22c5 s\u22122 for all objects. \nAt the equator, \ng = 9.78 m \u22c5 s\u22122 and at the poles g = 9.83 m \u22c5 s\u22122 . This difference is primarily due to the \nearth\u2019s rotation, which introduces an apparent (fictitious) repulsive force that affects the \ndetermination of g as given in Equation (8.2.2) and also flattens the spherical shape of \nEarth (the distance from the center of Earth is larger at the equator than it is at the poles \nby about 26.5 km ). Both the magnitude and the direction of the gravitational force also \nshow variations that depend on local features to an extent that's useful in prospecting for \noil, investigating the water table, navigating submerged submarines, and as well as many \nother practical uses. Such variations in g can be measured with a sensitive spring \nbalance. Local variations have been much studied over the past two decades in attempts \nto discover a proposed \u201cfifth force\u201d which would fall off faster than the gravitational \nforce that falls off as the inverse square of the distance between the objects. \n8.2.4 Electric Charge and Coulomb\u2019s Law", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_159", "text": "balance. Local variations have been much studied over the past two decades in attempts \nto discover a proposed \u201cfifth force\u201d which would fall off faster than the gravitational \nforce that falls off as the inverse square of the distance between the objects. \n8.2.4 Electric Charge and Coulomb\u2019s Law \nMatter has properties other than mass. Matter can also carry one of two types of observed \nelectric charge, positive and negative. Like charges repel, and opposite charges attract \neach other. The unit of charge in the SI system of units is called the coulomb [C]. \nThe smallest unit of \u201cfree\u201d charge known in nature is the charge of an electron or \nproton, which has a magnitude of \ne = 1.602 \u00d7 10\u221219 C . \n(8.2.4) \n8-7", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_160", "text": "It has been shown experimentally that charge carried by ordinary objects is quantized in \nintegral multiples of the magnitude of this free charge. The electron carries one unit of \nnegative charge ( qe = \u2212e ) and the proton carries one unit of positive charge ( qp = +e ). In \nan isolated system, the charge stays constant; in a closed system, an amount of \nunbalanced charge can neither be created nor destroyed. Charge can only be transferred \nfrom one object to another. \nConsider two point-like objects with charges q1 and q2 , separated by a distance \nin vacuum. By experimental observation, the two objects repel each other if they are \nr1, 2 \nboth positively or negatively charged (Figure 8.4a). They attract each other if they are \noppositely charged (Figure 8.5b). The force exerted on object 2 due to the interaction \nbetween objects 1 and 2 is given by Coulomb's Law, \nF\n\uf072 E \nq1 q2\n= k \n\u02c6 \n(8.2.5)\n1, 2 \ne \n2 r1, 2 \nr1, 2 \n\uf072\nwhere \u02c6 = \uf072 / \nis a unit vector directed from object 1 to object 2, and in SI units, \nr1,2 \nr1,2 \nr1,2 \n2 \u22c5 C\u22122\nke = 8.9875 \u00d7 109 N \u22c5 m\n, as illustrated in the Figure 8.5a. This law was derived \nempirically by Charles Augustin de Coulomb in the late 18th century. \nFigure 8.5 (a) and 8.5 (b) Coulomb interaction between two charges \nExample 8.1 Coulomb\u2019s Law and the Universal Law of Gravitation \nShow that both Coulomb\u2019s Law and the Universal Law of Gravitation satisfy Newton\u2019s \nThird Law. \nSolution: To see this, interchange 1 and 2 in the Universal Law of Gravitation to find the \nforce on object 1 due to the interaction between the objects. The only quantity to change \nsign is the unit vector \n8-8", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_161", "text": ".\nmany examples in which the constraint forces cannot be determined. \ninclined plane \n(a) \n(b) ' \nbead \nrotating rod \n\u02c6 = \u2212 \u02c6 \n(8.2.6)\nr2,1 \nr1, 2 . \nThen \nF\n\uf072 G \nm2 m1 \nm1 m2 \n= \u2212F\n\uf072 G\n= \u2212G \n\u02c6 = G \n\u02c6 \n(8.2.7)\n2,1 \n2 r2,1 \n2 r1,2 \n1,2 . \nr2,1 \nr1, 2 \nCoulomb\u2019s Law also satisfies Newton\u2019s Third Law since the only quantity to change sign \nis the unit vector, just as in the case of the Universal Law of Gravitation. \n8.3 Constraint Forces \nKnowledge of all the external and internal forces acting on each of the objects in a system \nand applying Newton\u2019s Second Law to each of the objects determine a set of equations of \nmotion. These equations of motion are not necessarily independent due to the fact that the \nmotion of the objects may be limited by equations of constraint. In addition there are \nforces of constraint that are determined by their effect on the motion of the objects and \nare not known beforehand or describable by some force law. For example: an object \nsliding down an inclined plane is constrained to move along the surface of the inclined \nplane (Figure 8.6a) and the surface exerts a contact force on the object; an object that \nslides down the surface of a sphere until it falls off experiences a contact force until it \nloses contact with the surface (Figure 8.6b); gas particles in a sealed vessel are \nconstrained to remain inside the vessel and therefore the wall must exert force on the gas \nmolecules to keep them inside the vessel (8.6c); and a bead constrained to slide outward \nalong a rotating rod is acting on by time dependent forces of the rod on the bead (Figure \n8.6d). We shall develop methods to determine these constraint forces although there are \n(c) \n(d) '  \n8-9", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_162", "text": "Figure 8.6 Constrained motions: (a) particle sliding down inclined plane, (b) particles \nsliding down surface of sphere, (c) gas molecules in a sealed vessel, and (d) bead sliding \non a rotating rod \n8.3.1 Contact Forces \nPushing, lifting and pulling are contact forces that we experience in the everyday world. \nRest your hand on a table; the atoms that form the molecules that make up the table and \nyour hand are in contact with each other. If you press harder, the atoms are also pressed \ncloser together. The electrons in the atoms begin to repel each other and your hand is \npushed in the opposite direction by the table. \nAccording to Newton\u2019s Third Law, the force of your hand on the table is equal in \nmagnitude and opposite in direction to the force of the table on your hand. Clearly, if you \npush harder the force increases. Try it! If you push your hand straight down on the table, \nthe table pushes back in a direction perpendicular (normal) to the surface. Slide your \nhand gently forward along the surface of the table. You barely feel the table pushing \nupward, but you do feel the friction acting as a resistive force to the motion of your hand. \nThis force acts tangential to the surface and opposite to the motion of your hand. Push \ndownward and forward. Try to estimate the magnitude of the force acting on your hand. \n!\n! \nThe force of the table acting on your hand, FC \u2261 C , is called the contact force. \n!\n! \nThis force has both a normal component to the surface, C\u22a5\u2261 N , called the normal force, \n!\n! \nand a tangential component to the surface, C \u2261 f , called the friction force (Figure 8.6).\n\" \nC \nN \nC \nf \nC \nFigure 8.6 Normal and tangential components of the contact force \nThe contact force, written in terms of its component forces, is therefore \n! \nC =  \n! \nC \u22a5 +  \n! \nC \" \u2261  \n! \nN +  \n!\nf .  \n(8.3.1)  \n\uf072 \nAny force can be decomposed into component vectors so the normal component, N , and \n! \nthe tangential component, f , are not independent forces but the vector components of the \ncontact force, perpendicular and parallel to the surface of contact. The contact force is a \ndistributed force acting over all the points of contact between your hand and the surface.  \n8-10", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_163", "text": "For most applications we shall treat the contact force as acting at single point but \nprecaution must be taken when the distributed nature of the contact force plays a key role \nin constraining the motion of a rigid body. \nIn Figure 8.7, the forces acting on your hand are shown. These forces include the \n\uf072\n! \ncontact force, C , of the table acting on your hand, the force of your forearm, Fforearm , \nacting on your hand (which is drawn at an angle indicating that you are pushing down on \n! \nyour hand as well as forward), and the gravitational interaction, Fg , between the earth \nand your hand. \nC \nFg\nFforearm \nFigure 8.7 Forces on hand when moving towards the left \nOne point to keep in mind is that the magnitudes of the two components of the contact \nforce depend on how hard you push or pull your hand and in what direction, a \ncharacteristic of constraint forces, in which the components are not specified by a force \nlaw but dependent on the particular motion of the hand. \nExample 8.2 Normal Component of the Contact Force and Weight \nHold a block in your hand such that your hand is at rest (Figure 8.8). You can feel the \n\u201cweight\u201d of the block against your palm. But what exactly do we mean by \u201cweight\u201d? \nFg = mg\n.\nN\nFigure 8.8 Block resting in hand \nFigure 8.9 Forces on block \n8-11", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_164", "text": "There are two forces acting on the block as shown in Figure 8.9. One force is the \n!\n!\ngravitational force between the earth and the block, and is denoted by Fg = mg . The \nother force acting on the block is the contact force between your hand and the block. \nBecause our hand is at rest, this contact force on the block points perpendicular to the \n! \nsurface, and hence has only a normal component, N . Let N denote the magnitude of the \nnormal force. Because the object is at rest in your hand, the vertical acceleration is zero. \nTherefore Newton\u2019s Second Law states that \n! \nN +  \n! \nF g =  \n!\n0 .  \n(8.3.2)  \nChoose the positive direction to be upwards and then in terms of vertical components we \nhave that \nN \u2212 mg = 0 . \n(8.3.3) \nwhich can be solved for the magnitude of the normal force \nN = mg . \n(8.3.4) \nWhen we talk about the \u201cweight\u201d of the block, we often are referring to the effect \nthe block has on a scale or on the feeling we have when we hold the block. These effects \nare actually effects of the normal force. We say that a block \u201cfeels lighter\u201d if there is an \nadditional force holding the block up. For example, you can rest the block in your hand, \nbut use your other hand to apply a force upwards on the block to make it feel lighter in \nyour supporting hand. \nThe word \u201cweight,\u201d is often used to describe the gravitational force that Earth \nexerts on an object. We shall always refer to this force as the gravitational force instead \nof \u201cweight.\u201d When you jump in the air, you feel \u201cweightless\u201d because there is no normal \nforce acting on you, even though Earth is still exerting a gravitational force on you; \nclearly, when you jump, you do not turn gravity off! \nThis example may also give rise to a misconception that the normal force is \nalways equal to the mass of the object times the magnitude of the gravitational \nacceleration at the surface of the earth. The normal force and the gravitational force are \ntwo completely different forces. In this particular example, the normal force is equal in \nmagnitude to the gravitational force and directed in the opposite direction because the \nobject is at rest. The normal force and the gravitational force do not form a Third Law \ninteraction pair of forces. In this example, our system is just the block and the normal \nforce and gravitational force are external forces acting on the block.", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_165", "text": "magnitude to the gravitational force and directed in the opposite direction because the \nobject is at rest. The normal force and the gravitational force do not form a Third Law \ninteraction pair of forces. In this example, our system is just the block and the normal \nforce and gravitational force are external forces acting on the block. \nLet\u2019s redefine our system as the block, your hand, and Earth. Then the normal \nforce and gravitational force are now internal forces in the system and we can now \nidentify the various interaction pairs of forces. We explicitly introduce our interaction \n! \npair notation to enable us to identify these interaction pairs: for example, let FE\ng \n,B denote \nthe gravitational force on the block due to the interaction with Earth. The gravitational \n8-12", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_166", "text": "! \nforce on Earth due to the interaction with the block is denoted by FB\ng \n,E , and these two \n!\n! \nforces form an interaction pair. By Newton\u2019s Third Law, FE\ng \n,B = \u2212FB\ng \n,E . Note that these \ntwo forces are acting on different objects, the block and Earth. The contact force on the \n! \nblock due to the interaction between the hand and the block is then denoted by N H ,B . \n! \nN B H\n, \n! \n!  \nThe force of the block on the hand, which we denote by N B,H , satisfies \n= \u2212N H ,B . \nBecause we are including your hand as part of the system, there are two additional forces \n! \nacting on the hand. There is the gravitational force on your hand FE\ng \n,H , satisfying \n!\n! \n! \ng\ng \ng\nFE,H = \u2212FH ,E , where FH ,E is the gravitational force on Earth due to your hand. Finally \n! \nthere is the force of your forearm holding your hand up, which we denote FF ,H . Because \nwe are not including the forearm in our system, this force is an external force to the \nsystem. The forces acting on your hand are shown in diagram on your hand is shown in \nFigure 8.10, and the just the interaction pairing of forces acting on Earth is shown in \nFigure 8.11 (we are not representing all other external forces acting on the Earth). \nFF ,H\nN B,H\nFE,H\ng\nFH\ng \n,E \nFB\ng \n,E \nFigure 8.10 Free-body force diagram on \nFigure 8.11 Gravitational forces on\nhand \nearth due to object and hand \n8.3.2 Kinetic and Static Friction \nWhen a block is pulled along a horizontal surface or sliding down an inclined plane there \nis a lateral force resisting the motion. If the block is at rest on the inclined plane, there is \nstill a lateral force resisting the motion. This resistive force is known as dry friction, and  \nthere are two distinguishing types when surfaces are in contact with each other. The first \ntype is when the two objects are moving relative to each other; the friction in that case is \ncalled kinetic friction or sliding friction. When the two surfaces are non-moving but \nthere is still a lateral force as in the example of the block at rest on an inclined plane, the \nforce is called, static friction. \n8-13", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_167", "text": "Leonardo da Vinci was the first to record the results of measurements on kinetic \nfriction over a twenty-year period between 1493\u20134 and about 1515. Based on his \n\uf072 \nmeasurements, the force of kinetic friction, f k , between two surfaces, he identified two \nkey properties of kinetic friction. The magnitude of kinetic friction is proportional to the \nnormal force between the two surfaces, \nN , \n(8.3.5)\nfk = \u00b5k \nwhere \u00b5k is called the coefficient of kinetic friction. The second result is rather \nsurprising in that the magnitude of the force is independent of the contact surface. \nConsider two blocks of the same mass, but different surface areas. The force necessary to \nmove the blocks at a constant speed is the same. The block in Figure 8.12a has twice the \ncontact area as the block shown in Figure 8.12b, but when the same external force is \napplied to either block, the blocks move at constant speed. These results of da Vinci were \nrediscovered by Guillaume Amontons and published in 1699. The third property that \nkinetic friction is independent of the speed of moving objects (for ordinary sliding \nspeeds) was discovered by Charles Augustin Coulomb. \nF\nF \nf k\nf k \n(a) \n(b) \nFigure 8.12 (a) and (b): kinetic friction is independent of the contact area \n! \nThe kinetic friction on surface 2 moving relative to surface 1 is denoted by, \nk . The\nf1,2 \ndirection of the force is always opposed to the relative direction of motion of surface 2 \nrelative to the surface 1. When one surface is at rest relative to our choice of reference \n! \nframe we will denote the friction force on the moving object by f k . \nThe second type of dry friction, static friction occurs when two surfaces are static \nrelative to each other. Because the static friction force between two surfaces forms a third \n! \nlaw interaction pair, will use the notation \ns to denote the static friction force on surface \nf1,2 \n2 due to the interaction between surfaces 1 and 2. Push your hand forward along a \nsurface; as you increase your pushing force, the frictional force feels stronger and \nstronger. Try this! Your hand will at first stick until you push hard enough, then your \nhand slides forward. The magnitude of the static frictional force, f s , depends on how \nhard you push. \nIf you rest your hand on a table without pushing horizontally, the static friction is \nzero. As you increase your push, the static friction increases until you push hard enough", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_168", "text": "hand slides forward. The magnitude of the static frictional force, f s , depends on how \nhard you push. \nIf you rest your hand on a table without pushing horizontally, the static friction is \nzero. As you increase your push, the static friction increases until you push hard enough \nthat your hand slips and starts to slide along the surface. Thus the magnitude of static \n8-14", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_169", "text": "friction can vary from zero to some maximum value, ( f ) \n, when the pushed object \ns max \nbegins to slip, \n0 \u2264 f \u2264 ( f ) \n. \n(8.3.6)\ns \ns max \nIs there a mathematical model for the magnitude of the maximum value of static \nfriction between two surfaces? Through experimentation, we find that this magnitude is, \nlike kinetic friction, proportional to the magnitude of the normal force \n( f ) \n= \u00b5 N . \n(8.3.7)\ns max \ns \nHere the constant of proportionality is \u00b5s , the coefficient of static friction. This constant \nis slightly greater than the constant \u00b5k associated with kinetic friction, \u00b5s > \u00b5k . This \nsmall difference accounts for the slipping and catching of chalk on a blackboard, \nfingernails on glass, or a violin bow on a string. \nThe direction of static friction on an object is always opposed to the direction of \nthe applied force (as long as the two surfaces are not accelerating). In Figure 8.13a, an \n!\n\uf072 \nexternal force, F , is applied the left and the static friction, f s , is shown pointing to the \n! \nright opposing the external force. In Figure 8.13b, the external force, F , is directed to the \n! \nright and the static friction, f s , is now pointing to the left. \nf s \nf s \nF\nF \n(a) \n(b) \nFigure 8.13 (a) and (b): External forces and the direction of static friction. \nAlthough the force law for the maximum magnitude of static friction resembles \nthe force law for sliding friction, there are important differences: \n1. The direction and magnitude of static friction on an object always depends on the \ndirection and magnitude of the applied forces acting on the object, where the magnitude \nof kinetic friction for a sliding object is fixed. \n2. The magnitude of static friction has a maximum possible value. If the magnitude of the \napplied force along the direction of the contact surface exceeds the magnitude of the \nmaximum value of static friction, then the object will start to slip (and be subject to \nkinetic friction.) We call this the just slipping condition. \n8-15", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_170", "text": "8.4 Free-body Force Diagram \n8.4.1 System \nWhen we try to describe forces acting on a collection of objects we must first take care to \nspecifically define the collection of objects that we are interested in, which define our \nsystem. Often the system is a single isolated object but it can consist of multiple objects. \nBecause force is a vector, the force acting on the system is a vector sum of the individual \n\uf072 \nF\n1\n2 \n\uf072 \nF \nforces acting on the system\n\uf072 \nF \n+\u22c5\u22c5\u22c5 \n(8.4.1) \n=  +  \nA free-body force diagram is a representation of the sum of all the forces that act \non a single system. We denote the system by a large circular dot, a \u201cpoint\u201d. (Later on in \nthe course we shall see that the \u201cpoint\u201d represents the center of mass of the system.) We \nrepresent each force that acts on the system by an arrow (indicating the direction of that \nforce). We draw the arrow at the \u201cpoint\u201d representing the system. For example, the forces \nthat regularly appear in free-body diagram are contact forces, tension, gravitation, \nfriction, pressure forces, spring forces, electric and magnetic forces, which we shall \nintroduce below. Sometimes we will draw the arrow representing the actual point in the \nsystem where the force is acting. When we do that, we will not represent the system by a \n\u201cpoint\u201d in the free-body diagram. \nSuppose we choose a Cartesian coordinate system, then we can resolve the force \ninto its component vectors \n\uf072 \nF = F \u02c6i + F \u02c6j+ F k\u02c6\n(8.4.2)\nx \ny \nz \nEach one of the component vectors is itself a vector sum of the individual \ncomponent vectors from each contributing force. We can use the free-body force diagram \nto make these vector decompositions of the individual forces. For example, the x \u00ad\ncomponent of the force is \nFx = F1,x + F2,x + \u22c5\u22c5\u22c5. \n(8.4.3) \n8.4.5 Modeling \nOne of the most central and yet most difficult tasks in analyzing a physical interaction is \ndeveloping a physical model. A physical model for the interaction consists of a \ndescription of the forces acting on all the objects. The difficulty arises in deciding which \nforces to include. For example in describing almost all planetary motions, the Universal \nLaw of Gravitation was the only force law that was needed. There were anomalies, for \nexample the small shift in Mercury\u2019s orbit. These anomalies are interesting because they", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_171", "text": "forces to include. For example in describing almost all planetary motions, the Universal \nLaw of Gravitation was the only force law that was needed. There were anomalies, for \nexample the small shift in Mercury\u2019s orbit. These anomalies are interesting because they \nmay lead to new physics. Einstein corrected Newton\u2019s Law of Gravitation by introducing \nGeneral Relativity and one of the first successful predictions of the new theory was the \nperihelion precession of Mercury\u2019s orbit. On the other hand, the anomalies may simply \n\u001b-16", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_172", "text": "be due to the complications introduced by forces that are well understood but \ncomplicated to model. When objects are in motion there is always some type of friction \npresent. Air friction is often neglected because the mathematical models for air resistance \nare fairly complicated even though the force of air resistance substantially changes the \nmotion. Static or kinetic friction between surfaces is sometimes ignored but not always. \nThe mathematical description of the friction between surfaces has a simple expression so \nit can be included without making the description mathematically intractable. A good \nway to start thinking about the problem is to make a simple model, excluding \ncomplications that are small order effects. Then we can check the predictions of the \nmodel. Once we are satisfied that we are on the right track, we can include more \ncomplicated effects. \n8.5 Tension in a Rope \n8.5.1 Definition of Tension in a Rope \nLet\u2019s return to our example of the very light rope (object 2 with m2 ! 0 ) that is attached \n! \nto a block (object 1) at the point B , and pulled by an applied force at point A , FA,2 \n(Figure 8.18a). \nB \n1 . \n2 \nA FA,2 \n. \nFigure 8.18a Massless rope pulling a block \nChoose a coordinate system with the \u02c6j -unit vector pointing upward in the normal \ndirection to the surface, and the \u02c6i -unit vector pointing in the positive x -direction, \n(Figure 8.18b). The force diagrams for the system consisting of the rope and block is \n! \nshown in Figure 8.19, and for the rope and block separately in Figure 8.20, where F2,1 is \n! \nthe force on the block (object 1) due to the rope (object 2), and F1,2 is the force on the \nrope due to the block. \nN \n\u02c6j \n\u02c6i \nB \n. .\n1\n2 . \nA FA,2 \n.\nf \nm2g\n0 \n+x \nm1g \nFigure 8.18b Forces acting on system consisting of block and rope \n8-17", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_173", "text": "The forces on the rope and the block must each sum to zero. Because the rope is not \naccelerating, Newton\u2019s Second Law applied to the rope requires that FA,2 \u2212 F1,2 = m2a \n(where we are using magnitudes for all the forces). \n\u02c6j \n\u02c6i\nN \n.\n2\nA FA,2 \n..\n1\nF2,1 \nF1,2 \nB . \n. \nf  \nm2g\n0 \nm1g \nFigure 8.19 Separate force diagrams for rope and block \nBecause we are assuming the mass of the rope is negligible therefore \n= 0; (massless rope) \n(8.5.1)\nFA,2 \u2212 F1,2 \n. \nIf we consider the case that the rope is very light, then the forces acting at the ends of the \nrope are nearly horizontal. Then if the rope-block system is moving at constant speed or \nat rest, Newton\u2019s Second Law is now \n= 0; (constant speed or at rest) . \n(8.5.2)\nFA,2 \u2212 F1,2 \nNewton\u2019s Second Law applied to the block in the +\u02c6i -direction requires that F2,1 \u2212 f = 0. \nNewton\u2019s Third Law, applied to the block-rope interaction pair requires that F1,2 = F2,1 . \nTherefore \nFA,2 = F1,2 = F2,1 = f . \n(8.5.3) \nThus the applied pulling force is transmitted through the rope to the block since it has the \nsame magnitude as the force of the rope on the block. In addition, the applied pulling \nforce is also equal to the friction force on the block. \nHow do we define \u201ctension\u201d at some point in a rope? Suppose make an imaginary slice of \nthe rope at a point P , a distance xP from point B , where the rope is attached to the \nblock. The imaginary slice divides the rope into two sections, labeled L (left) and R \n(right), as shown in Figure 8.20. \nA\nB\nP \nL \nR\n. \n. \n. \nimaginary slice\nxP \nFigure 8.20 Imaginary slice through the rope \n8-18", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_174", "text": "There is now a Third Law pair of forces acting between the left and right sections of the \n! \nrope. Denote the force acting on the left section by FR,L (xP ) , and the force acting on the \n! \nright section by \n) . Newton\u2019s Third Law requires that the forces in this interaction\nFL,R (xP \npair are equal in magnitude and opposite in direction. \n! \n! \nFR,L (xP ) = \u2212FL,R (xP ) \n(8.5.4) \n! \nThe force diagram for the left and right sections are shown in Figure 8.21 where \nis\nF1,L \nthe force on the left section of the rope due to the block-rope interaction. (We had \n! \npreviously denoted that force by \n). Now denote the force on the right section of the \nF1,2 \n! \nrope side due to the pulling force at the point A by FA,R , (which we had previously \n! \ndenoted by \n).\nFA,2 \nL \nR\n. \n. \n. \nA\nP\nB \n. \nP \nxP \nFR,L (xP ) \nFL,R (xP )\nF1,L \nFA,R \nFigure 8.21 Force diagram for the left and right sections of rope \nThe tension T (xP ) at a point P in rope lying a distance x from one the left end of the \nrope, is the magnitude of the action -reaction pair of forces acting at the point P , \n! \n! \n. \n(8.5.5)\nT (xP ) = \n=\nFR,L (xP ) \nFL,R (xP ) \nFor a rope of negligible mass, under tension, as in the above case, (even if the rope is \naccelerating) the sum of the horizontal forces applied to the left section and the right \nsection of the rope are zero, and therefore the tension is uniform and is equal to the \napplied pulling force, \nT \n(8.5.6)\n= FA,R . \nExample 8.3 Tension in a Massive Rope \n. \nA \n. \nB \n1 \n2 \nFA,R \nFigure 8.22a Massive rope pulling a block \n8-19", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_175", "text": "Consider a block of mass m1 that is lying on a horizontal surface. The coefficient of \nkinetic friction between the block and the surface is \u00b5k . A uniform rope of mass m2 and \nlength d is attached to the block. The rope is pulled from the side opposite the block \n! \nwith an applied force of magnitude \n= FA,2 . Because the rope is now massive, the \nFA,2 \npulling force makes an angle \u03c6 with respect to the horizontal in order to balance the \ngravitational force on the rope, (Figure 8.22a). Determine the tension in the rope as a \nfunction of distance x from the block. \nSolution: In the following analysis, we shall assume that the angle \u03c6 is very small and \ndepict the pulling and tension forces as essentially acting in the horizontal direction even \nthough there must be some small vertical component to balance the gravitational forces. \nThe key point to realize is that the rope is now massive and we must take in to account \nthe inertia of the rope when applying Newton\u2019s Second Law. Consider an imaginary slice \nthrough the rope at a distance x from the block (Figure 8.22b), dividing the rope into two \nsections. The right section has length d \u2212 x and mass \n= (m2 / d)(d \u2212 x) . The left \nmR \nsection has length x and mass mL = (m2 / d)(x) . \nL \nR\n. \n. A\nB \nimaginary slice \nd x\nx \nO \n+ x \nFigure 8.22b Imaginary slice through the rope \nThe free body force diagrams for the two sections of the rope are shown in Figure 8.22c, \nwhere T (x) is the tension in the rope at a distance x from the block, and \nF1,L = \n! \nF1,L \u2261 \n! \nF1,2 is the magnitude of the force on the left-section of the rope due to the \nrope-block interaction. \nL \nR\n. \n. \n. \nA\nB \n. \nx \nT (x) \n\u02c6i\n\u02c6j \nFA, \nT (x)\nF1,L \n.\n. \nmR g\nmL g \nR \nFigure 8.22c Force diagram for the left and right sections of rope \nApply Newton\u2019s Second Law to the right section of the rope yielding \n8-20", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_176", "text": "m\n\u2212 T (x) = mR \n= \n2 (d \u2212 x)aR , \n(8.5.7)\nFA,R \naR \nd \nwhere aR is the x -component of the acceleration of the right section of the rope. Apply \nNewton\u2019s Second Law to the left slice of the rope yielding \n= (m2 / d)x aL , \n(8.5.8)\nT (x) \u2212 F1,L = mLaL \nwhere aL is the x -component of the acceleration of the left piece of the rope. \nN \n\u02c6j \n\u02c6i\nFL,1 \n1.\nfk \nm1g \nFigure 8.23 Force diagram on sliding block \nThe force diagram on the block is shown in Figure 8.23. Newton\u2019s Second Law on the \nblock in the +\u02c6i -direction is FL,1 \u2212 fk = m1a1 and in the +\u02c6j -direction is N \u2212 m1g = 0 . The \nkinetic friction force acting on the block is fk = \u00b5kN = \u00b5km1g . Newton\u2019s Second Law on \nthe block in the +\u02c6i -direction becomes \n\u2212 \u00b5km1g = m1a1 , \n(8.5.9)\nFL,1 \nNewton\u2019s Third Law for the block-rope interaction is given by FL,1 = F1,L . Eq. (8.5.8) \nthen becomes \nT (x) \u2212 (\u00b5km1g + m1a1) = (m2 / d)xaL . \n(8.5.10) \nBecause the rope and block move together, the accelerations are equal which we denote \nby the symbol a \u2261 a1 = aL . Then Eq. (8.5.10) becomes \nT (x) = \u00b5km1g + (m1 + (m2 / d)x)a . \n(8.5.11) \nThis result is not unexpected because the tension is accelerating both the block and the \nleft section and is opposed by the frictional force. \nAlternatively, the force diagram on the system consisting of the rope and block is shown \nin Figure 8.24. \n8-21", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_177", "text": ". \nA\nB \n2 \n1 \nm1g \nN \nfk \n. \n.\n. \nm2 g \nFA, R \nFigure 8.24 Force diagram on block-rope system \nNewton\u2019s Second Law becomes \nFA,R \u2212 \u00b5k m1g = (m2 + m1)a \n(8.5.12) \nSolve Eq. (8.5.12) for FA,R and substitute into Eq. (8.5.7), and solve for the tension \nyielding Eq. (8.5.11). \nExample 8.4 Tension in a Suspended Rope \nA uniform rope of mass M and length L is suspended from a ceiling (Figure 8.25). The \nmagnitude of the acceleration due to gravity is g . (a) Find the tension in the rope at the \nupper end where the rope is fixed to the ceiling. (b) Find the tension in the rope as a \nfunction of the distance from the ceiling. (c) Find an equation for the rate of change of the \ntension with respect to distance from the ceiling in terms of M , L , and g . \ng \nL \nrope of \nmass M \n\u02c6j \n+ y \nFigure 8.25 Rope suspended from ceiling Figure 8.26 Coordinate system for \nsuspended rope \nSolution: (a) Begin by choosing a coordinate system with the origin at the ceiling and the \npositive y -direction pointing downward (Figure 8.26). In order to find the tension at the \nupper end of the rope, choose as a system the entire rope. The forces acting on the rope \nare the force at y = 0 holding the rope up, T ( y = 0) , and the gravitational force on the \nentire rope. The free-body force diagram is shown in Figure 8.27. \n8-22", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_178", "text": "T ( y = 0) \n\u02c6j \nMg \nFigure 8.27 Force diagram on rope \nBecause the acceleration is zero, Newton\u2019s Second Law on the rope is Mg \u2212 T ( y = 0) = 0 . \nTherefore the tension at the upper end is T ( y = 0) = Mg . \n(b) Recall that the tension at a point is the magnitude of the action-reaction pair of forces \nacting at that point. Make an imaginary slice in the rope a distance y from the ceiling \nseparating the rope into an upper segment 1, and lower segment 2 (Figure 8.28a). Choose \nthe upper segment as a system with mass m1 = ( M / L) y . The forces acting on the upper \nsegment are the gravitational force, the force T ( y = 0) holding the rope up, and the \ntension T ( y) at the point y , that is pulling the upper segment down. The free-body force \ndiagram is shown in Figure 8.28b. \nT ( y = 0) \n(a) \n(b) \ny \ny \n1 \n\u02c6j\n1 \nm1g\n\u02c6j \nT ( y) \nL\ny \n2 \nFigure 8.28 (a) Imaginary slice separates rope into two pieces. (b) Free-body force \ndiagram on upper piece of rope \nApply Newton\u2019s Second Law to the upper segment: m1g + T ( y) \u2212 T ( y = 0) = 0 . Therefore \nthe tension at a distance y from the ceiling is T ( y) = T ( y = 0) \u2212 m1g . Because \n= ( M / L) y is the mass of the segment piece and Mg is the tension at the upper end,\nm1 \nNewton\u2019s Second Law becomes \n8-23", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_179", "text": "T ( y) = Mg(1\u2212 y / L) \n(8.5.13) \nAs a check, we note that when y = L , the tension T ( y = L) = 0 , which is what we expect \nbecause there is no force acting at the lower end of the rope. \n(c) Differentiate Eq. (8.5.13) with respect to y yielding \ndT = \u2212( M / L)g . \n(8.5.14)\ndy \nThe rate that the tension is changing at a constant rate with respect to distance from the \ntop of the rope. \n8.5.2 Continuous Systems and Newton\u2019s Second Law as a Differential Equations \nWe can determine the tension at a distance y from the ceiling in Example 8.4, by an \nalternative method, a technique that will generalize to many types of \u201ccontinuous \nsystems\u201d. Choose a coordinate system with the origin at the ceiling and the positive y \u00ad\ndirection pointing downward as in Figure 8.25. Consider as the system a small element of \nthe rope between the points y and y + \u0394y . This small element has length \u0394y , The small \nelement has mass \u0394m = ( M / L)\u0394y and is shown in Figure 8.29. \n\u02c6j \ny \ny \ny + y \ny = L \nm \nFigure 8.29 Small mass element of the rope \nThe forces acting on the small element are the tension, T ( y) at y directed upward, the \ntension T ( y + \u0394y) at y + \u0394y directed downward, and the gravitational force \u0394mg \ndirected downward. The tension T ( y + \u0394y) is equal to the tension T ( y) plus a small \ndifference \u0394T , \nT ( y + \u0394y) = T ( y) + \u0394T . \n(8.5.15) \n8-24", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_180", "text": "y)\n)\n \n \n \n \n \n \n \n \n \n \n \n \n \n  \n  \n \n  \n \n  \n \n \n  \n \nThe small difference in general can be positive, zero, or negative. The free body force \ndiagram is shown in Figure 8.30. \nmg\nT(y\nT(\ny +\ny\ny\ny\n\u02c6j\n+\nT\nFigure 8.30 Free body force diagram on small mass element \nNow apply Newton\u2019s Second Law to the small element \n\u0394mg + T ( y) \u2212 (T ( y) + \u0394T ) = 0 \n(8.5.16) \nThe difference in the tension is then \u0394T = \u2212\u0394mg . We now substitute our result for the \nmass of the element \u0394m = ( M / L)\u0394y , and find that that \n\u0394T = \u2212( M / L)\u0394yg . \n(8.5.17) \nDivide through by \u0394y , yielding \u0394T / \u0394y = \u2212( M / L)g . Now take the limit in which the \nlength of the small element goes to zero, \u0394y \u2192 0 , \n\u0394T\nlim \n= \u2212( M / L)g . \n(8.5.18)\n\u0394y\u21920 \u0394y \nRecall that the left hand side of Eq. (8.5.18) is the definition of the derivative of the \ntension with respect to y , and so we arrive at Eq. (8.5.14), \ndT = \u2212( M / L)g .\ndy \nWe can solve the differential equation, Eq. (8.5.14), by a technique called separation of \nvariables. We rewrite the equation as dT = \u2212( M / L)gdy and integrate both sides. Our \nintegral will be a definite integral in which we integrate a \u2018dummy\u2019 integration variable \ny\u2032 from y\u2032 = 0 to y\u2032 = y and the corresponding T \u2032 from T \u2032 = T ( y = 0) to T \u2032 = T ( y) : \nT \u2032=T ( y) \ny\u2032= y\n\u222b \ndT \u2032 = \u2212( M / L)g\u222by\u2032=0 dy\u2032 . \n(8.5.19) \nT '=T ( y=0) \n8-25", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_181", "text": "After integration and substitution of the limits, we have that \nT ( y) \u2212 T ( y = 0) = \u2212( M / L)gy . \n(8.5.20) \nUs the fact that tension at the top of the rope is T ( y = 0) = Mg and find that \nT ( y) = Mg(1\u2212 y / L) \nin agreement with our earlier result, Eq. (8.5.13). \n8.6 Drag Forces in Fluids \nWhen a solid object moves through a fluid it will experience a resistive force, called the \ndrag force, opposing its motion. The fluid may be a liquid or a gas. This force is a very \ncomplicated force that depends on both the properties of the object and the properties of \nthe fluid. The force depends on the speed, size, and shape of the object. It also depends \non the density, viscosity and compressibility of the fluid. \nFor objects moving in air, the air drag is \nstill quite complicated but for rapidly \nTable 8.1 Drag Coefficients \nmoving objects the resistive force is \nroughly proportional to the square of the \nspeed v , the cross-sectional area A of \nthe object in a plane perpendicular to the \nmotion, the density \u03c1 of the air, and \nindependent of the viscosity of the air. \nTraditional the magnitude of the air drag \nfor rapidly moving objects is written as \nFdrag = 2\n1 CD A\u03c1v2 . \n(8.6.1) \nThe coefficient CD is called the drag \ncoefficient, a dimensionless number that \nis a property of the object. Table 8.1 \nlists the drag coefficient for some simple \nshapes, (each of these objects has a \nReynolds number of order 104 ). \nSphere\nHalf-sphere\nCone\nCube\nAngled cube\nLong cylinder\nShort cylinder\nStreamlined\nbody\nStreamlined\nhalf-body\nShape\nDrag coefficient\n0.47\n0.42\n0.50\n1.05\n0.80\n0.82\n1.15\n0.04\n0.09\nThe above model for air drag does not extend to all fluids. An object dropped in \noil, molasses, honey, or water will fall at different rates due to the different viscosities of \nthe fluid. For very low speeds, the drag force depends linearly on the speed and is also \nproportional to the viscosity \u03b7 of the fluid. For the special case of a sphere of radius R , \n8-26", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_182", "text": "the drag force law can be exactly deduced from the principles of fluid mechanics and is \ngiven by \n!\n!\nFdrag = \u22126\u03c0\u03b7Rv \n(sphere) . \n(8.6.2) \nThis force law is known as Stokes\u2019 Law. The coefficient of viscosity \u03b7 has SI units of \n\u22121 \u22c5s\n[N \u22c5m\u22122 \u22c5s] = [Pa \u22c5s] = [kg \u22c5m\n\u22121] ; a cgs unit called the poise is often encountered . \nSome typical coefficients of viscosity are listed in Table 8.2. \nTable 8.2: Coefficients of viscosity \nfluid \nTemperature, \n0 C \nCoefficient of viscosity \u03b7 ; [kg \u22c5m\u22121 \u22c5s\u22121] \nAcetone \n25 \n3.06 \u00d710\u22124 \nAir \n15 \n1.81\u00d710\u22125 \nBenzene \n25 \n6.04 \u00d710\u22124 \nBlood \n37 \n(3\u2212 4) \u00d710\u22123 \nCastor oil \n25 \n0.985 \nCorn Syrup \n25 \n1.3806 \nEthanol \n25 \n1.074 \u00d710\u22123 \nGlycerol \n20 \n1.2 \nMethanol \n25 \n5.44 \u00d710\u22124 \nMotor oil (SAE 10W) \n20 \n6.5\u00d710\u22122 \nOlive Oil \n25 \n8.1\u00d710\u22122 \nWater \n10 \n1.308 \u00d710\u22123 \nWater \n20 \n1.002 \u00d710\u22123 \nWater \n60 \n0.467 \u00d710\u22123 \nWater \n100 \n0.28 \u00d710\u22123 \nThis law can be applied to the motion of slow moving objects in a fluid, for example: \nvery small water droplets falling in a gravitational field, grains of sand settling in water, \nor the sedimentation rate of molecules in a fluid. In the later case, If we model a molecule \nas a sphere of radius R , the mass of the molecule is proportional to R3 and the drag force \nis proportion to R , therefore different sized molecules will have different rates of \nacceleration. \nThis is the basis for the design of measuring devices that separate \nmolecules of different molecular weights. \nIn many physical situations the force on an object will be modeled as depending on the \nobject\u2019s velocity. We have already seen static and kinetic friction between surfaces \nmodeled as being independent of the surfaces\u2019 relative velocity. Common experience \n(swimming, throwing a Frisbee) tells us that the frictional force between an object and a \n8-27", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_183", "text": "fluid can be a complicated function of velocity. Indeed, these complicated relations are \nan important part of such topics as aircraft design. \nExample 8.5 Drag Force at Low Speeds \nh \ng \nmarble: \nmass m and \nradius R \nolive oil \nFigure 8.31 Example 8.5 \nA spherical marble of radius R and mass m is released from rest and falls under the \ninfluence of gravity through a jar of olive oil of viscosity \u03b7 . The marble is released from \nrest just below the surface of the olive oil, a height h from the bottom of the jar. The \ngravitational acceleration is g (Figure 8.31). Neglect any force due to the buoyancy of \n!v \nthe olive oil. (i) Determine the velocity of the marble as a function of time, (ii) what is the \n!v\u221e\nmaximum possible velocity \n=  (t = \u221e) (terminal velocity), that the marble can obtain, \n(iii) determine an expression for the viscosity of olive oil \u03b7 in terms of g , m, R , and \n! \nv\u221e = \n, (iv) determine an expression for the position of the marble from just below the \nsurface of the olive oil as a function of time. \nSolution: Choose positive y -direction downwards with the origin at the initial position of \nthe marble as shown in Figure 8.32(a). \nv\u221e \ny(t) \n\u02c6j \nh \n+ y \nO \nv(t) \n(a) \n\u02c6j \nO \n. \n+ y \nmg \nFdrag \n(b) \nFigure 8.32 (a) Coordinate system for marble; (b) free body force diagram on marble \nThere are two forces acting on the marble: the gravitational force, and the drag force \nwhich is given by Eq. (8.6.2). The free body diagram is shown in the Figure 8.32(b). \nNewton\u2019s Second Law is then \ndv \nmg \u2212 6\u03c0\u03b7Rv = m \n, \n(8.6.3)\ndt \n8-28", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_184", "text": "where v is the y -component of the velocity of the marble. Let \u03b3 = 6\u03c0\u03b7R / m ; the SI \nunits \u03b3 are [s\u22121]. Then Eq. (8.6.3) becomes \ndv \ng \u2212\u03b3 v = \n, \n(8.6.4)\ndt \nSuppose the object has an initial y -component of velocity v(t = 0) = 0 . We shall solve \nEq. (8.6.3) using the method of separation of variables. The differential equation may be \nrewritten as \ndv \n= \u2212\u03b3 dt . \n(8.6.5)\n(v \u2212 g / \u03b3 ) \nThe integral version of Eq. (8.6.5) is then \nv\u2032=v(t ) \nt\u2032=t\ndv\u2032 \n= \u2212\u03b3 \ndt\u2032\n\u222b\n\u222b\nv\u2032 \u2212 g / \u03b3 \n. \n(8.6.6)\nv\u2032=0 \nt\u2032=0 \nIntegrating both sides of Eq. (8.6.6) yields \n\u239b v(t) \u2212 g / \u03b3 \u239e \n\u23a0\u239f = \u2212\u03b3 t\nln \u239d\u239c \n\u2212g / \u03b3 \n. \n(8.6.7) \nRecall that eln x = x , therefore upon exponentiation of Eq. (8.6.7) yields \nv(t) \u2212 g / \u03b3 \n\u2212\u03b3 t\n= e \n. \n(8.6.8)\n\u2212g / \u03b3 \nThus the y -component of the velocity as a function of time is given by \ng \n\u2212\u03b3 t ) = mg \n\u2212(6\u03c0\u03b7R/m)t )\nv(t) = \n(1\u2212 e \n(1\u2212 e \n(8.6.9)\n.\n\u03b3 \n6\u03c0\u03b7R \nA plot of v(t) vs. t is shown in Figure 8.31 with parameters R = 5.00 \u00d710\u22123m , \n\u22121 \u22c5s\u22121 \n\u22121\n\u03b7 = 8.10 \u00d710\u22122 kg \u22c5 m\n, m = 4.08 \u00d710\u22123 kg , and g / \u03b3 = 1.87 m \u22c5s . \n8-29", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_185", "text": "v(t)\nt\n[m s 1]\n5\n1\n2\n[s]\n[s]\nFigure 8.33 Plot of y -component of the velocity v(t) vs. t for marble falling through \noil with g / \u03b3 = 1.87 m \u22c5s\u22121 . \n\u2212(6\u03c0\u03b7R/m)t\nFor large values of t , the term e \napproaches zero, and the marble reaches a \nterminal velocity \nmg \nv\u221e = v(t = \u221e) = \n(8.6.10)\n6\u03c0\u03b7R . \nThe coefficient of viscosity can then be determined from the terminal velocity by the \ncondition that \nmg \n\u03b7 = \n(8.6.11)\n.\n6\u03c0 Rvter \nLet \u03c1 m denote the density of the marble. The mass of the spherical marble is \nm = (4 / 3)\u03c1 mR3. The terminal velocity is then \nR2\n2\u03c1 m\ng\nv\u221e = \n(8.6.12)\n.\n9\u03b7 \nThe terminal velocity depends on the square of the radius of the marble, indicating that \nlarger marbles will reach faster terminal speeds. \nThe position of the marble as a function of time is given by the integral expression \nt\u2032=t \ny(t) \u2212 y(t = 0) = \u222b v(t\u2032)dt\u2032 \n(8.6.13)\n, \nt\u2032=0 \nwhich after substitution of Eq. (8.6.9) and integration using the initial condition that \ny(t = 0) = 0 , becomes \ng\ng \n\u2212\u03b3 t \u22121\ny(t) = \u03b3 t +\n(e\n) . \n(8.6.14)\n\u03b3 2 \n8-30", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_186", "text": "Example 8.6 Drag Forces at High Speeds \n!\nAn object of mass m at time t = 0 is moving rapidly with velocity v0 through a fluid of \ndensity \u03c1 . Let A denote the cross-sectional area of the object in a plane perpendicular \nto the motion. The object experiences a retarding drag force whose magnitude is given by \nEq. (8.6.1). Determine an expression for the velocity of the object as a function of time. \nSolution: Choose a coordinate system such that the object is moving in the positive x \u00ad\n!\ndirection, v = v\u02c6i . Set \u03b2 = (1/ 2)CD A\u03c1 . Newton\u2019s Second Law can then be written as \n2 \ndv\n\u2212\u03b2 v = \n. \n(8.6.15)\ndt \nAn integral version of Eq. (8.6.15) is then \nv\u2032=v(t ) \nt\u2032=t\ndv\u2032 \n\u222b v\u20322 = \u2212\u03b2 \u222b dt\u2032 . \n(8.6.16) \nt\u2032=0\nv\u2032=v0 \nIntegration yields \n\u239b 1\n1 \u239e\n\u2212 \n\u2212\n\u23a0\u239f = \u2212\u03b2t . \n(8.6.17)\n\u239d\u239c v(t) \nv0 \nAfter some algebraic rearrangement the x -component of the velocity as a function of \ntime is given by \nv0\n1 \nv(t) =\n= \nv0 , \n(8.6.18)\n1+ v0\u03b2t \n1+ t / \u03c4 \nwhere \u03c4 = 1/ v0\u03b2 . A plot of v(t) vs. t is shown in Figure 8.34 with initial conditions \nv0 \n\u22121\n= 20 m \u22c5s\nand \u03b2 = 0.5 s\u22121 . \nv(t)\nm s 1]\nt\n[s]\n20\n5\n1\n2\nFigure 8.34 Plot of v(t) vs. t for damping force Fdrag = CD A\u03c1v\n2 \n8-31", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_187", "text": "8.7 Worked Examples \nExample 8.7 Staircase \nAn object of mass m at time t = 0 has speed v0 . It slides a distance s along a horizontal \nfloor and then off the top of a staircase (Figure 8.35). The coefficient of kinetic friction \nbetween the object and the floor is \u00b5k . The object strikes at the far end of the third stair. \nEach stair has a rise of h and a run of d . Neglect air resistance and use g for the \ngravitational constant. (a) What is the distance s that the object slides along the floor? \nFigure 8.35 Object falling down a staircase \nSolution: There are two distinct stages to the object\u2019s motion, the initial horizontal \nmotion and then free fall. The given final position of the object, at the far end of the third \nstair, will determine the horizontal component of the velocity at the instant the object left \nthe top of the stairs. This in turn can be used to determine the time the object decelerated \nalong the floor, and hence the distance traveled on the floor. The given quantities are m , \nv0 , \u00b5k , g , h and d . \nFor the horizontal motion, choose coordinates with the origin at the initial position of the \nblock. Choose the positive \u02c6i -direction to be horizontal, directed to the left in Figure 8.35, \nand the positive \u02c6j -direction to be vertical (up). The forces on the object are gravity \n\uf072\n\uf072 \n\uf072\n\u02c6 \nN = N \u02c6 \nf \u02c6\nmg = \u2212mg j , the normal force \nj and the kinetic frictional force fk = \u2212 k i . The \n\uf072\n\uf072\ncomponents of the vectors in Newton\u2019s Second Law, F = ma , are \n\u2212 fk = max \n(8.6.19)\nN \u2212 m g = may . \n8-32", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_188", "text": "The object does not move in the y -direction; ay = 0 and thus from the second expression \nin (8.6.19), N = m g . The magnitude of the frictional force is then f = \u00b5 N = \u00b5 mg , and \nk\nk \nk \nthe first expression in (8.6.19) gives the x -component of acceleration as ax = \u2212\u00b5k g . \nBecasue the acceleration is constant the x -component of the velocity is given by \nv ( )t = v + a t , \n(8.6.20)\nx \n0 \nx \nwhere v0 is the x -component of the velocity of the object when it just started sliding. \nThe displacement is given by \nx(t) \u2212 x0 \nt + 1 a t2. \n(8.6.21)\n= v0 \nx\n2 \nDenote the time the block just leaves the landing by t1, where x(t1) = s , and the speed \njust when it reaches the landing v (t1) = vx,1 . The initial speed is v0 and x0 = 0 . Using the \nx \ninitial and final conditions, and the value of the acceleration, Eq. (8.6.21) becomes \n1\n2\n\u2212 \ngt1 . \n(8.6.22)\ns = v0 t1 \n\u00b5k\n2 \nSolve Eq. (8.6.20) for the time the block reaches the edge of the landing, \nv \n\u2212 v \nx,1 \u2212 v0 \nv0 \nx,1 \n=\n= \n. \n(8.6.23)\nt1 \ng \ng\n\u2212\u00b5k \n\u00b5k \nSubstituting Eq. (8.6.23) into Eq. (8.6.22) yields \n2\n\u239b \n\u2212 v\n\u239e \n\u239b \n\u2212 v\n\u239e \n1 \nv0 \nx,1 \ng\n\u00b5k \nv0 \nx,1 \ng\n\u00b5k \n\u2212 \n(8.6.24) \ns = v0 \n\u00b5kg\n2\n\u239c\u239d \n\u239f\u23a0 \n\u239c\u239d \n\u239f\u23a0  \nand after some algebra, we can rewrite Eq. (8.6.24) as \n2 \u2212 v 2\nv0 \nx,1 \ns = \n. \n(8.6.25)\ng\n2\u00b5k \nFrom the top of the stair to the far end of the third stair, the object is in free fall. Choose \nthe positive \u02c6i -direction to be horizontal, directed to the left in Figure 8.35, and the \npositive \u02c6j -direction to be vertical (up) and now choose the origin at the top of the stairs,", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_189", "text": "the positive \u02c6i -direction to be horizontal, directed to the left in Figure 8.35, and the \npositive \u02c6j -direction to be vertical (up) and now choose the origin at the top of the stairs, \nwhere the object first goes into free fall. The components of acceleration are ax = 0 , \n8-33", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_190", "text": "a = \u2212g , the initial x -component of velocity is v , the initial y -component of velocity \ny\nx,1 \nis vy,0 = 0 , the initial x -position is x0 = 0 and the initial y -position is y0 = 0 . Reset \nt = 0 when the object just leaves the landing. Let t2 denote the instant the object hits the \nstair, where y(t2) = \u22123h and x(t2) = 3d . The equations describing the object\u2019s position \nand speed at time t = t2 are \n) = 3d = v \n(8.6.26)\nx(t2 \nx,1 t2 \n1\n2\ny(t2) = \u22123h = \u2212 gt2 . \n(8.6.27)\n2 \nSolve Eq. (8.6.26) for t2 to yield \nt2 = v \n3d . \n(8.6.28) \nx,1 \nSubstitute Eq. (8.6.28) into Eq. (8.6.27) and eliminate the variable t2 , \n1\n9d 2 \n3h = g \n2 . \n(8.6.29)\n2 vx,1 \nEq. (8.6.29) can now be solved for the square of the horizontal component of the velocity, \n3gd 2 \nvx,1 \n2 = \n. \n(8.6.30)\n2h \nNow substitute Eq. (8.6.30) into Eq. (8.6.25) to determine the distance the object traveled \non the landing, \nv0\n2 \u2212 (3gd 2 / 2h)\ns = \n. \n(8.6.31)\n2\u00b5k g \nExample 8.8 Cart Moving on a Track \nC \nB  \nFigure 8.36 A falling block will accelerate a cart on a track via the pulling force of the \nstring. The force sensor measures the tension in the string. \n8-34", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_191", "text": "Consider a cart that is free to slide along a horizontal track (Figure 8.36). A force is \napplied to the cart via a string that is attached to a force sensor mounted on the cart, \nwrapped around a pulley and attached to a block on the other end. When the block is \nreleased the cart will begin to accelerate. The force sensor and cart together have a mass \n, and the suspended block has mass mB . Neglect the small mass of the string and \nmC \npulley, and assume the string is inextensible. The coefficient of kinetic friction between \nthe cart and the track is \u00b5k . Determine (i) the acceleration of the cart, and (ii) the tension \nin the string. \nSolution: In general, we would like to draw free-body diagrams on all the individual \nobjects (cart, sensor, pulley, rope, and block) but we can also choose a system consisting \nof two (or more) objects knowing that the forces of interaction between any two objects \nwill cancel in pairs by Newton\u2019s Third Law. In this example, we shall choose the \nsensor/cart as one free-body, and the block as the other free-body. The free-body force \ndiagram for the sensor/cart is shown in Figure 8.37. \nC .\nfk \nN \nmC g \nTR,C \n\u02c6i \n\u02c6j \nFigure 8.37 Force diagram on sensor/cart with a vector decomposition of the contact \nforce into horizontal and vertical components \n!\nThere are three forces acting on the sensor/cart: the gravitational force mC g , the pulling \n! \nforce TR,C of the rope on the force sensor, and the contact force between the track and the \ncart. In Figure 8.34, we decompose the contact force into its two components, the kinetic \n!\n! \nfrictional force fk = \u2212 fk\u02c6i and the normal force, N = N \u02c6j . \n\uf072\nThe cart is only accelerating in the horizontal direction with a = a \n\u02c6i , so the \nC \nC,x \ncomponent of the force in the vertical direction must be zero, aC,y = 0 . We can now apply \nNewton\u2019s Second Law in the horizontal and vertical directions and find that \n\u02c6i : T \n\u2212 f = m a \n(8.6.32)\nR,C \nk \nC\nC,x \n\u02c6j : N \u2212 mC g = 0 . \n(8.6.33) \nFrom Eq. (8.6.33), we conclude that the normal component is \nN \ng . \n(8.6.34)\n= mC \n8-35", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_192", "text": "We use Equation (8.6.34) for the normal force to find that the magnitude of the kinetic \nfrictional force is \nN = \u00b5k\ng . \n(8.6.35)\nfk = \u00b5k \nmC \nThen Equation (8.6.32) becomes \ng = mC \n. \n(8.6.36)\nTR,C \u2212 \u00b5k mC \naC,x \nThe force diagram for the block is shown in Figure 8.38. The two forces acting on\n!\n!\nthe block are the pulling force TR,B of the string and the gravitational force mBg . We now \napply Newton\u2019s Second Law to the block and find that \n\u02c6jB : mBg \u2212 TR,B = mBaB,y . \n(8.6.37) \nB. \nTR,B \nmBg \n\u02c6jB \nFigure 8.38 Forces acting on the block \nIn Equation (8.6.37), the symbol aB, y represents the component of the acceleration with \nsign determined by our choice of downward direction for the unit vector \u02c6jB . Note that we \nmade a different choice of direction for the unit vector in the vertical direction in the free-\nbody diagram for the block shown in Figure 8.37. Each free-body diagram has an \nindependent set of unit vectors that define a sign convention for vector decomposition of \nthe forces acting on the free-body and the acceleration of the free-body. In our example, \nwith the unit vector pointing downwards in Figure 8.38, if we solve for the component of \nthe acceleration and it is positive, then we know that the direction of the acceleration is \ndownwards. \nThere is a second subtle way that signs are introduced with respect to the forces \nacting on a free-body. In our example, the force between the string and the block acting \non the block points upwards, so in the vector decomposition of the forces acting on the \nblock that appears on the left-hand side of Equation (8.6.37), this force has a minus sign \n! \nand the quantity TR,B = \u2212TR,B \u02c6jB where TR,B is assumed positive. \nOur assumption that the mass of the rope and the mass of the pulley are negligible \nenables us to assert that the tension in the rope is uniform and equal in magnitude to the \nforces at each end of the rope, \n8-36", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_193", "text": "TR,B = TR,C \u2261 T . \n(8.6.38) \nWe also assumed that the string is inextensible (does not stretch). This implies that the \nrope, block, and sensor/cart all have the same magnitude of acceleration, \n\u2261 a . \n(8.6.39)\naC,x = aB, y \nUsing Equations (8.6.38) and (8.6.39), we can now rewrite the equation of motion for the \nsensor/cart, Equation (8.6.36), as \ng = mCa , \n(8.6.40)\nT \u2212 \u00b5k mC \nand the equation of motion (8.6.37) for the block as \nmB g \u2212 T = mBa . \n(8.6.41) \nWe have only two unknowns T and a , so we can now solve the two equations (8.6.40) \nand (8.6.41) simultaneously for the acceleration of the sensor/cart and the tension in the \nrope. We first solve Equation (8.6.40) for the tension \nT \ng + mCa \n(8.6.42)\n= \u00b5k mC \nand then substitute Equation (8.6.42) into Equation (8.6.41) and find that \ng + mCa) = mBa . \n(8.6.43)\nmB g \u2212 (\u00b5k mC \nWe can now solve Equation (8.6.43) for the acceleration, \ng \u2212 \u00b5k \ng\nmB \nmC\na = \n. \n(8.6.44)\nmC + mB \nSubstitution of Equation (8.6.44) into Equation (8.6.42) gives the tension in the string, \nT = \u00b5 m g + m a \nk\nC \nC \nmB g \u2212 \u00b5k \ng\nmC \n(8.6.45)\n= \u00b5k mC g + mC \nmC + mB \nmCmB\n+ 1) \ng.\n= (\u00b5k \nmC + mB \nIn this example, we applied Newton\u2019s Second Law to two objects, one a \ncomposite object consisting of the sensor and the cart, and the other the block. We \nanalyzed the forces acting on each object and also any constraints imposed on the \n8-37", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_194", "text": "acceleration of each object. We used the force laws for kinetic friction and gravitation on \neach free-body system. The three equations of motion enable us to determine the forces \nthat depend on the parameters in the example: the tension in the rope, the acceleration of \nthe objects, and normal force between the cart and the table. \nExample 8.9 Pulleys and Ropes Constraint Conditions \nConsider the arrangement of pulleys and blocks shown in Figure 8.39. The pulleys are \nassumed massless and frictionless and the connecting strings are massless and \ninextensible. Denote the respective masses of the blocks as m1 , m2 and m3 . The upper \npulley in the figure is free to rotate but its center of mass does not move. Both pulleys \nhave the same radius R . (a) How are the accelerations of the objects related? (b) Draw \nforce diagrams on each moving object. (c) Solve for the accelerations of the objects and \nthe tensions in the ropes. \n2 \n3 \nP\n1 \nFigure 8.39 Constrained pulley system \nSolution: (a) Choose an origin at the center of the upper pulley. Introduce coordinate \nfunctions for the three moving blocks, y1, y2 and y3 . Introduce a coordinate function \nyP for the moving pulley (the pulley on the lower right in Figure 8.40). Choose \ndownward for positive direction; the coordinate system is shown in the figure below then. \nstring A \ny1 \nyP y3 y2 \n1 \n2 \n3 \nP. \n\u02c6j \nstring B \nFigure 8.40 Coordinated system for pulley system \nThe length of string A is given by \n8-38", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_195", "text": "l = y + y + \u03c0 R \n(8.6.46)\nA \n1 \nP \nwhere \u03c0 R is the arc length of the rope that is in contact with the pulley. This length is \nconstant, and so the second derivative with respect to time is zero, \n2\n2\n2\nd l \nd y \nd y \nA \n1 \nP\n0 = \n2 = \n2 +\n= ay,1 + ay P . \n(8.6.47)\ndt\ndt \ndt 2\n, \nThus block 1 and the moving pulley\u2019s components of acceleration are equal in magnitude \nbut opposite in sign, \nay P = \u2212ay,1 . \n(8.6.48)\n, \nThe length of string B is given by \nlB = ( y3 \u2212 yP ) + ( y2 \u2212 yP ) + \u03c0 R = y3 + y2 \u2212 2yP + \u03c0 R \n(8.6.49) \nwhere \u03c0 R is the arc length of the rope that is in contact with the pulley. This length is \nalso constant so the second derivative with respect to time is zero, \n2\n2\n2 \n2\nd l \nd y \nd y \nd y \nB \n2\n3 \nP\n0 =\n= \n+ \n\u2212 2 \n= ay,2 + a \n\u2212 2ay P . \n(8.6.50)\n2\n2\n2 \n2 \ny,3 \n,\ndt\ndt\ndt \ndt \nWe can substitute Equation (8.6.48) for the pulley acceleration into Equation (8.6.50) \nyielding the constraint relation between the components of the acceleration of the three \nblocks, \n0 = ay,2 + ay,3 + 2ay,1 . \n(8.6.51) \nb) Free-body Force diagrams: the forces acting on block 1 are: the gravitational force \n!\n! \nm1g and the pulling force TA,1 of string A acting on the block 1. Denote the magnitude \nof this force by TA . Because the string is assumed to be massless and the pulley is \nassumed to be massless and frictionless, the tension TA in the string is uniform and equal \nin magnitude to the pulling force of the string on the block. The free-body diagram on \nblock 1 is shown in Figure 8.41(a). \n.\n1 . \n.\n2 \n3 \nP . \n(a) \n(b) \n(c) \n(d) \n\u02c6j \nTA,1 \nm1g \nTB,2 \nm2g \nTB,3 \nm3g TB,P \nTB,P \nTA,P", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_196", "text": "block 1 is shown in Figure 8.41(a). \n.\n1 . \n.\n2 \n3 \nP . \n(a) \n(b) \n(c) \n(d) \n\u02c6j \nTA,1 \nm1g \nTB,2 \nm2g \nTB,3 \nm3g TB,P \nTB,P \nTA,P \nFigure 8.41 Free-body force diagram on (a) block 1; (b) block 2; (c) block 3; (d) pulley \n8-39", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_197", "text": "Newton\u2019s Second Law applied to block 1 is then \n\u02c6j: m1g \u2212 TA = m1 ay,1 . \n(8.6.52) \n!\nThe forces on the block 2 are the gravitational force m2g and string B holding the block, \n! \nTB,2 , with magnitude TB . The free-body diagram for the forces acting on block 2 is \nshown in Figure 8.41(b). Newton\u2019s second Law applied to block 2 is \n\u02c6j: m2 g \u2212 TB = m2 ay,2 . \n(8.6.53) \n!\nThe forces on the block 3 are the gravitational force m3g and string holding the block, \n! \nTB,3 , with magnitude equal to TB because pulley P has been assumed to be both \nfrictionless and massless. The free-body diagram for the forces acting on block 3 is \nshown in Figure 8.41(c). Newton\u2019s second Law applied to block 3 is \n\u02c6j: m3g \u2212 TB = m3 ay,3 . \n(8.6.54) \n\uf072\n\uf072 \nThe forces on the moving pulley P are the gravitational force mP g = 0 (the pulley is \n! \nassumed massless); string B pulls down on the pulley on each side with a force, TB,P , \n! \nwhich has magnitude TB . String A holds the pulley up with a force TA,P with the \nmagnitude TA equal to the tension in string A . The free-body diagram for the forces \nacting on the moving pulley is shown in Figure 8.41(d). Newton\u2019s second Law applied to \nthe pulley is \n\u02c6j: 2TB \u2212TA = mP ay P = 0 . \n(8.6.55)\n, \nBecause the pulley is assumed to be massless, we can use this last equation to determine \nthe condition that the tension in the two strings must satisfy, \n2TB = TA \n(8.6.56) \nWe are now in position to determine the accelerations of the blocks and the tension in the \ntwo strings. We record the relevant equations as a summary. \n0 = ay,2 + ay,3 + 2ay,1 \n(8.6.57) \nm g \u2212T = m a \n(8.6.58)\n1 \nA \n1 \ny,1 \nm g \u2212T = m a \n(8.6.59)\n2 \nB \n2 \ny,2 \nm g \u2212T = m a \n(8.6.60)\n3 \nB \n3 \ny,3 \n2TB = TA .", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_198", "text": "(8.6.58)\n1 \nA \n1 \ny,1 \nm g \u2212T = m a \n(8.6.59)\n2 \nB \n2 \ny,2 \nm g \u2212T = m a \n(8.6.60)\n3 \nB \n3 \ny,3 \n2TB = TA . \n(8.6.61) \n8-40", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_199", "text": "There are five equations with five unknowns, so we can solve this system. We shall first \nuse Equation (8.6.61) to eliminate the tension TA in Equation (8.6.58), yielding \nm g \u2212 T = m a . \n(8.6.62)\n1\n2 B \n1 \ny,1 \nWe now solve Equations (8.6.59), (8.6.60) and (8.6.62) for the accelerations, \nTB\nay,2 = g \u2212 \n(8.6.63)\nm2 \nTB\nay,3 = g \u2212 \n(8.6.64)\nm3 \n2TB\nay,1 = g \u2212 \n. \n(8.6.65)\nm1 \nWe now substitute these results for the accelerations into the constraint equation, \nEquation (8.6.57), \nT\nT \n4T \n\u239b 1\n1\n4 \u239e\nB\nB \nB\n0 = g \u2212 \n+ g \u2212 \n+ 2g \u2212 \n= 4g \u2212TB \u239c \n+\n+ \n\u239f . \n(8.6.66)\nm\nm\nm \nm\nm\nm\n2\n3 \n1 \n\u239d \n2 \n3\n1 \u23a0 \nWe can now solve this last equation for the tension in string B , \n4g \n4g m m m 3\nTB =\n= \n1 \n2 \n. \n(8.6.67)\n\u239b 1\n1\n4 \u239e \nm m + m m + 4m m \n1\n3\n1\n2 \n2\n3\n+\n+\n\u239c\n\u239f \nm\nm\nm\n\u239d \n2 \n3\n1 \u23a0 \nFrom Equation (8.6.61), the tension in string A is \n8g m m m\n1\n2\n3\nTA = 2TB = \n. \n(8.6.68)\nm m + m m + 4m m \n1\n3\n1\n2 \n2\n3 \nWe find the acceleration of block 1 from Equation (8.6.65), using Equation (8.6.67) for \nthe tension in string B, \n2T \n8g m m \nm m + m m \u2212 4m m \nB \n2\n3 \n1\n3\n1\n2 \n2\n3\nay,1 = g \u2212 \n= g \u2212 \n= g \n. (8.6.69)\nm \nm m + m m + 4m m \nm m + m m + 4m m \n1 \n1\n3\n1\n2 \n2\n3 \n1\n3\n1\n2 \n2\n3 \nWe find the acceleration of block 2 from Equation (8.6.63), using Equation (8.6.67) for \nthe tension in string B, \n8-41", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_200", "text": "T \n4g m m \n\u22123m m + m m + 4m m \nB \n1\n3 \n1\n3\n1\n2 \n2\n3\nay,2 = g \u2212 \n= g \u2212 \n= g \n. (8.6.70)\nm \nm m \n1 \n+ 4m m\n2 \nm m \n1\n3 + m m2 + 4m m3\n3 + m m \n2\n1\n2\n3 \n1\n2 \nSimilarly, we find the acceleration of block 3 from Equation (8.6.64), using Equation \n(8.6.67) for the tension in string B, \nT \n4 g m m \nm m \u2212 3m m + 4m m \n1\n3\n1\n2 \n2\n3\nB \n1\n2\nay,3 = g \u2212 \n= g \u2212 \n= g \n. (8.6.71)\nm \nm m + m m + 4m m \nm m + m m + 4m m \n3 \n1\n3\n1\n2 \n2\n3 \n1\n3\n1\n2 \n2\n3 \nAs a check on our algebra we note that \n2a + a \n+ a \n= \n1, y \n2, y \n3, y \nm m + m m \u2212 4m m \n\u22123m m + m m + 4m m \nm m \u2212 3m m + 4m m \n1\n3\n1\n2 \n2\n3 \n1\n3\n1\n2 \n2\n3 \n1\n3 \n1\n2 \n2\n3\n2g \n+ g \n+ g\nm m + m m + 4m m \nm m + m m + 4m m \nm m + m m + 4m m \n1\n3\n1\n2 \n2\n3 \n1\n3\n1\n2 \n2\n3 \n1\n3\n1\n2 \n2\n3 \n= 0. \nExample 8.10 Accelerating Wedge \nwedge \nblock of mass m \nA \nFigure 8.42 Block on accelerating wedge \n! \nA 45o wedge is pushed along a table with constant acceleration A according to an \nobserver at rest with respect to the table. A block of mass m slides without friction down \nthe wedge (Figure 8.42). Find its acceleration with respect to an observer at rest with \nrespect to the table. Write down a plan for finding the magnitude of the acceleration of \nthe block. Make sure you clearly state which concepts you plan to use to calculate any \nrelevant physical quantities. Also clearly state any assumptions you make. Be sure you \ninclude any free-body force diagrams or sketches that you plan to use. \nSolution: Choose a coordinate system for the block and wedge as shown in Figure 8.43. \n\uf072 \nThen A = A \n\u02c6i where A \nis the x-component of the acceleration of the wedge.\nx,w\nx,w \n8-42", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_201", "text": "wedge \nblock of mass m \nxb \n\u02c6j \n\u02c6i \nA\nx w \nyb \nl  \nFigure 8.43 Coordinate system for block on accelerating wedge \nWe shall apply Newton\u2019s Second Law to the block sliding down the wedge. Because the \nwedge is accelerating, there is a constraint relation between the x - and y - components \nof the acceleration of the block. In order to find that constraint we choose a coordinate \nsystem for the wedge and block sliding down the wedge shown in the figure below. We \nshall find the constraint relationship between the components of the accelerations of the \nblock and wedge by a geometric argument. From the figure above, we see that \nyb\ntan\u03c6 = \n. \n(8.6.72)\nl \u2212 (xb \u2212 xw ) \nTherefore \n\u2212 x ))tan\u03c6 . \n(8.6.73)\nyb = (l \u2212 (xb \nw \nIf we differentiate Eq. (8.6.73) twice with respect to time noting that \n2\nd l = 0 \n(8.6.74)\ndt 2 \nwe have that \nd 2 \n\u239b d 2 \nd 2\ny \nx\nx \u239e \nb \nb\nw\n= \u2212\u239c \n\u2212 \n\u239f tan\u03c6 . \n(8.6.75)\ndt2 \n\u239d dt2 \ndt2 \u23a0 \nTherefore \n\u2212 A )tan\u03c6 \n(8.6.76)\nab, y = \u2212(ab,x\nx,w \nwhere \nd 2 x\nAx,w = \ndt2 \nw . \n(8.6.77) \nWe now draw a free-body force diagram for the block (Figure 8.44). Newton\u2019s Second \nLaw in the \u02c6i - direction becomes \nN sin\u03c6 = mab,x . \n(8.6.78) \n8-43", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_202", "text": "and the \u02c6j-direction becomes \nN cos\u03c6 \u2212 mg = mab,y \n(8.6.79) \nN \n\u02c6j \n\u02c6i\n. \nmg \nFigure 8.44 Free-body force diagram on block \nWe can solve for the normal force from Eq. (8.6.78): \nmab,x\nN = \n(8.6.80)\nsin\u03c6 \nWe now substitute Eq. (8.6.76) and Eq. (8.6.80) into Eq. (8.6.79) yielding \nmab,x cos\u03c6 \u2212 mg = m(\u2212(a \n\u2212 A ) tan\u03c6) . \n(8.6.81)\nb,x\nw,x\nsin\u03c6 \nWe now clean this up yielding \nmab,x (cotan \u03c6 + tan\u03c6) = m(g + Aw,x tan\u03c6) \n(8.6.82) \nThus the x-component of the acceleration is then \ng + A tan\u03c6 \nab,x = \nw,x \n(8.6.83)\ncotan \u03c6 + tan\u03c6 \nFrom Eq. (8.6.76), the y -component of the acceleration is then \n\u239b g + Aw,x tan\u03c6\n\u239e \n\u2212 A ) tan\u03c6 = \u2212\n\u2212 A \u239f tan\u03c6 . \n(8.6.84)\nab, y = \u2212(ab,x \nw,x \n\u239c \nw,x\n\u239d cotan \u03c6 + tan\u03c6\n\u23a0 \nThis simplifies to \nAw,x \u2212 g tan\u03c6 \nab, y = \n(8.6.85)\ncotan \u03c6 + tan\u03c6 \n8-44", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_203", "text": "\uf06f \n\uf06f\n\uf06f\nWhen \u03c6 = 45 , cotan 45 = tan 45 = 1, and so Eq. (8.6.83) becomes \ng + A \n= \nw,x \n(8.6.86)\nab,x \n2 \nand Eq. (8.6.85) becomes \nA \u2212 g\n= \n. \n(8.6.87)\nab, y \n2 \nThe magnitude of the acceleration is then \n\u239b \n\u239d\u239c\n2 \n2\n\u239e \n\u239b \n\u239e \ng + Aw,x \nAw,x \u2212 g\na = ab,x \n2\n2 + ab,y \n(8.6.88) \n+ \n= \n\u239f\u23a0 \n\u239c\u239d \n\u239f\u23a0 \n2  \n2  \n\u239b \n\u239d\n\u239c\n\u23a0\n\u239f\n2 + A2\ng\nw,x \u239e  \na = \n. \n2  \nExample 8.11: Capstan \nA device called a capstan is used aboard ships in order to control a rope that is under \ngreat tension. The rope is wrapped around a fixed drum of radius R , usually for several \nturns (Figure 8.45 shows about three fourths turn as seen from overhead). The load on the \nrope pulls it with a force TA , and the sailor holds the other end of the rope with a much \nsmaller force TB . The coefficient of static friction between the rope and the drum is \u00b5s . \nThe sailor is holding the rope so that it is just about to slip. Show that TB = TAe\n\u2212 \u00b5s\u03b8BA , \nwhere \u03b8BA is the angle subtended by the rope on the drum. \nFigure 8.45 Capstan \nFigure 8.46 Small slice of rope \n8-45", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_204", "text": "Solution: We begin by considering a small slice of rope of arc length R \u0394\u03b8 , shown in the \nFigure 8.46. We choose unit vectors for the force diagram on this section of the rope and \nindicate them on Figure 8.47. The right edge of the slice is at angle \u03b8 and the left edge of \nthe slice is at \u03b8 + \u0394\u03b8 . The angle edge end of the slice makes with the horizontal is \u0394\u03b8 / 2 . \nThere are four forces acting on this section of the rope. The forces are the normal force \nbetween the capstan and the rope pointing outward, a static frictional force and the \ntensions at either end of the slice. The rope is held at the just slipping point, so if the load \nexerts a greater force the rope will slip to the right. Therefore the direction of the static \nfrictional force between the capstan and the rope, acting on the rope, points to the left. \nThe tension on the right side of the slice is denoted by T (\u03b8) \u2261 T , while the tension on the \nleft side of the slice is denoted by T (\u03b8 + \u0394\u03b8) \u2261 T + \u0394T . Does the tension in this slice \nfrom the right side to the left, increase, remain the same, or decrease? The tension \ndecreases because the load on the left side is less than the load on the right side. Note that \n\u0394T < 0 . \nFigure 8.47 Free-body force diagram on small slice of rope \nThe vector decomposition of the forces is given by \n\u02c6i : T cos(\u0394\u03b8 / 2) \u2212 fs \u2212 (T + \u0394T )cos(\u0394\u03b8 / 2) \n(8.6.89) \n\u02c6j: \u2212T sin(\u0394\u03b8 / 2) + N \u2212 (T + \u0394T )sin(\u0394\u03b8 / 2) . \n(8.6.90) \nFor small angles \u0394\u03b8 , cos(\u0394\u03b8 / 2) \u2245 1 and sin(\u0394\u03b8 / 2) \u2245 \u0394\u03b8 / 2 . Using the small angle \napproximations, the vector decomposition of the forces in the x -direction (the +\u02c6i \u00ad\ndirection) becomes \nT cos(\u0394\u03b8 / 2) \u2212 f \u2212 (T + \u0394T )cos(\u0394\u03b8 / 2) \uf03b T \u2212 f \u2212 (T + \u0394T )\ns\ns \n(8.6.91) \n= \u2212 f \u2212\u0394T \n. \ns \nBy the static equilibrium condition the sum of the x -components of the forces is zero, \n\u2212 f s \u2212\u0394T = 0 .", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_205", "text": "s\ns \n(8.6.91) \n= \u2212 f \u2212\u0394T \n. \ns \nBy the static equilibrium condition the sum of the x -components of the forces is zero, \n\u2212 f s \u2212\u0394T = 0 . \n(8.6.92) \n8-46", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_206", "text": "The vector decomposition of the forces in the y -direction (the +\u02c6j -direction) is \napproximately \n\u2212T sin(\u0394\u03b8 / 2) + N \u2212 (T + \u0394T )sin(\u0394\u03b8 / 2) \uf03b \u2212T \u0394\u03b8 / 2 + N \u2212 (T + \u0394T )\u0394\u03b8 / 2 (8.6.93) \n= \u2212T \u0394\u03b8 + N \u2212\u0394T \u0394\u03b8 / 2 \n. \nIn the last equation above we can ignore the terms proportional to \u0394T \u0394\u03b8 because these \nare the product of two small quantities and hence are much smaller than the terms \nproportional to either \u0394T or \u0394\u03b8 . The vector decomposition in the y -direction becomes \n\u2212T \u0394\u03b8 + N . \n(8.6.94) \nStatic equilibrium implies that this sum of the y -components of the forces is zero, \n\u2212T \u0394\u03b8 + N = 0 . \n(8.6.95) \nWe can solve this equation for the magnitude of the normal force \nN = T \u0394\u03b8 . \n(8.6.96) \nThe just slipping condition is that the magnitude of the static friction attains its maximum \nvalue \nf = ( f ) \n= \u00b5 N . \n(8.6.97)\ns \ns\nmax \ns \nWe can now combine the Equations (8.6.92) and (8.6.97) to yield \n\u0394T = \u2212\u00b5sN . \n(8.6.98) \nNow substitute the magnitude of the normal force, Equation (8.6.96), into Equation \n(8.6.98), yielding \n\u2212\u00b5sT\u0394\u03b8 \u2212\u0394T = 0 . \n(8.6.99) \nFinally, solve this equation for the ratio of the change in tension to the change in angle, \n\u0394T = \u2212\u00b5 T . \n(8.6.100)\n\u0394\u03b8 \ns \nThe derivative of tension with respect to the angle \u03b8 is defined to be the limit \ndT \n\u0394T\n\u2261 lim \n, \n(8.6.101)\nd\u03b8 \n\u0394\u03b8 \u21920 \u0394\u03b8 \nand Equation (8.6.100) becomes \n8-47", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_207", "text": "dT = \u2212\u00b5 T . \n(8.6.102)\nd\u03b8 \ns \nThis is an example of a first order linear differential equation that shows that the rate of \nchange of tension with respect to the angle \u03b8 is proportional to the negative of the \ntension at that angle \u03b8 . This equation can be solved by integration using the technique of \nseparation of variables. We first rewrite Equation (8.6.102) as \ndT = \u2212\u00b5s d\u03b8 . \n(8.6.103)\nT \nIntegrate both sides, noting that when \u03b8 = 0 , the tension is equal to force of the load TA , \nand when angle \u03b8 = \u03b8 , \nB the sailor applies to the rope,\nA B the tension is equal to the force T \nT =TB \n\u03b8 =\u03b8BA\ndT = \u2212 \n\u00b5 d\u03b8 . \n(8.6.104)\n\u222b\n\u222b \ns\nT\nT =TA \n\u03b8 =0 \nThe result of the integration is \n\u239b TB \u239e \nln \n= \u2212\u00b5 \n(8.6.105)\n\u239d\u239c T\ns \u03b8BA . \nA \u23a0\u239f \nNote that the exponential of the natural logarithm \n\u239b\n\u239b \n\u239e\u239e\nTB \nTB\nexp ln \u239c\n\u239f = \n, \n(8.6.106)\n\u239c\n\u239f \nT\nT\n\u239d\n\u239d A \u23a0\u23a0 \nA \nso exponentiating both sides of Equation (8.6.105) yields \n\u2212 \u00b5\nTB = e\ns \u03b8BA ; \n(8.6.107)\nTA \nthe tension decreases exponentially, \nTB = TA e\n\u2212 \u00b5s\u03b8BA , \n(8.6.108) \nBecause the tension decreases exponentially, the sailor need only apply a small force to \nprevent the rope from slipping. \nExample 8.12 Free Fall with Air Drag \nConsider an object of mass m that is in free fall but experiencing air resistance. The \nmagnitude of the drag force is given by Eq. (8.6.1), where \u03c1 is the density of air, A is \n8-48", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_208", "text": "the cross-sectional area of the object in a plane perpendicular to the motion, and CD is \nthe drag coefficient. Assume that the object is released from rest and very quickly attains \nspeeds in which Eq. (8.6.1) applies. Determine (i) the terminal velocity, and (ii) the \nvelocity of the object as a function of time. \nSolution: Choose positive y -direction downwards with the origin at the initial position \nof the object as shown in Figure 8.48(a). \ny(t) \n\u02c6j \nh \n+ y \nO \nv(t) \n(a) \n\u02c6j \nO \n. \n+ y \nmg \nFdrag \n(b) \nFigure 8.48 (a) Coordinate system for marble; (b) free body force diagram on marble \nThere are two forces acting on the object: the gravitational force, and the drag force \nwhich is given by Eq. (8.6.1). The free body diagram is shown in the Figure 8.48(b). \nNewton\u2019s Second Law is then \n2 \ndv \nmg \u2212 (1/ 2)CD A\u03c1v = m \n, \n(8.6.109)\ndt \nSet \u03b2 = (1/ 2)CD A\u03c1 . Newton\u2019s Second Law can then be written as \n2 \ndv \nmg \u2212\u03b2 v = m \n. \n(8.6.110)\ndt \nInitially when the object is just released with v = 0 , the air drag is zero and the \nacceleration dv / dt is maximum. As the object increases its velocity, the air drag \nbecomes larger and dv / dt decreases until the object reaches terminal velocity and \ndv / dt = 0 . Set dv / dt = 0 in Eq. (8.6.15) and solve for the terminal velocity yielding. \nv\u221e = mg \n\u03b2 = \n2mg \nCD A\u03c1 . \n(8.6.111)  \nValues for the magnitude of the terminal velocity is shown in Table 8.3 for a variety of \nobjects with the same drag coefficient CD = 0.5. \n8-49", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_209", "text": "Table 8.3 Terminal Velocities for Different Sized Objects with CD = 0.5 \nObject \nMass m (kg) \nArea A (m2 ) \nTerminal Velocity v\u221e (m \u22c5s\u22121 ) \nRain drop \n4 \u00d710\u22126 \n3\u00d710\u22126 \n6.5 \nHailstone \n4 \u00d710\u22123 \n3\u00d710\u22124 \n20 \nOsprey \n20 \n2.5\u00d710\u22121 \n50 \nHuman Being \n7.5 \u00d7101 \n6 \u00d710\u22121 \n60 \nIn order to integrate Eq. (8.6.15), we shall apply the technique of separation of variables \nand integration by partial fractions. First rewrite Eq. (8.6.15) as \n\u2212\u03b2 \ndv \ndv \n\u239b \n1\n1 \n\u239e\ndt = \n=\n= \u2212 \n+ \n\u23a0\u239f dv . (8.6.112)\nm \n\u239b 2 \u2212 mg \u239e (v2 \u2212 v\u221e \n2 ) \u239d\u239c 2v\u221e (v + v\u221e )\n2v\u221e (v \u2212 v\u221e )\nv\n\u239d\u239c \n\u03b2 \u23a0\u239f \nAn integral expression of Eq. (8.6.112) is then \nv\u2032=v(t ) \nv\u2032=v(t ) \nt\u2032=t\ndv\u2032 \ndv\u2032\n\u03b2\n\u2212 \n+\n= \u2212 \ndt\u2032 . \n(8.6.113)\n\u222b\n\u222b \n\u222b \nv\u2032=0 2v\u221e (v\u2032 + v\u221e ) \nv\u2032=0 2v\u221e (v\u2032 \u2212 v\u221e ) \nm t\u2032=0 \nIntegration yields \nv\u2032=v(t ) \nv\u2032=v(t ) \nt\u2032=t\ndv\u2032 \ndv\u2032\n\u03b2\n\u2212 \n+\n= \u2212 \ndt\u2032\n\u222b\n\u222b \n\u222b\n2v\u221e (v\u2032 + v\u221e )\n2v\u221e (v\u2032 \u2212 v\u221e ) \nm \nv\u2032=0 \nv\u2032=0 \nt\u2032=0 \n. \n(8.6.114)\n1 \u239b\n\u239b v(t) + v\u221e\u239e \n\u239b v\u221e\u2212 v(t)\u239e\u239e \n\u03b2 \n\u239c\u2212 ln \n\u23a0\u239f + ln \nt\n2v\u221e\u239d \n\u239d\u239c \nv\u221e \n\u239d\u239c \nv\u221e \n\u23a0\u239f \u239f\u23a0 \n= \u2212 m \nAfter some algebraic manipulations, Eq. (8.6.114) can be rewritten as \n\u239b v\u221e\u2212 v(t)\u239e \n2v\u221e\u03b2\nln \n\u23a0\u239f = \u2212 \nt \n(8.6.115)\n\u239d\u239c v(t) + v\u221e \nm \nExponentiate Eq. (8.6.115) yields \n\u239b v\u221e\u2212 v(t)\u239e\n\u2212 2v\u221e\u03b2 t . \n(8.6.116)\n\u239d\u239c v(t) + v\u221e \u23a0\u239f = e", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_210", "text": "(8.6.115)\n\u239d\u239c v(t) + v\u221e \nm \nExponentiate Eq. (8.6.115) yields \n\u239b v\u221e\u2212 v(t)\u239e\n\u2212 2v\u221e\u03b2 t . \n(8.6.116)\n\u239d\u239c v(t) + v\u221e \u23a0\u239f = e\nm \nAfter some algebraic rearrangement the y -component of the velocity as a function of \ntime is given by \n8-50", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_211", "text": "2v\u221e\u03b2\n\u239b\n\u2212 \nt \u239e \n1\u2212 e \n\u239b v\u221e\u03b2 \u239e \nv(t) = v\u221e\u239c \nm \u239f = v\u221e tan h\nt \n\u23a0\u239f . \n(8.6.117)\n2v\n\u239c \n\u2212\n\u221e\u03b2 t \u239f \n\u239d\u239c m \nm\n\u239d 1+ e \n\u23a0 \nv\u221e\u03b2 \n\u03b2 \nmg \n\u03b2g \n(1/ 2)CD A\u03c1g\nwhere \n= \n.\n= \n= \nm\nm \n\u03b2 \nm\nm \n8-51", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_212", "text": "Chapter 9 Circular Motion Dynamics \n9.1 Introduction Newton\u2019s Second Law and Circular Motion ................................. 1  \n9.2 Universal Law of Gravitation and the Circular Orbit of the Moon .................. 1  \n9.2.1 Universal Law of Gravitation ......................................................................... 2  \n9.2.2 Kepler\u2019s Third Law and Circular Motion..................................................... 4  \n9.3 Worked Examples Circular Motion...................................................................... 5  \nExample 9.1 Geosynchronous Orbit ....................................................................... 5  \nExample 9.2 Double Star System............................................................................. 6", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_213", "text": "Example 9.3 Rotating Objects ................................................................................. 9  \nExample 9.4 Tension in a Rope.............................................................................. 11  \nExample 9.5 Object Sliding in a Circular Orbit on the Inside of a Cone .......... 12  \nExample 9.6 Coin on a Rotating Turntable.......................................................... 14  \nAppendix 9A The Gravitational Field of a Spherical Shell of Matter ....................... 16", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_214", "text": "Chapter 9 Equation Chapter 9 Section 1Circular Motion  \nDynamics  \nI shall now recall to mind that the motion of the heavenly bodies is \ncircular, since the motion appropriate to a sphere is rotation in a circle.1\nNicholas Copernicus \n9.1 Introduction Newton\u2019s Second Law and Circular Motion \nWe have already shown that when an object moves in a circular orbit of radius r with\n\uf072\nangular velocity \u03c9 , it is most convenient to choose polar coordinates to describe the \nposition, velocity and acceleration vectors. In particular, the acceleration vector is given \nby \n\uf072 \n\u239bd\u03b8 \u239e \n2 \nd 2\u03b8 \na(t) = \u2212r \nr\u02c6(t) + r \n\u03b8\u02c6(t) .\n(9.1.1)\n\u239d\u239c dt \u23a0\u239f\ndt2 \n\uf072\n\uf072\nThen Newton\u2019s Second Law, F = ma , can be decomposed into radial ( r\u02c6 -) and tangential \n( \u03b8\u02c6 -) components \n\u239bd\u03b8 \u239e \n2 \nF r = \u2212mr \n(circular motion) , \n(9.1.2)\n\u239d\u239cdt \u23a0\u239f\nd 2\u03b8\nF\u03b8 = mr \n(circular motion) . \n(9.1.3)\ndt2 \nFor the special case of uniform circular motion, d 2\u03b8 / dt2 = 0 , and so the sum of the \ntangential components of the force acting on the object must therefore be zero, \nF\u03b8 = 0 (uniform circular motion) . \n(9.1.4) \n9.2 Universal Law of Gravitation and the Circular Orbit of the Moon \nAn important example of (approximate) circular motion is the orbit of the Moon around \nthe Earth. We can approximately calculate the time T the Moon takes to complete one \ncircle around the earth (a calculation of great importance to early lunar calendar systems, \nwhich became the basis for our current model.) Denote the distance from the moon to the \ncenter of the earth by Re, m . \n1 Dedicatory Letter to Pope Paul III. \n9-1", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_215", "text": "Because the Moon moves nearly in a circular orbit with angular speed \u03c9 = 2\u03c0 / T \nit is accelerating towards the Earth. The radial component of the acceleration (centripetal \nacceleration) is \n4\u03c0 2 R \nar = \u2212 \ne, m . \n(9.2.1)\nT 2 \nAccording to Newton\u2019s Second Law, there must be a centripetal force acting on the \nMoon directed towards the center of the Earth that accounts for this inward acceleration. \n9.2.1 Universal Law of Gravitation \nNewton\u2019s Universal Law of Gravitation describes the gravitational force between two \nbodies 1 and 2 with masses m1 and m2 respectively. This force is a radial force (always \npointing along the radial line connecting the masses) and the magnitude is proportional to \nthe inverse square of the distance that separates the bodies. Then the force on object 2 \ndue to the gravitational interaction between the bodies is given by, \n\uf072 \nm1 m2\n= \u2212G \n\u02c6 \n(9.2.2)\nF1,2 \n2 r1, 2 , \nr1, 2 \nwhere \nis the distance between the two bodies and r\u02c6 \nis the unit vector located at the \nr1,2 \n1,2 \nposition of object 2 and pointing from object 1 towards object 2. The Universal \n\u221211 \n2 \n\u22122\nGravitation Constant is G = 6.67\u00d710 N m \u22c5 kg\n\u22c5 \n. Figure 9.1 shows the direction of the \nforces on bodies 1 and 2 along with the unit vector r\u02c61,2 . \nFigure 9.1 Gravitational force of interaction between two bodies \nNewton realized that there were still some subtleties involved. First, why should \nthe mass of the Earth act as if it were all placed at the center? Newton showed that for a \nperfect sphere with uniform mass distribution, all the mass may be assumed to be located \nat the center. (This calculation is difficult and can be found in Appendix 9A to this \n9-2", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_216", "text": "chapter.) We assume for the present calculation that the Earth and the Moon are perfect \nspheres with uniform mass distribution. \nSecond, does this gravitational force between the Earth and the Moon form an \naction-reaction Third Law pair? When Newton first explained the Moon\u2019s motion in \n1666, he had still not formulated the Third Law, which accounted for the long delay in \nthe publication of the Principia. The link between the concept of force and the concept of \nan action-reaction pair of forces was the last piece needed to solve the puzzle of the effect \nof gravity on planetary orbits. Once Newton realized that the gravitational force between \nany two bodies forms an action-reaction pair, and satisfies both the Universal Law of \nGravitation and his newly formulated Third Law, he was able to solve the oldest and \nmost important physics problem of his time, the motion of the planets. \nThe test for the Universal Law of Gravitation was performed through \nexperimental observation of the motion of planets, which turned out to be resoundingly \nsuccessful. For almost 200 years, Newton\u2019s Universal Law was in excellent agreement \nwith observation. A sign of more complicated physics ahead, the first discrepancy only \noccurred when a slight deviation of the motion of Mercury was experimentally confirmed \nin 1882. The prediction of this deviation was the first success of Einstein\u2019s Theory of \nGeneral Relativity (formulated in 1915). \nWe can apply this Universal Law of Gravitation to calculate the period of the \nMoon\u2019s orbit around the Earth. The mass of the Moon is m1 = 7.36\u00d71022 kg and the mass \nof the Earth is m2 = 5.98\u00d71024 kg . \nThe distance from the Earth to the Moon is \nR e, m = 3.82 \u00d7 108 m . We show the force diagram in Figure 9.2. \nFigure 9.2 Gravitational force of moon \nNewton\u2019s Second Law of motion for the radial direction becomes \n4\u03c0 2 R\nm1 m2\n\u2212G \ne, m . \n(9.2.3)\n2 \n= \u2212m1\nR \nT 2 \ne, m \nWe can solve this equation for the period of the orbit, \n9-3", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_217", "text": "2 \n3 \ne,m \n2 \n4 R\nT \nG m \n\u03c0 \n= \n. \n(9.2.4)  \nSubstitute the given values for the radius of the orbit, the mass of the earth, and the \nuniversal gravitational constant. The period of the orbit is \nT = \n4\u03c0 2 (3.82 \u00d7108 m)3 \n(6.67 \u00d710\u221211 N \u22c5 m2 \u22c5 kg\u22122 )(5.98 \u00d71024 kg) = 2.35 \u00d7106 s . \n(9.2.5) \nThis period is given in days by \nT = (2.35\u00d7106 s) \n1 day \n8.64 \u00d7104 \n\u239b \n\u239d\u239c \ns \n\u239e \n\u23a0\u239f = 27.2 days. \n(9.2.6) \nThis period is called the sidereal month because it is the time that it takes for the Moon to \nreturn to a given position with respect to the stars. \nThe actual time T1 between full moons, called the synodic month (the average \nperiod of the Moon\u2019s revolution with respect to the earth and is 29.53 days, it may range \nbetween 29.27 days and 29.83 days), is longer than the sidereal month because the Earth \nis traveling around the Sun. So for the next full moon, the Moon must travel a little \nfarther than one full circle around the Earth in order to be on the other side of the Earth \nfrom the Sun (Figure 9.3). \nFigure 9.3: Orbital motion between full moons \nTherefore the time T1 between consecutive full moons is approximately T1 \uf03b T + \u0394T \nwhere \u0394T \uf03b T / 12 = 2.3 days . So T1 \uf03b 29.5 days . \n9.2.2 Kepler\u2019s Third Law and Circular Motion \nThe first thing that we notice from the above solution is that the period does not depend \non the mass of the Moon. We also notice that the square of the period is proportional to \nthe cube of the distance between the Earth and the Moon, \n9-4", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_218", "text": "4\u03c0 2R3 \n2 \ne,m \nT = \n. \n(9.2.7)\nG m2 \nThis is an example of Kepler\u2019s Third Law, of which Newton was aware. This \nconfirmation was convincing evidence to Newton that his Universal Law of Gravitation \nwas the correct mathematical description of the gravitational force law, even though he \nstill could not explain what \u201ccaused\u201d gravity. \n9.3 Worked Examples Circular Motion \nExample 9.1 Geosynchronous Orbit \nA geostationary satellite goes around the earth once every 23 hours 56 minutes and 4 \nseconds, (a sidereal day, shorter than the noon-to-noon solar day of 24 hours) so that its \nposition appears stationary with respect to a ground station. The mass of the earth is \nm e = 5.98 \u00d7 1024 kg . The mean radius of the earth is R e = 6.37 \u00d7 106 m . The universal \nconstant of gravitation is G = 6.67 \u00d7 10\u221211 N \u22c5 m2 \u22c5 kg\u22122 . What is the radius of the orbit of a \ngeostationary satellite? Approximately how many earth radii is this distance? \nSolution: The satellite\u2019s motion can be modeled as uniform circular motion. The \ngravitational force between the earth and the satellite keeps the satellite moving in a \ncircle (In Figure 9.4, the orbit is close to a scale drawing of the orbit). The acceleration of \nthe satellite is directed towards the center of the circle, that is, along the radially inward \ndirection. \nFigure 9.4 Geostationary satellite orbit (close to a scale drawing of orbit). \nChoose the origin at the center of the earth, and the unit vector r\u02c6 along the radial \ndirection. This choice of coordinates makes sense in this problem since the direction of \nacceleration is along the radial direction. \n9-5", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_219", "text": "\uf072\n\uf072\nLet r be the position vector of the satellite. The magnitude of r (we denote it as rs ) is \nthe distance of the satellite from the center of the earth, and hence the radius of its \ncircular orbit. Let \u03c9 be the angular velocity of the satellite, and the period is T = 2\u03c0 / \u03c9 . \nThe acceleration is directed inward, with magnitude rs\u03c9 2 ; in vector form, \n\uf072 a = \u2212rs\u03c9 2\u02c6r . \n(9.3.1) \nApply Newton\u2019s Second Law to the satellite for the radial component. The only force in \nthis direction is the gravitational force due to the Earth, \n\uf072 \nF \n= \u2212m \u03c9 2r r\u02c6 . \n(9.3.2)\ngrav \ns\ns \nThe inward radial force on the satellite is the gravitational attraction of the earth, \nm me\ns\n\u2212G \nr\u02c6 = \u2212m \u03c9 2r r\u02c6 . \n(9.3.3)\n2 \ns\ns\nrs \nEquating the r\u02c6 components, \nm me\ns\nG \n= m \u03c9 2r . \n(9.3.4)\n2 \ns\ns\nrs \nSolving for the radius of orbit of the satellite r s , \n1/ 3 \n\u239b G me \u239e \nrs = \u239c\n\u239f . \n(9.3.5)\n\u239d\u03c9 2 \u23a0 \nThe period T of the satellite\u2019s orbit in seconds is 86164 s and so the angular speed is \n2\u03c0 \n2\u03c0\n\u22125 \u22121\n\u03c9 \n= \n\u00d7 \ns . \n(9.3.6)\n=\n= 7.2921 10 \nT \n86164 s \nUsing the values of \u03c9, G and me in Equation (9.3.5), we determine rs , \n7\nrs = 4.22\u00d710 m = 6.62 Re . \n(9.3.7) \nExample 9.2 Double Star System \nConsider a double star system under the influence of gravitational force between the \nstars. Star 1 has mass m1 and star 2 has mass m2 . Assume that each star undergoes \nuniform circular motion such that the stars are always a fixed distance s apart (rotating \ncounterclockwise in Figure 9.5). What is the period of the orbit? \n9-6", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_220", "text": "Figure 9.5 Two stars undergoing circular orbits about each other \nSolution: Because the distance between the two stars doesn\u2019t change as they orbit about \neach other, there is a central point where the lines connecting the two objects intersect as \nthe objects move, as can be seen in the figure above. (We will see later on in the course \nthat central point is the center of mass of the system.) Choose radial coordinates for each \nstar with origin at that central point. Let r\u02c61 be a unit vector at Star 1 pointing radially \n\uf072\naway from the center of mass. The position of object 1 is then r\nr\u02c6 , where r1 is the \n1 = r1 1 \ndistance from the central point. Let r\u02c62 be a unit vector at Star 2 pointing radially away \n\uf072\nfrom the center of mass. The position of object 2 is then \nr\u02c6 , where \nis the \nr2 = r2 \n2 \nr2 \ndistance from the central point. Because the distance between the two stars is fixed we \nhave that \ns = r1 + r2 . \nThe coordinate system is shown in Figure 9.6 \n9-7\n\nFigure 9.6 Coordinate system for double star orbits \nThe gravitational force on object 1 is then \n\uf076 \nGm1m2\n= \u2212 \nr\u02c61.\nF2,1 \n2s \nThe gravitational force on object 2 is then \n\uf076 \nGm1m2\n= \u2212 \nr\u02c6\nF1,2 \n2\n2 . \ns \nThe force diagrams on the two stars are shown in Figure 9.7. \nFigure 9.7 Force diagrams on objects 1 and 2 \nLet \u03c9 denote the magnitude of the angular velocity of each star about the central point.\n\uf076\n\uf076\nThen Newton\u2019s Second Law, F = m a , for Star 1 in the radial direction r\u02c6 is\n1\n1\n1 \n1 \nm1 m2 \n\u03c9 2\n\u2212G \n2 \n.\n= \u2212m1 r1\ns \nWe can solve this for r1 , \nm \nr = G \n2 .\n1 \n2\n2\n\u03c9 s \n\uf076\n\uf076\nNewton\u2019s Second Law, F = m a , for Star 2 in the radial direction \u02c6 is\n2 \n2\n2 \nr2 \n9-8", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_221", "text": "m1 m2 \n\u03c9 2\n\u2212G \nr2\n.\n2 \n= \u2212m2\ns \nWe can solve this for r2 , \nm1\nr2 = G \n2 2 .\n\u03c9 s \nBecause s , the distance between the stars, is constant \n)\nm2 \nm1\n(m2 + m1\ns = r1 \n= G \n+ G \n= G\n+ r2 \n2\n2 \n2 \n.\n\u03c9 2s \n\u03c9 2s \n\u03c9 2s \nThus the magnitude of the angular velocity is \n)\u239e\n1 2 \n\u239b \n(m2 + m1\n\u03c9 = G \n3\n,\n\u239d\u239c \ns \n\u23a0\u239f \nand the period is then \n2\u03c0\n\u239b \n4\u03c0 2s3 \n\u239e\n1 2 \nT =\n= \n. \n(9.3.8)\n\u03c9 \n\u239d\u239c G(m2 + m1)\u23a0\u239f \nNote that both masses appear in the above expression for the period unlike the expression \nfor Kepler\u2019s Law for circular orbits. Eq. (9.2.7). The reason is that in the argument \nleading up to Eq. (9.2.7), we assumed that m1 << m2 , this was equivalent to assuming that \nthe central point was located at the center of the Earth. If we used Eq. (9.3.8) instead we \nwould find that the orbital period for the circular motion of the Earth and moon about \neach other is \nT = \n4\u03c0 2 (3.82 \u00d7108 m)3 \n(6.67 \u00d710\u221211 N \u22c5m2 \u22c5kg\u22122 )(5.98 \u00d71024 kg +7.36 \u00d71022 kg) = 2.33\u00d7106 s , \nwhich is 1.43 \u00d7 104 s = 0.17 d shorter than our previous calculation. \nExample 9.3 Rotating Objects \nTwo objects 1 and 2 of mass \nand \nare whirling around a shaft with a constant \nm1 \nm2 \nangular velocity \u03c9 . The first object is a distance d from the central axis, and the second \nobject is a distance 2d from the axis (Figure 9.8). You may ignore the mass of the strings \nand neglect the effect of gravity. (a) What is the tension in the string between the inner \n9-9", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_222", "text": "object and the outer object? (b) What is the tension in the string between the shaft and the \ninner object? \nFigure 9.8 Objects attached to a rotating shaft \nSolution: We begin by drawing separate force diagrams, Figure 9.9a for object 1 and \nFigure 9.9b for object 2. \n(a) \n(b)  \nFigure 9.9 (a) and 9.9 (b) Free-body force diagrams for objects 1 and 2  \n\uf076\n\uf076\nNewton\u2019s Second Law, F = m a , for the inner object in the radial direction is\n1 \n1\n1 \nr\nd \u03c9 2 .\n\u02c6 : T2 \u2212 T1 = \u2212m1 \n\uf076\n\uf076\nNewton\u2019s Second Law, F = m a , for the outer object in the radial direction is\n2 \n2\n2 \nr\u02c6 : \u2212 T2 = \u2212m2 2d \u03c9 2. \nThe tension in the string between the inner object and the outer object is therefore \n2d \u03c9 2 .\nT2 = m2 \n9-10", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_223", "text": "Using this result for T2 in the force equation for the inner object yields \n2d \u03c9 2 \u2212 T1 \nd \u03c9 2 ,\nm2 \n= \u2212m1 \nwhich can be solved for the tension in the string between the shaft and the inner object \n= d \u03c9 2(m1 + 2m2 ) .\nT1 \nExample 9.4 Tension in a Rope \nA uniform rope of mass mand length L is attached to shaft that is rotating at constant \nangular velocity \u03c9 . Find the tension in the rope as a function of distance from the shaft. \nYou may ignore the effect of gravitation. \nSolution: Divide the rope into small pieces of length \u0394r , each of mass \u0394m = (m / L)\u0394r . \nConsider the piece located a distance r from the shaft (Figure 9.10). \nFigure 9.10 Small slice of rotating rope \nThe radial component of the force on that piece is the difference between the tensions \nevaluated at the sides of the piece, Fr = T (r + \u0394r) \u2212 T (r) , (Figure 9.11). \nFigure 9.11 Free-body force diagram on small slice of rope \n= \u2212r\u03c9 2\nThe piece is accelerating inward with a radial component ar \n. Thus Newton\u2019s \nSecond Law becomes \nF = \u2212\u0394m\u03c9 2r \nr \n(9.3.9)\nT (r + \u0394r) \u2212 T (r) = \u2212(m / L)\u0394r r\u03c9 2 . \nDenote the difference in the tension by \u0394T = T (r + \u0394r) \u2212 T (r) . After dividing through by \n\u0394r , Eq. (9.3.9) becomes \n9-11", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_224", "text": "\u0394T = \u2212(m / L) r\u03c9 2 . \n(9.3.10)\n\u0394r \nIn the limit as \u0394r \u2192 0 , Eq. (9.3.10) becomes a differential equation, \ndT = \u2212(m / L)\u03c9 2r . \n(9.3.11)\ndr \nFrom this, we see immediately that the tension decreases with increasing radius. We \nshall solve this equation by integration \nr \u2032= r \nT (r) \u2212 T (L) = \u222b \ndT dr\u2032 = \u2212(m\u03c9 2 / L)\u222b\nr r\u2032 dr\u2032 \ndr\u2032 \nL \nr \u2032= L \n= \u2212(m\u03c9 2 / 2L)(r 2 \u2212 L2) \n(9.3.12) \n= (m\u03c9 2 / 2L)(L2 \u2212 r 2). \nWe use the fact that the tension, in the ideal case, will vanish at the end of the rope, \nr = L . Thus, \nT (r) = (m\u03c9 2 / 2L)(L2 \u2212 r 2). \n(9.3.13) \nThis last expression shows the expected functional form, in that the tension is largest \nclosest to the shaft, and vanishes at the end of the rope. \nExample 9.5 Object Sliding in a Circular Orbit on the Inside of a Cone \nConsider an object of mass m that slides without friction on the inside of a cone moving \nin a circular orbit with constant speed v0 . The cone makes an angle \u03b8 with respect to a \nvertical axis. The axis of the cone is vertical and gravity is directed downwards. The apex \nhalf-angle of the cone is \u03b8 as shown in Figure 9.12. Find the radius of the circular path \nand the time it takes to complete one circular orbit in terms of the given quantities and g . \nFigure 9.12 Object in a circular orbit on inside of a cone \n9-12", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_225", "text": "Solution: Choose cylindrical coordinates as shown in the above figure. Choose unit \nvectors r\u02c6 pointing in the radial outward direction and k\u02c6 pointing upwards. The force \ndiagram on the object is shown in Figure 9.13. \nFigure 9.13 Free-body force diagram on object \nThe two forces acting on the object are the normal force of the wall on the object and the \ngravitational force. Then Newton\u2019s Second Law in the r\u02c6 -direction becomes \n\u2212mv2 \n\u2212N cos\u03b8 = \nr \nand in the k\u02c6 -direction becomes \nN sin\u03b8 \u2212 mg = 0. \nThese equations can be re-expressed as \n2v\nN cos\u03b8 = m r \nN sin\u03b8 = m g . \nWe can divide these two equations, \nN sin\u03b8 \nmg \n= \nN cos\u03b8 \nmv2 / r \nyielding \nr g \ntan\u03b8 = \n2 . \nv \nThis can be solved for the radius, \nr = v2 \ntan\u03b8 . \ng \nThe centripetal force in this problem is the vector component of the contact force that is \npointing radially inwards, \nFcent = N cos\u03b8 = m g cot\u03b8 , \n9-13", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_226", "text": "where N sin\u03b8 = m g has been used to eliminate N in terms of m , g and \u03b8 . The radius \nis independent of the mass because the component of the normal force in the vertical \ndirection must balance the gravitational force, and so the normal force is proportional to \nthe mass. \nExample 9.6 Coin on a Rotating Turntable \nA coin of mass m (which you may treat as a point object) lies on a turntable, exactly at \nthe rim, a distance R from the center. The turntable turns at constant angular speed \u03c9 \nand the coin rides without slipping. Suppose the coefficient of static friction between the \nturntable and the coin is given by \u00b5 . Let g be the gravitational constant. What is the \nmaximum angular speed \u03c9max such that the coin does not slip? \nFigure 9.14 Coin on Rotating Turntable \nSolution: The coin undergoes circular motion at constant speed so it is accelerating \ninward. The force inward is static friction and at the just slipping point it has reached its \nmaximum value. We can use Newton\u2019s Second Law to find the maximum angular speed \n\u03c9max . We choose a polar coordinate system and the free-body force diagram is shown in \nthe figure below. \nFigure 9.15 Free-body force diagram on coin \nThe contact force is given by \n\uf072 \n\uf072\n\uf072 \nC = N + f s = N k\u02c6 \u2212 f s r\u02c6 . \n(9.3.14) \nThe gravitational force is given by \n\uf072 \nF grav = \u2212mg k\u02c6 . \n(9.3.15) \nNewton\u2019s Second Law in the radial direction is given by \n9-14", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_227", "text": "\u2212 f s = \u2212m R\u03c9 2 . \n(9.3.16) \nNewton\u2019s Second Law, Fz = ma z , in the z-direction, noting that the disc is static hence \na z = 0 , is given by \n0\nN \u2212 mg = . \n(9.3.17) \nThus the normal force is \nN = mg . \n(9.3.18) \nAs \u03c9 increases, the static friction increases in magnitude until at \u03c9 = \u03c9max and static \nfriction reaches its maximum value (noting Eq. (9.3.18)). \n( f ) \n= \u00b5N = \u00b5mg . \n(9.3.19)\ns max \nAt this value the disc slips. Thus substituting this value for the maximum static friction \ninto Eq. (9.3.16) yields \n2\n\u00b5mg = mR\u03c9max . \n(9.3.20) \nWe can now solve Eq. (9.3.20) for maximum angular speed \u03c9max such that the coin does \nnot slip \ng \nR \n\u00b5\n\u03c9  \n.  \n(9.3.21) \n=  \nmax \n9-15", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_228", "text": "Appendix 9A The Gravitational Field of a Spherical Shell of  \nMatter  \nWhen analyzing gravitational interactions between uniform spherical bodies we assumed \nwe could treat each sphere as a point-like mass located at the center of the sphere and \nthen use the Universal Law of Gravitation to determine the force between the two point-\nlike objects. We shall now justify that assumption. For simplicity we only need to \nconsider the interaction between a spherical object and a point-like mass. We would like \nto determine the gravitational force on the point-like object of mass \ndue to the \nm1 \ngravitational interaction with a solid uniform sphere of mass m2 and radius R . In order \nto determine the force law we shall first consider the interaction between the point-like \nobject and a uniform spherical shell of mass ms and radius R . We will show that: \n1)  The gravitational force acting on a point-like object of mass m1 located a distance \nr > R from the center of a uniform spherical shell of mass ms and radius R is the \nsame force that would arise if all the mass of the shell were placed at the center of \nthe shell. \n2)  The gravitational force on an object of mass m1 placed inside a spherical shell of \nmatter is zero. \nThe force law summarizes these results: \n\uf072  \nF s,1(r) = \n\u23a7\n\u23aa\u23a8 \n\u23aa\u23a9 \n\u2212G  msm1 r\u02c6, \nr > R \nr\uf072 \n2 \n,  \n0, \nr < R \nwhere r\u02c6 is the unit vector located at the position of the object and pointing radially away \nfrom the center of the shell. \nFor a uniform spherical distribution of matter, we can divide the sphere into thin shells. \nThen the force between the point-like object and each shell is the same as if all the mass \nof the shell were placed at the center of the shell. Then we add up all the contributions of \nthe shells (integration), the spherical distribution can be treated as point-like object \nlocated at the center of the sphere justifying our assumption. \nThus it suffices to analyze the case of the spherical shell. We shall first divide the shell \ninto small area elements and calculate the gravitational force on the point-like object due \nto one element of the shell and then add the forces due to all these elements via \nintegration. \n9-16", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_229", "text": "We begin by choosing a coordinate system. Choose our z -axis to be directed from the \n\uf072\ncenter of the sphere to the position of the object, at position r = z k\u02c6 , so that z \u2265 0. \n(Figure 9A.1 shows the object lying outside the shell with z > R ). \nFigure 9A.1 Object lying outside shell with z > R . \nChoose spherical coordinates as shown in Figure 9A.2. \nFigure 9A.2 Spherical coordinates \nFor a point on the surface of a sphere of radius r = R , the Cartesian coordinates are \nrelated to the spherical coordinates by \nx = Rsin\u03b8 cos\u03c6, \ny = Rsin\u03b8 sin\u03c6, \n(9.A.1)  \nz = Rcos\u03b8,  \n9-17\n\nwhere 0 \u2264\u03b8 \u2264\u03c0 and 0 \u2264\u03c6 \u2264 2\u03c0 . \nNote that the angle \u03b8 in Figure 9A.2 and Equations (9.A.1) is not the same as that in \nplane polar coordinates or cylindrical coordinates. The angle \u03b8 is known as the co\u00ad\nlatitude, the complement of the latitude. We now choose a small area element shown in \nFigure 9A.3. \nFigure 9A.3 Infinitesimal area element \nThe infinitesimal area element on the surface of the shell is given by \nda = R2 sin\u03b8d\u03b8d\u03c6 . \nThen the mass dm contained in that element is \ndm = \u03c3 da = \u03c3 R2 sin\u03b8 \u03b8 \u03c6 .\nd d \nwhere \u03c3 is the surface mass density given by \n\u03c3 = ms / 4\u03c0 R2 . \n\uf072 \nThe gravitational force \non the object of mass m1 that lies outside the shell due to \nFdm, m1 \nthe infinitesimal piece of the shell (with mass dm ) is shown in Figure 9A.4. \n9-18", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_230", "text": "Figure 9A.4 Force on a point-like object due to piece of shell \nThe contribution from the piece with mass dm to the gravitational force on the object of \nmass m1 that lies outside the shell has a component pointing in the negative k\u02c6 -direction \nand a component pointing radially away from the z -axis. By symmetry there is another \nmass element with the same differential mass dm\u2032 = dm on the other side of the shell \nwith same co-latitude \u03b8 but with \u03c6 replaced by \u03c6 \u00b1\u03c0 ; this replacement changes the sign \nof x and y in Equations (9.A.1) but leaves z unchanged. This other mass element \nproduces a gravitational force that exactly cancels the radial component of the force \npointing away from the z -axis. Therefore the sum of the forces of these differential mass \nelements on the object has only a component in the negative k\u02c6 -direction (Figure 9A.5) \nFigure 9A.5 Symmetric cancellation of components of force \n9-19\n\nTherefore we need only the z -component vector of the force due to the piece of the shell \non the point-like object. \nFigure 9A.6 Geometry for calculating the force due to piece of shell. \nFrom the geometry of the set-up (Figure 9A.6) we see that \n\uf072 \nm1dm\n(dF ) \u2261 dF k\u02c6 = \u2212G \ncos\u03b1k\u02c6 . \ns,1 z\nz \n2r s1 \nThus \nm1dm \nGm sm1 cos\u03b1 sin\u03b8d\u03b8d\u03c6\ndF z = \u2212G \ncos\u03b1 = \u2212 \n. \n(9.A.2)\n2\n4\u03c0 \n2\nr \nr \ns1 \ns1 \nThe integral of the force over the surface is then \n\u03b8 =\u03c0 \u03c6 = 2\u03c0\n\u03b8 =\u03c0 \u03c6 =2\u03c0\nGm m \ndmcos\u03b1 \ns \n1 \ncos\u03b1 sin\u03b8d\u03b8d\u03c6\nF = \u2212Gm \n= \u2212 \n. \n(9.A.3)\nz \n1 \u222b\u222b \n\u222b\u222b\nr 2 \n4\u03c0 \nr 2 \n\u03b8 =0 \u03c6 =0 \ns1 \n\u03b8 =0 \u03c6 =0 \ns1 \nThe \u03c6 -integral is straightforward yielding \nGm \n\u03b8=\u03c0 cos\u03b1 sin\u03b8d\u03b8\ns m1\nF = \u2212 \n. \n(9.A.4)\nz \n\u222b \n2\n2 \nr\n\u03b8=0 \ns1 \nFrom Figure 9A.6 we can use the law of cosines in two different ways \n9-20", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_231", "text": "2 = R2 + z\nrs1 \n2 \u2212 2 R z cos\u03b8 \n(9.A.5)\nR2 \n2 + r 2 \u2212 2r\n= z\nz cos\u03b1. \ns1 \ns,1 \nDifferentiating the first expression in (9.A.5), with R and z constant yields, \n2r dr = 2 R z sin\u03b8 d\u03b8 . \n(9.A.6)\ns,1 \ns,1 \nHence \nr \nsin\u03b8 d\u03b8 = \ns,1 dr \n(9.A.7)\ns,1 .\nR z \nand from the second expression in (9.A.5) we have that \ncos\u03b1 = \n1 \u23a1\u23a3(z2 \u2212 R2) + r 2 \u23a4\u23a6 . \n(9.A.8)\n2 zr s,1 \ns1 \nWe now have everything we need in terms of r s,1 . \nFor the case when z > R , rs,1 varies from z \u2212 R to z + R . Substituting Equations (9.A.7) \nand (9.A.8) into Eq. (9.A.3) and using the limits for the definite integral yields \nGm \n\u03b8 =\u03c0 cos\u03b1 sin\u03b8\nFz = \u2212 \n2 \nsm1 \u222b \n2 \nd\u03b8 \nr\n\u03b8 =0 \ns,1 \nGm m1 1 \nz+R 1\n1 r dr \n= \u2212 \ns \n\u23a1(z2 \u2212 R2 )+ r 2 \u23a4 \ns,1 \ns,1 \n(9.A.9)\n2\n2\n2 z \u222bz\u2212 R r \u23a3 \ns,1 \u23a6 r\nR z \ns,1 \ns,1 \nGm m1\n1 \n\u23a1 \n2 \u2212 R2 \nz+ R drs,1 \nz+ R \n\u23a4 \ns\n= \u2212\n\u23a2(z \n) \u222b \n+ \u222b \ndr s,1 \u23a5.\n2\n2\n2 R z2 \u23a2 \nz\u2212 R r s,1 \nz\u2212 R \n\u23a5\n\u23a3\n\u23a6 \nNo tables should be needed for these; the result is \nGm \n\u23a1 (z\n) \n\u23a4\nz+R\n2 \u2212 R2 \nsm1\n1\nF = \u2212 \n\u23a2\u2212 \n+ r \u23a5 \nz\ns,1 \n2\n2 R z2 \u23a2 \nr \n\u23a5\n\u23a3 \ns,1 \n\u23a6z\u2212R \nGm \n1 \n= \u2212 \nsm1 \n\u23a1\u23a3\u2212(z \u2212 R)+(z + R)+ 2 R\u23a4\u23a6 \n(9.A.10)\n2\n2 R z2 \nGm sm1\n= \u2212 \n2\n. \nz \n9-21", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_232", "text": "For the case when z < R , rs,1 varies from R \u2212 z to R + z . Then the integral is \n2 \u2212 R2 \nR+z \nGm m1\n1 \n\u23a1 (z\n) \n\u23a4\ns\nF = \u2212 \n\u23a2\u2212 \n+ r \u23a5 \nz\ns,1 \n2\n2 R z2 \u23a2 \nrs,1 \n\u23a5\n\u23a3\n\u23a6R\u2212z \nGm sm1\n1 \n= \u2212 \n\u23a1\u23a3\u2212(z \u2212 R) \u2212 (z + R)+ 2 z\u23a4\u23a6 \n(9.A.11)\n2\n2 R z2 \n= 0. \nSo we have demonstrated the proposition that for a point-like object located on the z \u00ad\naxis a distance z from the center of a spherical shell, the gravitational force on the point \nlike object is given by \n\uf072\n0 \ns,1 \nz\nF\n\u23a7\n\u23aa\u23a8  \nm m1\ns \nk\u02c6,\n\uf072  \n\u2212G  \nz > R\n(r) = \n2 \n.  \n\u23aa\u23a9 \n, \nz < R \nThis proves the result that the gravitational force inside the shell is zero and the \ngravitational force outside the shell is equivalent to putting all the mass at the center of \nthe shell. \n9-22", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_233", "text": "Chapter 10 Momentum, System of Particles, and Conservation of  \nMomentum  \n10.1 Introduction........................................................................................................... 1  \n10.2 Momentum (Quantity of Motion) and Average Impulse .................................. 1  \nExample 10.1 Impulse for a Non-Constant Force.................................................. 3  \n10.3 External and Internal Forces and the Change in Momentum of a System ..... 4  \n10.4 System of Particles ................................................................................................ 5  \n10.5 Center of Mass ...................................................................................................... 6", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_234", "text": "Example 10.2 Center of Mass of the Earth-Moon System .................................... 7  \nExample 10.3 Center of Mass of a Rod ................................................................... 8  \n10.6 Translational Motion of the Center of Mass .................................................... 10  \nExample 10.4 Forces on a Baseball Bat ................................................................ 10  \n10.7 Constancy of Momentum and Isolated Systems .............................................. 11  \n10.8 Momentum Changes and Non-isolated Systems .............................................. 12  \n10.9 Worked Examples............................................................................................... 12", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_235", "text": "10.9.1 Problem Solving Strategies ......................................................................... 12  \nExample 10.5 Exploding Projectile ....................................................................... 14  \nExample 10.6 Landing Plane and Sandbag .......................................................... 17", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_236", "text": "Chapter 10 Momentum, System of Particles, and Conservation  \nof Momentum \nLaw II: The change of motion is proportional to the motive force \nimpressed, and is made in the direction of the right line in which that force \nis impressed. \nIf any force generates a motion, a double force will generate double the \nmotion, a triple force triple the motion, whether that force is impressed \naltogether and at once or gradually and successively. And this motion \n(being always directed the same way with the generating force), if the body \nmoved before, is added or subtracted from the former motion, according as \nthey directly conspire with or are directly contrary to each other; or \nobliquely joined, when they are oblique, so as to produce a new motion \ncompounded from the determination of both. 1 \nIsaac Newton Principia \n10.1 Introduction \nWhen we apply a force to an object, through pushing, pulling, hitting or otherwise, we \nare applying that force over a discrete interval of time, \u0394t . During this time interval, the \napplied force may be constant, or it may vary in magnitude or direction. Forces may also \nbe applied continuously without interruption, such as the gravitational interaction \nbetween the earth and the moon. In this chapter we will investigate the relationship \nbetween forces and the time during which they are applied, and in the process learn about \nthe quantity of momentum, the principle of conservation of momentum, and its use in \nsolving a new set of problems involving systems of particles. \n10.2 Momentum (Quantity of Motion) and Average Impulse \n!\nConsider a point-like object (particle) of mass m that is moving with velocity v with \n\uf072\nrespect to some fixed reference frame. The quantity of motion or the momentum, p , of \nthe object is defined to be the product of the mass and velocity \n!\n! \np = mv . \n(10.2.1) \nMomentum is a reference frame dependent vector quantity, with direction and magnitude. \nThe direction of momentum is the same as the direction of the velocity. The magnitude of \nthe momentum is the product of the mass and the instantaneous speed. \n\u22121]\nUnits: In the SI system of units, momentum has units of [kg m s\n\u22c5\n\u22c5 \n. There is no special \nname for this combination of units. \n1 Isaac Newton. Mathematical Principles of Natural Philosophy. Translated by Andrew Motte (1729). \nRevised by Florian Cajori. Berkeley: University of California Press, 1934. p. 13. \n10-1", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_237", "text": "! \nDuring a time interval \u0394t , a non-uniform force F is applied to the particle. Because we \nare assuming that the mass of the point-like object does not change, Newton\u2019s Second \nLaw is then \nd\nd(m \n= m \n!v \n!v\n!a \n!\nF \n) . \n(10.2.2)\n= m \n= \ndt \ndt \nBecause we are assuming that the mass of the point-like object does not change, the \nSecond Law can be written as \n! \nF = d! p \ndt . \n(10.2.3) \nThe impulse of a force acting on a particle during a time interval [t,t + \u0394t] is defined as \nthe definite integral of the force from t to t + \u0394t , \nt\u2032=t+\u0394t\n! \nI =  \u222b  \n!\nF (t\u2032) dt\u2032 . \n(10.2.4) \nt\u2032=t \nThe SI units of impulse are [N \u22c5m] = [kg\u22c5m\u22c5s\u22121] which are the same units as the units of \nmomentum. \nApply Newton\u2019s Second Law in Eq. (10.2.4) yielding \nt\u2032=t+\u0394t \nt\u2032=t+\u0394t ! \np\u2032=p(t+\u0394t )\n!\n! \ndp \n!\n!\n!\n! \n!\n!\nI = \nF(t\u2032) dt\u2032 = \ndt\u2032 = \ndp\u2032 = p(t + \u0394t) \u2212 p(t) = \u0394p . \n(10.2.5)\n\u222b \n\u222b\n\u222b\ndt\u2032 \n!\n!\nt\u2032=t\nt\u2032=t \np\u2032=p(t ) \nEq. (10.2.5) represents the integral version of Newton\u2019s Second Law: the impulse applied \nby a force during the time interval [t,t + \u0394t] is equal to the change in momentum of the \nparticle during that time interval. \nThe average value of that force during the time interval \u0394t is given by the integral \nexpression \nt\u2032=t+\u0394t\n! \n1 \n! \nF ave = \u0394t \u222b F(t\u2032)dt\u2032 . \n(10.2.6) \nt\u2032=t \nThe product of the average force acting on an object and the time interval over which it is \napplied is called the average impulse, \n!\n! \nI \n= F \u0394t . \n(10.2.7)\nave \nave \n10-2", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_238", "text": "Multiply each side of Eq. (10.2.6) by \u0394t resulting in the statement that the average \nimpulse applied to a particle during the time interval [t,t + \u0394t] is equal to the change in \nmomentum of the particle during that time interval, \n!\n!\nI ave = \u0394p. \n(10.2.8) \nExample 10.1 Impulse for a Non-Constant Force \nSuppose you push an object for a time \u0394t = 1.0 s in the +x -direction. For the first half of \nthe interval, you push with a force that increases linearly with time according to \n! \n\u22121\nF(t) = bt \u02c6i, 0 \u2264 t \u2264 0.5s  with  b = 2.0 \u00d7101 N \u22c5s . \n(10.2.9) \nThen for the second half of the interval, you push with a linearly decreasing force, \n! \nF(t) = (d \u2212 bt)\u02c6i, 0.5s \u2264 t \u2264 1.0s  with d = 2.0 \u00d7101 N \n(10.2.10) \nThe force vs. time graph is shown in Figure 10.3. What is the impulse applied to the \nobject? \nFigure 10.3 Graph of force vs. time \nSolution: We can find the impulse by calculating the area under the force vs. time curve. \nSince the force vs. time graph consists of two triangles, the area under the curve is easy to \ncalculate and is given by \n\uf072 \n\u23a1 1\n1 \n\u23a4\nI = \u23a2 (b\u0394t / 2)(\u0394t / 2) + (b\u0394t / 2)(\u0394t / 2) \u23a5\u02c6i\n2\n2\n\u23a3\n\u23a6 \n(10.2.11)\n1 b(\u0394t)2 \u02c6i = 1 \n= \n(2.0 \u00d7101 N \u22c5s\u22121)(1.0s)2 \u02c6i = (5.0N \u22c5s)\u02c6i.\n4\n4 \n10-3", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_239", "text": "10.3 External and Internal Forces and the Change in Momentum of a \nSystem \nSo far we have restricted ourselves to considering how the momentum of an object \nchanges under the action of a force. For example, if we analyze in detail the forces acting \non the cart rolling down the inclined plane (Figure 10.4), we determine that there are \n\uf072 \nthree forces acting on the cart: the force Fspring, cart the spring applies to the cart; the \n\uf072 \ngravitational interaction Fearth, cart between the cart and the earth; and the contact force \n\uf072 \nFplane, cart between the inclined plane and the cart. If we define the cart as our system, then \neverything else acts as the surroundings. We illustrate this division of system and \nsurroundings in Figure 10.4. \n\uf072 \nF \n\uf072 \nF \n\uf072 \nF \nFigure 10.4 A diagram of a cart as a system and its surroundings \nThe forces acting on the cart are external forces. We refer to the vector sum of these \nexternal forces that are applied to the system (the cart) as the external force, \nspring, cart \nearth, cart \nplane, cart . \n\uf072 \nF ext \n(10.3.1) \n+  \n+ \n=  \n\uf072p \nThen Newton\u2019s Second Law applied to the cart, in terms of impulse, is \n\u0394 \n\uf072\nF  \n\uf072\nI \nt f\u222bt0 \next dt \u2261 \n(10.3.2) \n=  \n. \nsys \nsys \nLet\u2019s extend our system to two interacting objects, for example the cart and the spring. \nThe forces between the spring and cart are now internal forces. Both objects, the cart and \nthe spring, experience these internal forces, which by Newton\u2019s Third Law are equal in \nmagnitude and applied in opposite directions. So when we sum up the internal forces for \nthe whole system, they cancel. Thus the sum of all the internal forces is always zero, \n\uf072\n\uf072 \nFint = 0. \n(10.3.3) \nExternal forces are still acting on our system; the gravitational force, the contact force \nbetween the inclined plane and the cart, and also a new external force, the force between \nthe spring and the force sensor. The force acting on the system is the sum of the internal \nand the external forces. However, as we have shown, the internal forces cancel, so we \nhave that \n10-4", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_240", "text": "\uf072 \nF =  \n\uf072 \nF ext + \n\uf072 \nF int =  \n\uf072 \nF ext . \n(10.3.4) \n10.4 System of Particles \nSuppose we have a system of N particles labeled by the index i = 1, 2, 3, \uf04c , N . The \nith \nforce on the \nparticle is \nj= N \next +\ni \n\u2211\ni\ni, j \nj=1, j\u2260i \n\uf072 \nith \nith \nIn this expression Fj,i is the force on the \nparticle due to the interaction between the \njth \n\uf072\n\uf072 \n\uf072\nF \nith \non the \nparticle, \nj=N\n\uf072\n\uf072 \nFi \nint = \u2211 Fj,i . \n(10.4.2) \nj=1, j\u2260i \n\uf072\nF \n\uf072\nF\ni\ni \n\uf072\nF \nj\ni\nj \n\u2260\nand \nparticles. We sum over all particles with \nsince a particle cannot exert a \nF\n0\nforce on itself (equivalently, we could define \n), yielding the internal force acting \n= \ni i, \nThe force acting on the system is the sum over all i particles of the force acting on each \n\uf072\nF \nparticle, \n(10.4.1) \n=  \n. \nj= N \n\u2211\n\uf072\nF  \ni=N \ni=N \ni=N\n\uf072\nF  \n\uf072\nF \n\u2211  \u2211  \n\u2211 \next \next . \n(10.4.3)\n+ \n=  \n=  \n=  \nj,i \n\uf072 \nF \ni=1 \ni=1 \ni=1 j=1, j\u2260i \nNote that the double sum vanishes, \ni= N\nj = N \uf072\n\uf072\n\u2211\u2211 Fj,i = 0 , \n(10.4.4) \ni=1 j=1, j \u2260 i \nbecause all internal forces cancel in pairs, \ni, j \n\uf072 \nF  \n\uf072\n0 \n(10.4.5) \n+  \n=  . \nj,i \nThe force on the ith particle is equal to the rate of change in momentum of the ith \nparticle, \n\uf072 \nd \uf072 pi\n= \n. \n(10.4.6)\nFi \ndt \nWhen can now substitute Equation (10.4.6) into Equation (10.4.3) and determine that that \nthe external force is equal to the sum over all particles of the momentum change of each \nparticle, \n\uf072 \ni=\nFext \nN d p\uf072 \ni\n= \u2211 \n. \n(10.4.7) \ni=1 dt \n10-5", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_241", "text": "The momentum of the system is given by the sum \ni= N \uf072\n\uf072 psys = \u2211pi ; \n(10.4.8) \ni=1 \nmomenta add as vectors. We conclude that the external force causes the momentum of \nthe system to change, and we thus restate and generalize Newton\u2019s Second Law for a \nsystem of objects as \n\uf072 \nd p \uf072 \nFext \nsys \n= \n. \n(10.4.9)\ndt \nIn terms of impulse, this becomes the statement \n\u0394\uf072p \n\uf072\nF  \n\uf072\nI \n\u222bt\nt f \next dt \u2261 \n(10.4.10) \n=  \n. \nsys \n0 \n10.5 Center of Mass \nConsider two point-like particles with masses m1 and m2 . Choose a coordinate system \n\uf072\n\uf072\nwith a choice of origin such that body 1 has position r1 and body 2 has position r2 \n(Figure 10.5). \nFigure 10.5 Center of mass coordinate system. \n\uf072 \nThe center of mass vector, Rcm , of the two-body system is defined as \n\uf072\n\uf072\n\uf072 \nm r + m r\n1 1 \n2\n2\nRcm = \n. \n(10.5.1)\nm + m\n1\n2 \nWe shall now extend the concept of the center of mass to more general systems. Suppose \nwe have a system of N particles labeled by the index i = 1, 2, 3,\uf04c , N . Choose a \nith \n\uf072\ncoordinate system and denote the position of the \nparticle as ri . The mass of the system \nis given by the sum \n10-6", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_242", "text": "i=N \nm sys = \u2211mi \n(10.5.2) \ni=1 \nand the position of the center of mass of the system of particles is given by \n\uf072 \n1 i= N \n\uf072\nR cm = \n. \n(10.5.3)\n\u2211mi ri\nm sys i=1 \n(For a continuous rigid body, each point-like particle has mass dm and is located at the \n\uf072\nposition r\u2032 . The center of mass is then defined as an integral over the body, \ndmr \uf072\u2032 \n\uf072 \n\u222b \nRcm = \nbody \n. \n(10.5.4)\n\u222b dm \nbody \nExample 10.2 Center of Mass of the Earth-Moon System \nThe mean distance from the center of the earth to the center of the moon is \nrem = 3.84\u00d7108 m . The mass of the earth is me = 5.98 \u00d7 1024 kg and the mass of the moon \nis mm = 7.34\u00d71022 kg . The mean radius of the earth is re = 6.37 \u00d7 106 m . The mean radius \nof the moon is rm = 1.74\u00d7106 m . Where is the location of the center of mass of the earth-\nmoon system? Is it inside the earth\u2019s radius or outside? \nSolution: The center of mass of the earth-moon system is defined to be \n\uf072 \ni=N\n1 \n\uf072 \n1 \n\uf072\n\uf072\nR =\n= \n(m r + m r ) . \n(10.5.5)\ncm \n\u2211 miri \ne e \nm m\nm \ni=1 \nm + m \nsys \ne\nm \nChoose an origin at the center of the earth and a unit vector \u02c6i pointing towards the moon, \n\uf072\n\uf072 \nthen re = 0 . The center of mass of the earth-moon system is then \n\uf072\n\uf072 \n1 \n\uf072\n\uf072 \nm r \nm r \nm em \nm em \u02c6\nRcm = \n(mere + mmrm ) =\n= \ni \n(10.5.6)\nme + mm \nme + mm \nme + mm \n22 \n8\n\uf072 \n(7.34\u00d710 kg)(3.84\u00d710 m) \u02c6 \n6 \n\u02c6\nRcm = \ni = 4.66\u00d710 m i \n(10.5.7)\n24 \n22\n(5.98\u00d710 kg + 7.34\u00d710 kg) \nThe earth\u2019s mean radius is r = 6.37 \u00d7 106 m so the center of mass of the earth-moon \ne \nsystem lies within the earth. \n10-7", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_243", "text": "Example 10.3 Center of Mass of a Rod \nA thin rod has length L and mass M . \n0\n2\n(x) = \nx\n= M / L \nL2 \nM\nM \nL\nL \na) uniform rod \nb) non-uniform rod \nFigure 10.6 a) Uniform rod and b) non-uniform rod \n(a) Suppose the rod is uniform (Figure 10.6a). Find the position of the center of mass \nwith respect to the left end of the rod. (b) Now suppose the rod is not uniform (Figure \n10.6b) with a linear mass density that varies with the distance x from the left end \naccording to \n\u03bb0\n2\n\u03bb(x) = L2 x \n(10.5.8) \nwhere \u03bb0 is a constant and has SI units [kg \u22c5 m-1]. Find \u03bb0 and the position of the center \nof mass with respect to the left end of the rod. \nSolution: (a) Choose a coordinate system with the rod aligned along the x -axis and the \norigin located at the left end of the rod. The center of mass of the rod can be found using \nthe definition given in Eq. (10.5.4). In that expression dm is an infinitesimal mass \n\uf072\nelement and r is the vector from the origin to the mass element dm (Figure 10.6c). \n= M / L \ndm = dx \nx \n+x \nx = 0 \nx = L \nFigure 10.6c Infinitesimal mass element for rod  \nChoose an infinitesimal mass element dm located a distance x\u2032 from the origin. In this  \nproblem x\u2032 will be the integration variable. Let the length of the mass element be dx\u2032 . \nThen \ndm = \u03bb dx\u2032 \n(10.5.9) \n\uf072\nThe vector r = x\u2032 \u02c6i . The center of mass is found by integration \n10-8", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_244", "text": "\uf072 \n1 \n\uf072 \n1 x \n1\n2 x\u2032=L \n1\n2 \nL\n\u02c6 \n\u02c6\n\u02c6 \ncm \n\u222b \n\u2032\n\u2032 \u02c6 \n\u2032\nR\n= \nr dm = \n\u222b x dx i = \nx \ni = \n(L \u2212 0) i = \ni . (10.5.10)\nx\u2032=0\n2L \n2\nM body \nL x\u2032 =0\n2L \n(b) For a non-uniform rod (Figure 10.6d), \n2\n(x ) = L2\n0 x \nx \n+x \nx = 0 \nx = L\ndm = (x )dx \nFigure 10.6d Non-uniform rod \nthe mass element is found using Eq. (10.5.8) \n\u03bb0\n2\ndm = \u03bb( ) \nx\u2032 dx\u2032 = \u03bb = \nx\u2032 dx\u2032 . \n(10.5.11)\nL2 \n\uf072\nThe vector r = x\u2032 \u02c6i . The mass is found by integrating the mass element over the length of \nthe rod \nx=L\nx=L \nx\u2032=L \n\u03bb\n\u03bb\n\u03bb0\n2 \n\u03bb0\n3 \n0\n3 \n0\nM = \ndm = \u222b\u03bb( )\nx\u2032 dx\u2032 = \nx\u2032 dx\u2032 = \nx\u2032 \n= \n2 (L \u2212 0) = \nL . (10.5.12)\n\u222b \n2 \u222b \n2 \nx\u2032=0\n3L \n3\nbody \nx\u2032 =0 \nL\nx\u2032 =0\n3L \nTherefore \n3M\n\u03bb0 = \n(10.5.13)\nL \nThe center of mass is again found by integration \n\uf072 \n1 \n\uf072 \n3 \n3 \ncm = \n\u222b r dm = \n\u222b \nx \n\u03bb(x\u2032)x dx\u2032 \u02c6i = \n\u222b \nx\nx\u20323 dx\u2032 \u02c6i\nR \n\u2032 \nM body \n\u03bb0 L x\u2032=0 \nL3 \nx\u2032=0 \n(10.5.14)\n\uf072 \n3 \nx\u2032= \n3\n3\nL \u02c6i =\nR = \nx\u20324 \n4L3 (L4 \u2212 0) \u02c6i = \nL \u02c6i. \ncm \n4L3 \nx\u2032=0 \n4 \n. \n10-9", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_245", "text": "10.6 Translational Motion of the Center of Mass \nThe velocity of the center of mass is found by differentiation, \nsys \nm \n\uf072p\n\uf072\nV \n1 i=N \n\u2211mi \n\uf072v\n.  \n(10.6.1) \n=  \n=  \ni\ncm \nm \ni=1\nsys \nsys \nThe momentum is then expressed in terms of the velocity of the center of mass by \n\uf072\n\uf072 \np \n= m V . \n(10.6.2)\nsys \nsys \ncm \nWe have already determined that the external force is equal to the change of the \nmomentum of the system (Equation (10.4.9)). If we now substitute Equation (10.6.2) into \nEquation (10.4.9), and continue with our assumption of constant masses mi , we have that \nd\nsys \n\uf072p \n\uf072 \nV \n\uf072\nA  \n\uf072\nF \nd\next \n(10.6.3) \n= m \ncm = m\n=  \n, \ndt \ndt\nsys \nsys \ncm \n\uf072\n\uf072 \nwhere A , the derivative with respect to time of V , is the acceleration of the center of \ncm \ncm\nmass. From Equation (10.6.3) we can conclude that in considering the linear motion of \nthe center of mass, the sum of the external forces may be regarded as acting at the center \nof mass. \nExample 10.4 Forces on a Baseball Bat \nSuppose you push a baseball bat lying on a nearly frictionless table at the center of mass, \n! \nposition 2, with a force F (Figure 10.7). Will the acceleration of the center of mass be \ngreater than, equal to, or less than if you push the bat with the same force at either end, \npositions 1 and 3 \n1\n2 \n3 \ncm \nF \nF \nF \nFigure 10.7 Forces acting on a baseball bat \nSolution: The acceleration of the center of mass will be equal in the three cases. From \nour previous discussion, (Equation (10.6.3)), the acceleration of the center of mass is \nindependent of where the force is applied. However, the bat undergoes a very different \nmotion if we apply the force at one end or at the center of mass. When we apply the force \n10-10", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_246", "text": "at the center of mass all the particles in the baseball bat will undergo linear motion \n(Figure 10.7a). \ncm \nF \ncm translates, \nno rotation about cm \nFigure 10.7a Force applied at center of mass \nWhen we push the bat at one end, the particles that make up the baseball bat will no \nlonger undergo a linear motion even though the center of mass undergoes linear motion. \nIn fact, each particle will rotate about the center of mass of the bat while the center of \nmass of the bat accelerates in the direction of the applied force (Figure 10.7b). \ncm \nF\ncm translates and rotates \nFigure 10.7b Force applied at end of bat \n10.7 Constancy of Momentum and Isolated Systems \nSuppose we now completely isolate our system from the surroundings. When the external \nforce acting on the system is zero, \n\uf072\n\uf072 \nFext = 0 . \n(10.7.1) \nthe system is called an isolated system. For an isolated system, the change in the \nmomentum of the system is zero, \n\uf072 \n\u0394\uf072 psys = 0 \n(isolated system) , \n(10.7.2) \ntherefore the momentum of the isolated system is constant. The initial momentum of our \nsystem is the sum of the initial momentum of the individual particles, \n\uf072p\n= m1\nsys,i \n\uf072v1,i + m2 \n\uf072v2,i + \u22c5\u22c5\u22c5 . \n(10.7.3)  \nThe final momentum is the sum of the final momentum of the individual particles, \n\uf072p \n\uf072v \n\uf072v\n= m1\nsys, f \n1, f + m2 \n2, f + \u22c5\u22c5\u22c5. \n(10.7.4)  \nNote that the right-hand-sides of Equations. (10.7.3) and (10.7.4) are vector sums. \n10-11", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_247", "text": "When the external force on a system is zero, then the initial momentum of the \nsystem equals the final momentum of the system, \n\uf072\n\uf072 \np \n= p \n(10.7.5)\nsys,i \nsys, f . \n10.8 Momentum Changes and Non-isolated Systems \nSuppose the external force acting on the system is not zero, \n\uf072\n\uf072 \nFext \u2260 0. \n(10.8.1) \nand hence the system is not isolated. By Newton\u2019s Third Law, the sum of the force on the \nsurroundings is equal in magnitude but opposite in direction to the external force acting \non the system,\n\uf072 \nF sur = \u2212  \n\uf072 \nF ext .  \n(10.8.2)  \nIt\u2019s important to note that in Equation (10.8.2), all internal forces in the surroundings sum \nto zero. Thus the sum of the external force acting on the system and the force acting on \nthe surroundings is zero,\n\uf072 \nF sur + \n\uf072 \nF ext =  \n\uf072\n0 \n(10.8.3)  \nacting on a system \n. \n\uf072 \nFext \nWe have already found (Equation (10.4.9)) that the external force \n\uf072p \nis equal to the rate of change of the momentum of the system. Similarly, the force on the \nsurrounding is equal to the rate of change of the momentum of the surroundings. \nTherefore the momentum of both the system and surroundings is always conserved. \nFor a system and all of the surroundings that undergo any change of state, the \nchange in the momentum of the system and its surroundings is zero, \n\u0394\n\uf072p \n\uf072\n0 \n+ \u0394  \n(10.8.4) \n=  . \nsys \nsur \nEquation (10.8.4) is referred to as the Principle of Conservation of Momentum. \n10.9 Worked Examples \n10.9.1 Problem Solving Strategies \nWhen solving problems involving changing momentum in a system, we shall employ our \ngeneral problem solving strategy involving four basic steps: \n1. Understand \u2013 get a conceptual grasp of the problem. \n2. Devise a Plan - set up a procedure to obtain the desired solution. \n3. Carry our your plan \u2013 solve the problem! \n10-12", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_248", "text": "4. Look Back \u2013 check your solution and method of solution. \nWe shall develop a set of guiding ideas for the first two steps. \n1. Understand \u2013 get a conceptual grasp of the problem \nThe first question you should ask is whether or not momentum is constant in some \nsystem that is changing its state after undergoing an interaction. First you must identify \nthe objects that compose the system and how they are changing their state due to the \ninteraction. As a guide, try to determine which objects change their momentum in the \ncourse of interaction. You must keep track of the momentum of these objects before and \nafter any interaction. Second, momentum is a vector quantity so the question of whether \nmomentum is constant or not must be answered in each relevant direction. In order to \ndetermine this, there are two important considerations. You should identify any external \nforces acting on the system. Remember that a non-zero external force will cause the \n\uf072p \nmomentum of the system to change, (Equation (10.4.9) above), \n\uf072 \nd p \uf072 \nFext \nsys \n= \n. \n(10.9.1)\ndt \nEquation (10.9.1) is a vector equation; if the external force in some direction is zero, then \nthe change of momentum in that direction is zero. In some cases, external forces may act \nbut the time interval during which the interaction takes place is so small that the impulse \nis small in magnitude compared to the momentum and might be negligible. Recall that \nthe average external impulse changes the momentum of the system \n= \u0394 \n\uf072\nI  \n\uf072 \nF ext \u0394tint \n(10.9.2) \n=  \n. \nsys \n\uf072\n\uf072 \nIf the interaction time is small enough, the momentum of the system is constant, \u0394p \u2192 0 . \nIf the momentum is not constant then you must apply either Equation (10.9.1) or \nEquation (10.9.2). If the momentum of the system is constant, then you can apply \nEquation (10.7.5), \n\uf072\n\uf072 \np \n= p \n. \n(10.9.3)\nsys, i \nsys, f \nIf there is no net external force in some direction, for example the x -direction, the \ncomponent of momentum is constant in that direction, and you must apply \np \n= p \n(10.9.4)\nsys, x,i \nsys, x, f \n2. Devise a Plan - set up a procedure to obtain the desired solution \nDraw diagrams of all the elements of your system for the two states immediately before \nand after the system changes its state. Choose symbols to identify each mass and velocity", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_249", "text": "sys, x,i \nsys, x, f \n2. Devise a Plan - set up a procedure to obtain the desired solution \nDraw diagrams of all the elements of your system for the two states immediately before \nand after the system changes its state. Choose symbols to identify each mass and velocity \nin the system. Identify a set of positive directions and unit vectors for each state. Choose \n10-13", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_250", "text": "your symbols to correspond to the state and motion (this facilitates an easy interpretation, \nfor example (v \nrepresents the x -component of the velocity of object 1 in the initial \nx,i )1 \nstate and (v \nrepresents the x -component of the velocity of object 1 in the final state). \nx, f )1 \nDecide whether you are using components or magnitudes for your velocity symbols. \nSince momentum is a vector quantity, identify the initial and final vector components of \nthe momentum. We shall refer to these diagrams as momentum flow diagrams. Based on \nyour model you can now write expressions for the initial and final momentum of your \nsystem. As an example in which two objects are moving only in the x -direction, the \ninitial x -component of the momentum is \np \n= m1(v \n+ m2(v \n+\uf04c. \n(10.9.5)\nsys, x,i\nx,i )1 \nx,i )2 \nThe final x -component of the momentum is \npsys, x, f = m1(vx, f )1 + m2(vx, f )2 +\uf04c. \n(10.9.6) \nIf the x -component of the momentum is constant then \np \n= p \n(10.9.7)\nsys, x,i \nsys, x, f . \nWe can now substitute Equations (10.9.5) and (10.9.6) into Equation (10.9.7), yielding \n(v \n(v \n+\uf04c \n(v \n(v \n+\uf04c. \n(10.9.8)\nm1 \nx,i )1 + m2 \nx,i )2 \n= m1 \nx, f )1 + m2 \nx, f )2 \nEquation (10.9.8) can now be used for any further analysis required by a particular \nproblem. For example, you may have enough information to calculate the final velocities \nof the objects after the interaction. If so then carry out your plan and check your solution, \nespecially dimensions or units and any relevant vector directions. \nExample 10.5 Exploding Projectile \nAn instrument-carrying projectile of mass \naccidentally explodes at the top of its \nm1 \ntrajectory. The horizontal distance between launch point and the explosion is \n. The \nxi \nprojectile breaks into two pieces that fly apart horizontally. The larger piece, m3, has \nthree times the mass of the smaller piece, m2 . To the surprise of the scientist in charge, \nthe smaller piece returns to earth at the launching station. Neglect air resistance and", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_251", "text": ". The \nxi \nprojectile breaks into two pieces that fly apart horizontally. The larger piece, m3, has \nthree times the mass of the smaller piece, m2 . To the surprise of the scientist in charge, \nthe smaller piece returns to earth at the launching station. Neglect air resistance and \neffects due to the earth\u2019s curvature. How far away, x3, f , from the original launching point \ndoes the larger piece land? \n10-14", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_252", "text": "parabolic orbit\nm1 \nx = 0 \nxi \nv1 \n\u02c6i \nx = 0 \nx = 0 \nxi \nm2 \nm3 \nv2 \nv3 \nxi \nx3, f \nFigure 10.8 Exploding projectile trajectories \nSolution: We can solve this problem two different ways. The easiest approach is utilizes \nthe fact that the external force is the gravitational force and therefore the center of mass \nof the system follows a parabolic trajectory. From the information given in the problem \nm = m / 4 and m = 3m / 4 . Thus when the two objects return to the ground the center \n2\n1 \n3\n1 \nof mass of the system has traveled a distance Rcm = 2xi . We now use the definition of \ncenter of mass to find where the object with the greater mass hits the ground. Choose an \norigin at the starting point. The center of mass of the system is given by \n\uf072\n\uf072\n\uf072 \nm r + m r\n2 2 \n3 3 \nR \n= \n.\ncm \nm + m\n2\n3 \n! \nSo when the objects hit the ground R cm = 2xi \u02c6i , the object with the smaller mass returns \n\uf072\n\uf072 \n!\nto the origin, r2 = 0 , and the position vector of the other object is r3 = x3, f \u02c6i . So using \nthe definition of the center of mass, \n(3m1 / 4)x3, f \u02c6i \n(3m1 / 4)x3, f \u02c6i \n3\n\u02c6i = \n=\n= \n\u02c6i .\n2xi \nx3, f\nm1 / 4 + 3m1 / 4 \nm1\n4 \nTherefore \n8 \n=\nx3, f \nxi .\n3 \n10-15", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_253", "text": "Note that the neither the vertical height above ground nor the gravitational acceleration g \nentered into our solution. \nAlternatively, we can use conservation of momentum and kinematics to find the distance \ntraveled. Because the smaller piece returns to the starting point after the collision, the \nvelocity of the smaller piece immediately after the explosion is equal to the negative of \nthe velocity of original object immediately before the explosion. Because the collision is \ninstantaneous, the horizontal component of the momentum is constant during the \ncollision. We can use this to determine the speed of the larger piece after the collision. \nThe larger piece takes the same amount of time to return to the ground as the projectile \noriginally takes to reach the top of the flight. We can therefore determine how far the \nlarger piece traveled horizontally. \nWe begin by identifying various states in the problem. \nInitial state, time t0 = 0 : the projectile is launched. \nState 1 time t1: the projectile is at the top of its flight trajectory immediately before the \n! \nexplosion. The mass is m1 and the velocity of the projectile is v1 = v1\u02c6i . \nState 2 time t2 : immediately after the explosion, the projectile has broken into two \npieces, one of mass m2 moving backwards (in the negative x -direction) with velocity \n!\n! \n= \u2212v1 . The other piece of mass m3 is moving in the positive x -direction with velocity\nv2 \n! v3 = v3\u02c6i , (Figure 10.8). \nState 3: the two pieces strike the ground at time t f \n, one at the original launch site \n= 2t1 \nand the other at a distance \nfrom the launch site, as indicated in Figure 10.8. The \nx3, f \npieces take the same amount of time to reach the ground \u0394t = t1 because both pieces are \nfalling from the same height as the original piece reached at time t1, and each has no \ncomponent of velocity in the vertical direction immediately after the explosion. The \nmomentum flow diagram with state 1 as the initial state and state 2 as the final state are \nshown in the upper two diagrams in Figure 10.8. \nThe initial momentum at time t1 immediately before the explosion is \n! sys (t1 \n! \np \n) = m1 v1 . \n(10.9.9) \nThe momentum at time t2 immediately after the explosion is \n10-16", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_254", "text": "!psys (t2 ) = m2 \n!v2 + m3 \n!v3 = \u2212  1  \n4  m1 \n!v1 +  3  \n4  m1 \n!v3 \n(10.9.10)  \nDuring the duration of the instantaneous explosion, impulse due to the external \ngravitational force may be neglected and therefore the momentum of the system is \nconstant. In the horizontal direction, we have that \n1  \n3  \nm1 \n!v\nm1 \n!v\nm1\n!v3 . \n(10.9.11) \n= \u2212  \n+ \n1 \n1\n4  \n4  \nEquation (10.9.11) can now be solved for the velocity of the larger piece immediately \nafter the collision, \n! v3 = 5 \n3 \n! v1 . \n(10.9.12) \nThe larger piece travels a distance \nx3, f = v3 t1 = 5 \n3v1 t1 = 5 \n3 xi . \n(10.9.13) \nTherefore the total distance the larger piece traveled from the launching station is \n5\n8\n+\n= xi , \n(10.9.14)\nx f = xi \nxi\n3\n3 \nin agreement with our previous approach. \nExample 10.6 Landing Plane and Sandbag \nFigure 10.9 Plane and sandbag \nA light plane of mass 1000 kg makes an emergency landing on a short runway. With its \nengine off, it lands on the runway at a speed of 40 m s-1 . A hook on the plane snags a \n\u22c5 \ncable attached to a 120 kg sandbag and drags the sandbag along. If the coefficient of \nfriction between the sandbag and the runway is \u00b5k = 0.4 , and if the plane\u2019s brakes give \nan additional retarding force of magnitude 1400 N , how far does the plane go before it \ncomes to a stop? \n10-17", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_255", "text": "Solution: We shall assume that when the plane snags the sandbag, the collision is \ninstantaneous so the momentum in the horizontal direction remains constant, \np = p \n(10.9.15)\nx,i\nx,1 . \nWe then know the speed of the plane and the sandbag immediately after the collision. \nAfter the collision, there are two external forces acting on the system of the plane and \nsandbag, the friction between the sandbag and the ground and the braking force of the \nrunway on the plane. So we can use the Newton\u2019s Second Law to determine the \nacceleration and then one-dimensional kinematics to find the distance the plane traveled \nsince we can determine the change in kinetic energy. \nThe momentum of the plane immediately before the collision is \n\uf072 pi = m v p,i \u02c6i \n(10.9.16)\np \nThe momentum of the plane and sandbag immediately after the collision is \n\uf072 p = (m + m )v \n\u02c6i \n(10.9.17)\n1 \np \ns\np,1 \nBecause the x - component of the momentum is constant, we can substitute Eqs. \n(10.9.16) and (10.9.17) into Eq. (10.9.15) yielding \nm v \n= (m + m )v \n(10.9.18)\np\np,i \np\ns\np,1 . \nThe speed of the plane and sandbag immediately after the collision is \nm v \np\np,i\nvp,1 = \n(10.9.19)\nm + m \np\ns \nThe forces acting on the system consisting of the plane and the sandbag are the normal \nforce on the sandbag, \n\uf072 \nN \n= N \u02c6j , \n(10.9.20)\ng,s\ng ,s \nthe frictional force between the sandbag and the ground \n\uf072 \nf = \u2212 f \u02c6i = \u2212\u00b5 N \u02c6i , \n(10.9.21)\nk\nk \nk\ng ,s \nthe braking force on the plane \n\uf072 \nF \n= \u2212F \u02c6i , \n(10.9.22)\ng , p\ng, p \nand the gravitational force on the system, \n10-18", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_256", "text": "\uf072\n(m + m )g = \u2212(m + m )g\u02c6j. \n(10.9.23)\np\ns \np\ns \nNewton\u2019s Second Law in the \u02c6i -direction becomes \n\u2212 F \n= (m + m )a . \n(10.9.24)\ng , p \u2212 fk \np \ns \nx \nIf we just look at the vertical forces on the sandbag alone then Newton\u2019s Second Law in \nthe \u02c6j-direction becomes \nN \u2212 msg = 0 . \nThe frictional force on the sandbag is then \n\uf072 \nf = \u2212\u00b5 N \u02c6i = \u2212\u00b5 m g\u02c6i . \n(10.9.25)\nk \nk\ng ,s \nk\ns \nNewton\u2019s Second Law in the \u02c6i -direction becomes \n\u2212F \n\u2212 \u00b5km g = (m + m )a . \ng, p \ns \np\ns\nx \nThe x -component of the acceleration of the plane and the sand bag is then \n\u2212 F \n\u2212 \u00b5km g \ng , p\ns\nax = \n(10.9.26)\nm + m \np\ns \nWe choose our origin at the location of the plane immediately after the collision, \nxp (0) = 0 . Set t = 0 immediately after the collision. The x -component of the velocity of \nthe plane immediately after the collision is v \n= v \nSet t = t f when the plane just \nx,0 \np,1 . \ncomes to a stop. Because the acceleration is constant, the kinematic equations for the \nchange in velocity is \nv \n) \u2212 v \n= a t f . \nx, f (t f \np,1 \nx \nWe can solve this equation for t = t f , where v \n) = 0 \nx, f (t f \n= \u2212v \n/ a t .\nt f \np,1 \nx \nThen the position of the plane when it first comes to rest is \n1 \n2\n1 v2 \np,1 \nx (t ) \u2212 x (0) = v\nt + a t = \u2212 \n. \n(10.9.27)\np\nf\np \np,1 f \nx f\n2\n2 a x \n10-19", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_257", "text": "Then using xp (0) = 0 and substituting Eq. (10.9.26) into Eq. (10.9.27) yields \n1 (m + m )v2 \np \ns\np,1 \nxp (t f ) = \n. \n(10.9.28)\n2 (F \n+ \u00b5 m g)\ng , p \nk\ns \nWe now use the condition from conservation of the momentum law during the collision, \nEq. (10.9.19) in Eq. (10.9.28) yielding \n2\n2\nm v \np\np,i\nx (t ) = \n. \n(10.9.29)\np \nf \n2(m + m )(F \n+ \u00b5 m g)\np \ns\ng, p \nk\ns \nSubstituting the given values into Eq. (10.9.28) yields \n-1 )2\n(1000 kg)2(40 m \u22c5 s \nx \n) =\n= 3.8 \u00d7 102 m . (10.9.30)\np (t f \n2(1000 kg + 120 kg)(1400 N + (0.4)(120 kg)(9.8m \u22c5 s-2 )) \n10-20", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_258", "text": "Chapter 11 Reference Frames \n11.1 Introduction........................................................................................................... 1  \n11.2 Galilean Coordinate Transformations................................................................ 1  \n11.2.1 Relatively Inertial Reference Frames and the Principle of Relativity ...... 2  \n11.3 Law of Addition of Velocities: Newtonian Mechanics....................................... 3  \n11.4 Worked Examples................................................................................................. 3  \nExample 11.1 Relative Velocities of Two Moving Planes...................................... 3  \nExample 11.2 Relative Motion and Polar Coordinates ......................................... 5", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_259", "text": "Example 11.2 Relative Motion and Polar Coordinates ......................................... 5  \nExample 11.3 Recoil in Different Frames ............................................................... 7", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_260", "text": "1 \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n  \n \n \n \n \n \n \n \n \n \n  \n \n \n \n                      \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n  \n \n \n  \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n  \n  \n \n \n \n \n \n \n  \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n                                                   \n  \n \n \n \n \n \n \n \n \n \n \n \nChapter 11 Reference Frames \nExamples of this sort, together with the unsuccessful attempts to discover any \nmotion of the earth relatively to the \u201clight medium\u201d suggest that the phenomena \nof electromagnetism as well as mechanics possess no properties corresponding to \nthe idea of absolute rest. They suggest rather that, \u2026, the same laws of \nelectrodynamics and optics will be valid for all frames of reference for which the \nequations of mechanics hold good. We will raise this conjecture (the purport of \nwhich will hereafter be called the \u201cPrinciple of Relativity\u201d) to the status of a \npostulate, and also introduce another postulate, \u2026, namely that light is always \npropagated in empty space with a definite velocity c, which is independent of the \nstate of motion of the emitting body. 1 \nAlbert Einstein \n11.1 Introduction \nIn order to describe physical events that occur in space and time such as the motion of \nbodies, we introduced a coordinate system. Its spatial and temporal coordinates can now \nspecify a space-time event. In particular, the position of a moving body can be described \nby space-time events specified by its space-time coordinates. You can place an observer \nat the origin of coordinate system. The coordinate system with your observer acts as a \nreference frame for describing the position, velocity, and acceleration of bodies. The \nposition vector of the body depends on the choice of origin (location of your observer) \nbut the displacement, velocity, and acceleration vectors are independent of the location of \nthe observer. \nYou can always choose a second reference frame that is moving with respect to \nthe first reference frame. Then the position, velocity and acceleration of bodies as seen by \nthe different observers do depend on the relative motion of the two reference frames. The \nrelative motion can be described in terms of the relative position, velocity, and \nacceleration of the observer at the origin, O , in reference frame S with respect to a \nsecond observer located at the origin, O\u2032 , in reference frame S\u2032 . \n11.2 Galilean Coordinate Transformations \n\uf072 \nLet the vector R point from the origin of frame S to the origin of reference \nframe S\u2032. Suppose an object is located at a point 1. Denote the position vector of the \n\uf072\nobject with respect to origin of reference frame S by r . Denote the position vector of the \n\uf072\nobject with respect to origin of reference frame S\u2032 by r\u2032 (Figure 11.1). \n1 A. Einstein, Zur Elektrodynamik begetter K\u00f6rper, (On the\t\r \u00a0Electrodynamics of Moving", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_261", "text": "\uf072\nobject with respect to origin of reference frame S\u2032 by r\u2032 (Figure 11.1). \n1 A. Einstein, Zur Elektrodynamik begetter K\u00f6rper, (On the\t\r \u00a0Electrodynamics of Moving \nBodies), Ann. Physik, 17, 891 (1905); translated\t\r \u00a0by W. Perrett\t\r \u00a0and G.B. Jeffrey,\t\r \u00a0\n19223, in The Principle of Relativity, Dover,\t\r \u00a0New York. \n11-1", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_262", "text": "path of moving object in referenece frame S  \nR \nr \nS \nS \nr \nFigure 11.1 Two reference frames S and S\u2032. \nThe position vectors are related by \n! \n!\n! \nr\u2032 = r \u2212 R \n(11.2.1) \nThese coordinate transformations are called the Galilean Coordinate \nTransformations. They enable the observer in frame S to predict the position vector in \nframe S\u2032, based only on the position vector in frame S and the relative position of the \norigins of the two frames. \nThe relative velocity between the two reference frames is given by the time \n\uf072 \nderivative of the vector R , defined as the limit as of the displacement of the two origins \ndivided by an interval of time, as the interval of time becomes infinitesimally small, \n\uf072\n\uf072 \ndR\nV = \n. \n(11.2.2)\ndt \n11.2.1 Relatively Inertial Reference Frames and the Principle of Relativity \nIf the relative velocity between the two reference frames is constant, then the relative \nacceleration between the two reference frames is zero, \n\uf072\n\uf072\n\uf072\ndV\nA =\n= 0. \n(11.2.3)\ndt \nWhen two reference frames are moving with a constant velocity relative to each other as \nabove, the reference frames are called relatively inertial reference frames. \nWe can reinterpret Newton\u2019s First Law \nLaw 1: Every body continues in its state of rest, or of uniform motion in a right \nline, unless it is compelled to change that state by forces impressed upon it. \nas the Principle of Relativity: \n11-2", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_263", "text": "In relatively inertial reference frames, if there is no net force impressed on an \nobject at rest in frame S, then there is also no net force impressed on the object in \nframe S\u2032 . \n11.3 Law of Addition of Velocities: Newtonian Mechanics \nSuppose the object in Figure 11.1 is moving; then observers in different reference \nframes will measure different velocities. Denote the velocity of the object in frame S by\n\uf072\n\uf072 \n\uf072\n\uf072 \nv = dr dt , and the velocity of the object in frame S\u2032 by v\u2032 = dr\u2032 dt\u2032 . Since the derivative \nof the position is velocity, the velocities of the object in two different reference frames \nare related according to \ndr \uf072\u2032 \ndr \uf072 \ndR \n\uf072 \n= \n\u2212 \n, \n(11.3.1)\ndt\u2032 \ndt \ndt \n\uf072 \n\uf072\n\uf072 \nv\u2032 = v \u2212 V. \n(11.3.2) \nThis is called the Law of Addition of Velocities. \n11.4 Worked Examples \nExample 11.1 Relative Velocities of Two Moving Planes \nAn airplane A is traveling northeast with a speed of vA = 160 m \u22c5 s-1 . A second airplane \nB is traveling southeast with a speed of vB = 200 m \u22c5 s-1 . (a) Choose a coordinate system \n\uf072\n\uf072\nand write down an expression for the velocity of each airplane as vectors, vA and v B . \n\uf072\n\uf072\nCarefully use unit vectors to express your answer. (b) Sketch the vectors vA and v B on \nyour coordinate system. (c) Find a vector expression that expresses the velocity of aircraft \nA as seen from an observer flying in aircraft B. Calculate this vector. What is its \nmagnitude and direction? Sketch it on your coordinate system. \nSolution: From the information given in the problem we draw the velocity vectors of the \nairplanes as shown in Figure 11.2a. \n(a) \n(b) \nFigure 11.2 (a): Motion of two planes \nFigure 11.2 (b): Coordinate System \n11-3", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_264", "text": "An observer at rest with respect to the ground defines a reference frame S . Choose a \ncoordinate system shown in Figure 11.2b. According to this observer, airplane A is \n\uf072\nmoving with velocity vA = vA cos\u03b8A\u02c6i + vA sin\u03b8A \u02c6j , and airplane B is moving with \n\uf072\nvelocity vB = vB cos\u03b8B\u02c6i + vB sin\u03b8B \u02c6j. According to the information given in the problem \nairplane A flies northeast so \u03b8A = \u03c0 / 4 and airplane B flies southeast east so \u03b8B = \u2212\u03c0 / 4 . \n\uf072 \n-1 )\u02c6j\nThus vA = (80 2 m \u22c5 s-1 )\u02c6i + (80 2 m \u22c5 s-1 )\u02c6j and v\uf072 \nB = (100 2 m \u22c5 s-1 )\u02c6i \u2212 (100 2 m \u22c5 s \nConsider a second observer moving along with airplane B, defining reference frame S\u2032 . \nWhat is the velocity of airplane A according to this observer moving in airplane B ? The \nvelocity of the observer moving along in airplane B with respect to an observer at rest on \nthe \nground \nis \njust \nthe \nvelocity \nof \nairplane \nB \nand \nis \ngiven \nby\n\uf072\n\uf072\nV = vB = vB cos\u03b8B\u02c6i + vB sin\u03b8B \u02c6j . Using the Law of Addition of Velocities, Equation \n(11.3.2), the velocity of airplane A with respect to an observer moving along with \nAirplane B is given by \n\uf072\n\uf072\n\uf072 \nv\u2032 \nA = vA \u2212 V = (vA cos\u03b8A \u02c6i + vA sin\u03b8A \u02c6j) \u2212 (vB cos\u03b8B\u02c6i + vB sin\u03b8B \u02c6j) \n= (vA cos\u03b8A \u2212 vB cos\u03b8B)\u02c6i + (vA sin\u03b8A \u2212 vB sin\u03b8B)\u02c6j \n-1 ))\u02c6j.\n= ((80 2 m \u22c5 s-1 ) \u2212 (100 2 m \u22c5 s-1 ))\u02c6i + ((80 2 m \u22c5 s-1 ) + (100 2 m \u22c5 s\n(11.4.1) \n-1 )\u02c6j\n= \u2212(20 2 m \u22c5 s-1 )\u02c6i + (180 2 m \u22c5 s \n= v\u2032 \u02c6i + v\u2032 \u02c6j\nAx \nAy \nFigure 11.3 shows the velocity of airplane A with respect to airplane B in reference frame \nS\u2032 . \nFigure 11.3 Airplane A as seen from observer in airplane B \nThe magnitude of velocity of airplane A as seen by an observer moving with airplane B \nis given by \n11-4", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_265", "text": "\uf072 \n2 + \n2)1/ 2 \n-1 )2)1/2 \n-1 \nvA \u2032 = (v\u2032 \nv\u2032 \n= ((\u221220 2 m \u22c5 s-1 )2 + (180 2 m \u22c5 s \n= 256 m \u22c5 s . \n(11.4.2)\nAx \nAy \nThe angle of velocity of airplane A as seen by an observer moving with airplane B is \ngiven by, \n\u03b8\u2032 = tan\u22121(v\u2032 / v\u2032 ) = tan\u22121((180 2 m \u22c5 s-1 ) / (\u221220 2 m \u22c5 s-1 )) \nA \nAy \nAx \n. \n(11.4.3) \n= tan\u22121(\u22129) = 180\uf06f \u2212 83.7\uf06f = 96.3\uf06f \nExample 11.2 Relative Motion and Polar Coordinates \nBy relative velocity we mean velocity with respect to a specified coordinate system. (The \nterm velocity, alone, is understood to be relative to the observer\u2019s coordinate system.) (a)\n\uf072\nA point is observed to have velocity v A relative to coordinate system A . What is its \nvelocity relative to coordinate system B , which is displaced from system A by distance \n\uf072\n\uf072 \nR ? ( R can change in time.) (b) Particles a and b move in opposite directions around a \ncircle with the magnitude of the angular velocity \u03c9 , as shown in Figure 11.4. At t = 0 \n\uf072\nthey are both at the point r = l\u02c6j , where l is the radius of the circle. Find the velocity of \na relative to b . \nFigure 11.5 Particles a and b \nmoving relative to each other\nFigure 11.4 Particles a and b \nmoving relative to each other \n11-5", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_266", "text": "Solution: (a) The position vectors are related by \n\uf072r\nB\nA \n\uf072r =  \u2212  \n\uf072 \nR .  \n(11.4.4)  \nThe velocities are related by the taking derivatives, (law of addition of velocities Eq. \n(11.3.2)) \n\uf072v B =  \uf072v A \u2212  \n\uf072 \nV .  \n(11.4.5)  \n(b) Let\u2019s choose two reference frames; frame B is centered at particle b, and frame A is \ncentered at the center of the circle in Figure 11.5. Then the relative position vector \nbetween the origins of the two frames is given by \n\uf072 \nR = l r\u02c6 . \n(11.4.6) \nThe position vector of particle a relative to frame A is given by \n\uf072 = l r\u02c6\u2032 . \n(11.4.7)\nrA \nThe position vector of particle b in frame B can be found by substituting Eqs. (11.4.7) \nand (11.4.6) into Eq. (11.4.4), \n\uf072\n\uf072\n\uf072 \n\u2212 R = l r\u02c6\u2032 \u2212 l r\u02c6 . \n(11.4.8)\nrB = rA \nWe can decompose each of the unit vectors r\u02c6 and r\u02c6\u2032 with respect to the Cartesian unit \nvectors \u02c6i and \u02c6j (see Figure 11.5), \nr\u02c6 = \u2212 sin\u03b8 \u02c6i + cos\u03b8 \u02c6j \n(11.4.9) \nr\u02c6\u2032 = sin\u03b8 \u02c6i + cos\u03b8 \u02c6j. \n(11.4.10) \nThen Eq. (11.4.8) giving the position vector of particle b in frame B becomes \n\uf072 rB = l r\u02c6\u2032 \u2212 l r\u02c6 = l (sin\u03b8 \u02c6i + cos\u03b8 \u02c6j) \u2212 l (\u2212 sin\u03b8 \u02c6i + cos\u03b8 \u02c6j) = 2l sin\u03b8 \u02c6i . (11.4.11) \nIn order to find the velocity vector of particle a in frame B (i.e. with respect to particle b), \ndifferentiate Eq. (11.4.11) \n\uf072 \nd\nd\u03b8 \nv B = \n(2l sin\u03b8) \u02c6i = (2l cos\u03b8)\n\u02c6i = 2\u03c9l cos\u03b8 \u02c6i . \n(11.4.12)\ndt \ndt \n11-6", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_267", "text": "Example 11.3 Recoil in Different Frames \nA person of mass m1 is standing on a cart of mass m2 . Assume that the cart is free to \nmove on its wheels without friction. The person throws a ball of mass m3 at an angle of \n\u03b8 with respect to the horizontal as measured by the person in the cart. The ball is thrown \nwith a speed v0 with respect to the cart (Figure 11.6). (a) What is the final velocity of the \nball as seen by an observer fixed to the ground? (b) What is the final velocity of the cart\nas seen by an observer fixed to the ground? (c) With respect to the horizontal, what angle\nthe fixed observer see the ball leave the cart? \nFigure 11.6 Recoil of a person on cart due to thrown ball \nSolution: a), b) Our reference frame will be that fixed to the ground. We shall take as our \ninitial state that before the ball is thrown (cart, ball, throwing person stationary) and our \nfinal state that after the ball is thrown. We are assuming that there is no friction, and so \nthere are no external forces acting in the horizontal direction. The initial x -component of \nthe total momentum is zero, \ntotal \npx,0 = 0 . \n(11.4.13) \nAfter the ball is thrown, the cart and person have a final momentum \n\uf072 \n\u02c6i\np f ,cart = \u2212(m2 + m1)v f ,cart \n(11.4.14) \nas measured by the person on the ground, where v f ,cart is the speed of the person and cart. \n(The person\u2019s center of mass will move with respect to the cart while the ball is being \nthrown, but since we\u2019re interested in velocities, not positions, we need only assume that\nthe person is at rest with respect to the cart after the ball is thrown.) \nThe ball is thrown with a speed v0 and at an angle \u03b8 with respect to the horizontal as \nmeasured by the person in the cart. Therefore the person in the cart throws the ball with \nvelocity \n\uf072 v\u2032 f ,ball = v0 cos\u03b8 \u02c6i + v0 sin\u03b8 \u02c6j . \n(11.4.15). \n11-7", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_268", "text": "Because the cart is moving in the negative x -direction with speed v f ,cart just as the ball \nleaves the person\u2019s hand, the x -component of the velocity of the ball as measured by an \nobserver on the ground is given by \n. \n(11.4.16)\nvxf , ball = v0 cos\u03b8 \u2212 v f ,cart \nThe ball appears to have a smaller x -component of the velocity according to the observer\non the ground. The velocity of the ball as measured by an observer on the ground is \n\uf072 v \n= (v cos\u03b8 \u2212 v \n) \u02c6i + v sin\u03b8 \u02c6j. \n(11.4.17)\nf , ball \n0 \nf ,cart \n0 \nThe final momentum of the ball according to an observer on the ground is \n\uf072 p f , ball = m3 \u23a3\u23a1(v0 cos\u03b8 \u2212 v f ,cart ) \u02c6i + v0 sin\u03b8 \u02c6j\u23a6\u23a4 . \n(11.4.18) \nThe momentum flow diagram is shown in (Figure 11.7). \nFigure 11.7 Momentum flow diagram for recoil \nBecause the x -component of the momentum of the system is constant, we have that \n0 = ( p \n) \n+ ( p\nx, f cart \nx, f )ball \n(11.4.19) \n= \u2212(m + m )v \n+ m (v cos\u03b8 \u2212 v \n). \n2\n1 \nf , cart \n3 \n0 \nf ,cart \nWe can solve Equation (11.4.19) for the final speed and velocity of the cart as measured\nby an observer on the ground, \nm v cos\u03b8 \nv f ,cart = \n3 0 \n, \n(11.4.20)\nm + m + m\n2\n1\n3 \n\uf072 \nm v cos\u03b8\n\u02c6 \n3\n0\n\u02c6\nv f , cart = v f , cart i = \ni . \n(11.4.21)\nm + m + m\n2\n1\n3 \n11-8", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_269", "text": "Note that the y -component of the momentum is not constant because as the person is\nthrowing the ball he or she is pushing off the cart and the normal force with the ground \nexceeds the gravitational force so the net external force in the y -direction is non-zero. \nSubstituting Equation (11.4.20) into Equation (11.4.17) gives \n\uf072 v f ,ball = (v0 cos\u03b8 \u2212 v f ,cart ) \u02c6i + v0 sin\u03b8 \u02c6j \n(11.4.22) \n= \nm1 + m2\n(v0 cos\u03b8 )\u02c6i + (v0 sin\u03b8 ) \u02c6j. \nm1 + m2 + m3 \n\uf072\nAs a check, note that in the limit m3 << m1 + m2 , v f ,ball has speed v0 and is directed at an \nangle \u03b8 above the horizontal; the fact that the much more massive person-cart \ncombination is free to move doesn\u2019t affect the flight of the ball as seen by the fixed \nobserver. Also note that in the unrealistic limit m >> m1 \nthe ball is moving at a \n+ m2 \nspeed much smaller than v0 as it leaves the cart. \nc) The angle \u03c6 at which the ball is thrown as seen by the observer on the ground is given \nby \n)\n(v f ,ball y\nv0 sin\u03b8\n\u03c6 = tan\u22121 \n= tan\u22121 \n(v f ,ball )x \n\u23a1\u23a3(m1 + m2) / (m1 + m2 + m3)\u23a4\u23a6v0 cos\u03b8 \n(11.4.23)\n\u239e\n\u23a4 \n= tan\u22121 \u23a1\n\u23a2 \n\u239b m1 + m2 + m3 \n\u23a0\u239f tan\u03b8 \u23a5. \n\u23a2 \u239d\u239c \n\u23a3 \nm1 + m2 \n\u23a6\u23a5 \nFor arbitrary values for the masses, the above expression will not reduce to a simplified \nform. However, we can see that tan\u03c6 > tan\u03b8 for arbitrary masses, and that in the limit \n<< m1 + m2 , \u03c6 \u2192\u03b8 and in the unrealistic limit m3 >> m1 + m2 , \u03c6 \u2192\u03c0 / 2 . Can you \nm3 \nexplain this last odd prediction? \n11-9", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_270", "text": "Chapter 12 Momentum and the Flow of Mass \nPreface: The Challenger Flight.................................................................................... 1  \n12.1 Introduction........................................................................................................... 2  \n12.1.1 Transfer of Material into an Object, but no Transfer of Momentum ...... 3  \n12.1.2 Transfer of Material Out of an Object, but no Transfer of Momentum .. 3  \n12.1.3 Transfer of Material Impulses Object Via Transfer of Momentum......... 4  \n12.1.4 Material Continually Ejected From Object results in Recoil of Object ... 4  \n12.2 Worked Examples................................................................................................. 5  \nExample 12.1 Filling a Coal Car.............................................................................. 5", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_271", "text": "Example 12.2 Emptying a Freight Car ................................................................... 6  \nExample 12.3 Filling a Freight Car ......................................................................... 9  \nExample 12.4 Boat and Fire Hose ......................................................................... 12  \n12.3 Rocket Propulsion............................................................................................... 18  \n12.3.1 Rocket Equation in Gravity-free Space ..................................................... 21  \nExample 12.4 Single-Stage Rocket ........................................................................ 23", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_272", "text": "Example 12.5 Two-Stage Rocket ........................................................................... 23  \n12.3.2 Rocket in a Constant Gravitational Field:................................................. 24", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_273", "text": "Chapter 12 Momentum and the Flow of Mass \nEven though the release was pulled, the rocket did not rise at first, but the \nflame came out, and there was a steady roar. After a number of seconds it \nrose, slowly until it cleared the flame, and then at express-train speed, \ncurving over to the left, and striking the ice and snow, still going at a \nrapid rate. It looked almost magical as it rose, without any appreciably \ngreater noise or flame, as if it said, \u201cI've been here long enough; I think \nI'll be going somewhere else, if you don't mind.\u201d 1 \nRobert Goddard \nPreface: The Challenger Flight \nWhen the Rogers Commission in 1986 investigated the Challenger disaster, a \ncommission member, physicist Richard Feynman, made an extraordinary demonstration \nduring the hearings. \n\u201cHe (Feynman) also learned that rubber used to seal the solid rocket booster joints using \nO-rings, failed to expand when the temperature was at or below 32 degrees F (0 degrees \nC). The temperature at the time of the Challenger liftoff was 32 degrees F. Feynman \nnow believed that he had the solution, but to test it, he dropped a piece of the O-ring \nmaterial, squeezed with a C-clamp to simulate the actual conditions of the shuttle, into a \nglass of ice water. Ice, of course, is 32 degrees F. At this point one needs to understand \nexactly what role the O-rings play in the solid rocket booster (SRB) joints. When the \nmaterial in the SRB start to heat up, it expands and pushes against the sides of the SRB. \nIf there is an opening in a joint in the SRB, the gas tries to escape through that opening \n(think of it like water in a tea kettle escaping through the spout.) This leak in the \nChallenger's SRB was easily visible as a small flicker in a launch photo. This flicker \nturned into a flame and began heating the fuel tank, which then ruptured. When this \nhappened, the fuel tank released liquid hydrogen into the atmosphere where it exploded. \nAs Feynman explained, because the O-rings cannot expand in 32 degree weather, the gas \nfinds gaps in the joints, which led to the explosion of the booster and then the shuttle \nitself.\u201d2 \nIn the Report of the Presidential Commission on the Space Shuttle Challenger Accident \n(1986), Appendix F - Personal observations on the reliability of the Shuttle, Feynman \nwrote", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_274", "text": "finds gaps in the joints, which led to the explosion of the booster and then the shuttle \nitself.\u201d2 \nIn the Report of the Presidential Commission on the Space Shuttle Challenger Accident \n(1986), Appendix F - Personal observations on the reliability of the Shuttle, Feynman \nwrote \nThe Challenger flight is an excellent example. \u2026 The O-rings of the Solid Rocket \nBoosters were not designed to erode. Erosion was a clue that something was wrong. \nErosion was not something from which safety can be inferred. There was no way, without \n1 describing the first rocket flight using liquid propellants at Aunt Effie's farm, 17 March 1926. \n2 http://www.fotuva.org/online/frameload.htm?/online/challenger.htm. \n12-1", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_275", "text": "full understanding, that one could have confidence that conditions the next time might not \nproduce erosion three times more severe than the time before. Nevertheless, officials \nfooled themselves into thinking they had such understanding and confidence, in spite of \nthe peculiar variations from case to case. A mathematical model was made to calculate \nerosion. This was a model based not on physical understanding but on empirical curve \nfitting. To be more detailed, it was supposed a stream of hot gas impinged on the O-ring \nmaterial, and the heat was determined at the point of stagnation (so far, with reasonable \nphysical, thermodynamic laws). But to determine how much rubber eroded it was \nassumed this depended only on this heat by a formula suggested by data on a similar \nmaterial. A logarithmic plot suggested a straight line, so it was supposed that the erosion \nvaried as the .58 power of the heat, the .58 being determined by a nearest fit. At any rate, \nadjusting some other numbers, it was determined that the model agreed with the erosion \n(to depth of one-third the radius of the ring). There is nothing much so wrong with this as \nbelieving the answer! Uncertainties appear everywhere. How strong the gas stream might \nbe was unpredictable, it depended on holes formed in the putty. Blow-by showed that the \nring might fail even though not, or only partially eroded through. The empirical formula \nwas known to be uncertain, for it did not go directly through the very data points by \nwhich it was determined. There were a cloud of points some twice above, and some twice \nbelow the fitted curve, so erosions twice predicted were reasonable from that cause alone. \nSimilar uncertainties surrounded the other constants in the formula, etc., etc. When using \na mathematical model careful attention must be given to uncertainties in the model. \u2026 \nIn any event this has had very unfortunate consequences, the most serious of which is to \nencourage ordinary citizens to fly in such a dangerous machine, as if it had attained the \nsafety of an ordinary airliner. The astronauts, like test pilots, should know their risks, and \nwe honor them for their courage. Who can doubt that McAuliffe was equally a person of \ngreat courage, who was closer to an awareness of the true risk than NASA management \nwould have us believe? Let us make recommendations to ensure that NASA officials deal \nin a world of reality in understanding technological weaknesses and imperfections well \nenough to be actively trying to eliminate them. \u2026. For a successful technology, reality", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_276", "text": "great courage, who was closer to an awareness of the true risk than NASA management \nwould have us believe? Let us make recommendations to ensure that NASA officials deal \nin a world of reality in understanding technological weaknesses and imperfections well \nenough to be actively trying to eliminate them. \u2026. For a successful technology, reality \nmust take precedence over public relations, for nature cannot be fooled. 3 \n12.1 Introduction \nSo far we have restricted ourselves to considering systems consisting of discrete objects \nor point-like objects that have fixed amounts of mass. We shall now consider systems in \nwhich material flows between the objects in the system, for example we shall consider \ncoal falling from a hopper into a moving railroad car, sand leaking from railroad car fuel, \ngrain moving forward into a railroad car, and fuel ejected from the back of a rocket, In \neach of these examples material is continuously flows into or out of an object. We have \nalready shown that the total external force causes the momentum of a system to change, \n3 R. P. Feynman, Appendix F - Personal observations on the reliability of the Shuttle, Report of the \nPRESIDENTIAL COMMISSION on the Space Shuttle Challenger Accident (1986), \nhttp://history.nasa.gov/rogersrep/genindex.htm. \n12-2", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_277", "text": "\uf072 \nd p \uf072 \ntotal \nsystem \nF \n= \n. \n(12.2.1)\next \ndt \nWe shall analyze how the momentum of the constituent elements our system change over \na time interval [t,t + \u0394t], and then consider the limit as \u0394t \u2192 0 . We can then explicit \ncalculate the derivative on the right hand side of Eq. (12.2.1) and Eq. (12.2.1) becomes \n\uf072p\nd \n\u0394 \n(t + \u0394t) \u2212 \n(t)\nsystem \nsystem \nsystem \nsystem \next \n\uf072p \n\uf072p \n\uf072p\n\uf072\nFtotal \n= lim \n= lim \n(12.2.2) \n=  \n. \ndt \n\u0394t \n\u0394t\n\u0394t\u21920 \n\u0394t\u21920 \nWe need to be very careful how we apply this generalized version of Newton\u2019s Second \nLaw to systems in which mass flows between constituent objects. In particular, when we \nisolate elements as part of our system we must be careful to identify the mass \u0394m of the \nmaterial that continuous flows in or out of an object that is part of our system during the \ntime interval \u0394t under consideration. \nWe shall consider four categories of mass flow problems that are characterized by the \nmomentum transfer of the material of mass \u0394m . \n12.1.1 Transfer of Material into an Object, but no Transfer of Momentum \nConsider for example rain falling vertically downward with speed u into car of mass m \nmoving forward with speed v . A small amount of falling rain \u0394mr has no component of \nmomentum in the direction of motion of the car. There is a transfer of rain into the car but \nno transfer of momentum in the direction of motion of the car (Figure 12.1). \nu \nm r\nu \nv\nfrictionless\nrain\nv\nfrictionless\nFigure 12.1 Transfer of rain mass into the car but no transfer of momentum in direction \nof motion \n12.1.2 Transfer of Material Out of an Object, but no Transfer of Momentum \nThe material continually leaves the object but it does not transport any momentum away \nfrom the object in the direction of motion of the object (Figure 12.2). Consider an ice \nskater gliding on ice at speed v holding a bag of sand that is leaking straight down with \nrespect to the moving skater. The sand continually leaves the bag but it does not transport \nany momentum away from the bag in the direction of motion of the object. In Figure \n12.2, sand of mass \u0394ms leaves the bag. \n12-3", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_278", "text": "Figure 12.2 Transfer of mass out of object but no transfer of momentum in direction of \nmotion \n12.1.3 Transfer of Material Impulses Object Via Transfer of Momentum \nSuppose a fire hose is used to put out a fire on a boat of mass mb . Assume the column of \nwater moves horizontally with speed u . The incoming water continually hits the boat \npropelling it forward. During the time interval \u0394t , a column of water of mass \u0394ms will \nhit the boat that is moving forward with speed v increasing it\u2019s speed (Figure 12.3). \nFigure 12.3 Transfer of mass of water increases speed of boat \n12.1.4 Material Continually Ejected From Object results in Recoil of Object \nWhen fuel of mass \u0394mf is ejected from the back of a rocket with speed u relative to the \nrocket, the rocket of mass mr recoils forward. Figure 12.4a shows the recoil of the rocket \nin the reference frame of the rocket. The rocket recoils forward with speed \u0394vr . In a \nreference frame in which the rocket is moving forward with speed vr , then the speed after \nrecoil is v + \u0394v . The speed of the backwardly ejected fuel is u \u2212 v (Figure 12.4b).\nr\nr \nr \n(a) \n(b) \n12-4", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_279", "text": "Figure 12.4 Transfer of mass out of rocket provides impulse on rocket in (a) reference \nframe of rocket, (b) reference frame in which rocket moves with speed vr \nWe must carefully identify the momentum of the object and the material transferred at \n\uf072\ntime t in order to determine p \n(t) . We must also identify the momentum of the object \nsystem \n\uf072\nand the material transferred at time t + \u0394t in order to determine p \n(t + \u0394t) as well. \nsystem \nRecall that when we defined the momentum of a system, we assumed that the mass of the \nsystem remain constant. Therefore we cannot ignore the momentum of the transferred \nmaterial at time t + \u0394t even though it may have left the object; it is still part of our \nsystem (or at time t even though it has not flowed into the object yet). \n12.2 Worked Examples \nExample 12.1 Filling a Coal Car \nAn empty coal car of mass m0 starts from rest under an applied force of magnitude F . \nAt the same time coal begins to run into the car at a steady rate b from a coal hopper at \nrest along the track (Figure 12.5). Find the speed when a mass mc of coal has been \ntransferred. \nFigure 12.5 Filling a coal car \nSolution: We shall analyze the momentum changes in the horizontal direction, which we \ncall the x -direction. Because the falling coal does not have any horizontal velocity, the \nfalling coal is not transferring any momentum in the x -direction to the coal car. So we \nshall take as our system the empty coal car and a mass mc of coal that has been \ntransferred. Our initial state at t = 0 is when the coal car is empty and at rest before any \ncoal has been transferred. The x -component of the momentum of this initial state is zero, \npx (0) = 0. \n(12.3.1) \n12-5", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_280", "text": "Our final state at t = t is when all the coal of mass m = bt has been transferred into the\nf \nc\nf \ncar that is now moving at speed v f . The x -component of the momentum of this final \nstate is \np \n+ m \n+ bt f \n. \n(12.3.2)\nx (t f ) = (m0 \nc )v f = (m0\n)v f \nThere is an external constant force Fx = F applied through the transfer. The momentum \nprinciple applied to the x -direction is \nt f \nF dt = \u0394p = \n) \u2212 p (0) . \n(12.3.3)\n\u222b x \nx\npx (t f \nx \n0 \nBecause the force is constant, the integral is simple and the momentum principle becomes \nFt f \n+ bt f \n. \n(12.3.4)\n= (m0 \n)v f \nSo the final speed is \nFt f\n= \n. \n(12.3.5)\nv f \n(m0 + bt f ) \nExample 12.2 Emptying a Freight Car \nA freight car of mass mc contains sand of mass ms . At t = 0 a constant horizontal force \nof magnitude F is applied in the direction of rolling and at the same time a port in the \nbottom is opened to let the sand flow out at the constant rate b = dms / dt . Find the speed \nof the freight car when all the sand is gone (Figure 12.6). Assume that the freight car is at \nrest at t = 0. \nFigure 12.6 Emptying a freight car \nSolution: Choose the positive x -direction to point in the direction that the car is moving.  \nChoose for the system the amount of sand in the fright car at time t , mc (t) . At time t ,  \n12-6", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_281", "text": "!\nthe car is moving with velocity v c (t) = vc (t)\u02c6i . The momentum diagram for the system at \ntime t is shown in the diagram on the left in Figure 12.7. \ntime \ntime \nv c (t) \nv c (t) + v c \nt + t\nt \nm c (t) + m c \nm c (t) \nm s \nv c (t) + v c \nFigure 12.7 Momentum diagram at time t and at time t + \u0394t \nThe momentum of the system at time t is given by \n! psys (t) = m c (t)! v c (t) . \n(12.3.6) \nDuring the time interval [ ,t t \nt]\n+ \u0394 , an amount of sand of mass \ns\n\u0394m leaves the freight car \nand the mass of the freight car changes by m (t + \u0394t) = m (t) + \u0394m , where \u0394m = \u2212\u0394m . \nc \nc\nc \nc\ns \nAt \nthe \nend \nof \nthe \ninterval \nthe \ncar \nis \nmoving \nwith \nvelocity\n!  \nv c (t + \u0394t) = !  \nv c (t) + \u0394\nc \n!v = (vc (t) + \u0394v c )\u02c6i . The momentum diagram for the system at time \nt + \u0394t is shown in the diagram on the right in Figure 12.7. The momentum of the system \nat time t + \u0394t is given by \n! \n!\n! \np\n(t + \u0394t) = (\u0394m + m (t) + \u0394m )(v ! (t) + \u0394v ) = m (t)(v ! (t) + \u0394v ) .(12.3.7)\nsys \ns\nc \nc \nc \nc \nc \nc \nc \n!\n!\nNote that the sand that leaves the car is shown with velocity v c (t) + \u0394v c . This implies \nthat all the sand leaves the car with the velocity of the car at the end of the interval. This \nis an approximation. Because the sand leaves continuous, the velocity will vary from \n! \n!\n! \nv (t) to v (t) + \u0394v \nbut so does the change in mass of the car and these two \nc \nc\nc \ncontributions to the system\u2019s moment exactly cancel. The change in momentum of the \nsystem is then \n\u0394p \n! \n!  \nsys =  psys (t + \u0394t) \u2212!  \npsys (t) = mc (t)( !  \nv c (t) + \u0394!  \nv c ) \u2212 mc (t)!", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_282", "text": "system is then \n\u0394p \n! \n!  \nsys =  psys (t + \u0394t) \u2212!  \npsys (t) = mc (t)( !  \nv c (t) + \u0394!  \nv c ) \u2212 mc (t)!  \nv c (t) = mc (t)\u0394v !  \nc .(12.3.8)  \n! \nThroughout the interval a constant force F = F\u02c6i is applied to the system so the \nmomentum principle becomes \n12-7", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_283", "text": "!\n!\n(t + \u0394t) \u2212 \n\u0394! \n\u0394t \nvc \n(t) d! v c .\ndt \n(12.3.9)  \n!  \n= lim \n= lim mc (t) \n= m\n\u0394t \nc\n\u0394t\u21920 \n\u0394t\u21920 \nF \n(t)\nsys \nsys \nBecause the motion is one-dimensional, Eq. (12.3.9) written in terms of x -components \nbecomes \ndv c\nF = m (t) \n. \n(12.3.10)\nc \ndt \nDenote by initial mass of the car by mc,0 = mc + ms where mc is the mass of the car and \nis the mass of the sand in the car at t = 0 . The mass of the sand that has left the car at \nms \ntime t is given by \nt dm \nt \ns\nm (t) = \ndt = bdt = bt . \n(12.3.11)\ns \n\u222b\n\u222b \n0 dt \n0 \nThus \nm (t) = m \u2212 bt = m + m \u2212 bt . \n(12.3.12) \np \nc\nc,0 \nc\ns \nTherefore Eq. (12.3.10) becomes \ndv c\nF = (m + m \u2212 bt) \n. \n(12.3.13)\nc \ns\ndt \np \nThis equation can be solved for the x -component of the velocity at time t , vc (t) , (which \nin this case is the speed) by the method of separation of variables. Rewrite Eq. (12.3.13) \nas \nFdt \ndv c = \n. \n(12.3.14)\n(mc + ms \u2212 bt) \nThen integrate both sides of Eq. (12.3.14) with the limits as shown \nvc \u2032 =v (t )\nc \nt\u2032=t \nFdt\u2032 \ndv\u2032 = \n. \n(12.3.15)\n\u222b \nc \n\u222b m + m \u2212 bt\u2032 \nv\u2032 =0 \nt\u2032=0 \nc \ns\nc \nIntegration yields the speed of the car as a function of time \nt\u2032= t\nF \n\u239b m + m \u2212 bt \u239e \nF \n\u239b \nm + m \n\u239e\nF \nc\ns \nc\ns\nvc (t) = \u2212 \nln(m c + ms \u2212 bt\u2032) \n= \u2212 \nln \nln \n\u23a0\u239f . (12.3.16)\nb \nt\u2032= 0 \nb \n\u239d\u239c \nm c + m s \n\u23a0\u239f = b \n\u239d\u239c m c + m s \u2212 bt \n12-8", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_284", "text": "In writing Eq. (12.3.16), we used the property that ln(a) \u2212 ln(b) = ln(a / b) and therefore \nln(a / b) = \u2212 ln(b / a) . \nNote \nthat \nm + m \u2265 m + m \u2212 bt , \nso \nthe \nterm \nc\ns \nc\ns \n\u239b \nm + m \n\u239e \nln \nc\ns \n\u23a0\u239f \u2265 0 , and the speed of the car increases as we expect.\n\u239d\u239c m + m \u2212 bt \nc\ns \nExample 12.3 Filling a Freight Car \nGrain is blown into car A from car B at a rate of b kilograms per second. The grain \nleaves the chute vertically downward, so that it has the same horizontal velocity, u as car \nB , (Figure 12.8). Car A is initially at rest before any grain is transferred in and has mass \nAt the moment of interest, car A has mass \nand speed v . Determine an \nmA,0 . \nmA \nexpression for the speed car A as a function of time t . \nFigure 12.8 Filling a freight car \nSolution: Choose positive x -direction to the right in the direction the cars are moving. \nDefine the system at time t to be the car and grain that is already in it, which together has \nmass \n(t) , and the small amount of material of mass \u0394m that is blown into car A\nmA \ng \nduring the time interval [t,t + \u0394t] . At time that is moving with x -component of the \n!\nvelocity vA . At time t , car A is moving with velocity v A (t) = vA (t)\u02c6i , and the material \n!\nblown into car is moving with velocity u = u\u02c6i At time t + \u0394t , car A is moving with \n!\n!\nvelocity v (t) + \u0394v = (vA (t) + \u0394vA )\u02c6i , and the mass of car A is mA (t + \u0394t) = mA (t) + \u0394mA ,\nA \nA \nwhere \u0394mA = \u0394m g . The momentum diagram for times t and for t + \u0394t is shown in \nFigure 12.9. \ntime \ntime t + t\nt \n\u02c6i \nmA (t) \nm g \nu \nv A (t) \nA \nv A (t) + v A \nA \nmA (t) + mA \nFigure 12.9 Momentum diagram at times t and t + \u0394t \n12-9", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_285", "text": "The momentum at time t is \n! \nP (t) = mA (t)!v A (t) + \u0394mg \n!u . \n(12.3.17) \nsys \nThe momentum at time t + \u0394t is \n(t) + \u0394 \nThere are no external forces acting on the system in the x -direction and the external \nforces acting on the system perpendicular to the motion sum to zero, so the momentum \n!v \nprinciple becomes \n!\n! \n! \nP (t + \u0394t) \u2212 P (t)\nsys \nsys \n0 = lim \n. \n(12.3.19)\n\u0394t\u21920 \n\u0394t \nUsing the results above (Eqs. (12.3.17) and (12.3.18), the momentum principle becomes \n! \nP \n!v\n(t + \u0394t) = (mA (t) + \u0394mA )(  \n) . \n(12.3.18) \nA \nA\nsys \n!\n0  \n!u \n!v \n!v\n)( \n(t) + \u0394 \n(t)\n(t) + \u0394m )\n(mA (t) + \u0394mA \nA \nA ) \u2212 (mA \nA \ng \n!v \nlim \n(12 3 20)\n= \n.\n. . \n\u0394t\u21920 \n\u0394t \n!v \n!u \n!v \n!v \nwhich after using the condition that \u0394mA = \u0394mg and some rearrangement becomes \n(t)\u0394 \n(\n(t) \u2212 ) \n\u0394\nmA \nA \n\u0394mA \nA \n\u0394mA\nA \n!\n0 = lim \n+ lim \n+ lim \n. \n(12.3.21) \n\u0394t \n\u0394t \n\u0394t\n\u0394t\u21920 \n\u0394t\u21920 \n\u0394t\u21920 \n!v\n\u0394!\nIn the limit as , the product \u0394mA v A is a second order differential (the product of two \n\u0394!\nfirst order differentials) and the term \u0394mA v A / \u0394t approaches zero, therefore the \nmomentum principle yields the differential equation \nd\nA \n!\n0 \ndmA !v\n!u\n(t) \n( \n(t) \u2212) . \n(12.3.22) \n+ \n= mA \nA\ndt \ndt \nThe x -component of Eq. (12.3.22) is then \ndvA \ndmA\n0 = mA (t) \n+ \n(t) \u2212 u) . \n(12.3.23)\n(vA\ndt \ndt \nRearranging terms and using the fact that the material is blown into car A at a constant \nrate b \u2261 dmA / dt , we have that the rate of change of the x -component of the velocity of \ncar A is given by \ndvA (t) \ndt", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_286", "text": "(vA\ndt \ndt \nRearranging terms and using the fact that the material is blown into car A at a constant \nrate b \u2261 dmA / dt , we have that the rate of change of the x -component of the velocity of \ncar A is given by \ndvA (t) \ndt \n= b(u \u2212 vA (t)) \nmA (t) \n. \n(12.3.24) \n12-10", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_287", "text": "We cannot directly integrate Eq. (12.3.24) with respect to dt because the mass of the car \nA is a function of time. In order to find the x -component of the velocity of car A we \nneed to know the relationship between the mass of car A and the x -component of the \nvelocity of the car A . There are two approaches. In the first approach we separate \nvariables in Eq. (12.3.24) where we have suppressed the dependence on t in the \nexpressions for mA and vA yielding \ndvA \ndmA\n= \n, \n(12.3.25)\nu \u2212 vA \nmA \nwhich becomes the integral equation \nv\u2032 A = vA (t ) \n\u2032 \nm \u2032 A = mA (t ) \n\u2032\ndvA \ndmA\n= \n, \n(12.3.26)\n\u222b\n\u222b \n\u2032\n\u2032 \nv\u2032 A =0 u \u2212 vA \nm\u2032 A = mA,0 mA \nwhere mA,0 is the mass of the car before any material has been blown in. After integration \nwe have that \nu \n(t)\nmA\nln \n= ln \n. \n(12.3.27)\nu \u2212 vA (t) \nmA,0 \nExponentiate both side yields \nu \nmA (t)\n= \n. \n(12.3.28)\n(t)\nu \u2212 vA \nmA,0 \nWe can solve this equation for the x -component of the velocity of the car \nmA (t) \u2212 mA,0 \n(t) = \nu . \n(12.3.29)\nvA \n(t)\nmA \nBecause the material is blown into the car at a constant rate b \u2261 dmA / dt , the mass of the \ncar as a function of time is given by \n+ bt . \n(12.3.30)\nmA (t) = mA,0 \nTherefore substituting Eq. (12.3.30) into Eq. (12.3.29) yields the x -component of the \nvelocity of the car as a function of time \nbt\n(t) = \nu . \n(12.3.31)\nvA \n+ bt\nmA,0 \nIn a second approach, we substitute Eq. (12.3.30) into Eq. (12.3.24) yielding \n12-11", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_288", "text": "dvA \nb(u \u2212 vA )\n= \n. \n(12.3.32)\ndt \nm \n+ bt\nA,0 \nSeparate variables in Eq. (12.3.32): \ndvA \nbdt \n= \n, \n(12.3.33)\nu \u2212 v\nm \n+ bt\nA\nA,0 \nwhich then becomes the integral equation \nvA \u2032 =vA (t ) \nt \u2032 =t\ndvA \ndt\u2032 \n= \n. \n(12.3.34)\n\u222b \n\u2032\n\u2032\n\u222b \n\u2032 =0 u \u2212 v\nt \u2032 =0 m\n+ bt\u2032 \nvA \nA \nA,0 \nIntegration yields \nu\nm \n+ bt \nln \n= ln \nA,0 \n. \n(12.3.35)\nu \u2212 v (t) \nm\nA\nA,0 \nAgain exponentiate both sides resulting in \nu \nm\n+ bt \n= \nA,0 \n. \n(12.3.36)\nu \u2212 v (t) \nm\nA\nA,0 \nAfter some algebraic manipulation we can find the speed of the car as a function of time \nbt\n(t) = \nu . \n(12.3.37)\nvA \n+ bt\nmA,0 \nin agreement with Eq. (12.3.31). \nCheck result: \nWe can rewrite Eq. (12.3.37) as \n(mA,0 + bt)vA (t) = btu , \n(12.3.38) \nwhich illustrates the point that the momentum of the system at time t is equal to the \nmomentum of the grain that has been transferred to the system during the interval [0,t]. \nExample 12.4 Boat and Fire Hose \nA burning boat of mass m0 is initially at rest. A fire fighter stands on a bridge and sprays \nwater onto the boat. The water leaves the fire hose with a speed u at a rate \u03b1 (measured \nin kg s-1 ). Assume that the motion of the boat and the water jet are horizontal, that \n\u22c5 \ngravity does not play any role, and that the river can be treated as a frictionless surface. \nAlso assume that the change in the mass of the boat is only due to the water jet and that \nall the water from the jet is added to the boat, (Figure 12.10). \n12-12", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_289", "text": "Figure 12.10 Example 12.4 \na)  In a time interval [ ,t t + \u0394t] , an amount of water \u0394m hits the boat. Choose a \nsystem. Is the total momentum constant in your system? Write down a differential \nequation that results from the analysis of the momentum changes inside your \nsystem. \nb)  Integrate the differential equation you found in part a), to find the velocity ( )\nv m \nas a function of the increasing mass m of the boat, m0 , and u . \nSolution: Let\u2019s take as our system the boat, the amount of water of mass \u0394m that enters \nw \nthe boat during the time interval [ ,t t + \u0394t] and whatever water is in the boat at time t . \nThe water from the fire hose has a speed u . Denote the mass of the boat (including some \nwater) at time t by mb \n(t) , and the speed of the boat by v \u2261 vb (t) . At time t + \u0394t the\n\u2261 mb \nspeed of the boat is v + \u0394v . Choose the positive x - direction in the direction that the boat \nis moving. Then the x -components of the momentum of the system at time t and t + \u0394t \nare shown in Figure 12.11. \nu \nv \nt + t \nt \nm w \nm w \nv + v \nmb \nmb \nFigure 12.11 Momentum diagrams for burning boat \nBecause we are assuming that the burning boat slides with negligible resistance and that \ngravity has a negligible effect on the arc of the water jet, there are no external forces \nacting on the system in the x -direction. Therefore the x -component of the momentum of \nthe system is constant during the interval [ ,t t + \u0394t] and so \n12-13", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_290", "text": "p t( \nt) \n( )\n+  \u0394\n\u2212p t\nx\nx\n0 = lim \n. \n(12.3.39)\nt \n0\n\u0394 \u2192 \n\u0394t \nUsing the information from the figure above, Eq. (12.3.39) becomes \n(m + \u0394m )(v +  \u0394  \u2212\u0394  \nv)\n( m u\n \n+ m v)\nb\nw \nw\nb\n0 = lim \n. \n(12.3.40)\nt \n0\n\u0394 \u2192 \n\u0394t \nEq. (12.3.40) simplifies to \n\u0394v \n\u0394mw \n\u0394mw\u0394v \n\u0394mw\n0 = lim mb \n+ lim \nv + lim \n\u2212lim \nu . \n(12.3.41)\n\u0394 \u2192\nt \n0 \n\u0394 \u2192\nt \n0 \nt\nt \n0 \n\u0394\nt \n0 \u0394t\n\u0394 \u2192\n\u0394t \n\u0394 \n\u0394 \u2192 \nt \nThe third term vanishes when we take the limit \u0394 \u2192\nt \n0 because it is of second order in \nthe infinitesimal quantities (in this case \nm\nv) and \nwhen dividing by \u0394t\n\u0394\n\u0394\nso \nthe\nw \nquantity is of first order and hence vanishes since both \u0394mw \u2192 0 and \u0394 \u2192\nv \n0. Eq. \n(12.3.41) becomes \n\u0394v \n\u0394mw \n\u0394mw\n0 = lim mb \n+ lim \nv \u2212lim \nu. \n(12.3.42)\nt \n0 \nt \n\u0394 \u21920\n\u0394 \u2192 \n\u0394 \u21920 \nt\n\u0394t \n\u0394t \n\u0394t \nWe now use the definition of the derivatives: \n\u0394v\ndv\n \n\u0394m\n dm\nw\nw\nlim \n= \n; lim \n= \n. \n(12.3.43)\nt\n0\n\u0394 \u2192\nt\n0 \u0394t \ndt\n\u0394 \u2192\n\u0394t \ndt \nin Eq. (12.3.42) to fund the differential equation describing the relation between the \nacceleration of the boat and the time rate of change of the mass of water entering the \nboat \ndv \ndmw\n0 = mb \n+ \n(v \u2212 u) . \n(12.3.44)\ndt \ndt \nThe mass of the boat is increasing due to the addition of the water. Let m\nt( ) denote the \nw\nmass of the water that is in the boat at time t . Then the mass of the boat can be written as \n+ m (t) , \n(12.3.45)\nmb (t) = m0 \nw \nwhere m0 is the mass of the boat before any water entered. Note we are neglecting the \neffect of the fire on the mass of the boat. Differentiating Eq. (12.3.45) with respect to \ntime yields \ndmb", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_291", "text": "mb (t) = m0 \nw \nwhere m0 is the mass of the boat before any water entered. Note we are neglecting the \neffect of the fire on the mass of the boat. Differentiating Eq. (12.3.45) with respect to \ntime yields \ndmb \ndmw\n= \n, \n(12.3.46)\ndt \ndt \nThen Eq. (12.3.44) becomes \ndv \ndmb\n0 = mb \n+ \n(v \u2212 u). \n(12.3.47)\ndt \ndt \n12-14", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_292", "text": "(b) We can integrate this equation through the separation of variable technique. Rewrite \nEq. (12.3.47) as (cancel the common factor dt ) \ndv \ndmb\n= \u2212 \n. \n(12.3.48)\nv\nu\n \nm\n\u2212 \nb \nWe can then integrate both sides of Eq. (12.3.48) with the limits as shown \nv(t ) \nmb (t )\ndv \ndmb\n= \u2212 \n(12.3.49)\n\u222b\n\u222b\nv \u2212 u \nv=0 \nm0 mb \nIntegration yields \n\u239b v(t) \u2212 u\u239e\n\u239b mb (t)\u239e\nln \n\u23a0\u239f = \u2212 ln \n(12.3.50)\n\u239d\u239c \n\u2212u \n\u239d\u239c m0 \u23a0\u239f \nRecall that ln( / )\nb a so Eq. (12.3.50) becomes\na b = \u2212 ln( / ) \n\u239b v(t) \u2212 u \u239e\n\u239b \n\u239e\nln \n\u23a0\u239f = ln \nm0 \n(12.3.51)\n\u239d\u239c \n\u2212u \n\u239d\u239c mb (t)\u23a0\u239f \nAlso recall that exp(ln( / )) \n/\na b  = a b and so exponentiating both sides of Eq. (12.3.51) \nyields \nv(t) \u2212 u\nm0\n= \n(12.3.52)\n\u2212u \n(t)\nmb \nSo the speed of the boat at time t can be expressed as \n\u239b\n\u239e\nm0\nv(t) = u 1\u2212 \n(12.3.53)\n\u239d\u239c \n(t)\u23a0\u239f\nmb \nCheck result: \nWe can rewrite Eq. (12.3.52) as \nmb (t)(v(t) \u2212 u) = \u2212m0u \u21d2 mb (t)v(t) = (mb (t) \u2212 m0)u . \n(12.3.54) \nRecall that the mass of the water that enters the car during the interval [0,t] is \nm (t) = mb \n. Therefore Eq. (12.3.54) becomes \nw \n(t) \u2212 m0 \n(t)v(t) = m (t)u . \n(12.3.55)\nmb \nw \n12-15", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_293", "text": "During the interaction between the jet of water and the boat, the water transfers an \namount of momentum m (t)u to the boat and car producing a momentum \n(t)v(t) . \nw \nmb \nBecause all the water that collides with the boat ends up in the boat, all the interaction \nforces between the jet of water and the boat are internal forces. The boat recoils forward \nand the water recoils backward and through collisions with the boat stays in the boat. \nTherefore if we choose as our system, all of the water that eventually ends up in the boat \nand the boat then the momentum principle states \np (t) = p (0) , \n(12.3.56)\nsys \nsys \nwhere p (0) = m (t)u is the momentum of all of the water that eventually ends up in the\nsys \nw \nboat. \nNote that the problem didn\u2019t ask to find the speed of the boat as a function t . We shall \nnow show how to find that. We begin by observing that \ndmb \ndmw\n= \n\u2260\u03b1 \n(12.3.57)\ndt \ndt \nwhere the constant \u03b1 is measured in kg s\u22c5 -1 and is specified as a given constant \naccording to the information in the problem statement. The reason is that \u03b1 is the rate \nthat the water is ejected from the hose but not the rate that the water enters the boat. \nu t \nm = u t \nFigure 12.12 Mass per unit length of water jet \nConsider a small amount of water that is moving with speed u that, in a time interval \u0394t , \nflows through a cross sectional area oriented perpendicular to the flow (see Figure 12.12). \nThe area is larger than the cross sectional area of the jet of water. The amount of water \nthat floes through the area element \u0394m = \u03bbu\u0394t , where \u03bb is the mass per unit length of \nthe jet and u\u0394t is the length of the jet that flows through the area in the interval \u0394t . The \nmass rate of water that flows through the cross sectional area element is then \n\u0394m\n\u03b1 =\n= \u03bbu . \n(12.3.58)\n\u0394t \n12-16", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_294", "text": "In the Figure 12.13 we consider a small length u\u0394t of the water jet that is just behind the \nboat at time t . During the time interval [ ,t t + \u0394t], the boat moves a distance v t .\n\u0394 \nu t \nv \nt \nv t \n(u v) t \nt + t \nFigure 12.13 Amount of water that enter boat in time interval [t,t + \u0394t] \nOnly a fraction of the length u\u0394t of water enters the boat and is given by \n\u03b1\n\u0394mw = \u03bb(u \u2212 v)\u0394t = \n(u \u2212 v)\u0394t \n(12.3.59)\nu \nDividing Eq. (12.3.59) through by \u0394t and taking limits we have that \ndmw \n\u0394mw \n\u03b1 \n= lim \n= \n(u \u2212 v) = \u03b1(1\u2212 v ) . \n(12.3.60)\ndt \n\u0394t\u21920 \u0394t\nu \nu \nSubstituting Eq. (12.3.53) and Eq. (12.3.46) into Eq. (12.3.60) yields \ndmb = \u03b1(1\u2212 v \nm0\n) = \u03b1 \n. \n(12.3.61)\ndt \nu \n(t)\nmb \nWe can integrate this equation by separating variables to find an integral expression for \nthe mass of the boat as a function of time \nmb (t ) \nt \ndt . \n(12.3.62)\n\u222b mb dmb = \u03b1 m0 \u222b \nt=0\nm0 \nWe can easily integrate both sides of Eq. (12.3.62) yielding \n12-17", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_295", "text": "1 (mb (t)2 \u2212 m0\n2 ) = \u03b1 mb,0t . \n(12.3.63)\n2 \nThe mass of the boat as a function of time is then \n\u03b1t\n(t) = m0 1+ 2 \n. \n(12.3.64)\nmb \nm0 \nWe now substitute Eq. (12.3.64) into Eq. (12.3.65)yielding the speed of the burning boat \nas a function of time \n\u239b\n\u239e \n\u239c\n\u239f \n\u239c \n1 \n\u239f\nv t( ) = u 1 \n(12.3.66)\n\u239f\n\u239c\u2212 \n\u03b1t\n\u239c \n1+ 2 \n\u239f\n\u239c \n\u239f\nm\n\u239d \nb,0 \u23a0 \n12.3 Rocket Propulsion \n!\nA rocket at time t = ti is moving with velocity v r ,i with respect to a fixed reference \nframe. During the time interval [ti ,t f ] the rocket continuously burns fuel that is \n!\ncontinuously ejected backwards with velocity u relative to the rocket. This exhaust \nvelocity is independent of the velocity of the rocket. The rocket must exert a force to \naccelerate the ejected fuel backwards and therefore by Newton\u2019s Third law, the fuel \nexerts a force that is equal in magnitude but opposite in direction accelerating the rocket \n!\nforward. The rocket velocity is a function of time, v (t) . Because fuel is leaving the \nr \nrocket, the mass of the rocket is also a function of time, mr (t) , and is decreasing at a rate \n! \ndm / dt . Let F\ndenote the total external force acting on the rocket. We shall use the \nr \next \n!\nmomentum principle, to determine a differential equation that relates dv r / dt , dmr / dt , \n!\n! \n! \nu , v (t) , and F , an equation known as the rocket equation.\next \nr \nWe shall apply the momentum principle during the time interval [t,t + \u0394t] with \u0394t taken \nto be a small interval (we shall eventually consider the limit that \u0394t \u2192 0 ), and ti < t < t f . \nDuring this interval, choose as our system the mass of the rocket at time t , \nm = m (t) = m \n(t) , \n(12.3.67)\nsys \nr \nr ,d + mf \nwhere mr ,d is the dry mass of the rocket and mf (t) is the mass of the fuel in the rocket at", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_296", "text": "m = m (t) = m \n(t) , \n(12.3.67)\nsys \nr \nr ,d + mf \nwhere mr ,d is the dry mass of the rocket and mf (t) is the mass of the fuel in the rocket at \ntime t . During the time interval [t,t + \u0394t], a small amount of fuel of mass \u0394mf (in the \n12-18", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_297", "text": "mr t  \n t \n   \n \n \n \n \n  \n \n  \n \n  \n \n \n \n  \n \n \n \n \n \n \n \n   \n  \n   \n   \n  \n \n \n   \n \n \n \n \n  \n \n  \n   \n  \n \n \n \n  \n  \n \n!\nlimit that \u0394t \u2192 0 , \u0394mf \u2192 0 ) is ejected backwards with velocity u to the rocket. Before \nthe fuel is ejected, it is traveling at the velocity of the rocket and so during the time \ninterval [t,t + \u0394t], the elected fuel undergoes a change in momentum and the rocket \n! \nrecoils forward. At time t + \u0394t the rocket has velocity v r (t + \u0394t) . Although the ejected \nfuel continually changes its velocity, we shall assume that the fuel is all ejected at the \ninstant t + \u0394t and then consider the limit as \u0394t \u2192 0 . Therefore the velocity of the ejected \nfuel with respect to the fixed reference frame is the vector sum of the relative velocity of\n!\n!\nthe fuel with respect to the rocket and the velocity of the rocket, u + v r (t + \u0394t) . Figure \n12.14 represents momentum diagrams for our system at time t and t + \u0394t relative to a \n!\nfixed inertial reference frame in which velocity of the rocket at time t is v (t). \nr \nv r (t)\n( ) \ntime \nmf \nu + v \nv r (t + t)\nr (t + t) \nm r (t) + m r \ntime t + \nFigure 12.14 Momentum diagrams for system at time t and t + \u0394t \nThe momentum of the system at time t is \n!psys (t) = mr (t)v!\nr (t) .  \n(12.3.68) \nNote that the mass of the system at time t  is \nm = m (t) . \n(12.3.69)\nsys \nr \nThe momentum of the system at time t + \u0394t is \nt \n!psys (t + \u0394t) = mr (t + \u0394t)!vr (t + \u0394t) + \u0394m f (!u + !vr (t + \u0394t)) , \n(12.3.70) \nwhere m (t + \u0394t) = m (t) + \u0394m . With this notation the mass of the system at time t + \u0394t \nr \nr\nr \nis given by \nmsys = mr (t + \u0394t) + \u0394m f = mr (t) + \u0394mr + \u0394m f . \n(12.3.71) \nBecause the mass of the system is constant, setting Eq. (12.3.69) equal to Eq. (12.3.71) \nrequires that \n12-19", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_298", "text": "\u0394mr = \u2212\u0394m f . \n(12.3.72) \nThe momentum of the system at time t + \u0394t (Eq. (12.3.70)) can be rewritten as \n!p\n!p \n(t + \u0394t) = (mr (t) + \u0394mr \n(t)\n(t + \u0394t) \u2212\u0394mr \n!vr \n)\n(t + \u0394t) \u2212\u0394mr \n!u \n!vr \n( !u + !vr (t + \u0394t)) \nsys \n, (12.3.73) \n(t + \u0394t) = mr\nsys \nWe can now apply Newton\u2019s Second Law in the form of the momentum principle, \n!\n!\n(\n( )\n(\n\u0394 ) \u2212\u0394\n) \u2212 \n( )\n( )\nt\nt + t\nt\nt\nm \nv \nm u \nm \nv \nr\nr\nr\nr\nr\nFext = lim \nv \nv \nr \nr \n!\n!  \n\u0394t\n\u0394t\u21920 \n.. \n(12.3.74) \n! \n! \n(t + \u0394t) \u2212 \n(t) \n\u0394m ! \n(t) lim\n\u0394t\u21920 \n\u2212 lim \nr\n= m \nu\n\u0394t \n\u0394t\nr \n\u0394t\u21920 \nWe now take the limit as \n!v\n!\nFext \nEq. (12.3.75) is known as the rocket equation.  \nSuppose the rocket is moving in the positive x -direction with an external force given by  \n!\n!\nF = F \n\u02c6i Then u = \u2212u\u02c6i , where u > 0 is the relative speed of the fuel and it is moving \next \next,x \n!\nin the negative x -direction, v = v \u02c6i . Then the rocket equation (Eq. (12.3.75)) becomes\nr\nr ,x \ndv \ndm\nr ,x\nr\nF \n= m (t) \n+ \nu . \n(12.3.76)\next,x\nr \ndt \ndt \nNote that the rate of decrease of the mass of the rocket, dmr / dt , is equal to the negative \nof the rate of increase of the exhaust fuel \ndm \ndmf\nr = \u2212 \n. \n(12.3.77)\ndt \ndt \nWe can rewrite Eq. (12.3.76) as \ndm \ndv \nr\nr ,x\nF \n\u2212 \nu = m (t) \n. \n(12.3.78)\next,x\nr\ndt \ndt \nThe second term on the left-hand-side of Eq. (12.3.78) is called the thrust \nd \ndm\n\u2212\n!u", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_299", "text": "dm \ndv \nr\nr ,x\nF \n\u2212 \nu = m (t) \n. \n(12.3.78)\next,x\nr\ndt \ndt \nThe second term on the left-hand-side of Eq. (12.3.78) is called the thrust \nd \ndm\n\u2212\n!u\n(t) \n(12.3.75) \nr \nr\n= m \n. \ndt \ndt\nr \n12-20", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_300", "text": "dmr \ndmf\n= \u2212 \nu = \nu . \n(12.3.79)\nFthrust,x \ndt \ndt \nNote that this is not an extra force but the result of the forward recoil due to the ejection \nof the fuel. Because we are burning fuel at a positive rate dmf / dt > 0 and the speed \nu > 0 , the direction of the thrust is in the positive x -direction. \n12.3.1 Rocket Equation in Gravity-free Space \nWe shall first consider the case in which there are no external forces acting on the \nsystem, then Eq. (12.3.78) becomes \ndm \ndv \nr\nr ,x\n\u2212 \nu = mr (t) \n. \n(12.3.80)\ndt \ndt \nIn order to solve this equation, we separate the variable quantities vr ,x (t) and mr (t) and \nmultiply both sides by dt yielding \ndm r\ndv = \u2212u \n. \n(12.3.81)\nr ,x \nmr (t) \nWe now integrate both sides of Eq. (12.3.81) with limits corresponding to the values of \nthe x -component of the velocity and mass of the rocket at times ti when the ejection of \nthe burned fuel began and the time t f when the process stopped, \nv\u2032 =v\nm\u2032 = m\nr ,x\nr ,x , f \nr\nr , f u\n\u222b \ndvr \u2032 ,x = \u2212\u222b m\u2032 dmr \u2032 . \n(12.3.82) \nv\u2032 = v\nm\u2032 = m\nr\nr ,x\nr ,x ,i \nr\nr ,i \nPerforming the integration and substituting in the values at the endpoints yields \n\u239b \n\u239e \nm r , f \nm r ,i \n\u2212 v \n= \u2212u ln \n(12.3.83) \n\u239c\u239d \n\u239f\u23a0  \nv \n.  \nr ,x, f \nr ,x,i \nBecause the rocket is losing fuel, m \n< m , we can rewrite Eq. (12.3.83) as \nr , f\nr ,i \n\u239b \n\u239e \nm r ,i \nm r , f \n\u2212 v \n= u ln \n(12.3.84) \n\u239c\n\u239d  \n\u239f\n\u23a0  \nv \n.  \nr ,x, f \nr ,x,i \n12-21", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_301", "text": "We note ln(m / m ) > 1. Therefore v \n> v \nas we expect. After a slight\nr ,i\nr , f\nr ,x, f\nr ,x,i , \nrearrangement of Eq. (12.3.84), we have an expression for the x -component of the \nvelocity of the rocket as a function of the mass mr of the rocket \n\u239b \n\u239e \nm r ,i \nm r , f \n+ u ln \n.  \n(12.3.85) \n\u239c\n\u239d  \n\u239f\n\u23a0  \nv \n= v \nr ,x, f \nr ,x,i \nLet\u2019s examine our result. First, let\u2019s suppose that all the fuel was burned and ejected. \nThen m \n\u2261 m \nis the final dry mass of the rocket (empty of fuel). The ratio\nr , f\nr ,d \nm \nR = \nr ,i \n(12.3.86)\nm r ,d \nis the ratio of the initial mass of the rocket (including the mass of the fuel) to the final dry \nmass of the rocket (empty of fuel). The final velocity of the rocket is then \nv \n= v \n+ u ln R . \n(12.3.87)\nr ,x, f\nr ,x,i \nThis is why multistage rockets are used. You need a big container to store the fuel. Once \nall the fuel is burned in the first stage, the stage is disconnected from the rocket. During \nthe next stage the dry mass of the rocket is much less and so R is larger than the single \nstage, so the next burn stage will produce a larger final speed then if the same amount of \nfuel were burned with just one stage (more dry mass of the rocket). In general rockets do \nnot burn fuel at a constant rate but if we assume that the burning rate is constant where \ndmf \ndm\nb =\n= \u2212 \nr \n(12.3.88)\ndt \ndt \nthen we can integrate Eq. (12.3.88) \nmr \u2032 = mr (t ) \nt \u2032=t \ndm\u2032 = \u2212b \ndt\u2032 \n(12.3.89)\n\u222b \nr \n\u222b \nm\u2032 = \nt \u2032=ti\nr mr ,i \nand find an equation that describes how the mass of the rocket changes in time \nm (t) = m \u2212 b(t \u2212 ti ) . \n(12.3.90)\nr\nr ,i \nFor this special case, if we set t f = t in Eq. (12.3.85), then the velocity of the rocket as a", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_302", "text": "m (t) = m \u2212 b(t \u2212 ti ) . \n(12.3.90)\nr\nr ,i \nFor this special case, if we set t f = t in Eq. (12.3.85), then the velocity of the rocket as a \nfunction of time is given by \n\u239b \n\u239e \nm r ,i \nm \u2212 bt \nr ,i \n+ u ln \n(12.3.91) \nv \n= v\n\u239c\u239d \n\u239f\u23a0  \n.  \nr ,x, f \nr ,x,i \n12-22", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_303", "text": "Example 12.4 Single-Stage Rocket \nBefore a rocket begins to burn fuel, the rocket has a mass of mr ,i = 2.81\u00d7 107kg, of \nwhich the mass of the fuel is mf ,i = 2.46 \u00d7 107kg . The fuel is burned at a constant rate \nwith total burn time is 510 s and ejected at a speed u = 3000 m/s relative to the rocket. If \nthe rocket starts from rest in empty space, what is the final speed of the rocket after all \nthe fuel has been burned? \nSolution: The dry mass of the rocket is m \n\u2261 m \u2212 m \n= 0.35 \u00d7 107 kg , hence \nr ,d\nr ,i\nf ,i \nR = m / m \n= 8.03. The final speed of the rocket after all the fuel has burned is\nr ,i\nr ,d \nv \n= \u0394v = u ln R = 6250 m/s . \n(12.3.92)\nr , f\nr \nExample 12.5 Two-Stage Rocket \nNow suppose that the same rocket in Example 12.4 burns the fuel in two stages ejecting \nthe fuel in each stage at the same relative speed. In stage one, the available fuel to burn is \nm \n= 2.03 \u00d7 107 kg with burn time 150 s . Then the empty fuel tank and accessories \nf ,1,i \nfrom stage one are disconnected from the rest of the rocket. These disconnected parts \nhave a mass m = 1.4 \u00d7 106kg . All the remaining fuel with mass is burned during the \nsecond stage with burn time of 360 s . What is the final speed of the rocket after all the \nfuel has been burned? \nSolution: The mass of the rocket after all the fuel in the first stage is burned is \nm r ,1,d = m \n\u2212 m\n= 0.78 \u00d7 107kg and R1 = m r ,1,i / mr ,1,d = 3.60 . The change in speed \nr ,1,i\nf ,1,i \nafter the first stage is complete is \n\u0394vr ,1 = u ln R1 = 3840 m/s . \n(12.3.93) \nAfter the empty fuel tank and accessories from stage one are disconnected from the rest \nof the rocket, the remaining mass of the rocket is m \n= 2.1\u00d7 106kg . The remaining fuel \nr ,2,d \nhas mass m \n= 4.3 \u00d7 106kg . The mass of the rocket plus the unburned fuel at the", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_304", "text": "of the rocket, the remaining mass of the rocket is m \n= 2.1\u00d7 106kg . The remaining fuel \nr ,2,d \nhas mass m \n= 4.3 \u00d7 106kg . The mass of the rocket plus the unburned fuel at the \nf ,2,i \nbeginning of the second stage is m \n= 6.4 \u00d7 106 kg . Then \n= m \n/ m \n= 3.05 . \nr ,2,i\nR2 \nr ,2,i\nr ,2,d \nTherefore the rocket increases its speed during the second stage by an amount \n\u0394vr ,2 = u ln R2 = 3340 m/s . \n(12.3.94) \nThe final speed of the rocket is the sum of the change in speeds due to each stage, \n12-23", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_305", "text": "= \u0394v = u ln R1 + u ln R2 = u ln(R1R2) = 7190 m/s , \n(12.3.95)\nv f \nr \nwhich is greater than if the fuel were burned in one stage. Plots of the speed of the rocket \nas a function time for both one-stage and two-stage burns are shown Figure 12.15. \nFigure 12.15 Plots of speed of rocket for both one-stage burn and two-stage burn \n12.3.2 Rocket in a Constant Gravitational Field: \nNow suppose that the rocket takes off from rest at time t = 0 in a constant gravitational \nfield then the external force is \nF\n\uf072 total \n\uf072 \n= m g . \n(12.3.96)\next \nr \nChoose the positive x -axis in the upward direction then Fext,x (t) = \u2212mr (t)g . Then the \nrocket equation (Eq. (12.3.75) becomes \ndm \ndv \nr\nr ,x\n\u2212m (t)g \u2212 \nu = m (t) \n. \n(12.3.97)\nr \ndt \nr\ndt \nMultiply both sides of Eq. (12.3.97) by dt , and divide both sides by mr (t) . Then Eq. \n(12.3.97) can be written as \ndm\ndvr ,x = \u2212gdt \u2212 \nr u . \n(12.3.98)\nmr (t) \nWe now integrate both sides \n12-24", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_306", "text": "v \n(t ) \nm (t )\nr ,x \nr\nt\ndm\u2032 \ndv\u2032 = \u2212u\nr \u2212 g dt\u2032 , \n(12.3.99)\n\u222b \nr ,x \n\u222b m\u2032\n\u222b \nv \n0 \nm\nr \n0\nr ,x ,i = \nr ,i \nwhere mr ,i is the initial mass of the rocket and the fuel. Integration yields \n\u239b \n\u239b\n\u239e\n\u239e \nm r (t) \nm r ,i \nm r ,i\nv r ,x (t) = \u2212u ln \n\u239f\u23a0  \n\u2212 gt = u ln \u239c\u239d  \n\u2212 gt . \n(12.3.100)\n\u239c\u239d \n\u239f\u23a0 \n(t)\nm r \nAfter all the fuel is burned at t = t f , the mass of the rocket is equal to the dry mass \nand so\nmr , f = mr ,d \nv r ,x (t f ) = u ln R \u2212 gt f . \n(12.3.101) \nThe first term on the right hand side is independent of the burn time. However the second \nterm depends on the burn time. The shorter the burn time, the smaller the negative \ncontribution from the third turn, and hence the rocket ends up with a larger final speed. \nSo the rocket engine should burn the fuel as fast as possible in order to obtain the \nmaximum possible speed. \n12-25", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_307", "text": "Chapter 13 The Concept of Energy and Conservation of Energy \n13.1 The Concept of Energy and Conservation of Energy........................................ 2  \n13.2 Kinetic Energy....................................................................................................... 5  \nExample 13.1 Change in Kinetic Energy of a Car ................................................. 5  \n13.3 Kinematics and Kinetic Energy in One Dimension ........................................... 6  \n13.3.1 Constant Accelerated Motion ....................................................................... 6  \n13.3.2 Non-constant Accelerated Motion ................................................................ 7", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_308", "text": "13.4 Work done by Constant Forces ........................................................................... 9  \nExample 13.2 Work Done by Static Fiction.......................................................... 10  \nExample 13.3 Work Done by Force Applied in the Direction of Displacement 11  \nExample 13.5 Work done by Gravity Near the Surface of the Earth ................ 12  \n13.5 Work done by Non-Constant Forces................................................................. 13  \nExample 13.6 Work done by the Spring Force .................................................... 14  \n13.6 Work-Kinetic Energy Theorem......................................................................... 15  \nExample 13.7 Gravity and the Work-Energy Theorem ...................................... 16", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_309", "text": "Example 13.7 Gravity and the Work-Energy Theorem ...................................... 16  \nExample 13.7 Final Kinetic Energy of Moving Cup............................................ 17  \n13.7 Power Applied by a Constant Force ................................................................. 17  \nExample 13.8 Gravitational Power for a Falling Object..................................... 18  \nExample 13.9 Power Pushing a Cup ..................................................................... 20  \n13.8 Work and the Scalar Product ............................................................................ 21  \n13.8.1 Scalar Product.............................................................................................. 21", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_310", "text": "13.8.2 Kinetic Energy and the Scalar Product ..................................................... 23  \n13.8.2 Work and the Scalar Product ..................................................................... 24  \n13.9 Work done by a Non-Constant Force Along an Arbitrary Path.................... 27  \n13.9.1 Work Integral in Cartesian Coordinates................................................... 28  \n13.9.2 Work Integral in Cylindrical Coordinates ................................................ 29  \n13.10 Worked Examples............................................................................................. 30  \nExample 13.11 Work Done in a Constant Gravitation Field .............................. 30  \nExample 13.12 Hooke\u2019s Law Spring-Body System.............................................. 31", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_311", "text": "Example 13.12 Hooke\u2019s Law Spring-Body System.............................................. 31  \nExample 13.13 Work done by the Inverse Square Gravitation Force ............... 32  \nExample 13.14 Work Done by the Inverse Square Electrical Force.................. 33  \n13.11 Work-Kinetic Energy Theorem in Three Dimensions .................................. 34", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_312", "text": "13.11.1 Instantaneous Power Applied by a Non-Constant Force for Three \nDimensional Motion................................................................................................ 35 \nAppendix 13A Work Done on a System of Two Particles....................................... 36  \n13-1", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_313", "text": "Chapter 13 Energy, Kinetic Energy, and Work \nAcceleration of the expansion of the universe is one of the most exciting \nand significant discoveries in physics, with implications that could \nrevolutionize theories of quantum physics, gravitation, and cosmology. \nWith its revelation that close to the three-quarters of the energy density of \nthe universe, given the name dark energy, is of a new, unknown origin and \nthat its exotic gravitational \u201crepulsion\u201d will govern the fate of the \nuniverse, dark energy and the accelerating universe becomes a topic not \njust of great interest to research physicists but to science students at all \nlevels. 1 \nEric Linder \n13.1 The Concept of Energy and Conservation of Energy \nThe transformation of energy is a powerful concept that enables us to describe a vast \nnumber of processes: \nFalling water releases stored gravitational potential energy, which can become the \nkinetic energy associated with a coherent motion of matter. The harnessed mechanical \nenergy can be used to spin turbines and alternators, doing work to generate electrical \nenergy, transmitted to consumers along power lines. When you use any electrical \ndevice, the electrical energy is transformed into other forms of energy. In a \nrefrigerator, electrical energy is used to compress a gas into a liquid. During the \ncompression, some of the internal energy of the gas is transferred to the random \nmotion of molecules in the outside environment. The liquid flows from a high-\npressure region into a low-pressure region where the liquid evaporates. During the \nevaporation, the liquid absorbs energy from the random motion of molecules inside of \nthe refrigerator. The gas returns to the compressor. \n\u201cHuman beings transform the stored chemical energy of food into various forms \nnecessary for the maintenance of the functions of the various organ system, tissues \n2\nand cells in the body.\u201d A person can do work on their surroundings \u2013 for example, by \npedaling a bicycle \u2013 and transfer energy to the surroundings in the form of increasing \nrandom motion of air molecules, by using this catabolic energy. \nBurning gasoline in car engines converts chemical energy, stored in the molecular \nbonds of the constituent molecules of gasoline, into coherent (ordered) motion of the \nmolecules that constitute a piston. With the use of gearing and tire/road friction, this \nmotion is converted into kinetic energy of the car; the automobile moves. \n1 \nEric Linder, Resource Letter: Dark Energy and the Accelerating Universe, Am.J.Phys.76: 197\u00ad\n204, 2008; p. 197. \n2", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_314", "text": "motion is converted into kinetic energy of the car; the automobile moves. \n1 \nEric Linder, Resource Letter: Dark Energy and the Accelerating Universe, Am.J.Phys.76: 197\u00ad\n204, 2008; p. 197. \n2 \nGeorge B. Benedek and Felix M.H. Villars, Physics with Illustrative Examples from Medicine and \nBiology, Volume 1: Mechanics, Addison-Wesley, Reading, 1973, p. 115-6. \n13-2", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_315", "text": "Stretching or compressing a spring stores elastic potential energy that can be released \nas kinetic energy. \nThe process of vision begins with stored atomic energy released as electromagnetic \nradiation (light), which is detected by exciting photoreceptors in the eye, releasing \nchemical energy. \nWhen a proton fuses with deuterium (a hydrogen atom with a neutron and proton for \na nucleus), helium-three is formed (with a nucleus of two protons and one neutron) \nalong with radiant energy in the form of photons. The combined internal energy of \nthe proton and deuterium are greater than the internal energy of the helium-three. This \ndifference in internal energy is carried away by the photons as light energy. \nThere are many such processes involving different forms of energy: kinetic energy, \ngravitational energy, thermal energy, elastic energy, electrical energy, chemical energy, \nelectromagnetic energy, nuclear energy and more. The total energy is always conserved \nin these processes, although different forms of energy are converted into others. \nAny physical process can be characterized by two states, initial and final, between \nwhich energy transformations can occur. Each form of energy E j , where \u201c j \u201d is an \narbitrary label identifying one of the N forms of energy, may undergo a change during \nthis transformation, \n\u0394E j \u2261 Efinal, j \u2212 Einitial, j . \n(13.1.1) \nConservation of energy means that the sum of these changes is zero, \nN \n\u0394E1 + \u0394E2 + \u22c5\u22c5\u22c5+ \u0394EN = \u2211\u0394E j = 0 . \n(13.1.2) \nj=1 \nTwo important points emerge from this idea. First, we are interested primarily in \nchanges in energy and so we search for relations that describe how each form of energy \nchanges. Second, we must account for all the ways energy can change. If we observe a \nprocess, and the sum of the changes in energy is not zero, either our expressions for \nenergy are incorrect, or there is a new type of change of energy that we had not \npreviously discovered. This is our first example of the importance of conservation laws in \ndescribing physical processes, as energy is a key quantity conserved in all physical \nprocesses. If we can quantify the changes of different forms of energy, we have a very \npowerful tool to understand nature. \nWe will begin our analysis of conservation of energy by considering processes \ninvolving only a few forms of changing energy. We will make assumptions that greatly \nsimplify our description of these processes. At first we shall only consider processes", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_316", "text": "powerful tool to understand nature. \nWe will begin our analysis of conservation of energy by considering processes \ninvolving only a few forms of changing energy. We will make assumptions that greatly \nsimplify our description of these processes. At first we shall only consider processes \nacting on bodies in which the atoms move in a coherent fashion, ignoring processes in \nwhich energy is transferred into the random motion of atoms. Thus we will initially \nignore the effects of friction. We shall then treat processes involving friction between \n13-3", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_317", "text": "consider rigid bodies. We will later return to processes in which there is an energy \ntransfer resulting in an increase or decrease in random motion when we study the First \nLaw of Thermodynamics. \nEnergy is always conserved but we often prefer to restrict our attention to a set of \nobjects that we define to be our system. The rest of the universe acts as the surroundings. \nWe illustrate this division of system and surroundings in Figure 13.1. \nFigure 13.1 A diagram of a system and its surroundings with boundary \nBecause energy is conserved, any energy that leaves the system must cross \nthrough the boundary and enter the surroundings. Consider any physical process in which \nenergy transformations occur between initial and final states. We assert that \nwhen a system and its surroundings undergo a transition from an initial \nstate to a final state, the change in energy is zero, \n\u0394E = \u0394Esystem + \u0394Esurroundings = 0 . \n(13.1.3) \nEq. (13.1.3) is called conservation of energy and is our operating definition for energy. \nWe will sometime refer to Eq. (13.1.3) as the energy principle. In any physical \napplication, we first identify our system and surroundings, and then attempt to quantify \nchanges in energy. In order to do this, we need to identify every type of change of energy \nin every possible physical process. When there is no change in energy in the surroundings \nthen the system is called a closed system, and consequently the energy of a closed system \nis constant. \n\u0394E \n= 0, \n(closed system) . \n(13.1.4)\nsystem \nIf we add up all known changes in energy in the system and surroundings and do \nnot arrive at a zero sum, we have an open scientific problem. By searching for the \nmissing changes in energy, we may uncover some new physical phenomenon. Recently, \none of the most exciting open problems in cosmology is the apparent acceleration of the \nexpansion of the universe, which has been attributed to dark energy that resides in space \nitself, an energy type without a clearly known source.3 \n3 http://www-supernova.lbl.gov/~evlinder/sci.html \n13-4", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_318", "text": "13.2 Kinetic Energy \nThe first form of energy that we will study is an energy associated with the coherent \nmotion of molecules that constitute a body of mass m; this energy is called the kinetic \nenergy (from the Greek word kinetikos which translates as moving). Let us consider a car \nmoving along a straight road (along which we will place the x -axis). For an observer at \n\uf072 \n\u02c6\nrest with respect to the ground, the car has velocity v = vx i . The speed of the car is the \nmagnitude of the velocity, v \u2261 \n.\nvx \nThe kinetic energy K of a non-rotating body of mass m moving with speed \nv is defined to be the positive scalar quantity \n1\n2\nK \u2261 mv \n(13.2.1)\n2 \nThe kinetic energy is proportional to the square of the speed. The SI units for kinetic \n2 \u22c5 s\nenergy are [kg \u22c5 m\n\u22122] . This combination of units is defined to be a joule and is denoted \n2 \u22c5 s\u22122\nby [J] , thus 1J \u2261 1 kg \u22c5 m\n. (The SI unit of energy is named for James Prescott \nJoule.) The above definition of kinetic energy does not refer to any direction of motion, \njust the speed of the body. \nLet\u2019s consider a case in which our car changes velocity. For our initial state, the \n!\ncar moves with an initial velocity v = v \u02c6i along the x -axis. For the final state (at some \ni \nx,i \nlater time), the car has changed its velocity and now moves with a final velocity \n\uf072 v f = vx, f \u02c6i . Therefore the change in the kinetic energy is \n1 \n2 \u2212 1\n2\n\u0394K = mv f \nmvi . \n(13.2.2)\n2\n2 \nExample 13.1 Change in Kinetic Energy of a Car \nSuppose car A increases its speed from 10 to 20 mph and car B increases its speed from \n50 to 60 mph. Both cars have the same mass m. (a) What is the ratio of the change of \nkinetic energy of car B to the change of kinetic energy of car A? In particular, which car \nhas a greater change in kinetic energy? (b) What is the ratio of the change in kinetic \nenergy of car B to car A as seen by an observer moving with the initial velocity of car A? \nSolution: (a) The ratio of the change in kinetic energy of car B to car A is \n13-5", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_319", "text": "1 \n)2 \u2212 1 \n)2 \n)2 \u2212 (vB,i )2\nm(vB, f \nm(vB,i\n\u0394KB \n2\n2\n(vB, f\n=\n= \n)2\n\u0394K A \n1 \n)2 \u2212 1 \n)2 \n(vA, f )2 \u2212 (vA,i \n2 m(vA, f \n2 m(vA,i \n(60 mph)2 \u2212 (50 mph)2 \n=\n= 11/ 3. \n(20 mph)2 \u2212 (10 mph)2 \nThus car B has a much greater increase in its kinetic energy than car A. \n(b) In a reference moving with the speed of car A , car A increases its speed from rest to \n10 mph and car B increases its speed from 40 to 50 mph. The ratio is now \n1 \n)2 \u2212 1 \n)2 \n)2\n\u0394KB \n2 m(vB, f \n2 m(vB,0 \n(vB, f )2 \u2212 (vB,0 \n=\n= \n\u0394K A \n1 \n)2 \u2212 1 \n)2\n(vA, f )2 \u2212 (vA,0 )2 \n2 m(vA, f \n2 m(vA,0 \n(50 mph)2 \u2212 (40 mph)2 \n=\n= 9. \n(10 mph)2 \nThe ratio is greater than that found in part a). Note that from the new reference frame \nboth car A and car B have smaller increases in kinetic energy. \n13.3 Kinematics and Kinetic Energy in One Dimension \n13.3.1 Constant Accelerated Motion \nLet\u2019s consider a constant accelerated motion of a rigid body in one dimension in which \nwe treat the rigid body as a point mass. Suppose at t = 0 the body has an initial x \u00ad\ncomponent of the velocity given by vx,i . If the acceleration is in the direction of the \ndisplacement of the body then the body will increase its speed. If the acceleration is \nopposite the direction of the displacement then the acceleration will decrease the body\u2019s \nspeed. The displacement of the body is given by \n\u0394x = v\nt + 1 a t2 . \n(13.3.1)\nx,i\nx\n2 \nThe product of acceleration and the displacement is \na \u0394x = a (v\nt + 1 a t2) . \n(13.3.2)\nx \nx\nx,i\nx\n2 \nThe acceleration is given by \n13-6", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_320", "text": "\u0394v \n(v \n\u2212 v )\nx\nx, f\nx,i\na =\n= \n. \n(13.3.3)\nx \n\u0394t\nt \nTherefore \n\u239b\n\u239e\n(vx, f \u2212 v x,i ) \n1 (v x, f \u2212 v x,i ) \nt2\na \u0394x = \nv\nt + \n\u239f . \n(13.3.4)\nx \n\u239c x,i\nt \n\u239d \n2 \nt \n\u23a0 \nEquation (13.3.4) becomes \n1 \n1\n2 \u2212 1\n2\na \u0394x = (vx, f \u2212 v x,i )(v ) + (v \n\u2212 v )(v x, f \u2212 vx,i ) = v\nv \n(13.3.5)\nx\nx,i \nx, f\nx,i \nx, f \nx,i .\n2 \n2\n2 \nIf we multiply each side of Equation (13.3.5) by the mass m of the object this \nkinematical result takes on an interesting interpretation for the motion of the object. We \nhave \n1 \n2\n1\n2\nma x \u0394x = 2 mv x, \u2212\nf m 2 vx,i = K f \u2212 Ki . \n(13.3.6) \nRecall that for one-dimensional motion, Newton\u2019s Second Law is Fx = ma x , for the \nmotion considered here, Equation (13.3.6) becomes \nF \u0394x = K \u2212 K . \n(13.3.7)\nx \nf\ni \n13.3.2 Non-constant Accelerated Motion \nIf the acceleration is not constant, then we can divide the displacement into N intervals \nindexed by j = 1 to N . It will be convenient to denote the displacement intervals by \u0394x j , \nthe corresponding time intervals by \u0394t j and the x -components of the velocities at the \nbeginning and end of each interval as vx, j\u22121 and vx, j . Note that the x -component of the \nvelocity at the beginning and end of the first interval j = 1is then v = v \nand the \nx,1 \nx,i \nvelocity at the end of the last interval, j = N is vx N = vx, j . Consider the sum of the \n, \nproducts of the average acceleration (a ) \nand displacement \u0394x j in each interval,\nx, j ave \nj= N \n\u2211(a )\n\u0394x j . \n(13.3.8)\nx, j ave \nj=1 \nThe average acceleration over each interval is equal to \n\u0394v \n(v \n\u2212 v )\nx, j\nx, j+1 \nx, j", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_321", "text": "x, j ave \nj= N \n\u2211(a )\n\u0394x j . \n(13.3.8)\nx, j ave \nj=1 \nThe average acceleration over each interval is equal to \n\u0394v \n(v \n\u2212 v )\nx, j\nx, j+1 \nx, j\n(a ) \n=\n= \n, \n(13.3.9)\nx, j ave \n\u0394t j \n\u0394t j \n13-7", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_322", "text": "and so the contribution in each integral can be calculated as above and we have that \n1 \n2 \u2212 1 \n2\n(a x, j )ave \u0394x j = 2 v x, j \n2 v x, j\u22121 . \n(13.3.10) \nWhen we sum over all the terms only the last and first terms survive, all the other terms \ncancel in pairs, and we have that \nj= N \n1\n2 \n1\n2\n\u2211(ax, j )ave \u0394x j = 2 vx, f \u2212 2 vx,i . \n(13.3.11) \nj=1 \nIn the limit as N \u2192\u221e and \u0394x j \u2192 0 for all j (both conditions must be met!), the limit of \nthe sum is the definition of the definite integral of the acceleration with respect to the \nposition, \nx=x f\nj=N \nlim \u2211(a x, j )ave \u0394x j \u2261\u222b a x ( x) dx . \n(13.3.12)\nN\u2192\u221e \nj=1\n\u0394x j \u21920 \nx=xi \nTherefore In the limit as N \u2192\u221e and \u0394x j \u2192 0 for all j , with v \n\u2192 v \n, Eq. (13.3.11)\nx, N \nx, f \nbecomes \nx=x f \n1\n2 \n2\n( x)dx = (v \n\u2212 v ) \n(13.3.13)\nx \nx, f \nx,i\n\u222b a \n2 \nx=xi \nThis integral result is consequence of the definition that ax \u2261 dvx / dt . The integral in Eq. \n(13.3.13) is an integral with respect to space, while our previous integral \nt =t f \na (t) dt = v \n\u2212 v \n(13.3.14)\n\u222b \nx \nx, f \nx,i . \nt =ti \nrequires integrating acceleration with respect to time. Multiplying both sides of Eq. \n(13.3.13) by the mass m yields \nx=x f \n1\n2 \n2\nma ( x) dx = m(v \n\u2212 v ) = K f \u2212 Ki . \n(13.3.15)\n\u222b \nx \n2 \nx, f\nx,i \nx=xi \nWhen we introduce Newton\u2019s Second Law in the form Fx = ma x , then Eq. (13.3.15) \nbecomes \nx= x f \nF (x) dx = K \u2212 K \n(13.3.16)\n\u222b \nx \nf\ni . \nx = xi \n13-8", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_323", "text": "The integral of the x -component of the force with respect to displacement in Eq.  \n(13.3.16) applies to the motion of a point-like object. For extended bodies, Eq. (13.3.16) \napplies to the center of mass motion because the external force on a rigid body causes the \ncenter of mass to accelerate. \n13.4 Work done by Constant Forces \nWe will begin our discussion of the concept of work by analyzing the motion of an object \nin one dimension acted on by constant forces. Let\u2019s consider the following example: push \na cup forward with a constant force along a desktop. When the cup changes velocity (and \nhence kinetic energy), the sum of the forces acting on the cup must be non-zero according \nto Newton\u2019s Second Law. There are three forces involved in this motion: the applied \n\uf072 \n\uf072\n\uf072\n\uf072 \n\uf072 \n\uf072\npushing force Fa ; the contact force C \u2261 N + fk ; and gravity Fg = mg . The force diagram \non the cup is shown in Figure 13.2. \nFigure 13.2 Force diagram for cup. \nLet\u2019s choose our coordinate system so that the +x -direction is the direction of the \nforward motion of the cup. The pushing force can then be described by \n\uf072 \nFa\na\n= Fx \u02c6i . \n(13.4.1) \nSuppose a body moves from an initial point \nto a final point \nso that the \nxi \nx f \ndisplacement of the point the force acts on is \u0394x \u2261 x f \n. The work done by a \n\u2212 xi \n\uf072 \nconstant force Fa = Fx\na\u02c6i acting on the body is the product of the component of \nthe force Fx\na and the displacement \u0394x , \nW a = Fx\na \u0394x . \n(13.4.2) \nWork is a scalar quantity; it is not a vector quantity. The SI unit for work is \n13-9", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_324", "text": "-2 \n2\n-2 \n[1 N m] = [1 kg \u22c5 m s \n\u22c5 \n][1 m] = [1 kg \u22c5 m \u22c5s ] = [1J]. \n(13.4.3)\n\u22c5 \nNote that work has the same dimension and the same SI unit as kinetic energy. Because \nour applied force is along the direction of motion, both Fx\na > 0 and \u0394x > 0. In this \nexample, the work done is just the product of the magnitude of the applied force and the \ndistance through which that force acts and is positive. In the definition of work done by a \nforce, the force can act at any point on the body. The displacement that appears in \nEquation (13.4.2) is not the displacement of the body but the displacement of the point of \napplication of the force. For point-like objects, the displacement of the point of \napplication of the force is equal to the displacement of the body. However for an \nextended body, we need to focus on where the force acts and whether or not that point of \napplication undergoes any displacement in the direction of the force as the following \nexample illustrates. \nExample 13.2 Work Done by Static Fiction \nSuppose you are initially standing and you start walking by pushing against the ground \nwith your feet and your feet do not slip. What is the work done by the static friction force \nacting on you? \nSolution: When you apply a contact force against the ground, the ground applies an \nequal and opposite contact force on you. The tangential component of this constant force \nis the force of static friction acting on you. Since your foot is at rest while you are \npushing against the ground, there is no displacement of the point of application of this \nstatic friction force. Therefore static friction does zero work on you while you are \naccelerating. You may be surprised by this result but if you think about energy \ntransformation, chemical energy stored in your muscle cells is being transformed into \nkinetic energy of motion and thermal energy. \nWhen forces are opposing the motion, as in our example of pushing the cup, the \nkinetic friction force is given by \n! \nF f = fk ,x \u02c6i = \u2212\u00b5kN \u02c6i = \u2212\u00b5kmg \u02c6i . \n(13.4.4) \nHere the component of the force is in the opposite direction as the displacement. The \nwork done by the kinetic friction force is negative, \nW f = \u2212\u00b5kmg\u0394x . \n(13.4.5) \nSince the gravitation force is perpendicular to the motion of the cup, the gravitational", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_325", "text": "Here the component of the force is in the opposite direction as the displacement. The \nwork done by the kinetic friction force is negative, \nW f = \u2212\u00b5kmg\u0394x . \n(13.4.5) \nSince the gravitation force is perpendicular to the motion of the cup, the gravitational \nforce has no component along the line of motion. Therefore the gravitation force does \nzero work on the cup when the cup is slid forward in the horizontal direction. The \nnormal force is also perpendicular to the motion, and hence does no work. \n13-10", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_326", "text": "We see that the pushing force does positive work, the kinetic friction force does \nnegative work, and the gravitation and normal force does zero work. \nExample 13.3 Work Done by Force Applied in the Direction of Displacement \nPush a cup of mass 0.2 kg along a horizontal table with a force of magnitude 2.0 N for a \ndistance of 0.5 m. The coefficient of friction between the table and the cup is \u00b5k = 0.10 . \nCalculate the work done by the pushing force and the work done by the friction force. \nSolution: The work done by the pushing force is \nW a = Fx\na \u0394x = (2.0 N)(0.5 m) = 1.0 J . \n(13.4.6) \nThe work done by the friction force is \nW f = \u2212\u00b5kmg\u0394x = \u2212(0.1)(0.2 kg)(9.8 m \u22c5s-2 )(0.5 m)= \u2212 0.10 J . \n(13.4.7) \nExample 13.4 Work Done by Force Applied at an Angle to the Direction of \nDisplacement \nSuppose we push the cup in the previous example with a force of the same magnitude but \nat an angle \u03b8 = 30o upwards with respect to the table. Calculate the work done by the \npushing force. Calculate the work done by the kinetic friction force. \nSolution: The force diagram on the cup and coordinate system is shown in Figure 13.3. \nFigure 13.3 Force diagram on cup. \nThe x -component of the pushing force is now \nFx \na = F a cos(\u03b8 ) = (2.0 N)(cos(30\uf06f )) = 1.7 N . \n(13.4.8) \nThe work done by the pushing force is \n13-11", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_327", "text": "W a = Fx\na \u0394x = (1.7 N)(0.5 m) = 8.7 \u00d710\u22121 J . \n(13.4.9) \nThe kinetic friction force is \n\uf072 \nF f = \u2212\u00b5kN \u02c6i . \n(13.4.10) \nIn this case, the magnitude of the normal force is not simply the same as the weight of the \ncup. We need to find the y -component of the applied force, \nFy\na = F a sin(\u03b8 ) = (2.0 N)(sin(30o ) = 1.0 N . \n(13.4.11) \nTo find the normal force, we apply Newton\u2019s Second Law in the y -direction, \nFy\na + N \u2212 mg = 0 . \n(13.4.12) \nThen the normal force is \na\nN = mg \u2212 Fy = (0.2 kg)(9.8 m \u22c5s\u22122) \u2212 (1.0 N) = 9.6 \u00d710\u22121 N . \n(13.4.13) \nThe work done by the kinetic friction force is \nW f = \u2212\u00b5kN \u0394x = \u2212(0.1)(9.6 \u00d710\u22121 N)(0.5 m) = 4.8 \u00d710\u22122 J . \n(13.4.14) \nExample 13.5 Work done by Gravity Near the Surface of the Earth \nConsider a point-like body of mass m near the surface of the earth falling directly \ntowards the center of the earth. The gravitation force between the body and the earth is \n\uf072\n\uf072\nnearly constant, Fgrav = mg . Let\u2019s choose a coordinate system with the origin at the \nsurface of the earth and the + y -direction pointing away from the center of the earth \nSuppose the body starts from an initial point yi and falls to a final point y f closer to the \nearth. How much work does the gravitation force do on the body as it falls? \nSolution: The displacement of the body is negative, \u0394y \u2261 y f \n< 0 . The gravitation \n\u2212 yi \nforce is given by \n\uf072\n\uf072\nFg = mg = Fy\ng \u02c6j = \u2212mg \u02c6j . \n(13.4.15) \nThe work done on the body is then \nW g = Fy\ng \u0394y = \u2212mg\u0394y . \n(13.4.16) \n13-12", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_328", "text": "For a falling body, the displacement of the body is negative, \u0394y \u2261 y f \n< 0 ; therefore \n\u2212 yi \nthe work done by gravity is positive, W g > 0 . The gravitation force is pointing in the \nsame direction as the displacement of the falling object so the work should be positive. \nWhen an object is rising while under the influence of a gravitation force, \n\u0394y \u2261 y f \n> 0 . The work done by the gravitation force for a rising body is negative, \n\u2212 yi \nW g < 0 , because the gravitation force is pointing in the opposite direction from that in \nwhich the object is displaced. \nIt\u2019s important to note that the choice of the positive direction as being away from the \ncenter of the earth (\u201cup\u201d) does not make a difference. If the downward direction were \nchosen positive, the falling body would have a positive displacement and the \ngravitational force as given in Equation (13.4.15) would have a positive downward \ncomponent; the product Fy\ng \u0394y would still be positive. \n13.5 Work done by Non-Constant Forces \nConsider a body moving in the x -direction under the influence of a non-constant force in \n\uf072 \nthe x -direction, F = Fx \u02c6i . The body moves from an initial position xi to a final position \n. In order to calculate the work done by a non-constant force, we will divide up the \nx f \ndisplacement of the point of application of the force into a large number N of small \njth \ndisplacements \u0394x j where the index j marks the \ndisplacement and takes integer \nvalues from 1 to N . Let (Fx, j )ave denote the average value of the x -component of the \njth \nforce in the displacement interval [x j\u22121, x j ] . For the \ndisplacement interval we \ncalculate the contribution to the work \nWj = (F x, j )ave \u0394x j \n(13.5.1) \nThis contribution is a scalar so we add up these scalar quantities to get the total work \nj=N\nj=N \nWN = \u2211Wj = \u2211(F x, j )ave \u0394x j . \n(13.5.2) \nj=1 \nj=1 \nThe sum in Equation (13.5.2) depends on the number of divisions N and the width of the \nintervals \u0394x j . In order to define a quantity that is independent of the divisions, we take \nthe limit as N \u2192\u221e and \n\u2192 0 for all j . The work is then\n\u0394x j \n13-13", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_329", "text": "x=x f\nj= N \nW = lim \u2211(Fx, j )ave \u0394x j = \u222b Fx (x) dx \n(13.5.3)\nN\u2192\u221e \nj=1 \nx=xi\n\u21920\n\u0394x j \nThis last expression is the definite integral of the x -component of the force with respect \nto the parameter x . In Figure 13.5 we graph the x -component of the force as a function \nof the parameter x . The work integral is the area under this curve between x = xi and \nx = x f . \nFigure 13.5 Plot of x -component of a sample force Fx (x) as a function of x . \nExample 13.6 Work done by the Spring Force \nConnect one end of an unstretched spring of length l0 with spring constant k to an object \nresting on a smooth frictionless table and fix the other end of the spring to a wall. Choose \nan origin as shown in the figure. Stretch the spring by an amount \nand release the \nxi \nobject. How much work does the spring do on the object when the spring is stretched by \nan amount x f ? \nxi \nx f\nl0 \nx = 0 \n\u02c6i \nl0 \n\u02c6i \nl0 \n\u02c6i \nx = 0 \nx = 0 \nFigure 13.6 Equilibrium, initial and final states for a spring \nSolution: We first begin by choosing a coordinate system with our origin located at the \nposition of the object when the spring is unstretched (or uncompressed). We choose the \u02c6i \nunit vector to point in the direction the object moves when the spring is being stretched. \nWe choose the coordinate function x to denote the position of the object with respect to \nthe origin. We show the coordinate function and free-body force diagram in the figure \nbelow. \n13-14", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_330", "text": "l0 \nx = 0 \n\u02c6i \nx \nx = 0 \n\u02c6i \nx \nF = F x \u02c6i = kx\u02c6i \nFigure 13.6a Spring force \nThe spring force on the object is given by (Figure 13.6a) \n! \nF = Fx\u02c6i = \u2212k x\u02c6i \n(13.5.4) \nIn Figure 13.7 we show the graph of the x -component of the spring force, Fx (x) , as a \nfunction of x . \nF x (x) \nxi\nx f \n+x \nF x (x) = k x \nFigure 13.7 Plot of spring force Fx (x) vs. displacement x \nThe work done is just the area under the curve for the interval xi to x f , \nx\u2032=x f \nx\u2032=x f \nW = \u222b Fx x\u2032)dx\u2032 = \u222b\u2212kx dx\u2032 = \u2212 1\n2 k(x2 \nf \u2212 xi \n2)\n( \n\u2032 \n(13.5.5) \nx\u2032=xi \nx\u2032=xi \nThis result is independent of the sign of \nand \nbecause both quantities appear as \nxi \nx f \nsquares. If the spring is less stretched or compressed in the final state than in the initial \nstate, then the absolute value, \n, and the work done by the spring force is positive.\n<\nx f \nxi \nThe spring force does positive work on the body when the spring goes from a state of \n\u201cgreater tension\u201d to a state of \u201clesser tension.\u201d \n13.6 Work-Kinetic Energy Theorem \n13-15", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_331", "text": "There is a direct connection between the work done on a point-like object and the change \nin kinetic energy the point-like object undergoes. If the work done on the object is non-\nzero, this implies that an unbalanced force has acted on the object, and the object will \nhave undergone acceleration. For an object undergoing one-dimensional motion the left \nhand side of Equation (13.3.16) is the work done on the object by the component of the \nsum of the forces in the direction of displacement, \nx=x f \n1 \n2 \u2212 1\n2\nW = \nF dx = mv f \n= \u0394K \n(13.6.1)\n\u222b \nx \n2\n2 mvi = K f \u2212 Ki \nx=xi \nWhen the work done on an object is positive, the object will increase its speed, and \nnegative work done on an object causes a decrease in speed. When the work done is zero, \nthe object will maintain a constant speed. In fact, the work-energy relationship is quite \nprecise; the work done by the applied force on an object is identically equal to the change \nin kinetic energy of the object. \nExample 13.7 Gravity and the Work-Energy Theorem \nSuppose a ball of mass m = 0.2 kg starts from rest at a height \n= 15 m above the \ny0 \nsurface of the earth and falls down to a height y f = 5.0 m above the surface of the earth. \nWhat is the change in the kinetic energy? Find the final velocity using the work-energy \ntheorem. \nSolution: As only one force acts on the ball, the change in kinetic energy is the work \ndone by gravity, \nW g = \u2212mg( y f \u2212 y0) \n(13.6.2) \n= (\u22122.0 \u00d710\u22121 kg)(9.8 m \u22c5s-2 )(5 m \u221215 m) = 2.0 \u00d7101 J. \nThe ball started from rest, vy,0 = 0 . So the change in kinetic energy is \n1\n2 \u2212 1\n2\n1 \n2\n\u0394K = mv \nmv \n= mv \n. \n(13.6.3)\ny, f\ny,0 \ny, f\n2\n2\n2 \nWe can solve Equation (13.6.3) for the final velocity using Equation (13.6.2) \nv y, f = 2\u0394K \nm = 2W g \nm = 2(2.0 \u00d7101 J) \n0.2 kg \n= 1.4 \u00d7101 m \u22c5s-1 . \n(13.6.4)", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_332", "text": "v y, f = 2\u0394K \nm = 2W g \nm = 2(2.0 \u00d7101 J) \n0.2 kg \n= 1.4 \u00d7101 m \u22c5s-1 . \n(13.6.4) \nFor the falling ball in a constant gravitation field, the positive work of the gravitation \nforce on the body corresponds to an increasing kinetic energy and speed. For a rising \n13-16", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_333", "text": "body in the same field, the kinetic energy and hence the speed decrease since the work \ndone is negative. \nExample 13.7 Final Kinetic Energy of Moving Cup \nA person pushes a cup of mass 0.2 kg along a horizontal table with a force of magnitude \n2.0 N at an angle of 30o with respect to the horizontal for a distance of 0.5 m as in \nExample 13.4. The coefficient of friction between the table and the cup is \u00b5k = 0.1. If the \ncup was initially at rest, what is the final kinetic energy of the cup after being pushed 0.5 \nm? What is the final speed of the cup? \nSolution: The total work done on the cup is the sum of the work done by the pushing \nforce and the work done by the friction force, as given in Equations (13.4.9) and \n(13.4.14), \nW a +W f = (F a \u2212 \u00b5k\nW = \nN )(x f \n)\nx \n\u2212 xi \n. \n(13.6.5) \n= (1.7 N \u2212 9.6 \u00d710\u22122 N)(0.5 m) = 8.0 \u00d710\u22121 J \nThe initial velocity is zero so the change in kinetic energy is just \n1\n2 \u2212 1\n1\n\u0394K = mv \nmv \n2 = mv \n2 . \n(13.6.6)\ny, f\ny,0 \ny, f\n2\n2\n2 \nThus the work-kinetic energy theorem, Eq.(13.6.1)), enables us to solve for the final \nkinetic energy, \n1\n2\n= mv f = \u0394K = W = 8.0 \u00d710\u22121 J . \n(13.6.7)\nK f \n2 \nWe can solve for the final speed, \nv y, f = \n2K f \nm = 2W \nm = 2(8.0 \u00d710\u22121 J) \n0.2 kg \n= 2.9 m \u22c5s-1 . \n(13.6.8) \n13.7 Power Applied by a Constant Force \n\uf072 \nSuppose that an applied force Fa acts on a body during a time interval \u0394t , and the \ndisplacement of the point of application of the force is in the x -direction by an amount \n\u0394x . The work done, \u0394W a , during this interval is \n\u0394W a = Fx\na \u0394x . \n(13.7.1) \nwhere Fx\na is the x -component of the applied force. (Equation (13.7.1) is the same as", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_334", "text": "\u0394x . The work done, \u0394W a , during this interval is \n\u0394W a = Fx\na \u0394x . \n(13.7.1) \nwhere Fx\na is the x -component of the applied force. (Equation (13.7.1) is the same as \nEquation (13.4.2).) \n13-17", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_335", "text": "The average power of an applied force is defined to be the rate at which work is \ndone, \n\u0394W a \nF a \u0394x\na \nx\na\nP =\n= \n= F v \n. \n(13.7.2)\nave \nx \nave,x\n\u0394t \n\u0394t \nThe average power delivered to the body is equal to the component of the force in the \ndirection of motion times the component of the average velocity of the body. Power is a \nscalar quantity and can be positive, zero, or negative depending on the sign of work. The \n-1].\nSI units of power are called watts [W] and [1 W] = [1 J \u22c5 s \nThe instantaneous power at time t is defined to be the limit of the average power \nas the time interval [t,t + \u0394t] approaches zero, \n\u0394W a \nF a \u0394x\nPa \nx \na \u239b\n\u0394x \u239e \na\n= lim \n= lim \n= F x \nlim \n\u23a0\u239f = Fx vx . \n(13.7.3)\n\u0394t\u21920 \u0394t \n\u0394t\u21920 \u0394t \n\u239d\u239c \u0394t\u21920 \u0394t \nThe instantaneous power of a constant applied force is the product of the component of \nthe force in the direction of motion and the instantaneous velocity of the moving object. \nExample 13.8 Gravitational Power for a Falling Object \nSuppose a ball of mass m = 0.2 kg starts from rest at a height \n= 15 m above the \ny0 \nsurface of the earth and falls down to a height y f = 5.0 m above the surface of the earth. \nWhat is the average power exerted by the gravitation force? What is the instantaneous \npower when the ball is at a height y f = 5.0 m above the surface of the Earth? Make a \ngraph of power vs. time. You may ignore the effects of air resistance. \nSolution: There are two ways to solve this problem. Both approaches require calculating \nthe time interval \u0394t for the ball to fall. Set t0 = 0 for the time the ball was released. We \ncan solve for the time interval \u0394t = t f that it takes the ball to fall using the equation for a \nfreely falling object that starts from rest, \n1\n2\n= \n\u2212 gt f . \n(13.7.4)\ny f \ny0 \n2 \nThus the time interval for falling is \n2\n2\nt = \n( y \u2212 y ) = \n(15 m \u2212 5 m) = 1.4 s . \n(13.7.5)\nf \n0 \nf \n-2 \ng \n9.8 m \u22c5 s \n13-18", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_336", "text": "First approach: we can calculate the work done by gravity, \nW g = \u2212mg( y f \u2212 y0 ) \n(13.7.6) \n= (\u22122.0 \u00d710\u22121 kg)(9.8 m \u22c5s-2 )(5 m \u221215 m) = 2.0 \u00d7101 J. \nThen the average power is \ng \n\u0394W \n2.0 \u00d7101 J\nP ave =\n= \n= 1.4 \u00d7101 W . \n(13.7.7)\n\u0394t \n1.4 s \nSecond Approach. We calculate the gravitation force and the average velocity. The \ngravitation force is \nFy \ng = \u2212mg = \u2212(2.0 \u00d710\u22121 kg)(9.8 m \u22c5s-2 ) = \u22122.0 N . \n(13.7.8) \nThe average velocity is \n\u0394y \n5 m \u221215 m \n-1 \nv \n=\n= \n= \u22127.0 m \u22c5s . \n(13.7.9)\nave,y \n\u0394t \n1.4 s \nThe average power is therefore \nPg = F g v \n= (\u2212mg)v \nave \ny \nave,y \nave,y \n(13.7.10) \n= (\u22122.0 N)(\u22127.0 m \u22c5s-1) = 1.4 \u00d7101 W. \nIn order to find the instantaneous power at any time, we need to find the instantaneous \nvelocity at that time. The ball takes a time t f = 1.4 s to reach the height y f = 5.0 m . The \nvelocity at that height is given by \nvy = \u2212gt f = \u2212(9.8 m \u22c5 s-2 )(1.4 s) = \u22121.4 \u00d7 101 m \u22c5 s-1 . \n(13.7.11) \nSo the instantaneous power at time t f = 1.4 s is \nPg = Fy \ng vy = (\u2212mg)(\u2212gt f ) = mg 2t f \n(13.7.12) \n= (0.2 kg)(9.8 m \u22c5s-2 )2(1.4 s) = 2.7 \u00d7101 W \nIf this problem were done symbolically, the answers given in Equation (13.7.11) and \nEquation (13.7.12) would differ by a factor of two; the answers have been rounded to two \nsignificant figures. \n13-19", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_337", "text": "The instantaneous power grows linearly with time. The graph of power vs. time is shown \nin Figure 13.8. From the figure, it should be seen that the instantaneous power at any \ntime is twice the average power between t = 0 and that time. \nFigure 13.8 Graph of power vs. time \nExample 13.9 Power Pushing a Cup \nA person pushes a cup of mass 0.2 kg along a horizontal table with a force of magnitude \n2.0 N at an angle of 30o with respect to the horizontal for a distance of 0.5 m , as in \nExample 13.4. The coefficient of friction between the table and the cup is \u00b5k = 0.1. What \nis the average power of the pushing force? What is the average power of the kinetic \nfriction force? \nSolution: We will use the results from Examples 13.4 and 13.7 but keeping extra \nsignificant figures in the intermediate calculations. The work done by the pushing force \nis \nW a = F a (x f \n) = (1.732 N)(0.50 m) = 8.660 \u00d710\u22121 J . \n(13.7.13)\nx \n\u2212 x0 \nThe final speed of the cup is vx, f = 2.860 m s\u22c5 -1 . Assuming constant acceleration, the \ntime during which the cup was pushed is \n2(x f \u2212 x0)\nt f =\n= 0.3496s . \n(13.7.14)\nvx, f \nThe average power of the pushing force is then, with \u0394t = t f , \n\u0394W a \nPa \n8.660 \u00d710\u22121 J \nave =\n= \n= 2.340 W , \n(13.7.15)\n\u0394t \n0.3496 s \nor 2 3W to two significant figures. The work done by the friction force is\n. \nW f = fk \n\u2212 x0)\n(x f \n(13.7.16)\n) = \u2212(9.6 \u00d710\u22122 N)(0.50 m) = \u2212(4.8 \u00d710\u22122 J). \n= \u2212\u00b5k N (x f \u2212 x0 \n13-20", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_338", "text": "The average power of kinetic friction is \n\u0394W f \nP f \n\u22124.8 \u00d710\u22122 J \nave =\n= \n= \u22121.4 \u00d710\u22121 W. \n(13.7.17)\n\u0394t \n0.3496 s \nThe time rate of change of the kinetic energy for a body of mass m moving in the x \u00ad\ndirection is \ndK \nd \u239b 1\n2 \u239e \ndvx\n= \nmv \n= m\nv = ma v . \n(13.7.18)\n\u239c \nx \u239f \nx \nx\nx\ndt \ndt \u239d 2 \n\u23a0 \ndt \nBy Newton\u2019s Second Law, Fx = max , and so Equation (13.7.18) becomes \ndK = Fxvx = P . \n(13.7.19)\ndt \nThe instantaneous power delivered to the body is equal to the time rate of change of the \nkinetic energy of the body. \n13.8 Work and the Scalar Product \nWe shall introduce a vector operation, called the scalar product or \u201cdot product\u201d that \ntakes any two vectors and generates a scalar quantity (a number). We shall see that the \nphysical concept of work can be mathematically described by the scalar product between \nthe force and the displacement vectors. \n13.8.1 Scalar Product \n\uf072\n\uf072 \nLet A and B be two vectors. Because any two non-collinear vectors form a plane, we \n\uf072\n\uf072 \ndefine the angle \u03b8 to be the angle between the vectors A and B as shown in Figure \n13.9. Note that \u03b8 can vary from 0 to \u03c0 . \nFigure 13.9 Scalar product geometry. \n\uf072\n\uf072 \n\uf072\n\uf072 \nThe scalar product A \u22c5 B of the vectors A and B is defined to be product of the \n\uf072\n\uf072 \nmagnitude of the vectors A and B with the cosine of the angle \u03b8 between the \ntwo vectors: \n13-21", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_339", "text": "\uf075\uf072 \uf075\uf072 \nA \u22c5 B = ABcos(\u03b8) , \n(13.8.1) \n\uf072\n\uf072 \n\uf072\n\uf072 \nwhere A =| A | and B =| B | represent the magnitude of A and B respectively. \nThe scalar product can be positive, zero, or negative, depending on the value of \ncos\u03b8 . The scalar product is always a scalar quantity. \nThe angle formed by two vectors is therefore \n\u239b  \n\u22121 \u239c\u239c\u239d \n\u03b8 = cos\n\uf072\nA\uf072\nA  \n\u22c5  \n\uf072\nB\uf072\nB  \n\u239e  \n\u239f\u239f . \n(13.8.2) \n\u23a0 \n\uf072 \nThe magnitude of a vector A is given by the square root of the scalar product of the \n\uf072 \nvector A with itself. \n\uf072 \nA = ( \n\uf072 \nA \u22c5  \n\uf072 \nA )1/ 2 . \n(13.8.3)  \nWe can give a geometric interpretation to the scalar product by writing the definition as \n\uf072\n\uf072 \nA \u22c5 B = ( Acos(\u03b8)) B . \n(13.8.4) \n\uf072 \nB in the direction of \nIn this formulation, the term Acos\u03b8 is the projection of the vector \n\uf072 \nB \nthe vector \n. This projection is shown in Figure 13.10a. So the scalar product is the \nNote that we could also write the scalar product as \n\uf072 \nproduct of the projection of the length of A in the direction of \n\uf072 \nB with t\n\uf072 \nB\nhe length of \n. \n\uf072 \uf072 \nA \u22c5 B = A(Bcos(\u03b8)) . \n(13.8.5) \n\uf072 \n\uf072 \n\uf072 \nB \nNow the term Bcos(\u03b8) is the projection of the vector B in the direction of the vector A \nas shown in Figure 13.10b. From this perspective, the scalar product is the product of the \n\uf072\n\uf072 \nprojection of the length of \nin the direction of A with the length of A . \n(a) \n(b) \n13-22", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_340", "text": "Figure 13.10 (a) and (b) Projection of vectors and the scalar product \nFrom our definition of the scalar product we see that the scalar product of two vectors \nthat are perpendicular to each other is zero since the angle between the vectors is \u03c0 / 2 \nand cos(\u03c0 / 2) = 0. \nWe can calculate the scalar product between two vectors in a Cartesian coordinates \n\uf072 \nsystem \nas \nfollows. \nConsider \ntwo \nvectors \nA = A \u02c6i + A \u02c6j + A k\u02c6\nand \nx \ny \nz \n\uf072 \nB = B \u02c6i + B \u02c6j + B k\u02c6 . Recall that \nx\ny\nz \n\u02c6i \u22c5 \u02c6i = \u02c6j\u22c5 \u02c6j = k\u02c6 \u22c5k\u02c6 = 1 \n(13.8.6)\n\u02c6i \u22c5 \u02c6j = \u02c6j\u22c5k\u02c6 = \u02c6i \u22c5k\u02c6 = 0. \n\uf072\n\uf072 \nThe scalar product between A and B is then \n\uf072\n\uf072 \nA \u22c5 B = A B + A B + A B . \n(13.8.7)\nx\nx \ny\ny \nz\nz \nThe time derivative of the scalar product of two vectors is given by \nd \ndt ( \n\uf072\nA \u22c5  \n\uf072\nB ) =  d \ndt (Ax Bx + Ay By + Az Bz ) \nd\nd\nd \nd\nd\nd \n= \n(Ax )Bx + \n(Ay )By + \n( Az )Bz + Ax \n( Bx ) + Ay \n( By ) + Az \n( Bz ) \n(13.8.8)\ndt\ndt\ndt \ndt\ndt\ndt \n=  \u239b  \n\u239d\u239c  \nd \ndt \n\uf072\nA \u239e  \n\u23a0\u239f \u22c5  \n\uf072\nB +  \n\uf072\nA \u22c5\u239b \n\u239d\u239c \nd \ndt \n\uf072\nB \u239e  \n\u23a0\u239f .  \n\uf072\n\uf072 \nIn particular when A = B , then the time derivative of the square of the magnitude of the \n\uf072 \nvector A is given by \nd\nd \nd \nd \nd\n( \n\uf072\nA \u22c5  \n\uf072\nA ) =  \n\uf072\nA \u239e  \n\u23a0\u239f \u22c5  \n\uf072\nA +  \n\uf072\nA \u22c5  \n\uf072\nA \n\uf072\nA \u239e  \n\u23a0\u239f \u22c5  \n\uf072\nA .  \n(13.8.9) \n\u239b  \n\u239b  \n\u239e\n\u239b \nA2 \n= 2 \n=  \n\u239d\u239c  \n\u239d\u239c  \n\u23a0\u239f  \n\u239d\u239c \ndt \ndt \ndt \ndt \ndt \n13.8.2 Kinetic Energy and the Scalar Product \nFor an object undergoing three-dimensional motion, the velocity of the object in \n\uf072\nCartesian components is given by v = v \u02c6i + v \u02c6j + v k\u02c6 . Recall that the magnitude of a \nx\ny\nz \nvector is given by the square root of the scalar product of the vector with itself, \nA \u2261  \n\uf072 \nA \u2261 (  \n\uf072 \nA \u22c5  \n\uf072 \nA )1/ 2 \nA2 + A2 + \n2)1/ 2 \nz \nA \n= (  \n. \n(13.8.10) \nx \ny \nTherefore the square of the magnitude of the velocity is given by the expression \n13-23", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_341", "text": "2 \u2261 (v \uf072\uf072\n2 + v2 + v2\nv\n\u22c5 v) = v\n. \n(13.8.11)\nx \ny\nz \nHence the kinetic energy of the object is given by \n1 \n\uf072\uf072\n1\n2 + v2 + v\nK = m(v \u22c5 v) = m(v\n2) . \n(13.8.12)\nx \ny\nz\n2\n2 \n13.8.2 Work and the Scalar Product \nWork is an important physical example of the mathematical operation of taking the scalar \nproduct between two vectors. Recall that when a constant force acts on a body and the \npoint of application of the force undergoes a displacement along the x -axis, only the \ncomponent of the force along that direction contributes to the work, \nW = Fx \u0394x . \n(13.8.13) \n\uf072 \nSuppose we are pulling a body along a horizontal surface with a force F . Choose \n\uf072 \ncoordinates such that horizontal direction is the x -axis and the force F forms an angle \n\u03b2 with the positive x -direction. In Figure 13.11 we show the force vector \n\uf072 \nF = Fx \u02c6i + Fy \u02c6j and the displacement vector of the point of application of the force \n\uf072\n\uf072\n\u0394x = \u0394x\u02c6i . Note that \u0394x = \u0394x\u02c6i is the component of the displacement and hence can be \ngreater, equal, or less than zero (but is shown as greater than zero in the figure for \n\uf072 \n\uf072\nclarity). The scalar product between the force vector F and the displacement vector \u0394x \nis \n\uf072\n\uf072 \n\u02c6\n\u02c6 \n\u02c6\nF \u22c5\u0394x = (F i + F j) (\u22c5\u0394x i) = F \u0394x . \n(13.8.14)\nx\ny \nx \nFigure 13.11 Force and displacement vectors \nThe work done by the force is then \n\uf072\n\uf072\n\u0394W = F \u22c5\u0394x . \n(13.8.15) \n13-24", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_342", "text": "In general, the angle \u03b2 takes values within the range \u2212\u03c0 \u2264\u03b2 \u2264\u03c0 (in Figure 13.11, \n\uf072 \n0 \u2264\u03b2 \u2264\u03c0 / 2 ). Because the x -component of the force is Fx = F cos(\u03b2) where F =| F | \n\uf072 \ndenotes the magnitude of F , the work done by the force is \n\uf072\n\uf072\nW = F \u22c5\u0394x = (F cos( ))\u0394x .\n\u03b2 \n(13.8.16) \nExample 13.10 Object Sliding Down an Inclined Plane \nAn object of mass m = 4.0 kg , starting from rest, slides down an inclined plane of length \nl = 3.0 m . The plane is inclined by an angle of \u03b8 = 300 to the ground. The coefficient of \nkinetic friction is \u00b5k = 0.2. (a) What is the work done by each of the three forces while \nthe object is sliding down the inclined plane? (b) For each force, is the work done by the \nforce positive or negative? (c) What is the sum of the work done by the three forces? Is \nthis positive or negative? \nSolution: (a) and (b) Choose a coordinate system with the origin at the top of the inclined \nplane and the positive x -direction pointing down the inclined plane, and the positive y \u00ad\ndirection pointing towards the upper right as shown in Figure 13.12. While the object is \nsliding down the inclined plane, three uniform forces act on the object, the gravitational \nforce which points downward and has magnitude Fg = mg , the normal force N which is \nperpendicular to the surface of the inclined plane, and the friction force which opposes \nthe motion and is equal in magnitude to \n= \u00b5kN . A force diagram on the object is \nfk \nshown in Figure 13.13. \nFigure 13.12 Coordinate system for \nFigure 13.13 Free-body force diagram\nobject sliding down inclined plane \nfor object \nIn order to calculate the work we need to determine which forces have a component in \nthe direction of the displacement. Only the component of the gravitational force along the \npositive x -direction Fgx = mg sin\u03b8 and the friction force are directed along the \ndisplacement and therefore contribute to the work. We need to use Newton\u2019s Second Law \n13-25", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_343", "text": "to determine the magnitudes of the normal force. Because the object is constrained to \nmove along the positive x -direction, ay = 0 , Newton\u2019s Second Law in the \u02c6j -direction \nN \u2212 mg cos\u03b8 = 0 . Therefore N = mg cos\u03b8 and the magnitude of the friction force is \nmg cos\u03b8 .\nfk = \u00b5k \nWith our choice of coordinate system with the origin at the top of the inclined plane and \nthe positive x -direction pointing down the inclined plane, the displacement of the object \n\uf072\nis given by the vector \u0394r = \u0394x \u02c6i (Figure 13.14). \nFigure 13.14 Force vectors and displacement vector for object \n\uf072 \nFg\nThe vector decomposition of the three forces are \n= mgsin\u03b8 \u02c6i \u2212 mgcos\u03b8\u02c6j , \n\uf072\n\uf072 \nF f = \u2212\u00b5kmgcos\u03b8\u02c6i , and FN = mgcos\u03b8\u02c6j . The work done by the normal force is zero \nbecause the normal force is perpendicular the displacement \nW N \n\uf072\n\uf072 \n= FN \u22c5\u0394r = mgcos\u03b8\u02c6j\u22c5l \u02c6i = 0. \nThen the work done by the friction force is negative and given by \nW f \n\uf072\n\uf072 \n= F f \u22c5\u0394r = \u2212\u00b5kmgcos\u03b8\u02c6i \u22c5l \u02c6i = \u2212\u00b5kmgcos\u03b8l < 0 . \nSubstituting in the appropriate values yields \nW f = \u2212\u00b5kmg cos\u03b8l = \u2212(0.2)(4.0kg)(9.8m \u22c5s-2 )(3.0m)(cos(30o )(3.0m) = \u221220.4 J . \nThe work done by the gravitational force is positive and given by \n\uf072 \nW g \n\uf072 \n= Fg \u22c5\u0394r = (mgsin\u03b8 \u02c6i \u2212 mgcos\u03b8\u02c6j) \u22c5l \u02c6i = mglsin\u03b8 > 0. \nSubstituting in the appropriate values yields \nW g = mglsin\u03b8 = (4.0kg)(9.8 m \u22c5s-2 )(3.0m)(sin(30o ) = 58.8 J . \n13-26", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_344", "text": "(c) The scalar sum of the work done by the three forces is then \nW g +W f\nW =\n= mgl(sin\u03b8 \u2212 \u00b5k cos\u03b8 ) \nW = (4.0kg)(9.8m \u22c5s-2 )(3.0m)(sin(30o ) \u2212 (0.2)(cos(30o )) = 38.4 J. \n13.9 Work done by a Non-Constant Force Along an Arbitrary Path \n\uf072 \nSuppose that a non-constant force F acts on a point-like body of mass m while the body \nis moving on a three dimensional curved path. The position vector of the body at time t \n\uf072\nwith respect to a choice of origin is r( )t . In Figure 13.15 we show the orbit of the body \n\uf072\nfor a time interval [t t ] moving from an initial position r \u2261 r \uf072 \n) at time t = ti to a \ni , f \ni \n(t = ti \n\uf072\n\uf072\nfinal position r \u2261 r \n) at time t = t f .\nf \n(t = t f \nFigure 13.15 Path traced by the motion of a body.  \nWe divide the time interval [ti ,t f ] into N smaller intervals with [t ,t ], j = 1,\u22c5\u22c5\u22c5, N \nj\u22121 \nj \n\uf072\n\uf072\n\u2261 \uf072 \n\u2261 \uf072\nwith tN = t f . Consider two position vectors rj \nr(t = t j ) and rj \u22121 \nr(t = t j \u22121 ) the \ndisplacement vector during the corresponding time interval as \u0394 \uf072rj =  \uf072rj \u2212 \uf072rj \u22121 . \n\uf072  \nLet F \ndenote the force acting on the body during the interval [t j\u22121,t j ] . The average force in this \n\uf072 \ninterval is (Fj ) \nand the average work \u0394Wj done by the force during the time interval \nave \n[t j\u22121,t j ] is the scalar product between the average force vector and the displacement \nvector, \n\uf072\n\uf072\n\u0394Wj = (Fj )ave \u22c5\u0394rj . \n(13.8.17) \nThe force and the displacement vectors for the time interval [t j\u22121,t j ] are shown in Figure \n\uf072 \n13.16 (note that the subscript \u201cave\u201d on (Fj )\nhas been suppressed).\nave \n13-27", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_345", "text": "Figure 13.16 An infinitesimal work element. \nWe calculate the work by adding these scalar contributions to the work for each \ninterval [t j\u22121,t j ] , for j = 1 to N , \nj = N\nj = N \n= \u2211\u0394Wj \u2211 ( \n\uf072\nF j )ave \u22c5\u0394\uf072  \nrj . \n(13.8.18) \nWN \n=  \nj =1 \nj =1 \nWe would like to define work in a manner that is independent of the way we \n\uf072\ndivide the interval, so we take the limit as N \u2192\u221e and \n\u2192 0 for all j . In this limit, \n\u0394rj \nas the intervals become smaller and smaller, the distinction between the average force \nand the actual force vanishes. Thus if this limit exists and is well defined, then the work \ndone by the force is \nj = N\nlim \u2211 (\nN \u2192\u221e \n\uf072\nF j )ave \u22c5\u0394\uf072rj = \u222bi \nf \uf072\nF \u22c5 d\uf072r .  \n(13.8.19) \nW  =  \n\uf072r\n\u0394 \nj =1\n\u21920\nj \nNotice that this summation involves adding scalar quantities. This limit is called the line \n\uf072 \n\uf072\nintegral of the force F . The symbol dr is called the infinitesimal vector line element. \n\uf072\nAt time t , dr is tangent to the orbit of the body and is the limit of the displacement \n\uf072\n\uf072 \n\uf072\nvector \u0394r = r(t + \u0394t) \u2212 r( )t as \u0394t approaches zero. In this limit, the parameter t does not \nappear in the expression in Equation (13.8.19). \nIn general this line integral depends on the particular path the body takes between \n\uf072\n\uf072\n\uf072\nthe initial position ri and the final position rf , which matters when the force F is non-\nconstant in space, and when the contribution to the work can vary over different paths in \nspace. We can represent the integral in Equation (13.8.19) explicitly in a coordinate \n\uf072\nsystem by specifying the infinitesimal vector line element dr and then explicitly \ncomputing the scalar product. \n13.9.1 Work Integral in Cartesian Coordinates \nIn Cartesian coordinates the line element is \n13-28", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_346", "text": "\uf072 \n\u02c6\n\u02c6\n\u02c6\ndr = dx i + dy j + dz k , \n(13.8.20) \nwhere dx , dy , and dz represent arbitrary displacements in the \u02c6i -, \u02c6j -, and k\u02c6 -directions \nrespectively as seen in Figure 13.17. \nFigure 13.17 A line element in Cartesian coordinates. \nThe force vector can be represented in vector notation by \n\uf072 \n\u02c6\n\u02c6 \n\u02c6\nF = F i + F j + F k . \n(13.8.21)\nx\ny\nz \nThe infinitesimal work is the sum of the work done by the component of the force times \nthe component of the displacement in each direction, \ndW = F dx + F dy + F dz . \n(13.8.22)\nx\ny\nz \nEq. (13.8.22) is just the scalar product \n\uf072\n\uf072\ndW = F \u22c5 dr = (F \u02c6i + F \u02c6j + F k\u02c6 )\u22c5(dx \u02c6i + dy \u02c6j + dz k\u02c6 )\nx\ny\nz \n, \n(13.8.23) \n= F dx + F dy + F dz \nx\ny\nz \nThe work is \n\uf072r= \uf072r \n\uf072r= \uf072r \n\uf072r= \uf072r \n\uf072r= \uf072r \n\uf072r= \uf072r\nf \uf072\nF \u22c5 d\uf072r =  \nf \nf \nf \nf \n\u222b \uf072r0 \n\u222b\n\u222b\n\u222b\n\u222b\nx\nz \n\uf072r \n\uf072r \n\uf072r \n\uf072r \ny \n0 \n0\n0\n0 \nW  =  \n(F dx + F dy + F dz) \nF dx + \nFydy + \nFzdz . \n(13.8.24)\n=  \nx \n\uf072r \n\uf072r \n\uf072r \n\uf072r \n\uf072r\n= \n= \n= \n= \n= \n13.9.2 Work Integral in Cylindrical Coordinates \nIn cylindrical coordinates the line element is \ndr \uf072 = dr r\u02c6 + rd\u03b8 \u03b8\u02c6 + dz k\u02c6 , \n(13.8.25) \nwhere dr , rd\u03b8 , and dz represent arbitrary displacements in the \ndirections respectively as seen in Figure 13.18. \n\u02c6r -, \u03b8\u02c6 -, and k\u02c6 -\n13-29", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_347", "text": "\uf072\nFigure 13.18 Displacement vector d s between two points \nThe force vector can be represented in vector notation by \n\uf072 \nF = Fr \u02c6r + F\u03b8 \u03b8\u02c6 + Fz k\u02c6 . \n(13.8.26) \nThe infinitesimal work is the scalar product \n\uf072\n\uf072\ndW = F \u22c5 dr = (Fr r\u02c6 + F\u03b8 \u03b8\u02c6 + Fz k\u02c6 )\u22c5(dr r\u02c6 + rd\u03b8 \u03b8\u02c6 + dz k\u02c6 ) \n(13.8.27) \n= Frdr + F\u03b8 rd\u03b8 + Fzdz. \nThe work is \n\uf072r= \uf072r \n\uf072r= \uf072r \n\uf072r= \uf072r \n\uf072r= \uf072r \n\uf072r= \uf072r\nf \uf072\nF \u22c5 d\uf072r =  \nf \nf \nf \nf \nW  =  \u222b  \n\u222b\n\u222b\n\u222b\n\u222b\n\u03b8\n\u03b8\nr\nz \n\uf072r \n\uf072r \n\uf072r \n\uf072r \ndr + F rd\u03b8 + F dz) = \nF dr + \nF rd\u03b8 + \nr \n0 \n0\n0\n0 \n(F  \nFzdz . (13.8.28) \n\uf072r \n\uf072r \n\uf072r0 \n\uf072r \n\uf072r \n\uf072r\n= \n= \n= \n= \n= \n13.10 Worked Examples \nExample 13.11 Work Done in a Constant Gravitation Field \nThe work done in a uniform gravitation field is a fairly straightforward calculation when \nthe body moves in the direction of the field. Suppose the body is moving under the \n\uf072 \ninfluence of gravity, F = \u2212mg \u02c6j along a parabolic curve. The body begins at the point \n(x0, y0) and ends at the point (x f , y f ) . What is the work done by the gravitation force on \nthe body? \n\uf072\nSolution: The infinitesimal line element dr is therefore \n\uf072\ndr = dx \u02c6i + dy \u02c6j. \n(13.9.1) \n13-30", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_348", "text": "The scalar product that appears in the line integral can now be calculated, \n\uf072\n\uf072\nF \u22c5 d r = \u2212mg \u02c6j\u22c5[dx \u02c6i + dy \u02c6j] = \u2212mgdy . \n(13.9.2) \nThis result is not surprising since the force is only in the y -direction. Therefore the only \nnon-zero contribution to the work integral is in the y -direction, with the result that \nr \ny= y\ny= y\nf \nf\nf\n\uf072 \nW\ndr \uf072 = \nF dy = \n\u2212mgdy = \u2212mg y f\n( \n\u2212 y ) . \n(13.9.3)\n= \u222b F \u22c5\n\u222b \ny \n\u222b \n0 \nr \ny= y\ny= y\n0 \n0\n0 \nIn this case of a constant force, the work integral is independent of path. \nExample 13.12 Hooke\u2019s Law Spring-Body System \nConsider a spring-body system lying on a frictionless horizontal surface with one end of \nthe spring fixed to a wall and the other end attached to a body of mass m (Figure 13.19). \nCalculate the work done by the spring force on body as the body moves from some initial \nposition to some final position. \nFigure 13.19 A spring-body system. \nSolution: Choose the origin at the position of the center of the body when the spring is \nrelaxed (the equilibrium position). Let x be the displacement of the body from the origin. \nWe choose the +\u02c6i unit vector to point in the direction the body moves when the spring is \nbeing stretched (to the right of x = 0 in the figure). The spring force on the body is then \ngiven by \n\uf072 \n\u02c6\n\u02c6\nF = Fx i = \u2212kx i . \n(13.9.4) \nThe work done by the spring force on the mass is \nx = x f \n1\n2 \u2212 x\nW\n= \u222b (\u2212kx) dx = \u2212 k(x \n2) . \n(13.9.5)\nspring \nf \n0\n2 \nx = x0 \n13-31", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_349", "text": "Example 13.13 Work done by the Inverse Square Gravitation Force \nConsider a body of mass m in moving in a fixed orbital plane about the sun. The mass of \nthe sun is ms . How much work does the gravitation interaction between the sun and the \nbody done on the body during this motion? \nSolution: Let\u2019s assume that the sun is fixed and choose a polar coordinate system with \nthe origin at the center of the sun. Initially the body is at a distance r0 from the center of \nthe sun. In the final configuration the body has moved to a distance rf < r0 from the \ncenter of the sun. The infinitesimal displacement of the body is given by\n\uf072\ndr = dr r\u02c6 + rd\u03b8 \u03b8\u02c6 . The gravitation force between the sun and the body is given by \n\uf072 \nGm m\ns\nF \n= F\n\u02c6r = \u2212 \n\u02c6r . \n(13.9.6)\ngrav \ngrav \n2r \nThe infinitesimal work done work done by this gravitation force on the body is given by \n\uf072\n\uf072\ndW = F \n\u22c5 dr = (F \nr\u02c6)\u22c5(dr r\u02c6 + rd\u03b8 \u03b8\u02c6) = F \ndr . \n(13.9.7)\ngrav \ngrav,r \ngrav,r \nTherefore the work done on the object as the object moves from ri to rf is given by the \nintegral \nrf \nrf \nrf\n\uf072 \n\u239b Gm m\u239e\n\uf072 \nsun \nW = \u222b F grav \u22c5 dr = \u222b Fgrav,r dr = \u222b \u239d\u239c \u2212 \n2 \n\u23a0\u239f dr . \n(13.9.8)\nr\nri \nri \nri \nUpon evaluation of this integral, we have for the work \nrf \n\u239b\n\u239e\nrf \u239b Gm m\u239e \nGm m \n1\n1\nsun \nsun \nW = \u222b\u2212 \n\u23a0\u239f dr = \n= Gm sun m\u239c \n\u2212\n\u239f . \n(13.9.9)\n2\n\u239d\u239c \nr\nr \n\u239d rf \nri \u23a0\nri \nri \nBecause the body has moved closer to the sun, rf < ri , hence 1/ rf > 1/ ri . Thus the work \ndone by gravitation force between the sun and the body, on the body is positive, \n\u239b \n\u239e \n1  \n1 \nW = Gm m \nsun \n\u2212 \n> 0 \n(13.9.10) \n\u239c\n\u239d  \n\u239f\n\u23a0 \nrf \nri \nWe expect this result because the gravitation force points along the inward radial \ndirection, so the scalar product and hence work of the force and the displacement is \n13-32", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_350", "text": "positive when the body moves closer to the sun. Also we expect that the sign of the work \nis the same for a body moving closer to the sun as a body falling towards the earth in a \nconstant gravitation field, as seen in Example 4.7.1 above. \nExample 13.14 Work Done by the Inverse Square Electrical Force \nLet\u2019s consider two point-like bodies, body 1 and body 2, with charges \nand\nq1 \nq2 \nrespectively interacting via the electric force alone. Body 1 is fixed in place while body 2 \nis free to move in an orbital plane. How much work does the electric force do on the body \n2 during this motion? \nSolution: The calculation in nearly identical to the calculation of work done by the \ngravitational inverse square force in Example 13.13. The most significant difference is \nthat the electric force can be either attractive or repulsive while the gravitation force is \nalways attractive. Once again we choose polar coordinates centered on body 2 in the \nplane of the orbit. Initially a distance \nseparates the bodies and in the final state a \nr0 \ndistance rf separates the bodies. The electric force between the bodies is given by \n\uf072 \n1 q1q2\n\u02c6 \nr\u02c6 = \nr\u02c6 . \n(13.9.11)\nFelec = Felec r = Felec,r \n2\n4\u03c0\u03b50 r \nThe work done by this electric force on the body 2 is given by the integral \nrf\nrf \nrf\n\uf072\n\uf072 \n1 \nq q \nW = F \n\u22c5 dr = F \ndr = \n1 2 dr . \n(13.9.12)\n\u222b elec \n\u222b elec,r \n\u222b \n2\n4\u03c0\u03b50\nr\nri\nri \nri \nEvaluating this integral, we have for the work done by the electric force \nrf \n1 q1q2\n1 q1q2 \nrf \n1 \n\u239b 1\n1 \u239e \nW = \ndr = \u2212 \n= \u2212 \n\u2212\n\u239f . \n(13.9.13)\nq1q2 \u239c\n\u222b \n2\n2\n4\u03c0\u03b5 0 r \n4\u03c0\u03b5 0 r \n4\u03c0\u03b5 \nr\nr \nri \n0 \n\u239d f\ni \u23a0\nri \nIf the charges have opposite signs, q1q2 < 0 , we expect that the body 2 will move closer \nto body 1 so rf < ri , and 1/ rf > 1/ ri . From our result for the work, the work done by \nelectrical force in moving body 2 is positive, \n1\n1\nW = \u2212 4\u03c0\u03b50 \nq1q2( 1 \u2212 ) > 0 . \n(13.9.14)\nrf \nri \nOnce again we see that bodies under the influence of electric forces only will naturally", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_351", "text": "electrical force in moving body 2 is positive, \n1\n1\nW = \u2212 4\u03c0\u03b50 \nq1q2( 1 \u2212 ) > 0 . \n(13.9.14)\nrf \nri \nOnce again we see that bodies under the influence of electric forces only will naturally \nmove in the directions in which the force does positive work. If the charges have the \n13-33", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_352", "text": "same sign, then q1q2 > 0 . They will repel with rf > ri and 1/ rf < 1/ ri . Thus the work is \nonce again positive: \n1 \n\u239b 1\n1 \u239e\nW = \u2212 \nq1q2 \u239c \n\u2212\n\u239f > 0 . \n(13.9.15)\n4\u03c0\u03b50 \n\u239d rf \nri \u23a0 \n13.11 Work-Kinetic Energy Theorem in Three Dimensions \nRecall our mathematical result that for one-dimensional motion \nf\nf \nf \nf\ndvx \ndx \n1 \n2\n1 \n2\nm a dx = m \ndx = m dv \n= m v dv = mv \n\u2212 mv . (13.11.1)\n\u222b x \n\u222b\n\u222b \nx \n\u222b x\nx \nx, f\nx, i\ndt \ndt \n2\n2\ni\ni \ni \ni \nUsing Newton\u2019s Second Law in the form Fx = ma x , we concluded that \nf \n1 \n2\n1 \n2\n\u222b F dx = mv \n\u2212 mv . \n(13.11.2)\nx\nx, f \nx,i \ni \n2\n2 \nEq. (13.11.2) generalizes to the y - and z -directions: \nf \n1 \n2\n1 \n2\n\u222b Fy dy = 2 mv y, f \u2212 2 mv y, i , \n(13.11.3) \ni \nf \n1 \n2\n1 \n2\nF dz = mv \n\u2212 mv . \n(13.11.4)\n\u222b z\nz, f \nz, i \ni \n2 \n2 \nAdding Eqs. (13.11.2), (13.11.3), and (13.11.4) yields \nf \n1 \n1\n2\n2\n2 \n2\n2\n2\n\u222b(Fx dx + Fy dy + Fz dz) = 2 m(vx, f + vy, f + vz, f ) \u2212 2 m(vx, i + vy, i + vz, i ) . \n(13.11.5) \ni \nRecall (Eq. (13.8.24)) that the left hand side of Eq. (13.11.5) is the work done by the \n\uf072 \nforce F on the object \nf\nf \nf \uf072 \nW = \u222b dW = \u222b(F dx + F dy + F dz) = \u222bF \u22c5 dr \uf072 \n(13.11.6)\nx \ny \nz \ni\ni \ni \nThe right hand side of Eq. (13.11.5) is the change in kinetic energy of the object \n1\n2 \u2212 1\n2 \n1 \n2 \n2 \n2 \n1\n2\n2\n2", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_353", "text": "(13.11.6)\nx \ny \nz \ni\ni \ni \nThe right hand side of Eq. (13.11.5) is the change in kinetic energy of the object \n1\n2 \u2212 1\n2 \n1 \n2 \n2 \n2 \n1\n2\n2\n2\n\u0394K \u2261 K f \u2212 Ki = mv f \nmv0 = m(vx, f + vy, f + vz, f ) \u2212 m(vx, i + vy, i + vz, i ) . (13.11.7)\n2\n2\n2 \n2 \n13-34", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_354", "text": "Therefore Eq. (13.11.5) is the three dimensional generalization of the work-kinetic \nenergy theorem \n\u222b \nf \nF \n\uf072 \n\u22c5 dr \uf072 = K f \u2212 Ki . \n(13.11.8) \ni \nWhen the work done on an object is positive, the object will increase its speed, and \nnegative work done on an object causes a decrease in speed. When the work done is zero, \nthe object will maintain a constant speed. \n13.11.1 Instantaneous Power Applied by a Non-Constant Force for Three \nDimensional Motion \nRecall that for one-dimensional motion, the instantaneous power at time t is defined to \nbe the limit of the average power as the time interval [t,t + \u0394t] approaches zero, \na (t)v\nP(t) = Fx \nx (t) . \n(13.11.9) \nA more general result for the instantaneous power is found by using the expression for \ndW as given in Equation (13.8.23), \n\uf072\ndW \nF \n\uf072 \n\u22c5 d r \uf072\n\uf072\nP =\n= \n= F \u22c5 v . \n(13.11.10)\ndt \ndt \nThe time rate of change of the kinetic energy for a body of mass m is equal to the power, \ndK \n1 \nd \ndt =  2 m dt ( \uf072v \u22c5 \uf072v) = m d\uf072v \ndt \u22c5 \uf072v = m \uf072a\u22c5 \uf072v =  \n\uf072\nF \u22c5 \uf072v = P .  \n(13.11.11)  \nwhere the we used Eq. (13.8.9), Newton\u2019s Second Law and Eq. (13.11.10). \n13-35", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_355", "text": "Appendix 13A Work Done on a System of Two Particles \nWe shall show that the work done by an internal force in changing a system of two \nparticles of masses m1 and m2 respectively from an initial state A to a final state B is \nequal to \nW c = 1 \u00b5(vB \n2 \u2212 vA \n2 ) \n(13.1.1)\n2 \nwhere vB \n2 is the square of the relative velocity in state B , vA \n2 is the square of the relative \nvelocity in state A , and \u00b5 = m1m2 / (m1 + m2) . \nConsider two bodies 1 and 2 and an interaction pair of forces shown in Figure 13A.1. \nFigure 13A.1 System of two bodies interacting \nWe choose a coordinate system shown in Figure 13A.2. \nFigure 13A.2 Coordinate system for two-body interaction \nNewton\u2019s Second Law applied to body 1 is \n\uf072 \nd 2\uf072  \nF2,1 = m1 dt\nr\n2 \n1 \n(13.1.2)  \nand applied to body 2 is \n13-36", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_356", "text": "\uf072 \nd 2\uf072  \nF1,2 = m2 dt\nr\n2 \n2 . \n(13.1.3)  \nDivide each side of Equation (13.1.2) by m1 , \n\uf072 \nd 2\uf072\nF2,1 \nr1\n= \n(13.1.4)\ndt2\nm1 \nand divide each side of Equation (13.1.3) by m2 , \n\uf072 \n\uf072r \nr\nF1,2 \nd 2\uf072 \n= \n2 . \n(13.1.5)\ndt2\nm2 \n\uf072r \nSubtract Equation (13.1.5) from Equation (13.1.4) yielding \n\uf072r\nd 2 \nd 2 \nd 2 \n2,1 \n1,2 \n1 \n2 \n2,1 \n\uf072 \nF \n\uf072 \nF \n\u2212  \n\u2212  \n(13.1.6) \n=  \n=  \n, \ndt2 \ndt2 \ndt2\nm1 \nm2 \n\uf072\n\uf072  \n2,1 \n1\n2 .  \nEquation (13.1.6) to obtain  \n\uf072r \n\uf072r \n\uf072r\nF2,1 = \u2212F1, 2 \nwhere \n\u2212  \nUse Newton\u2019s Third Law, \n\uf072\nF\nd 2 \n2,1 \nthe left hand side of \n=  \non \n\uf072r1 \ndt2 \n\uf072r\n\uf072r\nd 2 \nd 2 \n2 \n2,1 \ndt2 \ndt2 \n\u239b 1\n1 \u239e  \n\u2212  \n(13.1.7) \n+ \n\u23a0\u239f = \n=  \n\u239d\u239c  \n.  \nm\nm\n1 \n2 \n\uf072r \n\uf072\nF\n\uf072r \n\uf072\nF \n\uf072\nThe quantity d 2r1,2 / dt 2 is the relative acceleration of body 1 with respect to body 2. \nDefine \n1\n1 \n1\n\u2261 \n+ \n. \n(13.1.8)\n\u00b5 \nm\nm\n1\n2 \nThe quantity \u00b5 is known as the reduced mass of the system. Equation (13.1.7) now takes \nthe form \n\uf072 \nd 2r \uf072 \nF = \u00b5 \n2,1 . \n(13.1.9)\n2,1 \ndt2 \nThe work done in the system in displacing the two masses from an initial state A to a \nfinal state B is given by \n2,1 \n1 \n1,2 \n2 \nB \nB \nW  = \u222b  \n\u22c5 d + \u222b  \n\u22c5 d . \n(13.1.10)  \nA\nA \n13-37", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_357", "text": "Recall by the work energy theorem that the LHS is the work done on the system, \nB \nB \n=  \n2,1 \n1 \n1,2 \n2 \nA\nA \nFrom Newton\u2019s Third Law, the sum in Equation (13.1.10) becomes \n\uf072r \n\uf072\nF\n\uf072r \n\uf072\nF\u222b  \n\u222b \nW  \n\u22c5 d \n\u22c5 d \n= \u0394K . \n(13.1.11) \n+  \nB \nB \nB \nB\n\uf072r \n\uf072\nF \nA\nA\nA \nA \n\uf072r \n(13.1.12), \n\uf072r \n\uf072\nF \n2\n2\nd\nd\nd\n2,1 \n2,1 \n2,1 \n2,1 \n2,1 \n2,1 \n\uf072r \n\uf072r \n\uf072\nF\n\uf072r \n\uf072\nd\nwhere \nis the relative displacement of the two bodies. We can now substitute \nr2 1, \nNewton\u2019s Second Law, Equation (13 1 9) for the relative acceleration into Equation \n. .\n, \n\uf072r \n\uf072r\n2,1 \n1\n2,1 \n2\n2,1 \n1\n2\n2,1 \n2,1 \n\uf072r\n\uf072\nF \n\uf072\nF\u222b  \n\u222b  \n\u222b  \n\u222b \nW  \n\u22c5 d \n\u22c5 d \n\u22c5(d \u2212 d \n\u22c5 d \n(13.1.12) \n\u2212  \n) \n=  \n=  \n=  \n,  \n\uf072r \ndt2 \n\u239b  \n\u239e \nB \nB \nB \n\u222b  \n\u222b  \n\u222b \nW  \n\u22c5 d \n\u22c5 d \n\u239f dt\n\u22c5  \n(13.1.13) \n\u239c \n\u00b5  \n= \u00b5 \n=  \n=  \n, \ndt2 \ndt\n\u239d  \n\u23a0 \nA\nA \nA \n\uf072r \n\uf072r \n\uf072r \n\uf072r \nd\uf072 \n\uf072 \nr2,1 \nwhere we have used the relation between the differential elements d \n= \ndt . The \nr2,1 \ndt \nproduct rule for derivatives of the scalar product of a vector with itself is given for this \nd\nd \nd 2 \nd\n2,1 \n2,1 \n2,1 \n2,1 \ncase by \n\u239b \n\u239e \n\u239f\u23a0  \n1 d\n\u22c5 \n\u22c5 \n(13.1.14) \n\u239c\u239d  \n=  \n.  \ndt2\n2 dt \ndt\ndt \ndt \nSubstitute Equation (13.1.14) into Equation (13.1.13), which then becomes \n\u239b \n\u239c\u239d \nd\uf072 r2,1 \ndt \u22c5  \nd\uf072 r2,1 \ndt \n\u239e  \n\u222b  \nB 1 d \n2 dt\nW  = \u00b5  \ndt . \n(13.1.15)\n\u239f\u23a0 \nA \nEquation (13.1.15) is now the integral of an exact derivative, yielding \n\uf072r\nd 2,1 \ndt \u22c5  \n\uf072r\nd 2,1 \ndt \nB\n\u239b  \n\u239e \n1  \n1  \nB = 1 \n2 \u2212 vA \n2 ),\n\u00b5(vB\nA\n\u00b5 ( \uf072v2,1 \u22c5 \uf072v2,1 ) \nW  =  \n(13.1.16) \n\u00b5 \u239c\u239d \n\u239f\u23a0 \n=  \n2  \n2  \n2  \nA \n\uf072\nwhere \nis the relative velocity between the two bodies. It\u2019s important to note that in \nv2,1", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_358", "text": "2 \u2212 vA \n2 ),\n\u00b5(vB\nA\n\u00b5 ( \uf072v2,1 \u22c5 \uf072v2,1 ) \nW  =  \n(13.1.16) \n\u00b5 \u239c\u239d \n\u239f\u23a0 \n=  \n2  \n2  \n2  \nA \n\uf072\nwhere \nis the relative velocity between the two bodies. It\u2019s important to note that in \nv2,1 \nthe above derivation had we exchanged the roles of body 1 and 2 i.e. 1\u2192 2 and 2 \u2192 1, \nwe would have obtained the identical result because \n13-38", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_359", "text": "\u2212 \n\uf072 \nF2,1 \uf072r \n= \u2212 \n\uf072r2 \n= \u2212 \n\uf072r1 \n=  \n1,2 \n1,2 \nd 1,2 \n1,2 \n\uf072r \n\uf072r\n\uf072 \nF \n\uf072v \n2,1 \n) = \u2212d \n\uf072r\n1 \n\u2212 \n(13.1.17)  \n2 \n2,1 \nEquation (13.1.16) implies that the work done is the change in the kinetic energy of the \nsystem, which we can write in terms of the reduced mass and the change in the square of \nrelative speed of the two objects \n\u0394K = 1 \n2 \u2212 vA \n2 ) . \n(13.1.18)\n\u00b5(vB\n2 \n\uf072r \n\uf072r\n= d( \n= \u2212  2,1. \n\uf072v \n13-39", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_360", "text": "Chapter 14 Potential Energy and Conservation of Energy \n14.1 Conservation of Energy........................................................................................ 1  \n14.2 Conservative and Non-Conservative Forces ...................................................... 2  \n14.3 Changes in Potential Energies of a System......................................................... 5  \n14.3.1 Change in Potential Energy for Several Conservative Forces................... 8  \n14.4 Change in Potential Energy and Zero Point for Potential Energy................... 9  \n14.4.1 Change in Gravitational Potential Energy Near Surface of the Earth ..... 9  \n14.4.2 Hooke\u2019s Law Spring-Object System .......................................................... 11  \n14.4.3 Inverse Square Gravitation Force.............................................................. 12  \n14.5 Mechanical Energy and Conservation of Mechanical Energy ....................... 13", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_361", "text": "14.5 Mechanical Energy and Conservation of Mechanical Energy ....................... 13  \n14.5.1 Change in Gravitational potential Energy Near Surface of the Earth ... 13  \n14.6 Spring Force Energy Diagram........................................................................... 14  \nExample 14.1 Energy Diagram.............................................................................. 17  \n14.7 Change of Mechanical Energy for Closed System with Internal Non\u00ad\nconservative Forces..................................................................................................... 19  \n14.7.1 Change of Mechanical Energy for a Non-closed System.......................... 20  \n14.8 Dissipative Forces: Friction ............................................................................... 21", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_362", "text": "14.8.1 Source Energy .............................................................................................. 22  \n14.9 Worked Examples............................................................................................... 22  \nExample 14.2 Escape Velocity of Toro.................................................................. 22  \nExample 14.3 Spring-Block-Loop-the-Loop......................................................... 24  \nExample 14.4 Mass-Spring on a Rough Surface .................................................. 26  \nExample 14.5 Cart-Spring on an Inclined Plane.................................................. 27", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_363", "text": "Example 14.5 Cart-Spring on an Inclined Plane.................................................. 27  \nExample 14.6 Object Sliding on a Sphere............................................................. 29", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_364", "text": "Equation Chapter 8 Section 1 Chapter 14 Potential Energy and  \nConservation of Energy \nThere is a fact, or if you wish, a law, governing all natural phenomena \nthat are known to date. There is no exception to this law \u2014 it is exact as \nfar as we know. The law is called the conservation of energy. It states that \nthere is a certain quantity, which we call energy that does not change in \nthe manifold changes which nature undergoes. That is a most abstract \nidea, because it is a mathematical principle; it says that there is a \nnumerical quantity, which does not change when something happens. It is \nnot a description of a mechanism, or anything concrete; it is just a strange \nfact that we can calculate some number and when we finish watching \nnature go through her tricks and calculate the number again, it is the \n1\nsame. \nRichard Feynman \nSo far we have analyzed the motion of point-like objects under the action of forces using \nNewton\u2019s Laws of Motion. We shall now introduce the Principle of Conservation of \nEnergy to study the change in energy of a system between its initial and final states. In \nparticular we shall introduce the concept of potential energy to describe the effect of \nconservative internal forces acting on the constituent components of a system. \n14.1 Conservation of Energy \nRecall from Chapter 13.1, the principle of conservation of energy. When a system and its \nsurroundings undergo a transition from an initial state to a final state, the change in \nenergy is zero, \n\u0394E = \u0394E \n= 0 . \n(14.1.1)\nsystem + \u0394Esurroundings \nFigure 14.1 Diagram of a system and its surroundings \nWe shall study types of energy transformations due to interactions both inside and across \nthe boundary of a system. \nRichard P. Feynman, Robert B. Leighton, and Matthew Sands, The Feynman Lectures on Physics, \nVol. 1, p. 4.1. \n14-1\n1", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_365", "text": "14.2 Conservative and Non-Conservative Forces \nOur first type of \u201cenergy accounting\u201d involves mechanical energy. There are two types of \nmechanical energy, kinetic energy and potential energy. Our first task is to define what \nwe mean by the change of the potential energy of a system. \n\uf072 \nWe defined the work done by a force F , on an object, which moves along a path \n\uf072\n\uf072\nfrom an initial position ri to a final position rf , as the integral of the component of the \nforce tangent to the path with respect to the displacement of the point of contact of the \nforce and the object, \nW = \u222b F \n\uf072 \n\u22c5 dr \uf072 . \n(14.2.1) \npath \nDoes the work done on the object by the force depend on the path taken by the \nobject? \n(a) \n(b) \nFigure 14.2 (a) and (b) Two different paths connecting the same initial and final points \nFirst consider the motion of an object under the influence of a gravitational force near the \nsurface of the earth. Let\u2019s consider two paths 1 and 2 shown in Figure 14.2. Both paths \nbegin at the initial point (xi , yi ) = (0, yi ) and end at the final point (x f , y f ) = (x f ,0) . The \ngravitational force always points downward, so with our choice of coordinates, \n\uf072 \nF = \u2212mg \u02c6j . The infinitesimal displacement along path 1 (Figure 14.2a) is given by \n\uf072\ndr1 = dx1 \u02c6i + dy1 \u02c6j . The scalar product is then \n\uf072\n\uf072\nF \u22c5 dr1 = \u2212mg \u02c6j\u22c5 (dx1 \u02c6i + dy1 \u02c6j) = \u2212mgdy1. \n(14.2.2) \nThe work done by gravity along path 1 is the integral \n( x f ,0) \n\uf072 \n\u22c5 d\uf072\nW1 = \u222b F\nr = \u222b\u2212mgdy1 = \u2212mg(0 \u2212 yi ) = mgyi . \n(14.2.3) \npath 1 \n(0,yi ) \n14-2", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_366", "text": "Path 2 consists of two legs (Figure 14.2b), leg A goes from the initial point (0, yi ) \nto the origin (0,0) , and leg B goes from the origin (0,0) to the final point (x f ,0) . We \nshall calculate the work done along the two legs and then sum them up. The infinitesimal \n\uf072\ndisplacement along leg A is given by drA = dyA \u02c6j . The scalar product is then \n\uf072\n\uf072\nF \u22c5 drA = \u2212mg \u02c6j\u22c5 dyA \u02c6j = \u2212mgdyA . \n(14.2.4) \nThe work done by gravity along leg A is the integral \n(0,0) \n\uf072 \n\uf072\nWA = \u222b F \u22c5 drA = \u222b\u2212mgdyA = \u2212mg(0 \u2212 yi ) = mgyi . \n(14.2.5) \nleg A \n(0, yi ) \n\uf072\nThe infinitesimal displacement along leg B is given by drB = dxB \u02c6i . The scalar product is \nthen \n\uf072\n\uf072\nF \u22c5 drB = \u2212mg \u02c6j\u22c5 dxB \u02c6i = 0 . \n(14.2.6) \nTherefore the work done by gravity along leg B is zero, WB = 0 , which is no surprise \nbecause leg B is perpendicular to the direction of the gravitation force. Therefore the \nwork done along path 2 is equal to the work along path 1, \nW2 = WA + WB = mgyi = W1 . \n(14.2.7) \nNow consider the motion of an object on a surface with a kinetic frictional force \nbetween the object and the surface and denote the coefficient of kinetic friction by \u00b5k . \nLet\u2019s compare two paths from an initial point xi to a final point x f . The first path is a \nstraight-line path. Along this path the work done is just \n\uf072 \nW f \n\uf072 \n= \u222b F \u22c5 dr = \u222b F dx = \u2212\u00b5k N s1 \nN \u0394x < 0 , \n(14.2.8)\nx \n= \u2212\u00b5k \npath 1 \npath 1 \nwhere the length of the path is equal to the displacement, s1 = \u0394x . Note that the fact that \nthe kinetic frictional force is directed opposite to the displacement, which is reflected in \nthe minus sign in Equation (14.2.8). The second path goes past x f some distance and \nthem comes back to x f (Figure 14.3). Because the force of friction always opposes the \nmotion, the work done by friction is negative, \n\uf072 \nW f \n\uf072", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_367", "text": "the minus sign in Equation (14.2.8). The second path goes past x f some distance and \nthem comes back to x f (Figure 14.3). Because the force of friction always opposes the \nmotion, the work done by friction is negative, \n\uf072 \nW f \n\uf072 \n= \u222b F \u22c5 dr = \u222b Fx dx = \u2212\u00b5k N s2 < 0 . \n(14.2.9) \npath 2 \npath 2 \n14-3", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_368", "text": "The work depends on the total distance traveled s2 , and is greater than the displacement \ns2 > \u0394x . The magnitude of the work done along the second path is greater than the \nmagnitude of the work done along the first path. \nFigure 14.3 Two different paths from xi to x f . \nThese two examples typify two fundamentally different types of forces and their \ncontribution to work. The work done by the gravitational force near the surface of the \nearth is independent of the path taken between the initial and final points. In the case of \nsliding friction, the work done depends on the path taken. \nWhenever the work done by a force in moving an object from an initial \npoint to a final point is independent of the path, the force is called a \nconservative force. \n\uf072 \nThe work done by a conservative force Fc in going around a closed path is zero. Consider \nthe two paths shown in Figure 14.4 that form a closed path starting and ending at the \npoint A with Cartesian coordinates (1,0) . \nFigure 14.4 Two paths in the presence of a conservative force. \nThe work done along path 1 (the upper path in the figure, blue if viewed in color) from \npoint A to point B with coordinates (0,1) is given by \n14-4", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_369", "text": "B \uf072 \nW1 = \u222bF c (1)\u22c5 dr\uf072 \n1 . \n(14.2.10) \nA \nThe work done along path 2 (the lower path, green in color) from B to A is given by \nA \uf072 \nW2 = \u222bF c (2)\u22c5 dr\uf072 \n2 . \n(14.2.11) \nB \nThe work done around the closed path is just the sum of the work along paths 1 and 2, \nB \uf072 \nA \uf072 \nW = W1 +W2 = \u222bF c (1) \u22c5 dr \uf072 \n1 + \u222bF c (2)\u22c5 dr \uf072 \n2 . \n(14.2.12) \nA\nB \nIf we reverse the endpoints of path 2, then the integral changes sign, \nA \nB\n\uf072\nF  \n\uf072\nF \n\uf072r\n2\n2 \n\uf072r \nB\nA \nWe can then substitute Equation (14.2.13) into Equation (14.2.12) to find that the work \ndone around the closed path is \n\u222b  \n\u222b \nW2 \n(2) \u22c5 d \n(2) \u22c5 d \n(14.2.13) \n= \u2212 \n=  \n.  \nc \nc \nB \nB\n\uf072\nF  \n\uf072\nF \n\uf072r\n1\n2 \n\uf072r \nA\nA \nSince the force is conservative, the work done between the points A to B is independent \nof the path, so \nB \uf072 \nB \uf072 \n\u222b  \n\u222b \nW  \n(1)\u22c5 d \n(2)\u22c5 d \n(14.2.14) \n\u2212 \n=  \n.  \nc \nc \n\uf072  \n\uf072 \n\u222b \n= \u222b \nr\nF\nr \npath \nF\n2 .\nc \n1\nc \nA\nA \nWe now use path independence of work for a conservative force (Equation (14.2.15) in \nEquation (14.2.14)) to conclude that the work done by a conservative force around a \nclosed path is zero, \n\uf0d1\u222b F \n\uf072 \nW = \nc \u22c5 dr \uf072 = 0 . \n(14.2.16) \nclosed \n14.3 Changes in Potential Energies of a System \nConsider an object near the surface of the earth as a system that is initially given a \nvelocity directed upwards. Once the object is released, the gravitation force, acting as an \nexternal force, does a negative amount of work on the object, and the kinetic energy \ndecreases until the object reaches its highest point, at which its kinetic energy is zero. The \n(1) \u22c5 d\n(2) \u22c5 d\n(14.2.15)  \n14-5", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_370", "text": "gravitational force then does positive work until the object returns to its initial starting \npoint with a velocity directed downward. If we ignore any effects of air resistance, the \ndescending object will then have the identical kinetic energy as when it was thrown. All \nthe kinetic energy was completely recovered. \nNow consider both the earth and the object as a system and assume that there are \nno other external forces acting on the system. Then the gravitational force is an internal \nconservative force, and does work on both the object and the earth during the motion. As \nthe object moves upward, the kinetic energy of the system decreases, primarily because \nthe object slows down, but there is also an imperceptible increase in the kinetic energy of \nthe earth. The change in kinetic energy of the earth must also be included because the \nearth is part of the system. When the object returns to its original height (vertical distance \nfrom the surface of the earth), all the kinetic energy in the system is recovered, even \nthough a very small amount has been transferred to the Earth. \nIf we included the air as part of the system, and the air resistance as a non\u00ad\nconservative internal force, then the kinetic energy lost due to the work done by the air \nresistance is not recoverable. This lost kinetic energy, which we have called thermal \nenergy, is distributed as random kinetic energy in both the air molecules and the \nmolecules that compose the object (and, to a smaller extent, the earth). \nWe shall define a new quantity, the change in the internal potential energy of the \nsystem, which measures the amount of lost kinetic energy that can be recovered during an \ninteraction. \nWhen only internal conservative forces act in a closed system, the sum of \nthe changes of the kinetic and potential energies of the system is zero. \nConsider a closed system, \u0394Esys = 0 , that consists of two objects with masses m1 \nand m2 respectively. Assume that there is only one conservative force (internal force) \nthat is the source of the interaction between two objects. We denote the force on object 1 \n\uf072 \ndue to the interaction with object 2 by F2,1 and the force on object 2 due to the interaction \n\uf072 \nwith object 1 by F1,2 . From Newton\u2019s Third Law, \n\uf072\n\uf072 \n= \u2212 \n(14.3.1)\nF2,1 \nF1,2 . \nThe forces acting on the objects are shown in Figure 14.5. \nFigure 14.5 Internal forces acting on two objects \n14-6", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_371", "text": "Choose a coordinate system (Figure 14.6) in which the position vector of object 1 \n\uf072\n\uf072\nis given by r1 and the position vector of object 2 is given by r2 . The relative position of \n2,1 \n1 \n2 . \n\uf072r \n\uf072\n\uf072\ninteraction, object 1 is displaced by dr1 and object 2 is displaced by dr2 , so the relative \n2,1 \n2 . \n\uf072r \n\uf072r \n\uf072r \n\uf072r\nobject 1 with respect to object 2 is given by \n\u2212  \nDuring the course of the \n=  \n\uf072r1\ndisplacement of the two objects during the interaction is given by d \n= d \u2212 d \n\uf072r \n\uf072r \n\uf072r \n\uf072r \n\uf072\nF\n\uf072r \n\uf072\nF \n2,1\n1\n2 \nRecall that the change in the kinetic energy of an object is equal to the work done by the \nforces in displacing the object. For two objects displaced from an initial state A to a \nfinal state B , \n2 .\n2,1 \n1 \n1,2 \nFigure 14.6 Coordinate system for two objects with relative position vector \n\u2212 \n=  \nB \nB \n\u222b  \n\u222b \n\u0394K sys = \u0394K1 + \u0394K2 = W \n\u22c5 d \n\u22c5 d \n(14.3.2) \n+ \n=  \nc \nA\nA \n(In Equation (14.3.2), the labels \u201c A \u201d and \u201c B \u201d refer to initial and final states, not paths.) \nFrom Newton\u2019s Third Law, Equation (14.3.1), the sum in Equation (14.3.2) becomes \nB \nB \nB \nB\n\uf072r \n\uf072\nF\n\uf072r \n\uf072r \n\uf072r \n\uf072\nF \n\uf072\nF \n\uf072r \n\uf072r \n\uf072\nF \n\uf072\nF \n\uf072r \n\uf072\nF2,1 \n1 \n2,1 \n2 \n2,1 \n1 \n2 \n2,1 \n2,1 \nis the relative displacement of the two objects. Note that since \n2,1 \n2,1 \n1,2 \n1,2 \n\u222b  \n\u222b  \n\u222b  \n\u222b \n\u0394K \n= W  \n\u22c5 d \n\u22c5 d \n\u22c5(d \u2212 d \n\u22c5 d \n(14.3.3) \n\u2212  \n) \n=  \n=  \n=  \nsys \nc \nA\nA\nA \nA \n!r \n!r \n!r2,1 \n1\n2 \n\uf072\n\uf072 \nand dr \uf072 = \u2212d r \uf072\nF2,1 = \u2212F1, 2 \n2,1 \n1,2, \u222b \nwhere d \n= d \u2212 d \nB \nB \n\u222b \n\u22c5 d \n\u22c5 d\n=  \n.  \nA\nA \n14-7", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_372", "text": "Consider a system consisting of two objects interacting through a \n\uf072 \nconservative force. Let F2,1 denote the force on object 1 due to the \n2,1\n1\n2 \n\uf072r \n\uf072r \n\uf072r\ninteraction with object 2 and let d \n= d \u2212 d \nbe the relative \ndisplacement of the two objects. The change in internal potential energy \nof the system is defined to be the negative of the work done by the \nconservative force when the objects undergo a relative displacement from \nthe initial state A to the final state B along any displacement that \nchanges the initial state A to the final state B , \nB \nB \n2,1 \n2,1 \n1,2 \nA\nA \n\uf072\nF \nOur definition of potential energy only holds for conservative forces, because the \nwork done by a conservative force does not depend on the path but only on the initial and \n\uf072r \nfinal positions. Because the work done by the conservative force is equal to the change in \nkinetic energy, we have that \n\uf072\nF \n\u0394U \n= \u2212\u0394K , (closed system with no non-conservative forces) . \n(14.3.5)\nsys \nsys \nRecall that the work done by a conservative force in going around a closed path is \nzero (Equation (14.2.16)); therefore the change in kinetic energy when a system returns \nto its initial state is zero. This means that the kinetic energy is completely recoverable. \nIn the Appendix 13A: Work Done on a System of Two Particles, we showed that \nthe work done by an internal force in changing a system of two particles of masses m1 \nand m2 respectively from an initial state A to a final state B is equal to \n1 \n2 \u2212 v\nW =\n\u00b5 (vB \nA \n2 ) = \u0394K sys , \n(14.3.6)\n2 \nwhere vB \n2 is the square of the relative velocity in state B , vA \n2 is the square of the relative \nvelocity in state A , and \u00b5 = m1m2 / (m1 + m2) is a quantity known as the reduced mass of \nthe system. \n14.3.1 Change in Potential Energy for Several Conservative Forces \nWhen there are several internal conservative forces acting on the system we define a \nseparate change in potential energy for the work done by each conservative force, \n\u0394U \n= \u2212W = \u2212\u222b \nB \nF \n\uf072 \n\u22c5 dr \uf072 \ni . \n(14.3.7)\nsys, i \nc,i \nc, i \nA \n\uf072r1,2 \n\u222b  \n\u222b \n\u0394U \n= \u2212W  = \u2212  \n\u22c5 d \n= \u2212  \n\u22c5 d \n(14.3.4) \n.  \nsys \nc \n14-8", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_373", "text": "\uf072 \n\uf072\nwhere Fc, i is a conservative internal force and dri a change in the relative positions of \n\uf072 \nthe objects on which Fc, i when the system is changed from state A to state B . The work \ndone is the sum of the work done by the individual conservative forces, \nW \n+ \u22c5\u22c5\u22c5 . \n(14.3.8)\nc = Wc,1 + Wc,2 \nHence, the sum of the changes in potential energies for the system is the sum \n\u0394U sys = \u0394Usys,1 + \u0394Usys,2 + \u22c5\u22c5\u22c5 . \n(14.3.9) \nTherefore the change in potential energy of the system is equal to the negative of the \nwork done \nB \uf072\n\uf072\n\u0394U sys = \u2212W c = \u2212\u2211\u222bF c, i \u22c5 dri . \n(14.3.10) \ni\nA \nIf the system is closed (external forces do no work), and there are no non-conservative \ninternal forces then Eq. (14.3.5) holds. \n14.4 Change in Potential Energy and Zero Point for Potential Energy \nWe already calculated the work done by different conservative forces: constant gravity \nnear the surface of the earth, the spring force, and the universal gravitation force. We \nchose the system in each case so that the conservative force was an external force. In \neach case, there was no change of potential energy and the work done was equal to the \nchange of kinetic energy, \nW = \u0394K sys . \n(14.4.1)\next \nWe now treat each of these conservative forces as internal forces and calculate the change \nin potential energy of the system according to our definition \n\u0394U \n= \u2212W = \u2212\u222b\nB \nF \n\uf072 \n\u22c5 dr \uf072 . \n(14.4.2)\nsys \nc \nc \nA \nWe shall also choose a zero reference potential for the potential energy of the system, so \nthat we can consider all changes in potential energy relative to this reference potential. \n14.4.1 Change in Gravitational Potential Energy Near Surface of the Earth \nLet\u2019s consider the example of an object falling near the surface of the earth. Choose our \nsystem to consist of the earth and the object. The gravitational force is now an internal \nconservative force acting inside the system. The distance separating the object and the \n14-9", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_374", "text": "center of mass of the earth, and the velocities of the earth and the object specifies the \ninitial and final states. \nLet\u2019s choose a coordinate system with the origin on the surface of the earth and the + y \u00ad\ndirection pointing away from the center of the earth. Because the displacement of the \nearth is negligible, we need only consider the displacement of the object in order to \ncalculate the change in potential energy of the system. \nSuppose the object starts at an initial height yi above the surface of the earth and ends at \n\uf072 \nfinal height y f . The gravitational force on the object is given by Fg = \u2212mg \u02c6j , the \n\uf072\ndisplacement is given by \ndr = dy \u02c6j, and the scalar product is given by \n\uf072\n\uf072\nFg \u22c5 dr = \u2212mg \u02c6j\u22c5 dy\u02c6j = \u2212mgdy . The work done by the gravitational force on the object is \nthen \nyf \nyf\n\uf072 \n\uf072\nW g = \u222b Fg \u22c5 dr = \u222b\u2212mg dy = \u2212mg(yf \u2212 yi ) . \n(14.4.3) \nyi ) \nyi ) \nThe change in potential energy is then given by \n\u0394U g = \u2212W g = mg \u0394y = mg y \u2212 mg y \n(14.4.4)\nf\ni . \nWe introduce a potential energy function U so that \n\u0394U g \u2261 U g\nf \u2212 Ui\ng . \n(14.4.5) \nOnly differences in the function U g have a physical meaning. We can choose a zero \nreference point for the potential energy anywhere we like. We have some flexibility to \nadapt our choice of zero for the potential energy to best fit a particular problem. Because \nthe change in potential energy only depended on the displacement, \u0394y . In the above \nexpression for the change of potential energy (Eq. (14.4.4)), let y f = y be an arbitrary \npoint and yi = 0 denote the surface of the earth. Choose the zero reference potential for \nthe potential energy to be at the surface of the earth corresponding to our origin y = 0 , \nwith U g (0) = 0 . Then \n\u0394U g = U g ( y) \u2212 U g (0) = U g ( y) . \n(14.4.6) \nSubstitute yi = 0 , y f = y and Eq. (14.4.6) into Eq. (14.4.4) yielding a potential energy as \na function of the height y above the surface of the earth, \nU g ( y) = mgy, with U g ( y = 0) = 0 .", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_375", "text": "a function of the height y above the surface of the earth, \nU g ( y) = mgy, with U g ( y = 0) = 0 . \n(14.4.7) \n14-10", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_376", "text": "14.4.2 Hooke\u2019s Law Spring-Object System \nConsider a spring-object system lying on a frictionless horizontal surface with one end of \nthe spring fixed to a wall and the other end attached to an object of mass m (Figure \n14.7). The spring force is an internal conservative force. The wall exerts an external force \non the spring-object system but since the point of contact of the wall with the spring \nundergoes no displacement, this external force does no work. \nFigure 14.7 A spring-object system. \nChoose the origin at the position of the center of the object when the spring is \nrelaxed (the equilibrium position). Let x be the displacement of the object from the \norigin. We choose the +\u02c6i unit vector to point in the direction the object moves when the \nspring is being stretched (to the right of x = 0 in the figure). The spring force on a mass \n\uf072\n\uf072\nis then given by Fs = Fx\ns \u02c6i = \u2212kx \u02c6i . The displacement is dr = dx \u02c6i . The scalar product is \n\uf072\n\uf072\nF\u22c5 dr = \u2212kx \u02c6i \u22c5 dx \u02c6i = \u2212kx dx . The work done by the spring force on the mass is \nx=x f\nx=x f \uf072 \nW s \n\uf072 \n1\n1 \n1 \n= \u222b F\u22c5 dr = \u2212 2 \u222b\u2212 (\u2212kx) dx = \u2212 k(x f \n2 \u2212 xi \n2) . \n(14.4.8)\n2 \n2 \nx=xi \nx=xi \nWe then define the change in potential energy in the spring-object system in moving the \nobject from an initial position \nfrom equilibrium to a final position x f \nfrom\nxi \nequilibrium by \n1 \n2\n\u0394U s \n) = \u2212W s = k(x\n2) . \n(14.4.9)\n\u2261 U s(x f ) \u2212 U s(xi \nf \u2212 xi\n2 \nTherefore an arbitrary stretch or compression of a spring-object system from equilibrium \n= 0 to a final position x f = x changes the potential energy by \n\u0394U s = U s(x f ) \u2212 U s(0) = 1 k x2 . \n(14.4.10)\n2 \nxi \n14-11", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_377", "text": "For the spring-object system, there is an obvious choice of position where the potential \nenergy is zero, the equilibrium position of the spring- object, \nU s(0) \u2261 0 . \n(14.4.11) \nThen with this choice of zero reference potential, the potential energy as a function of the \ndisplacement x from the equilibrium position is given by \nU s(x) = 1 k x2, with U s(0) \u2261 0 . \n(14.4.12)\n2 \n14.4.3 Inverse Square Gravitation Force \nConsider a system consisting of two objects of masses m1 and m2 that are separated by a \ncenter-to-center distance \nA coordinate system is shown in the Figure 14.8. The\nr2,1 . \ninternal gravitational force on object 1 due to the interaction between the two objects is \ngiven by \n\uf072 G \nG m1 m2\n= \u2212 \n\u02c6r2,1 . \n(14.4.13)\nF2,1 \n2r2,1 \n\uf072 \n\u02c6\nThe displacement vector is given by dr\nr . So the scalar product is\n2,1 = dr2,1 \n2,1 \n\uf072 G \n\uf072 \nG m1 m2 \nG m1 m2\n\u22c5 d \n= \u2212 \n\u02c6 \u22c5 dr2,1 \u02c6 \n= \u2212 \ndr2,1 . \n(14.4.14)\nF2 1, \nr2,1 \n2 \nr2,1 \nr2,1 \n2\nr2,1 \nr2,1 \nFigure 14.8 Gravitational interaction \nUsing our definition of potential energy (Eq. (14.3.4)), we have that the change in the \ngravitational potential energy of the system in moving the two objects from an initial \nposition in which the center of mass of the two objects are a distance ri apart to a final \nposition in which the center of mass of the two objects are a distance rf apart is given by \n14-12", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_378", "text": "B \uf072 \nf \nG m m \nG m m \nrf \nG m m \nG m m \n\u0394U G \nG \n\uf072 \n1\n2 \n1\n2 \n1\n2 \n1\n2\n= \u2212 \n+ \n. (14.4.15)\n= \u2212\n\u22c5 d \n= \u2212\u2212 \n= \u2212\n\u222bF2,1 \nr2,1 \n\u222b \n2 \ndr2,1 \nr\nA \nri \n2,1 \nr2,1 \nrf \nri\nri \nWe now choose our reference point for the zero of the potential energy to be at infinity, \n= \u221e , with the choice that U G (\u221e) \u2261 0 . By making this choice, the term 1/ r in the \nri \nexpression for the change in potential energy vanishes when \n= \u221e . The gravitational\nri \npotential energy as a function of the relative distance r between the two objects is given \nby \nG m1 m2\nU G (r) = \u2212 \n, with U G (\u221e) \u2261 0 . \n(14.4.16)\nr \n14.5 Mechanical Energy and Conservation of Mechanical Energy \nThe total change in the mechanical energy of the system is defined to be \nthe sum of the changes of the kinetic and the potential energies, \n\u0394E = \u0394K + \u0394U\n. \n(14.4.17)\nm \nsys \nsys \nFor a closed system with only conservative internal forces, the total change in the \nmechanical energy is zero, \n\u0394E = \u0394K + \u0394U \n= 0 . \n(14.4.18)\nm \nsys \nsys \nEquation (14.4.18) is the symbolic statement of what is called conservation of \nmechanical energy. Recall that the work done by a conservative force in going around a \nclosed path is zero (Equation (14.2.16)), therefore both the changes in kinetic energy and \npotential energy are zero when a closed system with only conservative internal forces \nreturns to its initial state. Throughout the process, the kinetic energy may change into \ninternal potential energy but if the system returns to its initial state, the kinetic energy is \ncompletely recoverable. We shall refer to a closed system in which processes take place \nin which only conservative forces act as completely reversible processes. \n14.5.1 Change in Gravitational potential Energy Near Surface of the Earth \nLet\u2019s consider the example of an object of mass mo falling near the surface of the earth \n(mass me ). Choose our system to consist of the earth and the object. The gravitational \nforce is now an internal conservative force acting inside the system. The initial and final \nstates are specified by the distance separating the object and the center of mass of the", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_379", "text": "(mass me ). Choose our system to consist of the earth and the object. The gravitational \nforce is now an internal conservative force acting inside the system. The initial and final \nstates are specified by the distance separating the object and the center of mass of the \nearth, and the velocities of the earth and the object. The change in kinetic energy between \nthe initial and final states for the system is \n\u0394K \n= \u0394K + \u0394K , \n(14.4.19)\nsys \ne\no \n14-13", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_380", "text": "\u239b 1 \n\u239e\n\u239b 1 \n\u239e\n)2 \u2212 1 \n)2 \n)2 \u2212 1 \n)2\n\u0394K sys = \u239d\u239c m e(ve, f \nm e(ve,i \u23a0\u239f + \u239d\u239c mo (vo, f \nmo (vo,i \u23a0\u239f . \n(14.4.20)\n2\n2 \n2\n2 \nThe change of kinetic energy of the earth due to the gravitational interaction between the \nearth and the object is negligible. The change in kinetic energy of the system is \napproximately equal to the change in kinetic energy of the object, \n1 \n)2 \u2212 1 \n)2\n\u0394K \n\u2245\u0394K = m (v \nm (v \n. \n(14.4.21)\nsys \no \no\no, f \no\no,i\n2\n2 \nWe now define the mechanical energy function for the system \nE = K +U g = 1 \n)2 + m gy, with U g (0) = 0 , \n(14.4.22)\nm \nm o (vb \no\n2 \nwhere K is the kinetic energy and U g is the potential energy. The change in mechanical \nenergy is then \n\u0394E \u2261 Em, f \u2212 E \n= (K + U g ) \u2212 (K + U g ) . \n(14.4.23)\nm \nm, i \nf \nf \ni \ni \nWhen the work done by the external forces is zero and there are no internal non\u00ad\nconservative forces, the total mechanical energy of the system is constant, \nE \n= E \n(14.4.24)\nm, f\nm, i , \nor equivalently \n(14.4.25)\n(K f + U f ) = (Ki + Ui ). \n14.6 Spring Force Energy Diagram \n\uf072 \nThe spring force on an object is a restoring force Fs = Fx\ns \u02c6i = \u2212k x \u02c6i where we choose a \ncoordinate system with the equilibrium position at xi = 0 and x is the amount the spring \nhas been stretched (x > 0) or compressed (x < 0) from its equilibrium position. We \ncalculate the potential energy difference Eq. (14.4.9) and found that \nU s (x) \u2212 U s (xi ) = \u2212\u222b\nx Fx\ns dx = 1 k(x2 \u2212 xi \n2) . \n(14.5.1)\nxi \n2 \nThe first fundamental theorem of calculus states that \nx\u2032= x dU\nU (x) \u2212 U (xi ) = \ndx\u2032 . \n(14.5.2)\n\u222bx\u2032= xi dx\u2032 \n14-14", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_381", "text": "Comparing Equation (14.5.1) with Equation (14.5.2) shows that the force is the negative \nderivative (with respect to position) of the potential energy, \nF s \ndU s (x)\n= \u2212 \n. \n(14.5.3)\nx \ndx \nChoose the zero reference point for the potential energy to be at the equilibrium position, \nU s (0) \u2261 0 . Then the potential energy function becomes \nU s (x) = 1 k x2 . \n(14.5.4)\n2 \nFrom this, we obtain the spring force law as \ns \ndU s (x) \nd \u239b 1 \n\u239e\nF = \u2212 \n= \u2212 \nk x2 \n\u23a0\u239f = \u2212k x . \n(14.5.5)\nx \ndx \ndx \u239d\u239c 2 \nIn Figure 14.9 we plot the potential energy function U s (x) for the spring force as \nfunction of x with U s (0) \u2261 0 (the units are arbitrary). \nFigure 14.9 Graph of potential energy function as function of x for the spring. \nThe minimum of the potential energy function occurs at the point where the first \nderivative vanishes \ndU s (x) \ndx \n= 0 . \n(14.5.6) \nFrom Equation (14.5.4), the minimum occurs at x = 0 , \n0 = dU s (x) \ndx \n= k x . \n(14.5.7) \n14-15", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_382", "text": "Because the force is the negative derivative of the potential energy, and this derivative \nvanishes at the minimum, we have that the spring force is zero at the minimum x = 0 \nagreeing with our force law, Fx\ns \n= 0 .\n= \u2212k x \nx =0\nx =0 \nThe potential energy function has positive curvature in the neighborhood of a \nminimum equilibrium point. If the object is extended a small distance x > 0 away from \nequilibrium, the slope of the potential energy function is positive, dU ( )\nx dx > 0 , hence \nthe component of the force is negative because Fx = \u2212 dU ( )\nx dx < 0 . Thus the object \nexperiences a restoring force towards the minimum point of the potential. If the object is \ncompresses with x < 0 then dU ( )\nx dx < 0 , hence the component of the force is positive, \nFx = \u2212 dU ( )\nx dx > 0 , and the object again experiences a restoring force back towards the \nminimum of the potential energy as in Figure 14.10. \nFigure 14.10 Stability diagram for the spring force. \nThe mechanical energy at any time is the sum of the kinetic energy \n( ) and the \nK x \npotential energy U s (x) \nEm = K(x) + U s (x) . \n(14.5.8) \nSuppose our spring-object system has no loss of mechanical energy due to dissipative \nforces such as friction or air resistance. Both the kinetic energy and the potential energy \nare functions of the position of the object with respect to equilibrium. The energy is a \nconstant of the motion and with our choice of U s (0) \u2261 0 , the energy can be either a \npositive value or zero. When the energy is zero, the object is at rest at the equilibrium \nposition. \nIn Figure 14.10, we draw a straight horizontal line corresponding to a non-zero \npositive value for the energy Em on the graph of potential energy as a function of x . The \nenergy intersects the potential energy function at two points {\u2212x \n, x } with x \n> 0 .\nmax \nmax \nmax \nThese points correspond to the maximum compression and maximum extension of the \nspring, which are called the turning points. The kinetic energy is the difference between \nthe energy and the potential energy, \n14-16", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_383", "text": "K(x) = Em \u2212 U s (x) . \n(14.5.9) \nAt the turning points, where Em = U s (x) , the kinetic energy is zero. Regions where the \nkinetic energy is negative, x < \u2212x \nor x > x \nare called the classically forbidden \nmax \nmax \nregions, which the object can never reach if subject to the laws of classical mechanics. In \nquantum mechanics, with similar energy diagrams for quantum systems, there is a very \nsmall probability that the quantum object can be found in a classically forbidden region. \nExample 14.1 Energy Diagram \nThe potential energy function for a particle of mass m , moving in the x -direction is \ngiven by \n3\n2 \u239e\n\u239b \u239b x \u239e\n\u239b x \u239e \nU (x) = \u2212U1 \u239c\n\u2212\n\u239f , \n(14.5.10)\n\u239c \u239d\u239c \n\u239d\u239c \n\u239d x1 \u23a0\u239f \nx1 \u23a0\u239f \u239f\u23a0 \nwhere U1 and x1 are positive constants and U (0) = 0 . (a) Sketch \nx \nas a function \nU ( ) /U1 \nof x / x1. (b) Find the points where the force on the particle is zero. Classify them as \nstable or unstable. Calculate the value of U (x) / U1 at these equilibrium points. (c) For \nenergies E that lies in 0 < E < (4 / 27)U1 find an equation whose solution yields the \nturning points along the x-axis about which the particle will undergo periodic motion. (d) \nSuppose E = (4 / 27)U1 and that the particle starts at x = 0 with speed v0 . Find v0 . \nSolution: a) Figure 14.11 shows a graph of U (x) vs. x , with the choice of values \n= 1.5 m ,\nx1 \n= 27 / 4 J , and E = 0.2 J .\nU1 \nFigure 14.11 Energy diagram for Example 14.1 \nb) The force on the particle is zero at the minimum of the potential which occurs at \n14-17", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_384", "text": "\u239b \n\u239e \n\u239b \n\u239e \n\u239b \n\u239e  \n(x) = \u2212 dU \ndx (x) = U1 \n3  \n2  \nx2 \u2212\nF  \n= 0 \n(14.5.11) \n\u239c\n\u239d  \n\u239f\n\u23a0  \nx\n\u239c\u239d \n\u239f\u23a0 \n\u239c\u239d \n\u239f\u23a0 \n3 \n2\nx \nx1 \nx1 \nwhich becomes \nx2 = (2x1 / 3)x . \n(14.5.12) \nWe can solve Eq. (14.5.12) for the extrema. This has two solutions \nx = (2x1 / 3) and x = 0 . \n(14.5.13) \nThe second derivative is given by \n\u239b \n\u239e \n\u239b \n\u239b \n\u239e \n\u239e \nd 2U \ndx2 (x) = \u2212U1 \n6  \n2  \n\u239f\u23a0  \nx \u2212\u239c\u239d  \n. \n(14.5.14) \n\u239c\n\u239d  \n\u239f\n\u23a0 \n\u239c\u239d \n\u239f\u23a0 \n3 \n2\nx1 \nx1 \nEvaluating the second derivative at x = (2x1 / 3) yields a negative quantity \n\u239b \n\u239e \n\u239b \n\u239e \n\u239b \n\u239e \nd 2U \ndx2 (x = (2x1 / 3)) = \u2212U1 \n2U \n2x1\n6  \n2  \n1\n\u2212 \n= \u2212  \n< 0 , \n(14.5.15) \n\u239c\n\u239d  \n\u239f\n\u23a0 \n\u239c\u239d \n\u239f\u23a0 \n\u239c\u239d \n\u239f\u23a0 \n3 \n2 \n2\n3 \nx1 \nx1 \nx1 \nindicating the solution x = (2x1 / 3) represents a local maximum and hence is an unstable point. \nAt x = (2x1 / 3) , the potential energy is given by the value U ((2x1 / 3)) = (4 / 27)U1 . Evaluating \nthe second derivative at x = 0 yields a positive quantity \n\u239b \n\u239e \n\u239b \n\u239b \n\u239e \n\u239e \nd 2U \ndx2 (x = 0) = \u2212U1 \n2U \n6  \n2  \n1\n\u239f\u23a0  \n0 \u2212 \u239c\u239d  \n> 0 , \n(14.5.16) \n\u239c\n\u239d  \n\u239f\n\u23a0  \n= \n\u239c\u239d \n\u239f\u23a0 \n3 \n2 \n2\nx1 \nx1 \nx1 \nindicating the solution x = 0 represents a local minimum and is a stable point. At the local \nminimum x = 0 , the potential energy U (0) = 0 . \nc) Consider a fixed value of the energy of the particle within the range \n4U1\nU (0) = 0 < E < U (2x1 / 3) = \n. \n(14.5.17)\n27 \nIf the particle at any time is found in the region x \n/ 3, where x and xb are the \na < x < xb < 2x1 \na \nturning points and are solutions to the equation \n14-18", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_385", "text": "\u239b\u239b x \u239e \n3 \n\u239b x \u239e \n2 \u239e \nE = U (x) = \u2212U1 \u239c\n\u2212\n\u239f . \n(14.5.18)\n\u239c \u239d\u239c \n\u239d\u239c \n\u239d x1 \u23a0\u239f \nx1 \u23a0\u239f \u239f\u23a0 \nthen the particle will undergo periodic motion between the values x \n. Within \na < x < xb \nthis region xa < x < xb , the kinetic energy is always positive because K(x) = E \u2212U (x) . \nThere is another solution x to Eq. (14.5.18) somewhere in the region x \n/ 3. If the \nc\nc > 2x1 \nparticle at any time is in the region x > xc then it at any later time it is restricted to the \nregion xc < x < +\u221e . \nFor E > U (2x1 / 3) = (4 / 27)U1, Eq. (14.5.18) has only one solution xd . For all values of \nx > xd , the kinetic energy is positive, which means that the particle can \u201cescape\u201d to \ninfinity but can never enter the region x < xd . \nFor E < U (0) = 0 , the kinetic energy is negative for the range \u2212\u221e < x < xe where xe \nsatisfies Eq. (14.5.18) and therefore this region of space is forbidden. \n(d) If the particle has speed v0 at x = 0 where the potential energy is zero, U (0) = 0 , the \nenergy of the particle is constant and equal to kinetic energy \n1\n2\nE = K(0) = \nmv0 . \n(14.5.19)\n2 \nTherefore \n1\n2\n(4 / 27)U1 = \nmv0 , \n(14.5.20)\n2 \nwhich we can solve for the speed \n8U1 / 27m .\nv0 = \n(14.5.21) \n14.7 Change of Mechanical Energy for Closed System with Internal \nNon-conservative Forces \nConsider a closed system (energy of the system is constant) that undergoes a \ntransformation from an initial state to a final state by a prescribed set of changes. \nWhenever the work done by a force in moving an object from an initial point to a \nfinal point depends on the path, the force is called a non-conservative force. \n14-19", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_386", "text": "Suppose the internal forces are both conservative and non-conservative. The work W \ndone by the forces is a sum of the conservative work W c , which is path-independent, and \nthe non-conservative work W nc , which is path-dependent, \nW = W c + W nc . \n(14.6.1) \nThe work done by the conservative forces is equal to the negative of the change in the \npotential energy \n\u0394U = \u2212W c . \n(14.6.2) \nSubstituting Equation (14.6.2) into Equation (14.6.1) yields \nW = \u2212\u0394U + W nc . \n(14.6.3) \nThe work done is equal to the change in the kinetic energy, \nW = \u0394K . \n(14.6.4) \nSubstituting Equation (14.6.4) into Equation (14.6.3) yields \n\u0394K = \u2212\u0394U + W nc . \n(14.6.5) \nwhich we can rearrange as \nW nc = \u0394K + \u0394U . \n(14.6.6) \nWe can now substitute Equation (14.6.4) into our expression for the change in the \nmechanical energy, Equation (14.4.17), with the result \nW nc = \u0394Em . \n(14.6.7) \nThe mechanical energy is no longer constant. The total change in energy of the system is \nzero, \n\u0394E \n= \u0394E \u2212 W = 0 . \n(14.6.8)\nsystem \nm \nnc \nEnergy is conserved but some mechanical energy has been transferred into non\u00ad\nrecoverable energy W nc . We shall refer to processes in which there is non-zero non\u00ad\nrecoverable energy as irreversible processes. \n14.7.1 Change of Mechanical Energy for a Non-closed System \nWhen the system is no longer closed but in contact with its surroundings, the change in \nenergy of the system is equal to the negative of the change in energy of the surroundings \n(Eq. (14.1.1)), \n14-20", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_387", "text": "\u0394Esystem = \u2212\u0394Esurroundings \n(14.6.9) \nIf the system is not isolated, the change in energy of the system can be the result of \nexternal work done by the surroundings on the system (which can be positive or negative) \nW\n= \u222b \nB \nF \n\uf072 \n\u22c5 dr \uf072 . \n(14.6.10)\next \next \nA \nThis work will result in the system undergoing coherent motion. Note that Wext > 0 if \nwork is done on the system ( \u0394Esurroundings < 0 ) and W < 0 if the system does work on the \next \nsurroundings ( \u0394Esurroundings > 0 ). If the system is in thermal contact with the surroundings, \nthen energy can flow into or out of the system. This energy flow due to thermal contact is \noften denoted by Q with the convention that Q > 0 if the energy flows into the system \n( \u0394Esurroundings < 0 ) and Q < 0 if the energy flows out of the system ( \u0394Esurroundings > 0 ). Then \nEq. (14.6.9) can be rewritten as \nW ext + Q = \u0394E sys \n(14.6.11) \nEquation (14.6.11) is also called the first law of thermodynamics. \nThis will result in either an increase or decrease in random thermal motion of the \nmolecules inside the system, There may also be other forms of energy that enter the \nsystem, for example radiative energy. \nSeveral questions naturally arise from this set of definitions and physical \nconcepts. Is it possible to identify all the conservative forces and calculate the associated \nchanges in potential energies? How do we account for non-conservative forces such as \nfriction that act at the boundary of the system? \n14.8 Dissipative Forces: Friction \nSuppose we consider an object moving on a rough surface. As the object slides it slows \ndown and stops. While the sliding occurs both the object and the surface increase in \ntemperature. The increase in temperature is due to the molecules inside the materials \nincreasing their kinetic energy. This random kinetic energy is called thermal energy. \nKinetic energy associated with the coherent motion of the molecules of the object has \nbeen dissipated into kinetic energy associated with random motion of the molecules \ncomposing the object and surface. \nIf we define the system to be just the object, then the friction force acts as an \nexternal force on the system and results in the dissipation of energy into both the block", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_388", "text": "been dissipated into kinetic energy associated with random motion of the molecules \ncomposing the object and surface. \nIf we define the system to be just the object, then the friction force acts as an \nexternal force on the system and results in the dissipation of energy into both the block \nand the surface. Without knowing further properties of the material we cannot determine \nthe exact changes in the energy of the system. \n14-21", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_389", "text": "Friction introduces a problem in that the point of contact is not well defined \nbecause the surface of contact is constantly deforming as the object moves along the \nsurface. If we considered the object and the surface as the system, then the friction force \nis an internal force, and the decrease in the kinetic energy of the moving object ends up as \nan increase in the internal random kinetic energy of the constituent parts of the system. \nWhen there is dissipation at the boundary of the system, we need an additional model \n(thermal equation of state) for how the dissipated energy distributes itself among the \nconstituent parts of the system. \n14.8.1 Source Energy \nConsider a person walking. The frictional force between the person and the ground does \nno work because the point of contact between the person\u2019s foot and the ground undergoes \nno displacement as the person applies a force against the ground, (there may be some \nslippage but that would be opposite the direction of motion of the person). However the \nkinetic energy of the object increases. Have we disproved the work-energy theorem? The \nanswer is no! The chemical energy stored in the body tissue is converted to kinetic \nenergy and thermal energy. Because the person-air-ground can be treated as a closed \nsystem, we have that \n0 = \u0394E sys = \u0394Echemical + \u0394Ethermal + \u0394Emechanical , (closed system) . \n(14.7.1) \nIf we assume that there is no change in the potential energy of the system, then \n= \u0394K . Therefore some of the internal chemical energy has been transformed \n\u0394Emechanical \ninto thermal energy and the rest has changed into the kinetic energy of the system, \n\u2212\u0394E \n= \u0394E \n+ \u0394K . \n(14.7.2)\nchemical \nthermal \n14.9 Worked Examples \nExample 14.2 Escape Velocity of Toro \nThe asteroid Toro, discovered in 1964, has a radius of about R = 5.0km and a mass of \nabout mt = 2.0 \u00d71015 kg . Let\u2019s assume that Toro is a perfectly uniform sphere. What is the \nescape velocity for an object of mass m on the surface of Toro? Could a person reach \nthis speed (on earth) by running? \nSolution: The only potential energy in this problem is the gravitational potential energy. \nWe choose the zero point for the potential energy to be when the object and Toro are an \ninfinite distance apart, U G (\u221e) \u2261 0 . With this choice, the potential energy when the object", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_390", "text": "this speed (on earth) by running? \nSolution: The only potential energy in this problem is the gravitational potential energy. \nWe choose the zero point for the potential energy to be when the object and Toro are an \ninfinite distance apart, U G (\u221e) \u2261 0 . With this choice, the potential energy when the object \nand Toro are a finite distance r apart is given by \nGmt m\nU G (r) = \u2212 \n(14.8.1)\nr \n14-22", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_391", "text": "with U G (\u221e) \u2261 0 . The expression escape velocity refers to the minimum speed necessary \nfor an object to escape the gravitational interaction of the asteroid and move off to an \ninfinite distance away. If the object has a speed less than the escape velocity, it will be \nunable to escape the gravitational force and must return to Toro. If the object has a speed \ngreater than the escape velocity, it will have a non-zero kinetic energy at infinity. The \ncondition for the escape velocity is that the object will have exactly zero kinetic energy at \ninfinity. \nWe choose our initial state, at time ti , when the object is at the surface of the asteroid \nwith speed equal to the escape velocity. We choose our final state, at time t f , to occur \nwhen the separation distance between the asteroid and the object is infinite. \nThe initial kinetic energy is \n= (1/ 2)mv \n2 . The initial potential energy is\nKi \nesc \n= \u2212Gm m / R , and so the initial mechanical energy is\nUi \nt \n1\n2 \nGmt m \n= mv \u2212 \n. \n(14.8.2)\nEi = Ki +Ui \nesc \n2 \nR \nThe final kinetic energy is K f = 0 , because this is the condition that defines the escape \nvelocity. The final potential energy is zero, U f = 0 because we chose the zero point for \npotential energy at infinity. The final mechanical energy is then \n= 0 . \n(14.8.3)\nE f = K f + U f \nThere is no non-conservative work, so the change in mechanical energy is zero \n0 = W = \u0394E \n(14.8.4)\nnc \nm = E f \u2212 Ei . \nTherefore \n\u239b 1\n2 \nGmt m\u239e\n0 = \u2212 \nmv esc \u2212 \n(14.8.5)\n\u239d\u239c 2 \nR \n\u23a0\u239f . \nThis can be solved for the escape velocity, \n(14.8.6) \n= 7.3 m \u22c5s\u22121. \nv esc = 2Gmt \nR \n= 2(6.67 \u00d710\u221211 N \u22c5m2 \u22c5kg\u22122 )(2.0 \u00d71015 kg) \n(5.0 \u00d7103 m) \n14-23", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_392", "text": "Considering that Olympic sprinters typically reach velocities of 12 m \u22c5 s\u22121, this is an easy \nspeed to attain by running on earth. It may be harder on Toro to generate the acceleration \nnecessary to reach this speed by pushing off the ground, since any slight upward force \nwill raise the runner\u2019s center of mass and it will take substantially more time than on \nearth to come back down for another push off the ground. \nExample 14.3 Spring-Block-Loop-the-Loop \nA small block of mass m is pushed against a spring with spring constant k and held in \nplace with a catch. The spring is compressed an unknown distance x (Figure 14.12). \nWhen the catch is removed, the block leaves the spring and slides along a frictionless \ncircular loop of radius r . When the block reaches the top of the loop, the force of the \nloop on the block (the normal force) is equal to twice the gravitational force on the mass. \n(a) Using conservation of energy, find the kinetic energy of the block at the top of \nthe loop. (b) Using Newton\u2019s Second Law, derive the equation of motion for the block \nwhen it is at the top of the loop. Specifically, find the speed vtop in terms of the \ngravitation constant g and the loop radius r . (c) What distance was the spring \ncompressed? \nFigure 14.12 Initial state for spring-block-loop-the-loop system \nSolution: a) Choose for the initial state the instant before the catch is released. The initial \nkinetic energy is Ki = 0 . The initial potential energy is non-zero, Ui = (1/ 2)k x2 . The \ninitial mechanical energy is then \n1\nEi = Ki + Ui = \nk x2 . \n(14.8.7)\n2 \nChoose for the final state the instant the block is at the top of the loop. The final kinetic \nenergy is K f = (1/ 2)mv2 ; the block is in motion with speed v . The final potential \ntop \ntop \nenergy is non-zero, U f = (mg)(2R) . The final mechanical energy is then \n= 2mgR + 1\n2 mv2 . \n(14.8.8)\nE f = K f + U f \ntop \nBecause we are assuming the track is frictionless and neglecting air resistance, there is no \nnon- conservative work. The change in mechanical energy is therefore zero, \n14-24", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_393", "text": "0 = W nc = \u0394Em = E f \u2212 Ei . \n(14.8.9) \nMechanical energy is conserved, E f \n, therefore\n= Ei \n1 \n2\n1\n2mgR + mvtop = k x2 . \n(14.8.10)\n2\n2 \nFrom Equation (14.8.10), the kinetic energy at the top of the loop is \n1 \n2\n1 \nmvtop = k x2 \u2212 2mgR . \n(14.8.11)\n2\n2 \nb) At the top of the loop, the forces on the block are the gravitational force of magnitude \nmg and the normal force of magnitude N , both directed down. Newton\u2019s Second Law \nin the radial direction, which is the downward direction, is \n2\nmv \n\u2212mg \u2212 N = \u2212 \ntop . \n(14.8.12)\nR \nIn this problem, we are given that when the block reaches the top of the loop, the force of \nthe loop on the block (the normal force, downward in this case) is equal to twice the \nweight of the block, N = 2mg . The Second Law, Eq. (14.8.12), then becomes \n2\nmv \n3mg = \ntop . \n(14.8.13)\nR \nWe can rewrite Equation (14.8.13) in terms of the kinetic energy as \n3 \n1\n2\nmg R = mvtop . \n(14.8.14)\n2\n2 \nThe speed at the top is therefore \nvtop = 3mg R . \n(14.8.15) \nc) Combing Equations (14.8.11) and (14.8.14) yields \n7 mg R = 1 k x2 . \n(14.8.16)\n2\n2 \nThus the initial displacement of the spring from equilibrium is \n14-25", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_394", "text": "7mg R \nx \nk \n= \n.  \n(14.8.17)  \nExample 14.4 Mass-Spring on a Rough Surface \nA block of mass m slides along a horizontal table with speed v0 . At x = 0 it hits a \nspring with spring constant k and begins to experience a friction force. The coefficient of \nfriction is variable and is given by \u00b5 = bx , where b is a positive constant. Find the loss \nin mechanical energy when the block first momentarily comes to rest. \nFigure 14.13 Spring-block system \nSolution: From the model given for the frictional force, we could find the non\u00ad\nconservative work done, which is the same as the loss of mechanical energy, if we knew \nthe position x f where the block first comes to rest. The most direct (and easiest) way to \nfind x f is to use the work-energy theorem. The initial mechanical energy is Ei = mvi \n2 / 2 \nand the final mechanical energy is E f = k x 2 \nf / 2 (note that there is no potential energy \nterm in \nand no kinetic energy term in \n). The difference between these two \nEi  \nE f \nmechanical energies is the non-conservative work done by the frictional force, \nx=x\nx=x\nx=x\nf\nf  \nf \nW = \nF dx = \n\u2212F \ndx = \n\u2212\u00b5 N dx \nnc  \n\u222b \nnc \n\u222b \nfriction \n\u222b \nx=0 \nx=0 \nx=0 \n(14.8.18) \nx \n1\n2\n= \u2212 \nf b xmg dx = \u2212 bmg x f .\n\u222b0\n2 \nWe then have that \nW = \u0394E \nnc \nm \nW  \n(14.8.19)\nnc = E f \u2212 Ei \n1 \n2\n1 \n1\n2\n\u2212 bmg x f = k x 2 \u2212\nf \nmvi .\n2 \n2\n2 \nSolving the last of these equations for x2 \nf yields \n14-26", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_395", "text": "mv0\n2 \nx2 = \n. \n(14.8.20)\nf\nk + bmg \nSubstitute Eq. (14.8.20) into Eq. (14.8.18) gives the result that \nbmg \nmv 2 \nmv 2 \u239b \nk \u239e\n\u22121 \nWnc = \u2212 \n0 \n= \u2212 \n0 \u239c1+ \n\u239f . \n(14.8.21)\n2 k + bmg \n2 \u239d \nbmg \u23a0 \nIt is worth checking that the above result is dimensionally correct. From the model, the \nparameter b must have dimensions of inverse length (the coefficient of friction \u00b5 must \nbe dimensionless), and so the product bmg has dimensions of force per length, as does \nthe spring constant k ; the result is dimensionally consistent. \nExample 14.5 Cart-Spring on an Inclined Plane \nAn object of mass m slides down a plane that is inclined at an angle \u03b8 from the \nhorizontal (Figure 14.14). The object starts out at rest. The center of mass of the cart is a \ndistance d from an unstretched spring that lies at the bottom of the plane. Assume the \nspring is massless, and has a spring constant k . Assume the inclined plane to be \nfrictionless. (a) How far will the spring compress when the mass first comes to rest? (b) \nNow assume that the inclined plane has a coefficient of kinetic friction \u00b5k . How far will \nthe spring compress when the mass first comes to rest? The friction is primarily between \nthe wheels and the bearings, not between the cart and the plane, but the friction force may \nbe modeled by a coefficient of friction \u00b5k . (c) In case (b), how much energy has been \nlost to friction? \nFigure 14.14 Cart on inclined plane \nSolution: Let x denote the displacement of the spring from the equilibrium position. \nChoose the zero point for the gravitational potential energy U g (0) = 0 not at the very \nbottom of the inclined plane, but at the location of the end of the unstretched spring. \nChoose the zero point for the spring potential energy where the spring is at its \nequilibrium position, U s (0) = 0 . \na) Choose for the initial state the instant the object is released (Figure 14.15). The initial \nkinetic energy is Ki = 0 . The initial potential energy is non-zero, Ui = mg d sin\u03b8 . The \ninitial mechanical energy is then \n14-27", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_396", "text": "= mg d sin\u03b8 \n(14.8.22)\nEi = Ki + Ui \nChoose for the final state the instant when the object first comes to rest and the spring is \ncompressed a distance x at the bottom of the inclined plane (Figure 14.16). The final \nkinetic energy is K f = 0 since the mass is not in motion. The final potential energy is \nnon-zero, \n= k x2 / 2 \u2212 x mg sin\u03b8 . Notice that the gravitational potential energy is \nU f \nnegative because the object has dropped below the height of the zero point of \ngravitational potential energy. \nFigure 14.15 Initial state \nFigure 14.16 Final state \nThe final mechanical energy is then \n= 1 k x2 \u2212 x mg sin\u03b8 . \n(14.8.23)\nE f = K f + U f \n2 \nBecause we are assuming the track is frictionless and neglecting air resistance, there is no \nnon- conservative work. The change in mechanical energy is therefore zero, \n0 = W nc = \u0394E m = E f \u2212 Ei . \n(14.8.24) \nTherefore \nd mg sin\u03b8 = 1 \n2 k x2 \u2212 x mg sin\u03b8 . \n(14.8.25) \nThis is a quadratic equation in x , \nx2 \u2212 2mg sin\u03b8 \nk \nx \u2212 2d mg sin\u03b8 \nk \n= 0 . \n(14.8.26) \nIn the quadratic formula, we want the positive choice of square root for the solution to \nensure a positive displacement of the spring from equilibrium, \n14-28", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_397", "text": "1+ 2(k d / mg)sin\u03b8 ). \nmg sin\u03b8\n\u239b m2 g 2 sin2\u03b8 \n2d mg sin\u03b8 \u239e\n1 2 \nx =\n+ \n+ \n\u239d\u239c \nk 2 \n\u23a0\u239f\nk \nk \n(14.8.27) \nmg \n= \n(sin\u03b8 +\nk \n(What would the solution with the negative root represent?) \nb) The effect of kinetic friction is that there is now a non-zero non-conservative work \ndone on the object, which has moved a distance, d + x , given by \nW \n(d + x) = \u2212\u00b5k N (d + x) = \u2212\u00b5k mg cos\u03b8(d + x) . \n(14.8.28)\nnc = \u2212 fk \nNote the normal force is found by using Newton\u2019s Second Law in the perpendicular direction \nto the inclined plane, \nN \u2212 mg cos\u03b8 = 0 . \n(14.8.29) \nThe change in mechanical energy is therefore \nW = \u0394E \n(14.8.30)\nnc \nm = E f \u2212 Ei , \nwhich becomes \n\u239b 1 \n\u239e\n\u2212\u00b5k mg cos\u03b8(d + x) = \nk x2 \u2212 x mg sin\u03b8\u23a0\u239f \u2212 d mg sin\u03b8 . \n(14.8.31)\n\u239d\u239c 2 \nEquation (14.8.31) simplifies to \n\u239b 1 \n\u239e\n0 = \nk x2 \u2212 x mg(sin\u03b8 \u2212 \u00b5k cos\u03b8 )\u23a0\u239f \u2212 d mg(sin\u03b8 \u2212 \u00b5k cos\u03b8 ) . \n(14.8.32)\n\u239d\u239c 2 \nThis is the same as Equation (14.8.25) above, but with sin\u03b8 \u2192 sin\u03b8 \u2212 \u00b5k cos\u03b8 . The \nmaximum displacement of the spring is when there is friction is then \nmg \nx = \n((sin\u03b8 \u2212 \u00b5k cos\u03b8 ) + 1+ 2(k d / mg)(sin\u03b8 \u2212 \u00b5k cos\u03b8 )) . \n(14.8.33)\nk \n. \nc) The energy lost to friction is given by W \nmg cos\u03b8(d + x) , where x is given in \nnc = \u2212\u00b5k \npart b). \nExample 14.6 Object Sliding on a Sphere \nA small point like object of mass m rests on top of a sphere of radius R . The object is \nreleased from the top of the sphere with a negligible speed and it slowly starts to slide \n(Figure 14.17). Let g denote the gravitation constant. (a) Determine the angle \u03b81 with \n14-29", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_398", "text": "respect to the vertical at which the object will lose contact with the surface of the sphere. \n(b) What is the speed v1 of the object at the instant it loses contact with the surface of the \nsphere. \nFigure 14.17 Object sliding on surface of sphere \nSolution: We begin by identifying the forces acting on the object. There are two forces \nacting on the object, the gravitation and radial normal force that the sphere exerts on the \nparticle that we denote by N . We draw a free-body force diagram for the object while it \nis sliding on the sphere. We choose polar coordinates as shown in Figure 14.18. \nFigure 14.18 Free-body force diagram on object \nThe key constraint is that when the particle just leaves the surface the normal force is \nzero, \nN (\u03b81) = 0 , \n(14.8.34) \nwhere \u03b81 denotes the angle with respect to the vertical at which the object will just lose \ncontact with the surface of the sphere. Because the normal force is perpendicular to the \ndisplacement of the object, it does no work on the object and hence conservation of \nenergy does not take into account the constraint on the motion imposed by the normal \nforce. In order to analyze the effect of the normal force we must use the radial component \nof Newton\u2019s Second Law, \n2v\nN \u2212 mg cos\u03b8 = \u2212m \n. \n(14.8.35)\nR \nThen when the object just loses contact with the surface, Eqs. (14.8.34) and (14.8.35) \nrequire that \n14-30", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_399", "text": "v1\n2 \nmg cos\u03b81 = m \n. \n(14.8.36)\nR \nwhere v1 denotes the speed of the object at the instant it loses contact with the surface of \nthe sphere. Note that the constrain condition Eq. (14.8.36) can be rewritten as \nmgRcos\u03b81 = mv1\n2 . \n(14.8.37) \nWe can now apply conservation of energy. Choose the zero reference point U = 0 for \npotential energy to be the midpoint of the sphere. \nIdentify the initial state as the instant the object is released (Figure 14.19). We can \nneglect the very small initial kinetic energy needed to move the object away from the top \nof the sphere and so Ki = 0 . The initial potential energy is non-zero, Ui = mgR . The \ninitial mechanical energy is then \n= mgR . \n(14.8.38)\nEi = Ki + Ui \nFigure 14.19 Initial state \nFigure 14.20 Final state \nChoose for the final state the instant the object leaves the sphere (Figure 14.20). The final \nkinetic energy is K f = mv1\n2 / 2 ; the object is in motion with speed v1 . The final potential \nenergy is non-zero, U f = mgRcos\u03b81 . The final mechanical energy is then \n1\nE f = K f + U f = \nmv1 \n2 + mgRcos\u03b81 . \n(14.8.39)\n2 \nBecause we are assuming the contact surface is frictionless and neglecting air resistance, \nthere is no non-conservative work. The change in mechanical energy is therefore zero, \n0 = W = \u0394E \n(14.8.40)\nnc \nm = E f \u2212 Ei . \nTherefore \n1 mv1\n2 + mgRcos\u03b81 = mgR . \n(14.8.41)\n2 \n14-31\n\nWe now solve the constraint condition Eq. (14.8.37) into Eq. (14.8.41) yielding \n1 mgRcos\u03b81 + mgRcos\u03b81 = mgR . \n(14.8.42)\n2 \nWe can now solve for the angle at which the object just leaves the surface \n= cos\u22121(2 / 3) . \n(14.8.43)\n\u03b81 \nWe now substitute this result into Eq. (14.8.37) and solve for the speed \n= 2gR / 3 . \n(14.8.44)\nv1 \n14-32", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_400", "text": "Chapter 15 Collision Theory \n15.1 Introduction........................................................................................................... 1  \n15.2 Reference Frames and Relative Velocities.......................................................... 1  \n15.2.1 Relative Velocities .......................................................................................... 3  \n15.2.2 Center-of-mass Reference Frame................................................................. 4  \n15.2.3 Kinetic Energy in the Center-of-Mass Reference Frame........................... 5  \n15.2.4 Change of Kinetic Energy and Relatively Inertial Reference Frames...... 5", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_401", "text": "15.2.4 Change of Kinetic Energy and Relatively Inertial Reference Frames...... 5  \n15.3 Characterizing Collisions ..................................................................................... 7  \n15.4 One-Dimensional Collisions Between Two Objects........................................... 7  \n15.4.1 One Dimensional Elastic Collision in Laboratory Reference Frame........ 7  \n15.4.2 One-Dimensional Collision Between Two Objects \u2013 Center-of-Mass  \nReference Frame ..................................................................................................... 11  \n15.5 Worked Examples............................................................................................... 12  \nExample 15.1 Elastic One-Dimensional Collision Between Two Objects.......... 12  \nExample 15.2 The Dissipation of Kinetic Energy in a Completely Inelastic", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_402", "text": "Example 15.1 Elastic One-Dimensional Collision Between Two Objects.......... 12  \nExample 15.2 The Dissipation of Kinetic Energy in a Completely Inelastic  \nCollision Between Two Objects ............................................................................. 13  \nExample 15.3 Bouncing Superballs ....................................................................... 14  \n15.6 Two Dimensional Elastic Collisions .................................................................. 17  \n15.6.1 Two-dimensional Elastic Collision in Laboratory Reference Frame...... 17  \nExample 15.5 Elastic Two-dimensional collision of identical particles.............. 20  \nExample 15.6 Two-dimensional elastic collision between particles of equal mass  \n................................................................................................................................... 21  \nExample 15.7 Two dimensional collision between particles of unequal mass ... 22", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_403", "text": "Example 15.7 Two dimensional collision between particles of unequal mass ... 22  \n15.7 Two-Dimensional Collisions in Center-of-Mass Reference Frame ................ 24  \n15.7.1 Two-Dimensional Collision in Center-of-Mass Reference Frame........... 24  \n15.7.2 Scattering in the Center-of-Mass Reference Frame ................................. 25  \nExample 15.8 Scattering in the Lab and CM Frames ......................................... 26", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_404", "text": "Chapter 15 Collision Theory \nDespite my resistance to hyperbole, the LHC [Large Hadron Collider] \nbelongs to a world that can only be described with superlatives. It is not \nmerely large: the LHC is the biggest machine ever built. It is not merely \ncold: the 1.9 kelvin (1.9 degrees Celsius above absolute zero) temperature \nnecessary for the LHC\u2019s supercomputing magnets to operate is the coldest \nextended region that we know of in the universe\u2014even colder than outer \nspace. The magnetic field is not merely big: the superconducting dipole \nmagnets generating a magnetic field more than 100,000 times stronger than \nthe Earth\u2019s are the strongest magnets in industrial production ever made. \nAnd the extremes don\u2019t end there. The vacuum inside the proton-containing \ntubes, a 10 trillionth of an atmosphere, is the most complete vacuum over \nthe largest region ever produced. The energy of the collisions are the highest \never generated on Earth, allowing us to study the interactions that occurred \nin the early universe the furthest back in time.1 \nLisa Randall \n15.1 Introduction \nWhen discussing conservation of momentum, we considered examples in which two \nobjects collide and stick together, and either there are no external forces acting in some \ndirection (or the collision was nearly instantaneous) so the component of the momentum \nof the system along that direction is constant. We shall now study collisions between \nobjects in more detail. In particular we shall consider cases in which the objects do not \nstick together. The momentum along a certain direction may still be constant but the \nmechanical energy of the system may change. We will begin our analysis by considering \ntwo-particle collision. We introduce the concept of the relative velocity between two \nparticles and show that it is independent of the choice of reference frame. We then show \nthat the change in kinetic energy only depends on the change of the square of the relative \nvelocity and therefore is also independent of the choice of reference frame. We will then \nstudy one- and two-dimensional collisions with zero change in potential energy. In \nparticular we will characterize the types of collisions by the change in kinetic energy and \nanalyze the possible outcomes of the collisions. \n15.2 Reference Frames and Relative Velocities \n\uf072 \nWe shall recall our definition of relative inertial reference frames. Let R be the \nvector from the origin of frame S to the origin of reference frame S\u2032 . Denote the", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_405", "text": "analyze the possible outcomes of the collisions. \n15.2 Reference Frames and Relative Velocities \n\uf072 \nWe shall recall our definition of relative inertial reference frames. Let R be the \nvector from the origin of frame S to the origin of reference frame S\u2032 . Denote the \n1 Randall, Lisa, Knocking on Heaven's Door: How Physics and Scientific Thinking Illuminate the Universe \nand the Modern World, Ecco, 2011. \n15-1", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_406", "text": "\uf072\nposition vector of the jth particle with respect to the origin of reference frame S by rj \njth \nand similarly, denote the position vector of the \nparticle with respect to the origin of \n\uf072\nreference frame S\u2032 by r\u2032 \nj (Figure 15.1). \nS \nrj \njth particle \nrj \nS \nR \nFigure 15.1 Position vector of jth particle in two reference frames. \nThe position vectors are related by\n\uf072rj \n\uf072 \nj \n= r\u2032 +  \n\uf072  \nR .  \n(15.2.1)  \nThe relative velocity (call this the boost velocity) between the two reference frames is \ngiven by \n\uf072\n\uf072 \ndR\nV = \n. \n(15.2.2)\ndt \nAssume the boost velocity between the two reference frames is constant. Then, the \nrelative acceleration between the two reference frames is zero, \n\uf072\n\uf072\n\uf072\ndV\nA =\n= 0 . \n(15.2.3)\ndt \nWhen Eq. (15.2.3) is satisfied, the reference frames S and S\u2032 are called relatively \ninertial reference frames. \nSuppose the jth particle in Figure 15.1 is moving; then observers in different \nreference frames will measure different velocities. Denote the velocity of jth particle in \n\uf072\n\uf072\nframe S by v j = drj / dt , and the velocity of the same particle in frame S\u2032 by \n\uf072\n\uf072\nv\u2032 j = drj \u2032 / dt . Taking derivative, the velocities of the particles in two different reference \nframes are related according to \n\uf072\n\uf072\n\uf072 \nv j = v\u2032 j + V . \n(15.2.4) \n15-2", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_407", "text": "15.2.1 Relative Velocities \nConsider two particles of masses m1 and m2 interacting via some force (Figure 15.2). \nFigure 15.2 Two interacting particles  \nChoose a coordinate system (Figure 15.3) in which the position vector of body 1 is given  \n\uf072\n\uf072\nby r1 and the position vector of body 2 is given by r2 . The relative position of body 1 \n\uf072 \n\uf072\n\uf072\nwith respect to body 2 is given by r1 2 = r1 \u2212 r2 . \n, \n\uf072\n\uf072 \nFigure 15.3 Coordinate system for two bodies. \n\uf072\nDuring the course of the interaction, body 1 is displaced by dr1 and body 2 is displaced \n\uf072\nby dr2 , so the relative displacement of the two bodies during the interaction is given by \n\uf072 \n\uf072\n\uf072\ndr1 2 = dr1 \u2212 dr2 . The relative velocity between the particles is\n, \n\uf072 \n\uf072\n\uf072\n\uf072 \ndr1 2,\ndr1 \ndr2 \n\uf072\n\uf072 \nv1 2 =\n= \n\u2212 \n= v1 \u2212 v2 . \n(15.2.5)\n, \ndt \ndt \ndt \nWe shall now show that the relative velocity between the two particles is independent of \nthe choice of reference frame providing that the reference frames are relatively inertial. \n\uf072\nThe relative velocity v\u2032 in reference frame S\u2032 can be determined from using Eq. \n12 \n(15.2.4) to express Eq. (15.2.5) in terms of the velocities in the reference frame S\u2032 , \n= (v \n\uf072 \n\uf072  \nv1, 2 = v1 \u2212 \uf072  \nv2 \n\uf072\n1 \u2032 + V) \u2212 (v2 \n\uf072 \n\uf072 \n\uf072 \u2032v1 \u2212 v\u2032 \n2 =  \uf072 \n\u2032  +  V) =  \n\u2032 \n(15.2.6)\nv1, 2 \n15-3", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_408", "text": "and is equal to the relative velocity in frame S . \nFor a two-particle interaction, the relative velocity between the two \nvectors is independent of the choice of relatively inertial reference frames. \n15.2.2 Center-of-mass Reference Frame \n\uf072\nLet r cm be the vector from the origin of frame S to the center-of-mass of the \nsystem of particles, a point that we will choose as the origin of reference frame S cm , \ncalled the center-of-mass reference frame. Denote the position vector of the jth particle \n\uf072\nwith respect to origin of reference frame S by \nand similarly, denote the position\nrj \njth  \n\uf072\nvector of the \nparticle with respect to origin of reference frame S \nby \n(Figure\ncm \nr\u2032 \nj \n15.4). \nS cm \nr cm \nrj \njth particle \nrj \nS ! \nFigure 15.4 Position vector of jth particle in the center-of-mass reference frame. \nThe position vector of the jth particle in the center-of-mass frame is then given by \n\uf072\n\uf072 \u2212 \uf072 \nr\u2032 \nr .  \n(15.2.7)\nj = rj \ncm \nThe velocity of the jth particle in the center-of-mass reference frame is then given by \n\uf072\n\uf072 \u2212 \uf072 \nv\u2032 j = v j \nv cm .  \n(15.2.8) \nThere are many collision problems in which the center-of-mass reference frame is the \nmost convenient reference frame to analyze the collision. \nConsider a system consisting of two particles, which we shall refer to as particle 1 and \nparticle 2. We can use Eq. (15.2.8) to determine the velocities of particles 1 and 2 in the \ncenter-of-mass, \n\uf072\n\uf072 \n\uf072\n\uf072\n\uf072 \n\uf072 \nm1\n+ m2\nm2  \n\u00b5 \uf072 \nv1 \u2032 = v1 \u2212 v cm = v1 \u2212  v1 \nv2 = \n(v\uf072 \n1, \u2212 v\uf072 \n2 ) = \nv1, 2 . \n(15.2.9)\nm1 + m2 \nm1 + m2 \nm1 \n15-4", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_409", "text": "where \uf072v12 =  \uf072v1 \u2212 \uf072v2 is the relative velocity of particle 1 with respect to particle 2 . A \nsimilar result holds for particle 2 : \n\uf072\n\uf072 \n\uf072\n\uf072\n\uf072 \n\uf072 \nm1\n+ m2\nm1  \n\uf072 \nv\u2032 \n2 = v2 \u2212 v cm = v2 \u2212  v1 \nv2 = \u2212 \n(v\uf072 \n1 \u2212 v\uf072 \n2 ) = \u2212 \u00b5 v1, 2 . \n(15.2.10)\nm1 + m2 \nm1 + m2 \nm2 \nThe momentum of the system the center-of-mass reference frame is zero as we expect, \n\uf072 \n\uf072\n\uf072 \n\uf072\n\uf072 \nm1 \u2032 + m2v\u2032 = \u00b5 \n\u2212 \u00b5\n= 0 .  \n(15.2.11)\nv1 \n2 \nv12 \nv12 \n15.2.3 Kinetic Energy in the Center-of-Mass Reference Frame \nThe kinetic energy in the center of mass reference frame is given by \n1 \n\uf072\n\uf072 \n1 \n\uf072\n\uf072\nK = \n\u2032 \u22c5 v\u2032 +  \nv\u2032 \u22c5 v\u2032 \n(15.2.12)\ncm \nm1v1 \n1 \nm2 \n2 \n2 .\n2\n2 \nWe now use Eqs. (15.2.9) and (15.2.10) to rewrite the kinetic energy in terms of the\n\uf072 \n\uf072\n\uf072\nrelative velocity v\u2032 \n\u2032 \u2212 v\u2032 \n2 ,\n12 = v1 \n\u239b  \n\u239e\u239b  \n\u239e \n1 \n\u239b  \n\u239e\u239b \n\u23a0\u239f\u22c5 \u2212 \u00b5 \n\u239e \n1 \n\uf072v \n\uf072v \n\uf072v \n\uf072v\n\u00b5 \nm1 \n\u00b5 \nm1 \n\u2212 \u00b5 \nK  \n\u23a0\u239f \u22c5  \n\u239d\u239c  \n\u23a0\u239f + \nm1 \nm2 \n=  \n\u239d\u239c  \n\u239d\u239c  \n\u239d\u239c  \n\u23a0\u239f \n1, 2 \n1, 2 \n1, 2 \n1, 2 \n2  \n2 \ncm \nm2 \nm2 \n. \n(15.2.13) \n\u239b 1\n1 \u239e \n1  \n1 \n\uf072v \n\uf072v\n2 \n2\n\u22c5  \n+ \n\u00b5  \n\u00b5v1, 2 \nm2 \u23a0\u239f =\n=  \n\u239d\u239c  m1 \n1, 2 \n1, 2 \n2  \n2  \nwhere we used the fact that we defined the reduced mass by \n1\n1 \n1\n\u2261 \n+ \n.  \n(15.2.14)\n\u00b5 \nm1 \nm2 \n15.2.4 Change of Kinetic Energy and Relatively Inertial Reference Frames \nThe kinetic energy of the two particles in reference frame S is given by \n1 \n2 + 1 \n2\n= \n.  \n(15.2.15)\nKS \nm1v1 \nm2v2\n2\n2 \nWe can take the scalar product of Eq. (15.2.8) to rewrite Eq. (15.2.15) as \n15-5", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_410", "text": "1 \n\uf072\n\uf072 \n\uf072\n\uf072 \n1 \n\uf072\n\uf072 \n\uf072\n\uf072 \n= m1\n\u2032 + v \n\u2032 + v ) + m2(v\u2032 + v )\u22c5(v\u2032 + v )\nKS \n(v1 \ncm )\u22c5(v1 \ncm \n2 \ncm \n2 \ncm \n2\n2 \n. \n(15.2.16)\n1\n1 \n1 \n\uf072\n\uf072\n2 + (m1 \n\uf072 \n= \n\u2032 2+ \n\u2032 2+ \n)v \nv\u2032 )\u22c5 v\nm1v1 \nm2v2 \n(m1 + m2 \ncm \nv1 \u2032 + m2\n2 \ncm \n2\n2 \n2 \nThe last term is zero due to the fact that the momentum of the system in the center of \nmass reference frame is zero (Eq. (15.2.11)). Therefore Eq. (15.2.16) becomes \n1\n1\n1 \n2\n= \n\u2032 2+ \n\u2032 2+\n+ m2)v \n. \n(15.2.17)\nKS \nm1v1 \nm2v2 \n(m1 \ncm \n2\n2 \n2 \nThe first two terms correspond to the kinetic energy in the center of mass frame, thus the \nkinetic energies in the two reference frames are related by \n= K + 1 \n+ m2)v 2 . \n(15.2.18)\nKS \ncm \n(m1 \ncm \n2 \nWe now use Eq. (15.2.13) to rewrite Eq. (15.2.18) as \n1\n2 + 1 \n2\n=\n+ m2)v \n(15.2.19)\nKS \n2 \u00b5v1, 2 \n2 (m1 \ncm \nEven though kinetic energy is a reference frame dependent quantity, because the second \nterm in Eq. (15.2.19) is a constant, the change in kinetic energy in either reference frame \nis equal to \n1\n2 \n2\n\u0394K = \u00b5(( \n) \u2212 ( \n) ) . \n(15.2.20)\nv1, 2 \nv1, 2 \n2 \nf\ni \nThis generalizes to any two relatively inertial reference frames because the relative \nvelocity is a reference frame independent quantity, \nthe change in kinetic energy is independent of the choice of relatively \ninertial reference frames. \nWe showed in Appendix 13A that when two particles of masses m1 and m2 interact, the \nwork done by the interaction force is equal to \n1\n2 \n2\nW = 2 \u00b5(( \n) \u2212 ( \n) ) . \n(15.2.21)\nv1, 2 \nv1, 2 \nf\ni \nHence we explicitly verified that for our two-particle system \nW = \u0394Ksys . \n(15.2.22) \n15-6", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_411", "text": "15.3 Characterizing Collisions \nIn a collision, the ratio of the magnitudes of the initial and final relative velocities is \ncalled the coefficient of restitution and denoted by the symbol e , \nvB\ne = \n. \n(15.2.23)\nvA \nIf the magnitude of the relative velocity does not change during a collision, e = 1, then \nthe change in kinetic energy is zero, (Eq. (15.2.21)). Collisions in which there is no \nchange in kinetic energy are called elastic collisions, \n\u0394K = 0, elastic collision . \n(15.2.24) \nIf the magnitude of the final relative velocity is less than the magnitude of the initial \nrelative velocity, e < 1, then the change in kinetic energy is negative. Collisions in which \nthe kinetic energy decreases are called inelastic collisions, \n\u0394K < 0, inelastic collision . \n(15.2.25) \nIf the two objects stick together after the collision, then the relative final velocity is zero, \ne = 0 . Such collisions are called totally inelastic. The change in kinetic energy can be \nfound from Eq. (15.2.21), \n1 \n2 \n1 m1m2\n2\n\u0394K = \u2212 \n= \u2212 \nvA , totally inelastic collision . \n(15.2.26)\n\u00b5 vA\n2\n2 m1 + m2 \nIf the magnitude of the final relative velocity is greater than the magnitude of the initial \nrelative velocity, e > 1, then the change in kinetic energy is positive. Collisions in which \nthe kinetic energy increases are called superelastic collisions, \n\u0394K > 0, superelastic collision . \n(15.2.27) \n15.4 One-Dimensional Collisions Between Two Objects \n15.4.1 One Dimensional Elastic Collision in Laboratory Reference Frame \nConsider a one-dimensional elastic collision between two objects moving in the x \u00ad\ndirection. One object, with mass m1 and initial x -component of the velocity v1x,i , \ncollides with an object of mass m2 and initial x -component of the velocity v2x,i . The \nscalar components v1x,i and v1x,i can be positive, negative or zero. No forces other than \nthe interaction force between the objects act during the collision. After the collision, the \n15-7", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_412", "text": "final x -component of the velocities are v1x, f and v2 x, f . We call this reference frame the \n\u201claboratory reference frame\u201d. \nFigure 15.5 One-dimensional elastic collision, laboratory reference frame  \nFor the collision depicted in Figure 15.5, \n> 0 , \n< 0, \n< 0 , and \n> 0. \nv1x,i \nv2x,i \nv1x, f \nv2 x, f \nBecause there are no external forces in the x -direction, momentum is constant in the x \u00ad\ndirection. Equating the momentum components before and after the collision gives the \nrelation \n+ m2\n= m1\n+ m2\n(15.3.1)\nm1v1x, i \nv2 x, i \nv1x, f \nv2x, f . \nBecause the collision is elastic, kinetic energy is constant. Equating the kinetic energy \nbefore and after the collision gives the relation \n1\n1 \n1\n1\n2\n2\n2 \n2\n+\n=\n+ \n(15.3.2)\nm1v1x,i \nm2v2x,i \nm1v1x, f \nm2v2 x, f\n2\n2 \n2\n2 \nRewrite these Eqs. (15.3.1) and (15.3.2) as \n) = m2 \n) \n(15.3.3)\nm1(v1x,i \u2212 v1x, f \n(v2 x, f \u2212 v2x,i \n2\n2 \n2\n2\n) = m2 \n) . \n(15.3.4)\nm1(v1x,i \u2212 v1x, f \n(v2 x, f \u2212 v2x,i \nEq. (15.3.4) can be written as \n) = m2 \n) . \n(15.3.5)\nm1(v1x,i \u2212 v1x, f )(v1x,i + v1x, f \n(v2 x, f \u2212 v2x,i )(v2 x, f + v2x,i \nDivide Eq. (15.3.4) by Eq. (15.3.3), yielding \n(15.3.6)\nv1x,i + v1x, f = v2 x,i + v2x, f . \nEq. (15.3.6) may be rewritten as \n\u2212 v1x, f . \n(15.3.7)\nv1x,i \u2212 v2 x,i = v2x, f", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_413", "text": "Eq. (15.3.6) may be rewritten as \n\u2212 v1x, f . \n(15.3.7)\nv1x,i \u2212 v2 x,i = v2x, f \nRecall that the relative velocity between the two objects is defined to be \n\uf072vrel \u2261\uf072v\n\u2261 \uf072v \u2212 \uf072v2 . \n(15.3.8)\n1,2 \n1 \n15-8", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_414", "text": "where we used the superscript \u201crel\u201d to remind ourselves that the velocity is a relative \nvelocity (and to simplify our notation). Thus vrel \nis the initial x -component\nx,i = v1x,i \u2212 v2 x,i \nof the relative velocity, and vrel \nis the final x -component of the relative \nx, f = v1x, f \u2212 v2 x, f \nvelocity. Therefore Eq. (15.3.7) states that during the interaction the initial relative \nvelocity is equal to the negative of the final relative velocity \n\uf072rel \n\uf072 rel \n= \u2212v , (1\u2212 dimensional energy-momentum prinicple) . \n(15.3.9)\nvi \nf \nConsequently the initial and final relative speeds are equal. We shall call this relationship \nbetween the relative initial and final velocities the one-dimensional energy-momentum \nprinciple because we have combined these two principles to realize this result. The \nenergy-momentum principle is independent of the masses of the colliding particles. \nAlthough we derived this result explicitly, we have already shown that the change in \nkinetic energy for a two-particle interaction (Eq. (15.2.20)), in our simplified notation is \ngiven by \n1 \nrel )2 \nf \u2212 (vrel )i \n2)\n\u0394K =\n\u00b5((v \n(15.3.10)\n2 \nTherefore for an elastic collision where \u0394K = 0 , the square of the relative speed remains \nconstant \nrel )2 \nrel )2\n(v\nf = (v\ni . \n(15.3.11) \nFor a one-dimensional collision, the magnitude of the relative speed remains constant but \nthe direction changes by 180\uf06f . \nWe can now solve for the final x -component of the velocities, \nand v2 x, f , as\nv1x, f \nfollows. Eq. (15.3.7) may be rewritten as \n\u2212 v2x,i . \n(15.3.12)\nv2 x, f = v1x, f + v1x,i \nNow substitute Eq. (15.3.12) into Eq. (15.3.1) yielding \nm1\n+ m2\n= m1\n+ m2 \n) . \n(15.3.13)\nv1x,i \nv2x,i \nv1x, f \n(v1x, f + v1x,i \u2212 v2 x,i", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_415", "text": "m1\n+ m2\n= m1\n+ m2 \n) . \n(15.3.13)\nv1x,i \nv2x,i \nv1x, f \n(v1x, f + v1x,i \u2212 v2 x,i \nSolving Eq. (15.3.13) for v1x, f involves some algebra and yields \n\u2212 m2\n2 m2\nm1\n=\n+ \nv2x,i . \n(15.3.14)\nv1x, f \nv1x,i\nm1 + m2 \nm1 + m2 \n15-9", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_416", "text": "To find v2 x, f , rewrite Eq. (15.3.7) as \n+ v2x,i .  \n(15.3.15)\nv1x, f = v2x, f \u2212 v1x,i \nNow substitute Eq. (15.3.15) into Eq. (15.3.1) yielding \n+ m2\n= m1 \n+ m2\n. \n(15.3.16)\nm1v1x,i \nv2 x,i \n(v2x, f \u2212 v1x,i + v2x,i )v1x, f \nv2x, f \nWe can solve Eq. (15.3.16) for v2 x, f and determine that \n\u2212 m1\n2 m1\nm2 \n.  \n(15.3.17)\nv2 x, f = v2x,i \n+ v1x,i\nm2 + m1 \nm2 + m1 \nConsider what happens in the limits m1 >> m2 in Eq. (15.3.14). Then \n+ \n;  \n(15.3.18)\nv1x, f \u2192 v1x,i  \n2 m2v2 x,i\nm1 \nthe more massive object\u2019s velocity component is only slightly changed by an amount \nproportional to the less massive object\u2019s x -component of momentum. Similarly, the less \nmassive object\u2019s final velocity approaches \n\u2192\u2212v2x,i  \n\u2212 v2 x,i . \n(15.3.19)\nv2 x, f  \n+ 2v1x,i = v1x,i + v1x,i \nWe can rewrite this as \n= vrel . \n(15.3.20)\nv2 x, f \u2212 v1x,i = v1x,i \u2212 v2x,i \nx,i \ni.e. the less massive object \u201crebounds\u201d with the same speed relative to the more massive \nobject which barely changed its speed. \nIf the objects are identical, or have the same mass, Eqs. (15.3.14) and (15.3.17) become \n= v2x,i , \n;  \n(15.3.21)\nv1x, f \nv2x, f = v1x,i \nthe objects have exchanged x -components of velocities, and unless we could somehow \ndistinguish the objects, we might not be able to tell if there was a collision at all. \n15-10", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_417", "text": "15.4.2 One-Dimensional Collision Between Two Objects \u2013 Center-of-Mass Reference \nFrame \nWe analyzed the one-dimensional elastic collision (Figure 15.5) in Section 15.4.1 in the \nlaboratory reference frame. Now let\u2019s view the collision from the center-of-mass (CM) \nframe. The x -component of velocity of the center-of-mass is \nm1 v1x,i + m2 v2x,i\nvx,cm = \n. \n(15.3.22)\nm1 + m2 \nWith respect to the center-of-mass, the x -components of the velocities of the objects are \nm2\n\u2032\n\u2212 v \n)\nv1x,i = v1x,i \nx,cm = (v1x,i \u2212 v2x,i m1 + m2 \n(15.3.23)\nm1\n\u2032\n\u2212 v \n)\n.\nv2 x,i = v2x,i \nx,cm = (v2x,i \u2212 v1x,i  m1 + m2  \nIn the CM frame the momentum of the system is zero before the collision and hence the \nmomentum of the system is zero after the collision. For an elastic collision, the only way \nfor both momentum and kinetic energy to be the same before and after the collision is \neither the objects have the same velocity (a miss) or to reverse the direction of the \nvelocities as shown in Figure 15.6. \nFigure 15.6 One-dimensional elastic collision in center-of-mass reference frame \nIn the CM frame, the final x -components of the velocities are \nm2\n\u2032\n\u2032 \n)\nv1x, f = \u2212v1x,i = (v2x,i \u2212 v1x,i m1 + m2 \n(15.3.24)\nm1\n\u2032\n\u2032 \n)\n.\nv2x, f = \u2212v2x,i = (v2 x,i \u2212 v1x,i m1 + m2 \nThe final x -components of the velocities in the \u201claboratory frame\u201d are then given by \n15-11", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_418", "text": "\u2032 \n+ v\nv1x, f = v1x, f \nx,cm \nm2 \nm1 v1x,i + m2 v2x,i\n) \n+  \n(15.3.25)\n= (v2x,i \u2212 v1x,i  m1 + m2 \nm1 + m2  \n\u2212 m2\n2 m2\nm1\n= v1x,i \n+ v2x,i\nm1 + m2 \nm1 + m2 \nas in Eq. (15.3.14) and a similar calculation reproduces Eq. (15.3.17). \n15.5 Worked Examples \nExample 15.1 Elastic One-Dimensional Collision Between Two Objects \n\u02c6i \n1 \n2\n\u02c6i\nv1,i = v1,x,i \nv2,i = 0  initial state \nm2 = 2m1 \n\u02c6i final state\n\u02c6i\nv1, f = v1,x, f \nv2, f = v2,x, f\n\u02c6i \nm2 = 2m1 \n1 \n2 \nFigure 15.7 Elastic collision between two non-identical carts \nConsider the elastic collision of two carts along a track; the incident cart 1 has mass m1 \nand moves with initial speed v1,i . The target cart has mass m2 = 2 m1 and is initially at \nrest, \n= 0, (Figure 15.7). Immediately after the collision, the incident cart has final\nv2,i \nspeed v1, f and the target cart has final speed v2, f . Calculate the final x -component of the \nvelocities of the carts as a function of the initial speed v1,i . \nSolution The momentum flow diagram for the objects before (initial state) and after \n(final state) the collision are shown in Figure 15.7. We can immediately use our results \nabove with m2 = 2 m1 and v2,i = 0 . The final x -component of velocity of cart 1 is given \nby Eq. (15.3.14), where we use v1x,i = v1,i \n15-12", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_419", "text": "v1x, f = \u2212 1 \n3 v1,i . \n(15.4.1) \nThe final x -component of velocity of cart 2 is given by Eq. (15.3.17) \nv2 x, f = 2 \n3 v1,i . \n(15.4.2) \nExample 15.2 The Dissipation of Kinetic Energy in a Completely Inelastic Collision\nBetween Two Objects \n\u02c6i \n= 0\nv1,i \nv2,i \ninitial state \n1 \n2 \n\u02c6i \nfinal state \nv f \n1 \n2 \nFigure 15.7b Inelastic collision between two non-identical carts \nAn incident cart of mass m1 and initial speed v1, i collides completely inelastically with a \ncart of mass m2 that is initially at rest (Figure 15.7b). There are no external forces acting \non the objects in the direction of the collision. Find \u0394K / Kinitial = (Kfinal \u2212 Kinitial ) / Kinitial . \nSolution: In the absence of any net force on the system consisting of the two carts, the \nmomentum after the collision will be the same as before the collision. After the collision \nthe carts will move in the direction of the initial velocity of the incident cart with a \ncommon speed v f found from applying the momentum condition \nm1v1, i = (m1 + m2)vf \u21d2 \nm1 \n(15.4.3)\n=\nvf \nv1, i . \nm1 + m2 \nThe initial relative speed is vi \nrel \n. The final relative velocity is zero because the carts\n= v1, i \nstick together so using Eq. (15.2.26), the change in kinetic energy is \n1 \nrel )2 \n1 m1m2\n2\n\u0394K = \u2212 \u00b5(vi \n= \u2212 \nv1, i . \n(15.4.4)\n2\n2 m1 + m2 \n15-13", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_420", "text": "The ratio of the change in kinetic energy to the initial kinetic energy is then \n\u0394K / Kinitial = \u2212 \nm2 \nm1 + m2 \n. \n(15.4.5) \nAs a check, we can calculate the change in kinetic energy via \n\u0394K = (K f \u2212 Ki ) = 1 \n2 (m1 + m2 )vf \n2 \u2212 1 \n2 v1, i \n2 \n= 1 \n2 (m1 + m2 ) \nm1 \nm1 + m2 \n\u239b \n\u239d\u239c \n\u239e \n\u23a0\u239f \n2 \nv1, i \n2 \u2212 1 \n2 v1, i \n2 \n= \nm1 \nm1 + m2 \n\u22121\n\u239b \n\u239d\u239c \n\u239e \n\u23a0\u239f \n1 \n2 m1v1, i \n2\n\u239b \n\u239d\u239c \n\u239e \n\u23a0\u239f = \u2212 1 \n2 \nm1m2 \nm1 + m2 \nv1, i \n2 . \n(15.4.6) \nin agreement with Eq. (15.4.4). \nExample 15.3 Bouncing Superballs \n1 \n2 \ng \nM2 >> M1 \nFigure 15.8b Two superballs dropping \nConsider two balls that are dropped from a height hi above the ground, one on top of the \nother (Figure 15.8). Ball 1 is on top and has mass M1, and ball 2 is underneath and has \nmass M2 with M2 >> M1 . Assume that there is no loss of kinetic energy during all \ncollisions. Ball 2 first collides with the ground and rebounds. Then, as ball 2`starts to \nmove upward, it collides with the ball 1 which is still moving downwards (figure below \nleft). How high will ball 1 rebound in the air? Hint: consider this collision as seen by an \nobserver moving upward with the same speed as the ball 2 has after it collides with \nground. What speed does ball 1 have in this reference frame after it collides with the ball \n2? \n15-14", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_421", "text": "Solution \nThe system consists of the two balls and the earth. There are five special states for this \nmotion shown in the figure below. \npart a)  \nInitial State: the balls are released from rest at a height hi above the ground.  \nState A: the balls just reach the ground with speed v = \n. This follows from \na \n2ghi \n= 0 \u21d2\u0394K = \u2212\u0394U . Thus (1 / 2)mv 2 \u2212 0 = \u2212mg\u0394h = mghi \u21d2 v =\n\u0394Emech \na \na \n2ghi . \nState B: immediately before the collision of the balls. Ball 2 has collided with the ground \nand reversed direction with the same speed, va , but ball 1 is still moving downward with \nspeed va . \nState C: immediately after the collision of the balls. Because we are assuming that \n>> m1, ball 2 does not change its speed as a result of the collision so it is still moving \nm2 \nupward with speed va . As a result of the collision, ball 1 moves upward with speed vb . \nFinal State: ball 1 reaches a maximum height hf = vb \n2 / 2g above the ground. This again \nfollows from \u0394K = \u2212\u0394U \u21d2 0 \u2212 (1/ 2)mv2 = \u2212mg\u0394h = \u2212mgh \u21d2 h = v2 / 2g .\nb \nf\nf\nb \nChoice of Reference Frame: \nAs indicated in the hint above, this collision is best analyzed from the reference frame of \nan observer moving upward with speed va , the speed of ball 2 just after it rebounded with \n15-15", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_422", "text": "the ground. In this frame immediately, before the collision, ball 1 is moving downward \nwith a speed vb \u2032 that is twice the speed seen by an observer at rest on the ground (lab \nreference frame). \nva \u2032 = 2va \n(15.4.7) \nThe mass of ball 2 is much larger than the mass of ball 1, m2 >> m1. This enables us to \nconsider the collision (between States B and C) to be equivalent to ball 1 bouncing off a \nhard wall, while ball 2 experiences virtually no recoil. Hence ball 2 remains at rest in the \nreference frame moving upwards with speed va with respect to observer at rest on \nground. Before the collision, ball 1 has speed va \u2032 = 2va . Since there is no loss of kinetic \nenergy during the collision, the result of the collision is that ball 1 changes direction but \nmaintains the same speed, \n\u2032 = 2v . \n(15.4.8)\nvb \na \nHowever, according to an observer at rest on the ground, after the collision ball 1 is \nmoving upwards with speed \n= 2v + v = 3v . \n(15.4.9)\nvb \na\na\na \nWhile rebounding, the mechanical energy of the smaller superball is constant (we \nconsider the smaller superball and the Earth as a system) hence between State C and the \nFinal State, \n\u0394K + \u0394U = 0 . \n(15.4.10) \nThe change in kinetic energy is \n\u0394K = \u2212 1 m1(3va )2 . \n(15.4.11)\n2 \nThe change in potential energy is \n\u0394U \ng hf . \n(15.4.12)\n= m1 \nSo the condition that mechanical energy is constant (Equation (15.4.10)) is now \n\u2212 1 m1(3v1a )2 + m1 g hf = 0 . \n(15.4.13)\n2 \nWe can rewrite Equation (15.4.13) as \n1 \n)2\nm1 g hf = 9 m1( v a . \n(15.4.14)\n2 \n15-16", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_423", "text": "Recall that we can also use the fact that the mechanical energy doesn\u2019t change between \nthe Initial State and State A yielding an equation similar to Eq. (15.4.14), \n1 \n)2\ng hi = m1( v \n. \n(15.4.15)\nm1 \na\n2 \nNow substitute the expression for the kinetic energy in Eq. (15.4.15) into Eq. (15.4.14) \nyielding \ng hf \ng hi . \n(15.4.16)\nm1 \n= 9 m1 \nThus ball 1 reaches a maximum height \n= 9 hi . \n(15.4.17)\nhf \n15.6 Two Dimensional Elastic Collisions \n15.6.1 Two-dimensional Elastic Collision in Laboratory Reference Frame \nConsider the elastic collision between two particles in which we neglect any external \nforces on the system consisting of the two particles. Particle 1 of mass m1 is initially \n\uf072\nmoving with velocity \nand collides elastically with a particle 2 of mass \nthat is\nv1, i\nm2 \ninitially at rest. We shall refer to the reference frame in which one particle is at rest, \u2018the \ntarget\u2019, as the laboratory reference frame. After the collision particle 1 moves with \n\uf072\n\uf072\nvelocity v1, f and particle 2 moves with velocity v2, f , (Figure 15.9). The angles \u03b81, f \nand \u03b82, f that the particles make with the positive forward direction of particle 1 are \nFigure 15.9 Two-dimensional collision in laboratory reference frame \ncalled the laboratory scattering angles. \n1, f \n2, f \n1 \n1 \n2 \n2 \nv1, i \nv1, f \nv2, f \n\uf072\nGenerally the initial velocity \nof particle 1 is known and we would like to determine \nv1, i \n\uf072\n\uf072\nthe final velocities \nand \n, which requires finding the magnitudes and directions \nv1, f \nv2, f \n15-17", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_424", "text": "of each of these vectors, v1, f , v2, f , \u03b81, f , and \u03b82, f . These quantities are related by the two \nequations describing the constancy of momentum, and the one equation describing \nconstancy of the kinetic energy. Therefore there is one degree of freedom that we must \nspecify in order to determine the outcome of the collision. In what follows we shall \nexpress our results for v1, f , v2, f , and \u03b82, f in terms of v1, i and \u03b81, f . \nThe components of the total momentum \uf072psys \ni = m1 \n\uf072v1,i + m2 \n\uf072v2,i in the initial state are given \nby \npsys \nx,i = m1v1,i \n(15.5.1)\nsys \npy,i = 0. \nThe components of the momentum \uf072psys \nf = m1 \n\uf072v1, f + m2 \n\uf072v2, f in the final state are given by \nsys \np\nv1, f cos\u03b81, f\nv2, f cos\u03b82, f\nx, f = m1 \n+ m2 \n(15.5.2) \npsys \ny, f = m1 v1, f sin\u03b81, f \u2212 m2 v2, f sin\u03b82, f . \nThere are no any external forces acting on the system, so each component of the total \nmomentum remains constant during the collision, \nsys \nsys \n= \n(15.5.3)\npx,i \npx, f \nsys \nsys \np \n= p \n. \n(15.5.4)\ny,i\ny, f \nEqs. (15.5.3) and (15.5.4) become \nm1 v1,i = m1 v1, f cos\u03b81, f + m2 v2, f cos\u03b82, f , \n(15.5.5) \n0 = m1 v1, f sin\u03b81, f \u2212 m2 v2, f sin\u03b82, f . \n(15.5.6) \nThe collision is elastic and therefore the system kinetic energy of is constant \nsys \nsys . \n(15.5.7)\nKi \n= K f \nUsing the given information, Eq. (15.5.7) becomes \n1\n2\n1 \n2 \n1 \n2\n=\n+ \n. \n(15.5.8)\nm1v1,i\nm1v1, f\nm2v2, f\n2\n2 \n2", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_425", "text": "sys . \n(15.5.7)\nKi \n= K f \nUsing the given information, Eq. (15.5.7) becomes \n1\n2\n1 \n2 \n1 \n2\n=\n+ \n. \n(15.5.8)\nm1v1,i\nm1v1, f\nm2v2, f\n2\n2 \n2 \nRewrite the expressions in Eqs. (15.5.5) and (15.5.6) as \nm2v2, f cos\u03b82, f = m1(v1,i \u2212 v1, f cos\u03b81, f ), \n(15.5.9) \n15-18", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_426", "text": ".\nm2v2, f sin\u03b82, f = m1v1, f sin\u03b81, f \n(15.5.10) \nSquare each of the expressions in Eqs. (15.5.9) and (15.5.10), add them together and use \nthe identity cos2 \u03b8 + sin2 \u03b8 = 1 yielding \n2 \nm1\n2\n2 \n2\n= \n) . \n(15.5.11)\nv2, f \n2 (v1,i \u2212 2v1,iv1, f cos\u03b81, f + v1, f\nm2 \nSubstituting Eq. (15.5.11) into Eq. (15.5.8) yields \n1 \n2\n1 \n2\n1 m1\n2\n2 \n2  \n2 m1v1,i = 2 m1v1, f + \n(v1,i \u2212 2v1,i v1, f cos\u03b81, f + v1, f ) . \n(15.5.12) \n2 m2 \nEq. (15.5.12) simplifies to \n\u239b\n\u239e \n\u239b\n\u239e\nm1\n2 \nm1 \nm1\n2\n0 = 1+ \n\u23a0\u239f v1, f \u2212 \n2v1,i v1, f cos\u03b81, f \u2212 1\u2212 \n\u23a0\u239f v1,i , \n(15.5.13)\n\u239d\u239c \n\u239d\u239c\nm2 \nm2 \nm2 \nLet \u03b1 = m1 / m2 then Eq. (15.5.13) can be written as \n2 \n2\n0 = (1+ \u03b1 )v1, f \u2212 2\u03b1v1,i v1, f cos\u03b81, f \u2212 (1\u2212\u03b1 )v1, i , \n(15.5.14) \nThe solution to this quadratic equation is given by \n1/2 \n\u03b1 2\n2 \n2\n\u03b1v1,i cos\u03b81, f \u00b1(\nv1,i cos2 \u03b81, f + (1\u2212\u03b1 )v1,i )\n= \n. \n(15.5.15)\nv1, f \n(1+ \u03b1 ) \nDivide the expressions in Eq. (15.5.9), yielding \nv2, f sin\u03b82, f \nv1, f sin\u03b81, f\n= \n. \n(15.5.16)\nv2, f cos\u03b82, f \nv1,i \u2212 v1, f cos\u03b81, f \nEq. (15.5.16) simplifies to \nv1, f sin\u03b81, f\ntan\u03b82, f = \n.", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_427", "text": "= \n. \n(15.5.16)\nv2, f cos\u03b82, f \nv1,i \u2212 v1, f cos\u03b81, f \nEq. (15.5.16) simplifies to \nv1, f sin\u03b81, f\ntan\u03b82, f = \n. \n(15.5.17)\nv1,i \u2212 v1, f cos\u03b81, f \nThe relationship between the scattering angles in Eq. (15.5.17) is independent of the \nmasses of the colliding particles. Thus the scattering angle for particle 2 is \n15-19", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_428", "text": "0 \n \n \n \n \n \n  \n \n \n  \n \n \n \n \n \n1\nv1, i\nv1, f\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n  \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n  \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n  \n  \n \n \n  \n \n \n \n \n  \n \n \n \n \n  \n \n\u239b \nsin\u03b81, f \n\u239e\nv1, f\n= tan\u22121 \u239c\n\u239f \n(15.5.18)\n\u03b82, f \n\u239d v1,i \u2212 v1, f cos\u03b81, f \u23a0 \nWe can now use Eq. (15.5.10) to find an expression for the final velocity of particle 1 \nv1, f sin\u03b81, f\n= \n. \n(15.5.19)\nv2, f \n\u03b1 sin\u03b82, f \nExample 15.5 Elastic Two-dimensional collision of identical particles \n1 \n2 \n2 \nv2, f \n1, f = 30 \n2, f \n\u02c6i \n\u02c6j \nFigure 15.10 Momentum flow diagram for two-dimensional elastic collision \nObject 1 with mass \nis initially moving with a speed \n= 3.0m \u22c5s\u22121 and collides \nm1 \nv1,i \nelastically with object 2 that has the same mass, m2 = m1, and is initially at rest. After the \ncollision, object 1 moves with an unknown speed v1, f at an angle \u03b81, f with respect to its \ninitial direction of motion and object 2 moves with an unknown speed \n, at an \nv2, f \nunknown angle \u03b82, f (as shown in the Figure 15.10). Find the final speeds of each of the \nobjects and the angle \u03b82, f . \nSolution: Because the masses are equal, \u03b1 = 1. We are given that v1,i = 3.0 m \u22c5s\u22121 and \n\u03b81, f = 30o . Hence Eq. (15.5.14) reduces to \nv1, f = v1,i cos\u03b81, f = (3.0 m \u22c5s\u22121)cos30\uf06f = 2.6 m \u22c5s\u22121 . \n(15.5.20) \nSubstituting Eq. (15.5.20) in Eq. (15.5.17) yields \n15-20", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_429", "text": "1\nv1, i\nv1, f\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n  \n \n \n \n \n  \n  \n \n \n \n \n \n \n \n \n \n \n  \n  \n \n \n \n  \n  \n  \n \n \n \n\u239b \nsin\u03b81, f \n\u239e\nv1, f\n= tan\u22121\n\u03b82, f  \n\u239c\n\u239f\n\u239d v1,i \u2212 v1, f cos\u03b81, f \u23a0 \n\u239b \n(2.6 m \u22c5s\u22121)sin(30\uf06f ) \n\u239e \n= tan\u22121  \n(15.5.21)\n\u03b82, f \n\u239d\u239c 3.0 m \u22c5s\u22121 \u2212 (2.6 m \u22c5s\u22121)cos(30\uf06f )\u23a0\u239f \n= 60\uf06f . \nThe above results for v1, f and \u03b82,f may be substituted into either of the expressions in \nEq. (15.5.9), or Eq. (15.5.11), to find v2, f = 1.5m \u22c5 s\u22121 . Eq. (15.5.11) also has the solution \nv2, f = 0 , which would correspond to the incident particle missing the target completely. \nBefore going on, the fact that \u03b81, f +\u03b82, f = 90\uf06f , that is, the objects move away from the \ncollision point at right angles, is not a coincidence. A vector derivation is presented in \nExample 15.6. We can see this result algebraically from the above result. Substituting \nEq. (15.5.20) v1, f = v1,i cos\u03b81, f in Eq. (15.5.17) yields \ncos\u03b81, f sin\u03b81, f\n= \n= tan(90\uf06f \u2212\u03b81, f ) ; \n(15.5.22)\ntan\u03b82, f \n2 \n= cot\u03b81, f\n1\u2212 cos\u03b81, f \nshowing that \u03b81, f +\u03b82, f = 90\uf06f , the angles \u03b81, f and \u03b82, f are complements. \nExample 15.6 Two-dimensional elastic collision between particles of equal mass \nShow that the equal mass particles emerge from a two-dimensional elastic collision at \nright angles by making explicit use of the fact that momentum is a vector quantity. \n1 \n2 \n2 \nv2, f \n1, f \n2, f \n\u02c6i \n\u02c6j \nFigure 15.11 Elastic scattering of identical particles \n15-21", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_430", "text": "Solution: Choose a reference frame in which particle 2 is initially at rest (Figure 15.11). \nThere are no external forces acting on the two objects during the collision (the collision \nforces are all internal), therefore momentum is constant \n\uf072\n\uf072\nsys \nsys \n= p f , \n(15.5.23)\npi \nwhich becomes \nm1 \n\uf072v \n\uf072v \n\uf072v\n(15.5.24) \n+ m1 \n= m1 \n. \n1, i \n1, f \n2, f \nEq. (15.5.24) simplifies to\n\uf072v\n=  \uf072v \n\uf072v\n(15.5.25) \n+  \n. \n1,i \n1, f \n2, f \nRecall the vector identity that the square of the speed is given by the dot product \n\uf072\uf072 \nv \u22c5 v = v2 . With this identity in mind, we take the dot product of each side of Eq. \n(15.5.25) with itself, \n\uf072v\n\u22c5 \uf072v\n= ( \uf072v \n\uf072v \n\uf072v\nf\n1, \n\uf072v\n) \u22c5(  \n) \n+  \n+ \n1,i \n1,i \n1, f \n2, f \n2, f \n(15.5.26) \n=  \uf072v \n\uf072v \n\uf072v\n\uf072v \n\uf072v\n\u22c5 \uf072v\n\u22c5  \n+ 2  \n\u22c5  \n+  \n2, f .\n1, f \n1, f \n1, f \n2, f \n2, f \nThis becomes \n2\n2 \n2\n+ 2\uf072 \n\u22c5 \uf072 \n. \n(15.5.27)\nv1,i = v1, f \nv1, f v2, f + v2, f \nRecall that kinetic energy is the same before and after an elastic collision, and the masses \nof the two objects are equal, so constancy of energy, (Eq. (15.3.2)) simplifies to \n2\n2 \n2 . \n(15.5.28)\nv1,i = v1, f + v2, f \nComparing Eq. (15.5.27) to Eq. (15.5.28), we see that \n\uf072\n\uf072 \nv1, f \u22c5 v2, f = 0 . \n(15.5.29) \nThe dot product of two nonzero vectors is zero when the two vectors are at right angles to \neach other justifying our claim that the collision particles emerge at right angles to each \nother. \nExample 15.7 Two dimensional collision between particles of unequal mass \nParticle 1 of mass m1 , initially moving in the positive x -direction (to the right in the \nfigure below) with speed v1,i , collides with particle 2 of mass m = m /3, which is \n2\n1", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_431", "text": "other. \nExample 15.7 Two dimensional collision between particles of unequal mass \nParticle 1 of mass m1 , initially moving in the positive x -direction (to the right in the \nfigure below) with speed v1,i , collides with particle 2 of mass m = m /3, which is \n2\n1 \ninitially moving in the opposite direction (Figure 15.12) with an unknown speed v2,i . \nAssume that the total external force acting on the particles is zero. Do not assume the \ncollision is elastic. After the collision, particle 1 moves with speed v1, f \n/ 2 in the \n= v1,i \nnegative y -direction. After the collision, particle 2 moves with an unknown speed v2, f , \n15-22", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_432", "text": "at an angle \u03b82, f = 45o with respect to the positive x -direction. (i) Determine the initial \nspeed v2,i of particle 2 and the final speed v2, f of particle 2 in terms of v1,i . (ii) Is the \ncollision elastic? \n\u02c6i \n\u02c6j \nbefore \nafter \nm1 = m \n1 \n2 \n1 \n2 \n= 45 \nm2 = m / 3 \nv1,i \nv1, f = v1,i / 2 \nv2, f \nv2,i \nFigure 15.12 Two-dimensional collision between particles of unequal mass  \nSolution: We choose as our system the two particles. We are given that v1, f \n/ 2 . We \n= v1,i \napply the two momentum conditions, \n/ 3)v2,i \n/ 3) v2, f ( 2 / 2) \n(15.5.30)\nm1 v1,i \u2212 (m1 \n= (m1 \n0 = m1 \n/ 3) v2, f ( 2 / 2) . \n(15.5.31)\nv1, f \u2212 (m1 \nSolve Eq. (15.5.31) for v2, f \n= \n(15.5.32)\nv2, f = 3 2v1, f \n3 2 v1,i\n2 \nSubstitute Eq. (15.5.32) into Eq. (15.5.30) and solve for v2,i \n=(3/ 2)v1,i . \n(15.5.33)\nv2,i \nThe initial kinetic energy is then \n1\n2\n1 \n2\n7\n2\n=\n+ \n/ 3)v2,i = \n. \n(15.5.34)\nKi \nm1v1,i \n(m1 \nm1v1,i\n2\n2 \n8 \nThe final kinetic energy is \n1 \n1 \n1\n3\n7\n2 \n2 \n2\n2\n2\n= m1 \n+ m2 \n= m1 \n+ m1 \n= m1 \n. \n(15.5.35)\nK f \nv1, f\nv2, f\nv1,i\nv1,i\nv1,i\n2 \n2 \n8\n4\n8 \nComparing our results, we see that kinetic energy is constant so the collision is elastic. \n15-23", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_433", "text": "15.7 Two-Dimensional Collisions in Center-of-Mass Reference Frame \n15.7.1 Two-Dimensional Collision in Center-of-Mass Reference Frame \nConsider the elastic collision between two particles in the laboratory reference frame\n\uf072\n(Figure 15.9). Particle 1 of mass m1 is initially moving with velocity v1, i and collides \nelastically with a particle 2 of mass m2 that is initially at rest. After the collision the \n\uf072\n\uf072\nparticle 1 moves with velocity v1, f and particle 2 moves with velocity v2, f . In section \n15.7.1 we determined how to find v1, f , v2, f , and \u03b82, f in terms of v1, i and \u03b82, f . We shall \nnow analyze the collision in the center-of-mass reference frame, which is boosted form \nthe laboratory frame by the velocity of center-of-mass given by \n\uf072v\nm1 1, i\n\uf072v\n=  \ncm \nm1 + m2 \n. \n(15.5.36)  \nBecause we assumed that there are no external forces acting on the system, the center-of\u00ad\nmass velocity remains constant during the interaction. \n1 \n1 \n2 \n2 \nv2, f \nv1, f \nv1,i \nv2,i \ncm \nFigure 15.13 Two-dimensional elastic collision in center-of-mass reference frame \nRecall the velocities of particles 1 and 2 in the center-of-mass frame are given by \n(Eq.,(15.2.9) and (15.2.10)). In the center-of-mass reference frame the velocities of the \ntwo incoming particles are in opposite directions, as are the velocities of the two outgoing \nparticles after the collision (Figure 15.13). The angle \u0398 cm between the incoming and \noutgoing velocities is called the center-of-mass scattering angle. \n15-24", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_434", "text": "15.7.2 Scattering in the Center-of-Mass Reference Frame \n\uf072\nConsider a collision between particle 1 of mass m1 and velocity \nand particle 2 of\nv1,i \nmass m2 at rest in the laboratory frame. Particle 1 is scattered elastically through a \nscattering angle \u0398 in the center-of-mass frame. The center-of-mass velocity is given by \n\uf072 \n\uf072 \nm1v1,i\nv cm = \n. \n(15.5.37)\nm1 + m2 \nIn the center-of-mass frame, the momentum of the system of two particles is zero \n\uf072\n\uf072\n\uf072\n\uf072 \n\uf072\n0 = m1\n+ m2v\u2032 = m1\n+ m2v\u2032 \n. \n(15.5.38)\nv1, \u2032 i \n2,i \nv1, \u2032 f \n2, f \nTherefore \n\uf072 \nm2 \uf072 \n= \u2212 \nv\u2032 . \n(15.5.39)\nv1, \u2032 i \n2,i\nm1 \n\uf072 \nm2 \uf072 \n= \u2212 \nv\u2032 \n(15.5.40)\nv1, \u2032 f \n2, f\nm1 \nThe energy condition in the center-of-mass frame is \n1\n2 + 1 \n2\n1\n2\n1 \n2\n=\n+ \n. \n(15.5.41)\nm1v1, \u2032 i\nm2v2, \u2032 i\nm1v1, \u2032 f\nm2v2, \u2032 f\n2\n2 \n2\n2 \nSubstituting Eqs. (15.5.39) and (15.5.40) into Eq. (15.5.41) yields \n\u2032 i = \u2032 f . \n(15.5.42)\nv1, \nv1, \n(we are only considering magnitudes). Therefore \n\u2032 i = \n. \n(15.5.43)\nv2, \nv2, \u2032 f \nBecause the magnitude of the velocity of a particle in the center-of-mass reference frame \nis proportional to the relative velocity of the two particles, Eqs. (15.5.42) and (15.5.43) \nimply that the magnitude of the relative velocity also does not change \n\uf072 \n\uf072 \n, \n(15.5.44)\nv\u2032 \n= v\u2032 \n1, 2, i \n1, 2, f \nverifying our earlier result that for an elastic collision the relative speed remains the \nsame, (Eq. (15.2.20)). However the direction of the relative velocity is rotated by the \ncenter-of-mass scattering angle \u0398 cm . This generalizes the energy-momentum principle to \ntwo dimensions. Recall that the relative velocity is independent of the reference frame, \n15-25", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_435", "text": "1\nv1, i\nv1, f\n     \n \n \n \n \n \n \n \n  \n  \n  \n \n \n \n \n \n \n \n  \n \n \n \n    \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n  \n \n  \n \n \n \n  \n \n \n  \n  \n \n\uf072\n\uf072\n\u2212 \uf072 \n\u2212 \uf072 \n= \nv\u2032 \n(15.5.45)\nv1, i \nv2, i \nv1, \u2032 i \n2, i \n\uf072\n\uf072 \nIn the laboratory reference frame \n= 0 , hence the initial relative velocity is\nv2,i \n\uf072\n\uf072\n\uf072\n\u2032 \n=\n= \n, and the velocities in the center-of-mass frame of the particles are then\nv1, 2, i \nv1, 2, i \nv1, i \n\uf072 \n\u00b5 \uf072 \n= \n(15.5.46)\nv1, \u2032 i \nv1, i\nm1 \n\uf072 \n= \u2212 \u00b5 \uf072 \nv\u2032 \n. \n(15.5.47)\n2, i \nv1, i\nm2 \nTherefore the magnitudes of the final velocities in the center-of-mass frame are \n\u00b5\n\u00b5\n\u00b5\n=\n= \n\u2032 \n=\n= \n(15.5.48)\nv1, \u2032 f\nv1, \u2032 i\nv1, 2, i\nv1, 2, i\nv1, i . \nm1 \nm1 \nm1 \n\u00b5\n\u00b5\n\u00b5\n=\n= \n\u2032 \n=\n= \n(15.5.49)\nv2, \u2032 f\nv2, \u2032 i\nv1, 2, i\nv1, 2, i\nv1, i . \nm2 \nm2 \nm2 \nExample 15.8 Scattering in the Lab and CM Frames \n\uf072\nParticle 1 of mass \nand velocity \nby a particle of mass m2 at rest in the laboratory \nm1 \nv1,i \nframe is scattered elastically through a scattering angle \u0398 in the center of mass frame, \n(Figure 15.14). Find (i) the scattering angle of the incoming particle in the laboratory \nframe, (ii) the magnitude of the final velocity of the incoming particle in the laboratory \nreference frame, and (iii) the fractional loss of kinetic energy of the incoming particle. \nv1,i \nv1, f \n\u02c6j \n\u02c6i \n1\n1 \ncm \n1, f \n2 \n1\n2 v2,i \n2\n2, f \ncm \n2 \nv2, f \nv2, f \nFigure 15.14 Scattering in the laboratory and center-of-mass reference frames \n15-26", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_436", "text": "Solution: \ni) In order to determine the center-of-mass scattering angle we use the transformation law \nfor velocities \n\uf072\n\uf072 \n\u2212 \uf072 \n= \nv \n. \n(15.5.50)\nv1, \u2032 f \nv1, f \ncm \nIn Figure 15.15 we show the collision in the center-of-mass frame along with the \nlaboratory frame final velocities and scattering angles. \n\u02c6i \n1 \nv1, f \nv1,i \n1, f \nv cm \nv1, f \nv2,i \ncm \n2 \nv cm \nv2, f \n2, f \ncm \n\u02c6j \n2\n1 \nv2, f \nFigure 15.15 Final velocities of colliding particles \nVector decomposition of Eq. (15.5.50) yields \nv1, f cos\u03b81, i = v1, \u2032 f cos\u0398 cm \u2212 v cm , \n(15.5.51)  \nv1, f sin\u03b81, i = v1, \u2032 f sin\u0398 cm . \n(15.5.52)  \nwhere we choose as our directions the horizontal and vertical Divide Eq. (15.5.52) by  \n(15.5.51) yields \nsin\u03b81, i \nsin\u0398\nv1, f\nv1, \u2032 f \ncm \ntan\u03b81, i =\n= \n(15.5.53)\n\u2032 f cos\u0398\n\u2212 v\nv1, f cos\u03b81, i\nv1, \ncm \ncm \nBecause \u2032 i = \n, we can rewrite Eq. (15.5.53) as\nv1, \nv1, \u2032 f \nv1, \u2032 i sin\u0398 cm \ntan\u03b81, i = \n(15.5.54)\n\u2032 i cos\u0398\n\u2212 v\nv1, \ncm \ncm \nWe now substitute Eqs. (15.5.48) and v \n/ (m1 \n) into Eq. (15.5.54) yielding\ncm = m1v1, i \n+ m2 \n15-27", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_437", "text": "sin\u0398\nm2 \ncm \ntan\u03b81, i = \n. \n(15.5.55)\ncos\u0398\n\u2212 m1\ncm \n/ m2 \nThus in the laboratory frame particle 1 scatters by an angle \n\u239b\n\u239e\nm2 \ncm \n\u03b81, i = tan\u22121 \nsin\u0398 \n. \n(15.5.56)\n\u239d\u239c cos\u0398 cm \u2212 m1 / m2 \u23a0\u239f \nii) We can calculate the square of the final velocity in the laboratory frame \n\uf072v1, f \u22c5 \uf072v1, f = ( \uf072v1, \u2032 f + \uf072  \nv cm ) \u22c5( \uf072v1, \u2032 f + \uf072vcm ) . \n(15.5.57)  \nwhich becomes \n2\n2 + 2v \uf072\n\uf072 \n2 \n2\nv1, f = v1, \u2032 f \n1, \u2032 f \u22c5 v cm + vcm = v1, \u2032 f \n2 + 2v1, \u2032 f vcm cos\u0398 cm + vcm . \n(15.5.58) \nWe use the fact that \n\u2032 f =\n= (\u00b5 / m1 \n= (\u00b5 / m1 \n= (m2 / m1 \nto rewrite \nv1, \nv1, \u2032 i \n)v1,2, i \n)v1, i \n+ m2 )v1, i \nEq. (15.5.58) as \n\u239b\n\u239e \n2 \n2 \nm2 \nm2m1 \nm1\n2 \n2\nv1, f = \n\u239d\u239c \n+ m2 \u23a0\u239f v1, i \n2 + 2 \n+ m2)2 v1, i cos\u0398 cm + \n)2 v1, i . \n(15.5.59)\nm1 \n(m1 \n(m1 + m2 \nThus \n2 + 2m2\n2 1/2 \ncos\u0398\n(m2 \nm1 \ncm + m1 )\n= \n. \n(15.5.60)\nv1, f\nv1, i\nm1 + m2 \n(iii) The fractional change in the kinetic energy of particle 1 in the laboratory frame is \ngiven by \n2 + 2m2\n2\nK1, f \u2212 K1, i\nv1, f \n2 \u2212 v1, i \n2 \nm2 \nm1cos\u0398 + m1\n2m2m1(cos\u0398\n\u22121) \ncm \ncm \n=\n= \n\u22121 = \n.(15.5.61)\n2 \n)2 \n)2\nK1, i\nv1, i \n(m1 + m2\n(m1 + m2 \nWe can also determine the scattering angle \u0398 cm in the center-of-mass reference frame \nfrom the scattering angle \u03b81, i of particle 1 in the laboratory. We now rewrite the \nmomentum relations as", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_438", "text": "2 \n)2 \n)2\nK1, i\nv1, i \n(m1 + m2\n(m1 + m2 \nWe can also determine the scattering angle \u0398 cm in the center-of-mass reference frame \nfrom the scattering angle \u03b81, i of particle 1 in the laboratory. We now rewrite the \nmomentum relations as \nv1, f cos\u03b81, i + v cm = v1, \u2032 f cos\u0398 cm , \n(15.5.62) \nv1, f sin\u03b81, i = v1, \u2032 f sin\u0398 cm . \n(15.5.63) \nIn a similar fashion to the above argument, we have that \n15-28", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_439", "text": "v1, f sin\u03b81, f\ntan\u0398 = \n. \n(15.5.64)\ncm \nv1, f cos\u03b81, f + v cm \nRecall from our analysis of the collision in the laboratory frame that if we specify one of \nthe four parameters v1, f , v2, f , \u03b81, f , or v1, f , then we can solve for the other three in terms \nof the initial parameters \nand \nWith that caveat, we can use Eq. (15.5.64) to\nv1, i\nv2, i . \ndetermine \u0398 . \ncm \n15-29", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_440", "text": "Chapter 16 Two Dimensional Rotational Kinematics \n16.1 Introduction........................................................................................................... 1  \n16.2 Fixed Axis Rotation: Rotational Kinematics...................................................... 1  \n16.2.1 Fixed Axis Rotation........................................................................................ 1  \n16.2.2 Angular Velocity and Angular Acceleration ............................................... 2  \n16.2.3 Sign Convention: Angular Velocity and Angular Acceleration ................ 4  \n16.2.4 Tangential Velocity and Tangential Acceleration....................................... 4", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_441", "text": "16.2.4 Tangential Velocity and Tangential Acceleration....................................... 4  \nExample 16.1 Turntable ........................................................................................... 5  \n16.3 Rotational Kinetic Energy and Moment of Inertia............................................ 6  \n16.3.1 Rotational Kinetic Energy and Moment of Inertia..................................... 6  \n16.3.2 Moment of Inertia of a Rod of Uniform Mass Density............................... 7  \nExample 16.2 Moment of Inertia of a Uniform Disc.............................................. 8  \n16.3.3 Parallel Axis Theorem ................................................................................. 10  \n16.3.4 Parallel Axis Theorem Applied to a Uniform Rod ................................... 11", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_442", "text": "16.3.4 Parallel Axis Theorem Applied to a Uniform Rod ................................... 11  \nExample 16.3 Rotational Kinetic Energy of Disk ................................................ 11  \n16.4 Conservation of Energy for Fixed Axis Rotation............................................. 12  \nExample 16.4 Energy and Pulley System.............................................................. 12  \nExample 16.5 Physical Pendulum.......................................................................... 14  \nAppendix 16A: Proof of the Parallel Axis Theorem.................................................... 18", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_443", "text": "Chapter 16 Two Dimensional Rotational Kinematics \nMost galaxies exhibit rising rotational velocities at the largest measured \nvelocity; only for the very largest galaxies are the rotation curves flat. Thus \nthe smallest SC\u2019s (i.e. lowest luminosity) exhibit the same lack of Keplerian \nvelocity decrease at large R as do the high-luminosity spirals. The form for \nthe rotation curves implies that the mass is not centrally condensed, but \nthat significant mass is located at large R. The integral mass is increasing \nat least as fast as R. The mass is not converging to a limiting mass at the \nedge of the optical image. The conclusion is inescapable than non-\nluminous matter exists beyond the optical galaxy. 1 \nVera Rubin \n16.1 Introduction \nThe physical objects that we encounter in the world consist of collections of atoms that \nare bound together to form systems of particles. When forces are applied, the shape of \nthe body may be stretched or compressed like a spring, or sheared like jello. In some \nsystems the constituent particles are very loosely bound to each other as in fluids and \ngasses, and the distances between the constituent particles will vary. We shall begin by \nrestricting ourselves to an ideal category of objects, rigid bodies, which do not stretch, \ncompress, or shear. \nA body is called a rigid body if the distance between any two points in the body \ndoes not change in time. Rigid bodies, unlike point masses, can have forces applied at \ndifferent points in the body. Let\u2019s start by considering the simplest example of rigid body \nmotion, rotation about a fixed axis. \n16.2 Fixed Axis Rotation: Rotational Kinematics \n16.2.1 Fixed Axis Rotation \nA simple example of rotation about a fixed axis is the motion of a compact disc in \na CD player, which is driven by a motor inside the player. In a simplified model of this \nmotion, the motor produces angular acceleration, causing the disc to spin. As the disc is \nset in motion, resistive forces oppose the motion until the disc no longer has any angular \nacceleration, and the disc now spins at a constant angular velocity. Throughout this \nprocess, the CD rotates about an axis passing through the center of the disc, and is \nperpendicular to the plane of the disc (see Figure 16.1). This type of motion is called \nfixed-axis rotation. \n1V.C. Rubin, W.K. Jr. Ford, N Thonnard, Rotational properties of 21 SC galaxies with a large range of", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_444", "text": "perpendicular to the plane of the disc (see Figure 16.1). This type of motion is called \nfixed-axis rotation. \n1V.C. Rubin, W.K. Jr. Ford, N Thonnard, Rotational properties of 21 SC galaxies with a large range of \nluminosities and radii, from NGC 4605 /R = 4kpc/ to UGC 2885 /R = 122 kpc/, Astrophysical Journal, Part \n1, vol. 238, June 1, 1980, p. 471-487. \n16-\u0014", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_445", "text": "\u0014 \n \n \n \n \n \n \n \n  \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n  \n \n \n \n \n \n \n \n \n \n \n \nFigure 16.1 Rotation of a compact disc about a fixed axis. \nWhen we ride a bicycle forward, the wheels rotate about an axis passing through \nthe center of each wheel and perpendicular to the plane of the wheel (Figure 16.2). As \nlong as the bicycle does not turn, this axis keeps pointing in the same direction. This \nmotion is more complicated than our spinning CD because the wheel is both moving \n\uf072\n(translating) with some center of mass velocity, vcm , and rotating with an angular speed \n\u03c9 . \nFigure 16.2 Fixed axis rotation and center of mass translation for a bicycle wheel. \nWhen we turn the bicycle\u2019s handlebars, we change the bike\u2019s trajectory and the \naxis of rotation of each wheel changes direction. Other examples of non-fixed axis \nrotation are the motion of a spinning top, or a gyroscope, or even the change in the \ndirection of the earth\u2019s rotation axis. This type of motion is much harder to analyze, so \nwe will restrict ourselves in this chapter to considering fixed axis rotation, with or \nwithout translation. \n16.2.2 Angular Velocity and Angular Acceleration \nFor a rigid body undergoing fixed-axis rotation, we can divide the body up into small \nvolume elements with mass \u0394mi . Each of these volume elements is moving in a circle of \nradius ri about the axis of rotation (Figure 16.3). \n16-2", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_446", "text": "\u0014 \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n  \n \n \n \n \n \n \n \n \n  \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n  \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n  \n \n \n \n \n \n \n  \nFigure 16.3 Coordinate system for fixed-axis rotation. \nWe will adopt the notation implied in Figure 16.3, and denote the vector from the axis to \n\uf072\n\uf072 \nr\nthe point where the mass element is located as \n, with magnitude r = \n. Suppose the \nri\ni \ni \nfixed axis of rotation is the z -axis. Introduce a right-handed coordinate system for an \nangle \u03b8 in the plane of rotation and the choice of the positive z -direction perpendicular \nto that plane of rotation. Recall our definition of the angular velocity vector. The angular \nvelocity vector is directed along the z -axis with z -component equal to the time \nderivative of the angle \u03b8 , \n\u03c9\uf072 = d\u03b8 k\u02c6 = \u03c9 k\u02c6 . \n(16.1.1)\ndt \nz \nThe angular velocity vector for the mass element undergoing fixed axis rotation with \n\u03c9 z > 0 is shown in Figure 16.4. Because the body is rigid, all the mass elements will have \n\uf072\n\uf072\nthe same angular velocity \u03c9 and hence the same angular acceleration \u03b1 . If the bodies \ndid not have the same angular velocity, the mass elements would \u201ccatch up to\u201d or \u201cpass\u201d \neach other, precluded by the rigid-body assumption. \nFigure 16.4 Angular velocity vector for a mass element for fixed axis rotation \nIn a similar fashion, all points in the rigid body have the same angular acceleration, \n\uf072 \nd 2\u03b8\n\u03b1 = \nk\u02c6 = \u03b1 z k\u02c6 . \n(16.1.2)\ndt2 \n16-3", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_447", "text": "The angular acceleration vector is shown in Figure 16.5. \nFigure 16.5 Angular acceleration vector for a rigid body rotating about the z -axis \n16.2.3 Sign Convention: Angular Velocity and Angular Acceleration \nFor rotational problems we shall always choose a right-handed cylindrical coordinate \nsystem. If the positive z -axis points up, then we choose \u03b8 to be increasing in the \ncounterclockwise direction as shown in Figures 16.4 and 16.5. If the rigid body rotates in \nthe counterclockwise direction, then the z -component of the angular velocity is positive, \n\u03c9 z = d\u03b8 / dt > 0 . The angular velocity vector then points in the +k\u02c6 -direction as shown in \nFigure 16.4. If the rigid body rotates in the clockwise direction, then the z -component of \nthe angular velocity angular velocity is negative, \u03c9 z = d\u03b8 / dt < 0 . The angular velocity \nvector then points in the \u2212k\u02c6 -direction. \nIf the rigid body increases its rate of rotation in the counterclockwise (positive) \ndirection \nthen \nthe \nz -component \nof \nthe \nangular \nacceleration \nis \npositive, \n\u03b1 z \u2261 d 2\u03b8 dt2 = d\u03c9 z / dt > 0 . The angular acceleration vector then points in the +k\u02c6 \u00ad\ndirection as shown in Figure 16.5. If the rigid body decreases its rate of rotation in the \ncounterclockwise (positive) direction then the z -component of the angular acceleration is \nnegative, \u03b1 z = d 2\u03b8 / dt2 = d\u03c9 z / dt < 0 . The angular acceleration vector then points in the \n\uf072\n\uf072\n\u2212k\u02c6 -direction. To phrase this more generally, if \u03b1 and \u03c9 point in the same direction, the \nbody is speeding up, if in opposite directions, the body is slowing down. This general \nresult is independent of the choice of positive direction of rotation. Note that in Figure \n16.1, the CD has the angular velocity vector points downward (in the \u2212k\u02c6 -direction). \n16.2.4 Tangential Velocity and Tangential Acceleration \nBecause the small element of mass, \u0394mi , is moving in a circle of radius ri with angular \n\uf072\nvelocity \u03c9 = \u03c9 z k\u02c6 , the element has a tangential velocity component \n\u03c9 . \n(16.1.3)\nv\u03b8 , i = ri \nz \n16-4", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_448", "text": "If the magnitude of the tangential velocity is changing, the mass element undergoes a \ntangential acceleration given by \n\u03b1 . \n(16.1.4)\na\u03b8 , i = ri \nz \nRecall that the mass element is always accelerating inward with radial component given \nby \n2v\u03b8 , i \n2\na \n= \u2212 \n= \u2212r \u03c9 . \n(16.1.5)\nr , i \ni \nz \nri  \nExample 16.1 Turntable \n1\nA turntable is a uniform disc of mass 1.2 kg and a radius 1.3 10 cm . The turntable is \n\u00d7 \nspinning initially in a counterclockwise direction when seen from above at a constant rate \nof f0 = 33 cycles min\u22121 (33 rpm ). The motor is turned off and the turntable slows to a \n\u22c5 \nstop in 8.0 s . Assume that the angular acceleration is constant. (a) What is the initial \nangular velocity of the turntable? (b) What is the angular acceleration of the turntable? \nSolution: (a) Choose a coordinate system shown in Figure 16.6. \nFigure 16.6 Coordinate system for turntable \nInitially, the disc is spinning with a frequency \n\u239b \ncycles \u239e\u239b1min \u239e\nf0 = \u239c33 \n\u239f\u239c \n\u22c5\n\u22121 = 0.55 Hz , \n(16.1.6)\n\u239f = 0.55 cycles s \n\u239d \nmin \u23a0\u239d 60 s \u23a0 \nso the initial angular velocity has magnitude \n\u239b \nradian \u239e\u239b \ncycles \u239e\n\u03c90 = 2\u03c0 f0 = \u239c 2\u03c0 \n\u239f\u239c 0.55 \n\u239f = 3.5 rad s\u22c5\n\u22121 . \n(16.1.7)\n\u239d \ncycle \u23a0\u239d \ns \n\u23a0 \nThe angular velocity vector points in the +k\u02c6 -direction as shown above. \n16-5", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_449", "text": "(b) The final angular velocity is zero, so the component of the angular acceleration is \n\u0394\u03c9 z \n\u03c9 f \u2212\u03c90 \n\u22123.5 rad \u22c5 s\u22121 \n\u22122\n\u03b1 z =\n=\n= \n= \u22124.3 \u00d7 10\u22121 rad \u22c5 s . \n(16.1.8)\n\u0394t \nt f \u2212 t0 \n8.0 s \nThe z -component of the angular acceleration is negative, the disc is slowing down and so \nthe angular acceleration vector then points in the \u2212 \u02c6k -direction as shown in Figure 16.7. \nFigure 16.7 Angular acceleration vector for turntable \n16.3 Rotational Kinetic Energy and Moment of Inertia \n16.3.1 Rotational Kinetic Energy and Moment of Inertia \nWe have already defined translational kinetic energy for a point object as K = (1/ 2)mv2 ; \nwe now define the rotational kinetic energy for a rigid body about its center of mass. \nFigure 16.8 Volume element undergoing fixed-axis rotation about the z -axis that passes \nthrough the center of mass. \nChoose the z -axis to lie along the axis of rotation passing through the center of mass. As \nin Section 16.2.2, divide the body into volume elements of mass \u0394mi (Figure 16.8). Each \nindividual mass element \u0394mi undergoes circular motion about the center of mass with z \u00ad\n16-6", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_450", "text": "component of angular velocity \u03c9 cm in a circle of radius r \n. Therefore the velocity of \ncm, i \n\uf072\neach element is given by v \n= r \u03c9\n\u03b8\u02c6 . The rotational kinetic energy is then\ncm, i \ncm, i \ncm \n1 \n2\n1 \n2\n2\nK cm, i = 2 \u0394mi v cm, i = 2 \u0394mir cm, i\u03c9 cm . \n(16.2.1) \nWe now add up the kinetic energy for all the mass elements, \ni=N\ni=N \u239b\u2211 \n1 \n\u239e\nK \n= lim \nK \n= lim \n\u0394mir 2\n2 \ncm \n\u2211 \ncm, i \n\u2211\u239d\u239c \ncm, i \u23a0\u239f \u03c9 cm \ni\u2192\u221e \ni\u2192\u221e \n2\ni=1 \ni=1 \ni\n\u0394mi\u21920 \n\u0394mi\u21920 \n(16.2.2)\n\u239b\n\u239e\n1 \n2\n2\n= \n,\ndmrdm \u239f\u03c9 cm \n\u239c 2 \u222b\n\u239d \nbody \n\u23a0 \nwhere dm is an infinitesimal mass element undergoing a circular orbit of radius rdm \nabout the axis passing through the center of mass. \nThe quantity \n2\nI = \ndmrdm . \n(16.2.3)\ncm \n\u222b \nbo dy \nis called the moment of inertia of the rigid body about a fixed axis passing \nthrough the center of mass, and is a physical property of the body. The SI units for \nmoment of inertia are \u23a1kg \u22c5 m2 \u23a4\u23a6 .\n\u23a3 \nThus \n\u239b\n\u239e\n1\n1\n2\n2 \n2\nK cm = \u239c 2 \u222b dmrdm \u239f\u03c9 cm \u2261 \nIcm \u03c9 cm . \n(16.2.4)\n2\n\u239d \nbody \n\u23a0 \n16.3.2 Moment of Inertia of a Rod of Uniform Mass Density \nConsider a thin uniform rod of length L and mass m . In this problem, we will calculate \nthe moment of inertia about an axis perpendicular to the rod that passes through the \ncenter of mass of the rod. A sketch of the rod, volume element, and axis is shown in \nFigure 16.9. Choose Cartesian coordinates, with the origin at the center of mass of the \nrod, which is midway between the endpoints since the rod is uniform. Choose the x -axis \nto lie along the length of the rod, with the positive x -direction to the right, as in the \nfigure. \n16-7", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_451", "text": "Figure 16.9 Moment of inertia of a uniform rod about center of mass. \nIdentify an infinitesimal mass element dm = \u03bb dx , located at a displacement x from the \ncenter of the rod, where the mass per unit length \u03bb = m / L is a constant, as we have \nassumed the rod to be uniform. When the rod rotates about an axis perpendicular to the \nrod that passes through the center of mass of the rod, the element traces out a circle of \nradius rdm = x . We add together the contributions from each infinitesimal element as we \ngo from x = \u2212L 2 to x = L 2 . The integral is then \nL/2 \nL/ 2 \n2\nI \n= \u222b r\ndm = \u03bb\u222b\u2212 \n(x2) dx = \u03bb x3 \ncm \ndm \nL/ 2 \n3\nbo dy \n\u2212 L/ 2 \n(16.2.5) \nm (L / 2)3 \nm (\u2212L / 2)3 \n1 \n= \n\u2212 \n= \nm L2.\nL \n3 \nL \n3 \n12 \nBy using a constant mass per unit length along the rod, we need not consider variations in \nthe mass density in any direction other than the x - axis. We also assume that the width is \nthe rod is negligible. (Technically we should treat the rod as a cylinder or a rectangle in \nthe \u00ad\nx y plane if the axis is along the z - axis. The calculation of the moment of inertia in \nthese cases would be more complicated.) \nExample 16.2 Moment of Inertia of a Uniform Disc \nA thin uniform disc of mass M and radius R is mounted on an axle passing through the \ncenter of the disc, perpendicular to the plane of the disc. Calculate the moment of inertia \nabout an axis that passes perpendicular to the disc through the center of mass of the disc \nSolution: As a starting point, consider the contribution to the moment of inertia from the \nmass element dm show in Figure 16.10. Let r denote the distance form the center of \nmass of the disc to the mass element. \nFigure 16.10 Infinitesimal mass element and coordinate system for disc. \n16-8", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_452", "text": "Choose cylindrical coordinates with the coordinates (r,\u03b8) in the plane and the z -axis \nperpendicular to the plane. The area element \nda = r dr d\u03b8 \n(16.2.6) \nmay be thought of as the product of arc length r d\u03b8 and the radial width dr . Since the \ndisc is uniform, the mass per unit area is a constant, \ndm\nm \nM\ntotal \n\u03c3 =\n= \n= \n. \n(16.2.7)\n\u03c0 R2\nda \nArea \nTherefore the mass in the infinitesimal area element as given in Equation (16.2.6), a \ndistance r from the axis of rotation, is given by \nM\ndm = \u03c3 r dr d\u03b8 = \nr dr d\u03b8 . \n(16.2.8)\n\u03c0 R2 \nWhen the disc rotates, the mass element traces out a circle of radius rdm = r ; that is, the \ndistance from the center is the perpendicular distance from the axis of rotation. The \nmoment of inertia integral is now an integral in two dimensions; the angle \u03b8 varies from \n\u03b8 = 0 to \u03b8 = 2\u03c0 , and the radial coordinate r varies from r = 0 to r = R . Thus the limits \nof the integral are \n2 \nM\nr \n\u03b8 = \nI cm = \u222b rdm dm = \n\u03c0 R2 \u222b \n= R \u222b\u03b8 =0\n2\u03c0 r3 d\u03b8 dr . \n(16.2.9)\nr =0 \nbody \nThe integral can now be explicitly calculated by first integrating the \u03b8 -coordinate \nM\nr = R \n\u03b8 = 2\u03c0 \nM\nr = R \n2 M\nr = R\n\u239e\nI cm = \n\u03c0 R2 \u222b \n\u239b\u222b \nd\u03b8\u23a0 r3dr = \n\u03c0 R2 \u222b \n2\u03c0r3dr = \nR2 \u222b \nr3dr \n(16.2.10)\nr =0 \u239d\n\u03b8 =0 \nr =0 \nr =0 \nand then integrating the r -coordinate, \nr = R \n2 M\nr = R \n2 M r 4 \n2 M R4\n1\nI \n= \nr3dr = \n=\n= \nMR2 . \n(16.2.11)\ncm \nR2 \u222br =0 \nR2 4 \nR2\n4 \n2 \nr =0 \nRemark: Instead of taking the area element as a small patch da = r dr d\u03b8 , choose a ring \nof radius r and width dr . Then the area of this ring is given by", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_453", "text": "MR2 . \n(16.2.11)\ncm \nR2 \u222br =0 \nR2 4 \nR2\n4 \n2 \nr =0 \nRemark: Instead of taking the area element as a small patch da = r dr d\u03b8 , choose a ring \nof radius r and width dr . Then the area of this ring is given by \ndaring = \u03c0(r + dr)2 \u2212\u03c0r 2 = \u03c0r 2 + 2\u03c0r dr + \u03c0 (dr)2 \u2212\u03c0r 2 = 2\u03c0r dr + \u03c0(dr)2 . (16.2.12) \n16-9", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_454", "text": "In the limit that dr \u2192 0 , the term proportional to (dr)2 can be ignored and the area is \nda = 2\u03c0rdr . This equivalent to first integrating the d\u03b8 variable \n\u03b8 = 2\u03c0 \n\u239e\ndaring = r dr \u239b\n\u239d \u222b\u03b8 =0 d\u03b8\u23a0 = 2\u03c0r dr . \n(16.2.13) \nThen the mass element is \nM\ndm \n= \u03c3 da \n= \n2\u03c0r dr . \n(16.2.14)\nring \nring \n\u03c0 R2 \nThe moment of inertia integral is just an integral in the variable r , \n2\u03c0 M\nr = R \n1\nI cm = \u222b (r\u22a5 )2 dm = \n\u03c0 R2 \nr3dr = \nMR2 . \n(16.2.15)\n\u222br =0\n2\nbody \n16.3.3 Parallel Axis Theorem \nConsider a rigid body of mass m undergoing fixed-axis rotation. Consider two parallel \naxes. The first axis passes through the center of mass of the body, and the moment of \ninertia about this first axis is Icm . The second axis passes through some other point S in \nthe body. Let dS ,cm denote the perpendicular distance between the two parallel axes \n(Figure 16.11). \nFigure 16.11 Geometry of the parallel axis theorem. \nThen the moment of inertia IS about an axis passing through a point S is related to Icm \nby \nIS = I\n+ m dS \n2 \n. \n(16.2.16)\ncm \n,cm \n16-10", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_455", "text": "16.3.4 Parallel Axis Theorem Applied to a Uniform Rod \nLet point S be the left end of the rod of Figure 16.9. Then the distance from the center of \nmass to the end of the rod is dS ,cm = L / 2 . The moment of inertia IS = Iend about an axis \npassing through the endpoint is related to the moment of inertia about an axis passing \nthrough the center of mass, Icm = (1/12) m L2 , according to Equation (16.2.16), \n1 \n2\n1\n2\n1\n2\nIS = \nm L + m L = m L . \n(16.2.17)\n12\n4 \n3 \nIn this case it\u2019s easy and useful to check by direct calculation. Use Equation (16.2.5) but \nwith the limits changed to x\u2032 = 0 and x\u2032 = L , where x\u2032 = x + L / 2 , \nIend = \u222b r\u22a5 \n2 dm = \u03bb\u222b\nL x\u20322 dx\u2032 \n0 \nbody \nL \n(16.2.18) \nx\u20323 \nm ( )\nL 3 \nm (0)3\n1 \n2\n= \u03bb \n= \n\u2212 \n= m L . \n3 0 \nL 3 \nL 3\n3 \nExample 16.3 Rotational Kinetic Energy of Disk \nA disk with mass M and radius R is spinning with angular speed \u03c9 about an axis that \npasses through the rim of the disk perpendicular to its plane. The moment of inertia about \ncm is Icm = (1/ 2)mR2 . What is the kinetic energy of the disk? \nSolution: The parallel axis theorem states the moment of inertia about an axis passing \nperpendicular to the plane of the disc and passing through a point on the edge of the disc \nis equal to \nI \n= I + mR2 . \n(16.2.19)\nedge \ncm \nThe moment of inertia about an axis passing perpendicular to the plane of the disc and \npassing through the center of mass of the disc is equal to Icm = (1/ 2)mR2 . Therefore \nIedge = (3/ 2)mR2 . \n(16.2.20) \nThe kinetic energy is then \nK = (1/ 2)Iedge\u03c9 2 = (3/ 4)mR2\u03c9 2 . \n(16.2.21) \n16-11", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_456", "text": "16.4 Conservation of Energy for Fixed Axis Rotation \nConsider a closed system ( \u0394E \n= 0 ) under action of only conservative internal forces.\nsystem \nThen the change in the mechanical energy of the system is zero \n\u0394E \n) = 0 . \n(16.3.1)\nm = \u0394U + \u0394K = (U f + K f ) \u2212 (Ui + Ki \nFor fixed axis rotation with a component of angular velocity \u03c9 about the fixed axis, the \nchange in kinetic energy is given by \n1 \u03c9 2 \n1\n\u0394K \u2261 K f \n= \nIS\n\u2212 IS\u03c9 i \n2 ,  \n(16.3.2)\n\u2212 Ki \nf\n2\n2 \nwhere S is a point that lies on the fixed axis. Then conservation of energy implies that \n1\n2 \n1\n2\n+ IS\n=\n+ IS \n(16.3.3)\nU f \n\u03c9 f \nUi \n\u03c9 i\n2\n2 \nExample 16.4 Energy and Pulley System \nA wheel in the shape of a uniform disk of radius R and mass mp is mounted on a \nfrictionless horizontal axis. The wheel has moment of inertia about the center of mass \nI \n= (1/ 2)m R2 . A massless cord is wrapped around the wheel and one end of the cord is \ncm \np \nattached to an object of mass m2 that can slide up or down a frictionless inclined plane. \nThe other end of the cord is attached to a second object of mass m1 that hangs over the \nedge of the inclined plane. The plane is inclined from the horizontal by an angle \u03b8 \n(Figure 16.12). Once the objects are released from rest, the cord moves without slipping \naround the disk. Calculate the speed of block 2 as a function of distance that it moves \ndown the inclined plane using energy techniques. Assume there are no energy losses due \nto friction and that the rope does not slip around the pulley \nFigure 16.12 Pulley and blocks  \nFigure 16.13 Coordinate system for \npulley and blocks \n16-12", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_457", "text": "Solution: Define a coordinate system as shown in Figure 16.13. Choose the zero for the \ngravitational potential energy at a height equal to the center of the pulley. In Figure 16.14 \nillustrates the energy diagrams for the initial state and a dynamic state at an arbitrary time \nwhen the blocks are sliding. \nFigure 16.14 Energy diagrams for initial state and dynamic state at arbitrary time \nThen the initial mechanical energy is \nEi = Ui = \u2212m1gy1, i \u2212 m2 gx2,i sin\u03b8 . \n(16.3.4) \nThe mechanical energy, when block 2 has moved a distance \nd = x2 \n(16.3.5)\n\u2212 x2, i \nis given by \n1\n2 + 1 \n2 + 1 \u03c9 2\nE = U + K = \u2212m1\nsin\u03b8 + \n. \n(16.3.6)\ngy1 \u2212 m2 gx2 \nm1v1 \nm2v2 \nIP\n2\n2 \n2 \nThe rope connects the two blocks, and so the blocks move at the same speed \nv \u2261 v1 = v2 . \n(16.3.7) \nThe rope does not slip on the pulley; therefore as the rope moves around the pulley the \ntangential speed of the rope is equal to the speed of the blocks \nvtan = R\u03c9 = v . \n(16.3.8) \nEq. (16.3.6) can now be simplified \n1 \u239b \nIP \u239e \n2\nE = U + K = \u2212m gy \u2212 m gx sin\u03b8 + \nm + m + \n. \n(16.3.9)\n1\n1\n2\n2 \n1\n2 \nR2 \u23a0\u239f v \n2 \u239d\u239c \nBecause we have assumed that there is no loss of mechanical energy, we can set Ei = E \nand find that \n16-13", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_458", "text": "3\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n  \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n  \n \n \n \n \n  \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n  \n \n \n \n \n  \n \n \n \n \n  \n \n \n \n \n \n \n \n  \n \n  \n \n \n \n \n \n \n \n  \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n  \n  \n  \n \n \n \n \n \n \n \n1 \u239b \nIP \u239e \n2\n\u2212m1gy1, i \u2212 m2 gx2, i sin\u03b8 = \u2212m1gy1 \u2212 m2 gx2 sin\u03b8 + \nm1 + m2 + R2 \u23a0\u239f v , (16.3.10)\n2 \u239d\u239c \nwhich simplifies to \n1 \u239b\n\u239e\nIP \n2\n\u2212m1g( y1,0 \u2212 y1) + m2 g(x2 \u2212 x2,0 )sin\u03b8 = \nm1 + m2 + R2 \u23a0\u239f v . \n(16.3.11)\n2 \u239d\u239c \nWe finally note that the movement of block 1 and block 2 are constrained by the \nrelationship \nd = x2 \n= \n\u2212 y1 . \n(16.3.12)\n\u2212 x2, i\ny1, i \nThen Eq. (16.3.11) becomes \n1 \u239b \nIP \u239e \n2\ngd(\u2212m1 + m2 sin\u03b8) = \nm1 + m2 + \nR2 \u23a0\u239f v . \n(16.3.13)\n2 \u239d\u239c \nWe can now solve for the speed as a function of distance d = x2 \nthat block 2 has \n\u2212 x2, i \ntraveled down the incline plane \n2gd(\u2212m1 \nsin\u03b8)\nv =\n+ m2 \n. \n(16.3.14)\n(m1 + m2 + (IP / R2)) \nIf we assume that the moment of inertial of the pulley is Icm \np \n= (1/ 2)m R2 , then the speed \nbecomes \n(16.3.15) \nExample 16.5 Physical Pendulum \nA physical pendulum consists of a uniform rod of mass m1 pivoted at one end about the \npoint S . The rod has length l1 and moment of inertia I1 about the pivot point. A disc of \nmass \nand radius \nwith moment of inertia I \nabout its center of mass is rigidly \nm2 \nr2 \ncm \nattached a distance l2 from the pivot point. The pendulum is initially displaced to an \nangle \u03b8i and then released from rest. (a) What is the moment of inertia of the physical \npendulum about the pivot point S ? (b) How far from the pivot point is the center of mass \nof the system? (c) What is the angular speed of the pendulum when the pendulum is at \nthe bottom of its swing? \nv = \n2gd(\u2212m1 + m2 sin\u03b8) \nm1 + m2 + (1 / 2)mP\n( \n) \n. \n16-14", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_459", "text": "Figure 16.15 Rod and with fixed disc pivoted about the point S \nSolution: a) The moment of inertia about the pivot point is the sum of the moment of \ninertia of the rod, given as I1 , and the moment of inertia of the disc about the pivot point.  \nThe moment of inertia of the disc about the pivot point is found from the parallel axis \ntheorem, \nI \n= I + m l2 . \n(16.3.16)\ndisc \ncm \n2\n2 \nThe moment of inertia of the system consisting of the rod and disc about the pivot point \nS is then \nI = I + I \n= I + I + m l2 . \n(16.3.17)\nS \n1 \ndisc \n1 \ncm \n2\n2 \nThe center of mass of the system is located a distance from the pivot point \n/ 2) + m2\nm1(l1 \nl2\nl cm = \n. \n(16.3.18)\nm1 + m2 \nb) We can use conservation of mechanical energy, to find the angular speed of the \npendulum at the bottom of its swing. Take the zero point of gravitational potential energy \nto be the point where the bottom of the rod is at its lowest point, that is, \u03b8 = 0 . The \ninitial state energy diagram for the rod is shown in Figure 16.16a and the initial state \nenergy diagram for the disc is shown in Figure 16.16b. \n16-15", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_460", "text": "6\n \n \n \n \n \n \n \n \n \n \n \n \n \n  \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n  \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n        \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n(a) \n(b) \nFigure 16.16 (a) Initial state energy diagram for rod (b) Initial state energy diagram for \ndisc \nThe initial mechanical energy is then \nl1\n= \n\u2212 cos\u03b8i \ncos\u03b8i ) , \n(16.3.19)\nEi \nUi = m1 g (l1 \n) + m2 g (l1 \u2212 l2\n2 \nAt the bottom of the swing, \u03b8 f = 0 , and the system has angular velocity \u03c9 f . The \nmechanical energy at the bottom of the swing is \nl1 \n1\n2\ng \n\u2212 l2) + IS\u03c9 f , \n(16.3.20)\nE f = U f + K f = m1 \n+ m2 g(l1\n2\n2 \nwith IS as found in Equation (16.3.17). There are no non-conservative forces acting, so \nthe mechanical energy is constant therefore equating the expressions in (16.3.19) and \n(16.3.20) we get that \nl1 \nl1\n1 \u03c9 2\n\u2212 cos\u03b8i \ncos\u03b8i \ng \n) + \n(16.3.21)\nm1 g (l1 \n) + m2 g (l1 \u2212 l2 \n) = m1 \n+ m2 g(l1 \u2212 l2 \nIS\nf ,\n2 \n2\n2 \nThis simplifies to \n\u239b\n\u239e \n1\nm1 l1 \nl2 \u23a0\u239f g (1\u2212 cos\u03b8i ) = \n2 \n(16.3.22)\n+ m2 \nIS\u03c9 f ,\n\u239d\u239c 2\n2 \nWe now solve for \u03c9 f (taking the positive square root to insure that we are calculating \nangular speed) \n16-16", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_461", "text": "\u239e\n2\u239b m1 l1 + m2 l2 \u23a0\u239f g (1\u2212 cos\u03b8i )\n\u239d\u239c 2\n\u03c9 f = \n, \n(16.3.23)\nIS \nFinally we substitute in Eq.(16.3.17) in to Eq. (16.3.23) and find \n\u239e\n2\u239b m1 l1 + m2 l2 \u23a0\u239f g (1\u2212 cos\u03b8i )\n\u239d\u239c 2 \n= \n. \n(16.3.24)\n\u03c9 f \n2\n+ I + m2\nI1 \ncm \nl2 \nNote that we can rewrite Eq. (16.3.22), using Eq. (16.3.18) for the distance between the \ncenter of mass and the pivot point, to get \n1\n2\n+ m2)l g (1\u2212 cos\u03b8i ) = \nIS\u03c9 f , \n(16.3.25)\n(m1 \ncm \n2 \nWe can interpret this equation as follows. Treat the system as a point particle of mass \nlocated at the center of mass l . Take the zero point of gravitational potential\nm1 + m2 \ncm \nenergy to be the point where the center of mass is at its lowest point, that is, \u03b8 = 0 . Then \n)l g (1\u2212 cos\u03b8i ) , \n(16.3.26)\nEi = (m1 + m2 \ncm\n1 \n2\n= \n. \n(16.3.27)\nE f \nIS\u03c9 f\n2 \nThus conservation of energy reproduces Eq. (16.3.25). \n16-17", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_462", "text": "8\n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n  \n \n \n \n \n \n \n  \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n  \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n  \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n  \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n  \n \n  \n \n \n \n \n \n \n  \n \n \n \n \n  \n \n \n \n \n \n \n \n \n \n \n \n \n \n  \n \n \n \n \n \nAppendix 16A: Proof of the Parallel Axis Theorem \nIdentify an infinitesimal volume element of mass dm . The vector from the point S to the \n\uf072\n\uf072\nmass element is \n, \n, the vector from the center of mass to the mass element is rdm , and \nrS dm \n\uf072\nthe vector from the point S to the center of mass is rS ,cm . \n\uf072r \n\uf072r \n\uf072r \nFigure 16A.1 Geometry of the parallel axis theorem. \nS , dm \nS ,cm \ndm . \nFrom Figure 16A.1, we see that \n(16.A.1) \n=  \n+  \n\uf072\n\uf072\nThe notation gets complicated at this point. The vector r \nhas a component vector r\ndm \n\uf050, dm \n\uf072\nparallel to the axis through the center of mass and a component vector r\u22a5 , dm \nperpendicular to the axis through the center of mass. The magnitude of the perpendicular \ncomponent vector is \n\uf072 \n(16.A.2)\nr cm, \u22a5 , dm = r\u22a5 , dm . \n\uf072\n\uf072\nThe vector rS dm has a component vector \nparallel to the axis through the point S \n, \nrS ,\uf050, dm \n\uf072\nand a component vector rS ,\u22a5,dm perpendicular to the axis through the point S . The \nmagnitude of the perpendicular component vector is \n\uf072 r \n= rS ,\u22a5,dm . \n(16.A.3)\nS ,\u22a5,dm \n\uf072\n\uf072\nThe vector rS ,cm has a component vector rS ,\uf050,cm parallel to both axes and a perpendicular \n\uf072\ncomponent vector rS ,\u22a5,cm that is perpendicular to both axes (the axes are parallel, of \ncourse). The magnitude of the perpendicular component vector is \n\uf072 r \n= dS ,cm . \n(16.A.4)\nS ,\u22a5,cm \n16-18", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_463", "text": "Equation (16.A.1) is now expressed as two equations, \nS ,\u22a5, dm \nS ,\u22a5,cm \n\u22a5,dm \nS ,\uf050,dm \nS ,\uf050,cm \n\uf072r \n\uf072r \n\uf072r \n\uf072r \n\uf072r\n+ \n=  \n(16.A.5) \n\uf072r\n+ \n=  \n\uf050,dm . \nAt this point, note that if we had simply decided that the two parallel axes are parallel to \nthe z -direction, we could have saved some steps and perhaps spared some of the notation \nwith the triple subscripts. However, we want a more general result, one valid for cases \nwhere the axes are not fixed, or when different objects in the same problem have different \naxes. For example, consider the turning bicycle, for which the two wheel axes will not be \nparallel, or a spinning top that precesses (wobbles). Such cases will be considered in \nlater on, and we will show the general case of the parallel axis theorem in anticipation of \nuse for more general situations. \nThe moment of inertia about the point S is \n\uf072r \n= \n)2 . \n(16.A.6)\nIS \n\u222b dm(rS , \u22a5,dm \nbody \n\uf072r \nFrom (16.A.5) we have \nS , \u22a5,dm \nS , \u22a5,dm \nS , \u22a5,cm \n)2\n(rS , \u22a5,dm \n\u22c5 \n=  \n\uf072r \n\uf072r \n\uf072rS , \u22a5,cm \n\u22a5,dm \n2 \n)2 + 2\n= dS ,cm + (r\u22a5, dm \n= (  \n) \u22c5 (  \n) \n(16.A.7) \n+  \n+  \n\uf072r \n\u22a5,dm \n\uf072rS , \u22a5,cm \n\u22a5,dm. \nThus we have for the moment of inertia about S , \n\uf072\n\uf072 \n= \u222b dm dS \n2 \n+ \n)2 + 2 \u222b dm(r \n\u22c5 r \n) . \n(16.A.8)\nIS \n,cm \n\u222b dm(r\u22a5,dm \nS ,\u22a5,cm \n\u22a5,dm \nbody \nbody \nbody \nIn the first integral in Equation (16.A.8), rS ,\u22a5,cm = dS ,cm is the distance between the \nparallel axes and is a constant. Therefore we can rewrite the integral as \n2 \ndm \n2 \n. \n(16.A.9)\ndS ,cm \u222b \n= m dS ,cm \nbody \nThe second term in Equation (16.A.8) is the moment of inertia about the axis through the \ncenter of mass, \n)2\nI \n= \u222b dm (r\u22a5,dm \n. \n(16.A.10)\ncm \nbo dy \n\u22c5  \n\uf072r", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_464", "text": "dS ,cm \u222b \n= m dS ,cm \nbody \nThe second term in Equation (16.A.8) is the moment of inertia about the axis through the \ncenter of mass, \n)2\nI \n= \u222b dm (r\u22a5,dm \n. \n(16.A.10)\ncm \nbo dy \n\u22c5  \n\uf072r \n16-19", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_465", "text": "0\n \n \n  \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n \n    \n \n    \n     \n  \n \n \n \n \n \n \n  \n \n \n \n \n\uf072\nThe third integral in Equation (16.A.8) is zero. To see this, note that the term rS ,\u22a5,cm is a \nconstant and may be taken out of the integral, \nS ,\u22a5,cm \n\u22a5,dm \nS ,\u22a5,cm \nbody \nbody \n\uf072\nThe integral \u222b dm r\u22a5,dm is the perpendicular component of the position of the center of \nbo dy \n\uf072 \nmass with respect to the center of mass, and hence 0 , with the result that \n2 \u222b dm (r \uf072 \n\u22c5 r \uf072 \n) = 0 . \n(16.A.12)\nS ,\u22a5,cm \n\u22a5,dm \nbody \nThus, the moment of inertia about S is just the sum of the first two integrals in \n\uf072r \nEquation (16 A 8)\n. .\nI = I + md 2 \n, \n(16.A.13)\nS \ncm \nS ,cm \n\uf072r \nproving the parallel axis theorem. \n\uf072r \n\uf072r\n2 \u222b dm ( \n\u22c5  \n) =  \n\u22c5 2 \u222b dm\n(16.A.11) \n\u22a5,dm \n16-20", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_466", "text": "Chapter 17 Two Dimensional Rotational Dynamics \ntorque, n. \na. The twisting or rotary force in a piece of mechanism (as a \nmeasurable quantity); the moment of a system of forces producing \nrotation. \nOxford English Dictionary \n17.1 Introduction \nA body is called a rigid body if the distance between any two points in the body \ndoes not change in time. Rigid bodies, unlike point masses, can have forces applied at \ndifferent points in the body. For most objects, treating as a rigid body is an idealization, \nbut a very good one. In addition to forces applied at points, forces may be distributed \nover the entire body. Forces that are distributed over a body are difficult to analyze; \nhowever, for example, we regularly experience the effect of the gravitational force on \nbodies. Based on our experience observing the effect of the gravitational force on rigid \nbodies, we shall demonstrate that the gravitational force can be concentrated at a point in \n\uf072\nthe rigid body called the center of gravity, which for small bodies (so that g may be \ntaken as constant within the body) is identical to the center of mass of the body. \nLet\u2019s consider a rigid rod thrown in the air (Figure 17.1) so that the rod is \n\uf072\nspinning as its center of mass moves with velocity vcm . We have explored the physics of \ntranslational motion; now, we wish to investigate the properties of rotational motion \nexhibited in the rod\u2019s motion, beginning with the notion that every particle is rotating \nabout the center of mass with the same angular (rotational) velocity. \nFigure 17.1 The center of mass of a thrown rigid rod follows a parabolic trajectory while \nthe rod rotates about the center of mass. \nWe can use Newton\u2019s Second Law to predict how the center of mass will move. \nBecause the only external force on the rod is the gravitational force (neglecting the action \nof air resistance), the center of mass of the body will move in a parabolic trajectory. \n17-1", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_467", "text": "How was the rod induced to rotate? In order to spin the rod, we applied a torque \nwith our fingers and wrist to one end of the rod as the rod was released. The applied \ntorque is proportional to the angular acceleration. The constant of proportionality is the \nmoment of inertia. When external forces and torques are present, the motion of a rigid \nbody can be extremely complicated while it is translating and rotating in space. \nIn order to describe the relationship between torque, moment of inertia, and \nangular acceleration, we will introduce a new vector operation called the vector product \nalso know as the \u201ccross product\u201d that takes any two vectors and generates a new vector. \nThe vector product is a type of \u201cmultiplication\u201d law that turns our vector space (law for \naddition of vectors) into a vector algebra (a vector algebra is a vector space with an \nadditional rule for multiplication of vectors). \n17.2 Vector Product (Cross Product) \n\uf072\n\uf072 \nLet A and B be two vectors. Because any two non-parallel vectors form \n\uf072 \na plane, we denote the angle \u03b8 to be the angle between the vectors A and \n\uf072 \n\uf072\n\uf072 \nB as shown in Figure 17.2. The magnitude of the vector product A \u00d7 B \n\uf072\n\uf072 \nof the vectors A and B is defined to be product of the magnitude of the \n\uf072\n\uf072 \nvectors A and B with the sine of the angle \u03b8 between the two vectors, \n\uf072 \nA \u00d7  \n\uf072 \nB =  \n\uf072 \nB  \n\uf072 \nA \nsin(\u03b8) . \n(17.2.1) \nThe angle \u03b8 between the vectors is limited to the values 0 \u2264\u03b8 \u2264\u03c0 \nensuring that sin( ) \u2265 0 .\n\u03b8 \nFigure 17.2 Vector product geometry. \n\uf072 \nThe direction of the vector product is defined as follows. The vectors A \n\uf072 \nand B form a plane. Consider the direction perpendicular to this plane. \nThere are two possibilities: we shall choose one of these two (the one \n\uf072\n\uf072 \nshown in Figure 17.2) for the direction of the vector product A \u00d7 B using \na convention that is commonly called the \u201cright-hand rule\u201d. \n17-2", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_468", "text": "17.2.1 Right-hand Rule for the Direction of Vector Product \n\uf072\n\uf072 \nso that the tails are touching. Then draw \n\uf072 \nB \nThe first step is to redraw the vectors A and B \n\uf072 \nan arc starting from the vector A and finishing on the vector \n. Curl your right fingers \nthe same way as the arc. Your right thumb points in the direction of the vector product \n\uf072\n\uf072 \nA \u00d7 B (Figure 17.3). \nFigure 17.3 Right-Hand Rule. \n\uf072\n\uf072 \nYou should remember that the direction of the vector product A \u00d7 B is perpendicular to \n\uf072\n\uf072 \nthe plane formed by A and B . We can give a geometric interpretation to the magnitude \nof the vector product by writing the magnitude as \n\uf072 \nA \u00d7  \n\uf072 \nB =  \n\uf072 \nA ( \n\uf072 \nB sin\u03b8 ). \n(17.2.2) \n\uf072\n\uf072 \nThe vectors A and B form a parallelogram. The area of the parallelogram is equal to the \nheight times the base, which is the magnitude of the vector product. In Figure 17.4, two \n\uf072 \nB \ndifferent representations of the height and base of a parallelogram are illustrated. As \n\uf072 \ndirection perpendicular to the vector B . We could also write the magnitude of the vector \n\uf072  \nB\ndepicted in Figure 17.4a, the term \nsin\u03b8 is the projection of the vector \nin the \nproduct as\n\uf072 \nA \u00d7  \n\uf072 \nB = ( \n\uf072 \nA sin\u03b8 )\n\uf072 \nB .  \n(17.2.3)  \n\uf072 \n\uf072 \nA\nThe term \nsin\u03b8 is the projection of the vector A in the direction perpendicular to the \nvector \n\uf072 \nB as shown in Figure 17.4(b). The vector product of two vectors that are parallel \n(or anti-parallel) to each other is zero because the angle between the vectors is 0 (or \u03c0 ) \nand sin(0) = 0 (or sin( ) \n\u03c0 = 0 ). Geometrically, two parallel vectors do not have a unique \ncomponent perpendicular to their common direction. \n17-3", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_469", "text": "(a)  \n(b) \nFigure 17.4 Projection of (a) \n\uf072 \nB perpendicular to \n\uf072 \nA , (b) of \n\uf072 \nA perpendicular to \n\uf072 \nB \n17.2.2 Properties of the Vector Product \n(1)  The vector product is anti-commutative because changing the order of the vectors \nchanges the direction of the vector product by the right hand rule: \n\uf072\n\uf072 \n\uf072\n\uf072 \nA \u00d7 B = \u2212B \u00d7 A .  \n(17.2.4) \n\uf072  \n(2) The vector product between a vector c A where c is a scalar and a vector \n\uf072 \nB is \n\uf072\n\uf072 \n\uf072\n\uf072 \nc A \u00d7 B = c (A \u00d7 B) . \n(17.2.5) \nSimilarly, \n\uf072 \n\uf072 \n\uf072\n\uf072 \nA \u00d7 c B = c (A \u00d7 B) .  \n(17.2.6) \n\uf072\n\uf072  \n\uf072 \n(3)  The vector product between the sum of two vectors A and B with a vector C is \n\uf072 \n\uf072 \n\uf072 \n\uf072\n\uf072\n\uf072\n\uf072 \n(A + B)\u00d7C = A \u00d7C + B\u00d7C \n(17.2.7) \nSimilarly, \n\uf072  \n\uf072\n\uf072 \n\uf072\n\uf072\n\uf072\n\uf072 \nA \u00d7 (B + C) = A \u00d7 B + A \u00d7C .  \n(17.2.8) \n17.2.3 Vector Decomposition and the Vector Product: Cartesian Coordinates \nWe first calculate that the magnitude of vector product of the unit vectors \u02c6i and \u02c6j : \n\u02c6 \u02c6 \n\u02c6 \u02c6\n| i \u00d7 j | | || |sin( \n= i\nj \n\u03c0 / 2) = 1,  \n(17.2.9) \n\u02c6\n\u02c6\nbecause the unit vectors have magnitude | | | |\ni = j = 1 and sin( \u03c0 / 2) = 1. By the right hand \nrule, the direction of \u02c6i \u00d7 \u02c6j is in the +k\u02c6 as shown in Figure 17.5. Thus \u02c6i \u00d7 \u02c6j = k\u02c6 . \n17-4", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_470", "text": "Figure 17.5 Vector product of \u02c6i \u00d7 \u02c6j \nWe note that the same rule applies for the unit vectors in the y and z directions, \n\u02c6\n\u02c6 \n\u02c6\n\u02c6\n\u02c6\n\u02c6\nj\u00d7k = i, k \u00d7 i = j . \n(17.2.10) \nBy the anti-commutatively property (1) of the vector product, \n\u02c6 \u02c6 \n\u02c6\n\u02c6\n\u02c6 \n\u02c6\nj\u00d7 i = \u2212k, i \u00d7k = \u2212j \n(17.2.11) \nThe vector product of the unit vector \u02c6i with itself is zero because the two unit vectors are \nparallel to each other, (sin(0) = 0 ), \n\u02c6 \u02c6 \n\u02c6 \u02c6\n| i \u00d7 i | | || | sin(0) \ni\ni \n= 0 . \n(17.2.12)\n= \nThe vector product of the unit vector \u02c6j with itself and the unit vector k\u02c6 with itself are \nalso zero for the same reason, \n\u02c6 \u02c6 \n\u02c6\n\u02c6 = 0 . \n(17.2.13)\n\u00d7\nj j = 0, k \u00d7k \nWith these properties in mind we can now develop an algebraic expression for the vector \nproduct in terms of components. Let\u2019s choose a Cartesian coordinate system with the \n\uf072 \nvector B pointing along the positive x-axis with positive x-component Bx . Then the \n\uf072\n\uf072 \nvectors A and B can be written as \n\uf072 \n\u02c6\n\u02c6 \n\u02c6\nA = A i + A j + A k \n(17.2.14)\nx\ny\nz \n\uf072 \n\u02c6\nB = Bx i , \n(17.2.15) \nrespectively. The vector product in vector components is \n\uf072\n\uf072 \n\u02c6\n\u02c6\n\u02c6 \n\u02c6\nA \u00d7 B = ( A i + A j + A k)\u00d7 B i . \n(17.2.16)\nx\ny\nz \nx \n17-5", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_471", "text": "This becomes, \n\uf072\n\uf072 \n\u02c6\n\u02c6\n\u02c6\n\u02c6 \n\u02c6\n\u02c6\nA \u00d7 B = (A i \u00d7 B i) + (A j\u00d7 B i) + (A k \u00d7 B i)\nx\nx \ny\nx\nz\nx \n\u02c6 \u02c6\n\u02c6 \u02c6 \n\u02c6\n\u02c6\n= A B (i \u00d7 i) + A B (j\u00d7 i) + A B (k \u00d7 i) . \n(17.2.17)\nx\nx \ny\nx\nz\nx \n= \u2212A B k\u02c6 + A B \u02c6j\ny\nx \nz\nx \nThe vector component expression for the vector product easily generalizes for arbitrary \nvectors \n\uf072 \nA = A \u02c6i + A \u02c6j+ A k\u02c6 \n(17.2.18)\nx \ny \nz \n\uf072 \n\u02c6\n\u02c6\n\u02c6\nB = B i + B j + B k , \n(17.2.19)\nx\ny\nz \nto yield \n\uf072\n\uf072 \n\u02c6\n\u02c6 \n\u02c6\nA \u00d7 B = (A B \u2212 A B ) i + (A B \u2212 A B ) j + (A B \u2212 A B ) k . \n(17.2.20)\ny\nz\nz\ny \nz\nx\nx\nz \nx\ny\ny\nx \n17.2.4 Vector Decomposition and the Vector Product: Cylindrical Coordinates \nRecall the cylindrical coordinate system, which we show in Figure 17.6. We have chosen \ntwo directions, radial and tangential in the plane, and a perpendicular direction to the \nplane. \nFigure 17.6 Cylindrical coordinates  \nThe unit vectors are at right angles to each other and so using the right hand rule, the  \nvector product of the unit vectors are given by the relations \n\u02c6r \u00d7 \u02c6\u03b8 = \u02c6k \n(17.2.21) \n\u02c6\u03b8 \u00d7 \u02c6k = \u02c6r \n(17.2.22) \n\u02c6k \u00d7 \u02c6r = \u02c6\u03b8 . \n(17.2.23) \nBecause the vector product satisfies \n\uf072 \nA \u00d7  \n\uf072 \nB = \u2212  \n\uf072 \nB \u00d7  \n\uf072 \nA , we also have that \n\u03b8\u02c6 \u00d7 r\u02c6 = \u2212k\u02c6 \n(17.2.24) \n17-6", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_472", "text": "k\u02c6 \u00d7 \u03b8\u02c6 = \u2212r\u02c6 \n(17.2.25) \n\u02c6 = \u2212\u03b8\u02c6\nr\u02c6 \u00d7 k\n. \n(17.2.26) \nFinally \n\uf072 \nr\u02c6 \u00d7 r\u02c6 = \u03b8\u02c6 \u00d7 \u03b8\u02c6 = k\u02c6 \u00d7 k\u02c6 = 0 . \n(17.2.27) \nExample 17.1 Vector Products \n\uf072\n\uf072\n\u02c6 \n\u02c6\n\u02c6 \n\u02c6\n\u02c6\n\u02c6\nGiven two vectors, A = 2 i + \u22123 j + 7 k and B = 5i + j + 2k , find \n\uf072 \nA \u00d7  \n\uf072 \nB .  \nSolution: \n\uf072\n\uf072 \nA \u00d7 B = ( A B \u2212 A B ) \u02c6i + ( A B \u2212 A B ) \u02c6j + ( A B \u2212 A B ) k\u02c6 \ny\nz\nz\ny \nz\nx\nx\nz \nx\ny\ny\nx \n= ((\u22123)(2) \u2212 (7)(1)) \u02c6i + ((7)(5) \u2212 (2)(2)) \u02c6j+ ((2)(1) \u2212 (\u22123)(5)) k\u02c6 \n= \u221213 \u02c6i + 31\u02c6j+17 k\u02c6. \nExample 17.2 Law of Sines \nFor \nthe \ntriangle\n\uf072 \nB \nshown \nin \nFigure \n17.7a, \nprove \nthe \nlaw \nof \nsines, \n\uf072 \nA / sin\u03b1 = \n/ sin \u03b2 = \n\uf072 \nC / sin\u03b3 , using the vector product. \nFigure 17.7 (a) Example 17.2 \nFigure 17.7 (b) Vector analysis \n\uf072\n\uf072 \n\uf072 \nSolution: Consider the area of a triangle formed by three vectors A , B , and C , where \n\uf072\n\uf072\n\uf072 \n\uf072\n\uf072\n\uf072 \nA + B + C = 0 \n(Figure \n17.7b). \nBecause \nA + B + C = 0 , \nwe \nhave \nthat \n\uf072 \n\uf072\n\uf072\n\uf072 \n\uf072\n\uf072\n\uf072\n\uf072 \n\uf072\n\uf072 \n\uf072\n\uf072 \n\uf072\n\uf072 \n\uf072\n\uf072 \n0 = A \u00d7 (A + B + C) = A \u00d7 B + A \u00d7C . Thus A \u00d7 B = \u2212A \u00d7C or \n. From\nA \u00d7 B = A \u00d7C \n\uf072\n\uf072 \n\uf072 \n\uf072\n\uf072 \n\uf072 \uf072\n\uf072 \nFigure 17.7b we see that \nsin \u03b2 . Therefore\nA \u00d7 B = A B sin \u03b3 and A \u00d7C = A C \n\uf072 \n\uf072 \uf072 \n\uf072\n\uf072 \n\uf072 \nsin \u03b2 , and hence \n/ sin \u03b3 . A similar argument shows that \nA B sin \u03b3 = A C \nB / sin \u03b2 = C \n\uf072\n\uf072 \n/ sin\u03b1 proving the law of sines.\nB / sin \u03b2 = A \nExample 17.3 Unit Normal \n\uf072\n\uf072\n\u02c6\n\u02c6\n\u02c6 \n\u02c6\n\u02c6\n\u02c6\nFind a unit vector perpendicular to A = i + j \u2212 k and B = \u22122i \u2212 j + 3k . \n17-7", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_473", "text": "\uf072\n\uf072 \n\uf072\n\uf072  \nSolution: The vector product A \u00d7 B is perpendicular to both A and B . Therefore the \n\uf072\n\uf072\n\uf072\n\uf072 \n\uf072\n\uf072 \nunit vectors n\u02c6 = \u00b1A \u00d7 B / \nare perpendicular to both A and B . We first calculate\nA \u00d7 B \n\uf072\n\uf072 \nA \u00d7 B = ( A B \u2212 A B ) \u02c6i + ( A B \u2212 A B ) \u02c6j+ ( A B \u2212 A B ) k\u02c6 \ny\nz\nz\ny \nz\nx\nx\nz \nx\ny\ny\nx \n= ((1)(3) \u2212 (\u22121)(\u22121)) \u02c6i + ((\u22121)(2) \u2212 (1)(3)) \u02c6j + ((1)(\u22121) \u2212 (1)(2)) k\u02c6 \n= 2 \u02c6i \u2212 5 \u02c6j\u2212 3 k\u02c6. \nWe now calculate the magnitude \n\uf072\n\uf072 \n= (22 + 52 + 32)1/2 = (38)1/2 \nA \u00d7 B \n. \nTherefore the perpendicular unit vectors are \n\uf072\n\uf072 \n\uf072\n\uf072 \nn\u02c6 = \u00b1A \u00d7 B / A \u00d7 B = \u00b1(2 \u02c6i \u2212 5 \u02c6j\u2212 3 k\u02c6 ) / (38)1/2 . \nExample 17.4 Volume of Parallelepiped \n\uf072\n\uf072 \nShow that the volume of a parallelepiped with edges formed by the vectors A , B , and \n\uf072 \n\uf072\n\uf072\n\uf072 \nC is given by A \u22c5(B\u00d7C) . \nSolution: The volume of a parallelepiped is given by area of the base times height. If the \n\uf072\n\uf072 \nbase is formed by the vectors B and C , then the area of the base is given by the \n\uf072\n\uf072 \n\uf072\n\uf072 \n\uf072\n\uf072 \nmagnitude of B \u00d7C. The vector B \u00d7C = \nn\u02c6 where n\u02c6 is a unit vector perpendicular \nB\u00d7C \nto the base (Figure 17.8). \nFigure 17.8 Example 17.4 \n\uf072 \nThe projection of the vector A along the direction n\u02c6 gives the height of the \n\uf072 \nparallelepiped. This projection is given by taking the dot product of A with a unit vector \n\uf072 \n\u22c5 \u02c6\nand is equal to A n = height . Therefore \n\uf072\n\uf072\n\uf072 \n\uf072 \n\uf072\n\uf072 \n\uf072\n\uf072 \uf072 \nA \u22c5 (B \u00d7 C) = A \u22c5 ( B \u00d7 C )n\u02c6 = ( B \u00d7 C )A \u22c5 n\u02c6 = (area)(height) = (volume) . \n17-8", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_474", "text": "Example 17.5 Vector Decomposition \n\uf072 \nLet A be an arbitrary vector and let n\u02c6 be a unit vector in some fixed direction. Show \n\uf072\n\uf072 \n\uf072 \nthat A = (A \u22c5 n\u02c6)n\u02c6 + (n\u02c6 \u00d7 A) \u00d7 n\u02c6 . \n\uf072\n\uf072 \nSolution: Let A = An\u02c6 + A\u22a5e\u02c6 where A is the component A in the direction of n\u02c6 , e\u02c6 is\n\uf050\n\uf050 \n\uf072 \nthe direction of the projection of A in a plane perpendicular to n\u02c6 , and A\u22a5 is the \n\uf072 \n\uf072 \n\u02c6 \u02c6 \ncomponent of A in the direction of e\u02c6 . Because e n\n\u22c5 = 0 , we have that A \u22c5 n\u02c6 = A\uf050 . Note \nthat \n\uf072 \nn\u02c6 \u00d7 A = n\u02c6 \u00d7 ( A\uf050n\u02c6 + A\u22a5 e\u02c6) = n\u02c6 \u00d7 A\u22a5 e\u02c6 = A\u22a5 (n\u02c6 \u00d7 e\u02c6) . \nThe unit vector n\u02c6 \u00d7e\u02c6 lies in the plane perpendicular to n\u02c6 and is also perpendicular to e\u02c6 . \nTherefore (n\u02c6 \u00d7 e\u02c6) \u00d7 n\u02c6 is also a unit vector that is parallel to e\u02c6 (by the right hand rule. So \n\uf072 \n(n\u02c6 \u00d7 A) \u00d7 n\u02c6 = A\u22a5 e\u02c6 . Thus \n\uf072 \n\uf072\n\uf072 \nA = A\uf050n\u02c6 + A\u22a5e\u02c6 = (A \u22c5n\u02c6)n\u02c6 + (n\u02c6 \u00d7 A) \u00d7 n\u02c6 . \n17.3 Torque \n17.3.1 Definition of Torque about a Point \nIn order to understand the dynamics of a rotating rigid body we will introduce a new \n\uf072\n\uf072 \n\uf072\nquantity, the torque. Let a force FP with magnitude F = \nact at a point P . Let \n, \nFP \nrS P \n\uf072 r\nbe the vector from the point S to a point P , with magnitude r = \n. The angle \n,\nS P \n\uf072\n\uf072\nbetween the vectors \n, and FP is \u03b8 with [0 \u2264\u03b8 \u2264\u03c0 ] (Figure 17.9).\nrS P \nFigure 17.9 Torque about a point S due to a force acting at a point P \n\uf072 \nThe torque about a point S due to force FP acting at P , is defined by \n\uf072\n\uf072 \n\uf072 \n\u00d7 FP . \n(17.2.28)\n\u03c4S = rS , P \n17-9", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_475", "text": "\uf072 \nThe magnitude of the torque about a point S due to force FP acting at P , \nis given by \n\uf072\n\u2261 \n= r F sin\u03b8 . \n(17.2.29)\n\u03c4 S \n\u03c4S \nThe SI units for torque are [N \u22c5 m]. The direction of the torque is perpendicular to the \n\uf072\n\uf072\nplane formed by the vectors \n, \nand FP (for [0 < \u03b8 < \u03c0 ] ), and by definition points in \nrS P \nthe direction of the unit normal vector to the plane n\u02c6 RHR as shown in Figure 17.10. \nFigure 17.10 Vector direction for the torque  \nFigure 17.11 shows the two different ways of defining height and base for a  \n\uf072\n\uf072\nparallelogram defined by the vectors \n, \nand FP .\nrS P \nFigure 17.11 Area of the torque parallelogram. \n\uf072 \nLet r = r sin\u03b8 and let F = F sin\u03b8 be the component of the force F \nthat is \n\u22a5\n\u22a5 \nP \nperpendicular to the line passing from the point S to P . (Recall the angle \u03b8 has a range \nof values 0 \u2264\u03b8 \u2264\u03c0 so both r\u22a5\u2265 0 and F\u22a5\u2265 0 .) Then the area of the parallelogram \n\uf072\n\uf072\ndefined by \n, and FP is given by\nrS P \nArea = \u03c4 = r F = r F = r F sin\u03b8 . \n(17.2.30)\nS \n\u22a5\n\u22a5 \n17-10", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_476", "text": "We can interpret the quantity r\u22a5 as follows. \nFigure 17.12 The moment arm about the point S and line of action of force passing \nthrough the point P \n\uf072 \nWe begin by drawing the line of action of the force FP . This is a straight line passing \n\uf072 \nthrough P , parallel to the direction of the force FP . Draw a perpendicular to this line of \naction that passes through the point S (Figure 17.12). The length of this perpendicular, \n\uf072 \nr\u22a5 = r sin\u03b8 , is called the moment arm about the point S of the force FP . \nYou should keep in mind three important properties of torque: \n\uf072\n\uf072\n1. The torque is zero if the vectors \n, \nand FP are parallel (\u03b8 = 0) or anti-parallel\nrS P \n(\u03b8 = \u03c0 ). \n2.  Torque is a vector whose direction and magnitude depend on the choice of a point \nS about which the torque is calculated. \n3.  The direction of torque is perpendicular to the plane formed by the two vectors, \n\uf072 \n\uf072 r\nFP and r = \n(the vector from the point S to a point P ).\n,\nS P \n17.3.2 Alternative Approach to Assigning a Sign Convention for Torque \n\uf072 \n\uf072\nIn the case where all of the forces Fi and position vectors \n, are coplanar (or zero), we \nri P \ncan, instead of referring to the direction of torque, assign a purely algebraic positive or \nnegative sign to torque according to the following convention. We note that the arc in \nFigure 17.13a circles in counterclockwise direction. (Figures 17.13a and 17.13b use the \nsimplifying assumption, for the purpose of the figure only, that the two vectors in \n\uf072 \n\uf072\nquestion, FP and \n, are perpendicular. The point S about which torques are calculated \nrS P \nis not shown.) \n17-11", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_477", "text": "(a)  \n(b) \nFigure 17.13 (a) Positive torque out of plane, (b) positive torque into plane \nWe can associate with this counterclockwise orientation a unit normal vector n\u02c6 \naccording to the right-hand rule: curl your right hand fingers in the counterclockwise \ndirection and your right thumb will then point in the n\u02c6 1 direction (Figure 17.13a). The \narc in Figure 17.13b circles in the clockwise direction, and we associate this orientation \nwith the unit normal n\u02c6 2 . \nIt\u2019s important to note that the terms \u201cclockwise\u201d and \u201ccounterclockwise\u201d might be \n\uf072 \n\uf072\ndifferent for different observers. For instance, if the plane containing FP and rS P is \n, \nhorizontal, an observer above the plane and an observer below the plane would disagree \non the two terms. For a vertical plane, the directions that two observers on opposite sides \nof the plane would be mirror images of each other, and so again the observers would \ndisagree. \n1.  Suppose we choose counterclockwise as positive. Then we assign a positive sign \nfor the component of the torque when the torque is in the same direction as the \n\uf072\n\uf072 \n\uf072 \n\uf072\n\uf072 r\nunit normal n\u02c6 , i.e. \u03c4 = r\n\u00d7 F = + \nn\u02c6 1 , (Figure 17.13a).\n1 \nS \nS ,P \nP \nS ,P FP \n2.  Suppose we choose clockwise as positive. Then we assign a negative sign for the \ncomponent of the torque in Figure 17.13b because the torque is directed opposite \n\uf072\n\uf072 \n\uf072 \nn\u02c6\nr\nto the unit normal n\u02c6 , i.e. \u03c4 = r\uf072 \n\u00d7 F\n\uf072 \n= \u2212\n2 \nS \nS ,P \nP \nS ,P FP \n2 . \nExample 17.6 Torque and Vector Product \n\uf072\n\uf072  \n\u02c6\n\u02c6\nConsider two vectors rP ,F = x\u02c6i with x > 0 and F = F i + F k with F > 0 and F > 0 .\nx\nz\nx \nz \n\uf072\n\uf072 \nCalculate the torque rP,F \u00d7 F . \nSolution: We calculate the vector product noting that in a right handed choice of unit \n\uf072 \nvectors, \u02c6i \u00d7\u02c6i = 0 and \u02c6i \u00d7k\u02c6 = \u2212 \u02c6j, \n\uf072\n\uf072 \nrP,F \u00d7 F = x\u02c6i \u00d7 (Fx \u02c6i + Fzk\u02c6) = (x\u02c6i \u00d7 Fx \u02c6i) + (x\u02c6i \u00d7 Fzk\u02c6) = \u2212xFz \u02c6j . \n17-12", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_478", "text": "Because x > 0 and Fz > 0 , the direction of the vector product is in the negative y \u00ad\ndirection. \nExample 17.7 Calculating Torque \nIn Figure 17.14, a force of magnitude F is applied to one end of a lever of length L. What \nis the magnitude and direction of the torque about the point S? \nFigure 17.14 Example 17.7 \nFigure 17.15 Coordinate system \nSolution: Choose units vectors such that \u02c6i \u00d7 \u02c6j = k\u02c6 , with \u02c6i pointing to the right and \u02c6j \n\uf072\n\uf072 \n\uf072 \npointing up (Figure 17.15). The torque about the point S is given by \n\u00d7 F ,\n\u03c4S = rS , F \n\uf072\n\uf072 \nwhere rSF = Lcos\u03b8\u02c6i + Lsin\u03b8\u02c6j and F = \u2212F\u02c6j then \n\uf072\u03c4S = (Lcos\u03b8 \u02c6i + Lsin\u03b8 \u02c6j) \u00d7 \u2212F \u02c6j = \u2212FLcos\u03b8 k\u02c6 . \nExample 17.8 Torque and the Ankle \nA person of mass m is crouching with their weight evenly distributed on both tiptoes.  \nThe free-body force diagram on the skeletal part of the foot is shown in Figure 17.16. The \n\uf072 \nnormal force N acts at the contact point between the foot and the ground. In this position, \n\uf072 \n\uf072 \nF\nthe tibia acts on the foot at the point S with a force F of an unknown magnitude F = \nand makes an unknown angle \u03b2 with the vertical. This force acts on the ankle a \nhorizontal distance s from the point where the foot contacts the floor. The Achilles \n\uf072 \nT\ntendon also acts on the foot and is under considerable tension with magnitude T \u2261 \nand \nacts at an angle \u03b1 with the horizontal as shown in the figure. The tendon acts on the \nankle a horizontal distance b from the point S where the tibia acts on the foot. You may \nignore the weight of the foot. Let g be the gravitational constant. Compute the torque \nabout the point S due to (a) the tendon force on the foot; (b) the force of the tibia on the \nfoot; (c) the normal force of the floor on the foot. \n17-13", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_479", "text": "Figure 17.16 Force diagram and coordinate system for ankle  \nSolution: (a) We shall first calculate the torque due to the force of the Achilles tendon on  \n\uf072 \n\u02c6\n\u02c6\nthe ankle. The tendon force has the vector decomposition T = T cos\u03b1i + T sin\u03b1 j. \nFigure 17.17 Torque diagram for tendon \nFigure 17.18 Torque diagram for normal \nforce on ankle \n\uf072\nforce on ankle rS ,N \n\uf072 \n\u02c6\nThe vector from the point S to the point of action of the force is given by r , = \u2212bi\nS T \n\uf072 \n(Figure 17.17). Therefore the torque due to the force of the tendon T on the ankle about \nthe point S is then \n\uf072\n\uf072\n\uf072 \n\u03c4 S ,T = rS ,T \u00d7 T = \u2212b\u02c6i \u00d7 (T cos\u03b1\u02c6i + T sin\u03b1\u02c6j) = \u2212bT sin\u03b1 k\u02c6 . \n(b) The torque diagram for the normal force is shown in Figure 17.18. The vector from \nthe point S to the point where the normal force acts on the foot is given by \n17-14", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_480", "text": "\uf072 rS ,N = (s\u02c6i \u2212 h\u02c6j) . Because the weight is evenly distributed on the two feet, the normal force \non one foot is equal to half the weight, or N = (1/ 2)mg . The normal force is therefore \n\uf072 \ngiven by N = N \u02c6j = (1/ 2)mg \u02c6j. Therefore the torque of the normal force about the point \nS is \n\uf072\n\uf072\n\u03c4 S N\n, = r , \u00d7 N \u02c6j = (s\u02c6i \u2212 h\u02c6j)\u00d7 N \u02c6j = s N k\u02c6 = (1/ 2) s mg k\u02c6 .\nS N \n\uf072 \n(c) The force F that the tibia exerts on the ankle will make no contribution to the torque \nabout this point S since the tibia force acts at the point S and therefore the vector \n\uf072 \n= 0 \n\uf072 \n.\nrS F \n, \n17.4 Torque, Angular Acceleration, and Moment of Inertia \n17.4.1 Torque Equation for Fixed Axis Rotation \nFor fixed-axis rotation, there is a direct relation between the component of the torque \nalong the axis of rotation and angular acceleration. Consider the forces that act on the \nrotating body. Generally, the forces on different volume elements will be different, and \n\uf072 \nso we will denote the force on the volume element of mass \u0394mi by Fi . Choose the z \u00ad\naxis to lie along the axis of rotation. Divide the body into volume elements of mass \u0394mi . \nLet the point S denote a specific point along the axis of rotation (Figure 17.19). Each \nvolume element undergoes a tangential acceleration as the volume element moves in a \n\uf072 r\ncircular orbit of radius ri = \nabout the fixed axis.\ni \nFigure 17.19: Volume element undergoing fixed-axis rotation about the z -axis. \nThe vector from the point S to the volume element is given by \n\uf072\n\uf072 \nr \n= z k\u02c6 + r = z k\u02c6 + r r\u02c6 \n(17.3.1)\nS , i\ni\ni\ni \ni \n17-15", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_481", "text": "where zi is the distance along the axis of rotation between the point S and the volume \n\uf072 \nelement. The torque about S due to the force Fi acting on the volume element is \ngiven by \n\uf072 \n\uf072\n\uf072 \n= \n\u00d7 Fi . \n(17.3.2)\n\u03c4S , i \nrS , i \nSubstituting Eq. (17.3.1) into Eq. (17.3.2) gives \n\uf072\n\uf072\u03c4S , i = (zi k\u02c6 + ri r\u02c6) \u00d7 Fi . \n(17.3.3) \nFor fixed-axis rotation, we are interested in the z -component of the torque, which must \nbe the term \n\uf072\n\uf072 \n(\u03c4S , i )z = (ri r\u02c6 \u00d7 Fi )z \n(17.3.4) \n\uf072 \nbecause the vector product zi k\u02c6 \u00d7 Fi must be directed perpendicular to the plane formed \n\uf072\n\u02c6\nby the vectors k and Fi , hence perpendicular to the z -axis. The force acting on the \nvolume element has components \n\uf072 \nFi = F r , i r\u02c6 + F\u03b8 , i \u03b8\u02c6 + F z, i k\u02c6 . \n(17.3.5) \nThe z -component Fz i, of the force cannot contribute a torque in the z -direction, and so \nsubstituting Eq. (17.3.5) into Eq. (17.3.4) yields \n\uf072\n(\n) \n\u02c6 \u00d7 (F \u02c6 \n\u03b8)) . \n(17.3.6)\n\u03c4S , i\nz = (ri r\nr ,i r + F\u03b8 ,i \u02c6 \nz \nFigure 17.20 Tangential force acting on a volume element. \nThe radial force does not contribute to the torque about the z -axis, since \n\uf072 \nr\u02c6 \u00d7 F r\u02c6 = 0 . \n(17.3.7)\nri \nr , i \n17-16", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_482", "text": "So, we are interested in the contribution due to torque about the z -axis due to the \ntangential component of the force on the volume element (Figure 17.20). The component \nof the torque about the z -axis is given by \n\uf072 \n\u02c6\n(\n) \nr\u02c6 \u00d7 F\u03b8 ,i \u03b8) \n(17.3.8)\n\u03c4S , i\nz = (ri \nz = ri F\u03b8 ,i . \nThe z -component of the torque is directed upwards in Figure 17.20, where \nis\nF\u03b8 , i \npositive (the tangential force is directed counterclockwise, as in the figure). Applying \nNewton\u2019s Second Law in the tangential direction, \n(17.3.9)\nF\u03b8 , i = \u0394mi a\u03b8 , i . \nUsing our kinematics result that the tangential acceleration is a\u03b8 , i \n\u03b1 , where \u03b1 is\n= r i \nz \nz \nthe z -component of angular acceleration, we have that \n\u03b1 . \n(17.3.10)\nF\u03b8 , i = \u0394mi ri \nz \nFrom Eq. (17.3.8), the component of the torque about the z -axis is then given by \n\uf072\n(\u03c4S , i )z = r i F\u03b8 , i = \u0394mi ri \n2\u03b1 z . \n(17.3.11) \nThe component of the torque about the z -axis is the summation of the torques on all the \nvolume elements, \ni=N\ni=N\n\uf072\n\uf072\n(\u03c4S )z = \u2211(\u03c4S , i )z = \u2211r\u22a5, i F\u03b8 , i \ni=1 \ni=1 \n(17.3.12)\ni=N \n= \u2211\u0394miri \n2\u03b1 z . \ni=1 \nBecause each element has the same z -component of angular acceleration, \u03b1 z , the \nsummation becomes \n\uf072 \n\u239b i= N \n2 \u239e\n(\u03c4S )z = \u2211\u0394mi ri \u23a0\u239f \u03b1 z . \n(17.3.13)\n\u239d\u239c \ni=1 \nRecalling our definition of the moment of inertia, (Chapter 16.3) the z -component of the \ntorque is proportional to the z -component of angular acceleration, \n\u03c4 S ,z = IS \u03b1 z , \n(17.3.14) \n17-17", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_483", "text": "and the moment of inertia, IS , is the constant of proportionality. The torque about the \npoint S is the sum of the external torques and the internal torques \n\uf072\u03c4 S =  \uf072\u03c4 S \next + \uf072\u03c4 S \nint .  \n(17.3.15)  \nThe external torque about the point S is the sum of the torques due to the net external \nforce acting on each element \ni= N \ni= N\n\uf072\n\uf072 \n\uf072\n\uf072 \next \next \next\n\u03c4S = \u2211\u03c4S ,i = \u2211rS ,i \u00d7 Fi . \n(17.3.16) \ni=1 \ni=1 \nThe internal torque arise from the torques due to the internal forces acting between pairs \n\uf072r \nof elements \n\u03c4\n\u03c4\n\u2211\u03c4 \n\u2211 S ,i \n\uf072\n\uf072\n\uf072\nj= N \nj = N\nN \ni= N \ni= N\n\uf072\nF \n\u2211  \n\u2211  \n\u2211 \nint \nint \nint \n(17.3.17) \n\u00d7 \n=  \n=  \n=  \n. \nS \nS , j \nS , j, i \nj, i \n\uf072 \nF\n\uf072r \ni=1 \ni=1  j=1 \ni=1 j =1 \nj\u2260 i\nj \u2260 i \n\uf072\n\uf072 \n\uf072r \nWe know by Newton\u2019s Third Law that the internal forces cancel in pairs, Fj,i = \u2212Fi, j , and \nhence the sum of the internal forces is zero \n\uf072 \ni= N j = N \uf072\n\uf072 \n0 = \u2211\u2211 Fj, i .  \n(17.3.18) \n\uf072 \ni=1  j=1 \nj\u2260 i \nDoes the same statement hold about pairs of internal torques? Consider the sum of \ninternal torques arising from the interaction between the ith and jth particles \n\u03c4\n\u03c4 \nS ,i\nS , j\ni, j . \n\uf072 \nF \nint \nint \n(17.3.19) \n+  \n\u00d7  \n+  \n\u00d7 \n=  \nS , j, i \nS ,i, j \nj, i \nBy the Newton\u2019s Third Law this sum becomes \nr \n\uf072\n\uf072 \n\uf072\n\uf072\n\u03c4\n\u03c4 \nS ,i \n\uf072 \nint \nint\nF \n= (  \n\u2212  \n) \u00d7  j, i . \n(17.3.20) \n+ \nr ,\nS j\nS , j, i \nS , i, j \n\uf072r \n\uf072r \n\uf072\n\uf072 \n\uf072\n\uf072\nIn the Figure 17.21, the vector r \u2212 r \npoints from the jth element to the ith element. If\nS ,i \nS , j \nthe internal forces between a pair of particles are directed along the line joining the two \nparticles then the torque due to the internal forces cancel in pairs. \n\u03c4\n\u03c4 \nS ,i\nS , j \n\uf072 \nF  \n\uf072\n0 \nint \nint \n= (  \n\u2212  \n) \u00d7  \n(17.3.21) \n+  \n=  . \nS , j, i \nS , i, j \nj, i \n17-18", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_484", "text": "Figure 17.21 The internal force is directed along the line connecting the ith and jth \nparticles  \nThis is a stronger version of Newton\u2019s Third Law than we have so far since we have \nadded the additional requirement regarding the direction of all the internal forces between \npairs of particles. With this assumption, the torque is just due to the external forces \n\uf072\n\uf072 ext . \n(17.3.22)\n\u03c4S = \u03c4S \nThus Eq. (17.3.14) becomes \next )\n(\u03c4 S \n\u03b1 , \n(17.3.23)\nz = IS \nz \nThis is very similar to Newton\u2019s Second Law: the total force is proportional to the \nacceleration, \n\uf072\n\uf072\nF = ma . \n(17.3.24) \nwhere the mass, m, is the constant of proportionality. \n17.4.2 Torque Acts at the Center of Gravity \nSuppose a rigid body in static equilibrium consists of N particles labeled by the index \ni = 1, 2, 3, ..., N . Choose a coordinate system with a choice of origin O such that mass mi has \n\uf072 \n\uf072\n\uf072\nposition ri . Each point particle experiences a gravitational force Fgravity,i = mi g . The total \ntorque about the origin is then zero (static equilibrium condition), \n\uf072r \n\uf072\nF\n\uf072r\n\uf072\n\uf072\u03c4\n\u03c4\ni \ngravity,i\ni \ni=1 \ni=1 \ni=1 \n\uf072\nIf the gravitational acceleration g is assumed constant, we can rearrange the summation \n\uf072\n\uf072\nin Eq. (17.3.25) by pulling the constant vector g out of the summation ( g appears in \neach term in the summation), \n17-19 \ni=N \ni= N \ni=N\n\uf072\n0 \n\uf072g\n\u2211  \n\u2211  \n\u2211  \n(17.3.25) \n\u00d7  \n\u00d7 mi \n=  \n=  \n=  \n=  . \nO \nO,i", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_485", "text": "i=N \n= \u2211  \u00d7 mi \n\uf072g =  \u239b i=N \n\u239e \uf072g =  \n\uf072\n0 .  \n(17.3.26) \n\u2211  \n\u23a0\u239f \u00d7 \nmi\n\u239d\u239c \n\u03c4O\ni \ni \nWe now use our definition of the center of the center of mass, Eq. (10.5.3), to rewrite Eq. \n(17.3.26) as \n\uf072r \n\uf072r \n\u03c4\nri \n\uf072 \n\uf072\ni=1 \ni=1 \ni=N \n\u2211 \n\uf072\nR  \n\uf072\nR  \n\uf072\n0 \n\uf072 \n\uf072g \n\uf072g \n\uf072g\nMT \n\u00d7 MT \n(17.3.27) \n\u00d7 mi \n\u00d7 \n=  \n=  \n=  \n=  . \nO \ncm \ncm \ni=1 \nThus the torque due to the gravitational force acting on each point-like particle is \nequivalent to the torque due to the gravitational force acting on a point-like particle of \nmass M T located at a point in the body called the center of gravity, which is equal to the \n\uf072\ncenter of mass of the body in the typical case in which the gravitational acceleration g is \nconstant throughout the body. \nExample 17.9 Turntable \n1\nThe turntable in Example 16.1, of mass 1.2 kg and radius 1.3 10 cm , has a moment of \n\u00d7 \ninertia IS = \n\u00d7\n\u22122 kg m 2 about an\n1.01 10 \n\u22c5 \naxis through the center of the turntable and \nperpendicular to the turntable. The turntable is spinning at an initial constant frequency \n= 33cycles \u22c5 min\u22121 . The motor is turned off and the turntable slows to a stop in 8.0 s \nfi \ndue to frictional torque. Assume that the angular acceleration is constant. What is the \nmagnitude of the frictional torque acting on the turntable? \nSolution: We have already calculated the angular acceleration of the turntable in \nExample 16.1, where we found that \n\u0394\u03c9 z \n\u03c9 f \u2212\u03c9 i \n\u22123.5 rad \u22c5s\u22121 \n\u22122\n\u03b1 z =\n=\n= \n= \u22124.3\u00d710\u22121 rad \u22c5s\n(17.3.28)\n\u0394t \n8.0 s \nt f \u2212 ti \nand so the magnitude of the frictional torque is \n\u03c4 fric \n\u22122)\n\u03b1 = (1.01\u00d710\u22122 kg \u22c5 m2)(4.3\u00d710\u22121 rad \u22c5s\n= IS\nz \nz \n(17.3.29) \n= 4.3\u00d710\u22123 N \u22c5 m. \nExample 17.10 Pulley and blocks \nA pulley of mass mp , radius R , and moment of inertia about its center of mass Icm , is", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_486", "text": "= IS\nz \nz \n(17.3.29) \n= 4.3\u00d710\u22123 N \u22c5 m. \nExample 17.10 Pulley and blocks \nA pulley of mass mp , radius R , and moment of inertia about its center of mass Icm , is \nattached to the edge of a table. An inextensible string of negligible mass is wrapped \naround the pulley and attached on one end to block 1 that hangs over the edge of the table \n(Figure 17.22). The other end of the string is attached to block 2 that slides along a table. \n17-20", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_487", "text": "The coefficient of sliding friction between the table and the block 2 is \u00b5k . Block 1 has \nmass m1 and block 2 has mass m2 , with m1 > \u00b5km2 . At time t = 0 , the blocks are \nreleased from rest and the string does not slip around the pulley. At time t = t1 , block 1 \nhits the ground. Let g denote the gravitational constant. (a) Find the magnitude of the \nacceleration of each block. (b) How far did the block 1 fall before hitting the ground? \nFigure 17.22 Example 17.10 \nFigure 17.23 Torque diagram for pulley \nSolution: The torque diagram for the pulley is shown in the figure below where we \nchoose k\u02c6 pointing into the page. Note that the tensions in the string on either side of the \npulley are not equal. The reason is that the pulley is massive. To understand why, \nremember that the difference in the magnitudes of the torques due to the tension on either \nside of the pulley is equal to the moment of inertia times the magnitude of the angular \n\uf072\nacceleration, which is non -zero for a massive pulley. So the tensions cannot be equal. \nFrom our torque diagram, the torque about the point O at the center of the pulley is given \n\u03c4 \nby\n\uf072 \n\uf072 \n\uf072 \n\uf072  \n)k\u02c6\nT1 \nT = R(T \u2212 T\n2 \n1 \n2 \n(17.3.30) \n\u00d7  +  \n\u00d7 \n=  rO,1 \nrO,2 \n. \nO \nTherefore the torque equation (17.3.23) becomes \n\u2212 T2) = I \u03b1 . \n(17.3.31)\nR(T1 \nz \nz \nThe free body force diagrams on the two blocks are shown in Figure 17.23. \n(a) \n(b)  \nFigure 17.23 Free-body force diagrams on (a) block 2, (b) block 1  \n17-21", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_488", "text": "Newton\u2019s Second Law on block 1 yields \nm1g \u2212 T1 = m1ay1 . \n(17.3.32) \nNewton\u2019s Second Law on block 2 in the \u02c6j direction yields \nN \u2212 m2 g = 0 . \n(17.3.33) \nNewton\u2019s Second Law on block 2 in the \u02c6i direction yields \n= m2ax2 . \n(17.3.34)\nT2 \u2212 fk \nThe kinetic friction force is given by \nN \ng \n(17.3.35)\nfk = \u00b5k \n= \u00b5km2 \nTherefore Eq. (17.3.34) becomes \n\u2212 \u00b5k\ng = m2ax 2 . \n(17.3.36)\nT2 \nm2 \nBlock 1 and block 2 are constrained to have the same acceleration so \na \u2261 a x1 = a x2 . \n(17.3.37) \nWe can solve Eqs. (17.3.32) and (17.3.36) for the two tensions yielding \nT = m g \u2212 m a , \n(17.3.38)\n1\n1 \n1 \ng + m2a . \n(17.3.39)\nT2 = \u00b5km2 \nAt point on the rim of the pulley has a tangential acceleration that is equal to the \nacceleration of the blocks so \na = a\u03b8 = R\u03b1 z . \n(17.3.40) \nThe torque equation (Eq. (17.3.31)) then becomes \nI\nT1 \u2212 T2 = \nR\nz \n2 a . \n(17.3.41) \nSubstituting Eqs. (17.3.38) and (17.3.39) into Eq. (17.3.41) yields \nI \nm1g \u2212 m1a \u2212 (\u00b5k m2 g + m2a) = \nR\nz \n2 a , \n(17.3.42) \nwhich we can now solve for the accelerations of the blocks \n17-22", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_489", "text": "m1g \u2212 \u00b5k\ng\nm2\na = \n. \n(17.3.43)\n+ I / R2\nm1 + m2 \nz \nBlock 1 hits the ground at time t1, therefore it traveled a distance \n\u239b \n\u239e \nm1g \u2212 \u00b5k\n1  \nm g\n2 \nt1 \n2 .  \n(17.3.44) \ny = \n1\n\u239c\u239d \n\u239f\u23a0 \n/ R2\n2  \n+ I\nm1 + m2 \nz \nExample 17.11 Experimental Method for Determining Moment of Inertia \nA steel washer is mounted on a cylindrical rotor of radius r = 12.7 mm . A massless \nstring, with an object of mass m = 0.055 kg attached to the other end, is wrapped around \nthe side of the rotor and passes over a massless pulley (Figure 17.24). Assume that there \nis a constant frictional torque about the axis of the rotor. The object is released and falls. \nAs the object falls, the rotor undergoes an angular acceleration of magnitude \u03b11. After \nthe string detaches from the rotor, the rotor coasts to a stop with an angular acceleration \n\u2212\nof magnitude \u03b12 . Let g = 9.8 m s\u22c5 \n2 denote the gravitational constant. Based on the data \nin the Figure 17.25, what is the moment of inertia IR of the rotor assembly (including the \nwasher) about the rotation axis? \nFigure 17.24 Steel washer, rotor, pulley, \nFigure 17.26 Graph of angular speed vs.\nand hanging object \ntime for falling object \nSolution: We begin by drawing a force-torque diagram (Figure 17.26a) for the rotor and \na free-body diagram for hanger (Figure 17.26b). (The choice of positive directions are \n\uf072\nindicated on the figures.) The frictional torque on the rotor is then given by \u03c4 f = \u2212\u03c4 f k\u02c6 \nwhere we use \u03c4 f as the magnitude of the frictional torque. The torque about the center of \n\uf072\nthe rotor due to the tension in the string is given by \u03c4T = rT k\u02c6 where r is the radius of \n17-23", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_490", "text": "\uf072\nthe rotor. The angular acceleration of the rotor is given by \u03b11 = \u03b11 k\u02c6 and we expect that \n> 0 because the rotor is speeding up.\n\u03b11 \n(a) \n(b) \nFigure 17.26 (a) Force-torque diagram on rotor and (b) free-body force diagram on \nhanging object \nWhile the hanger is falling, the rotor-washer combination has a net torque due to the \ntension in the string and the frictional torque, and using the rotational equation of motion, \nTr \u2212\u03c4 = I \u03b1 . \n(17.4.1)\nf\nR 1 \nWe apply Newton\u2019s Second Law to the hanger and find that \nmg \u2212T = ma1 = m\u03b11r , \n(17.4.2) \nwhere a1 \nhas been used to express the linear acceleration of the falling hanger to \n= r\u03b11 \nthe angular acceleration of the rotor; \nthat is, the string does not stretch. Before \nproceeding, it might be illustrative to multiply Eq. (17.4.2) by r and add to Eq. (17.4.1) \nto obtain \nmgr \u2212\u03c4 f = (IR + mr 2)\u03b11 . \n(17.4.3) \nEq. (17.4.3) contains the unknown frictional torque, and this torque is determined by \nconsidering the slowing of the rotor/washer after the string has detached. \nFigure 17.27 Torque diagram on rotor when string has detached \n17-24", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_491", "text": "The torque on the system is just this frictional torque (Figure 17.27), and so \n= IR\n(17.4.4)\n\u2212\u03c4 f \n\u03b12 \nNote that in Eq. (17.4.4), \u03c4 f > 0 and \u03b12 < 0 . Subtracting Eq. (17.4.4) from Eq. (17.4.3) \neliminates \u03c4 f , \nmgr = mr\n\u2212\u03b12) . \n(17.4.5)\n2\u03b11 + IR (\u03b11 \nWe can now solve for IR yielding \nmr(g \u2212 r\u03b11)\n= \n. \n(17.4.6)\nIR \n\u03b11 \u2212\u03b1 2 \nFor a numerical result, we use the data collected during a trial run resulting in the graph \nof angular speed vs. time for the falling object shown in Figure 17.25. The values for \u03b11 \nand \u03b12 can be determined by calculating the slope of the two straight lines in Figure \n17.28 yielding \n= (96rad \u22c5s\u22121) / (1.15s) = 83 rad \u22c5s\u22122,\n\u03b11 \n= \u2212(89rad \u22c5s\u22121) / (2.85s) = \u2212 31rad \u22c5s\u22122.\n\u03b1 2 \nInserting these values into Eq. (17.4.6) yields \nIR = 5.3 10 \n\u00d7\n\u22125 kg \u22c5 m2 . \n(17.4.7) \n17.5 Torque and Rotational Work \nWhen a constant torque \u03c4 s,z is applied to an object, and the object rotates through an \nangle \u0394\u03b8 about a fixed z -axis through the center of mass, then the torque does an \namount of work \u0394W = \u03c4 S ,z \u0394\u03b8 on the object. By extension of the linear work-energy \ntheorem, the amount of work done is equal to the change in the rotational kinetic energy \nof the object, \n1 \n\u03c9 2 \n1 \n2\nW = \nI \n\u2212 I \n= K \n\u2212 K \n(17.4.8)\nrot \n2\ncm \nf \n2 cm \u03c9 i \nrot, f \nrot,i . \nThe rate of doing this work is the rotational power exerted by the torque, \ndW \n\u0394W\nd\u03b8\nrot \nrot \nProt \u2261 \n= lim \n= \u03c4 S ,z \u03c9 z . \n(17.4.9)\n= \u03c4 S ,z\ndt \n\u0394t\u21920 \u0394t \ndt \n17-25", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_492", "text": "17.5.1 Rotational Work \nConsider a rigid body rotating about an axis. Each small element of mass \u0394mi in the rigid \nbody is moving in a circle of radius (rS , i )\u22a5 about the axis of rotation passing through the \npoint S . Each mass element undergoes a small angular displacement \u0394\u03b8 under the \n\uf072 \naction of a tangential force, F\u03b8 , i = F\u03b8 , i \u03b8\u02c6 , where \u03b8\u02c6 is the unit vector pointing in the \ntangential direction (Figure 17.20). The element will then have an associated \n\uf072\ndisplacement vector for this motion, \u0394r \n= r \u0394\u03b8 \u03b8\u02c6 and the work done by the tangential \nS , i\ni \nforce is \n\uf072 \n\u0394Wrot,i = F\u03b8 ,i \u22c5\u0394r \uf072 \nS ,i = (F\u03b8 ,i \u03b8\u02c6) \u22c5(ri\u0394\u03b8 \u03b8\u02c6) = riF\u03b8 ,i\u0394\u03b8 . \n(17.4.10) \nRecall the result of Eq. (17.3.8) that the component of the torque (in the direction along \n\uf072 \nthe axis of rotation) about S due to the tangential force, F\u03b8 , i , acting on the mass element \n\u0394mi is \n(\u03c4 S ,i )z = ri F\u03b8 ,i , \n(17.4.11) \nand so Eq. (17.4.10) becomes \n\u0394Wrot, i = (\u03c4 S ,i )z \u0394\u03b8 . \n(17.4.12) \nSumming over all the elements yields \nWrot = \u2211\u0394Wrot, i = \u2211((\u03c4 S ,i )z )\u0394\u03b8 = \u03c4 S ,z \u0394\u03b8 , \n(17.4.13) \ni\ni \nthe rotational work is the product of the torque and the angular displacement. In the limit \nof small angles, \u0394\u03b8 \u2192 d\u03b8 , \u0394W \u2192 dW and the differential rotational work is\nrot \nrot \ndWrot = \u03c4 S ,zd\u03b8 . \n(17.4.14) \nWe can integrate this amount of rotational work as the angle coordinate of the rigid body \nchanges from some initial value \u03b8 = \u03b8i to some final value \u03b8 = \u03b8 f , \nWrot = \u222b dWrot \n\u03b8 f \u03c4 S ,z d\u03b8 . \n(17.4.15)\n= \u222b\u03b8i \n17.5.2 Rotational Work-Kinetic Energy Theorem \nWe will now show that the rotational work is equal to the change in rotational kinetic \nenergy. We begin by substituting our result from Eq. (17.3.14) into Eq. (17.4.14) for the", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_493", "text": "= \u222b\u03b8i \n17.5.2 Rotational Work-Kinetic Energy Theorem \nWe will now show that the rotational work is equal to the change in rotational kinetic \nenergy. We begin by substituting our result from Eq. (17.3.14) into Eq. (17.4.14) for the \ninfinitesimal rotational work, \ndW \n\u03b1 d\u03b8 . \n(17.4.16)\nrot = IS \nz \n17-26", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_494", "text": "Recall that the rate of change of angular velocity is equal to the angular acceleration, \n\u03b1 \u2261 d\u03c9 dt and that the angular velocity is \u03c9 z \u2261 d\u03b8 dt . Note that in the limit of small \nz\nz \ndisplacements, \nd\u03c9 z\nd\u03b8\nd\u03b8 = d\u03c9 \n= d\u03c9 \u03c9 . \n(17.4.17)\nz \nz\nz\ndt \ndt \nTherefore the infinitesimal rotational work is \nd\u03c9 z\nd\u03b8\ndW \n\u03b1 \nd\u03c9 \nd\u03c9 \u03c9 . \n(17.4.18)\nrot = IS \nz d\u03b8 = IS \nd\u03b8 = IS \nz \n= IS \nz\nz\ndt \ndt \nWe can integrate this amount of rotational work as the angular velocity of the rigid body \nchanges from some initial value \u03c9 = \u03c9 \nto some final value \u03c9 = \u03c9 \nz\nz,i \nz\nz, f , \n\u03c9 z , f \n1 \n2\n1 \n2\n= \u222b dW\n= \u222b \n\u03c9\n\u03c9\n\u03c9 \n(17.4.19)\nWrot \nrot \nIS d\u03c9 z \nz = \nIS \nz, f \u2212 IS \nz,i . \n\u03c9 z ,i \n2\n2 \nWhen a rigid body is rotating about a fixed axis passing through a point S in the body, \nthere is both rotation and translation about the center of mass unless S is the center of \nmass. If we choose the point S in the above equation for the rotational work to be the \ncenter of mass, then \n1 \n2\n1 \n2\nW = \nI \u03c9\n\u2212 I \u03c9 \n= K \n\u2212 K \n\u2261\u0394K . \n(17.4.20)\nrot \ncm \ncm, f \ncm \ncm,i \nrot, f \nrot,i \nrot \n2\n2 \nNote that because the z -component of the angular velocity of the center of mass appears \nas a square, we can just use its magnitude in Eq. (17.4.20). \n17.5.3 Rotational Power \nThe rotational power is defined as the rate of doing rotational work, \ndW\nProt \u2261 \nrot . \n(17.4.21)\ndt \nWe can use our result for the infinitesimal work to find that the rotational power is the \nproduct of the applied torque with the angular velocity of the rigid body, \ndWrot \nd\u03b8\nProt \u2261 \n= \u03c4 S ,z \n= \u03c4 S ,z \u03c9 . \n(17.4.22)\ndt\ndt \nz \n17-27", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_495", "text": "Example 17.12 Work Done by Frictional Torque \nA steel washer is mounted on the shaft of a small motor. The moment of inertia of the \nmotor and washer is I0 . The washer is set into motion. When it reaches an initial \nangular velocity \u03c90 , at t = 0 , the power to the motor is shut off, and the washer slows \ndown during a time interval \u0394t1 = t until it reaches an angular velocity of \u03c9 at time t . \na \na\na \nAt that instant, a second steel washer with a moment of inertia Iw is dropped on top of \nthe first washer. Assume that the second washer is only in contact with the first washer.  \nThe collision takes place over a time \u0394tint \n\u2212 t after which the two washers and rotor \n= tb \na \nrotate with angular speed \u03c9b . Assume the frictional torque on the axle (magnitude \u03c4 f ) is \nindependent of speed, and remains the same when the second washer is dropped. (a) \nWhat angle does the rotor rotate through during the collision? (b) What is the work done \nby the friction torque from the bearings during the collision? (c) Write down an equation \nfor conservation of energy. Can you solve this equation for \u03c9b ? (d) What is the average \nrate that work is being done by the friction torque during the collision? \nSolution: We begin by solving for the frictional torque during the first stage of motion \nwhen the rotor is slowing down. We choose a coordinate system shown in Figure 17.29. \nFigure 17.29 Coordinate system for Example 17.12 \nThe component of average angular acceleration is given by \n\u03c9a \u2212\u03c90\n\u03b11 =\n< 0 .\nta \nWe can use the rotational equation of motion, and find that the frictional torque satisfies \n\u239b\u03c9a \u2212\u03c90 \u239e\n\u2212\u03c4 f = I0 \u239d\u239c \n\u0394t1 \n\u23a0\u239f . \n17-28", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_496", "text": "During the collision, the component of the average angular acceleration of the rotor is \ngiven by \n\u03c9b \u2212\u03c9a\n\u03b12 =\n< 0 .\n(\u0394tint ) \nThe angle the rotor rotates through during the collision is (analogous to linear motion \nwith constant acceleration) \n1\n2 \n1 \u239b\u03c9b \u2212\u03c9a \u239e \n2\n1\n\u0394\u03b82 = \u03c9a \u0394tint + 2 \u03b12 \u0394tint = \u03c9a \u0394tint + \n\u23a0\u239f \u0394tint = (\u03c9b + \u03c9a )\u0394tint > 0 .\n2 \u239d\u239c \u0394tint \n2 \nThe non-conservative work done by the bearing friction during the collision is \nW f ,b = \u2212\u03c4 f \u0394\u03b8rotor = \u2212\u03c4 f \n1 (\u03c9a + \u03c9b )\u0394tint .\n2 \nUsing our result for the frictional torque, the work done by the bearing friction during the \ncollision is \n1 \n\u239b\u03c9a \u2212\u03c90 \u239e\nW f ,b = \nI0 \n+ \u03c9b )\u0394tint < 0 .\n2 \n\u239d\u239c \n\u0394t1 \n\u23a0\u239f (\u03c9a \nThe negative work is consistent with the fact that the kinetic energy of the rotor is \ndecreasing as the rotor is slowing down. Using the work energy theorem during the \ncollision the kinetic energy of the rotor has deceased by \nW f ,b = 1 (I0 + Iw )\u03c9b \n2 \u2212 1 I0\u03c9a \n2 .\n2\n2 \nUsing our result for the work, we have that \n1 \n\u239b\u03c9a \u2212\u03c90 \u239e \n1 \n1\n2\nI0 \n\u23a0\u239f (\u03c9a + \u03c9b )\u0394tint = (I0 + Iw )\u03c9b \n2 \u2212 I0\u03c9a .\n2 \n\u239d\u239c \n\u0394t1 \n2\n2 \nThis is a quadratic equation for the angular speed \u03c9b of the rotor and washer \nimmediately after the collision that we can in principle solve. However remember that we \nassumed that the frictional torque is independent of the speed of the rotor. Hence the best \npractice would be to measure \u03c90 , \u03c9a , \u03c9b , \u0394t1 , \u0394tint , I0 , and Iw and then determine \nhow closely our model agrees with conservation of energy. The rate of work done by the \nfrictional torque is given by \nW f ,b \n1 \n\u239b\u03c9a \u2212\u03c90 \u239e\nPf =\n= \nI0 \n+ \u03c9b ) < 0 . \n\u0394tint \n2 \n\u239d\u239c \n\u0394t1 \n\u23a0\u239f (\u03c9a \n17-29", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_497", "text": "Chapter 18 Static Equilibrium \n18.1 Introduction Static Equilibrium.......................................................................... 1  \n18.2 Lever Law.............................................................................................................. 2  \nExample 18.1 Lever Law.......................................................................................... 4  \n18.3 Generalized Lever Law ........................................................................................ 5  \n18.4 Worked Examples................................................................................................. 7", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_498", "text": "Example 18.2 Suspended Rod.................................................................................. 7  \nExample 18.3 Person Standing on a Hill............................................................... 10  \nExample 18.4 The Knee.......................................................................................... 11  \nAppendix 18A The Torques About any Two Points are Equal for a Body in Static  \nEquilibrium ................................................................................................................. 16", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_499", "text": "Chapter 18 Static Equilibrium  \nThe proof of the correctness of a new rule can be attained by the repeated \napplication of it, the frequent comparison with experience, the putting of it to the \ntest under the most diverse circumstances. This process, would in the natural \ncourse of events, be carried out in time. The discoverer, however hastens to reach \nhis goal more quickly. He compares the results that flow from his rule with all the \nexperiences with which he is familiar, with all older rules, repeatedly tested in \ntimes gone by, and watches to see if he does not light on contradictions. In this \nprocedure, the greatest credit is, as it should be, conceded to the oldest and most \nfamiliar experiences, the most thoroughly tested rules. Our instinctive \nexperiences, those generalizations that are made involuntarily, by the irresistible \nforce of the innumerable facts that press upon us, enjoy a peculiar authority; and \nthis is perfectly warranted by the consideration that it is precisely the elimination \nof subjective caprice and of individual error that is the object aimed at.1 \nErnst Mach \n18.1 Introduction Static Equilibrium \nWhen the vector sum of the forces acting on a point-like object is zero then the object \nwill continue in its state of rest, or of uniform motion in a straight line. If the object is in \nuniform motion we can always change reference frames so that the object will be at rest. \nWe showed that for a collection of point-like objects the sum of the external forces may \nbe regarded as acting at the center of mass. So if that sum is zero the center of mass will \ncontinue in its state of rest, or of uniform motion in a straight line. We introduced the \nidea of a rigid body, and again showed that in addition to the fact that the sum of the \nexternal forces may be regarded as acting at the center of mass, forces like the \ngravitational force that acts at every point in the body may be treated as acting at the \ncenter of mass. However for an extended rigid body it matters where the force is applied \nbecause even though the sum of the forces on the body may be zero, a non-zero sum of \ntorques on the body may still produce angular acceleration. In particular for fixed axis \nrotation, the torque along the axis of rotation on the object is proportional to the angular \nacceleration. It is possible that sum of the torques may be zero on a body that is not \nconstrained to rotate about a fixed axis and the body may still undergo rotation. We will", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_500", "text": "rotation, the torque along the axis of rotation on the object is proportional to the angular \nacceleration. It is possible that sum of the torques may be zero on a body that is not \nconstrained to rotate about a fixed axis and the body may still undergo rotation. We will \nrestrict ourselves to the special case in which in an inertial reference frame both the \ncenter of mass of the body is at rest and the body does not undergo any rotation, a \ncondition that is called static equilibrium of an extended object. \nThe two sufficient and necessary conditions for a rigid body to be in static \nequilibrium are: \n1 Ernst Mach, The Science of Mechanics: A Critical and Historical Account of Its \nDevelopment, translated by Thomas J. McCormack, Sixth Edition with Revisions through \nthe Ninth German Edition, Open Court Publishing, Illinois. \n18-1", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_501", "text": "(1) The sum of the forces acting on the rigid body is zero, \n\uf072 \nF\n1\n2 \n\uf072 \nF \n\uf072 \nF =  +  +\u22c5\u22c5\u22c5=  \n\uf072\n0 .  \n(18.1.1)  \n(2) The vector sum of the torques about any point S in a rigid body is zero, \n\uf072\n\uf072\n\uf072 \n\uf072 \n+\n+\u22c5\u22c5\u22c5= 0 . \n(18.1.2)\n\u03c4S = \u03c4S ,1 \n\u03c4S ,2 \n18.2 Lever Law \nLet\u2019s consider a uniform rigid beam of mass mb balanced on a pivot near the center of \nmass of the beam. We place two objects 1 and 2 of masses m1 and m2 on the beam, at \ndistances d1 and d2 respectively from the pivot, so that the beam is static (that is, the \nbeam is not rotating. See Figure 18.1.) We shall neglect the thickness of the beam and \ntake the pivot point to be the center of mass. \nFigure 18.1 Pivoted Lever \nLet\u2019s consider the forces acting on the beam. The earth attracts the beam \ndownward. This gravitational force acts on every atom in the beam, but we can \n\uf072\nsummarize its action by stating that the gravitational force mb g is concentrated at a point \nin the beam called the center of gravity of the beam, which is identical to the center of \n\uf072 \nmass of the uniform beam. There is also a contact force Fpivot between the pivot and the \nbeam, acting upwards on the beam at the pivot point. The objects 1 and 2 exert normal \n\uf072\n\uf072 \n\uf072\n\uf072 \nforces downwards on the beam, N1,b \u2261 N1, and N2,b \u2261 N2 , with magnitudes N1, and N2 , \nrespectively. Note that the normal forces are not the gravitational forces acting on the \nobjects, but contact forces between the beam and the objects. (In this case, they are \nmathematically the same, due to the horizontal configuration of the beam and the fact that \nall objects are in static equilibrium.) The distances d1 and d2 are called the moment arms \n\uf072\n\uf072 \nwith respect to the pivot point for the forces N1 and N2 , respectively. The force diagram \n\uf072 \non the beam is shown in Figure 18.2. Note that the pivot force Fpivot and the force of \n\uf072\ngravity mb g each has a zero moment arm about the pivot point. \n18-2", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_502", "text": "Figure 18.2 Free-body diagram on beam \nBecause we assume the beam is not moving, the sum of the forces in the vertical \ndirection acting on the beam is therefore zero, \nFpivot \u2212 mb g \u2212 N1 \u2212 N2 = 0 . \n(18.2.1) \nThe force diagrams on the objects are shown in Figure 18.3. Note the magnitude of the \nnormal forces on the objects are also N1 and N2 since these are each part of an action\u00ad\n\uf072\n\uf072 \n\uf072\n\uf072 \nreaction pair, N1, b = \u2212Nb,1 , and N2, b = \u2212Nb, 2 . \n(a) \n(b) \nFigure 18.3 Free-body force diagrams for each body. \nThe condition that the forces sum to zero is not sufficient to completely predict the \nmotion of the beam. All we can deduce is that the center of mass of the system is at rest \n(or moving with a uniform velocity). In order for the beam not to rotate the sum of the \ntorques about any point must be zero. In particular the sum of the torques about the pivot \npoint must be zero. Because the moment arm of the gravitational force and the pivot force \nis zero, only the two normal forces produce a torque on the beam. If we choose out of the \npage as positive direction for the torque (or equivalently counterclockwise rotations are \npositive) then the condition that the sum of the torques about the pivot point is zero \nbecomes \nd N \u2212 d N = 0 . \n(18.2.2)\n2\n2 \n1\n1 \nThe magnitudes of the two torques about the pivot point are equal, a condition known as \nthe lever law. \n18-3", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_503", "text": "Lever Law: A beam of length l is balanced on a pivot point that is placed \ndirectly beneath the center of mass of the beam. The beam will not \nundergo rotation if the product of the normal force with the moment arm \nto the pivot is the same for each body, \nd N \n1\n1 = d N \n2 \n2 . \n(18.2.3) \nExample 18.1 Lever Law \nSuppose a uniform beam of length l = 1.0 m and mass mB = 2.0 kg is balanced on a pivot \npoint, placed directly beneath the center of the beam. We place body 1 with mass \nm1 = 0.3 kg a distance d1 = 0.4 m to the right of the pivot point, and a second body 2 \nwith m2 = 0.6 kg a distance d2 to the left of the pivot point, such that the beam neither \n\uf072 \ntranslates nor rotates. (a) What is the force Fpivot that the pivot exerts on the beam? (b) \nWhat is the distance d2 that maintains static equilibrium? \nSolution: a) By Newton\u2019s Third Law, the beam exerts equal and opposite normal forces \nof magnitude N1 on body 1, and N2 on body 2. The condition for force equilibrium \napplied separately to the two bodies yields \n\u2212 m g = 0 , \n(18.2.4)\nN1\n1 \nN2 \n2 \n(18.2.5)\n\u2212 m g = 0 . \nThus the total force acting on the beam is zero, \nFpivot \u2212 (mb + m1 + m2)g = 0 , \n(18.2.6) \nand the pivot force is \n)g\nFpivot = (mb + m1 + m2 \n(18.2.7) \n= (2.0 kg+ 0.3 kg+ 0.6 kg)(9.8 m\u22c5s\u22122) = 2.8 \u00d7101 N. \nb) We can compute the distance d2 from the Lever Law, \nd1 N1 \nd1 m1g\nd1 m1 \n(0.4 m)(0.3 kg) \nd2 =\n=\n=\n= \n= 0.2 m . \n(18.2.8)\ng \n0.6 kg \nN2 \nm2 \nm2 \n18-4", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_504", "text": "18.3 Generalized Lever Law \n\uf072\n\uf072 \nWe can extend the Lever Law to the case in which two external forces F1 and F2 are \nacting on the pivoted beam at angles \u03b81 and \u03b82 with respect to the horizontal as shown in \nthe Figure 18.4. Throughout this discussion the angles will be limited to the range \n[0 \u2264\u03b81,\u03b82 \u2264\u03c0 ] . We shall again neglect the thickness of the beam and take the pivot point \nto be the center of mass. \nFigure 18.4 Forces acting at angles to a pivoted beam. \n\uf072\n\uf072 \nThe forces F1 and F2 can be decomposed into separate vectors components respectively \n\uf072\n\uf072\n\uf072\n\uf072 \n\uf072\n\uf072 \n( \n, \n) and ( \n, \n) , where \nand \nare the horizontal vector projections of \nF1,\uf050 F1, \u22a5 \nF2,\uf050 F2, \u22a5 \nF1,\uf050 \nF2,\uf050 \n\uf072 \nthe two forces with respect to the direction formed by the length of the beam, and F1,\u22a5 \n\uf072 \nand F2,\u22a5 are the perpendicular vector projections respectively to the beam (Figure 18.5), \nwith \n\uf072\n\uf072 \n\uf072 \nF = F + F \n(18.3.1)\n1 \n1,\uf050 \n1,\u22a5 , \n\uf072\n\uf072 \n\uf072 \nF = F + F \n(18.3.2)\n2 \n2,\uf050 \n2,\u22a5 . \nFigure 18.5 Vector decomposition of forces. \nThe horizontal components of the forces are \nF1,\uf050 = F1 cos\u03b81 , \nF2,\uf050 = \u2212F2 cos\u03b82 , \n(18.3.3) \n(18.3.4) \n18-5", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_505", "text": "where our choice of positive horizontal direction is to the right. Neither horizontal force \ncomponent contributes to possible rotational motion of the beam. The sum of these \nhorizontal forces must be zero, \nF cos\u03b8 \u2212 F cos\u03b8 = 0 . \n(18.3.5)\n1\n1\n2 \n2 \nThe perpendicular component forces are \nF1,\u22a5 = F1 sin\u03b81, \n(18.3.6) \nF2,\u22a5 = F2 sin\u03b82 , \n(18.3.7) \nwhere the positive vertical direction is upwards. The perpendicular components of the \nforces must also sum to zero, \nFpivot \u2212 mbg + F1 sin\u03b81 + F2 sin\u03b82 = 0 . \n(18.3.8) \nOnly the vertical components F1,\u22a5 and F2,\u22a5 of the external forces are involved in the \nlever law (but the horizontal components must balance, as in Equation (18.3.5), for \nequilibrium). Then the Lever Law can be extended as follows. \nGeneralized Lever Law A beam of length l is balanced on a pivot point \nthat is placed directly beneath the center of mass of the beam. Suppose a \n\uf072 \nforce F1 acts on the beam a distance d1 to the right of the pivot point. A \n\uf072 \nsecond force F2 acts on the beam a distance d2 to the left of the pivot \npoint. The beam will remain in static equilibrium if the following two \nconditions are satisfied: \n1) \nThe total force on the beam is zero, \n2) \nThe product of the magnitude of the perpendicular component of \nthe force with the distance to the pivot is the same for each force, \nF \nF \n. \n(18.3.9)\n= d2\nd1 \n1,\u22a5 \n2,\u22a5 \nThe Generalized Lever Law can be stated in an equivalent form, \nd F sin\u03b8 = d F sin\u03b8 . \n(18.3.10)\n1\n1\n1\n2\n2 \n2 \nWe shall now show that the generalized lever law can be reinterpreted as the statement \nthat the vector sum of the torques about the pivot point S is zero when there are just two \n\uf072\n\uf072 \nforces F1 and F2 acting on our beam as shown in Figure 18.6. \n18-6", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_506", "text": "Figure 18.6 Force and torque diagram. \nLet\u2019s choose the positive z -direction to point out of the plane of the page then torque \npointing out of the page will have a positive z -component of torque (counterclockwise \nrotations are positive). From our definition of torque about the pivot point, the magnitude \n\uf072 \nof torque due to force F1 is given by \n\u03c4 \n= d F sin\u03b8 . \n(18.3.11)\nS ,1 \n1\n1 \n1 \nFrom the right hand rule this is out of the page (in the counterclockwise direction) so the \ncomponent of the torque is positive, hence, \n) \nsin\u03b81. \n(18.3.12)\n(\u03c4 S ,1 z = d1F1 \n\uf072 \nThe torque due to F2 about the pivot point is into the page (the clockwise direction) and \nthe component of the torque is negative and given by \n) \nsin\u03b82 . \n(18.3.13)\n(\u03c4 S ,2 z = \u2212d2 F2 \nThe z -component of the torque is the sum of the z -components of the individual torques \nand is zero, \n)\n)\n) = d1\nsin\u03b81 \nsin\u03b82 = 0 , \n(18.3.14)\n(\u03c4 S , total z = (\u03c4 S ,1 z + (\u03c4 S ,2 z \nF1 \n\u2212 d2 F2 \nwhich is equivalent to the Generalized Lever Law, Equation (18.3.10), \nd F\n1 1 sin\u03b81 = d F \n2 2 sin\u03b82 . \n18.4 Worked Examples \nExample 18.2 Suspended Rod \nA uniform rod of length l = 2.0 m and mass m = 4.0 kg is hinged to a wall at one end \nand suspended from the wall by a cable that is attached to the other end of the rod at an \n18-7", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_507", "text": "angle of \u03b2 = 30o to the rod (see Figure 18.7). Assume the cable has zero mass. There is a \ncontact force at the pivot on the rod. The magnitude and direction of this force is \nunknown. One of the most difficult parts of these types of problems is to introduce an \nangle for the pivot force and then solve for that angle if possible. In this problem you will \nsolve for the magnitude of the tension in the cable and the direction and magnitude of the \npivot force. (a) What is the tension in the cable? (b) What angle does the pivot force \nmake with the beam? (c) What is the magnitude of the pivot force? \nFigure 18.7 Example 18.2 \nFigure 18.8 Force and torque diagram. \nSolution: a) The force diagram is shown in Figure 18.8. Take the positive \u02c6i -direction to \nbe to the right in the figure above, and take the positive \u02c6j -direction to be vertically \n\uf072 \n\u02c6\nupward. The forces on the rod are: the gravitational force m g = \u2212m g j , acting at the \n\uf072 \ncenter of the rod; the force that the cable exerts on the rod, T = T (\u2212cos\u03b2 \u02c6i + sin \u03b2 \u02c6j) , \n\uf072 \nacting at the right end of the rod; and the pivot force F \n= F(cos\u03b1 \u02c6i + sin\u03b1 \u02c6j), acting at \npivot \nthe left end of the rod. If 0 < \u03b1 < \u03c0 / 2 , the pivot force is directed up and to the right in \nthe figure. If 0 > \u03b1 > \u2212\u03c0 / 2 , the pivot force is directed down and to the right. We have \nno reason, at this point, to expect that \u03b1 will be in either of the quadrants, but it must be \nin one or the other. \nFor static equilibrium, the sum of the forces must be zero, and hence the sums of the \ncomponents of the forces must be zero, \n0 = \u2212T cos\u03b2 + F cos\u03b1 \n(18.4.1)\n0 = \u2212mg + T sin\u03b2 + F sin\u03b1. \nWith respect to the pivot point, and taking positive torques to be counterclockwise, the \ngravitational force exerts a negative torque of magnitude mg(l / 2) and the cable exerts a \npositive torque of magnitude T l sin \u03b2 . The pivot force exerts no torque about the pivot. \nSetting the sum of the torques equal to zero then gives", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_508", "text": "gravitational force exerts a negative torque of magnitude mg(l / 2) and the cable exerts a \npositive torque of magnitude T l sin \u03b2 . The pivot force exerts no torque about the pivot. \nSetting the sum of the torques equal to zero then gives \n0 = T l sin \u03b2 \u2212 mg(l / 2) \nmg \n(18.4.2)\nT = \n.\n2sin\u03b2 \n18-8", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_509", "text": "This result has many features we would expect; proportional to the weight of the rod and \ninversely proportional to the sine of the angle made by the cable with respect to the \nhorizontal. Inserting numerical values gives \n\u22122)\nmg \n(4.0kg)(9.8m \u22c5s\nT =\n= \n= 39.2N. \n(18.4.3)\n2sin\u03b2 \n2sin30\uf06f \nThere are many ways to find the angle \u03b1 . Substituting Eq. (18.4.2) for the tension into \nboth force equations in Eq. (18.4.1) yields \nF cos\u03b1 = T cos\u03b2 = (mg / 2)cot \u03b2 \n(18.4.4)\nF sin\u03b1 = mg \u2212 T sin \u03b2 = mg / 2. \nIn Eq. (18.4.4), dividing one equation by the other, we see that tan\u03b1 = tan \u03b2 , \u03b1 = \u03b2 . \nThe horizontal forces on the rod must cancel. The tension force and the pivot force act \nwith the same angle (but in opposite horizontal directions) and hence must have the same \nmagnitude, \nF = T = 39.2N . \n(18.4.5) \nAs an alternative, if we had not done the previous parts, we could find torques about the \npoint where the cable is attached to the wall. The cable exerts no torque about this point \nand the y -component of the pivot force exerts no torque as well. The moment arm of the \nx -component of the pivot force is l tan \u03b2 and the moment arm of the weight is l / 2 . \nEquating the magnitudes of these two torques gives \nl\nF cos\u03b1 l tan \u03b2 = mg ,\n2 \nequivalent to the first equation in Eq. (18.4.4). Similarly, evaluating torques about the \nright end of the rod, the cable exerts no torques and the x -component of the pivot force \nexerts no torque. The moment arm of the y -component of the pivot force is l and the \nmoment arm of the weight is l / 2 . Equating the magnitudes of these two torques gives \nl\nF sin\u03b1 l = mg ,\n2 \nreproducing the second equation in Eq. (18.4.4). The point of this alternative solution is \nto show that choosing a different origin (or even more than one origin) in order to remove", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_510", "text": "l\nF sin\u03b1 l = mg ,\n2 \nreproducing the second equation in Eq. (18.4.4). The point of this alternative solution is \nto show that choosing a different origin (or even more than one origin) in order to remove \nan unknown force from the torques equations might give a desired result more directly. \n18-9", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_511", "text": "Example 18.3 Person Standing on a Hill \nA person is standing on a hill that is sloped at an angle of \u03b1 with respect to the \nhorizontal (Figure 18.9). The person\u2019s legs are separated by a distance d , with one foot \nuphill and one downhill. The center of mass of the person is at a distance h above the \nground, perpendicular to the hillside, midway between the person\u2019s feet. Assume that the \ncoefficient of static friction between the person\u2019s feet and the hill is sufficiently large that \nthe person will not slip. (a) What is the magnitude of the normal force on each foot? (b) \nHow far must the feet be apart so that the normal force on the upper foot is just zero? \nThis is the moment when the person starts to rotate and fall over. \nFigure 18.10 Free-body force diagram\nFigure 18.9 Person standing on hill  \nfor person standing on hill  \nSolution: The force diagram on the person is shown in Figure 18.10. Note that the \ncontact forces have been decomposed into components perpendicular and parallel to the \nhillside. A choice of unit vectors and positive direction for torque is also shown. \nApplying Newton\u2019s Second Law to the two components of the net force, \n\u02c6j: N1 + N2 \u2212 mg cos\u03b1 = 0 \n(18.4.6) \n\u02c6i : f1 + f2 \u2212 mg sin\u03b1 = 0 . \n(18.4.7) \nThese two equations imply that \nN1 + N2 = mg cos\u03b1 \n(18.4.8) \nf1 + f2 = mg sin\u03b1 . \n(18.4.9) \nEvaluating torques about the center of mass, \nd\nh( f1 + f2) + (N2 \u2212 N1) \n= 0 . \n(18.4.10)\n2 \n18-10", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_512", "text": "Equation (18.4.10) can be rewritten as \n2h( f1 + f2)\n= \n. \n(18.4.11)\nN1 \u2212 N2 \nd \nSubstitution of Equation (18.4.9) into Equation (18.4.11) yields \n2 ( \nh mg sin\u03b1 )\nN1 \u2212 N2 = \n. \n(18.4.12)\nd \nWe can solve for N1 by adding Equations (18.4.8) and (18.4.12), and then dividing by 2, \nyielding \n1 \nh mg sin\u03b1 ) \nh \n\u239e\n( \n\u239b 1\nN1 = mg cos\u03b1 +\n= mg \u239c cos\u03b1 + sin\u03b1 \u239f . \n(18.4.13)\n2 \nd \n\u239d 2 \nd \n\u23a0 \nSimilarly, we can solve for N2 by subtracting Equation (18.4.12) from Equation (18.4.8) \nand dividing by 2, yielding \n\u239b 1 \nh \n\u239e\nN2 = mg \u239c cos\u03b1 \u2212 sin\u03b1 \u239f . \n(18.4.14)\n\u239d 2 \nd \n\u23a0 \nThe normal force N2 as given in Equation (18.4.14) vanishes when \n1 \nh \ncos\u03b1 = \nsin\u03b1 , \n(18.4.15)\n2 \nd \nwhich can be solved for the minimum distance between the legs, \nd = 2 (tan\nh \n\u03b1 ) . \n(18.4.16) \nIt should be noted that no specific model for the frictional force was used, that is, no \ncoefficient of static friction entered the problem. The two frictional forces f1 and f2 \nwere not determined separately; only their sum entered the above calculations. \nExample 18.4 The Knee \nA man of mass m = 70kg is about to start a race. Assume the runner\u2019s weight is equally \ndistributed on both legs. The patellar ligament in the knee is attached to the upper tibia \n\uf072 \nand runs over the kneecap. When the knee is bent, a tensile force, T , that the ligament \nexerts on the upper tibia, is directed at an angle of \u03b8 = 40\u00b0 with respect to the horizontal. \n\uf072 \nThe femur exerts a force F on the upper tibia. The angle, \u03b1 , that this force makes with \nthe vertical will vary and is one of the unknowns to solve for. Assume that the ligament is \nconnected a distance, d = 3.8cm , directly below the contact point of the femur on the \n18-11", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_513", "text": "1 \ntibia. The contact point between the foot and the ground is a distance s = 3.6 10 cm\n\u00d7 \nfrom the vertical line passing through contact point of the femur on the tibia. The center \nof mass of the lower leg lies a distance x = \n\u00d7 \n1\n1.8 10 cm from this same vertical line. \nSuppose the mass mL of the lower leg is a 1/10 of the mass of the body (Figure 18.11). (a) \n\uf072 \nFind the magnitude T of the force T of the patellar ligament on the tibia. (b) Find the \n\uf072 \ndirection (the angle \u03b1 ) of the force F of the femur on the tibia. (c) Find the magnitude \n\uf072 \nF of the force F of the femur on the tibia. \nFigure 18.11 Example 18.4 \nFigure 18.12 Torque-force diagram for \nknee \nSolutions: a) Choose the unit vector \u02c6i to be directed horizontally to the right and \u02c6j \ndirected vertically upwards. The first condition for static equilibrium, Eq. (18.1.1), that \nthe sum of the forces is zero becomes \n\u02c6i : \u2212 F sin\u03b1 + T cos\u03b8 = 0. \n(18.4.17) \n\u02c6j : N \u2212 F cos\u03b1 + T sin\u03b8 \u2212 (1 / 10)mg = 0. \n(18.4.18) \nBecause the weight is evenly distributed on the two feet, the normal force on one foot is \nequal to half the weight, or \nN = (1/ 2)mg ; \n(18.4.19) \nEquation (18.4.18) becomes \n\u02c6j : (1/ 2)mg \u2212 F cos\u03b1 + T sin\u03b8 \u2212 (1/ 10)mg = 0 . \n(18.4.20)\n(2 / 5)mg \u2212 F cos\u03b1 + T sin\u03b8 = 0. \n18-12", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_514", "text": "The torque-force diagram on the knee is shown in Figure 18.12. Choose the point of \naction of the ligament on the tibia as the point S about which to compute torques. Note \n\uf072 \nthat the tensile force, T , that the ligament exerts on the upper tibia will make no \ncontribution to the torque about this point S . This may help slightly in doing the \ncalculations. Choose counterclockwise as the positive direction for the torque; this is the \n\uf072 \npositive k\u02c6 - direction. Then the torque due to the force F of the femur on the tibia is \n\uf072 \n\uf072\n\uf072 \n\u03c4\n= r \u00d7 F = d \u02c6j \u00d7 (\u2212 F sin\u03b1 \u02c6i \u2212 F cos\u03b1\u02c6j) = d F sin\u03b1 k\u02c6 . \n(18.4.21)\nS ,1 \nS ,1 \nThe torque due to the mass of the leg is \n\uf072\n\uf072\n\u03c4 \n= r \n\u00d7 (\u2212mg / 10)\u02c6j = (\u2212x \u02c6i \u2212 y \u02c6j) \u00d7 (\u2212mg / 10)\u02c6j = (1/ 10)x mg k\u02c6 . (18.4.22)\nS ,2 \nS ,2 \nL \n\u03c4 \nThe torque due to the normal force of the ground is \n\u03c4\n\uf072\n\uf072\n\u03c4 \n= r \u00d7 N \u02c6j = (\u2212s\u02c6i \u2212 y \u02c6j) \u00d7 N \u02c6j = \u2212s N k\u02c6 = \u2212(1/ 2)s mg k\u02c6 . \n(18.4.23)\nS \nS \nN\n,3 \n,3 \n\u03c4\n(In Equations (18.4.22) and (18.4.23), yL and yN are the vertical displacements of the \n\u03c4\npoint where the weight of the leg and the normal force with respect to the point S ; as can \nbe seen, these quantities do not enter directly into the calculations.) The condition that the \nsum of the torques about the point S vanishes, Eq. (18.1.2), \n\uf072\n\uf072\n\uf072\n\uf072\n\uf072 \n0 , \n(18.4.24) \n+  \n+ \n=  \n=  \nS , total \nS ,1 \nS ,2 \nS ,3 \nbecomes \n\uf072 \nd F sin\u03b1 k\u02c6 + (1/ 10)x mg k\u02c6 \u2212 (1/ 2)s mg k\u02c6 = 0 . \n(18.4.25) \nThe three equations in the three unknowns are summarized below: \n\u2212 F sin\u03b1 + T cos\u03b8 = 0 \n(2 / 5)mg \u2212 F cos\u03b1 + T sin\u03b8 = 0 \n(18.4.26) \nd F sin\u03b1 + (1/ 10)x mg \u2212 (1/ 2)s mg = 0.", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_515", "text": "\u2212 F sin\u03b1 + T cos\u03b8 = 0 \n(2 / 5)mg \u2212 F cos\u03b1 + T sin\u03b8 = 0 \n(18.4.26) \nd F sin\u03b1 + (1/ 10)x mg \u2212 (1/ 2)s mg = 0. \nThe horizontal force equation, the first in (18.4.26), implies that \nF sin\u03b1 = T cos\u03b8 . \n(18.4.27) \nSubstituting this into the torque equation, the third equation of (18.4.26), yields \ndT cos\u03b8 + (1/ 10)x mg \u2212 s(1/ 2)mg = 0 . \n(18.4.28) \n18-13", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_516", "text": "Note that Equation (18.4.28) is the equation that would have been obtained if we had \nchosen the contact point between the tibia and the femur as the point about which to \ndetermine torques. Had we chosen this point, we would have saved one minor algebraic \n\uf072 \nstep. We can solve this Equation (18.4.28) for the magnitude T of the force T of the \npatellar ligament on the tibia, \ns(1/ 2)mg \u2212 (1/ 10)x mg \nT = \n. \n(18.4.29)\nd cos\u03b8 \nInserting numerical values into Equation (18.4.29), \n(3.6\u00d7 10\u22121m)(1/2) \u2212 (1/10)(1.8\u00d7 10\u22121m) \n\u22122)\nT = (70kg)(9.8m \u22c5 s \n(3.8\u00d7 10\u22122m)cos(40\u00b0) \n(18.4.30) \n= 3.8 \u00d7 103 N. \n\uf072 \nb) We can now solve for the direction \u03b1 of the force F of the femur on the tibia as \nfollows. Rewrite the two force equations in (18.4.26) as \nF cos\u03b1 = (2 / 5)mg + T sin\u03b8 \n(18.4.31)\nF sin\u03b1 = T cos\u03b8. \nDividing these equations yields \nF cos\u03b1 \n(2 / 5)mg + T sin\u03b8 \n= cotan\u03b1 = \n, \n(18.4.32)\nF sin\u03b1 \nT cos\u03b8 \nAnd so \n\u239b (2 / 5)mg + T sin\u03b8 \u239e\n\u03b1 = cotan\u22121 \n\u239d\u239c \nT cos\u03b8 \n\u23a0\u239f \n(18.4.33)\n\u239b (2 / 5)(70kg)(9.8m \u22c5 s\u22122) + (3.4 \u00d7 103 N)sin(40\u00b0)\u239e\n\u03b1 = cotan\u22121 \n\u23a0\u239f = 47\u00b0. \n\u239d\u239c \n(3.4 \u00d7 103N)cos(40\u00b0) \nc) We can now use the horizontal force equation to calculate the magnitude F of the \n\uf072 \nforce of the femur F on the tibia from Equation (18.4.27), \n(3.8 \u00d7 103 N)cos(40\u00b0)\nF =\n= 4.0 \u00d7 103 N . \n(18.4.34)\nsin(47\u00b0) \nNote you can find a symbolic expression for \u03b1 that did not involve the intermediate", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_517", "text": "(3.8 \u00d7 103 N)cos(40\u00b0)\nF =\n= 4.0 \u00d7 103 N . \n(18.4.34)\nsin(47\u00b0) \nNote you can find a symbolic expression for \u03b1 that did not involve the intermediate \nnumerical calculation of the tension. This is rather complicated algebraically; basically, \nthe last two equations in (18.4.26) are solved for F and T in terms of \u03b1 , \u03b8 and the \n18-14", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_518", "text": "other variables (Cramer\u2019s Rule is suggested) and the results substituted into the first of \n(18.4.26). The resulting expression is \n(s / 2 \u2212 x / 10)sin(40\u00b0) + ((2d / 5)cos(40\u00b0)) \ncot\u03b1 = \n(s / 2 \u2212 x / 10)cos(40\u00b0) \n(18.4.35)\n2d / 5 \n= tan(40\u00b0) + s / 2 \u2212 x / 10 \nwhich leads to the same numerical result, \u03b1 = 47\u00b0 . \n18-15\n\nAppendix 18A The Torques About Any Two Points are Equal for a  \nBody in Static Equilibrium  \nWhen the net force on a body is zero, the torques about any two points are equal. To \nshow this, consider any two points A and B . Choose a coordinate system with origin O \n\uf072\n\uf072\nand denote the constant vector from A to B by \n, . Suppose a force Fi is acting at the \nrA B \n\uf072\npoint rO,i . The vector from the point A to the point where the force acts is denoted by \n\uf072 , and the vectors from the point B to the point where the force acts is denoted by \nrA, i \n\uf072 rB, i . \n\uf072r \n\uf072r \n\uf072r \nFigure 18A.1 Location of body i with respect to the points A and B . \nIn Figure 18A.1, the position vectors satisfy \nA, i\nA, B\nB, i . \n(18.A.1) \n+ \n=  \nThe sum of the torques about the point A is given by \n\uf072\ni= N \uf072\n\uf072 \n\u03c4 A = \u2211rA,i \u00d7 Fi . \n(18.A.2) \ni=1 \nThe sum of the torques about the point B is given by \ni= N \n\uf072\n\uf072\n\uf072\n\u03c4 B = \u2211rB,i \u00d7 Fi . \n(18.A.3) \ni=1 \nWe can now substitute Equation (18.A.1) into Equation (18.A.2) and find that \n\uf072\nF\n\uf072r \n\uf072\nF\n\uf072r \n\uf072\nF\n\uf072r \n\uf072r \n\uf072\nF\n\uf072r\n\uf072\u03c4 \nA, i\ni \nA, B\nB, i\ni \nA, B\ni \nB, i\ni \ni=1 \ni=1 \ni=1 \ni=1 \n\uf072\nIn the next-to-last term in Equation (18.A.4), the vector \n, \nis constant and so may be \nrA B \ntaken outside the summation, \ni= N \ni= N \ni= N \ni= N \n\u2211  \n\u2211  \n\u2211  \n\u2211 \n(  \n) \u00d7  \n(18.A.4) \n\u00d7  \n+  \n\u00d7  +  \n\u00d7 \n=  \n=  \n=  \n. \nA \n18-16", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_519", "text": "i=N \n\uf072 \ni=N \uf072\n\u2211r \uf072 \n\u00d7 F = r \uf072 \n, \u00d7\u2211Fi . \n(18.A.5)\nA B\n, \ni \nA B \ni=1 \ni=1 \nWe are assuming that there is no net force on the body, and so the sum of the forces on \nthe body is zero, \ni= N \uf072\n\u2211Fi = 0 \n\uf072 \n. \n(18.A.6) \ni=1 \nTherefore the torque about point A , Equation (18.A.2), becomes \n\uf072 \ni=N \uf072 \n\uf072\n\uf072\n\u03c4 A = \u2211rB,i \u00d7 Fi = \u03c4 B . \n(18.A.7) \ni=1 \nFor static equilibrium problems, the result of Equation (18.A.7) tells us that it does not \nmatter which point we use to determine torques. In fact, note that the position of the \nchosen origin did not affect the result at all. Choosing the point about which to calculate \ntorques (variously called \u201c A \u201d, \u201c B \u201d, \u201c S \u201d or sometimes \u201cO \u201d) so that unknown forces do \nnot exert torques about that point may often greatly simplify calculations. \n18-17", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_520", "text": "Chapter 19 Angular Momentum \n19.1 Introduction........................................................................................................... 2  \n19.2 Angular Momentum about a Point for a Particle.............................................. 3  \n19.2.1 Angular Momentum for a Point Particle..................................................... 3  \n19.2.2 Right-Hand-Rule for the Direction of the Angular Momentum ............... 4  \nExample 19.1 Angular Momentum: Constant Velocity ........................................ 5  \nExample 19.2 Angular Momentum and Circular Motion..................................... 6  \nExample 19.3 Angular Momentum About a Point along Central Axis for Circular  \nMotion ........................................................................................................................ 6", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_521", "text": "19.3 Torque and the Time Derivative of Angular Momentum about a Point for a Particle  \n........................................................................................................................................ 8  \n19.4 Conservation of Angular Momentum about a Point ......................................... 9  \nExample 19.4 Meteor Flyby of Earth.................................................................... 10  \n19.5 Angular Impulse and Change in Angular Momentum ................................... 12  \n19.6 Angular Momentum of a System of Particles .................................................. 13  \nExample 19.5 Angular Momentum of Two Particles undergoing Circular Motion", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_522", "text": "Example 19.5 Angular Momentum of Two Particles undergoing Circular Motion  \n................................................................................................................................... 14  \nExample 19.6 Angular Momentum of a System of Particles about Different Points  \n................................................................................................................................... 16  \n19.7 Angular Momentum and Torque for Fixed Axis Rotation ............................. 17  \nExample 19.6 Circular Ring................................................................................... 20  \n19.8 Principle of Conservation of Angular Momentum .......................................... 21", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_523", "text": "19.8 Principle of Conservation of Angular Momentum .......................................... 21  \nExample 19.7 Collision Between Pivoted Rod and Object.................................. 21  \n19.9 External Angular Impulse and Change in Angular Momentum ................... 26  \nExample 19.8 Angular Impulse on Steel Washer................................................. 26", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_524", "text": "Chapter 19 Angular Momentum \nThe situation, in brief, is that newtonian physics is incapable of predicting \nconservation of angular momentum, but no isolated system has yet been \nencountered experimentally for which angular momentum is not conserved. We \nconclude that conservation of angular momentum is an independent physical law, \nand until a contradiction is observed, our physical understanding must be guided \nby it. 1 \nDan Kleppner \n19.1 Introduction \nWhen we consider a system of objects, we have shown that the external force, acting at the center \nof mass of the system, is equal to the time derivative of the total momentum of the system,\n\uf072 \n(19.1.1)\n= \n. \n\uf072r\n\uf072 \ndp\nFext \nsys \ndt \nWe now introduce the rotational analog of Equation (19.1.1). We will first introduce the concept \n\uf072\nof angular momentum for a point-like particle of mass m with linear momentum p about a point \nS \nS , defined by the equation\n\uf072 \nL \n\uf072p\n(19.1.2) \n\u00d7 \n=  \n, \nS \n\uf072\nwhere rS is the vector from the point S to the particle. We will show in this chapter that the \ntorque about the point S acting on the particle is equal to the rate of change of the angular \nmomentum about the point S of the particle, \n\uf072\u03c4S = d \n\uf072 \nLS \ndt . \n(19.1.3) \nEquation (19.1.3) generalizes to any body undergoing rotation.  \nWe shall concern ourselves first with the special case of rigid body undergoing fixed axis rotation  \n\uf072\nabout the z-axis with angular velocity \u03c9 = \u03c9 z k\u02c6 . We divide up the rigid body into N elements \n\uf072\nlabeled by the index i , i = 1,2,\u2026N , the ith element having mass mi and position vector rS i, . The \nrigid body has a moment of inertia IS about some point S on the fixed axis, (often taken to be the \n\uf072 \nz-axis, but not always) which rotates with angular velocity \u03c9 about this axis. The angular \nmomentum is then the vector sum of the individual angular momenta, \n1 Kleppner, Daniel, An Introduction to Mechanics (1973), p. 307. \n2", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_525", "text": "\uf072\nL  \ni=N \uf072\nL  \ni=N \n\u00d7 \uf072pi \n(19.1.4) \n\u2211  \n\u2211 \n=  \n=  \nS \nS ,i \ni=1 \ni=1 \nS ,i \nWhen the rotation axis is the z-axis the z-component of the angular momentum, LS ,z , about the \npoint S is then given by \nLS ,z = IS \u03c9 z . \n(19.1.5) \nWe shall show that the z-component of the torque about the point S , \u03c4 S ,z , is then the time \nderivative of the z-component of angular momentum about the point S , \ndLS ,z\nd\u03c9 z\n= \n\u03b1 . \n(19.1.6)\n\u03c4 S ,z \n= IS \n= IS \nz\ndt \ndt \n\uf072r \n19.2 Angular Momentum about a Point for a Particle \n19.2.1 Angular Momentum for a Point Particle \n\uf072\nConsider a point-like particle of mass m moving with a velocity v (Figure 19.1) with momentum \n\uf072\n\uf072 \np = mv . \n.\nS \n. \nrS \nm \np \nFigure 19.1 A point-like particle and its angular momentum about S . \n\uf072\nConsider a point S located anywhere in space. Let rS denote the vector from the point S to the \nlocation of the object. \n\uf072 \nDefine the angular momentum LS about the point S of a point-like particle as the \nvector product of the vector from the point S to the location of the object with the \nmomentum of the particle, \n\uf072 \n\uf072\n\uf072\nLS = rS \u00d7p . \n(19.2.1) \n2 \u22c5s\nThe derived SI units for angular momentum are [kg \u22c5 m\n\u22121] = [N \u22c5m \u22c5s] = [J \u22c5s]. There is no \nspecial name for this set of units. \nBecause angular momentum is defined as a vector, we begin by studying its magnitude and \ndirection. The magnitude of the angular momentum about S is given by \n3", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_526", "text": "\uf072 \n\uf072 \uf072 \n= r p sin\u03b8 , \n(19.2.2)\nLS \nS \n\uf072\nwhere \u03b8 is the angle between the vectors and p , and lies within the range [0 \u2264\u03b8 \u2264\u03c0 ] Analogous \nto the magnitude of torque, there are two ways to determine the magnitude of the angular \nmomentum about S . \n.S \n.\nrS \np \nrS \nS \nrS\n.\nmoment arm \n. \np \npS \nFigure 19.2 (a) Moment arm. \n(b) Perpendicular component of momentum. \nDefine the moment arm, rS \n\u22a5 , (Figure 19.2 (a)), as the perpendicular distance from the point S to \nthe line defined by the direction of the momentum. Then \n\uf072 \nrS \n\u22a5 = \nsin\u03b8 . \n(19.2.3)\nrS \nHence the magnitude of the angular momentum is the product of the moment arm with the \nmagnitude of the momentum, \n\uf072 \n\uf072 p . \n(19.2.4)\nLS = rS \n\u22a5 \nAlternatively, let Error! Objects cannot be created from editing field codes. denote the \nmagnitude of the component of the momentum perpendicular to the line defined by the direction \n\uf072\nof the vector rS . From the geometry shown in Figure 19.2 (b), \n\uf072 p sin\u03b8 . \n(19.2.5)\npS \n\u22a5 = \nThus the magnitude of the angular momentum is the product of the distance from S to the \nparticle with pS \n\u22a5 , \n\uf072 \n\uf072 \n= \n\u22a5 . \n(19.2.6)\nLS \nrS pS \n19.2.2 Right-Hand-Rule for the Direction of the Angular Momentum \nWe shall define the direction of the angular momentum about the point S by a right hand rule. \n\uf072\n\uf072\nDraw the vectors rS and p so their tails are touching. Then draw an arc starting from the vector \n\uf072\n\uf072 \nrS and finishing on the vector p . (There are two such arcs; choose the shorter one.) This arc is \neither in the clockwise or counterclockwise direction. Curl the fingers of your right hand in the \nsame direction as the arc. Your right thumb points in the direction of the angular momentum. \n4", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_527", "text": "\u22c5 \n  \n  \n\u22c5 \n \n \n \n \n \n  \n  \n \n \n \n \n \n \n  \n \n  \n \n \n \n \n  \n \n \n \n \n  \n \n \n   \n \n \n \n    \n     \n          \np\nLS = rS \nrS \np \n.\n.\nS \nFigure 19.3 The right hand rule for determining the direction of angular momentum about S .  \nRemember that, as in all vector products, the direction of the angular momentum about S is  \n\uf072\n\uf072\nperpendicular to the plane formed by rS and p . \nExample 19.1 Angular Momentum: Constant Velocity \nA particle of mass m = 2.0 kg moves as shown in Figure 19.4 with a uniform velocity \n\uf072 \n\u22121 \u02c6 \n\u22121 \u02c6\nv = 3.0 m s\ni + 3.0 m s\nj . At time t , the particle passes through the point (2.0 m, 3.0 m) . Find \nthe direction and the magnitude of the angular momentum about the point S (the origin) at time \nt . \n.\nS . \nrS \nm \np \n+ x \n+ y \n\u02c6i \n\u02c6j \n\u02c6k\nFigure 19.4 Example 19.4 \nSolution: Choose Cartesian coordinates with unit vectors shown in the figure above. The vector \n\uf072\nfrom the point S to the location of the particle is rS = 2.0 m \u02c6i + 3.0 m \u02c6j . The angular momentum \n\uf072 \nvector LO of the particle about the origin S is given by: \n\uf072 \n\uf072\n\uf072\n\uf072 \n\uf072\nLS = rS \u00d7 p = rS \u00d7 mv \n= (2.0m \u02c6i + 3.0m \u02c6j) \u00d7 (2kg)(3.0m \u22c5s\u22121\u02c6i + 3.0m \u22c5s\u22121\u02c6j) \n= 0 +12kg \u22c5 m2 \u22c5s\u22121 k\u02c6 \u221218kg \u22c5m2 \u22c5s\u22121(\u2212k\u02c6 ) + 0 \n\uf072 \n2 \u22c5s\u22121 k\n= \u22126kg \u22c5 m\n\u02c6. \n\uf072\n\uf072\n\uf072\n\uf072\n\uf072 \n\uf072\n\uf072\n\uf072\n\uf072\n\uf072\n\uf072 \nIn the above, the relations i \u00d7 j = k, j \u00d7 i = \u2212k, i \u00d7 i = j \u00d7 j = 0 were used. \n5", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_528", "text": "Example 19.2 Angular Momentum and Circular Motion \nA particle of mass m moves in a circle of radius R about the z -axis in the x-y plane defined by \n\uf072 \n\u02c6\nz = 0 with angular velocity \u03c9 = \u03c9 z k , \u03c9 z > 0 , (Figure 19.5). Find the magnitude and the direction \n\uf072 \nof the angular momentum LS relative to the point S lying at the center of the circular orbit, (the \norigin). \n. \np \n\u02c6\n\u02c6\n\u02c6k\nS . \nLS \n+ z \n= \nz \u02c6k \nrS = R \u02c6r \nr \nFigure 19.5 Example 19.2 \n\uf072\nSolution: The velocity of the particle is given by v = R\u03c9 z \u03b8\u02c6 . The vector from the center of the \n\uf072\ncircle (the point S ) to the object is given by rS = R r\u02c6 . The angular momentum about the center of \nthe circle is the vector product \n\uf072\n\uf072\n\uf072\n\uf072\n\uf072 \n\uf072\nLS = rS \u00d7 p = rS \u00d7 mv = Rmv k\u02c6 = RmR\u03c9 z k\u02c6 = mR2\u03c9 z k\u02c6 = IS\u03c9 . \n\uf072 \nThe magnitude is \n= mR2\u03c9 z , and the direction is in the + k\u02c6 -direction. For the particle, the \nLS \nmoment of inertia about the z -axis is IS = mR2 , therefore the angular momentum about S is \n\uf072\n\uf072 \n= IS\u03c9 .\nLS \n\uf072 \nThe fact that LS is in the same direction as the angular velocity is due to the fact that the point S \nlies on the plane of motion. \nExample 19.3 Angular Momentum About a Point along Central Axis for Circular Motion \n\uf072\nA particle of mass m moves in a circle of radius R with angular velocity \u03c9 = \u03c9 z k\u02c6 , \u03c9 z > 0 , \nabout the z - axis in a plane parallel to but a distance h above the x-y plane (Figure 19.6). Find \n\uf072 \nthe magnitude and the direction of the angular momentum LS relative to the point S (the origin). \n6", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_529", "text": "k\u02c6\n= \n+ z\nz \n\u02c6\np. \n\u02c6 \nk \n\u02c6r \nh \nR  \nS \n+ x  \nFigure 19.6 Example 19.3 \n\uf072 \nSolution: The easiest way to calculate LS is to use cylindrical coordinates. We begin by writing \n\uf072\n\uf072\nthe two vectors rS and p in polar coordinates. We start with the vector from point S (the origin) \n\uf072\nto the location of the moving object, rS = Rr\u02c6 + hk\u02c6 . The momentum vector is tangent to the \n\uf072\n\uf072\ncircular orbit so p = mv = mR\u03c9 z \u03b8\u02c6 . Using the fact that r\u02c6 \u00d7 \u03b8\u02c6 = k\u02c6 and k\u02c6 \u00d7 \u03b8\u02c6 = \u2212r\u02c6 , the angular \nmomentum about point S is \n\uf072 \n\uf072\n\uf072\nLS = rS \u00d7 p = (Rr\u02c6 + hk\u02c6 ) \u00d7 mR\u03c9 z \u03b8\u02c6 = mR2\u03c9 z k\u02c6 \u2212 hmR\u03c9 z r\u02c6 \nk\u02c6\n= \nz . \n+ z \n+ x\nS \nrS \np \nLS \nR \nh \nFigure 19.7 Angular momentum about the point S \n\uf072 \nThe magnitude of LS is given by \n\uf072 \n)2)1/2 \n(h2 + R2)1/2 \n= ((mR2\u03c9 )2 + (hmR\u03c9 \n= mR\u03c9\nLS \nz\nz \nz \n\uf072 \nThe direction of LS is given by (Figure 19.7) \nLS ,z\nR\n\u2212 \n=\n= tan\u03c6\nLS ,r\nh \n7", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_530", "text": "We also present a geometric argument. Suppose the particle has coordinates (x, y,h) . The \n\uf072rS \n\uf072 \nL \n\uf072p . The vectors r \uf072\n\uf072\nangular momentum about the origin is given by \nand\n\u00d7 \n=  \np are\nS \nS \nperpendicular to each other so the angular momentum is perpendicular to the plane formed by\n\uf072\nthose two vectors. Recall that the speed v = R\u03c9 z . Suppose the vector rS forms an angle \u03c6 \n\uf072 \nwith the z -axis. Then LS forms an angle \u03c6 with respect to the x \u2212 y plane as shown in the \n\uf072 \nfigure above. The magnitude of LS is \n\uf072rS \n\uf072 \nL \n\uf072v = (h2 + R2)1/2 mR\u03c9\n=  \nm\nS \nz \n\uf072 \nThe magnitude of LS is constant, but its direction is changing as the particle moves in a \ncircular orbit about the z -axis, sweeping out a cone as shown in Figure 19.8. We draw the \n\uf072 \nvector LS at the origin because it is defined at that point. \nk\u02c6\n= \nz\n+ z \nLS \nS \n+ x \ndt \n\uf072rS \n\uf072 \nL\nFigure 19.8 Direction of angular momentum about the point S sweeps out a cone \nThe important point to keep in mind regarding this calculation is that for any point along the \nz -axis not at the center of the circular orbit of a single particle, the angular momentum about \nthat point does not point along the z -axis but it is has a non-zero component in the x \u2212 y \nplane (or in the \u2212r\u02c6 direction if you use polar coordinates). The z -component of the angular \nmomentum about any point along the z -axis is independent of the location of that point along \nthe axis. \n19.3 Torque and the Time Derivative of Angular Momentum about a Point \nfor a Particle \nWe will now show that the torque about a point S is equal to the time derivative of the \nangular momentum about S , \n\uf072 \n\uf072 \ndLS\n= \n. \n(19.3.1)\n\u03c4S \nTake the time derivative of the angular momentum about S , \nd \nS \ndt =  d \ndt ( \n\u00d7 \uf072p) .  \n(19.3.2)  \n19-8", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_531", "text": "In this equation we are taking the time derivative of a vector product of two vectors. There are \ntwo important facts that will help us simplify this expression. First, the time derivative of the \nvector product of two vectors satisfies the product rule, \nd \nS\n\uf072r\nS\nS \n\uf072r \n\uf072 \nL\nd \ndt S \n\uf072rS \n\uf072p\n\u239b  \n\u239e\n\u239b  \n\u239e \n\u239b  \n\u239e  \n\u239b  \n\u239e \nd \nd\n\uf072p \n\uf072p\n(  \n) =  \n(19.3.3) \n\u00d7  \n\u23a0\u239f \u00d7  \u239f + \n\u23a0 \n\u00d7 \n=  \n\u239c\u239d \n\u239d\u239c  \n\u23a0\u239f .  \n\u239d\u239c  \n\u239d\u239c  \n\u23a0\u239f \ndt \ndt \ndt \nSecond, the first term on the right hand side vanishes, \n\uf072\ndrS \n\uf072\n\uf072 \n\uf072\n\uf072 \n\u00d7p = v \u00d7 m v = 0 . \n(19.3.4)\ndt \nThe rate of angular momentum change about the point S is then \nd \ndt \n\uf072 \nL\n\uf072p\nd\n\uf072rS \n\uf072rS \nS \n\uf072 \nL\nFrom Newton\u2019s Second Law, the force on the particle is equal to the derivative of the linear \nmomentum, \n\uf072\n\uf072 \ndp\nF = \n. \n(19.3.6)\ndt \nTherefore the rate of change in time of angular momentum about the point S is \nd \nS \n(19.3.5) \n\u00d7 \n=  \n. \ndt \n\uf072\nF \n(19.3.7) \n\u00d7 \n=  \n. \ndt \n\uf072 \nRecall that the torque about the point S due to the force F acting on the particle is \n\uf072 \n\uf072\n\uf072 \n\u03c4S = rS \u00d7 F . \n(19.3.8) \nCombining the expressions in (19.3.7) and (19.3.8), it is readily seen that the torque about the \npoint S is equal to the rate of change of angular momentum about the point S , \n\uf072 \n\uf072 \ndLS\n= \n. \n(19.3.9)\n\u03c4S \ndt \n19.4 Conservation of Angular Momentum about a Point \n19-9", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_532", "text": "So far we have introduced two conservation principles, showing that energy is constant for \nclosed systems (no change in energy in the surroundings) and linear momentum is constant \nisolated system. The change in mechanical energy of a closed system is \nWnc = \u0394Em = \u0394K + \u0394U , \n(closed system) . \n(19.3.10) \nIf the non-conservative work done in the system is zero, then the mechanical energy is \nconstant, \n0 = W \n= \u0394K + \u0394U , \n(closed system) . \n(19.3.11)\nnc = \u0394Emechanical \nThe conservation of linear momentum arises from Newton\u2019s Second Law applied to systems, \n\uf072\nF ext \nN d\n= \u2211 dt \n\uf072p\nd \n=  \ni \ndt \n\uf072psys \n(19.3.12)  \ni=1 \nThus if the external force in any direction is zero, then the component of the momentum of the \nsystem in that direction is a constant. For example, if there are no external forces in the x \u00ad\nand y -directions then \n\uf072\n0  \n\uf072\nF ext ) \nd \uf072p\n= (  \n(  \n) \n=  \nx \ndt \nsys x \n(19.3.13)\n\uf072\n0  \n\uf072\nF ext ) \nd \uf072p\n= (  \n(  \n) \n=  \n. \ny \ndt \nsys y \n\uf072 \nL \nWe can now use our relation between torque about a point S and the change of the angular \nmomentum about S , Eq. (19.3.9), to introduce a new conservation law. Suppose we can find \na point S such that torque about the point S is zero, \n\uf072 \n\uf072\n\uf072 \ndLS\n0 =\n= \n, \n(19.3.14)\n\u03c4S \ndt \nthen the angular momentum about the point S is a constant vector, and so the change in \n\u0394 \nangular momentum is zero,\n\uf072 \nL  \n\uf072 \nL  \n\uf072\n0 \n\u2261  \n\u2212  \n(19.3.15) \n=  . \nS \nS , f \nS ,i \nThus when the torque about a point S is zero, the final angular momentum about S is equal \nto the initial angular momentum, \n\uf072 \n\uf072 \nLS , f = LS ,i . \n(19.3.16) \nExample 19.4 Meteor Flyby of Earth \nA meteor of mass m is approaching Earth as shown in the figure. The radius of Earth is RE .  \nThe mass of Earth is \n. Assume that the meteor started very far away from Earth with  \nM E \n19-10", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_533", "text": "speed vi and at a perpendicular distance h from the axis of symmetry of the orbit. At some \nlater time the meteor just grazes Earth (Figure 19.9). You may ignore all other gravitational \nforces except due to Earth. Find the distance h . Hint: What quantities are constant for this \norbit? \n. \n. \nh \nvi ( ) \nv f ( ) \n. \nv(RE ) \nE (M E ,RE ) \nFigure 19.9 Meteor flyby of earth \nSolution: In this problem both energy and angular momentum about the center of Earth are \nconstant (see below for justification). \nThe meteor\u2019s mass is so much small than the mass of Earth that we will assume that the \nearth\u2019s motion is not affected by the meteor. We\u2019ll also need to neglect any air resistance \nwhen the meteor approaches Earth. Choose the center of Earth, (point S ) to calculate the \ntorque and angular momentum. The force on the meteor is \n\uf072 G \nGmM E\n= \u2212 \nr\u02c6\nFE,m \n2r\n\uf072\nThe vector from the center of Earth to the meteor is \n= rr\u02c6 . The torque about S is zero \nrS \nbecause they two vectors are anti-parallel \n\uf072\n\uf072 \nGmM E \n\uf072\n\uf072 \nG\n\u00d7 \n= rr\u02c6 \u00d7 \u2212 \nr\u02c6 = 0\n\u03c4S = rS \nFE,m \n2r \nTherefore the angular momentum about the center of Earth is a constant. \nThe initial angular momentum is \n\uf072\n\uf072 \n\uf072\nLS ,i = rS ,i \u00d7 mvi = (xi \u02c6i + h\u02c6j) \u00d7 mvi \u02c6i = \u2212hmvik\u02c6 \nWhen the meteor just grazes Earth, the angular momentum is \n\uf072\n\uf072 \n\uf072\nLS , E = rS , E \u00d7 mv p = RE \u02c6i \u00d7 mv p (\u2212\u02c6j) = \u2212REmv p k\u02c6 \n19-11", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_534", "text": "where we have used vp for the speed of the meteor at its nearest approach to Earth. The \nconstancy of angular momentum requires that \nmvih = mv pRE \nIn order to solve for h , we need to find vp . Because we are neglecting all forces on the \nmeteor other than Earth\u2019s gravity, mechanical energy is constant, and \n1 \n2\n1 \n2 \u2212 GmM E\n= mv \n,\nmvi \np\n2\n2 \nRE \nwhere we have taken the meteor to have speed vi at a distance \u201cvery far away from Earth\u201d to \nmean that we neglect any gravitational potential energy in the meteor-Earth system, when \nr \u2192\u221e , U (r) = \u2212GmM E / r \u2192 0 . From the angular momentum condition, vp = vih / RE and \ntherefore the energy condition can be rewritten as \n2 \n2 \u239b h \u239e \n2 \n\u2212 2GM E\nvi = vi \u239d\u239c RE \u23a0\u239f \nRE \nwhich we solve for the impact parameter h \nh = RE \n2 + 2GM E RE \nvi \n2 \n. \n\uf072\nL \n\uf072 \nL\n\uf072 \n19.5 Angular Impulse and Change in Angular Momentum \n\uf072 \n\u2212 ti\nIf there is a total applied torque \u03c4S about a point S over an interval of time \u0394t = t f \n, then \nthe torque applies an angular impulse about a point S , given by \n\uf072\n\uf072 \n= \nt f \ndt . \n(19.4.1)\nJS \u222b\u03c4S\nti \n\uf072\n\uf072total /\nBecause \u03c4S = d LS \ndt , the angular impulse about S is equal to the change in angular \nmomentum about S , \nd\n\u03c4S\nS \n= \u0394 \n\uf072\nJ  \n\uf072\nL  \n\uf072\nL \nt f\u222bti \nt f\u222bti \ndt \ndt \n\u2212  \n(19.4.2) \n=  \n=  \n=  \nS ,i .\nS \nS \nS , f\ndt \n19-12", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_535", "text": "This result is the rotational analog to linear impulse, which is equal to the change in \nmomentum, \ndt = \u222bti \n= \u0394\uf072p \n\uf072p\nd\nt f\u222bti \nt f \n=  \n=  \n\uf072\nI \n\uf072\nF \ndt\n\uf072p f \u2212 \uf072pi . \n(19.4.3) \ndt \n19.6 Angular Momentum of a System of Particles \nWe now calculate the angular momentum about the point S associated with a system of N \njth \npoint particles. Label each individual particle by the index j , j = 1,2,\uf04c, N . Let the \n\uf072\nparticle have mass mj and velocity v j . The momentum of an individual particle is then \n\uf072\n\uf072 \n\uf072 \np j = mj v j . Let rS , j be the vector from the point S to the jth particle, and let \u03b8 j be the angle \n\uf072\n\uf072\nbetween the vectors r \nand p (Figure 19.10).\nS , j \nj \n\uf072r \nFigure 19.10 System of particles \n\uf072 \nThe angular momentum LS , j of the jth particle is \nS , j \n\uf072 \nL \n\uf072p\n(19.5.1) \n\u00d7 \n=  \nj .\nS , j \n\uf072r \n\uf072\nL \nThe angular momentum for the system of particles is the vector sum of the individual angular \nmomenta, \nsys \nS , j \nj=N \n\u2211 \nj=N \n\u2211\n\uf072\nL \n\uf072p\n(19.5.2) \n\u00d7 \n=  \n=  \n. \nS \nS , j \nj \n\uf072 \nL \nj=1 \nj=1 \nThe change in the angular momentum of the system of particles about a point S is given by \nsys \nd \nd\nd\nS\nS , j\nj \n\uf072r \ndt \n\uf072p \ndt \n\u239b  \n\u239e \nj= \n\u2211 \nN \nj=N\n\uf072\nL \nd\n\uf072rS , j \n/ dt , the first term in the parentheses \n\uf072p\n(19.5.3) \n\u00d7  +  \n\u00d7 \n=  \n= \u2211\u239c\u239d \n.\n\u239f\u23a0\nS , j \nj\ndt \ndt j=1 \nj=1 \n\uf072 \n= d\uf072 rS , j\nBecause the velocity of the jth particle is vS , j \nvanishes (the cross product of a vector with itself is zero because they are parallel to each \nother) \n19-13", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_536", "text": "d S , j \ndt \n\uf072r \n\u00d7 \uf072p j =  \uf072vS , j \u00d7 mj \n\uf072vS , j = 0 . \n(19.5.4)  \n\uf072\n\uf072\nSubstitute Eq. (19.5.4) and Fj = dp j / dt into Eq. (19.5.3) yielding \nd\nS \nd\nj \nS , j\nS , j \n\uf072r \n\uf072p\n\uf072r \n\uf072 \nL\n\u239b  \n\u239e \nj=N \n\u2211  \nj=N\n\u2211 \n\uf072\nF  \nsys \ndt \n(  \n) .  \n(19.5.5) \n\u00d7  \n\u00d7 \n=  \n= \n\u239c\u239d \n\u239f\u23a0 \nj\ndt\nj=1 \nj=1 \nBecause \nj= \nj=1 \u2211 \nN \nj= \nj=1 \u2211 \nN\n\uf072  \n\uf072\n\uf072\n\uf072\u03c4 \nWe have already shown in Chapter 17.4 that when we assume all internal forces are directed \n\u03c4\nalong the line connecting the two interacting objects then the internal torque about the point \nS is zero, \n\u03c4\n\u2211\n\uf072\n\uf072 \n\u03c4int \nS = 0 . \n(19.5.7) \nEq. (19.5.6) simplifies to \nj= \nj=\n\uf072rS j,\n\u2211 \nj 1\n= \n(\uf072  \n)  \nint\next +\nFj \n(19.5.6) \n\u00d7 \nrS , j \n=  \n=  \nS , j \nS \nS \nN \nN\n\uf072\nF \n\uf072\u03c4 \n\uf072\u03c4 \n( \n)  \next . \n(19.5.8)\n\u00d7 \n=  \n=  \nj \nS , j \nS \nj=1 \nTherefore Eq. (19.5.5) becomes \n\uf072 \n\uf072 \nd sys \next \nLS\n= \n. \n(19.5.9)\n\u03c4S \ndt \nThe external torque about the point S is equal to the time derivative of the angular \nmomentum of the system about that point. \nExample 19.5 Angular Momentum of Two Particles undergoing Circular Motion \nTwo identical particles of mass m move in a circle of radius R , with angular velocity\n\uf072\u03c9 = \u03c9 z k\u02c6 , \u03c9 z > 0, \u03c9 about the z -axis in a plane parallel to but a distance h above the x-y \nplane. The particles are located on opposite sides of the circle (Figure 19.11). Find the \nmagnitude and the direction of the angular momentum about the point S (the origin). \nk\u02c6\n= \nz\n+ z \np2 .\n2 \n.\n1 \nh \nR \nS \n+ x \nFigure 19.11 Example 19.5 \np1 \n19-14", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_537", "text": "19.3 \nSolution: The angular momentum about the origin is the sum of the contributions from each \nobject. The calculation of each contribution will be identical to the calculation in Example \n. \n\u02c6k \n+ z \n+ x\nS \n. \nrS ,1 \np1\np2 \n1\n2 \n\u02c6r1 \n\u02c6 \n1 \n\u02c6k\nLS ,1 \nR \nh \nFigure 19.12 Angular momentum of particle 1 about origin \n\u02c6k  \nk\u02c6 \n+ z  \n.\n2 \n\u02c6 \n1. p1\n\u02c6r2 \np2 \n2 \nh\nrS ,2 \nLS ,2 \nS \n+ x \nR \nFigure 19.13 Angular momentum of particle 2 about origin \nFor particle 1 (Figure 19.12), the angular momentum about the point S is \n\uf072 \n\uf072\n\uf072\nLS ,1 = rS ,1 \u00d7 p1 = (Rr\u02c61 + hk\u02c6 ) \u00d7 mR\u03c9 z \u03b8\u02c6\n1 = mR2\u03c9 z k\u02c6 \u2212 hmR\u03c9 z r\u02c61 . \nFor particle 2, (Figure 19.13), the angular momentum about the point S is \n\uf072 \n\uf072\n\uf072\nLS ,2 = rS ,2 \u00d7 p2 = (Rr\u02c62 + hk\u02c6 ) \u00d7 mR\u03c9 z \u03b8\u02c6\n2 = mR2\u03c9 z k\u02c6 \u2212 hmR\u03c9 z r\u02c62 . \nBecause the particles are located on opposite sides of the circle, r\u02c61 = \u2212r\u02c62 . The vector sum only \npoints along the z -axis and is equal to \n\uf072\n\uf072 \n\uf072 \nLS = LS ,1 + LS ,2 = 2mR2\u03c9 z k\u02c6 . \n(19.5.10) \nThe two angular momentum vectors are shown in Figure 19.14. \n19-15", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_538", "text": "\u02c6k \n+ z .\n. \n1\n2 \nLS ,2\nLS ,1 \nLS \nS \n+ x  \np1\np2 \nFigure 19.14 Angular momentum about the point S of both particles and their sum \nThe moment of inertia of the two particles about the z -axis is given by IS = 2mR2 . Therefore \n\uf072\n\uf072 \n= IS\u03c9 . The important point about this example is that the two objects are symmetrically \nLS \ndistributed with respect to the z -axis (opposite sides of the circular orbit). Therefore the \n\uf072 \n2\nangular momentum about any point S along the z -axis has the same value LS = 2mr \u03c9k\u02c6 , \n\uf072r \nwhich is constant in magnitude and points in the + z -direction. This result generalizes to any \nrigid body in which the mass is distributed symmetrically about the axis of rotation. \nExample 19.6 Angular Momentum of a System of Particles about Different Points \nConsider a system of N particles, and two points A and B (Figure 19.15). The angular \nmomentum of the jth particle about the point A is given by \nA,j \n\uf072 \nL \n\uf072v\n(19.5.11) \n\u00d7 mj \n=  \nj .\nA,j \nFigure 19.15 Vector triangle relating position of object and points A and B \nThe angular momentum of the system of particles about the point A is given by the sum \n\uf072\nL  \nN \uf072\nL  \nN \uf072rA,j \nj=1\nj=1 \nThe angular momentum about the point B can be calculated in a similar way and is given by \n\uf072v\n\u2211  \n\u2211  \n(19.5.12) \n\u00d7 mj \n=  \n=  \nA \nA,j \nj \n19-16", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_539", "text": "\uf072\nL  \nN \uf072\nL  \nN \n\u00d7 mj \n\uf072v j . \n(19.5.13) \n\u2211  \n\u2211 \n=  \n=  \nB \nB,j \nB,j \nA,j \nB,j \nA,B . \nWe can substitute Eq. (19.5.14) into Eq. (19.5.12) yielding \n\uf072r \n\uf072r \n\uf072r \n\uf072r \nj=1\nj=1 \nFrom Figure 19.15, the vectors \n(19.5.14) \n+ \n=  \nN \nN \nN\n\uf072\nL \n\uf072r\nB, j + A,B ) \u00d7 mj \nB, j \u00d7 mj \nA,B \u00d7 mj \n\uf072\nThe first term in Eq. (19.5.15) is the angular momentum about the point B . The vector rA,B is \n\uf072r \na constant and so can be pulled out of the sum in the second term, and Eq. (19 5 15) becomes\n. . \n\uf072rA,B \u00d7 \n\uf072r\n\uf072r\n\uf072v\n\uf072v\n\uf072v\n\u2211( \nj=1 \n\u2211  \n\u2211  \n(19.5.15) \n+ \n=  \n=  \n. \nA \nj \nj \nj\nj=1 \nj=1 \n\u2211 \nN\n\uf072\nL \n\uf072\nL \n\uf072v\n(19.5.16) \nB + \nmj \n= \nA \nj\nj=1 \nThe sum in the second term is the momentum of the system \n\uf072 \n= \nN \n\uf072 . \n(19.5.17)\npsys \u2211mjv j\nj=1 \nTherefore the angular momentum about the points A and B are related by \n\uf072rA,B \u00d7 \nThus if the momentum of the system is zero, the angular momentum is the same about any \npoint. \n\uf072\nL \n\uf072\nL \n\uf072p\n(19.5.18) \nB +\n= \nA \nsys \n\uf072\nL A =  \n\uf072\nL B , \n( \uf072psys =  \n\uf072\n0 ) . \n(19.5.19)  \nIn particular, the momentum of a system of particles is zero by definition in the center of mass \n\uf072\n\uf072 \nreference frame because in that reference frame psys = 0 . Hence the angular momentum is the \nsame about any point in the center of mass reference frame. \n19.7 Angular Momentum and Torque for Fixed Axis Rotation \nWe have shown that, for fixed axis rotation, the component of torque that causes the angular \nvelocity to change is the rotational analog of Newton\u2019s Second Law, \n19-17", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_540", "text": "\uf072\n\uf072\n\u03c4ext \nS \n\u03b1 . \n(19.5.20)\n= IS \nWe shall now see that this is a special case of the more general result \n\uf072 \nd \uf072 \next \nsys\n= \n. \n(19.5.21)\n\u03c4S \nLS\ndt \nConsider a rigid body rotating about a fixed axis passing through the point S and take the \nfixed axis of rotation to be the z -axis. Recall that all the points in the rigid body rotate about \n\uf072\nthe z -axis with the same angular velocity \u03c9 = (d\u03b8 / dt)k\u02c6 = \u03c9 z k\u02c6 . In a similar fashion, all \n\uf072\npoints in the rigid body have the same angular acceleration, \u03b1 = (d 2\u03b8 / dt2) k\u02c6 = \u03b1 z k\u02c6 . Let the \npoint S lie somewhere along the z -axis. \nAs before, the body is divided into individual elements. We calculate the contribution of each \nelement to the angular momentum about the point S , and then sum over all the elements. The \nsummation will become an integral for a continuous body. \nEach individual element has a mass \u0394mj and is moving in a circle of radius rS \n\u22a5 \n, j about the axis \n\uf072\nof rotation. Let rS , j be the vector from the point S to the element. The momentum of the \n\uf072\nelement, p j , is tangent to this circle (Figure 19.16). \n= \nz\u02c6k\nfixed axis \nof rotation \n+ z \n. .\nmj \nS \nrS , j \nrS , j\n. \nrS , j \n\u02c6 \n\u02c6k\n\u02c6r \nrigid body \np j \nFigure 19.16 Geometry of instantaneous rotation. \nThe angular momentum of the jth element about the point S is given by\n\uf072rS , j \n\uf072\nvector rS , j can be decomposed into parallel and perpendicular components with respect to the \n\uf072 \n\uf072\n\uf072 \n\uf072 \n\uf072\uf050\n\uf050 \n\u22a5\n\u22a5 \n\u22a5\naxis of rotation r \n= r + r \n(Figure 19.16), where \n= \nand \n\uf050 = \n. The\nS , j\nS , j\nS , j \nrS , j \nrS , j \nrS , j \nrS , j \n\uf072 \n\u22a5\nmomentum is given by p \n\u03c9 \u03b8\u02c6 . Then the angular momentum about the point S is\nj = \u0394mj rS , j\nz \n\uf072 \nL \n\uf072p j . The\n\u00d7 \n=  \nS , j \n19-18", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_541", "text": "\uf072 \n\uf072\n\uf072 \n\u22a5 \n\uf050 \n\u22a5\nk\n\u03c9 \u03b8\u02c6)\nLS , j = rS , j \u00d7 p j = (rS , jr\u02c6 + rS , j \u02c6 ) \u00d7(\u0394mj rS , j\nz \n. \n(19.5.22) \n= \u0394mj (rS \n\u22a5 \n, j )2\u03c9 z k\u02c6 \u2212\u0394mjrS \n\uf050 \n, j rS \n\u22a5 \n, j\u03c9 z r\u02c6 \nIn the last expression in Equation (19.5.22), the second term has a direction that is \nperpendicular to the z -axis. Therefore the z -component of the angular momentum about the \n\uf072\n\uf072\npoint S , (LS , j )z , arises entirely from the second term, rS \n\u22a5 \n, j \u00d7p j . Therefore the z -component \nof the angular momentum about S is \n\u22a5 )2\u03c9\n(LS , j )z = \u0394mj (rS , j\nz . \n(19.5.23) \nThe z -component of the angular momentum of the system about S is the summation over all \nthe elements, \nsys \n\u22a5\nLS , z = \u2211(LS , j )z =\u2211\u0394mj (rS , j )2\u03c9 z . \n(19.5.24) \nj\nj \nFor a continuous mass distribution the summation becomes an integral over the body, \nsys = \ndm (rdm )2\u03c9 , \n(19.5.25)\nLS , z \n\u222b \nz \nbody \nwhere rdm is the distance form the fixed z -axis to the infinitesimal element of mass dm . The \nmoment of inertia of a rigid body about a fixed z -axis passing through a point S is given by \nan integral over the body \n)2\n= \n. \n(19.5.26)\nIS \n\u222b dm (rdm \nbo dy \nThus the z -component of the angular momentum about S for a fixed axis that passes through \nS in the z -direction is proportional to the z -component of the angular velocity, \u03c9 z , \nLsys \nS , z \n\u03c9 . \n(19.5.27)\n= IS \nz \nFor fixed axis rotation, our result that torque about a point is equal to the time derivative of \nthe angular momentum about that point, \n\uf072ext \nd \uf072 sys \n= \n, \n(19.5.28)\n\u03c4S \nLS\ndt \ncan now be resolved in the z -direction, \ndLsys \nS , z\next \nd\nd\u03c9 z\nd 2\u03b8 \n=\n= \n\u03c9 ) = IS \n\u03b1 , \n(19.5.29)\n\u03c4 S ,z \n(IS \nz \n= IS \n= IS \nz\ndt\ndt \ndt \ndt2", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_542", "text": "\u03c4S \nLS\ndt \ncan now be resolved in the z -direction, \ndLsys \nS , z\next \nd\nd\u03c9 z\nd 2\u03b8 \n=\n= \n\u03c9 ) = IS \n\u03b1 , \n(19.5.29)\n\u03c4 S ,z \n(IS \nz \n= IS \n= IS \nz\ndt\ndt \ndt \ndt2 \n19-19", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_543", "text": "in agreement with our earlier result that the z -component of torque about the point S is equal \nto the product of moment of inertia about IS , and the z -component of the angular \nacceleration, \u03b1 z . \nExample 19.6 Circular Ring \nA circular ring of radius R , and mass M is rotating about the z -axis in a plane parallel to \nbut a distance h above the x-y plane. The z -component of the angular velocity is \u03c9 z (Figure \n\uf072 \n19.17). Find the magnitude and the direction of the angular momentum LS along at any point \nS on the central z -axis. \n\u02c6 \n. \nR\nM \n= \nzk \n+ z \n. \nLS\nh \nS \n+ x  \nFigure 19.17 Example 19.6 \nSolution: Use the same symmetry argument as we did in Example 19.5. The ring can be \nthought of as made up of pairs of point like objects on opposite sides of the ring each of mass \nm (Figure 19.18). \n+ z \n+ x\nS \nR \nh \n= \nz\u02c6k\nM \nLS \n. \n. \nFigure 19.18 Ring as a sum of pairs of symmetrically distributed particles \nEach pair has a non-zero z-component of the angular momentum taken about any point S \n\uf072 \n\uf072\n\uf072 \nLpair \nS \npair R2\u03c9\nalong the z -axis, \n= LS ,1 + LS ,2 = 2mR2\u03c9 z k\u02c6 = m\nz k\u02c6 . The angular momentum of the \nring about the point S is then the sum over all the pairs \n\uf072 \npair R2\u03c9\n= \u2211 m \nk\u02c6 = MR2\u03c9 k\u02c6 . \n(19.5.30)\nLS \nz \nz \npairs \n19-20", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_544", "text": "Recall that the moment of inertia of a ring is given by \n)2 = MR2\n= \ndm (rdm \n. \n(19.5.31)\nIS \n\u222b \nbody \nFor the symmetric ring, the angular momentum about S points in the direction of the angular \nvelocity and is equal to \n= IS\u03c9 k\u02c6 . \n(19.5.32)\nLS \nz \n19.8 Principle of Conservation of Angular Momentum \nConsider a system of particles. We begin with the result that we derived in Section 19.7 that \nthe torque about a point S is equal to the time derivative of the angular momentum about that \npoint S , \n\uf072 \n\uf072 \nd sys \n\uf072 \nL \next \nLS\n= \n. \n(19.5.33)\n\u03c4S \ndt \nWith this assumption, the torque due to the external forces is equal to the rate of change of the \nangular momentum \n\uf072 sys \n\uf072 \nL\n\uf072ext \ndLS\n= \n. \n(19.5.34)\n\u03c4S \ndt \nPrinciple of Conservation of Angular Momentum \n\uf072\n\uf072 \nL \nIf the external torque acting on a system is zero, then the angular momentum of \nthe system is constant. So for any change of state of the system the change in \nangular momentum is zero \n\u0394 \n\uf072\n0 \n\u2212 (  \n(19.5.35) \n= \nsys \u2261 (\nS \nsys ) f\nS \nsys )i\nS \n.  \nEquivalently the angular momentum is constant \n\uf072\n\uf072 \nsys ) f \nsys )i . \n(19.5.36)\n(LS \n= (LS \nSo far no isolated system has been encountered such that the angular momentum is not \nconstant so our assumption that internal torques cancel is pairs can be taken as an \nexperimental observation. \nExample 19.7 Collision Between Pivoted Rod and Object \nA point-like object of mass m1 moving with constant speed vi strikes a rigid uniform rod of \nlength l and mass m2 that is hanging by a frictionless pivot from the ceiling. Immediately \n19-21", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_545", "text": "after striking the rod, the object continues forward but its speed decreases to vi / 2 (Figure \n19.19). The moment of inertia of the rod about its center of mass is Icm = (1/ 12)m2l2 . Gravity \nacts with acceleration g downward. (a) For what value of vi will the rod just touch the \nceiling on its first swing? (b) For what ratio m2 / m1 will the collision be elastic? \nfrictionless pivot \nl\nm2\ng\nm2 \nm1g \nm1 \nvi \nvi / 2 \nFigure 19.19 Example 19.7 \nSolution: We begin by identifying our system, which consists of the object and the uniform \nrod. We identify three states; an initial state i : immediately before the collision, state a : \nimmediately after the collision, and state f : the instant the rod touches the ceiling when the \nfinal angular speed is zero. We would like to know if any of our fundamental quantities: \nmomentum, energy, and angular momentum, are constant during these state changes, state i \nto state a , state a to state f . \n. \nA \n. \nm1g \nm2g \nFpivot ,2 \nF1,2\nF2,1 \n.S \nA \ncm \nFigure 19.20 Free-body force diagrams on particle and rod \nWe start with the transition from state i to state a . The pivot force holding the rod to the \nceiling is an external force acting at the pivot point S . There is also the gravitational force \nacting at the center of mass of the rod and on the object. There are also internal forces due to \nthe collision of the rod and the object at point A (Figure 19.20). \n19-22", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_546", "text": "The external force means that momentum is not constant. The point of action of the external \npivot force is fixed and so does no work. However, we do not know whether or not the \ncollision is elastic and so we cannot assume that mechanical energy is constant. Choose the \npivot point S as the point about which to calculate torque, then the torque diagrams are \nshown in Figure 19.21. \nA \n. \nm1g \nm2g \nFpivot ,2 \nF1,2\nF2,1 \n.S\n.S \nrS ,cm \ncm \nrS ,A \nA \nrS ,A \n\uf072g \n\uf072r \n\uf072r \n\uf072 \nF\n\uf072r \nFigure 19.21 Torque diagrams on particle and rod with torque calculated about pivot point S \n\uf072 \nF \nThe torque on the system about the pivot S is then the sum of terms \n\uf072r\n\u03c4sys \nS \nS ,S \npivot,2 \nS ,A \n1,2 \nS ,A \n2,1 \nS ,cm \nS , A \u00d7 m1 \n\uf072 \nF\n\uf072r \n\uf072\n\uf072g\n.(19.5.37) \n\u00d7  \n+  \n\u00d7  \n+  \n\u00d7  \n+  \n\u00d7 m2 + \n=  \n\uf072\n\uf072 \nThe external pivot force does not contribute any torque because rS ,S = 0 . The internal forces \n\uf072\n\uf072 \nbetween the rod and the object are equal in magnitude and opposite in direction, F1,2 = \u2212F2,1 \n(Newton\u2019s Third Law), and so their contributions to the torque add to zero. If the collision is \n\uf072\n\uf072\ninstantaneous then the gravitational force is parallel to \nand \nso the two gravitational \nrS ,cm \nrS ,A \n\uf072\n\uf072 \nsys \ntorques are zero. Therefore the torque on the system about the pivot point is zero, \u03c4S = 0 . \nThus the angular momentum about the pivot point is constant, \n\uf072\n\uf072 \nsys \nsys \nLS ,i = LS ,a . \n(19.5.38) \n.\nl \nS\n\u02c6j \n.\n.\n.S \nk\u02c6\n= \na \na\n\u02c6i \nk\u02c6 \ncm \nrS ,A \nstate i \nstate a\nrS ,A \nvi / 2\nvi\nA \nA  \nFigure 19.22 Angular momentum diagram \n19-23", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_547", "text": "In order to calculate the angular momentum we draw a diagram showing the momentum of \nthe object and the angular speed of the rod in (Figure 19.22). The angular momentum about S \nimmediately before the collision is \n\uf072\n\uf072\nLsys \nS , i = rS ,1 \u00d7 m1v\uf072 \ni = l(\u2212\u02c6j) \u00d7 m1vi \u02c6i = lm1vik\u02c6 . \nThe angular momentum about S immediately after the collision is \n\uf072\n\uf072\n\uf072 \n\uf072\nLsys \nS , a = rS ,1 \u00d7 m1vi / 2 + IS \u03c9 a = l(\u2212\u02c6j) \u00d7 m1(vi / 2)\u02c6i + IS\u03c9 a k\u02c6 = (lm1vi / 2) k\u02c6 + IS\u03c9 a k\u02c6 . \nTherefore the condition that the angular momentum about S is constant during the collision \nbecomes \nlm1vik\u02c6 = (lm1vi / 2 + IS\u03c9 a )k\u02c6 . \nWe can solve for the angular speed immediately after the collision \nlm1vi\n\u03c9 = \n. \na \n2IS \nBy the parallel axis theorem the moment of inertial of a uniform rod about the pivot point is \nl2 \nl2\n(l / 2)2 + I = (1/ 4)m2l2 + (1/12)m2\n= (1/ 3)m2\n. \n(19.5.39)\nIS = m2 \ncm \nTherefore the angular speed immediately after the collision is \n3m1vi\n= \n. \n(19.5.40)\n\u03c9 2 \n2m2l \n.\n.S\nS \na = \na\u02c6k\n. \n. \nstate a \nl / 2 % \nU g = 0\ncm \nstate f \nA\nA \nFigure 19.23 Energy diagram for transition from state a to state f . \nFor the transition from state a to state f , we know that the gravitational force is conservative \nand the pivot force does no work so mechanical energy is constant. \n19-24", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_548", "text": "mech \nmech \nE a \n= E f \nWe draw an energy diagram only for the rod because the kinetic energy for the particle is not \nchanging between states a and f , (Figure 19.23), with a choice of zero for the potential \nenergy at the center of mass. The mechanical energy of the rod and particle immediately after \nthe collision is \nmech \n1 \u03c9 2 + 1\nE \n= \n/ 2)2 . \na \nIS\na \nm1(vi\n2\n2 \nUsing our results for the moment of inertia IS (Eq. (19.5.39)) and \u03c9 2 (Eq. (19.5.40)), we \nhave that \n2 \n2\n2 \nmech \n1 \nl2 \u239b 3m1vi \u239e \n1\n3m1 vi \n1\nE \n= (1/ 3)m2\n+ \n/ 2)2 =\n+ m1(vi / 2)2 . (19.5.41)\na\nm1(vi\n2 \n\u239d\u239c 2m2l \u23a0\u239f \n2\n8m2\n2 \nThe mechanical energy when the rod just reaches the ceiling when the final angular speed is \nzero is then \nmech \n1 \n= m2 g(l / 2) + m1\n/ 2)2 .\nE f \n(vi\n2 \nThen the condition that the mechanical energy is constant becomes \n3m1\n2vi \n2\n1 \n1\n+ \n/ 2)2 \ng(l / 2) + \n/ 2)2 . \n(19.5.42)\nm1(vi \n= m2 \nm1(vi\n8m2\n2 \n2 \nWe can now solve Eq. (19.5.42) for the initial speed of the object \nm2\n4gl . \n(19.5.43)\nvi = m1\n3 \nWe now return to the transition from state i to state a . and determine the constraint on the \nmass ratio in order for the collision to be elastic. The mechanical energy before the collision is \nmech \n1\n2\n= \n. \n(19.5.44)\nEi \nm1vi\n2 \nIf we impose the condition that the collision is elastic then \nmech \nmech \n= E \n. \n(19.5.45)\nEi \na \nSubstituting Eqs. (19.5.41) and (19.5.44) into Eq. (19.5.45) yields \n19-25", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_549", "text": "1  \n2\n3m1\n2vi \n2\n1 \nm1\n=\n+ m1\n/ 2)2 .\nvi  \n(vi\n2\n8m2\n2 \nThis simplifies to \n3 \n2\n3m1\n2vi \n2 \n=\nm1vi\n8 \n8m2 \nHence we can solve for the mass ratio necessary to ensure that the collision is elastic if the \nfinal speed of the object is half it\u2019s initial speed \nL\nm2 = 1.  \n(19.5.46)\nm1 \nL \nNotice that this mass ratio is independent of the initial speed of the object. \nL \n19.9 External Angular Impulse and Change in Angular Momentum \nDefine the external angular impulse about a point S applied as the integral of \nthe external torque about S \nt f\n\uf072\n\uf072\next \u2261 \next dt .  \n(19.5.47)\nJS  \n\u222b\u03c4S \nti \nThen the external angular impulse about S is equal to the change in angular momentum \n\uf072 sys \nsys \nsys \nt f \nt f\n\uf072 \n\uf072\n\uf072 \nd\n\uf072 \next \u2261\nS \n\u222b \next dt =\n\u03c4S \n\u222b \nS dt\nJ \n\u2212  \n(19.5.48) \n=  \n. \nS , f \nS ,i\ndt\nti \nti \nNotice that this is the rotational analog to our statement about impulse and momentum, \n\uf072p\nd sys \ndt dt =  \uf072psys, f \u2212 \uf072psys,i . \n(19.5.49)  \nt f \next \u2261\nS \n\u222b \n\uf072\nF \nt f \next dt = \u222b \n\uf072\nI  \nti \nti \nExample 19.8 Angular Impulse on Steel Washer \nA steel washer is mounted on the shaft of a small motor. The moment of inertia of the motor \nand washer is I0 . The washer is set into motion. When it reaches an initial angular speed \u03c90 , \nat t = 0 , the power to the motor is shut off, and the washer slows down until it reaches an \nangular speed of \u03c9 a at time ta . At that instant, a second steel washer with a moment of inertia \nw is dropped on top of the first washer. Assume that the second washer is only in contact \n19-26 \nI", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_550", "text": "with the first washer. The collision takes place over a time \n\u2212 t . Assume the \n\u0394tint = tb \na \nfrictional torque on the axle is independent of speed, and remains the same when the second \nwasher is dropped. The two washers continue to slow down during the time interval \nuntil they stop at time t = t f . (a) What is the angular acceleration while the \n\u0394t2 = t f \u2212 tb \nwasher and motor are slowing down during the interval \u0394t1 = ta ? (b) Suppose the collision is \nnearly instantaneous, \u0394tint \n\u2212 t ) \uf03b 0 . What is the angular speed \u03c9b of the two washers \n= (tb \na \nimmediately after the collision is finished (when the washers rotate together)? \nwasher \nmotor \n= \nz\u02c6k\nFigure 19.24 Example 19.8 \nNow suppose the collision is not instantaneous but that the frictional torque is independent of \nthe speed of the rotor. (c) What is the angular impulse during the collision? (d) What is the \nangular velocity \u03c9b of the two washers immediately after the collision is finished (when the \nwashers rotate together)? (e) What is the angular deceleration \u03b12 after the collision? \nSolution: a) The angular acceleration of the motor and washer from the instant when the \npower is shut off until the second washer was dropped is given by \n\u03c9 \u2212\u03c90\n= \na \n< 0. \n(19.5.50)\n\u03b11 \n\u0394t1 \n(b) If the collision is nearly instantaneous, then there is no angular impulse and therefore the \nz -component of the angular momentum about the rotation axis of the motor remains constant \n0 = \u0394L \n\u2212 I0\u03c9 . \n(19.5.51)\nz = Lf ,z \u2212 L0,z = (I0 + I w )\u03c9b \na \nWe can solve Eq. (19.5.51) for the angular speed \u03c9b of the two washers immediately after the \ncollision is finished \n19-27", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_551", "text": "I0\n= \n\u03c9 . \n(19.5.52)\n\u03c9b\na\n+ I\nI0 \nw \n(c) The angular acceleration found in part a) is due to the frictional torque in the motor. \nmotor \nf \nwasher \noverhead view \nFigure 19.25 Frictional torque in the motor \nLet \uf072\u03c4 f = \u2212\u03c4 f \u02c6k where \u03c4 f is the magnitude of the frictional torque (Figure 19.25) then \n\u2212\u03c4 f = I0\u03b11 = I0 (\u03c9 a \u2212 \u03c90 ) \n\u0394t1 \n. \n(19.5.53) \nDuring the collision with the second washer, the frictional torque exerts an angular impulse \n(pointing along the z -axis in the figure), \n\u0394tint \nJz = \u2212\u222b\ntb \u03c4 f dt = \u2212\u03c4 f \u0394tint = I0(\u03c9 a \u2212\u03c90) \n. \n(19.5.54)\nta \n\u0394t1 \n(d) The z -component of the angular momentum about the rotation axis of the motor changes \nduring the collision, \n\u0394L \n+ I \n\u03c9 . \n(19.5.55)\nz = Lf ,z \u2212 L0,z = (I0 \nw )\u03c9 b \u2212 I0 \na \nThe change in the z -component of the angular momentum is equal to the z -component of the \nangular impulse \nJz = \u0394Lz . \n(19.5.56) \nThus, equating the expressions in Equations (19.5.54) and (19.5.55), yields \n\u239b\n\u239e\n\u0394tint \n(\u03c9 \u2212\u03c90) \n\u2212 (I0)\u03c9 . \n(19.5.57)\nI0 \na \n= (I0 + I w )\u03c9b \na\n\u239d\u239c \u0394t1 \u23a0\u239f \n19-28", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_552", "text": "Solve Equation (19.5.57) for the angular velocity immediately after the collision, \n\u239b \n\u239b\n\u239e\n\u239e \n\u03c9b = \nI0 \n\u239c (\u03c9 a \u2212\u03c90) \u0394tint \n\u23a0\u239f + \u03c9 a \u239f . \n(19.5.58)\n(I0 + I w ) \u239d \n\u239d\u239c\u0394t1 \n\u23a0 \nIf there were no frictional torque, then the first term in the brackets would vanish, and the \nsecond term of Eq. (19.5.58) would be the only contribution to the final angular speed. \n(e) The final angular acceleration \u03b12 is given by \n0 \u2212\u03c9b \n\u239b \n\u239b\n\u239e\n\u239e\nI0 \n\u0394tint \n\u03b12 =\n= \u2212\n\u239c (\u03c9 a \u2212\u03c90) \n\u23a0\u239f + \u03c9 a \u239f . \n(19.5.59)\n\u0394t2\n(I0 + I w )\u0394t2 \u239d \n\u239d\u239c\u0394t1 \n\u23a0 \n19-29", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_553", "text": "Chapter 20 Rigid Body: Translation and Rotational Motion Kinematics \nfor Fixed Axis Rotation \n20.1 Introduction........................................................................................................... 1  \n20.2 Constrained Motion: Translation and Rotation ................................................ 1  \n20.2.1 Rolling without slipping ................................................................................ 5  \nExample 20.1 Bicycle Wheel Rolling Without Slipping ........................................ 6  \nExample 20.2 Cylinder Rolling Without Slipping Down an Inclined Plane........ 9  \nExample 20.3 Falling Yo-Yo .................................................................................. 10", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_554", "text": "Example 20.4 Unwinding Drum ............................................................................ 11  \n20.3 Angular Momentum for a System of Particles Undergoing Translational and  \nRotational .................................................................................................................... 12  \nExample 20.5 Earth\u2019s Motion Around the Sun.................................................... 15  \n20.4 Kinetic Energy of a System of Particles............................................................ 17  \n20.5 Rotational Kinetic Energy for a Rigid Body Undergoing Fixed Axis Rotation  \n...................................................................................................................................... 18", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_555", "text": "Appendix 20A Chasles\u2019s Theorem: Rotation and Translation of a Rigid Body ... 19", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_556", "text": "Chapter 20 Rigid Body: Translation and Rotational Motion  \nKinematics for Fixed Axis Rotation  \nHence I feel no shame in asserting that this whole region engirdled by the \nmoon, and the center of the earth, traverse this grand circle amid the rest \nof the planets in an annual revolution around the sun. Near the sun is the \ncenter of the universe. Moreover, since the sun remains stationary, \nwhatever appears as a motion of the sun is really due rather to the motion \nof the earth.1 \nCopernicus \n20.1 Introduction \nThe general motion of a rigid body of mass m consists of a translation of the center of \n\uf072 \nmass with velocity V cm and a rotation about the center of mass with all elements of the \n\uf072\nrigid body rotating with the same angular velocity \u03c9 cm . We prove this result in Appendix \nA. Figure 20.1 shows the center of mass of a thrown rigid rod follows a parabolic \ntrajectory while the rod rotates about the center of mass. \nFigure 20.1 The center of mass of a thrown rigid rod follows a parabolic trajectory while \nthe rod rotates about the center of mass. \n20.2 Constrained Motion: Translation and Rotation \nWe shall encounter many examples of a rolling object whose motion is constrained. For \nexample we will study the motion of an object rolling along a level or inclined surface \nand the motion of a yo-yo unwinding and winding along a string. We will examine the \nconstraint conditions between the translational quantities that describe the motion of the \ncenter of mass, displacement, velocity and acceleration, and the rotational quantities that \ndescribe the motion about the center of mass, angular displacement, angular velocity and \nangular acceleration. We begin with a discussion about the rotation and translation of a \nrolling wheel. \n1Nicolaus Copernicus, De revolutionibus orbium coelestium (On the Revolutions of the Celestial Spheres), \nBook 1 Chapter 10. \n20-1", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_557", "text": "Figure 20.2 Rolling Wheel \nConsider a wheel of radius R is rolling in a straight line (Figure 20.2). The center of \n\uf072 \nmass of the wheel is moving in a straight line at a constant velocity V cm . Let\u2019s analyze \nthe motion of a point P on the rim of the wheel. \n\uf072\nLet v P denote the velocity of a point P on the rim of the wheel with respect to reference \n\uf072\nframe O at rest with respect to the ground (Figure 20.3a). Let v\u2032 P denote the velocity of \nthe point P on the rim with respect to the center of mass reference frame Ocm moving \n\uf072 \nwith velocity V cm with respect to at O (Figure 20.3b). (You should review the definition \nof the center of mass reference frame in Chapter 15.2.1.) We can use the law of addition \nof velocities (Eq.15.2.4) to relate these three velocities, \n\uf072\n\uf072\n\uf072 \n= v\u2032 + V . \n(20.2.1)\nv P \nP \ncm \nLet\u2019s choose Cartesian coordinates for the translation motion and polar coordinates for \nthe motion about the center of mass as shown in Figure 20.3. \n(a) \n(b) \nFigure 20.3 (a) reference frame fixed to ground, (b) center of mass reference frame \nThe center of mass velocity in the reference frame fixed to the ground is given by \n\uf072\nVcm = Vcm \u02c6i . \n(20.2.2) \n20-2", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_558", "text": "where Vcm is the speed of the center of mass. The position of the center of mass in the \nreference frame fixed to the ground is given by \n\uf072\nRcm (t) = (Xcm, 0 + Vcmt)\u02c6i , \n(20.2.3) \nwhere Xcm, 0 is the initial x -component of the center of mass at t = 0 . The angular \nvelocity of the wheel in the center of mass reference frame is given by \n\uf072\u03c9 \n= \u03c9 k\u02c6 . \n(20.2.4)\ncm \ncm \nwhere \u03c9 cm is the angular speed. The point P on the rim is undergoing uniform circular \nmotion with the velocity in the center of mass reference frame given by \n\uf072 \n\u02c6\nv\u2032 P = R\u03c9 cm\u03b8 . \n(20.2.5) \n\uf072\nIf we want to use the law of addition of velocities then we should express v\u2032 P = R\u03c9 cm\u03b8\u02c6 in \nCartesian coordinates. Assume that at t = 0 , \u03b8(t = 0) = 0 i.e. the point P is at the top of \nthe wheel at t = 0 . Then the unit vectors in polar coordinates satisfy (Figure 20.4) \nr\u02c6 = sin\u03b8\u02c6i \u2212 cos\u03b8\u02c6j . \n(20.2.6)\n\u03b8\u02c6 = cos\u03b8\u02c6i + sin\u03b8 \u02c6j \nTherefore the velocity of the point P on the rim in the center of mass reference frame is \ngiven by \nv\uf072\u2032 P = R\u03c9 cm\u03b8\u02c6 = R\u03c9 cm (cos\u03b8\u02c6i \u2212 sin\u03b8\u02c6j) . \n(20.2.7) \nFigure 20.4 Unit vectors \nNow substitute Eqs. (20.2.2) and (20.2.7) into Eq. (20.2.1) for the velocity of a point P \non the rim in the reference frame fixed to the ground \n\uf072 v P = R\u03c9 cm (cos\u03b8\u02c6i + sin\u03b8\u02c6j) +V cm \u02c6i \n. \n(20.2.8) \n= (V + R\u03c9 cos\u03b8 )\u02c6i + R\u03c9 sin\u03b8\u02c6j\ncm \ncm \ncm \n20-3", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_559", "text": "The point P is in contact with the ground when \u03b8 = \u03c0 . At that instant the velocity of a \npoint P on the rim in the reference frame fixed to the ground is \n\uf072 v (\u03b8 = \u03c0 ) = (V \u2212 R\u03c9 )\u02c6i . \n(20.2.9)\nP \ncm \ncm \nWhat velocity does the observer at rest on the ground measure for the point on the rim \nwhen that point is in contact with the ground? In order to understand the relationship \nbetween Vcm and \u03c9 cm , we consider the displacement of the center of mass for a small \ntime interval \u0394t (Figure 20.5). \nFigure 20.5 Displacement of center of mass in ground reference frame. \nFrom Eq. (20.2.3) the x -component of the displacement of the center of mass is \n\u0394Xcm =Vcm \u0394t . \n(20.2.10) \nThe point P on the rim in the center of mass reference frame is undergoing circular \nmotion (Figure 20.6). \nFigure 20.6: Small displacement of point on rim in center of mass reference frame. \n20-4", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_560", "text": "In the center of mass reference frame, the magnitude of the tangential displacement is \ngiven by the arc length subtended by the angular displacement \u0394\u03b8 = \u03c9cm \u0394t , \n\u0394s = R\u0394\u03b8 = R\u03c9cm \u0394t . \n(20.2.11) \nCase 1: if the x -component of the displacement of the center of mass is equal to the arc \nlength subtended by \u0394\u03b8 , then the wheel is rolling without slipping or skidding, rolling \nwithout slipping for short, along the surface with \n\u0394Xcm = \u0394s . \n(20.2.12) \nSubstitute Eq. (20.2.10) and Eq. (20.2.11) into Eq. (20.2.12) and divide through by \u0394t . \nThen the rolling without slipping condition becomes \nV = R\u03c9 , \n(rolling without slipping). \n(20.2.13)\ncm \ncm \nCase 2: if the x -component of the displacement of the center of mass is greater than the \narc length subtended by \u0394\u03b8 , then the wheel is skidding along the surface with \n\u0394Xcm > \u0394s . \n(20.2.14) \nSubstitute Eqs. (20.2.10) and (20.2.11) into Eq. (20.2.14) and divide through by \u0394t , then \nV > R\u03c9 , \n(skidding) . \n(20.2.15)\ncm \ncm \nCase 3: if the x -component of the displacement of the center of mass is less than the arc \nlength subtended by \u0394\u03b8 , then the wheel is slipping along the surface with \n\u0394Xcm < \u0394s . \n(20.2.16) \nArguing as above the slipping condition becomes \nV < R\u03c9 , \n(slipping) . \n(20.2.17)\ncm \ncm \n20.2.1 Rolling without slipping \nWhen a wheel is rolling without slipping, the velocity of a point P on the rim is zero \nwhen it is in contact with the ground. In Eq. (20.2.9) set \u03b8 = \u03c0 , \n\uf072\n\uf072 \nv P (\u03b8 = \u03c0 ) = (V cm \u2212 R\u03c9 cm )\u02c6i = (R\u03c9 cm \u2212 R\u03c9 cm )\u02c6i = 0 . \n(20.2.18) \nThis makes sense because the velocity of the point P on the rim in the center of mass \nreference frame when it is in contact with the ground points in the opposite direction as", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_561", "text": "(20.2.18) \nThis makes sense because the velocity of the point P on the rim in the center of mass \nreference frame when it is in contact with the ground points in the opposite direction as \nthe translational motion of the center of mass of the wheel. The two velocities have the \n20-5", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_562", "text": "same magnitude so the vector sum is zero. The observer at rest on the ground sees the \ncontact point on the rim at rest relative to the ground. \nThus any frictional force acting between the tire and the ground on the wheel is static \nfriction because the two surfaces are instantaneously at rest with respect to each other. \nRecall that the direction of the static frictional force depends on the other forces acting on \nthe wheel. \nExample 20.1 Bicycle Wheel Rolling Without Slipping \nConsider a bicycle wheel of radius R that is rolling in a straight line without slipping. \nThe velocity of the center of mass in a reference frame fixed to the ground is given by \n\uf072 \nvelocity Vcm . A bead is fixed to a spoke a distance b from the center of the wheel \n(Figure 20.7). (a) Find the position, velocity, and acceleration of the bead as a function of \ntime in the center of mass reference frame. (b) Find the position, velocity, and \nacceleration of the bead as a function of time as seen in a reference frame fixed to the \nground. \nFigure 20.8 Coordinate system for bead\nFigure 20.7 Example 20.1 \nin center of mass reference frame \nSolution: a) Choose the center of mass reference frame with an origin at the center of the \nwheel, and moving with the wheel. Choose polar coordinates (Figure 20.8). The z \u00ad\ncomponent of the angular velocity \u03c9 cm = d\u03b8 / dt > 0 . Then the bead is moving uniformly \nin a circle of radius r = b with the position, velocity, and acceleration given by \n\uf072\n\uf072 \n\uf072 \n2\nrb \u2032 = b r\u02c6, \nv\u2032 \nb = b\u03c9 cm \u03b8\u02c6, a\u2032 \nb = \u2212b\u03c9 cm r\u02c6 . \n(20.2.19) \nBecause the wheel is rolling without slipping, the velocity of a point on the rim of the \nwheel has speed v\u2032 \nP = R\u03c9 cm . This is equal to the speed of the center of mass of the wheel \nVcm , thus \nV = R\u03c9 . \n(20.2.20)\ncm \ncm \nNote that at t = 0 , the angle \u03b8 = \u03b80 = 0 . So the angle grows in time as \n20-6", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_563", "text": "\u03b8(t) = \u03c9 t = (V / R)t . \n(20.2.21)\ncm \ncm \nThe velocity and acceleration of the bead with respect to the center of the wheel are then \n\uf072 \nbV \n\uf072 \nbV 2 \ncm \ncm \nv\u2032 \nb = \n\u03b8\u02c6, a\u2032 \nb = \u2212 R2 r\u02c6 . \n(20.2.22)\nR \nb) Define a second reference frame fixed to the ground with choice of origin, Cartesian \ncoordinates and unit vectors as shown in Figure 20.9. \nFigure 20.9 Coordinates of bead in reference frame fixed to ground \nThen the position vector of the center of mass in the reference frame fixed to the ground \nis given by \n\uf072 \nR (t) = X \n\u02c6i + R \u02c6j = V t \u02c6i + R \u02c6j. \n(20.2.23)\ncm \ncm \ncm \nThe relative velocity of the two frames is the derivative \n\uf072 \nd \n\uf072 \ndXcm \nRcm \nVcm =\n= \n\u02c6i = Vcm \u02c6i . \n(20.2.24)\ndt \ndt \nBecause the center of the wheel is moving at a uniform speed the relative acceleration of \nthe two frames is zero, \n\uf072\nA \nd \n\uf072 \nV  \ncm \n=\n= \n\uf072\n0 .  \n(20.2.25) \ncm \ndt \nDefine the position, velocity, and acceleration in this frame (with respect to the ground) \nby \n\uf072\n\uf072 \n\uf072 \nrb (t) = xb (t) \u02c6i + yb (t) \u02c6j, \nvb (t) = vb,x (t) \u02c6i + vb,y (t) \u02c6j, a(t) = ab,x (t) \u02c6i + ab,y (t) \u02c6j . (20.2.26) \nThen the position vectors are related by \n20-7", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_564", "text": "\uf072\n\uf072 \n\uf072\n(t) = R (t) + rb \u2032(t) . \n(20.2.27)\nrb \ncm \nIn order to add these vectors we need to decompose the position vector in the center of \nmass reference frame into Cartesian components, \n\uf072 rb \u2032(t) = b r\u02c6(t) = bsin\u03b8(t) \u02c6i + bcos\u03b8(t) \u02c6j . \n(20.2.28) \nThen using the relation \u03b8(t) = (V cm / R)t , Eq. (20.2.28) becomes \n\uf072\n\uf072 \n\uf072 \nrb (t) = R cm (t) + rb \u2032(t) = (V cm t \u02c6i + R \u02c6j) + (bsin\u03b8(t) \u02c6i + bcos\u03b8(t) \u02c6j) \n. \n(20.2.29) \n= (V t + bsin((V / R)t)) \u02c6i +( R + bcos((V / R)t)) \u02c6j\ncm \ncm \ncm \nThus the position components of the bead with respect to the reference frame fixed to the \nground are given by \n(t) = V t + bsin((V / R)t) \n(20.2.30)\nxb \ncm \ncm \n(t) = R + bcos((V / R)t) . \n(20.2.31)\nyb \ncm \nA plot of the y -component vs. the x -component of the position of the bead in the \nreference frame fixed to the ground is shown in Figure 20.10 below using the values \nV cm = 5 m \u22c5 s-1 , R = 0.25 m , and b = 0.125 m . We can differentiate the position vector in \nthe reference frame fixed to the ground to find the velocity of the bead \n\uf072  \nvb (t) = d\uf072rb \ndt (t) = d (V t + bsin((V / R)t)) \u02c6i + d (R + bcos((V / R)t) )\u02c6j ,\ncm \ncm \ncm \n(20.2.32) \ndt \ndt \n\uf072 v (t) = (V + (b / R)V cos((V / R)t)) \u02c6i \u2212 ((b / R)V sin((V / R)t) )\u02c6j. \n(20.2.33)\nb \ncm \ncm \ncm \ncm \nFigure 20.10 Plot of the y -component vs. the x -component of the position of the bead \n20-8", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_565", "text": "Alternatively, we can decompose the velocity of the bead in the center of mass reference \nframe into Cartesian coordinates \n\uf072v\u2032 (t) = (b / R)V (cos((V / R)t) \u02c6i \u2212 sin((V / R)t) \u02c6j) . \n(20.2.34)\nb \ncm \ncm \ncm \nThe law of addition of velocities is then \n\uf072 \n\uf072\n\uf072\n(t) = V + v\u2032 (t) , \n(20.2.35)\nvb \ncm \nb \n\uf072 v (t) = V \u02c6i + (b / R)V (cos((V / R)t) \u02c6i \u2212 sin((V / R)t) \u02c6j) , \n(20.2.36)\nb \ncm \ncm \ncm \ncm \n\uf072 v (t) = (V \n+ (b / R)V cos((V / R)t)) \u02c6i \u2212 (b / R)sin((V / R)t) \u02c6j, (20.2.37)\nb \ncm \ncm \ncm \ncm \nin agreement with our previous result. The acceleration is the same in either frame so \n\uf072\n\uf072 \n2\n(t) = a\u2032 = \u2212(b / R2)V r\u02c6 = \u2212(b / R2)V 2 (sin((V / R)t) \u02c6i + cos((V / R)t) \u02c6j). \n(20.2.38)\nab \nb \ncm \ncm \ncm \ncm \nWhen the bead is at the rim of the wheel, b = R , then the position of the bead in the \nreference frame fixed to the ground is given by \n\uf072 rb (t) = (V cm t + Rsin((V cm / R)t)) \u02c6i + R(1+ cos((V cm / R)t))) \u02c6j . \n(20.2.39) \nThe path traced out by the bead in the reference frame fixed to the ground is called a \ncycloid. \nExample 20.2 Cylinder Rolling Without Slipping Down an Inclined Plane \nA uniform cylinder of outer radius R and mass M with moment of inertia about the \ncenter of mass I cm = (1/ 2) M R2 starts from rest and rolls without slipping down an \nincline tilted at an angle \u03b2 from the horizontal. The center of mass of the cylinder has \ndropped a vertical distance h when it reaches the bottom of the incline. Let g denote the \ngravitational constant. What is the relation between the component of the acceleration of", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_566", "text": "incline tilted at an angle \u03b2 from the horizontal. The center of mass of the cylinder has \ndropped a vertical distance h when it reaches the bottom of the incline. Let g denote the \ngravitational constant. What is the relation between the component of the acceleration of \nthe center of mass in the direction down the inclined plane and the component of the \nangular acceleration into the page of Figure 20.11? \n20-9", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_567", "text": "Figure 20.11 Example 20.2 \nSolution: We begin by choosing a coordinate system for the translational and rotational \nmotion as shown in Figure 20.12. \nFigure 20.12 Coordinate system for rolling cylinder \nFor a time interval \u0394t , the displacement of the center of mass is given by \n\uf072 \n\u0394R (t) = \u0394X \n\u02c6i . The arc length due to the angular displacement of a point on the rim \ncm \ncm \nduring the time interval \u0394t is given by \u0394s = R\u0394\u03b8 . The rolling without slipping condition \nis \n\u0394X \n= R\u0394\u03b8 . \ncm \nIf we divide both sides by \u0394t and take the limit as \u0394t \u2192 0 then the rolling without \nslipping condition show that the x -component of the center of mass velocity is equal to \nthe magnitude of the tangential component of the velocity of a point on the rim \n\u0394X cm \n\u0394\u03b8\nV = lim \n= lim R \n= R\u03c9 . \ncm \ncm \n\u0394t\u21920 \u0394t \n\u0394t\u21920 \n\u0394t \nSimilarly if we differentiate both sides of the above equation, we find a relation between \nthe x -component of the center of mass acceleration is equal to the magnitude of the \ntangential component of the acceleration of a point on the rim \ndV \nd\u03c9 \ncm \ncm \nA =\n= R \n= R\u03b1 . \ncm \ncm \ndt \ndt \nExample 20.3 Falling Yo-Yo \nA Yo-Yo of mass m has an axle of radius b and a spool of radius R (Figure 20.13a). Its \nmoment of inertia about the center of mass can be taken to be I = (1/ 2)mR2 (the \nthickness of the string can be neglected). The Yo-Yo is released from rest. What is the \nrelation between the angular acceleration about the center of mass and the linear \nacceleration of the center of mass? \n20-10", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_568", "text": "Solution: Choose coordinates as shown in Figure 20.13b. \nFigure 20.13a Example 20.3 \nFigure 20.13b Coordinate system for \nYo-Yo \nConsider a point on the rim of the axle at a distance r = b from the center of mass. As the \nyo-yo falls, the arc length \u0394s = b\u0394\u03b8 subtended by the rotation of this point is equal to \nlength of string that has unraveled, an amount \u0394l . In a time interval \u0394t , b\u0394\u03b8 = \u0394l . \nTherefore b\u0394\u03b8 / \u0394t = \u0394l / \u0394t . Taking limits, noting that, V \n= dl / dt , we have that \ncm, y \nb\u03c9 \n= V \n. Differentiating a second time yields b\u03b1 \n= A \n.\ncm \ncm, y \ncm \ncm, y \nExample 20.4 Unwinding Drum \nDrum A of mass m and radius R is suspended from a drum B also of mass m and \nradius R , which is free to rotate about its axis. The suspension is in the form of a \nmassless metal tape wound around the outside of each drum, and free to unwind (Figure \n20.14). Gravity acts with acceleration g downwards. Both drums are initially at rest. \nFind the initial acceleration of drum A , assuming that it moves straight down. \nFigure 20.14 Example 20.4 \n20-11", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_569", "text": "Solution: The key to solving this problem is to determine the relation between the three \nkinematic quantities \u03b1 A , \u03b1 B , and aA , the angular accelerations of the two drums and the \nlinear acceleration of drum A . Choose the positive y -axis pointing downward with the \norigin at the center of drum B . After a time interval \u0394t , the center of drum A has \nundergone a displacement \u0394y . An amount of tape \u0394lA \nhas unraveled from drum \n= R\u0394\u03b8 A \nA , and an amount of tape \u0394lB \nhas unraveled from drum B . Therefore the \n= R\u0394\u03b8B \ndisplacement of the center of drum A is equal to the total amount of tape that has \nunwound from the two drums, \u0394y = \u0394lA \n. Dividing through by \u0394t\n+ \u0394lB = R\u0394\u03b8 A + R\u0394\u03b8B \nand taking the limit as \u0394t \u2192 0 yields \ndy \nd\u03b8 A\nd\u03b8B\n= R \n+ R \n.\ndt\ndt \ndt \nDifferentiating a second time yields the desired relation between the angular \naccelerations of the two drums and the linear acceleration of drum A , \nd 2 y\nd 2\u03b8 A\nd 2\u03b8B\n= R \n+ R \ndt2 \ndt2 \ndt2 \naA, y = R\u03b1 A + R\u03b1 B . \n20.3 Angular Momentum for a System of Particles Undergoing \nTranslational and Rotational \nWe shall now show that the angular momentum of a body about a point S can be \ndecomposed into two vector parts, the angular momentum of the center of mass (treated \nas a point particle) about the point S , and the angular momentum of the rotational \nmotion about the center of mass. \nConsider a system of N particles located at the points labeled i = 1,2,\uf04c, N . The angular \nmomentum about the point S is the sum \n\uf072r \n\uf072\nLtotal \nS , i \ni=1 \ni=1 \n\uf072\nwhere rS i, is the vector from the point S to the ith particle (Figure 20.15) satisfying \n\uf072\n\uf072 \n\uf072 \nr = r \n+ r \n(20.3.2)\nS , i \nS , cm \ncm,i , \n\uf072 \n\uf072\n\uf072 \nv \n= V + v \n(20.3.3)\nS ,i \ncm \ncm,i , \n\uf072\n\uf072 \nwhere \n= V . We can now substitute both Eqs. (20.3.2) and (20.3.3) into Eq. \nvS , cm \ncm \n(20.3.1) yielding \nN \uf072\nL  \nN\n\uf072v\n\u2211  \n\u2211 \n= (  \n) , \n(20.3.1) \n\u00d7 mi \n=  \nS \nS ,i \ni \n20-12", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_570", "text": "\uf072 \nN \n\uf072 \nLtotal \n\uf072\n\uf072 \n\uf072 \n= \u2211(r\n+ r \n) \u00d7 m (V + v \n) . \n(20.3.4)\nS \nS , cm \ncm,i \ni \ncm \ncm,i  \ni=1  \nFigure 20.15 Vector Triangle \n\uf072r \nWhen we expand the expression in Equation (20.3.4), we have four terms, \nS ,cm \nS ,cm \n\uf072r\ntotal = \n\uf072\nLS \n\uf072\nV  \nN \nN\n\uf072v\n\u2211  \n\u2211 \n(  \n)  \n(  \n) \n\u00d7 mi \n+  \n\u00d7 mi\ncm,i \ncm \ni=1 \ni=1 \n(20.3.5) \nN \nN\n\uf072\nV \n( \uf072  \nr cm,i \u00d7 mi \n\uf072v cm,i )  + \n\uf072 \n\u2211  \n\u2211 (  \n). \n+  \n\u00d7 mi\nr cm,i \ncm \nth \ni\nmass and not on the location of the \nparticle. Therefore in the first term in the above \n\uf072\nL\n\uf072r \ni=1 \ni=1 \n\uf072r \n\uf072\nThe vector \nis a constant vector that depends only on the location of the center of \nrS , cm \n\uf072 \nequation, \ncan be taken outside the summation. Similarly, in the second term the \nrS , cm \n\uf072 \nvelocity of the center of mass V cm is the same for each term in the summation, and may \nbe taken outside the summation, \ntotal \nS ,cm \nS ,cm \n\uf072\nV  \nN \nN\n\u239b  \n\u239e  \n\u239b  \n\u239e \n\uf072v\n\u2211  \n\u2211 \n\u00d7  \n\u23a0\u239f +  \n\u00d7 \nmi \nmi \n=  \n\u239d\u239c  \n\u239d\u239c  \n\u23a0\u239f \nS \ncm,i \ncm \ni=1 \ni=1 \n(20.3.6) \nN \nN\n\uf072\nV \n\u239e \n\u239b \n\uf072 \n\uf072 \n\uf072v\n\u2211(  \n\u2211 \ncm,i ) \n+  \n\u00d7 mi \n+  \n\u23a0\u239f \u00d7 \nr cm,i \nmir cm,i\n\u239d\u239c  \n.  \ncm \ni=1 \ni=1 \nThe first and third terms in Eq. (20.3.6) are both zero due to the fact that \nN \n\uf072\n\u2211mir cm,i = 0 \ni=1 \n(20.3.7)\nN \n\uf072\n\u2211miv cm,i = 0. \ni=1 \nN \n\uf072\nWe first show that \u2211m r \nis zero. We begin by using Eq. (20.3.2),\ni cm,i  \ni=1  \n20-13", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_571", "text": "N \nN \ncm,i  \ni=1 \ni=1  \n\uf072r\n\u2211  \n\u2211 \n(mi \n)  \n(\n(mi \n\u2212  \n))  \n=  \ni \nS ,cm \ni\nS ,cm \ni\nS ,cm \nSubstitute the definition of the center of mass (Eq. 10.5.3) into Eq. (20.3.8) yielding \n\uf072r \n\uf072r \n\uf072r \n\uf072r \n\uf072r\n\uf072r\n(20.3.8) \nN \nN \nN\n\u239b  \n\u239e \n\u2211  \n\u2211(mi \n\u2211  \ntotal \n\u2212  \n)  \n\u2212 m\nmi \nmi \n=  \n=  \n\u239d\u239c  \n\u23a0\u239f  \n. \ni=1 \ni=1 \ni=1 \nN \n\u2211 (mi \n\uf072  \nr cm,i )  =  \nN \n\u2211  \n1 \nN\n\uf072  \nm r\ni i =  \n\uf072\n0 .  \n(20.3.9) \n\uf072  \nm r\ni i \u2212 mtotal \n\u2211 \ntotal \nm\ni=1 \ni=1 \ni=1 \nN \n\uf072\nThe vanishing of \u2211m v \n= 0 follows directly from the definition of the center of mass \ni\ncm,i  \ni=1  \nframe, that the momentum in the center of mass is zero. Equivalently the derivative of \nEq. (20.3.9) is zero. We could also simply calculate and find that \n\uf072 \n\uf072v \n\uf072 \nV \n\u2211mi \n= \u2211 ( \n\u2212  \n) \nv \nmi\ncm,i \ni \ncm \ni\ni\n\uf072\nV \n\uf072v\n\u2211  \n\u2211 \n\u2212 \nmi \nmi \n=  \ni \n(20.3.10) \ncm \ni \ni\n\uf072\nV  \n\uf072\nV \ntotal \ntotal \n\u2212 \n= m \nm \ncm \ncm \n=  \n\uf072\n0 .  \nr \nL \nWe can now simplify Eq. (20.3.6) for the angular momentum about the point S using the \nN \n\uf072 \nV \n\uf072 \nfact that, mT = \u2211mi , and psys = mT \ncm (in reference frame O ): \ni=1 \n\uf072total \n\uf072 \nN \uf072 \nS , cm \n\uf072  \n\uf072 \n\u2211(  \n) . \n(20.3.11) \nsys +\n\u00d7  \n\u00d7 mi \n= \np\nr cm,i\nv\nS \ncm,i \ni=1 \n\uf072\n\uf072 \n\uf072\nConsider the first term in Equation (20.3.11), \n\u00d7 p ; the vector \nis the vector\nrS ,cm \nsys \nrS ,cm \nfrom the point S to the center of mass. If we treat the system as a point-like particle of \nmass mT located at the center of mass, then the momentum of this point-like particle is \n\uf072\n\uf072 \np\nV . Thus the first term is the angular momentum about the point S of this \nsys = mT \ncm \n\u201cpoint-like particle\u201d, which is called the orbital angular momentum about S , \nL\n\uf072orbital \nS \n\uf072\n\uf072\n\u00d7 p\n. \n(20.3.12)\n= rS ,cm \nsys \nfor the system of particles. \n20-14", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_572", "text": "N \nConsider the second term in Equation (20.3.11), \u2211(r \uf072 \n\u00d7 m v \uf072 \n) ; the quantity inside \ncm,i \ni\ncm,i \ni=1 \nthe summation is the angular momentum of the ith particle with respect to the origin in \nthe center of mass reference frame Ocm (recall the origin in the center of mass reference \nframe is the center of mass of the system), \n\uf072\n\uf072 \n\uf072\nL \n= r \nv \n(20.3.13)\ncm,i \ncm,i \u00d7 mi \ncm,i . \nHence the total angular momentum of the system with respect to the center of mass in the \ncenter of mass reference frame is given by \n\uf072 \nN \uf072 \nN \uf072  \nr cm,i \u00d7 mi \n\uf072v cm,i ) . \n(20.3.14) \n\u2211  \n\u2211( \nspin \nL\nL \n=  \n=  \ncm,i\ncm \ni=1 \ni=1 \na vector quantity we call the spin angular momentum. Thus we see that the total angular \nmomentum about the point S is the sum of these two terms, \n\uf072\n\uf072 \n\uf072 \nLtotal \norbital + Lspin \n= L\n. \n(20.3.15)\nS \nS \ncm \nThis decomposition of angular momentum into a piece associated with the translational \nmotion of the center of mass and a second piece associated with the rotational motion \nabout the center of mass in the center of mass reference frame is the key conceptual \nfoundation for what follows. \nExample 20.5 Earth\u2019s Motion Around the Sun \nThe earth, of mass me = 5.97\u00d71024 kg and (mean) radius Re = 6.38\u00d7106 m , moves in a \nnearly circular orbit of radius r s,e = 1.50 \u00d71011 m around the sun with a period \nTorbit = 365.25 days , and spins about its axis in a period Tspin = 23 hr 56 min , the axis \ninclined to the normal to the plane of its orbit around the sun by 23.5\u00b0 (in Figure 20.16, \nthe relative size of the earth and sun, and the radius and shape of the orbit are not \nrepresentative of the actual quantities). \nFigure 20.16 Example 20.5 \n20-15", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_573", "text": "If we approximate the earth as a uniform sphere, then the moment of inertia of the earth \nabout its center of mass is \nIcm = 2 m R e\n2 . \n(20.3.16)\n5\ne \nIf we choose the point S to be at the center of the sun, and assume the orbit is circular, \nthen the orbital angular momentum is \n\uf072 \nLorbital \nS \n\uf072\n\uf072 \n\u02c6\n\u02c6\n\u00d7 p\n= r r\u02c6 \u00d7 m v \u03b8 = r m v k . \n(20.3.17)\n= rS ,cm \nsys \ns,e \ne\ncm \ns,e \ne\ncm \nThe velocity of the center of mass of the earth about the sun is related to the orbital \nangular velocity by \nvcm = rs,e \u03c9orbit , \n(20.3.18) \nwhere the orbital angular speed is \n2\u03c0 \n2\u03c0 \n=\n=\n\u03c9orbit \nTorbit \n(365.25 d)(8.640 \u00d7 104 s \u22c5 d\u22121) \n(20.3.19) \n= 1.991\u00d7 10\u22127 rad \u22c5 s\u22121. \nThe orbital angular momentum about S is then \n\uf072orbital = m r \nk\nLS \ne \n2 \u03c9orbit \u02c6 \ns,e \n= (5.97 \u00d7 1024 kg)(1.50 \u00d7 1011 m)2(1.991\u00d7 10\u22127 rad \u22c5 s\u22121) k\u02c6 \n(20.3.20) \n2 \u22c5 s\n= (2.68 \u00d7 1040 kg \u22c5 m\n\u22121) k\u02c6. \nThe spin angular momentum is given by \nL\n\uf072spin \n\uf072 \n2 \n= I \u03c9\n= m R \nn\u02c6 , \n(20.3.21)\ncm \ncm \nspin \n5 \ne\ne\n2 \u03c9spin \nwhere n\u02c6 is a unit normal pointing along the axis of rotation of the earth and \n2\u03c0 \n2\u03c0\n\u22125 \n\u22121\n\u03c9spin =\n= \n\u00d7\n\u22c5 \n. \n(20.3.22)\n= 7.293 10\nrad s\nT \n8.616\u00d7104 s\nspin \nThe spin angular momentum is then \n\uf072spin \nL cm = 2 (5.97 \u00d7 1024 kg)(6.38 \u00d7 106 m)2(7.293 \u00d7 10\u22125 rad \u22c5 s\u22121) n\u02c6 \n5 \n(20.3.23) \n2 \u22c5 s\n= (7.10 \u00d7 1033 kg \u22c5 m\n\u22121) n\u02c6. \n20-16", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_574", "text": "The ratio of the magnitudes of the orbital angular momentum about S to the spin angular \nmomentum is greater than a million, \norbital \n2 \u03c9\nm r\nr 2 T\nL \ne s,e \norbit \n5 s,e \nspin \nS \n= \n=\n= 3.77\u00d7106 , \n(20.3.24)\nspin \n2\n2\nL \n(2 / 5)m R \u03c9 \n2 R T\ncm \ne\ne \nspin \ne \norbit \nas this ratio is proportional to the square of the ratio of the distance to the sun to the \nradius of the earth. The angular momentum about S is then \n\uf072total \n2\n\u02c6\n2 \n2\nL\n= m r \u03c9 \nk + m R \u03c9 \nn\u02c6 . \n(20.3.25)\nS \ne s,e \norbit \ne \ne \nspin \n5 \nThe orbit and spin periods are known to far more precision than the average values used \nfor the earth\u2019s orbit radius and mean radius. Two different values have been used for one \n\u201cday;\u201d in converting the orbit period from days to seconds, the value for the solar day, \nT \n= 86,400s was used. In converting the earth\u2019s spin angular frequency, the sidereal\nsolar \nday, Tsidereal = Tspin = 86,160s was used. The two periods, the solar day from noon to noon \nand the sidereal day from the difference between the times that a fixed star is at the same \nplace in the sky, do differ in the third significant figure. \n20.4 Kinetic Energy of a System of Particles \nith \n\uf072\nConsider a system of particles. The \nparticle has mass mi and velocity vi with respect \nto a reference frame O . The kinetic energy of the system of particles is given by \nK = \u2211 \n1 \n2 \n1 \u2211mi \n=\nmi vi\n2\n2 \n\uf072vi \u22c5\uf072vi \ni \ni \n(20.4.1)  \n+  \n\uf072\nV \n+  \n\uf072\nV \n1 \u2211mi (\n2 \n\uf072 \n)\u22c5( \uf072v\n). \nv\n=  \ncm,i \ncm,i\ncm \ncm \ni \nwhere Equation 15.2.6 has been used to express \uf072 vi in terms of \uf072 v cm,i \n\uf072 \nand Vcm . Expanding \nthe last dot product in Equation (20.4.1), \n!\n!\n!\n1 \u2211mi\n2 \n! \n! \n! \nV \u22c5 V  \nV \nK =  \n( \n\u22c5  \n+ 2 \n\u22c5  \n) \n+ \nv \nv \nv \ncm,i \ncm,i \ncm,i\ncm \ncm \ncm \ni \n! \n! \n! \n1  \n2  \n1  \n2  \n! \n! \n! \n\u2211  \n\u2211mi (  \n) + \u2211 \nV\nV  \nV \n(20.4.2) \n(  \n\u22c5  \n) +  \n\u22c5  \n\u22c5 \nmi v \nv \nmiv\n=  \ncm,i \ncm,i \ncm,i\ncm \ncm \ncm \ni \ni\ni \n1\n1 \n\u239b", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_575", "text": "cm,i\ncm \ncm \ncm \ni \n! \n! \n! \n1  \n2  \n1  \n2  \n! \n! \n! \n\u2211  \n\u2211mi (  \n) + \u2211 \nV\nV  \nV \n(20.4.2) \n(  \n\u22c5  \n) +  \n\u22c5  \n\u22c5 \nmi v \nv \nmiv\n=  \ncm,i \ncm,i \ncm,i\ncm \ncm \ncm \ni \ni\ni \n1\n1 \n\u239b\n\u2211 2\n2 \n2 \ncm,i \n! \n\u239e ! \n\u2211  \n\u239d\u239c\u2211m \n\u23a0\u239f \u22c5 \ni \ni \nV  \nV \n+  \n+ \n=  \nmiv \nmi \nv \n.  \ncm,i\n2  \ncm \ncm \ni \n20-17", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_576", "text": "The last term in the third equation in (20.4.2) vanishes as we showed in Eq. (20.3.7). \nThen Equation (20.4.2) reduces to \n1 \n2\n1 \n2\nK = \u2211 miv cm,i + \u2211mi V cm \ni 2\n2 i \n(20.4.3)\n1 \n2 \n1 \ntotal V 2\n= \u2211 miv \n+ m \n. \ncm,i \ncm \ni 2\n2 \nWe interpret the first term as the sum of the individual kinetic energies of the particles of \nthe system in the center of mass reference frame O cm and the second term as the kinetic \nenergy of the center of mass motion in reference frame O . \nAt this point, it\u2019s important to note that no assumption was made regarding the mass \nelements being constituents of a rigid body. Equation (20.4.3) is valid for a rigid body, a \ngas, a firecracker (but K is certainly not the same before and after detonation), and the \nsixteen pool balls after the break, or any collection of objects for which the center of \nmass can be determined. \n20.5 Rotational Kinetic Energy for a Rigid Body Undergoing Fixed Axis \nRotation \n\uf072\nThe rotational kinetic energy for the rigid body, using v \n= (r \n) \u03c9\n\u03b8\u02c6 , simplifies to\ncm, i \ncm, i \u22a5 \ncm \nK = 1 I \u03c9 2 . \n(20.5.1)\nrot \ncm \ncm \n2 \nTherefore the total kinetic energy of a translating and rotating rigid body is \n1\n2\n1 \n2\n= K \n+ K = mV + I \u03c9 . \n(20.5.2)\nKtotal \ntrans \nrot \ncm \ncm \ncm \n2\n2 \n20-18", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_577", "text": "Appendix 20A Chasles\u2019s Theorem: Rotation and Translation of a Rigid \nBody \nWe now return to our description of the translating and rotating rod that we first \nconsidered when we began our discussion of rigid bodies. We shall now show that the \nmotion of any rigid body consists of a translation of the center of mass and rotation about \nthe center of mass. \nWe shall demonstrate this for a rigid body by dividing up the rigid body into point-like \nconstituents. Consider two point-like constituents with masses m1 and m2 . Choose a \n\uf072\ncoordinate system with a choice of origin such that body 1 has position r1 and body 2 \n\uf072\nhas position r2 (Figure 20A.1). The relative position vector is given by \n1,2 \n1 \n\uf072r \n\uf072r \n\uf072r .\n2\n\u2212  \n(20.A.1) \n=  \nFigure 20A.1 Two-body coordinate system. \n\uf072 \nRecall we defined the center of mass vector, R cm , of the two-body system as \n\uf072r \n\uf072r\nm1\n1 + m2 2 \ncm \nm1 + m2 \nIn Figure 20A.2 we show the center of mass coordinate system. \n\uf072\nR \n(20.A.2) \n=  \n.  \nFigure 20A.2 Position coordinates with respect to center of mass \n20-19", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_578", "text": "The position vector of the object 1 with respect to the center of mass is given by \nm1 1 + m2\n2 \nm2 \nm1 + m2 \nm1 + m2 \n\uf072r \n\uf072r\n\uf072r1\ncm \n\uf072\nR \ncm,1 \n1 \n1\n2 \n1,2 , \nwhere \n\uf072r \nm m \n\u00b5 = \n1 \n2 , \n(20.A.4) \nr \nm + m\n1\n2 \n\uf072r \n\uf072r \nis the reduced mass. In addition, the relative position vector between the two objects is \n\uf072r \nindependent of the choice of reference frame, \n\uf072r\n12 \n1 \n2 \ncm,1 \ncm,2 \ncm,1 \ncm,2 \ncm,1,2 . \n\uf072r \n\uf072r \n\uf072r \n\uf072r \nr \n\uf072r \n\uf072r \n\u00b5 \nm1 \n\u2212  \n\u2212  \n(  \u2212  ) =  \n(20.A.3) \n=  \n=  \n=  \n\uf072 \n\uf072 \nR  \n\uf072 \nR \n\uf072 \n\u2212  = (  \n) \u2212 (  \n) =  \n\u2212  \n(20.A.5) \n+  \n+ \n=  \n=  \ncm \ncm \nBecause the center of mass is at the origin in the center of mass reference frame, \n\uf072r\nm\n+ m\n1 cm,1 \n2 cm,2 \n\uf072r\n\uf072\n0 \n(20.A.6) \n=  .  \nm1 + m2 \nTherefore \n\uf072\n\uf072 \n(20.A.7)\nm1rcm,1 = \u2212m2rcm,2 \n\uf072 \n\uf072 \n. \n(20.A.8)\n= m2\nm1 rcm,1 \nrcm,2 \nThe displacement of object 1 about the center of mass is given by taking the derivative of \nEq. (20.A.3), \nd\uf072 rcm,1 = \u00b5 \nm1 \nd\uf072 r1,2 . \n(20.A.9) \n\uf072r \n\uf072r \n\uf072r \nA similar calculation for the position of object 2 with respect to the center of mass yields \nfor the position and displacement with respect to the center of mass \n\u00b5 \ncm,2 \n1,2 , \n\uf072\nR \n\u2212  \n= \u2212  \n(20.A.10) \n=  2 \ncm \nm2 \n\uf072 \n\u00b5 \n\uf072\nd \n= \u2212 \nd \n(20.A.11) \nrcm,2 \nr1,2 . \nm2 \nLet i = 1,2 . An arbitrary displacement of the ith object is given respectively by \n\uf072\n\uf072 \n\uf072 \ndr = dr \n+ dR , \n(20.A.12)\ni \ncm,i \ncm \n20-20", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_579", "text": "\uf072\nwhich is the sum of a displacement about the center of mass dr cm,i and a displacement of \n\uf072 \nthe center of mass dR cm . The displacement of objects 1 and 2 are constrained by the \ncondition that the distance between the objects must remain constant since the body is \nrigid. In particular, the distance between objects 1 and 2 is given by \n1,2 \n1\n2 \n1\n2 \n\uf072r \n\uf072r \n\uf072r \n\uf072r \n\uf072r \n2 = (  \u2212  ) \u22c5 (  \u2212  ).  \n(20.A.13)  \nBecause this distance is constant we can differentiate Eq. (20.A.13), yielding the rigid \nbody condition that \n1\n2 \n1\n2 \n1,2 \n\uf072r \n20A.1. Translation of the Center of Mass \n\uf072r\n\uf072r \nThe condition (Eq. (20.A.14)) can be satisfied if the relative displacement vector between \n\uf072r\n\uf072r \n1,2 \n1\n2 \n\uf072r \n\uf072r \n\uf072r \n\uf072r1,2 \n\u2212  ) \u22c5 (d \u2212 d \n\u22c5 d\n0 = 2(  \n) = 2  \n(20.A.14)  \nthe two objects is zero,\n\uf072\n0 \nd \n= d \u2212 d \n(20.A.15) \n=  .  \n\uf072r \n\uf072r \nThis implies, using, Eq. (20.A.9) and Eq. (20.A.11), that the displacement with respect to \nthe center of mass is zero, \ncm,1 \ncm,2 \n\uf072\n0 \nd \n= d \n(20.A.16) \n=  .  \nThus by Eq. (20.A.12), the displacement of each object is equal to the displacement of \nthe center of mass, \n\uf072 \nd\uf072 = dR , \n(20.A.17)\nri \ncm \nwhich means that the body is undergoing pure translation. \n20A.2 Rotation about the Center of Mass \n\uf072r \n\uf072r \n\uf072r1,2 \n1\n2 \nterms of the center of mass coordinates. Using Eq. (20.A.9), the rigid body condition (Eq. \n(20.A.14)) becomes \n0 = 2 \u00b5 \nm1 \n\uf072 r1,2 \u22c5 d\uf072 rcm,1 . \n(20.A.18) \nBecause the relative position vector between the two objects is independent of the choice \nof reference frame (Eq. (20.A.5)), the rigid body condition Eq. (20.A.14) in the center of \nmass reference frame is then given by \n\uf072\n\uf072\n0 = 2r \n\u22c5 dr \n(20.A.19)\ncm,1,2 \ncm,1 . \n\uf072\n0 \nNow suppose that d \n= d \u2212 d \u2260  . The rigid body condition can be expressed in", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_580", "text": "mass reference frame is then given by \n\uf072\n\uf072\n0 = 2r \n\u22c5 dr \n(20.A.19)\ncm,1,2 \ncm,1 . \n\uf072\n0 \nNow suppose that d \n= d \u2212 d \u2260  . The rigid body condition can be expressed in \n20-21", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_581", "text": "This condition is satisfied if the relative displacement is perpendicular to the line passing \nthrough the center of mass, \n\uf072\n\uf072\n\u22a5 dr \n(20.A.20)\nrcm,1,2 \ncm,1 . \n\uf072\n\uf072\nBy a similar argument, rcm,1,2 \u22a5 drcm,2 . In order for these displacements to correspond to a \nrotation about the center of mass, the displacements must have the same angular \ndisplacement. \nFigure 20A.3 Infinitesimal angular displacements in the center of mass reference frame \nIn Figure 20A.3, the infinitesimal angular displacement of each object is given by \ndr\uf072 \ncm,1 \nd\u03b81 = \uf072 rcm,1 \ndr\uf072 \ncm,2 \nd\u03b82 = \uf072 rcm,2 \n, \n(20.A.21)  \n. \n(20.A.22)  \nFrom Eq. (20.A.9) and Eq. (20.A.11), we can rewrite Eqs. (20.A.21) and (20.A.22) as \nd\u03b81 = \u00b5 \nm1 \nd\uf072 r1,2 \n\uf072 rcm,1 \n, \nd\u03b82 = \u00b5 \nm2 \nd\uf072 r1,2 \n\uf072 rcm,2 \n. \nRecall that in the center of mass reference frame m1 \n\uf072 rcm,1 = m2 \n\uf072 rcm,2 \nand hence the angular displacements are equal, \n(20.A.23) \n(20.A.24) \n(Eq. (20.A.8)) \n20-22\n\nd\u03b81 = d\u03b82 = d\u03b8 . \n(20.A.25) \n\uf072\nTherefore the displacement of the ith object dri differs from the displacement of the \n\uf072 \ncenter of mass dR cm by a vector that corresponds to an infinitesimal rotation in the center \nof mass reference frame \n\uf072\n\uf072 \n\uf072\ndr = dR + dr \n(20.A.26)\ni \ncm \ncm,i . \nWe have shown that the displacement of a rigid body is the vector sum of the \ndisplacement of the center of mass (translation of the center of mass) and an infinitesimal \nrotation about the center of mass. \n20-23", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_582", "text": "Chapter 23 Simple Harmonic Motion \n23.1 Introduction: Periodic Motion............................................................................. 1  \n23.1.1 Simple Harmonic Motion: Quantitative ...................................................... 1  \n23.2 Simple Harmonic Motion: Analytic .................................................................... 3  \n23.2.1 General Solution of Simple Harmonic Oscillator Equation ...................... 6  \nExample 23.1: Phase and Amplitude ...................................................................... 7  \nExample 23.2: Block-Spring System ..................................................................... 10  \n23.3 Energy and the Simple Harmonic Oscillator ................................................... 11", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_583", "text": "23.3 Energy and the Simple Harmonic Oscillator ................................................... 11  \n23.3.1 Simple Pendulum: Force Approach ........................................................... 13  \n23.3.2 Simple Pendulum: Energy Approach ........................................................ 16  \n23.4 Worked Examples............................................................................................... 18  \nExample 23.3: Rolling Without Slipping Oscillating Cylinder........................... 18  \nExample 23.4: U-Tube ............................................................................................ 19", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_584", "text": "23.5 Damped Oscillatory Motion............................................................................... 21  \n23.5.1 Energy in the Underdamped Oscillator..................................................... 24  \n23.6 Forced Damped Oscillator ................................................................................. 26  \n23.6.1 Resonance ..................................................................................................... 27  \n23.6.2 Mechanical Energy ...................................................................................... 30  \nExample 23.5: Time-Averaged Mechanical Energy ............................................ 30", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_585", "text": "Example 23.5: Time-Averaged Mechanical Energy ............................................ 30  \n23.6.3 The Time-averaged Power .......................................................................... 34  \n23.6.4 Quality Factor .............................................................................................. 35  \n23.7 Small Oscillations................................................................................................ 36  \nExample 23.6: Quartic Potential ........................................................................... 39  \nExample 23.7: Lennard-Jones 6-12 Potential....................................................... 41", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_586", "text": "Appendix 23A: Solution to Simple Harmonic Oscillator Equation ....................... 42  \nAppendix 23B: Complex Numbers............................................................................ 45  \nAppendix 23C: Solution to the Underdamped Simple Harmonic Oscillator ........ 48  \nAppendix 23D: Solution to the Forced Damped Oscillator Equation.................... 50", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_587", "text": "Chapter 23 Simple Harmonic Motion \n\u2026Indeed it is not in the nature of a simple pendulum to provide equal and \nreliable measurements of time, since the wide lateral excursions often \nmade may be observed to be slower than more narrow ones; however, we \nhave been led in a different direction by geometry, from which we have \nfound a means of suspending the pendulum, with which we were \npreviously unacquainted, and by giving close attention to a line with a \ncertain curvature, the time of the swing can be chosen equal to some \ncalculated value and is seen clearly in practice to be in wonderful \nagreement with that ratio. As we have checked the lapses of time \nmeasured by these clocks after making repeated land and sea trials, the \neffects of motion are seen to have been avoided, so sure and reliable are \nthe measurements; now it can be seen that both astronomical studies and \nthe art of navigation will be greatly helped by them\u2026 1 \nChristian Huygens \n23.1 Introduction: Periodic Motion \nThere are two basic ways to measure time: by duration or periodic motion. Early clocks \nmeasured duration by calibrating the burning of incense or wax, or the flow of water or \nsand from a container. Our calendar consists of years determined by the motion of the \nsun; months determined by the motion of the moon; days by the rotation of the earth; \nhours by the motion of cyclic motion of gear trains; and seconds by the oscillations of \nsprings or pendulums. In modern times a second is defined by a specific number of \nvibrations of radiation, corresponding to the transition between the two hyperfine levels \nof the ground state of the cesium 133 atom. \nSundials calibrate the motion of the sun through the sky, including seasonal \ncorrections. A clock escapement is a device that can transform continuous movement into \ndiscrete movements of a gear train. The early escapements used oscillatory motion to stop \nand start the turning of a weight-driven rotating drum. Soon, complicated escapements \nwere regulated by pendulums, the theory of which was first developed by the physicist \nChristian Huygens in the mid 17th century. The accuracy of clocks was increased and the \nsize reduced by the discovery of the oscillatory properties of springs by Robert Hooke.  \nBy the middle of the 18th century, the technology of timekeeping advanced to the point \nthat William Harrison developed timekeeping devices that were accurate to one second in \na century. \n23.1.1 Simple Harmonic Motion: Quantitative", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_588", "text": "size reduced by the discovery of the oscillatory properties of springs by Robert Hooke.  \nBy the middle of the 18th century, the technology of timekeeping advanced to the point \nthat William Harrison developed timekeeping devices that were accurate to one second in \na century. \n23.1.1 Simple Harmonic Motion: Quantitative \n1 Christian Huygens, The Pendulum Clock or The Motion of Pendulums Adapted to Clocks By Geometrical \nDemonstrations, tr. Ian Bruce, p. 1. \n23-1", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_589", "text": "One of the most important examples of periodic motion is simple harmonic \nmotion (SHM), in which some physical quantity varies sinusoidally. Suppose a function \nof time has the form of a sine wave function, \ny(t) = Asin(2\u03c0t / T ) \n(23.1.1) \nwhere A > 0 is the amplitude (maximum value). The function y(t) varies between A \nand \u2212 A, because a sine function varies between +1 and \u22121. A plot of y t( ) vs. time is \nshown in Figure 23.1. \nFigure 23.1 Sinusoidal function of time \nThe sine function is periodic in time. This means that the value of the function at \ntime t will be exactly the same at a later time t\u2032 = t + T , where T is the period. That the \nsine function satisfies the periodic condition can be seen from \n\u23a1 2\u03c0\n\u23a4 \n\u23a1 2\u03c0\n\u23a4 \n\u23a1 2\u03c0 \u23a4\ny t( + T ) = Asin \n(t + T ) = Asin \nt + 2\u03c0 = Asin \nt = y t( ) . (23.1.2)\n\u23a2 \n\u23a5\n\u23a2 \n\u23a5\n\u23a2\n\u23a5\n\u23a3 T \n\u23a6\n\u23a3 T \n\u23a6\n\u23a3 T \u23a6 \nThe frequency, f , is defined to be \nf \u2261 1/T . \n(23.1.3) \n\u23a1\u22121\nThe SI unit of frequency is inverse seconds, s \u23a6\u23a4 , or hertz [Hz]. The angular frequency\n\u23a3 \nof oscillation is defined to be \n\u2261 2\u03c0 / T = 2\u03c0 f , \n(23.1.4)\n\u03c90 \nand is measured in radians per second. (The angular frequency of oscillation is denoted \nby \u03c90 to distinguish from the angular speed \u03c9 = d\u03b8 / dt .) One oscillation per second, \n1 Hz , corresponds to an angular frequency of 2\u03c0 rad s\u22c5\n\u22121 . (Unfortunately, the same \n23-2", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_590", "text": "symbol \u03c9 is used for angular speed in circular motion. For uniform circular motion the \nangular speed is equal to the angular frequency but for non-uniform motion the angular \nspeed is not constant. The angular frequency for simple harmonic motion is a constant by \ndefinition.) We therefore have several different mathematical representations for \nsinusoidal motion \ny(t) = Asin(2\u03c0t / T ) = Asin(2\u03c0 f t) = Asin(\u03c90t) . \n(23.1.5) \n23.2 Simple Harmonic Motion: Analytic \nOur first example of a system that demonstrates simple harmonic motion is a spring-\nobject system on a frictionless surface, shown in Figure 23.2 \nFigure 23.2 Spring-object system \nThe object is attached to one end of a spring. The other end of the spring is attached to a \nwall at the left in Figure 23.2. Assume that the object undergoes one-dimensional motion. \nThe spring has a spring constant k and equilibrium length leq . Choose the origin at the \nequilibrium position and choose the positive x -direction to the right in the Figure 23.2. \nIn the figure, x > 0 corresponds to an extended spring, and x < 0 to a compressed spring. \nDefine x t( ) to be the position of the object with respect to the equilibrium position. The \nforce acting on the spring is a linear restoring force, Fx = \u2212k x (Figure 23.3). The initial \nconditions are as follows. The spring is initially stretched a distance l0 and given some \ninitial speed v0 to the right away from the equilibrium position. The initial position of the \nstretched spring from the equilibrium position (our choice of origin) is x0 = (l0 \u2212 leq ) > 0 \nand its initial x -component of the velocity is v \n> 0. \nx,0 = v0 \n23-3", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_591", "text": "Figure 23.3 Free-body force diagram for spring-object system \nNewton\u2019s Second law in the x -direction becomes \n2\nd x \n\u2212k x = m \n2 . \n(23.2.1)\ndt \nThis equation of motion, Eq. (23.2.1), is called the simple harmonic oscillator equation \n(SHO). Because the spring force depends on the distance x , the acceleration is not \nconstant. Eq. (23.2.1) is a second order linear differential equation, in which the second \nderivative of the dependent variable is proportional to the negative of the dependent \nvariable, \n2\nd x \nk \n2 = \u2212 \nx . \n(23.2.2)\ndt \nm \nIn this case, the constant of proportionality is k / m , \nEq. (23.2.2) can be solved from energy considerations or other advanced techniques but \ninstead we shall first guess the solution and then verify that the guess satisfies the SHO \ndifferential equation (see Appendix 22.3.A for a derivation of the solution). \nWe are looking for a position function x(t) such that the second time derivative position \nfunction is proportional to the negative of the position function. Since the sine and cosine \nfunctions both satisfy this property, we make a preliminary ansatz (educated guess) that \nour position function is given by \nx(t) = Acos((2\u03c0 / T )t) = Acos(\u03c90 t) , \n(23.2.3) \nwhere \u03c90 is the angular frequency (as of yet, undetermined). \nWe shall now find the condition that the angular frequency \u03c90 must satisfy in order to \ninsure that the function in Eq. (23.2.3) solves the simple harmonic oscillator equation, Eq. \n(23.2.1). The first and second derivatives of the position function are given by \n23-4", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_592", "text": "dx \nAsin(\u03c90t)\n= \u2212\u03c90\ndt \n(23.2.4)\nd 2 x = \u2212\u03c90\n2 Acos(\u03c90 t) = \u2212\u03c90\n2 x. \ndt2 \nSubstitute the second derivative, the second expression in Eq. (23.2.4), and the position \nfunction, Equation (23.2.3), into the SHO Equation (23.2.1), yielding \n\u2212\u03c90\n2 Acos(\u03c90 t) = \u2212 k Acos(\u03c90 t) . \n(23.2.5)\nm \nEq. (23.2.5) is valid for all times provided that \n\u03c90 = \nk \nm . \n(23.2.6) \nThe period of oscillation is then \nT = 2\u03c0 \n\u03c90 \n= 2\u03c0 m \nk . \n(23.2.7) \nOne possible solution for the position of the block is \n\u239b \n\u239e \nk \nx(t) = Acos \n(23.2.8) \nt\n\u239c\n\u239d  \n\u239f\n\u23a0  \n,  \nm \nand therefore by differentiation, the x -component of the velocity of the block is \n\u239b \n\u239e \nk \nk \nv x (t) = \u2212 \nAsin \n(23.2.9) \nt\n\u239c\n\u239d  \n\u239f\n\u23a0  \n. \nm \nm \nNote that at t = 0 , the position of the object is x0 \u2261 x t( = 0) = A since cos(0) = 1 and the \nvelocity is v \n\u2261 v (t = 0) = 0 since sin(0) = 0 . The solution in (23.2.8) describes an \nx,0 \nx \nobject that is released from rest at an initial position A = x0 but does not satisfy the initial \nvelocity condition, v (t = 0) = v \n\u2260 0 . We can try a sine function as another possible\nx\nx,0 \nsolution, \n\u239b k \u239e \nx t( ) = Bsin \nt . \n(23.2.10)\n\u239c \n\u239f \nm\n\u239d \n\u23a0 \nThis function also satisfies the simple harmonic oscillator equation because \n23-5", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_593", "text": "d 2 \n\u239b \n\u239e\nx\nk \nk \n2\n= \u2212 \nBsin \nx , \n(23.2.11)\n\u239c \nt\u239f = \u2212\u03c90\ndt2 \nm \n\u239d m \u23a0 \nwhere \u03c90 = k / m . The x -component of the velocity associated with Eq. (23.2.10) is \ndx \nk\n\u239b \n\u239e \nk \nv x (t) = \nBcos \n.  \n(23.2.12) \nt\n\u239c\n\u239d  \n\u239f\n\u23a0  \n=  \ndt \nm \nm \nThe proposed solution in Eq. (23.2.10) has initial conditions x0 \u2261 x t( = 0) = 0 and \nv \n\u2261 v (t = 0) = ( k / m)B , thus B = vx,0 / k / m . This solution describes an object that \nx,0 \nx \nis initially at the equilibrium position but has an initial non-zero x -component of the \nvelocity, vx,0 \u2260 0 . \n23.2.1 General Solution of Simple Harmonic Oscillator Equation \nSuppose x1( )t and x2( )t are both solutions of the simple harmonic oscillator equation, \nd 2 \nk \nx1(t) = \u2212 \nx1(t)\ndt2 \nm \n(23.2.13)\nd 2 \nk \nx2(t) = \u2212 \nx2(t). \ndt2 \nm \nThen the sum x t( ) = x1( )t + x2( )t of the two solutions is also a solution. To see this, \nconsider \nd 2 \nd 2 \nd 2 \nd 2\nx(t) \nx1(t) \nx2(t)\n= \n(x1(t) + x2(t)) =\n+ \n. \n(23.2.14)\ndt2 \ndt2 \ndt2 \ndt2 \nUsing the fact that x1( )t and x2( )t both solve the simple harmonic oscillator equation \n(23.2.13), we see that \nd 2 \nk\nk \nk\nx t( ) = \u2212 \nx ( )t + \u2212 \nx ( )t = \u2212 (x ( )t + x ( )t )\n2 \n1\n2 \n1\n2\ndt \nm\nm\nm \n(23.2.15)\nk \n= \u2212 \nx t( ). \nm \nThus the linear combination x t( ) = x1( )t + x2( )t is also a solution of the SHO equation, \nEq. (23.2.1). Therefore the sum of the sine and cosine solutions is the general solution,", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_594", "text": "x t( ). \nm \nThus the linear combination x t( ) = x1( )t + x2( )t is also a solution of the SHO equation, \nEq. (23.2.1). Therefore the sum of the sine and cosine solutions is the general solution, \nx(t) = C cos(\u03c90 t) + Dsin(\u03c90 t) , \n(23.2.16) \n23-6", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_595", "text": "where the constant coefficients C and D depend on a given set of initial conditions \nx \u2261 x t( = 0) and v \n\u2261 v (t = 0) where \nand v \nare constants. For this general \n0 \nx,0 \nx\nx0 \nx,0 \nsolution, the x -component of the velocity of the object at time t is then obtained by \ndifferentiating the position function, \ndx \nv (t) =\n= \u2212\u03c90C sin(\u03c90 \nD cos(\u03c90 t) . \n(23.2.17)\nx \nt) + \u03c90\ndt \nTo find the constants C and D , substitute t = 0 into the Eqs. (23.2.16) and (23.2.17). \nBecause cos(0) = 1 and sin(0) = 0 , the initial position at time t = 0 is \n\u2261 x(t = 0) = C . \n(23.2.18)\nx0 \nThe x -component of the velocity at time t = 0 is \nv \n= v (t = 0) = \u2212\u03c90C sin(0) + \u03c90 Dcos(0) = \u03c90 D . \n(23.2.19)\nx,0 \nx \nThus \nv \nC = x0 and D = \nx,0 . \n(23.2.20)\n\u03c90 \nThe position of the object-spring system is then given by \n\u239b \n\u239e \n\u239b \n\u239e \nk \nk\nv x,0 \nk / m \nx(t) = x0 cos \nsin \n(23.2.21) \nt +  \nt\n\u239c\n\u239d  \n\u239f\n\u23a0  \n\u239c\n\u239d  \n\u239f\n\u23a0 \nm \nm \nand the x -component of the velocity of the object-spring system is \n\u239b \n\u239b\n\u239e\n\u239e \nk \nk \nk \nv x (t) = \u2212 \nx0 sin \n(23.2.22) \nt\u239f\n\u23a0  \n+ v x,0 cos \u239c\n\u239d  \nt\n\u239c\n\u239d  \n\u239f\n\u23a0  \n.  \nm \nm \nm \nAlthough we had previously specified x > 0 and v \n> 0 , Eq. (23.2.21) is seen to be a \n0 \nx,0 \nvalid solution of the SHO equation for any values of x0 and v x,0 . \nExample 23.1: Phase and Amplitude \nShow that x(t) = C cos\u03c90t + Dsin\u03c90t = Acos(\u03c90t +\u03c6) , where A = (C 2 + D2)1 2 > 0 , and \n\u03c6 = tan\u22121(\u2212D / C). \n23-7", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_596", "text": "Solution: Use the identity Acos(\u03c90t + \u03c6) = Acos(\u03c90t)cos(\u03c6) \u2212 Asin(\u03c90t)sin(\u03c6) . Thus \nC cos(\u03c90t) + Dsin(\u03c90t) = Acos(\u03c90t)cos(\u03c6) \u2212 Asin(\u03c90t)sin(\u03c6) . Comparing coefficients \nwe see that C = Acos\u03c6 and D = \u2212 Asin\u03c6 . Therefore \n(C 2 + D2)1 2 \nA2\n= A2(cos2 \u03c6 + sin2 \u03c6) = \n. \nWe choose the positive square root to ensure that A > 0 , and thus \nA = (C 2 + D2)1 2 \n(23.2.23) \nsin\u03c6\n\u2212 D / A\nD \ntan\u03c6 =\n= \n= \u2212 \n, \ncos\u03c6 \nC / A\nC \n\u03c6 = tan\u22121(\u2212D / C) . \n(23.2.24) \nThus the position as a function of time can be written as \nx(t) = Acos(\u03c90t +\u03c6) . \n(23.2.25) \nIn Eq. (23.2.25) the quantity \u03c90t +\u03c6 is called the phase, and \u03c6 is called the phase \nconstant. Because cos(\u03c90t + \u03c6) varies between +1 and \u22121 , and A > 0 , A is the \namplitude defined earlier. We now substitute Eq. (23.2.20) into Eq. (23.2.23) and find \nthat the amplitude of the motion described in Equation (23.2.21), that is, the maximum \nvalue of x t( ) , and the phase are given by \nA = x0 \n2 + (v x,0 / \u03c90 )2 . \n(23.2.26)  \n\u03c6 = tan\u22121(\u2212v \n/ \u03c9 x ) . \n(23.2.27)\nx,0 \n0\n0 \nA plot of x t( ) vs. t is shown in Figure 23.4a with the values A = 3 , T = \u03c0 , and \n\u03c6 = \u03c0 / 4 . Note that x(t) = Acos(\u03c90t + \u03c6) takes on its maximum value when \ncos(\u03c90t + \u03c6) = 1 . This occurs when \u03c90t + \u03c6 = 2\u03c0 n where n = 0, \u00b1 1, \u00b1 2,\u22c5\u22c5\u22c5 . The", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_597", "text": "cos(\u03c90t + \u03c6) = 1 . This occurs when \u03c90t + \u03c6 = 2\u03c0 n where n = 0, \u00b1 1, \u00b1 2,\u22c5\u22c5\u22c5 . The \nmaximum value associated with n = 0 occurs when \u03c90t + \u03c6 = 0 or t = \u2212\u03c6 / \u03c90 . For the \ncase shown in Figure 23.4a where \u03c6 = \u03c0 / 4 , this maximum occurs at the instant \nt = \u2212T / 8 . Let\u2019s plot x(t) = Acos(\u03c90t + \u03c6) vs. t for \u03c6 = 0 (Figure 23.4b). For \u03c6 > 0 , \nFigure 23.4a shows the plot x(t) = Acos(\u03c90t + \u03c6) vs. t . Notice that when \u03c6 > 0 , x(t) is \nshifted to the left compared with the case \u03c6 = 0 (compare Figures 23.4a with 23.4b). The \nfunction x(t) = Acos(\u03c90t + \u03c6) with \u03c6 > 0 reaches its maximum value at an earlier time \nthan the function x(t) = Acos(\u03c90t) . The difference in phases for these two cases is \n(\u03c90t +\u03c6) \u2212\u03c90t = \u03c6 and \u03c6 is sometimes referred to as the phase shift. When \u03c6 < 0 , the \n23-8", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_598", "text": "function x(t) = Acos(\u03c90t + \u03c6) reaches its maximum value at a later time t = T / 8 than \nthe function x(t) = Acos(\u03c90t) as shown in Figure 23.4c. \n(a) \n(b) \n(c) \nFigure 23.4 Phase shift of x(t) = Acos(\u03c90t + \u03c6) (a) to the left by \u03c6 = \u03c0 / 4 , (b) no shift \n\u03c6 = 0, (c) to the right \u03c6 = \u2212\u03c0 / 4 \n23-9", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_599", "text": "Example 23.2: Block-Spring System \nA block of mass m is attached to a spring with spring constant k and is free to slide \nalong a horizontal frictionless surface. At t = 0 , the block-spring system is stretched an \namount x > 0 from the equilibrium position and is released from rest, v \n= 0 . What is\n0 \nx,0 \nthe period of oscillation of the block? What is the velocity of the block when it first \ncomes back to the equilibrium position? \nSolution: The position of the block can be determined from Eq. (23.2.21) by substituting \nthe initial conditions x > 0 , and v \n= 0 yielding\n0 \nx,0 \n\u239b k \u239e \nx t( ) = x0 cos\u239c \nt , \n(23.2.28)\n\u239f \nm\n\u239d \n\u23a0 \nand the x -component of its velocity is given by Eq. (23.2.22), \n\u2212\nk\n\u239b \n\u239e \nk \nv x (t) = \nx0 sin \n.  \n(23.2.29) \nt\n\u239c\n\u239d  \n\u239f\n\u23a0 \nm \nm \nThe angular frequency of oscillation is \u03c90 = k / m and the period is given by \nEq. (23.2.7), \n2\u03c0 \nm\nT =\n= 2\u03c0 \n. \n(23.2.30)\nk\n\u03c90 \nThe block first reaches equilibrium when the position function first reaches zero. This \noccurs at time t1 satisfying \nk \n\u03c0\n\u03c0 \nm\nT\nt1 = \n, t1 = \n= \n. \n(23.2.31)\nm \n2\n2 \nk \n4 \nThe x -component of the velocity at time t1 is then \n\u239b \n\u239e\nk\nk\nk \nk \nvx (t1) = \u2212 \nsin\u239c \nsin(\u03c0 / 2) = \u2212 \n(23.2.32)\nt1\u239f = \u2212\nx0 \nx0 \nx0 = \u2212\u03c90 x0\nm\nm\nm \nm\n\u239d \n\u23a0 \nNote that the block is moving in the negative x -direction at time t1; the block has moved \nfrom a positive initial position to the equilibrium position (Figure 23.4(b)). \n23-10", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_600", "text": "v \n23.3 Energy and the Simple Harmonic Oscillator \nLet\u2019s consider the block-spring system of Example 23.2 in which the block is initially \nstretched an amount x0 > 0 from the equilibrium position and is released from rest, \nx,0 = 0 . We shall consider three states: state 1, the initial state; state 2, at an arbitrary \ntime in which the position and velocity are non-zero; and state 3, when the object first \ncomes back to the equilibrium position. We shall show that the mechanical energy has \nthe same value for each of these states and is constant throughout the motion. Choose the \nequilibrium position for the zero point of the potential energy. \nState 1: all the energy is stored in the object-spring potential energy, U1 = (1/ 2) k x0\n2 . The \nobject is released from rest so the kinetic energy is zero, K1 = 0 . The total mechanical \nenergy is then \nE1 = U1 = 1 k x0\n2 . \n(23.3.1)\n2 \nState 2: at some time t , the position and x -component of the velocity of the object are \ngiven by \nx(t) = x0 cos \nk \nm t \n\u239b \n\u239d\n\u239c\n\u239e \n\u23a0\n\u239f\nk \nm x0 sin \nk \nm t \n\u239b \n\u239d\n\u239c\n(23.3.2) \n\u239e  \n(t) = \u2212\n\u239f\n\u23a0  \nv \n. \nx \nThe kinetic energy is \n1 \n2\n1\n2\n2 \u239b k \u239e\nK2 = mv = k x0 sin \nt , \n(23.3.3)\n\u239c \n\u239f \n2\n2 \nm\n\u239d \n\u23a0 \nand the potential energy is \n1\n2\n1\n2 \n2 \u239b k \u239e\nU2 = k x = k x0 cos \nt . \n(23.3.4)\n\u239c \n\u239f \n2\n2 \nm\n\u239d \n\u23a0 \nThe mechanical energy is the sum of the kinetic and potential energies \n1 \n2 + 1 \n= mv \nk x2\nE2 = K2 +U2 \nx\n2 \n2 \n\u239b \n\u239e \nk \nk\n\u239b \n\u239b \n\u239e \n\u239e \n1 k x0\n2 \n2\n\u239f\u23a0  \n+ sin2 \u239c\u239d \n(23.3.5) \nt \nt\n\u239c\n\u239d  \n\u239f\n\u23a0  \n\u239c\u239d \n\u239f\u23a0  \n=  \ncos\n2  \nm \nm \n1\n2\n= k x0 ,\n2 \n23-11", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_601", "text": "2 \u03c90\nwhere we used the identity that cos\nt + sin2 \u03c90t = 1 , \nand that \u03c90 = k / m (Eq. \n(23.2.6)). \nThe mechanical energy in state 2 is equal to the initial potential energy in state 1, so the \nmechanical energy is constant. This should come as no surprise; we isolated the object-\nspring system so that there is no external work performed on the system and no internal \nnon-conservative forces doing work. \nFigure 23.5 State 3 at equilibrium and in motion \nState 3: now the object is at the equilibrium position so the potential energy is zero, \nU3 = 0 , and the mechanical energy is in the form of kinetic energy (Figure 23.5). \n1\n2\nE3 = K3 = mveq . \n(23.3.6)\n2 \nBecause the system is closed, mechanical energy is constant, \n= E3. \n(23.3.7)\nE1 \nTherefore the initial stored potential energy is released as kinetic energy, \n1\n2\n1 \n2\nk x0 = mveq , \n(23.3.8)\n2\n2 \nand the x -component of velocity at the equilibrium position is given by \nk \nv \n= \u00b1 \nx0 . \n(23.3.9)\nx,eq \nm \nNote that the plus-minus sign indicates that when the block is at equilibrium, there are \ntwo possible motions: in the positive x -direction or the negative x -direction. If we take \nx0 > 0 , then the block starts moving towards the origin, and v x,eq will be negative the first \ntime the block moves through the equilibrium position. \n23-12", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_602", "text": "We can show more generally that the mechanical energy is constant at all times as \nfollows. The mechanical energy at an arbitrary time is given by \n1 \n2 + 1\nE = K +U = mv \nk x2 . \n(23.3.10)\n2 \nx \n2 \nDifferentiate Eq. (23.3.10) \nd 2\ndE \ndvx \ndx \n\u239b \nx \n\u239e \n= mv x \n+ k x \n= vx m \n+ k x . \n(23.3.11)\ndt \ndt\ndt \n\u239d\u239c \ndt2 \n\u23a0\u239f \nNow substitute the simple harmonic oscillator equation of motion, (Eq. (23.2.1) ) into Eq. \n(23.3.11) yielding \ndE = 0 , \n(23.3.12)\ndt \ndemonstrating that the mechanical energy is a constant of the motion. \n23.3.1 Simple Pendulum: Force Approach \nA pendulum consists of an object hanging from the end of a string or rigid rod pivoted \nabout the point P . The object is pulled to one side and allowed to oscillate. If the object \nhas negligible size and the string or rod is massless, then the pendulum is called a simple \npendulum. Consider a simple pendulum consisting of a massless string of length l and a \npoint-like object of mass m attached to one end, called the bob. Suppose the string is \nfixed at the other end and is initially pulled out at an angle \u03b80 from the vertical and \nreleased from rest (Figure 23.6). Neglect any dissipation due to air resistance or frictional \nforces acting at the pivot. \nFigure 23.6 Simple pendulum \nLet\u2019s choose polar coordinates for the pendulum as shown in Figure 23.7a along with the \nfree-body force diagram for the suspended object (Figure 23.7b). The angle \u03b8 is defined \nwith respect to the equilibrium position. When \u03b8 > 0 , the bob is has moved to the right, \nand when \u03b8 < 0 , the bob has moved to the left. The object will move in a circular arc \ncentered at the pivot point. The forces on the object are the tension in the string \n23-13", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_603", "text": "\uf072 \n\uf072\nT = \u2212T r\u02c6 and gravity mg . \nThe gravitation force on the object has r\u02c6 - and \u03b8\u02c6 \u00ad\ncomponents given by \n\uf072 \nmg = mg(cos\u03b8 r\u02c6 \u2212 sin\u03b8 \u03b8\u02c6) . \n(23.3.13) \nFigure 23.7 (a) Coordinate system \nFigure 23.7 (b) free-body force diagram \nOur concern is with the tangential component of the gravitational force, \nF\u03b8 = \u2212mgsin\u03b8 . \n(23.3.14) \nThe sign in Eq. (23.3.14) is crucial; the tangential force tends to restore the pendulum to \nthe equilibrium value \u03b8 = 0 . If \u03b8 > 0 , F\u03b8 < 0 and if \u03b8 < 0 , F\u03b8 > 0 , where we are that \nbecause the string is flexible, the angle \u03b8 is restricted to the range \u2212\u03c0 / 2 < \u03b8 < \u03c0 / 2 . (For \nangles \u03b8 > \u03c0 / 2 , the string would go slack.) In both instances the tangential component \nof the force is directed towards the equilibrium position. The tangential component of \nacceleration is \na\u03b8 = l\u03b1 = l d 2\u03b8 . \n(23.3.15)\ndt 2 \nNewton\u2019s Second Law, F\u03b8 = ma\u03b8 , yields \n\u2212mgl sin\u03b8 = ml2 d 2\u03b8 . \n(23.3.16)\ndt 2 \nWe can rewrite this equation is the form \nd 2\u03b8 \ng\n= \u2212 sin\u03b8 . \n(23.3.17)\ndt 2 \nl \nThis is not the simple harmonic oscillator equation although it still describes periodic \nmotion. In the limit of small oscillations, sin\u03b8 \u2245\u03b8 , Eq. (23.3.17) becomes \n23-14", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_604", "text": "d 2\u03b8 \ng\n\u2245\u2212 \u03b8. \n(23.3.18)\ndt 2 \nl \nThis equation is similar to the object-spring simple harmonic oscillator differential \nequation \nd 2 x\nk \n= \u2212 \nx . \n(23.3.19)\ndt 2 \nm \nBy comparison with Eq. (23.2.6) the angular frequency of oscillation for the pendulum is \napproximately \n\u03c9 0 \uf03b g\nl , \n(23.3.20) \nwith period \n2\u03c0 \nl . \n(23.3.21)\nT = \n\uf03b 2\u03c0\n\u03c9 0 \ng \nThe solutions to Eq. (23.3.18) can be modeled after Eq. (23.2.21). With the initial \nconditions that the pendulum is released from rest, d\u03b8 (t = 0) = 0 , at a small angle \ndt \n\u03b8(t = 0) = \u03b80 , the angle the string makes with the vertical as a function of time is given by \n\u239b 2\u03c0 \u239e \n\u239b g \u239e\n\u03b8(t) = \u03b80 cos(\u03c9 0 t) = \u03b80 cos \nt\u23a0\u239f = \u03b80 cos \nt \n(23.3.22)\n\u23a0\u239f .\n\u239d\u239c T \n\u239d\u239c l \nThe z -component of the angular velocity of the bob is \n\u239b \n\u239e\nd\u03b8 \ng\ng\n\u03c9 (t) = \n(t) = \u2212 \nsin \u239c \nt\u239f . \n(23.3.23)\nz \n\u03b80\ndt\nl \nl\n\u239d \n\u23a0 \nKeep in mind that the component of the angular velocity \u03c9 z = d\u03b8 / dt changes with time \nin an oscillatory manner (sinusoidally in the limit of small oscillations). The angular \nfrequency \u03c9 0 is a parameter that describes the system. The z -component of the angular \nvelocity \u03c9 z (t) , besides being time-dependent, depends on the amplitude of oscillation \u03b80 . \nIn the limit of small oscillations, \u03c90 does not depend on the amplitude of oscillation. \nThe fact that the period is independent of the mass of the object follows algebraically \nfrom the fact that the mass appears on both sides of Newton\u2019s Second Law and hence \ncancels. Consider also the argument that is attributed to Galileo: if a pendulum, \nconsisting of two identical masses joined together, were set to oscillate, the two halves \nwould not exert forces on each other. So, if the pendulum were split into two pieces, the \n23-15", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_605", "text": "pieces would oscillate the same as if they were one piece. This argument can be \nextended to simple pendula of arbitrary masses. \n23.3.2 Simple Pendulum: Energy Approach \nWe can use energy methods to find the differential equation describing the time evolution \nof the angle \u03b8 . When the string is at an angle \u03b8 with respect to the vertical, the \ngravitational potential energy (relative to a choice of zero potential energy at the bottom \nof the swing where \u03b8 = 0 as shown in Figure 23.8) is given by \nU = mgl(1\u2212 cos\u03b8 ) \n(23.3.24) \nThe \u03b8 -component of the velocity of the object is given by v\u03b8 = l(d\u03b8 / dt) so the kinetic \nenergy is \nK =  1\n2 mv2 =  1\n2 m\n\u239e\u239f\u23a0 \n\u239b \ndt\nd\u03b8\nl\u239c\u239d\n2 \n.  \n(23.3.25)  \nFigure 23.8 Energy diagram for simple pendulum \nThe mechanical energy of the system is then \n1 \n\u239b l d\u03b8 \u239e \n2 \nE = K + U =\n+ mgl (1 \u2212 cos\u03b8 ) . \n(23.3.26)\n2 m\u239d\u239c dt \u23a0\u239f \nBecause we assumed that there is no non-conservative work (i.e. no air resistance or \nfrictional forces acting at the pivot), the energy is constant, hence \ndE \n1 \nd 2\u03b8 \nd\u03b8\n0 =\n= 2 m 2l2 d\u03b8 \n+ mgl sin\u03b8\ndt \ndt dt 2 \ndt \n(23.3.27)\n\u239b d 2\u03b8 \ng \n\u239e \n= ml2 d\u03b8 \n+ sin\u03b8\u23a0\u239f .\ndt \u239d\u239c dt 2 \nl \n23-16", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_606", "text": "There are two solutions to this equation; the first one d\u03b8 / dt = 0 is the equilibrium \nsolution. That the z -component of the angular velocity is zero means the suspended \nobject is not moving. The second solution is the one we are interested in \nd 2\u03b8 \ng\n+ sin\u03b8 = 0 , \n(23.3.28)\ndt 2 \nl \nwhich is the same differential equation (Eq. (23.3.16)) that we found using the force \nmethod. \nWe can find the time t1 that the object first reaches the bottom of the circular arc by \nsetting \u03b8(t1) = 0 in Eq. (23.3.22) \n\u239b g \u239e\n0 = \u03b80 cos \nt1 \u23a0\u239f . \n(23.3.29)\n\u239d\u239c l \nThis zero occurs when the argument of the cosine satisfies \n\u03c0\nt1 = \n(23.3.30)\n2 .\ng\nl \nThe z -component of the angular velocity at time t1 is therefore \nd\u03b8 \ndt (t1) = \u2212 \n\u03b80 sin\u239b \n\u239d\u239c \nt1 \n\u239e \n\u23a0\u239f = \u2212 \n\u03b80 sin \u239b\u03c0 \u239e \n\u239d\u239c \u23a0\u239f = \u2212\n2  \ng\nl \ng\nl \ng\nl \ng\nl \u03b80 . \n(23.3.31)  \nNote that the negative sign means that the bob is moving in the negative \u03b8\u02c6 -direction \nwhen it first reaches the bottom of the arc. The \u03b8 -component of the velocity at time t1 is \ntherefore \n= l d\u03b8 \ng \n\u239b g \u239e \n\u239b\u03c0 \u239e\n\u03b80 sin \nt1 \u23a0\u239f = \u2212 lg \u03b80 sin \n\u23a0\u239f = \u2212 lg \u03b80 .(23.3.32)\nv\u03b8 (t1) \u2261 v1\n(t1) = \u2212l \n\u239d\u239c\ndt \nl \n\u239d\u239c l \n2 \nWe can also find the components of both the velocity and angular velocity using energy \nmethods. When we release the bob from rest, the energy is only potential energy \n0\nE = U0 = mgl(1 \u2212 cos\u03b80 ) \u2245 mgl \u03b8 2 \n, \n(23.3.33)\n2 \nwhere we used the approximation that cos\u03b80 \u2245 1\u2212\u03b80\n2 / 2 . When the bob is at the bottom \nof the arc, the only contribution to the mechanical energy is the kinetic energy given by \n23-17", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_607", "text": "K1 = 2 \n1 mv1 \n2 . \n(23.3.34) \nBecause the energy is constant, we have that U0 = K1 or \nmgl \u03b8 \n2 \n2 \n0 = 1 \n2 \n2\nmv1 . \n(23.3.35) \nWe can solve for the \u03b8 -component of the velocity at the bottom of the arc \nv\u03b8 ,1 = \u00b1 gl \u03b80 . \n(23.3.36) \nThe two possible solutions correspond to the different directions that the motion of the \nbob can have when at the bottom. The z -component of the angular velocity is then \nd\u03b8 (t1) = \ndt \nv1 \nl = \u00b1 g\nl \u03b80 , \n(23.3.37)  \nin agreement with our previous calculation. \nIf we do not make the small angle approximation, we can still use energy techniques to \nfind the \u03b8 -component of the velocity at the bottom of the arc by equating the energies at \nthe two positions \n1\n2\nmgl (1 \u2212 cos\u03b80 ) = 2 mv1 , \n(23.3.38) \n(23.3.39) \n23.4 Worked Examples \nExample 23.3: Rolling Without Slipping Oscillating Cylinder \nAttach a solid cylinder of mass M and radius R to a horizontal massless spring with \nspring constant k so that it can roll without slipping along a horizontal surface. At time t , \nthe center of mass of the cylinder is moving with speed Vcm and the spring is compressed \nv\u03b8 , 1 = \u00b1 2gl 1\u2212 cos\u03b80\n( \n) . \na distance x from its equilibrium length. What is the period of simple harmonic motion \nfor the center of mass of the cylinder? \nFigure 23.9 Example 23.3 \n23-18", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_608", "text": "Solution: At time t , the energy of the rolling cylinder and spring system is \n2\n1\n2  \n1\n2  \n1\n2  \n\u239b d\u03b8 \u239e\nMvcm \n\u239d\u239c \n\u23a0\u239f \nwhere x is the amount the spring has compressed, Icm = (1/ 2)MR2 , and because it is \nrolling without slipping \n2 \nd\u03b8 \ndt = Vcm \nR . \n(23.4.2) \nTherefore the energy is \nkx2\nE =  \nIcm \n(23.4.1) \n+  \n+  \n. \ndt \n2\n1\n2  \n1\n4  \n1\n2  \n3\n4  \n1\n2  \nVcm \n\u239b\u239c\u239d  \n\u239e\u239f\u23a0  \n2\nMVcm\nMVcm \nThe energy is constant (no non-conservative force is doing work on the system) so \ndE \n3 \ndVcm \n1 \ndx \n3 M d 2 x\n0 =\n= \n2MVcm \n+ 2 k2x \n= Vcm (\ndt 2 + kx) \n(23.4.4)\ndt \n4 \ndt \ndt \n2 \n2 \nBecause Vcm is non-zero most of the time, the displacement of the spring satisfies a \nsimple harmonic oscillator equation \nd 2 x \ndt 2 + 2k \n3M x = 0 . \n(23.4.5) \nHence the period is \nT = 2\u03c0 \n\u03c9 0 \n= 2\u03c0 \n. \n(23.4.6) \nExample 23.4: U-Tube \nA U-tube open at both ends is filled with an incompressible fluid of density \u03c1 . The \ncross-sectional area A of the tube is uniform and the total length of the fluid in the tube \nis L . A piston is used to depress the height of the liquid column on one side by a distance \nx0 , (raising the other side by the same distance) and then is quickly removed (Figure \n23.10). What is the angular frequency of the ensuing simple harmonic motion? Neglect \nany resistive forces and at the walls of the U-tube. \nMR2 \nkx2 \nkx2\nE =  \n(23.4.3) \n+  \n+  \n+ \n=  \n. \nR  \n3M \n2k \n23-19", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_609", "text": "Figure 23.10 Example 23.4 \nFigure 23.11 Energy diagram for water \nSolution: We shall use conservation of energy. First choose as a zero for gravitational \npotential energy in the configuration where the water levels are equal on both sides of the \ntube. When the piston on one side depresses the fluid, it rises on the other. At a given \ninstant in time when a portion of the fluid of mass \u0394m = \u03c1 Ax is a height x above the \nequilibrium height (Figure 23.11), the potential energy of the fluid is given by \nU = \u0394mgx = (\u03c1 Ax)gx = \u03c1 Agx2 . \n(23.4.7) \nAt that same instant the entire fluid of length L and mass m = \u03c1 AL is moving with \nspeed v , so the kinetic energy is \nK = 1 mv 2 = 1 \u03c1 ALv2 . \n(23.4.8)\n2\n2 \nThus the total energy is \nE = K +U = 1 \u03c1 ALv2 + \u03c1 Agx2 . \n(23.4.9)\n2 \nBy neglecting resistive force, the mechanical energy of the fluid is constant. Therefore \ndE\ndv \ndx\n0 =\n= \u03c1 ALv \n+ 2\u03c1 Agx \n. \n(23.4.10)\ndt\ndt \ndt \nIf we just consider the top of the fluid above the equilibrium position on the right arm in \nFigure 23.13, we rewrite Eq. (23.4.10) as \ndE \ndvx \ndx\n0 =\n= \u03c1 ALv \n+ 2\u03c1 Agx \n, \n(23.4.11)\ndt \nx dt \ndt \nwhere vx = dx / dt . We now rewrite the energy condition using dvx / dt = d 2 x / dt2 as \n23-20", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_610", "text": "d 2\n\u239b \nx \n\u239e\n0 = vx \u03c1 A L \n+ 2gx . \n(23.4.12)\n\u239d\u239c \ndt2 \n\u23a0\u239f \nThis condition is satisfied when vx = 0 , i.e. the equilibrium condition or when \nd 2 x\n0 = L \n+ 2gx . \n(23.4.13)\ndt2 \nThis last condition can be written as \nd 2 x \n2g\n= \u2212 \nx . \n(23.4.14)\ndt2 \nL \nThis last equation is the simple harmonic oscillator equation. Using the same \nmathematical techniques as we used for the spring-block system, the solution for the \nheight of the fluid above the equilibrium position is given by \nx(t) = Bcos(\u03c90t) + C sin(\u03c90t) , \n(23.4.15) \nwhere \n2g\n\u03c90 = \n(23.4.16)\nL \nis the angular frequency of oscillation. The x -component of the velocity of the fluid on \nthe right-hand side of the U-tube is given by \ndx(t)\nv (t) = \nBsin(\u03c90t) + \u03c90C cos(\u03c90t) . \n(23.4.17)\nx \n= \u2212\u03c90\ndt \nThe coefficients B and C are determined by the initial conditions. At t = 0 , the height of \nthe fluid is x t( = 0) = B = x0 . At t = 0 , the speed is zero so vx (t = 0) = \u03c90C = 0 , hence \nC = 0 . The height of the fluid above the equilibrium position on the right hand-side of \nthe U-tube as a function of time is thus \n\u239b \n\u239e \n2g\nx(t) = x0 cos \n.  \n(23.4.18) \nt\n\u239c\u239d \n\u239f\u23a0 \nL  \n23.5 Damped Oscillatory Motion \nLet\u2019s now consider our spring-block system moving on a horizontal frictionless surface \nbut now the block is attached to a damper that resists the motion of the block due to \nviscous friction. This damper, commonly called a dashpot, is shown in Figure 23.13. The \nviscous force arises when objects move through fluids at speeds slow enough so that \nthere is no turbulence. When the viscous force opposes the motion and is proportional to \nthe velocity, so that \n23-21", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_611", "text": "\uf072 \n\uf072 \n= \u2212bv , \n(23.5.1)\nfvis \nthe dashpot is referred to as a linear dashpot. The constant of proportionality b depends \non the properties of the dashpot. \nFigure 23.12 Spring-block system connected to a linear dashpot \nChoose the origin at the equilibrium position and choose the positive x -direction to the \nright in the Figure 23.13. Define x(t) to be the position of the object with respect to the \nequilibrium position. The x -component of the total force acting on the spring is the sum \nof the linear restoring spring force, and the viscous friction force (Figure 23.13), \ndx\nF = \u2212k x \u2212 b \n(23.5.2)\nx \ndt \nFigure 23.13 Free-body force diagram for spring-object system with linear dashpot \nNewton\u2019s Second law in the x -direction becomes \n\u2212k x \u2212 b dx \ndt = m d 2 x \ndt2 . \n(23.5.3) \nWe can rewrite Eq. (23.5.3) as \nd 2 x \ndt2 + b \nm \ndx \ndt + k \nm x = 0 . \n(23.5.4) \nWhen (b / m)2 < 4k / m , the oscillator is called underdamped, and the solution to Eq. \n(23.5.4) is given by \nx(t) = x m e\u2212\u03b1t cos(\u03b3 t +\u03c6) \n(23.5.5) \n23-22", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_612", "text": "= (k / m \u2212 (b / 2m)2)1 2 \nwhere \u03b3 \nis the angular frequency of oscillation, \u03b1 = b 2m is a \nparameter that measured the exponential decay of the oscillations, x m is a constant and \u03c6 \nis the phase constant. Recall the undamped oscillator has angular frequency \n= (k / m)1 2 , so the angular frequency of the underdamped oscillator can be expressed \n\u03c90 \nas \n2 \u2212\u03b1 2)1 2 \n\u03b3 \n. \n(23.5.6)\n= (\u03c90 \nIn Appendix 23B: Complex Numbers, we introduce complex numbers and use them to \nsolve Eq.(23.5.4) in Appendix 23C: Solution to the Underdamped Simple Harmonic \nOscillator Equation. \nThe x -component of the velocity of the object is given by \n\u2212\u03b1t\nvx (t) = dx dt = (\u2212\u03b3 x sin(\u03b3 t +\u03c6) \u2212\u03b1 x cos(\u03b3 t +\u03c6))e \n. \n(23.5.7)\nm\nm \nThe position and the x -component of the velocity of the object oscillate but the \namplitudes of the oscillations decay exponentially. In Figure 23.14, the position is plotted \nas a function of time for the underdamped system for the special case \u03c6 = 0 . For that case \nx(t) = x m e\u2212\u03b1t cos(\u03b3 t) . \n(23.5.8) \nand \n\u2212\u03b1t\nvx (t) = dx dt = (\u2212\u03b3 x sin(\u03b3 t) \u2212\u03b1 x cos(\u03b3 t))e \n. \n(23.5.9)\nm \nm \nFigure 23.14 Plot of position x(t) of object for underdamped oscillator with \u03c6 = 0 \nBecause the coefficient of exponential decay \u03b1 = b 2m is proportional to the b, we see \nthat the position will decay more rapidly if the viscous force increases. We can introduce \na time constant \n\u03c4 = 1 \u03b1 = 2m / b . \n(23.5.10) \n23-23", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_613", "text": "When t = \u03c4 , the position is \nx(t = \u03c4 ) = x m cos(\u03b3\u03c4 )e\u22121 . \n(23.5.11) \nThe envelope of exponential decay has now decreases by a factor of e\u22121 , i.e. the \namplitude can be at most x m e\u22121 . During this time interval [0,\u03c4 ] , the position has \nundergone a number of oscillations. The total number of radians associated with those \noscillations is given by \n= (k / m \u2212 (b / 2m)2)1 2 (2m / b).\n\u03b3\u03c4 \n(23.5.12) \nThe closest integral number of cycles is then \nn = \u23a1\u23a3\u03b3\u03c4 / 2\u03c0 \u23a4\u23a6 = \u23a1\u23a3(k / m \u2212 (b / 2m)2)1 2 (m / \u03c0b)\u23a4\u23a6 . \n(23.5.13) \nIf the system is very weakly damped, such that (b / m)2 << 4k / m , then we can \napproximate the number of cycles by \n(k / m)1 2 (m / \u03c0b)\nn = \u23a1\u23a3\u03b3\u03c4 \n\u23a4\u23a6 = \u23a1\u23a3\u03c90(m / \u03c0 b)\u23a4\u23a6 , \n(23.5.14)\n2\u03c0 \u23a4\u23a6 \uf03b \u23a1\u23a3 \n= (k / m)1 2 \nwhere \u03c90 \nis the angular frequency of the undamped oscillator. \nWe define the quality, Q , of this oscillating system to be proportional to the number of \nintegral cycles it takes for the exponential envelope of the position function to fall off by \na factor of e\u22121. The constant of proportionality is chosen to be \u03c0 . Thus \nQ = n\u03c0 . \n(23.5.15) \nFor the weakly damped case, we have that \nQ \uf03b \u03c90(m / b) . \n(23.5.16) \n23.5.1 Energy in the Underdamped Oscillator \n= (k / m \u2212 (b / 2m)2)1 2 \nFor the underdamped oscillator, (b / m)2 < 4k / m , \u03b3 \n, and \n\u03b1 = b 2m . Let\u2019s choose t = 0 such that the phase shift is zero \u03c6 = 0 . The stored energy \nin the system will decay due to the energy loss due to dissipation. The mechanical energy \nstored in the potential and kinetic energies is then given by \n1 kx2 + 1\n2\nE = \nmv . \n(23.5.17)\n2\n2 \n23-24", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_614", "text": "where the position and the x -component of the velocity are given by Eqs. (23.5.8) and \n(23.5.9). The mechanical energy is then \n1\n2 \n\u22122\u03b1t + 1 \n\u22122\u03b1t\nE = kx cos2(\u03b3 t)e\nm(\u2212\u03b3 x sin(\u03b3 t) \u2212\u03b1 x cos(\u03b3 t))\n2 e \n. \n(23.5.18)\nm \nm\nm\n2\n2 \nExpanding this expression yields \n1 \n2 \n\u22122\u03b1t + 1 \n\u22122\u03b1t\nE = (k + m\u03b1 2)x cos 2(\u03b3 t)e\u22122\u03b1t + m\u03b3\u03b1 x 2 sin(\u03b3 t)cos(\u03b3 t)e\nm\u03b3 2 x 2 sin2(\u03b3 t)e \n(23.5.19)\nm\nm \nm\n2\n2 \nThe kinetic energy, potential energy, and mechanical energy are shown in Figure 23.15. \nFigure 23.15 Kinetic, potential and mechanical energy for the underdamped oscillator \nThe stored energy at time t = 0 is \nE(t = 0) = 1 \n2 (k + m\u03b1 2 )x m \n2 \n(23.5.20) \nThe mechanical energy at the conclusion of one cycle, with \u03b3 T = 2\u03c0 , is \nE(t = T ) = 1 \n2 (k + m\u03b1 2 )x m \n2e\u22122\u03b1T \n(23.5.21) \nThe change in the mechanical energy for one cycle is then \nE(t = T ) \u2212 E(t = 0) = \u2212 1 \n2 (k + m\u03b1 2 )x m \n2 (1\u2212 e\u22122\u03b1T ) . \n(23.5.22) \n23-25", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_615", "text": "= b2\nRecall that \u03b1 2 \n4m2 . Therefore \nE(t = T ) \u2212 E(t = 0) = \u2212 1 (k + b2 4m)x m\n2(1\u2212 e\u22122\u03b1T ) . \n(23.5.23)\n2 \nWe can show (although the calculation is lengthy) that the energy dissipated by the \nviscous force over one cycle is given by the integral \nT \uf072 \n2\n\uf072 \n\u239b \nb2 \u239e x m\n= \n\u22c5 v dt = \u2212k + \n(1\u2212 e\u22122\u03b1t ) . \n(23.5.24)\nEdis \u222bFvis \n0 \n\u239d\u239c \n4m\u23a0\u239f 2 \nBy comparison with Eq. (23.5.23), the change in the mechanical energy in the \nunderdamped oscillator during one cycle is equal to the energy dissipated due to the \nviscous force during one cycle. \n23.6 Forced Damped Oscillator \nLet\u2019s drive our damped spring-object system by a sinusoidal force. Suppose that the x \u00ad\ncomponent of the driving force is given by \nFx (t) = F0 cos(\u03c9t) , \n(23.6.1) \nwhere F0 is called the amplitude \n(maximum value) and \u03c9 is the driving angular \nfrequency. The force varies between F0 and \u2212F0 because the cosine function varies \nbetween +1 and \u22121. Define x(t) to be the position of the object with respect to the \nequilibrium position. The x -component of the force acting on the object is now the sum \ndx\nF \ncos(\u03c9t) \u2212 kx \u2212 b \n. \n(23.6.2)\nx = F0 \ndt \nNewton\u2019s Second law in the x -direction becomes \nd 2\ndx \nx\ncos(\u03c9t) \u2212 kx \u2212 b \n= m \n. \n(23.6.3)\nF0 \ndt \ndt2 \nWe can rewrite Eq. (23.6.3) as \nd 2 x \ndx\ncos(\u03c9t) = m \n+ b \n+ kx . \n(23.6.4)\nF0 \ndt2 \ndt \n23-26", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_616", "text": "We derive the solution to Eq. (23.6.4) in Appendix 23E: Solution to the forced Damped \nOscillator Equation. The solution to is given by the function \nx(t) = x0 cos(\u03c9t +\u03c6) , \n(23.6.5) \nwhere the amplitude x0 is a function of the driving angular frequency \u03c9 and is given by \nF0 / m\n(\u03c9 ) = \n. \n(23.6.6)\nx0 \n1/2 \n2 \u2212\u03c9 2)2\n((b / m)2\u03c9 2 + (\u03c90 \n) \nThe phase constant \u03c6 is also a function of the driving angular frequency \u03c9 and is given \nby \n\u239b (b / m)\u03c9 \u239e \n\u03c6(\u03c9 ) = tan\u22121 \u239c \n2 \u239f\n\u239d \u03c9 2 \u2212 \u03c90 \u23a0 \n. \n(23.6.7) \nIn Eqs. (23.6.6) and (23.6.7) \nk\n\u03c90 = \n(23.6.8)\nm \nis the natural angular frequency associated with the undriven undamped oscillator. The x \n-component of the velocity can be found by differentiating Eq. (23.6.5), \ndx \nv (t) = \nsin(\u03c9t +\u03c6) , \n(23.6.9)\nx \n(t) = \u2212\u03c9 x0\ndt \nwhere the amplitude x0(\u03c9 ) is given by Eq. (23.6.6) and the phase constant \u03c6(\u03c9 ) is given \nby Eq. (23.6.7). \n23.6.1 Resonance \nWhen b / m << 2\u03c90 we say that the oscillator is lightly damped. For a lightly-damped \ndriven oscillator, after a transitory period, the position of the object will oscillate with the \nsame angular frequency as the driving force. The plot of amplitude x0(\u03c9 ) vs. driving \nangular frequency \u03c9 for a lightly damped forced oscillator is shown in Figure 23.16. If \nthe angular frequency is increased from zero, the amplitude of the x0(\u03c9 ) will increase \nuntil it reaches a maximum when the angular frequency of the driving force is the same \nas the natural angular frequency, \u03c90 , associated with the undamped oscillator. This is \ncalled resonance. When the driving angular frequency is increased above the natural \nangular frequency the amplitude of the position oscillations diminishes. \n23-27", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_617", "text": "Figure 23.16 Plot of amplitude x0(\u03c9 ) vs. driving angular frequency \u03c9 for a lightly \ndamped oscillator with b / m << 2\u03c90 \nWe can find the angular frequency such that the amplitude x0(\u03c9 ) is at a maximum by \nsetting the derivative of Eq. (23.6.6) equal to zero, \nd\nF0(2\u03c9 ) \n((b / m)2 \u2212 2(\u03c90\n2 \u2212\u03c9 2)) \n0 = \n(\u03c9 ) = \u2212 \n. \n(23.6.10)\nx0 \n3/2 \ndt \n2m \n2 \u2212\u03c9 2)2\n((b / m)2\u03c9 2 + (\u03c90 \n) \nThis vanishes when \n\u03c9 = (\u03c90\n2 \u2212 (b / m)2 / 2)1/2 . \n(23.6.11) \nFor the lightly-damped oscillator, \u03c90 >> (1/ 2)b / m , and so the maximum value of the \namplitude occurs when \n= (k / m)1/2 . \n(23.6.12)\n\u03c9 \uf03b \u03c90 \nThe amplitude at resonance is then \nF0\nx0(\u03c9 = \u03c90) = \n(lightly damped) . \n(23.6.13)\nb\u03c90 \nThe plot of phase constant \u03c6(\u03c9 ) vs. driving angular frequency \u03c9 for a lightly damped \nforced oscillator is shown in Figure 23.17. \n23-28", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_618", "text": "Figure 23.17 Plot of phase constant \u03c6(\u03c9 ) vs. driving angular frequency \u03c9 for a lightly \ndamped oscillator with b / m << 2\u03c90 \nThe phase constant at resonance is zero, \n\u03c6(\u03c9 = \u03c90) = 0 . \n(23.6.14) \nAt resonance, the x -component of the velocity is given by \ndx \nF0\nvx (t) = \n(t) = \u2212 \nsin(\u03c90t) \n(lightly damped) . \n(23.6.15)\ndt \nb \nWhen the oscillator is not lightly damped ( b / m \uf03b \u03c90 ), the resonance peak is shifted to \nthe left of \u03c9 = \u03c90 as shown in the plot of amplitude vs. angular frequency in Figure \n23.18. The corresponding plot of phase constant vs. angular frequency for the non-lightly \ndamped oscillator is shown in Figure 23.19. \nFigure 23.18 Plot of amplitude vs. angular frequency for lightly-damped driven oscillator \nwhere b / m \uf03b \u03c90 \n23-29", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_619", "text": "Figure 23.19 Plot of phase constant vs. angular frequency for lightly-damped driven \noscillator where b / m \uf03b \u03c90 \n23.6.2 Mechanical Energy \nThe kinetic energy for the driven damped oscillator is given by \nK(t) = 1 mv 2(t) = 1 m\u03c9 2 x0\n2 sin2(\u03c9t +\u03c6) . \n(23.6.16)\n2\n2 \nThe potential energy is given by \n1 \n1\n2\nU (t) = kx2(t) = kx0 cos2(\u03c9t +\u03c6) . \n(23.6.17)\n2\n2 \nThe mechanical energy is then \n1 \n1\n1 m\u03c9 2 \n1\nE(t) = mv2(t) + kx2(t) = \nx0\n2 sin2(\u03c9t +\u03c6) + kx0 \n2 cos2(\u03c9t +\u03c6) .(23.6.18)\n2\n2\n2 \n2 \nExample 23.5: Time-Averaged Mechanical Energy \nThe period of one cycle is given by T = 2\u03c0 / \u03c9 . Show that \n1 T \n1\n(i) \n\u222b sin2(\u03c9t +\u03c6) dt = \n, \n(23.6.19)\nT 0\n2 \n1 T \n(ii) \n\u222b cos2(\u03c9t +\u03c6)dt = 1 , \n(23.6.20)\nT 0\n2 \nT \n1 \nT \n\u222b \n0 \n(iii) \nsin(\u03c9t)cos(\u03c9t) dt = 0 . \n(23.6.21) \n23-30", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_620", "text": "Solution: (i) We use the trigonometric identity \nsin2(\u03c9t +\u03c6)) = 1 (1\u2212 cos(2(\u03c9t +\u03c6)) \n(23.6.22)\n2 \nto rewrite the integral in Eq. (23.6.19) as \n1 \nT \nT \n\u222bsin2(\u03c9t +\u03c6)) dt = 1 \u222b(1\u2212 cos(2(\u03c9t +\u03c6))dt \n(23.6.23)\nT 0\n2T 0 \nIntegration yields \nT =2\u03c0 /\u03c9\n1 \nT \n1 \u239b sin(2(\u03c9t +\u03c6)) \u239e\n\u222b(1\u2212 cos(2(\u03c9t +\u03c6)) dt = \n\u2212\n2T 0 \n2 \u239d\u239c \n2\u03c9 \n\u23a0\u239f \nT =0 \n(23.6.24) \n1 \u239b sin(4\u03c0 + 2\u03c6) sin(2\u03c6)\u239e \n1 \n= \n\u2212\n\u2212 \n,\n\u23a0\u239f = \n2 \u239d\u239c \n2\u03c9 \n2\u03c9 \n2 \nwhere we used the trigonometric identity that \nsin(4\u03c0 + 2\u03c6) = sin(4\u03c0 )cos(2\u03c6) + sin(2\u03c6)cos(4\u03c0 ) = sin(2\u03c6) , \n(23.6.25) \nproving Eq. (23.6.19). \n(ii) We use a similar argument starting with the trigonometric identity that \ncos2(\u03c9t +\u03c6)) = 1 (1+ cos(2(\u03c9t +\u03c6)) . \n(23.6.26)\n2 \nThen \n1 \nT \nT \n\u222bcos2(\u03c9t +\u03c6)) dt = 1 \u222b(1+ cos(2(\u03c9t +\u03c6)) dt . \n(23.6.27)\nT 0\n2T 0 \nIntegration yields \nT =2\u03c0 /\u03c9\n1 \nT \n1 \u239b sin(2(\u03c9t +\u03c6)) \u239e\n(1+ cos(2(\u03c9t +\u03c6)) dt =\n+\n\u222b\n2T 0\n2 \u239d\u239c \n2\u03c9 \n\u23a0\u239f \nT =0 \n(23.6.28) \n1 \u239b sin(4\u03c0 + 2\u03c6) sin(2\u03c6)\u239e \n1 \n=\n+ \n\u2212 \n.\n\u23a0\u239f = \n2 \u239d\u239c \n2\u03c9 \n2\u03c9 \n2 \n(iii) We first use the trigonometric identity that \n23-31", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_621", "text": "1\nsin(\u03c9t)cos(\u03c9t) = sin(\u03c9t) . \n(23.6.29)\n2 \nThen \n1 \nT \nT \nsin(\u03c9t)cos(\u03c9t)dt = \nsin(\u03c9t) dt\n\u222b \n1 \u222b\nT 0 \nT 0 \n(23.6.30)\nT \n1\n1 cos(\u03c9t)\n= \u2212 \n= \u2212 \n(1\u22121) = 0.\nT \n2\u03c9 \n0\n2\u03c9T \nThe values of the integrals in Example 23.5 are called the time-averaged values. We \ndenote the time-average value of a function f (t) over one period by \n1 T \nf \u2261\n\u222b f (t) dt . \n(23.6.31)\nT 0 \nIn particular, the time-average kinetic energy as a function of the angular frequency is \ngiven by \n1 m\u03c9 2 \n2\nK(\u03c9 ) = \nx0 . \n(23.6.32)\n4 \nThe time-averaged potential energy as a function of the angular frequency is given by \n1\n2\nU (\u03c9 ) = kx0 . \n(23.6.33)\n4 \nThe time-averaged value of the mechanical energy as a function of the angular frequency \nis given by \n1 m\u03c9 2 \n2 + 1\n1\nE(\u03c9 ) = \nx0 \nkx0 \n2 = (m\u03c9 2 + k)x0 \n2 . \n(23.6.34)\n4 \n4\n4 \nWe now substitute Eq. (23.6.6) for the amplitude into Eq. (23.6.34) yielding \n2 \n2 + \u03c9 2)\nF0\n(\u03c90\n= \n. \n(23.6.35)\nE(\u03c9 ) \n4m \u239b (b / m)2\u03c9 2 + \n2 \u2212\u03c9 2\n2 \u239e \n\u239d \n(\u03c90 \n) \u23a0 \nA plot of the time-averaged energy versus angular frequency for the lightly-damped case \n( b / m << 2\u03c90 ) is shown in Figure 23.20. \n23-32", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_622", "text": "Figure 23.20 Plot of the time-averaged energy versus angular frequency for the \nlightly-damped case ( b / m << 2\u03c90 ) \nWe can simplify the expression for the time-averaged energy for the lightly-damped case \nby observing that the time-averaged energy is nearly zero everywhere except where \n, (see Figure 23.20). We first substitute \u03c9 = \u03c90 everywhere in Eq. (23.6.35)\n\u03c9 = \u03c90 \n2 \u2212\u03c9 2\nexcept the term \u03c90 \nthat appears in the denominator, yielding \n2 \n2)\nF0\n(\u03c90\n= \n. \n(23.6.36)\nE(\u03c9 ) \n2m \u239b (b / m)2\u03c90\n2 +(\u03c90\n2 \u2212\u03c9 2 )\n2 \u239e \n\u239d\n\u23a0 \nWe can approximate the term \n2 \u2212\u03c9 2\n\u03c90 \n\u2212\u03c9 )(\u03c90 + \u03c9 ) \uf03b 2\u03c90 \n\u2212\u03c9 ) \n(23.6.37)\n= (\u03c90 \n(\u03c90 \nThen Eq. (23.6.36) becomes \nF0\n2 \n1\nE(\u03c9 ) = \n(lightly damped) . \n(23.6.38)\n2m ((b / m)2 + 4(\u03c90 \u2212\u03c9 )2 ) \nThe right-hand expression of Eq. (23.6.38) takes on its maximum value when the \ndenominator has its minimum value. By inspection, this occurs when \u03c9 = \u03c90 . \nAlternatively, to find the maximum value, we set the derivative of Eq. (23.6.35) equal to \nzero and solve for \u03c9 , \n23-33", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_623", "text": "0 = d \nd F0\n2 \n1\nE(\u03c9 ) = \nd\u03c9 \nd\u03c9 2m ((b / m)2 + 4(\u03c90 \u2212\u03c9 )2 ) \n. \n(23.6.39)\n4F0\n2 \n\u2212\u03c9 )\n= \n(\u03c90 \nm \n\u2212\u03c9 )2 2\n((b / m)2 + 4(\u03c90 \n) \nThe maximum occurs when occurs at \u03c9 = \u03c90 and has the value \nmF0\n2 \nE(\u03c90) = \n(underdamped) . \n(23.6.40)\n2b2 \n23.6.3 The Time-averaged Power \nThe time-averaged power delivered by the driving force is given by the expression \n1 \nT \n1 \nT\nF0\n2\u03c9 cos(\u03c9t)sin(\u03c9t +\u03c6)\nP(\u03c9 ) = \nv dt = \u2212 \ndt , \n(23.6.41)\nT \u222b \n0 \nFx\nx \nT \u222b \n0 m (b / m)2\u03c9 2 + (\u03c90\n2 \u2212\u03c9 2)2 1/2 \n(\n) \nwhere we used Eq. (23.6.1) for the driving force, and Eq. (23.6.9) for the x -component \nof the velocity of the object. We use the trigonometric identity \nsin(\u03c9t +\u03c6) = sin(\u03c9t)cos(\u03c6) + cos(\u03c9t)sin(\u03c6) \n(23.6.42) \nto rewrite the integral in Eq. (23.6.41) as two integrals \n1 \nT\nF0\n2\u03c9 cos(\u03c9t)sin(\u03c9t)cos(\u03c6)\nP(\u03c9 ) = \u2212 \ndt\nT \u222b \n2 \u2212\u03c9 2)2 1/2 \n0 m((b / m)2\u03c9 2 + (\u03c90 \n) \n(23.6.43)\n1 \nT\nF0\n2\u03c9 cos2(\u03c9t)sin(\u03c6)\n\u2212 \ndt.\nT \u222b \n2 \u2212\u03c9 2)2 1/2 \n0 m((b / m)2\u03c9 2 + (\u03c90 \n) \nUsing the time-averaged results from Example 23.5, we see that the first term in Eq. \n(23.6.43) is zero and the second term becomes \nF0\n2\u03c9 sin(\u03c6)\n= \n(23.6.44)\nP(\u03c9 ) \n2 \u2212\u03c9 2)2 1/2 \n2m((b / m)2\u03c9 2 + (\u03c90 \n)", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_624", "text": "(23.6.43) is zero and the second term becomes \nF0\n2\u03c9 sin(\u03c6)\n= \n(23.6.44)\nP(\u03c9 ) \n2 \u2212\u03c9 2)2 1/2 \n2m((b / m)2\u03c9 2 + (\u03c90 \n) \nFor the underdamped driven oscillator, we make the same approximations in Eq. \n(23.6.44) that we made for the time-averaged energy. In the term in the numerator and the \n23-34", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_625", "text": "term on the left in the denominator, we set \u03c9 \uf03b \u03c90 , and we use Eq. (23.6.37) in the term \non the right in the denominator yielding \nF0\n2 sin(\u03c6)\nP(\u03c9 ) = \n1/2 \n(underdamped) . \n(23.6.45) \n2m((b / m)2 + 2(\u03c90 \u2212\u03c9 )) \nThe time-averaged power dissipated by the resistive force is given by \nT \nT\nT\n1 \n1\n1 \nF0\n2\u03c9 2 sin2(\u03c9t +\u03c6)dt\n(\u03c9 ) = T \u222b(F x )dis vx dt = \u2212 T \u222b bv x \n2 dt = T \u222b \n2 (b / m)2\u03c9 2 + (\u03c90\n2 \u2212\u03c9 2)2\nPdis  \n0 \n0 \n0 m (\n) \n,(23.6.46)\nF0\n2\u03c9 2dt \n= \n.  \n2m2 ((b / m)2\u03c9 2 + (\u03c90\n2 \u2212\u03c9 2)2 )  \nwhere we used Eq. (23.5.1) for the dissipative force, Eq. (23.6.9) for the x -component of \nthe velocity of the object, and Eq. (23.6.19) for the time-averaging. \n23.6.4 Quality Factor \nThe plot of the time-averaged energy vs. the driving angular frequency for the \nunderdamped oscullator has a width, \u0394\u03c9 (Figure 23.20). One way to characterize this \nwidth is to define \u0394\u03c9 = \u03c9 \n, where \u03c9 \u00b1 are the values of the angular frequency such \n+ \u2212\u03c9 \u2212 \nthat time-averaged energy is equal to one half its maximum value \n1 \nmF0\n2 \nE(\u03c9 \u00b1 ) = \nE(\u03c90) = \n. \n(23.6.47)\n4b2\n2 \nThe quantity \u0394\u03c9 is called the line width at half energy maximum also known as the \nresonance width. We can now solve for \u03c9 \u00b1 by setting \nF0\n2 \n1 \nmF0\n2 \nE(\u03c9 \u00b1 ) =\n= \n, \n(23.6.48)\n2m ((b / m)2 + 4(\u03c90 \u2212\u03c9 \u00b1 )2 ) \n4b2 \nyielding the condition that \n)2\n(b / m)2 = 4(\u03c90 \u2212\u03c9 \u00b1 \n. \n(23.6.49) \nTaking square roots of Eq. (23.6.49) yields", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_626", "text": "4b2 \nyielding the condition that \n)2\n(b / m)2 = 4(\u03c90 \u2212\u03c9 \u00b1 \n. \n(23.6.49) \nTaking square roots of Eq. (23.6.49) yields \n\uf06d(b / 2m) = \u03c90 \u2212\u03c9 \u00b1 . \n(23.6.50) \n23-35", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_627", "text": "Therefore \n\u00b1 (b / 2m) . \n(23.6.51)\n\u03c9 \u00b1 = \u03c90 \nThe half-width is then \n\u0394\u03c9 = \u03c9 \n+ (b / 2m)) \u2212 (\u03c90 \u2212 (b / 2m)) = b / m . \n(23.6.52)\n+ \u2212\u03c9 \u2212 = (\u03c90 \nWe define the quality Q of the resonance as the ratio of the resonant angular frequency \nto the line width, \n\u03c90 \n\u03c90\nQ =\n= \n. \n(23.6.53)\n\u0394\u03c9 \nb / m \nFigure 23.21 Plot of time-averaged energy vs. angular frequency for different values of \nb / m \nIn Figure 23.21 we plot the time-averaged energy vs. angular frequency for several \ndifferent values of the quality factor Q = 10, 5, and 3. Recall that this was the same result \nthat we had for the quality of the free oscillations of the damped oscillator, Eq. (23.5.16) \n(because we chose the factor \u03c0 in Eq. (23.5.16)). \n23.7 Small Oscillations \nAny object moving subject to a force associated with a potential energy function that is \nquadratic will undergo simple harmonic motion, \n1 \n)2\nU (x) = U0 + k(x \u2212 x \n. \n(23.7.1)\n2 \neq \nwhere k is a \u201cspring constant\u201d, xeq is the equilibrium position, and the constant U0 just \ndepends on the choice of reference point xref for zero potential energy, U (xref ) = 0 , \n23-36", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_628", "text": "1 \n)2\n+ \n\u2212 x \n. \n(23.7.2)\n0 = U (xref ) = U0 \n2 k(xref \neq \nTherefore the constant is \n1 \n)2\nU0 = \u2212 2 k(xref \u2212 xeq . \n(23.7.3) \nThe minimum of the potential x0 corresponds to the point where the x -component of the \nforce is zero, \ndU \n\u2212 x ) = 0 \u21d2 x0 = x , \n(23.7.4)\n= 2k(x0 \neq \neq \ndx x = x0 \ncorresponding to the equilibrium position. Therefore the constant is U (x0) = U0 and we \nrewrite our potential function as \n1 \n)2\nU (x) = U (x0) + k(x \u2212 x0 . \n(23.7.5)\n2 \nNow suppose that a potential energy function is not quadratic but still has a minimum at \nx0 . For example, consider the potential energy function \n3\n2 \u239e\n\u239b \u239b x \u239e\n\u239b x \u239e \nU (x) = \u2212U1 \u239c\n\u2212\n\u239f , \n(23.7.6)\n\u239c \u239d\u239c \n\u239d\u239c \n\u239d x1 \u23a0\u239f \nx1 \u23a0\u239f \u239f\u23a0 \n(Figure 23.22), which has a stable minimum at x0 . \nFigure 23.22 Potential energy function with stable minima and unstable maxima \nWhen the energy of the system is very close to the value of the potential energy at the \nminimum U (x0) , we shall show that the system will undergo small oscillations about the \n23-37", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_629", "text": "minimum value x0 . We shall use the Taylor formula to approximate the potential \nfunction as a polynomial. We shall show that near the minimum x0 , we can approximate \nthe potential function by a quadratic function similar to Eq. (23.7.5) and show that the \nsystem undergoes simple harmonic motion for small oscillations about the minimum x0 . \nWe begin by expanding the potential energy function about the minimum point using the \nTaylor formula \ndU \n1 d 2U \n)2 + 1 d 3U\nU (x) = U (x0) + \n(x \u2212 x0) + \n(x \u2212 x0 \n(x \u2212 x0)3 + \u22c5\u22c5\u22c5 (23.7.7)\ndx \n2! dx2 \n3! dx3 \nx=x0 \nx=x0 \nx=x0 \n1 d 3U \n)3\nwhere \n(x \u2212 x0)3 is a third order term in that it is proportional to (x \u2212 x0 \n, and \n3! dx3 \nx=x0 \nd 3U \nd 2U \ndU \n, dx2 \nand \nare constants. If x0 is the minimum of the potential \ndx3 \ndx\n, \nx=x0\nx=x0 \nx=x0 \nenergy, then the linear term is zero, because \ndU \n= 0 \n(23.7.8)\ndx x=x0 \nand so Eq. ((23.7.7)) becomes \n1 d 2U \n)2 + 1 d 3U\nU (x) \uf03b U (x0) + \n(x \u2212 x0 \n(x \u2212 x0)3 + \u22c5\u22c5\u22c5 \n(23.7.9)\n2 dx2 \n3! dx3 \nx=x0 \nx=x0 \nFor small displacements from the equilibrium point such that \nis sufficiently small, \nthe third order term and higher order terms are very small and can be ignored. Then the \npotential energy function is approximately a quadratic function, \nx \u2212 x0 \n1 d 2U \n)2 \n1 \n)2\nU (x) \uf03b U (x0) + \n(x \u2212 x0 \n= U (x0) + \n(x \u2212 x0 \n(23.7.10)\nkeff \n2 dx2 \n2 \nx=x0 \nwhere we define keff , the effective spring constant, by \n\u2261 \n. \n(23.7.11)\nkeff \nd\ndx\n2U \n2 \nx=x0 \n23-38", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_630", "text": "Because the potential energy function is now approximated by a quadratic function, the \nsystem will undergo simple harmonic motion for small displacements from the minimum \nwith a force given by \nF x = \u2212 dU \ndx = \u2212keff (x \u2212 x0 ) . \n(23.7.12) \nAt x = x0 , the force is zero \nF x (x0 ) = dU \ndx (x0 ) = 0 . \n(23.7.13) \nWe can determine the period of oscillation by substituting Eq. (23.7.12) into Newton\u2019s \nSecond Law \nx\n(x \u2212 x0 \n(23.7.14)\n\u2212keff \n) = meff \nd\ndt\n2 \n2 \nwhere meff is the effective mass. For a two-particle system, the effective mass is the \nreduced mass of the system. \nm1m2\n= \n(23.7.15)\nmeff \n\u2261 \u00b5red , \nm1 + m2 \nEq. (23.7.14) has the same form as the spring-object ideal oscillator. Therefore the \nangular frequency of small oscillations is given by \n\u03c90 = \nkeff \nmeff \n= \nd 2U \ndx2 \nx=x0 \nmeff \n. \n(23.7.16)  \nExample 23.6: Quartic Potential \nA system with effective mass m has a potential energy given by \n\u239b\n\u239b x \u239e \n2 \n\u239b x \u239e \n4 \u239e \nU (x) = \n\u239c\u22122 \n+ \n\u239f , \n(23.7.17)\nU0 \u239c \n\u239d\u239c \n\u239d\u239c \n\u239d \nx0 \u23a0\u239f \nx0 \u23a0\u239f \u239f\u23a0 \nwhere U0 and x0 are positive constants and U (0) = 0 . (a) Find the points where the \nforce on the particle is zero. Classify these points as stable or unstable. Calculate the \nvalue of U (x) / U0 at these equilibrium points. (b) If the particle is given a small \ndisplacement from an equilibrium point, find the angular frequency of small oscillation. \nSolution: (a) A plot of U (x) / U0 as a function of x / x0 is shown in Figure 23.23. \n23-39", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_631", "text": "Figure 22.23 Plot of U (x) / U0 as a function of x / x0 \nThe force on the particle is zero at the minimum of the potential energy, \n\u239b \n2\n4 \n\u239e\ndU \n\u239b 1 \u239e\n\u239b 1 \u239e \n0 =\n= \n\u239c\u22124 \nx + 4 \nx\nU0\ndx \n\u239c\u239d \n\u239d\u239c x0 \u23a0\u239f \n\u239d\u239c x0 \u23a0\u239f \n3 \n\u239f\u23a0\n\u239f \n(23.7.18) \n\u239b 1 \u239e \n2 \u239b\n\u239b x \u239e \n2 \u239e \n= \u22124U0 x \n\u239c1\u2212 \n\u239f\u21d2 x2 = x0\n2 and x = 0. \n\u239d\u239c \n\u239d\u239c\nx0 \u23a0\u239f \u239c\u239d \nx0 \u23a0\u239f \u239f\u23a0 \nThe equilibrium points are at x = \u00b1x0 which are stable and x = 0 which is unstable. The \nsecond derivative of the potential energy is given by \n\u239b\n\u239b 1 \u239e \n2 \n\u239b 1 \u239e \n4 \n\u239e\nd 2U \n2 \u239f\n\u239c\u22124 \n+12 \nx\n. \n(23.7.19)\ndx2 = U0 \n\u239d\u239c \n\u239d\u239c\n\u239c\u239d \nx0 \u23a0\u239f \nx0 \u23a0\u239f \n\u239f\u23a0 \nIf the particle is given a small displacement from x = x0 then \n\u239b \n2\n4 \n\u239e\nd 2U \n\u239b 1 \u239e\n\u239b 1 \u239e \n2 \u239f \n8\n\u239c\u22124 \n+12 \n. \n(23.7.20)\n= U0 \nx0 = U0\n2\ndx2 \n\u239c \n\u239d\u239c x0 \u23a0\u239f \n\u239d\u239c x0 \u23a0\u239f \n\u239f \nx0\nx=x0 \n\u239d\n\u23a0 \n(b) The angular frequency of small oscillations is given by \n. \n(23.7.21)\n\u03c90 = d 2U \ndx2 \nx=x0 \nmx0 \n/ m = 8U0 \n2 \n23-40", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_632", "text": "Example 23.7: Lennard-Jones 6-12 Potential \nA commonly used potential energy function to describe the interaction between two atoms is the \nLennard-Jones 6-12 potential \n/ r)12 \u2212 2(r0\nU (r) = U0 \u23a1\u23a3(r0\n/ r)6 \u23a4\u23a6 ; r > 0 , \n(23.7.22) \nwhere r is the distance between the atoms. Find the angular frequency of small oscillations \nabout the stable equilibrium position for two identical atoms bound to each other by the Lennard-\nJones interaction. Let m denote the effective mass of the system of two atoms. \nSolution: The equilibrium points are found by setting the first derivative of the potential \nenergy equal to zero, \n\u23a1 \n6 \n\u23a4\ndU \n12 \u221213 +12r0 \n6r \u22127\n6 \u22127 \nr0\n0 =\n= U0 \u23a1\u23a3\u221212r0 r\n\u23a4\u23a6 = U012r0 r \u23a2\u2212\u239b \u239e +1\u23a5 . \n(23.7.23)\ndr \n\u239d\u239c r \u23a0\u239f\n\u23a2\u23a3 \n\u23a5\u23a6 \nThe equilibrium point occurs when r = r0 . The second derivative of the potential energy \nfunction is \n12 \n6 \u22128\nd\ndr\n2U \n2 = U0 \u23a1\u23a3+(12)(13)r0 r \u221214 \u2212 (12)(7)r0 r \u23a4\u23a6 . \n(23.7.24) \nEvaluating this at r = r0 yields \nd 2U \n= 72U0\n\u22122 . \n(23.7.25)\nr0\ndr 2 \nr=r0 \nThe angular frequency of small oscillation is therefore \n\u03c90 = d 2U \ndr 2 \nr=r0 \n/ m = 72U0 / mr0 \n2 . \n(23.7.26)  \n23-41", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_633", "text": "Appendix 23A: Solution to Simple Harmonic Oscillator Equation \nIn our analysis of the solution of the simple harmonic oscillator equation of motion, \nEquation (23.2.1), \n2\nd x \n\u2212k x = m \n, \n(23.A.1)\ndt 2 \nwe assumed that the solution was a linear combination of sinusoidal functions, \nx(t) = Acos(\u03c90 t) + Bsin(\u03c90 t) , \n(23.A.2) \nwhere \u03c90 = k / m . We shall now derive Eq. (23.A.2).  \nAssume that the mechanical energy of the spring-object system is given by the constant  \nE . Choose the reference point for potential energy to be the unstretched position of the \nspring. Let x denote the amount the spring has been compressed ( x < 0 ) or stretched \n( x > 0 ) from equilibrium at time t and denote the amount the spring has been \ncompressed or stretched from equilibrium at time t = 0 by x t( = 0) \u2261 x0 . Let vx = dx / dt \ndenote the x -component of the velocity at time t and denote the x -component of the \nvelocity at time t = 0 by v (t = 0) \u2261 v \nThe constancy of the mechanical energy is then \nx \nx,0 . \nexpressed as \n1\n2\n1 \n2\nE = K +U = k x + mv . \n(23.A.3)\n2\n2 \nWe can solve Eq. (23.A.3) for the square of the x -component of the velocity, \n2\n2E\nk \n2\n2E \u239b \nk \n2 \u239e \nv = \n\u2212 \nx = \n1\u2212 \nx \u23a0\u239f . \n(23.A.4)\nx \nm\nm \nm \u239d\u239c \n2E \nTaking square roots, we have \ndx \n2E\nk \n2\n= \n1\u2212 \nx . \n(23.A.5)\ndt \nm \n2E \n(why we take the positive square root will be explained below). \nLet a1 \u2261 2E / m and a2 \u2261 k / 2E . It\u2019s worth noting that a1 has dimensions of velocity \nand w has dimensions of [length]\u22122 . Eq. (23.A.5) is separable, \n23-42", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_634", "text": "dx \n2\n= a1 \nx\n1\u2212 a2\ndt \n(23.A.6)\ndx \ndt. \n2 = a1 \nx\n1\u2212 a2 \nWe now integrate Eq. (23.A.6), \ndx \n= \u222b dt . \n(23.A.7)\na1 \nThe integral on the left in Eq. (23.A.7) is well known, and a derivation is presented here. \n1\u2212 a1 x2\n\u222b \nWe make a change of variables cos\u03b8 = \nx with the differentials d\u03b8 and dx related\na2 \nby \u2212sin\u03b8 d\u03b8 = a2 dx . The integration variable is \n\u03b8 = cos\u22121 ( a2 x) . \n(23.A.8) \nEq. (23.A.7) then becomes \na1 dt . \n(23.A.9) \nThis is a good point at which to check the dimensions. The term on the left in Eq. \n\u2212sin\u03b8 d\u03b8 \n1\u2212cos2 \u03b8\n\u222b \n= \na2\n\u222b \n(23.A.9) is dimensionless, and the product \non the right has dimensions of inverse \na2 a1 \ntime, [length]\u22121[length \u22c5 time\u22121] = [time\u22121] , so \ndt is dimensionless. Using the \na2 a1 \n2 \u03b8\ntrigonometric identity 1\u2212cos\n= sin\u03b8 , Eq. (23.A.9) reduces to \ndt . \n(23.A.10)\n\u222b d\u03b8 = \u2212\u222b a2 a1 \nAlthough at this point in the derivation we don\u2019t know that \n, which has \na2 a1 \ndimensions of frequency, is the angular frequency of oscillation, we\u2019ll use some foresight \nand make the identification \nk \n2E\nk\n\u2261 \n= \n= \n, \n(23.A.11)\n\u03c90 \na2 a1 \n2E\nm \nm \nand Eq. (23.A.10) becomes \n\u03b8 \nt \nd\u03b8 = \u2212 \ndt . \n(23.A.12)\n\u222b \n\u222b\u03c90 \n\u03b8 \nt=0\n=\u03b80 \n23-43", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_635", "text": "2E \nk \nAfter integration we have \nt , \n(23.A.13)\n\u03b8 \u2212\u03b80 = \u2212\u03c90 \n\u22121 \na2\nwhere \u03b80 \u2261 \u2212\u03c6 is the constant of integration. Because \u03b8 = cos ( \nx(t)) , Eq. (23.A.13) \nbecomes \n\u22121\ncos ( \nx(t)\nt +\u03c6) . \n(23.A.14)\na2 \n) = \u2212(\u03c90 \nTake the cosine of each side of Eq. (23.A.14), yielding \n1\n2E \nx(t) = \ncos(\u2212(\u03c90 t +\u03c6)) = \ncos(\u03c90 t +\u03c6) . \n(23.A.15)\nk\na2 \nAt t = 0, \n\u2261 x(t = 0) = \ncos\u03c6 . \n(23.A.16)\nx0 \nThe x -component of the velocity as a function of time is then \ndx(t)\n2E \nv x (t) = \nsin(\u03c90 t + \u03c6) . \n(23.A.17)\n= \u2212\u03c90\ndt \nk \nAt t = 0, \n2E \nv \n\u2261 v (t = 0) = \u2212\u03c90 \nsin\u03c6 . \n(23.A.18)\nx,0 \nx \nk \nWe can determine the constant \u03c6 by dividing the expressions in Eqs. (23.A.18) and \n(23.A.16), \nv \n\u2212 \nx,0 = tan\u03c6. \n(23.A.19)\n\u03c90 x0 \nThus the constant \u03c6 can be determined by the initial conditions and the angular \nfrequency of oscillation, \n\u239b \nv \n\u239e \n\u03c6 = tan\u22121 \u239c\u2212 \nx,0 \u239f . \n(23.A.20)\n\u239d\n\u03c90 x0 \u23a0 \nUse the identity \ncos(\u03c90t +\u03c6) = cos(\u03c90t)cos(\u03c6) \u2212 sin(\u03c90t)sin(\u03c6) \n(23.A.21) \nto expand Eq. (23.A.15) yielding \n23-44", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_636", "text": "2E \n2E \nx(t) = \ncos(\u03c90t)cos(\u03c6) \u2212 \nsin(\u03c90t)sin(\u03c6) , \n(23.A.22)\nk\nk \nand substituting Eqs. (23.A.16) and (23.A.18) into Eq. (23.A.22) yields \nv \nx(t) = x cos\u03c9 t + \nx,0 sin\u03c9 t , \n(23.A.23)\n0\n0 \n0\n\u03c90 \nagreeing with Eq. (23.2.21). \nSo, what about the missing \u00b1 that should have been in Eq. (23.A.5)? Strictly speaking, \nwe would need to redo the derivation for the block moving in different directions.  \nMathematically, this would mean replacing \u03c6 by \u03c0 \u2212\u03c6 (or \u03c6 \u2212\u03c0 ) when the block\u2019s \nvelocity changes direction. Changing from the positive square root to the negative and \nchanging \u03c6 to \u03c0 \u2212\u03c6 have the collective action of reproducing Eq. (23.A.23). \nAppendix 23B: Complex Numbers \nA complex number z can be written as a sum of a real number x and a purely imaginary \nnumber iy where i = \nz = x + iy . \n(23.B.1) \nThe complex number can be represented as a point in the x-y plane as show in Figure \n23B.1. \n\u22121 , \nFigure 23B.1 Complex numbers \nThe complex conjugate z of a complex number z is defined to be \nThe modulus of a complex number is \nz = x \u2212 iy . \n(23.B.2) \nz = (zz )1 2 = ((x + iy)(x \u2212 iy))1 2 = (x2 + y2 )1 2 . \n(23.B.3) \n23-45", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_637", "text": "z\nwhere we used the fact that i2 = \u22121. The modulus \nrepresents the length of the ray \nfrom the origin to the complex number z in Figure 23B.1. Let \u03c6 denote the angle that \nthe ray with the positive x -axis in Figure 23B.1. Then \nx = z cos\u03c6 , \n(23.B.4) \ny = z sin\u03c6 . \n(23.B.5) \nHence the angle \u03c6 is given by \n\u03c6 = tan\u22121( y / x) . \n(23.B.6) \nThe inverse of a complex number is then \n1 \nz\nx \u2212 iy \n=\n= \n. \n(23.B.7)\n2 + y\nz \nzz \n(x\n2) \nThe modulus of the inverse is the inverse of the modulus; \n1 \n1\n1 \n=\n= \n. \n(23.B.8)\n2 + y2)1 2 \nz \n(x\nz \nThe sum of two complex numbers, z1 \n+ iy1 and z2 \n+ iy2 , is the complex number\n= x1 \n= x2 \n(23.B.9)\nz3 = z1 + z2 = (x1 + x2) + i( y1 + y2) = x3 + iy3 , \nwhere x3 \n+ x2 , \n= \nWe can represent this by the vector sum in Figure \n= x1 \ny3 \ny1 + y2 . \n23B.2, \nFigure 23B.2 Sum of two complex numbers \nThe product of two complex numbers is given by \n23-46", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_638", "text": "+ iy1\n+ iy2) = (x1\nz3 = z1z2 = (x1 \n)(x2 \nx2 \u2212 y1 y2) + i(x1 y2 + x2 y1) = x3 + iy3 , (23.B.10) \nwhere x3 = x1\ny2 , and y3 \ny1 .\nx2 \u2212 y1 \n= x1 y2 + x2 \nOne of the most important identities in mathematics is the Euler formula, \nei\u03c6 = cos\u03c6 + isin\u03c6 . \n(23.B.11) \nThis identity follows from the power series representations for the exponential, sine, and \ncosine functions, \n\u03c6 2 \n\u03c63 \n\u03c6 4 \n\u03c65 \ni\u03c6 \nn=\u221e 1 (i\u03c6)n\ne = \u2211 \n= 1+ i\u03c6 \u2212 \n\u2212 i \n+\n+ i \n..., \n(23.B.12) \nn=0 n! \n2 \n3! \n4! \n5! \n\u03c6 2 \n\u03c6 4 \ncos\u03c6 = 1\u2212 \n+ \n\u2212 ..., \n(23.B.13)\n2 \n4! \n\u03c63 \n\u03c65 \nsin\u03c6 = \u03c6 \u2212 \n+ \n\u2212 .... \n(23.B.14)\n3! \n5! \nWe define two projection operators. The first one takes the complex number ei\u03c6 and \ngives its real part, \nRe ei\u03c6 = cos\u03c6 . \n(23.B.15) \nThe second operator takes the complex number ei\u03c6 and gives its imaginary part, which is \nthe real number \nIm ei\u03c6 = sin\u03c6 . \n(23.B.16) \nA complex number z = x + iy can also be represented as the product of a modulus z and \na phase factor ei\u03c6 , \nz = z ei\u03c6 . \n(23.B.17) \nThe inverse of a complex number is then \n1\n1 \n1 \n= \ni\u03c6 = \ne\u2212 i\u03c6 , \n(23.B.18)\nz \nz\nz e \nwhere we used the fact that \n1 \n\u2212 i\u03c6 \ni\u03c6 = e . \n(23.B.19)\ne \ni\u03c61\nz\nIn terms of modulus and phase, the sum of two complex numbers, z1 = \ne \nand\n1 \ni\u03c62\nz = z e , is\n2 \n23-47 \n2", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_639", "text": "i\u03c61\n= \ne + \ne\ni\u03c62 . \n(23.B.20)\nz1 + z2 \nz1 \nz2 \nA special case of this result is when the phase angles are equal, \u03c61 \n, then the sum\n= \u03c62 \nz1 + z2 has the same phase factor e\ni\u03c61 as z1 and z2 , \ni\u03c61\ni\u03c61 \ni\u03c61 =\n= \ne + \ne \n( \n+ \n)e . \n(23.B.21)\nz1 + z2 \nz1 \nz2 \nz1 \nz2 \ni\u03c62\nz \nz\nThe product of two complex numbers, z1 = \ne\ni\u03c61 , and z2 = \ne \nis\n1 \n2 \ni\u03c61+\u03c62\ni\u03c61 \ni\u03c62 =\nz1z2 = \ne \ne \ne \n. \n(23.B.22)\nz1 \nz2 \nz1 z2 \nWhen the phases are equal, the product does not have the same factor as z1 and z2 , \ni\u03c61\ni\u03c61\nz1z2 = \ne \ne = \ne\ni2\u03c61 . \n(23.B.23)\nz1 \nz2 \nz1 z2 \nAppendix 23C: Solution to the Underdamped Simple Harmonic \nOscillator \nConsider the underdamped simple harmonic oscillator equation (Eq. (23.5.4)), \nd 2 x \nb dx \nk\n+\n+ \nx = 0 . \n(23.C.1)\ndt2 \nm dt \nm \nWhen (b / m)2 < 4k / m , we show that the equation has a solution of the form \nx(t) = x m e\u2212\u03b1t cos(\u03b3 t +\u03c6) . \n(23.C.2) \nSolution: Let\u2019s suppose the function x(t) has the form \nzt )\nx(t) = ARe(e \n(23.C.3) \nwhere z is a number (possibly complex) and A is a real number. Then \ndx = zAezt \n(23.C.4)\ndt \nd 2 x \n2 Aezt \n= z \n(23.C.5)\ndt2 \n23-48", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_640", "text": "We now substitute Eqs. (23.C.3), (23.C.4), and (23.C.5), into Eq. (23.C.1) resulting in \n2 Aezt + b \nk Aezt \nz\nzAezt +\n= 0 . \n(23.C.6)\nm\nm \nCollecting terms in Eq. (23.C.6) yields \n\u239b 2 + b\nk \n\u23a0\u239f\n\u239e Aezt \nz\nz +\n= 0 \n(23.C.7)\n\u239d\u239c \nm\nm \nThe condition for the solution is that \n2 + b\nk \nz\nz +\n= 0 . \n(23.C.8)\nm\nm \nThis quadratic equation has solutions \n\u2212(b / m) \u00b1 ((b / m)2 \u2212 4k / m)1 2 \nz = \n. \n(23.C.9)\n2 \nWhen (b / m)2 < 4k / m , the oscillator is called underdamped, and we have two solutions \nfor z , however the solutions are complex numbers. Let \n= (k / m \u2212 (b / 2m)2)1 2 ;\n\u03b3 \n(23.C.10) \nand \n\u03b1 = b 2m . \n(23.C.11) \n. \nRecall that the imaginary number i = \u22121 . The two solutions are then z1 \n\u03b3\n= \u2212\u03b1 + i t and \nz2 = \u2212\u03b1 \u2212 i t \n\u03b3 . Because our system is linear, our general solution is a linear combination \nof these two solutions, \n\u2212\u03b1 +i\u03b3 t + A2\n\u2212\u03b1\u2212i\u03b3 t \ni\u03b3 t + A2\n\u2212i\u03b3 t )e\u2212\u03b1t\nx(t) = A1e\ne \n= ( A1e\ne\n, \n(23.C.12) \nwhere A1 and A2 are constants. We shall transform this expression into a more familiar \nequation involving sine and cosine functions with help from the Euler formula, \n\u00b1i\u03b3 t\ne \n= cos(\u03b3 t) \u00b1 isin(\u03b3 t). \n(23.C.13) \nTherefore we can rewrite our solution as \n\u2212\u03b1t\nx(t) = ( A1(cos(\u03b3 t) + isin(\u03b3 t)) + A2(cos(\u03b3 t) \u2212 isin(\u03b3 t)) )e \n. \n(23.C.14) \n23-49", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_641", "text": "A little rearrangement yields \n\u2212\u03b1t\nx(t) = (( A1 + A2)cos(\u03b3 t) + i( A1 \u2212 A2)sin(\u03b3 t))e \n. \n(23.C.15) \nDefine two new constants C = \nand D = i( A1 \u2212 A2). Then our solution looks like\nA1 + A2 \nx(t) = (C cos(\u03b3 t) + Dsin(\u03b3 t))e\u2212\u03b1t . \n(23.C.16) \nRecall from Example 23.5 that we can rewrite \nC cos(\u03b3 t) + Dsin(\u03b3 t) = x m cos(\u03b3 t +\u03c6) \n(23.C.17) \n, \nwhere \nx m = (C 2 + D2)1 2 , and \u03c6 = tan\u22121(D / C) . \nThen our general solution for the underdamped case (Eq. (23.C.16)) can be written as \nx(t) = x m e\u2212\u03b1t cos(\u03b3 t +\u03c6) . \n(23.C.18) \nThere are two other possible cases which we shall not analyze: when (b / m)2 > 4k / m , a \ncase referred to as overdamped, and when (b / m)2 = 4k / m , a case referred to as \ncritically damped. \nAppendix 23D: Solution to the Forced Damped Oscillator Equation \nWe shall now use complex numbers to solve the differential equation \nd 2 x \ndx\nF0 cos(\u03c9t) = m \n+ b \n+ kx . \n(23.D.1)\ndt2 \ndt \nWe begin by assuming a solution of the form \nx(t) = x0 cos(\u03c9t +\u03c6) . \n(23.D.2) \nwhere the amplitude x0 and the phase constant \u03c6 need to be determined. We begin by \ndefining the complex function \ni(\u03c9t+\u03c6 )\nz(t) = x0e \n. \n(23.D.3) \nOur desired solution can be found by taking the real projection \n23-50", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_642", "text": "x(t) = Re(z(t)) = x0 cos(\u03c9t +\u03c6) . \n(23.D.4) \nOur differential equation can now be written as \nd 2 \ni\u03c9t \nz \ndz\nF0e \n= m \n+ b \n+ kz . \n(23.D.5)\ndt2 \ndt \nWe take the first and second derivatives of Eq. (23.D.3), \ndz \ni(\u03c9t+\u03c6 )\n(t) = i\u03c9 x0e \n= i\u03c9 z . \n(23.D.6)\ndt \nd 2 z \ni(\u03c9t+\u03c6 ) = \u2212\u03c9 2\n(t) = \u2212\u03c9 2 x0e\nz . \n(23.D.7)\ndt2 \nWe substitute Eqs. (23.D.3), (23.D.6), and (23.D.7) into Eq. (23.D.5) yielding \ni\u03c9t\ni(\u03c9t+\u03c6 )\nF0e \n= (\u2212\u03c9 2m + bi\u03c9 + k)z = (\u2212\u03c9 2m + bi\u03c9 + k)x0e \n. \n(23.D.8) \nWe divide Eq. (23.D.8) through by ei\u03c9t and collect terms using yielding \nF0 / m \nx0ei\u03c6 = \n. \n(23.D.9)\n((\u03c90\n2 \u2212\u03c9 2) + i(b / m)\u03c9 ) \nwhere we have used \u03c90\n2 = k / m . Introduce the complex number \nz1 = (\u03c90\n2 \u2212\u03c9 2) + i(b / m)\u03c9 . \n(23.D.10) \nThen Eq. (23.D.9) can be written as \ni\u03c6 \nF0\nx0e = \n. \n(23.D.11)\nmy \nMultiply the numerator and denominator of Eq. (23.D.11) by the complex conjugate \n2 \u2212\u03c9 2) \u2212 i(b / m)\u03c9 yielding\nz1 = (\u03c90 \n((\u03c90\n2 \u2212\u03c9 2) \u2212 i(b / m)\u03c9 )\ni\u03c6 \nF0 z1 \nF0\nx0e =\n= \n\u2261 u + iv . \n(23.D.12)\nmz1\nm ((\u03c90\n2 \u2212\u03c9 2)2 + (b / m)2\u03c9 2)\nz1 \nwhere \n2 \u2212\u03c9 2)\nF0\n(\u03c90\nu = \n,", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_643", "text": "F0 z1 \nF0\nx0e =\n= \n\u2261 u + iv . \n(23.D.12)\nmz1\nm ((\u03c90\n2 \u2212\u03c9 2)2 + (b / m)2\u03c9 2)\nz1 \nwhere \n2 \u2212\u03c9 2)\nF0\n(\u03c90\nu = \n, \n(23.D.13)\nm ((\u03c90\n2 \u2212\u03c9 2)2 + (b / m)2\u03c9 2) \n23-51", "source": "Meca_Notes", "format": "pdf"}
-{"id": "Meca_Notes_chunk_644", "text": "v = \u2212 F0 \nm \n(b / m)\u03c9 \n((\u03c90 \n2 \u2212 \u03c9 2 )2 + (b / m)2\u03c9 2 ) . \n(23.D.14) \nTherefore the modulus x0 is given by \nx0 = (u2 + v2 )1/2 = \nF0 / m \n((\u03c90 \n2 \u2212 \u03c9 2 )2 + (b / m)2\u03c9 2 ) , \n(23.D.15) \nand the phase is given by \n\u03c6 = tan\u22121(v / u) = \u2212(b / m)\u03c9 \n(\u03c90 \n2 \u2212 \u03c9 2 ) . \n(23.D.16) \n23-52", "source": "Meca_Notes", "format": "pdf"}